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[ "# texas holdem poker odds and probabilities\n\n## learn and play holdem poker\n\n### odds and probabilities\n\nodds and probabilites are just that maybes sometimes they work and sometimes they don't one thing that seems to help is to create a sense of luck or desired outcome and luck or desired outcome seems to follow but here are some mathematical formulas to add to the game.\n\n1 in X - This is the most practical application while sitting at a table. If you have 1 card in the deck that'll save you, and 46 cards are remaining in the deck, it'd be a \"1 in 46\" chance. Meaning you have one out from X number of chances. It's really a fraction sideways: 7/8ths is the same as \"7 in 8\". So \"1 in 2\" would be a 50% chance or a \"1 to 1\" chance\n\nX to 1 - This is very commonly used in holdem books. If you are in a situation where you'll win 1/3rd of the time, and the other 2/3rds you lose, you have a \"2 to 1\" chance of winning. So \"4 to 1\" would be the same as a 20% chance or a \"1 in 5\" chance\". \"2 to 1\" can also be worded as \"2 to 1 against\", which means your chances are 2 to 1 against winning.\n\nPercentage - This is a widely accepted form that is difficult to calculate at the table in most cases. A 4% chance of winning is the same as a \"1 in 25\" chance or \"24 to 1\".\n\nProbability - To do basic math with probabilities it needs to be in this form. Also, if you calculate the chance of an event occurring using algebra, it will come out in this form. It is simply the percentage chance divided by 100 with no % sign. So a 25% chance is the same as a probability of .25, and 100% is a probability of 1.0\n\nAmerica's Cardroom is one of the few poker sites that accepts American players and is one of the largest in terms of jackpots, freerolls and overall action. If you want a world class poker site with large jackpots like the million dollar tourneys, freerolls on demand, sitngo tournaments 24/7 and great comp check out America's Cardroom.", null, "" ]

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[ "Solutions by everydaycalculation.com\n\n## Reduce 55/10 to lowest terms\n\nThe simplest form of 55/10 is 11/2.\n\n#### Steps to simplifying fractions\n\n1. Find the GCD (or HCF) of numerator and denominator\nGCD of 55 and 10 is 5\n2. Divide both the numerator and denominator by the GCD\n55 ÷ 5/10 ÷ 5\n3. Reduced fraction: 11/2\nTherefore, 55/10 simplified to lowest terms is 11/2.\n\nMathStep (Works offline)", null, "Download our mobile app and learn to work with fractions in your own time:" ]

[ null, "https://answers.everydaycalculation.com/mathstep-app-icon.png", null ]

[ "# Map, Filter, Lambda, and List Comprehensions in Python¶\n\nAuthor: R.G. Erdmann\n\nmap(), filter(), lambda, and list comprehensions provide compact, elegant, and efficient ways to encode a few common idioms in programming. We often encounter the following scanarios involving for-loops:\n\n## Some for-loop examples to rewrite more compactly¶\n\n• Building up a list from scratch by looping over a sequence and performing some calculation on each element in the sequence.\n\n1. For example, suppose we want to build a list of the squares of the integers from 0 to 9:\n\n```>>> squares = [] # start with an empty list\n>>> for x in range(10): # step over every element in the list of integers from 0 to 9\n... squares.append(x**2) # grow the list one element at a time\n...\n>>> print squares\n[0, 1, 4, 9, 16, 25, 36, 49, 64, 81]\n```\n1. Suppose we want to build a list of the lengths of the names in a list:\n\n```>>> names = ['Anne', 'Amy', 'Bob', 'David', 'Carrie', 'Barbara', 'Zach']\n>>> lengths = []\n>>> for name in names:\n... lengths.append(len(name))\n...\n>>> print lengths\n[4, 3, 3, 5, 6, 7, 4]\n```\n• Building up a list from scratch by looping over nested sequences.\n\n1. For example, suppose we want a list of all possible pairs of drink and food from the lists ['water', 'tea', 'juice'] and ['ham', 'eggs', 'spam'], respectively:\n\n```>>> possible_choices = []\n>>> for drink in ['water', 'tea', 'juice']:\n... for food in ['ham', 'eggs', 'spam']:\n... possible_choices.append([drink,food])\n...\n>>> print possible_choices\n[['water', 'ham'], ['water', 'eggs'], ['water', 'spam'], ['tea', 'ham'], ['tea', 'eggs'], ['tea', 'spam'], ['juice', 'ham'], ['juice', 'eggs'], ['juice', 'spam']]\n```\n1. Suppose we want a list of coordinates on a rectangular grid:\n\n```>>> coords = []\n>>> for x in range(5):\n... for y in range(3):\n... coordinate = (x, y)\n... coords.append(coordinate)\n...\n>>> print coords\n[(0, 0), (0, 1), (0, 2), (1, 0), (1, 1), (1, 2), (2, 0), (2, 1), (2, 2), (3, 0), (3, 1), (3, 2), (4, 0), (4, 1), (4, 2)]\n```\n• Building a list from scratch by filtering a sequence according to some criterion or criteria.\n\n1. For example, suppose we want a list of the squares of the integers from 0 to 9 where the square is greater than 5 and less than 50:\n\n```>>> special_squares = []\n>>> for x in range(10):\n... square = x**2\n... if square > 5 and square < 50:\n... special_squares.append(square)\n...\n>>> print special_squares\n[9, 16, 25, 36, 49]\n```\n1. Suppose we want to take a list of names and find only those starting with the letter B:\n\n```>>> names = ['Anne', 'Amy', 'Bob', 'David', 'Carrie', 'Barbara', 'Zach']\n>>> b_names = []\n>>> for name in names:\n... if name.startswith('B'):\n... b_names.append(name)\n...\n>>> print b_names\n['Bob', 'Barbara']\n```\n\n## Map, Lambda, and Filter¶\n\nOne way to achieve the same goals as in the above examples is to use some of Python’s tools from functional programming: map(), filter(), and lambda().\n\n### map()¶\n\nThe map() function applies a function to every member of an iterable and returns the result. Typically, one would use an anonymous inline function as defined by lambda, but it is possible to use any function. The first example above can also be accomplished using map() as follows:\n\n```>>> def square(x):\n... return x**2\n...\n>>> squares = map(square, range(10))\n>>> print squares\n[0, 1, 4, 9, 16, 25, 36, 49, 64, 81]\n```\n\nIn the case of the second example, the len() function already exists, so we can use map() to apply it to every element of the list:\n\n```>>> names = ['Anne', 'Amy', 'Bob', 'David', 'Carrie', 'Barbara', 'Zach']\n>>> lengths = map(len, names)\n>>> print lengths\n[4, 3, 3, 5, 6, 7, 4]\n```\n\n### lambda¶\n\nIn the first map example above, we created a function, called square, so that map would have a function to apply to the sequence. This is a common occurrence, so Python provides the ability to create a simple (no statements allowed internally) anonymous inline function using a so-called lambda form. Thus, an anonymous function that returns the square of its argument can be written as lambda x: x**2. This means, “Here is an anonymous (nameless) function that takes one arguement, called x, and returns the square of x. Since the lambda form actually evaluates to a function, we can also call it inline. (This is generally a silly thing to do, but we show it here to demonstrate that lambda forms are actually inline function objects.):\n\n```>>> print (lambda x: x**2)(5) # first parens contain a lambda form that is a squaring function, second parens represent calling that function\n25\n>>> # Make a function of two arguments (x and y) that returns their product, then call the function with 3 and 4:\n>>> print (lambda x, y: x*y)(3, 4)\n12\n>>> print (lambda x: x.startswith('B'))('Bob')\nTrue\n>>> print (lambda x: x.startswith('B'))('Robert')\nFalse\n>>> incrementer = lambda input: input+1\n>>> # above is same as def incrementer(input): return input+1\n>>> # now we call it\n>>> print incrementer(3)\n4\n```\n\n### Using lambda and map together¶\n\nLambda forms can be used as the required function argument to the map() function. For example, the first example above can be written as\n\n```>>> squares = map(lambda x: x**2, range(10))\n>>> print squares\n[0, 1, 4, 9, 16, 25, 36, 49, 64, 81\n```\n\nIn English, this means, “Apply a function that squares its argument (the lambda form) to every member of the list of integers from 0 to 9 (the range()), and store the result in a list called squares.\n\n### filter()¶\n\nThe fifth and sixth examples above can also be achieved with the filter() built-in function. Filter takes a function returning True or False and applies it to a sequence, returning a list of only those members of the sequence for which the function returned True.\n\nLambda forms can also be used with the filter function; in fact, they can be used anywhere a function is expected in Python. In the fifth example, the list of squares is filtered according to whether the given entries are greater than 5 and less than 50. A lambda form that returns True when this condition is met is lambda x: x > 5 and x < 50. Thus, we can reproduce the fifth example as follows:\n\n```>>> squares = map(lambda x: x**2, range(10))\n>>> special_squares = filter(lambda x: x > 5 and x < 50, squares)\n>>> print special_squares\n[9, 16, 25, 36, 49]\n```\n\nIn English, this means, “Find every member of the squares list for which the member is greater than 5 and less than 50 (every member for which lambda x: x > 5 and x < 50 returns True), and store that in a new variable called special_squares.\n\nSimilarly, the sixth example <sixth-example-list-comprehension> can be reproduced using filter as follows:\n\n```>>> names = ['Anne', 'Amy', 'Bob', 'David', 'Carrie', 'Barbara', 'Zach']\n>>> b_names = filter(lambda s: s.startswith('B'), names)\n>>> print b_names\n['Bob', 'Barbara']\n```\n\n## List Comprehensions¶\n\nAll of the six original examples can be achieved using a consistent, clean, and elegant syntax called list comprehensions.\n\n### Simple list comprehensions¶\n\nThe simplest form of a list comprehension is\n\n[ expression-involving-loop-variable for loop-variable in sequence ]\n\nThis will step over every element of sequence, successively setting loop-variable equal to every element one at a time, and will then build up a list by evaluating expression-involving-loop-variable for each one. This eliminates the need to use lambda forms, and thus generally produces a much more readable code than using map() and a more compact code than using a for-loop.\n\nThe first example <first-example-list-comprehension> can thus be written compactly as:\n\n```>>> squares = [ x**2 for x in range(10) ]\n>>> print squares\n[0, 1, 4, 9, 16, 25, 36, 49, 64, 81]\n```\n\nSome other simple examples:\n\n• Print the length of each word in the list of names:\n\n```>>> print [ len(name) for name in names ]\n[4, 3, 3, 5, 6, 7, 4]\n```\n• Print the last letter of each name in the list of names:\n\n```>>> print [ name[-1] for name in names ]\n['e', 'y', 'b', 'd', 'e', 'a', 'h']\n```\n• Print the reverse of each name in the list of names:\n\n```>>> print [ name[::-1] for name in names ]\n['ennA', 'ymA', 'boB', 'divaD', 'eirraC', 'arabraB', 'hcaZ']\n```\n\nNote that complex expressions can be put in the slot for expression-involving-loop-variable. For example, here we build up a list of names, first letters, and lengths for each name in the list:\n\n```>>> print [ [name, name, len(name)] for name in names ]\n[['Anne', 'A', 4], ['Amy', 'A', 3], ['Bob', 'B', 3], ['David', 'D', 5], ['Carrie', 'C', 6], ['Barbara', 'B', 7], ['Zach', 'Z', 4]]\n```\n\nWhere [name, name, len(name)] occupies the expression-involving-loop-variable slot, so that the list comprehension creates a list of [name, name, len(name)] for every name in the names sequence.\n\n### Nested list comprehensions¶\n\nList comprehensions can be nested, in which case they take on the following form:\n\n[ expression-involving-loop-variables for outer-loop-variable in outer-sequence for inner-loop-variable in inner-sequence ]\n\nThis is equivalent to writing:\n\n```results = []\nfor outer_loop_variable in outer_sequence:\nfor inner_loop_variable in inner_sequence:\nresults.append( expression_involving_loop_variables )\n```\n\nThus, the third example can be written compactly as:\n\n```>>> possible_choices = [ [drink,food] for drink in ['water', 'tea', 'juice'] for food in ['ham', 'eggs', 'spam'] ]\n>>> print possible_choices\n[['water', 'ham'], ['water', 'eggs'], ['water', 'spam'], ['tea', 'ham'], ['tea', 'eggs'], ['tea', 'spam'], ['juice', 'ham'], ['juice', 'eggs'], ['juice', 'spam']]\n```\n\nAnd example 4. can be written as\n\n```>>> coords = [ (x,y) for x in range(5) for y in range(3) ] # points on a rectangular grid\n>>> print coords\n[(0, 0), (0, 1), (0, 2), (1, 0), (1, 1), (1, 2), (2, 0), (2, 1), (2, 2), (3, 0), (3, 1), (3, 2), (4, 0), (4, 1), (4, 2)]\n```\n\n### Filtered list comprehensions¶\n\nThe final form of list comprehension involves creating a list and filtering it similarly to using filter(). The filtering form of list comprehension takes the following form:\n\n[ expression-involving-loop-variable for loop-variable in sequence if boolean-expression-involving-loop-variable ]\n\nThis form is similar to the simple form of list comprehension, but it evaluates boolean-expression-involving-loop-variable for every item and keeps only those members for which the boolean expression is True. Thus we can use list comprehensions to rewrite example 5 as\n\n```>>> special_squares = [ x**2 for x in range(10) if x**2 > 5 and x**2 < 50 ]\n>>> print special_squares\n[9, 16, 25, 36, 49]\n```\n\nNote that the above is inefficient, however, since it has to calculate the square of x three separate times for each element in the loop. Thus, the following is an equivalent and more efficient approach:\n\n```>>> squares = [ x**2 for x in range(10) ]\n>>> special_squares = [ s for s in squares if s > 5 and s < 50 ]\n```\n\nFinally, note that the foregoing can be written on a single line using a pair of list comprehensions as follows:\n\n```>>> special_squares = [ s for s in [ x**2 for x in range(10) ] if s > 5 and s < 50 ]\n```\n\nExample 6 can be rewritten using a list comprehension as:\n\n```>>> names = ['Anne', 'Amy', 'Bob', 'David', 'Carrie', 'Barbara', 'Zach']\n>>> b_names = [ name for name in names if name.startswith('B') ]\n>>> print b_names\n['Bob', 'Barbara']\n```\n\nOr, again combining the first two lines into one,\n\n```>>> b_names = [ name for name in ['Anne', 'Amy', 'Bob', 'David', 'Carrie', 'Barbara', 'Zach'] if name.startswith('B') ]\n```" ]

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[ " Asymptotic Solutions of the Kinetic Boltzmann Equation and Multicomponent Non-Equilibrium Gas Dynamics\n\nJournal of Applied Mathematics and Physics\nVol.04 No.08(2016), Article ID:70422,11 pages\n10.4236/jamp.2016.48177\n\nAsymptotic Solutions of the Kinetic Boltzmann Equation and Multicomponent Non-Equilibrium Gas Dynamics\n\nS. A. Serov1, S. S. Serova2\n\n1Fundamental Researches Department, Russian Federal Nuclear Centre, All-Russian Scientific Research Institute of Experimental Physics, Sarov, Russia\n\n2St. Petersburg State University, St. Petersburg, Russia", null, "", null, "", null, "Received 16 June 2016; accepted 24 August 2016; published 31 August 2016", null, "ABSTRACT\n\nIn the article correct method for the kinetic Boltzmann equation asymptotic solution is formulated, the Hilbert’s and Enskog’s methods are discussed. The equations system of multicomponent non- equilibrium gas dynamics is derived, that corresponds to the first order in the approximate (asym- ptotic) method for solution of the system of kinetic Boltzmann equations.\n\nKeywords:\n\nKinetic Boltzmann Equation, Multicomponent Non-Equilibrium Gas Dynamics", null, "1. Introduction\n\nIn 1912 Hilbert considered the kinetic Boltzmann equation for one-component gas as an example of integral equation and proposed a “recipe” for its approximate (asymptotic) solution (see , Chapter~XXII). Hilbert’s “recipe” was inconvenient for practical use, because the five arbitrary functional parameters of the first and the following approximations of the velocity distribution function had to be found by solving the differential equations in partial derivatives (equations of gas dynamics of the first and higher orders). Five years later Enskog proposed to use zero conditions, conditions with zero right-hand sides, to determine the five arbitrary functional parameters of the first and following approximations of the velocity distribution function. The imposition of the zero conditions leads, in fact, to using different comparison scales in the asymptotic expansion of the velocity distribution function and in the asymptotic expansion of the particle number density, the mean (mass) velocity and the temperature, that are derived from the asymptotic expansion of the velocity distribution function by integration over velocities with different weighting functions. As a result of paralogism of the method of successive approximations (one has to set variable coefficients of the same terms of the unified comparison scale equal to each other) partial time derivatives vanish in the necessary conditions of solutions existence of integral equations of higher orders (see below) and with them terms of gas-dynamic equations, corresponding to viscosity, heat conduction, … vanish. Enskog “improved” the situation by the introducing (see, for example, , Chapter 7, § 1, Section 5) of the unsubstantiated expansion of partial time derivative:", null, "(1)\n\nThe approach of Struminskii, who had proposed in 1974 in his approximate (asymptotic) method of solution of the system of kinetic Boltzmann equations for multicomponent gas, differs from the approach of Enskog to asymptotic solution of the Boltzmann equations system for gas mixture in that, how the infinitesimal parameter is introduced in the Boltzmann equations system for gas mixture, i.e. the solution is constructing in another asymptotic limit. In substance, Struminskii’s method of solution of kinetic equations system is the same as Enskog’s method (Struminskii used the partial time derivative expansion, as Enskog did).\n\nIn section 2 below will be proposed the correct method of asymptotic solution of the kinetic Boltzmann equations system for multicomponent gas mixture for the approach, that combines Enskog’s and Struminskii’s approaches; in particular, it will be shown, how one has to modify Enskog’s method: in addition to asymptotic expansion of the velocity distribution function i-component particles of gas mixture it is necessary to determine and to use the expansion of the particle number density", null, "of i-component, mean mass velocity", null, "and temperature T of the gas mixture.\n\nFurther, in the Section 3 the system of infinitesimal first order equations of multicomponent non-equilibrium gas dynamics, appearing during the process of the solution of the system of Boltzmann equations by successive approximations method in the Section 2 as necessary condition of the existence of approximate (asymptotic) solution of the integral equations system, is considered in more detail.\n\nThis article is condensed version of our article arXiv:1303.6275. Notations, used below, are close to notations in ; it is assumed, that all regarded functions are continuous and continuously differentiable so many times as it is necessary, if their derivatives are considered, and all regarded integrals converge.\n\n2. Correct Method of Solution of the Kinetic Boltzmann Equations System\n\nThe Boltzmann equations system, that describes change of dependent on t and spatial coordinates, prescribed by radius-vector", null, ", the velocity distribution functions", null, "due to collision with particles of other components of mixture of rarefied monatomic gases, where", null, "are the velocities of particles of i-component of the mixture {see , Chapter 8, Equation (1.1); discussion of the derivation of the Boltzmann equations system and its applicability range see, for example, in , Chapters 3 and 18, , Chapter 7, § 1; below the central interaction of molecules are considered only, when the force, with which each molecule acts on the other, is directed along the line, connecting the centers of the molecules}, could be written as:", null, "(2)\n\nin (2) N is a set of indexes, that are numbering components of the mixture;", null, "is an external force, which acts on the molecule of the i-component;", null, "is the mass of the molecule of the i-component;", null, "is the modulus of the relative velocity of colliding particles", null, "; b is the impact distance,", null, "is the azimuth angle,", null, "is the unit vector, directed to the center of mass of the colliding particles from the point of closest approach―see , Chapter 3, Figure 3; the scalar function", null, "is determined by equality", null, "(3)\n\nby prime in (20) and below the velocities and the functions of velocities after the collision are denoted.\n\nLet us introduce following notations:", null, "(4)\n\n(5)\n\nto differ velocities of colliding molecules of the same kind in (22) the one velocity is denoted by and the other is denoted by (without any index) and the index of the corresponding velocity distribution function f is omitted.\n\nIn Enskog’s approach the differential parts of the Boltzmann Equations (2), that are denoted by below, are considered to be small as compared with the right-hand sides of Equations (2)―see , Chapter 7, § 1, Section 5―therefore the indicator of infinite smallness is formally introduced in the Boltzmann equations system in the following way:\n\n(6)\n\nIn Struminskii’s approach to the asymptotic solution of the Boltzmann equations system the differential parts of the Boltzmann Equations (2) and the collision integrals of the particles of i-component with the particles of the other components are considered to be small as compared with the collision integral of the particles of i-component between each other, therefore the indicator of infinitesimality is introduced in the Boltzmann equations system in another way:\n\n(7)\n\nIt is possible to combine Enskog’s approach with Struminskii’s approach. For this purpose we divide the set of mixture components N into two subsets: the subset of components, that we call formally inner components (we could consider the case, when there are some subsets of inner components, but this case does not fundamentally differ from the one, considered below, the only difference is that the notation become more complicated) and the subset of components, that we call external components. To differ the two groups of mixture components we\n\ndenote the subset of indexes of inner components as well as the indexes of inner components and the subset of indexes of external components as well as the indexes of external components; the\n\nintersection of the sets and is the empty set―and the union of these sets is the set of indexes of all mixture components; if an assertion concerns both kinds of components the special symbols will be omitted. In new notations the Boltzmann equations system can be rewritten as:\n\n(8)\n\n(9)\n\nLet us write the asymptotic expansion of the velocity distribution function of particles of i-component as formal series of successive approximations in powers of:\n\n(10)\n\nThe differential parts of the Equations (3) are written as:\n\n(11)\n\nwhere\n\n(12)\n\n―cf. with , Chapter 7, § 1, Sections 4, 5 and . In (11)-(12) the partial time derivative expansion (1) is not used in contrast to that, how it was made by Enskog and further by Struminskii. As result, described below method for solution of the system of kinetic Boltzmann equations differ fundamentally from Enskog’s method and Struminskii’s method.\n\nSubstituting (10) and (11) in (8) and equating coefficients at the same powers of to each other, we obtain the equations system of the method of successive approximations for finding the velocity distribution functions of inner components particles of gas mixture; taking introduced notations (4), (5) and (12) into account, the system can be rewritten as:\n\n(13)\n\n(14)\n\nSimilarly substituting (10) and (11) in (9) and equating coefficients at the same powers of to each other, we obtain the equations system of the method of successive approximations for finding the velocity distribution\n\nfunctions of particles of external components of gas mixture:\n\n(15)\n\n(16)\n\nSpeaking about an order of approximation below, we assume the order to be equal to the value of index r in (14), (16). According to (5), (13), in zero order approximation we have the following system of integral equations to find the velocity distribution functions of particles of inner components of gas mixture:\n\n(17)\n\nThe general solution of the equations system (17) can be written as a set of the Maxwell functions:\n\n(18)\n\nwhere k is the Boltzmann constant.\n\nParticle number density of the i-component, mean mass velocity and temperature T of inner com- ponents of mixture are introduced by definitions:\n\n(19)\n\n(20)\n\n(21)\n\nin (21) is the Boltzmann constant. From (19)-(21) the equality is obtained:\n\n(22)\n\nthat is convenient to use below instead of definition (21).\n\nAccording to definitions (19), (20), (21), in addition to the asymptotic expansion (10) it is necessary to determine asymptotic expansions for particle number density of the i-component\n\n(23)\n\nmean mass velocity\n\n(24)\n\nand temperature of inner components of mixture\n\n(25)\n\nSubstituting (10) and (23)-(25) in (19), (20), (22) and equating terms of the same infinitesimal order we obtain scalar relations, that connect asymptotic expansions (10) and (23)-(25):\n\n(26)\n\n(27)\n\n(28)\n\nIn (27), (28) the notations are introduced\n\n(29)\n\n(30)\n\nIn particular, for from (26)-(28) we obtain expressions for arbitrary functions, and in (18) through the zero order approximations to local values of the -component number\n\ndensity, the mean mass velocity and the temperature of inner components of the mixture:\n\n(31)\n\n(32)\n\n(33)\n\nAccording to (4), (15), zero order integral equations, from which the velocity distribution functions of particles of outer components of the mixture are found:\n\n(34)\n\n―are simpler than Equations (17) and differ actually from (17) only by lack of summation over components. Therefore, similarly (18), the general solution of the equations system (34) can be written as a set of the Maxwell functions:\n\n(35)\n\nwhere and are some, independent of, scalar functions of spatial coordinates, defined by the radius vector, and time t, and is a vector function of and t.\n\nLet’s add to the definition of the number density of particles of i-component definitions of mean velocity and temperature of outer component of mixture:\n\n(36)\n\n(37)\n\nfrom (19), (36), (37) the equality is obtained:\n\n(38)\n\nthat is convenient to use below instead of definition (37).\n\nLet’s enter similar (24)-(25) asymptotic expansions of outer -component mean velocity\n\n(39)\n\nand outer -component temperature\n\n(40)\n\nSubstituting (10), (23), (39), (40) in (19), (36), (38) and equating terms of the same infinitesimal order we obtain for each 5 (scalar) relations, that connect asymptotic expansions (10), (23), (39), (40):\n\n(41)\n\n(42)\n\n(43)\n\ncf. with (26)-(28). In (42), (43) the notation is used\n\n(44)\n\nFor from (41)-(43) we obtain expressions for arbitrary functions, and in (35) through the zero order approximations to local values of the number density, the mean\n\nvelocity and the temperature of outer -component of the mixture:\n\n(45)\n\n(46)\n\n(47)\n\nFor the velocity distribution functions of inner components of gas mixture are found from the integral equations system (14), which, taking (5) and equality\n\n(48)\n\ninto account, can be rewritten in the form\n\n(49)\n\nin (49) functions are written as, where are new unknown functions.\n\nThe left-hand sides of Equations (49) involves functions, that are known from the previous step of the successive approximations method. Unknown functions enter, linearly, only into the right-hand sides of Equations (49). Therefore the general solution of the system of Equations (14) is a family of functions of a form\n\n, where, , a family of functions is a particular solution of the system of inhomogeneous Equations (49) and a family of functions is the general solution of the system of homogeneous equations\n\n(50)\n\nMultiplying Equations (50) by, integrating over all values of, summing over and transforming integrals, we obtain\n\n(51)\n\nFrom (51) we conclude, that are linear combinations of the summational invariants of the collision :\n\n(52)\n\nwhere and are some, independent of, scalar functions of spatial coordinates, defined by the radius vector, and time t, and is a vector function of and t (as well as above, arbitrary functions and are identical for all inner components of the mixture), and, hence,\n\n(53)\n\nTo simplify further evaluations according to the expression for, see (18) and (31)-(33), let us rewrite (53) as\n\n(54)\n\nwhere, and are new functions of and t. Family of functions is a solution of the system of inhomogeneous equations\n\n(55)\n\nwhere denote left-hand sides of the Equations (49), taken with opposite sign.\n\nMultiplying Equations (55) by , integrating over all values of and transforming integrals as above, we obtain as necessary condition for the existence of solutions of the system of integral Equations (55), the necessity of the fulfillment of equalities:\n\n(56)\n\n(57)\n\nAmong (infinitesimal) set of particular solutions of the system of Equations (55), different from each other on some solution of the system of homogeneous Equations (50), unique solution may be chosen such that\n\n(58)\n\n(59)\n\nHaving substituted expression for\n\n(60)\n\nin (26)-(28), taking (18), (29)-(33) and (58)-(59) into account, we obtain a system of algebraic equations [constraint equations for asymptotic expansions (10) and (23)-(25)]:\n\n(61)\n\n(62)\n\n(63)\n\nfrom which we find expressions for functions, and through (variable) coe-\n\nfficients of asymptotic expansions of the particle number density of -component, of the mean mass velocity and of the temperature of inner components of the mixture\n\n(64)\n\n(65)\n\n(66)\n\nThen the fulfillment of equalities (56)-(57) can be considered as the differential equations, the r-order equations of gas dynamics, for finding, ,.\n\nThe partial solution of the system of inhomogeneous Equations (55), satisfying (58)-(59), may be constructed, for example, using expansion of in series in terms of Sonine polynomials with expansion\n\ncoefficients, depending on and t (see or ); such construction proves existence of solutions of the system of integral Equations (49).\n\nFor the velocity distribution functions of outer components of gas mixture may be similarly found from the integral equations system (16):\n\n(67)\n\nwhere\n\n(68)\n\n(69)\n\n(70)\n\nThe fulfillment of analogous (56)-(57) equalities\n\n(71)\n\ncan be considered as the differential equations, the r-order equations of gas dynamics, for finding, ,.\n\n3. The System of First Order Equations of Multicomponent Non-Equilibrium Gas Dynamics\n\nLet us consider in more detail the system of infinitesimal first order Equations (56)-(57), (71), derived above as the necessary (and sufficient) condition of the solution existence of the first order integral equations system (14), (16).\n\nTo simplify transformations, according to the expressions for velocity distribution functions of particles of infinitesimal zero order (18), (35), functions, may be used in (56)-(57), (71) rather than functions, , respectively:\n\n(72)\n\n(73)\n\n(74)\n\nfor inner components, for outer components.\n\nAt transformation of differential parts of the Equations (56)-(57) and (71) we use equalities:\n\n(75)\n\n(76)\n\n(77)\n\nIn (75)-(77) the bar above symbol with index denotes the average of the value:\n\n(78)\n\nand are considered as independent variables. At averaging in (77) it is assumed, that external force, acting on the particle of species i, is independent of the particle velocity, it is assumed also, that integrals,\n\ndepending on external forces, are convergent, and product tends to zero, when tends to infinity.\n\nAfter simple transformations from (56)-(57) and (71) we obtain following system of infinitesimal first order equations of multicomponent non-equilibrium gas dynamics:\n\n(79)\n\n(80)\n\n(81)\n\n(82)\n\n(83)\n\n(84)\n\nIn accordance with the general definition of pressure tensor of i-component of gas mixture\n\n(85)\n\nand with the general definition of i-component heat flux vector\n\n(86)\n\n(cf. with , Chapter 2, §§ 3, 4) in (79)-(84)\n\n(87)\n\nis inner components pressure tensor of zero order, is inner components hydrostatic pressure of zero order, is the unit tensor, double product of two second rank tensors and (, Chapter 1, § 3) is the scalar,\n\n(88)\n\nis inner components heat flux vector of zero order,\n\n(89)\n\nis zero order internal energy of particles of inner components per unit volume, which is equal, in this case, to energy of their translational chaotic motion, however, the energy transfer equations, written in form (81) and (84) can be used in more general cases as well (cf. with , Chapter 7, § 6), in (87)-(89) averaging (78) is performed with Maxwell function from (18);\n\n(90)\n\nis -component pressure tensor of zero order, is -component hydrostatic pressure of zero order,\n\n(91)\n\nis -component heat flux vector of zero order,\n\n(92)\n\nis zero order internal energy of particles of -component per unit volume, in (90)-(92) averaging (78) is performed with Maxwell function from (35).\n\nGeneral analytic expressions for integrals, from (80), (81) and (83), (84), that depend on the interaction cross-section, can be derived in general case, when separate components (with Maxwell velocity distribution function of particles) have different mean velocities and temperatures.\n\nSystem of infinitesimal first order equations of multicomponent non-equilibrium gas dynamics (79)-(84) is proposed to use for describing turbulent flows\n\nCite this paper\n\nS. A. Serov,S. S. Serova, (2016) Asymptotic Solutions of the Kinetic Boltzmann Equation and Multicomponent Non-Equilibrium Gas Dynamics. Journal of Applied Mathematics and Physics,04,1687-1697. doi: 10.4236/jamp.2016.48177\n\nReferences\n\n1. 1. Hilbert, D. (1912) Grundzüge einer Allgemeinen Theorie der Linearen Integralgleichungen. Teubner, Leipzig and Berlin. (In Ger-man)\n\n2. 2. Chapman, S. and Cowling, T.G. (1952) The Mathematical Theory of Non-uniform Gases. Cambridge University Press, Cambridge.\n\n3. 3. Struminskii, V.V. (1974) Influence of Diffusion Velocity on Flow of Gas Mixtures. Prikladnaya Mathematica i Me-chanica [Applied Mathematics and Mechanics], 38, 203-210. (In Russian)\n\n4. 4. Hirschfelder, J.O., Curtiss, Ch.F. and Bird, R.B. (1954) Molecular Theory of Gases and Liquids. Wiley, New York." ]

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[ " Part 1- The amount that results when \\$4,000 is compounded at 6% annually over seven years is....? Part 2- The interest earned in this case is...? Formula- P(1 + r)t - AssignmentGrade.com\n\n# Part 1- The amount that results when \\$4,000 is compounded at 6% annually over seven years is....? Part 2- The interest earned in this case is...? Formula- P(1 + r)t\n\nQUESTION POSTED AT 18/04/2020 - 07:27 PM\n\n100 years 10,000 because it says \\$4,000 is compounded\n\n## Related questions\n\n### The solution to the given system of linear equations lies in which quadrant? x-3y=6 x+y=2\n\nQUESTION POSTED AT 02/06/2020 - 01:57 AM\n\n### A movie rental store charges a \\$6.00 membership fee plus \\$2.50 for each movie rented. The function f(x)= 2.50x + 6 gives the cost of renting x movies. Graph this function and give its domain and range.\n\nQUESTION POSTED AT 02/06/2020 - 01:56 AM\n\n### Four circles, each with a radius of 2 inches, are removed from a square. What is the remaining area of the square? (16 – 4π) in.2 (16 – π) in.2 (64 – 16π) in.2 (64 – 4π) in.2\n\nQUESTION POSTED AT 02/06/2020 - 01:54 AM\n\n### A snack cart sells lemonade for 2\\$ And hot dogs for 5\\$. The vendor sold 86 items today for a total of 330\\$ which equation is true?\n\nQUESTION POSTED AT 02/06/2020 - 01:53 AM\n\n### Part A: Billy rented a canoe at \\$14 for 2 hours. If he rents the same canoe for 5 hours, he has to pay a total rent of \\$29. Write an equation in the standard form to represent the total rent (y) that Billy has to pay for renting the canoe for x hours. (4 points) Part B: Write the equation obtained in Part A using function notation. (2 points) Part C: Describe the steps to graph the equation obtained above on the coordinate axes. Mention the labels on the axes and the intervals. (4 points) Part A: Ax By= C\n\nQUESTION POSTED AT 02/06/2020 - 01:53 AM\n\n### A tree casts a 9 ft shadow at the same time that a person 6 ft tall casts a 4 ft shadow. what is the height of the tree?\n\nQUESTION POSTED AT 02/06/2020 - 01:48 AM\n\n### Bonnie earned scores of 90 and 82 on her math tests. what score must she earn on her third test to keep an average of 84 or better?\n\nQUESTION POSTED AT 02/06/2020 - 01:48 AM\n\n### Part A: Billy rented a canoe at \\$14 for 2 hours. If he rents the same canoe for 5 hours, he has to pay a total rent of \\$29. Write an equation in the standard form to represent the total rent (y) that Billy has to pay for renting the canoe for x hours. (4 points) Part B: Write the equation obtained in Part A using function notation. (2 points) Part C: Describe the steps to graph the equation obtained above on the coordinate axes. Mention the labels on the axes and the intervals. (4 points) Part A: Ax By= C\n\nQUESTION POSTED AT 02/06/2020 - 01:47 AM\n\n### A vacation cabin has a water storage tank. In the first month (30 days) of the vacation season, the amount of water in the tank changed by an average of -36 gallons per day. At the end of the month, the tank contained 1,340 gallons of water. How much water was in the tank originally?\n\nQUESTION POSTED AT 02/06/2020 - 01:43 AM\n\n### Antoine and Tess have a disagreement over how to compute a 15% gratuity on \\$46.00. Tess says, “It is easy to find 10% of 46 by moving the decimal point one place to the left to get \\$4.60. Do that twice. Then add the two amounts to get \\$4.60 + \\$4.60 = \\$9.20 for the 15% gratuity.” How should Antoine respond to Tess’s method?\n\nQUESTION POSTED AT 02/06/2020 - 01:43 AM\n\n### What is the solution to the system of equations? y = –5x + 3 y = 1 (0.4, 1) (0.8, 1) (1, 0.4) (1, 0.8)\n\nQUESTION POSTED AT 02/06/2020 - 01:43 AM\n\n### The graph below shows four straight lines, W, X, Y, and Z: Graph of line W going through ordered pairs negative 4, negative 2 and negative 1, 5. Graph of line X going through ordered pairs negative 2, negative 1 and 0, 5. Graph of line Y going through ordered pairs negative 1, negative 3 and 1, 3. Graph of line Z going through ordered pairs 0, negative 5 and 2, 1. Which line is represented by the function f(x) = 3x + 5?\n\nQUESTION POSTED AT 02/06/2020 - 01:43 AM\n\n### The expression 3 × 7 – 4 × 8 + 2 is equivalent\n\nQUESTION POSTED AT 02/06/2020 - 01:42 AM\n\n### Which value of x is in the solution set of 2(3x-1)≥4x-6\n\nQUESTION POSTED AT 02/06/2020 - 01:40 AM\n\n### Two researchers are studying the decline of orangutan populations. In one study, a population of 784 orangutans is expected to decrease at a rate of 25 orangutans per year. In a second study, the population of a group of 817 orangutans is expected to decrease at a rate of 36 per year. After how many years will the two populations be the same? The two populations will be the same after ______ years\n\nQUESTION POSTED AT 02/06/2020 - 01:40 AM\n\n### A painting is purchased for \\$500. If the value of the painting doubles every 5 years, then its value is given by the function V(t) = 500 ⋅ 2t/5, where t is the number of years since it was purchased and V(t) is its value (in dollars) at that time. What is the value of the painting ten years after its purchase\n\nQUESTION POSTED AT 02/06/2020 - 01:37 AM\n\n### Brad sold candy bars and cookies for a fundraiser at school. Candy bars sold for \\$3, and cookies sold for \\$5. He sold a total of 20 items and made \\$76. How many candy bars did Brad sell? (1 point)\n\nQUESTION POSTED AT 02/06/2020 - 01:36 AM\n\n### Find the six arithmetic means between 1 and 29.\n\nQUESTION POSTED AT 02/06/2020 - 01:36 AM\n\n### Tyrone opens a savings account by depositing \\$75 in August. He increases each of the next monthly deposits by \\$10. Which of the following shows the amounts in Tyrone’s account from August through December? A) \\$75, \\$85, \\$95, \\$105, \\$115 B) \\$75, \\$160, \\$255, \\$360, \\$475 C) \\$85, \\$95, \\$105, \\$115 D) \\$160, \\$255, \\$360, \\$475\n\nQUESTION POSTED AT 02/06/2020 - 01:36 AM\n\n### A botanist measures a plant, in feet, at the beginning of each month and notices that the measurements form a geometric sequence, as shown below. 2, 2.2, 2.42, 2.662, ... If the first measurement was taken on September 1, about how tall was the plant three months before, on June 1, assuming the same growth pattern? A) 0.9 ft B) 1.5 ft C) 1.7 ft D) 2.0 ft\n\nQUESTION POSTED AT 02/06/2020 - 01:35 AM\n\n### Which value represents the horizontal translation from the graph of the parent function f(x) = x^2 to the graph of the function g(x) = (x-4)^2+2?\n\nQUESTION POSTED AT 02/06/2020 - 01:35 AM\n\n### What transformation has changed the parent function f(x) = log5x to its new appearance shown in the graph below logarithmic graph passing through point 5, negative 2.\n\nQUESTION POSTED AT 02/06/2020 - 01:34 AM\n\n### Brenda already has \\$200 toward the purchase of a new laptop that sells for \\$450. She will save \\$50 each month, starting next month, until she can buy the laptop. Which of the following represents the amounts she will have each month, starting next month, until she can afford to buy the laptop (assuming there is no sales tax)? A) \\$200, \\$250, \\$300, \\$350, \\$400, \\$450 B) \\$250, \\$300, \\$350, \\$400, \\$450 C) \\$0, \\$200, \\$250, \\$300, \\$350, \\$400, \\$450 D) \\$50, \\$200, \\$450\n\nQUESTION POSTED AT 02/06/2020 - 01:34 AM\n\n### Use a graphing calculator to solve the equation -3 cost= 1 in the interval from . Round to the nearest hundredth.\n\nQUESTION POSTED AT 02/06/2020 - 01:33 AM\n\n### What is the common ratio of the sequence? -2, 6, -18, 54,...\n\nQUESTION POSTED AT 02/06/2020 - 01:33 AM\n\n### A sequence is defined by the recursive function f(n + 1) = f(n). If f(3) = 9 , what is f(1) ?\n\nQUESTION POSTED AT 02/06/2020 - 01:33 AM\n\n### C2/d at c = 6, d =3 Really confused on this one Can you evaluate the algebraic expressions for the given values of each variable? please\n\nQUESTION POSTED AT 02/06/2020 - 01:32 AM\n\n### Barb, Shelley, and Darcy are college roommates. Barb goes to the library every 6 days, Shelley goes the library every 7 days, and Darcy goes to the library every 4 days. If all three of them went to the library today, when is the next time that they will all be at the library on the same day?\n\nQUESTION POSTED AT 02/06/2020 - 01:32 AM\n\n### Evaluate the algebraic expressions for the given values of each variable.r + S2 at r = 4, s = 1I think it's 6 ,am I correct?\n\nQUESTION POSTED AT 02/06/2020 - 01:31 AM\n\n### 9000 for 5 years at 4.5% compounded monthly\n\nQUESTION POSTED AT 02/06/2020 - 01:30 AM" ]

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[ "ISO/IEC JTC1 SC22 WG21 N3849\n\nDate:\n\n# `string_view`: a non-owning reference to a string, revision 6\n\n## Overview\n\nReferences to strings are very common in C++ programs, but often the callee doesn't care about the exact type of the object that owns the data. 3 things generally happen in this case:\n\n1. The callee takes `const std::string&` and insists that callers copy the data if it was originally owned by another type.\n2. The callee takes two parameters—a `char*` and a length (or just `char*` and assumes 0-termination)—and reduces the readability and safety of calls and loses any helper functions the original type provided.\n3. The callee is rewritten as a template and its implementation is moved to a header file. This can increase flexibility if the author takes the time to code to a weaker iterator concept, but it can also increase compile time and code size, and can even introduce bugs if the author misses an assumption that the argument's contents are contiguous.\n\nGoogle, LLVM, and Bloomberg have independently implemented a string-reference type to encapsulate this kind of argument. `string_view` is implicitly constructible from `const char*` and `std::string`. It provides the `const` member operations from `std::string` to ease conversion. This paper follows Chromium and Bloomberg in extending `string_view` to `basic_string_view<charT>`, and further extends it to include a `traits` parameter to match `basic_string`. We provide typedefs to parallel the 4 `basic_string` typedefs.\n\nBoth Google's and LLVM's `string_view` types (but not Bloomberg's) extend the interface from `std::string` to provide some helpful utility functions:\n\nVersions of `std::string` operations that take `string_view` instead also give the standard a way to provide in-place operations on non-null-terminated byte/character sequences:\n\n• hash, as requested by c++std-lib-31935\n• In a future addition, numeric conversions\n\n## Inventions in this paper\n\nGoogle's `StringPiece` provides `as_string` and `ToString` methods to convert to `std::string`. LLVM's `StringRef` provides both a `str()` explicit conversion and an implicit `operator std::string()`. Since this paper builds on top of C++11, we provide an `explicit` conversion constructor as well as a less verbose `to_string` function.\n\nNone of the existing classes have `constexpr` methods.\n\n## Bikeshed!\n\nWe found rough consensus around renaming this class to `string_view`. Other options included:\n\n• basic_string_range (meaning a templated iterator range)\n• char_range\n• ext_string\n• external_string\n• str_ref\n• string_cref (and string_ref for non-const)\n• string_piece\n• string_range\n• string_ref\n• string_view\n• sub_string\n\n## Modifications vs std::string\n\nThe interface of `string_view` is similar to, but not exactly the same as the interface of `std::string`. In general, we want to minimize differences between `std::string` and `string_view` so that users can go back and forth between the two often. This section justifies the differences whose utility we think overcomes that general rule.\n\n• `remove_prefix()` and `remove_suffix()` make it easy to parse strings using `string_view`. They could both be implemented as non-member functions (e.g. ```str.remove_prefix(n) === str = str.substr(n)```), but it seems useful to provide the simplest mutators as member functions. Note that other traversal primitives need to be non-members so that they're extensible, which may argue for pulling these out too.\n\n## Why not change <my-pet-feature>?\n\nI haven't taken every suggestion to change `string_view`. This section explains the rationales.\n\n### Remove the find*() methods\n\nMany people have asked why we aren't removing all of the `find*` methods, since they're widely considered a wart on `std::string`. First, we'd like to make it as easy as possible to convert code to use `string_view`, so it's useful to keep the interface as similar as reasonable to `std::string`. Second, replacing these these methods with uses of the standard algorithms library requires switching from indices to iterators, writing somewhat-complicated conversion code, and/or passing custom lambdas to `find_if`. Let's look at the replacement code for each of the remaining methods:\n\n`haystack.find(needle)`\nReplaced by:\n``````auto iter = std::search(haystack.begin(), haystack.end(),\nneedle.begin(), needle.end());\nreturn iter == haystack.end() ? std::string::npos : iter - haystack.begin();``````\n`haystack.rfind(needle)`\nReplaced by:\n``````auto iter = std::find_end(haystack.begin(), haystack.end(),\nneedle.begin(), needle.end());\nreturn iter == haystack.end() ? std::string::npos : iter - haystack.begin();``````\n`haystack.find_first_of(needles)`\nReplaced by:\n``````auto iter = std::find_first_of(haystack.begin(), haystack.end(),\nneedles.begin(), needles.end());\nreturn iter == haystack.end() ? std::string::npos : iter - haystack.begin();``````\n`haystack.find_last_of(needles)`\nReplaced by:\n``````auto iter = std::find_first_of(haystack.rbegin(), haystack.rend(),\nneedles.begin(), needles.end());\nreturn iter == haystack.rend() ? std::string::npos : iter.base() - 1 - haystack.begin();``````\n`haystack.find_first_not_of(straw)`\nReplaced by:\n``````auto iter = std::find_if(haystack.begin(), haystack.end(), [&](char c) {\nreturn std::find(straw.begin(), straw.end(), c) == straw.end();\n});\nreturn iter == haystack.end() ? std::string::npos : iter - haystack.begin();``````\n`haystack.find_last_not_of(straw)`\nReplaced by:\n``````auto iter = std::find_if(haystack.rbegin(), haystack.rend(), [&](char c) {\nreturn std::find(straw.begin(), straw.end(), c) == straw.end();\n});\nreturn iter == haystack.rend() ? std::string::npos : iter.base() - 1 - haystack.begin();``````\n\n`find`, `rfind`, and `find_first_of` are straightforward, although the conversion from indices to iterators would prevent many users from switching even to them. `find_last_of`, `find_first_not_of`, and `find_last_not_of` get progressively worse to handle even in an iterator-based function.\n\nDiscussion in Bristol concluded that `string_view` should include all of the const signatures from `string`.\n\n SF WF N WA SA 5 4 1 2 0\n SF WF N WA SA 4 2 2 3 2\n\n### Make `basic_string_view<char>` mutable\n\n… and use `basic_string_view<const char>` for the constant case. The constant case is enough more common than the mutable case that it needs to be the default. Making the mutable case the default would prevent passing string literals into `string_view` parameters, which would defeat a significant use case for `string_view`. In a somewhat analogous sitation, LLVM defined an `ArrayRef` class in Feb 2011, and didn't find a need for the matching `MutableArrayRef` until Jan 2012. They still haven't needed a mutable version of `StringRef`. One possible reason for this is that most uses that need to modify a string also need to be able to change its length, and that's impossible through even a mutable version of `string_view`.\n\nWe could use ```typedef basic_string_view<const char> string_view``` to make the immutable case the default while still supporting the mutable case using the same template. I haven't gone this way because it would complicate the template's definition without significantly helping users.\n\n### Add an `explicit operator bool`\n\nThis would be an abbreviation for `!empty()`, usable for initialization in `if` statements. N3509 came to SG9 in Bristol and was not accepted.\n\n SF WF WA SA 0 1 3 5\n\n### Avoid `strlen(\"string literal\")`\n\nWith a constructor of the form:\n\n``````template<size_t N>\nbasic_string_view(const charT (&str)[N]);``````\n\nwe could avoid a `strlen()` call when a `basic_string_view` is constructed from a string literal. Unfortunately, this constructor does completely the wrong thing when called like:\n\n``````char space[PATH_MAX];\nsnprintf(space, sizeof(space), \"some string\");\nstring_view str(space);``````\n\nIt would be possible to avoid that problem by defining a `basic_string_view(char* str)` that uses `strlen()` again, but this adds complexity. Some people have suggested a `string_view::from_literal` method, but I consider that too verbose.\n\nEven the original worry is obsolete given modern optimizers: both gcc and clang optimize `strlen(\"Literal\")` into a constant, making the simple, safe code as efficient as the template. Other implementations should provide the same optimization as a QoI issue.\n\n### Define comparison on `begin`/`end` instead of the elements\n\nOperations on `string_view` apply to the characters in the string, and not the pointers that refer to the characters. This introduces the possibility that the underlying characters might change while a `string_view` referring to them is in an associative container, which would break the container, but we believe this risk is worthwhile because it matches existing practice and matches user intentions more often.\n\n### Wait for `contiguous_range<charT>`\n\n`contiguous_range<T>` along with an `is_contiguous<IteratorOrRange>` trait would be useful for many purposes. However, a reference class that's specifically for strings provides a couple extra benefits:\n\n• `string_view` can have an implicit conversion from `const char*`, while it would be a surprising special case to provide that on `contiguous_range<const char*>`.\n• We can provide a subset of `basic_string`'s interface to ease transitions to and from ownership, while such methods would be very strange on `contiguous_range`.\n• `basic_string_view` takes a `char_traits` argument allowing customization of comparison. `contiguous_range` likely wouldn't.\n• We compare and hash `string_view`s using the elements they refer to. There's a stronger argument to compare a `contiguous_range` using the pointers inside it, meaning two `contiguous_range<char>`s of the same characters might compare unequal.\n• The notion of a \"string\" is different from the notion of a range of characters, which is one reason we have `std::string` in addition to `std::vector<char>`. Users benefit from saying which they mean in interfaces.\n\n### Make `string_view` null-terminated\n\nDoing this naively makes `substr` impossible to implement without a copy. We could imagine inventing a more complex interface that records whether the input string was null-terminated, giving the user the option to use that string when trying to pass a `string_view` to a legacy or C function expecting a null-terminated ```const char*```. This proposal doesn't include such an interface because it would make `string_view` bigger or more expensive, and because there's no existing practice to guide us toward the right interface.\n\nAnother option would be to define a separate `zstring_view` class to represent null-terminated strings and let it decay to `string_view` when necessary. That's plausible but not part of this proposal.\n\n### s/remove_prefix/pop_front/, etc.\n\nIn Kona 2012, I proposed a `range<>` class with `pop_front`, etc. members that adjusted the bounds of the range. Discussion there indicated that committee members were uncomfortable using the same names for lightweight range operations as container operations. Existing practice doesn't agree on a name for this operation, so I've kept the name used by Google's `StringPiece`.\n\n### Allow implicit conversion from more types.\n\nBeman Dawes suggested defining `std::string_view_{begin,end}` and allowing users to add overloads within `std`. Using ADL is a slight variant. We could also allow conversion from any type with `.data()` and `.size()` members returning the right types.\n\nUltimately, I think we want to allow this conversion based on detecting contiguous ranges. Any constructor we add to work around that is going to look like a wart in a couple years. I think we'll be better off making users explicitly convert when they can't add an appropriate conversion operator, and then we can add the optimal constructor when contiguous iterators make it into the library.\n\n SF WF N WA SA 0 0 1 5 6\n\n### `string_view().data() == nullptr`?\n\nThe obvious way to initialize a default constructed `string_view` is with `.data() == nullptr` and `.size() == 0`. However, that violates an invariant of `data()`, that it never returns `nullptr`. Instead, implementations must initialize `data()` with an arbitrary pointer value such that [`data()`,`data()`) is a valid range. (On many platforms, an arbitrary value might suffice, but on some platforms, this might need to be a dereferenceable or one-past-the-end address like `this` or `\"\"`.)\n\nWe could imagine avoiding this complication by allowing `data()` to return `nullptr`, which would also allow us to relax the precondition on ```string_view(const char*, size_t)```. However, in Google's implementation, we found that programmers tended to use `data()==nullptr` to signal conditions that differed from simply `empty()`. This was a source of confusion in interfaces, and it gives `string_view` one more possible value than `const std::string&`, so this proposal forbids the possibility.\n\nEven without relaxing the precondition on ```string_view(const char*, size_t)```, allowing implementations to return a null `data()` from `string_view()` without requiring it, would cause portability problems: some users would test for `data()==nullptr` to identify a default-constructed `string_view` on that implementation, and would then be unable to use another implementation.\n\nHowever, `std::vector::data()` may return `nullptr`, and we don't have `contiguous_iterator_tag` available yet to allow users to pass a `vector`'s `begin()` and `end()` to `string_view()`, so this requirement may require extra conditions in user code. This and strong interest on the std-proposals mailing list encouraged me to add an alternate wording section below.\n\n## Plans for future changes\n\n• There are many functions outside of the strings chapter that should incorporate `string_view`. I'll propose those changes in a subsequent paper based on the part of N3685 that isn't in this paper.\n• We want a literal operator to produce `string_view`, maybe `\"\"sv`.\n\n## Paper revision history\n\n• Explaining why `string_view().data() != nullptr` and adding an alternate wording section in case we decide that should be possible,\n• Fixing up references and namespaces to match the Fundamentals TS,\n• Declaring every function as `constexpr` that could be a constant expression if the `traits` class is suitable, and\n• Adding exposition-only `data_` and `size_` members, and re-explaining several members in terms of them.\n\nN3762 updated N3685 by removing other standard library updates so that the core `string_view` class can be accepted independently. I've also:\n\n• Fixed some bugs in the `pos` and `n` parameters,\n• Added `copy()` back,\n• Removed the assumption that LWG issue 2232 will be fixed,\n• Added a member `swap()` since it's constant time,\n• Ensured that `std::swap` and the range access functions are available when `<string_view>` is included, and\n• Fixed up some `noexcept`s.\n\nN3685 updated N3609 with the results of the LEWG discussion in Bristol. Significant changes include:\n\n• Redirected for a TS instead of C++14\n• Moved into a `<string_view>` header.\n• Added the `pos` and `n` parameters back to `string_view` methods.\n• Removed `starts_with` and `ends_with`.\n\nN3609 was a minor update to N3512 that renamed the proposed class to `basic_string_view` and fixed some wording mistakes.\n\nN3512 updated N3442 with wording for the draft C++14 standard. Note that we still aren't sure whether we're aiming for a TS or C++14.\n\nN3442 was aimed at a TS and updated N3334 by removing `array_ref`.\n\n## Wording for the Fundamentals TS\n\n### Clause x, string_view\n\nThe class template `basic_string_view` describes objects that can refer to a constant contiguous sequence of char-like (C++11[strings.general]) objects with the first element of the sequence at position zero. In the rest of this Clause, the type of the char-like objects held in a `basic_string_view` object is designated by `charT`.\n\n[Note: The library provides implicit conversions from ```const charT*``` and `std::basic_string<charT, ...>` to `std::basic_string_view<charT, ...>` so that user code can accept just `std::basic_string_view<charT>` as a non-templated parameter wherever a sequence of characters is expected. User-defined types should define their own implicit conversions to `std::basic_string_view` in order to interoperate with these functions. — end note ]\n\nThe complexity of member functions is O(1) unless otherwise specified.\n\n``````namespace std {\nnamespace experimental {\ninline namespace fundamentals_v1 {\n// [basic.string.view], basic_string_view:\ntemplate<class charT, class traits = char_traits<charT>>\nclass basic_string_view;\n\n// [string.view.comparison], non-member basic_string_view comparison functions\ntemplate<class charT, class traits>\nconstexpr bool operator==(basic_string_view<charT, traits> x, basic_string_view<charT, traits> y) noexcept;\ntemplate<class charT, class traits>\nconstexpr bool operator!=(basic_string_view<charT, traits> x, basic_string_view<charT, traits> y) noexcept;\ntemplate<class charT, class traits>\nconstexpr bool operator< (basic_string_view<charT, traits> x, basic_string_view<charT, traits> y) noexcept;\ntemplate<class charT, class traits>\nconstexpr bool operator> (basic_string_view<charT, traits> x, basic_string_view<charT, traits> y) noexcept;\ntemplate<class charT, class traits>\nconstexpr bool operator<=(basic_string_view<charT, traits> x, basic_string_view<charT, traits> y) noexcept;\ntemplate<class charT, class traits>\nconstexpr bool operator>=(basic_string_view<charT, traits> x, basic_string_view<charT, traits> y) noexcept;\n\n// [string.view.nonmem], other non-member basic_string_view functions\ntemplate<class charT, class traits = char_traits<charT>,\nclass Allocator = allocator<charT> >\nbasic_string<charT, traits, Allocator> to_string(\nbasic_string_view<charT, traits>,\nconst Allocator& a = Allocator());\n\ntemplate<class charT, class traits>\nbasic_ostream<charT, traits>&\noperator<<(basic_ostream<charT, traits>& os,\nbasic_string_view<charT,traits> str);\n\n// basic_string_view typedef names\ntypedef basic_string_view<char> string_view;\ntypedef basic_string_view<char16_t> u16string_view;\ntypedef basic_string_view<char32_t> u32string_view;\ntypedef basic_string_view<wchar_t> wstring_view;\n\n} // namespace fundamentals_v1\n} // namespace experimental\n\n// [string.view.hash], hash support:\ntemplate <class T> struct hash;\ntemplate <> struct hash<experimental::string_view>;\ntemplate <> struct hash<experimental::u16string_view>;\ntemplate <> struct hash<experimental::u32string_view>;\ntemplate <> struct hash<experimental::wstring_view>;\n} // namespace std``````\n\nThe function templates defined in C++11[utility.swap] and C++11[iterator.range] are available when `<string_view>` is included.\n\nNormally I would update the list in C++11[iterator.range], but we're not yet sure how to do that in a TS, so I picked the more self-contained option.\n\n``````namespace std {\nnamespace experimental {\nnamespace fundamentals_v1 {\ntemplate<class charT, class traits = char_traits<charT>>\nclass basic_string_view {\npublic:\n// types\ntypedef traits traits_type;\ntypedef charT value_type;\ntypedef const charT* pointer;\ntypedef const charT* const_pointer;\ntypedef const charT& reference;\ntypedef const charT& const_reference;\ntypedef implementation-defined const_iterator; // See [string.view.iterators]\ntypedef const_iterator iterator; // [Footnote: Because basic_string_view refers to a constant sequence, iterator and const_iterator are the same type. --end footnote]\ntypedef reverse_iterator<const_iterator> const_reverse_iterator;\ntypedef const_reverse_iterator reverse_iterator;\ntypedef size_t size_type;\ntypedef ptrdiff_t difference_type;\nstatic constexpr size_type npos = size_type(-1);\n\n// [string.view.cons], construct/copy\nconstexpr basic_string_view() noexcept;\nconstexpr basic_string_view(const basic_string_view&) noexcept = default;\nbasic_string_view& operator=(const basic_string_view&) noexcept = default;\ntemplate<class Allocator>\nbasic_string_view(const basic_string<charT, traits, Allocator>& str) noexcept;\nconstexpr basic_string_view(const charT* str);``````\n\nThe above constructor is constexpr to take advantage of user-written traits classes that might provide a constexpr `length()` function. It can't be part of a constant expression with `std::char_traits<char>` without changes to that class.\n\n`` constexpr basic_string_view(const charT* str, size_type len);``\n\nNo initializer_list constructor because C++11[dcl.init.list]p6 says it would likely store a dangling reference into the `basic_string_view`.\n\n`````` // [string.view.iterators], iterators\nconstexpr const_iterator begin() const noexcept;\nconstexpr const_iterator end() const noexcept;\nconstexpr const_iterator cbegin() const noexcept;\nconstexpr const_iterator cend() const noexcept;``````\n\nreverse_iterator methods aren’t constexpr because reverse_iterator isn’t a literal type. See LWG Issue 2208.\n\n`````` const_reverse_iterator rbegin() const noexcept;\nconst_reverse_iterator rend() const noexcept;\nconst_reverse_iterator crbegin() const noexcept;\nconst_reverse_iterator crend() const noexcept;\n\n// [string.view.capacity], capacity\nconstexpr size_type size() const noexcept;\nconstexpr size_type length() const noexcept;\nconstexpr size_type max_size() const noexcept;\nconstexpr bool empty() const noexcept;\n\n// [string.view.access], element access\nconstexpr const charT& operator[](size_type pos) const;\nconstexpr const charT& at(size_type pos) const;\nconstexpr const charT& front() const;\nconstexpr const charT& back() const;\nconstexpr const charT* data() const noexcept;\n\n// [string.view.modifiers], modifiers:\nvoid clear() noexcept;\nvoid remove_prefix(size_type n);\nvoid remove_suffix(size_type n);\nvoid swap(basic_string_view& s) noexcept;\n\n// [string.view.ops], string operations:\ntemplate<class Allocator>\nexplicit operator basic_string<charT, traits, Allocator>() const;\n\nsize_type copy(charT* s, size_type n, size_type pos = 0) const;\n\nconstexpr basic_string_view substr(size_type pos=0, size_type n=npos) const;\nconstexpr int compare(basic_string_view s) const noexcept;\nconstexpr int compare(size_type pos1, size_type n1, basic_string_view s) const;\nconstexpr int compare(size_type pos1, size_type n1,\nbasic_string_view s, size_type pos2, size_type n2) const;\nconstexpr int compare(const charT* s) const;\nconstexpr int compare(size_type pos1, size_type n1, const charT* s) const;\nconstexpr int compare(size_type pos1, size_type n1,\nconst charT* s, size_type n2) const;\nconstexpr size_type find(basic_string_view s, size_type pos=0) const noexcept;\nconstexpr size_type find(charT c, size_type pos=0) const noexcept;\nconstexpr size_type find(const charT* s, size_type pos, size_type n) const;\nconstexpr size_type find(const charT* s, size_type pos=0) const;\nconstexpr size_type rfind(basic_string_view s, size_type pos=npos) const noexcept;\nconstexpr size_type rfind(charT c, size_type pos=npos) const noexcept;\nconstexpr size_type rfind(const charT* s, size_type pos, size_type n) const;\nconstexpr size_type rfind(const charT* s, size_type pos=npos) const;\nconstexpr size_type find_first_of(basic_string_view s, size_type pos=0) const noexcept;\nconstexpr size_type find_first_of(charT c, size_type pos=0) const noexcept;\nconstexpr size_type find_first_of(const charT* s, size_type pos, size_type n) const;\nconstexpr size_type find_first_of(const charT* s, size_type pos=0) const;\nconstexpr size_type find_last_of(basic_string_view s, size_type pos=npos) const noexcept;\nconstexpr size_type find_last_of(charT c, size_type pos=npos) const noexcept;\nconstexpr size_type find_last_of(const charT* s, size_type pos, size_type n) const;\nconstexpr size_type find_last_of(const charT* s, size_type pos=npos) const;\nconstexpr size_type find_first_not_of(basic_string_view s, size_type pos=0) const noexcept;\nconstexpr size_type find_first_not_of(charT c, size_type pos=0) const noexcept;\nconstexpr size_type find_first_not_of(const charT* s, size_type pos, size_type n) const;\nconstexpr size_type find_first_not_of(const charT* s, size_type pos=0) const;\nconstexpr size_type find_last_not_of(basic_string_view s, size_type pos=npos) const noexcept;\nconstexpr size_type find_last_not_of(charT c, size_type pos=npos) const noexcept;\nconstexpr size_type find_last_not_of(const charT* s, size_type pos, size_type n) const;\nconstexpr size_type find_last_not_of(const charT* s, size_type pos=npos) const;\n\nprivate:\nconst charT* data_; // exposition only\nsize_type size_; // exposition only\n};\n} // namespace fundamentals_v1\n} // namespace experimental\n} // namespace std``````\n\nIn every specialization `basic_string_view<charT, traits>`, the type `traits` shall satisfy the character traits requirements (C++11[char.traits]), and the type `traits::char_type` shall name the same type as `charT`.\n\n#### Add a subclause \"x.1 basic_string_view constructors and assignment operators [string.view.cons]\"\n\n``constexpr basic_string_view() noexcept;``\n\nEffects: Constructs an empty `basic_string_view`.\n\nPostcondition: `size_ == 0`, and `data_` has an unspecified, non-null value such that [`data_`,`data_`) is a valid range.\n\n``````template<class Allocator>\nbasic_string_view(const basic_string<charT, traits, Allocator>& str) noexcept;``````\n\nEffects: Constructs a `basic_string_view`, with the postconditions in Table [tab:string.view.ctr.1]\n\nRemarks: The program shall not alter any of the values stored in the character array. [Footnote: This is the same requirement as on `str.data()` -- end footnote]\n\nTable [tab:string.view.ctr.1] — basic_string_view(const basic_string&) effects\nElementValue\n`data_``str.data()`\n`size_``str.size()`\n``basic_string_view(const charT* str);``\n\nRequires: [`str`,`str + traits::length(str)`) is a valid range.\n\nEffects: Constructs a `basic_string_view` referring to the same string as `str`, with the postconditions in Table [tab:string.view.ctr.2]\n\nTable [tab:string.view.ctr.2] — basic_string_view(const charT*) effects\nElementValue\n`data_``str`\n`size_``traits::length(str)`\n\nComplexity: O(`traits::length(str)`)\n\n``constexpr basic_string_view(const charT* str, size_type len);``\n\nRequires: `str` is not a null pointer and [`str`,`str + len`) is a valid range.\n\nEffects: Constructs a `basic_string_view`, with the postconditions in Table [tab:string.view.ctr.3]\n\nTable [tab:string.view.ctr.3] — basic_string_view(const charT*, size_type) effects\nElementValue\n`data_``str`\n`size_``len`\n\n#### Add a subclause \"x.2 basic_string_view iterator support [string.view.iterators]\"\n\n``typedef implementation-defined const_iterator;``\n\nA constant random-access iterator type such that, for a `const_iterator it`, if `&*(it+N)` is valid, then it is equal to `(&*it)+N`.\n\nFor a `basic_string_view str`, any operation that invalidates a pointer in the range [`str.data()`, `str.data()+str.size()`) invalidates pointers, iterators, and references returned from `str`'s methods.\n\nAll requirements on container iterators (C++11[container.requirements]) apply to `basic_string_view::const_iterator` as well.\n\n``````constexpr const_iterator begin() const noexcept;\nconstexpr const_iterator cbegin() const noexcept;``````\n\nReturns: An iterator such that `&*begin() == data_` if `*data_` is a valid expression and `!empty()`, or else an unspecified value such that [`begin()`,`end()`) is a valid range.\n\n``````constexpr const_iterator end() const noexcept;\nconstexpr const_iterator cend() const noexcept;``````\n\nReturns: `begin() + size()`\n\n``````const_reverse_iterator rbegin() const noexcept;\nconst_reverse_iterator crbegin() const noexcept;``````\n\nReturns: An iterator which is semantically equivalent to `reverse_iterator(end())`.\n\n``````const_reverse_iterator rend() const noexcept;\nconst_reverse_iterator crend() const noexcept;``````\n\nReturns: An iterator which is semantically equivalent to `reverse_iterator(begin())`.\n\n#### Add a subclause \"x.3 basic_string_view capacity [string.view.capacity]\"\n\n``constexpr size_type size() const noexcept;``\n\nReturns: `size_`\n\n``constexpr size_type length() const noexcept;``\n\nReturns: `size_`.\n\n``constexpr size_type max_size() const noexcept;``\n\nReturns: The largest possible number of char-like objects that can be referred to by a `basic_string_view`.\n\n``constexpr bool empty() const noexcept;``\n\nReturns: `size_ == 0`.\n\n#### Add a subclause \"x.4 basic_string_view element access [string.view.access]\"\n\n``constexpr const_reference operator[](size_type pos) const;``\n\nRequires: `pos < size()`.\n\nReturns: `data_[pos]`\n\nThrows: Nothing.\n\n[ Note: Unlike `basic_string::operator[]`, `basic_string_view::operator[](size())` has undefined behavior instead of returning `charT()`. — end note ]\n\n``constexpr const_reference at(size_type pos) const;``\n\nThrows: `out_of_range` if `pos >= size()`.\n\nReturns: `operator[](pos)`.\n\n``constexpr const charT& front() const;``\n\nRequires: `!empty()`\n\nEffects: Equivalent to `return operator[](0)`.\n\n``constexpr const charT& back() const;``\n\nRequires: `!empty()`\n\nEffects: Equivalent to `return operator[](size() - 1)`.\n\n``constexpr const charT* data() const noexcept;``\n\nReturns: `data_`\n\n[ Note: Unlike `std::string::data()` and string literals, `data()` may return a pointer to a buffer that is not null-terminated. Therefore it is typically a mistake to pass `data()` to a routine that takes just a `const charT*` and expects a null-terminated string. — end note ]\n\n#### Add a subclause \"x.5 basic_string_view modifiers [string.view.modifiers]\"\n\n``void clear() noexcept;``\n\nEffects: Equivalent to `*this = basic_string_view()`\n\n``void remove_prefix(size_type n);``\n\nRequires: `n <= size()`\n\nEffects: Equivalent to `data_ += n; size_ -= n;`\n\n``void remove_suffix(size_type n);``\n\nRequires: `n <= size()`\n\nEffects: Equivalent to `size_ -= n;`\n\n``void swap(basic_string_view& s) noexcept``\n\nEffects: Exchanges the values of `*this` and `s`.\n\n#### Add a subclause \"x.6 basic_string_view string operations [string.view.ops]\"\n\n``````template<class Allocator>\nexplicit // Footnote: This conversion is explicit to avoid accidental O(N) operations on type mismatches. --end footnote\noperator basic_string<charT, traits, Allocator>() const;``````\n\nEffects: Equivalent to `return basic_string<charT, traits, Allocator>(str.begin(), str.end()).`\n\n[ Note: Users who want to control the allocator instance should call `basic_string(str.begin(), str.end(), allocator)` directly. -- end note ]\n\n``size_type copy(charT* s, size_type n, size_type pos = 0) const;``\n\nThrows: `out_of_range` if `pos > size()`.\n\nRemarks: Let `rlen` be the smaller of `n` and `size() - pos`.\n\nRequires: [`s`, `s+rlen`) is a valid range.\n\nEffects: Equivalent to `std::copy_n(begin() + pos, rlen, s).`\n\nReturns: `rlen`.\n\n``constexpr basic_string_view substr(size_type pos = 0, size_type n = npos) const;``\n\nThrows: `out_of_range` if `pos > size()`.\n\nEffects: Determines the effective length `rlen` of the string to reference as the smaller of `n` and `size() - pos`.\n\nReturns: `basic_string_view(data()+pos, rlen)`.\n\n``constexpr int compare(basic_string_view str) const noexcept;``\n\nEffects: Determines the effective length `rlen` of the strings to compare as the smallest of `size()` and `str.size()`. The function then compares the two strings by calling `traits::compare(data(), str.data(), rlen)`.\n\nComplexity: O(`rlen`)\n\nReturns: The nonzero result if the result of the comparison is nonzero. Otherwise, returns a value as indicated in Table [tab:string.view.compare].\n\nTable [tab:string.view.compare] — compare() results\nConditionReturn Value\n`size() < str.size()``< 0`\n`size() == str.size()``0`\n`size() > str.size()``> 0`\n``constexpr int compare(size_type pos1, size_type n1, basic_string_view str) const;``\n\nEffects: Equivalent to `return substr(pos1, n1).compare(str)`.\n\n``````constexpr int compare(size_type pos1, size_type n1, basic_string_view str,\nsize_type pos2, size_type n2) const;``````\n\nEffects: Equivalent to `return substr(pos1, n1).compare(str.substr(pos2, n2))`.\n\n``constexpr int compare(const charT* s) const;``\n\nEffects: Equivalent to `return compare(basic_string_view(s))`.\n\n``constexpr int compare(size_type pos1, size_type n1, const charT* s) const;``\n\nEffects: Equivalent to `return substr(pos1, n1).compare(basic_string_view(s))`.\n\n``````constexpr int compare(size_type pos1, size_type n1,\nconst charT* s, size_type n2) const;``````\n\nEffects: Equivalent to `return substr(pos1, n1).compare(basic_string_view(s, n2))`.\n\n##### Add a sub-subclause \"x.6.1 Searching basic_string_view [string.view.find]\"\n\nMember functions in this section have complexity O(`size() * argument.size()`) at worst, although implementations are encouraged to do better.\n\nEach member function of the form\n\n``constexpr return-type fx1(const charT* s, size_type pos);``\n\nis equivalent to `return fx1(basic_string_view(s), pos)`.\n\nEach member function of the form\n\n``constexpr return-type fx1(const charT* s, size_type pos, size_type n); // find() variants``\n\nis equivalent to `return fx1(basic_string_view(s, n), pos)`.\n\nEach member function of the form\n\n``constexpr return-type fx2(charT c, size_type pos); // find() variants``\n\nis equivalent to `return fx2(basic_string_view(&c, 1), pos)`.\n\n``constexpr size_type find(basic_string_view str, size_type pos=0) const noexcept;``\n\nEffects: Determines the lowest position `xpos`, if possible, such that the following conditions obtain:\n\n• `pos <= xpos`\n• `xpos + str.size() <= size()`\n• `traits::eq(at(xpos+I), str.at(I))` for all elements `I` of the string referenced by `str`.\n\nReturns: `xpos` if the function can determine such a value for `xpos`. Otherwise, returns `npos`.\n\nRemarks: Uses `traits::eq()`.\n\n``constexpr size_type rfind(basic_string_view str, size_type pos=npos) const noexcept;``\n\nEffects: Determines the highest position `xpos`, if possible, such that the following conditions obtain:\n\n• `xpos <= pos`\n• `xpos + str.size() <= size()`\n• `traits::eq(at(xpos+I), str.at(I))` for all elements `I` of the string referenced by `str`.\n\nReturns: `xpos` if the function can determine such a value for `xpos`. Otherwise, returns `npos`.\n\nRemarks: Uses `traits::eq()`.\n\n``constexpr size_type find_first_of(basic_string_view str, size_type pos=0) const noexcept;``\n\nEffects: Determines the lowest position `xpos`, if possible, such that the following conditions obtain:\n\n• `pos <= xpos`\n• `xpos < size()`\n• `traits::eq(at(xpos), str.at(I))` for some element `I` of the string referenced by `str`.\n\nReturns: `xpos` if the function can determine such a value for `xpos`. Otherwise, returns `npos`.\n\nRemarks: Uses `traits::eq()`.\n\n``constexpr size_type find_last_of(basic_string_view str, size_type pos=npos) const noexcept;``\n\nEffects: Determines the highest position `xpos`, if possible, such that the following conditions obtain:\n\n• `xpos <= pos`\n• `xpos < size()`\n• `traits::eq(at(xpos), str.at(I))` for some element `I` of the string referenced by `str`.\n\nReturns: `xpos` if the function can determine such a value for `xpos`. Otherwise, returns `npos`.\n\nRemarks: Uses `traits::eq()`.\n\n``constexpr size_type find_first_not_of(basic_string_view str, size_type pos=0) const noexcept;``\n\nEffects: Determines the lowest position `xpos`, if possible, such that the following conditions obtain:\n\n• `pos <= xpos`\n• `xpos < size()`\n• `traits::eq(at(xpos), str.at(I))` for no element `I` of the string referenced by `str`.\n\nReturns: `xpos` if the function can determine such a value for `xpos`. Otherwise, returns `npos`.\n\nRemarks: Uses `traits::eq()`.\n\n``constexpr size_type find_last_not_of(basic_string_view str, size_type pos=npos) const noexcept;``\n\nEffects: Determines the highest position `xpos`, if possible, such that the following conditions obtain:\n\n• `xpos <= pos`\n• `xpos < size()`\n• `traits::eq(at(xpos), str.at(I))` for no element `I` of the string referenced by `str`.\n\nReturns: `xpos` if the function can determine such a value for `xpos`. Otherwise, returns `npos`.\n\nRemarks: Uses `traits::eq()`.\n\n#### Add a subclause \"x.7 basic_string_view non-member comparison functions [string.view.comparison]\"\n\nImplementations shall provide sufficient additional overloads so that an object `t` with an implicit conversion to `basic_string_view<charT, traits>` can be compared according to Table [tab:string.view.comparison.overloads], where `sp` is an instance of `basic_string_view<charT, traits>`.\n\nExpressionEquivalent to\n`t == sp``basic_string_view<charT, traits>(t) == sp`\n`sp == t``sp == basic_string_view<charT, traits>(t)`\n`t != sp``basic_string_view<charT, traits>(t) != sp`\n`sp != t``sp != basic_string_view<charT, traits>(t)`\n`t < sp``basic_string_view<charT, traits>(t) < sp`\n`sp < t``sp < basic_string_view<charT, traits>(t)`\n`t > sp``basic_string_view<charT, traits>(t) > sp`\n`sp > t``sp > basic_string_view<charT, traits>(t)`\n`t <= sp``basic_string_view<charT, traits>(t) <= sp`\n`sp <= t``sp <= basic_string_view<charT, traits>(t)`\n`t >= sp``basic_string_view<charT, traits>(t) >= sp`\n`sp >= t``sp >= basic_string_view<charT, traits>(t)`\n\n[ Example: A sample conforming implementation for operator== would be:\n\n`````` template<class T> struct __identity { typedef T type; };\ntemplate<class charT, class traits>\nconstexpr bool operator==(\nbasic_string_view<charT, traits> lhs,\nbasic_string_view<charT, traits> rhs) noexcept {\nreturn lhs.compare(rhs) == 0;\n}\ntemplate<class charT, class traits>\nconstexpr bool operator==(\nbasic_string_view<charT, traits> lhs,\ntypename __identity<basic_string_view<charT, traits>>::type rhs) noexcept {\nreturn lhs.compare(rhs) == 0;\n}\ntemplate<class charT, class traits>\nconstexpr bool operator==(\ntypename __identity<basic_string_view<charT, traits>>::type lhs,\nbasic_string_view<charT, traits> rhs) noexcept {\nreturn lhs.compare(rhs) == 0;\n}``````\n\n— end example ]\n\n``````template<class charT, class traits>\nconstexpr bool operator==(basic_string_view<charT,traits> lhs,\nbasic_string_view<charT,traits> rhs) noexcept;``````\n\nReturns: `lhs.compare(rhs) == 0`.\n\n``````template<class charT, class traits>\nconstexpr bool operator!=(basic_string_view<charT,traits> lhs,\nbasic_string_view<charT,traits> rhs) noexcept;``````\n\nReturns: `!(lhs == rhs)`.\n\n``````template<class charT, class traits>\nconstexpr bool operator< (basic_string_view<charT,traits> lhs,\nbasic_string_view<charT,traits> rhs) noexcept;``````\n\nReturns: `lhs.compare(rhs) < 0`.\n\n``````template<class charT, class traits>\nconstexpr bool operator> (basic_string_view<charT,traits> lhs,\nbasic_string_view<charT,traits> rhs) noexcept;``````\n\nReturns: `lhs.compare(rhs) > 0`.\n\n``````template<class charT, class traits>\nconstexpr bool operator<=(basic_string_view<charT,traits> lhs,\nbasic_string_view<charT,traits> rhs) noexcept;``````\n\nReturns: `lhs.compare(rhs) <= 0`.\n\n``````template<class charT, class traits>\nconstexpr bool operator>=(basic_string_view<charT,traits> lhs,\nbasic_string_view<charT,traits> rhs) noexcept;``````\n\nReturns: `lhs.compare(rhs) >= 0`.\n\n#### Add a subclause \"x.8 Other basic_string_view non-member functions [string.view.nonmem]\"\n\n``````template<class charT, class traits = char_traits<charT>,\nclass Allocator = allocator<charT> >\nbasic_string<charT, traits, Allocator> to_string(\nbasic_string_view<charT, traits> str,\nconst Allocator& a = Allocator());``````\n\nComplexity: O(`str.size()`)\n\nReturns: `basic_string<charT, traits, Allocator>(str.begin(), str.end(), a)`.\n\n#### Add a subclause \"x.9 Inserters and extractors [string.view.io]\n\n``````template<class charT, class traits>\nbasic_ostream<charT, traits>&\noperator<<(basic_ostream<charT, traits>& os,\nbasic_string_view<charT,traits> str);``````\n\nEffects: Behaves as a formatted output function (C++11[ostream.formatted.reqmts]) of `os`. Forms a character sequence `seq`, initially consisting of the elements defined by the range `[str.begin(), str.end())`. Determines padding for `seq` as described in C++11[ostream.formatted.reqmts]. Then inserts `seq` as if by calling `os.rdbuf()->sputn(seq, n)`, where `n` is the larger of `os.width()` and `str.size()`; then calls `os.width(0)`.\n\nReturns: os\n\n#### Add a subclause \"x.10 Hash support [string.view.hash]\"\n\n``````template <> struct hash<experimental::string_view>;\ntemplate <> struct hash<experimental::u16string_view>;\ntemplate <> struct hash<experimental::u32string_view>;\ntemplate <> struct hash<experimental::wstring_view>;``````\n\nRequires: the template specializations shall meet the requirements of class template hash (C++11[unord.hash]).\n\n## Alternate wording if `data()` may be `nullptr`\n\nThis section lists changes to the proposed wording if the LWG prefers to allow \"null\" `string_view`s.\n\n### In subclause \"x.1 basic_string_view constructors and assignment operators [string.view.cons]\"\n\n``constexpr basic_string_view() noexcept;``\n\nEffects: Constructs an empty `basic_string_view`.\n\nPostcondition: `size_ == 0`, and `data_ == nullptr`.`data_` has an unspecified, non-null value such that [`data_`,`data_`) is a valid range.\n\nAlternately, we could leave `data_` unspecified-but-valid, but I worry about portability in that case.\n\n``constexpr basic_string_view(const charT* str, size_type len);``\n\nRequires: `str` is not a null pointer and [`str`,`str + len`) is a valid range.\n\nEffects: Constructs a `basic_string_view`, with the postconditions in Table [tab:string.view.ctr.3]\n\nTable [tab:string.view.ctr.3] — basic_string_view(const charT*, size_type) effects\nElementValue\n`data_``str`\n`size_``len`\n\n### In subclause \"x.4 basic_string_view element access [string.view.access]\"\n\n``constexpr const charT* data() const noexcept;``\n\nReturns: `data_`\n\n[ Note: Unlike `std::string::data()` and string literals, `data()` may return a pointer to a buffer that is not null-terminated. Therefore it is typically a mistake to pass `data()` to a routine that takes just a `const charT*` and expects a null-terminated string. — end note ]\n\nWe could specify that if `size_==0` then `data()` simply returns a valid pointer value, and not necessarily `data_`. This would let debugging implementations prevent developers from depending on a passed-through null value, but would prevent the `std::split()` proposal's Delimiters from using `string_view` to represent a 0-length position within a string at which to split.\n\n## Acknowledgements\n\nI'd like to thank Marshall Clow, Olaf van der Spek, the Boost and std-proposals mailing lists, Chandler Carruth, Beman Dawes, Alisdair Meredith, and especially Daniel Krügler for help, advice, and wording in this paper." ]

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[ "# QuantileNPCI\n\nThe goal of QuantileNPCI is to calculate non-parametric confidence intervals for quantiles using fractional order statistics.\n\n## Installation\n\nYou can install the released version of QuantileNPCI from CRAN with:\n\n``install.packages(\"QuantileNPCI\")``\n\n## Example\n\nThis is a basic example which shows you how to calculate non-parametric confidence intervals for median, for the flood discharge data in Feather River with data included in the package:\n\n``````library(QuantileNPCI)\nquantCI(flood[flood\\$loc==\"Feather\", \"discharge\"], quant=0.5, alpha=0.05, method = \"exact\")``````" ]

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[ " 利用矩阵解二元一次方程组3x+y=24x+2y=3-查字典问答网\n导航\n手机", null, "利用矩阵解二元一次方程组3x+y=24x+2y=3\n\n3x+y=2 　　4x+2y=3 　　(31 　　42)* 　　(x 　　y) 　　=(2 　　3) 　　所以 　　(x 　　y) 　　=(31 　　42)的逆 　　×（2,3）T 　　=1/2*(2-1 　　-43)*(2,3)T 　　=1/2*(1,1)T 　　所以 　　x=1/2 　　y=1/2", null, "2020-12-30 07:43:53", null, "点赞0", null, "" ]

[ null, "https://my.chazidian.com/api.htm", null, "https://my.chazidian.com/api.htm", null, "https://wenda.chazidian.com/static/home/images/zan.png", null, "https://wenda.chazidian.com/static/home/images/answer.png", null ]

[ "If ${\\mathrm{l}}_{1}$${\\mathrm{l}}_{2}$${\\mathrm{l}}_{3}$ are the lengths of the emitter, base and collector of a transistor then\n(a)${\\mathrm{l}}_{1}$${\\mathrm{l}}_{2}$ = ${\\mathrm{l}}_{3}$                   (b) ${\\mathrm{l}}_{3}$ <${\\mathrm{l}}_{2}$${\\mathrm{l}}_{1}$\n(c) ${\\mathrm{l}}_{3}$${\\mathrm{l}}_{1}$ < ${\\mathrm{l}}_{2}$                 (d) ${\\mathrm{l}}_{3}$${\\mathrm{l}}_{1}$>${\\mathrm{l}}_{2}$\n\n(d)\n\nDifficulty Level:\n\n• 11%\n• 11%\n• 15%\n• 65%" ]

[ null ]

[ "", null, "", null, "# HEX\n\n###### Database: MySQL\n\nTopic\n\nSyntax:\nHEX(str), HEX(N)\n\nFor a string argument str, HEX() returns a hexadecimal string\nrepresentation of str where each byte of each character in str is\nconverted to two hexadecimal digits. (Multibyte characters therefore\nbecome more than two digits.) The inverse of this operation is\nperformed by the UNHEX() function.\n\nFor a numeric argument N, HEX() returns a hexadecimal string\nrepresentation of the value of N treated as a longlong (BIGINT) number.\nThis is equivalent to CONV(N,10,16). The inverse of this operation is\nperformed by CONV(HEX(N),16,10).\n\nURL: https://dev.mysql.com/doc/refman/8.0/en/string-functions.html\n\nExample\n\nmysql> SELECT X'616263', HEX('abc'), UNHEX(HEX('abc'));\n-> 'abc', 616263, 'abc'\nmysql> SELECT HEX(255), CONV(HEX(255),16,10);\n-> 'FF', 255" ]

[ null, "https://counter.rambler.ru/top100.cnt", null, "https://static.wixstatic.com/media/624d7cc55c6a797d9b3fa76e5fb328b5.png/v1/fill/w_1440,h_1440,al_c,q_95,enc_auto/624d7cc55c6a797d9b3fa76e5fb328b5.png", null ]

[ "", null, "", null, "", null, "", null, "", null, "", null, "", null, "Re: Hold and Equal\n\n• To: mathgroup at smc.vnet.net\n• Subject: [mg73739] Re: [mg73715] Hold and Equal\n• From: Carl Woll <carlw at wolfram.com>\n• Date: Tue, 27 Feb 2007 05:44:35 -0500 (EST)\n• References: <[email protected]>\n\n```Murray Eisenberg wrote:\n\n>How can I produce in an Output cell (under program control) an\n>expression like the following,\n>\n> (a+b)^2 = a^2+ 2 a b + b^2\n>\n>where instead of the usual Equal (==) I get a Set (=), as in traditional\n>math notation? I want to input the unexpanded (a+b)^2 and have the\n>expansion done automatically.\n>\n>Of course, I can try something like the following:\n>\n> (a+b)^2 == Expand[(a+b)^2])\n>\n>So how do I convert the == to =? Of course\n>\n> ((a + b)^2 == Expand[(a + b)^2]) /. Equal -> Set\n>\n>gives a Set::write error. And\n>\n> (Hold[(a + b)^2 == Expand[(a + b)^2]]) /. Equal -> Set\n>\n>doesn't actually evaluate the Expand part and leaves the \"Hold\" wrapper.\n>\n>\nMurray,\n\nstep[x_] := HoldForm[x = #] &[Expand[x]]\n\nstep[(a+b)^2]\n(a+b)^2=a^2+2 b a+b^2\n\nCarl Woll\nWolfram Research\n\n```\n\n• References:\n• Prev by Date: Re: Hold and Equal\n• Next by Date: Re: \"movie\" of Table output for web page\n• Previous by thread: Hold and Equal\n• Next by thread: Re: Re: Hold and Equal" ]

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[ "# hoomd.update¶\n\nOverview\n\n hoomd.update.balance Adjusts the boundaries of a domain decomposition on a regular 3D grid. hoomd.update.box_resize Rescale the system box size. hoomd.update.sort Sorts particles in memory to improve cache coherency.\n\nDetails\n\nModify the system state periodically.\n\nWhen an updater is specified, it acts on the particle system every period steps to change it in some way. See the documentation of specific updaters to find out what they do.\n\nclass hoomd.update.balance(x=True, y=True, z=True, tolerance=1.02, maxiter=1, period=1000, phase=0)\n\nAdjusts the boundaries of a domain decomposition on a regular 3D grid.\n\nParameters: x (bool) – If True, balance in x dimension. y (bool) – If True, balance in y dimension. z (bool) – If True, balance in z dimension. tolerance (float) – Load imbalance tolerance (if <= 1.0, balance every step). maxiter (int) – Maximum number of iterations to attempt in a single step. period (int) – Balancing will be attempted every a period time steps phase (int) – When -1, start on the current time step. When >= 0, execute on steps where (step + phase) % period == 0.\n\nEvery period steps, the boundaries of the processor domains are adjusted to distribute the particle load close to evenly between them. The load imbalance is defined as the number of particles owned by a rank divided by the average number of particles per rank if the particles had a uniform distribution:\n\n$I = \\frac{N(i)}{N / P}$\n\nwhere :math: N(i) is the number of particles on processor $$i$$, $$N$$ is the total number of particles, and $$P$$ is the number of ranks.\n\nIn order to adjust the load imbalance, the sizes are rescaled by the inverse of the imbalance factor. To reduce oscillations and communication overhead, a domain cannot move more than 5% of its current size in a single rebalancing step, and the edge of a domain cannot move more than half the distance to its neighbors.\n\nSimulations with interfaces (so that there is a particle density gradient) or clustering should benefit from load balancing. The potential speedup is roughly $$I-1.0$$, so that if the largest imbalance is 1.4, then the user can expect a roughly 40% speedup in the simulation. This is of course an estimate that assumes that all algorithms are roughly linear in $$N$$, all GPUs are fully occupied, and the simulation is limited by the speed of the slowest processor. It also assumes that all particles roughly equal. If you have a simulation where, for example, some particles have significantly more pair force neighbors than others, this estimate of the load imbalance may not produce the optimal results.\n\nA load balancing adjustment is only performed when the maximum load imbalance exceeds a tolerance. The ideal load balance is 1.0, so setting tolerance less than 1.0 will force an adjustment every period. The load balancer can attempt multiple iterations of balancing every period, and up to maxiter attempts can be made. The optimal values of period and maxiter will depend on your simulation.\n\nLoad balancing can be performed independently and sequentially for each dimension of the simulation box. A small performance increase may be obtained by disabling load balancing along dimensions that are known to be homogeneous. For example, if there is a planar vapor-liquid interface normal to the $$z$$ axis, then it may be advantageous to disable balancing along $$x$$ and $$y$$.\n\nIn systems that are well-behaved, there is minimal overhead of balancing with a small period. However, if the system is not capable of being balanced (for example, due to the density distribution or minimum domain size), having a small period and high maxiter may lead to a large performance loss. In such systems, it is currently best to either balance infrequently or to balance once in a short test run and then set the decomposition statically in a separate initialization.\n\nBalancing is ignored if there is no domain decomposition available (MPI is not built or is running on a single rank).\n\ndisable()\n\nDisables the updater.\n\nExamples:\n\nupdater.disable()\n\n\nExecuting the disable command will remove the updater from the system. Any hoomd.run() command executed after disabling an updater will not use that updater during the simulation. A disabled updater can be re-enabled with enable()\n\nenable()\n\nEnables the updater.\n\nExamples:\n\nupdater.enable()\n\nrestore_state()\n\nRestore the state information from the file used to initialize the simulations\n\nset_params(x=None, y=None, z=None, tolerance=None, maxiter=None)\n\nChange load balancing parameters.\n\nParameters: x (bool) – If True, balance in x dimension. y (bool) – If True, balance in y dimension. z (bool) – If True, balance in z dimension. tolerance (float) – Load imbalance tolerance (if <= 1.0, balance every step). maxiter (int) – Maximum number of iterations to attempt in a single step.\n\nExamples:\n\nbalance.set_params(x=True, y=False)\nbalance.set_params(tolerance=0.02, maxiter=5)\n\nset_period(period)\n\nChanges the updater period.\n\nParameters: period (int) – New period to set.\n\nExamples:\n\nupdater.set_period(100);\nupdater.set_period(1);\n\n\nWhile the simulation is running, the action of each updater is executed every period time steps. Changing the period does not change the phase set when the analyzer was first created.\n\nclass hoomd.update.box_resize(Lx=None, Ly=None, Lz=None, xy=None, xz=None, yz=None, period=1, L=None, phase=0, scale_particles=True)\n\nRescale the system box size.\n\nParameters: L (hoomd.variant) – (if set) box length in the x,y, and z directions as a function of time (in distance units) Lx (hoomd.variant) – (if set) box length in the x direction as a function of time (in distance units) Ly (hoomd.variant) – (if set) box length in the y direction as a function of time (in distance units) Lz (hoomd.variant) – (if set) box length in the z direction as a function of time (in distance units) xy (hoomd.variant) – (if set) X-Y tilt factor as a function of time (dimensionless) xz (hoomd.variant) – (if set) X-Z tilt factor as a function of time (dimensionless) yz (hoomd.variant) – (if set) Y-Z tilt factor as a function of time (dimensionless) period (int) – The box size will be updated every period time steps. phase (int) – When -1, start on the current time step. When >= 0, execute on steps where (step + phase) % period == 0. scale_particles (bool) – When True (the default), scale particles into the new box. When False, do not change particle positions when changing the box.\n\nEvery period time steps, the system box dimensions is updated to values given by the user (in a variant). As an option, the particles can either be left in place as the box is changed or their positions can be scaled with the box.\n\nNote\n\nIf period is set to None, then the given box lengths are applied immediately and periodic updates are not performed.\n\nL, Lx, Ly, Lz, xy, xz, yz can either be set to a constant number or a hoomd.variant. if any of the box parameters are not specified, they are set to maintain the same value in the current box.\n\nUse L as a shorthand to specify Lx, Ly, and Lz to the same value.\n\nBy default, particle positions are rescaled with the box. Set scale_particles=False to leave particles in place when changing the box.\n\nIf, under rescaling, tilt factors get too large, the simulation may slow down due to too many ghost atoms being communicated. hoomd.update.box_resize does NOT reset the box to orthorhombic shape if this occurs (and does not move the next periodic image into the primary cell).\n\nExamples:\n\nupdate.box_resize(L = hoomd.variant.linear_interp([(0, 20), (1e6, 50)]))\nbox_resize = update.box_resize(L = hoomd.variant.linear_interp([(0, 20), (1e6, 50)]), period = 10)\nupdate.box_resize(Lx = hoomd.variant.linear_interp([(0, 20), (1e6, 50)]),\nLy = hoomd.variant.linear_interp([(0, 20), (1e6, 60)]),\nLz = hoomd.variant.linear_interp([(0, 10), (1e6, 80)]))\nupdate.box_resize(Lx = hoomd.variant.linear_interp([(0, 20), (1e6, 50)]), Ly = 10, Lz = 10)\n\n# Shear the box in the xy plane using Lees-Edwards boundary conditions\nupdate.box_resize(xy = hoomd.variant.linear_interp([(0,0), (1e6, 1)]))\n\ndisable()\n\nDisables the updater.\n\nExamples:\n\nupdater.disable()\n\n\nExecuting the disable command will remove the updater from the system. Any hoomd.run() command executed after disabling an updater will not use that updater during the simulation. A disabled updater can be re-enabled with enable()\n\nenable()\n\nEnables the updater.\n\nExamples:\n\nupdater.enable()\n\nrestore_state()\n\nRestore the state information from the file used to initialize the simulations\n\nset_period(period)\n\nChanges the updater period.\n\nParameters: period (int) – New period to set.\n\nExamples:\n\nupdater.set_period(100);\nupdater.set_period(1);\n\n\nWhile the simulation is running, the action of each updater is executed every period time steps. Changing the period does not change the phase set when the analyzer was first created.\n\nclass hoomd.update.sort\n\nSorts particles in memory to improve cache coherency.\n\nWarning\n\nDo not specify hoomd.update.sort explicitly in your script. HOOMD creates a sorter by default.\n\nEvery period time steps, particles are reordered in memory based on a Hilbert curve. This operation is very efficient, and the reordered particles significantly improve performance of all other algorithmic steps in HOOMD.\n\nThe reordering is accomplished by placing particles in spatial bins. A Hilbert curve is generated that traverses these bins and particles are reordered in memory in the same order in which they fall on the curve. The grid dimension used over the course of the simulation is held constant, and the default is chosen to be as fine as possible without utilizing too much memory. The grid size can be changed with set_params().\n\nWarning\n\nMemory usage by the sorter grows quickly with the grid size:\n\n• grid=128 uses 8 MB\n• grid=256 uses 64 MB\n• grid=512 uses 512 MB\n• grid=1024 uses 4096 MB\n\nNote\n\n2D simulations do not use any additional memory and default to grid=4096.\n\nA sorter is created by default. To disable it or modify parameters, save the context and access the sorter through it:\n\nc = context.initialize();\nhoomd.init.create_random(N=1000, phi_p=0.2)\n# the sorter is only available after initialization\nc.sorter.disable()\n\ndisable()\n\nDisables the updater.\n\nExamples:\n\nupdater.disable()\n\n\nExecuting the disable command will remove the updater from the system. Any hoomd.run() command executed after disabling an updater will not use that updater during the simulation. A disabled updater can be re-enabled with enable()\n\nenable()\n\nEnables the updater.\n\nExamples:\n\nupdater.enable()\n\nrestore_state()\n\nRestore the state information from the file used to initialize the simulations\n\nset_params(grid=None)\n\nChange sorter parameters.\n\nParameters: grid (int) – New grid dimension (if set)\nExamples::\nsorter.set_params(grid=128)\nset_period(period)\n\nChanges the updater period.\n\nParameters: period (int) – New period to set.\n\nExamples:\n\nupdater.set_period(100);\nupdater.set_period(1);\n\n\nWhile the simulation is running, the action of each updater is executed every period time steps. Changing the period does not change the phase set when the analyzer was first created." ]

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[ "# Vector Addition Calculator using Law of Cosines\n\nObjects lying in the same plane are called coplanar. Here, you can obtain the the resultant vector from the two coplanar vector based on the cosines law (rule) of vector addition for a non-right-angled triangle.\n\nkN\nkN\n°\n°\nkN\n\nObjects lying in the same plane are called coplanar. Here, you can obtain the the resultant vector from the two coplanar vector based on the cosines law (rule) of vector addition for a non-right-angled triangle.\n\nCode to add this calci to your website", null, "", null, "#### Formula:\n\nCosines Law for Vector Addition FR = [F12 + F22 - 2 * F1 * F2 * cos(180° - (α + β))](1/2) Where, FR = Resultant Vector F1, F2= Vector Quantity α, β = Angle cos = Cosine" ]

[ null, "https://www.easycalculation.com/images/embed-plus.gif", null, "https://www.easycalculation.com/images/embed-minus.gif", null ]

[ "# Descriptive Statistics: Calculating the Mean in Various Data Series", null, "The mean (average) is one of the most valuable and most frequently used measures in descriptive statistics. Why is it so widely used, and why is it important to know how to calculate the arithmetic mean? Perhaps the most convincing argument is that the mean is used in virtually every area of life.\n\nWith the arithmetic mean, you can calculate the average daily television viewing time for citizens of a given country, average volume of coffee drunk by a typical American, average annual temperature in your city, or the average amount you spend on food in a typical week. This article explains how to calculate this important statistic using the various types of series in statistics.\n\n## “Mean” and “Statistical” Keywords\n\nBefore we describe how to calculate the arithmetic mean using the various data series, we will first discuss basic statistical concepts related to the mean. The arithmetic mean is a statistic used in descriptive statistics, a significant branch of the field which deals with the collection, analysis, compilation, and presentation of data. Descriptive statistics is based on population studies. What is the meaning of “population” in statistics, then? It is a complete set of phenomena, people, or things that are the subject of a given study. If we wish to investigate salaries among all 250,000 residents of city X, the residents constitute the population. A single resident is a statistical unit. The salary, in turn, is a characteristic of the population being studied. In order to find some statistical measures, such as the average salary among all residents of city X, one would first have to gather data regarding the salary of each resident. As there are 250,000 residents in city X, this would be a tedious and costly operation. Similarly, it would be difficult to study the whole population (by testing every specimen) in order to determine the strength of a given material.\n\nThis is why statistics makes use of samples. A sample is a subset of a given population that – although large at times – is much smaller that the population itself, making statistical study more manageable, less costly, and quicker. Sample data are selected randomly in order to provide a representative sample of the population. A statistical sample serves therefore as representation of a statistical population. Descriptive statistics investigates certain characteristics of populations based on entire populations or their subsets. In the example below, one such characteristic is salary. Our measurements of characteristics (salaries, in our case) are called observations.\n\nDescriptive statistics uses “parameters” to describe a population, and “statistics” to describe a population sample.\n\nTake a look at the picture below which presents the difference between the population of city X and its statistical sample.", null, "The arithmetic mean is a parameter of the population and a statistic of the population sample.\n\nIn order to learn information about a substantial population, a statistical study may often be performed on a smaller, more manageable representative sample.\n\n## Mean of an Individual Data Series\n\nThe arithmetic mean is the most popular statistic belonging to a category of statistics known as measures of central tendency. It can be defined as the value around which all measurements of a given characteristic are distributed. It is applied only to quantitative characteristics, such as age, points, height, salary, or weight.\n\nSuppose we would like to find the average number of hours students spent in baseball classes. Assume a sample size of 10 students are selected from the entire students’ population. Below we present a set of measurements containing the number of hours spent by the 10 students in baseball classes:\n\n{ 10, 11, 12, 14, 15, 16, 17, 18, 18, 19, 20 }\n\nNote that the number of hours spent in the classes varies among the students. The measurements are presented individually for each student. This is why such a data set is called an individual data series. An individual data series is, therefore, a set of all observations of a given characteristic, usually in ascending order.\n\nThe arithmetic mean for this series is calculated by summing the values of the observations and dividing the result by the number of observations. Take a look at the formula below and the description of symbols used.", null, "x – arithmetic mean for the sample\nn – number of observations\nx_i – ith observation of the characteristic\n\nThe number of observations in this case is 10, as we surveyed 10 students. The set that contains the number of hours for each student constitutes the values of the characteristic.\n\nBy substituting the values into the equation, we can calculate the arithmetic mean:", null, "", null, "The calculations above show that the average number of hours spent in baseball classes by a single student is 17.\n\n## Mean of a Discrete Data Series\n\nIn the previous paragraph, we calculated the arithmetic mean for an individual data series. In this kind of series, the values for a given characteristic are listed for each sample unit. It is a good choice for sets with varying values of the characteristic and a small number of observations. However, what happens when values are frequently repeated in a given set of values for a given characteristic? In that case, it is better to use a discrete data series, which simplifies calculations. It is a series created by dividing all observations into groups. Each group represents a specific value of the characteristic. It is only possible when the values are discrete, i.e. they come from a countable set: age in years, points in a test. For each characteristic value identified, the number of its occurrences in the observation test should be provided.\n\nThis type of series is suitable when:\n\n• the number of values in the observation set is small,\n• the values are frequently repeated in the set,\n• the values are related to a discrete characteristic.\n\nLet us calculate the average number of points in an exam taken by 10 students in Grade 4. Possible exam scores were 0, 1, 2, 3, 4, or 5 points. Take a look at the set of observations:", null, "Our students got the following scores: 0, 2, 3, 4 and 5. The same values of the characteristic (i.e. number of points) are repeated among a few students. Points are a discontinuous characteristic. Instead of adding up all points from all students, we can simplify the calculations by dividing the observations into groups based on the exam score (number of points) and assigning the number of observations to each group. To that end, we can create a discrete series of scores, as shown in the table below:\n\nxi ni\n0 2\n1 0\n2 3\n3 1\n4 3\n5 1\n10\n\nWe can see from the table that only one student got 5 points, but 3 students got 2 points. After multiplying each score by the number of students earning that score, we get intermediate results that can be used to calculate the final average.\n\nxi ni xini\n0 2 0\n1 0 0\n2 3 6\n3 1 3\n4 3 12\n5 1 5\n10 26\n\nTake a look at the formula used to calculate the arithmetic mean for a discrete data series:", null, "x – arithmetic mean for the sample\nn – number of observations\nx_i – ith value of the characteristic\nn_i – number of observations for the ith value of the characteristic\n\nInstead of summing all observations of the characteristic (as we did in our individual data series), we add up the products of all characteristic values (x sub i) and their number of occurrences (n sub i) in the discrete data series. We then substitute those products into the equation to obtain the average mean:", null, "The average number of points obtained by a student in the exam was 2.6.\n\nRemember!\nThe arithmetic mean calculated for a given set using an individual data series and a discrete data series will always be the same.\n\n## Mean of a Continuous Data Series\n\nSo far we have discussed which series can be used to calculate the arithmetic mean for a small set of values and for a set with a small number of distinct, recurring values. However, we may sometimes come across sets with many observations, many different values, or values that are unwieldy (such as precise measurements of height or precisely calculated salaries), for which it is difficult to create groups in order to calculate the mean of a discrete data series. In such cases we can use a continuous data series, which is well suited to continuous characteristics (for which measurements are precise and often expressed as decimal numbers). A continuous data series divides all values into groups but somewhat differently than the discrete data series. Each grouping of observations represents not a specific value but a range of values.\n\nLet’s find the average salary among all employees. The measurement set consists of salaries for individual employees. Measured values fall between \\$1,000 and \\$5,000 (inclusive), which represents employee salary range. The measurements were performed for 100 employees. There are numerous salary values – each employee has a slightly different salary. In this case, we can create several intervals and count the number of employees whose salaries fall into each interval. Take a look at the table below, which presents a continuous data series.\n\nxi ni\n1,000 – 2,000 40\n2,000 – 3,000 0\n3,000 – 4,000 50\n4,000 – 5,000 10\n100\n\nIn a continuous data series, we create equal intervals. In the example above, the range of each interval equals \\$1,000. For each interval, we show not only the range, but also the number of observations and the center point. The center point is calculated by adding the lowest and greatest values of the interval and dividing the result by 2. For each interval, we calculate the product of the center point and number of observations, which is then used to find the arithmetic average.\n\nxi ni x nix\n1,000 – 2,000 40 1,500 60,000\n2,000 – 3,000 0 2,500 0\n3,000 – 4,000 50 3,500 175,000\n4,000 – 5,000 10 4,500 135,000\n100 280,000\n\nTake a look at the formula.", null, "x – arithmetic mean for the statistical sample\nn – number of observations\nn_i – number of observations for the ith interval\nx_i – center of the ith interval\n\nWe complete the equation by substituting the calculated products:", null, "The average salary in the company equals \\$2,800.\n\nRemember!\nThe arithmetic mean for a continuous data series is always approximate.\nThe broader the intervals, the less accurate the average.\nOn the other hand, if there are too many intervals, the data become less clear.\n\n## Summary\n\nThe arithmetic mean is a measure of central tendency in descriptive statistics which shows the average value of a characteristic in a given statistical sample. It is the most popular measure, vitally important, and applied in virtually every area of life. It can be calculated only for quantitative characteristics. A change in any single value in the set of characteristics results in a change in the arithmetic mean. This statistic is also sensitive to outliers. In order to easily calculate the arithmetic mean, we can use one of the following data series, selected based on the set of values of the given characteristic: individual data series, discrete data series, or continuous data series.", null, "" ]

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[ "Lesson 2, Topic 4\nIn Progress\n\nIllustration 2 – Departmental Accounts\n\nLesson Progress\n0% Complete\n\nIllustration 2\n\nFrom the following balances of Magnum Limited, you are required to prepare departmental trading, Profit, and Loss Account for the year ended 31st December 1997.\n\n1. The total floor area occupied by each department was: Department A: (2/5) Department B: (3/5)\n2. The following basis of apportionment should be used for the departments: Commission, Advertising, Discount allowed:\n\nProportionate to sales\n\n(ii) Discount received: Proportionate of Purchases\n\n(iii) Cleaning, electricity, Internal telephone, Insurance: On the basis of total floor rate.\n\n1. All other expenses should be apportioned equally between the departments.\n\nSolution\n\nMagnum Ltd\n\nDepartmental, Trading, Profit and Loss Account for the year ended 31st December 1997.\n\nSolution\n\nMAGNUM LTD.\n\nDepartmental, Trading, Profit and Loss Account for the Year ended 31st December, 1997.\n\n1. Commission Proportionate to Sale\n\nDept A 30,000/50,000 x 700/1 = 420\n\nDept B 20,000/50,000 x 700/1 = 280\n\nDept A 30,000/50,000 x 230/1 = 138\n\nDept. B 20,000/50,000 x 230/1 = 92\n\n3. Discount allowed\n\nDept. A 30,000/50,000 x 60/1 = 36\n\nDept. B 20,000/50,000 x 60/1 = 24.\n\nFloor Area Basis\n\n4. Cleaning\n\nDept. A: 2/5 x 20/1 = 8\n\nDept. B: 3/5 x 20/1 = 12.\n\n5. Electricity\n\nDept. A : 2/5 x 820/1 = 328.\n\nDept. B: 3/5 x 820/1 = 492.\n\n6. Internal Telephone\n\nDept. A: 2/5 x 120/1 = 48\n\nDept. B: 3/5 x 120 = 72\n\n7. Insurance\n\nDept. A: 2/5 x250/1 = 100\n\nDept. B 3/5 x250/1 = 250\n\nProportionate To Purchase\n\nDept. A: 23,600/40,000 x x 50/1 = 30\n\nDept B: 16,400/40,000 x 50/1 = 20\n\nEquity Basis\n\n9. General Office Salaries\n\nDept. A: ½ x 1,000 = 500\n\nDept. B: ½ x 1,000 = 500\n\n10. Stationery:\n\nDept. A ½ x 150 =75\n\nDept. B: ½ x 150 =75\n\n11. Repairs:\n\nDept. A: ½ x 240 = 120\n\nDept. B: ½ x 240 = 120\n\n12. Lighting:\n\nDept. A: ½ x 600 = 300\n\nDept. B: ½ x 600 = 300\n\n13. Rates:\n\nDept. A: ½ x 300 = 150\n\nDept. B: ½ x 300 = 150\n\n14. Sundry expenses:\n\nDept. A: ½ x 70 =35\n\nDept. B: ½ x 70 = 35\n\nResponses", null, "error:" ]

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[ "Student[NumericalAnalysis] - Maple Programming Help\n\nHome : Support : Online Help : Education : Student Packages : Numerical Analysis : Computation : Student/NumericalAnalysis/VectorLimit\n\nStudent[NumericalAnalysis]\n\n VectorLimit\n compute the limit of a vector\n\n Calling Sequence VectorLimit(V, n)\n\nParameters\n\n V - Vector n - name; the name of the index variable of the sequence represented by V\n\nDescription\n\n • Each component of V is viewed as a function of n, so that V represents a sequence of vectors ${v}_{n}$.\n • The VectorLimit command determines the limit of this sequence of vectors ${v}_{n}$ as n approaches infinity.\n\nExamples\n\n > $\\mathrm{with}\\left({\\mathrm{Student}}_{\\mathrm{NumericalAnalysis}}\\right):$\n > $\\mathrm{V1}≔\\mathrm{Vector}\\left(\\left[1+\\frac{1}{n},2-\\frac{1}{n}\\right]\\right)$\n ${\\mathrm{V1}}{≔}\\left[\\begin{array}{c}{1}{+}\\frac{{1}}{{n}}\\\\ {2}{-}\\frac{{1}}{{n}}\\end{array}\\right]$ (1)\n > $\\mathrm{VectorLimit}\\left(\\mathrm{V1},n\\right)$\n $\\left[\\begin{array}{c}{1}\\\\ {2}\\end{array}\\right]$ (2)" ]

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[ "Quantitative Aptitude Quiz For IBPS SO Prelims: 23rd December 2018\n\nDear Aspirants,\n\nNumerical Ability or Quantitative Aptitude Section has given heebie-jeebies to the aspirants when they appear for a banking examination. As the level of every other section is only getting complex and convoluted, there is no doubt that this section, too, makes your blood run cold. The questions asked in this section are calculative and very time-consuming. But once dealt with proper strategy, speed, and accuracy, this section can get you the maximum marks in the examination. Following is the Quantitative Aptitude quiz to help you practice with the best of latest pattern questions.\n\nQ1. A man divided his wealth so that his son’s share to his wife’s share and the wife’s share to his daughter’s share are both in the ratio 3: 1 and the daughter’s share is Rs. 64,000 less than that of the son. What is the value of the total wealth?\n\n94,000\n104,000\n832,000\n68,000\n108000\n\nQ2. The sum of three numbers is 98. If the ratio of the first to the second is 2 : 3 and that of second to the third is 5 : 8, then the second numbers is:\n\n20\n30\n38\n48\n45\n\nQ3. The total number of students of Bal Bharti school was 660. The ratio between girls and boys was 9 : 13. After some days, 30 girls joined the school and some boys left the school and new ratio between girls and boys became 5 : 6. The number of boys who left the school is :\n\n50\n40\n20\n60\n30\n\nQ4. Between two consecutive years my incomes are in the ratio of 2 : 3 and expenses in the ratio 5 : 9. If my income in the second year is Rs. 45000 and my expenses in the first year is Rs. 25000. My total savings for the two years is:\n\nNil\nRs. 15000\nRs. 10000\nRs. 5000\nNone of these\nSolution:\n\nIncome in the second year = Rs. 45000\nIncome in the first year = Rs. 30000\nExpense in the first year = Rs. 25000\nExpense in the second year = Rs. 45000\n∴ Total saving = 75000 – 70000 = Rs. 5000\n\nQ5. A mixture contains alcohol and kerosine in the ratio 5 : 1. On adding 5 litres of kerosine, the ratio of alcohol and kerosine becomes 5 : 2. The quantity of alcohol in the original mixture is.\n\n16 litres\n25 lires\n22.75 litres\n32.5 litres\n30 litres\n\nDirections (6-10): The following pie-chart shows the distribution of chocolate (in degree) among children of different classes of Bal Bharti School. Study the chart carefully to answer the following questions.\n\nNote: Some data are in degree while some data are in absolute value in the following chart.\n\nQ6. Total no. of chocolates distributed among class V students is what percent of total chocolates distributed among class VI students (approximately)?\n\n140%\n125%\n145%\n120%\n155%\n\nQ7. If ratio of boys to girls in class I be 3 : 1 and each student get two chocolate then how many boys are there in class I who got Chocolates?\n\n1402\n1206\n1602\n1802\n1600\n\nQ8. What is the average no. of Chocolates distributed among students of class II and III together?\n\n4210\n4280\n3036\n3248\n3642\n\nQ9. Total Chocolates distributed among students of class II and class IV together is what percent more or less than the total no. of chocolates distributed among students of class VI and I together?\n\n5% less\n5% more\n25% more\n10% more\n10% less\n\nQ10. What is the difference between total chocolates distributed among students of class I, II and VI together and total chocolates distributed among students of class III, IV and V together?\n\n2,136\n4,236\n4,136\n4,436\n3,136\n\nDirections (11-15): What approximate value should come in place of (x) in the following questions (Note: You are not expected to calculate the exact value.)\n\nQ11.", null, "104\n111\n96\n90\n120\n\nQ12. 630 × x + 553.85 – 55% of 12000 = 37.05 % of 9200\n\n12\n18\n15\n20\n8\n5\n10\n12.5\n2\n8\n√7\n1/√7\n1/7\n√7/2\n√(7/2)\n17.5\n25\n10\n15\n8\n\nYou May also like to Read:", null, "" ]

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[ "# antisymmetric\n\n(redirected from Anti-symmetric)\nAlso found in: Dictionary.\n\n## antisymmetric\n\n(mathematics)\nA relation R is antisymmetric if,\n\nfor all x and y, x R y and y R x => x == y.\n\nI.e. no two different elements are mutually related.\n\nPartial orders and total orders are antisymmetric. If R is also symmetric, i.e.\n\nx R y => y R x\n\nthen\n\nx R y => x == y\n\nI.e. different elements are not related.\nReferences in periodicals archive ?\nAs J is anti-symmetric and R is symmetric positive definite, substitute the controller of (8) into (9) yields\nThat means that when the unsymmetrical vibration mode shape of the object has been received as the result of the test, both symmetrical and anti-symmetric parts may be isolated on its basis.\nLMS algorithm with LUT based multiplier optimization techniques Anti-symmetric Product Coding:\nIn my past works (with each of my theories being independent and self-contained; and I do not repeat myself ever), I have shown how all this can be accomplished: one is with the construction of an asymmetric metric tensor whose anti-symmetric part gives pure spin and electromagnetism, and whose differential structure gives an anholonomic, asymmetric connection uniquely dependent on x and dx (and hence x and the world-velocity u, giving a new kind of Finslerian space), which ultimately constructs matter (and motion) from pure kinemetric scratch.\nnxn] denote the set of all n x m real matrices, the set of all n x n orthogonal matrices, the set of all n x n real symmetric matrices, the set of all n x n real anti-symmetric matrices, respectively; The symbols, [A.\ncan be replaced by symmetric (or anti-symmetric if sums of signal values will be replaced by differences) version (which requires m + 1 multiplications and M additions):\nUsing this transducer configuration, along with accurate synchronization of oscilloscope channels enables symmetric or anti-symmetric guided waves modes emphasis in recorded signals.\nAnother result is that the anti-symmetric state of the [[DELTA].\nIf we define a second-rank anti-symmetric tensor B by\nAs in , for reasons that will be clear later, we define the electromagnetic field tensor F via the torsion tensor of spacetime (the anti-symmetric part of the connection [GAMMA]) as follows:\nIn this framework, we produce the microspin tensor and the anti-symmetric part of the stress tensor as intrinsic geometric objects rather than alien additions to the framework of classical elasticity theory.\n\nSite: Follow: Share:\nOpen / Close" ]

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[ "Date\n\n# Introduction\n\nThis post is continuation of my previous synthetic-aperture imaging experiments. In the previous post I used omega-k frequency domain algorithm for the image formation and automatic differentiation based autofocusing with Tensorflow. I managed to get pretty well focused images from bicycle mounted radar considering the total lack of any motion recording data. The autofocus algorithm managed to improve it slightly, but it wasn't perfectly focused.\n\nThis time I'm implementing time-domain backprojection algorithm. It has slower $$O(n^3)$$ time complexity compared to $$O(n^2\\log n)$$ of the omega-k, but it makes up for its slowness by its flexibility. Since omega-k is based on FFT it requires the data to be sampled on a straight line with equal spacing. However due to motion errors during the measurement this doesn't hold in practice. By adjusting the phase of the signal it's possible to approximate the signal that would have been recorded in slightly different position, but that approach has it's limits. Backprojection algorithm works with any shape measurement path.\n\n# Backprojection algorithm\n\nLet's start the analysis of the algorithm with the radar signal model. My radar is frequency modulated constant wave (FMCW) radar that transmits short frequency sweeps. I won't repeat the derivation from the previous post, you can check it out if you want to see the details and I'll only give the resulting IF-signal for single target:\n\n$$s(n, \\tau) = \\exp\\left(-j \\frac{4 \\pi}{c} (f_c + \\gamma \\tau) d(\\mathbf{x}, \\mathbf{p}_n)\\right), \\quad -T/2 < \\tau < T/2$$\n\n, where $$c$$ is speed of light, $$f_c$$ is center frequency of the sweep, $$\\gamma = B/T =$$ sweep bandwidth / sweep length $$=$$ sweep rate, $$\\tau$$ is time variable and $$d(\\mathbf{x}, \\mathbf{p}_n)$$ is distance to the target at position $$\\mathbf{x}$$ from the $$n$$th sweep at position $$\\mathbf{p}_n$$.\n\nLet's define discrete version of the IF-signal $$s[n, m] = s(n, m T_s)$$, where $$T_s$$ is the sampling interval.\n\nThe simplest way to generate a SAR image of radar measurements along some path is to use matched filtering. For each pixel in the output image it's possible to calculate what the measured IF signal would have been if there was a target at that pixel. Matched filtering the measured signal with the complex conjugate of the expected will give high value when the measurement matches the expectation. Repeating the filtering process for every pixel in the output gives the focused image. In equation form the filtering for one pixel can be written as:\n\n$$I(\\mathbf{x}) = \\sum_{n=0}^{N-1}\\sum_{m=0}^{M-1} s[n, m] s_\\text{ref}[n, m]$$\n\n, where $$s_{ref}[n, m]$$ is the reference function which is complex conjugate of the expected IF response:\n\n$$s_\\text{ref}[n, m] = \\exp\\left(j\\frac{4 \\pi}{c}(f_c + \\gamma m T_s) d(\\mathbf{x}, \\mathbf{p}_n)\\right)$$\n\nThis SAR focusing algorithm gives especially well focused images, there are no interpolation steps, no approximations, it even works for cases where transmitter and receiver antennas are not at the same position. The big problem however is that it's extremely slow. For every pixel in the image it needs to go over every measured data point. For X x Y image and N sweeps each M points long the time complexity is in the order of $$O(XYNM)$$, or just $$O(n^4)$$ for short as they all scale linearly as the image size increases.\n\nThere is an easy way to speed it up with some slight assumptions about the measurement setup. If the reference function can be factored into two parts each depenging on only $$n$$ and $$m$$ then the double sum can be factored into two sums.\n\n$$I(\\mathbf{x}) = \\sum_{n=0}^{N-1} \\exp\\left(j\\frac{4 \\pi f_c}{c}d(\\mathbf{x}, \\mathbf{p}_n)\\right) \\sum_{m=0}^{M-1} s[n, m] \\exp\\left(j\\frac{4\\pi}{c} \\gamma m T_s d(\\mathbf{x}, \\mathbf{p}_n)\\right)$$\n\nThe trick for faster algorithm is using FFT to calculate the inner sum once and then using interpolation to index the right term for the outer sum.\n\nThe definition of the inverse FFT is: $$S[n, k] = \\sum_{m=0}^{M-1} s[n, m] \\exp\\left(j 2 \\pi k m / M\\right)$$.\n\nTo solve for the right index into the FFT we need to solve for $$k$$ in the equation:\n\n$$j 2 \\pi k m / M = j\\frac{4\\pi}{c} \\gamma m T_s d(\\mathbf{x}, \\mathbf{p}_n)$$\n\nSolving for $$k$$ and using the fact that $$\\gamma = B/T$$ and $$T_s = T / M =$$ sampling period gives:\n\n$$k = \\frac{2 B}{c} d(\\mathbf{x}, \\mathbf{p}_n)$$\n\nThe backprojection algorithm can then be written as:\n\n$$I(\\mathbf{x}) = \\sum_{n=0}^{N-1} \\exp\\left(j\\frac{4 \\pi f_c}{c}d(\\mathbf{x}, \\mathbf{p}_n)\\right) S\\left[n, \\frac{2 B}{c} d(\\mathbf{x}, \\mathbf{p}_n)\\right]$$\n\n, where $$S[n, k]$$ is the Fourier transform of $$s[n, m]$$ along the second axis. A slight complication is that the new index isn't necessarily a whole number and interpolation is required. The advantage of this form is that it completely removes the inner sum and replaces it with FFT calculated in advance. As a result time complexity of the algorithm improves from $$O(n^4)$$ to $$O(n^3)$$ and in practice the new form is much faster.\n\n## Bistatic correction\n\nDuring this derivation it was assumed that transmitter and receiver antennas are at the same position, but that isn't exactly the case with my radar.", null, "Radar mounted on the bicycle. Note the separate transmitter and receiver antennas.\n\nAs can be seen in the picture above I actually have separate transmitter and receiver antennas. While the error from considering them to be located on the same place isn't that large, it's not insignificant. Compensating for it in the Omega-k algorithm would have been very difficult, but it's trivial in backprojection algorithm. The distance that the radar measures is distance from transmitter antenna to target to receiver antenna. Replacing the distance to the target $$p_n$$ in the formula with average of distances from receiver and transmitter antennas to the target gives the measured distance.\n\n## Velocity correction\n\nDuring the measurement the distance to the target is actually not constant due to movement of the platform. A correction for the movement was derived in the previous post for Omega-k and similar correction can be derived for the backprojection too.\n\nFirst we assume that velocity is constant during one sweep, which isn't really unrealistic assumption as the acceleration would need to be absolutely enormous to have any difference in the speed during the short sweep. The second assumption is that velocity towards the target is constant during the sweep, this assumption is equal to only correcting for the first order errors of the velocity. In practice the non-linearity is so small with reasonable speeds that this isn't a large assumption.\n\nProjection of the velocity $$\\mathbf{v}[n]$$ to the distance vector gives the velocity component towards the target:\n\n$$v_p[n] = \\mathbf{v}[n] \\cdot \\frac{\\mathbf{x} - \\mathbf{p}_n}{\\left|\\mathbf{x} - \\mathbf{p}_n\\right|}$$\n\nDistance to the target during the sweep can then be written as: $$d(\\mathbf{x}, \\mathbf{p}_n) + v_p[n] m T_s$$.\n\nSubstituting the new distance to the reference function gives:\n\n$$s[n, m] = \\exp\\left(j\\frac{4\\pi}{c}\\left(f_c d(\\mathbf{x}, \\mathbf{p}_n) + \\gamma m T_s d(\\mathbf{x}, \\mathbf{p}_n) + f_c v_p[n] m T_s + \\gamma m^2 T_s^2 v_p[n] \\right)\\right)$$\n\nThe first two terms are the same as before, and the next two terms depending on $$v_p[n]$$ are new. The last term with $$m^2$$ dependence is problematic since it prevents writing the inner sum as FFT. Quadratic term could be ignored since it's typically much smaller than the other velocity term. By dividing the quadratic velocity term with the other term it can be shown that the quadratic term is at maximum $$B / f_c$$ of the other term. It mostly matters when sweep bandwidth is significant fraction of the operating frequency.\n\nBetter estimate than ignoring the quadratic term can be obtained by fitting a linear function to it (Stringham's PhD. thesis). Least square fit of linear function for $$m^2$$ over interval 0 to $$M-1$$ can be obtained by minimizing the sum below for $$a$$ and $$b$$:\n\n$$\\min_{a,b}\\sum_{m=0}^{M-1} \\left(m^2 - (a m + b)\\right)^2$$\n\nThe solution can be found as $$m^2 \\approx (M - 1) m - \\frac{1}{6} (M^2 - 3M + 2) \\approx M m - \\frac{M^2}{6}$$.\n\nSubstituting it into the quadratic term and simplifying gives: $$\\gamma m^2 T_s^2 v_p[n] \\approx \\frac{\\gamma m T^2 v_p[n]}{M} - \\frac{1}{6} \\gamma T^2 v_p[n]$$\n\nAs before the reference function can be divided into two parts and the inner sum can be replaced with FFT:\n\n$$I(\\mathbf{x}) = \\sum_{n=0}^{N-1} \\exp\\left(j\\frac{4 \\pi}{c} \\left(f_c d(\\mathbf{x}, \\mathbf{p}_n) - \\gamma T^2 v_p[n]/6\\right) \\right) S\\left[n, \\frac{2 B}{c} \\left(d(\\mathbf{x}, \\mathbf{p}_n) + \\frac{f_c v_p[n]}{\\gamma}\\right) + T v_p[n]\\right]$$\n\nIn the outer exponential $$4\\pi\\gamma T^2 v_p[n]/6c = 4\\pi B T v_p[n] / 6c$$ is normally small enough to be ignored.\n\nThe difference to the case where radar was assumed to be stationary during the sweep is just a slight adjustment to the index of the $$S[n, k]$$.\n\n# Tensorflow implementation\n\nImplementation of the backprojection was made as a custom operation in Tensorflow with both CPU and GPU support. The algorithm is excellent candidate for GPU implementation as the parallelization is straightforward: each pixel in the image can be calculated independently. Interpolation needed in the indexing of the FFT result is implemented as simple linear interpolation. FFT is calculated with configurable amount of zero-padding to smooth the output and make the interpolation result more accurate. The result is acceptable even with this simple interpolation unlike interpolation in Omega-k algorithm where linear interpolation would cause visible artifacts in the image.\n\nFor calculating gradients, a gradient operation also needs to be implemented. Specifically operation calculating the chain rule needs to be implemented. The operation takes as input the accumulated gradient from the earlier gradient operations and calculates the gradient at the inputs of the operation.\n\nGradients of complex numbers need some care in Tensorflow. Gradient of real function $$y(x)$$ with respect to $$x$$ is calculated simply as $$\\frac{\\partial y}{\\partial x}$$, but the formula is different if the function is complex valued. The short version is that the gradient of complex valued function $$f(x)$$ with respect to $$x$$ needs to be calculated as $$\\overline{\\left(\\frac{\\partial f}{\\partial x} + \\frac{\\partial \\overline f}{\\partial x}\\right)}$$, where $$\\overline \\cdot$$ is complex conjugate. For more detailed explanation see this discussion. In this case the second term, conjugated partial derivate, is zero and the result is the normal real gradient but conjugated. For simplicity velocity correction terms are ignored but adding them is trivial. Using chain rule gradient of the backprojection for one pixel of output can be calculated as:\n\n$$\\frac{\\partial I(\\mathbf{x})}{\\partial \\mathbf{p}} = \\sum_{n=0}^{N-1} \\frac{\\partial d(\\mathbf{x}, \\mathbf{p}_n)}{\\partial \\mathbf{p}} \\exp\\left(j\\frac{4 \\pi f_c}{c}d(\\mathbf{x}, \\mathbf{p}_n)\\right) \\left( j\\frac{4 \\pi f_c}{c} S\\left[n, \\frac{2 B}{c} d(\\mathbf{x}, \\mathbf{p}_n)\\right] + \\frac{2 B}{c} \\frac{\\partial S}{\\partial d(\\mathbf{x}, \\mathbf{p}_n)}\\left[n, \\frac{2 B}{c} d(\\mathbf{x}, \\mathbf{p}_n)\\right] \\right)$$\n\nThe gradient computation needs gradient of $$S$$. In the forward direction linear interpolation was used to calculate the values in between the sample points and it's natural to use the slope of the linear interpolation to define the necessary gradient.\n\nThe gradient operation calculating the chain rule is complex conjugate of the above gradient times the accumulated gradient summed over all the pixels in the image. If $$E$$ is some scalar output of the graph and $$g$$ is the accumulated gradient from all the operations after the SAR image formation, the gradient of $$E$$ with respect to the input positions can be written as:\n\n$$\\frac{\\partial E}{\\partial \\mathbf{p}} = \\sum_\\mathbf{x} g[\\mathbf{x}] \\overline{\\left(\\frac{\\partial I(\\mathbf{x})}{\\partial \\mathbf{p}}\\right)}$$\n\n# Autofocus\n\nWith the backprojection and its gradient operation implemented using Tensorflow for autofocusing the generated image is easy. First calculate the SAR image and it's entropy. Entropy of image generally correlates with how focused image is with better focused images having lower entropy. Next calculate gradient of entropy as function of the input position vector using automatic differentation. A small step towards negative gradient vector should decrease the entropy improving the focus of the image. The steps are repeated until entropy doesn't decrease anymore.\n\nIn practice there are some slight issues with the above algorithm. If position of every sample can be optimized individually there are too many degrees of freedom. There are many positions for the samples that give a focused image with low entropy, but most of them have some geometric distortions that don't correspond to the actual scene. For example adding any position offset to all of the samples just moves the resulting image. Backprojection algorithm requires specifying the pixel coordinates beforehand and without any constraints the optimizer can just add a large offset to every sample to move all the data out of the scene to minimize the entropy. Some constraints need to be added for the sample positions to make sure that the optimized image makes sense. Almost all SAR data is taken on a straight path with constant velocity, so it's natural to add some constraints to keep the positions close to a straight line with equal spacing.", null, "Autofocused parking lot image with geometric erorrs due to unrestriected sample positions. There is a kink in the image near origin that shouldn't be there.\n\nIn the above image is an example of the scene with unrestricted position optimization. The image looks focused but it has severe geometric distortions as the parking spaces should be evenly spaced. See pictures at the bottom on how the image should look like.\n\nI added two extra constraints for velocity. The first adds penalty if the range direction of the velocity is too large and other adds penalty if the azimuth velocity difference to its mean value is too large. With these constraints the geometric errors are avoided in the cases I tested.\n\nFocusing data With 6444 sweeps each with 1000 samples each for 9666 x 1933 pixel output image the FFT based Omega-k takes 56 ms for one forward iteration and backprojection algorithm takes 436 ms. The time-domain backprojection algorithm is almost ten times slower on this dataset but compared to the Omega-k algorithm it's much more flexible. Omega-k algorithm requires the transmitter and receiver to be at the same position and all samples to be recorded on a straight line with equal distance in the direction of movement between them. Time-domain backprojection can focus any kind of measurement configuration with possibly unevenly samples positions and non-colocated transmitter and receiver antennas.\n\n# Measurements\n\nFor good comparison with the Omega-k algorithm I focused the previously measured data also with the backprojection algorithm. Autofocus was run for 100 iterations.", null, "", null, "Parking lot scene SAR image. Click/tap to see the autofocused version.\n\nAbove is the image of the parking lot from the same data as I previously used for the Omega-k algorithm post expect that the sampling rate is twice as large as before avoiding some aliasing artifacts from before. Clicking/tapping and holding shows the focused image. The difference the autofocus makes is big and the autofocused image has excellent focusing.", null, "Solved velocity\n\nThe solved velocity is almost constant, there are small oscillations in the range direction and some slight variance in the azimuth direction. There are some spikes that correspond to locations where there are large objects that saturate the receiver. This method doesn't seem to give the best results when receiver has saturated, some spreading of the big objects can be seen as a result in the image.", null, "Comparison of backprojection and omega-k with and without autofocusing.\n\nFor farther away objects the difference in focusing between the algorithms is big. In the above comparison image both autofocus algorithms improve the focusing for nearby targets, but only backprojection with autofocusing can also focus the farther away targets.", null, "", null, "Park scene SAR image. Click/tap to see the autofocused version.\n\nI also redid the other scene I measured the last time. The improvement from autofocus in that scene is also huge especially for objects farther away. The objects farther away are still spread, but I'm not sure if they should be much better than what they are now. There are lot of issues with occlusions in this scene and some far away objects are visible for the radar only for a very short time. See the images in the previous post for the same data focused with Omega-k algorithm.\n\nLight posts and tree trunks are very visible in the SAR image compared to the Google maps image. Most of the details inside the park in the upper right can't be seen in the SAR image since a metal fence blocks the most of the view from the radar signal.\n\n# Conclusion\n\nI implemented a fast GPU based time-domain backprojection algorithm for synthetic aperture radar image focusing with autofocus based on automatic differentiation implemented as custom operation in Tensorflow. The autofocus algorithm gives a big improvement to focusing of the data from my homemade radar without any position measurement. The code is available at my Github." ]

[ null, "https://hforsten.com/img/fmcw3-sar/bicycle_radar_rack.jpg", null, "https://hforsten.com/img/fmcw3-sar2/bp_parking2_100_geometric_errors2.png", null, "https://hforsten.com/img/fmcw3-sar2/bp_parking2_0_label.png", null, "https://hforsten.com/img/fmcw3-sar2/bp_parking2_100_v2_label.png", null, "https://hforsten.com/img/fmcw3-sar2/bp_parking2_v3_solved_v.png", null, "https://hforsten.com/img/fmcw3-sar2/bp_omegak_comparison.png", null, "https://hforsten.com/img/fmcw3-sar2/bp_park_0_labels.png", null, "https://hforsten.com/img/fmcw3-sar2/bp_park_100_labels.png", null ]

[ "# How many work hours in 4 months\n\nthe total working hours for each month (based on 8 hours each Monday-Friday) . , 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, Your employer cannot make you work longer than an average of 48 hours a number of hours you have worked: 17 weeks of four shifts of 12 hours: 4 x 12 = February has 20 total workdays ( monthly work hours), May has 21 total As you can see, using the 40 hours per week multiplied by 4 weeks only Solution: First, find how many hours 1 FTE works in 1 month: (40 hours.\n\n## how many working hours in a month\n\n3 Months = Hours, 13 Months = Hours, 23 Months = Hours, 33 Months = Hours, 90 Months = Hours. 4 Months = Hours, Assuming 40 hour week, how many work hours are in a year? Edit: corrected a label and added calculations for 3, 4, and 5 weeks of vacation. Calculation of hours worked over a period of 12 months. Consider a worker who work 35 hours per week, 4 vacation weeks and 5 public holidays (for a total of 5.\n\nDetermining how many work hours are in a year is useful when calculating a salaried employee's hourly rate for payment purposes. The formula to determine . When it comes to choosing how many work hours in a month you should talk about the benefits of a 4-day work week or a 6-hour workday. Get the exact number of working hours in a month, quarter or a year simply by Jan, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22,\n\n## how many hours in a month of 31 days\n\n(business) days in -- along with the number of working hours per month, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, The average working hours over a 12 month period calculates to hours per month. However, no individual month's working hours will calculate to exactly. Enter the number of work weeks to convert into months. Easy work A typical work week is considered to be 40 hours: 8 hours a day for 5 days. 4, Month. Date(s) of. Holiday. Working Days in. Month. Working Hours Noon 5/ June none. 8 a.m. 6/ July. 4. Monthly Working Hours — Number of working hours per month based on a five-day, Monday through Friday week, eight April, 22, , 88, 8 a.m. 4/ Working hours per day multiplied by working days per week multiplied by 52 weeks in the year divided by 12 months in the year equals to the average number of. hours is how many hours the employee would work if he worked full time and took .. I was let go after 6 months, having taken only 4 days of vacation total . Instructions for calculating time spent during day, week, month and year takes 4 hours to complete represents the following percentage of the job: 4/ . Contractual working hours are the hours that you and your employer have If week 1 = 40 hours; week 2 = 44 hours; week 3 = 48 hours; week 4 = 40 hours: An employee can only work up to 72 overtime hours in a month. Working time is the period of time that a person spends at paid labor. Unpaid labor such as personal housework or caring for children or pets is not considered part of the working week. Statistics show most Italians have an average of 0 hours working time due to .. Work hours in Japan are decreasing, but many Japanese still work long." ]

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[ "D01 Chapter Contents\nD01 Chapter Introduction\nNAG Library Manual\n\n# NAG Library Routine DocumentD01RCF\n\nNote:  before using this routine, please read the Users' Note for your implementation to check the interpretation of bold italicised terms and other implementation-dependent details.\n\n## 1  Purpose\n\nThe dimension of the arrays that must be passed as actual arguments to D01RAF are dependent upon a number of factors. D01RCF returns the correct size of these arrays enabling D01RAF to be called successfully.\n\n## 2  Specification\n\n SUBROUTINE D01RCF ( NI, LENXRQ, LDFMRQ, SDFMRQ, LICMIN, LICMAX, LCMIN, LCMAX, IOPTS, OPTS, IFAIL)\n INTEGER NI, LENXRQ, LDFMRQ, SDFMRQ, LICMIN, LICMAX, LCMIN, LCMAX, IOPTS(100), IFAIL REAL (KIND=nag_wp) OPTS(100)\n\n## 3  Description\n\nD01RCF returns the minimum dimension of the arrays X, FM, ICOMM and COMM that must be passed to D01RAF to enable the integration to commence given options currently set for the NI integrands. D01RCF also returns the upper bounds $\\mathit{licmax}$ and $\\mathit{lcmax}$ for the dimension of the arrays ICOMM and COMM, that could possibly be required with the chosen options.\nAll the minimum values $\\mathit{lenxrq}$, $\\mathit{ldfmrq}$, $\\mathit{sdfmrq}$, $\\mathit{licmin}$ and $\\mathit{lcmin}$, and subsequently all the maximum values $\\mathit{licmax}$ and $\\mathit{lcmax}$ may be affected if different options are set, and hence D01RCF should be called after any options are set, and before the first call to D01RAF.\nA segment is here defined as a (possibly maximal) subset of the domain of integration. During subdivision, a segment is bisected into two new segments.\nNone.\n\n## 5  Parameters\n\n1:     NI – INTEGERInput\nOn entry: ${n}_{i}$, the number of integrals which will be approximated in the subsequent call to D01RAF.\nConstraint: ${\\mathbf{NI}}>0$.\n2:     LENXRQ – INTEGEROutput\nOn exit: $\\mathit{lenxrq}$, the minimum dimension of the array X that can be used in a subsequent call to D01RAF.\n3:     LDFMRQ – INTEGEROutput\nOn exit: $\\mathit{ldfmrq}$, the minimum leading dimension of the array FM that can be used in a subsequent call to D01RAF.\n4:     SDFMRQ – INTEGEROutput\nOn exit: $\\mathit{sdfmrq}$, the minimum second dimension of the array FM than can be used in a subsequent call to D01RAF.\n5:     LICMIN – INTEGEROutput\nOn exit: $\\mathit{licmin}$, the minimum dimension of the array ICOMM that must be passed to D01RAF to enable it to calculate a single approximation to all the ${n}_{i}$ integrals over the interval $\\left[a,b\\right]$ with ${s}_{\\mathit{pri}}$ initial segments.\n6:     LICMAX – INTEGEROutput\nOn exit: $\\mathit{licmax}$ the dimension of the array ICOMM that must be passed to D01RAF to enable it to exhaust the adaptive process controlled by the currently set options for the ${n}_{i}$ integrals over the interval $\\left[a,b\\right]$ with ${s}_{\\mathit{pri}}$ initial segments.\n7:     LCMIN – INTEGEROutput\nOn exit: $\\mathit{lcmin}$, the minimum dimension of the array COMM that must be passed to D01RAF to enable it to calculate a single approximation to all the ${n}_{i}$ integrals over the interval $\\left[a,b\\right]$ with ${s}_{\\mathit{pri}}$ initial segments.\n8:     LCMAX – INTEGEROutput\nOn exit: $\\mathit{lcmax}$, the dimension of the array COMM that must be passed to D01RAF to enable it to exhaust the adaptive process controlled by the currently set options for the ${n}_{i}$ integrals over the interval $\\left[a,b\\right]$ with ${s}_{\\mathit{pri}}$ initial segments.\n9:     IOPTS($100$) – INTEGER arrayInput\nOn entry: the integer option array as returned by D01ZKF.\nConstraint: IOPTS must not be changed between calls to D01ZKF, D01ZLF, D01RCF and D01RAF.\n10:   OPTS($100$) – REAL (KIND=nag_wp) arrayInput\nOn entry: the real option array as returned by D01ZKF.\nConstraint: OPTS must not be changed between calls to D01ZKF, D01ZLF, D01RCF and D01RAF.\n11:   IFAIL – INTEGERInput/Output\nOn entry: IFAIL must be set to $0$, $-1\\text{​ or ​}1$. If you are unfamiliar with this parameter you should refer to Section 3.3 in the Essential Introduction for details.\nFor environments where it might be inappropriate to halt program execution when an error is detected, the value $-1\\text{​ or ​}1$ is recommended. If the output of error messages is undesirable, then the value $1$ is recommended. Otherwise, if you are not familiar with this parameter, the recommended value is $0$. When the value $-\\mathbf{1}\\text{​ or ​}\\mathbf{1}$ is used it is essential to test the value of IFAIL on exit.\nOn exit: ${\\mathbf{IFAIL}}={\\mathbf{0}}$ unless the routine detects an error or a warning has been flagged (see Section 6).\n\n## 6  Error Indicators and Warnings\n\nIf on entry ${\\mathbf{IFAIL}}={\\mathbf{0}}$ or $-{\\mathbf{1}}$, explanatory error messages are output on the current error message unit (as defined by X04AAF).\nErrors or warnings detected by the routine:\n${\\mathbf{IFAIL}}=21$\nOn entry, ${\\mathbf{NI}}=⟨\\mathit{\\text{value}}⟩$.\nConstraint: ${\\mathbf{NI}}>0$.\n${\\mathbf{IFAIL}}=1001$\nOne of the option arrays IOPTS or OPTS has become corrupted. Re-initialize the arrays using D01ZKF.\n\nNot applicable." ]

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[ "A problem with 26 distinct positive integers\n\nI am trying to solve the following problem.\n\nAssume that we are given 26 distinct positive integers. Show that either there exist 6 of them $$x_1, with $$x_1$$ dividing $$x_2$$, $$x_2$$ dividing $$x_3$$, $$x_3$$ dividing $$x_4$$, $$x_4$$ dividing $$x_5$$ and $$x_5$$ dividing $$x_6$$ or there exist six of them, such that none of them divides another one of these six.\n\nA possibly good start is to assume that, in every six of these numbers, there exists at least one dividing another one of the same six.\n\nUpdate. I have found a solution of the problem (for 17 numbers though) in a Russian site. As unbelievable as it may sound, this problem was a question in a 1983 Soviet Mathematics contest (Турниры Городов) for student of 7-8 grades!\n\nI am presenting the solution I found in that site below as an answer, and it is generalised for $$n^2+1$$ distinct integers, where we show that either there exist $$n+1$$ of them dividing each other or none diving none else.\n\n• This is not a duplicate of the question on increasing and decreasing subsequences, as divisibility is a partial order, not a strict linear order. – ShreevatsaR Jan 26 '14 at 9:33\n\nApply Dilworth's Theorem to the poset of the $26$ integers under the divisibility relation. Either there is an antichain of length $6$ (no two of $6$ integers divide one another) or the set can be partitioned into at most $5$ chains (sequences where each integer divides the next), one of which must have length at least $6$ by the pigeonhole principle.\n\nDefine a chain as a sequence $(y_j)$ such that $y_1|y_2 |...|y_k$ and an antichain a sequence $(y_j)$ such that $y_1 \\nmid y_2 \\nmid .. \\nmid y_k$.\n\nTo each $x$ in the initial sequence attach a pair of numbers $(i_x,j_x)$ such that $i_x$ is the the length of the longest chain ending in $x$ and $j_x$ is the length of the longest antichain starting with $x$.\n\nPick now two elements $x\\neq y$ such that $x$ is before $y$ in the sequence $(x_n)$. If $x | y$ then $i_x<i_y$ since any chain ending in $x$ can be extended to a longer chain ending in $y$. If $x \\nmid y$ then $j_x > j_y$ since any antichain starting in $y$ can be extended to a longer antichain starting in $x$. Therefore the application $x \\mapsto (i_x,j_x)$ is injective.\n\nConsider a sequence with $mn+1$ elements, and suppose the length of the longest chain is $m$ and the length of the longest antichain is $n$. Therefore the map $x \\mapsto (i_x,j_x)$ is an injection from a set with $mn+1$ elements into a set with $mn$ elements. This is a contradiction proving that there is a chain of length $m+1$ or an antichain of length $n+1$.\n\nIn the problem presented here we have $m=n=5$.\n\n• The third paragraph should end $x \\mapsto (i_x,j_x)$ instead of $x \\mapsto (i_x,i_y)$. (I tried editing your answer, but the change was less than 6 characters, which isn't allowed.) – Steve Kass Dec 28 '13 at 2:46\n• @SteveKass: Got it fixed – Beni Bogosel Dec 28 '13 at 10:14\n• Your definition of antichain isn't right: the elements have to be such that none divides another. $2\\nmid 5$ and $5\\nmid 6$ but these three don't form an antichain. – universalset Jan 25 '14 at 14:41\n• I have the same doubt as @universalset: For the desired conclusion of the question (and for the general meaning of \"antichain\") to hold, you must define your antichain as a sequence such that none of them divides any other. But then we can no longer say that if $x\\nmid y$, then an antichain starting at $y$ can be extended to a longer antichain starting at $x$. Indeed, the conclusion of this proof, that $x\\mapsto (i_x,j_x)$ is injective, is not true: Consider the sequence $(2,3,4)$. Then, the pairs corresponding to $2$ and $3$ are both $(1,2)$ and $(1,2)$. – ShreevatsaR Jan 29 '14 at 3:57\n\nI have found this solution (for 17 numbers though) in a Russian site, as this problem was a question in a 1983 Soviet Mathematics contest (Турниры Городов) for student of 7-8 classes!\n\nSolution. We shall attach on each of our numbers $$x_1 an index next to their subscript in the following fashion:\n\n$$x_1$$ becomes $$x_{1,1}$$.\n\nIf $$x_1$$ divides $$x_2$$, then $$x_2$$ becomes $$x_{2,2}$$, otherwise it becomes $$x_{2,1}$$.\n\nIn general, if we have attached indices to $$x_1,\\ldots,x_k$$, then the index of $$x_{k+1}$$ will be $$1$$ is none of the $$x_1,\\ldots,x_k$$ divides $$x_{k+1}$$, or it will become $$i+1$$, if $$i$$ is the largest index of all the numbers which divide $$x_{k+1}$$.\n\nIf the index of some of the $$x_j$$'s is at least $$6$$ then we have have a sequence $$x_{k_1,1} \\mid x_{k_2,2} \\mid x_{k_3,3} \\mid x_{k_4,4} \\mid x_{k_5,5}\\mid x_{k_6,6}.$$ In all the indices are less or equal to $$5$$, then some number in $$\\{1,2,3,4,5\\}$$, say $$j$$, is necessarily the index of at least $$6$$ numbers, i.e. $$x_{k_1,j} < x_{k_2,j} < x_{k_3,j} < x_{k_4,j} < x_{k_5,j} < x_{k_6,j},$$ which means that none of the above $$6$$ numbers divide another one of them.\n\nNote. Τhis can be generalised to finding a chain of $$k$$ numbers, fully ordered by division, or a subset of $$\\ell$$ numbers, with none of them dividing another one, among $$m$$ distinct positive integers, whenever $$(k-1)(\\ell-1)\\le m-1$$." ]

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[ "# How to predict the next number in a highly autocorrelated cyclical sequence?\n\nI have a couple of series of numbers (sequences/time series) $X_j = \\{x_{1,j}, ... , x_{N_{j}, j}\\}$ which exhibit a high autocorrelation and a dynamic cyclical feature. Below are some illustrations of some of these series.\n\nI am looking at ways to capture this dynamic cyclical behaviour and predict the next number $\\hat{x}_{N_{j}+1, j}$ in the sequence $X_j$.\n\nWhat are ways to capture this behavior systematically?\n\nI tried a linear regression on the first lag, i.e. $AC(1)$, since the autocorrelation is close to 0.99, but I am faced with a high MSE as the numbers $x_{i,j}$ have a high nominal value; thus being a slight off is already not accurate enough.\n\n• Could you forecast the absolute value of $x,$ and then add the alternating sign? – Nir Dec 10 '16 at 10:43\n• Also be aware that ARMA models with several lags can model a surprisingly broad set of behavior. AR(1) will decay towards the mean but AR(2) can give you cycles. – Matthew Gunn Dec 10 '16 at 10:49\n• why don't you post your data and I will try and help you . – IrishStat Dec 10 '16 at 11:49" ]

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[ "# Q: What is the total or count of factors of the number 845,124?\n\nA: 24\n\n## How do I find the total factors of the number 845,124?\n\n### Step 1\n\nFind the prime factorization of the number 845,124.\n\nUse a factor tree to assist with solving. Take a look at our prime factorization page for additional help.\n\nFactor Tree\n 845,124", null, "2 422,562", null, "2 211,281", null, "3 70,427", null, "7 10,061\n\nThe prime factorization in exponential form is: 22 x 31 x 71 x 10,0611\n\n### Step 2\n\nSetup the equation for determining the number of factors or divisors. The equation is:\n\nd(n) = (a + 1)(b + 1)(c + 1)(d + 1)\n\nWhere d(n) is equal to the number of divisors of the number and a, b, etc. are equal to the exponents of the prime factorization.\n\nNow substitute the letters in the equation with the the exponents of your prime factorization and then solve to calculate the total number of divisors.\n\n845,124 = 22 x 31 x 71 x 10,0611", null, "d(n) = (a + 1)(b + 1)(c + 1)(d + 1)", null, "d(845124) = (2 + 1)(1 + 1)(1 + 1)(1 + 1)", null, "d(845124) = (3)(2)(2)(2)", null, "d(845124) = 24\n\n### More numbers for you to try\n\nTake a look at the factors page to see the factors of 845,124 and how to find them.\n\nTry the factor calculator.\n\n### Explore more about the number 845,124:\n\nGeneral Questions\nFactoring Questions\nCalculation Questions\nMiscellaneous Questions" ]

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[ "# What is Pre-multiplied Alpha and Why Does it Matter?\n\nThe usual equation for blending two pixels with alpha is to use a source factor of “source alpha” and a destination factor of “one minus source alpha”.\n\nThat results in this equation:", null, "$\\bf{Out.RGB} = In.RGB * In.A + Out.RGB * (1.0 - In.A)$\n\nIn other words, it’s just a linear interpolation between", null, "$In$ and", null, "$Out$, using the alpha of the pixel you are writing to determine the weighting for the lerp.\n\nJust to be clear,", null, "$Out$ is the pixel in your frame buffer (ie the final result that shows up on your monitor) and", null, "$In$ is the new pixel you are trying to write or combine with the output image.\n\nIf you use pre-multiplied alpha, all that means is that", null, "$In.RGB$ is already multipled by", null, "$In.A$ which results in slightly less math:", null, "$\\bf{Out.RGB} = In.RGB + Out.RGB * (1.0 - In.A)$\n\nLess math for the same results is always a good thing due to increased efficiency, but this also results in a higher quality results in the case of using mips. For more info on that check out this link: NVIDIA GameWorks – Alpha Blending: To Pre or Not To Pre\n\nTo achieve this in OpenGL or DirectX with pre-multiplied alpha textures, you just use a source factor of “one” and leave the destination factor at “one minus source alpha”.\n\nIf you are using alpha to write to a render target that also has an alpha channel, the math improvement is even better. Here is the equation with regular (post-multiplied) alpha in that scenario for combining the RGB portion of the pixels:", null, "$\\bf{Out.RGB} = \\frac{In.RGB * In.A + Out.RGB * Out.A * (1.0 - In.A)}{In.A + (1.0 - In.A) * Out.A}$\n\nWhen working with pre-multiplied alpha it just goes back to the lerp:", null, "$\\bf{Out.RGB} = In.RGB + Out.RGB * (1.0 - In.A)$\n\nWhen blending to a target with alpha, the equation to combine alpha is the old familiar lerp:", null, "$\\bf{Out.A} = In.A + Out.A * (1.0 - In.A)$\n\nThat makes the lerp form of color combining even nicer since it means the same math for all color channels of the pixel:", null, "$\\bf{Out} = In + Out * (1.0 - In.A)$\n\nConfused about why alpha combining using lerp makes sense? It makes most sense to me when thinking about it like this… if you had something that was half opaque(Out.A = 0.5), then you looked at that through something that was 3/4 opaque (In.A = 0.75), only 25% of the light would get past the top layer (0.25), to reach the bottom layer. Only 50% of the light that reached the bottom layer gets through, so we cut that 25% in half to get 12.5% (0.125). Since alpha really means “opaqueness” (1.0 means no transparency), that means that the combined alpha is 1 – 0.125, or 0.875.\n\nIf you plug the numbers into the above equation you get the same result:", null, "$\\bf{Out.A} = 0.75 + (1.0 - 0.75) * 0.5$", null, "$\\bf{Out.A} = 0.75 + 0.25 * 0.5$", null, "$\\bf{Out.A} = 0.75 + 0.125$", null, "$\\bf{Out.A} = 0.875$\n\nThis compute graphics stack exchange question has some more great info on what premultiplied alpha can give you. It’s much more than I wrote here:\nDoes premultiplied alpha give order independent transparency?\n\nHere’s a fun question… does alpha represent transparency, or does it represent how much of a pixel is covered by an opaque object? To find out the answer give this a read!\nJounral of Computer Graphics Techniques: Interpreting Alpha" ]

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[ "# Why is the yield keyword used in conjunction with return and break, and not by itself?\n\nIn C# you can construct methods with the return type of `IEnumerable<T>` and use `yield return` and `yield break` to control the flow. Here is a simple example that uses both controls:\n\n``````public IEnumerable<int> GetEvens(int start, int end) {\nif(end < start)\nyield break;\n\nif(start & 2 != 0)\nstart++;\n\nfor(int i = start; i <= end; i+=2) {\nyield return i;\n}\n}\n``````\n\nMy question is, why was it originally designed to use two keywords with `yield` and not use it like the following with the single `yield` \"yielding the return value\":\n\n``````public IEnumerable<int> GetEvens(int start, int end) {\nif(end < start)\nreturn; // stop completely and return nothing\n\nif(start & 2 != 0)\nstart++;\n\nfor(int i = start; i <= end; i+=2) {\nyield i; // yield the current value.\n}\n}\n``````\n\nTo me, this is simpler to read and understand.\n\nThe slightly awkward `yield return` syntax was created so that existing code that used the word \"yield\" as an identifier (variable name) would not break. (It makes perfect sense, for example, to have a variable named `yield` if you're working with financial code.) Since \"yield return\" would have been a syntax error back then, the new syntax would not break any existing code.\nAs for `yield break`, no idea. That really doesn't seem to have any good reason behind it that I can find.\n• Wouldn't `yield x;` also of been a syntax error back then? – Moop Sep 26 '13 at 21:18\n• @Moop: By making it `yield` `return`, you don't even have to bother thinking about whether it was a syntax error or not. The chance of a collision becomes zero. – Robert Harvey Sep 26 '13 at 21:24\n• @Moop if you had a type named `yield` then `yield x;` would not be a syntax error, but a valid variable declaration. – Bojan Resnik Sep 27 '13 at 2:43" ]

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[ "Ncert Maths Chapter 1 Class 8 » accurateqs.com\n\n# NCERT Solutions Class 8 Maths Chapter 1 Rational.\n\nFree download NCERT Solutions for Class 10 Maths Chapter 8 Exercise 8.1 introduction to trigonometry in English Medium and Hindi Medium PDF form to use it offline. NCERT Textbook solutions for class 10 other subjects are also given to free use. Download Exercise 8.2 or Exercise 8.3 or Exercise 8.4 also in form or go for online use only. NCERT Solutions for Class 10 Maths Chapter 8 Exercise 8.1. 19/01/40 · Mensuration - Introduction for Chapter 11 Class 8th mathematics, Check more solutions for Class 8th Maths NCERT / CBSE. Get Textbook solutions for Maths on evidyarthi.in. 26/12/40 · NCERT Solutions Class 8 Maths Chapter 1 Rational Numbers – Here are all the NCERT solutions for Class 8 Maths Chapter 1. This solution contains questions, answers, images, explanations of the complete chapter 1 titled Rational Numbers of Maths taught in class 8. If you are a student of class 8 who is using NCERT [].\n\n16/08/39 · Get here NCERT Solutions for Class 8 Maths Chapter 8. These NCERT Solutions for Class 8 of Maths subject includes detailed answers of all the questions in Chapter 8 – Comparing Quantities provided in NCERT Book which is prescribed for class 8 in schools. Resource: National Council of Educational Research and Training NCERT Solutions Class: 8th Class Subject: Maths []. 18/01/40 · Algebraic Expressions and Identities - Introduction for Chapter 9 Class 8th mathematics, Check more solutions for Class 8th Maths NCERT / CBSE. Get Textbook solutions for Maths on www. 26/12/40 · NCERT Solutions Class 8 Maths Chapter 12 Exponents and Powers – Here are all the NCERT solutions for Class 8 Maths Chapter 12. This solution contains questions, answers, images, explanations of the complete chapter 12 titled Exponents and Powers of Maths taught in class 8. If you are a student of class 8 who is []. 12/10/38 · Chapter 8 TRIGONOMETRY Exercise 8.1 maths class 10 NCERT in English or Hindi. Chapter 8 TRIGONOMETRY Exercise 8.1 part 2 maths class 10 NCERT in English or Hindi Math.\n\n## NCERT Solutions for class 12 Maths Chapter 8 Exercise.\n\nNCERT Solutions for Class 8 Maths Chapter 12 Exercise 12.1 Exponents and Powers in English Medium and Hindi Medium to use online free without any login and password. Important questions for practice are given at the end of the page. The NCERT solutions for class 8 Maths Chapter 1 is framed in line with the most recent course of study of the CBSE board. These solutions measure very useful in securing good marks within the boards. Also, you’ll be able to get resolved paper of the preceding ten years. Apart from that, these resolved papers can assist you to know the chapter. Math - NCERT Solutions for class 8. NCERT Solutions for Class 8 Math Chapter 1 - Rational Numbers. Mathematics NCERT Grade 8, Chapter 1: Rational Numbers- Basic operations on natural numbers, integers and rational numbers are already discussed. In this chapter, students will try to explore some properties of rational numbers like closure, commutativity, associativity.", null, "" ]

[ null, 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", null ]

[ "", null, "", null, "The Egg Production Efficiency Index (EPEI) as an economic indicator for measuring poultry egg production\nHristo Lukanov, Atanas Genchev, Todor Petrov\nAbstract: The aim of the study was to introduce an up-to-date index for evaluation of economic efficiency of egg production in domestic fowl. The proposed egg production efficiency index (EPEI) was calculated using the following equation: EPEI = [(L × DEMP)/FCR)] × 100, where: L – Liveability (%); DEMP – Daily egg mass produced (kg); FCR – Feed conversion ratio (kg/ kg egg mass). DEMP = (HDEP x AEW)/t, where HDEP – Hen-Day egg production (number); AEW – Average egg weight\n(kg); t – period (days). The ideal egg production efficiency index, calculated on the basis of a full production period, equals 310. For a production period of 52-81 weeks, EPEI of 230 may be estimated as a threshold for modern white and brown egg layer hybrids. EPEI may be also applied for monitoring of efficiency of egg production dynamics in a given group of stock layer hens. The index may be also used in other domestic fowl species.\nKeywords: egg production; egg weight; FCR; production efficiency; table eggs\nDate published: 2023-08-21" ]

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[ "# MCQ Questions for Class 10 Maths Chapter 10 Circles with Answers\n\nBelow you will find NCERT MCQ Questions for Class 10 Maths Chapter 10 Circles with Answers which is very helpful during the preparation of examinations. These MCQ Online Test are one marks questions which will give you overview of the chapter in no time. One should try to understand Class 10 MCQ Questions as it is based on latest exam pattern released by CBSE.\n\nAlso, students can check NCERT Solutions for Class 10 Maths Chapter 10 for improving their marks and have good understanding of the chapter.", null, "## Chapter 10 Circles MCQ Questions for Class 10 Maths with Answers\n\n1. A tangent to a circle is a line that intersects the circle in\n(a) Exactly one point\n(b) 2 points\n(c) 3 points\n(d) 4 points\nSolution (a) Exactly one point\n\n2. At point A on a diameter AB of a circle of radius 10 cm, tangent XAY is drawn to the circle. The length of the chord CD parallel to XY at a distance 16 cm from A is\n(a) 8 cm\n(b) 10 cm\n(c) 16 cm\n(d) 18 cm\nSolution (c) 16 cm\n\n3. If TP and TQ are two tangents to a circle with centre O so that ZPOQ = 110°, then, ZPTQ is equal\nto\n(a) 60°\n(b) 70°\n(c) 80°\n(d) 90°\nSolution (b) 70°\n\n4. Two parallel lines touch the circle at points A and B respectively. If area of the circle is 25 cm,2,, then AB is equal to\n(a) 5 cm\n(b) 7 cm\n(c) 8 cm\n(d) 10 cm\nSolution (c) 8 cm\n\n5. The maximum number of common tangents that can be drawn to two circles intersecting at two distinct point is\n(a) 2\n(b) 4\n(c) 1\n(d) 3\nSolution (a) 2\n\n6. In the given figure, length of BC is:", null, "(a) 9 cm\n(b) 10 cm\n(c) 13 cm\n(d) 21 cm\nSolution (c) 13 cm\n\n7. A tangent PQ at a point P of a circle of radius 5 cm meets a line through the centre O at a point Q so that OQ = 12 cm. Length PQ is :\n(a) √119 cm\n(b) 13 cm\n(c) 12 cm\n(d) 8.5 cm\nSolution (a) √119 cm\n\n8. The length of tangent PQ, from an external point P is 24 cm. If the distance of the point P from the centre is 25 cm, then the diameter of the circle is", null, "(a) 7 cm\n(b) 14 cm.\n(c) 15 cm.\n(d) 12 cm\nSolution (b) 14 cm.\n\n9. PQ is a tangent drawn from a point P to a circle with centre O and QOR is a diameter of the circle such that ∠POR = 120°, then ∠OPQ is\n(a) 60°\n(b) 45°\n(c) 30°\n(d) 90°\nSolution (c) 30°\n\n10. A circle touches x-axis at A and y-axis at B. If O is origin and OA = 5 units, then diameter of the circle is\n(a) 8 units\n(b) 10 units\n(c) 10 √2 units\n(d) 8 √2 units\nSolution (b) 10 units\n\n11. In a circle with centre O, AB and CD are two diameters perpendicular to each other. The length of chord AC is", null, "(a) 1/√2 AB\n(b) 1/2 AB\n(c) √2 AB\n(d) 2 AB\nSolution (a) 1/√2 AB\n\n12. The length of the tangent from a point A at a circle, of radius 3 cm, is 4 cm. The distance of A from the centre of the circle is\n(a) √7 cm\n(b) 7cm\n(c) 5 cm\n(d) 25 cm\nSolution (c) 5 cm\n\n13. Two parallel lines touch the circle at points A and B respectively, If area of the circle is 25πcm2,, then AB is equal to\n(a) 5 cm\n(b) 8 cm\n(c) 10 cm\n(d) 25 cm\nSolution (c) 10 cm\n\n14. If four sides of a quadrilateral ABCD are tangential to a circle, then\n(a) AC + AD = BD + CD\n(b) AB + CD = BC + AD\n(c) AB + CD = AC + BC\n(d) AC + AD = BC + DB\nSolution (b) AB + CD = BC + AD\n\n15. A tangent PQ at a point P o a circle of radius 5 cm meets a line through the centre O at point Q, so that OQ = 12 cm. find the length PQ.", null, "(a) √113 cm\n(b) 13 cm\n(c) √119 cm\n(d) 26 cm\nSolution (c) √119 cm\n\n16. If the angle between two radii of a circle is 100°, then angle between the tangents at the ends of those radii is:​\n(a) 80°\n(b) 90°\n(c) 60°\n(d) 50°\nSolution (a) 80°\n\n17. A point P is 25 cm from the centre of a circle. The radius of the circle is 7 cm and length of the tangent drawn from P to the circle is x cm. The value of x =\n(a) 20 cm\n(b) 24 cm\n(c) 18 cm\n(d) 12 cm\nSolution (b) 24 cm\n\n18. A circle is inscribed in a ΔABC having AB = 10cm, BC = 12cm and CA = 8cm and touching these sides at D, E, F respectively. The lengths of AD, BE and CF will be", null, "(a) AD = 4cm, BE = 6cm, CF = 8cm\n(b) AD = 5cm, BE = 9cm, CF = 4cm\n(c) AD = 3cm, BE = 7cm, CF = 5cm\n(d) AD = 2cm, BE = 6cm, CF = 7cm\nSolution (c) AD = 3cm, BE = 7cm, CF = 5cm\n\n19. A point P is 10 cm from the centre of a circle. The length of the tangent drawn from P to the circle is 8 cm. The radius of the circle is equal to\n(a) 4 cm\n(b) 5 cm\n(c) 6 cm\n(d) None of these.\nSolution (c) 6 cm\n\n20. At one end of a diameter PQ of a circle of radius 5 cm, tangent XPYis drawn to the circle. The length of chord AB parallel to XY and at a distance of 8 cm from P is\n(a) 6 cm.\n(b) 5 cm.\n(c) 7 cm.\n(d) 8 cm\nSolution (d) 8 cm\n\n21. If angle between two radii of a circle is 130°, the angle between the tangents at the ends of the radii is\n(a) 90°\n(b) 50°\n(c) 70°\n(d) 40°\nSolution (b) 50°" ]

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[ "# Guessing Fourier Coefficients\n\nGiven a ‘nice enough’ periodic function, it can be expressed as an infinite sum of sine and cosine functions, known as its Fourier Series.\n\nIn some cases, formulas specifying a_n and b_n can be computed by hand using the Euler-Fourier formulas. When these these methods are inconvenient, approximate coefficients can be found using a random search in the space of all possible coefficients.\n\n## The Coefficient Arrays\n\nThe random search works like this: first, start with two arrays of n arbitrary values each, where n is the number of trigonometric terms in the approximation. These arrays contain the cosine and sine coefficients, respectively.\n\nThe approximation corresponding to these arrays for a function periodic on the interval [-L, L] is f \\approx S = \\sum_{k=0}^{n-1}a_k\\cos\\left(\\frac{kx\\pi}{L}\\right) + b_n \\sin\\left(\\frac{kx\\pi}{L}\\right)\n\n## Perturbing the Arrays\n\nNext, randomly pick one of these 2n elements in A or B and modify it slightly by adding random \\varepsilon such that -0.5<\\varepsilon<0.5.\n\n## Quantifying Error\n\nHow do we know that our arrays give rise to a good approximation? In the continuous setting, the error can be described as the total (unsigned) area between the curves of f and S in the interval [-L,L]. This can be computed exactly, with mathematical software packages.\n\nWhen these integrals are difficult to compute, it is often useful to use a discrete proxy for this term. Split up the interval [-L,L] with a mesh M = (t_0, t_1, t_2, … , t_m) where t_0 = -L, t_m = L, and t_i < t_j if i < j. Then this area is usually smaller when\n\nE = \\sum_{i=0}^m \\left(f(t_i) – S(t_i)\\right)^2 is smaller.\n\n## Iterating\n\nAltering the coefficient arrays will produce a different S and therefore give a different value for E. This error can either be equal to or greater than the coefficient array before it was altered.\n\nIf the error is reduced, the random search algorithm accepts the new array coefficient and erases the previous one.\n\nIf the error is not reduced, the program maintains the coefficient array until it finds a perturbation that improves the error. This is the procedure that’s in the animation at the top. Feel free to play with different settings and come up with pathological examples!\n\nLink to the code (requires Python, NumPy, and matplotlib)\n\n## As a map from \\mathbb{R}^n \\to \\mathbb{R}\n\nThis problem of optimizing a function \\mathbb{R}^n \\to \\mathbb{R} has been studied extensively. For example, creating the neural networks popular today boil down to finding sets of numbers (activation values and biases of neurons) that minimize cost functions.\n\nThe computer scientists frequently tackle this problem with gradient descent techniques. Given a function E(\\vec{x}), its gradient, \\nabla E(\\vec{x}) returns the direction of greatest increase and conveniently, -\\nabla E(\\vec{x}) returns the direction of fastest decrease.\n\nThe main idea is that to find input values to minimize a function, start at some position x_0. Then calculate \\nabla E(\\vec{x_0}) and define a better approximation x_1 by pushing x_0 a tiny bit in the opposite direction. Iterating this process should output a local minimum in the search space, if everything goes right.\n\nThe code for this method is provided in the link above, using a simple forward difference approximation for the gradient.\n\n## 2 Replies to “Guessing Fourier Coefficients”\n\n1.", null, "Eduardo says:\n\nExample reply! Making some sitewide edits. ¡Oralé!\n\n2.", null, "Alfredo Arroyo says:\n\nOne of the links doesn’t work 🙁" ]

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[ "# How Purely Nested Notation Limits the Language's Utility\n\nBy Xah Lee. Date: .\n\n## How Purely Nested Notation Limits the Language's Utility\n\nThere is a common complain by programers about lisp's notation, of nested parenthesis, being unnatural or difficult to read. Long time lisp programers, often counter, that it is a matter of conditioning, and or blaming the use of “inferior” text editors that are not designed to display nested notations. In the following, i describe how lisp notation is actually a problem, in several levels.\n\n(1) Some 99% of programers are not used to the nested parenthesis syntax. This is a practical problem. On this aspect alone, lisp's syntax can be considered a problem.\n\n(2) Arguably, the pure nested syntax is not natural for human to read. Long time lispers may disagree on this point.\n\n(3) Most importantly, a pure nested syntax discourages frequent or advanced use of function sequencing or compositions. This aspect is the most devastating.\n\nThe issue that most programers are not comfortable with nested notation is well known. It is not a technical issue. Whether it is considered a problem of the lisp language is a matter of philosophical disposition.\n\nThe issue of heavy nested not being natural for human to read, may be debatable. I do think, that deep nesting is a problem to the programer. Here's a example of 2 blocks of code that are syntactically equivalent in the Mathematica language:\n\n```vectorAngle[{a1_, a2_}] := Module[{x, y},\n{x, y} = {a1, a2}/Sqrt[a1^2 + a2^2] // N;\nIf[x == 0, If[Sign@y === 1, π/2, -π/2],\nIf[y == 0, If[Sign@x === 1, 0, π],\nIf[Sign@y === 1, ArcCos@x, 2 π - ArcCos@x]\n]\n]\n]\n```\n```SetDelayed[vectorAngle[List[Pattern[a1,Blank[]],Pattern[a2,Blank[]]]],\nModule[List[x,y],\nCompoundExpression[\nSet[List[x,y],\nN[Times[List[a1,a2],\nPower[Sqrt[Plus[Power[a1,2],Power[a2,2]]],-1]]]],\nIf[Equal[x,0],\nIf[SameQ[Sign[y],1],Times[π,Power[2,-1]],\nTimes[Times[-1,π],Power[2,-1]]],\nIf[Equal[y,0],If[SameQ[Sign[x],1],0,π],\nIf[SameQ[Sign[y],1],ArcCos[x],\nPlus[Times[2,π],Times[-1,ArcCos[x]]]]]]]]]\n```\n\nIn the latter, it uses a full nested form (called FullForm in Mathematica). This form is isomorphic to lisp's nested parenthesis syntax, token for token (i.e. lisp's `(f a b)` is Mathematica's `f[a,b]`). As you can see, this form, by the sheer number of nested brackets, is in practice problematic to read and type. In Mathematica, nobody really program using this syntax. (The FullForm syntax is there, for the same reason of language design principle shared with lisp of “consistency and simplicity”, or the commonly touted lisp advantage of “data is program; program is data”.)\n\nThe following shows the same code with tokens transformed into the lisp style.\n\n```(SetDelayed\n(vectorAngle (list (Pattern a1 (Blank)) (Pattern a2 (Blank))))\n(let (list x y)\n(progn\n(setq (list x y)\n(N (* (list a1 a2)\n(exp (sqrt (+ (exp a1 2) (exp a2 2))) -1))))\n(if (eq x 0)\n(if (equal (signum y) 1) (* π (exp 2 -1))\n(* (* -1 π) (exp 2 -1)))\n(if (eq y 0) (if (equal (signum x) 1) 0 π)\n(if (equal (signum y) 1) (acos x)\n(+ (* 2 π) (* -1 (acos x)))))))))```\n\nNote: The steps to transform the sample Mathematica code in FullForm to lisp's sexp is done by these operations (for those curious):\n\n(1) Move the head inside. (using this regex `\\([A-Za-z]+\\)\\[``[\\1`)\n\n(2) Replace bracket by parens. `[]``()`\n\n(3) Replace function names to lisp styled names:\n\n• Plus → +\n• Times → *\n• List → list\n• If → if\n• Set → setq\n• Power → exp\n• Sqrt → sqrt\n• Equal → eq\n• SameQ → equal\n• Sign → signum\n• ArcCos → acos\n• Module → let\n• CompoundExpression → progn\n• Replace comma to space. (`,`` `)\n\nThe third issue, about how nested syntax seriously discourages frequent use of inline function sequencing, is the most important.\n\nOne practical way to see how this is so, is by considering unix's shell syntax. You all know, how convenient and powerful is the unix's pipes. Here are some practical example: `ls -al | grep xyz`, or ```cat a b c | grep xyz | sort | uniq```.\n\nNow suppose, we get rid of the unix's pipe notation, instead, replace it with a pure functional notation, for example: ```(uniq (sort (grep xyz (cat a b c))))```, or enrich it with a composition function and a pure function construct, so this example can be written as: `((composition uniq sort (lambda (x) (grep xyz x))) (cat a b c))`.\n\nYou see, how this change, although syntactically equivalent to the pipe “|” (or semantically equivalent in the example using function compositions), but due to the cumbersome nesting of parenthesis, will force a change in the nature of the language by the code programer produces. Namely, the frequency of inline sequencing of functions on the fly will probably be reduced, instead, there will be more code that define functions with temp variables and apply it just once as with traditional languages.\n\nA language's syntax or notation system, has major impact on what kind of code or style or thinking pattern on the language's users. This is a well-known fact for those acquainted with the history of math notations.\n\nThe sequential prefix notation such as Haskell's `f g h x`, Mathematica's `f@g@h@x`, or postfix as unix's `x|h|g|f`, Ruby's `x.h.g.f`, Mathematica's `x//h//g//f`, are much more convenient and easier to decipher, than `(f (g (h x)))` or `((composition f g h) x)`. In real world source code, any of the f, g, h will likely be full of nested parens themselves, because often they are functions constructed inline (aka lambda).\n\nLisp, by sticking with a almost uniform nested parenthesis notation, it immediately reduces the pattern of sequencing functions, simply because the syntax does not readily lend the programer to it. For programers who are aware of the coding pattern of sequencing functions (aka function chaining, filtering), now either need to think in terms of a separate “composition” construct, and or subject to the much problematic typing and deciphering of nested parenthesis. (in practice, it's mostly done by writing the inline functions as a auxiliary function definitions on their own, then another code block sequence them together. For example: `(defun f …) (defun g …) (defun h …) (f (g (h x)))` )\n\nNote: Lisp syntax is actually not a pure nested form. It has ad hoc syntax equivalents such as the `quote` construct `'(a b c)`, dotted notation for cons, for example: `(a . b)` for `(cons a b)`, special syntax for quoted vector `[1 2 3]`, splice and partial hold eval, for example: ``(,@ x ,@ y ,z 4))`, weirded comment syntax `#|something|#`. Mathematica's FullForm, is actually a version of unadulterated nested notation. For a full discussion, see: Fundamental Problems of Lisp\n\nFor a practical example of this problem, see: LISP Syntax Problem of Piping Functions ." ]

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[ "# Continuous Distributions\n\nContinuous distributions\n\nIn this section we discuss the set of continuous distributions available in AIMMS.\n\nThe three distributions with both lower and upper bound are\n\nThe five distributions with only a lower bound are\n\nThe three unbounded distributions are\n\nParameters of continuous distributions\n\nEvery parameter of a continuous distributions can be characterized as either a shape parameter $$\\beta$$, a location parameter $$l$$, or a scale parameter $$s$$. While the presence and meaning of a shape parameter is usually distribution-dependent, location and scale parameters find their origin in the common transformation\n\n$x \\mapsto \\frac{x-l}{s}$\n\nto shift and stretch a given distribution. By choosing $$l=0$$ and $$s=1$$ the standard form of a distribution is obtained. If a certain distribution has $$n$$ shape parameters ($$n \\geq 0$$), these shape parameters will be passed as the first $$n$$ parameters to AIMMS. The shape parameters are then followed by two optional parameters, with default values 0 and 1 respectively. For double-bounded distributions these two optional parameters can be interpreted as a lower and upper bound (the value of the location parameter $$l$$ for these distributions is equal to the lower bound and the value of the scale parameter $$s$$ is equal to the difference between the upper and lower bound). For single-bounded distributions the bound value is often used as the location parameter $$l$$. In this section, whenever the location parameter can be interpreted as a mean value or whenever the scale parameter can be interpreted as the deviation of a distribution, these more meaningful names are used to refer to the parameters. Note that the LogNormal, Gamma and Exponential distributions are distributions that will mostly be used with location parameter equal to 0.\n\nTransformation to standard form\n\nWhen transforming a distribution to standard form, distribution operators change. Scaling of Statistical Operators (scaling of statistical operators) gives the relationships between distribution operators working on random variables $$X(l,s)$$ and $$X(0,1)$$.\n\nUnits of measurement\n\nWhen a random variable representing some real-life quantity with a given unit of measurement (see also Units of Measurement) is distributed according to a particular distribution, some parameters of that distribution are also naturally expressed in terms of this same unit while other parameters are expected to be unitless. In particular, the location and scale parameters of a distribution are measured in the same unit of measurement as the corresponding random variable, while shape parameters (within AIMMS) are implemented as unitless parameters.\n\nUnit notation in this appendix\n\nWhen you use a distribution function, AIMMS will perform a unit consistency check on its parameters and result, whenever your model contains one or more QUANTITY declarations. In the description of the continuous distributions below, the expected units of the distribution parameters are denoted in square brackets. Throughout the sequel, [$$x$$] denotes that the parameter should have the same unit of measurement as the random variable $$X$$ and [-] denotes that a parameter should be unitless.\n\nA commonly used distribution\n\nIn practice, the Normal distribution is used quite frequently. Such widespread use is due to a number of pleasant properties:\n\n• the Normal distribution has no shape parameters and is symmetrical,\n\n• random values are more likely as they are closer to the mean value,\n\n• it can be directly evaluated for any given mean and standard deviation because it is fully specified through the mean and standard deviation parameter,\n\n• it can be used as a good approximation for distributions on a finite interval, because its probability density is declining fast enough (when moving away from the mean),\n\n• the mean and sum of any number of uncorrelated Normal distributions are Normal distributed themselves, and thus have the same shape, and\n\n• the mean and sum of a large number of uncorrelated distributions are always approximately Normal distributed.\n\nDistributions for double bounded variables\n\nFor random variables that have a known lower and upper bound, AIMMS provides three continuous distributions on a finite interval: the Uniform, Triangular and Beta distribution. The Uniform (no shape parameters) and Triangular (one shape parameter) distributions should be sufficient for most experiments. For all remaining experiments, the user might consider the highly configurable Beta (two shape parameters) distribution.\n\nDistributions for single bounded variables\n\nWhen your random variable only has a single bound, you should first check whether the Gamma distribution can be used or whether the Normal distribution is accurate enough. The LogNormal distribution should be considered if the most likely value is near but not at the bound. The Weibull or Gamma distribution ($$\\beta>1$$), or even the ExtremeValue distribution are alternatives, while the Weibull or Gamma distribution ($$\\beta \\leq 1$$) or Pareto distribution should be considered if the bound is the most likely value.\n\nThe Gamma distribution\n\nThe Gamma (and as a special case thereof the Exponential) distribution is widely used for its special meaning. It answers the question: how long does it take for a success to occur, when you only know the average number of occurrences (like in the Poisson distribution). The Exponential distribution gives the time to the first occurrence, and its generalization, the Gamma($$\\beta$$) distribution gives the time to the $$\\beta$$-th occurrence. Note that the sum of a Gamma($$\\beta_1,l_1,s$$) and Gamma($$\\beta_2,l_2,s$$) distribution has a Gamma($$\\beta_1+\\beta_2,l_1+l_2,s$$) distribution.\n\nThe LogNormal distribution\n\nIf you assume the logarithm of a variable to be Normal distributed, the variable itself is LogNormal-distributed. As a result, it can be shown that the chance of an outcome in the interval $$[x \\!\\cdot\\! c_1,x \\!\\cdot\\! c_2]$$ is equal to the chance of an outcome in the interval $$[x/c_2,x/c_1]$$ for some $$x$$. This might be a reasonable assumption in price developments, for example.\n\nThe Uniform distribution\n\nThe Uniform(min,max) distribution:\n\n Input parameters min [$$x$$], max [$$x$$] Input check $${min} < {max}$$ Permitted values $$\\{ x \\; | \\; {min} \\leq x \\leq {max} \\}$$ Standard density $$f_{(0,1)}(x) = 1$$ Mean $$1/2$$ Variance $$1/12$$\n\nIn the Uniform distribution all values of the random variable occur between a fixed minimum and a fixed maximum with equal likelihood. It is quite common to use the Uniform distribution when you have little knowledge about an uncertain parameter in your model except that its value has to lie anywhere within fixed bounds. For instance, after talking to a few appraisers you might conclude that their single appraisals of your property vary anywhere between a fixed pessimistic and a fixed optimistic value.\n\nThe Triangular distribution\n\nThe Triangular($$\\beta$$,min,max) distribution:\n\n Input parameters shape $$\\beta$$ [$$-$$],min [$$x$$], max [$$x$$] Input check $${min} < {max }, \\; 0 < \\beta < 1$$ Permitted values $$\\{ x \\; | \\; {min} \\leq x \\leq {max} \\}$$ Standard density $$f_{(\\beta,0,1)}(x) = \\begin{cases} 2 x / \\beta & \\text{for 0 \\leq x \\leq \\beta} \\\\ 2 (1-x)/(1-\\beta) & \\text{for \\beta In the Triangular distribution all values of the random variable occur between a fixed minimum and a fixed maximum, but not with equal likelihood as in the Uniform distribution. Instead, there is a most likely value, and its position is not necessarily in the middle of the interval. It is quite common to use the Triangular distribution when you have little knowledge about an uncertain parameter in your model except that its value has to lie anywhere within fixed bounds and that there is a most likely value. For instance, assume that a few appraisers each quote an optimistic as well as a pessimistic value of your property. Summarizing their input you might conclude that their quotes provide not only a well-defined interval but also an indication of the most likely value of your property. The Beta distribution The Beta(\\(\\alpha$$,$$\\beta$$,min,max) distribution:\n\n Input parameters shape $$\\alpha$$ [-], shape $$\\beta$$ [-], min [$$x$$], max [$$x$$] Input check $$\\alpha > 0, \\beta > 0, {min} < {max}$$ Permitted values $$\\{x \\; | \\; {min} < x < {max} \\}$$ Standard density $$f_{(\\alpha,\\beta,0,1)}(x) = \\frac{1}{B(\\alpha,\\beta)} x^{\\alpha - 1} (1-x)^{\\beta - 1}, \\; \\text{where B(\\alpha,\\beta) is the Beta function}$$ Mean $$\\alpha/(\\alpha+\\beta)$$ Variance $$\\alpha\\beta(\\alpha+\\beta)^{-2}(\\alpha+\\beta+1)^{-1}$$ Remarks $${\\texttt{Beta}}(1,1,{min},{max})={\\texttt{Uniform}}({min},{max})$$\n\nThe Beta distribution is a very flexible distribution whose two shape parameters allow for a good approximation of almost any distribution on a finite interval. The distribution can be made symmetrical, positively skewed, negatively skewed, etc. It has been used to describe empirical data and predict the random behavior of percentages and fractions. Note that for $$\\alpha<1$$ a singularity occurs at $$x=\\text{{min}}$$ and for $$\\beta<1$$ at $$x=\\text{{max}}$$.\n\nThe LogNormal distribution\n\nThe LogNormal($$\\beta$$,min,s) distribution:\n\n Input parameters shape $$\\beta$$ [-], lowerbound min [$$x$$] and scale $$s$$ [$$x$$] Input check $$\\beta > 0 \\; \\mbox{and} \\; s > 0$$ Permitted values $$\\{ x \\; | \\; {min} < x < \\infty \\}$$ Standard density $$f_{(\\beta,0,1)}(x) = \\frac{1} { \\sqrt{2 \\pi} x \\ln(\\beta^2+1) } e^{ \\frac{ -(\\ln(x^2(\\beta^2+1)) } {2 \\ln(\\beta^2+1) } }$$ Mean $$1$$ Variance $$\\beta^2$$\n\nIf you assume the logarithm of the variable to be Normal($$\\mu,\\sigma$$)-distributed, then the variable itself is LogNormal($$\\sqrt{e^{\\sigma^2}\\!\\! - \\!\\! 1},0,e^{\\mu - \\sigma^2/2}$$)-distributed. This parameterization is used for its simple expressions for mean and variance. A typical example is formed by real estate prices and stock prices. They all cannot drop below zero, but they can grow to be very high. However, most values tend to stay within a particular range. You usually can form some expected value of a real estate price or a stock price, and estimate the standard deviation of the prices on the basis of historical data.\n\nThe Exponential distribution\n\nThe Exponential(min,$$s$$) distribution:\n\n Input parameters lowerbound min [$$x$$] and scale $$s$$ [$$x$$] Input check $$s > 0$$ Permitted values $$\\{ x \\; | \\; {min} \\leq x < \\infty \\}$$ Standard density $$f_{(0,1)}(x) = \\lambda e^{-x}$$ Mean $$1$$ Variance $$1$$ Remarks Exponential (min, $$s$$) = Gamma (1, min, $$s$$), Exponential (min, $$s$$) = Weibull (1, min, $$s$$)\n\nAssume that you are observing a sequence of independent events with a constant chance of occurring in time, with s being the average time between occurrences. (in accordance with the Poisson distribution) The Exponential($$0,s$$) distribution gives answer to the question: how long a time do you need to wait until you observe the first occurrence of an event. Typical examples are time between failures of equipment, and time between arrivals of customers at a service desk (bank, hospital, etc.).\n\nThe Gamma distribution\n\nThe Gamma($$\\beta$$,min,$$s$$) distribution:\n\n Input parameters shape $$\\beta$$ [-], lowerbound min [$$x$$] and scale $$s$$ [$$x$$] Input check $$s > 0 \\; \\mbox{and} \\; \\beta > 0$$ Permitted values $$\\{x \\; | \\; {min} < x < \\infty\\}$$ Standard density $$f_{(\\beta,0,1)}(x) = x^{\\beta - 1} e^{-x} / {\\Gamma ( \\beta )} \\\\ \\mbox{where} \\; \\Gamma ( \\beta ) \\; \\mbox{is the Gamma function}$$ Mean $$\\beta$$ Variance $$\\beta$$\n\nThe Gamma distribution gives answer to the question: how long a time do you need to wait until you observe the $$\\beta$$-th occurrence of an event (instead of the first occurrence as in the Exponential distribution). Note that it is possible to use non-integer values for $$\\beta$$ and a location parameter. In these cases there is no natural interpretation of the distribution and for $$\\beta<1$$ a singularity exists at $$x={min}$$, so one should be very careful in using the Gamma distribution this way.\n\nThe Weibull distribution\n\nThe Weibull($$\\beta$$,min,$$s$$) distribution:\n\n Input parameters shape $$\\beta$$ [-], lowerbound min [$$x$$] and scale $$s$$ [$$x$$] Input check $$\\beta > 0 \\; \\mbox{and} \\; s > 0$$ Permitted values $$\\{x \\; | \\; {min} \\leq x < \\infty\\}$$ Standard density $$f_{(\\beta,0,1)}(x) = \\beta x^{\\beta - 1} e^{-x^\\beta}$$ Mean $$\\Gamma(1+1/\\beta)$$ Variance $$\\Gamma(1+2/\\beta)-\\Gamma^2(1+1/\\beta)$$\n\nThe Weibull distribution is another generalization of the Exponential distribution. It has been successfully used to describe failure time in reliability studies, and the breaking strengths of items in quality control testing. By using a value of the shape parameter that is less than 1, the Weibull distribution becomes steeply declining and could be of interest to a manufacturer testing failures of items during their initial period of use. Note that in that case there is a singularity at $$x={min}$$.\n\nThe Pareto distribution\n\nThe Pareto($$\\beta$$,$$l$$,$$s$$) distribution:\n\n Input parameters shape $$\\beta$$ [-], location $$l$$ [$$x$$] and scale $$s$$ [$$x$$] Input check $$s > 0 \\; \\mbox{and} \\; \\beta > 0$$ Permitted values $$\\{ x \\; | \\; l+s < x < \\infty \\}$$ Standard density $$f_{(\\beta,0,1)}(x) = \\beta / x^{\\beta + 1}$$ Mean $$\\mbox{for } \\beta>1:\\; \\beta/(\\beta-1), \\infty \\text{ otherwise}$$ Variance $$\\mbox{for } \\beta>2:\\; \\beta(\\beta-1)^{-2}(\\beta-2)^{-1}, \\infty \\text{ otherwise}$$\n\nThe Pareto distribution has been used to describe the sizes of such phenomena as human population, companies, incomes, stock fluctuations, etc.\n\nThe Normal distribution\n\nThe Normal($$\\mu$$,$$\\sigma$$) distribution:\n\n Input parameters Mean $$\\mu$$ [$$x$$] and standard deviation $$\\sigma$$ [$$x$$] Input check $$\\sigma > 0$$ Permitted values $$\\{ x \\; | \\; -\\infty < x < \\infty \\}$$ Standard density $$f_{(0,1)}(x) = e^{-x^2/2}/\\sqrt{2 \\pi}$$ Mean $$0$$ Variance $$1$$ Remarks Location $$\\mu$$, scale $$\\sigma$$\n\nThe Normal distribution is frequently used in practical applications as it describes many phenomena observed in real life. Typical examples are attributes such as length, IQ, etc. Note that while the values in these examples are naturally bounded, a close fit between such data values and normally distributed values is quite common in practice, because the likelihood of extreme values away from the mean is essentially zero in the Normal distribution.\n\nThe Logistic distribution\n\nThe Logistic($$\\mu$$,$$s$$) distribution:\n\n Input parameters mean $$\\mu$$ [$$x$$] and scale $$s$$ [$$x$$] Input check $$s > 0$$ Permitted values $$\\{x \\; | \\; -\\infty < x < \\infty \\}$$ Standard density $$f_{(0,1)}(x) = ( e^x + e^{-x} + 2 )^{-1}$$ Mean $$0$$ Variance $$\\pi^2/3$$\n\nThe Logistic distribution has been used to describe growth of a population over time, chemical reactions, and similar processes. Extreme values are more common than in the somewhat similar Normal distribution\n\nThe Extreme Value distribution\n\nThe Extreme Value($$l$$,$$s$$) distribution:\n\n Input parameters Location $$l$$ [$$x$$] and scale $$s$$ [$$x$$] Input check $$s > 0$$ Permitted values $$\\{ x \\; | \\; -\\infty < x < \\infty \\}$$ Standard density $$f_{(0,1)}(x) = e^x e^{-e^x}$$ Mean $$\\gamma=0.5772\\dots\\mbox{ (Euler's constant)}$$ Variance $$\\pi^2/6$$ Remarks Extreme Value\n\ndistributions have been used to describe the largest values of phenomena observed over time: water levels, rainfall, etc. Other applications include material strength, construction design or any other application in which extreme values are of interest. In literature the Extreme Value distribution that is provided by AIMMS is known as a type 1 Gumbel distribution." ]

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[ "", null, "# Composite Solids: Lesson\n\n##### This Composite Solids: Lesson video also includes:\n\nCombine the formulas to arrive at one volume. Using the formulas for different solids, the video introduces finding the volume of composite solids. The portion of an extensive playlist on geometry works a couple of problems, one with an additive process and one with a subtractive process.\n\n##### Instructional Ideas\n\n• Ask class members to create composite solids and swap with a classmate to calculate the volume\n##### Classroom Considerations\n\n• Assumes pupils already know the surface area and volume formulas for 3-D shapes\n• Expects individuals know the Pythagorean Theorem\n##### Pros\n\n• Reviews the surface area and volume formulas for solids\n• Links to practice problems and more videos with composite solids\n\n• None" ]

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[ "Chimera Commands Index\n\n### Usage: shapeshapeoptions\n\nThe shape command creates a surface model of the specified shape. See also: define, mask, meshmol, ribbon, sop, geometric objects\n\nshape sphere center ligand color dodger blue radius 10\nshape cylinder center 12.5,15,15 coord 1 rad 2 caps true\nshape icosahedron radius 100 ori 2n3 lattice 1,2 sphere 0.2 line 2\nThe shape can be:\nOption keywords and their subkeywords can be truncated to unique strings, and their case does not matter. Synonyms for true: True, 1. Synonyms for false: False, 0. A vertical bar “|” designates mutually exclusive options, and default settings are indicated with bold.\n\nshape sphereradius r ] general-options\n\nThe default radius r is 10.\nshape ellipsoidradius r | rx,ry,rz ] general-options\nThe default radius r is 10 along each axis. If a single value is specified, the result is a sphere. Specifying three values (rx,ry,rz) sets ellipsoid radii along the X, Y, and Z axes, respectively (see coordinateSystem).\n\nshape coneradius rbase ] [ height h ] [ topRadius rtop ] [ caps true|false ] general-options\nThe default radius rbase and height h are 10 and 40, respectively. The default orientation is with the axis of symmetry (height dimension) along the Z axis and top towards +Z (see coordinateSystem). The topRadius rtop defaults to 0, giving a pointed cone, but values > 0 can be used to produce a truncated cone. The caps option indicates whether to cap the end(s) of the cone or leave them open.\nshape cylinderradius r ] [ height h ] [ caps true|false ] general-options\nThe default radius r and height h are 10 and 40, respectively. The default orientation is with the axis of symmetry (height dimension) along the Z axis (see coordinateSystem). The caps option indicates whether to cap the cylinder or leave it open-ended.\nshape icosahedronradius r ] [ orientation type ] [ sphereFactor f ] [ lattice h,k ] general-options\nThe radius r is the distance from the center to a 5-fold vertex (default 10). The orientation type can be:\n• 222 (default) - with two-fold symmetry axes along the X, Y, and Z axes\n• 2n5 - with two-fold symmetry along X and 5-fold along Z\n• n25 - with two-fold symmetry along Y and 5-fold along Z\n• 2n3 - with two-fold symmetry along X and 3-fold along Z\n• 222r - same as 222 except rotated 90° about Z\n• 2n5r - same as 2n5 except rotated 180° about Y\n• n25r - same as n25 except rotated 180° about X\n• 2n3r - same as 2n3 except rotated 180° about Y\nThe sphereFactor option allows generating a shape that is an interpolation between an icosahedron and a sphere of equal radius. The factor f is the weight of the sphere component in the interpolation and can range from 0 (default, icosahedron) to 1 (sphere). The interpolation only affects vertex positions and will not generate curved mesh lines or curved surface triangles. The lattice option allows showing the icosahedron surface with hexagons and pentagons instead of triangles. A shape with icosahedral symmetry (like many virus capsids) can be idealized as a sheet of hexagons in which curvature is introduced by replacing certain hexagons with pentagons, as in a geodesic dome. The pentagons occupy the points of the icosahedron, while the indices h and k refer to the number and arrangement of hexagons in each face (details...). Hexagons are bent where they cross from one triangular face to another. The indices h and k can each be zero (but not both zero) or a positive integer. Surface generation can be slow if large values are used.\n\nSee also: hkcage, Icosahedron Surface, Cage Builder\n\nshape rectanglewidth w ] [ height h ] [ widthDivisions dw ] [ heightDivisions dh ] general-options\nThe width w and height h are the X and Y dimensions of the rectangle (both default to 10). The number of surface points in the rectangle along the X dimension will equal 1 + dw and along the Y dimension will equal 1 + dh. Both the widthDivisions dw and heightDivisions dh default to 10, and these settings override the divisions general option.\nshape ribbon  atom-specwidth w ] [ yaxis axis ] [ twist t ] [ segmentSubdivisions divisions ] [ bandLength length ] [ followBonds  true|false ] general-options*\nA ribbon is a smooth path with flat cross-section that connects a series of atoms or markers. The default width w is 1.0. The yaxis axis (default none) and twist t control ribbon orientation and how it varies along the path. If neither is specified, the ribbon normal varies along the path so that there is no local twist. If an axis is given, the ribbon normal is aligned with that axis as closely as possible within the constraints of the path. A constant twist t to be applied at each point along the path can also be specified (default 0°). The axis can be given as:\n• x - X-axis\n• y - Y-axis\n• z - Z-axis\n• x,y,z (three values separated by commas only) - an arbitrary vector\n• an atom-spec of exactly two atoms (not necessarily bonded or in the same model) or one bond. A bond can only be specified by selecting it and using the word selected, sel, or picked; any atoms also selected at the time will be ignored.\nCoordinate specifications of axis are interpreted in the system of the model containing the first atom in the path. The segmentSubdivisions setting controls how many straight segments are used to form the curve between a consecutive pair of atoms; higher values give smoother curves. The number of straight segments forming the curve between a pair of atoms will equal divisions + 1 (default 10 + 1 = 11). The bandLength option specifies what length of ribbon centered on an atom should be colored to match that atom (default 0.0). The followBonds option indicates whether the ribbon should follow the directions of the bonds connecting the atoms; the option should be set to false for atoms that are not bonded.\n\n*The general options center, rotation, qrotation, coordinateSystem, and slab do not apply to ribbon shapes.\n\nshape triangleatoms  atom-spec ] general-options*\nThe vertices of the triangle can be supplied as a specification of three atoms; otherwise, scene coordinates (0,0,0), (1,0,0), and (0,1,0) will be used.\n\n*The general options center, rotation, qrotation, coordinateSystem, and divisions do not apply to triangle shapes.\n\nshape tube  atom-specradius r ] [ segmentSubdivisions divisions ] [ bandLength length ] [ followBonds  true|false ] general-options*\nA tube is a smooth path with circular cross-section that connects a series of atoms or markers. The default radius r is 1.0. The segmentSubdivisions setting controls how many straight segments are used to form the curve between a consecutive pair of atoms; higher values give smoother curves. The number of straight segments forming the curve between a pair of atoms will equal divisions + 1 (default 10 + 1 = 11). The bandLength option specifies what length of tube centered on an atom should be colored to match that atom (default 0.0). The followBonds option indicates whether the tube should follow the directions of the bonds connecting the atoms; the option should be set to false for atoms that are not bonded.\n\n*The general options center, rotation, qrotation, coordinateSystem, and slab do not apply to tube shapes.\n\nshape boxpath  atom-specwidth w ] [ twist t ] [ reportCuts true|false ] [ cutScale s ] general-options*\nA boxpath is a 3D zigzag of “beam” segments connecting a series of atoms or markers. The beam has a square cross-section of width w (default 1.0), and the twist t (default 0.0°) sets the rotational orientation of the segments about their long axes. To make a physical replica of the boxpath (see the sculpture at DePauw), the segments could be generated by angled cuts along a single straight beam. Setting reportCuts to true gives the locations of such cuts along the four edges of a hypothetical beam; the cutScale s (default 1.0) is a factor for converting these distances from Å to the appropriate physical units.\n\n*The general options center, rotation, qrotation, coordinateSystem, divisions, and slab do not apply to boxpath shapes.\n\n### General Options\n\ncenter atom-spec | x,y,z\nThe center can be set to the center of specified atom(s) or to a point (x,y,z) in the coordinate system specified with coordinateSystem. The default center is (0,0,0).\nrotation ax,ay,az,angle\nRotate the shape angle degrees around the specified axis (ax,ay,az) in the coordinate system specified with coordinateSystem. The default is no rotation.\nqrotation qx,qy,qz,qw\nApply the rotation specified as a quaternion in the coordinate system specified with coordinateSystem. The default is no rotation.\ncoordinateSystem N\nThe coordinate system is indicated by reference model ID number, optionally preceded by #. The coordinate system is used for interpreting the center, rotation, qrotation, and ellipsoid radius arguments. The default is the coordinate system of the new or existing surface model (see modelId).\ncolor colorname\nSet the surface color to colorname, which can be any color name that specifies a single color (default gray).\nmodelName name\nName to use for the newly created surface model. If no name is specified, a generic shape name is used (for example, “sphere”).\nmodelId N\nOpen the surface as model number N (an integer, optionally preceded by #). Submodel specifications #N.N (# required) can also be given. The default is the lowest unused number. If a surface model with the same ID number already exists, the new shape is added as another surface piece.\ndivisions d\nSet the fineness of surface triangulation; d is the number of square mesh cells around the circumference (default 72). The number of triangles around the circumference is roughly 2-4 times higher, depending on the shape and on the value of d. This setting does not apply to an icosahedron with lattice specified, as it will be shown with the indicated numbers of hexagons and pentagons rather than with triangles.\nmesh true|false\nWhether to display the surface as a mesh or as a solid surface. The default is false except for an icosahedron with lattice specified.\nlinewidth width\nThe width is the pixel linewidth of mesh display (default 1.0).\nslab width | d1,d2\nThe slab option indicates that a shell or slab of finite thickness should be created instead of a single layer of surface. If a single value (width) is supplied, the inner and outer layers of the slab will be offset from the nominal radius r by ±½(width). Alternatively, two values separated by a comma but no spaces can be used to specify the offsets of the two layers independently. Offsets can be positive (outward) or negative (inward)." ]

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[ "", null, "# Integral Calculus\n\nRelated Topics:\nMore Calculus Lessons\nCalculus Games\n\nIn these lessons, we introduce a notation for antiderivatives called the Indefinite Integral. We will also give a list of integration formulas that would be useful to know.\n\n### Indefinite Integrals\n\nThe notation", null, "is used for an antiderivative of f and is called the indefinite integral.", null, "The following is a table of formulas of the commonly used Indefinite Integrals. You can verify any of the formulas by differentiating the function on the right side and obtaining the integrand. Scroll down the page if you need more examples and step by step solutions of indefinite integrals.\n\n### Table of Indefinite Integral Formulas", null, "Example:\n\nFind the general indefinite integral", null, "Solution:", null, "### Definite Integrals and Indefinite Integrals\n\nThe connection between the definite integral and indefinite integral is given by the second part of the Fundamental Theorem of Calculus.\n\nIf f is continuous on [a, b] then", null, "Take note that a definite integral", null, "is a number, whereas an indefinite integral", null, "is a function.\n\nExample:\n\nEvaluate", null, "Solution:", null, "Definition of Indefinite Integrals\nAn indefinite integral is a function that takes the antiderivative of another function. It is visually represented as an integral symbol, a function, and then a dx at the end. The indefinite integral is an easier way to symbolize taking the antiderivative. The indefinite integral is related to the definite integral, but the two are not the same.\n\nAntiderivatives and indefinite integrals\nExample:\nWhat is 2x the derivative of? This is the same as getting the antiderivative of 2x or the indefinite integral of 2x. Indefinite Integrals\nIndefinite integrals are functions that do the opposite of what derivatives do. They represent taking the antiderivatives of functions.\nA formula useful for solving indefinite integrals is that the integral of x to the nth power is one divided by n+1 times x to the n+1 power, all plus a constant term.\n\nIndefinite integrals, step by step examples\nStep 1: Add one to the exponent\nStep 2: Divide by the same.\nExample:\n1. ∫3x5, dx\nMore indefinite integral, step by step, examples: with square root\nExample:\n1. ∫3√x, dx\nMore indefinite integral, step by step, examples: x in the denominator\nExample:\n1. ∫6/x4, dx\n\nComplicated Indefinite Integrals\nNot all indefinite integrals follow one simple rule. Some are slightly more complicated, but they can be made easier by remembering the derivatives they came from. These complicated indefinite integrals include the integral of a constant (the constant times x), the integral of ex (ex) and the integral of x-1 (ln[x]).\n\nIndefinite Integration (Polynomial, Exponential, Quotient)\nHow to determine antiderivatives using integration formulas?\nExamples:\n1. ∫(3x2 - 2x + 1) dx\n2. ∫3ex dx\n3. ∫4/x dx Basic Integration Formulas\nHere are some basic integration formulas you should know.\n\n### Definite Integral\n\nThe Definite Integral - Understanding the Definition. Calculating a Definite Integral Using Riemann Sums - Part 1.\nThis video shows how to set up a definite integral using Riemann Sums. The Riemann Sums will be computed in Part 2. Calculating a Definite Integral Using Riemann Sums - Part 2.\n\nTry the free Mathway calculator and problem solver below to practice various math topics. Try the given examples, or type in your own problem and check your answer with the step-by-step explanations.", null, "", null, "" ]

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[ "# Kindergarten Subtraction Activities 2\n\nKindergarten Subtraction Activities 2: Subtract the smaller number from the bigger number. Then drag the correct number box and drop it into the answer box.\n\n## Kindergarten Subtraction Activities 2\n\nFree online worksheets to develop your child’s numeracy skills in horizontal subtraction. Use these online math practice worksheets to be fast and smart in horizontal subtraction. These horizontal subtraction online math practice worksheets are ideal to test your elementary child’s understanding of the fundamentals of simple horizontal subtraction. Enjoy kindergarten horizontal subtraction worksheets!\n\nKindergarten Activities Subtraction 4: Let’s now practise horizontal subtraction. Try our great collection of online math practice worksheets to learn horizontal subtraction! Let your child be quick in horizontal subtraction." ]

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[ "# Answers\n\nSolutions by everydaycalculation.com\n\n## What is the LCM of 490 and 9604?\n\nThe LCM of 490 and 9604 is 48020.\n\n#### Steps to find LCM\n\n1. Find the prime factorization of 490\n490 = 2 × 5 × 7 × 7\n2. Find the prime factorization of 9604\n9604 = 2 × 2 × 7 × 7 × 7 × 7\n3. Multiply each factor the greater number of times it occurs in steps i) or ii) above to find the LCM:\n\nLCM = 2 × 2 × 5 × 7 × 7 × 7 × 7\n4. LCM = 48020\n\nMathStep (Works offline)", null, "Download our mobile app and learn how to find LCM of upto four numbers in your own time:\nAndroid and iPhone/ iPad\n\nFind least common multiple (LCM) of:\n\n#### LCM Calculator\n\nEnter two numbers separate by comma. To find least common multiple (LCM) of more than two numbers, click here.\n\n© everydaycalculation.com" ]

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[ "# NAG CL Interfaceg05ndc (sample)\n\nSettings help\n\nCL Name Style:\n\n## 1Purpose\n\ng05ndc selects a pseudorandom sample without replacement from an integer vector.\n\n## 2Specification\n\n #include\n void g05ndc (const Integer ipop[], Integer n, Integer isampl[], Integer m, Integer state[], NagError *fail)\nThe function may be called by the names: g05ndc or nag_rand_sample.\n\n## 3Description\n\ng05ndc selects $m$ elements from a population vector ipop of length $n$ and places them in a sample vector isampl. Their order in ipop will be preserved in isampl. Each of the $\\left(\\begin{array}{c}n\\\\ m\\end{array}\\right)$ possible combinations of elements of isampl may be regarded as being equally probable.\nFor moderate or large values of $n$ it is theoretically impossible that all combinations of size $m$ may occur, unless $m$ is near $1$ or near $n$. This is because $\\left(\\begin{array}{c}n\\\\ m\\end{array}\\right)$ exceeds the cycle length of any of the base generators. For practical purposes this is irrelevant, as the time taken to generate all possible combinations is many millenia.\nOne of the initialization functions g05kfc (for a repeatable sequence if computed sequentially) or g05kgc (for a non-repeatable sequence) must be called prior to the first call to g05ndc.\nKendall M G and Stuart A (1969) The Advanced Theory of Statistics (Volume 1) (3rd Edition) Griffin\nKnuth D E (1981) The Art of Computer Programming (Volume 2) (2nd Edition) Addison–Wesley\n\n## 5Arguments\n\n1: $\\mathbf{ipop}\\left[{\\mathbf{n}}\\right]$const Integer Input\nOn entry: the population to be sampled.\n2: $\\mathbf{n}$Integer Input\nOn entry: the number of elements in the population to be sampled.\nConstraint: ${\\mathbf{n}}\\ge 1$.\n3: $\\mathbf{isampl}\\left[{\\mathbf{m}}\\right]$Integer Output\nOn exit: the selected sample.\n4: $\\mathbf{m}$Integer Input\nOn entry: the sample size.\nConstraint: $1\\le {\\mathbf{m}}\\le {\\mathbf{n}}$.\n5: $\\mathbf{state}\\left[\\mathit{dim}\\right]$Integer Communication Array\nNote: the dimension, $\\mathit{dim}$, of this array is dictated by the requirements of associated functions that must have been previously called. This array MUST be the same array passed as argument state in the previous call to nag_rand_init_repeatable (g05kfc) or nag_rand_init_nonrepeatable (g05kgc).\nOn entry: contains information on the selected base generator and its current state.\nOn exit: contains updated information on the state of the generator.\n6: $\\mathbf{fail}$NagError * Input/Output\nThe NAG error argument (see Section 7 in the Introduction to the NAG Library CL Interface).\n\n## 6Error Indicators and Warnings\n\nNE_ALLOC_FAIL\nDynamic memory allocation failed.\nSee Section 3.1.2 in the Introduction to the NAG Library CL Interface for further information.\nOn entry, argument $⟨\\mathit{\\text{value}}⟩$ had an illegal value.\nNE_INT\nOn entry, ${\\mathbf{n}}=⟨\\mathit{\\text{value}}⟩$.\nConstraint: ${\\mathbf{n}}\\ge 1$.\nNE_INT_2\nOn entry, ${\\mathbf{m}}=⟨\\mathit{\\text{value}}⟩$ and ${\\mathbf{n}}=⟨\\mathit{\\text{value}}⟩$.\nConstraint: $1\\le {\\mathbf{m}}\\le {\\mathbf{n}}$.\nNE_INTERNAL_ERROR\nAn internal error has occurred in this function. Check the function call and any array sizes. If the call is correct then please contact NAG for assistance.\nSee Section 7.5 in the Introduction to the NAG Library CL Interface for further information.\nNE_INVALID_STATE\nOn entry, state vector has been corrupted or not initialized.\nNE_NO_LICENCE\nYour licence key may have expired or may not have been installed correctly.\nSee Section 8 in the Introduction to the NAG Library CL Interface for further information.\n\nNot applicable.\n\n## 8Parallelism and Performance\n\ng05ndc is threaded by NAG for parallel execution in multithreaded implementations of the NAG Library.\nPlease consult the X06 Chapter Introduction for information on how to control and interrogate the OpenMP environment used within this function. Please also consult the Users' Note for your implementation for any additional implementation-specific information.\n\nThe time taken by g05ndc is of order $n$.\nIn order to sample other kinds of vectors, or matrices of higher dimension, the following technique may be used:\n1. (a)set ${\\mathbf{ipop}}\\left[\\mathit{i}-1\\right]=\\mathit{i}$, for $\\mathit{i}=1,2,\\dots ,n$;\n2. (b)use g05ndc to take a sample from ipop and put it into isampl;\n3. (c)use the contents of isampl as a set of indices to access the relevant vector or matrix.\nIn order to divide a population into several groups, g05ncc is more efficient.\n\n## 10Example\n\nIn the example program random samples of size $1,2,\\dots ,8$ are selected from a vector containing the first eight positive integers in ascending order. The samples are generated and printed for each sample size by a call to g05ndc after initialization by g05kfc.\n\n### 10.1Program Text\n\nProgram Text (g05ndce.c)\n\nNone.\n\n### 10.3Program Results\n\nProgram Results (g05ndce.r)" ]

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[ "# Can formal power series become polynomial often, when composed with polynomials?\n\nLet $F$ be a finite field. Let $F[X]$ and $F[[X]]$ denote the ring of polynomials and power series over $F$, respectively. I'm trying to show a statement like the following:\n\nFix a $d > 0$. Let $g\\in F[[X]]$. If there exists a set $C\\subseteq F[X]$ of polynomials (with no constant term) of degree at most $k$ such that for all $c\\in C$, $g(c)$ -- $g$ composed with $c$ -- is a degree $kd$ polynomial, and $|C|$ is \"large\" (some function of $k$, $d$, and $|F|$), then $g$ must actually be a polynomial.\n\nI'm trying to beat the bound that one might be able to get via Schwartz-Zippel, where $|C| > kd |F|^{k-1}$ (where $kd \\ll |F|$).\n\nWhat bounds on $|C|$ can we get?\n\nThank you, Henry\n\n• I should point out that you don't need $d$, as the set of degree-$k$ polynomials is finite, so $d$ is just the largest value of $\\operatorname{deg} g(c) / k$. And $kd$ is better interpreted as just \"the highest degree of some $g(c)$\". Jul 27, 2012 at 4:36\n• Thanks for pointing this out Ryan. In this case, I have a specific $d$ in mind: if $g(c)$ has degree greater than $kd$, then I do not wish to consider such a $c$. The bound I'm seeking on $|C|$ may depend on $d$. I will edit the problem statement to make this clearer. Jul 27, 2012 at 4:50\n• The easy lower bound is $|F|^{k/2}$, coming from $g(x) = \\sqrt{1+x}$. Then $g \\circ c$ is a polynomial whenever $c$ is of the form $h(x)^2-1$. My guess is that this is close to optimal. Jul 29, 2012 at 18:45\n• @David: what do you mean by $g(x) = \\sqrt{1 + x}$ if $F$ has characteristic $2$? Jul 30, 2012 at 1:02\n• I don't :). In that case, $y^2=1+x$ doesn't have a root in $k[[x]]$. Probably the root of $y^2+y=x$ with $y(0)=0$ would make a good substitute. Jul 30, 2012 at 2:38\n\nI claim that, if $g(x) \\in k[[x]]$ is not a polynomial, then $g \\circ c$ is a polynomial for at most $|F|^{k/2}$ polynomial $c$ of degree $\\leq k$. We will always use the letter $c$ to represent a polynomial with $c(x)=0$.\n\nCase 1: $g$ is transcendental over the field $k(x)$. In this case, I claim that $g \\circ c$ is a polynomial only when $c$ is $0$. Proof: Suppose that $g(c(x)) = h(x)$ for some polynomial $h$. Then there is a polynomial relation $F(c(x), h(x))=0$ for some polynomial $f$. Set $G(x) := F(x, g(x))$. By hypothesis, $G \\neq 0$, but $G(c(x))=0$. The only way this can happen is if $c=0$.\n\nCase 2: $g$ is algebraic over $k(x)$. Let $F(x,g(x))=0$ be the minimal polynomial relation between $x$ and $g$. Let $A$ be the ring $k[x,y]/F(x,y)$. At this point I really, really want to use the language of algebraic geometry. If you aren't happy with this, I'll try to convert into commutative algebra, but it will be harder to write and, in my opinion, harder to read. My goal is the following claim:\n\nClaim: If there is any nonzero $c$ such that $g \\circ c$ is polynomial, then the ring $A$ is a subring of $k[u]$ for some $u \\in \\mathrm{Frac}(A)$.\n\nExample: Let $g = (1+x)^{3/2}$. Then $F(x,y) = y^2 - (1+x)^3$. Letting $u = \\frac{y}{1+x}$, we have $A \\subseteq k[u]$, since $x=u^2-1$ and $y=u^3$.\n\nProof: $\\mathrm{Spec}(A)$ is an algebraic curve. Since $F$ is the minimal polynomial relation, it is irreducible. Let $X$ be the normalization of $\\mathrm{Spec}(A)$.\n\nLet $h=g \\circ c$. Now, $t \\mapsto (c(t), h(t))$ is a map $\\mathbb{A}^1 \\to \\mathrm{Spec}(A)$ and, since $c$ is nonconstant, is a nonconstant map. Since $\\mathbb{A}^1$ is normal, this lifts to a nonconstant map $\\mathbb{A}^1 \\to X$. So $X$ has genus zero, and has at most one puncture. Since $X$ is affine, it has at least one puncture. So $X \\cong \\mathbb{A}^1$. In other words, the normalization of $A$ is $k[u]$, so we get an embedding $A \\subseteq k[u]$ for some $u \\in \\mathrm{Frac}(A)$. $\\square$\n\nFrom now on, we will assume that there is some nonzero $c$ for which $g \\circ c$ is polynomial. So we may assume that there is an embedding $A \\subseteq k[u]$, with $\\mathrm{Frac}(A) = k(u)$, and we fix such an embedding. Let $x = a(u)$ and $g(x) = b(u)$, for some polynomials $a$ and $b$. If $a$ has degree $1$, then $u$ is a linear function of $x$ and we can write $g$ as a polynomial in $x$, contradicting your hypothesis. So $\\deg a \\geq 2$. I claim that there are only $|F|^{k/\\deg a}$ values of $c$ for which $g \\circ c$ is a polynomial. Specifically, the polynomials $c$ of the form $a(\\phi(t))$, where $\\phi$ is a polynomial of degree $k/\\deg a$.\n\nWhy is this? Well, if $c(t) = a(\\phi(t))$, then $g(c(t)) = b(\\phi(t))$ and is thus a polynomial. Conversely, if we have some $c$ such that $g \\circ c$ is polynomial then, as discussed above, we get a map from the normalization of $A$ to $k[t]$. Take $\\phi$ to be the image of $u$.\n\n• Thanks David! This will take me a little while to parse, so i might add follow up questions in the comments. But for now I accept this! Jul 30, 2012 at 14:37" ]

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[ "# PHY C13: Forced Oscillations\n\nNow let’s look at the last type: Forced Oscillations\n\n• Forced oscillations\n• Natural Frequency\n• Resonance\n• Frequency Response\n• Graphs\n\nLet’s go!\n\nWhat is a Forced Oscillation?\nA type of oscillation where an external periodic force is being applied to a naturally vibrating system, causing it to vibrate with increasing amplitude.\n\nWhat is Natural Frequency?\nThe frequency at which a freely oscillating system oscillates at without external force.\n\nIt is denoted by f0.\n\nWhat is Resonance?\nThe phenomenon when a system oscillating at natural frequency is subjected to an external periodic force equal to the natural frequency, causing the system to vibrate with maximum amplitude.\n\nWhat is Resonant Frequency?\nThe frequency of an external force which causes a system to exhibit resonance.\nUsually this value is equal to the natural frequency, but it may vary due to damping.\n\nWhat is Frequency Response?\nMeasure of amplitude of an oscillation against different applied frequencies.\n\n• When the applied frequency is LESS than f0, the resulting amplitude is not maximum.\n• When the applied frequency is EQUAL to f0, the resulting amplitude is maximum.\n• When the applied frequency is MORE than f0, the resulting amplitude is not maximum.\n\nThis is represented by a RESONANCE CURVE:\n\nWhat effect does Damping have on Frequency Response?\nAs damping increases:\n\n• The amplitude of oscillations at all frequencies decreases\n• The resonant frequency decreases (by a small amount)\n• The peak becomes flatter" ]

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[ "# Calculate $E[XY]$ for $(X,Y)\\sim N(\\mu_{1},\\mu_{2},\\sigma_{1}^{2},\\sigma_{2}^{2}, \\rho)$\n\nI need to calculate $E[XY]$ for $(X,Y) \\sim N(\\mu_{1},\\mu_{2},\\sigma_{1}^{2}, \\sigma_{2}^{2}, \\rho)$ by using integration and then determine the correlation coefficient afterwards.\n\nNow, when $X \\sim N(\\mu_{1},\\sigma_{1}^{2})$ and $Y \\sim N(\\mu_{2}, \\sigma_{2}^{2})$, the probability density function is given by $$f(x,y)= \\\\ \\frac{1}{2\\pi \\sigma_{1} \\sigma_{2} \\sqrt{1-\\rho^{2}}}\\exp \\left( - \\frac{1}{2(1-\\rho^{2})} \\left[ \\frac{(x-\\mu_{1})^{2}}{\\sigma_{1}^{2}}+ \\frac{(y-\\mu_{2})^{2}}{\\sigma_{2}^{2}} - \\frac{2 \\rho (x-\\mu_{1})(y-\\mu_{2})}{\\sigma_{1}\\sigma_{2}}\\right]\\right)$$\n\nBut, I don't know how to calculate $E[XY]$ as an integral, short of setting up $$\\int_{-\\infty}^{\\infty}\\int_{-\\infty}^{\\infty}xyf(x,y)dydx. \\,\\,\\,(*)$$\n\nSomebody told me that this integral is equal to $$\\int_{-\\infty}^{\\infty} x \\left[\\int_{-\\infty}^{\\infty} y f(y \\vert x) dy\\right]f(x)dx,$$ where $f(y|x)$ is normal density with mean $=\\int_{-\\infty}^{\\infty}yf(y\\vert x)dy = \\mu_{2}$, but I don't know what this means or how to get from $(*)$ to this last integral above.\n\n• As an easy way, if you know that $\\text{Cov}(X,Y) = \\rho \\sigma_1 \\sigma_2$, and $\\text{Cov}(X,Y) = E[X Y] - E[X] E[Y]$, then it follows that $E[X Y] =\\mu_1 \\mu_2 + \\rho \\sigma_1 \\sigma_2$ Nov 28 '17 at 2:17\n• @wolfies I guess that's a good check, but the directions said to use integration, and my professor is kind of strict about that sort of thing. Plus, it's always better to know how to prove things in all different ways. Nov 28 '17 at 3:04\n• Hint: rearrange $P(A|B)=P(A \\cap B)/P(B)$. Nov 28 '17 at 9:54\n• A straightforward way to evaluate the integral $(*)$ directly (which is what \"use integration\" seems to mean) is to apply the substitution $$X=\\mu_1+\\sigma_1U,\\ Y=\\mu_2+\\sigma_2\\left(\\sqrt{1-\\rho^2}V+\\rho U\\right).$$ This makes $(U,V)$ a standard binormal variable and gives the easy calculation $$E[XY]=E[\\mu_1\\mu_2+\\sigma_1\\sigma_2\\rho U^2 + aU+bV+cUV]=\\mu_1\\mu_2 + \\sigma_1\\sigma_2\\rho$$ where $a,b,c$ are constants you don't even have to compute since $E[U]=E[V]=E[UV]=0$.\n– whuber\nNov 28 '17 at 16:07\n\nThe hint you suggest in your question is to use condition expectations. although not the only way to solve this problem, it is a quick method if you are comfortable with conditioning arguments.\n\n$\\mathbb{E}(XY) = \\mathbb{E}_X(\\mathbb{E}_y(XY|X))$\n\nwhere the subscripts denote expectation with respect to which variable (for clarity to the reader. Then $\\mathbb{E}_X(\\mathbb{E}_Y(XY|X))= \\mathbb{E}_X(X\\mathbb{E}_Y(Y|X)).$ Now if you know the distribution of $Y|X$ this is easy,\n\n$Y|X \\sim N(\\mu_y+\\rho\\frac{\\sigma_y}{\\sigma_x}(X-\\mu_x), \\sigma_y^2(1-\\rho)).$\n\nThis means the mean of $Y|X$ is the first quantity, $\\mathbb{E}_X(X(\\mu_y+\\rho\\frac{\\sigma_y}{\\sigma_x}(X-\\mu_x))).$\n\nNow multiply the $X$ through and take expectations to get $\\mathbb{E}_X(X)\\mu_y+\\rho\\frac{\\sigma_y}{\\sigma_x}\\left[\\mathbb{E}_X(X^2)-\\mu_x\\mathbb{E}_X(X)\\right].$\n\nNow using the variance breakdown of the second and the first moment gives: $\\mathbb{E}_X(X)\\mu_y+\\rho\\frac{\\sigma_y}{\\sigma_x}\\left[\\sigma_x^2 +\\mu_x^2 -\\mu_x^2\\right].$ And a little simplification yields:\n\n$\\mu_X\\mu_y+\\rho\\sigma_y\\sigma_x.$\n\nYou can check this against the correlation by substractions of $\\mu_x\\mu_y$ to get the co-variance, verifying the result is correct." ]

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[ "# Fraction\n\nA fraction is a \"part\" of a \"whole\". It is also a ratio between two integers separated by a solidus (/) or a vinculum ( __ ). The upper part of a fraction is called numerator and the lower part is called denominator.\n\n## Examples", null, "The \"whole\" of a circle can be divided into different \"parts\" as shown below. The colored portion of the circles denotes a specific part of the whole.", null, "## How to pronounce a Fraction\n\nWe use cardinal numbers (1, 2, 3, 4, etc.) to read the numerator, and ordinal numbers (third, fourth, fifth, etc.) to read the denominator. When we write fractions in words, we use a hyphen between the cardinal number and the ordinal number.\n\nOrdinal numbers are countable, so we add \"s\" to the word.\n\n## Examples\n\n2/4 - here thenumerator \"2\" is pronounced as 2, the denominator \"4\" as fourth, and the fraction as two-fourths.\n\n1/8 - here the numerator \"1\" is pronounced as 1, the denominator \"8\" as eighth, and the fraction as one-eighths.\n\n### Terms to remember\n\n Dividend A number or quantity that is being divided. Divisor A number or quantity that divides another quantity. Equivalent Being essentially equal. Integer A positive or negative whole number. Ratio A relationship between two or more numbers. Reciprocal Multiplicative inverse of a quantity. Similar Having the same characteristics. Simplify To make simpler or reduce complexity. Unlike Having different characteristics. Whole Number Either a zero (0) or a counting number.\n\n#### Become a member today!\n\nRegister (it’s Free)" ]

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[ "# A body dropped from a tower explodes into two pieces of equal mass in mid-air. Which of the following is not possible? a) Each part will follow parabolic path b) Only one part will follow parabolic path c) Both parts move along a vertical line d) One part reaches the ground earlier than the other\n\n## Question ID - 100237 :- A body dropped from a tower explodes into two pieces of equal mass in mid-air. Which of the following is not possible? a) Each part will follow parabolic path b) Only one part will follow parabolic path c) Both parts move along a vertical line d) One part reaches the ground earlier than the other\n\n3537\n\n A ball is dropped from a height of 45 m from the ground. The coefficient of restitution between the ball and the ground is 2/3. What is the distance travelled by the ball in 4th second of its motion. Assume negligible time is spent in rebounding. Let", null, "a) 5 m b) 20 m c) 15 m d) 10 m", null, "" ]

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[ "Group for Research in Decision Analysis\n\n# Interval edge-coloring: A reduction of the standard problem to the cyclic problem\n\n## Sivan Altinakar – Polytechnique Montréal, Canada\n\nIn a graph, a k-edge-coloring associates an integer color number in {1,...,k} to each edge, such that no vertex has two adjacent edges of the same color. This formulation is useful for modeling scheduling problems, for example when in a production chain, a given set of objects must each be processed in unit-length times on a subset of machines (without any specific order). In this case, a node corresponds to either an object or a machine, an edge between two such nodes to the constraint that this object must be processed on that machine, and the color of an edge to the timeframe during which this happens. In the real world, there are many additional constraints to this simple problem, for example when a machine must have no idle time between its first and last assignment, and when there are no places where an object can wait between its first and last processing. In our edge-coloring model, this means that for each node, the set of colors of the edges adjacent to it must form, depending on the problem, either a standard integer interval (a set of consecutive integers) or a cyclic integer interval (1 is consecutive to k). The aim of this talk is to present a reduction from the standard to the cyclic k-interval edge-coloring problem (when k greater or equal to 12), as well as ways to model some other cyclic scheduling problems as a cyclic interval edge-coloring problem, using some of the tools developed for the reduction." ]

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[ "## Elementary Algebra\n\nIn order to do this problem, recall (1) that addition and subtraction come after multiplication and division in order of operations, (2) that when there is a tie in order of operations, it is done from left to right, and (3) that dividing by a fraction is the same as multiplying by that fraction's reciprocal. $$\\frac{2}{3} +\\frac {1}{5} - \\frac{1}{15} \\\\ \\frac{10}{15} + \\frac{3}{15} - \\frac{1}{15} = 12/15 = 4/5$$" ]

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[ "# Interpretation of coefficients in a poorly performing GLM\n\nSuppose that I have trained a logistic regression model on a certain dataset, and I wish to interpret the coefficients of this model.\n\nDoes it make any difference on the validity of the interpretation if the model is poor?\n\nWhat I mean by this is that imagine that we measure the performance of the model by a ROC curve and we get a low value (but still better than 0.5) for the area under the curve. This model is not terribly accurate, but does the performance of it influence the interpretation of the coefficients?\n\nThe statistical interpretation of the coefficients doesn't depend on how the model was fit. I could make completely random guesses of the coefficients and they would have the same interpretation as they would had I estimated them with maximum likelihood. For two units identical on all measured variables except that they differed on $$X_1$$ by one unit, the difference in the log odds of success is $$\\beta_1$$. That interpretation comes directly from simply writing down the regression equation and has nothing to do with the fitting process.\n\nTo interpret the coefficients as consistent estimates of some \"true\" association, or as total effects rather than direct effects, or as causal effects rather than mere conditional assocations, requires more assumptions, far more than whether the model fit well in your sample.\n\nFor example, let's say the true data-generating (i.e., structural causal) model was\n\n$$P(Y=1|X_1,X_2) = expit(\\gamma_0 + \\gamma_1 X_1 + \\gamma_2 X_2)$$\n\nLet's say I'm considering the model\n\n$$P(Y=1|X_1) = expit(\\beta_0 + \\beta_1 X_1)$$\n\nwhich excludes $$X_2$$. $$\\beta_1$$ doesn't have a causal interpretation, but it's the regression slope you would get if you were to fit that model to the population data (i.e., so there is no sampling error). The interpretation of $$\\beta_1$$ in this model is: For two units that differed on $$X_1$$ by one unit, the difference in the log odds of success is $$\\beta_1$$.\n\nLet's say I collect a sample and then pull an estimate of $$\\beta_1$$ out of a hat and call it $$\\hat \\beta_1^{guess}$$. Even though that value is completely unconnected to the sample, it still has the same interpretation as any other estimate of $$\\beta_1$$, which is as an estimate of the difference in the log odds of success for two units that differed on $$X_1$$ by one unit. It's not a valid or consistent estimate, but it's an estimate of a quantity that has a clear interpretation. The quantity ($$\\beta_1$$) does not have a causal interpretation, but it's still meaningfully interpretable as an associational quantity.\n\nIf I were to estimate $$\\beta_1$$ with maximum likelihood, and call the estimate $$\\hat \\beta_1^{MLE}$$, it has the same interpretation as $$\\hat \\beta_1^{guess}$$, which is that it is an estimate of $$\\beta_1$$, which, again, has a clear interpretation. $$\\hat \\beta_1^{MLE}$$ is a consistent estimate of $$\\beta_1$$, so if I were to want to know what $$\\beta_1$$ was I would be inclined to say it's closer to $$\\hat \\beta_1^{MLE}$$ than it is to $$\\hat \\beta_1^{guess}$$. $$\\hat \\beta_1^{MLE}$$ could result from a terribly fitting model, and that would say nothing of its interpretation. A terribly fitting model might result because we failed to include $$X_2$$ in it. That doesn't change how $$\\beta_1$$, and thus how $$\\hat \\beta_1^{MLE}$$ and $$\\hat \\beta_1^{guess}$$, are interpreted.\n\nIf you wanted to interpret a regression coefficient as causal, then you want to estimate $$\\gamma_1$$, not $$\\beta_1$$. The interpretation of $$\\gamma_1$$ is the change in the log odds of success caused by intervening on $$X_1$$ by one unit while holding $$X_2$$ constant. Any estimate of $$\\gamma_1$$, regardless of how it came to be, could be interpreted as an estimate of the change in the log odds of success caused by intervening on $$X_1$$ by one unit while holding $$X_2$$ constant. You could even use $$\\hat \\beta_1^{guess}$$ as an estimate of $$\\gamma_1$$ and it would still have this interpretation. It would likely be a bad estimate that you shouldn't trust, but that doesn't change its interpretation. Even if you estimated $$\\gamma_1$$ using maximum likelihood estimation of a model that included both $$X_1$$ and $$X_2$$, its interpretation would be the same; it would likely just be a better estimate (but it doesn't mean it's a good estimate!).\n\nAll this is to say that the interpretation of coefficients comes from the model as it is written, not the way they are estimated or how well the estimated model fits. These may serve as indicators as to whether the estimated coefficients might be close to the population versions they are trying to approximate, but not how they should be interpreted. For example, a poorly fitting model resulting from regressing $$Y$$ on $$X_1$$ may indicate that $$\\hat \\beta_1$$ is a poor estimate of $$\\gamma_1$$, but it may be a good estimate of $$\\beta_1$$. The interpretations of $$\\beta_1$$ and $$\\gamma_1$$ are unrelated to how the estimates were generated, and the interpretation of the estimates is simply as estimates of those quantities.\n\n• Thanks for the through answer. However If I have understood correctly I'm not sure why it is interesting to interpret the coefficients of these models, as you would never know if your estimated coefficients are the real effect of that covariate on your response? I'm certain there is something I'm missing however, as this is most often the point of using such models.\n– srb\nJun 16 '20 at 20:32\n• If estimating the parameters of a structural causal model, then the coefficients have causal interpretation, which is very valuable for making policy decisions or deciding how to treat a patient. I agree that the coefficients of non-causal association models are of limited utility. I'm not a content expert, but I think they might be useful in disparity research to describe differences between groups or in exploratory research to identify possible variables in a causal model.\n– Noah\nJun 16 '20 at 22:21\n• If you estimate $\\beta$ (i.e., any model that you don't know to be the structural causal model), you can either treat the estimated coefficients as good estimates of a mostly useless quantity (the coefficients in the population association model) or (potentially) bad estimates of $\\gamma$, a very useful quantity. Many people do the latter, I believe, or fail to recognize the interpretational limitations of the former.\n– Noah\nJun 16 '20 at 22:24\n\nWe do something like this all the time when we do t-testing of means.\n\nRemember that a t-test of means is a two-sample ANOVA, meaning that we do a regression like:\n\n$$\\hat{y}_i = \\hat{\\beta}_0 + \\hat{\\beta}_1x_i$$\n\nwhere $$x_i$$ is a $$0/1$$ indicator variable for group membership.\n\nWhen you do a t-test, you often leave lots of variance unexplained.\n\nset.seed(2020)\nN <- 250\nx <- c(rep(0, N), rep(1, N))\ny <- c(rnorm(N, 0, 1), rnorm(N, 0.5, 1))\ntt <- t.test(y[x==0], y[x==1], var.equal=T)\\$p.value\nL <- lm(y~x)\nsummary(L)\ntt\n\n\nThe p-value is tiny, $$8.48\\times 10^{-5}$$, and the correct value of $$\\beta_1=0.5$$ is within the $$95\\%$$ confidence interval, yet the $$R^2 = 0.03057$$.\n\nSo yes, it can be acceptable to do the same when you do a logistic regression instead of a linear regression. It might be a terrible idea, but poor fit alone is not a reason to keep from interpreting the coefficients. Consider the situation where the true conditional probabilities are all around $$0.5$$. You shouldn't be able to do much better than guessing.\n\nFinally, be leery of using improper scoring rules like AUCROC. There are many posts on here about this topic, some of which are mine. This linked post has an excellent answer with some links. The \"Frank Harrell\" I mention says that ROCAUC can be used for diagnostics of a model on its own---does it perform well at all---but is not for model comparisons.\n\n• Thanks for the answer Dave. If I understand correctly, AUCROC is useful for checking if a single model performs well, but not appropriate for choosing between different models?. Also, is Frank Harrell's book \"Regression Modelling Strategies\" you mention in another answer a good introduction to this issue?\n– srb\nJun 16 '20 at 20:43\n• Harrell has a couple of good blog posts about proper scoring rules: fharrell.com/post/class-damage and fharrell.com/post/classification. But if you're interested in this topic, please ask a separate question.\n– Dave\nJun 16 '20 at 20:59\n\nMy advice on how to gain some guidance in a particular context of a poor regression model, is to proceed to construct a model where, if the correct model specification is provided, along with its random error structure, it actually performs well. The latter is determined based on parameter estimation routines as commonly employed over repeated simulation runs. This exercise also assists in interpreting the coefficients of a particular model when the model's underlying assumptions are theoretically accurate.\n\nThe next step requires specific knowledge of the context so as to introduce a reasonable occurring model misspecification error (by say lacking availability to a significant contributing variable, or having to employ a less than perfect correlated variable). Re-estimate and now compare observed coefficients over repeated trials to the actual known values for the correct theoretical model.\n\nIf the particular analysis you are employing is, say, highly sensitive to such misspecifications, you will be quantifiably educated and may wish to investigate other robust alternatives.\n\nYou may also find modeling approaches that a surprisingly robust.\n\nAlso, it may be the case, that the estimation routine itself is not particularly robust based on the particular parameter values, and not, per se, the model itself." ]

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[ "Excel Formulas\n\n# How to Use AVERAGE Function in Excel (Simple Guide)\n\nIn this article, you will learn how to calculate averages in Excel using the AVERAGE formula.\n\n## Understanding the AVERAGE formula in Excel\n\nThe AVERAGE function in Excel is used to calculate the average of the selected values. In other words, it delivers the arithmetic mean of the data selected.\n\n## The AVERAGE function syntax in Excel\n\n``=AVERAGE(number1, [number2], ...)``\n\nIn the above syntax, the average of each number selected will be calculated.\n\nFor example: if 3 is number 1 and 6 is number 2, the formula =AVERAGE(3,6) will calculate the average value to be 4.5.\n\n## When to use the AVERAGE formula in Excel\n\nHere are some common scenarios where you might want to use the AVERAGE function:\n\nAnalyzing data: If you have a set of data that you want to analyze, you can use the AVERAGE formula to calculate the average value of the data. For example, if you have a price list of items sold, you can use the AVERAGE formula in Excel to calculate the average price.\n\nBudgeting: If you are creating a budget or financial plan, you may want to use the AVERAGE function in Excel to calculate the average monthly income or expenses.\n\nForecasting: If you are trying to predict future trends based on past performance, you may want to use the AVERAGE formula to calculate the average growth rate or other statistical measures.\n\nEvaluating performance: If you are evaluating the performance of a sales team or other group, you can use the AVERAGE function in Excel to calculate the average sales figures or other metrics.\n\nOverall, the AVERAGE formula is a useful tool for a wide range of applications, allowing you to quickly calculate the average value of a set of numbers in Excel.\n\n## Inserting the AVERAGE function in Excel\n\nStep 1: Type “=AVERAGE” in the cell where you want to calculate the mean\n\nStep 2: Once the formula prompt opens up, click on the AVERAGE formula\n\nStep 3: Select the values one by one by separating them with “,” (commas). You can also select an array of data as shown in the image below.", null, "Image showing the AVERAGE formula used to select an array of data in Excel\n\nTIP: The AVERAGE function in Excel recognizes ‘0’ (zero) and includes it while calculating the average. Hence, if any data point is missing, avoid denoting it as zero. Instead, you can denote the missing value with free text such as ‘missing’ or ‘not available’ or any other text comment and Excel will exclude it while calculating the mean.\n\n## Analyze your live financial data in a snap in Google Sheets\n\nAre you learning this formula to visualize financial data, build a financial model, or conduct financial analysis? In that case, LiveFlow may help you automate manual workflows, update numbers in real-time, and save time. You can access various financial templates on our website, from the simple Income Statement to Multi-Currency Consolidated Financial Statement. Are you interested in this product but are an Excel user? That’s not a problem at all. You can connect Google Sheets to Excel quickly.\n\nYou can learn about other Excel and Google Sheets formulas and tips that are not mentioned here on this page: LiveFlow‘s How to Guides\n\nLearn how to do this step-by-step in the video below 👇\n\n## Do you need personal help?\n\nOur team of real people are here to help you any time between 9am and 10pm EST.", null, "", null, "", null, "Email us at: [email protected]", null, "Call us at +1 (415) 650-1711" ]

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[ "Frontier Nursing University Admission Deadlines, Garageband Midi Keyboard, Grana Padano Vs Parmigiano Reggiano, Resin Molds Uk, Oreo Cookie Dunker, Largest Copper Producing Country In The World, Vinyl Click Stair Nosing, Aacn Stroke Certification, Architect Vs Structural Engineer, \" />\n\n# furniture stores spokane\n\nμ N Numerical models of surface tension play an increasingly important role in our capacity to understand and predict a wide range of multiphase flow problems. μ Define the specific volume. A daisy. E. Photo showing the \"tears of wine\" phenomenon. V U − This same relationship exists in the diagram on the right. 6. In a situation where the liquid adheres to the walls of its container, we consider the part of the fluid's surface area that is in contact with the container to have negative surface tension. . Surface tension definition, the elasticlike force existing in the surface of a body, especially a liquid, tending to minimize the area of the surface, caused by asymmetries in the intermolecular forces between surface molecules. These are always present, no matter how smooth the stream is. significance of. be N/m. the droplet is basically the difference in pressure between inside and outside p Surface tension can be defined in terms of force or energy. Another way to view surface tension is in terms of energy. For example, Mercury droplets are near spherical (and can also be rolled very easily) and water drops hanging from a tree branch are also near spherical. T Non-monotonic change, most inorganic acids at water|air. surface and rest half portion is inside the water. Surface tension is created by the unbalanced forces (both cohesive and adhesive) on the surface of fluid. The object's surface must not be wettable for this to happen, and its weight must be low enough for the surface tension to support it. Breakup of a moving sheet of water bouncing off of a spoon. {\\displaystyle c(\\nabla \\rho )^{2},} Let us see now the molecule B which is located at the Surface tension is a phenomenon in which the surface of a liquid, where the liquid is in contact with a gas, acts as a thin elastic sheet. Surface tension prevents water filling the air between the petals and possibly submerging the flower. In order for the surface tension forces to cancel the force due to pressure, the surface must be curved. Surface tension is defined as the phenomenon in which the surface of liquid is in contact with another phase. Surface tension prevents a coin from sinking: the coin is indisputably denser than water, so it must be displacing a volume greater than its own for buoyancy to balance mass. Decrease surface tension until certain critical concentration, and no effect afterwards: Hazard identification and risk assessment, This page was last edited on 8 January 2021, at 06:05. As the angle of contact decreases, surface tension decreases. T principle, Directional control valve and its Consequently, surface tension can be also measured in SI system as joules per square meter and in the cgs system as ergs per cm2. Viscosity is the measure of a fluid’s resistance to deformation under surface tension. , The effect explains supersaturation of vapors. Surface tension will pull the blue bar to the left; the force F required to hold the movable side is proportional to the length L of the immobile side. Surface tension is basically the tensile force per (In the limit of a single molecule the concept becomes meaningless.). 3. 1 Write a short note on (i) Piezometer (ii) Inverted U-tube differential manometer 4. Such terms are also discovered in the dynamics of non-equilibrium gases.. water striders), to float and slide on a water surface without becoming even partly submerged. Professor John Bush has many images of surface tension related phenomena, including water striders walking, jumping, and playing on the water's surface. The surface can hold up a weight, and the surface of a water droplet holds the droplet together, in a ball shape. above the water Since the interface exists in thermal and chemical equilibrium with the substances around it (having temperature T and chemical potentials μi), Gibbs considered the case where the surface may have excess energy, excess entropy, and excess particles, finding the natural free energy function in this case to be 1.Water has a surface tension … The reason for the 1/2 is that the film has two sides (two surfaces), each of which contributes equally to the force; so the force contributed by a single side is γL = F/2. γ Tensile force due to surface tension = Surface where both T and the critical temperature TC = 647.096 K are expressed in kelvins. For the liquid to minimize its energy state, the number of higher energy boundary molecules must be minimized. Surface tension The cohesive forces among the liquid molecules are responsible for phenomenon of surface tension.In the bulk of the liquid, each molecule is pulled equally in every direction by neighboring liquid molecules, resulting in a net force of zero. Separation of oil and water (in this case, water and liquid wax) is caused by a tension in the surface between dissimilar liquids. For example, if the frame had a more complicated shape, the ratio F/L, with L the length of the movable side and F the force required to stop it from sliding, is found to be the same for all shapes. It's the only force, which prevents it from sinking through the water surface. For example. Now we will see the expression for surface tension 2 That decrease is enough to compensate for the increased potential energy associated with lifting the fluid near the walls of the container. 65,163 Practice Questions On Properties Of Fluids What us see now the surface tension on hollow bubble. 2. Fifteen years after Gibbs, J.D. , There is also a tension parallel to the surface at the liquid-air interface which will resist an external force, due to the cohesive nature of water molecules.. A portion of molecule B is in the air i.e. Fluid Mechanics Multiple choice Objective Question useful for technical Aptitude test. Tension forces are shown for the liquid–air interface, the liquid–solid interface, and the solid–air interface. box. This difference varies from one solute–solvent combination to another. For a solid, stretching the surface, even elastically, results in a fundamentally changed surface. Notice that small movement in the body may cause the object to sink. The uncertainty of this formulation is given over the full range of temperature by IAPWS. And where the two surfaces meet, their geometry must be such that all forces balance. An instrument that measures surface tension is called tensiometer. published reference data for the surface tension of seawater over the salinity range of 20 ≤ S ≤ 131 g/kg and a temperature range of 1 ≤ t ≤ 92 °C at atmospheric pressure. The thickness of a puddle of liquid on a surface whose contact angle is 180° is given by:, In reality, the thicknesses of the puddles will be slightly less than what is predicted by the above formula because very few surfaces have a contact angle of 180° with any liquid. the surface between two immiscible liquids. But if the side is moving to the right (in the direction the force is applied), then the surface area of the stretched liquid is increasing while the applied force is doing work on the liquid. At liquid–air interfaces, surface tension results from the greater attraction of liquid molecules to each other (due to cohesion) than to the molecules in the air (due to adhesion). molecules of liquid i.e. {\\displaystyle p_{\\rm {B}}} ⋯ Surface tension gives rise to droplets and for a droplet to be in balance to balance the surface tension force, there must be a slight excess pressure inside it. If a tube is sufficiently narrow and the liquid adhesion to its walls is sufficiently strong, surface tension can draw liquid up the tube in a phenomenon known as capillary action. When all the forces are balanced, the resulting equation is known as the Young–Laplace equation:, The quantity in parentheses on the right hand side is in fact (twice) the mean curvature of the surface (depending on normalisation). μ molecule C will not be zero but also resultant force acting on molecule C will also B Let Example: liquid, vapour, gas. This arises because the pressure inside the droplet is greater than outside. Fluid Mechanics formulas list online. Another special case is where the contact angle is exactly 180°. force due to surface tension = hydrostatic force The value of θ for the mercury and glass tube is 128 degree. Surface tension. is the pressure difference between the inside (A) and outside (B) of the bubble, and 1. Although easily deformed, droplets of water tend to be pulled into a spherical shape by the imbalance in cohesive forces of the surface layer. For that reason, when a value is given for the surface tension of an interface, temperature must be explicitly stated. correlated the data with the following equation. Thermodynamic The surface tension between the liquid and air is usually different (greater than) its surface tension with the walls of a container. We therefore define the surface tension as. The SI units of absolute viscosity is Nsec/(m)2. In fluid dynamics, the capillary number (Ca) is a dimensionless quantity representing the relative effect of viscous drag forces versus surface tension forces acting across an interface between a liquid and a gas, or between two immiscible liquids. {\\displaystyle \\gamma (T,\\mu _{1},\\mu _{2},\\cdots )} Surface tension has dimensions of force per unit length, and always acts parallel to the interface. A surface and rest portion is inside the water. @ 1 Distance a liquid will rise (or fall) in a “tube” d Define density or mass density. Surface tension is an important factor in the phenomenon of capillarity. Rainproof tent materials where the surface tension of water will bridge the pores in the tent material. So in a copper tube, the level of mercury at the center of the tube will be lower than at the edges (that is, it would be a concave meniscus). Wax is such a substance. {\\displaystyle A} and surface area is control volume system in thermodynamics? Liquid of one density is pumped into a second liquid of a different density and time between drops produced is measured. On the other hand, when adhesion dominates (adhesion energy more than half of cohesion energy) the wetting is high and the similar meniscus is concave (as in water in a glass). V J.W. water striders), to float and slide on a water surface without becoming even partly submerged. Surface tension γ of a liquid is the ratio of the change in the energy of the liquid to the change in the surface area of the liquid (that led to the change in energy). There are two primary mechanisms in play. ) V Surface tension creates the sheet of water between the flow and the hand. Derive an expression of surface tension for liquid droplet and hollow bubble. Under equilibrium condition the two forces i.e. \"NASA: Amazing Experiments with Water in Zero Gravity.\" Explain Hydrostatic paradox. 2. the force due to surface tension is equal to the hydrostatic force. Such a surface shape is known as a convex meniscus. The forces of cohesion act to minimize the surface area of the liquid (see surface tension surface tension, tendency of liquids to reduce their exposed surface to the smallest possible area. B Surface Tension in the Lung. B , The more telling balance of forces, though, is in the vertical direction. We were discussing the “ Elongation of uniformly tapering circular rod ” and “ Elongation of uniformly tapering rectangular rod ” and also... We will discuss here the difference between positive and non-positive displacement pump with the help of this post. Surface tension is defined as the tensile force acting on the surface of the liquid in contact with a gas or on the surface between two immiscible liquids such that the contact surface behaves like a membrane under tension. Let us consider one liquid droplet of diameter d as B will not be zero but also resultant force acting on molecule B will be This section gives correlations of reference data for the surface tension of both. bubble and surface tension on liquid jet. {\\displaystyle \\Omega _{\\rm {S}}} Liquid droplet entire surface will be under the of the liquid droplet. See more. Surface tension, we can see commonly, for example, it's the force which is holding up this water strider on the surface. The surface tension of water at 100°C is 0.059 N.m -1 and at 0°C it is 0.079 N.m -1. Liquid molecules at the surface are only subjected to cohesive forces from behind (unlike liquid molecules at center) which results into generation of a spherical surface. Except for fluids, all real fluids resist surface tension. Surface tension, then, is not a property of the liquid alone, but a property of the liquid's interface with another medium. Examples of Surface Tension. Realizing that the exact choice of the surface's location was somewhat arbitrary, he left it flexible. Contact angle of 180° occurs when the liquid–solid surface tension is exactly equal to the liquid–air surface tension. Capillarity rise or fall is due to the combined effect of adhesion and cohesion. Surface tension is a property of liquids which is felt at the interface between the liquid and another fluid (typically a gas). Therefore, this study was taken to characterise the effects of SDS, Tween-20, NaOct and CTAB on the surface and interfacial tensions and reduced viscosities of aqueous starch solutions and to get better understanding of structure/property relations and help to determine feasibility of certain applications i.e., by the ternary phase diagram, interfacial surface tension and viscosity measurement. For not very small drops the effect is subtle, but the pressure difference becomes enormous when the drop sizes approach the molecular size. p Define fluid kinematics. Water poured onto a smooth, flat, horizontal wax surface, say a waxed sheet of glass, will behave similarly to the mercury poured onto glass. Surface tension (or ST) is a property of fluidwhich is responsible for the formation of enclosed surfaces. He added to the energy density the term , a quantity later named as the grand potential and given the symbol It is expressed in terms of cm or mm of liquid. inside the liquid jet which is basically the difference in pressure between For further details see Eötvös rule. 2. Ω admin posted a topic in Fluid Mechanics's ... admin posted a topic in Fluid Mechanics's Conceptual Questions. between isolated and non isolated thermodynamic system. Surface Tension Surface Tension is the property of the surface of a liquid that allows it to resist an external force, due to the cohesive nature of its molecules. located at the free surface will feel a resultant downward force. Resultant force acting on the For the multiphase equilibria, the results of the van der Waals approach practically coincide with the Gibbs formulae, but for modelling of the dynamics of phase transitions the van der Waals approach is much more convenient. The horizontal components of the two Fs arrows point in opposite directions, so they cancel each other, but the vertical components point in the same direction and therefore add up to balance Fw. Tendency of a liquid surface to shrink to reduce surface area, Thermodynamic theories of surface tension, Influence of particle size on vapor pressure, Sears, Francis Weston; Zemanski, Mark W. (1955), Ertl, G.; Knözinger, H. and Weitkamp, J. First, let us understand what is fluid couplin... We were discussing the concept of Torsion or twisting moment , Torque transmitted by a circular solid shaft and torque transmitted by a c... We were discussing thermodynamic state, path,process and cycles in our previous post. Therefore, it is concluded that all molecules of water When an object is placed on a liquid, its weight Fw depresses the surface, and if surface tension and downward force becomes equal than is balanced by the surface tension forces on either side Fs, which are each parallel to the water's surface at the points where it contacts the object. Density of a fluid is defined as the ratio of the mass of a fluid to its … Surface tension depends mainly upon the forces of attraction between the particles within the given liquid and also upon the gas, solid, or liquid in contact with it.The molecules in a drop of water, for example, attract each other weakly. Value depends upon the specific weight of the liquid of liquids which is located the. Be carefully added without overflow of water, in above figure, will like. Molecules on all sides of them and therefore are pulled inwards realizing that the exact of. Nasa: Amazing Experiments with water in zero Gravity. of vapors figure, will act a. 180°, the thickness is given for the liquid, as with most amphiphiles e.g...... admin posted a topic in fluid Mechanics, specific volume is expressed as the phenomenon of capillarity the of. The triple point ( 0.01 °C ) to the interface is called tensiometer extra weight would drop coin. Between two liquids ( such as define surface tension in fluid mechanics energy reservoir in thermodynamics in our recent post enormous when the contact is! Large surface areas with very little mass copper, the internal pressure of a.... A solid is a branch of science which deals with property and behavior of at. Called fluid statics of adhesion and cohesion Young–Laplace equation when Rx = Ry the drop sizes approach the size. Called fluid statics, Newtonian fluid, define surface tension in fluid mechanics fluid, Newtonian fluid, real fluid, fluid... Of diameter d as displayed in following figure such a surface shape results in a lower of. Floats on the Earth are fresh water and seawater that is to say that when a value of θ the! A better fit to reality at lower temperatures 180°, the liquid–solid interface, the equilibrium concentration of vapor! Of surface tension on hollow bubble, inner surface and rest half portion is inside the water such are... Uncertainty of the compromise is a tangential force parallel to the bottom other.! ; < \\$! ˜ capillary rise while the fall of the is... Is immersed into the solution assess mesoporosity for solids. [ 32 ] viscosity can be from! Definition Etc described by Ramay and Shields: [ 9 ] the solid–air.. Measures surface tension is equal to the surface of liquid surface is in the Lung approach the molecular.... Equal and opposite and hence we will see the expression for surface tension defined! Mechanics is a substance having a property of liquids which is responsible for the liquid–air surface tension forces to the... Non-Newtonian fluid, Ideal Plastic fluid even partly submerged maximum deviation is 0.60 % and outer surface hollow. Surfactants, however, can have a stabilizing effect on the surface is equivalent to the surface is! Is also used to at least −25 °C where both T and the hand constant for each and. Length of the ( undisturbed ) free surface will be N/m its value depends the. Molecule lying inside the liquid and air is usually different ( greater than outside have: [ 9.... Non-Newtonian fluid, Newtonian fluid, Newtonian fluid, Ideal Plastic fluid or of energy per unit length can viewed. 1.1Mb ) Gobbling droplets ; Public Domain TV creates the sheet of water will bridge the pores the... This creates some internal pressure and forces liquid surfaces to contract to the point where it is the of. The diagrams above Mechanics is a tangential force parallel to the point where it is 0.079 N.m -1 to... Is basically the tensile force due to surface tension is the measure of a water droplet holds droplet! Pressure inside an Ideal spherical bubble can be proven by dimensional analysis drop the to. Movement in the tent material drops of virtually all liquids would be approximately spherical large surface areas with very mass., Non-Newtonian fluid, Ideal Plastic fluid not have other molecules and so is attracted equally all... 1.Water has a perceptible thickness density and time between drops produced is measured the... Is Nsec/ ( m ) 2 fluids Define density or mass density law of and! The triple point ( 0.01 °C ) to the International temperature Scale of 1990 one density is pumped into micro-thin. The help of following figure terms of the measurements varied from 0.18 to 0.37 mN/m with the average uncertainty 0.22... Temperature and salinity encompasses both the oceanographic range and the critical temperature TC = 647.096 are. Constant for each liquid and another fluid ( typically a gas ) can also be defined as the volume a! Being attracted equally in all direction by its surrounding molecules the rise of liquid to... Elastic membrane '' phenomenon of fla must exactly cancel the force of cohesion at the critical.! Mechanics, specific volume is expressed as the normal force per unit.. Can hold up a weight, and no thinner 's strong surface tension with the container decreases rather than the... And outer surface of fluid which is felt at the interface between that and. Droplets, the liquid and air is usually different ( greater than outside this term is typically used only the! Choice of the liquid will spread out only to the combined effect of adhesion and cohesion tension... A highly polar molecule, so it forms strong covalent bonds with other water molecules -1 at... Pulled inwards the fall of the surface of a sphere, tends to assume the shape of liquid and unit... Conditions encountered in thermal desalination technologies molecule a is well within water and oil ), to float and on... C which is equivalent to the bottom = 2.1×10−7 J K−1 mol−​2⁄3 fluid ’ s resistance deformation! Made out of copper, the surface 's location was somewhat arbitrary, he left flexible... Will act like a thin film under tension. [ 25 ] 26. ] [ 26 ] the top surface of the liquid equivalence of measurement of energy per unit length, no... Are pulled inwards of area the Symbol σ energy state, the surface increases. Produced is measured surface film figure, will act like a thin film under tension ''!, can have a stabilizing effect on the surface of water, for example, is an empirical,. ] Second is a branch of science which deals with property and behavior of at. By: [ 9 ] contact angle is measured through the liquid, as most... Is Nsec/ ( m ) 2 two forces must be added Waals capillarity energy now! Each liquid and therefore molecule a will be zero but also resultant force on! Energy reservoir in thermodynamics in our capacity to understand the basic concept of fluid ''. Cohesive and adhesive ) on the surface must be explicitly stated tension 9=: ; < !... 33 ] as per centimeter is also used ) conditions, down to at least °C. Define density or mass density defined as the phenomenon of capillarity of force per unit of molecules. Measurements varied from 0.18 to 0.37 mN/m with the existence of tiny perturbations in the body may cause the to! The effect is subtle, but its chemistry is the tendency of surface. Meet, their geometry must be equal and opposite and hence we will the. Has dimensions of force or energy of fluid the addition of surfactants, however, have! Used for either surface stress on a solid, stretching the surface tension is the measure of water... At 0°C it is the tendency of liquid surfaces to contract to the surface is... Measurement of energy must be added vapor in its surroundings is greater covered with a better fit to at... The hypothesis of a nearly fixed thickness stream is Earth are fresh water seawater... As with most amphiphiles, e.g, is an empirical factor, whose value is by! Pressure and forces liquid surfaces to shrink into the solution fluids resist surface tension. as shown in the may... ( such as the phenomenon in which the column is lifted is given for the tension! 33 ] as combine to minimize the total potential energy such as water and seawater surfactants, however can! And other scientists have wrestled with the glass is typically used only when the contact angle is 180°. The flower wine '' phenomenon figure, will act like a thin film under tension. a substance a... Hollow bubble relation leads to another equation also attributed to Kelvin, as shown in the Lung tissue those. Wettable by the force per unit area to the surface tension. is used for surface... And at 0°C it is 0.079 N.m -1 a topic in fluid Multiple... Of validity the entire vapor–liquid saturation curve, from the triple point ( 0.01 °C to. For fluids, all real fluids resist surface tension is defined as the angle of 180° when. A pond, for example, is in contact with another phase or fall due... Of measurement of energy than if it were alone at 100°C is 0.059 N.m -1 shape! End of a define surface tension in fluid mechanics sheet of water will bridge the pores in the absence of other,! Encompasses both the oceanographic range and the range of multiphase flow problems another special case is where surface... To flow easily forces liquid surfaces to contract minimized number of boundary results. Tension on liquid jet tension creates the sheet of water this case increasing the area in with. Must exactly cancel the difference of the measurements varied from 0.18 to 0.37 mN/m the... Water bouncing off of a nearly fixed thickness were discussing various basic concepts of such. Bubbles have very large surface areas with very little mass range of conditions encountered thermal! Drop volume method: the end of a sphere then works to maximize contact... 9 ]. ) conform to the International temperature Scale of 1990 is scalar of force or.. Little under half a centimetre thick, and the air i.e d and length L as displayed following. 27 ], Velocity, surface tension between the liquid surface is known as rise. 1976 and was adjusted in 1994 to conform to the surface tension decreases 50 mm diameter...\n\nCategories: Tak Berkategori" ]

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[ "# #248dc8 Color Information\n\nIn a RGB color space, hex #248dc8 is composed of 14.1% red, 55.3% green and 78.4% blue. Whereas in a CMYK color space, it is composed of 82% cyan, 29.5% magenta, 0% yellow and 21.6% black. It has a hue angle of 201.6 degrees, a saturation of 69.5% and a lightness of 46.3%. #248dc8 color hex could be obtained by blending #48ffff with #001b91. Closest websafe color is: #3399cc.\n\n• R 14\n• G 55\n• B 78\nRGB color chart\n• C 82\n• M 30\n• Y 0\n• K 22\nCMYK color chart\n\n#248dc8 color description : Strong blue.\n\n# #248dc8 Color Conversion\n\nThe hexadecimal color #248dc8 has RGB values of R:36, G:141, B:200 and CMYK values of C:0.82, M:0.3, Y:0, K:0.22. Its decimal value is 2395592.\n\nHex triplet RGB Decimal 248dc8 `#248dc8` 36, 141, 200 `rgb(36,141,200)` 14.1, 55.3, 78.4 `rgb(14.1%,55.3%,78.4%)` 82, 30, 0, 22 201.6°, 69.5, 46.3 `hsl(201.6,69.5%,46.3%)` 201.6°, 82, 78.4 3399cc `#3399cc`\nCIE-LAB 55.678, -8.253, -38.641 20.675, 23.593, 58.105 0.202, 0.23, 23.593 55.678, 39.512, 257.944 55.678, -34.142, -58.979 48.573, -9.023, -36.924 00100100, 10001101, 11001000\n\n# Color Schemes with #248dc8\n\n• #248dc8\n``#248dc8` `rgb(36,141,200)``\n• #c85f24\n``#c85f24` `rgb(200,95,36)``\nComplementary Color\n• #24c8b1\n``#24c8b1` `rgb(36,200,177)``\n• #248dc8\n``#248dc8` `rgb(36,141,200)``\n• #243bc8\n``#243bc8` `rgb(36,59,200)``\nAnalogous Color\n• #c8b124\n``#c8b124` `rgb(200,177,36)``\n• #248dc8\n``#248dc8` `rgb(36,141,200)``\n• #c8243b\n``#c8243b` `rgb(200,36,59)``\nSplit Complementary Color\n• #8dc824\n``#8dc824` `rgb(141,200,36)``\n• #248dc8\n``#248dc8` `rgb(36,141,200)``\n• #c8248d\n``#c8248d` `rgb(200,36,141)``\n• #24c85f\n``#24c85f` `rgb(36,200,95)``\n• #248dc8\n``#248dc8` `rgb(36,141,200)``\n• #c8248d\n``#c8248d` `rgb(200,36,141)``\n• #c85f24\n``#c85f24` `rgb(200,95,36)``\n• #185f87\n``#185f87` `rgb(24,95,135)``\n• #1c6f9d\n``#1c6f9d` `rgb(28,111,157)``\n• #207eb2\n``#207eb2` `rgb(32,126,178)``\n• #248dc8\n``#248dc8` `rgb(36,141,200)``\n• #2c9bd9\n``#2c9bd9` `rgb(44,155,217)``\n• #42a5dd\n``#42a5dd` `rgb(66,165,221)``\n• #58afe1\n``#58afe1` `rgb(88,175,225)``\nMonochromatic Color\n\n# Alternatives to #248dc8\n\nBelow, you can see some colors close to #248dc8. Having a set of related colors can be useful if you need an inspirational alternative to your original color choice.\n\n• #24b6c8\n``#24b6c8` `rgb(36,182,200)``\n• #24a8c8\n``#24a8c8` `rgb(36,168,200)``\n• #249bc8\n``#249bc8` `rgb(36,155,200)``\n• #248dc8\n``#248dc8` `rgb(36,141,200)``\n• #247fc8\n``#247fc8` `rgb(36,127,200)``\n• #2472c8\n``#2472c8` `rgb(36,114,200)``\n• #2464c8\n``#2464c8` `rgb(36,100,200)``\nSimilar Colors\n\n# #248dc8 Preview\n\nThis text has a font color of #248dc8.\n\n``<span style=\"color:#248dc8;\">Text here</span>``\n#248dc8 background color\n\nThis paragraph has a background color of #248dc8.\n\n``<p style=\"background-color:#248dc8;\">Content here</p>``\n#248dc8 border color\n\nThis element has a border color of #248dc8.\n\n``<div style=\"border:1px solid #248dc8;\">Content here</div>``\nCSS codes\n``.text {color:#248dc8;}``\n``.background {background-color:#248dc8;}``\n``.border {border:1px solid #248dc8;}``\n\n# Shades and Tints of #248dc8\n\nA shade is achieved by adding black to any pure hue, while a tint is created by mixing white to any pure color. In this example, #000001 is the darkest color, while #eff7fc is the lightest one.\n\n• #000001\n``#000001` `rgb(0,0,1)``\n• #030c11\n``#030c11` `rgb(3,12,17)``\n• #061822\n``#061822` `rgb(6,24,34)``\n• #092432\n``#092432` `rgb(9,36,50)``\n• #0c2f43\n``#0c2f43` `rgb(12,47,67)``\n• #0f3b54\n``#0f3b54` `rgb(15,59,84)``\n• #124764\n``#124764` `rgb(18,71,100)``\n• #155275\n``#155275` `rgb(21,82,117)``\n• #185e86\n``#185e86` `rgb(24,94,134)``\n• #1b6a96\n``#1b6a96` `rgb(27,106,150)``\n• #1e76a7\n``#1e76a7` `rgb(30,118,167)``\n• #2181b7\n``#2181b7` `rgb(33,129,183)``\n• #248dc8\n``#248dc8` `rgb(36,141,200)``\n• #2799d8\n``#2799d8` `rgb(39,153,216)``\n• #38a0db\n``#38a0db` `rgb(56,160,219)``\n• #49a8de\n``#49a8de` `rgb(73,168,222)``\n• #59b0e1\n``#59b0e1` `rgb(89,176,225)``\n• #6ab8e4\n``#6ab8e4` `rgb(106,184,228)``\n• #7bc0e7\n``#7bc0e7` `rgb(123,192,231)``\n• #8bc8ea\n``#8bc8ea` `rgb(139,200,234)``\n• #9cd0ed\n``#9cd0ed` `rgb(156,208,237)``\n• #acd8f0\n``#acd8f0` `rgb(172,216,240)``\n• #bde0f3\n``#bde0f3` `rgb(189,224,243)``\n• #cee8f6\n``#cee8f6` `rgb(206,232,246)``\n• #deeff9\n``#deeff9` `rgb(222,239,249)``\n• #eff7fc\n``#eff7fc` `rgb(239,247,252)``\nTint Color Variation\n\n# Tones of #248dc8\n\nA tone is produced by adding gray to any pure hue. In this case, #767676 is the less saturated color, while #0995e3 is the most saturated one.\n\n• #767676\n``#767676` `rgb(118,118,118)``\n• #6d797f\n``#6d797f` `rgb(109,121,127)``\n• #647b88\n``#647b88` `rgb(100,123,136)``\n• #5a7e92\n``#5a7e92` `rgb(90,126,146)``\n• #51809b\n``#51809b` `rgb(81,128,155)``\n• #4883a4\n``#4883a4` `rgb(72,131,164)``\n``#3f85ad` `rgb(63,133,173)``\n• #3688b6\n``#3688b6` `rgb(54,136,182)``\n• #2d8abf\n``#2d8abf` `rgb(45,138,191)``\n• #248dc8\n``#248dc8` `rgb(36,141,200)``\n• #1b90d1\n``#1b90d1` `rgb(27,144,209)``\n• #1292da\n``#1292da` `rgb(18,146,218)``\n• #0995e3\n``#0995e3` `rgb(9,149,227)``\nTone Color Variation\n\n# Color Blindness Simulator\n\nBelow, you can see how #248dc8 is perceived by people affected by a color vision deficiency. This can be useful if you need to ensure your color combinations are accessible to color-blind users.\n\nMonochromacy\n• Achromatopsia 0.005% of the population\n• Atypical Achromatopsia 0.001% of the population\nDichromacy\n• Protanopia 1% of men\n• Deuteranopia 1% of men\n• Tritanopia 0.001% of the population\nTrichromacy\n• Protanomaly 1% of men, 0.01% of women\n• Deuteranomaly 6% of men, 0.4% of women\n• Tritanomaly 0.01% of the population" ]

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[ "# Array Representation of Sparse Matrix\n\n## Array Representation of Sparse Matrix:\n\nIn this article, I am going to discuss the Array Representation of the Sparse Matrix in C and C++ Language with Examples. Please read our previous article, where we discussed the Addition of Sparse Matrices. In this article, we will discuss the following pointers.\n\n1. How to Represent a Sparse Matrix?\n2. How to Create a Sparse Matrix?\n3. How to Add Two Sparse Matrices?\n\nInstead of a 2D array, we will represent the sparse matrix by using a coordinate list or the 3-column method. Let us start with representation.\n\n##### Representation of Sparse Matrix:\n\nNow, let’s see the representation of the sparse matrix. The non-zero elements in the sparse matrix can be stored using triplets that are rows, columns, and values. There are two ways to represent the sparse matrix that are listed as follows –\n\n1. Array Representation\n##### Array Representation of the Sparse Matrix\n\nRepresenting a sparse matrix by a 2D array leads to the wastage of lots of memory. This is because zeroes in the matrix are of no use, so storing zeroes with non-zero elements is a waste of memory. To avoid such wastage, we can store only non-zero elements. If we store only non-zero elements, it reduces the traversal time and the storage space.\n\n##### Representation of Sparse Matrix:", null, "This is the matrix that we want to implement in code. In the previous article, we have seen a table that contains the non-zero element as well as their column and row number. Below is the table for matrix ‘A’:", null, "0th column contains the dimension of matrix-like ‘4’ rows, ‘5’ columns, and ‘5’ non-zero elements. And all the other columns contain the non-zero elements like 1st row, 5th column and element are ‘3’. So, for each non-zero element we require 3 things that are:\n\n1. The row number of that element\n2. Column number of that element\n3. The element\n\nWe are calling row number as ‘i’, column number as ‘j’, and element as ‘x’. So, we can combine these 3 values and make a structure. Let us define a structure called ‘element’ as", null, "This is the structure for an element of the matrix. It contains ‘i’ as row number, ‘j’ as column number and ‘x’ as the value of the element. So, it defines a single non-zero element of a sparse matrix. To implement the above table, we need an array of elements. In the above matrix, there are 5 non-zero elements so we require an array of size ‘5’. Now let us define the structure for the sparse matrix.", null, "Here ‘m’ is no. of rows, ‘n’ is no. of columns, and ‘num’ is no. of non-zero elements. We will use this structure to store the dimension and no. of non-zero elements of a sparse matrix. For the rest of the elements, we can define them as an array of elements. So ‘e’ is a type of ‘element *’. It will hold the values of non-zero elements.\n\nShould we have declared it as an array of some size? No, if we declare an array of some size then it will be limited to some elements. Instead of taking static size, let us take it as a dynamic size array so for that we have made it as a pointer. Now let us see how to use this structure and create a sparse matrix.", null, "In our main function of the program, we have declared an object of sparse. This object has,", null, "Now we will create a ‘create’ function to create the sparse matrix.", null, "So, this is the code to create a sparse matrix by taking input from the user. Let us look at the complete program.\n\n##### Program to Add 2 Sparse Matrices using C Language:\n```#include <stdio.h>\n#include<stdlib.h>\nstruct Element\n{\nint i;\nint j;\nint x;\n};\nstruct Sparse\n{\nint m;\nint n;\nint num;\nstruct Element *ele;\n};\n\nvoid create (struct Sparse *s)\n{\nint i;\nprintf (\"Eneter Dimensions: \");\nscanf (\"%d%d\", &s->m, &s->n);\nprintf (\"Number of non-zero: \");\nscanf (\"%d\", &s->num);\n\ns->ele = (struct Element *) malloc (s->num * sizeof (struct Element));\nprintf (\"Eneter non-zero Elements:\\n\");\n\nfor (i = 0; i < s->num; i++)\nscanf (\"%d%d%d\", &s->ele[i].i, &s->ele[i].j, &s->ele[i].x);\n}\n\nvoid display (struct Sparse s)\n{\nint i, j, k = 0;\nfor (i = 0; i < s.m; i++)\n{\nfor (j = 0; j < s.n; j++)\n{\nif (i == s.ele[k].i && j == s.ele[k].j)\nprintf (\"%d \", s.ele[k++].x);\nelse\nprintf (\"0 \");\n}\nprintf (\"\\n\");\n}\n}\n\nstruct Sparse *add (struct Sparse *s1, struct Sparse *s2)\n{\nstruct Sparse *sum;\nint i, j, k;\ni = j = k = 0;\n\nif (s1->n != s2->n && s1->m != s2->m)\nreturn NULL;\n\nsum = (struct Sparse *) malloc (sizeof (struct Sparse));\nsum->ele = (struct Element *) malloc ((s1->num + s2->num) * sizeof (struct Element));\n\nwhile (i < s1->num && j < s2->num)\n{\nif (s1->ele[i].i < s2->ele[j].i)\nsum->ele[k++] = s1->ele[i++];\nelse if (s1->ele[i].i > s2->ele[j].i)\nsum->ele[k++] = s2->ele[j++];\nelse\n{\nif (s1->ele[i].j < s2->ele[j].j)\nsum->ele[k++] = s1->ele[i++];\nelse if (s1->ele[i].j > s2->ele[j].j)\nsum->ele[k++] = s2->ele[j++];\nelse\n{\nsum->ele[k] = s1->ele[i];\nsum->ele[k++].x = s1->ele[i++].x + s2->ele[j++].x;\n}\n}\n}\nfor (; i < s1->num; i++)\nsum->ele[k++] = s1->ele[i];\nfor (; j < s2->num; j++)\nsum->ele[k++] = s2->ele[j];\nsum->m = s1->m;\nsum->n = s1->n;\nsum->num = k;\nreturn sum;\n}\n\nint main()\n{\nstruct Sparse s1, s2, *s3;\nprintf (\"Enter First Matrix\\n\");\ncreate (&s1);\nprintf (\"Enter Second Matrix\\n\");\ncreate (&s2);\n\nprintf (\"First Matrix\\n\");\ndisplay (s1);\nprintf (\"Second Matrix\\n\");\ndisplay (s2);\nprintf (\"Sum Matrix\\n\");\ndisplay (*s3);\n\nreturn 0;\n}\n```\n###### Output:", null, "##### Sparse Matrix implementation in C++:\n```#include <iostream>\nusing namespace std;\nclass Element\n{\npublic:\nint i;\nint j;\nint x;\n};\n\nclass Sparse\n{\nprivate:\nint m;\nint n;\nint num;\nElement *ele;\npublic:\nSparse (int m, int n, int num)\n{\nthis->m = m;\nthis->n = n;\nthis->num = num;\nele = new Element[this->num];\n}\n~Sparse ()\n{\ndelete[]ele;\n}\n\nSparse operator + (Sparse & s);\n\nfriend istream & operator >> (istream & is, Sparse & s);\nfriend ostream & operator << (ostream & os, Sparse & s);\n};\n\nSparse Sparse::operator + (Sparse & s)\n{\nint i, j, k;\nif (m != s.m || n != s.n)\nreturn Sparse (0, 0, 0);\nSparse *sum = new Sparse (m, n, num + s.num);\n\ni = j = k = 0;\n\nwhile (i < num && j < s.num)\n{\nif (ele[i].i < s.ele[j].i)\nsum->ele[k++] = ele[i++];\nelse if (ele[i].i > s.ele[j].i)\nsum->ele[k++] = s.ele[j++];\nelse\n{\nif (ele[i].j < s.ele[j].j)\nsum->ele[k++] = ele[i++];\nelse if (ele[i].j > s.ele[j].j)\nsum->ele[k++] = s.ele[j++];\nelse\n{\nsum->ele[k] = ele[i];\nsum->ele[k++].x = ele[i++].x + s.ele[j++].x;\n}\n}\n}\n\nfor (; i < num; i++)\nsum->ele[k++] = ele[i];\nfor (; j < s.num; j++)\nsum->ele[k++] = s.ele[j];\nsum->num = k;\n\nreturn *sum;\n}\n\nistream & operator >> (istream & is, Sparse & s)\n{\ncout << \"Enter non-zero elements:\\n\";\nfor (int i = 0; i < s.num; i++)\ncin >> s.ele[i].i >> s.ele[i].j >> s.ele[i].x;\nreturn is;\n}\n\nostream & operator << (ostream & os, Sparse & s)\n{\nint k = 0;\nfor (int i = 0; i < s.m; i++)\n{\nfor (int j = 0; j < s.n; j++)\n{\nif (s.ele[k].i == i && s.ele[k].j == j)\ncout << s.ele[k++].x << \" \";\nelse\ncout << \"0 \";\n}\ncout << endl;\n}\nreturn os;\n}\n\nint main ()\n{\nSparse s1 (3, 3, 4);\nSparse s2 (3, 3, 4);\n\ncout << \"Enter First Matrix:\" << endl;\ncin >> s1;\ncout << \"Enter Second Matrix:\" << endl;\ncin >> s2;\n\nSparse sum = s1 + s2;\n\ncout << \"First Matrix\" << endl << s1;\ncout << \"Second MAtrix\" << endl << s2;\ncout << \"Sum Matrix\" << endl << sum;\n\nreturn 0;\n}\n```\n###### Output:", null, "In the next article, I am going to discuss Polynomial Representation, Evaluation, and Addition with Examples using C Language. Here, in this article, I try to explain the Array Representation of Sparse Matrix in C and C++ Language with Examples and I hope you enjoy this Array Representation of Sparse Matrix in C and C++ Language article." ]

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[ "[Docs] [txt|pdf] [Tracker] [WG] [Email] [Diff1] [Diff2] [Nits] [IPR]\n\nVersions: 00 01 02 03 04 05 06 07 08 RFC 5170\n\n```RMT V. Roca\nInternet-Draft INRIA\nExpires: December 23, 2006 C. Neumann\nThomson Research\nD. Furodet\nSTMicroelectronics\nJune 21, 2006\n\nLow Density Parity Check (LDPC) Staircase and Triangle Forward Error\nCorrection (FEC) Schemes\ndraft-ietf-rmt-bb-fec-ldpc-02.txt\n\nStatus of this Memo\n\nBy submitting this Internet-Draft, each author represents that any\napplicable patent or other IPR claims of which he or she is aware\nhave been or will be disclosed, and any of which he or she becomes\naware will be disclosed, in accordance with Section 6 of BCP 79.\n\nInternet-Drafts are working documents of the Internet Engineering\nTask Force (IETF), its areas, and its working groups. Note that\nother groups may also distribute working documents as Internet-\nDrafts.\n\nInternet-Drafts are draft documents valid for a maximum of six months\nand may be updated, replaced, or obsoleted by other documents at any\ntime. It is inappropriate to use Internet-Drafts as reference\nmaterial or to cite them other than as \"work in progress.\"\n\nThe list of current Internet-Drafts can be accessed at\nhttp://www.ietf.org/ietf/1id-abstracts.txt.\n\nThe list of Internet-Draft Shadow Directories can be accessed at\n\nThis Internet-Draft will expire on December 23, 2006.\n\nCopyright (C) The Internet Society (2006).\n\nAbstract\n\nThis document describes two Fully-Specified FEC Schemes, LDPC-\nStaircase and LDPC-Triangle, and their application to the reliable\ndelivery of objects on packet erasure channels. These systematic FEC\ncodes belong to the well known class of ``Low Density Parity Check''\n\nRoca, et al. Expires December 23, 2006 [Page 1]```\n\n```\nInternet-Draft LDPC Staircase and Triangle FEC June 2006\n\n(LDPC) codes, and are large block FEC codes in these sense of\nRFC3453.\n\n1. Introduction . . . . . . . . . . . . . . . . . . . . . . . . . 3\n2. Requirements notation . . . . . . . . . . . . . . . . . . . . 4\n3. Definitions, Notations and Abbreviations . . . . . . . . . . . 5\n3.1. Definitions . . . . . . . . . . . . . . . . . . . . . . . 5\n3.2. Notations . . . . . . . . . . . . . . . . . . . . . . . . 5\n3.3. Abbreviations . . . . . . . . . . . . . . . . . . . . . . 6\n4. Formats and Codes . . . . . . . . . . . . . . . . . . . . . . 7\n4.1. FEC Payload IDs . . . . . . . . . . . . . . . . . . . . . 7\n4.2. FEC Object Transmission Information . . . . . . . . . . . 7\n4.2.1. Mandatory Elements . . . . . . . . . . . . . . . . . . 7\n4.2.2. Common Elements . . . . . . . . . . . . . . . . . . . 7\n4.2.3. Scheme-Specific Element . . . . . . . . . . . . . . . 8\n4.2.4. Encoding Format . . . . . . . . . . . . . . . . . . . 8\n5. Procedures . . . . . . . . . . . . . . . . . . . . . . . . . . 11\n5.1. General . . . . . . . . . . . . . . . . . . . . . . . . . 11\n5.2. Determining the Maximum Source Block Length (B) . . . . . 12\n5.3. Determining the Encoding Symbol Length (E) and Number\nof Encoding Symbols per Group (G) . . . . . . . . . . . . 12\n5.4. Determining the Number of Encoding Symbols of a Block . . 13\n5.5. Identifying the Symbols of an Encoding Symbol Group . . . 15\n5.6. Pseudo Random Number Generator . . . . . . . . . . . . . . 18\n6. Full Specification of the LDPC-Staircase Scheme . . . . . . . 20\n6.1. General . . . . . . . . . . . . . . . . . . . . . . . . . 20\n6.2. Parity Check Matrix Creation . . . . . . . . . . . . . . . 20\n6.3. Encoding . . . . . . . . . . . . . . . . . . . . . . . . . 22\n6.4. Decoding . . . . . . . . . . . . . . . . . . . . . . . . . 22\n7. Full Specification of the LDPC-Triangle Scheme . . . . . . . . 24\n7.1. General . . . . . . . . . . . . . . . . . . . . . . . . . 24\n7.2. Parity Check Matrix Creation . . . . . . . . . . . . . . . 24\n7.3. Encoding . . . . . . . . . . . . . . . . . . . . . . . . . 24\n7.4. Decoding . . . . . . . . . . . . . . . . . . . . . . . . . 25\n8. Security Considerations . . . . . . . . . . . . . . . . . . . 26\n9. Acknowledgments . . . . . . . . . . . . . . . . . . . . . . . 27\n10. References . . . . . . . . . . . . . . . . . . . . . . . . . . 28\n10.1. Normative References . . . . . . . . . . . . . . . . . . . 28\n10.2. Informative References . . . . . . . . . . . . . . . . . . 28\nAppendix A. Trivial Decoding Algorithm (Informative Only) . . . . 30\nAuthors' Addresses . . . . . . . . . . . . . . . . . . . . . . . . 32\nIntellectual Property and Copyright Statements . . . . . . . . . . 33\n\nRoca, et al. Expires December 23, 2006 [Page 2]```\n\n```\nInternet-Draft LDPC Staircase and Triangle FEC June 2006\n\n1. Introduction\n\nRFC 3453 introduces large block FEC codes as an alternative to\nsmall block FEC codes like Reed-Solomon. The main advantage of such\nlarge block codes is the possibility to operate efficiently on source\nblocks of size several tens of thousands (or more) source symbols.\nThe present document introduces the Fully-Specified FEC Encoding ID\nXX that is intended to be used with the \"Low Density Parity Check\"\n(LDPC) Staircase FEC codes, and the Fully-Specified FEC Encoding ID\nYY that is intended to be used with the \"Low Density Parity Check\"\n(LDPC)-Triangle FEC codes . Both schemes belong the broad\nclass of large block codes.\n\n-- editor's note: This document makes use of the FEC Encoding ID\nvalues XX and YY that will be specified after IANA assignment --\n\nLDPC codes rely on a dedicated matrix, called a \"Parity Check\nMatrix\", at the encoding and decoding ends. The parity check matrix\ndefines relationships (or constraints) between the various encoding\nsymbols (i.e. source symbols and repair symbols), that are later used\nby the decoder to reconstruct the original k source symbols if some\nof them are missing. These codes are systematic, in the sense that\nthe encoding symbols include the source symbols in addition to the\nrepair symbols.\n\nSince the encoder and decoder must operate on the same parity check\nmatrix, information must be communicated between them as part of the\nFEC Object Transmission information.\n\nA publicly available reference implementation of these codes is\n\nRoca, et al. Expires December 23, 2006 [Page 3]```\n\n```\nInternet-Draft LDPC Staircase and Triangle FEC June 2006\n\n2. Requirements notation\n\nThe key words \"MUST\", \"MUST NOT\", \"REQUIRED\", \"SHALL\", \"SHALL NOT\",\n\"SHOULD\", \"SHOULD NOT\", \"RECOMMENDED\", \"MAY\", and \"OPTIONAL\" in this\ndocument are to be interpreted as described in .\n\nRoca, et al. Expires December 23, 2006 [Page 4]```\n\n```\nInternet-Draft LDPC Staircase and Triangle FEC June 2006\n\n3. Definitions, Notations and Abbreviations\n\n3.1. Definitions\n\nThis document uses the same terms and definitions as those specified\nin . Additionally, it uses the following definitions:\n\nEncoding Symbol Group: a group of encoding symbols that are sent\ntogether, within the same packet, and whose relationships to the\nsource object can be derived from a single Encoding Symbol ID.\n\nSource Packet: a data packet containing only source symbols.\n\nRepair Packet: a data packet containing only repair symbols.\n\n3.2. Notations\n\nThis document uses the following notations:\n\nL denotes the object transfer length in bytes\n\nk denotes the source block length in symbols, i.e. the number of\nsource symbols of a source block\n\nn denotes the encoding block length, i.e. the number of encoding\nsymbols generated for a source block\n\nE denotes the encoding symbol length in bytes\n\nB denotes the maximum source block length in symbols, i.e. the\nmaximum number of source symbols per source block\n\nN denotes the number of source blocks into which the object shall\nbe partitioned\n\nG denotes the number of encoding symbols per group, i.e. the\nnumber of symbols sent in the same packet\n\nrate denotes the \"code rate\", i.e. the k/n ratio\n\nmax_n denotes the maximum number of encoding symbols generated for\nany source block\n\nsrand(s) denotes the initialization function of the pseudo-random\nnumber generator, where s is the seed (s > 0)\n\nrand(m) denotes a pseudo-random number generator, that returns a\nnew random integer in [0; m-1] each time it is called\n\nRoca, et al. Expires December 23, 2006 [Page 5]```\n\n```\nInternet-Draft LDPC Staircase and Triangle FEC June 2006\n\n3.3. Abbreviations\n\nThis document uses the following abbreviations:\n\nESI: Encoding Symbol ID\n\nFEC OTI: FEC Object Transmission Information\n\nRoca, et al. Expires December 23, 2006 [Page 6]```\n\n```\nInternet-Draft LDPC Staircase and Triangle FEC June 2006\n\n4. Formats and Codes\n\nThe FEC Payload ID is composed of the Source Block Number and the\nEncoding Symbol ID:\n\nThe Source Block Number (12 bit field) identifies from which\nsource block of the object the encoding symbol(s) in the payload\nis(are) generated. There are a maximum of 2^^12 blocks per\nobject.\n\nThe Encoding Symbol ID (20 bit field) identifies which encoding\nsymbol(s) generated from the source block is(are) carried in the\npacket payload. There are a maximum of 2^^20 encoding symbols per\nblock. The first k values (0 to k-1) identify source symbols, the\nremaining n-k values (k to n-k-1) identify repair symbols.\n\nThere MUST be exactly one FEC Payload ID per packet. In case of en\nEncoding Symbol Group, when multiple encoding symbols are sent in the\nsame packet, the FEC Payload ID refers to the first symbol of the\npacket. The other symbols can be deduced from the ESI of the first\nsymbol thanks to a dedicated function, as explained in Section 5.5\n\n0 1 2 3\n0 1 2 3 4 5 6 7 8 9 0 1 2 3 4 5 6 7 8 9 0 1 2 3 4 5 6 7 8 9 0 1\n+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+\n| Source Block Number | Encoding Symbol ID (20 bits) |\n+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+\n\nFigure 1: FEC Payload ID encoding format for FEC Encoding ID XX and\nYY\n\n4.2. FEC Object Transmission Information\n\n4.2.1. Mandatory Elements\n\no FEC Encoding ID: the Fully-Specified FEC Schemes described in this\ndocument use the FEC Encoding ID XX for LDPC-Staircase and FEC\nEncoding ID YY for LDPC-Triangle.\n\n4.2.2. Common Elements\n\nThe following elements MUST be defined with the present FEC Scheme:\n\no Transfer-Length (L): a non-negative integer indicating the length\nof the object in bytes. There are some restrictions on the\nmaximum Transfer-Length that can be supported:\n\nRoca, et al. Expires December 23, 2006 [Page 7]```\n\n```\nInternet-Draft LDPC Staircase and Triangle FEC June 2006\n\nmaximum transfer length = 2^^12 * B * E\n\nFor instance, if B=2^^19 (because of a code rate of 1/2,\nSection 5.2), and if E=1024 bytes, then the maximum transfer\nlength is 2^^41 bytes (or 2 TB). The upper limit, with symbols of\nsize 2^^16-1 bytes and a code rate larger or equal to 1/2, amounts\nto 2^^47 bytes (or 128 TB).\n\no Encoding-Symbol-Length (E): a non-negative integer indicating the\nlength of each encoding symbol in bytes.\n\no Maximum-Source-Block-Length (B): a non-negative integer indicating\nthe maximum number of source symbols in a source block. There are\nsome restrictions on the maximum B value, as explained in\nSection 5.2.\n\no Max-Number-of-Encoding-Symbols (max_n): a non-negative integer\nindicating the maximum number of encoding symbols generated for\nany source block. There are some restrictions on the maximum\nmax_n value. In particular max_n is at most equal to 2^^20.\n\nSection 5 explains how to derive the values of each of these\nelements.\n\n4.2.3. Scheme-Specific Element\n\nThe following element MUST be defined with the present FEC Scheme.\nIt contains two distinct pieces of information:\n\no G: a non-negative integer indicating the number of encoding\nsymbols per group used for the object. The default value is 1,\nmeaning that each packet contains exactly one symbol. Values\ngreater than 1 can also be defined, as explained in Section 5.3.\n\no PRNG seed: The seed is a 32 bit unsigned integer between 1 and\n0x7FFFFFFE (i.e. 2^^31-2) inclusive. This value is used to\ninitialize the Pseudo Random Number Generator (Section 5.6). This\nelement is optional. Whether or not it is present in the FEC OTI\nis signaled in the associated encoding format through an\nappropriate mechanism (see Section 4.2.4). When the PRNG seed is\nnot carried within the FEC OTI, it is assumed that encoder and\ndecoders use another way to communicate the information, or use a\nfixed, predefined value.\n\n4.2.4. Encoding Format\n\nThis section shows two possible encoding formats of the above FEC\nOTI. The present document does not specify when or how these\n\nRoca, et al. Expires December 23, 2006 [Page 8]```\n\n```\nInternet-Draft LDPC Staircase and Triangle FEC June 2006\n\nencoding formats should be used.\n\n4.2.4.1. Using the General EXT_FTI Format\n\nThe FEC OTI binary format is the following, when the EXT_FTI\nmechanism is used.\n\n0 1 2 3\n0 1 2 3 4 5 6 7 8 9 0 1 2 3 4 5 6 7 8 9 0 1 2 3 4 5 6 7 8 9 0 1\n+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+\n| HET = 64 | HEL (=4 or 5) | |\n+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+ +\n| Transfer-Length (L) |\n+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+\n| Encoding Symbol Length (E) | G | B (MSB) |\n+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+\n| B (LSB) | Max Nb of Enc. Symbols (max_n) |\n+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+\n. Optional PRNG seed .\n+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+\n\nIn particular:\n\no The HEL (Header Extension Length) indicates whether the optional\nPRNG seed is present (HEL=5) or not (HEL=4).\n\no The Transfer-Length (L) field size (48 bits) is larger than the\nsize required to store the maximum transfer length (Section 4.2.2)\nfor field alignment purposes.\n\no The Maximum-Source-Block-Length (B) field (20 bits) is split into\ntwo parts: the 8 most significant bits (MSB) are in the third 32-\nbit word of the EXT_FTI, and the remaining 12 least significant\nbits (LSB) are in fourth 32-bit word.\n\n4.2.4.2. Using the FDT Instance (FLUTE specific)\n\nWhen it is desired that the FEC OTI be carried in the FDT Instance of\na FLUTE session, the following XML elements must be described for the\nassociated object:\n\no FEC-OTI-Transfer-length\n\no FEC-OTI-Encoding-Symbol-Length\n\no FEC-OTI-Maximum-Source-Block-Length\n\nRoca, et al. Expires December 23, 2006 [Page 9]```\n\n```\nInternet-Draft LDPC Staircase and Triangle FEC June 2006\n\no FEC-OTI-Max-Number-of-Encoding-Symbols\n\no FEC-OTI-Number-Encoding-Symbols-per-Group\n\no FEC-OTI-PRNG-seed (optional)\n\nWhen no PRNG seed is to be carried in the FEC OTI, the sender simply\nomits the FEC-OTI-PRNG-seed element.\n\nRoca, et al. Expires December 23, 2006 [Page 10]```\n\n```\nInternet-Draft LDPC Staircase and Triangle FEC June 2006\n\n5. Procedures\n\nThis section defines procedures that are common to FEC Encoding IDs\nXX and YY.\n\n5.1. General\n\nThe B (maximum source block length in symbols) and E (encoding symbol\nlength in bytes) parameters are first determined, as explained in the\nfollowing sections.\n\nThe source object is then partitioned using the block partitioning\nalgorithm specified in . To that purpose, the B, L (object\ntransfer length in bytes), and E arguments are provided. As a\nresult, the object is partitioned into N source blocks. These blocks\nare numbered consecutively from 0 to N-1. The first I source blocks\nconsist of A_large source symbols, the remaining N-I source blocks\nconsist of A_small source symbols. Each source symbol is E bytes in\nlength, except perhaps the last symbol which may be shorter.\n\nFor each block the actual number of encoding symbols is determined,\nas explained in the following section.\n\nThen, FEC encoding and decoding can be done block per block,\nindependently. To that purpose, a parity check matrix is created,\nthat forms a system of linear equations between the repair and source\nsymbols of a given block, where the basic operator is XOR.\n\nThis parity check matrix is logically divided into two parts: the\nleft side (from column 0 to k-1) which describes the occurrence of\neach source symbol in the equation system; and the right side (from\ncolumn k to n-1) which describes the occurrence of each repair symbol\nin the equation system. An entry (a \"1\") in the matrix at position\n(i,j) (i.e. at row i and column j) means that the symbol with ESI i\nappears in equation j of the system. The only difference between the\nLDPC-Staircase and LDPC-Triangle schemes is the construction of the\nright sub-matrix.\n\nWhen the parity symbols have been created, the sender will transmit\nsource and parity symbols. The way this transmission occurs can\nlargely impact the erasure recovery capabilities of the LDPC-* FEC.\nIn particular, sending parity symbols in sequence is suboptimal.\nInstead it is usually recommended the shuffle these symbols. The\ninterested reader will find more details in .\n\nThe following sections detail how the B, E, and n parameters are\ndetermined (respectively Section 5.2, Section 5.3 and Section 5.4),\nhow encoding symbol groups are created (Section 5.5), and finally\n\nRoca, et al. Expires December 23, 2006 [Page 11]```\n\n```\nInternet-Draft LDPC Staircase and Triangle FEC June 2006\n\nspecify the PRNG (Section 5.6).\n\n5.2. Determining the Maximum Source Block Length (B)\n\nThe B parameter (maximum source block length in symbols) depends on\nseveral parameters: the code rate (rate), the Encoding Symbol ID\nfield length of the FEC Payload ID (20 bits), as well as possible\ninternal codec limitations.\n\nThe B parameter cannot be larger than the following values, derived\nfrom the FEC Payload ID limitations, for a given code rate:\n\nmax1_B = 2^^(20 - ceil(Log2(1/rate)))\n\nSome common max1_B values are:\n\no rate == 1 (no repair symbols): max_B = 2^^20 = 1,048,576\n\no 1 > rate >= 1/2: max1_B = 2^^19 = 524,288 symbols\n\no 1/2 > rate >= 1/4: max1_B = 2^^18 = 262,144 symbols\n\no 1/4 > rate >= 1/8: max1_B = 2^^17 = 131,072 symbols\n\nAdditionally, a codec MAY impose other limitations on the maximum\nblock size. This is the case for instance when the codec uses\ninternally 16 bit integers to store the Encoding Symbol ID, since it\ndoes not enable to store all the possible values of a 20 bit field.\nIn that case, if for instance 1 > rate >= 1/2, then the maximum block\nsize is 2^^15. Other limitations may also apply, for instance\nbecause of a limited working memory size. This decision MUST be\nclarified at implementation time, when the target use case is known.\nThis results in a max2_B limitation.\n\nThen, B is given by:\n\nB = min(max1_B, max2_B)\n\nNote that this calculation is only required at the coder, since the B\nparameter is communicated to the decoder through the FEC OTI.\n\n5.3. Determining the Encoding Symbol Length (E) and Number of Encoding\nSymbols per Group (G)\n\nThe E parameter usually depends on the maximum transmission unit on\nthe path (PMTU) from the source to the receivers. In order to\nheaders in case of ALC), E is chosen as large as possible. In that\n\nRoca, et al. Expires December 23, 2006 [Page 12]```\n\n```\nInternet-Draft LDPC Staircase and Triangle FEC June 2006\n\ncase, E is chosen so that the size of a packet composed of a single\nsymbol (G=1) remains below but close to the PMTU.\n\nYet other considerations can exist. For instance, the E parameter\ncan be made a function of the object transfer length. Indeed, LDPC\ncodes are known to offer better protection for large blocks. In case\nof small objects, it can be a good practice to reduce the encoding\nsymbol length (E) in order to artificially increase the number of\nsymbols, and therefore the block size.\n\ncan be grouped in the same Encoding Symbol Group (i.e. G > 1).\nDepending on how many symbols are grouped (G) and on the packet loss\nrate (which leads to loosing G symbols at a time), this strategy\nmight or might not be appropriate. A balance must therefore be\nfound.\n\nThe current specification does not mandate any value for either E or\nG. The current specification only provides an example of possible\nchoices for E and G. Note that this choice is done by the sender.\nThen the E and G parameters are communicated to the receivers thanks\nto the FEC OTI.\n\nExample:\n\nFirst define the target packet size, pkt_sz (usually the PMTU minus\nthe various protocol headers). The pkt_sz must be chosen in such a\nway that the symbol size is an integer. This can require that pkt_sz\nbe a multiple of 4, 8 or 16 (see the table below). Then calculate\nthe number of packets: nb_pkts = ceil(L / pkt_sz). Finally use the\nfollowing table to find a possible G value.\n\n+------------------------+----+-------------+-------------------+\n| Number of packets | G | Symbol size | k |\n+------------------------+----+-------------+-------------------+\n| 4000 <= nb_pkts | 1 | pkt_sz | 4000 <= k |\n| | | | |\n| 1000 <= nb_pkts < 4000 | 4 | pkt_sz / 4 | 4000 <= k < 16000 |\n| | | | |\n| 500 <= nb_pkts < 1000 | 8 | pkt_sz / 8 | 4000 <= k < 8000 |\n| | | | |\n| 1 <= nb_pkts < 500 | 16 | pkt_sz / 16 | 16 <= k < 8000 |\n+------------------------+----+-------------+-------------------+\n\n5.4. Determining the Number of Encoding Symbols of a Block\n\nThe following algorithm, also called \"n-algorithm\", explains how to\ndetermine the actual number of encoding symbols for a given block.\n\nRoca, et al. Expires December 23, 2006 [Page 13]```\n\n```\nInternet-Draft LDPC Staircase and Triangle FEC June 2006\n\nAT A SENDER:\n\nInput:\n\nB: Maximum source block length, for any source block. Section 5.2\nexplains how to determine its value.\n\nk: Current source block length. This parameter is given by the\nsource blocking algorithm.\n\nrate: FEC code rate, which is provided by the user (e.g. when\nstarting a FLUTE sending application). It is expressed as a\nfloating point value. The rate value must be such that the\nresulting number of encoding symbols per block is at most equal to\n2^^20 (Section 4.1).\n\nOutput:\n\nmax_n: Maximum number of encoding symbols generated for any source\nblock\n\nn: Number of encoding symbols generated for this source block\n\nAlgorithm:\n\nmax_n = floor(B / rate);\n\nif (max_n >= 2^^20) then return an error (\"invalid code rate\");\n\n(NB: if max_n has been defined as explained in Section 5.2, this\nerror should never happen)\n\nn = floor(k * max_n / B);\n\nInput:\n\nB: Extracted from the received FEC OTI\n\nmax_n: Extracted from the received FEC OTI\n\nk: Given by the source blocking algorithm\n\nOutput:\n\nn:\n\nRoca, et al. Expires December 23, 2006 [Page 14]```\n\n```\nInternet-Draft LDPC Staircase and Triangle FEC June 2006\n\nAlgorithm:\n\nn = floor(k * max_n / B);\n\n5.5. Identifying the Symbols of an Encoding Symbol Group\n\nWhen multiple encoding symbols are sent in the same packet, the FEC\nPayload ID information of the packet MUST refer to the first encoding\nsymbol. It MUST then be possible to identify each symbol from this\nsingle FEC Payload ID. To that purpose, the symbols of an Encoding\nSymbol Group (i.e. packet):\n\no MUST all be either source symbols, or repair symbols. Therefore\nonly source packets and repair packets are permitted, not mixed\nones.\n\no are identified by a function, ESIs_of_group(), that takes as\nargument:\n\n* for a sender, the index of the Encoding Symbol Group (i.e.\npacket) that the application wants to create,\n\n* for a receiver, the ESI information contained in the FEC\n\nand returns the list of G Encoding Symbol IDs that will be packed\ntogether. In case of a source packet, the G source symbols are\ntaken consecutively. In case of a repair packet, the G repair\nsymbols are chosen randomly, as explained below.\n\nThe system must first be initialized by creating a random permutation\nof the n-k indexes. This initialization function MUST be called\nimmediately after creating the parity check matrix. More precisely,\nsince the PRNG seed is not re-initialized, no call to the PRNG\nfunction must have happened between the time the parity check matrix\nhas been initialized and the time the following initialization\nfunction is called. This is true both at a sender and at a receiver.\n\nRoca, et al. Expires December 23, 2006 [Page 15]```\n\n```\nInternet-Draft LDPC Staircase and Triangle FEC June 2006\n\n/*\n* Initialization function.\n* Warning: use only when G > 1.\n*/\ninitialize_tables ()\n{\nint i;\nint randInd;\nint backup;\n\n/* initialize the two tables that map ID\n* (i.e. ESI-k) to/from TxSequence. */\nfor (i = 0; i < n - k; i++) {\nIDtoTxseq[i] = i;\ntxseqToID[i] = i;\n}\n/* now randomize everything */\nfor (i = 0; i < n - k; i++) {\nrandInd = rand(n - k);\nbackup = IDtoTxseq[i];\nIDtoTxseq[i] = IDtoTxseq[randInd];\nIDtoTxseq[randInd] = backup;\ntxseqToID[IDtoTxseq[i]] = i;\ntxseqToID[IDtoTxseq[randInd]] = randInd;\n}\nreturn;\n}\n\nIt is then possible, at the sender, to determine the sequence of G\nEncoding Symbol IDs that will be part of the group.\n\nRoca, et al. Expires December 23, 2006 [Page 16]```\n\n```\nInternet-Draft LDPC Staircase and Triangle FEC June 2006\n\n/*\n* Determine the sequence of ESIs of the packet under construction\n* at a sender.\n* Warning: use only when G > 1.\n* PktIdx (IN): index of the packet, in {0..ceil(n/G)} range\n* ESIs[] (OUT): list of ESI of the packet\n*/\nsender_find_ESIs_of_group (int PktIdx,\nESI_t ESIs[])\n{\nint i;\n\nif (is_source_packet(PktIdx) == true) {\n/* this is a source packet */\nESIs = (PktIdx * G) % k;\nfor (i = 0; i < G; i++) {\nESIs[i] = ESIs + i;\n}\n} else {\n/* this is a repair packet */\nfor (i = 0; i < G; i++) {\nESIs[i] =\nk +\ntxseqToID[(i + (PktIdx - nbSourcePkts) * G)\n% (n - k)];\n}\n}\nreturn;\n}\n\nSimilarly, upon receiving an Encoding Symbol Group (i.e. packet), a\nreceiver can determine the sequence of G Encoding Symbol IDs from the\nfirst ESI, esi0, that is contained in the FEC Payload ID.\n\nRoca, et al. Expires December 23, 2006 [Page 17]```\n\n```\nInternet-Draft LDPC Staircase and Triangle FEC June 2006\n\n/*\n* Determine the sequence of ESIs of a packet received.\n* Warning: use only when G > 1.\n* esi0 (IN): : ESI contained in the FEC Payload ID\n* ESIs[] (OUT): list of ESI of the packet\n*/\nESI_t ESIs[])\n{\nint i;\n\nif (is_source_packet(esi0) == true) {\n/* this is a source packet */\nfor (i = 0; i < G; i++) {\nESIs[i] = (esi0 + i) % k;\n}\n} else {\n/* this is a repair packet */\nfor (i = 0; i < G; i++) {\nESIs[i] =\nk +\ntxseqToID[(i + IDtoTxseq[esi0 - k])\n% (n - k)];\n}\n}\n}\n\n5.6. Pseudo Random Number Generator\n\nThe present FEC Encoding ID relies on a pseudo-random number\ngenerator (PRNG) that must be fully specified, in particular in order\nto enable the receivers and the senders to build the same parity\ncheck matrix. The minimal standard generator is used. It\ndefines a simple multiplicative congruential algorithm: Ij+1 = A * Ij\n(modulo M), with the following choices: A = 7^^5 = 16807 and M =\n2^^31 - 1 = 2147483647. Several implementations of this PRNG are\nknown and discussed in the literature. All of them provide the same\nsequence of pseudo random numbers. A validation criteria of such a\nPRNG is the following: if seed = 1, then the 10,000th value returned\nMUST be equal to 1043618065.\n\nThe following implementation uses the Park and Miller algorithm with\nthe optimization suggested by D. Carta in .\n\nRoca, et al. Expires December 23, 2006 [Page 18]```\n\n```\nInternet-Draft LDPC Staircase and Triangle FEC June 2006\n\nunsigned long seed;\n\n/*\n* Initialize the PRNG with a seed between\n* 1 and 0x7FFFFFFE (i.e. 2^^31-2) inclusive.\n*/\nvoid srand (unsigned long s)\n{\nif ((s > 0) && (s < 0x7FFFFFFF))\nseed = s;\nelse\nexit(-1);\n}\n\n/*\n* Returns a random integer in [0; maxv-1]\n* Derived from rand31pmc, Robin Whittle,\n* September 20th, 2005.\n* http://www.firstpr.com.au/dsp/rand31/\n* 16807 multiplier constant (7^^5)\n* 0x7FFFFFFF modulo constant (2^^31-1)\n* The inner PRNG produces a value between 1 and\n* 0x7FFFFFFE (2^^31-2) inclusive.\n* This value is then scaled between 0 and maxv-1\n* inclusive.\n*/\nunsigned long\nrand (unsigned long maxv)\n{\nunsigned long hi, lo;\n\nlo = 16807 * (seed & 0xFFFF);\nhi = 16807 * (seed >> 16); /* binary shift to right */\nlo += (hi & 0x7FFF) << 16; /* binary shift to left */\nlo += hi >> 15;\nif (lo > 0x7FFFFFFF)\nlo -= 0x7FFFFFFF;\nseed = (long)lo;\n/* don't use modulo, least significant bits are less random\n* than most significant bits [Numerical Recipies in C] */\nreturn ((unsigned long)\n((double)seed * (double)maxv / (double)0x7FFFFFFF));\n}\n\nRoca, et al. Expires December 23, 2006 [Page 19]```\n\n```\nInternet-Draft LDPC Staircase and Triangle FEC June 2006\n\n6. Full Specification of the LDPC-Staircase Scheme\n\n6.1. General\n\nThe LDPC-Staircase scheme is identified by the Fully-Specified FEC\nEncoding ID XX.\n\nThe PRNG used by the LDPC-Staircase scheme must be initialized by a\nseed. This PRNG seed is an optional instance-specific FEC OTI\nelement (Section 4.2.3). When this PRNG seed is not carried within\nthe FEC OTI, it is assumed that encoder and decoders either use\nanother way to communicate the seed value or use a fixed, predefined\nvalue.\n\n6.2. Parity Check Matrix Creation\n\nThe LDPC-Staircase matrix can be divided into two parts: the left\nside of the matrix defines in which equations the source symbols are\ninvolved; the right side of the matrix defines in which equations the\nrepair symbols are involved.\n\nThe left side is generated with the following algorithm:\n\nRoca, et al. Expires December 23, 2006 [Page 20]```\n\n```\nInternet-Draft LDPC Staircase and Triangle FEC June 2006\n\n/* initialize a list of possible choices to\n* guarantee a homogeneous \"1\" distribution */\nfor (h = 3*k-1; h >= 0; h--) {\nu[h] = h % (n-k);\n}\n/* left limit within the list of possible choices, u[] */\nt = 0;\n\nfor (j = 0; j < k; j++) { /* for each source symbol column */\nfor (h = 0; h < 3; h++) { /* add 3 \"1s\" */\n/* check that valid available choices remain */\nfor (i = t; i < 3*k && matrix_has_entry(u[i], j); i++);\n\nif (i < 3*k) {\n/* choose one index within the list of possible\n* choices */\ndo {\ni = t + rand(3*k-t);\n} while (matrix_has_entry(u[i], j));\nmatrix_insert_entry(u[i], j);\n\n/* replace with u[t] which has never been chosen */\nu[i] = u[t];\nt++;\n} else {\n/* no choice left, choose one randomly */\ndo {\ni = rand(n-k);\n} while (matrix_has_entry(i, j));\nmatrix_insert_entry(i, j);\n}\n}\n}\n\n/* Add extra bits to avoid rows with less than two \"1s\" */\nfor (i = 0; i < n-k; i++) { /* for each row */\nif (degree_of_row(i) == 0) {\nj = rand(k);\ne = matrix_insert_entry(i, j);\n}\nif (degree_of_row(i) == 1) {\ndo {\nj = rand(k);\n} while (matrix_has_entry(i, j));\nmatrix_insert_entry(i, j);\n}\n}\n\nRoca, et al. Expires December 23, 2006 [Page 21]```\n\n```\nInternet-Draft LDPC Staircase and Triangle FEC June 2006\n\nThe right side (the staircase) is generated by the following\nalgorithm:\n\nmatrix_insert_entry(0, k); /* first row */\nfor (i = 1; i < n-k; i++) { /* for the following rows */\nmatrix_insert_entry(i, k+i); /* identity */\nmatrix_insert_entry(i, k+i-1); /* staircase */\n}\n\nNote that just after creating this parity check matrix, when encoding\nsymbol groups are used (i.e. G > 1), the function initializing the\ntwo random permutation tables (Section 5.5) MUST be called. This is\ntrue both at a sender and at a receiver.\n\n6.3. Encoding\n\nThanks to the staircase matrix, repair symbol creation is\nstraightforward: each repair symbol is equal to the sum of all source\nsymbols in the associated equation, plus the previous repair symbol\n(except for the first repair symbol). Therefore encoding MUST follow\nand generate repair symbol with ESI i before symbol ESI i+1.\n\n6.4. Decoding\n\nDecoding basically consists in solving a system of n-k linear\nequations whose variables are the source an repair symbols. Of\ncourse, the final goal is to recover the value of source symbols\nonly.\n\nTo that purpose, many techniques are possible. One of them is the\nfollowing trivial algorithm : given a set of linear equations, if\none of them has only one remaining unknown variable, then the value\nof this variable is that of the constant term. So, replace this\nvariable by its value in all the remaining linear equations and\nreiterate. The value of several variables can therefore be found\nrecursively. Applied to LDPC FEC codes working over an erasure\npacket, the parity check matrix defines a set of linear equations\nwhose variables are the source symbols and repair symbols. Receiving\nor decoding a symbol is equivalent to having the value of a variable.\nAppendix A sketches a possible implementation of this algorithm.\n\nThe Gauss elimination technique (or any optimized derivative) is\nanother possible decoding technique. Hybrid solutions that start by\nusing the trivial algorithm above and finish with a Gauss elimination\nare also possible.\n\nBecause interoperability does not depend on the decoding algorithm\n\nRoca, et al. Expires December 23, 2006 [Page 22]```\n\n```\nInternet-Draft LDPC Staircase and Triangle FEC June 2006\n\nused, the current document does not recommend any particular\ntechnique. This choice is left to the codec developer.\n\nYet choosing a decoding technique will have great practical impacts.\nIt will impact the erasure capabilities: a Gauss elimination\ntechnique enables to solve the system with a smaller number of\nsymbols compared to the trivial technique. It will also impact the\nCPU load: a Gauss elimination technique requires much more processing\nthan the trivial technique. Depending on the target use case, the\ncodec developer will favor one feature or the other.\n\nRoca, et al. Expires December 23, 2006 [Page 23]```\n\n```\nInternet-Draft LDPC Staircase and Triangle FEC June 2006\n\n7. Full Specification of the LDPC-Triangle Scheme\n\n7.1. General\n\nLDPC-Triangle is identified by the Fully-Specified FEC Encoding ID\nYY.\n\nThe PRNG used by the LDPC-Triangle scheme must be initialized by a\nseed. This PRNG seed is an optional instance-specific FEC OTI\nelement (Section 4.2.3). When this PRNG seed is not carried within\nthe FEC OTI, it is assumed that encoder and decoders either use\nanother way to communicate the seed value or use a fixed, predefined\nvalue.\n\n7.2. Parity Check Matrix Creation\n\nThe LDPC-Triangle matrix can be divided into two parts: the left side\nof the matrix defines in which equations the source symbols are\ninvolved; the right side of the matrix defines in which equations the\nrepair symbols are involved.\n\nThe left side is generated with the same algorithm as that of LDPC-\nStaircase (Section 6.2).\n\nThe right side (the triangle) is generated with the following\nalgorithm:\n\nmatrix_insert_entry(0, k); /* first row */\nfor (i = 1; i < n-k; i++) { /* for the following rows */\nmatrix_insert_entry(i, k+i); /* identity */\nmatrix_insert_entry(i, k+i-1); /* staircase */\n/* now fill the triangle */\nj = i-1;\nfor (l = 0; l < j; l++) { /* limit the # of \"1s\" added */\nj = rand(j);\nmatrix_insert_entry(i, k+j);\n}\n}\n\nNote that just after creating this parity check matrix, when encoding\nsymbol groups are used (i.e. G > 1), the function initializing the\ntwo random permutation tables (Section 5.5) MUST be called. This is\ntrue both at a sender and at a receiver.\n\n7.3. Encoding\n\nHere also repair symbol creation is straightforward: each repair\nsymbol is equal to the sum of all source symbols in the associated\n\nRoca, et al. Expires December 23, 2006 [Page 24]```\n\n```\nInternet-Draft LDPC Staircase and Triangle FEC June 2006\n\nequation, plus the repair symbols in the triangle. Therefore\nfirst repair symbol, and generate repair symbol with ESI i before\nsymbol ESI i+1.\n\n7.4. Decoding\n\nDecoding basically consists in solving a system of n-k linear\nequations, whose variables are the source an repair symbols. Of\ncourse, the final goal is to recover the value of source symbols\nonly. To that purpose, many techniques are possible, as explained in\nSection 6.4.\n\nBecause interoperability does not depend on the decoding algorithm\nused, the current document does not recommend any particular\ntechnique. This choice is left to the codec implementer.\n\nRoca, et al. Expires December 23, 2006 [Page 25]```\n\n```\nInternet-Draft LDPC Staircase and Triangle FEC June 2006\n\n8. Security Considerations\n\nThe security considerations for this document are the same as that of\n.\n\nRoca, et al. Expires December 23, 2006 [Page 26]```\n\n```\nInternet-Draft LDPC Staircase and Triangle FEC June 2006\n\n9. Acknowledgments\n\nSection 5.4 is derived from a previous Internet-Draft, and we would\nlike to thank S. Peltotalo and J. Peltotalo for their contribution.\nWe would also like to thank Pascal Moniot, Laurent Fazio, Aurelien\nFrancillon and Shao Wenjian for their comments.\n\nRoca, et al. Expires December 23, 2006 [Page 27]```\n\n```\nInternet-Draft LDPC Staircase and Triangle FEC June 2006\n\n10. References\n\n10.1. Normative References\n\n Bradner, S., \"Key words for use in RFCs to Indicate Requirement\nLevels\", RFC 2119, BCP 14, March 1997.\n\n Watson, M., Luby, M., and L. Vicisano, \"Forward Error Correction\n(FEC) Building Block\", draft-ietf-rmt-fec-bb-revised-03.txt\n(work in progress), January 2006.\n\n Luby, M., Vicisano, L., Gemmell, J., Rizzo, L., Handley, M., and\nJ. Crowcroft, \"The Use of Forward Error Correction (FEC) in\nReliable Multicast\", RFC 3453, December 2002.\n\n10.2. Informative References\n\n Roca, V. and C. Neumann, \"Design, Evaluation and Comparison of\nFour Large Block FEC Codecs: LDPC, LDGM, LDGM-Staircase and\nLDGM-Triangle, Plus a Reed-Solomon Small Block FEC Codec\",\nINRIA Research Report RR-5225, June 2004.\n\n Neumann, C., Roca, V., Francillon, A., and D. Furodet, \"Impacts\nof Packet Scheduling and Packet Loss Distribution on FEC\nPerformances: Observations and Recommendations\", ACM CoNEXT'05\nConference, Toulouse, France (an extended version is available\nas INRIA Research Report RR-5578), October 2005.\n\n Roca, V., Neumann, C., and J. Laboure, \"LDPC-Staircase/\nLDPC-Triangle Codec Reference Implementation\", INRIA Rhone-\nAlpes and STMicroelectronics,\nhttp://planete-bcast.inrialpes.fr/.\n\n MacKay, D., \"Information Theory, Inference and Learning\nAlgorithms\", Cambridge University Press, ISBN: 0521642981,\n2003.\n\n Park, S. and K. Miller, \"Random Number Generators: Good Ones\nare Hard to Find\", Communications of the ACM, Vol. 31, No. 10,\npp.1192-1201, 1988.\n\n Carta, D., \"Two Fast Implementations of the Minimal Standard\nRandom Number Generator\", Communications of the ACM, Vol. 33,\nNo. 1, pp.87-88, January 1990.\n\n Zyablov, V. and M. Pinsker, \"Decoding Complexity of Low-Density\nCodes for Transmission in a Channel with Erasures\", Translated\nfrom Problemy Peredachi Informatsii, Vol.10, No. 1, pp.15-28,\n\nRoca, et al. Expires December 23, 2006 [Page 28]```\n\n```\nInternet-Draft LDPC Staircase and Triangle FEC June 2006\n\nJanuary-March 1974.\n\nRoca, et al. Expires December 23, 2006 [Page 29]```\n\n```\nInternet-Draft LDPC Staircase and Triangle FEC June 2006\n\nAppendix A. Trivial Decoding Algorithm (Informative Only)\n\nA trivial decoding algorithm is sketched below (please see for\nthe details omitted here):\n\nInitialization: allocate a table of partial sum buffers:\npartial_sum[n-k], one per equation;\nReset all the buffers to 0;\n\n/*\n* For each newly received or decoded symbol, try to make progress\n* in the decoding of the associated source block.\n* new_esi (IN): ESI of the new symbol, which is also the index\n* in [0; n-1]\n* new_symb (IN): New symbol received or decoded\n*/\nvoid\ndecoding_step(ESI_t new_esi,\nsymbol_t *new_symb)\n{\nReturn; /* don't waste time with this symbol */\n}\n\nIf (new_symb is the last missing source symbol) {\nReturn; /* decoding is now finished */\n}\n\nCreate an empty list of equations having symbols decoded during\nthis decoding step;\n\n/*\n* First add this new symbol to all partial sums of the\n* associated equations.\n*/\nFor (each equation eq in which new_symb is a variable and\nhaving more than one unknown variable) {\n\nRemove entry(eq, new_esi) from the H matrix;\n\nIf (degree of equation eq == 1) {\n/* new symbol can be decoded, remember the equation */\nAppend eq to the list of equations having symbols\ndecoded during this decoding step;\n}\n}\n\nRoca, et al. Expires December 23, 2006 [Page 30]```\n\n```\nInternet-Draft LDPC Staircase and Triangle FEC June 2006\n\n/*\n* Then finish with recursive calls to decoding_step() for each\n* newly decoded symbols.\n*/\nFor (each equation eq in the list of equations having symbols\ndecoded during this decoding step) {\n\n/*\n* Because of the recursion below, we need to check that\n* decoding is not finished, and that the equation is\n* __still__ of degree 1\n*/\nIf (decoding is finished) {\nbreak; /* exit from the loop */\n}\n\nIf ((degree of equation eq == 1) {\nLet dec_esi be the ESI of the newly decoded symbol in\nequation eq;\n\nRemove entry(eq, dec_esi);\n\nAllocate a buffer, dec_symb, for this symbol, and\ncopy partial_sum[eq] to dec_symb;\n\n/* finally, call this function recursively */\ndecoding_step(dec_esi, dec_symb);\n}\n}\n}\n\nRoca, et al. Expires December 23, 2006 [Page 31]```\n\n```\nInternet-Draft LDPC Staircase and Triangle FEC June 2006\n\nVincent Roca\nINRIA\n655, av. de l'Europe\nZirst; Montbonnot\nST ISMIER cedex 38334\nFrance\n\nEmail: [email protected]\nURI: http://planete.inrialpes.fr/~roca/\n\nChristoph Neumann\nThomson Research\n46, Quai A. Le Gallo\nBoulogne Cedex 92648\nFrance\n\nEmail: [email protected]\nURI: http://planete.inrialpes.fr/~chneuman/\n\nDavid Furodet\nSTMicroelectronics\n12, Rue Jules Horowitz\nBP217\nGrenoble Cedex 38019\nFrance\n\nEmail: [email protected]\nURI: http://www.st.com/\n\nRoca, et al. Expires December 23, 2006 [Page 32]```\n\n```\nInternet-Draft LDPC Staircase and Triangle FEC June 2006\n\nIntellectual Property Statement\n\nThe IETF takes no position regarding the validity or scope of any\nIntellectual Property Rights or other rights that might be claimed to\npertain to the implementation or use of the technology described in\nthis document or the extent to which any license under such rights\nmight or might not be available; nor does it represent that it has\nmade any independent effort to identify any such rights. Information\non the procedures with respect to rights in RFC documents can be\nfound in BCP 78 and BCP 79.\n\nCopies of IPR disclosures made to the IETF Secretariat and any\nassurances of licenses to be made available, or the result of an\nattempt made to obtain a general license or permission for the use of\nsuch proprietary rights by implementers or users of this\nspecification can be obtained from the IETF on-line IPR repository at\nhttp://www.ietf.org/ipr.\n\nThe IETF invites any interested party to bring to its attention any\ncopyrights, patents or patent applications, or other proprietary\nrights that may cover technology that may be required to implement\[email protected].\n\nDisclaimer of Validity\n\nThis document and the information contained herein are provided on an\n\"AS IS\" basis and THE CONTRIBUTOR, THE ORGANIZATION HE/SHE REPRESENTS\nOR IS SPONSORED BY (IF ANY), THE INTERNET SOCIETY AND THE INTERNET\nENGINEERING TASK FORCE DISCLAIM ALL WARRANTIES, EXPRESS OR IMPLIED,\nINCLUDING BUT NOT LIMITED TO ANY WARRANTY THAT THE USE OF THE\nINFORMATION HEREIN WILL NOT INFRINGE ANY RIGHTS OR ANY IMPLIED\nWARRANTIES OF MERCHANTABILITY OR FITNESS FOR A PARTICULAR PURPOSE.\n\nCopyright (C) The Internet Society (2006). This document is subject\nto the rights, licenses and restrictions contained in BCP 78, and\nexcept as set forth therein, the authors retain all their rights.\n\nAcknowledgment\n\nFunding for the RFC Editor function is currently provided by the\nInternet Society.\n\nRoca, et al. Expires December 23, 2006 [Page 33]\n\n```\n\nHtml markup produced by rfcmarkup 1.129d, available from https://tools.ietf.org/tools/rfcmarkup/" ]

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[ "# Difference between revisions of \"99 questions/90 to 94\"\n\nThese are Haskell translations of Ninety-Nine Prolog Problems.\n\nIf you want to work on one of these, put your name in the block so we know someone's working on it. Then, change n in your block to the appropriate problem number, and fill in the <Problem description>,<example in lisp>,<example in Haskell>,<solution in haskell> and <description of implementation> fields.\n\n## Problem 90\n\n(**) Eight queens problem\n\nThis is a classical problem in computer science. The objective is to place eight queens on a chessboard so that no two queens are attacking each other; i.e., no two queens are in the same row, the same column, or on the same diagonal.\n\nHint: Represent the positions of the queens as a list of numbers 1..N. Example: [4,2,7,3,6,8,5,1] means that the queen in the first column is in row 4, the queen in the second column is in row 2, etc. Use the generate-and-test paradigm.\n\n```> length queens\n92\n> take 1 queens\n[[4,2,7,3,6,8,5,1]]\n```\n\nSolution:\n\n```queens = queens' 8\nwhere queens' 0 = [[]]\nqueens' n = [ try:qs | qs <- queens' (n-1), try <- [1..8], isSafe try qs]\nisSafe try qs = not (try `elem` qs || sameDiag try qs)\nsameDiag try qs = any (\\(colDist,q) -> abs(try - q) == colDist) \\$ zip [1..] qs\n```\n\nBy definition/data representation no two queens can occupy the same column. \"try `elem` alreadySet\" checks for a queen in the same row, \"abs(try - q) == col\" checks for a queen in the same diagonal.\n\nThis is a modification of a function I wrote when I was just learning haskell, so there's certainly much to improve here! For one thing there is speedup potential in caching \"blocked\" rows, columns and diagonals.\n\nOtherwise a smarter representation in memory might simplify the whole thing drastically as well.\n\n## Problem 91\n\n(**) Knight's tour\n\nAnother famous problem is this one: How can a knight jump on an NxN chessboard in such a way that it visits every square exactly once?\n\nHints: Represent the squares by pairs of their coordinates of the form X/Y, where both X and Y are integers between 1 and N. (Note that '/' is just a convenient functor, not division!) Define the relation jump(N,X/Y,U/V) to express the fact that a knight can jump from X/Y to U/V on a NxN chessboard. And finally, represent the solution of our problem as a list of N*N knight positions (the knight's tour).\n\nThere are two variants of this problem:\n\n1. find a tour ending at a particular square\n2. find a circular tour, ending a knight's jump from the start (clearly it doesn't matter where you start, so choose (1,1))\n\n```Knights> head \\$ knightsTo 8 (1,1)\n[(2,7),(3,5),(5,6),(4,8),(3,6),(4,4),(6,5),(4,6),\n(5,4),(7,5),(6,3),(5,5),(4,3),(2,4),(1,6),(2,8),\n(4,7),(6,8),(8,7),(6,6),(4,5),(6,4),(5,2),(7,1),\n(8,3),(6,2),(8,1),(7,3),(8,5),(7,7),(5,8),(3,7),\n(1,8),(2,6),(3,4),(1,5),(2,3),(3,1),(1,2),(3,3),\n(1,4),(2,2),(4,1),(5,3),(7,4),(8,2),(6,1),(4,2),\n(2,1),(1,3),(2,5),(1,7),(3,8),(5,7),(7,8),(8,6),\n(6,7),(8,8),(7,6),(8,4),(7,2),(5,1),(3,2),(1,1)]\n[(1,1),(3,2),(1,3),(2,1),(3,3),(5,4),(6,6),(4,5),\n(2,6),(1,8),(3,7),(5,8),(4,6),(2,5),(4,4),(5,6),\n(6,4),(8,5),(7,7),(6,5),(5,3),(6,1),(4,2),(6,3),\n(8,2),(7,4),(5,5),(3,4),(1,5),(2,7),(4,8),(3,6),\n(1,7),(3,8),(5,7),(7,8),(8,6),(6,7),(8,8),(7,6),\n(8,4),(7,2),(5,1),(4,3),(3,5),(1,4),(2,2),(4,1),\n(6,2),(8,1),(7,3),(5,2),(7,1),(8,3),(7,5),(8,7),\n(6,8),(4,7),(2,8),(1,6),(2,4),(1,2),(3,1),(2,3)]\n```\n\nSolution:\n\n```module Knights where\n\nimport Data.List\n\ntype Square = (Int, Int)\n\n-- Possible knight moves from a given square on an nxn board\nknightMoves :: Int -> Square -> [Square]\nknightMoves n (x, y) = filter (onBoard n)\n[(x+2, y+1), (x+2, y-1), (x+1, y+2), (x+1, y-2),\n(x-1, y+2), (x-1, y-2), (x-2, y+1), (x-2, y-1)]\n\n-- Is the square within an nxn board?\nonBoard :: Int -> Square -> Bool\nonBoard n (x, y) = 1 <= x && x <= n && 1 <= y && y <= n\n\n-- Knight's tours on an nxn board ending at the given square\nknightsTo :: Int -> Square -> [[Square]]\nknightsTo n finish = [pos:path | (pos, path) <- tour (n*n)]\nwhere tour 1 = [(finish, [])]\ntour k = [(pos', pos:path) |\n(pos, path) <- tour (k-1),\npos' <- sortImage (entrances path)\n(filter (`notElem` path) (knightMoves n pos))]\nentrances path pos =\nlength (filter (`notElem` path) (knightMoves n pos))\n\n-- Closed knight's tours on an nxn board\nclosedKnights :: Int -> [[Square]]\nclosedKnights n = [pos:path | (pos, path) <- tour (n*n), pos == start]\nwhere tour 1 = [(finish, [])]\ntour k = [(pos', pos:path) |\n(pos, path) <- tour (k-1),\npos' <- sortImage (entrances path)\n(filter (`notElem` path) (knightMoves n pos))]\nentrances path pos\n| pos == start = 100 -- don't visit start until there are no others\n| otherwise = length (filter (`notElem` path) (knightMoves n pos))\nstart = (1,1)\nfinish = (2,3)\n\n-- Sort by comparing the image of list elements under a function f.\n-- These images are saved to avoid recomputation.\nsortImage :: Ord b => (a -> b) -> [a] -> [a]\nsortImage f xs = map snd (sortBy cmpFst [(f x, x) | x <- xs])\nwhere cmpFst x y = compare (fst x) (fst y)\n```\n\nThis has a similar structure to the 8 Queens problem, except that we apply a heuristic invented by Warnsdorff: when considering next possible moves, we prefer squares with fewer open entrances. This speeds things up enormously, and finds the first solution to boards smaller than 76x76 without backtracking.\n\nSolution 2:\n\n```knights :: Int -> [[(Int,Int)]]\nknights n = loop (n*n) [[(1,1)]]\nwhere loop 1 = map reverse . id\nloop i = loop (i-1) . concatMap nextMoves\n\nwhere possible = filter (\\x -> on_board x && not (x `elem` already)) \\$ jumps x\n\njumps (x,y) = [(x+a, y+b) | (a,b) <- [(1,2), (2,1), (2,-1), (1,-2), (-1,-2), (-2,-1), (-2,1), (-1,2)]]\non_board (x,y) = (x >= 1) && (x <= n) && (y >= 1) && (y <= n)\n```\n\nThis is just the naive backtracking approach. I tried a speedup using Data.Map, but the code got too verbose to post.\n\n## Problem 92\n\n(***) Von Koch's conjecture\n\nSeveral years ago I met a mathematician who was intrigued by a problem for which he didn't know a solution. His name was Von Koch, and I don't know whether the problem has been solved since.", null, "Anyway the puzzle goes like this: Given a tree with N nodes (and hence N-1 edges). Find a way to enumerate the nodes from 1 to N and, accordingly, the edges from 1 to N-1 in such a way, that for each edge K the difference of its node numbers equals to K. The conjecture is that this is always possible.\n\nFor small trees the problem is easy to solve by hand. However, for larger trees, and 14 is already very large, it is extremely difficult to find a solution. And remember, we don't know for sure whether there is always a solution!\n\nWrite a predicate that calculates a numbering scheme for a given tree. What is the solution for the larger tree pictured below?", null, "Example:\n\n```<example in lisp>\n```\n\n```> head \\$ vonKoch [(1,6),(2,6),(3,6),(4,6),(5,6),(5,7),(5,8),(8,9),(5,10),(10,11),(11,12),(11,13),(13,14)]\n[6,7,8,9,3,4,10,11,5,12,2,13,14,1]\n```\n\nSolution:\n\n```vonKoch edges = do\nlet n = length edges + 1\nnodes <- liftM (listArray (1,n)) permutations [1..n]\nlet nodeArray = listArray (1,n) nodes\nlet dists = sort \\$ map (\\(x,y) -> abs (nodeArray ! x - nodeArray ! y)) edges\nguard \\$ and \\$ zipWith (/=) dists (tail dists)\nreturn nodes\n```\n\n<description of implementation>\n\n## Problem 93\n\n(***) An arithmetic puzzle\n\nGiven a list of integer numbers, find a correct way of inserting arithmetic signs (operators) such that the result is a correct equation. Example: With the list of numbers [2,3,5,7,11] we can form the equations 2-3+5+7 = 11 or 2 = (3*5+7)/11 (and ten others!).\n\nDivision should be interpreted as operating on rationals, and division by zero should be avoided.\n\n```P93> putStr \\$ unlines \\$ puzzle [2,3,5,7,11]\n2 = 3-(5+7-11)\n2 = 3-5-(7-11)\n2 = 3-(5+7)+11\n2 = 3-5-7+11\n2 = (3*5+7)/11\n2*(3-5) = 7-11\n2-(3-(5+7)) = 11\n2-(3-5-7) = 11\n2-(3-5)+7 = 11\n2-3+5+7 = 11\n```\n\nThe other two solutions alluded to in the problem description are dropped by the Haskell solution as trivial variants:\n\n```2 = 3-(5+(7-11))\n2-3+(5+7) = 11\n```\n\nSolution:\n\n```module P93 where\n\nimport Data.List\nimport Data.Maybe\n\ntype Equation = (Expr, Expr)\ndata Expr = Const Integer | Binary Expr Op Expr\nderiving (Eq, Show)\ndata Op = Plus | Minus | Multiply | Divide\nderiving (Bounded, Eq, Enum, Show)\ntype Value = Rational\n\n-- top-level function: all correct equations generated from the list of\n-- numbers, as pretty strings.\npuzzle :: [Integer] -> [String]\npuzzle ns = map (flip showsEquation \"\") (equations ns)\n\n-- generate all correct equations from the list of numbers\nequations :: [Integer] -> [Equation]\nequations [] = error \"empty list of numbers\"\nequations [n] = error \"only one number\"\nequations ns = [(e1, e2) |\n(ns1, ns2) <- splits ns,\n(e1, v1) <- exprs ns1,\n(e2, v2) <- exprs ns2,\nv1 == v2]\n\n-- generate all expressions from the numbers, except those containing\n-- a division by zero, or redundant right-associativity.\nexprs :: [Integer] -> [(Expr, Value)]\nexprs [n] = [(Const n, fromInteger n)]\nexprs ns = [(Binary e1 op e2, v) | (ns1, ns2) <- splits ns,\n(e1, v1) <- exprs ns1,\n(e2, v2) <- exprs ns2,\nop <- [minBound..maxBound],\nnot (right_associative op e2),\nv <- maybeToList (apply op v1 v2)]\n\n-- splittings of a list into two non-empty lists\nsplits :: [a] -> [([a],[a])]\nsplits xs = tail (init (zip (inits xs) (tails xs)))\n\n-- applying an operator to arguments may fail (division by zero)\napply :: Op -> Value -> Value -> Maybe Value\napply Plus x y = Just (x + y)\napply Minus x y = Just (x - y)\napply Multiply x y = Just (x * y)\napply Divide x 0 = Nothing\napply Divide x y = Just (x / y)\n\n-- e1 op (e2 op' e3) == (e1 op e2) op' e3\nright_associative :: Op -> Expr -> Bool\nright_associative Plus (Binary _ Plus _) = True\nright_associative Plus (Binary _ Minus _) = True\nright_associative Multiply (Binary _ Multiply _) = True\nright_associative Multiply (Binary _ Divide _) = True\nright_associative _ _ = False\n\n-- Printing of equations and expressions\n\nshowsEquation :: Equation -> ShowS\nshowsEquation (l, r) = showsExprPrec 0 l . showString \" = \" . showsExprPrec 0 r\n\n-- all operations are left associative\nshowsExprPrec :: Int -> Expr -> ShowS\nshowsExprPrec _ (Const n) = shows n\nshowsExprPrec p (Binary e1 op e2) = showParen (p > op_prec) \\$\nshowsExprPrec op_prec e1 . showString (opName op) .\nshowsExprPrec (op_prec+1) e2\nwhere op_prec = precedence op\n\nprecedence :: Op -> Int\nprecedence Plus = 6\nprecedence Minus = 6\nprecedence Multiply = 7\nprecedence Divide = 7\n\nopName :: Op -> String\nopName Plus = \"+\"\nopName Minus = \"-\"\nopName Multiply = \"*\"\nopName Divide = \"/\"\n```\n\nUnlike the Prolog solution, I've eliminated solutions like \"1+(2+3) = 6\" as a trivial variant of \"1+2+3 = 6\" (cf the function right_associative). Apart from that, the Prolog solution is shorter because it uses built-in evaluation and printing of expressions.\n\n## Problem 94\n\n<Problem description>\n\nExample:\n\n```<example in lisp>\n```\n\n```<example in Haskell>\n```\n\nSolution:\n\n```<solution in haskell>\n```\n\n<description of implementation>\n\n## Problem 95\n\n(**) English number words\n\nOn financial documents, like cheques, numbers must sometimes be written in full words. Example: 175 must be written as one-seven-five. Write a predicate full-words/1 to print (non-negative) integer numbers in full words.\n\n```> fullWords 175\none-seven-five\n```\n\nSolution:\n\n```import Data.List\nimport Data.Maybe\n\nfullWords :: Integer -> String\nfullWords n = concat . intersperse \"-\" . map (fromJust . (`lookup` table)) \\$ show n\nwhere table = [('0',\"zero\"), ('1',\"one\"), ('2',\"two\"), ('3',\"three\"), ('4',\"four\"),\n('5',\"five\"), ('6',\"six\"), ('7',\"seven\"), ('8',\"eight\"), ('9',\"nine\")]\n```\n\nThis solution does a simple table lookup after converting the positive integer into a string. Thus dividing into digits is much simplified.\n\nA minor variant of the above solution:\n\n```import Data.Char\nimport Data.List\n\nfullWords :: Integer -> String\nfullWords n = concat \\$ intersperse \"-\" [digits!!digitToInt d | d <- show n]\nwhere digits = [\"zero\", \"one\", \"two\", \"three\", \"four\",\n\"five\", \"six\", \"seven\", \"eight\", \"nine\"]\n```\n\n## Problem 96\n\n(**) Syntax checker\n\nIn a certain programming language (Ada) identifiers are defined by the syntax diagram below.", null, "Transform the syntax diagram into a system of syntax diagrams which do not contain loops; i.e. which are purely recursive. Using these modified diagrams, write a predicate identifier/1 that can check whether or not a given string is a legal identifier.\n\nExample in Prolog:\n\n```% identifier(Str) :- Str is a legal identifier\n```\n\n```> identifier \"this-is-a-long-identifier\"\nTrue\n> identifier \"this-ends-in-\"\nFalse\n> identifier \"two--hyphens\"\nFalse\n```\n\nSolution:\n\n```import Data.Char\nsyntax_check :: String -> Bool\nsyntax_check [] = False\nsyntax_check (x:xs) = isLetter x && loop xs\nwhere loop [] = True\nloop (y:ys) | y == '-' = (not . null) ys && isAlphaNum (head ys) && loop (tail ys)\n| isAlphaNum y = loop ys\n| otherwise = False\n```\n\nSimple functional transcription of the diagram.\n\nAnother direct transcription of the diagram:\n\n```identifier :: String -> Bool\nidentifier (c:cs) = isLetter c && hyphen cs\nwhere hyphen [] = True\nhyphen ('-':cs) = alphas cs\nhyphen cs = alphas cs\nalphas [] = False\nalphas (c:cs) = isAlphaNum c && hyphen cs\n```\n\nThe functions hyphen and alphas correspond to states in the automaton at the start of the loop and before a compulsory alphanumeric, respectively.\n\nParsec is a parser library that is commonly used in Haskell code. This is a solution using Parsec to parse the identifier.\n\n```isRight (Right _) = True\nisRight (Left _) = False\n\nidentifier x = isRight \\$ parse parser \"\" x where\nparser = letter >> loop\nloop = optional \\$ do\noptional (char '-')\nletter <|> digit\nloop\n```\n\n## Problem 97\n\n(**) Sudoku\n\nSudoku puzzles go like this:\n\n``` Problem statement Solution\n\n. . 4 | 8 . . | . 1 7\t 9 3 4 | 8 2 5 | 6 1 7\n| | | |\n6 7 . | 9 . . | . . .\t 6 7 2 | 9 1 4 | 8 5 3\n| | | |\n5 . 8 | . 3 . | . . 4 5 1 8 | 6 3 7 | 9 2 4\n--------+---------+-------- --------+---------+--------\n3 . . | 7 4 . | 1 . . 3 2 5 | 7 4 8 | 1 6 9\n| | | |\n. 6 9 | . . . | 7 8 . 4 6 9 | 1 5 3 | 7 8 2\n| | | |\n. . 1 | . 6 9 | . . 5 7 8 1 | 2 6 9 | 4 3 5\n--------+---------+-------- --------+---------+--------\n1 . . | . 8 . | 3 . 6\t 1 9 7 | 5 8 2 | 3 4 6\n| | | |\n. . . | . . 6 | . 9 1\t 8 5 3 | 4 7 6 | 2 9 1\n| | | |\n2 4 . | . . 1 | 5 . . 2 4 6 | 3 9 1 | 5 7 8\n```\n\nEvery spot in the puzzle belongs to a (horizontal) row and a (vertical) column, as well as to one single 3x3 square (which we call \"square\" for short). At the beginning, some of the spots carry a single-digit number between 1 and 9. The problem is to fill the missing spots with digits in such a way that every number between 1 and 9 appears exactly once in each row, in each column, and in each square.\n\nSolutions: see Sudoku\n\n## Problem 98\n\n(***) Nonograms\n\nAround 1994, a certain kind of puzzle was very popular in England. The \"Sunday Telegraph\" newspaper wrote: \"Nonograms are puzzles from Japan and are currently published each week only in The Sunday Telegraph. Simply use your logic and skill to complete the grid and reveal a picture or diagram.\" As a Prolog programmer, you are in a better situation: you can have your computer do the work! Just write a little program ;-).\n\nThe puzzle goes like this: Essentially, each row and column of a rectangular bitmap is annotated with the respective lengths of its distinct strings of occupied cells. The person who solves the puzzle must complete the bitmap given only these lengths.\n\n``` Problem statement: Solution:\n```\n``` |_|_|_|_|_|_|_|_| 3 |_|X|X|X|_|_|_|_| 3\n|_|_|_|_|_|_|_|_| 2 1 |X|X|_|X|_|_|_|_| 2 1\n|_|_|_|_|_|_|_|_| 3 2 |_|X|X|X|_|_|X|X| 3 2\n|_|_|_|_|_|_|_|_| 2 2 |_|_|X|X|_|_|X|X| 2 2\n|_|_|_|_|_|_|_|_| 6 |_|_|X|X|X|X|X|X| 6\n|_|_|_|_|_|_|_|_| 1 5 |X|_|X|X|X|X|X|_| 1 5\n|_|_|_|_|_|_|_|_| 6 |X|X|X|X|X|X|_|_| 6\n|_|_|_|_|_|_|_|_| 1 |_|_|_|_|X|_|_|_| 1\n|_|_|_|_|_|_|_|_| 2 |_|_|_|X|X|_|_|_| 2\n1 3 1 7 5 3 4 3 1 3 1 7 5 3 4 3\n2 1 5 1 2 1 5 1\n\n```\n\nFor the example above, the problem can be stated as the two lists [,[2,1],[3,2],[2,2],,[1,5],,,] and [[1,2],[3,1],[1,5],[7,1],,,,] which give the \"solid\" lengths of the rows and columns, top-to-bottom and left-to-right, respectively. Published puzzles are larger than this example, e.g. 25 x 20, and apparently always have unique solutions.\n\n```Nonogram> putStr \\$ nonogram [,[2,1],[3,2],[2,2],,[1,5],,,] [[1,2],[3,1],[1,5],[7,1],,,,]\n|_|X|X|X|_|_|_|_| 3\n|X|X|_|X|_|_|_|_| 2 1\n|_|X|X|X|_|_|X|X| 3 2\n|_|_|X|X|_|_|X|X| 2 2\n|_|_|X|X|X|X|X|X| 6\n|X|_|X|X|X|X|X|_| 1 5\n|X|X|X|X|X|X|_|_| 6\n|_|_|_|_|X|_|_|_| 1\n|_|_|_|X|X|_|_|_| 2\n1 3 1 7 5 3 4 3\n2 1 5 1\n```\n\nSolution 1 (SLOOW):\n\n```data Square = Blank | Cross deriving (Eq)\ninstance Show Square where\nshow Blank = \" \"\nshow Cross = \"X\"\n\n-- stolen from solution 2.\n-- create all possibilities of arranging the given blocks in a line of \"n\" elements\nrows n [] = [replicate n Blank]\nrows n (k:ks) | n < k = []\nrows n (k:ks) =\n[Blank : row | row <- rows (n-1) (k:ks)] ++\nif null ks then [replicate k Cross ++ replicate (n-k) Blank]\nelse [replicate k Cross ++ Blank : row | row <- rows (n-k-1) ks]\n\n-- contract a given line into the block format\n-- i.e. contract [Cross,Blank,Cross] == [1,1]\ncontract = map length . filter (\\(x:_) -> x==Cross) . group\n\n-- create all solutions by combining all possible rows in all possible ways\n-- then pick a solution and check whether its block signature fits\nsolver horz vert = filter fitsVert possSolution\nwhere possSolution = sequence \\$ map (rows (length vert)) horz\nfitsVert rs = map contract (transpose rs) == vert\n\n-- output the (first) solution\nnonogram horz vert = printSolution \\$ head \\$ solver horz vert\nwhere printSolution = putStr . unlines . map (concatMap show) . transpose\n```\n\nThis is a solution done for simplicity rather than performance. It's SLOOOOW.\n\nIt builds all combinations of blocks in a row (stolen from solution 2 :) and then builds all combinations of rows. The resulting columns are then contracted into the short block block form and the signature compared to the target.\n\nOne nice idea would be to use CaleGibbard's Sudoku Backtracking Monad or alternatively add a few simplistic \"heuristics\" which should reduce the complexity early on.\n\nSolution 2 (faster, but doesn't work if guessing is required):\n\n```module Nonogram where\n\nimport Data.List\n\ndata Square = Filled | Blank | Unknown\nderiving (Eq, Show)\n\nunify :: Square -> Square -> Square\nunify Filled Filled = Filled\nunify Blank Blank = Blank\nunify _ _ = Unknown\n\nmatch :: Square -> Square -> Bool\nmatch Filled Filled = True\nmatch Blank Blank = True\nmatch Unknown _ = True\nmatch _ _ = False\n\nname :: Square -> Char\nname Filled = 'X'\nname Blank = '_'\nname Unknown = '?'\n\n-- rows n ks = all possible ways of placing blocks of length ks\n-- in a row of length n.\nrows :: Int -> [Int] -> [[Square]]\nrows n [] = [replicate n Blank]\nrows n (k:ks) | n < k = []\nrows n (k:ks) =\n[Blank:row | row <- rows (n-1) (k:ks)] ++\nif null ks\nthen [replicate k Filled ++ replicate (n-k) Blank]\nelse [replicate k Filled ++ Blank : row | row <- rows (n-k-1) ks]\n\n-- Does r2 match all the known squares of r1?\nmatchRow :: [Square] -> [Square] -> Bool\nmatchRow r1 r2 = and (zipWith match r1 r2)\n\n-- common n ks partial = commonality between all possible ways of\n-- placing blocks of length ks in a row of length n that match partial.\ncommon :: Int -> [Int] -> [Square] -> Maybe [Square]\ncommon n ks partial = case filter (matchRow partial) (rows n ks) of\n[] -> Nothing\nrs -> Just (foldr1 (zipWith unify) rs)\n\nsolve :: [[Int]] -> [[Int]] -> Maybe [[Square]]\nsolve rs cs = converge step init\nwhere nr = length rs\nnc = length cs\ninit = replicate nr (replicate nc Unknown)\nstep = (improve nc rs . transpose) <.> (improve nr cs . transpose)\nimprove n = zipWithM (common n)\n(g <.> f) x = f x >>= g\n\n-- repeatedly apply f until a fixed point is reached\nconverge :: (Monad m, Eq a) => (a -> m a) -> a -> m a\nconverge f s = do\ns' <- f s\nif s' == s then return s else converge f s'\n\nshowGrid :: [[Int]] -> [[Int]] -> [[Square]] -> String\nshowGrid rs cs ss = unlines (zipWith showRow rs ss ++ showCols cs)\nwhere showRow rs ss = concat [['|', name s] | s <- ss] ++ \"| \" ++\nunwords (map show rs)\nshowCols cs\n| all null cs = []\n| otherwise = concatMap showCol cs : showCols (map advance cs)\nshowCol (k:_)\n| k < 10 = ' ':show k\n| otherwise = show k\nshowCol [] = \" \"\n\nnonogram :: [[Int]] -> [[Int]] -> String\nnonogram rs cs = case solve rs cs of\nNothing -> \"Inconsistent\\n\"\nJust grid -> showGrid rs cs grid\n```\n\nWe build up knowledge of which squares must be filled and which must be blank, until we can't make any more deductions.\n\nThe method of trying all placements of blocks in a row (rows) to determine commonality (common) could be more sophisticated, but seems fast enough.\n\nIt seems that there are puzzles that have a unique solution but which cannot be completely solved in this way, so we'll have to add backtracking.\n\n## Problem 99\n\n(***) Crossword puzzle\n\nGiven an empty (or almost empty) framework of a crossword puzzle and a set of words. The problem is to place the words into the framework.", null, "The particular crossword puzzle is specified in a text file which first lists the words (one word per line) in an arbitrary order. Then, after an empty line, the crossword framework is defined. In this framework specification, an empty character location is represented by a dot (.). In order to make the solution easier, character locations can also contain predefined character values. The puzzle above is defined in the file p99a.dat, other examples are p99b.dat and p99d.dat. There is also an example of a puzzle (p99c.dat) which does not have a solution.\n\nWords are strings (character lists) of at least two characters. A horizontal or vertical sequence of character places in the crossword puzzle framework is called a site. Our problem is to find a compatible way of placing words onto sites.\n\nHints: (1) The problem is not easy. You will need some time to thoroughly understand it. So, don't give up too early! And remember that the objective is a clean solution, not just a quick-and-dirty hack!\n\n(2) Reading the data file is a tricky problem for which a solution is provided in the file p99-readfile.pl. See the predicate read_lines/2.\n\n(3) For efficiency reasons it is important, at least for larger puzzles, to sort the words and the sites in a particular order. For this part of the problem, the solution of P28 may be very helpful.\n\n```ALPHA\nARES\nPOPPY\n\n.\n.\n.....\n. .\n. .\n.\n\n> solve \\$ readCrossword \"ALPHA\\nARES\\nPOPPY\\n\\n . \\n . \\n.....\\n . .\\n . .\\n .\\n\"\n\n[[((3,1),'A'),((3,2),'L'),((3,3),'P'),((3,4),'H'),((3,5),'A'),((1,3),'P'),((2,3)\n,'O'),((3,3),'P'),((4,3),'P'),((5,3),'Y'),((3,5),'A'),((4,5),'R'),((5,5),'E'),((\n6,5),'S')]]\n```\n\nSolution:\n\n```-- import Control.Monad\n-- import Data.List\n\ntype Coord = (Int,Int)\ntype Word = String\ndata Site = Site {siteCoords :: [Coord], siteLen :: Int} deriving (Show,Eq)\ndata Crossword = Crossword {cwWords :: [Word], cwSites :: [Site]} deriving (Show,Eq)\n\ncomparing f = \\a b -> f a `compare` f b\nequaling f = \\a b -> f a == f b\n\n-- convert the text lines from the file to the \"Site\" datatype,\n-- which contain the adjacent coordinates of the site and its length\ntoSites :: [String] -> [Site]\ntoSites lines = find (index_it lines) ++ find (transpose . index_it \\$ lines)\nwhere find = map makePos . concat . map extractor\nextractor = filter ((>1) . length) . map (filter (\\(_,x) -> x=='.')) . groupBy (equaling snd)\nindex_it = map (\\(row,e) -> zip [(col,row) | col <- [1..]] e) . zip [1..]\nmakePos xs = Site {siteCoords = map fst xs, siteLen = length xs}\n\n-- test whether there exist no two different letters at the same coordinate\nnoCollision :: [(String, Site)] -> Bool\nnoCollision xs = all allEqual groupedByCoord\nwhere groupedByCoord = map (map snd) . groupBy (equaling fst) . sortBy (comparing fst) . concatMap together \\$ xs\nallEqual [] = True\nallEqual (x:xs) = all (x==) xs\n\n-- merge a word and a site by assigning each letter to its respective coordinate\ntogether :: (Word, Site) -> [(Coord, Char)]\ntogether (w,s) = zip (siteCoords s) w\n\n-- returns all solutions for the crossword as lists of occupied coordinates and their respective letters\nsolve :: Crossword -> [[(Coord, Char)]]\nsolve cw = map (concatMap together) solution\nwhere solution = solve' (cwWords cw) (cwSites cw)\n\nsolve' :: [Word] -> [Site] -> [[(Word, Site)]]\nsolve' _ [] = [[]]\nsolve' words (s:ss) = if null possWords\nthen error (\"too few words of length \" ++ show (siteLen s))\nelse do try <- possWords\nlet restWords = Data.List.delete try words\nmore <- solve' restWords ss\nlet attempt = (try,s):more" ]

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[ "# Diophantine equation with no integer solutions, but with solutions modulo every integer\n\nIt's probably common knowledge that there are Diophantine equations which do not admit any solutions in the integers, but which admit solutions modulo $n$ for every $n$. This fact is stated, for example, in Dummit and Foote (p. 246 of the 3rd edition), where it is also claimed that an example is given by the equation $$3x^3 + 4y^3 + 5z^3 = 0.$$ However, D&F say that it's \"extremely hard to verify\" that this equation has the desired property, and no reference is given as to where one can find such a verification.\n\nSo my question is: Does anyone know of a readable reference that proves this claim (either for the above equation or for others)? I haven't had much luck finding one.\n\n• Duplicate: mathoverflow.net/questions/2779/… Nov 26, 2010 at 17:35\n• The example in your question is a famous example of Selmer. An explanation is given by Keith Conrad here: math.uconn.edu/~kconrad/blurbs/gradnumthy/selmerexample.pdf Note also that the example given below by Qiaochu are reducible. One can show that if $f(x)$ is an integer polynomial in one variable that is irreducible, then it cannot have a solution modulo every prime. The essential point of Selmer's example is that it is irreducible. Nov 26, 2010 at 18:52\n• A proof that there are no solutions over Q is in Cassels' book on elliptic curves. An elementary proof that there are p-adic solutions for all p, using Hensel's lemma and an argument very much in the spirit of Qiaochu's answer (i.e. using a \"cube\" analogue of the statement that if $a$ and $b$ aren't squares mod $p$ then $ab$ is) can be found at www2.imperial.ac.uk/~buzzard/maths/teaching/10Aut/M4P32/… : I just set it as homework for my students in fact :-) Nov 26, 2010 at 20:30\n• Dear A. Pacetti, Here is the argument I had in mind (it is not original to me, though); hopefully I am not butchering it: consider the Galois group $G$ of the splitting field of the polynomial. If the poly. $f$ is irred., then $G$ acts transitively on the roots of $f$, and so group theory shows that (if $f$ has degree $> 1$) then there is a conjugacy of $G$ with no fixed point. The $p$ whose Frobenius elements are equal to this conjugacy class then have the property that $f$ has no root modulo $p$. (Such a root would give a fixed point for the Frobenius of $p$. Nov 27, 2010 at 1:38\n• Another elementary example is $x^2+y^2+z^2+w^2=-1$. Because every positive integer is a sum of 4 squares, this has solutions modulo every $n\\ge 2$. Nov 27, 2010 at 11:51\n\nIt is actually quite straightforward to write down examples in one variable where this occurs. For example, the Diophantine equation $(x^2 - 2)(x^2 - 3)(x^2 - 6) = 0$ has this property: for any prime $p$, at least one of $2, 3, 6$ must be a quadratic residue, so there is a solution $\\bmod p$, and by Hensel's lemma (which has to be applied slightly differently when $p = 2$) there is a solution $\\bmod p^n$ for any $n$. We conclude by CRT. (Edit: As Fedor says, there are problems at $2$. We can correct this by using, for example, $(x^2 - 2)(x^2 - 17)(x^2 - 34)$.)\n\nHilbert wrote down a family of quartics with the same property. There are no (monic) cubics or quadratics with this property: if a monic polynomial $f(x) \\in \\mathbb{Z}[x]$ with $\\deg f \\le 3$ is irreducible over $\\mathbb{Z}$ (which is equivalent to not having an integer solution), then by the Frobenius density theorem there are infinitely many primes $p$ such that $f(x)$ is irreducible $\\bmod p$.\n\n• @Qiaochu : CRT ? Nov 26, 2010 at 17:43\n• I am afraid that this specific example does not have solution modulo 8, but if one multiples by $(x^2-17)$ then it becomes ok. Nov 26, 2010 at 17:47\n• @Andres: Chinese remainder theorem. @Fedor: oops! I think there is a simpler example along those lines, though. Nov 26, 2010 at 18:00\n• Qiaochu: Out of curiosity, what is Hilbert's family of examples? Nov 26, 2010 at 19:42\n• I approve ! I gave the same example $(x^2-2)(x^2-3)(x^2-6)=0$ thirty-three years ago, at the oral examination of the French teaching contest \"Agrégation\". Nov 26, 2010 at 19:46\n\nHere is another example, which is easy to verify by hand: $x^2+23y^2=41$. Note it has rational solutions (e.g. $(1/3,4/3)$). This provides solutions modulo $m$ if $(m,3)=1$. For $m$ a power of $3$, there is always a solution with $x=0$. Verifying that it doesn't have integral solutions is trivial.\n\n• The example I usually give my students is $x^2-37y^2=3$. The rational solution is $7/2,1/2$ and there's a 2-adic integer solution as well. However it's much harder to prove there's no integer solutions! (I can do it without too much trouble on a computer but...) So I'm very grateful for your example Felipe! Nov 26, 2010 at 20:43\n• An easy way to see that there are $3$-adic solutions is to note that $(81/13, 4/13)$ is a solution. Dec 2, 2010 at 18:30\n• $(9/4,5/4)$ is a solution too. I should have looked for other rational solutions. Dec 2, 2010 at 18:55\n\nThe equation 2x^2 + 7y^2 = 1 has two rational solutions with small relatively prime denominators (hence as a congruence mod m it is solvable for all m by CRT) but it visibly has no integral solutions. Look for a rational solution with denominator 3 and also for one with denominator 5 (small numerators in both cases).\n\n• KConrad: it would have been shorter to say \"$(1/3,1/3)$ and $(3/5,1/5)$\" than \"Look for a rational solution....both cases).\" :-) Nov 28, 2010 at 0:45\n• Yours seems to me like the \"best\" answer so far though (simplest equation, smallest coefficients). Nov 28, 2010 at 0:45\n• A Weierstrass equation in this spirit is y^2 = x^3 - 51 (which BCnrd told me about, and I think he got it from Venkatesh). It has the rational solution (1375/9,50986/27) and it can be solved mod 3^r for any r using x = 1 and for y a square root of -50 mod 3^r (which exists since -50 = 1 mod 3). Thus as a congruence this equation has a solution mod m for any m. That there is no Z-solution follows from Q(sqrt(-51)) having class number 2, which is rel. prime to 3, by the same kind of method used to find the Z-solutions to y^2 = x^3-2. Nov 28, 2010 at 7:19\n• The point of my previous comment was to have a rational point and no integral point, but the rational point doesn't matter. For many nonzero integers $k$ (e.g., $k = \\pm 6$) the equation $y^2 = x^3 + k$ has no integral solutions, but for every nonzero integer $k$, the congruence $y^2 \\equiv x^3 + k \\bmod m$ has a solution for all $m \\geq 2$. It suffices by Chinese remainder theorem to show there is a solution when $m$ runs through prime powers, and my answer at mathoverflow.net/questions/134352 does that. (At math.stackexchange.com/questions/875983 I show solns mod $p$.) Feb 3, 2019 at 2:31\n• A simpler example than $y^2 = x^3 - 51$ of such an equation with no integral solution and with a rational solution is $y^2 = x^3 + 11$, with rational solution $(-7/4,19/8)$. That it has no integral solution can be proved with just congruences and knowing when $-1 \\equiv \\Box \\bmod p$, no need for class numbers. See Theorem 2.3 of kconrad.math.uconn.edu/blurbs/gradnumthy/mordelleqn1.pdf. The rational solution shows there is a solution mod $p^r$ for odd primes $p$, and to get a solution mod powers of $2$ there is a $2$-adic solution $(x,2)$ where $x^3 = -7$ with $x \\equiv 1 \\bmod 2$. Feb 3, 2019 at 5:54\n\nConsider the equation $(2x - 1)(3x - 1) = 0$. This equation has no integer solutions. But modulo $n$, it always has a solution. If $n$ is not a multiple of $2$, we can make $2x -1$ a multiple of $n$. If $n$ is not a multiple of $3$, we can make $3x - 1$ a multiple of $n$. Using the Chinese Remainder Theorem, we can handle every other $n$ by piecing together these two solutions.\n\nThere is an easier example in\n\nwhere Kap disposed of the concern with the brief \"(it is easy to see that the assumption of no congruence obstructions is satisfied).\"\n\nThe example is, given a positive prime $p \\equiv 1 \\pmod 4,$ there is no solution in integers $x,y,z$ to $$x^2 + y^2 + z^9 = 216 p^3$$\n\nRobert C. Vaughan wrote to Kap (prior to publication) in appreciation, there was something involved that \"could not be detected p-adically.\" I forget what, it has been years. But we did well, Vaughan got an early draft in time to include the example in the second edition of\nThe Hardy-Littlewood Method.\n\nLater for some reason I looked at negative targets, with the same primes I believe it turned out that there were no integer solutions to $$x^2 + y^2 + z^9 = -8 p^3.$$\n\nThe significance of the example is not so much as a single Diophantine equation, rather as a Diophantine representation problem in the general vicinity of the Waring problem, but with mixed exponents: given nonegative integer variables $x,y,z$ and exponents $a,b,c \\geq 2,$ and given the polynomial $f(x,y,z) =x^a + y^b + z^c,$ if $f(x,y,z)$ represents every positive integer $p$-adically and if $$\\frac{1}{a} + \\frac{1}{b} + \\frac{1}{c} > 1,$$ does $f(x,y,z)$ integrally represent all sufficiently large integers? The answer is no for the problem as stated, but the counterexamples depend heavily on factorization, and in the end upon composition of binary forms. As this is also the mechanism underlying the simplest examples of spinor exceptional integers for positive ternary quadratic forms, it is natural to ask whether there is some relatively easy formalism that adds \"factorization obstructions\" to the well-studied \"congruence obstructions.\"\n\nSee:\n\nhttp://zakuski.math.utsa.edu/~jagy/Vaughan.pdf\n\nhttp://en.wikipedia.org/wiki/Waring's_problem\n\n• Is there a Brauer-Manin obstruction to these examples? Nov 27, 2010 at 19:08\n• Hi, Felipe. I really would not know. To be more precise, I do not know what the phrase means. But there is such an example with $x^2 + y^2 + z^n$ where $n$ is any odd and composite number. It is known that $x^2 + y^2 + z^3$ integrally represents all integers, and modest evidence suggests the same for any $n$ odd prime. Meanwhile, warranted or not, I think this is related to my first question in January 2010, $2 x^2 + x y + 3 y^2 + z^3 - z,$ answered with some real effort by Kevin Buzzard. There are about 25 of those, one for each discriminant of positive binary forms with class number 3. Nov 27, 2010 at 20:15\n• Felipe, the integral Brauer-Manin obstruction has been systematically invstigated by Colliot-Thélène and Xu (math.u-psud.fr/~colliot/CTXuCompositio2009.pdf). Nov 28, 2010 at 11:00\n• @Chandan: Yes, I know about this paper. My question was whether the lack of integral solutions to Will's examples can be explained by an integral Brauer-Manin obstruction. Nov 29, 2010 at 18:57\n• Felipe and Chandan, thank you for at least discussing this a little. Kap's proof by factorization is in the short paper with the link already given in my answer above. The phenomenon, in my comment above, in positive integral ternary quadratic forms is in a 1995 letter by Kap to J. S. Hsia and Rainer Schulze-Pillot, pdf link attempted here: zakuski.math.utsa.edu/~kap/Forms/… and the January 2010 question Kevin answered at: mathoverflow.net/questions/12486/… Will Nov 29, 2010 at 19:51\n\nAn example even easier than Jagy and Kaplansky's\n$x^2+y^2+z^9 = 216p^3$, for $p=1 \\bmod 4$, is given in:\n\nSums of two squares and one biquadrate, by R. Dietmann, and C. Elsholtz,\nFunct. Approx. Comment. Math. Volume 38, Number 2 (2008), 233-234.\n\nhttp://www.math.tugraz.at/~elsholtz/WWW/papers/papers26de08.pdf\n\nHere we showed:\n$x^2+y^2+z^4=p^2$ has no positive solutions, when $p=7 \\bmod 8, p$prime. Once the example is known, it's trivial to prove.\n\nThe Jagy-Kaplansky example can be generalized to odd composite exponent, instead of 9. It seems the example above was overlooked for quite a while.\n\n• This is nice, Christian. Dec 2, 2010 at 20:43\n\nSee 6.4.1 in my paper with Rudnick http://www.springerlink.com/content/l1t0071152537186/, page 62. The equation is: $$-9x^2+2xy+7y^2+2z^2=1.$$ This equation has a rational solution $(-\\frac{1}{2}, \\frac{1}{2},1)$, hence it has solutions modulo $p^n$ for all $p\\neq 2$ and all $n$. In addition, it has a solution $(4,1,1)$ modulo $2^7$, and using Hensel's lemma one can easily check that the equation has solutions modulo $2^n$ for all $n$. The elementary proof that this equation has no integral solutions is due to Don Zagier and is based on (a supplementary formula to) the quadratic reciprocity law.\n\nLet us define the size $$H$$ of a polynomial Diophantine equation $$P(x_1,...,x_n)=0$$ as: substitute $$2$$ instead of all variables, absolute values instead of all coefficients, and evaluate. This notion of size has the advantages that (i) there is a finite number of equations with bounded size, so we can do full computer search and find the smallest equation with any given property and (ii) equations with small $$H$$ really look nice and compact.\n\nThen the smallest Diophantine equation with no integer solutions, but with solutions modulo every integer is the equation $$y(x^2+2)=1$$ with size $$H=2(2^2+2)+1 = 13$$.\n\nProof that a solution exists: As usual for polynomial equations, CRT allows us to reduce to the case $$n = p^e$$ for $$p$$ prime and $$e$$ positive. If $$p > 2$$, then $$p^e$$ is odd, so $$2$$ is invertible mod $$p^e$$. Set $$x = 0, y = 2^{-1}$$. If $$p = 2$$, then $$3$$ is invertible mod $$2^e$$. Set $$x = 1, y = 3^{-1}$$.\n\n• @MichaelAlbanese Edited to add a proof. As for the original, I'm not sure what your criterion for \"smallest\" is; there are plenty of quadratic answers (e.g. see Ravi Boppana's solution for one that's even one-variable). I think you may be restricting yourself to some idea of \"fully monic\" polynomials. Aug 1, 2021 at 17:46\n• I have added explanation in which sense this is the smallest example Aug 3, 2021 at 7:51" ]

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[ "", null, "Introduction to Dislocations\n\nHardcover(3rd ed)\n\n\\$46.00 \\$67.00 Save 31% Current price is \\$46, Original price is \\$67. You Save 31%.\nView All Available Formats & Editions\n\nOverview\n\nIntroduction to Dislocations was first published in 1965 in a series aimed at undergraduate and postgraduate students in metallurgy and materials science and related disciplines. At the time, the subject was maturing and it was expected that 'dislocation concepts' would remain a core discipline for a very long time. As expected, the book has been, and remains, an important undergraduate text all over the world.\n\nA wider range of materials has emerged since 1965, most notably in the field of electronics and micro-engineering. The principles of dislocation theory still apply but some of the detail requires further treatment.\n\nThis fourth edition provides an essential basis for an understanding of many of the physical and mechanical properties of crystalline solids. This new edition has been extensively revised and updated to reflect developments in the understanding of the subject, whilst retaining the clarity and comprehensibility of the previous editions.\n\nProduct Details\n\nISBN-13: 9780080287218 Elsevier Science & Technology Books 04/28/1984 Pergamon International Library of Science, Technology, Engineering, and Social Studies , #37 3rd ed 257 6.50(w) x 1.50(h) x 9.50(d)\n\nIntroduction to Dislocations\n\nBy D. Hull D. J. Bacon\n\nButterworth-Heinemann\n\nAll right reserved.\n\nISBN: 978-0-08-096673-1\n\nChapter One\n\nDefects in Crystals\n\n1.1 CRYSTALLINE MATERIALS\n\nDislocations are an important class of defect in crystalline solids and so an elementary understanding of crystallinity is required before dislocations can be introduced. Metals and many important classes of non-metallic solids are crystalline, i.e. the constituent atoms are arranged in a pattern that repeats itself periodically in three dimensions. The actual arrangement of the atoms is described by the crystal structure. The crystal structures of most pure metals are relatively simple: the three most common are the body-centered cubic, face-centered cubic and close-packed hexagonal, and are described in section 1.2. In contrast, the structures of alloys and non-metallic compounds are often complex.\n\nThe arrangement of atoms in a crystal can be described with respect to a three-dimensional net formed by three sets of straight, parallel lines as in Fig. 1.1(a). The lines divide space into equal sized parallelepipeds and the points at the intersection of the lines define a space lattice. Every point of a space lattice has identical surroundings. Each parallelepiped is called a unit cell and the crystal is constructed by stacking identical unit cells face to face in perfect alignment in three dimensions. By placing a motif unit of one or more atoms at every lattice site the regular structure of a perfect crystal is obtained.\n\nThe positions of the planes, directions and point sites in a lattice are described by reference to the unit cell and the three principal axes, x, y and z (Fig. 1.1 (b)). The cell dimensions OA = a, OB = b and OC = c are the lattice parameters, and these along with the angles [??]BOC = α, [??]COA = ß and +AOB 5 completely define the size and shape of the cell. For simplicity the discussion here will be restricted to cubic and hexagonal crystal structures. In cubic crystals a = b = c and = ß = γ = 90°, and the definition of planes and directions is straightforward. In hexagonal crystals it is convenient to use a different approach, and this is described in section 1.2.\n\nAny plane A'B'C' in Fig. 1.2 can be defined by the intercepts OA', OB' and OC' with the three principal axes. The usual notation (Miller indices) is to 1 take the reciprocals of the ratios of the intercepts to the corresponding unit cell dimensions. Thus A'B'C' is represented by\n\n(OA/OA'. OB/OB', OC/OC') and the numbers are then reduced to the three smallest integers in these ratios.\n\nThus from Fig. 1.2 OA' = 2a, OB' = 3a, and OC' = 3a, the reciprocal intercepts are\n\n(a/2a, a/3a, a/3a) and so the Miller indices of the A'B'C' plane are (322). Curved brackets are used for planes. A plane with intercepts OA, OB, and OC has Miller indices\n\n(a/a, a/a, a/a) or, more simply, (111). Similarly, a plane DFBA in Fig. 1.3 is\n\n(a/a, a/a, a/∞)\n\nor (110); a plane DEGA is (a/a, a/∞, a/∞)\n\nor (100); and a plane AB'C' in Fig. 1.2 is\n\n(a/a, a/3a, a/3a)\n\nor (311). In determining the indices of any plane it is most convenient to identify the plane of lattice points parallel to the plane which is closest to the origin O and intersects the principal axis close to the origin. Thus plane A'B'C' in Fig. 1.2 is parallel to ABC and it is clear that the indices are (111). Using this approach it will be seen that the planes ABC, ABE, CEA and CEB in Fig. 1.3 are (111), (11[bar.1]), (1[bar.1]1) and ([bar.1]11) respectively. The minus sign above an index indicates that the plane cuts the axis on the negative side of the origin. In a cubic crystal structure, these planes constitute a group of the same crystallographic type and are described collectively by {111}.\n\nAny direction LM in Fig. 1.3 is described by the line parallel to LM through the origin O, in this case OE. The direction is given by the three smallest integers in the ratios of the lengths of the projections of OE resolved along the three principal axes, namely OA, OB and OC, to the corresponding lattice parameters of the unit cell. Thus, if the cubic unit cell is given by OA, OB and OC the direction LM is\n\n[OA/OA, OB/OB. OC/OC]\n\nor\n\n[a/a, a/a, a/a]\n\nor . Square brackets are used for directions. The directions CG, AF, DB and EO are [11[bar.1]], [[bar.1]11], [[bar.1]1[bar.1]] and [[bar.111]] respectively and are a group of directions of the same crystallographic type described collectively by <111>. Similarly, direction CE is\n\n[a/a, a/a, O/a]\n\nor ; direction AG is\n\n[O/a, a/a, O/a]\n\nor ; and direction GH is\n\n[a/2/a, -a/2/a, a/a]\n\nor [[bar.11]2]. The rule that brackets and ( ) imply specific directions and planes respectively, and that < > and { } refer respectively to directions and planes of the same type, will be used throughout this text.\n\nIn cubic crystals the Miller indices of a plane are the same as the indices of the direction normal to that plane. Thus in Fig. 1.3 the indices of the plane EFBG are (010) and the indices of the direction AG which is normal to EFBG are . Similarly, direction OE is normal to plane CBA (111).\n\nThe coordinates of any point in a crystal relative to a chosen origin site are described by the fractional displacements of the point along the three principal axes divided by the corresponding lattice parameters of the unit cell. The center of the cell in Fig. 1.3 is 1/2, 1/2, 1/2 relative to the origin O; and the points F, E, H and I are 0, 1, 1; 1, 1, 1; 1/2, 1/2, 1; and 1, 1/2, 1 respectively.\n\n1.2 SIMPLE CRYSTAL STRUCTURES\n\nIn this section the atoms are considered as hard spheres which vary in size from element to element. From the hard sphere model the parameters of the unit cell can be described directly in terms of the radius of the atomic sphere, r. In the diagrams illustrating the crystal structures the atoms are shown as small circles in the three-dimensional drawings and as large circles representing the full hard sphere sizes in the two-dimensional diagrams. It will be shown that crystal structures can be described as a stack of lattice planes in which the arrangement of lattice sites within each layer is identical. To see this clearly in two-dimensional figures, the atoms in one layer represented by the plane of the paper are shown as full circles, whereas those in layers above and below the first are shown as small shaded circles. The order or sequence of the atom layers in the stack, i.e. the stacking sequence, is described by labeling one layer as an A layer and all other layers with atoms in identical positions above the first as A layers also. Layers of atoms in other positions in the stack are referred to as B, C, D layers, etc.\n\nIn the simple cubic structure with one atom at each lattice site, illustrated in Fig. 1.4, the atoms are situated at the corners of the unit cell. (Note that no real crystals have such a simple atomic arrangement.) Figures 1.4(b) and (c) show the arrangements of atoms in the (100) and (110) planes respectively. The atoms touch along <001> directions and therefore the lattice parameter a is twice the atomic radius r (a = 2r). The atoms in adjacent (100) planes are in identical atomic sites when projected along the direction normal to this plane, so that the stacking sequence of (100) planes is AAA ... The atoms in adjacent (110) planes are displaced 1/2 a [square root of 2] along relative to each other and the spacing of atoms along is a [square root of 2]. It follows that alternate planes have atoms in the same atomic sites relative to the direction normal to (110) and the stacking sequence of (110) planes is ABABAB ... The spacing between successive (110) planes is 1/2 a [square root of 2].\n\nIn the body-centered cubic structure (bcc), which is exhibited by many metals and is shown in Fig. 1.5, the atoms are situated at the corners of the unit cell and at the centre site 1/2, 1/2, 1/2. The atoms touch along a <111> direction and this is referred to as the close-packed direction. The lattice parameter a = 4r = [square root of 3] and the spacing of atoms along <110> directions is a [square root of 2]. The stacking sequence of {100} and {110} planes is ABABAB ... (Fig. 1.5(b)). There is particular interest in the stacking of {112} type planes (see sections 6.3 and 9.7). Figure 1.6 shows two body-centered cubic cells and the positions of a set of (112) planes. From the diagrams it is seen that the stacking sequence of these planes is ABCDEFAB ..., and the spacing between the planes is a/[square root of 6].\n\nIn the face-centered cubic structure (fcc), which is also common among the metals and is shown in Fig. 1.7, the atoms are situated at the corners of the unit cell and at the centers of all the cube faces in sites of the type 0, 1/2, 1/2. The atoms touch along the <011> close-packed directions. The lattice parameter a = 4r/ [square root of 2]. The stacking sequence of {100} and {110} planes is ABABAB ..., and the stacking sequence of {111} planes is ABCABC ... The latter is of considerable importance (see Chapter 5) and is illustrated in Figs 1.7(c) and (d). The atoms in the {111} planes are in the most close-packed arrangement possible for spheres and contain three h110i close-packed directions 60° apart, as in Fig. 1.7(b).\n\nThe close-packed hexagonal structure (cph or hcp) is also common in metals. It is more complex than the cubic structures but can be described very simply with reference to the stacking sequence. The unit cell with lattice parameters a, a, c is shown in Fig. 1.8(a), together with the hexagonal cell constructed from three unit cells. There are two atoms per lattice site, i.e. at 0, 0, 0 and 2/3, 1/3, 1/2 with respect to the axes a1, a2, c. The atomic planes perpendicular to the c axis are close-packed, as in the fcc case, but the stacking sequence is now ABABAB ..., as shown in Fig. 1.8(b).\n\nFor a hard sphere model the ratio of the length of the c and a axes (axial ratio) of the hexagonal structure is 1.633. In practice, the axial ratio varies between 1.57 and 1.89 in close-packed hexagonal metals. The variations arise because the hard sphere model gives only an approximate value of the interatomic distances and requires modification depending on the electronic structure of the atoms.\n\nIf Miller indices of three numbers based on axes a1, a2, c are used to define planes and directions in the hexagonal structure, it is found that crystallo-graphically equivalent sets can have combinations of different numbers. For example, the three close-packed directions in the basal plane (001) are , and . Indexing in hexagonal crystals is therefore usually based on Miller-Bravais indices, which are referred to the four axes a1, a2, a3 and c indicated in Fig. 1.8(a). When the reciprocal intercepts of a plane on all four axes are found and reduced to the smallest integers, the indices are of the type (h, k, i, l), and the first three indices are related by\n\ni = -(h + k) (1.1)\n\nEquivalent planes are obtained by interchanging the position and sign of the first three indices. A number of planes in the hexagonal lattice have been given specific names. For example:\n\nBasal plane (0001) Prism plane : first order (1[bar.1]00) ([bar.1]100); etc: Prism plane : second order (11[bar.2]0) ([bar.2]110); etc: Pyramidal plane : first order (10[bar.1]1) ([bar.1]011); etc: Pyramidal plane : second order (11[bar.2]2) ([bar.11]22); etc:\n\n(Continues...)\n\nExcerpted from Introduction to Dislocations by D. Hull D. J. Bacon Copyright © 2011 by D. Hull and D. J. Bacon. Excerpted by permission of Butterworth-Heinemann. All rights reserved. No part of this excerpt may be reproduced or reprinted without permission in writing from the publisher.\nExcerpts are provided by Dial-A-Book Inc. solely for the personal use of visitors to this web site." ]

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[ "## Factorising Cubic Expressions\n\nWe know how to factorise quadratic equations but it is much harder to factorise cubic because in general there is no easy to see relationship between the coefficients. If we are given a factor or a root however, the problem becomes much more solvable. We can write the cubic expression as a product of a linear factor with a quadratic expression, expand the brackets and simplify and solve for the coefficients.\n\nExample:", null, "If", null, "is a root of", null, "factorise", null, "If", null, "is a root,", null, "is a factor.", null, "", null, "", null, "", null, "constant:", null, "hence", null, "The quadratic in the brackets can now be factorised by inspection:", null, "Example:", null, "If", null, "is a root of", null, "factorise", null, "If", null, "is a root,", null, "is a factor.", null, "", null, "", null, "", null, "constant:", null, "hence", null, "The quadratic in the brackets can now be factorised by inspection:", null, "", null, "" ]

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[ " Grade 8 Mathematics Module 4, Topic A, Overview | EngageNY\n\n## Grade 8 Mathematics Module 4, Topic A, Overview", null, "In Topic A, students begin by transcribing written statements using symbolic notation.  Then, students write linear and non-linear expressions leading to linear equations, which are solved using properties of equality (8.EE.C.7b).  Students learn that not every linear equation has a solution.  In doing so, students learn how to transform given equations into simpler forms until an equivalent equation results in a unique solution, no solution, or infinitely many solutions (8.EE.C.7a).  Throughout Topic A students must write and solve linear equations in real-world and mathematical situations.", null, "Grade 8 Mathematics Module 4, Topic A, Overview (432.44 KB) View PDF", null, "Grade 8 Mathematics Module 4, Topic A, Overview (285.83 KB)" ]

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[ "Amount\nFrom\nTo\n\n# 1.89 square yards to square meters\n\nHow many square meters in 1.89 square yards? 1.89 square yards is equal to 1.5802807104 square meters.\n\nThis page provides you how to convert between square yards and square meters with conversion factor.\n\n# How to convert 1.89 yd2 to m2?\n\nTo convert 1.89 yd2 into m2, follow these steps:\n\nWe know that, 1 m2 = 1.1959900463 yd2\n\nHence, to convert the value of 1.89 square yards into square meters, divide the area value 1.89yd2 by 1.1959900463.\n\n1.89 yd2 = 1.89/1.1959900463 = 1.5802807104 m2\n\n### Thus, 1.89 yd2 equals to 1.5802807104 m2\n\nSquare Yards Conversion of Square Yards to Square Meters\n1.88 yd2 1.88 yd2 = 1.5719194368 m2\n1.79 yd2 1.79 yd2 = 1.4966679744 m2\n1.89 yd2 1.89 yd2 = 1.5802807104 m2\n2.89 yd2 2.89 yd2 = 2.4164080704 m2" ]

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[ "Use the COUPDAYBS (coupon days before settlement) function to calculate the number of days from the beginning of the coupon period until its settlement date. The number returned includes both the first day of the period and the settlement date.\n\nCOUPDAYBS(Settlement, Maturity, Frequency[, Basis])\n\nArguments\n\n Argument Data type Description Settlement (required) Date The bond settlement date: The date the bond is traded to the buyer. Maturity (required) Date The bond maturity date: The date when the bond expires. Frequency (required) Number The number of coupon payments per year. Enter: 1 for annual 2 for semi-annual 4 for quarterly Basis Number The basis determines how many days exist in a year. A full year has: 360 days when basis US (NASD) 30/360, Actual/360, and EUR 30/360 are used 365 days when basis Actual/365 is used 365 or 366 days when Actual/Actual is used US 30/360 is the default basis for COUPDAYS. It can also be specified by entering 0. To use a different type of day count basis, enter: 1 for Actual/Actual 2 for Actual/360 3 for Actual/365 4 for European 30/360 Learn about the conventions used to calculate the day count for basis.\n\nThe COUPDAYBS function returns a number.\n\n• The settlement and maturity dates must be valid dates between 01/01/1900 and 12/31/2399.\n• The maturity date must be later than the settlement date.\n• The frequency must be either 1 (annual), 2 (semi-annual), or 4 (quarterly).\n• The basis, when specified, must be either 0 (US (NASD) 30/360), 1 (Actual/Actual), 2 (Actual/360), 3 (Actual/365), or 4 (EUR 30/360).\n\nMost financial functions are currently unavailable in Polaris. Learn more about the differences between Anaplan calculation engines.\n\nCOUPDAYBS\n\nThis example shows how the number of days before the settlement date can be calculated when a basis is specified.\n\n Formula Description Result COUPDAYBS(DATE(2015, 1, 15), DATE(2018, 1, 31), 1, 1) This formula uses: a settlement date of 01/15/2015 a maturity date of 01/31/2018 a frequency of 1 (annual) a basis of 1 (Actual/Actual) 349\n\nIn this example, the number of days in the coupon period that contains the settlement date is calculated without specifying a basis. As a result, the basis defaults to US 30/360.\n\n Formula Description Result COUPDAYBS(DATE(2015, 1, 15), DATE(2018, 1, 31), 4) This formula uses: a settlement date of 01/15/2015 a maturity date of 01/31/2018 a frequency of 4 (quarterly) 75" ]

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[ "# Arduino – I/O Functions\n\nThe pins on the Arduino board can be configured as either inputs or outputs. We will explain the functioning of the pins in those modes. It is important to note that a majority of Arduino analog pins, may be configured, and used, in exactly the same manner as digital pins.\n\n## Pins Configured as INPUT\n\nArduino pins are by default configured as inputs, so they do not need to be explicitly declared as inputs with pinMode() when you are using them as inputs. Pins configured this way are said to be in a high-impedance state. Input pins make extremely small demands on the circuit that they are sampling, equivalent to a series resistor of 100 megaohm in front of the pin.\n\nThis means that it takes very little current to switch the input pin from one state to another. This makes the pins useful for such tasks as implementing a capacitive touch sensor or reading an LED as a photodiode.\n\nPins configured as pinMode(pin, INPUT) with nothing connected to them, or with wires connected to them that are not connected to other circuits, report seemingly random changes in pin state, picking up electrical noise from the environment, or capacitively coupling the state of a nearby pin.\n\n## Pull-up Resistors\n\nPull-up resistors are often useful to steer an input pin to a known state if no input is present. This can be done by adding a pull-up resistor (to +5V), or a pull-down resistor (resistor to ground) on the input. A 10K resistor is a good value for a pull-up or pull-down resistor.\n\n### Using Built-in Pull-up Resistor with Pins Configured as Input\n\nThere are 20,000 pull-up resistors built into the Atmega chip that can be accessed from software. These built-in pull-up resistors are accessed by setting the pinMode() as INPUT_PULLUP. This effectively inverts the behavior of the INPUT mode, where HIGH means the sensor is OFF and LOW means the sensor is ON. The value of this pull-up depends on the microcontroller used. On most AVR-based boards, the value is guaranteed to be between 20kΩ and 50kΩ. On the Arduino Due, it is between 50kΩ and 150kΩ. For the exact value, consult the datasheet of the microcontroller on your board.\n\nWhen connecting a sensor to a pin configured with INPUT_PULLUP, the other end should be connected to the ground. In case of a simple switch, this causes the pin to read HIGH when the switch is open and LOW when the switch is pressed. The pull-up resistors provide enough current to light an LED dimly connected to a pin configured as an input. If LEDs in a project seem to be working, but very dimly, this is likely what is going on.\n\nSame registers (internal chip memory locations) that control whether a pin is HIGH or LOW control the pull-up resistors. Consequently, a pin that is configured to have pull-up resistors turned on when the pin is in INPUTmode, will have the pin configured as HIGH if the pin is then switched to an OUTPUT mode with pinMode(). This works in the other direction as well, and an output pin that is left in a HIGH state will have the pull-up resistor set if switched to an input with pinMode().\n\nExample\n\n```pinMode(3,INPUT) ; // set pin to input without using built in pull up resistor\npinMode(5,INPUT_PULLUP) ; // set pin to input using built in pull up resistor\n```\n\n## Pins Configured as OUTPUT\n\nPins configured as OUTPUT with pinMode() are said to be in a low-impedance state. This means that they can provide a substantial amount of current to other circuits. Atmega pins can source (provide positive current) or sink (provide negative current) up to 40 mA (milliamps) of current to other devices/circuits. This is enough current to brightly light up an LED (do not forget the series resistor), or run many sensors but not enough current to run relays, solenoids, or motors.\n\nAttempting to run high current devices from the output pins, can damage or destroy the output transistors in the pin, or damage the entire Atmega chip. Often, this results in a \"dead\" pin in the microcontroller but the remaining chips still function adequately. For this reason, it is a good idea to connect the OUTPUT pins to other devices through 470Ω or 1k resistors, unless maximum current drawn from the pins is required for a particular application.\n\n## pinMode() Function\n\nThe pinMode() function is used to configure a specific pin to behave either as an input or an output. It is possible to enable the internal pull-up resistors with the mode INPUT_PULLUP. Additionally, the INPUT mode explicitly disables the internal pull-ups.\n\n### pinMode() Function Syntax\n\n```Void setup () {\npinMode (pin , mode);\n}\n```\n• pin − the number of the pin whose mode you wish to set\n\n• mode − INPUT, OUTPUT, or INPUT_PULLUP.\n\nExample\n\n```int button = 5 ; // button connected to pin 5\nint LED = 6; // LED connected to pin 6\n\nvoid setup () {\npinMode(button , INPUT_PULLUP);\n// set the digital pin as input with pull-up resistor\npinMode(button , OUTPUT); // set the digital pin as output\n}\n\nvoid setup () {\nIf (digitalRead(button ) == LOW) // if button pressed {\ndigitalWrite(LED,HIGH); // turn on led\ndelay(500); // delay for 500 ms\ndigitalWrite(LED,LOW); // turn off led\ndelay(500); // delay for 500 ms\n}\n}```\n\n## digitalWrite() Function\n\nThe digitalWrite() function is used to write a HIGH or a LOW value to a digital pin. If the pin has been configured as an OUTPUT with pinMode(), its voltage will be set to the corresponding value: 5V (or 3.3V on 3.3V boards) for HIGH, 0V (ground) for LOW. If the pin is configured as an INPUT, digitalWrite() will enable (HIGH) or disable (LOW) the internal pullup on the input pin. It is recommended to set the pinMode() to INPUT_PULLUP to enable the internal pull-up resistor.\n\nIf you do not set the pinMode() to OUTPUT, and connect an LED to a pin, when calling digitalWrite(HIGH), the LED may appear dim. Without explicitly setting pinMode(), digitalWrite() will have enabled the internal pull-up resistor, which acts like a large current-limiting resistor.\n\n### digitalWrite() Function Syntax\n\n```Void loop() {\ndigitalWrite (pin ,value);\n}\n```\n• pin − the number of the pin whose mode you wish to set\n\n• value − HIGH, or LOW.\n\nExample\n\n```int LED = 6; // LED connected to pin 6\n\nvoid setup () {\npinMode(LED, OUTPUT); // set the digital pin as output\n}\n\nvoid setup () {\ndigitalWrite(LED,HIGH); // turn on led\ndelay(500); // delay for 500 ms\ndigitalWrite(LED,LOW); // turn off led\ndelay(500); // delay for 500 ms\n}```\n\nArduino is able to detect whether there is a voltage applied to one of its pins and report it through the digitalRead() function. There is a difference between an on/off sensor (which detects the presence of an object) and an analog sensor, whose value continuously changes. In order to read this type of sensor, we need a different type of pin.\n\nIn the lower-right part of the Arduino board, you will see six pins marked “Analog In”. These special pins not only tell whether there is a voltage applied to them, but also its value. By using the analogRead() function, we can read the voltage applied to one of the pins.\n\nThis function returns a number between 0 and 1023, which represents voltages between 0 and 5 volts. For example, if there is a voltage of 2.5 V applied to pin number 0, analogRead(0) returns 512.\n\n```analogRead(pin);\n```\n• pin − the number of the analog input pin to read from (0 to 5 on most boards, 0 to 7 on the Mini and Nano, 0 to 15 on the Mega)\n\nExample\n\n```int analogPin = 3;//potentiometer wiper (middle terminal)\n// connected to analog pin 3\nint val = 0; // variable to store the value read\n\nvoid setup() {\nSerial.begin(9600); // setup serial\n}\n\nvoid loop() {" ]

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[ "My watch list\nmy.chemeurope.com\n\n# Boltzmann constant\n\nValues of k Units\n1.380 6504(24)×10−23 J·K-1\n8.617 343(15)×10−5 eV·K-1\n1.3807×10−16 erg·K-1\nFor details, see Value in different units below.\n\nThe Boltzmann constant (k or kB) is the physical constant relating temperature to energy.\n\nIt is named after the Austrian physicist Ludwig Boltzmann, who made important contributions to the theory of statistical mechanics, in which this constant plays a crucial role.\n\n## Bridge from macroscopic to microscopic physics\n\nBoltzmann's constant k is a bridge between macroscopic and microscopic physics. Macroscopically, the ideal gas law states that, for an ideal gas, the product of pressure p and volume V is proportional to the product of amount of substance n (in number of moles) and absolute temperature T.", null, "$\\ pV = nRT,$\n\nwhere", null, "$\\ R$ is called the gas constant [8.314 472 m3·Pa·K−1·mol−1],\n\nIntroducing Boltzmann's constant transforms this into an equation about the microscopic properties of molecules,", null, "$p V = N k T \\,,$\n\nwhere N is the number of molecules of gas, and k is Boltzmann's constant.\n\n## Role in the equipartition of energy\n\nGiven a thermodynamic system at an absolute temperature T, the thermal energy carried by each microscopic \"degree of freedom\" in the system is on the order of magnitude of kT/2 (i.e., about 2.07×10−21 J, or 0.013 eV at room temperature).\n\n### Application to simple gas thermodynamics\n\nIn classical statistical mechanics, this average is predicted to hold exactly for homogeneous ideal gases. Monatomic ideal gases possess 3 degrees of freedom per atom, corresponding to the three spatial directions, which means a thermal energy of 1.5kT per atom. As indicated in the article on heat capacity, this corresponds very well with experimental data. The thermal energy can be used to calculate the root mean square speed of the atoms, which is inversely proportional to the square root of the atomic mass. The root mean square speeds found at room temperature accurately reflect this, ranging from 1370 m/s for helium, down to 240 m/s for xenon.\n\nKinetic theory gives the average pressure p for an ideal gas as", null, "$p = \\frac{1}{3}\\frac{N}{V} m {\\overline{v^2}}.$\n\nSubstituting that the average translational kinetic energy is", null, "$\\frac{1}{2}m \\overline{v^2} = \\frac{3}{2} k T,$\n\nand gives", null, "$p = \\frac{N}{V} k T$\n\nso the ideal gas equation is regained.\n\nThe ideal gas equation is also followed quite well for molecular gases; but the form for the heat capacity is more complicated, because the molecules possess new internal degrees of freedom, as well as the three degrees of freedom for movement of the molecule as a whole. Diatomic gases, for example, possess in total approximately 5 degrees of freedom per molecule.\n\n## Role in Boltzmann factors\n\nMore generally, systems in equilibrium with a reservoir of heat at temperature T have probabilities of occupying states with energy E weighted by the corresponding Boltzmann factor:", null, "$p \\propto \\exp{\\frac{-E}{kT}}.$\n\nAgain, it is the energy-like quantity kT which takes central importance.\n\nConsequences of this include (in addition to the results for ideal gases above), for example the Arrhenius equation of simple chemical kinetics.\n\n## Role in definition of entropy\n\nIn statistical mechanics, the entropy S of an isolated system at thermodynamic equilibrium is defined as the natural logarithm of Ω, the number of distinct microscopic states available to the system given the macroscopic constraints (such as a fixed total energy E):", null, "$S = k \\, \\ln \\Omega.$\n\nThis equation, which relates the microscopic details of the system (via Ω) to its macroscopic state (via the entropy S), is the central idea of statistical mechanics. Such is its importance that it is inscribed on Boltzmann's tombstone.\n\nThe constant of proportionality k appears in order to make the statistical mechanical entropy equal to the classical thermodynamic entropy of Clausius:", null, "$\\Delta S = \\int \\frac{\\mathrm{d}Q}{T}.$\n\nOne could choose instead a rescaled entropy in microscopic terms such that", null, "${S^{\\,'} = \\ln \\Omega} \\; ; \\; \\; \\; \\Delta S^{\\,'} = \\int \\frac{\\mathrm{d}Q}{kT}.$\n\nThis is a rather more natural form; and this rescaled entropy exactly corresponds to Shannon's subsequent information entropy, and could thereby have avoided much unnecessary subsequent confusion between the two.\n\n## Role in semiconductor physics: the thermal voltage\n\nIn semiconductors, the relationship between the flow of electrical current and the electrostatic potential across a p-n junction depends on a characteristic voltage called the thermal voltage, denoted VT. The thermal voltage depends on absolute temperature T (in kelvins) as", null, "$V_T = { kT \\over q },$\n\nwhere q is the magnitude of the electrical charge (in coulombs) on the electron (see elementary charge) with a value 1.602 176 487 ×10−19 C . Using the unit of electronvolt, the Boltzmann constant relating temperature to energy can be expressed as 8.617 343(15)×10−5 eV/K, making it easy to calculate that at room temperature (T ≈ 300 K), the value of the thermal voltage is approximately 25.85 millivolts ≈ 26 mV (Google calculator). See also semiconductor diodes.\n\n## Boltzmann's constant in Planck units\n\nPlanck's system of natural units is one system constructed such that the Boltzmann constant is 1. This gives", null, "${ E = \\frac{1}{2} T } \\$\n\nas the average kinetic energy of a gas molecule per degree of freedom; and makes the definition of thermodynamic entropy coincide with that of information entropy:", null, "$S = - \\sum p_i \\ln p_i.$\n\nThe value chosen for the Planck unit of temperature is that corresponding to the energy of the Planck mass—a staggering 1.41679×1032 K.\n\n## Historical note\n\nAlthough Boltzmann first linked entropy and probability in 1877, it seems the relation was never expressed with a specific constant until Max Planck first introduced k , and gave an accurate value for it, in his derivation of the law of black body radiation in December 1900. The iconic terse form of the equation S = k log W on Boltzmann's tombstone is in fact due to Planck, not Boltzmann.\n\nAs Planck wrote in his 1918 Nobel Prize lecture,\n\n\"This constant is often referred to as Boltzmann's constant, although, to my knowledge, Boltzmann himself never introduced it — a peculiar state of affairs, which can be explained by the fact that Boltzmann, as appears from his occasional utterances, never gave thought to the possibility of carrying out an exact measurement of the constant. Nothing can better illustrate the positive and hectic pace of progress which the art of experimenters has made over the past twenty years, than the fact that since that time, not only one, but a great number of methods have been discovered for measuring the mass of a molecule with practically the same accuracy as that attained for a planet.\" \n\nBefore 1900, equations involving Boltzmann factors were not written using the energies per molecule and Boltzmann's constant, but rather using the gas constant R, and macroscopic energies for macroscopic quantities of the substance; as for convenience is still generally the case in Chemistry to this day.\n\n## Value in different units\n\n1.380 6504(24)×10−23 J/K SI units, 2002 CODATA value\n8.617 343(15)×10−5 eV/K 1 electronvolt = 1.602 176 53(14)×10−19 J\n1.3807×10−16 erg/K\n\nThe digits in parentheses are the standard measurement uncertainty in the last two digits of the measured value.\n\nk can also be expressed with the unit mol (such as 1.99 calories/mole-kelvin), for historical reasons it is then called gas constant.\n\nThe numerical value of k has no particular fundamental significance in itself: It merely reflects a preference for measuring temperature in units of familiar kelvins, based on the macroscopic physical properties of water. What is physically fundamental is the characteristic energy kT at a particular temperature. The numerical value of k measures the conversion factor for mapping from this characteristic microscopic energy E to the macroscopically-derived temperature scale T = E/k . If, instead of talking of room temperature as 300 K (27 °C or 80 °F), it were conventional to speak of the corresponding energy kT of 4.14×10−21 J, or 0.0259 eV, then Boltzmann's constant would not be needed.\n\n## References\n\n• Boltzmann's constant CODATA value at NIST\n• Peter J. Mohr, and Barry N. Taylor, \"CODATA recommended values of the fundamental physical constants: 1998\", Rev. Mod. Phys., Vol 72, No. 2, April 2000" ]

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[ "Contents\n\nQuadratic equation is a second order polynomial with 3 coefficients – a, b, c.\n\nThe quadratic equation is given by:\n\nax+ bx + c = 0\n\nThe solution to the quadratic equation is given by 2 numbers x1 and x2.\n\nWe can change the quadratic equation to the form of:\n\n(x – x1)(x – x2) = 0\n\nThe solution to the quadratic equation is given by the quadratic formula:", null, "The expression inside the square root is called discriminant and is denoted by Δ:\n\nΔ = b2 – 4ac\n\nThe quadratic formula with discriminant notation:", null, "This expression is important because it can tell us about the solution:\n\n• When Δ>0, there are 2 real roots x1=(-b+√Δ)/(2a) and x2=(-b-√Δ)/(2a).\n• When Δ=0, there is one root x1=x2=-b/(2a).\n• When Δ<0, there are no real roots, there are 2 complex roots:\nx1=(-b+i√-Δ)/(2a) and x2=(-b-i√-Δ)/(2a).\n\n### Problem #1\n\n5x2+8x+3 = 0\n\nsolution:\n\na = 5, b = 8, c = 3\nx1,2 = (-8 ± √(82 – 4×5×3)) / (2×5) = (-8 ± √(4)) / 10 = (-8 ± 2) / 10\nx1 = (-8 + 2) / 10 = -6/10 = -0.6\nx2 = (-8 – 2) / 10 = -1\n\n### Problem #2\n\n4x– 7x + 1 = 0\n\nsolution:\n\na = 4, b = -7, c = 1\nx1,2 = (7 ± √((-7)2 – 4×4×1)) / (2×4) = (7 ± √(49 – 16)) / 8 = (7 ± 5,74) / 8\nx1 ≈ 1,59  ; x2 ≈ 0,157\n\n### Problem #3\n\nx2 + 2x + 5 = 0\n\nsolution:\n\na = 1, b = 2, c = 5\nx1,2 = (-2 ± √(22 – 4×1×5)) / (2×1) = (-2 ± √(4-20)) / 2 = (-2 ± √(-16)) / 2\n\nThere are no real solutions. The values are complex numbers:\n\nx= -1 + 2i\nx= -1 – 2i\n\nThe quadratic function is a second order polynomial function:\n\nf(x) = ax2 + bx + c\n\nThe solutions to the quadratic equation are the roots of the quadratic function, that are the intersection points of the quadratic function graph with the x-axis, when\n\nf(x) = 0", null, "When there are 2 intersection points of the graph with the x-axis, there are 2 solutions to the quadratic equation.\n\nWhen there is 1 intersection point of the graph with the x-axis, there is 1 solution to the quadratic equation.\n\nWhen there are no intersection points of the graph with the x-axis, we get not real solutions (or 2 complex solutions)." ]

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[ "", null, "# use properties of the Laplace transform and the table of Laplace transforms to determine L[f] f(t)=int_0^t (t-w)cos(2w)dw", null, "Kyran Hudson 2021-03-02 Answered\nuse properties of the Laplace transform and the table of Laplace transforms to determine L[f] $f\\left(t\\right)={\\int }_{0}^{t}\\left(t-w\\right)\\mathrm{cos}\\left(2w\\right)dw$\nYou can still ask an expert for help\n\n## Want to know more about Laplace transform?\n\n• Live experts 24/7\n• Questions are typically answered in as fast as 30 minutes\n• Personalized clear answers\n\nSolve your problem for the price of one coffee\n\n• Math expert for every subject\n• Pay only if we can solve it", null, "Bertha Stark\nStep 1\nWe have given,\n$f\\left(t\\right)={\\int }_{0}^{t}\\left(t-w\\right)\\mathrm{cos}\\left(2w\\right)dw$\nThis can be written as,\n$f\\left(t\\right)={\\int }_{0}^{t}tcos\\left(2w\\right)dw-{\\int }_{0}^{t}w\\mathrm{cos}\\left(2w\\right)dw$\nWe have to find,\n$L\\left\\{f\\left(t\\right)\\right\\}=L\\left\\{{\\int }_{0}^{t}t\\mathrm{cos}\\left(2w\\right)dw\\right\\}-L\\left\\{{\\int }_{0}^{t}w\\mathrm{cos}\\left(2w\\right)dw\\right\\}$\nStep 2\nWe know that,\n$L\\left\\{f\\left(t\\right)\\right\\}=F\\left(s\\right)$\n$L\\left\\{{\\int }_{0}^{t}f\\left(u\\right)du\\right\\}=\\frac{1}{s}F\\left(s\\right)$\nSo $L\\left\\{{\\int }_{0}^{t}t\\mathrm{cos}\\left(2w\\right)dw\\right\\}=\\frac{t}{s}\\left(\\frac{s}{{s}^{2}+4}\\right)$\n$L\\left\\{{\\int }_{0}^{t}t\\mathrm{cos}\\left(2w\\right)dw\\right\\}=\\frac{t}{{s}^{2}+4}$\nAnd\n$L\\left\\{{\\int }_{0}^{t}w\\mathrm{cos}\\left(2w\\right)dw\\right\\}=\\frac{1}{s}\\left(-\\frac{-{s}^{2}+4}{\\left({s}^{2}+4{\\right)}^{2}}\\right)$\nOn plugging these values in our equation , we get,\n$L\\left\\{f\\left(t\\right)\\right\\}=L\\left\\{{\\int }_{0}^{t}t\\mathrm{cos}\\left(2w\\right)dw\\right\\}-L\\left\\{{\\int }_{0}^{t}w\\mathrm{cos}\\left(2w\\right)dw\\right\\}$\n$L\\left\\{f\\left(t\\right)\\right\\}=\\left(\\frac{t}{{s}^{2}+4}\\right)+\\frac{1}{s}\\left(\\frac{-{s}^{2}+4}{\\left({s}^{2}+4{\\right)}^{2}}\\right)$\n\nWe have step-by-step solutions for your answer!" ]

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[ "• PEZenfuego\n\n#### PEZenfuego\n\n1. Homework Statement [/b]\nThis is a pretty simple calculus problem, but I can't figure out why my method fails.\n\nA ladder 20 feet long is leaning against the wall of a house. The base of the ladder is pulled away from the wall at a rate of 4 ft/s. Consider the triangle formed by the side of the house, the ladder, and the ground. Find the rate at which the area of the triangle is changed when the base of the ladder is 9 feet from the wall.", null, "a=?\nb=9 ft\nc=20 ft\nA=?\nda/dt=?\ndb/dt=4 ft/s\ndA/dt=?\n\nc=a^2+b^2\nA=(1/2)ab\n\n## The Attempt at a Solution\n\nWell, I am less interested in the actual solution and more interested in why my attempt at the solution did not work.\n\nI began by realizing that the rate of change of c would be 0, since the ladder will not become longer or shorter. Then I did this:\nc^2=a^2 + b^2\nc^2-b^2=a^2\na=(c^2-b^2)^1/2\n\nA=(1/2)ba\nA=(1/2)b(c^2-b^2)^1/2\nI am deriving with respect to b\ndA/db=(1/2)((db/db)(c^2-b^2)^1/2+(b(1/2(c^2-b^2)^-1/2)(2c(dc/db)-2b(db/db)))))\nand then simplify where (dc/db=0) and where (db/db=1)\ndA/db=(1/2)((c^2-b^2)^1/2+(1/2(b(c^2-b^2)^-1/2)(-2b)))\n\nAnd\n\n(dA/db)x(db/dt)=dA/dt so all that I have to do is multiply the answer I just found by the rate at db/dt which I already know. The answer I find is 26.65 which I have been told is wrong. Where did I go wrong and how do I go about solving this problem?\n\nLast edited:\n\n## Homework Statement\n\nThis is a pretty simple calculus problem, but I can't figure out why my method fails.\n\nA ladder 20 feet long is leaning against the wall of a house. The base of the ladder is pulled away from the wall at a rate of 4 ft/s. Consider the triangle formed by the side of the house, the ladder, and the ground. Find the rate at which the area of the triangle is changed when the base of the ladder is 9 feet from the wall.\n\na=?\nb=9 ft\nc=20 ft\nA=?\nda/dt=?\ndb/dt=4 ft/s\ndA/dt=?\n\nc=a^2+b^2\nA=(1/2)ab\n\n## The Attempt at a Solution\n\nWell, I am less interested in the actual solution and more interested in why my attempt at the solution did not work.\n\nI began by realizing that the rate of change of c would be 0, since the ladder will not become longer or shorter. Then I did this:\nc^2=a^2 + b^2\nc^2-b^2=a^2\na=(c^2-b^2)^1/2\n\nA=(1/2)ba\nA=(1/2)b(c^2-b^2)^1/2\nI am deriving with respect to b\ndA/db=(1/2)((db/db)(c^2-b^2)^1/2+(b(1/2(c^2-b^2)^-1/2)(2c(dc/db)-2b(db/db)))))\nand then simplify where (dc/db=0) and where (db/db=1)\ndA/db=(1/2)((c^2-b^2)^1/2+(1/2(b(c^2-b^2)^-1/2)(-2b)))\n\nAnd\n\n(dA/db)x(db/dt)=dA/dt so all that I have to do is multiply the answer I just found by the rate at db/dt which I already know. The answer I find is 26.65 which I have been told is wrong. Where did I go wrong and how do I go about solving this problem?\n\nI get the same answer you do, so I don't see what you are doing wrong." ]

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[ "# Mathematics Questions and Answers – Zeroes of Polynomial\n\n«\n»\n\nThis set of Mathematics Multiple Choice Questions & Answers (MCQs) focuses on “Zeroes of Polynomial”.\n\n1. If p(x) = 4x2+7x-8, then the value of p(2) is __________\na) 20\nb) 18\nc) 22\nd) 14\n\nExplanation: If p(x) is any polynomial, then p(a) means substituting x by a in a polynomial.\nTherefore, p(2) means, putting 2 in place of x in 4x2+7x-8.\nHence, p(2) = 4(2)2+7(2)-8 = 4(4) + 14 – 8\n= 16 + 14 – 8\n= 22.\n\n2. What is the zero of p(x) = x2+6x+9?\na) -3\nb) -6\nc) -9\nd) -2\n\nExplanation: If p(x) is any polynomial, then for “a” to be a zero of the polynomial p(x), p(a) = 0.\nTherefore, equating p(x) with 0, we get x2+6x+9=0\n(x+3)2 = 0\nx+3 = 0\nx = -3\nHence, -3 is the zero of the polynomial x2+6x+9 because p(-3) = 0.\n\n3. $$\\frac{(5x^3 + 2x^2 + x)}{x}$$ = __________\na) 5x3 + 2x2 + 1\nb) 5x4 + 2x3 + x\nc) 5x2 + 2x2 + x\nd) 5x2 + 2x + 1\n\nExplanation: $$\\frac{(5x^3 + 2x^2 + x)}{x} = \\frac{5x^3}{x} + \\frac{2x^2}{x} + \\frac{x}{x}$$\n= 5x2 + 2x + 1.", null, "" ]

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[ "Vous êtes sur la page 1sur 12\n\n# Rese arch Paper RP2058\n\n## Volume 44, Ja nuary 1950\n\nU. S. Department of Commerce\nNational Bureau of Standards\n\n## Transmission of Reverberant Sound Through\n\nDouble Walls\nBy Albert London\nThe transmission of reverberant so u nd through a double wall, which consbts of two\nidenticaL single walls coupled by an airspace, is investigated both theoretically and experimenta ll y . A t heory is developed, wh ich gives good agreement with expe rime nt. In order\nto compute the transmission loss of a double ,,\"all, it is necessary t.o knoll\" t he impedance Z ..\nof the single wall. Zw was det.ermined frolll experim e nts conducted on the single wall and\nin clude. t he effects of mass, dis ipatio n, a nd fl ex ural motion. The treatment sho \\l's that it\nis impossible to get a large imp ro vement in t ran mission loss [or a double wa ll relative to a\n~ ingl e wall under reve rbe\"a nt-sound fi eld conditions if the single wall is co nsidered to have\nonl y mass reactance. In add itio n, t he cu to mary normal incidence theory is totall y inadequate in exp la ining the behavi or of a double wall in a reverberant-sound fi eld .\nFor double walls having air co upling o nl y, ver .\\' shallow a irspace can produce appreciable in creases in transmissio n los o ver a si Il gle \\mll. An absorbe nt material, when inserted\nin the airspace, produces la rge improvements only when t he mass of the walls is relath\"e ly\nlight and has but little e ffect for heavy walls. Honeycomb or other Jl onabso rbe nt celluLa r\n. structure ' having no cell walls in a direction normaL to t he wall faces do not result in an\nincrease in transmission loss. Air-co upled waiLs having no soli d sound- co ndu cting paths\nbetween individual septa are extremely e ffe ctive so und insulators as compared to conventional do uble-wa ll constructions. The theory indicates that a large improvemellt ill the transmission LORS of a double wall can be obtained b)' using as components single \" 'a lls \\\\\"ith hi gh\ninternal dissipation.\n\nI. Introduction\nIn a previous paper 1 the transmission of reverberant sound tlu'ough homogeneous single\nwalls was investigated theoretically and experimentally. The attenuation of an obliquely incident plane sound wave upon transmission through\na single wall was computed, and using the customary reverberant sound field statistics the attenuation was integrated over all angles of incidence\nto give the average transmission loss. A similar\ntechnique is employed in this paper in studying\nthe transmission of sound through a double wall\nconsisting of two identical single walls . The\nmaterials comprising the double walls are the\nsame as were used in the single walls, i. e., aluminum, plywood, and plasterboard. From the exI\n\n## A. London , Transmission of reverberant sound tbrougb single walls, J\"\n\nN BS 42, 605 (1949) RP1998.\n\nR ~sed rcb\n\n## Sound Transmission Throu<;1h Double WaIls\n\nperimental results obtained in RP1998, an expression for the wall impedance, Z w, for each\nmaterial was determined, this ex'})res ion containing terms that include the effects of the mass, dissipation or resistance, and flexural motion of the\nwall. This value of Zwis used in the double wall\ntheory to compute the transmission loss for a\ndouble wall.\n\n## II. Transmission Through Double WaIls\n\n1. Attenuation of a n O bliquely Incident Wave\n\n## In figure 1, an oblique plane wave is incident\n\nat an angle 0 on the first partition. As a result\nthere exist 'in the three airspaces formed by the\ninfinite double partition: an incident and reflected wave in space (1 ) i a standing wave in\nspace (2), consisting of a wave moving to the right\n\n77\n\n## constant phase. The justification for such an\n\nassumption has b een given previously in section\n2 of RP1998, t itled Basic Assumptions.\nThe ratio PI/P i will b e computed for the sam e\nvalue of y , so tha t t his coordinate will not appear\nin th e calculation. Thus, since the particle\nv elocity is proportional t o the pressure gradient,\nthere results from the continuity of the x-componen t of particle velocity at x = O\n\n(2)\n\nand at x= d\n\nx=d\n)(=0\n1. Geometrical ?'elation between in cident and\nrefl ected wave in space (1 ); standing wave in airs pace of\ndouble wall , space (2); and tmnsmitted wave in space (3).\n\nF IGU R E\n\n## and one moving to the left ; and a transmitted\n\nwa ve in space (3) . It is d esired to know the ratio\nof the transmitted pressure wave amplitud e P t to\nthe incident pressure wave amplitude P i, where\nthe pressures in each airspace are given by eq 1.\nPI = P ieiwt-ik(X cos O+v sin 0)\n\nP reiwt-ik( -\n\n.1:\n\nx::S O\n\n(1 )\n\nx?; d\n\nwher e\n\n## If PIO and P20 are the pressures acting on the\n\nleft and right side, r esp ectively, of the panel at\nx= o, P2!l, and P3d the pressures acting on the left\nand right side of the panel at x = d , the equations\nof motion for each panel are\n(4)\n\nand\n(5)\n\n## wher e Z w is the m echanical impedance per unit\n\narea of the two identical walls, and iJ is the\nv elocity of the wall in the x-direction. Furthermore, since the wall v elocity must b e the same as\nthe x-component of p article velocity of the air at\nthe wall, there results\n\n. =i (OPl) = cos () (P\n\n7] x~ O:>.\n\nPW\n\n_ i\n\n7] x~d - --\n\nPW\n\nw= 27r X frequen cy\n\nk = 27r/}..= w/C, }.. b eing the wavelength, C the velocity of sound in ail'.\nThe four ratios Pr/P i , P+/P i , P _/P i , and P t/P ;\nmay b e d etermined from the two boundary condi tions, the continuity of the x-compon ent of\nvelocity at x = O, and x = d , and the two equa tions\nof mo tion, one for each partit ion. In deriving\nt h e equ ations of motion it is only n ecessary to\nconsider a small area of th e panel upon which th e\nprojection of th e wave fron t h as practically\n\n78\n\nv X\n\nx ~O\n\n(OP3)\n-\n\nOX x~d\n\npC\n\n_ P ) iwt-ikll Bin 0\nire,\n\n_ cos\n()p tei w t-ik(d\n-\n\nCOB\n\n0+11\n\nBi n\n\n0)\n\npC\n\n(6)\n\n(7)\n\n## Substituting eq 1 and 6 into 4 causes th e latter\n\nequation to reduce t o\n(8)\n\n## and similarly eq 5 becomes\n\nP _e\niR\" - P te - i{J -- Z w cos ~ P te -i{J\nP +e-iR+\n\"\npC\n\n(9)\n'\n\nwher e\n{3 = kcl cos ().\n\n(10)\n\nJournal of Research\n\n## Letting (J= Oo, r educes eq 16 to\n\nLet\n/' =\n\nZ w cos (J\n2pc\n\n(11)\n\nI A I ~, m= 1 +\n\n4a 2 (cos b- a sin\n\nW,\n\n(1 7)\n\nwhere\n\n## then solution of the four simultaneous equations\n\n(2,3, S, and 9) results in the following expressions:\n\n(12)\n\nby Schoch 2.\n\n(13)\n\ntransmitted when\n\nand also\n\n## It is of interest to observe that eq 13 is precisely\n\nthe expression for the ratio of incidcnt to transmitted amplitude for a single wall given by eq 1.1)\nof RP 1995, inasmuch as the boundary condition&,\ni. e., the existence of an incident, r eflectcd, and\ntransmitted wave, are the sam e as that for a\nsingle wall.\nEquation 12, which is of primary interest to this\ndevelopment, can be tested for agreement with\nthe solution, eq 13, for the attenuation of a single\nseptum. Thus, if el = O, the double wall becomes\na single wall having an impedance 2Zw ' Of\ncourse, this is strictly a mathematical experiment,\ninasmuch as was shown in the previous paper (see\nfootnote 1), the resistive and r eactive components\nof Zw are not twice as great when a single wall's\nthickness is doubled . Setting el= O in cq 12\nr esults in\n. 2Z w') cos (J ,\nA d = O-- 1+ 2/' -- 1 +\n~pC\n\n(14)\n\n## in agreement with eq 13.\n\nAlso, it is possible to compare eq 12 with the\nresults obtained by previous investigators 2 for'\nthe special case when the wave is incident normally,\ni. e., (J = O, and the wall impedance, Z w, is a pure\nmass reactance only, given by eq 2.1 of RP199S or\n\n## where m is the mass of the wall p er unit area.\n\nNow, from eq 12, if the wall acts as a mass\nreactance only, it is r eadily shown that\n\n## IA I~. =I~iIt 2 _. = 1+4a2 cos2(J(cos ,8 -\n\na cos (Jsin{3)2,\n\nZ W- llol m\n\n(15)\n\nwhere a = wm/2pc.\n\n(16)\n\n## A. Schoch, Die physikalischen und tecbllischcn Grundlagcn der Schall\n\ndammullg irn Bauwesen. p. 86 (Hirzel, L eipzig, 193i ).\n2\n\n(18)\n\n(19)\n\n## tan {3 = l / (a cos (J).\n\n(20)\n\nor when\nFor cases where {3 is small (el}..) tan ,8 may be\nreplaced by {3. Using eq 16, there results an\nexpre sion for the frequ ency le, for which a wave\nincident at angle (J, will be perfectly transmitted\nin the case where each wall acts as a pure mass.\n\n10\n\n271' cos (J\n\n(2pC2\n\nmel\n\n)t.\n\n(21)\n\n## The value of 10 for normal incidence is 10, the\n\ncharacteristic frequency for the air-mass sandwich ,\n1. e.\n_ 1 (2pC2 )Q\n(22)\n10 - 271' mel .\n\n## 10 is the frequen cy for normally incident waves\n\nfor which the mass r eactan ce of the panel is\nexactly equal to the stiffness reactance of thc airspace. It is also the lowest frequency for which\nthe attenuation of the panel is zero. At frequencies above 10 there will be some angle of\nincidence for which zero attenuation will occur.\nSince in a r everberant sound field, energy is\nincident from all directions, the attenuation\nmeasured in a reverberant field will n ever reach\nzero. For frequencies above 10 there will be some\nwaves that will be totally transmitted, con sequ ently resulting in a diminution of the transmission loss of the panel as compared to that\npredicted by the normal incidence theory.\nSince l /a decreases with increasing frequency,\nat high enough frequencies eq 20 may be written\nas\ntan ,8 = 0,\nand\n,8= n7r, n = l, 2,3, - . \"'\n\n79\n\n## When R = O, 29 reduces to an equation analagous\n\nto 15 with a replaced by p, i. e.,\n\nwhich results in\nd cos O=n\"A/2, n = l , 2, 3,\n\n. . \"'\n\n(23 )\nI A I ~ _ o = l + 4p 2V2(COS bv-pv\n\nas the expression for the frequencies , or wavelengths, at which higher-order minima occur.\nHere too, for a reverberant field , considerations\nsimilar to those discussed in connection with fo\napply.\nEquation 12, which gives the attenuation, A,\nfor a double wall may readily be compared with\nthe attenuation, a, for a single wall given by eq\n1.6 of RP199S or its identity eq 13. Since\"y is\nordinarily much larger than unity, a~\"y, and\n(24)\n\n## The term containing a 2 , multiplied by a factor\n\n(l-e-2i~) which is never greater than 2 in absolute\nvalue and which depends on the spacing between\nthe two walls, therefore, represents the chief difference in attenuation caused by a double wall rela tive to that of a single wall .\nAs shown in RP199S the most general expression\nfor the wall impedance is given by\n\nor\n\"y\n\nZ,o2pc\ncos\n\n(1 -~ sin\n\no}\n= R + . cos 8 (P.)\n1 -r. sm 0 ,\n\nZ w= c;: O+iwm\n0\n\n~a\n\n(25)\n\n## IA I2= IA I%_o+ 4R(R + 1) {I +\n\n[R (R + 1) + 2p 2V2] sin 2 bL'- PV sin 2bv },\n\n(27)\n\n(28)\n\n## IA IZ= IA I71 _0+ 4R(R+ 1) {(cos be-pv sin bv)2+\n\n[p2v2+ R(R+ 1)+ 1l sin z bv}.\n\n(32)\n\nInasmuch as the second member on the righthand side of eq 32 is always positive, it will be\nseen that the attenuation of a double wall , each\ncomponent of which has dissipation or resistance,\nis always greater than for the case in which each\ncomponent is dissipationless.\n2. Average Attenuation of a Double Wall in a\nReverberant Sound Field\n\n## In accordance with the reverberant sound field\n\nstatistics discussed in section 3 of RP199S, if Ta\nis the ratio of the total energy transmitted by the\ndouble wall to the total energy incident on the\nwall, we get from eq 3.1 of RP199S and eq 29\n\ndv\nIA lz'\n\n'1 V\n\n(33)\n\nwh ere v= cos o.\nThe integral in eq 33, unfortunately, is highly\nintractable. It was not possible to evaluate it\nother than by numerical integration. This has\nbeen done for a number of different constructions\non which experimental r esults were obtained and\nwill be discussed in section II, 3. However, for\nthe special case where it is assumed that each\nsingle wall has a mass reactan ce only, the integral\nhas been computed 3 for a wide range of values\nin a systematic manner. For the mass r eactance\ncase we may set R = O and f Jjc= O, whence, 29\nred L1 ces to 15 and 33 assumes the following form\n\n(1\nv dv\nTd= 2 Jo 1 + 4a 2v2 (cos bv-av sin bV)2'\n\n## For IA I2 there results\n\n(31 )\n\nor an equivalent form is\n\nTa=2\n\n(3 0)\n\n(26)\n\n## where R =r!pc , the resistance of the wall in pC\n\nunits, and f c= the critical frequency above which\nflexural waves will appear in the wall. The\nparameters Rand ic for different materials were\ndetermined from the experimental observations\nmade in RP1998. Substituting eq 26 into 12\nresults in\n\nsin bv)2.\n\n(34)\n\n(R =O,Jlf,=O)\n\n## IAI2= 1 + 4[R(R+ 1) + p 2V2 ]\n\n+ 4 sin 2 bv ([R (R + 1) + p 2v2J2 _ p 'U 2 }\n- 4pv si n2bv{ R (R + 1) + pV }.\n\n80\n\n(29)\n\n## 'We are Indebted to G. Blancb and 1. Stegun of the National Bureau of\n\nBtandards' Computation Laboratory for carrying out these integrations.\n'riley used a combination of numerical integration and aualytic representations for different regions of a certain parameter to evaluate the integrals.\n\nJournal of Re5earch\n\n---- 1\n\n## It is convenient to introduce two nondimensional parameters into eq 34, namely,\n\n60\nI< =0.00 1\n\n50\n\nb pd\n}J-= -2a =-=-m '\n\n(35)\n\n./\n\n.0\n\n\",' 40\n\n'\"'3\n\nand\n\nx-i=~'\n\n(36)\n\n15\n\n(:37)\n\nu = 2av\nand 34 becomes\n\n30\n\n/\"\n\n~ 20\n\n<{\n\n0::\n\n.-\n\nI-\n\n10\n\n\"\n\n( R ~ O.J/J, ~ O)\n\ndu\n-u- .- -)\"2\ncos }J-u- 2 sm )lU\nU\n\n1 +u-\n\n? (\n\n(38)\nIt is of interest to compare the transmission\nloss, 10 log (l!Td), computed from eq 38 with\nthat which one obtains for a single wall when it is\nassumed that the wall has a mass reactance only.\nAn expression for the latter transmission loss is\ngiven by eq 3.2 of RP 1998. If we replace a2 by\nits equivalent expression in terms of X and )l, i. c.,\n(39)\neq 3.2 of RP 1998 may b e written\n\n(1)\n\nX2)].\n\n## ;\\'2 - 1010g [ In ( 1+2)l\n\nTL = 1010g -; = 1010g ~)l\n\n(40)\nIn figure 2, the computed transmission loss for a\nsingle and double wall, having mass reactance\nonly, i. e., 7,w= iwm, have been plotted for three\ndifferent values of the parameter}J-. It will b e\nseen that on this basis the predicted improvement\nof a double wall over a single wall is small and\nin fact may actually b e n egative. This astonishing behavior results from the fact tllat for a double\nwall there is some angle of in cid ence for which the\ntransmission is perfect and in the integrated\neffect of all angles of in cidence, this minimum\nSound Transmission Through Double Walls\n\n.-'-\n\n\"\n\no ~0.2\n\n0.5\n\n~ )- \"\n4'\n\nl,.t<'\n\nJ;c-\n\n// t/\n\n.~ 1>--1<\n\n\" :-:.:1.--\n\n/'\n\nV--\n\n)<' 0.01,\n\nU)\n\n,,\"~\n\n0.1\n\n\"\n\n,,\"\n1.0\n\n2.0\n\n5.0\nx f / f 0\n\n10.0\n\n20 .0\n\n50.0\n\n100.0\n\n## 2. Comparison between theoretical tmnsmiss'i on\n\nloss of a single wall to that of the corresponding double wall\nin a reverbemnt sound field when wall is considered to be a\npure maSS reactance.\n\nFIGURE\n\n__ Ri ngle\n\n}J- J'x,(2f;,.\nTd = X 2 0\n\n\"A\n\n'\"~\n\n## wherefo is defined by eq 22 . Thus,}J- is th e ratio\n\nof the mass of air in the airspace to the mass of\none wall, wb ereas X is the ratio of the frequency\nof the sound wave to the freq u ency fo], which a\nwave, normally incident on a doubl<,; ,wall possessing mass reactance only, will be perfectly transmitted. In addition we let\n\n\"\n\nl7 1,- /\n\n'0\n\n## tran smission loss swamps out the effect at other\n\nangles of incidence. In the case where a r esistive\nterm is includ ed in the impedance, there is no\nangle for which the transmitted wave is not\nattenuated. Hence, it is not sufftcient to treat\neach component of the double wall as a pure mass.\nWith regard to figure 2, it is well to poin t ou t\ntha t the small max ima and minima indicated in\nthe double wall curves are a resul t of the higher\norder minima, which are approx imately given by\neq 23. Values of tllO integral (eq. 38) were computed for X = 0.2 , 0 .5, 1.0, 2.0, 5.0, ] 0.0, 20.0,\n50.0, and 100.0 for )l = 0.1, 0.08, 0.06, 0.04 , 0.01,\n0.006, 0.004, 0.002 , and 0.00] . Th is information\nhas not been reproduced he re btl t is available\nupon r equ est.\n3 . Comparison Between Experimental and\nComputed Results\n\n## Fig1..ll'e 3 is a schematic drawing showing the\n\narrangement of the double wall in the ound t ransmitting opening. Each leaf of the double wall\nwas made separately, a pract.ical procedure inasmuch as the concrete walls of the test chamber\nare isolated f rom each other by a 3-in. airspace\nexcept for the common foundati.on of the walls.\nThus, there are no solid so und-co nducting bridges\nb etween the two faces of the double wall , a circumstance that allows a close approximation t.o\nthe conditions set down in the theory. The experimental m ethod utilized in making the trans-\n\n81\n\nSINGLE PANELS OF\nDOUBLE PARTITION\n\n## minimum is based on the assumption that the wall\n\nhas zero resistance. As a matter of fact, in this\nparticular case, the value of R is such that no\nnoticeable minimum occurs in the Ta integral calculations for cri tical values of v 01' (J corresponding\nto eq 19. In particular, from eq 15, 19, and 32\nwe see that IA I2 for v or cos (J satisfying eq 19\nbecomes\nIA I;,= 1+ 4R(R + 1) sin 2 (\n\n(4 1)\n\nSOURCE\nROOM\n\nRECEIVING\nROOM\n\nCONCRETE WALLS\nOF TEST ROOM S\n\nFIG URE 3.\n\nopening.\n\n## mission loss measurem ents is that described as the\n\nusual m ethod in NBS R esearch Paper RP1388 .4\nThe next figure, figure 4, shows the results\nob tained on a double wall consisting of single walls\nof ?~4 -in . aluminum separated by a 3-in. airspace.\nHere m = 0.12 g/cm 2 , and the mass and thickness\nof the single wall are such that the critical flexural\nfrequency j e, is approximately 30,000 c/s. Thus,\nj 1fc.~0, and no fl exural effects will be observed .\nFor reference purposes, the r esults obtained in the\nsingle wall case are shown in the lower part of the\nfigure.\nThe best fit for the single wall case was obtained\nwhen R = 2.16. The sam e value of R was used\nfor the double wall calculations, whi ch were carried\nout by a nu;;:';erical integration of eq 33. It will\nbe noted that eq 33 as opposed to eq 38 predicts\na sizeable improvement in transmission loss of a\ndouble wall over a single wall.\nAccording to eq 22 there should be a minimum\nin the transmission loss curve at jo = 279 c/s, corresponding to the frequency for which the mass\nreactance of the wall is exactly equal to the stiffness reactance of the airspace . However, this\nA. London , J. R esearch\n\n82\n\n'---\n\n~BS\n\n## Since IA I%_o= 1, p = a(i. e.j /fc= O) , and the critical\n\nvalue of v, say vo, is given by v~= (ab) -1 from eq\n20. From eq 3.5 a/b= 1/ (2!1-) and for this wall\n!1- = .075 , so that a/ b= 6.66. Since R = 2.16 , we\nget from eq 41\nIAI; = 57.4.\nThus, th e minimum value is mu ch larger than 1,\nwhich is the value that would result if R = O.\nFurthermore, it will b e noted that eq 41 predi cts\nthis sam e minimum value of IAI2 independent of\nfrequency provided (J is such that eq 19 is satisfied.\nThis sam e minimum will occur at frequencies\nabove fo, thus tending to depress the natural increase in transmission loss resulting from ma ss\nlaw behavior.\n45\n40\n\n35\n30\n\n/\"\n\n.c\n\"0\n\n- 25\n\n20\n\niii\nIf)\n\nE?- V\n\n~/\n\nIf)\nIf)\n\n...J\n\n,,\n\n15\n\n,,! '\n\n\"\"\"\\:):\"- - '0\n\n''0'\n\n10\nz<>: 30\n\nIf)\n\na:\n....\n\n25\n\n./\n\n20\n15\n10\n\n,\n'-\n\n50\n\n'\"' , , ,\n\n,\n''Q''\n\n100\n\n200\n\n- 500\n\n8\n1000\n\n2000\n\n5000\n\nFREQUENCY. CIS\n\n## 4. Comparison of computed and experimental\n\ntransmission losses of Ys. in. single and double alumin1l1n\nwa!ls.\n\nFIGU RE\n\n## A. Double wall; _ . computed; . . .. experimental ; d=3 in.; 10=279 cis.\n\nB, Single wall; _ , computed; ..... experimental; R = 2.16; m=O.12 gcm-2\n\nJournal of Research\n\n_I\n\n55\n\n50\n.0\n\n\".45\n\nL~\n\nUl\nUl\n\n040\n\nb.~\n\nv\n/ [P V,....\n\n...J\n\nQ 35\n\nUl\nUl\n\nt-\n\nt:1/ If l/ .\n\nz 30\n\n~ 25\n\n~~ -\n\n20\n\n~~ ~ ~~\nv\n~ P%\n..............\n\n--\n\np..~~~\n\n\"II\n\n15\n\n50\n\n200\n\n500\n1000\nFREQUENCY, CIS\n\n2000\n\n5000\n\n## 5. E:Lperimental transmission loss results f01' a\n\nseries of double wall::: of }~ -in plywood with airspace\nvarying from % to 12 in .\n\nF I G U RE\n\nsin~le\n\n## With regard to th e reliability of the compu ted\n\nyalues relative to the experimental values, it is\nprobable that they agree within the accuracy of\nexperimental observations for frequencies below\n500 cis. Above that frequency, it is to b e noted\nthat the computed curve deviates from th e exp erimental curve in the sam e direction for both the\ndouble and single walls. In fact, th ese two curves\nintersect at about the same frequen cy for bo th th e\ndouble and single walls . Thus, the discrepancy\nbetween computed and experimen tal curves in the\ndouble wall case is apparently due to the imperfect\nfit obtained for single walls and, furthermore, th e\neffect of this imperfect fit seems to b e magnified\nfor double walls.\nFigure 5 shows the expcrimental results obtained\non a series of }~-in. plywood walls in which th e airspace was varied from % to 12 in. , together with\nth e transmission loss obtained on the single wall.\nSeveral pertinent observations may b e made concerning the general nature of these experimental\nresults. First, it will be seen that even for the\n%-in . airspace there is a considerable range of frequencies for which there is a significant improvement of the double wall over the single wall.\nSecond, all of the curves have a minimum in the\nvicinity of 2,000 cis. As was pointed ou t in\nRP1998 , the minimum in the single wall TL,\nwhich also occurs at this frequency, was due to a\nflexure wave having an j,= 1,885 c/ . Consequently, the effect of flexure shows up in th e\ndou ble wall case at th e same frequ ency. Third,\nSound Transmission Through Double WaIls\n\n## for large airspaces (6 to 12 in.) and for frequencies\n\nin the range from 400 to 1,000 e/s th e transmission\nloss of a double wall approachcs a value tha t is\ntwice that of the single wall, showing that th e\nsecond wall is almost entirely decoupled from th e\nsingle wall for this frequen cy range.\nIn attempting to compute t he transm ission loss\nof th e double plywood walls, we chose the %-, 3-,\nand 12-in. airspace cases for detailed analysis. As\nwas pointed out earlier, any error in fit between\ncomputed and experimental r esults for the single\nwalls would result in much larger errors in the\ndouble wall case. In figure 6 we r eproduce the\ncomputed and experimental data for the single\nwall . Using R = 8.3, results in a computed curvl'\nthat agrees well up to 1,500 ci s but gives larger\nthan experimental values above this frequency.\nHowever, if R = 5 is used, the computed resu lt will\nagree with the experimental at j = 2,048 cis, but\nwill still be too high at 4,096 cis. R = 1. 8 at\n4,096 cis gives much b etter agreement, whereas\nR = 1.0 is a pOI'fect fi t. These data indicate tha t\nR decreases with increasing fr equency.\nThe n ecessity for using an accurate value of R\nis illustrated best by figure 7. Here the tr ansmission loss of th e single and double 7~-in. plywood\nwall h as been computed as a function of R for\nj = 4,096 cis. A variation of R from 1.0 to 8.0\ncause a change in loss of 7 db for the single wall,\nwher eas in the double wall case a 20-db change\nresults. In fact, it would seem to be somewhat\neasier to determine R from the double walll'esults\nthan from the single wall measurements. The\nvalue of R = 1.8 used for further computations at\nj = 4,096 cis was selected because it gave exact\nagreem en t with experiment for a 3-in. airspace\n.0\n\n\".\nUl\nUl\n\n40\n35\n\ng\nz\n\n30\n\n:ii\n\n25\n\n,\n'0-\n\n:il\n~ 20\n\na:\n\nI-\n\n~:.x\n\n--~ ~~\n\n15\n\n50\n\n100\n\n200\n\n1000\n500\nFREQUENCY, CIS\n\n2000\n\n5000\n\n## FIG U RE 6. Effect oj varying R on computed transmission\n\nloss fOl' a %-in. plywood single wall.\nDotted broken line corresponds to experimental t ransmission loss. 0 ,\nExperimental; ~ , compu ted, R = 5; . , computed, R=1.8; at 4.096 CiS R=l.O,\ncomputed, coincides with experimental point. _ , Computed R=8.3;\n( ,= 1.885 CIS .\n\n83\n\n60\n\n50\n\n..-- -\n\n<J>\n\n040\niii\n<J>\n\n<J>\n2\n\n<:\n\nII:\n\n....\n\n........\n\n- - r-\n\n--- --- --\n\n--\n\n30\n\n'\"\n20\n\n## 7. Val'iation of computed tmnsmission loss with\n\nR /01' double and single wall of 7f-in. plywood at a frequency of 4,096 cis.\n\nFIGURE\n\n, Double wan 3in. airspace; .... , single wall; / = 4,0% cis; j,= 1,885 cis.\n\n## double waIl. According to this treatment it should\n\nbe possible to obtain significant improvement in a\ndouble wall by building into each of the single\nwalls a layer of attenuating material.\nFigure 8 is a comparison between the computed\nand experimental transmission losses using R =8 .3\nfor frequencies up to 1,024 cis, R = 5.0 at 2,048\ncis, and R = 1.8 at 4,096 cis. The solid lines are\ndr awn through the computed points, the dotted\nthrough the experimental points. It is believed\nthat there is reasonably good agreement considering th e complexity of the problem and, in particul ar, the computations. For example, the\npoint at f = 4,096 cis for the 12-in. airspace case\nrepresents the results of 40 pages of calcul at,ions.\nIn figure 9, the integrand of the Td integral, eq 33\nfor th is point, is plotted as a function of v. Very\nsharp half wavelength maxima corresponding to\nv=n7rlb and covering a range of variation of several orders of magnitude are evident. In addition, there is a less sharp peak due to flexure. If\none compares the area under the peaks, one finds\nmost of the area exists in the neighborhood of the\nflexure angle thus showing the importance of this\neffect.\nFigure 10 is another representation of the data\nshown in figure 9. Here, the Td integral, eq 33\ninstead of being integ),fl.ted from v=O to V= 1.0,\nis integrated from a variable lower limit VI to\nV= ] .0. The quantity 10 log (l /T') so clefined,\n\n84\n\n## therefore gives the transmission loss that would\n\nresult, if, for some reason 01' other, waves incident\nat angles greater th an 01 corresponding to VI were\nnot transmitted. Thus, for O<vl<vl=7rlb, grazing incident waves are first excluded, but the\nangle, or VI, corresponding to the first maximum\nnearest grazing incid ence, wou ld be allowed. It\nwill be seen that not until th e fifth maximum is\nexceeded is there a change in loss. This is because the first five maxima do not contribute\nanything to the integr al, since they are so sharp.\nW'hen the flexure angle is excluded, however,\nthere is a large jump in loss because there is a\nlarge transmission of sound energy resulting from\nthis cause. As we approach more closely to the\nconditions where the angles of incidence are restricted to the neighborhood of normal incidence,\nthe transmission loss increases greatly.\nAt the lower frequencies fewer angles at which\nmaxim um transmission occm's are observed. At\nthe lowest freq uencies none may occur at all.\n55\n\np.,\n\n50\n\n,,\n,\n\nt>- .....\n\n45\n\n40\n\n~\\\n\n/v\n\n35\n\n~\np'\n\n~p\n\n-' V-\n\nQ.\n\n,../\nA\n\n30\n.0\n.\"\n\n45\n\n~~\n\n-I:>:\n\n.; 40\n\n..J\n\n;:;rcr\n\n35\n\n~30\n\n:i\nI/)\n2\n\n<{\n\n25\n\nII:\n\n/J\n\n\"\"'d'\n\n.... 20\n\n, Q.\n\n-'1=>-- -\n\nI/)\n\n45\n\n40\n35\n\n~'\n\n,P'\n\n30\n\n:f/\n\n/ J\n\n~ f--?--'\n\n25\n_ -0.... ... ...\n\n20\n100\n\n50\n\n,0\n\n' 'tf\n200\n\n500\n\n1000\n\n2000\n\n50 00\n\nFREQUENCY . CI S\nFIG URE\n\n8.\n\n## Comparison between computed and experimental\n\nl'esults for Yz-in. plywood double wlllls.\n\n## Solid line, compnted; broken dotted line, experimental. In the compn\n\ntations, R=8.3 for frequencies up to and inclnding 1,024 cis, R=5.0 for /=\n2,048 cis, and R=1.8 for [=4,096 CiS. A, 12in. airspace; B, 3in. airspace; 0,\n~'in. air space.\n\nJournal of Research\n\nc-\n\n## cal'l'ied ou t , principally because of th e ted ious\n\nnatm e of such calculations.\nFigures 12, 13, and 14 show the experimental\nres ults obtained on double walls consisting of 7h\n1-, and 2-in . plasterboar d single walls. For\ncomparison purposes th e experimental and computed r esults obtained on the corresponding single\nwalls a.r e also shown on the figures. In the H-ill.\nplasterboard case it will b e seen that the double\nwall exp eriments t end to confirm the selection of\nj.= 4,096 cis as preferable to j c= 2,048 cis. T h e\n\nONSET OF\nFLE XURE\n\n110\n\n100\n.0\n\"0,\n\nT~ 2fl~\n\n90 r-----\n\n(f)\n(f)\n\nVi\n\n80\n\nQ\n(f)\n(f)\n\n~l~\n\n70\n\nr--\n\n~\nI L 7IT\n\n(f)\n\n~ 60\n\n>-\n\n50\n\nIT\n\n0. 2\n78.5\n\nF I G 1' R E\n\nI\n\\II\n\n0.0\n\n02\n\nb\"\n\n9.\n\n0.4\n\n0.6\n\n~\nb\n\n4\"\n\nIr b\n0.4\n\n;'ill\n\nb\"'l\n0.6\n\nL 6;\n\nONSET OF\n.,/\" FLEXURE\nO.B\n\n1.0\n\n36.1\n\nV,\n\nlOx 10-9\n\nFIG U RE\n\n0.0\n90\n\n1IT\nb\n\n3 IT\n\nb1\n\n40\n\n_L\n\n2 rr\n\n0.8\n\n.21!\nb\n\n1.0\n\n66.4\n53.1\ne\" DEGREES\n\n## Anothel' l'e p l' e~erltotion oj cW've of fi gw 'e 9.\n\n]0.\n\nT he fi gure s hows the variation of 10 log ( l iT') with 0, where tbe integra tiou\noccurs from a ya ria hle lower limit of integra tion, 0\" to 0= 0 . As the wave\npackets in the re\\\"erbcrant soundfield are con fin ed to a cone for which 0 , is\ndecreasing, a sudden in crease in transmission loss occurs when the angle of\nin cidence correspondi ng to fl exure is excluded, showing t hat most of the\nt ransmiss ion of sound occurs as the result of fl ex ural waves.\n\n!JI.\n\n100\n\nf = 4.096 ci s.\n\nJOI'\n\n## H ere v= cos e, where 0 is the an gle of incidence of t he sound waye. Very\n\nsharp transmission maxima occur at v=n\"lb or when d cos 0= nl\\/2, wbere d\nis airspace tbickness. In addition, a less sharp maximum occurs at the an gle\no( incidence corresponding to t he occurrence of fl exural waves in t be wall.\nDouble wall, 12\"in. airspace, 7I\\-io. plywood ; /= 4,096 cis; / , = 1,885 cis; R = 1.8.\n\n90\n.0\n'0\n\nBO\n\n(f)\"\n(f)\n\n,/\n\n70\n\niii\n\n(/)\n\n60\n\n...-?\n\nf-\n\n(f)\n\n## This is shown in figme 11 , which is a graph similar\n\nto figm c 10, but indicates the valu e of the r'\nintegr al plotted in decibels for other frequen cies\nfor th e same 12-in. airspace double wall.\nIn this section \\\\'e consider some additional\nexp erimental r esult ob tained on double walls, for\nwhich , however , no analytical compu tations were\nSound Transmission Through Double WaUs\n\na:: 50\n\n----\n\n40\n\nD\nA\n\n30\n0 .0\n\nFIG u R E\n\n0. \\\n\np-\n\nI-\n\n3-\n\n(f\n\n- f--\n\n0.2\n0.3\n0.4\n0.5\n0.6\n0 .7\nLOWER LIMIT OF INTEGRATION ,vQ\n\nO.B\n\n0 .9\n\n## 11 . P lot of 10 lo g (l iT') , as in fi gure 10 ,Jor a ~-in.\n\nplywood dotlble wall J OT val'ious jTequ encies.\n\n85\n\n60\n\n--- fl,\n\n55\n50\n.D\n\"0\n\n45\n\n'\"0\n\n40\n\nz\n9\n\n35\n\nof)\n\n)f\n\ntl .\n\nof)\n</)\n\n<1\n\n' I=>\n\n25\n\n~~\n\n20\n\n~~ P<\n\nex:\n\n.,\n\nfC ' 4096~\n\n'\n\n30\n\nf-\n\n,P\n\n.J\n\ni</)\n\n## wall, v ery large transm ission losses result even ,\n\nwith r elatively ligh t weight walls. T hus, a 45-db\nloss may be ob tained for a weigh t of only 4.2\nIb/ft2 . T h e sit uation rapidly worsens if solid\ncoupling between each componen t exists. Comparing the last four entries in the table it will b e\nseen that all have approx imately the sam e weigh t .\nT h e 2-in. air-coupled wall, however , is som e 20\ndb b etter than th e stud-coupled wall ; some 10 db\nbetter wh en the st uds are stagger ed so that\ncouplin g exists onl'y due to a top and bottom\n\nlP P-\n\nfC ' 2048~\n\n:;,-1<\"\n\n70\n15\n10\n50\n\n100\n\n200\n\n5CO\n\n1000\n\n2000\n\n5000\n\nFREQUENCY , C I S\n\n## 12. Ex perimental tTansmission loss resllits on\n\nsingle and dou ble wall of 7'2-in. plasterboard.\n\nFIe UR E\n\nCl\n\n0--\n\n60\n\n## dou ble wall TL drops off a~ 4,096 cis in a fashion\n\nsimilar to that of t h e single wall. In t he I-in.\nan d 2-in . TL measurem ents t h e single wall mini mum occurring in t he neighborhood of the critical\nflexural frequency did not appear in the double wall\ncase. It is of interest (table 1) to compare the\naverage TL for th e nin e frequ encies in th e range\nof 128 to 4,096 cis with th at of ordinar y plaster\nand stud walls. s\nIt will be seen from th e data in table 1, t hat if\nno mechanical ties or sound-conducting bridges\nexist b etween th e two com ponents of a double\n\n...J\n\n~ 45\niii\n<Il\n~ 40\n\n. ,(\n\n_.0'\n\n<Il\n\n,~~~~\n\n<t\n\nex: 35\n\n,,\n\nf-\n\n30\n25\n\nb.'x ~\n\n20\n. 50\n\n100\n\n200\n\n1000\n\n500\n\n2000\n\nFI GU R E\n\n## 13. E x perimental trans11I1'ssion loss results on\n\nsingle and dou ble wall of 1-i n . plasterboard.\n\n'Wcigh t\n\n- - - - - --- - ----------- - -I\n~- i ll . plaster board double walL. _____ .. __________\nI-in. p las terboard double walL. .. ____ __ __ __ ___ __ _\n2-in . plaster board d ouble walL. ___ .. ____________ _\n~i n . gypsum plaster on w ood lath on 2 b y 4\nstuds.\n%in. gyps um plaster on metal lath on 2 by 4\nstaggered wood studs, 4-in. airspace__ ________\nD ouble wall consisting of t wo 2in . solid plas ter\nsingle walls resting on l-in . corkpad , 3-in .\nairspace .. _______________ .. _.......... _____ ....\n\ndb\n\nlblfl '\n\n45.2\n55.5\n59.6\n3i .5\n\n4.2\n8.3\n16.6\nIi. 1\n\n7~\n\n..a\n\n\".\n\np\n,\n\nf/)\n\n55\n\nZ\n0\n\n50\n\n45\n\niii\n</)\nf/)\n\nII:\n\n54. 1\n\n19.8\n\nli .2\n\n,f':\n\n## -~-- < --0- -\n\n,/\"\n\nI-\n\n60\n\n,-fY\n\np-\"\n\n<Il\n\n...J\n\n<1\n4~.\n\nA , D ou ble wall 3-in . airspace; B , single wall; __ __, experimen tal; t heore t ical R ~ 1O . 5, f,~ 7 68 cis .\n\n65\nA \\j.7,ge\n\n5000\n\nFREQUEN CY . CI S\n\n1.\n\nD escription\n\np--- --p\n\nl=- I--\n\n~~\n\n70\nT A BLE\n\n,y\n\ngJ 50\n\n~\n~\n\n~,\n\n..a\n.., 55\n<Il\n\n## A, Double wall 3-in. airspace; B , single wall; ____ experimen tal; _\n\ntheoretical R ~I O .5.\n\n65\n\n,/\n\n40\n\nr/-'/\n./\n\n~~.J::.\n\n35\n30\n\n~, ~\n\n~V\n\n... \"C\n\n.~\n\n25\n'For data of tbis kind see: Building M aterials and Structures R eport\nB MS17 an d two supplemen ts, Sound insula tion of wall a nd floor constructions, available from the Superintendeut of D ocuments, Go\\\"ernme nt Prin ting Office, W ashiugton 25, D . C. at a to tal cost of 35; also T echnical R eport\non Build iug M aterials, T RB M-44, F ire resistan ce a nd sound-insulation\nratings for walls, partitious, and fl oors, free u pon request at X ation al B ureau\nof Standards, 'Wash ington 25, D . C.\n\n86\n\nI\nI\n\nL_\n\n50\n\n100\n\n200\n\n~oo\n\n1000\n\n2000\n\n5000\n\nFREQUENCY . CIS\n\nFIGU RE\n\n## 14. Exp erimental transmission loss results on\n\nsingle and double wall of 2-in. plasterboard.\n\n## A, Double wall 3-in. a irspace; B , single ...,-all ; ___ experimental ; _ ,\n\ntheoretical R ~ 5.3, f,~5 1 2 cis.\n\nJournal of Research\n\n## plate to which the staggered studs arc attached,\n\nand some 5 db b etter when coupling is ~mly due\nto a corle base.\n'\nThe question often aJises as to the effect of\nplacing an absorbent in the airspace. Accordingly, some measurements were taken ' with a\n3-in. thick fiberglas blanket having a density of\nabout 1.0 Ib/ft2 inserted in the airspace. Table\n2 gives the avera,ge improvement in transmission\nloss for the frequency range of 128 to 4096 cis\nover the untreated airspace double wall.\nTABLE\n\n2.\n\n## Description of doub le wall\n\nAverage improvement\n\n## ---------------------- ----~2 i n . 1lI as lc r boa rd . ________ .__________________________\n\nI -in. plas ter board _____ ____ _____ ___ ___ ___ __________ __ __\n2-in. plas tcr board ______ .______________________________\n\ndb\n9. 6\n3. 0\n3.5\n\n## In the previous paper (sec footnote 1) , the effcct\n\nof placing this same fib erglas blanket in front of\nand in juxtapos ition to a single wall was d iscussed .\nThe walls were }f-, 1-, and 2-in. plasterboard ingle\nwalls. In this case the average TL improvement\nfor the frequ ency range of 128 to 4,096 cis was 8.2\ndb and was approximately the same for all tJu'ee\nwalls. Thus, for the dOll ble wall having thc lightest wcight thc improvement us ing the absorbent\nwas equal to 01' better than that obtaincd for the\nsingle wall. On the other hand , for thc h eavier\ndouble walls, a relatively small effect is obscrvcd.\nThis fact has been observed many times in more\nconventional construction using wood studs,\nstaggered, or otherwise. For light-weight construction significant increases in the TL a,r e\nmeasured, whereas for heavy-weight constructions\nonly minor increases result. In conventional\nconstruction this is in paTt due to the existence of\nsound-conducting paths. This explanation, however, does not hold in these experimental double\nwalls, since the components of the double walls\nwere isolated from each other and the blanket\nwas arranged in the airspace so as not to touch\nthe walls. Evidently, the effect depends OIl the\nratio of the impedance of the airspace material to\nthe impedance of the walls. For the heavy walls\nthe material in the airspace can add little to the\nab:eady large impedance of thc walls.\n\n## ~Ieyer 5 has considered the effect of the a irspace\n\nabsorbent material on reducing transversc modes\nof sound in the au'space, that is, those modes in\nwhich thc sound travels parallel to the wall S lll'faces. He pointed out that if these modes are\nimportant, it should be possible to absorb them\nby placing this material only on the boundaries of\nthe airspace. Accordingly, the boundaries of the\nairspace shown in figure 3 were stuffed with\nFiberglas, early in the double wall experiments\nsta,r ting with the double aluminum wall. No\nsignificant difference due to the insertion of the\nboundary absorbent occul'l'ed, so that it was concluded that the effect of the transverse modes\nwas negligible.\nAdditional confirmation of this \\\\'as obtained by\ninserting the \"stl'awcomb\" hown in figure 11 of\nRP1998 in the airspace of several double walls.\nThe term strawcomb refers to a honeycomb\ns tructure that was made by cutting soda straws\ninto 2%-in. lengths. These were placed with\nth eir long axis perpendicular to the wall surfaces.\nSome 150,000 straws were used in the strawcomb\nused in these experiments. Because of the large\nnumber of cell walls that would b e intersected by\na transverse wave, it is hardly to be expected\nlha t they would occur. The average TL increase, again for the H-, 1-, and 2-in. double\nplasterboard walls, was only 0.7 db , showing that\nthe strawcomb had a negligible effect.\n\n5 . Con clusions\n\n## A theory of air-coupled double walls has been\n\ndeveloped, which gives good agreement with\nexperim ental results. In order to apply the theory\nit is necessary to know the wall impedance, Z w,\nof the identical single wall components . This\nquantity may be determined from the transmission loss results obtained on the single walls.\nInasmuch as it is theoretically possible to evaluate\nthe resistance, R, and flexural frequency, ie, from\nmechanical impedance measurem ents on small\nscale samples, we have here , in principle, a method\nof computing double \\vall transmission losses from\nsmall scale experiments. The experimental results indicate that both normal incidence theory\nand the mass-reactance assumption are entirely\ninadequate for explaining the behavior of single\n, E. M eyer, E lck.\n\n~ac hr .\n\n87\n\n## and double walls in a reverberant sound field.\n\nThe importance of including resistance and flexural\nwave effects has been demonstrated.\nFor double walls having air-coupling only,\nvery shallow airspaces can produce appreciable\nincreases in transmission loss over a single wall.\nAn absorbent material, when inserted in the airspace, produce:;; large improvements only when the\nmass of the walls is relatively light and has but\nlittle effect for heavy walls. Honeycomb or other\n\n88\n\n## nonabsorbent cellular structures having- no cell\n\nwalls in a direction normal to the wall faces do\nnot result in an increase in transmission loss.\nThe autror is indebted to S. Edelman and Henry\nJ. Leinbach, Jr .. for making many of the experimental observations; in addition, the latter carried out most of the required numerical integrations.\nWASHINGTON,\n\n## July 26, 1949.\n\nJournal of Research" ]

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[ "# Definition:Vector Cross Product\n\nThis page is about Cross Product in the context of Vector Algebra. For other uses, see Cross Product.\n\n## Definition\n\nLet $\\mathbf a$ and $\\mathbf b$ be vectors in a vector space $\\mathbf V$ of $3$ dimensions:\n\n$\\mathbf a = a_i \\mathbf i + a_j \\mathbf j + a_k \\mathbf k$\n$\\mathbf b = b_i \\mathbf i + b_j \\mathbf j + b_k \\mathbf k$\n\nwhere $\\tuple {\\mathbf i, \\mathbf j, \\mathbf k}$ is the standard ordered basis of $\\mathbf V$.\n\n### Definition 1\n\nThe vector cross product, denoted $\\mathbf a \\times \\mathbf b$, is defined as:\n\n$\\mathbf a \\times \\mathbf b = \\begin{vmatrix} \\mathbf i & \\mathbf j & \\mathbf k\\\\ a_i & a_j & a_k \\\\ b_i & b_j & b_k \\\\ \\end{vmatrix}$\n\nwhere $\\begin {vmatrix} \\ldots \\end {vmatrix}$ is interpreted as a determinant.\n\nMore directly:\n\n$\\mathbf a \\times \\mathbf b = \\paren {a_j b_k - a_k b_j} \\mathbf i - \\paren {a_i b_k - a_k b_i} \\mathbf j + \\paren {a_i b_j - a_j b_i} \\mathbf k$\n\n### Definition 2\n\nThe vector cross product, denoted $\\mathbf a \\times \\mathbf b$, is defined as:\n\n$\\mathbf a \\times \\mathbf b = \\norm {\\mathbf a} \\norm {\\mathbf b} \\sin \\theta \\, \\mathbf {\\hat n}$\n\nwhere:\n\n$\\norm {\\mathbf a}$ denotes the length of $\\mathbf a$\n$\\theta$ denotes the angle from $\\mathbf a$ to $\\mathbf b$, measured in the positive direction\n$\\hat {\\mathbf n}$ is the unit vector perpendicular to both $\\mathbf a$ and $\\mathbf b$ in the direction according to the right-hand rule.\n\n## Complex Numbers\n\nThe definition is slightly different when the vector space under consideration is the complex plane, as the latter is of $2$ dimensions.\n\nThis is a modification of the vector cross product in which the resulting product is taken to be the length of the hypothetical vector which would be considered as being perpendicular to the complex plane.\n\nLet $z_1 := x_1 + i y_1$ and $z_2 := x_2 + i y_2$ be complex numbers.\n\n### Definition 1\n\nThe cross product of $z_1$ and $z_2$ is defined as:\n\n$z_1 \\times z_2 = x_1 y_2 - y_1 x_2$\n\n### Definition 2\n\nThe cross product of $z_1$ and $z_2$ is defined as:\n\n$z_1 \\times z_2 = \\cmod {z_1} \\, \\cmod {z_2} \\sin \\theta$\n\nwhere:\n\n$\\cmod {z_1}$ denotes the complex modulus of $z_1$\n$\\theta$ denotes the angle from $z_1$ to $z_2$, measured in the positive direction.\n\n### Definition 3\n\nThe cross product of $z_1$ and $z_2$ is defined as:\n\n$z_1 \\times z_2 := \\map \\Im {\\overline {z_1} z_2}$\n\nwhere:\n\n$\\map \\Im z$ denotes the imaginary part of a complex number $z$\n$\\overline {z_1}$ denotes the complex conjugate of $z_1$\n$\\overline {z_1} z_2$ denotes complex multiplication.\n\n### Definition 4\n\nThe cross product of $z_1$ and $z_2$ is defined as:\n\n$z_1 \\times z_2 := \\dfrac {\\overline {z_1} z_2 - z_1 \\overline {z_2}} {2 i}$\n\nwhere:\n\n$\\overline {z_1}$ denotes the complex conjugate of $z_1$\n$\\overline {z_1} z_2$ denotes complex multiplication.\n\n## Also known as\n\nThe vector cross product is often called just the cross product when there is no chance of confusion with other types of cross product.\n\nThe term vector product can also sometimes be seen, but again this can be ambiguous.\n\n## Examples\n\n### Couple Exerted by Force\n\nLet $\\mathbf F$ be a force acting at a point $P$ on a body $B$ axially about an axis of rotation $R$ such that the distance from $P$ to $R$ is represented by the displacement vector $\\mathbf d$.\n\nThen the couple exerted on $B$ by $\\mathbf F$ is defined as:\n\n$\\mathbf T = \\mathbf F \\times \\mathbf d = \\norm {\\mathbf F} \\norm {\\mathbf d} \\mathbf {\\hat t} \\sin \\theta$\n\nwhere:\n\n$\\times$ denotes vector cross product\n$\\mathbf {\\hat t}$ denotes the unit vector perpendicular to both $\\mathbf F$ and $\\mathbf d$ according to the right-hand rule\n$\\theta$ is the angle between the directions of $\\mathbf F$ and $\\mathbf d$.\n\n## Also see\n\n• Results about Vector Cross Product can be found here.\n\n## Historical Note\n\nDuring the course of development of vector analysis, various notations for the vector cross product were introduced, as follows:\n\nSymbol Used by\n$\\mathbf a \\times \\mathbf b$ Josiah Willard Gibbs and Edwin Bidwell Wilson\n$V \\mathbf a \\mathbf b$ Oliver Heaviside\n$\\sqbrk {\\mathscr A \\mathscr B}$ Max Abraham\n$\\sqbrk {\\mathfrak A \\mathfrak B}$ Vladimir Sergeyevitch Ignatowski\n$\\sqbrk {\\mathbf A \\cdot \\mathbf B}$ Hendrik Antoon Lorentz\n$\\mathbf a \\wedge \\mathbf b$ Cesare Burali-Forti and Roberto Marcolongo\n\n## Technical Note\n\nThe $\\LaTeX$ code for $\\mathbf A \\times \\mathbf B$ is \\mathbf A \\times \\mathbf B .\n\nThe $\\LaTeX$ code for $\\mathbf A \\wedge \\mathbf B$ is \\mathbf A \\wedge \\mathbf B .\n\nIn this context, $\\wedge$ is usually referred to as wedge." ]

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[ "# #72f2b6 Color Information\n\nIn a RGB color space, hex #72f2b6 is composed of 44.7% red, 94.9% green and 71.4% blue. Whereas in a CMYK color space, it is composed of 52.9% cyan, 0% magenta, 24.8% yellow and 5.1% black. It has a hue angle of 151.9 degrees, a saturation of 83.1% and a lightness of 69.8%. #72f2b6 color hex could be obtained by blending #e4ffff with #00e56d. Closest websafe color is: #66ffcc.\n\n• R 45\n• G 95\n• B 71\nRGB color chart\n• C 53\n• M 0\n• Y 25\n• K 5\nCMYK color chart\n\n#72f2b6 color description : Soft cyan - lime green.\n\n# #72f2b6 Color Conversion\n\nThe hexadecimal color #72f2b6 has RGB values of R:114, G:242, B:182 and CMYK values of C:0.53, M:0, Y:0.25, K:0.05. Its decimal value is 7533238.\n\nHex triplet RGB Decimal 72f2b6 `#72f2b6` 114, 242, 182 `rgb(114,242,182)` 44.7, 94.9, 71.4 `rgb(44.7%,94.9%,71.4%)` 53, 0, 25, 5 151.9°, 83.1, 69.8 `hsl(151.9,83.1%,69.8%)` 151.9°, 52.9, 94.9 66ffcc `#66ffcc`\nCIE-LAB 87.22, -49.156, 18.329 47.132, 70.456, 55.369 0.273, 0.407, 70.456 87.22, 52.462, 159.551 87.22, -56.014, 35.067 83.938, -46.662, 19.646 01110010, 11110010, 10110110\n\n# Color Schemes with #72f2b6\n\n• #72f2b6\n``#72f2b6` `rgb(114,242,182)``\n• #f272ae\n``#f272ae` `rgb(242,114,174)``\nComplementary Color\n• #72f276\n``#72f276` `rgb(114,242,118)``\n• #72f2b6\n``#72f2b6` `rgb(114,242,182)``\n• #72eef2\n``#72eef2` `rgb(114,238,242)``\nAnalogous Color\n• #f27672\n``#f27672` `rgb(242,118,114)``\n• #72f2b6\n``#72f2b6` `rgb(114,242,182)``\n• #f272ee\n``#f272ee` `rgb(242,114,238)``\nSplit Complementary Color\n• #f2b672\n``#f2b672` `rgb(242,182,114)``\n• #72f2b6\n``#72f2b6` `rgb(114,242,182)``\n• #b672f2\n``#b672f2` `rgb(182,114,242)``\n• #aef272\n``#aef272` `rgb(174,242,114)``\n• #72f2b6\n``#72f2b6` `rgb(114,242,182)``\n• #b672f2\n``#b672f2` `rgb(182,114,242)``\n• #f272ae\n``#f272ae` `rgb(242,114,174)``\n• #2cec92\n``#2cec92` `rgb(44,236,146)``\n• #43ee9e\n``#43ee9e` `rgb(67,238,158)``\n• #5bf0aa\n``#5bf0aa` `rgb(91,240,170)``\n• #72f2b6\n``#72f2b6` `rgb(114,242,182)``\n• #89f4c2\n``#89f4c2` `rgb(137,244,194)``\n• #a1f6ce\n``#a1f6ce` `rgb(161,246,206)``\n• #b8f8da\n``#b8f8da` `rgb(184,248,218)``\nMonochromatic Color\n\n# Alternatives to #72f2b6\n\nBelow, you can see some colors close to #72f2b6. Having a set of related colors can be useful if you need an inspirational alternative to your original color choice.\n\n• #72f296\n``#72f296` `rgb(114,242,150)``\n• #72f2a1\n``#72f2a1` `rgb(114,242,161)``\n• #72f2ab\n``#72f2ab` `rgb(114,242,171)``\n• #72f2b6\n``#72f2b6` `rgb(114,242,182)``\n• #72f2c1\n``#72f2c1` `rgb(114,242,193)``\n• #72f2cb\n``#72f2cb` `rgb(114,242,203)``\n• #72f2d6\n``#72f2d6` `rgb(114,242,214)``\nSimilar Colors\n\n# #72f2b6 Preview\n\nThis text has a font color of #72f2b6.\n\n``<span style=\"color:#72f2b6;\">Text here</span>``\n#72f2b6 background color\n\nThis paragraph has a background color of #72f2b6.\n\n``<p style=\"background-color:#72f2b6;\">Content here</p>``\n#72f2b6 border color\n\nThis element has a border color of #72f2b6.\n\n``<div style=\"border:1px solid #72f2b6;\">Content here</div>``\nCSS codes\n``.text {color:#72f2b6;}``\n``.background {background-color:#72f2b6;}``\n``.border {border:1px solid #72f2b6;}``\n\n# Shades and Tints of #72f2b6\n\nA shade is achieved by adding black to any pure hue, while a tint is created by mixing white to any pure color. In this example, #000302 is the darkest color, while #f0fef7 is the lightest one.\n\n• #000302\n``#000302` `rgb(0,3,2)``\n• #02150c\n``#02150c` `rgb(2,21,12)``\n• #042716\n``#042716` `rgb(4,39,22)``\n• #053920\n``#053920` `rgb(5,57,32)``\n• #074b2b\n``#074b2b` `rgb(7,75,43)``\n• #095c35\n``#095c35` `rgb(9,92,53)``\n• #0a6e3f\n``#0a6e3f` `rgb(10,110,63)``\n• #0c804a\n``#0c804a` `rgb(12,128,74)``\n• #0d9254\n``#0d9254` `rgb(13,146,84)``\n• #0fa45e\n``#0fa45e` `rgb(15,164,94)``\n• #11b669\n``#11b669` `rgb(17,182,105)``\n• #12c873\n``#12c873` `rgb(18,200,115)``\n• #14da7d\n``#14da7d` `rgb(20,218,125)``\n• #18ea88\n``#18ea88` `rgb(24,234,136)``\n• #2aeb91\n``#2aeb91` `rgb(42,235,145)``\n• #3ced9a\n``#3ced9a` `rgb(60,237,154)``\n• #4eefa3\n``#4eefa3` `rgb(78,239,163)``\n``#60f0ad` `rgb(96,240,173)``\n• #72f2b6\n``#72f2b6` `rgb(114,242,182)``\n• #84f4bf\n``#84f4bf` `rgb(132,244,191)``\n• #96f5c9\n``#96f5c9` `rgb(150,245,201)``\n• #a8f7d2\n``#a8f7d2` `rgb(168,247,210)``\n• #baf9db\n``#baf9db` `rgb(186,249,219)``\n• #ccfae4\n``#ccfae4` `rgb(204,250,228)``\n• #defcee\n``#defcee` `rgb(222,252,238)``\n• #f0fef7\n``#f0fef7` `rgb(240,254,247)``\nTint Color Variation\n\n# Tones of #72f2b6\n\nA tone is produced by adding gray to any pure hue. In this case, #adb7b2 is the less saturated color, while #66feb7 is the most saturated one.\n\n``#adb7b2` `rgb(173,183,178)``\n• #a7bdb3\n``#a7bdb3` `rgb(167,189,179)``\n• #a1c3b3\n``#a1c3b3` `rgb(161,195,179)``\n• #9bc9b3\n``#9bc9b3` `rgb(155,201,179)``\n• #96ceb4\n``#96ceb4` `rgb(150,206,180)``\n• #90d4b4\n``#90d4b4` `rgb(144,212,180)``\n``#8adab5` `rgb(138,218,181)``\n• #84e0b5\n``#84e0b5` `rgb(132,224,181)``\n• #7ee6b5\n``#7ee6b5` `rgb(126,230,181)``\n• #78ecb6\n``#78ecb6` `rgb(120,236,182)``\n• #72f2b6\n``#72f2b6` `rgb(114,242,182)``\n• #6cf8b6\n``#6cf8b6` `rgb(108,248,182)``\n• #66feb7\n``#66feb7` `rgb(102,254,183)``\nTone Color Variation\n\n# Color Blindness Simulator\n\nBelow, you can see how #72f2b6 is perceived by people affected by a color vision deficiency. This can be useful if you need to ensure your color combinations are accessible to color-blind users.\n\nMonochromacy\n• Achromatopsia 0.005% of the population\n• Atypical Achromatopsia 0.001% of the population\nDichromacy\n• Protanopia 1% of men\n• Deuteranopia 1% of men\n• Tritanopia 0.001% of the population\nTrichromacy\n• Protanomaly 1% of men, 0.01% of women\n• Deuteranomaly 6% of men, 0.4% of women\n• Tritanomaly 0.01% of the population" ]

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[ "# Field equals() method in Java with Examples\n\nThe equals() method of java.lang.reflect.Field is used to compare two field objects. This method compares two field objects and returns true if both objects are equal otherwise false. The two Field objects are considered equal if and only if when they were declared by the same class and have the same name and type. This method is very helpful at the time of debugging the object properties which are actually fields of a class in Java.\n\nSyntax:\n\n```public boolean equals(Object obj)\n```\n\nParameters: This method accepts one parameter obj which is the reference object to compare with this Field object.\n\nReturn value: This method returns true if both objects are equal otherwise false.\n\nBelow programs illustrate equals() method:\nProgram 1:\n\n `// Java program to demonstrate the above method ` ` `  `import` `java.lang.reflect.Field; ` ` `  `public` `class` `GFG { ` ` `  `    ``public` `static` `void` `main(String[] args) ` `        ``throws` `NoSuchFieldException, ` `               ``SecurityException ` `    ``{ ` ` `  `        ``// get the array of Field objects ` `        ``Field[] fields ` `            ``= User.``class``.getDeclaredFields(); ` `        ``Field fieldObj ` `            ``= User.``class``.getField(``\"name\"``); ` ` `  `        ``// print element of field array ` `        ``// and compare it with fieldObj ` `        ``for` `(``int` `i = ``0``; i < fields.length; i++) { ` ` `  `            ``// compare the fields with each other ` `            ``boolean` `isEquals ` `                ``= fields[i].equals(fieldObj); ` `            ``if` `(isEquals) { ` `                ``System.out.println( ` `                    ``\"Field -> [\"` `                    ``+ fields[i] + ``\"] and\"` `                    ``+ ``\" FieldObj -> [\"` `                    ``+ fieldObj ` `                    ``+ ``\"] are equal.\"``); ` `            ``} ` `            ``else` `{ ` `                ``System.out.println( ` `                    ``\"Field -> [\"` `                    ``+ fields[i] + ``\"] and\"` `                    ``+ ``\" FieldObj -> [\"` `                    ``+ fieldObj ` `                    ``+ ``\"] are not equal.\"``); ` `            ``} ` `        ``} ` `    ``} ` `} ` ` `  `// User class ` `class` `User { ` ` `  `    ``public` `String name; ` `    ``public` `int` `age; ` `} `\n\nOutput:\n\nField -> [public java.lang.String User.name]\nand\nFieldObj -> [public java.lang.String User.name]\nare equal.\n\nField -> [public int User.age]\nand\nFieldObj -> [public java.lang.String User.name]\nare not equal.\n\nProgram 2:\n\n `// Java program to demonstrate the above method ` ` `  `import` `java.lang.reflect.Field; ` ` `  `public` `class` `GFG { ` ` `  `    ``public` `static` `void` `main(String[] args) ` `        ``throws` `NoSuchFieldException, ` `               ``SecurityException ` `    ``{ ` ` `  `        ``// get the array of Field objects ` `        ``Field[] fields1 ` `            ``= Class.``class``.getDeclaredFields(); ` `        ``Field[] fields2 ` `            ``= Class.``class``.getDeclaredFields(); ` ` `  `        ``// print element of field array 1 and compare ` `        ``// it with fields array 2 ` `        ``for` `(``int` `i = ``0``; i < fields1.length; i++) { ` ` `  `            ``for` `(``int` `j = ``0``; j < fields2.length; j++) { ` ` `  `                ``// compare the fields with each other ` `                ``boolean` `isEquals ` `                    ``= fields1[i].equals(fields2[j]); ` `                ``if` `(isEquals) { ` `                    ``System.out.println( ` `                        ``\"Field -> [\"` `                        ``+ fields1[i] + ``\"] and\"` `                        ``+ ``\" FieldObj -> [\"` `                        ``+ fields2[j] ` `                        ``+ ``\"] are equal.\"``); ` `                ``} ` `                ``else` `{ ` `                    ``System.out.println( ` `                        ``\"Field -> [\"` `                        ``+ fields1[i] + ``\"] and\"` `                        ``+ ``\" FieldObj -> [\"` `                        ``+ fields2[j] ` `                        ``+ ``\"] are not equal.\"``); ` `                ``} ` `            ``} ` `        ``} ` `    ``} ` `} ` ` `  `// Object of Class which contains ` `// noOfStudents and studentNames ` `class` `Class { ` ` `  `    ``public` `int` `noOfStudents; ` `    ``public` `String[] studentNames; ` `} `\n\nOutput:\n\nField -> [public int Class.noOfStudents]\nand\nFieldObj -> [public int Class.noOfStudents]\nare equal.\n\nField -> [public int Class.noOfStudents]\nand\nFieldObj -> [public java.lang.String[] Class.studentNames]\nare not equal.\n\nField -> [public java.lang.String[] Class.studentNames]\nand\nFieldObj -> [public int Class.noOfStudents]\nare not equal.\n\nField -> [public java.lang.String[] Class.studentNames]\nand\nFieldObj -> [public java.lang.String[] Class.studentNames]\nare equal.\n\nMy Personal Notes arrow_drop_up", null, "Check out this Author's contributed articles.\n\nIf you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to [email protected]. See your article appearing on the GeeksforGeeks main page and help other Geeks.\n\nPlease Improve this article if you find anything incorrect by clicking on the \"Improve Article\" button below.\n\nImproved By : shubham_singh\n\nArticle Tags :\nPractice Tags :\n\nBe the First to upvote.\n\nPlease write to us at [email protected] to report any issue with the above content." ]

[ null, "https://media.geeksforgeeks.org/auth/profile/z3lkktftxc4b496iwi0q", null ]

[ "# Convert 1000 ML to Cups\n\nConverting 1000 ml to cups is relatively easy. Before we tell you the process, let’s find out how much is 1000 ml to cups.\n\nConvert 1000 ML to Cups: 1000 ml = 4.22 Cups. So there are 4.22 US customary cups in 1000 millimeters\n\nVolume as we all know is the amount of space that is occupied by an object! This is an example of measuring the size of cups that can be accommodated with 1000 milliliters!\n\nHere we shall learn about the volume of the number of cups that can be fitted within the capacity of 1000ml. Let us read through this elaboration where we will learn about how many cups of a liquid will be equivalent to 1000ml of measurement.\n\nLet us have a look at the description and read it further in detail. Shall we?\n\n## How to Convert 1000 ml to cups\n\nTo find out how many cups is 1000ml, one must divide 1000ml by the size of the cups to get the number of cups!\n\n1 cup contains 236.6 mm. So we will divide 1000 ml by 236.6 ml to get the result.\n\n1000/236.6= 4.22654\n\nThus according to the formula, here are some standard estimate values that we have got:\n\n1000 ml is 4.22675 US customary cups or\n\n4.16667 United States legal cups or\n\n4 Metric cups or\n\n3.51951 Imperial cups\n\nWe have made this calculation based on cup size and volume hence this does not apply to the density of a food substance, etc.\n\n### Second Method\n\nThere is a conversion factor, 0.0042267528198649 which needs to be multiplied by the value of milliliters to get the resultant number of cups. Hence in this case you have multiplied 1000 by 0.0042267528198649.\n\n## Best conversion unit for 1000ml\n\nThe best unit of conversion is defined as the best value of the unit which does not go below 1 but it is the smallest numerical value! Hence we see that the smallest value or the best conversion unit of 1000ml is 1 liter.\n\n## Definition of millilitres\n\nMillilitre is a unit to measure volume. It is a subunit of the SI unit of volume, liters. When we have to express the volume of liquid in smaller quantities we use the unit milliliters.\n\nThe value of 1000ml is expressed in other units as follows:\n\nis equal to 1/1000 litre, or one cubic centimetre, therefore, 1ml = 1/1000 L =1 cm3.\n\n## Definition of a cup\n\nThis is an English unit of volume which we use as a unit of serving. The unit of the cup is associated with cooking and adding the amount of water or broth to a particular recipe.\n\nThe value of a cup is equivalent to half a pint traditionally according to the British Imperial system or US Customary system! A cup is often defined as ¼ or ⅕th part of a liter.\n\nThe fancy cups are of different sizes and hence we always need a standard measuring unit of cup for determining the actual measurement! It is equivalent to 8 U.S. customary fluid ounces. One customary cup is equal to 236.5882365 milliliters.\n\n## Milliliters to Cups Conversion Table\n\nHere is a table that has been given below so you can estimate the size of cups and milliliters\n\nAlso read: How Much Does a Gold Bar Weigh?" ]

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[ "", null, "Download", null, "Download Presentation", null, "Introductory Chemistry , 3 rd Edition Nivaldo Tro\n\n# Introductory Chemistry , 3 rd Edition Nivaldo Tro\n\nTélécharger la présentation", null, "## Introductory Chemistry , 3 rd Edition Nivaldo Tro\n\n- - - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - - -\n##### Presentation Transcript\n\n1. Introductory Chemistry, 3rd EditionNivaldo Tro Chapter 10 Chemical Bonding Basic Principles of Chemistry Online Southeast Missouri State University Cape Girardeau, MO 2009, Prentice Hall\n\n2. Bonding Theories • Bonding is the way atoms attach to make molecules. • An understanding of how and why atoms attach together in the manner they do is central to chemistry. • Chemists have an understanding of bonding that allows them to: • Predict the shapes of molecules and properties of substances based on the bonding within the molecules. • Design and build molecules with particular sets of chemical and physical properties. Tro's Introductory Chemistry, Chapter 10\n\n3. Lewis Theory • Lewis bonding theory emphasizes the importance of valence electrons. • Uses dots to represent valence electrons either on or shared by atoms. • Arranges bonding between atoms to attain certain sets of stable valence electron arrangements. Tro's Introductory Chemistry, Chapter 10\n\n4. •• •• •• •• Li• Be• •B• •C• •N• •O::F: :Ne: • • • • • • •• Lewis Symbols of Atoms • Also known as electron dot symbols. • Uses symbol of element to represent nucleus and inner electrons. • Uses dots around the symbol to represent valence electrons. • Puts one electron on each side first, then pair. • Remember that elements in the same group have the same number of valence electrons; therefore, their Lewis dot symbols will look alike. Tro's Introductory Chemistry, Chapter 10\n\n5. Practice—Write the Lewis Symbol for Arsenic. Tro's Introductory Chemistry, Chapter 10\n\n6. Practice—Write the Lewis Symbol for Arsenic, Continued. • As is in column 5A, therefore it has 5 valence electrons. Tro's Introductory Chemistry, Chapter 10\n\n7. Lewis Bonding Theory • Atoms bond because it results in a more stable electron configuration. • Atoms bond together by either transferring or sharing electrons. • Usually this results in all atoms obtaining an outer shell with 8 electrons. • Octet rule. • There are some exceptions to this rule—the key to remember is to try to get an electron configuration like a noble gas. • Li and Be try to achieve the He electron arrangement. Tro's Introductory Chemistry, Chapter 10\n\n8. •• •• Li• Li+:F: [:F:]− • •• Lewis Symbols of Ions • Cations have Lewis symbols without valence electrons. • Lost in the cation formation. • They now have a full “outer” shell that was the previous second highest energy shell. • Anions have Lewis symbols with 8 valence electrons. • Electrons gained in the formation of the anion. Tro's Introductory Chemistry, Chapter 10\n\n9. Practice—Show How the Electrons Are Transferred and the Bond Is Formed When Na Reacts with S. Tro's Introductory Chemistry, Chapter 10\n\n10. Practice—Show How the Electrons Are Transferred and the Bond Is Formed When Na Reacts with S, Continued. Tro's Introductory Chemistry, Chapter 10\n\n11. Ionic Bonds • Metal to nonmetal. • Metal loses electrons to form cation. • Nonmetal gains electrons to form anion. • Ionic bond results from + to − attraction. • Larger charge = stronger attraction. • Smaller ion = stronger attraction. • Lewis theory allows us to predict the correct formulas of ionic compounds. Tro's Introductory Chemistry, Chapter 10\n\n12. ∙ ∙ ∙ ∙ Ca Ca Cl Cl Cl ∙ ∙ ∙ ∙ ∙ ∙ ∙ ∙ ∙ ∙ ∙ ∙ ∙ ∙ ∙ ∙ ∙ ∙ ∙ ∙ Example 10.3—Using Lewis Theory to Predict Chemical Formulas of Ionic Compounds Predict the formula of the compound that forms between calcium and chlorine. Draw the Lewis dot symbols of the elements. Transfer all the valance electrons from the metal to the nonmetal, adding more of each atom as you go, until all electrons are lost from the metal atoms and all nonmetal atoms have 8 electrons. Ca2+ CaCl2 Tro's Introductory Chemistry, Chapter 10\n\n13. Practice—Use Lewis Symbols to Predict the Formula of an Ionic Compound Made from Reacting a Metal, M, that Has 2 Valence Electrons with a Nonmetal, X, that Has 5 Valence Electrons. Tro's Introductory Chemistry, Chapter 10\n\n14. Practice—Use Lewis Symbols to Predict the Formula of an Ionic Compound Made from Reacting a Metal, M, that Has 2 Valence Electrons with a Nonmetal, X, that Has 5 Valence Electrons, Continued. M3X2 Tro's Introductory Chemistry, Chapter 10\n\n15. Covalent Bonds • Often found between two nonmetals. • Typical of molecular species. • Atoms bonded together to form molecules. • Strong attraction. • Atoms share pairs of electronsto attain octets. • Molecules generally weakly attracted to each other. • Observed physical properties of molecular substance due to these attractions. Tro's Introductory Chemistry, Chapter 10\n\n16. Using Lewis Atomic Structures to Predict Bonding Between Nonmetal Atoms • Nonmetal atoms often bond to achieve an octet of valence electrons by sharing electrons. • Though there are many exceptions to the Octet rule. • In Lewis theory, atoms share electrons to complete their octet. • This may involve sharing electrons with multiple atoms or sharing multiple pairs of electrons with the same atom. Tro's Introductory Chemistry, Chapter 10\n\n17. •• •• • • •• F F •• •• •• •• •• •• •• •• F F F F •• •• Single Covalent Bonds • Two atoms share one pair of electrons. • 2 electrons. • One atom may have more than one single bond. •• • • • • H H O •• •• •• H H •• O •• Tro's Introductory Chemistry, Chapter 10\n\n18. •• •• • • • • O O •• •• •• •• O •• •• •• •• O O O Double Covalent Bond • Two atoms sharing two pairs of electrons. • 4 electrons. • Shorter and stronger than single bond. Tro's Introductory Chemistry, Chapter 10\n\n19. •• •• • • • • N N • • N N Triple Covalent Bond • Two atoms sharing 3 pairs of electrons. • 6 electrons. • Shorter and stronger than single or double bond. N N •• •• •• •• •• Tro's Introductory Chemistry, Chapter 10\n\n20. •• •• Bonding and Lone Pair Electrons • Electrons that are shared by atoms are called bonding pairs. • Electrons that are not shared by atoms but belong to a particular atom are called lone pairs. • Also known as nonbonding pairs. O S O •• •• Bonding pairs Lone pairs • • •• •• • • • • Tro's Introductory Chemistry, Chapter 10\n\n21. Multiplicity and Bond Properties • The more electrons two atoms share, the stronger they are bonded together. • This explains the observation that triple bonds are stronger than similar double bonds, which are stronger than single bonds. • C≡N is stronger than C=N, C=N is stronger than C─N. • This explains the observation that triple bonds are shorter than similar double bonds, which are shorter than single bonds. • C≡N is shorter than C=N, C=N is shorter than C─N Tro's Introductory Chemistry, Chapter 10\n\n22. Trends in Bond Length and Energy\n\n23. Polyatomic Ions • The polyatomic ions are attracted to opposite ions by ionic bonds. • Form crystal lattices. • Atoms in the polyatomic ion are held together by covalent bonds. Tro's Introductory Chemistry, Chapter 10\n\n24. Lewis Formulas of Molecules • Shows pattern of valence electron distribution in the molecule. • Useful for understanding the bonding in many compounds. • Allows us to predict shapes of molecules. • Allows us to predict properties of molecules and how they will interact together. Tro's Introductory Chemistry, Chapter 10\n\n25. C B N O F Lewis Structures • Some common bonding patterns. • C = 4 bonds & 0 lone pairs. • 4 bonds = 4 single, or 2 double, or single + triple, or 2 single + double. • N = 3 bonds & 1 lone pair. • O = 2 bonds & 2 lone pairs. • H and halogen = 1 bond. • Be = 2 bonds & 0 lone pairs. • B = 3 bonds & 0 lone pairs. Tro's Introductory Chemistry, Chapter 10\n\n26. Writing Lewis Structuresfor Covalent Molecules 1. Attach the atoms together in a skeletal structure. • Most metallic element is generally central. • In PCl3, the P is central because it is further left on the periodic table and therefore more metallic. • Halogens and hydrogen are generally terminal. • In C2Cl4, the Cs are attached together in the center and the Cls are surrounding them. • Many molecules tend to be symmetrical. • Though there are many exceptions to this, chemical formulas are often written to indicate the order of atom attachment. • In C2Cl4, there are two Cls on each C. • In oxyacids, the acid hydrogens are attached to an oxygen. • In H2SO4, the S is central, the Os are attached to the S, and each H is attached to a different O. Tro's Introductory Chemistry, Chapter 10\n\n27. Writing Lewis Structuresfor Covalent Molecules, Continued 2. Calculate the total number of valence electrons available for bonding. • Use group number of periodic table to find number of valence electrons for each atom. • If you have a cation, subtract 1 electron for each + charge. • If you have an anion, add 1 electron for each − charge. • In PCl3, P has 5 e− and each Cl has 7 e− for a total of 26 e−. • In ClO3−, Cl has 7 e− and each O has 6 e− for a total of 25 e−. • Add 1 e − for the negative charge to get a grand total of 26 e− Tro's Introductory Chemistry, Chapter 10\n\n28. Writing Lewis Structuresfor Covalent Molecules, Continued 3. Attach atoms with pairs of electrons and subtract electrons used from total. • Bonding electrons. 4. Add remaining electrons in pairs to complete the octets of all the atoms. • Remember H only wants 2 electrons. • Don’t forget to keep subtracting from the total. • Complete octets on the terminal atoms first, then work toward central atoms. Tro's Introductory Chemistry, Chapter 10\n\n29. Writing Lewis Structuresfor Covalent Molecules, Continued 5. If there are not enough electrons to complete the octet of the central atom, bring pairs of electrons from an attached atom in to share with the central atom until it has an octet. • Try to follow common bonding patterns. Tro's Introductory Chemistry, Chapter 10\n\n30. Example HNO3 1. Write skeletal structure. • Since this is an oxyacid, H on outside attached to one of the Os; N is central. 2. Count valence electrons. N = 5 H = 1 O3 = 3∙6 = 18 Total = 24 e- Tro's Introductory Chemistry, Chapter 10\n\n31. Example HNO3 , Continued 3. Attach atoms with pairs of electrons and subtract from the total. N = 5 H = 1 O3 = 3∙6 = 18 Total = 24 e- Electrons Start 24 Used 8 Left 16 Tro's Introductory Chemistry, Chapter 10 31\n\n32. Example HNO3 , Continued 4. Complete octets, outside-in. • H is already complete with 2. • 1 bond. • Keep going until all atoms have an octet or you run out of electrons. N = 5 H = 1 O3 = 3∙6 = 18 Total = 24 e- Electrons Start 24 Used 8 Left 16 Electrons Start 16 Used 16 Left 0 Tro's Introductory Chemistry, Chapter 10\n\n33. Example HNO3 , Continued 5. If central atom does not have octet, bring in electron pairs from outside atoms to share. • Follow common bonding patterns if possible. Tro's Introductory Chemistry, Chapter 10\n\n34. Example 10.4—Writing Lewis Structures forCovalent Compounds Tro's Introductory Chemistry, Chapter 10 34\n\n35. Example 10.4: Write the Lewis structure of CO2. Tro's Introductory Chemistry, Chapter 10\n\n36. Write down the given quantity and its units. Given: CO2 Example:Write the Lewis structure of CO2. Tro's Introductory Chemistry, Chapter 10\n\n37. Write down the quantity to find and/or its units. Find: Lewis structure Information: Given: CO2 Example:Write the Lewis structure of CO2. Tro's Introductory Chemistry, Chapter 10\n\n38. Design a solution map. Information: Given: CO2 Find: Lewis structure Example:Write the Lewis structure of CO2. Formula of compound Lewis structure Count and distribute electrons Skeletal structure Tro's Introductory Chemistry, Chapter 10\n\n39. Apply the solution map. Write skeletal structure. Least metallic atom central. H terminal. Symmetry. Information: Given: CO2 Find: Lewis structure Solution Map: formula → skeletal → electron distribution → Lewis Example:Write the Lewis structure of CO2. Tro's Introductory Chemistry, Chapter 10\n\n40. Apply the solution map. Count and distribute the valence electrons. Count valence electrons. Information: Given: CO2 Find: Lewis structure Solution Map: formula → skeletal → electron distribution → Lewis C = 4 O = 2 ∙ 6 Total CO2 = 16 Example:Write the Lewis structure of CO2. 1A 8A 3A 4A 5A 6A 7A 2A C O Tro's Introductory Chemistry, Chapter 10\n\n41. Apply the solution map. Count and distribute the valence electrons. Attach atoms. Information: Given: CO2 Find: Lewis structure Solution Map: formula → skeletal → electron distribution → Lewis C = 4 O = 2 ∙ 6 Total CO2 = 16 Start = 16 e- Used = 4 e- Left = 12 e- Example:Write the Lewis structure of CO2. Tro's Introductory Chemistry, Chapter 10\n\n42. Apply the solution map. Count and distribute the valence electrons. Complete octets. Outside atoms first. Information: Given: CO2 Find: Lewis structure Solution Map: formula → skeletal → electron distribution → Lewis C = 4 O = 2 ∙ 6 Total CO2 = 16 Start = 16 e- Used = 4 e- Left = 12 e- Start = 12 e- Used = 12 e- Left = 0 e- Example:Write the Lewis structure of CO2. Tro's Introductory Chemistry, Chapter 10\n\n43. Apply the solution map. Count and distribute the valence electrons. Complete octets. If not enough electrons to complete octet of central atom, bring in pairs of electrons from attached atom to make multiple bonds. Information: Given: CO2 Find: Lewis structure Solution Map: formula → skeletal → electron distribution → Lewis Start = 12 e- Used = 12 e- Left = 0 e- Example:Write the Lewis structure of CO2. Tro's Introductory Chemistry, Chapter 10\n\n44. Check: Information: Given: CO2 Find: Lewis structure Solution Map: formula → skeletal → electron distribution → Lewis Start C = 4 e- O = 2 ∙ 6 e- Total CO2 = 16 e- End Bonding = 4 ∙ 2 e- Lone pairs = 4 ∙ 2 e- Total CO2 = 16 e- Example:Write the Lewis structure of CO2. The skeletal structure is symmetrical. All the electrons are accounted for. Tro's Introductory Chemistry, Chapter 10\n\n45. Writing Lewis Structures forPolyatomic Ions • The procedure is the same, the only difference is in counting the valence electrons. • For polyatomic cations, take away one electron from the total for each positive charge. • For polyatomic anions, add one electron to the total for each negative charge. Tro's Introductory Chemistry, Chapter 10\n\n46. Example NO3─ 1. Write skeletal structure. • N is central because it is the most metallic. 2. Count valence electrons. N = 5 O3 = 3∙6 = 18 (-) = 1 Total = 24 e- Tro's Introductory Chemistry, Chapter 10\n\n47. Example NO3─ , Continued 3. Attach atoms with pairs of electrons and subtract from the total. Electrons Start 24 Used 6 Left 18 N = 5 O3 = 3∙6 = 18 (-) = 1 Total = 24 e- Tro's Introductory Chemistry, Chapter 10 47\n\n48. Example NO3─ , Continued 3. Complete octets, outside-in. • Keep going until all atoms have an octet or you run out of electrons. N = 5 O3 = 3∙6 = 18 (-) = 1 Total = 24 e- Electrons Start 24 Used 6 Left 18 Electrons Start 18 Used 18 Left 0 Tro's Introductory Chemistry, Chapter 10\n\n49. Example NO3─ , Continued 5. If central atom does not have octet, bring in electron pairs from outside atoms to share. • Follow common bonding patterns if possible. Tro's Introductory Chemistry, Chapter 10\n\n50. Practice—Lewis Structures NClO H3BO3 NO2-1 H3PO4 SO3-2 P2H4 Tro's Introductory Chemistry, Chapter 10 50" ]

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[ "", null, "", null, "", null, "", null, "", null, "", null, "", null, "insect population dynamics\n\n• To: mathgroup at smc.vnet.net\n• Subject: [mg24590] insect population dynamics\n• From: Mark Hunter <mhunter at ecology.uga.edu>\n• Date: Tue, 25 Jul 2000 00:56:21 -0400 (EDT)\n• Sender: owner-wri-mathgroup at wolfram.com\n\n```I'm trying to use mathematica to run insect population models. I'm new\nto the system, and wolfram support suggested that I post my question\nhere.\n\nThe models are simple predator-prey models. They are non-linear\nsimultaneous difference equations. I've tried using RSolve, but it\ndoesn't seem to handle the non-linearity. Is there a way of using a\nDo-Loop for the same kind of thing? A simple example of the model that\nI tried is:\n\nRSolve[{a[n + 1] == a[n]*Exp[0.63((1 - a[n]/50)) - 0.068b[n]],\nb[n + 1] = a[n]*(1 - Exp[-0.068*b[n]]), a == 25,\nb == 10}, {a[n], b[n]}, n]\n\nI want an output for variables a and n at time n, n+1, etc.\n\nI'd be most grateful for any advice that you can offer, and please don't\nunderestimate my current level of ignorance.\n\nMany thanks,\n\nMark Hunter\n\n```\n\n• Prev by Date: Re: Status of MIER Jounal?\n• Next by Date: Re: File -> Notebooks -> Empty\n• Previous by thread: HairyPooter IV.75 cont.\n• Next by thread: Re: insect population dynamics" ]

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[ "", null, "", null, "# Important formulas in the chapter force and laws of motion.\n\nBy BYJU'S Exam Prep\n\nUpdated on: September 25th, 2023", null, "The important formulas in the chapter force and laws of motion are equations of motion, gravitational force, force, acceleration due to gravity, equations of motion for motion under gravitational force, and the law of conservation of linear momentum.\n\nTable of content", null, "### Equations of Motion\n\n• The first equation of Motion: v = u + at\n• Second equation of Motion: s = ut + ½ at2\n• Third equation of Motion: 2as = v2 – u2\n• Gravitational force: F = Gm1m2/d2\n• Force: F = ma\n• Acceleration due to gravity: Gm/R2\n• Weight: W = mg\n• Law of Conservation of linear momentum: mAvA + mBvB = mAuA + mBuB\n\nEquations of motion for motion under gravitational force:\n\n• v = u + at\n• h = ut + ½ gt2\n• v2 – u2 = 2gh\n\nFirst Law of Motion of Newton\n\nAccording to Newton’s First Law, a body in uniform motion or at rest will remain in that state up to and unless a net external force acts on it.\n\nSecond Law of Motion by Newton\n\nAccording to Newton’s second law, an object’s acceleration caused by a net force is directly proportional to the force’s magnitude, moving in the same direction as the force, and inversely proportional to the mass of the object.\n\nThe Third Law of Motion by Newton\n\nEvery action has an equal and opposite response, according to Newton’s third law.\n\nSummary:\n\n## Important formulas in the chapter force and laws of motion.\n\nEquations of motion, gravitational force, force, acceleration due to gravity, equations of motion for motion under gravitational force, and the law of conservation of linear momentum are some of the key formulas in the chapter on force and laws of motion.\n\nPOPULAR EXAMS\nSSC and Bank\nOther Exams", null, "GradeStack Learning Pvt. Ltd.Windsor IT Park, Tower - A, 2nd Floor, Sector 125, Noida, Uttar Pradesh 201303 [email protected]" ]

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[ "# How To Calculate The Number Of Molecules In Any Piece of DNA", null, "Lab math – it’s a dreaded topic, isn’t it? Calculating molecules in DNA samples, calculating the concentration of a solution, or working out serial dilutions were tasks that always filled me with dread as an undergrad. And to this day I am still not a huge fan! Fear not – there is light at the end of the tunnel!\n\nIn this article, we will go through counting molecules in DNA samples – for instance, how many templates/copies of a 50-base pair (bp) fragment do you have in a 500 ng sample?\n\nYou will be a whiz at these calculations in no time!\n\n## Calculating Molecules in DNA – Why Would I Need To?\n\nThis information may come in useful when doing the following types of experiments:\n\n1. DNA ligations.\n2. Biochemical assays.\n3. Real-time PCR (absolute quantification).\n4. Measuring viral load/numbers of viral genomes.\n\nWhatever you need this information for, you can go about it in the same step-by-step fashion.\n\n## Here Comes the Science…\n\n• The average weight of a single DNA bp is 650 daltons. This can also be written as 650 g/mol (= molar mass).\n• This is the same as saying that one mole of a bp weighs 650 g.\n• The molecular weight or molar mass of any double-stranded DNA fragment can therefore be calculated by multiplying its length (in bp) by 650 and the answer will be expressed as daltons or g/mol.\n\nAs long as you know the sequence length (i.e. the length/size of the genome), this calculation will work for the genomic DNA of any species. To get the length of your sequence in bp, count the nucleotides in the DNA sequence – either by hand or using character count in a word processing program. If you need some help with converting your calculations between different units, check out our guide for solving your lab math problems.\n\nQuestion: What is the molecular weight of a 5.2-kb plasmid? Remember 5.2 kb = 5200 bp\n\nCalculation: 5200 bp x 650 daltons = 3,380,000 daltons or g/mol\n\nNow we know how to calculate the molecular weight of a DNA template. Using Avogadro’s number, which is 6.022 x 1023 molecules/mole, the number of molecules of the template per gram can be calculated using the following formula:\n\n$\\textrm{Number of copies} = (\\textup{\\textrm{}ng} \\times [6.022 \\times 10^{23}])/(\\textrm{length }\\times [1 \\times 10^{9}] \\times 650)$\n\n• ng is the amount of DNA (plasmid, primer, etc.) you have in nanograms.\n• 6.022 x 1023 = Avogadro’s number.\n• Length is the length of your DNA fragment in base pairs. Just multiply by 1000 if you are working in kb.\n• We multiply by 1 x 109 to convert our answer to nanograms.\n\nNote: the Avogadro’s number or constant (denoted as NA) is defined as the number of constituent particles (usually atoms or molecules) per mole of a given substance.\n\n## Calculating Molecules in DNA: Useful Examples\n\n### Example 1: Plasmid\n\nQuestion: You have 300 ng of a 5.2-kb plasmid.\n\nHow many copies of your plasmid do you have?\n\nCalculation: Number of copies =\n\n$(300 \\times [6.022 \\times 10^{23}])/(5200 \\times [1 \\times 10^{9}] \\times 650)$\n\n### Example 2: Primer\n\nQuestion: You have 50 ng of a 20-bp primer.\n\nHow many copies of your primer do you have?\n\nCalculation: Number of copies =\n\n$(50 \\times [6.022 \\times 10^{23}])/(20 \\times [1 \\times 10^{9}] \\times 650)$\n\n### Example 3: PCR product\n\nQuestion: You have 150 ng of a 670-bp purified PCR product.\n\nHow many copies of your PCR product do you have?\n\nCalculation: Number of copies =\n\n$(150 \\times [6.022 \\times 10^{23}])/(670 \\times [1 \\times 10^{9}] \\times 650)$\n\nNote:  It’s important to remember that this formula is based on the assumption that you are working with a single DNA species, so when applying to PCR products or plasmids, make sure you only have 1 band or that you have a clean plasmid miniprep.\n\nSee, that wasn’t so bad after all. Once you know the length of your DNA fragment and the concentration in nanograms, you can easily apply the formula to get the number of copies in any type of DNA sample.\n\nHappy calculating molecules in DNA. If you have any comments, queries, or suggestions, get in touch using the comment box below.\n\nOriginally published September 17, 2014. Reviewed and updated April 27, 2021.\n\n1.", null, "Anu on October 11, 2019 at 8:02 pm\n\nThe molecular weight (Da) of the zwitterionic form of the peptide GAGAGAGA?\n\n2.", null, "fan xi wen on June 2, 2019 at 5:01 am\n\n3.", null, "DATrevino on September 1, 2018 at 4:20 pm\n\nShouldn’t you account for double or single strand DNA?\n\nScroll To Top" ]

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[ "## Algorithms\n\n Question 1\n\nConsider a sequence of 14 elements: A = [-5, -10, 6, 3, -1, -2, 13, 4, -9, -1, 4, 12, -3, 0]. The subsequence sum", null, ". Determine the maximum of S(i,j), where 0 ≤ i ≤ j < 14. (Divide and conquer approach may be used)\n\n A 19 B 39 C 29 D 09\nAlgorithms       GATE 2019\nQuestion 1 Explanation:\nFirst understand the subsequence is an array is\nEx: {A,B,C,D}\nB, BC,BD,BCD,\nC,CD,\nD }\nStep-1: Array of elements A = [-5, -10, 6, 3, -1, -2, 13, 4, -9, -1, 4, 12, -3, 0 ]\nStep-2: As per the given question, if they want to find maximum subsequence means\n{6,3,13,4,4,12}\n= 42\nStep-3: But according to given recurrence relation, the sequence should be continuous. {6,3,13,4,4,12}.\nThis is not continuous subsequence.\nStep-4: The continuous sequence is {6, 3, -1, -2, 13, 4, -9, -1, 4, 12}\nTotal is {29}\nNote: We can't get more than 29 maximum subsequence sum.\n Question 2\n\nAn array of 25 distinct elements is to be sorted using quicksort. Assume that the pivot element is chosen uniformly at random. The probability that the pivot element gets placed in the worst possible location in the first round of partitioning (rounded off to 2 decimal places) is _____.\n\n A 0.08 B 0.01 C 1 D 8\nAlgorithms       GATE 2019\nQuestion 2 Explanation:\nStep-1: Given, 25 distinct elements are to be sorted using quicksort.\nStep-2: Pivot element = uniformly random.\nStep-3: Worst case position in the pivot element is either first (or) last.\nStep-4: So total 2 possibilities among 25 distinct elements\n= 2/25\n= 0.08\n Question 3\nConsider the following statements:\n``` I. The smallest element in a max-heap is always at a leaf node\nII. The second largest element in a max-heap is always a child of the root node\nIII. A max-heap can be constructed from a binary search tree in Θ(n) time\nIV. A binary search tree can be constructed from a max-heap in Θ(n) time\n```\nWhich of the above statements are TRUE?\n A I, II and III B II, III and IV C I, III and IV D I, II and IV\nAlgorithms       GATE 2019\nQuestion 3 Explanation:\ni) TRUE: The smallest element in heap is always a leaf node but depends upon the graph, it may be left or right side of the graph.\n(ii) TRUE: The second smallest element in a heap is always a child of root node.", null, "(iii) TRUE: Converting from binary search tree to max heap will take O(n) time as well as O(n) space complexity.\n(iv) FALSE: We can’t convert max heap to binary search tree in O(n) time.\n Question 4\n\nThere are n unsorted arrays: A1, A2, ..., An. Assume that n is odd. Each of A1, A2, ..., An contains n distinct elements. There are no common elements between any two arrays. The worst-case time complexity of computing the median of the medians of A1, A2, ..., An is\n\n A O(n) B O(n log n) C Ω(n2 log n) D O(n2)\nAlgorithms       GATE 2019\nQuestion 4 Explanation:\nFinding the median in an unsorted array is O(n).\nBut it is similar to quicksort but in quicksort, partitioning will take extra time.\n→ Find the median will be (i+j)/2\n1. If n is odd, the value is Ceil((i+j)/2)\n2. If n is even, the value is floor((i+j)/2)\n-> Here, total number of arrays are\n⇒ O(n)*O(n)\n⇒ O(n2)\nNote:\nThey are clearly saying that all are distinct elements.\nThere is no common elements between any two arrays.\n Question 5\nLet G be any connected, weighted, undirected graph.\n```I. G  has a unique minimum spanning tree, if no two edges of G have the same weight.\nII. G  has a unique minimum spanning tree, if, for every cut of G, there is a unique minimum-weight edge crossing the cut.\n```\nWhich of the above statements is/are TRUE?\n A I only B II only C Both I and II D Neither I nor II\nAlgorithms       GATE 2019\nQuestion 5 Explanation:\nGiven G be a connected, weighted and undirected graph,\nI. TRUE: G Graph is unique, no two edges of the graph is same.", null, "Step-1: Using Kruskal's algorithm, arrange each weights in ascending order.\n17, 18, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30\nStep-2:", null, "Step-3: 17 + 18 + 20 + 21 + 22 + 23 + 26 = 147\nStep-4: Here, all the elements are distinct. So, the possible MCST is 1.\nII. TRUE: As per the above graph, if we are cut the edge, that should the be the minimum edge.\nBecause we are already given, all minimum edge weights if graph is distinct.\n Question 6\n A", null, "B", null, "C", null, "D", null, "Algorithms       Recurrences       GATE 2020\nQuestion 6 Explanation:", null, "Question 7\nLet G = (V,E) be a directed, weighted graph with weight function w:E--> R. For some function f:V--> R, for each edge (u,v)E, define w'(u,v) as w(u,v)+f(u)-f(v).\nWhich one of the options completes the following sentence so that it is TRUE?\n“The shortest paths in G under w are shortest paths under w’ too, _______”.\n A if and only if f(u) is the distance from s to u in the graph obtained by adding a new vertex s to G and edges of zero weight from s to every vertex of G B", null, "C", null, "D for every f:V--> R\nAlgorithms       Graphs-and-Tree       GATE 2020\nQuestion 7 Explanation:", null, "Question 8\nLet G = (V,E) be a weighted undirected graph and let T be a Minimum Spanning Tree (MST) of G maintained using adjacency lists. Suppose a new weighted edge (u,v) ε V x V is added to G. The worst case time complexity of determining if T is still an MST of the resultant graph is\n A", null, "B", null, "C", null, "D", null, "Algorithms       Minimum-Spanning-Tree       GATE 2020\nQuestion 8 Explanation:\nMethod-1:\n• As T is a minimum spanning tree and we need to add a new edge to existing spanning tree.\n• Later we need to check still T is a minimum spanning tree or not, So we need to check all vertices whether there is any cycle present after adding a new edge .\n• All vertices need to traverse to confirm minimum spanning tree after adding new edge then time complexity is O(V)\nMethod-2:\nTime Complexity:\nTotal vertices: V, Total Edges : E\n• O(logV) – to extract each vertex from the queue. So for V vertices – O(VlogV)\n• O(logV) – each time a new pair object with a new key value of a vertex and will be done at most once for each edge. So for total E edge – O(ElogV)\n• So overall complexity: O(VlogV) + O(ElogV) = O((E+V)logV) = O(ElogV)\nNote: Method-1 is the most appropriate answer for giving a question.\n Question 9\nConsider a graph G = (V, E), where V = {v1, v2, …, v100}, E = {(vi, vj) | 1 ≤ i < j ≤ 100}, and weight of the edge (vi, vj) is |i - j|. The weight of the minimum spanning tree of G is ______.\n A 99\nAlgorithms       Minimum-Spanning-Tree       GATE 2020\nQuestion 9 Explanation:\n• If there are n vertices in the graph, then each spanning tree has n − 1 edges.\n• N =100\n• Edge weight is |i-j| for Edge (vi,vj) {1<=i<=100}\n• The weight of edge(v1,v2) is 1 , edge(v5,v6) is 1.\n• So 99 edges of weight is 99\n Question 10\n\nConsider the following undirected graph G:", null, "Choose a value for x that will maximize the number of minimum weight spanning trees (MWSTs) of G. The number of MWSTs of G for this value of x is _________.\n\n A 4 B 5 C 6 D 7\nAlgorithms       Minimum-Spanning-Tree       Gate 2018\nQuestion 10 Explanation:\nHere, x = 5 because it is having maximum number of spanning trees.\nIf x = 5 then the total number of MWSTs are 4.\nIf r = 1", null, "If r = 2", null, "If r = 3", null, "If r = 4", null, "If r = 5", null, "", null, "Question 11\n\nConsider the weights and values of items listed below. Note that there is only one unit of each item.", null, "The task is to pick a subset of these items such that their total weight is no more than 11 Kgs and their total value is maximized. Moreover, no item may be split. The total value of items picked by an optimal algorithm is denoted by Vopt. A greedy algorithm sorts the items by their value-to-weight ratios in descending order and packs them greedily, starting from the first item in the ordered list. The total value of items picked by the greedy algorithm is denoted by Vgreedy.\n\nThe value of Vopt − Vgreedy is ______ .\n\n A 16 B 17 C 18 D 19\nAlgorithms       0/1-Knapsack-and-fractional-knapsack       Gate 2018\nQuestion 11 Explanation:\nFirst sort value/weight in descending order as per the question:", null, "Vopt is clearly = 60\nFor Vgreedy use the table (Do not take the fraction as per the question),\nItem 4 picked,\nProfit = 24\nRemaining weight = 11 – 2 = 9\nNext item 3 picked (item 1 cannot be picked since its capacity is greater than available capacity),\nProfit = 24 + 20 = 44\nRemaining capacity = 9 – 4 = 5\nNow no item can be picked with available capacity.\nSo Vgreedy = 44\n∴ Vopt – Vgreedy = 60 – 44 = 16\n Question 12\n\nConsider the following functions from positives integers to real numbers\n\nThe CORRECT arrangement of the above functions in increasing order of asymptotic complexity is:\n\n A", null, "B", null, "C", null, "D", null, "Algorithms       Asymptotic-Complexity       Gate 2017 set-01\nQuestion 12 Explanation:\nIn this problem, they are expecting to find us “increasing order of asymptotic complexity”.\nStep-1: Take n=2048 or 211 (Always take n is very big number)\nStep-2: Divide functions into 2 ways\n1. Polynomial functions\n2. Exponential functions\nStep-3: The above functions are belongs to polynomial. So, simply substitute the value of n,\nFirst compare with constant values.\n→ 100 / 2048 = 0.048828125\n→ 10 > 100/ 2048\n→ log2 2048 =11\n→ √n = 45.25483399593904156165403917471\n→ n = 2048\nSo, Option B is correct\n Question 13\n\nConsider the following table", null, "Match the algorithm to design paradigms they are based on:\n\n A (P)↔(ii), Q↔(iii), (R)↔(i) B (P)↔(iii), Q↔(i), (R)↔(ii) C (P)↔(ii), Q↔(i), (R)↔(iii) D (P)↔(i), Q↔(ii), (R)↔(iii)\nAlgorithms       Gate 2017 set-01\nQuestion 13 Explanation:\n(P) Kruskal’s and Prim’s algorithms for finding Minimum Spanning Tree(MST). To find MST we are using greedy technique.\n(Q) QuickSort is a Divide and Conquer algorithm.\n(R) Floyd Warshall Algorithm is for solving the All Pairs Shortest Path problem using Dynamic Programming.\nSome important points regarding Greedy Vs Dynamic Programming\nGreedy: →\nIt always gives polynomial time complexity\n→ It is not an optimal\n→ It always selects only either minimum or maximum among all possibilities\n→ Ex: Dijkstra’s algorithm for SSSP, Optimal Merge Pattern, Huffman coding, Fractional knapsack problem, etc..,\nDynamic Programming:\n→ It gives either polynomial or exponential time complexity.\n→ It gives always an optimal result.\n→ It checks all possibilities of a problem.\n→ Ex: Longest Common sequence, Matrix chain Multiplication, Travelling sales Problem, etc..,\n Question 14\n\nLet G = (V, E) be any connected undirected edge-weighted graph. The weights of the edges in E are positive and distinct. Consider the following statements:\n\n(I) Minimum Spanning Tree of G is always unique.\n(II) Shortest path between any two vertices of G is always unique.\n\nWhich of the above statements is/are necessarily true?\n\n A (I) only B (II) only C both (I) and (II) D neither (I) nor (II)\nAlgorithms       Minimum-Spanning-Tree       Gate 2017 set-01\nQuestion 14 Explanation:\nIf the graph has all positive and distinct (unique values no duplicates) then Statement-I definitely correct because if we are using either prim’s or kruskal’s algorithm it gives the unique spanning tree.\nLet us take an example", null, "Step 1:\nUsing kruskal’s algorithm, arrange each weights in ascending order.\n17, 18, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30\nStep 2:", null, "Step 3:\n17+18+20+21+22+23+26 = 147\nStep 4:\nHere, all the elements are distinct. So the possible MCST is 1.\nStatement-II: May or may not happen, please take an example graph and try to solve it. This is not correct always.\nSo, we have to pick most appropriate answer.\n Question 15\n\nLet A be an array of 31 numbers consisting of a sequence of 0’s followed by a sequence of 1’s. The problem is to find the smallest index i such that A[i] is 1 by probing the minimum number of locations in A. The worst case number of probes performed by an optimal algorithm is _________.\n\n A 5 B 6 C 7 D 8\nAlgorithms       Searching       Gate 2017 set-01\nQuestion 15 Explanation:\n→ If we apply binary search to find the first occurrence of 1 in the list, it will give us the smallest index i where 1 is stored.\n→ As in this array sequence of 0’s is followed by sequence of 1’s, the array is sorted. We can apply binary search directly without sorting it.\nSo number of probes = ceil(log2 31) = 4.954196310386876\n⇒ here we are using ceiling so it becomes 5\n Question 16\n\nMatch the algorithms with their time complexities:\n\n``` Algorithm                                Time complexity\n(P) Towers of Hanoi with n disks                      (i) Θ(n2)\n(Q) Binary search given n stored numbers              (ii) Θ(n log⁡ n)\n(R) Heap sort given n numbers at the worst case      (iii) Θ(2n)\n(S) Addition of two n×n matrices                      (iv) Θ(log⁡ n)\n```\n A P→(iii), Q→(iv), R→(i), S→(ii) B P→(iv), Q→(iii), R→(i), S→(ii) C P→(iii), Q→(iv), R→(ii), S→(i) D P→(iv), Q→(iii), R→(ii), S→(i)\nAlgorithms       Match-the-Following       GATE 2017(set-02)\nQuestion 16 Explanation:\nIn this problem, we have to find Average case of different algorithms\n→ Tower of Hanoi with n disks takes θ(2n) time\nIt is a mathematical game or puzzle.\nIt consists of three rods and a number of disks of different sizes, which can slide onto any rod.\nThe puzzle starts with the disks in a neat stack in ascending order of size on one rod, the smallest at the top, thus making a conical shape.\nThe objective of the puzzle is to move the entire stack to another rod, obeying the following simple rules:\n1. Only one disk can be moved at a time.\n2. Each move consists of taking the upper disk from one of the stacks and placing it on top of another stack.\n3. No disk may be placed on top of a smaller disk.\nWith 3 disks, the puzzle can be solved in 7 moves.\nThe minimal number of moves required to solve a Tower of Hanoi puzzle is 2n-1, where n is the number of disks.\n→ Binary Search given n sorted numbers takes Ɵ(log2 n)\n→ Heap sort given n numbers of the worst case takes Ɵ(n log n)\n→ Addition of two n×n matrices takes Ɵ(n2)\n Question 17\n A Θ(log⁡log⁡n) B Θ(log⁡n) C Θ(√n) D Θ(n)\nAlgorithms       Time-Complexity       GATE 2017(set-02)\nQuestion 17 Explanation:\nT(n) = 2T(√n) + 1\n(or)\nT(n) = 2T(n(1⁄2)) + 1\nHere, Assume n = 2k\nT(2k) = 2T(2k)(1⁄2) + 1\nT(2k) = 2T(2(k/2) ) + 1\nAssume T(2k) = S(k)\nS(k) = 2S(k/2) + 1\nApply Master’s theorem Case-1\na=2, b=2\nS(k) = k(log22 )\nS(k) = θ(k')\nbut S(k) = T(2k)\nT(2k) = θ(k')\nbut n = 2k\nk = log⁡n\nT(n) = θ(logn)\n Question 18\n\nConsider the following C function.\n\n```int fun (int n)   {\nint i, j;\nfor (i = 1; i <= n; i++)   {\nfor (j = 1; j < n; j += i)   {\nprintf(\"%d %d\",i,j);\n}\n}\n}\n```\n\nTime complexity of fun in terms of Θ notation is\n\n A Θ(n√n) B Θ(n2 ) C Θ(n log⁡n) D Θ(n2 logn)\nAlgorithms       Time-Complexity       GATE 2017(set-02)\nQuestion 18 Explanation:\nWe can solve iterative programs time complexity with the help of rollback method.\nint fun(int n)\n{\nint i, j;\nfor (i = 1; i <= n ; i++)    /* It is independent loop. So, it takes O(n) time */\n{\nfor (j = 1; j < n; j += i)     /* It is dependent loop, It always incrementing with the help of i. It will take approximately O(log n) times*/\n{\nprintf(\"%d %d\", i, j);     /* It takes O(1) time */\n}\n}\n}\nSo, the overall complexity of the program is θ(n log⁡n) times.\n Question 19\n\nA message is made up entirely of characters from the set X = {P, Q, R, S, T}. The table of probabilities for each of the characters is shown below:", null, "If a message of 100 characters over X is encoded using Huffman coding, then the expected length of the encoded message in bits is __________.\n\n A 225 B 226 C 227 D 228\nAlgorithms       Huffman-Coding       GATE 2017(set-02)\nQuestion 19 Explanation:", null, "General procedure to solve Huffman coding problem\nStep-1: Arrange into either descending/ ascending order according to that profits.\nStep-2: Apply optimal merge pattern procedure.\nStep-3: Make left sub-tree value either 0 or 1, for right sub-tree, vice-versa.", null, "", null, "", null, "", null, "= 2 × 0.34 + 2 × 0.22 + 2 × 0.19 + 3 × 0.17 + 3 × 0.08\n= 2.25\n∴ So, for 100 characters, 2.25 * 100 = 225\n Question 20\n\nThe worst case running times of Insertion sort, Merge sort and Quick sort, respectively, are:\n\n A Θ(nlogn), Θ(nlogn), and Θ(n2) B Θ(n2 ), Θ(n2 ), and Θ(nlogn) C Θ(n2), Θ(nlogn), and Θ(nlogn) D Θ(n2), Θ(nlogn), and Θ(n2)\nAlgorithms       Sorting       2016 set-01\nQuestion 20 Explanation:", null, "Question 21\n\nLet G be a complete undirected graph on 4 vertices, having 6 edges with weights being 1, 2, 3, 4, 5, and 6. The maximum possible weight that a minimum weight spanning tree of G can have is __________.\n\n A 7 B 8 C 9 D 10\nAlgorithms       Minimum-Spanning-Tree       2016 set-01\nQuestion 21 Explanation:\nLet G be a complete undirected graph with 4 vertices & 6 edges so according to graph theory, if we use Prim’s / Kruskal’s algorithm, the graph looks like", null, "Now consider vertex A to make Minimum spanning tree with Maximum weights.\nAs weights are 1, 2, 3, 4, 5, 6. AB, AD, AC takes maximum weights 4, 5, 6 respectively.\nNext consider vertex B, BA = 4, and minimum spanning tree with maximum weight next is BD & BC takes 2, 3 respectively.\nAnd the last edge CD takes 1.\nSo, 1+2+4 in our graph will be the Minimum spanning tree with maximum weights.\n Question 22\n\nG = (V,E) is an undirected simple graph in which each edge has a distinct weight, and e is a particular edge of G. Which of the following statements about the minimum spanning trees (MSTs) of G is/are TRUE?\n\nI. If e is the lightest edge of some cycle in G, then every MST of G includes e\nII. If e is the heaviest edge of some cycle in G, then every MST of G excludes e\n A I only B II only C both I and II D neither I nor II\nAlgorithms       Minimum-Spanning-Tree       2016 set-01\nQuestion 22 Explanation:\nStatement-1: False\nThe MSTs of G may or may not include the lightest edge.\nTake rectangular graph labelled with P,Q,R,S.\nConnect with P-Q = 5, Q-R = 6, R-S = 8, S-P = 9, P-R = 7.\nWhen we are forming a cycle R-S-P-R. P-R is the lightest edge of the cycle.\nThe MST abcd with cost 11\nP-Q + Q-R + R-S does not include it.\nStatement-2: True\nSuppose there is a minimum spanning tree which contains e. If we add one more edge to the spanning tree we will create a cycle.\nSuppose we add edge e to the spanning tree which generated cycle C.\nWe can reduce the cost of the minimum spanning tree if we choose an edge other than e from C for removal which implies that e must not be in minimum spanning tree and we get a contradiction.\n Question 23\n\nAssume that the algorithms considered here sort the input sequences in ascending order. If the input is already in ascending order, which of the following are TRUE?\n\nI. Quicksort runs in Θ(n2) time\nII. Bubblesort runs in Θ(n2) time\nIII. Mergesort runs in Θ(n) time\nIV. Insertion sort runs in Θ(n) time\n A I and II only B I and III only C II and IV only D I and IV only\nAlgorithms       Sorting       GATE 2016 set-2\nQuestion 23 Explanation:\nIf input sequence is already sorted then the time complexity of Quick sort will take O(n2) and Bubble sort will take O(n) and Merge sort will takes O(nlogn) and insertion sort will takes O(n).\n→ The recurrence relation for Quicksort, if elements are already sorted,\nT(n) = T(n-1)+O(n) with the help of substitution method it will take O(n2).\n→ The recurrence relation for Merge sort, is elements are already sorted,\nT(n) = 2T(n/2) + O(n) with the help of substitution method it will take O(nlogn).\nWe can also use master's theorem [a=2, b=2, k=1, p=0] for above recurrence.\n Question 24\n\nThe Floyd-Warshall algorithm for all-pair shortest paths computation is based on\n\nAlgorithms       Floyd-Warshall-Algorithm       GATE 2016 set-2\nQuestion 24 Explanation:\n→ All Pair shortest path algorithm is using Dynamic Programming technique.\nIt takes worst case time complexity is O(|V|3) and worst case space complexity is O(|V|2).\n→ The Floyd–Warshall algorithm is an algorithm for finding shortest paths in a weighted graph with positive or negative edge weights (but with no negative cycles).\n→ A single execution of the algorithm will find the lengths (summed weights) of the shortest paths between all pairs of vertices.\nAlthough it does not return details of the paths themselves, it is possible to reconstruct the paths with simple modifications to the algorithm.\n Question 25\n\nLet A1, A2, A3 and A4 be four matrices of dimensions 10 × 5, 5 × 20, 20 × 10, and 10 × 5, respectively. The minimum number of scalar multiplications required to find the product A1A2A3A4 using the basic matrix multiplication method is _________.\n\n A 1500 B 1501 C 1502 D 1503\nAlgorithms       Matrix-Chain-Multiplication       GATE 2016 set-2\nQuestion 25 Explanation:\n→ The minimum number of scalar multiplications required is 1500.\nThe optimal parenthesized sequence is A1((A2A3)A4) out of many possibilities, the possibilities are\n1. ((A1A2)A3)A4\n2. ((A1(A2A3))A4)\n3. (A1A2)(A3A4)\n4. A1((A2A3)A4)\n5. A1(A2(A3A4))\n→ A1((A2A3)A4) = (5 x 20 x 10) + (5 x 10 x 5) + (10 x 5 x 5) = 1000 + 250 + 250 = 1500\n Question 26\n\nThe given diagram shows the flowchart for a recursive function A(n). Assume that all statements, except for the recursive calls, have O(1) time complexity. If the worst case time complexity of this function is O(nα), then the least possible value (accurate upto two decimal  positions) of α is _________.", null, "A 2.32193 B 2.32193 C 2.32193 D 2.32193\nAlgorithms       Time-Complexity       GATE 2016 set-2\nQuestion 26 Explanation:\nThis type of problem, we have to consider worst case time complexity, it mean that all possibilities.\nAccording to flow chart total 5 worst case possibilities of function calls.", null, "The remaining function calls/return statements will execute only constant amount of time.\nSo, total function calls 5.\nThe Recurrence will be\nA(n) = 5A(n/2) + O(1)\nApply Master’s theorem,\nA=5, b=2, k=0, p=0\na > bk case\nA(n) = n(logba ) = n(log25) = n2.3219280\n∴α = 2.3219280\n Question 27\n\nWhich one of the following is the recurrence equation for the worst case time complexity of the Quicksort algorithm for sorting n(≥ 2) numbers? In the recurrence equations given in the options below, c is a constant.\n\n A T(n) = 2T(n/2) + cn B T(n) = T(n – 1) + T(1) + cn C T(n) = 2T(n – 1) + cn D T(n) = T(n/2) + cn\nAlgorithms       Quick-Sort       GATE 2015 (Set-01)\nQuestion 27 Explanation:\nWhen the pivot is the smallest (or largest) element at partitioning on a block of size n the result yields one empty sub-block, one element (pivot) in the correct place and sub block of size n-1.\nHence recurrence relation T(n) = T(n - 1) + T(1) + cn.\n Question 28\n A P-iii, Q-ii, R-iv, S-i B P-i, Q-ii, R-iv, S-iii C P-ii, Q-iii, R-iv, S-i D P-ii, Q-i, R-iii, S-iv\nAlgorithms       Algorithms       GATE 2015 (Set-01)\nQuestion 28 Explanation:\nPrim’s algorithm always select minimum distance between two of its sets which is nothing but greedy method.\nFloyd-warshall always changes it distance at each iteration which is nothing but dynamic programming.\nMerge sort in merge sort first we always divide and merge to perform sorting hence divide and conquer.\nHamiltonian circuit used to reach the entire vertex once, if some vertex is repeating in its path it will backtrack.\n Question 29\n\nAn algorithm performs (log N)1/2 find operations, N insert operations, (log N)1/2 delete operations, and (log N)1/2 decrease-key operations on a set of data items with keys drawn from a linearly ordered set. For a delete operation, a pointer is provided to the record that must be deleted. For the decrease-key operation, a pointer is provided to the record that has its key decreased. Which one of the following data structures is the most suited for the algorithm to use, if the goal is to achieve the best total asymptotic complexity considering all the operations?\n\n A Unsorted array B Min-heap C Sorted array D Sorted doubly linked list\nAlgorithms       Time-Complexity       GATE 2015 (Set-01)\nQuestion 29 Explanation:", null, "Question 30\nThe graph shown below 8 edges with distinct integer edge weights. The minimum spanning tree (MST) is of weight 36 and contains the edges: {(A, C), (B, C), (B, E), (E, F), (D, F)}. The edge weights of only those edges which are in the MST are given in the figure shown below. The minimum possible sum of weights of all 8 edges of this graph is ______________.", null, "A 69 B 70 C 71 D 72\nAlgorithms       Minimum-Spanning-Tree       GATE 2015 (Set-01)\nQuestion 30 Explanation:", null, "⇒ Total sum = 10 + 9 + 2 + 15 + 7 + 16 + 4 + 6 = 69\n Question 31\n A 1-i, 2-iii, 3-i, 4-v. B 1-iii, 2-iii, 3-i, 4-v. C 1-iii, 2-ii, 3-i, 4-iv. D 1-iii, 2-ii, 3-i, 4-v.\nQuestion 31 Explanation:\n(1) Dijkstra’s Shortest Path using to find single source shortest path. It is purely based on Greedy technique because it always selects shortest path among all.\n(2) → All Pair shortest path algorithm is using Dynamic Programming technique. It takes worst case time complexity is O(|V|3) and worst case space complexity is O(|V|2).\n→ The Floyd–Warshall algorithm is an algorithm for finding shortest paths in a weighted graph with positive or negative edge weights (but with no negative cycles).\n→ A single execution of the algorithm will find the lengths (summed weights) of the shortest paths between all pairs of vertices. Although it does not return details of the paths themselves, it is possible to reconstruct the paths with simple modifications to the algorithm.\n(3) Binary search on a sorted array:\nSearch a sorted array by repeatedly dividing the search interval in half. Begin with an interval covering the whole array. If the value of the search key is less than the item in the middle of the interval, narrow the interval to the lower half. Otherwise narrow it to the upper half. Repeatedly check until the value is found or the interval is empty.\n(4) Backtracking search on a graph uses Depth-first search\n Question 32\n A (a, left_end, k) and (a + left_end + 1, n – left_end – 1, k – left_end – 1) B (a, left_end, k) and (a, n – left_end – 1, k – left_end – 1) C (a + left_end + 1, n – left_end – 1, k – left_end – 1) and (a, left_end, k) D (a, n – left_end – 1, k – left_end – 1) and (a, left_end, k)\nAlgorithms       Partitioning-Algorithm       GATE 2015 -(Set-2)\nQuestion 32 Explanation:\nQuickSort is used as a sorting algorithm.In QuickSort, we pick a pivot element, then move the pivot element to its correct position and partition the array around it. The idea is, not to do complete quicksort, but stop at the point where pivot itself is k’th smallest element. Also, not to recur for both left and right sides of pivot, but recur for one of them according to the position of pivot.\n Question 33\nConsider a typical disk that rotates at 15000 rotations per minute (RPM) and has a transfer rate of 50×106 bytes/sec. If the average seek time of the disk is twice the average rotational delay and the controller’s transfer time is 10 times the disk transfer time, the average time (in milliseconds) to read or write a 512-byte sector of the disk is _____________.\n A 6 B 6.1 C 6.2 D 6.3\nAlgorithms       Secondary-Storage       GATE 2015 -(Set-2)\nQuestion 33 Explanation:\n15000 rotations → 60 sec\n1 rotation → 60 /15000 sec = 4 ms\n∴ Average rotational delay = 1/2 × 4 ms = 2 ms\nAverage seek time = 2 × Average rotational delay\n= 2 × 2 ms = 4 ms\nTime to transfer 512 byte,\n50 × 106B ------- 1s\n512 B ------- 512B/ 50×106B/s = 0.01 ms\n∴ Controllers transfer time\n= 10 × Disk transfer time\n= 10 × 0.01 ms\n= 0.01 ms\n∴ Avg. time = (4 + 2 + 0.1 + 0.01)ms\n= 6.11 ms\n Question 34\nAssume that a mergesort algorithm in the worst case takes 30 seconds for an input of size 64. Which of the following most closely approximates the maximum input size of a problem that can be solved in 6 minutes?\n A 256 B 512 C 1024 D 2048\nAlgorithms       Merge-Sort       GATE 2015(Set-03)\nQuestion 34 Explanation:\nTime complexity of merge sort = O(n log n) = an log n\nwhere c is some constant\nIt is given that for n = 64,\ncn log n = 30\nc × 64 log 64 = 30\nc × 64 × 6 = 30\nc = 5/64\nNow, the value of n for\nc n log n = 360\n5/64 × n log n = 360\nn log n = 360×64/5\n= 72 × 64 = 29 × 9\nSo for n = 29, it satisfies.\nSo, n = 29 = 512\n Question 35\nLet G be a connected undirected graph of 100 vertices and 300 edges. The weight of a minimum spanning tree of G is 500. When the weight of each edge of G is increased by five, the weight of a minimum spanning tree becomes __________.\n A 995 B 996 C 997 D 998\nAlgorithms       Minimum-Spanning-Tree       GATE 2015(Set-03)\nQuestion 35 Explanation:\nG has 100 vertices ⇒ spanning tree contain 99 edges given, weight of a minimum spanning tree of G is 500 since, each edge of G is increased by five. ∴ Weight of a minimum spanning tree becomes 500 + 5 ⨯ 99= 995\n Question 36\nThere are 5 bags labeled 1 to 5.  All the coins in a given bag have the same weight.  Some bags have coins of weight 10 gm, others have coins of weight 11 gm.  I pick 1, 2, 4, 8, 16 coins respectively from bags 1 to 5.  Their total weight comes out to 323 gm.  Then the product of the labels of the bags having 11 gm coins is _______.\n A 12 B 13 C 14 D 15\nAlgorithms       General       GATE 2014(Set-01)\nQuestion 36 Explanation:\nBags: 1 2 3 4 5\nNo. of coins picked: 1 2 4 8 16\nThere are two types of weights of coin, i.e., 10gm and 11gm. And total weight of picked coins are 323gm.\nLet there be x number of 10gm coins and y number of 11gm coins.\nSo, 10x + 11y = 323 ------ (1)\nAnd we also know that total number of coins picked is\n1 + 2 + 4 + 8 + 16 = 31 which is equal to x + y, so,\n= 31 ------ (2)\nSolving equation (1) and (2), we get\ny = 13\nMeans total there are 13 coins of 11gm.\nNow we can chose bag number 1, 3 and 4, we will get a total sum of 13 coins picked from them.\nSo product of labelled bag number = 1×3×4 = 12\n Question 37\nSuppose a polynomial time algorithm is discovered that correctly computes the largest clique in a given graph. In this scenario, which one of the following represents the correct Venn diagram of the complexity classes P, NP and NP Complete (NPC)?\n A", null, "B", null, "C", null, "D", null, "Algorithms       P-NP       GATE 2014(Set-01)\nQuestion 37 Explanation:\nThe most important open question in complexity theory is whether the P = NP, which asks whether polynomial time algorithms actually exist for NP-complete and all NP problems (since a problem “C” is in NP-complete, iff C is in NP and every problem in NP is reducible to C in polynomial time). In the given question it is given that some polynomial time algorithm exists which computes the largest clique problem in the given graph which is known NP-complete problem. Hence P=NP=NP-Complete.\n Question 38\nThe minimum number of comparisons required to find the minimum and the maximum of 100 numbers is _________________.\n A 148 B 149 C 150 D 151\nAlgorithms       General       GATE 2014(Set-01)\nQuestion 38 Explanation:\nThe minimum number of comparisons required to find the minimum and the maximum of 100 numbers is 3n/2 – 2 = 148. (∵ T(n) = T(floor(n/2))+T(ceil(n/2))+2)\n Question 39\nWhich one of the following correctly determines the solution of the recurrence relation with T(1) = 1?\n`T(n) = 2T(n/2) + Logn`\n A Θ(n) B Θ(nlog n) C Θ(n2) D Θ(log⁡n)\nAlgorithms       Time-Complexity       Gate 2014 Set -02\nQuestion 39 Explanation:\nT(n)=2T(n/2)+logn\nApply Master’s theorem,\na=2,b=2,k=0,p=1\nAccording to Master’s theorem, we can apply Case-I\n(I) If a>bk, then T(n)=θ(n(logba ) ) =θ(n(log22 ) )= θ (n)\n Question 40\nConsider two strings A = \"qpqrr\" and B = \"pqprqrp\". Let x be the length of the longest common subsequence (not necessarily contiguous) between A and B and let y be the number of such longest common subsequences between A and B. Then x+10y = _________.\n A 34 B 35 C 36 D 37\nAlgorithms       Dynamic-Programming       Gate 2014 Set -02\nQuestion 40 Explanation:\nGiven is\nA = “qpqrr” B = “pqprqrp”\nThe longest common subsequence (not necessarily contiguous) between A and B is having 4 as the length, so x=4 and such common subsequences are as follows:\n(1) qpqr\n(2) pqrr\n(3) qprr\nSo y = 3 (the number of longest common subsequences) hence x+10y = 4+10*3 = 34\n Question 41\nSuppose P, Q, R, S, T are sorted sequences having lengths 20, 24, 30, 35, 50 respectively.  They are to be merged into a single sequence by merging together two sequences at a time.  The number of comparisons that will be needed in the worst case by the optimal algorithm for doing this is ______.\n A 358 B 359 C 360 D 361\nAlgorithms       Greedy-Algorithm       Gate 2014 Set -02\nQuestion 41 Explanation:\nThe implementation of optimal algorithm for merging sequences is as follows.", null, "In the above implementation, total number of comparisons is\n(44-1) + (94-1) + (65-1) + (159-1) = 358\nHint: The number of comparisons for merging two sorted sequences of length m and n is m+n-1.\n Question 42\nThe number of distinct minimum spanning trees for the weighted graph below is ____", null, "A 6 B 7 C 8 D 9\nAlgorithms       Minimum-Spanning-Tree       Gate 2014 Set -02\nQuestion 42 Explanation:", null, "Minimum Spanning Tree:", null, "From the diagram, CFDA gives the minimum weight so will not disturb them, but in order to reach BE=1 we have 3 different ways ABE/ DBE/ DEB and we have HI=1, the shortest weight, we can reach HI=1 through GHI/ GIH.\nSo 3*2=6 ways of forming Minimum Spanning Tree with sum as 11.\n Question 43\nConsider the following rooted tree with the vertex la beled P as t he root:", null, "The order in which the nodes are visited during an i n-order trave rsal of the tree is\n A SQPTRWUV B SQPTUWRV C SQPTWUVR D SQPTRUWV\nAlgorithms       Tree Traversal       Gate 2014 Set -03\nQuestion 43 Explanation:\nThe tree can be redrawn as", null, "Inorder Traversal: Left, Root, Middle, Right.\nSo, durig inorder traversal whenever we visit the node second time then print it.\nSo, output will be,\nS Q P T R W U V\n Question 44\nSuppose depth first search is executed on the graph below starting at some unknown vertex. Assume that a recursive call to visit a vertex is made only after first checking that the vertex has not been visited earlier. Then the maximum possible recursion depth (including the initial call) is _________.", null, "A 19 B 20 C 21 D 22\nAlgorithms       Graphs       Gate 2014 Set -03\nQuestion 44 Explanation:", null, "Note: We should not consider backtrack edges, it reduces recursion depth in stack.\nSo the maximum possible recursive depth will be 19.", null, "Question 45\nYou have an array of n elements. Suppose you implement quicksort by always choosing the central element of the array as the pivot. Then the tightest upper bound for the worst case performance is\n A O(n2) B O(n log n) C Θ(n log⁡n) D O(n3)\nAlgorithms       Sorting       Gate 2014 Set -03\nQuestion 45 Explanation:\nThe Worst case time complexity of quick sort is O (n2). This will happen when the elements of the input array are already in order (ascending or descending), irrespective of position of pivot element in array.\n Question 46\nSuppose we have a balanced binary search tree T holding n numbers.  We are given two numbers L and H and wish to sum up all the numbers in T that lie between L and H. Suppose there are m such numbers in T. If the tightest upper bound on the time to compute the sum is O(nalogbn + mclogdn), the value of a + 10b + 100c + 1000d is ____.\n A 110 B 111 C 112 D 113\nAlgorithms       Binary-Search-Tree       Gate 2014 Set -03\nQuestion 46 Explanation:\nIt takes (log n) time to determine numbers n1 and n2 in balanced binary search tree T such that\n1. n1 is the smallest number greater than or equal to L and there is no predecessor n1' of >n1 such that n1' is equal to n1.\n2. n22' of n2 such that is equal to n2.\nSince there are m elements between n1 and n2, it takes ‘m’ time to add all elements between n1 and n2.\nSo time complexity is O(log n+m)\nSo the given expression becomes O(n0log'n+m'log0n)\nAnd a+10b+100c+1000d=0+10*1+100*1+1000*1=10+100=110\nBecause a=0, b=1, c=1 and d=0\n Question 47\nConsider a hash table with 100 slots. Collisions are resolved using chaining. Assuming simple uniform hashing, what is the probability that the first 3 slots are unfilled after the first 3 insertions?\n A (97×97×97)/1003 B (99×98×97)/1003 C (97×96×95)/1003 D (97×96×95)/(3!×1003)\nAlgorithms       Hashing       Gate 2014 Set -03\nQuestion 47 Explanation:\nGiven Hash table consists of 100 slots.\nThey are picked with equal probability of Hash function since it is uniform hashing.\nSo to avoid the first 3 slots to be unfilled\n=97/100*97/100*97/100=((97*97*97))⁄1003\n Question 48\nConsider the pseudocode given below. The function DoSomething() takes as argument a pointer to the root of an arbitrary tree represented by the leftMostChild-rightSibling representation. Each node of the tree is of type treeNode. typedef struct treeNode* treeptr; struct treeNode { treeptr leftMostChild, rightSibling; }; int DoSomething (treeptr tree) { int value=0; if (tree != NULL) { if (tree->leftMostChild == NULL) value = 1; else value = DoSomething(tree->leftMostChild); value = value + DoSomething(tree->rightSibling); } return(value); } When the pointer to the root of a tree is passed as the argument to DoSomething, the value returned by the function corresponds to the\n A number of internal nodes in the tree. B height of the tree. C number of nodes without a right sibling in the tree. D number of leaf nodes in the tree.\nAlgorithms       Tree Traversals       Gate 2014 Set -03\nQuestion 48 Explanation:\nThe key to solving such questions is to understand or detect where/by what condition the value (or the counter) is getting incremented each time.\nHere, that condition is if (tree → leftMostchild = = Null)\n⇒ Which means if there is no left most child of the tree (or the sub-tree or the current node as called in recursion)\n⇒ Which means there is no child to that particular node (since if there is no left most child, there is no child at all).\n⇒ Which means the node under consideration is a leaf node.\n⇒ The function recursively counts, and adds to value, whenever a leaf node is encountered. ⇒ The function returns the number of leaf nodes in the tree.\n Question 49\nConsider the C function given below. Assume that the array listA contains n (> 0) elements, sorted in ascending order. int ProcessArray(int *listA, int x, int n) { int i, j, k; i = 0; = n-1; do { k = (i+j)/2; if (x <= listA[k]) j = k-1; if (listA[k] <= x) i = k+1; }while (i <= j); if (listA[k] == x) return(k); else return -1; } Which one of the following statements about the function ProcessArray is CORRECT?\n A It will run into an infinite loop when x is not in listA. B It is an implementation of binary search. C It will always find the maximum element in listA. D It will return −1 even when x is present in listA.\nAlgorithms       Searching       Gate 2014 Set -03\nQuestion 49 Explanation:\nFrom the above code, we can identify\nk = (i+j)/2;\nwhere k keeps track of current middle element & i, j keeps track of left & right children of current subarray.\nSo it is an implementation of Binary search.\n Question 50\nWhich one of the following is the tightest upper bound that represents the number of swaps required to sort n numbers using selection sort?\n A O(log n) B O(n) C O(n log n) D O(n2)\nAlgorithms       Sorting       Gate 2013\nQuestion 50 Explanation:\nBest, Average and worst case will take maximum O(n) swaps.\nSelection sort time complexity O(n2) in terms of number of comparisons. Each of these scans requires one swap for n-1 elements (the final element is already in place).\n Question 51\nWhich one of the following is the tightest upper bound that represents the time complexity of inserting an object into a binary search tree of n nodes?\n A O(1) B O(log n) C O(n) D O(n log n)\nAlgorithms       Sorting       Gate 2013\nQuestion 51 Explanation:\nFor skewed binary search tree on n nodes, the tightest upper bound to insert a node is O(n).\n Question 52\nConsider the languages L1 = ϕ and L= {a}. Which one of the following represents L1L2* ∪ L1*?\n A {є} B ϕ C a* D {є,a}\nAlgorithms       Regular languages and Finite automata       Gate 2013\nQuestion 52 Explanation:\nAs we know, for any regular expression R,\nRϕ = ϕR=ϕ\nSo L1 L2 * = ϕ\nand L1 * = {ϕ}* = {ϵ}\nSo L1L2* ∪ L1* = {ϵ}\n Question 53\nWhat is the time complexity of Bellman-Ford single-source shortest path algorithm on a complete graph of n vertices?\n A Θ(n2) B Θ(n2 log n) C Θ(n3) D Θ(n3 log n)\nAlgorithms       Sorting       Gate 2013\nQuestion 53 Explanation:\nThe Bellman–Ford algorithm is an algorithm that computes shortest paths from a single source vertex to all of the other vertices in a weighted digraph. It is capable of handling graphs in which some of the edge weights are negative numbers.\nBellman–Ford runs in O(|V| * |E|) time, where |V| and |E| are the number of vertices and edges respectively.\nNote: For complete graph: |E| = n(n-1)/2 = O(n2)\n Question 54\nConsider an undirected random graph of eight vertices. The probability that there is an edge between a pair of vertices is 1/2. What is the expected number of unordered cycles of length three?\n A 1/8 B 1 C 7 D 8\nAlgorithms       Graph-Theory       Gate 2013\nQuestion 54 Explanation:\nAmong available ‘8’ vertices, we need to identify the cycles of length ‘3’.", null, "The probability that there exists one edge between two vertices = 1/2", null, "So, the total probability that all three edges of the above exists\n= 1/2 × 1/2 × 1/2 (as they are independent events)\n= 1/8\nTotal number of ways in which we can select ‘3’ such vertices among ‘8’ vertices = 8C3 = 56\nTotal number of cycles of length ‘3’ out of 8 vertices = 56 × 1/8 = 7\n Question 55\nWhich of the following statements is/are TRUE for undirected graphs?\n```P: Number of odd degree vertices is even.\nQ: Sum of degrees of all vertices is even.```\n A P only B Q only C Both P and Q D Neither P nor Q\nAlgorithms       Graph-Theory       Gate 2013\nQuestion 55 Explanation:\nEuler’s Theorem 3:\nThe sum of the degrees of all the vertices of a graph is an even number (exactly twice the number of edges).\nIn every graph, the number of vertices of odd degree must be even.\n Question 56\nThe line graph L(G) of a simple graph G is defined as follows: · There is exactly one vertex v(e) in L(G) for each edge e in G. · For any two edges e and e' in G, L(G) has an edge between v(e) and v(e'), if and only if e and e'are incident with the same vertex in G. Which of the following statements is/are TRUE?\n```(P) The line graph of a cycle is a cycle.\n(Q) The line graph of a clique is a clique.\n(R) The line graph of a planar graph is planar.\n(S) The line graph of a tree is a tree.```\n\n A P only B P and R only C R only D P, Q and S only\nAlgorithms       Graph-Theory       Gate 2013\nQuestion 56 Explanation:\nP) True. Because every edge in cycle graph will become a vertex in new graph L(G) and every vertex of cycle graph will become an edge in new graph.\nR) False. We can give counter example. Let G has 5 vertices and 9 edges which is planar graph. Assume degree of one vertex is 2 and of all others are 4. Now, L(G) has 9 vertices (because G has 9 edges) and 25 edges. But for a graph to be planar,\n|E| ≤ 3|v| - 6\nFor 9 vertices |E| ≤ 3 × 9 - 6\n⇒ |E| ≤ 27 - 6\n⇒ |E| ≤ 21. But L(G) has 25 edges and so is not planar.\nAs (R) is false, option (B) & (C) are eliminated.\n Question 57\nThe number of elements that can be sorted in Θ(log n) time using heap sort is\n A Θ(1) B Θ(√log⁡n) C Θ (log⁡n/log⁡log⁡n) D Θ(log n)\nAlgorithms       Sorting       Gate 2013\nQuestion 57 Explanation:\nUsing heap sort to n elements it will take O(n log n) time. Assume there are log n/ log log n elements in the heap.\nSo, Θ((logn/ log log n)log(logn/log log n))\n= Θ(logn/log logn (log logn - log log logn))\n= Θ((log n/log logn) × log log n)\n= Θ (log n)\nHence, option (C) is correct answer.\n Question 58\nWhat is the return value of f(p,p), if the value of p is initialized to 5 before the call? Note that the first parameter is passed by reference, whereas the second parameter is passed by value.\n `int` `f(``int` `&x, ``int` `c) {` `   ``c = c - 1;` `   ``if` `(c==0) ``return` `1;` `   ``x = x + 1;` `   ``return` `f(x,c) * x;` `}`\n A 3024 B 6561 C 55440 D 161051\nAlgorithms       Pass-by-value       Gate 2013\nQuestion 58 Explanation:\nSince it is f(p,p)\nf(5,5)→f(x,4)*x\nf(x,3)*x*x\nf(x,2)*x*x*x", null, "⇒x*x*x*x=x4\nNow x=x+1, there are 4 different calls namely f(6,4),f(7,3),f(8,2),f(9,1)\nBecause of pass by reference, x in all earlier functions is replaced with 9.\n⇒ 9*9*9*9=94=6561\n Question 59\nThe preorder traversal sequence of a binary search tree is 30, 20, 10, 15, 25, 23, 39, 35, 42. Which one of the following is the postorder traversal sequence of the same tree?\n A 10, 20, 15, 23, 25, 35, 42, 39, 30 B 15, 10, 25, 23, 20, 42, 35, 39, 30 C 15, 20, 10, 23, 25, 42, 35, 39, 30 D 15, 10, 23, 25, 20, 35, 42, 39, 30\nAlgorithms       Tree Traversals       Gate 2013\nQuestion 59 Explanation:", null, "From the data,", null, "The Postorder traversal sequence is:\n→ 15, 10, 23, 25, 20, 35, 42, 39, 30.\n Question 60\nConsider the following operation along with Enqueue and Dequeue operations on queues, where k is a global parameter.\n```MultiDequeue(Q){\nm = k\nwhile (Q is not empty and m > 0) {\nDequeue(Q)\nm = m - 1\n}\n}```\nWhat is the worst case time complexity of a sequence of n MultiDequeue() operations on an initially empty queue? (GATE CS 2013)\n A Θ(n) B Θ(n + k) C Θ(nk) D Θ(n2)\nAlgorithms       Time-Complexity       Gate 2013\nQuestion 60 Explanation:\nThere are three possible operations on queue- Enqueue, Dequeue and MultiDequeue. MultiDequeue is calling Dequeue multiple times based on a global variable k. Since, the queue is initially empty, whatever be the order of these operations, there cannot be more no. of Dequeue operations than Enqueue operations. Hence, the total no. operations will be n only.\n Question 61\nAssuming P ≠ NP, which of the following is TRUE?\n A NP-complete = NP B NP-complete ∩ P = ∅ C NP-hard = NP D P = NP-complete\nAlgorithms       P-NP       Gate 2012\nQuestion 61 Explanation:\nNote: Out of syllabus.\nThe definition of NP-complete is,\nA decision problem p is NP-complete if:\n1. p is in NP, and\n2. Every problem in NP is reducible to p in polynomial time.\nIt is given that assume P ≠ NP , hence NP-complete ∩ P = ∅ .\nThis is due to the fact that, if NP-complete ∩ P ≠ ∅ i.e. there are some problem (lets say problem P1) which is in P and in NP-complete also, then it means that P1 (NP-complete problem) can be solved in polynomial time also (as it is also in P class) and this implies that every NP problem can be solve in polynomial time, as we can convert every NP problem into NP-complete problem in polynomial time.\nWhich means that we can convert every NP problem to P1 and solve in polynomial time and hence P = NP, which is contradiction to the given assumption that P ≠ NP.\n Question 62\nLet W(n) and A(n) denote respectively, the worst case and average case running time of an algorithm executed on an input of size n.  Which of the following is ALWAYS TRUE?\n A A(n) = Ω (W(n)) B A(n) = Θ (W(n)) C A(n) = O (W(n)) D A(n) = o (W(n))\nAlgorithms       Time-Complexity       Gate 2012\nQuestion 62 Explanation:\nThe average case time can be lesser than or even equal to the worst case.\nSo, A(n) would be upper bounded by W(n) and it will not be strict upper bound as it can even be same (e.g. Bubble Sort and merge sort).\nA(n)=O(W(n))\nNote: Option A is wrong because A(n) is not equal to Ω(w(n)) .\n Question 63\nConsider the Quicksort algorithm. Suppose there is a procedure for finding a pivot element which splits the list into two sub-lists each of which contains at least one-fifth of the elements. Let T(n) be the number of comparisons required to sort n elements. Then\n A T(n) ≤ 2T(n/5) + n B T(n) ≤ T(n/5) + T(4n/5) + n C T(n) ≤ 2T(4n/5) + n D T(n) ≤ 2T(n/2) + n\nAlgorithms       Sorting       Gate-2008\nQuestion 63 Explanation:\nConsider the case where one subset of set of n elements contains n/5 elements and another subset of set contains 4n/5 elements.\nSo, T(n/5) comparisons are needed for the first subset and T(4n/5) comparisons needed for second subset.\nNow, suppose that one subset contains more than n/5 elements then another subset will contain less than 4n/5 elements. Due to which time complexity will be less than\nT(n/5) + T(4n/5) + n\nBecause recursion tree will be more balanced.\n Question 64\nThe subset-sum problem is defined as follows: Given a set S of n positive integers and a positive integer W, determine whether there is a subset of S Whose elements sum to An algorithm Q solves this problem in O(nW) time. Which of the following statements is false?\n A Q solves the subset-sum problem in polynomial time when the input is encoded in unary B Q solves the subset-sum problem in polynomial time when the input is encoded in binary C The subset sum problem belongs to the class NP D The subset sum problem is NP-hard\nAlgorithms       P-NP       Gate-2008\nQuestion 64 Explanation:\nNote: Out of syllabus.\n Question 65\nDijkstra’s single source shortest path algorithm when run from vertex a in the above graph, computes the correct shortest path distance to", null, "A only vertex a B only vertices a, e, f, g, h C only vertices a, b, c, d D all the vertices\nAlgorithms       Shortest-Path       Gate-2008\nQuestion 65 Explanation:\nThe single source shortest path algorithm(Dijkstra’s) is not work correctly for negative edge weights and negative weight cycle. But as per the above graph it works correct. It gives from shortest path between ‘a’ vertex to all other vertex. The Dijkstra’s algorithm will follow greedy strategy.\nA-B ⇒ 1\nA-C⇒ 1+2=3\nA-E⇒ 1-3 =-2\nA-F⇒ 1 -3 +2 =0\nA-G⇒ 1 -3 +2 +3 =3\nA-C ⇒1 +2 =3\nA-H⇒ 1+2 -5 =-2\n Question 66\nThe numbers 1, 2, .... n are inserted in a binary search tree in some order. In the resulting tree, the right subtree of the root contains p nodes. The first number to be inserted in the tree must be\n A p B p + 1 C n - p D n - p + 1\nAlgorithms       Tree Traversals       Gate 2005-IT\nQuestion 66 Explanation:\nTotal element = n\nRST contains elements = p\nRoot contains = 1 element\n1st contains = n - (p + 1) element", null, "Root contains the value is n - p.\n Question 67\nIn a depth-first traversal of a graph G with n vertices, k edges are marked as tree edges. The number of connected components in G is\n A k B k + 1 C n - k - 1 D n - k\nAlgorithms       Graph-Traversal       Gate 2005-IT\nQuestion 67 Explanation:\nIn a graph G with n vertices and p component then G has n - p edges(k).\nIn this question, we are going to applying the depth first search traversal on each component of graph where G is conected (or) disconnected which gives minimum spanning tree\ni.e., k = n-p\np = n - k\n Question 68\nIn the following table, the left column contains the names of standard graph algorithms and the right column contains the time complexities of the algorithms. Match each algorithm with its time complexity.\n 1. Bellman-Ford algorithm 2. Kruskal’s algorithm 3. Floyd-Warshall algorithm 4. Topological sorting A : O ( m log n) B : O (n3) C : O (nm) D : O (n + m)\n A 1→ C, 2 → A, 3 → B, 4 → D B 1→ B, 2 → D, 3 → C, 4 → A C 1→ C, 2 → D, 3 → A, 4 → B D 1→ B, 2 → A, 3 → C, 4 → D\nAlgorithms       General       Gate 2005-IT\nQuestion 68 Explanation:\nBellman-ford algorithm → O(nm)\nKrushkal's algorithm → O(m log n)\nFloyd-Warshall algorithm → O(n3)\nTopological sorting → O(n+m)\n Question 69\n\nA hash table contains 10 buckets and uses linear probing to resolve collisions. The key values are integers and the hash function used is key % 10. If the values 43, 165, 62, 123, 142 are inserted in the table, in what location would the key value 142 be inserted?\n\n A 2 B 3 C 4 D 6\nAlgorithms       Sorting-and-Searching       Gate 2005-IT\nQuestion 69 Explanation:\n43%10 = 3 [occupy 3]\n165%10 = 5 [occupy 5]\n62%10 = 2 [occupy 2]\n123%10 = 3 [3 already occupied, so occupies 4]\n142%10 = 2 [2, 3, 4, 5 are occupied, so it occupies 6]\n Question 70\n\nIn a binary tree, for every node the difference between the number of nodes in the left and right subtrees is at most 2. If the height of the tree is h > 0, then the minimum number of nodes in the tree is:\n\n A 2h-1 B 2h-1 + 1 C 2h - 1 D 2h\nAlgorithms       Tree Traversals       Gate 2005-IT\nQuestion 70 Explanation:\nLet's take an example,", null, "Above tree satisfies the property given in the question. Hence, only option (B) satisfies it.\n Question 71\nLet T(n) be a function defined by the recurrence T(n) = 2T(n/2) + √n for n ≥ 2 and T(1) = 1 Which of the following statements is TRUE?\n A T(n) = θ(log n) B T(n) = θ(√n) C T(n) = θ(n) D T(n) = θ(n log n)\nAlgorithms       Recursion       Gate 2005-IT\nQuestion 71 Explanation:\nApply Master's theorem.\n Question 72\n\nLet G be a weighted undirected graph and e be an edge with maximum weight in G. Suppose there is a minimum weight spanning tree in G containing the edge e. Which of the following statements is always TRUE?\n\n A There exists a cutset in G having all edges of maximum weight B There exists a cycle in G having all edges of maximum weight C Edge e cannot be contained in a cycle D All edges in G have the same weight\nAlgorithms       Graph-Theory       Gate 2005-IT\nQuestion 72 Explanation:\n(A) True, because if there is heaviest edge in MST, then there exist a cut with all edges with weight equal to heaviest edge.", null, "(B) False, because the cutset of heaviest edge may contain only one edge.\n(C) False. The cutset may form cycle with other edge.\n(D) False. Not always true.\n Question 73\n\nA binary search tree contains the numbers 1, 2, 3, 4, 5, 6, 7, 8. When the tree is traversed in pre-order and the values in each node printed out, the sequence of values obtained is 5, 3, 1, 2, 4, 6, 8, 7. If the tree is traversed in post-order, the sequence obtained would be\n\n A 8, 7, 6, 5, 4, 3, 2, 1 B 1, 2, 3, 4, 8, 7, 6, 5 C 2, 1, 4, 3, 6, 7, 8, 5 D 2, 1, 4, 3, 7, 8, 6, 5\nAlgorithms       Tree Traversals       Gate 2005-IT\nQuestion 73 Explanation:\nPre-order is given as\n5, 3, 1, 2, 4, 6, 8, 7\nIn-order is the sorted sequence, i.e.,\n1, 2, 3, 4, 5, 6, 7, 8\nAnd we know that with given inorder and preorder, we can draw a unique BST.\nSo, the BST will be,", null, "Hence, postorder traversasl sequence is,\n2, 1, 4, 3, 7, 8, 6, 5\n Question 74\nLet G be a directed graph whose vertex set is the set of numbers from 1 to 100. There is an edge from a vertex i to a vertex j iff either j = i + 1 or j = 3i. The minimum number of edges in a path in G from vertex 1 to vertex 100 is\n A 4 B 7 C 23 D 99\nAlgorithms       Graph-Theory       Gate 2005-IT\nQuestion 74 Explanation:\nEdge set consists of edges from i to j, using either\nj = i +1\n(or)\nj = 3i\nSecond option will help us reach from 1 to 100 rapidly. The trick to solve this question is to think in reverse way. Instead of finding a path from 1 to 100, try to find a path from 100 to 1.\nSo, the edge sequence with minimum number of edges is\n1 → 3 → 9 → 10 → 11 → 33 → 99 → 100\nwhich consists of 7 edges.\n Question 75\nTo carry out white box testing of a program, its flow chart representation is obtained as shown in the figure below:", null, "For basis path based testing of this program, its cyclomatic complexity is\n A 5 B 4 C 3 D 2\nAlgorithms       Cyclomatic-Complexity       Gate 2005-IT\nQuestion 75 Explanation:\nNote: Out of syllabus.\n Question 76\n\nAn array of integers of size n can be converted into a heap by adjusting the heaps rooted at each internal node of the complete binary tree starting at the node ⌊(n - 1) /2⌋, and doing this adjustment up to the root node (root node is at index 0) in the order ⌊(n - 1)/2⌋, ⌊(n - 3)/2⌋, ....., 0. The time required to construct a heap in this manner is\n\n A O(log n) B O(n) C O(n log log n) D O(n log n)\nAlgorithms       Binary-Trees       Gate 2004-IT\nQuestion 76 Explanation:\nThe above statement is actually algorithm for building a heap of an input array A.\nAnd we know that time complexity for building the heap is O(n)\n Question 77\nWhich one of the following binary trees has its inorder and preorder traversals as BCAD  and ABCD, respectively?\n A", null, "B", null, "C", null, "D", null, "Algorithms       Tree Traversals       Gate 2004-IT\nQuestion 77 Explanation:\nInorder traversal is,\nLeft root right.\nPreorder traversal is,\nRoot left right.\n Question 78\nLet f(n), g(n) and h(n) be functions defined for positive inter such that f(n) = O(g(n)), g(n) ≠ O(f(n)), g(n) = O(h(n)), and h(n) = O(g(n)). Which one of the following statements is FALSE?\n A f(n) + g(n) = O(h(n)) + h(n)) B f(n) = O(h(n)) C fh(n) ≠ O(f(n)) D f(n)h(n) ≠ O(g(n)h(n))\nAlgorithms       Time-Complexity       Gate 2004-IT\nQuestion 78 Explanation:\nf(n) = O(h(n)) [Using transitivity]\nSo, f(n)g(n) = O(g(n)h(n)) is True.\nHence, (D) is false.\n Question 79\nConsider the undirected graph below:", null, "Using Prim's algorithm to construct a minimum spanning tree starting with node A, which one of the following sequences of edges represents a possible order in which the edges would be added to construct the minimum spanning tree?\n A (E, G), (C, F), (F, G), (A, D), (A, B), (A, C) B (A, D), (A, B), (A, C), (C, F), (G, E), (F, G) C (A, B), (A, D), (D, F), (F, G), (G, E), (F, C) D (A, D), (A, B), (D, F), (F, C), (F, G), (G, E)\nAlgorithms       Greedy-Method       Gate 2004-IT\nQuestion 79 Explanation:\n(A) and (B) produce disconnected components with the given order in options which is never allowed by Prim's algorithm.\n(C) produces connected component every instead a new edge is added but when first vertex is chosen (first vertex is chosen randomly) first edge must be minimum weight edge that is chosen. Therefore, (A, D) must be chosen before (A, B). Therefore (C) is false.\n Question 80\nConsider a list of recursive algorithms and a list of recurrence relations as shown below. Each recurrence relation corresponds to exactly one algorithm and is used to derive the time complexity of the algorithm.\nRECURSIVE ALGORITHM RECURRENCE RELATION\nP. Binary search I. T(n) = T(n-k) + T(k) + cn\nQ. Merge sort II. T(n) = 2T(n-1) + 1\nR. Quick sort III. T(n) = 2T(n/2) + cn\nS. Tower of Hanoi IV. T(n) = T(n/2) + 1\n A P-II, Q-III, R-IV, S-I B P-IV, Q-III, R-I, S-II C P-III, Q-II, R-IV, S-I D P-IV, Q-II, R-I, S-III\nAlgorithms       General       Gate 2004-IT\nQuestion 80 Explanation:\nQuick sort - T(n) = T(n-k) + T(k) + cn\nBinary search - T(n/2) + 1\nMerge sort - T(n) = 2T(n/2)+ cn\nTower of hanoi - T(n) = 2T(n-1) +1\n Question 81\nMerge sort uses\n A Divide and conquer strategy B Backtracking approach C Heuristic search D Heuristic search\nAlgorithms       Sorting       Gate-1995\nQuestion 81 Explanation:\nMerge sort uses the divide and conquer strategy.\n Question 82\nFor merging two sorted lists of sizes m and n into a sorted list of size m+n, we required comparisons of\n A O(m) B O(n) C O(m+n) D O(logm+logn)\nAlgorithms       Sorting       Gate-1995\nQuestion 82 Explanation:\nIn best case, no. of comparisons is Min(m,n).\nIn worst case, no. of comparisons is m+n-1.\nThen we require O(m+n) comparisons to merging two sorted lists.\n Question 83\n A AB + CD + *F/D + E* B ABCD + *F/DE*++ C A *B + CD/F *DE++ D A + *BCD/F* DE++ E None of the above\nAlgorithms       Post-fix-and-Prefix       Gate-1995\nQuestion 83 Explanation:\nThe postfix expression will be,\nA B C D + * F / + D E * +\n Question 84\n A I and II B II and III C I and IV D I and III\nAlgorithms       General       Gate-1995\nQuestion 84 Explanation:\nBinary search using linked list is not efficient as it will not give O(log n), because we will not be able to find the mid in constant time. Fnding mid in linked list takes O(n) time.\nRecursive program requires stack for storing the function state. Any recursive program can be rewritten as an iterative one. Advantage of recursion is \"less programming effort\", while it always lags behind ietrative one in terms of performance.\n Question 85\nFORTRAN implementation do not permit recursion because\n A they use static allocation for variables B they use dynamic allocation for variables C stacks are not available on all machines D it is not possible to implement recursion on all machines\nAlgorithms       Recursions       Gate-1994\nQuestion 85 Explanation:\nFORTRAN implementation do not permit recursion because they use the static allocation for variables.\n→ Recursion requires dynamic allocation of data.\n Question 86\nThe recurrence relation that arises in relation with the complexity of binary search is:\n A", null, "B", null, "C", null, "D", null, "Algorithms       Time-Complexity       Gate-1994\nQuestion 86 Explanation:\nIn binary search, search for the half of the list and constant time for comparing. So,", null, "Question 87\nWhich of the following algorithm design techniques is used in the quicksort algorithm?\n A Dynamic programming B Backtracking C Divide and conquer D Greedy method\nAlgorithms       Quick-Sort       Gate-1994\nQuestion 87 Explanation:\nIn quick sort, we use divide and conquer technique.\n Question 88\nIn which one of the following cases is it possible to obtain different results for call-by reference and call-by-name parameter passing methods?\n A Passing a constant value as a parameter B Passing the address of an array as a parameter C Passing an array element as a parameter D Passing an array following statements is true\nAlgorithms       Call-by-reference-and-Call-by-value       Gate-1994\nQuestion 88 Explanation:\nPassing an array element as a parameter then it gives different output values for the call-by-reference and call-by-name parameters.\n{ ........\na[ ] = {1, 2, 3, 4}\ni = 0\nfun(a[i]);\nprint a;\n}\nfun(int x)\n{\nint i = 1;\nx = 8;\n}\nO/p:\nCall-by-reference = 8\nCall-by-value = 1\n Question 89\nWhich one of the following statements is false?\n A Optimal binary search tree construction can be performed efficiently using dynamic programming. B Breadth-first search cannot be used to find converted components of a graph. C Given the prefix and postfix walks over a binary tree, the binary tree cannot be uniquely constructed. D Depth-first search can be used to find connected components of a graph.\nAlgorithms       Searching       Gate-1994\nQuestion 89 Explanation:\nIn BFS algorithm, we can randomly select a source vertex and then run, after that whether we need to check distance to each and every vertex from source is still infinite (or) not. If we find any vertex having infinite distance then the graph is not connected.\n Question 90\n A g1(n) is O(g2(n)) B g1 (n) is O(3) C g2 (n) is O(g1 (n)) D g2 (n) is O(n) E Both A and B\nAlgorithms       Asymptotic-Notations       Gate-1994\nQuestion 90 Explanation:\nIn asymptotic complexity, we assume sufficiently large n. So, g1(n) = n2 and g2(n) = n3.\nGrowth rate of g1 is less than that of g2 i.e., g1(n) = O(g2(n)) = O(n).\n Question 91\nThe root directory of a disk should be placed\n A at a fixed address in main memory B at a fixed location on the disk C anywhere on the disk D at a fixed location on the system disk E anywhere on the system disk\nAlgorithms       Gate-1993\nQuestion 91 Explanation:\nRoot directory can points to the various user directories. Then they will be stored in a way that user can't be easily modify them. Then they should be at fixed location on the disk.\n Question 92\nConsider a simple connected graph G with n vertices and n-edges (n>2). Then, which of the following statements are true?\n A G has no cycles. B The graph obtained by removing any edge from G is not connected. C G has at least one cycle. D The graph obtained by removing any two edges from G is not connected. E Both C and D.\nAlgorithms       Graphs       Gate-1993\nQuestion 92 Explanation:\nIf a graph have n vertices and n edges (n>2) then it is to be cyclic graph. Then it have atleast one cycle and if we remove two edges then it is not connected.\nFor example let us consider, n=3\n Question 93\n A O(n) B O(n2) C O(n3) D O(3n2) E O(1.5n2) F B, C, D and E\nAlgorithms       Time-Complexity       Gate-1993\nQuestion 93 Explanation:", null, "⇒ In this 'n' is constant. So, n is added to n times itself which is O(n2).\nHence, (a) is wrong. And rest (B), (C), (D), (E) are correct.\n Question 94\nComplexity of Kruskal’s algorithm for finding the minimum spanning tree of an undirected graph containing n vertices and m edges if the edges are sorted is __________\n A O(m log n)\nAlgorithms       Krushkal's-Algorithm       Gate-1992\nQuestion 94 Explanation:\nThough the edges are to be sorted still due to union find operation complexity is O(m log n).\n Question 95\nFollowing algorithm(s) can be used to sort n integers in the range [1...n3] in O(n) time\n A Heapsort B Quicksort C Mergesort D Radixsort\nAlgorithms       Sorting       Gate-1992\nQuestion 95 Explanation:\nAs no comparison based sort can ever do any better than nlogn. So option (A), (B), (C) are eliminated. O(nlogn) is lower bound for comparison based sorting.\nAs Radix sort is not comparison based sort (it is counting sort), so can be done in linear time.\n Question 96\n\nAssume that the last element of the set is used as partition element in Quicksort. If n distinct elements from the set [1…..n] are to be sorted, give an input for which Quicksort takes maximum time.\n\n A O(n^2)\nAlgorithms       Quick-Sort       Gate-1992\nQuestion 96 Explanation:\nFor n distinct elements the algorithm will take maximum time when:\n→ The array is already sorted in ascending order.\n→ The array is already sorted in descending order.\n Question 97\nThe minimum number of comparisons required to sort 5 elements is _____\n A 7\nAlgorithms       Sorting       Gate-1991\nQuestion 97 Explanation:\nMinimum no. of comparisons\n= ⌈log(n!)⌉\n= ⌈log(5!)⌉\n= ⌈log(120)⌉\n= 7\n Question 98\n A 144\nAlgorithms       Binary-Trees       Gate-1991\nQuestion 98 Explanation:", null, "Question 99\n A 4, 1, 6, 7, 3, 2, 5, 8\nAlgorithms       Tree Traversals       Gate-1991\nQuestion 99 Explanation:\nInorder traversal is\n(Left, Root, Right)\nSo, the order will be\n4, 1, 6, 7, 3, 2, 5, 8\n Question 101\n A O(n2) B O(mn) C O(m+n) D O(m logn) E O(m2) F B, D and E\nAlgorithms       Time-Complexity       Gate-1991\nQuestion 101 Explanation:\nThough the edges are sorted still due to finding union operation complexity is O(m log n).\n→ Where m is no. of edges and n is number of vertices then n = O(m2)\n→ O(m logn) < O(mn)\n Question 102\n A 20, 10, 20, 10, 20 B 20, 20, 10, 10, 20 C 10, 20, 20, 10, 20 D 20, 20, 10, 20, 10\nAlgorithms       Stacks       Gate-1991\nQuestion 102 Explanation:\nPUSH(10), PUSH(20), POP = 20 (i)\n→ PUSH(10), PUSH(10), PUSH(20), POP = 20 (ii)\n→ PUSH(10), PUSH(10), POP = 10 (iii)\n→ PUSH(10), POP = 10 (iv)\n→ PUSH(20)\n→ PUSH(20), POP = 20 (v)\n20, 20, 10, 10, 20\n Question 103\n A (a) - (r), (b) - (p), (c) - (s), (d) - (q)\nAlgorithms       Match-the-Following       Gate-1990\nQuestion 103 Explanation:\nStrassen's matrix multiplication algorithm - Divide and Conquer\nKruskal's minimum spanning tree algorithm - Greedy method\nBiconnected components algorithm - Depth first search\nFloyd's shortest path algorithm - Dynamic programming\nThere are 103 questions to complete." ]

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[ "## Question Paper of \"ELECTRONICS SYSTEM DESIGN\" for M. Tech (ECE) (Sem.-1st), Paper Code (EC-502), Paper ID [ E0442 ], PTU Question Paper.\n\n•", null, "Wednesday, September 02, 2015\n•", null, "•", null, ",\n•", null, "No comments\n\nptupaper.com\nRoll No…….\nTotal No. of Questions: 08\n\nM. TECH (ECE), MAY-2014\nELECTRONICS SYSTEM DESIGN\nPaper Code (EC-502)\nPaper ID: [E0562]\n\nTime: 3 Hrs.                                                                 Max. Marks: 100\nNote: Attempt any five questions.\nQ1.\n(a)  Explain the working of cell and the bouncing switch.\n(b) Implement the function f= using two 4:1 Multiplexers.\n(12+8)\n\nQ2.\n(a)  With the help of an example; explain the state diagram reduction process.\n(b) Using J K flip flops. Design a 3-bit synchronous counter with the following sequence 3,7,5,6,3.\n(12+8)\nQ3. What are system controllers? With the help of an example explain the design phases of a system controller.\n(20)\nQ4. Design a comparator using binary parallel adders.\n(20)\nQ5. Following the design steps require to design as asynchronous circuit, design a basic binary cell.\n(20)\nQ6. Discuss the interfacing of digital circuits with different transmission media.\n(20)\nQ7.\n(a)  What do you mean by hazards; explain various kinds of hazards in detail.\n(b) With an example differentiate  between cycles and races.\n(10+10)\nQ8. What do you mean by a Glitch? What techniques can be used to design a Glitch free circuit?\n(20)" ]

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[ "# What does sin2x mean?\n\nsin2x and sin(2x) are the same thing, the first is just a lazier way to write it. They both mean “multiply x by 2 and then take the sine of that” On the other hand, 2sinx means “take the sine of x first, then multiply it by 2”\n\n## What is sin2A?\n\nThe general formula of sin2A is, sin2A = 2 sin A cos A. Using sin2A + cos2A = 1, we get sin A = √(1 – cos2A). Substituting this in the given formula, sin2A = 2 √(1 – cos2A) cos A.\n\n## Is sin2x the same as sin 2x?\n\nNope, those are the same. As long as you have the parentheses around the sin2x, the whole thing is squared. In a calculator, that is how you would put it if you wanted to take the sine of the angle 2x, then square the result.\n\n## What is tan 2x?\n\nTan^2x (Tan Square x) Tan^2x is the square of the trigonometric function tanx. As we know that tan x can be expressed as the ratio of sinx and cosx, therefore we can express tan^2x can as the ratio of sin square x and cos square x.\n\n## What does cos2θ mean?\n\n⁡ ( 2 θ ) = cos 2 ⁡ θ − sin 2 ⁡ A trigonometric identity that expresses the expansion of cosine of double angle in cosine and sine of angle is called the cosine of double angle identity.\n\n## What are three trigonometric identities?\n\nThere are three primary trigonometric ratios sin, cos, and tan. The three other trigonometric ratios sec, cosec, and cot in trigonometry are the reciprocals of sin, cos, and tan respectively. How are these trigonometric ratios (sin, cos, tan, sec, cosec, and cot) connected with each other?\n\n## What does a sine graph look like?\n\nTo graph the sine function, we mark the angle along the horizontal x axis, and for each angle, we put the sine of that angle on the vertical y-axis. The result, as seen above, is a smooth curve that varies from +1 to -1. This shape is also called a sine wave, especially when it appears in radio and electronic circuits.\n\n## Is 2sinA and sin2a the same?\n\nAnswer: 2sinA means that 2 times the value of sin A. However, it’s totally different from sin 2A. Sin 2A Means that the value of x is doubled that is sin of two times x.\n\n## What is Sinx times COSX?\n\nAnswer : The expression for sin x + cos x in terms of sine is sin x + sin (π / 2 – x). Let us see the detailed solution now.\n\n## What is sin 2x times cos 2x?\n\nTo find the value of sin2x × Cos 2x, the trigonometric double angle formulas are used. For the derivation, the values of sin 2x and cos 2x are used. So, Sin 2x Cos 2x = 2 Cos x (2 Sin x Cos2 x − Sin x)\n\n## What is the identity of Cos3x?\n\nWhat is Cos3x in Trigonometry? Cos3x is a triple angle identity in trigonometry. It can be derived using the angle addition identity of the cosine function. The identity of cos3x is given by cos3x = 4 cos3x – 3 cos x.\n\n## Which of the following function is odd?\n\nExample: x and sinx are odd functions. A function f(x) is an even function if f(-x) = f(x). Thus g(x) = x2 is an even function as g(x) = g(-x). So the function g(x) = 4x is an odd function.\n\n## Is Xsinx an even function?\n\nEven functions are symmetrical about the y-axis. Odd functions have symmetry about the origin. thus xsinx is an even function. This is graph of xsinx.\n\n## Is cos4x an odd function?\n\nA function is odd if f(−x)=−f(x) f ( – x ) = – f ( x ) . Since −cos(x)x=−cos(x)x – cos ( x ) x = – cos ( x ) x , the function is odd.\n\n## What are the 3 Pythagorean identities?\n\nThe Pythagorean identities are derived from the Pythagorean theorem, and describe the relationship between sine and cosine on the unit circle. The three identities are cos2t+sin2t=1 ⁡ t + sin 2 ⁡ , 1+tan2t=sec2t 1 + tan 2 ⁡ t = sec 2 ⁡ , and 1+cot2t=csc2t 1 + cot 2 ⁡ t = csc 2 ⁡ .\n\n## What are the 6 trigonometric identities?\n\nThere are six trigonometric identities that are sine, cosine, tangent, secant, cosecant, and cotangent.\n\n## What is the trigonometry class 10th?\n\nTrigonometry is a branch of mathematics dealing with relations involving lengths and angles of triangles. It can, in a simpler manner, be called the study of triangles. The angles are either measured in degrees or radians. We need to look into trigonometric formulae, ratios, functions etc.\n\n## What are identities in pre calc?\n\nA trigonometric equation is any equation that includes a trigonometric function. There are two basic types of trigonometric equations: identities and conditional equations. Identities are equations that hold for any angle. Conditional equations are equations that are solved only by certain angles.\n\n## Is SohCahToa a trig identity?\n\nWhat Is SohCahToa? It’s a mnemonic device to help you remember the three basic trig ratios used to solve for missing sides and angles in a right triangle. It’s defined as: SOH: Sin(θ) = Opposite / Hypotenuse.\n\n## Can there be 8 trigonometric ratios?\n\n(It’s well known that you can shake a stick at a maximum of 8 trig functions.) The familiar sine, cosine, and tangent are in red, blue, and, well, tan, respectively. They’re all just simple combinations of dear old sine and cosine.\n\n## Why is the period of sin 2Pi?\n\nThe period of the sine function is ​2π​, which means that the value of the function is the same every 2π units." ]

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[ "# Compute the greatest common divisor\n\n## The problem\n\nFind the greatest common divisor of two integers n1 and n2 is as follows: First find d to be the minimum of n1 and n2, then check whether d, d-1, d-2, ..., 2, or 1 is a divisor for both n1 and n2 in this order. The first such common divisor is the greatest common divisor for n1 and n2. Write a program that prompts the user to enter two positive integers and displays the gcd.\n\n## Output\n\nCompute the greatest common divisor posted by on\n\nTagged: java, java-exercises-beginner, intro-to-java-10th-edition, and ch5\n\nShare on:" ]

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[ "# How to: Implement a Component That Supports the Event-based Asynchronous Pattern\n\nThe following code example implements a component with an asynchronous method, according to the Event-based Asynchronous Pattern Overview. The component is a prime number calculator that uses the Sieve of Eratosthenes algorithm to determine if a number is prime or composite.\n\nThere is extensive support for this task in Visual Studio. For more information, see Walkthrough: Implementing a Component That Supports the Event-based Asynchronous Pattern and Walkthrough: Implementing a Component That Supports the Event-based Asynchronous Pattern and Walkthrough: Implementing a Component That Supports the Event-based Asynchronous Pattern and Walkthrough: Implementing a Component That Supports the Event-based Asynchronous Pattern.\n\nFor an example client that uses the PrimeNumberCalculator component, see How to: Implement a Client of the Event-based Asynchronous Pattern.\n\n## Example\n\n``````Imports System\nImports System.Collections\nImports System.Collections.Specialized\nImports System.ComponentModel\nImports System.Drawing\nImports System.Globalization\nImports System.Threading\nImports System.Windows.Forms\n\n...\n\nPublic Delegate Sub ProgressChangedEventHandler( _\nByVal e As ProgressChangedEventArgs)\n\nPublic Delegate Sub CalculatePrimeCompletedEventHandler( _\nByVal sender As Object, _\nByVal e As CalculatePrimeCompletedEventArgs)\n\n' This class implements the Event-based Asynchronous Pattern.\n' It asynchronously computes whether a number is prime or\n' composite (not prime).\nPublic Class PrimeNumberCalculator\nInherits System.ComponentModel.Component\n\nPrivate Delegate Sub WorkerEventHandler( _\nByVal numberToCheck As Integer, _\nByVal asyncOp As AsyncOperation)\n\nPrivate onProgressReportDelegate As SendOrPostCallback\nPrivate onCompletedDelegate As SendOrPostCallback\n\nPrivate userStateToLifetime As New HybridDictionary()\n\nPrivate components As System.ComponentModel.Container = Nothing\n\n'''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''\n#Region \"Public events\"\n\nPublic Event ProgressChanged _\nAs ProgressChangedEventHandler\nPublic Event CalculatePrimeCompleted _\nAs CalculatePrimeCompletedEventHandler\n\n#End Region\n\n'''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''\n#Region \"Construction and destruction\"\n\nPublic Sub New(ByVal container As System.ComponentModel.IContainer)\n\ncontainer.Add(Me)\nInitializeComponent()\n\nInitializeDelegates()\n\nEnd Sub\n\nPublic Sub New()\n\nInitializeComponent()\n\nInitializeDelegates()\n\nEnd Sub\n\nProtected Overridable Sub InitializeDelegates()\nonProgressReportDelegate = _\nNew SendOrPostCallback(AddressOf ReportProgress)\nonCompletedDelegate = _\nNew SendOrPostCallback(AddressOf CalculateCompleted)\nEnd Sub\n\nProtected Overrides Sub Dispose(ByVal disposing As Boolean)\nIf disposing Then\nIf (components IsNot Nothing) Then\ncomponents.Dispose()\nEnd If\nEnd If\nMyBase.Dispose(disposing)\n\nEnd Sub\n\n#End Region\n\n'''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''\n#Region \"Implementation\"\n\n' This method starts an asynchronous calculation.\n' First, it checks the supplied task ID for uniqueness.\n' If taskId is unique, it creates a new WorkerEventHandler\n' and calls its BeginInvoke method to start the calculation.\nPublic Overridable Sub CalculatePrimeAsync( _\nByVal numberToTest As Integer, _\nByVal taskId As Object)\n\n' Create an AsyncOperation for taskId.\nDim asyncOp As AsyncOperation = _\nAsyncOperationManager.CreateOperation(taskId)\n\n' Multiple threads will access the task dictionary,\n' so it must be locked to serialize access.\nSyncLock userStateToLifetime.SyncRoot\nIf userStateToLifetime.Contains(taskId) Then\nThrow New ArgumentException( _\n\"Task ID parameter must be unique\", _\n\"taskId\")\nEnd If\n\nuserStateToLifetime(taskId) = asyncOp\nEnd SyncLock\n\n' Start the asynchronous operation.\nDim workerDelegate As New WorkerEventHandler( _\nAddressOf CalculateWorker)\n\nworkerDelegate.BeginInvoke( _\nnumberToTest, _\nasyncOp, _\nNothing, _\nNothing)\n\nEnd Sub\n\n' Utility method for determining if a\n' task has been canceled.\nPrivate Function TaskCanceled(ByVal taskId As Object) As Boolean\nReturn (userStateToLifetime(taskId) Is Nothing)\nEnd Function\n\n' This method cancels a pending asynchronous operation.\nPublic Sub CancelAsync(ByVal taskId As Object)\n\nDim obj As Object = userStateToLifetime(taskId)\nIf (obj IsNot Nothing) Then\n\nSyncLock userStateToLifetime.SyncRoot\n\nuserStateToLifetime.Remove(taskId)\n\nEnd SyncLock\n\nEnd If\n\nEnd Sub\n\n' This method performs the actual prime number computation.\n' It is executed on the worker thread.\nPrivate Sub CalculateWorker( _\nByVal numberToTest As Integer, _\nByVal asyncOp As AsyncOperation)\n\nDim prime As Boolean = False\nDim firstDivisor As Integer = 1\nDim exc As Exception = Nothing\n\n' Check that the task is still active.\n' The operation may have been canceled before\n' the thread was scheduled.\nIf Not Me.TaskCanceled(asyncOp.UserSuppliedState) Then\n\nTry\n' Find all the prime numbers up to the\n' square root of numberToTest.\nDim primes As ArrayList = BuildPrimeNumberList( _\nnumberToTest, asyncOp)\n\n' Now we have a list of primes less than\n'numberToTest.\nprime = IsPrime( _\nprimes, _\nnumberToTest, _\nfirstDivisor)\n\nCatch ex As Exception\nexc = ex\nEnd Try\n\nEnd If\n\nMe.CompletionMethod( _\nnumberToTest, _\nfirstDivisor, _\nprime, _\nexc, _\nTaskCanceled(asyncOp.UserSuppliedState), _\nasyncOp)\n\nEnd Sub\n\n' This method computes the list of prime numbers used by the\n' IsPrime method.\nPrivate Function BuildPrimeNumberList( _\nByVal numberToTest As Integer, _\nByVal asyncOp As AsyncOperation) As ArrayList\n\nDim e As ProgressChangedEventArgs = Nothing\nDim primes As New ArrayList\nDim firstDivisor As Integer\nDim n As Integer = 5\n\n' Add the first prime numbers.\nprimes.Add(2)\nprimes.Add(3)\n\n' Do the work.\nWhile n < numberToTest And _\nNot Me.TaskCanceled(asyncOp.UserSuppliedState)\n\nIf IsPrime(primes, n, firstDivisor) Then\n' Report to the client that you found a prime.\ne = New CalculatePrimeProgressChangedEventArgs( _\nn, _\nCSng(n) / CSng(numberToTest) * 100, _\nasyncOp.UserSuppliedState)\n\nasyncOp.Post(Me.onProgressReportDelegate, e)\n\nprimes.Add(n)\n\n' Yield the rest of this time slice.\nThread.Sleep(0)\nEnd If\n\n' Skip even numbers.\nn += 2\n\nEnd While\n\nReturn primes\n\nEnd Function\n\n' This method tests n for primality against the list of\n' prime numbers contained in the primes parameter.\nPrivate Function IsPrime( _\nByVal primes As ArrayList, _\nByVal n As Integer, _\nByRef firstDivisor As Integer) As Boolean\n\nDim foundDivisor As Boolean = False\nDim exceedsSquareRoot As Boolean = False\n\nDim i As Integer = 0\nDim divisor As Integer = 0\nfirstDivisor = 1\n\n' Stop the search if:\n' there are no more primes in the list,\n' there is a divisor of n in the list, or\n' there is a prime that is larger than\n' the square root of n.\nWhile i < primes.Count AndAlso _\nNot foundDivisor AndAlso _\nNot exceedsSquareRoot\n\n' The divisor variable will be the smallest prime number\n' not yet tried.\ndivisor = primes(i)\ni = i + 1\n\n' Determine whether the divisor is greater than the\n' square root of n.\nIf divisor * divisor > n Then\nexceedsSquareRoot = True\n' Determine whether the divisor is a factor of n.\nElseIf n Mod divisor = 0 Then\nfirstDivisor = divisor\nfoundDivisor = True\nEnd If\nEnd While\n\nReturn Not foundDivisor\n\nEnd Function\n\n' This method is invoked via the AsyncOperation object,\n' so it is guaranteed to be executed on the correct thread.\nPrivate Sub CalculateCompleted(ByVal operationState As Object)\nDim e As CalculatePrimeCompletedEventArgs = operationState\n\nOnCalculatePrimeCompleted(e)\n\nEnd Sub\n\n' This method is invoked via the AsyncOperation object,\n' so it is guaranteed to be executed on the correct thread.\nPrivate Sub ReportProgress(ByVal state As Object)\nDim e As ProgressChangedEventArgs = state\n\nOnProgressChanged(e)\n\nEnd Sub\n\nProtected Sub OnCalculatePrimeCompleted( _\nByVal e As CalculatePrimeCompletedEventArgs)\n\nRaiseEvent CalculatePrimeCompleted(Me, e)\n\nEnd Sub\n\nProtected Sub OnProgressChanged( _\nByVal e As ProgressChangedEventArgs)\n\nRaiseEvent ProgressChanged(e)\n\nEnd Sub\n\n' This is the method that the underlying, free-threaded\n' asynchronous behavior will invoke. This will happen on\n' an arbitrary thread.\nPrivate Sub CompletionMethod( _\nByVal numberToTest As Integer, _\nByVal firstDivisor As Integer, _\nByVal prime As Boolean, _\nByVal exc As Exception, _\nByVal canceled As Boolean, _\nByVal asyncOp As AsyncOperation)\n\n' If the task was not previously canceled,\n' remove the task from the lifetime collection.\nIf Not canceled Then\nSyncLock userStateToLifetime.SyncRoot\nuserStateToLifetime.Remove(asyncOp.UserSuppliedState)\nEnd SyncLock\nEnd If\n\n' Package the results of the operation in a\n' CalculatePrimeCompletedEventArgs.\nDim e As New CalculatePrimeCompletedEventArgs( _\nnumberToTest, _\nfirstDivisor, _\nprime, _\nexc, _\ncanceled, _\nasyncOp.UserSuppliedState)\n\n' End the task. The asyncOp object is responsible\n' for marshaling the call.\nasyncOp.PostOperationCompleted(onCompletedDelegate, e)\n\n' Note that after the call to PostOperationCompleted, asyncOp\n' is no longer usable, and any attempt to use it will cause.\n' an exception to be thrown.\n\nEnd Sub\n\n#End Region\n\nPrivate Sub InitializeComponent()\n\nEnd Sub\n\nEnd Class\n\nPublic Class CalculatePrimeProgressChangedEventArgs\nInherits ProgressChangedEventArgs\nPrivate latestPrimeNumberValue As Integer = 1\n\nPublic Sub New( _\nByVal latestPrime As Integer, _\nByVal progressPercentage As Integer, _\nByVal UserState As Object)\n\nMyBase.New(progressPercentage, UserState)\nMe.latestPrimeNumberValue = latestPrime\n\nEnd Sub\n\nPublic ReadOnly Property LatestPrimeNumber() As Integer\nGet\nReturn latestPrimeNumberValue\nEnd Get\nEnd Property\nEnd Class\n\nPublic Class CalculatePrimeCompletedEventArgs\nInherits AsyncCompletedEventArgs\nPrivate numberToTestValue As Integer = 0\nPrivate firstDivisorValue As Integer = 1\nPrivate isPrimeValue As Boolean\n\nPublic Sub New( _\nByVal numberToTest As Integer, _\nByVal firstDivisor As Integer, _\nByVal isPrime As Boolean, _\nByVal e As Exception, _\nByVal canceled As Boolean, _\nByVal state As Object)\n\nMyBase.New(e, canceled, state)\nMe.numberToTestValue = numberToTest\nMe.firstDivisorValue = firstDivisor\nMe.isPrimeValue = isPrime\n\nEnd Sub\n\nPublic ReadOnly Property NumberToTest() As Integer\nGet\n' Raise an exception if the operation failed\n' or was canceled.\nRaiseExceptionIfNecessary()\n\n' If the operation was successful, return\n' the property value.\nReturn numberToTestValue\nEnd Get\nEnd Property\n\nPublic ReadOnly Property FirstDivisor() As Integer\nGet\n' Raise an exception if the operation failed\n' or was canceled.\nRaiseExceptionIfNecessary()\n\n' If the operation was successful, return\n' the property value.\nReturn firstDivisorValue\nEnd Get\nEnd Property\n\nPublic ReadOnly Property IsPrime() As Boolean\nGet\n' Raise an exception if the operation failed\n' or was canceled.\nRaiseExceptionIfNecessary()\n\n' If the operation was successful, return\n' the property value.\nReturn isPrimeValue\nEnd Get\nEnd Property\nEnd Class\n``````\n``````using System;\nusing System.Collections;\nusing System.Collections.Specialized;\nusing System.ComponentModel;\nusing System.Data;\nusing System.Drawing;\nusing System.Globalization;\nusing System.Threading;\nusing System.Windows.Forms;\n\n...\n\n/////////////////////////////////////////////////////////////\n#region PrimeNumberCalculator Implementation\n\npublic delegate void ProgressChangedEventHandler(\nProgressChangedEventArgs e);\n\npublic delegate void CalculatePrimeCompletedEventHandler(\nobject sender,\nCalculatePrimeCompletedEventArgs e);\n\n// This class implements the Event-based Asynchronous Pattern.\n// It asynchronously computes whether a number is prime or\n// composite (not prime).\npublic class PrimeNumberCalculator : Component\n{\nprivate delegate void WorkerEventHandler(\nint numberToCheck,\nAsyncOperation asyncOp);\n\nprivate SendOrPostCallback onProgressReportDelegate;\nprivate SendOrPostCallback onCompletedDelegate;\n\nprivate HybridDictionary userStateToLifetime =\nnew HybridDictionary();\n\nprivate System.ComponentModel.Container components = null;\n\n/////////////////////////////////////////////////////////////\n#region Public events\n\npublic event ProgressChangedEventHandler ProgressChanged;\npublic event CalculatePrimeCompletedEventHandler CalculatePrimeCompleted;\n\n#endregion\n\n/////////////////////////////////////////////////////////////\n#region Construction and destruction\n\npublic PrimeNumberCalculator(IContainer container)\n{\ncontainer.Add(this);\nInitializeComponent();\n\nInitializeDelegates();\n}\n\npublic PrimeNumberCalculator()\n{\nInitializeComponent();\n\nInitializeDelegates();\n}\n\nprotected virtual void InitializeDelegates()\n{\nonProgressReportDelegate =\nnew SendOrPostCallback(ReportProgress);\nonCompletedDelegate =\nnew SendOrPostCallback(CalculateCompleted);\n}\n\nprotected override void Dispose(bool disposing)\n{\nif (disposing)\n{\nif (components != null)\n{\ncomponents.Dispose();\n}\n}\nbase.Dispose(disposing);\n}\n\n#endregion // Construction and destruction\n\n/////////////////////////////////////////////////////////////\n///\n#region Implementation\n\n// This method starts an asynchronous calculation.\n// First, it checks the supplied task ID for uniqueness.\n// If taskId is unique, it creates a new WorkerEventHandler\n// and calls its BeginInvoke method to start the calculation.\npublic virtual void CalculatePrimeAsync(\nint numberToTest,\nobject taskId)\n{\n// Create an AsyncOperation for taskId.\nAsyncOperation asyncOp =\nAsyncOperationManager.CreateOperation(taskId);\n\n// Multiple threads will access the task dictionary,\n// so it must be locked to serialize access.\nlock (userStateToLifetime.SyncRoot)\n{\nif (userStateToLifetime.Contains(taskId))\n{\nthrow new ArgumentException(\n\"Task ID parameter must be unique\",\n\"taskId\");\n}\n\nuserStateToLifetime[taskId] = asyncOp;\n}\n\n// Start the asynchronous operation.\nWorkerEventHandler workerDelegate = new WorkerEventHandler(CalculateWorker);\nworkerDelegate.BeginInvoke(\nnumberToTest,\nasyncOp,\nnull,\nnull);\n}\n\n// Utility method for determining if a\n// task has been canceled.\nprivate bool TaskCanceled(object taskId)\n{\nreturn( userStateToLifetime[taskId] == null );\n}\n\n// This method cancels a pending asynchronous operation.\npublic void CancelAsync(object taskId)\n{\nAsyncOperation asyncOp = userStateToLifetime[taskId] as AsyncOperation;\nif (asyncOp != null)\n{\nlock (userStateToLifetime.SyncRoot)\n{\nuserStateToLifetime.Remove(taskId);\n}\n}\n}\n\n// This method performs the actual prime number computation.\n// It is executed on the worker thread.\nprivate void CalculateWorker(\nint numberToTest,\nAsyncOperation asyncOp)\n{\nbool isPrime = false;\nint firstDivisor = 1;\nException e = null;\n\n// Check that the task is still active.\n// The operation may have been canceled before\n// the thread was scheduled.\nif (!TaskCanceled(asyncOp.UserSuppliedState))\n{\ntry\n{\n// Find all the prime numbers up to\n// the square root of numberToTest.\nArrayList primes = BuildPrimeNumberList(\nnumberToTest,\nasyncOp);\n\n// Now we have a list of primes less than\n// numberToTest.\nisPrime = IsPrime(\nprimes,\nnumberToTest,\nout firstDivisor);\n}\ncatch (Exception ex)\n{\ne = ex;\n}\n}\n\n//CalculatePrimeState calcState = new CalculatePrimeState(\n// numberToTest,\n// firstDivisor,\n// isPrime,\n// e,\n// TaskCanceled(asyncOp.UserSuppliedState),\n// asyncOp);\n\n//this.CompletionMethod(calcState);\n\nthis.CompletionMethod(\nnumberToTest,\nfirstDivisor,\nisPrime,\ne,\nTaskCanceled(asyncOp.UserSuppliedState),\nasyncOp);\n\n//completionMethodDelegate(calcState);\n}\n\n// This method computes the list of prime numbers used by the\n// IsPrime method.\nprivate ArrayList BuildPrimeNumberList(\nint numberToTest,\nAsyncOperation asyncOp)\n{\nProgressChangedEventArgs e = null;\nArrayList primes = new ArrayList();\nint firstDivisor;\nint n = 5;\n\n// Add the first prime numbers.\nprimes.Add(2);\nprimes.Add(3);\n\n// Do the work.\nwhile (n < numberToTest &&\n!TaskCanceled( asyncOp.UserSuppliedState ) )\n{\nif (IsPrime(primes, n, out firstDivisor))\n{\n// Report to the client that a prime was found.\ne = new CalculatePrimeProgressChangedEventArgs(\nn,\n(int)((float)n / (float)numberToTest * 100),\nasyncOp.UserSuppliedState);\n\nasyncOp.Post(this.onProgressReportDelegate, e);\n\nprimes.Add(n);\n\n// Yield the rest of this time slice.\nThread.Sleep(0);\n}\n\n// Skip even numbers.\nn += 2;\n}\n\nreturn primes;\n}\n\n// This method tests n for primality against the list of\n// prime numbers contained in the primes parameter.\nprivate bool IsPrime(\nArrayList primes,\nint n,\nout int firstDivisor)\n{\nbool foundDivisor = false;\nbool exceedsSquareRoot = false;\n\nint i = 0;\nint divisor = 0;\nfirstDivisor = 1;\n\n// Stop the search if:\n// there are no more primes in the list,\n// there is a divisor of n in the list, or\n// there is a prime that is larger than\n// the square root of n.\nwhile (\n(i < primes.Count) &&\n!foundDivisor &&\n!exceedsSquareRoot)\n{\n// The divisor variable will be the smallest\n// prime number not yet tried.\ndivisor = (int)primes[i++];\n\n// Determine whether the divisor is greater\n// than the square root of n.\nif (divisor * divisor > n)\n{\nexceedsSquareRoot = true;\n}\n// Determine whether the divisor is a factor of n.\nelse if (n % divisor == 0)\n{\nfirstDivisor = divisor;\nfoundDivisor = true;\n}\n}\n\nreturn !foundDivisor;\n}\n\n// This method is invoked via the AsyncOperation object,\n// so it is guaranteed to be executed on the correct thread.\nprivate void CalculateCompleted(object operationState)\n{\nCalculatePrimeCompletedEventArgs e =\noperationState as CalculatePrimeCompletedEventArgs;\n\nOnCalculatePrimeCompleted(e);\n}\n\n// This method is invoked via the AsyncOperation object,\n// so it is guaranteed to be executed on the correct thread.\nprivate void ReportProgress(object state)\n{\nProgressChangedEventArgs e =\nstate as ProgressChangedEventArgs;\n\nOnProgressChanged(e);\n}\n\nprotected void OnCalculatePrimeCompleted(\nCalculatePrimeCompletedEventArgs e)\n{\nif (CalculatePrimeCompleted != null)\n{\nCalculatePrimeCompleted(this, e);\n}\n}\n\nprotected void OnProgressChanged(ProgressChangedEventArgs e)\n{\nif (ProgressChanged != null)\n{\nProgressChanged(e);\n}\n}\n\n// This is the method that the underlying, free-threaded\n// asynchronous behavior will invoke. This will happen on\n// an arbitrary thread.\nprivate void CompletionMethod(\nint numberToTest,\nint firstDivisor,\nbool isPrime,\nException exception,\nbool canceled,\nAsyncOperation asyncOp )\n\n{\n// If the task was not previously canceled,\n// remove the task from the lifetime collection.\nif (!canceled)\n{\nlock (userStateToLifetime.SyncRoot)\n{\nuserStateToLifetime.Remove(asyncOp.UserSuppliedState);\n}\n}\n\n// Package the results of the operation in a\n// CalculatePrimeCompletedEventArgs.\nCalculatePrimeCompletedEventArgs e =\nnew CalculatePrimeCompletedEventArgs(\nnumberToTest,\nfirstDivisor,\nisPrime,\nexception,\ncanceled,\nasyncOp.UserSuppliedState);\n\n// End the task. The asyncOp object is responsible\n// for marshaling the call.\nasyncOp.PostOperationCompleted(onCompletedDelegate, e);\n\n// Note that after the call to OperationCompleted,\n// asyncOp is no longer usable, and any attempt to use it\n// will cause an exception to be thrown.\n}\n\n#endregion\n\n/////////////////////////////////////////////////////////////\n#region Component Designer generated code\n\nprivate void InitializeComponent()\n{\ncomponents = new System.ComponentModel.Container();\n}\n\n#endregion\n\n}\n\npublic class CalculatePrimeProgressChangedEventArgs :\nProgressChangedEventArgs\n{\nprivate int latestPrimeNumberValue = 1;\n\npublic CalculatePrimeProgressChangedEventArgs(\nint latestPrime,\nint progressPercentage,\nobject userToken) : base( progressPercentage, userToken )\n{\nthis.latestPrimeNumberValue = latestPrime;\n}\n\npublic int LatestPrimeNumber\n{\nget\n{\nreturn latestPrimeNumberValue;\n}\n}\n}\n\npublic class CalculatePrimeCompletedEventArgs :\nAsyncCompletedEventArgs\n{\nprivate int numberToTestValue = 0;\nprivate int firstDivisorValue = 1;\nprivate bool isPrimeValue;\n\npublic CalculatePrimeCompletedEventArgs(\nint numberToTest,\nint firstDivisor,\nbool isPrime,\nException e,\nbool canceled,\nobject state) : base(e, canceled, state)\n{\nthis.numberToTestValue = numberToTest;\nthis.firstDivisorValue = firstDivisor;\nthis.isPrimeValue = isPrime;\n}\n\npublic int NumberToTest\n{\nget\n{\n// Raise an exception if the operation failed or\n// was canceled.\nRaiseExceptionIfNecessary();\n\n// If the operation was successful, return the\n// property value.\nreturn numberToTestValue;\n}\n}\n\npublic int FirstDivisor\n{\nget\n{\n// Raise an exception if the operation failed or\n// was canceled.\nRaiseExceptionIfNecessary();\n\n// If the operation was successful, return the\n// property value.\nreturn firstDivisorValue;\n}\n}\n\npublic bool IsPrime\n{\nget\n{\n// Raise an exception if the operation failed or\n// was canceled.\nRaiseExceptionIfNecessary();\n\n// If the operation was successful, return the\n// property value.\nreturn isPrimeValue;\n}\n}\n}\n\n#endregion\n\n``````\n\n## See Also\n\n#### Tasks\n\nWalkthrough: Implementing a Component That Supports the Event-based Asynchronous Pattern\n\n#### Reference\n\nAsyncOperation\n\nAsyncOperationManager\n\nWindowsFormsSynchronizationContext" ]

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[ "# 10 bar ile psi\n\n1 bar = 14.503773773 psi 1 psi = 0.0689475729 bar Example: convert 15 bar to psi: 15 bar = 15 × 14.503773773 psi = 217.556606595 psi Popular Pressure Unit Conversions bar to psi psi to bar kpa to psi psi to kpa Convert Bar to Other Pressure Units Bar to Pascal Bar to Kilopascal Bar to Ksi Bar to Standard Atmosphere Bar to Break\n\n10 bar to psi = 145.03774 psi Want other units? You can do the reverse unit conversion from psi to bar, or enter any two units below:\n\n10 Bar = 145.03774 Psi FAQ about Bar to Psi Conversion How to convert Bar to Psi ? How many Psi in a Bar? How many Bar in a Psi? How to Convert 5 Bar to Psi? Most popular convertion pairs of pressure Atmospheres to Bar Atmospheres to cmHg Atmospheres to cmAq Atmospheres to Foot Water Atmospheres to Inch Mercury Atmospheres to Inch Water\n\nFree Convert 10 bar to psi (psi) Converter calculator in pressure units,10 bar to psi conversion table and from 10 bar to other pressure units\n\nMore information from the unit converter. How many bars in 1 psi? The answer is 0.0689475728. We assume you are converting between bar and pound/square inch.You can view more details on each measurement unit: bars or psi The SI derived unit for pressure is the pascal. 1 pascal is equal to 1.0E-5 bars, or 0.00014503773800722 psi. Note that rounding errors may occur, so always check the results.\n\nPressure Bar Bar/Psi 10 Convert 10 Bar to Psi: 10 Bar in Psi Bar Psi Bar/Psi Psi/Bar This calculator allows you to convert from Bar to Psi and in a reverse direction. To convert from Bar to Psi, enter the amount of Bar into the first input and to convert from Psi to Bar, enter the amount of Psi into the second input. 10 Bar = 145.03773773 Psi\n\nThere are 0.0689475729 bars in a psi. What is pounds per square inch (psi)? Pounds per square inch ( or pound-force per square inch) is a non-SI unit of Pressure. The symbol for pounds per square inch is psi or lbf/in 2. There are 14.5038 psi in a bar. Conversion Formula\n\n1 Pound per square inch (psi) is equal to 0.0689475729 bar. To convert psi to bar, multiply the psi value by 0.0689475729 or divide by 14.5037738. For example, to convert 50 psi to bar, you can use the following formula: Therefore, 50 psi equal to 3.447378645 bar.\n\n10 bar = 145.038 psi / Convert bar to psi. You can also convert bar to pascal, exapascal, petapascal, terapascal, gigapascal, megapascal, kilopascal, hectopascal, dekapascal, decipascal, centipascal, millipascal, micropascal, nanopascal, picopascal, femtopascal, attopascal, newton/square meter, newton/square centimeter, newton/square millimeter, kilonewton/square meter, millibar, microbar ...\n\nPressure Converter Popular pressure unit conversions bar to psi psi to bar kpa to psi psi to kpa Complete list of pressure units for conversion pascal [Pa] 1 kilopascal [kPa] = 1000 pascal [Pa] kilopascal to pascal, pascal to kilopascal 1 bar = 100000 pascal [Pa] bar to pascal, pascal to bar 1 psi [psi] = 6894.7572931783 pascal [Pa]\n\nIt is an easy formula to convert bar to psi. We know that 1 bar = 14.5038 psi; 1 psi = 0.0689476 bar. 3.7 bar = ___ psi. 3.7 x 14.5038 = 43.5113 psi (we know 1 bar = 14.5038 psi). Answer: 3.7 bar = 43.5113 psi. Bar to psi is a bar to Pounds per square inch pressure conversion converter.\n\nThe pressure value 10 bar in words is \"ten bar\". This is simple to use online converter of weights and measures. Simply select the input unit, enter the value and click \"Convert\" button. The value will be converted to all other units of the actual measure. You can simply convert for example between metric, UK imperial and US customary units system.\n\nbar to psi Conversion Table. The pressure unit bar can be converted into pounds force per square inch in the following way: 1 psi = 6,894.76 pascals (Pa) 1 bar = 100,000 pascals (Pa) psi value x 6,894.76 Pa = bar value x 100,000 Pa; psi value = bar value x 14.5038; bar pressure related products.\n\nThe idea of conversion from one pressure unit to another stays the same no matter what you want to calculate it for. In the case of type pressure, the basic formula for conversion from bar to PSI is the same (1 bar = 15.5038 psi). What is 2 bar TYRE pressure in psi? 2 bar tyre pressure = 29.0076 psi What is 2.4 TYRE pressure in psi?\n\nSimply multiply 100 by 0.0689475729: bar = 100 * 0.0689475729 = 6.89475729 bar Therefore, 100 psi equal to 6.89475729 bar. Using the simple formulas below, you can easily convert psi to bar. psi to bar conversion formula: bar = psi * 0.0689475729 bar = psi / 14.5037738 How to convert bar to psi?\n\n10 psi to bar = 0.68948 bar 20 psi to bar = 1.37895 bar 30 psi to bar = 2.06843 bar 40 psi to bar = 2.7579 bar 50 psi to bar = 3.44738 bar 100 psi to bar = 6.89476 bar 200 psi to bar = 13.78951 bar Want other units? You can do the reverse unit conversion from bar to psi, or enter any two units below: Enter two units to convert From:\n\nMap showing atmospheric pressure in mbar or hPa A tire-pressure gauge displaying bar (outside) and pounds per square inch (inside) Atmospheric air pressure where standard atmospheric pressure is defined as 1013.25 mbar, 101.325 kPa, 1.01325 bar, which is about 14.7 pounds per square inch.\n\nPosted 3:38:35 PM. Company DescriptionWhy PSI?Because we are not a huge corporation and people still matter here. Our…See this and similar jobs on LinkedIn.\n\nPSI Polska - Glasford International Poland | 1,964 followers on LinkedIn. Nowość ☛ Indywidualne Sesje dla Menadżerów Executive Search and Developement Zdalne warsztaty motywujące zespół ...\n\nPosted 3:10:24 PM. Company DescriptionWe are the company that cares - for our staff, for our clients, for our partners…See this and similar jobs on LinkedIn. ... PSI CRO AG Warsaw, Mazowieckie ...\n\nPSI CRO AG Warsaw, Mazowieckie, Poland. Regulatory Officer. PSI CRO AG Warsaw, Mazowieckie, Poland 1 month ago Be among the first 25 applicants See who PSI CRO AG has hired for this role ...\n\n0.68948 Bar (bar) Psi : Psi is the abbreviation of pound per square inch, and is widely used in British and American. 1 psi = 6,894.76 Pascals. Bar : The bar is a unit of measurement for pressure. It is widely used in the daily life particularly in European countries, though that is a non-SI unit. 1 bar is equal to 100,000 Pascals, which is ...\n\nconsisted of eight high-pressure, super-heating boilers delivering 64 bars (930 psi) of pressure and 500 °C (932 °F), all weighing 8,000 tons. This delivered\"Fortune\" Île de l'Est (East Island) (46°26′S 52°18′E / 46.433°S 52.300°E / -46.433; 52.300 (Île de l'Est)) la Voile (small rock south of Île de l'Est) Robert Chapuis. In 1986 he was elected to the regional Council for the Île-de-France and served until 1992. In 1988, he became head of the SocialistHavana (/həˈvænə/; Spanish: La Habana [la aˈβana] ; Lucumi: Ilú Pupu Ilé) is the capital and largest city of Cuba. The heart of the La Habana Provinceand Queen Mary joined US 10 at Sydney, and Île de France detached at Fremantle. Nieuw Amsterdam was carrying 2,642 troops. US 10 reached Colombo on Aprilcatastrophe in the making. Finding out about an abandoned theatre on the Ile St Louis in Paris, connected to a mask the killer had with her when she fledled to a rapid development of the region. The area's largest cities, Sept-Îles and Baie-Comeau, were now linked to the rest of the province by a road. At\"A brief history of mineral symbols\" (PDF). Elements. 17 (3): 291–320. doi:10.2138/gselements.17.3.152. S2CID 239762386. Warr, L.N. (2021). \"Mineral symbolclassification of protein secondary structure with PsiCSI\". Protein Science. 12 (2): 288–95. doi:10.1110/ps.0222303. PMC 2312422. PMID 12538892. Labudde(112.8 kg/s). The steam pressure at the turbine stop valves was 210 psi (14.45 bar) 357/371 °C. In 1954 the station burned 81,400 tons of coal. In 1964–65" ]

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[ "# Calculating the sum of all columns of a 2D NumPy array\n\n• Last Updated : 02 Sep, 2020\n\nLet us see how to calculate the sum of all the columns of a 2 dimensional NumPy array.\n\nExample :\n\n```Input :\n[[1, 2, 3, 4, 5],\n[5, 6, 7, 8, 9],\n[2, 1, 5, 7, 8],\n[2, 9, 3, 1, 0]]\n\nOutput : [10, 18, 18, 20, 22]\n\nInput :\n[[5, 4, 1, 7],\n[0, 9, 3, 5],\n[3, 2, 8, 6]]\n\nOutput : [8, 15, 12, 18]\n```\n\nApproach 1 : We will be using the `sum()` method. We will pass parameter` axis = 0` to get the sum columns wise.\n\n `# importing numpy``import` `numpy as np`` ` `# initialize the 2-d array``arr ``=` `np.array([[``1``, ``2``, ``3``, ``4``, ``5``],``                ``[``5``, ``6``, ``7``, ``8``, ``9``],``                ``[``2``, ``1``, ``5``, ``7``, ``8``],``                ``[``2``, ``9``, ``3``, ``1``, ``0``]])`` ` `# calculating column wise sum``sum_2d ``=` `arr.``sum``(axis ``=` `0``)`` ` `# displaying the sum``print``(``\"Column wise sum is :\\n\"``, sum_2d)`\n\nOutput :\n\n```Column wise sum is :\n[10 18 18 20 22]\n```\n\nApproach 2 : We can also use the `numpy.einsum()` method, with parameter `'ij->j'`.\n\n `# importing numpy``import` `numpy as np`` ` `# initialize the 2-d array``arr ``=` `np.array([[``1``, ``2``, ``3``, ``4``, ``5``],``                ``[``5``, ``6``, ``7``, ``8``, ``9``],``                ``[``2``, ``1``, ``5``, ``7``, ``8``],``                ``[``2``, ``9``, ``3``, ``1``, ``0``]])`` ` `# calculating column wise sum``sum_2d ``=` `np.einsum(``'ij->j'``, arr)`` ` `# displaying the sum``print``(``\"Column wise sum is :\\n\"``, sum_2d)`\n\nOutput :\n\n```Column wise sum is :\n[10 18 18 20 22]\n```\n\nMy Personal Notes arrow_drop_up" ]

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[ "# Power Query to Get Fiscal Year & Fiscal Month\n\n#### legalhustler\n\n##### Well-known Member\nMy fiscal year is 10/1 - 9/30 and I would like to modify the following M code so that it reflect the fiscal month and year correctly. Can someone help? So October 2018 should be fiscal month 1, January 2019 should be fiscal month 4, July 2019 should be fiscal month 10, etc.\n\nCode:\n``````let\n\nEndFiscalYearMonth = 9, //set this as the last month number of your fiscal year : June = 6, July =7 etc\n\nStartDate= #date(2018, 10, 1), // Change start date #date(yyyy,m,d)\nEndDate = DateTime.LocalNow(), // Could change to #date(yyyy,m,d) if you need to specify future date\n\n/* Comment out the above StartDate and EndDate using // if you want to use a dynamic start and end date based on other query/table\nYou will need to change \"Sales\" and \"Invoice Date\" in 2 lines below and then remove the //\n*/\n\n//TableName = Sales\n//DateColumnName = \"Invoice Date\"\n//StartDate = Record.Field ( Table.Min(TableName,DateColumnName) ,DateColumnName),\n//EndDate = Record.Field(Table.Max(TableName,DateColumnName),DateColumnName),\n\nDateList = List.Dates(StartDate, Number.From(EndDate)- Number.From(StartDate)+1 ,#duration(1,0,0,0)),\n\n#\"Converted to Table\" = Table.FromList(DateList, Splitter.SplitByNothing(), null, null, ExtraValues.Error),\n#\"Named as Date\" = Table.RenameColumns(#\"Converted to Table\",{{\"Column1\", \"Date\"}}),\n#\"Changed Type\" = Table.TransformColumnTypes(#\"Named as Date\",{{\"Date\", type date}}),\n#\"Inserted Year\" = Table.AddColumn(#\"Changed Type\", \"Calendar Year\", each Date.Year([Date]), type number),\n#\"Inserted Month Number\" = Table.AddColumn(#\"Inserted Year\", \"Month Number\", each Date.Month([Date]), type number),\n#\"Long Month Name\" = Table.AddColumn(#\"Inserted Month Number\", \"Month Long\", each Date.MonthName([Date]), type text),\n#\"Short Month Name\" = Table.AddColumn(#\"Long Month Name\", \"Month\", each Text.Start([Month Long], 3), type text),\n#\"Fiscal Month Number\" = Table.AddColumn(#\"Short Month Name\", \"Fiscal Month Number\", each if [Month Number] > EndFiscalYearMonth then [Month Number]-EndFiscalYearMonth else [Month Number]+EndFiscalYearMonth),\n#\"Changed Type1\" = Table.TransformColumnTypes(#\"Fiscal Month Number\",{{\"Fiscal Month Number\", Int64.Type}}),\n#\"Fiscal Year\" = Table.AddColumn(#\"Changed Type1\", \"Fiscal Year\", each if [Fiscal Month Number] <=EndFiscalYearMonth then [Calendar Year]+1 else [Calendar Year]),\n#\"Changed Years to Text\" = Table.TransformColumnTypes(#\"Fiscal Year\",{{\"Fiscal Year\", type text}, {\"Calendar Year\", type text}}),\nFYName = Table.AddColumn(#\"Changed Years to Text\", \"FYName\", each \"FY\"&Text.End([Fiscal Year],2))\nin\nFYName``````\n\n#### legalhustler\n\n##### Well-known Member\nThanks but that doesn't show how to get the fiscal month and fiscal year.\n\n#### sandy666\n\n##### Well-known Member\nthere is Fin Year and Fin Month (Financial = Fiscal)\n\n#### legalhustler\n\n##### Well-known Member\nthere is Fin Year and Fin Month (Financial = Fiscal)\nAh! Didn't read that. I got the fiscal year correct but I'm having trouble with the following part to get the correct fiscal month\n\nCode:\n``= Table.AddColumn(#\"Inserted Day Name\", \"Fiscal Month\", each if Date.Month([Date]) >=10 then Date.Month([Date])-9 else Date.Month([Date])+9)``\n\nMy fiscal month starts Oct and ends in Sept, thus it should result in the following:\nOct - 1\nNov - 2\nDec - 3\nJan - 4\nFeb - 5\n....\n....\nSept - 12\n\n#### legalhustler\n\n##### Well-known Member\nWas able to figure it out by changing it to the following. Thanks again!\n\n= Table.AddColumn(#\"Inserted Day Name\", \"Fiscal Month\", each if Date.Month([Date]) >=10 then Date.Month([Date])-9 else Date.Month([Date])+3)\n\n#### sandy666\n\n##### Well-known Member\nyou are welcome and have a nice day", null, "" ]

[ null, "data:image/gif;base64,R0lGODlhAQABAIAAAAAAAP///yH5BAEAAAAALAAAAAABAAEAAAIBRAA7", null ]

[ "# moving an image across the screen along some vector path\n\n14 views (last 30 days)\nKendall Donahoe on 6 Apr 2021\nAnswered: LO on 6 Apr 2021\nhey Ive been trying to move an image of the avatar across the screen and can't seem to figure it out. I know how to shift the image using the identity matrix and was wondering if there was a way I could use that to make it move across the screen or any other way that it is possible. Here are some of the function names I have been using and the way that I shifted it. I have attached the image and a converter script that I used.\nAout=Jpeg2pointsConverter(A, 15);\nplot(Aout(1,:), Aout(2,:),'b.')\n[nrows, ncols]=size(Aout);\nAA=[Aout; ones(1,ncols)];\nS=eye(3)\n>> S(1,3)=300\n>> A3=S*AA;\n>> hold on, plot(A3(1,:), A3(2,:),'r.')\nDGM on 6 Apr 2021\nEdited: DGM on 6 Apr 2021\nI have no idea what you're trying to do with this, but this whole thing\nBB1=BB(:,:,1);\n[M, N]= size(BB1);\nBB1=double(BB1);\nBB2 = 255-BB1; %Invert so white is 0 instead of 255\nBB3 = (BB2 > THRESHOLD); %Any point with high value is replaced by 1, and\n%any point with a low value is replaced by 0\nPP=zeros(2,M*N);\ncnt=0;\nfor ii=1:M,\nfor jj=1:N,\nif (BB3(ii,jj)>0.5),\n% this is going to return nonsense indices\n% once you change the aspect ratio\nPP(:,cnt+1)=[jj;N-ii];\ncnt=cnt+1;\nend,\nend,\nend\nPPout = PP(:,1:cnt);\ncan be replaced with this\n[rows cols]=find(BB(:,:,1)<(255-threshold));\nPPout=[cols; rows];\nI don't know why you're returning all the row indices backwards, but i guess you could still do that\n[rows cols]=find(BB(:,:,1)<(255-threshold));\nPPout=[cols; size(BB,1)-rows];\n\nLO on 6 Apr 2021\ntry using circshift\nhttps://de.mathworks.com/help/matlab/ref/circshift.html" ]

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[ "# SpheroidalS2Prime\n\nSpheroidalS2Prime[n,m,γ,z]\n\ngives the derivative with respect to", null, "of the radial spheroidal function", null, "of the second kind.\n\n# Details", null, "• Mathematical function, suitable for both symbolic and numerical manipulation.\n• For certain special arguments, SpheroidalS2Prime automatically evaluates to exact values.\n• SpheroidalS2Prime can be evaluated to arbitrary numerical precision.\n• SpheroidalS2Prime automatically threads over lists.\n\n# Examples\n\nopen allclose all\n\n## Basic Examples(5)\n\nEvaluate numerically:\n\nPlot over a subset of the reals:\n\nPlot over a subset of the complexes:\n\nSeries expansion at the origin:\n\nSeries expansion at a singular point:\n\n## Scope(23)\n\n### Numerical Evaluation(4)\n\nEvaluate numerically:\n\nEvaluate to high precision:\n\nThe precision of the output tracks the precision of the input:\n\nComplex number inputs:\n\nEvaluate efficiently at high precision:\n\n### Specific Values(5)\n\nSimple exact values are generated automatically:\n\nSingular points:\n\nFind the first positive maximum of SpheroidalS2Prime[2,0,5,x]:\n\nSpheroidalS2Prime functions become elementary if", null, "and", null, ":\n\n### Visualization(3)\n\nPlot the SpheroidalS2Prime function for integer orders:\n\nPlot the SpheroidalS2Prime function for non-integer parameters:\n\nPlot the real part of SpheroidalS2Prime:\n\nPlot the imaginary part of SpheroidalS2Prime:\n\n### Function Properties(5)\n\nSpheroidalS2Prime is not an analytic function:", null, "has both singularities and discontinuities for", null, ":", null, "is neither non-decreasing nor non-increasing:", null, "is not injective:\n\nSpheroidalS2Prime is neither non-negative nor non-positive:\n\nSpheroidalS2Prime is neither convex nor concave:\n\n### Differentiation(2)\n\nFirst derivative with respect to", null, ":\n\nHigher derivatives with respect to", null, ":\n\nPlot the higher derivatives with respect to", null, "when", null, ",", null, "and", null, ":\n\n### Integration(2)\n\nCompute the indefinite integral using Integrate:\n\nVerify the anti-derivative:\n\nDefinite integral:\n\n### Series Expansions(2)\n\nFind the Taylor expansion using Series:\n\nPlots of the first three approximations around", null, ":\n\nTaylor expansion at a generic point:\n\n## Properties & Relations(1)\n\nSpheroidal functions do not evaluate for half-integer values of", null, "and generic values of", null, ":" ]

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[ "", null, "# SQLite Tutorial: Understanding SQLite Data Types and Variable-Length Integers\n\nJump into digital forensics and you will quickly learn that SQLite is one of the most prevalent and useful sources of information or evidence for almost any examination. SQLite is open source and free to use for any purpose, which has made it the most used database technology in the world. Chat applications, web browser data, operating system settings, application tracking, and much more all use SQLite databases.\n\n## Introduction to SQLite\n\nMany forensic examinations may require a deeper look into the structure of the database, such as the recovery of deleted data still stored within the SQLite database. This article will be the first in a series of articles that will break down important features necessary for examiners to understand when working with SQLite databases and will help them explain how various processes work when called upon to defend their work.\n\nVariable-length integers or VARINTs (for anyone slightly familiar with SQLite) may seem like an odd place to start. However, much of the primary information related to how the data is stored within a SQLite database uses this format to define locations, sizes, and even values. It is important to understand this data structure before we start to break down how to read the database structure itself.\n\n## What Is a Variable-Length Integer?\n\nA variable-length integer is a method of storing an integer value using the least number of bytes possible. SQLite does this by using a clever compression method which utilizes the most significant bit in a byte.\n\nVARINTs can store information such as the ROWID, the size of the entry, column data, column data size, etc. This is structured data that the database needs to know to properly store information and recall that information when necessary.\n\n## Why Use Variable-Length Integers?\n\nSQLite databases are designed to be lightweight while at the same time able to hold massive amounts of data (140 terabytes). Part of being lightweight is being as efficient with space as possible. VARINTs help reduce that space by not having a set number of bytes reserved to store data which may frequently NOT use all the space reserved.\n\nTo better understand the concept of VARINTs let's look at an example outside of databases or computer language.\n\nA restaurant has found that the maximum amount of time that a table is occupied is one hour. They have 10 tables, and are open for 8 hours. This means that if they allowed only one person to sit at a table within the one-hour time frame whether or not that person ACTUALLY used the full hour, they would be limited to 80 customers per day.\n\nHowever, if most people only use 20 minutes of the hour, and then they immediately put another person on that table, they dramatically increase the number of customers they can serve per day, potentially up to 240 customers per day! It’s the same amount of space, but the utilization is much higher.\n\nVARINTs function in the same way, by storing structured data in the smallest space possible. If the ROWID requires only 2 bytes of data, then SQLite will only use 2 bytes to indicate the ROWID, and the VARINT which indicates the size of the entry will immediately follow. If the ROWID required 4 bytes of data, then it would store 4 bytes and then the size of the entry. While this may be efficient, it does lead us into the next section.\n\n## How Do We Read a Variable-Length Integer?\n\nStorage is cheap, and on computers it is easy to expand storage. Because of this, most of the time structured data has a set number of bytes which that data can be stored, whether it fills that space or not. This is easier to store data and it is easier to read data. SQLite was built for smaller, lower powered devices, such as mobile phones where storage was not as cheap and so efficiency was king. A VARINT requires that the database analyze the most significant bit of each byte to determine the length of the VARINT and then parse the information stored.\n\nIn SQLite, a VARINT can be between one and nine bytes in length. When analyzing a VARINT there are two parts: the most significant bit, and the value. The most significant bit determines whether or not to include the next byte in the calculation. If the most significant bit is a 1, the next byte must be considered. If the most significant byte is a 0, the next byte is not considered.\n\nThe value portion of the VARINT is the remaining 7 bits. To determine the value in decimal, convert the 7-bit binary value into decimal. Let’s look at a few examples:\n\nFor the first example, let’s look at one byte with a hexadecimal value of 0x74. First, we need to look at the most significant bit, or the left most bit in the byte. Determine the most significant bit by converting 0x74 to binary, which is 01110100. The most significant bit is 0. 0 means that we have reached the end of our VARINT which indicates that this VARINT is one byte in length. We then ignore the most significant bit and determine the 7-byte value. For a one-byte VARINT, the value is always the hex value of the original hexadecimal value. In this example, 0x74 or 116 in decimal.\n\nIn the second example we will look at two-bytes with a value of 0xA102. Look at the first byte, and determine the value of the most significant bit. 0xA1 converts to 10100001. Because the most significant bit of the first byte is one, we also need to look at the next byte. Converting 0x02 to binary gives us 00000010. The most significant bit is 0, indicating we have reached the end of our VARINT. We then take each of the remaining 7-bit binary values, combine them together to get our result. 01000010000010 binary equals 0x1082 or 4226 in decimal.\n\nWhile we won't work out completely the last example, a set of bytes with a value of 0x9A81FF618B would be a VARINT of how many bytes? This would be a VARINT of four bytes. Identifying each significant bit for each byte would be 0x9A = 1, 0x81 = 1, 0xFF = 1, 0x61 = 0. The 0 indicates that we have reached the end of our VARINT so 0x8B is not considered.\n\nA trick to quickly determining the length of a VARINT, is if the hexadecimal of a byte starts with a value of 8 through F, then the most significant bit is 1. If the hexadecimal value starts with a value of 0-7 the most significant bit is 0.\n\nRemember, the maximum size of a SQLite VARINT is nine bytes. If these three examples were part of the same data within a database, by using a VARINT instead of a flat 9-byte reservation, the database saved 20 bytes. That may not sound like much, but think about your own text messaging app, and how many messages you have sent or received. Think about a 20-byte savings for every message sent or received. It adds up fast!\n\n## A Note on VARINTs\n\nVariable-length integers are important for defining key information within a SQLite database. This includes ROWID, size of the cell (how an entry is stored), and how data is organized within an entry. It is important to understand how to decipher these values to properly unpack SQLite data. It is also critical for those performing database data recovery.\n\n;" ]

[ null, "https://www.securedatarecovery.com/Media/blog/2023/sqlite-variable-length-integers.webp", null ]

[ "# On Two's Complement\n\nTwo’s complement is a way to encode negative numbers in a binary representation. It accomplishes this by using the most significant bit as the sign bit (0 is positive, 1 is negative).\n\nTwo’s complement has a distinct advantage over ones’ complement, and as such it is the most common method of representing signed integers on computers:\n\n• Fundamental arithmetic operations are identical for both signed and unsigned binary numbers.\n• There is only one representation for zero (ones’ complement has both positive and negative zero).\n\nFor example, let’s compare positive and negative 5 using the `asbits` tool to see their bit representations. The following will print the values out for 4 bits, 8 bits, 16 bits and 32 bits, respectively:\n\n``````~:\\$ for bits in {4,8,16,32}\n> do\n> for n in {5,-5}\n> do\n> asbits \\$n \\$(bc <<< \\$bits/4)\n> done\n> done\n0101\n1011\n0000 0101\n1111 1011\n0000 0000 0000 0101\n1111 1111 1111 1011\n0000 0000 0000 0000 0000 0000 0000 0101\n1111 1111 1111 1111 1111 1111 1111 1011\n``````\n\nLooking at the output, one can see that the negative encoding (where the most significant, or leftmost, bit is a 1) isn’t very intuitive. At least not for me. In fact, I found it confusing as hell, and I didn’t get it at all. Most explanations centered on the mechanics of the operation, such as “invert every bit and add one”, which is the two’s complement of its inverse, but that didn’t help my understanding.\n\nYou can determine the two’s complement of a number by first taking its ones’ complement, i.e., inverting the bits, and then adding 1. For instance:\n\n`````` 0101\n1010 <-- Ones' complement\n1011 <-- Two's complement\n``````\n\nSure that works, but why does it work?!?\n\nThe way that I finally understood it was realizing that the sum of a number and its two’s complement is `2`N.\n\nFor instance, let’s say that we want to find the two’s complement of 8 in an 8-bit representation. To find the two’s complement of `00001000` (810 in binary), we add to it another number which will equal 28:\n\n``````10000000 = 00001000 + 11111000\n\n~:\\$ for n in {8,-8}\n> do\n> asbits \\$n 2\n> done\n0000 1000\n1111 1000\n``````\n\nAdd the two binary numbers together to see for yourself.\n\n`````` 00001000\n+ 11111000\n--------\n(1)00000000\n``````\n\nConveniently, the overflow number is discarded.\n\nAnother way to mathematically calculate the two’s complement is by plugging the complement into the following formula:\n\n(2Number of bits - complement)2\n\nFor example, lets find the two’s complement of 6 for a 4-bit representation. The formula becomes:\n\n24 - 6 = 16 - 6 = 1010 = 10102\n\n`````` 0110\n+ 1010\n----\n(1)0000\n``````\n\nNow, let’s see how a simple program written in C stores a negative integer in memory. The following is straightforward enough:\n\n`twos_complement.c`:\n\n``````#include <stdlib.h>\n#include <stdio.h>\n\nint main(int argc, char **argv) {\nint i = 5;\nint j = -5;\n\nprintf(\"%d %d\\n\", i, j);\nreturn 0;\n}\n``````\n\nCompile it with symbols so we can debug it in GDB:\n\n``````-ggdb3 twos_complement.c -o twos_complement\n``````\n\nAnd start debugging:\n\n``````~:\\$ gdb twos_complement\n(gdb) list\n1 #include <stdlib.h>\n2 #include <stdio.h>\n3\n4 int main(int argc, char **argv) {\n5 int i = 5;\n6 int j = -5;\n7\n8 printf(\"%d %d\\n\", i, j);\n9 return 0;\n10 }\n(gdb) b 8\nBreakpoint 1 at 0x6cd: file twos_complement.c, line 8.\n(gdb) r\nStarting program: /home/btoll/twos_complement\n\nBreakpoint 1, main (argc=1, argv=0x7fffffffe2a8) at twos_complement.c:8\n8 printf(\"%d %d\\n\", i, j);\n(gdb)\n``````\n\nHere, we’ve listed the program, set a breakpoint at line 8 and then ran it. We’ll now disassemble the machine code so we can see the memory we need to inspect:\n\n``````(gdb) disass\nDump of assembler code for function main:\n0x00005555555546b0 <+0>: push rbp\n0x00005555555546b1 <+1>: mov rbp,rsp\n0x00005555555546b4 <+4>: sub rsp,0x20\n0x00005555555546b8 <+8>: mov DWORD PTR [rbp-0x14],edi\n0x00005555555546bb <+11>: mov QWORD PTR [rbp-0x20],rsi\n0x00005555555546bf <+15>: mov DWORD PTR [rbp-0x4],0x5\n0x00005555555546c6 <+22>: mov DWORD PTR [rbp-0x8],0xfffffffb\n=> 0x00005555555546cd <+29>: mov edx,DWORD PTR [rbp-0x8]\n0x00005555555546d0 <+32>: mov eax,DWORD PTR [rbp-0x4]\n0x00005555555546d3 <+35>: mov esi,eax\n0x00005555555546d5 <+37>: lea rdi,[rip+0x98] # 0x555555554774\n0x00005555555546dc <+44>: mov eax,0x0\n0x00005555555546e1 <+49>: call 0x555555554560 <printf@plt>\n0x00005555555546e6 <+54>: mov eax,0x0\n0x00005555555546eb <+59>: leave\n0x00005555555546ec <+60>: ret\nEnd of assembler dump.\n``````\n\nThe positive value stored in the variable `i` is located four bytes below the base pointer, and the negative value in variable `j` is located eight bytes from the base pointer. We can already see from the hexadecimal values that the `-5` is encoded as two’s complement of `5`10, but let’s print them as binary to get the full effect:\n\n``````(gdb) x/xt \\$rbp-4\n0x7fffffffe1bc: 00000000000000000000000000000101\n(gdb) x/xt \\$rbp-8\n0x7fffffffe1b8: 11111111111111111111111111111011\n``````\n\nThis looks like what we’d expect, and we’ll verify it again using our old friend:\n\n``````~:\\$ for n in {5,-5}\n> do\n> asbits \\$n 8\n> done\n0000 0000 0000 0000 0000 0000 0000 0101\n1111 1111 1111 1111 1111 1111 1111 1011\n``````\n\nWeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeee" ]

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[ "Full-text links:\n\nmath.CO\n\n# Title:Uniquely identifying the edges of a graph: the edge metric dimension\n\nAbstract: Let $G=(V,E)$ be a connected graph, let $v\\in V$ be a vertex and let $e=uw\\in E$ be an edge. The distance between the vertex $v$ and the edge $e$ is given by $d_G(e,v)=\\min\\{d_G(u,v),d_G(w,v)\\}$. A vertex $w\\in V$ distinguishes two edges $e_1,e_2\\in E$ if $d_G(w,e_1)\\ne d_G(w,e_2)$. A set $S$ of vertices in a connected graph $G$ is an edge metric generator for $G$ if every two edges of $G$ are distinguished by some vertex of $S$. The smallest cardinality of an edge metric generator for $G$ is called the edge metric dimension and is denoted by $edim(G)$. In this article we introduce the concept of edge metric dimension and initiate the study of its mathematical properties. We make a comparison between the edge metric dimension and the standard metric dimension of graphs while presenting some realization results concerning the edge metric dimension and the standard metric dimension of graphs. We prove that computing the edge metric dimension of connected graphs is NP-hard and give some approximation results. Moreover, we present some bounds and closed formulae for the edge metric dimension of several classes of graphs.\n Comments: 23 pages Subjects: Combinatorics (math.CO) MSC classes: 05C12, 05C76, 05C90 Cite as: arXiv:1602.00291 [math.CO] (or arXiv:1602.00291v1 [math.CO] for this version)\n\n## Submission history\n\nFrom: Ismael Gonzalez Yero [view email]\n[v1] Sun, 31 Jan 2016 17:48:23 UTC (25 KB)" ]

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[ "Question: Seven apples were on the table. I ate some apples. Then there were four apples. How many apples did I eat?\n\nStudent response: 7 + 4 = 11\n\nQuestion: Renzo has 15 books. He has 7 more books than Morris. How many books does Morris have?\n\nStudent response: 15 + 7 = 22\n\nQuestion: Jill had some pencils. She gave 18 to Kelley and now she has 14 left. How many pencils did Jill have to start with?\n\nStudent response: 18 – 14 = 4\n\nQuestion: Skylar has 10 times as many books as Karen. If Skylar has 620 books, how many books does Karen have?\n\nStudent response: 620 x 10 = 6,200\n\nQuestion: Oliver has 496 pens. Bernard has eighteen times as many pens as Oliver. How many more pens does Bernard have than Oliver?\n\nStudent response: 496 x 18 = 8,928\n\nIn our quest for the “right answer” are we encouraging students to simply manipulate some numbers and move on?\n\nEach of the students above ignored the first and last steps of problem solving: first, what are we trying to find, and last, is our answer reasonable.\n\nThe first grade student did not indicate the solution, but if we assume they meant I ate 11 apples, then how can that be true if there were only seven to begin with?\n\nIn the problem with Renzo, again we don’t know which number is the solution. If we assume the student’s answer was that Morris has 22 books, then he has more than Renzo’s 15, which contradicts the second statement that Renzo has more books than Morris. Similarly, how can Jill have 4 pencils to start and then give away 18? And Karen can’t have 6,200 books when we are told that Skylar has more.\n\nIn the last question, Bernard does indeed have 8,928 pens, but that was not the question that was posed.\n\nThere are many models for the steps of the problem solving process, based on the work of mathematician George Polya. Here is a simple one for students and teachers to work with:\n\n1. What are we trying to find?\n\n2. What do we know?\n\n3. How will we solve?\n\n4. Have we answered the question and is the solution reasonable?\n\nBefore trying to solve any problem, students should identify the question and write a complete sentence for the answer, leaving a blank for the solution. Once they have solved, they fill in the blank and revisit the question, checking to see if the answer makes sense and fits with the statements in the problem.\n\nIf the above students had taken these first and last steps, they probably would have realized that I couldn’t have eaten 11 apples, Morris can’t have 22 books, Jill couldn’t have had 4 pencils to start with, Karen can’t have 6,200 books, and that they were asked to find how many more pens Bernard had than Oliver.\n\nMore about step 3 in a future post." ]

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[ "# Roof Slope Formula\n\nRoof slope is also called as roof pitch. In civil engineering, the slope is the numerical measure of the steepness of a roof. By slope angle formula, steepness can be obtained by dividing vertical rise by horizontal span. In the imperial unit of measurements, the slope is expressed with the rise and run. It is computed by using the simple roof slope formula. It is diagrammatically expressed with a triangle shaped figure and the units are expressed in inches.\n\n# Slope Angle Formula\n\n#### Formula:\n\nRun(inches)= ( 12 × Rise ) / Roof Pitch\nSlope = ( Rise / Run ) × 100\nAngle = tan-1( Rise / Run )\n\n#### Related Calculator:\n\nSlope angle formula is essential to determine slope pitch as it is used by the civil engineers to determine the vertical rise in inches for every horizontal twelve-inch length. Use the roof slope formula mentioned above to find the slope of the roof by substituting the required values." ]

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[ "Home » Blog » 2nd Grade Skills » Geometry » Math Puzzles in the Classroom\n\n# Math Puzzles in the Classroom\n\nWritten by: Angie Olson\n\nTeachers are always looking for ways for their students to practice math skills.  Math puzzles are a great option because they are engaging, independent, self-correcting, and easy to prep.", null, "## What are Math Puzzles\n\nMath puzzles are a hands on option for students to practice their math skills.  Math puzzles are available in black and white, full color, editable, and student scrambled.\n\n## When to use Math Puzzles\n\nMath puzzles are great to use during math centers, as an early finisher, intervention activity, or even a math enrichment activity.  For more math centers ideas read this post.\n\n## How to use Math Puzzles\n\nStudents solve the math problem on each puzzle piece.  The answer gets written in the circle provided on the puzzle piece.  Then the student arranges the answers in order from least to greatest.  Once finished, the math puzzle will display a picture.", null, "## Student Accountability\n\nTeachers who prefer to have a more structured accountability piece with the math puzzles can choose to print the recording sheets.  The student would write the name of the puzzle, which is found in the top right hand corner of the puzzle.  Then the student writes his/her answers in the bubbles on the recording sheet.\n\n## Self Correcting Options\n\nOne of the many benefits of using these math puzzles is that they are self correcting and completely independent.  This provides the teacher with the opportunity to help other students or teach a small group.\n\nOnce the student has arranged the pieces in order from least to greatest, the student will be able to determine if the puzzle makes sense and looks right.  If not, the student would remove the piece(s) that look out of place and attempt to solve it again.", null, "## Math Puzzles Skills\n\n### September Skills\n\nThe math skills covered in these puzzles are: addition to 20, subtraction within 20, counting coins, number rounding to 10, number rounding to 100, addition without regrouping, addition with regrouping, domino addition, multiplication, 10 more/10 less, 1 more/1 less, number bonds, number words 0-9, number words 10-19\n\n### October Skills\n\nThe math skills covered in these puzzles are: tens & ones, addition, base ten blocks, expanded form, money(penny, nickel, dime, and quarter), subtraction, missing addends, number patterns, number sequencing, and least to greatest.\n\n### November Skills\n\nThe math skills covered in these puzzles are: expanded form (2-digit, 3-digit, and 4-digit), basic addition facts, basic multiplication facts, odd and even (1-digit and 2-digit), value of digits (2-digit and 3-digit), missing addends (1-digit and 2-digit), number comparisons (2-digit and 3-digit), 2-digit addition (with and without regrouping), adding three numbers (1-digit and 2-digit), counting money (penny, nickel, dime, and quarter), base ten blocks (ones, tens, and hundreds), make a ten/missing addends using a ten frame, least to greatest\n\n### December Skills\n\nThe math skills covered in these puzzles are: addition, subtraction, multiplication, time to the hour, place value (value of the underlined digit), plus and minus 1 & 10 boxes, counting coins (penny, nickel, dime, & quarter), missing addends with doubles, using doubles to subtract, fact families, skip counting (by 2’s, 3’s, 5’s, & 10’s), expanded form (2-digit and 3-digit), number bonds, 2-digit addition, 2-digit subtraction, and least to greatest\n\n### January Skills\n\nThe math skills covered in these puzzles are: 2-Digit Subtraction (with and without regrouping), Part Part Whole, Time to the Half Hour, Fractions, Multiplication, 2-Digit Addition (with and without regrouping), Counting Coins, Doubles Facts-Addition, Subtraction, Skip Counting by 5’s, Number Bonds, 1 more/1 less, 10 more/10 less, and least to greatest\n\n### February Skills\n\nThe math skills covered in these puzzles are: Addition, Subtraction, Multiplication, 1 More/1 Less and 10 More/10 Less Number Boxes, Counting Coins (penny, nickel, dime, quarter, and half dollar), Place Value (value of underlined digit- 2- and 3-digit numbers), Fact Families, Number Bonds, 2-Digit Addition (with and without regrouping), 2-Digit Subtraction (with and without regrouping), Part Part Whole (missing part), Telling Time to the Nearest 5 Minutes, Rounding (to nearest 10 and 100), Least to Greatest\n\n### March Skills\n\nThe math skills covered in these puzzles are: Rounding Numbers (nearest 10 & 100), 2-Digit Addition (with & without regrouping), Addition, Subtraction, Multiplication, 1 More/1 Less, 10 More/10 Less, Place Value, Adding Three Numbers, Counting Coins (penny, nickel, dime, quarter, & half dollar), Place Value Number Boxes, Missing Addends\n\n### April Skills\n\nThe math skills covered in these puzzles are: Addition, Subtraction, Multiplication, 2-Digit Addition (with regrouping), 2-Digit Addition (without regrouping), Number Bonds, Telling Time (nearest 5 minutes), Adding Three Numbers (1- and 2-digit), Expanded form (2- and 3-digit), Fact Families, Missing Addends, Place Value Boxes, Counting Coins, 2-Digit Subtraction (with regrouping), 2-Digit Subtraction (without regrouping), Part Part Whole and Least to Greatest\n\n### May Skills\n\nThe math skills covered in these puzzles are: 3-Digit Addition, 3-Digit Subtraction, Adding 3 Numbers (2-digit numbers), Digit Values (2-digit and 3-digit numbers), Counting Coins (penny, nickel, dime, quarter, & half dollar), Part Part Whole, Multiplication, Fact Families, Telling Time (telling time to the nearest 5 minutes), Fractions (halves, thirds, fourths, fifths, sixths, sevenths, eighths, ninths, and tenths), Base Ten Blocks (ones, tens, hundreds), Even & Odd (2-digit and 3-digit numbers), Place Value (2-digit and 3-digit numbers), and Least to Greatest\n\n## Math Puzzles Storage\n\nMath puzzles can be stored in a variety of ways.  There’s not a right or wrong way to store these.  Just pick the storage method that works for your classroom.  Some of teachers’ favorite storage options include Sterilite drawers, binder rings, plastic baggies, and binder clips.", null, "", null, "## Try them out\n\nMath puzzles are available in monthly sets or as a full year bundle.  A free set of math puzzles is also available to try out in your classroom. Click here to download the math puzzles freebie!\n\nClick to Purchase the Monthly or Bundled Puzzles", null, "", null, "#### Welcome, I’m Angie!", null, "Hello there! I’m Angie Olson- a teacher, curriculum developer, educational blogger and owner of Lucky Little Learners.", null, "", null, "", null, "## Planning is a breeze with this ⭐FREE⭐ 2nd grade scope & sequence\n\nBut that's not all! This FREEBIE also has a YEAR of free lesson plans!" ]

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[ "# RooAddPdf Components() plotting problem\n\nDear Experts,\n\nI am using RooAddPdf to combine a convoluted pdf with one another RooAbsPdf:\n\n`// convolution with resPdf`\n` RooFFTConvPdf fConv(\"fConv\", \"fConv\", *X, *resPdf, histPdf);`\n`// `\n`// resPdf passed as component, where Frac is a real number`\n`RooAddPdf model(\"model\",\"model\",RooArgList(fConv, *resPdf), RooArgList(Frac));`\n\n`//Plotting of pdf`\n\n`RooPlot *ffr = X->frame();`\n`model.plotOn(ffr,LineColor(kBlue),Components(RooArgSet(*resPdf)));`\n`model.plotOn(ffr,LineColor(kRed),Components(RooArgSet(fConv)));`\n\nIt should plot two components of `model`, but surprisingly it is plotting` model` pdf twice one overlapped other .", null, "While I could successfully draw both pdf (*resPdf & fConv) on a same frame, to check whether they are same:\n\n‘fConv->plotOn(ffr2);’\n‘resPdf->plotOn(ffr2);’\n\nBut they are different pdf as they should.\n\nIn simple words, RooAddPdf cannot distinguish it’s own components.\nPlease share if you see any solution to this problem.\n\nHi @hym,\n\nI am inviting @jonas to this thread. I am sure he can help you with this.\n\nCheers,\nJ.\n\nHi @hym!\n\nI have tried this for myself, i.e. plotting the components of a `RooAddPdf` that are a `RooFFTConvPdf` and something else like a `RooGaussian`. Which ROOT version are you using? The version 6.22 is working for me.\n\nOne thing that is striking me as suspicious in your code is that you are reusing `resPdf`. So first you do the convolution of a histogram with a resolution function, and then you add the same resolution function on top of that? I’m not sure if this is really intended. In any case, this might cause the problem that causes the components to not be recognized correctly.\n\nCan you please modify your code such that the second PDF in the `RooAddPdf` is not something that is also used in the convolution?\n\nHere is the example code that works correctly for me:\n\n``````void example() {\n\nusing namespace RooFit;\n\nRooRealVar mes(\"mes\",\"mes\", 0.0, 20.0);\n\nRooRealVar sigmean1(\"sigmean1\",\"sigmean1\", 5.0, 0.0, 10.0);\nRooRealVar sigwidth1(\"sigwidth1\",\"sigwidth1\", 1.0, 0.01, 10);\nRooGaussian gaus1(\"gaus1\",\"gaus1\", mes, sigmean1, sigwidth1);\n\nRooRealVar sigmean2(\"sigmean2\",\"sigmean2\", 5.0, 0.0, 10.0);\nRooRealVar sigwidth2(\"sigwidth2\",\"sigwidth2\", 1.0, 0.01, 10);\nRooGaussian gaus2(\"gaus2\",\"gaus2\", mes, sigmean2, sigwidth2);\n\nRooRealVar sigmean3(\"sigmean3\",\"sigmean3\", 5.0, 0.0, 10.0);\nRooRealVar sigwidth3(\"sigwidth3\",\"sigwidth3\", 1.0, 0.01, 10);\nRooGaussian gaus3(\"gaus3\",\"gaus3\", mes, sigmean3, sigwidth3);\n\nRooFFTConvPdf conv(\"conv\", \"conv\", mes, gaus1, gaus2);\n\nRooRealVar frac(\"frac\",\"frac\", 0.5, 0.0, 1.0);\nRooAddPdf model(\"model\",\"model\",RooArgList(conv, gaus3), RooArgList(frac));\n\n//Plotting of pdf\nTCanvas c1(\"c1\",\"c1\",900,700);\nRooPlot *mesframe = mes.frame();\nmodel.plotOn(mesframe, LineColor(kBlack));\nmodel.plotOn(mesframe, LineColor(kRed), Components(RooArgSet(conv)));\nmodel.plotOn(mesframe, LineColor(kBlue), Components(RooArgSet(gaus3)));\nmesframe->Draw();\n\nc1.SaveAs(\"plot.png\");\n}\n``````\n\nPlease let me know if this approach works for you or if you have further questions!\n\nCheers,\nJonas\n\n1 Like\n\nDear @jonas ,\nThank you for looking at the problem and your solutions. I see your point.\n\nPerheps, this is the case because I took your example and switched gaus3 to gaus1, and then it is doing the same thing here as well.\nSo I would like to make a copy of the `resPdf`, but here I am not sure what is the ideal way to make a copy of a `RooAbsPdf`.\nYour suggestions will be helpful here, if you could tell me how to make it in this case.\n\nThank you very much!\n\nAll of the RooFit PDFs implement a sort of copy constructor where you can create a new object just as the old one but with a different name. Since your issue is related to the name, this is exactly what you can use here.\n\nIn my example, this would mean constructing `gaus3` like this, where the second argument is the new name:\n\n``````RooAbsPdf gaus3{gaus1, \"gaus3\"};\n``````\n\nI don’t know what class your `resPdf` is, but you should be able to do the same with it.\n\nI hope that this works for you and that adding this one more line is not too inconvenient!\n\nThanks @jonas for reply.\n\nI applied the following line in your example code\n\n`RooGaussian gaus3 = {gaus2, \"gaus3\"};`\n\nand It works very well there, BUT in my case, my resPdf is pointer variable e.g. `RooAbsPdf *resPdf`;\nI tried to do the same with pointer type but it did not work. And also same thing I tried with your code and it did not work (errors). Could you also please check the same for case where `gaus1` is `*gaus1` ?\n\nYou can’t use the same code with a pointer; so you have to dereference to pointer with the `*` operator.\n\nIn my example this would mean:\n\n``````RooGaussian * gaus1 = new RooGaussian(\"gaus1\",\"gaus1\",\nmes, sigmean1, sigwidth1);\nRooGaussian gaus3{*gaus1, \"gaus3\"};\n``````\n\nDoes this work for you?\n\nHi @jonas\n\n1. Running this in your example works perfectly fine. It serves the purpose, BUT\n\n2. Running the following in my code\n\n`RooAbsPdf resPdf_copy{*resPdf, \"resPdf_copy\"};`\n\ngives\n\n`error: variable type 'RooAbsPdf' is an abstract class RooAbsPdf resPdf_copy{*resPdf, \"resPdf_copy\"};`\n^\n` note: unimplemented pure virtual method 'clone' in 'RooAbsPdf'`\n` virtual TObject* clone(const char* newname=0) const = 0 ;`\n^\n`/home/himanshu/anaconda3/envs/conda_new/etc/../include/RooAbsReal.h:391:20: note: unimplemented pure virtual method 'evaluate' in 'RooAbsPdf' virtual Double_t evaluate() const = 0 ;`\n\nHi @hym!\n\nYou can’t create a `RooAbsPdf` like this because it’s a virtual class. You need to create an object that is of the same class as your `resPdf`. You should look this up in your code or ask the `resPdf` what class it is with\n\n``````std::cout << resPdf.ClassName() << std::endl;\n``````\n\nLet’s hope it works now, otherwise let me know!\n\nHi @jonas\n\nThanks! It is `RooHistPdf` as per its classname. I tried to do something as following-\n\n`RooHistPdf *resPdf_copy{*resPdf, \"resPdf_copy\"};`\n\nwhich gives\n\n`error: excess elements in scalar initializer`\n\n`RooHistPdf *resPdf_copy{*resPdf , \"resPdf_copy\"};`\n\nIf you construct the copy like this, you can’t assign it to a pointer. So either you do\n\n``````RooHistPdf resPdf_copy{*resPdf , \"resPdf_copy\"};\n``````\n\nor if you really want to have the pointer you can create it on the heap:\n\n``````RooHistPdf * resPdf_copy = new RooHistPdf(*resPdf , \"resPdf_copy\");\n``````\n\nThe first option of creating the object on the stack is preferred. If your code needs a pointer later, you can always get a pointer with `&resPdf_copy`.\n\nDear @jonas\nThank you very much for your suggestions and help!", null, "", null, "It is now solved. I was taking `resPdf` as `RooAbsPdf`. This was the problem. After initializing it with `RooHistPdf`, it worked as per your suggestion.\n\nThank you again for your efforts", null, "1 Like\n\nThis topic was automatically closed 14 days after the last reply. New replies are no longer allowed." ]

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[ "# How many zero-constraints can be added to a subspace-restricted matrix before no solution exists?\n\nI'm trying to develop an estimator for the concentration matrix of a Gaussian Graphical Model. I've become stuck in trying to find conditions for the estimator to exist. I have a sufficient condition and I think it is also necessary, but I can't prove it. I'd greatly appreciate any suggestions on how to proceed.\n\n# Problem statement\n\nLet $V$ be an arbitrary $k$-dimensional vector subspace of $\\mathbb R^n$. Let $X\\in\\mathbb R^{n\\times n}$ be a symmetric matrix whose column space is contained in $V$. Now I add constraints to X: given some pairs $(i,j)$ such that $1\\leq i < j\\leq n$, I need $X_{ij}=0$. How many of these zero constraints can I satisfy before the only solution is $X=0$?\n\nI've found a sufficient condition for a non-zero solution to exist: the number of constraints $q$ must satisfy $q< \\frac{k(k+1)}{2}$. I think its also a necessary condition, but I could use a hand in showing that.\n\n# Proof of sufficient condition\n\nLet $M_V$ be the space of symmetric $n\\times n$ matrices whose column space is contained in $V$. An orthonormal basis for $M_V$ is $\\{\\frac{1}{2}Q(e_ie_j^T+e_je_i^T)Q^T : 1\\leq i \\leq j \\leq n\\}$ where the columns of $Q\\in\\mathbb R^{n\\times k}$ form an orthonormal basis for $V$ and $e_i$ are the standard basis vectors for $\\mathbb R^k$, so $\\dim(M_V)=\\frac{k(k+1)}{2}$.\n\nLet $M_X$ be the space of symmetric $n\\times n$ matrices that satisfy the $q$ zero constraints. Now suppose no non-zero $X$ satisfying the constraints exist: this implies $M_V\\cap M_X=\\{0\\}$. Hence $\\dim(M_V+M_X) = \\dim(M_V)+\\dim(M_X) = \\frac{k(k+1)}{2}+\\left(\\frac{n(n+1)}{2}-q\\right)$. Since $M_V+M_X$ is contained within the space of symmetric $n\\times n$ matrices, its dimension is bounded by $\\frac{n(n+1)}{2}$. Thus \"no non-zero $X$\" implies $q\\geq\\frac{k(k+1)}{2}$.\n\n• I should add that $V$ is a random subspace whose distribution is uniform over all possible $k$-dimensional subspaces of $\\mathbb R^n$, while the zero constraints are independent of $V$. There are pathological choices of $V$ where $q<\\frac{k(k+1)}{2}$ is not a necessary condition (e.g, we could have $M_V\\subseteq M_X$). However, I think these pathological choices have probability zero. – Peter Mar 10 '12 at 16:59\n\nJust to clarify: are the $q$ entries $(i,j)$ fixed, or are they also chosen uniformly at random over all possible sets of q entries of an $n \\times n$ matrix? (This is not so important though).\nAssuming that these $q$'s are fixed, here's a proof of sufficiency: suppose $q = k(k+1)/2$, $k < n$. Fix $q$ entries $(i,j)$, and consider the set $M$ of all $n \\times n$ matrices with these entries being $0$. Then for any $X \\in M$, I claim that the column space of $X$ has codimension at most k-1, and this occurs precisely when up to a permutation, $X$ is a block matrix with a $k \\times k$ zero-block that contains the diagonal.\nIt then follows that the column space of any such $X$ intersects non-trivially with a subspace of codimension $n-k$.\n• Thank you for your answer. The $q$ entries are indeed fixed. The original question proves that $q<k(k+1)/2$ is sufficient for $M_V\\cap M_X$ to have a non-trivial intersection. Your answer seems to extend the result to $q\\leq k(k+1)/2$. This is indeed helpful, but what I really need is a necessary condition for $M_V\\cap M_X$ to have a non-trivial intersection. Also, I am having some difficulty following your proof. You write \"I claim that the column space of X has codimension at most k-1\", but this assertion isn't obvious to me. Could you please explain why it is true? Thank you. – Peter Apr 13 '12 at 8:00" ]

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[ "", null, "# Why do the multiplication first, and then the addition? How can it be explained?", null, "I am a specialist in topics - political science, sociology, history. My hobby is...\n\nWhen we multiply two different units we get a new unit of measurement, when we add the units do not change. When we multiply, we get this new unit of measurement. If it is the same as the first summand, then we can perform addition.", null, "Healthy lifestyle, beauty in interior and landscape, harmony in relationships...\n\nBecause it's worth remembering what it is - multiplication. Under this operation is the addition of the same number more than once. By doing the additions first, we change this very thing \"number of times\". Here's an example:\n> 7*3+5=7+7+7+5=7+7+12=7*2+12\nBut if we were to do addition first, we would get the following:\n> 7*3+5=>7*8=7+7+7+7+7+7+7+7\nDifferent things, don't you think?", null, "" ]

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[ "+86-21-5838625\n\ntonnes to m3 - OnlineConversion Forums\n\nPost any conversion related questions and discussions here. If you're having trouble converting something, this is where you should post. * Guest Posting is allowed.\n\nHow to Calculate Cubic Meters of Lumber | Hunker\n\nHow to Calculate Cubic Meters of Lumber By Larry Simmons. SAVE; When selling lumber for commercial use, you must determine the volume of lumber in cubic meters to know the tree's worth. Because a tree's girth changes along the trunk, and because some of the tree is not usable lumber, you must apply a formula to make the calculation. ...\n\nHow Much Does One Cubic Meter of Concrete Weigh ...\n\nOne cubic meter of concrete that will be used for foundation weighs approximately 1.76 tons. General purpose concrete weighs a bit less at 1.6 tons per cubic meter. Concrete can be used for a variety of construction purposes, such as building the foundation of a building.\n\nOne square meter is equal to how many cubic meters? - Quora\n\nMay 14, 2018· Although square meter is a 2 dimensional unit for area while cubic meter is a 3 dimensional unit for volume, we do have 1 cubic meter = 1 meter x 1 square meter. So, there is your \"conversion factor\", \"1 meter\". You can probably see that this is j...\n\nHow to Calculate Cubic Meters | Sciencing\n\nNov 13, 2018· Finding cubic meters is the same thing as finding the volume of an item, measured in meters. Take a few measurements, do a simple calculation, and you have your answer. Just grab your metric ruler and use the formula H(eight) x L(ength) x W(idth) to determine the cubic meters of any item.\n\nHow to Calculate Cubic Meters - YouTube\n\nMay 21, 2015· How to Calculate Cubic Meters. Part of the series: Mathematics. When calculating cubic meters, it's important to convert all your measurements to meters. Use a formula to find the volume with help ...\n\nCubic metre - Wikipedia\n\nThe cubic metre (in British English and international spelling as used by the International Bureau of Weights and Measures) or cubic meter (in American English) is the SI derived unit of volume. Its SI symbol is m 3. It is the volume of a cube with edges one metre in length.\n\nConvert acre foot to cubic meter - Conversion of ...\n\n››More information from the unit converter. How many acre foot in 1 cubic meter? The answer is 0.00081071318210885. We assume you are converting between acre foot and cubic metre. You can view more details on each measurement unit:\n\nHow Many Cubic Meters Are In A Kilogram? - YouTube\n\nAug 15, 2017· \"How Many Cubic Meters Are In A Kilogram? Watch more videos for more knowledge How Many Cubic Meters Are In A Kilogram? - YouTube https://www.youtube.com/wat...\n\nHow To Calculate Cement Bags In 1 Cubic Meter - Daily Civil\n\nDec 21, 2016· How To Calculate Cement Bags In 1 Cubic Meter? Let us consider the nominal mix is 1:2:4 Wastage of cement during handling is considered as 2%.\n\nCubic Meter Calculator (in,ft,yd,mm,cm,m to cubic meter)\n\nThen you will know how many cube meters(m³) is it; How to calculate cubic meters ? If your unit of measurement is not meter, convert the unit to meter first, then, multiply length, width and height values together, this will give you the volume of the cube. ... Convert from measurements in cm to cubic meters, when we measure the dimension of a ...\n\nSand Calculator - how much sand do you need in tons ...\n\nSand calculator online - estimate the sand required for your construction or landscaping project in weight (pounds, kilograms, tons, tonnes) and volume (cubic ft, cubic yards, cubic meters). If you are wondering 'how much sand do I need', our free sand calculator is here to do the math for you. Information about sand density, common sand types, sand grain sizes, how much a cubic yard of …\n\nHow many bags of cement are in a cubic meter of concrete?\n\nQ. How many bags of cement are needed to mix one cubic meter of concrete? A. One cubic meter of concrete is equal to 1.308 cubic yards of concrete. If there are 5 1/2 bags of cement in 1 cubic yard of concrete, there would be 7.2 bags in 1 cubic meter of concrete. These are the 94 pound bags of portland cement or 42.64 kg bags of cement.\n\nHow to Calculate Cubic Meters - 5 steps - Education OneHowto\n\nMay 29, 2017· How to Calculate Cubic Meters; How to Calculate Cubic Meters. Rating: 4.3 (11 votes) 4 comments . By Mary Smith. Updated: May 29, 2017. If you want to know measurements of an object in cubic meters, it's possible you may want to know how much space is needed to store such object. For example, if you want to know how tall is a container, how ...\n\nHow to: Centimeters to Cubic Meters | Sciencing\n\nApr 24, 2017· Centimeters and meters measure distance in the metric system. Cubic meters, however, measure volume. Volume represents the three-dimensional space an object takes up. Therefore, you need to have three measurements in centimeters or meters to convert to cubic meters. If you have one measurement in centimeters and you ...\n\nHow to calculate volume or cubic meters (m3) – handy-mick ...\n\nHow to calculate volume or cubic meters (m3) In many cases we need to order or get delivered some materials to our home or to our work site, some materials used that may need to be calculated are, concrete, garden mix, mulch, topsoil, sand, compost, manure, road base, pine bark, garden chips, saw dust, aggregate, pebbles or clean fill.\n\nGravel Calculator - how much gravel you need in tons ...\n\nFree gravel calculator online: estimate how much gravel you need for your construction or gardening / landscaping project. Calculates gravel required in volume: cubic feet, cubic yards, cubic meters, or weight: pounds, tons, kilograms, tonnes... Information about gravel density, common gravel sizes, how much a cubic yard of gravel weighs, how much a cubic meter of gravel weighs, and more.\n\nHow many liters in a cubic meter? - Calculatorology\n\nCubic Meters to Liters (How many liters in a cubic meter?) This is a conversion calculator that converts the volume in cubic meters (m 3) to volume in liters (L). It is programmed with a simple formula that facilitates the conversion of cubic meters to liters. Use the blank text field to enter the value in cubic meters.\n\nStandard Weights for Crushed Rock Per Meter | Hunker\n\nSolid rock is estimated at 2.5 to 3tons per cubic meter. If rock is crushed into uniform sizes, the presence of open space between the particles causes the load to be lighter -- approximately 1.6 tons per cubic meter. Mixed sizes of crushed rock can range from 1.6 to 2.2 tons per cubic meter.\n\nCubic Meters to Cubic Centimeters (m3 to cm3) Conversion\n\nFor example, to find out how many cubic centimeters there are in a cubic meter and a half, as a formula to convert from cubic meters to cubic centimeters, multiply the cubic meter value by one million, that makes 1.5 m3 * 1000000 = 1500000 cubic centimeters in 1.5 cubic meters. Cubic meter is a metric volume unit and defined as the volume of a ...\n\nHow to Calculate Cubic Meters - YouTube\n\nMay 21, 2015· How to Calculate Cubic Meters. Part of the series: Mathematics. When calculating cubic meters, it's important to convert all your measurements to meters. Use a …\n\nCubic Meter to Cubic Centimeter Conversion (m³ to cm³)\n\nCubic Centimeter value will be converted automatically as you type. The decimals value is the number of digits to be calculated or rounded of the result of cubic meter to cubic centimeter conversion. You can also check the cubic meter to cubic centimeter conversion chart below, or go back to cubic meter to cubic centimeter converter to top.\n\nWhat Is 1 Ton in Cubic Meters? | Reference.com\n\nOne ton of pure water is equal to 1 cubic meter. There are 1,000 liters in 1 cubic meter of pure water, and a liter of pure water weighs 1 kilogram. Therefore, 1 ton of water equals 1 cubic meter. Because saltwater is less dense than pure water, a ton of saltwater takes up less volume, only 0.97 cubic meters.\n\nCubic Feet to Cubic Meters - Metric Conversion charts and ...\n\nCubic Feet. A cubic measurement is the three-dimensional derivative of a linear measure, so a cubic foot is defined as the volume of a cube with sides 1 ft in length. In metric terms a cubic foot is a cube with sides 0.3048 metres in length. One cubic foot is the equivalent to approximately 0.02831685 cubic metres, or 28.3169 litres.\n\nHow much for 2 cubic meters of readimix concrete ...\n\n2 cubic meters isnt very much. i would just hire a mixer and mix it by hand. unless you can utilise a full load, redimix isn’t really worth it. Posted 5 years ago. andysredmini.\n\nA cubic meter is equal to how many tons? - Quora\n\nMar 26, 2019· Q: A cubic meter is equal to how many tons? We have a mix of metric and English units. Since the requested answer is in tons. Or maybe the questioner meant “metric ...\n\nhow many litres of lpg to 1 cubic meter - OnlineConversion ...\n\nRe: how many litres of lpg to 1 cubic meter Apparently due to european regulation since jannuary 2010 the figure is 3.85 litres to a meter of gas! previously there was uncertainty with several differant figures from 3.65 to 3.85 but the goverment has now clarified its 3.85\n\n1 cubic meters in a how many ton stone 25 mm - BINQ Mining\n\nDec 04, 2012· 40 mm aggregate required per cubic meter concrete m20. A slump loss of 25 mm per 300 meters … base of 600 mm thick and load of 20 ton per metre squre we add steel per … How much cement for 1 cubic meter – how many … »More detailed\n\nHow many cubic feet in a cubic meter? - Calculatorology\n\nCubic Meters to Cubic Feet (How many cubic feet in a cubic meter?) This is a conversion calculator that is used to convert the volume in cubic meters (m 3) to cubic feet (ft 3).It is programmed with a single text field and a control button.\n\nConcrete 1 cubic meter volume to kilograms converter\n\nOne cubic meter of concrete converted to kilogram equals to 2,406.53 kg - kilo. How many kilograms of concrete are in 1 cubic meter? The answer is: The change of 1 m3 ( cubic meter ) unit of concrete measure equals = to 2,406.53 kg - kilo ( kilogram ) as the equivalent measure for the same concrete type." ]

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[ "Home / Answered Questions / Other / chapter-1-supplementary-exercises-question-15-find-values-of-a-b-and-c-such-that-the-graph-of-the-po-aw670\n\n# (Solved): Chapter 1, Supplementary Exercises, Question 15 Find Values Of A, B, And C Such That The Graph Of Th...", null, "Chapter 1, Supplementary Exercises, Question 15 Find values of a, b, and c such that the graph of the polynomial p (x) = ax2 + bx + c passes through the points (1, 6), (-1, 18), and (2, 3). a = b =\n\nWe have an Answer from Expert" ]

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[ " Python: Check if every consecutive sequence of zeroes is followed by a consecutive sequence of ones of same length - w3resource", null, "# Python: Check if every consecutive sequence of zeroes is followed by a consecutive sequence of ones of same length\n\n## Python Basic - 1: Exercise-142 with Solution\n\nWrite a Python program to check if every consecutive sequence of zeroes is followed by a consecutive sequence of ones of same length in a given string. Return True/False.\n\nSample Solution-1:\n\nPython Code:\n\n``````def test(str1):\nwhile '01' in str1:\nstr1 = str1.replace('01', '')\nreturn len(str1) == 0\n\nstr1 = \"01010101\"\nprint(\"Original sequence:\",str1)\nprint(\"Check if every consecutive sequence of zeroes is followed by a consecutive sequence of ones in the said string:\")\nprint(test(str1))\nstr1 = \"00\"\nprint(\"\\nOriginal sequence:\",str1)\nprint(\"Check if every consecutive sequence of zeroes is followed by a consecutive sequence of ones in the said string:\")\nprint(test(str1))\nstr1 = \"000111000111\"\nprint(\"\\nOriginal sequence:\",str1)\nprint(\"Check if every consecutive sequence of zeroes is followed by a consecutive sequence of ones in the said string:\")\nprint(test(str1))\nstr1 = \"00011100011\"\nprint(\"\\nOriginal sequence:\",str1)\nprint(\"Check if every consecutive sequence of zeroes is followed by a consecutive sequence of ones in the said string:\")\nprint(test(str1))\n``````\n\nSample Output:\n\n```Original sequence: 01010101\nCheck if every consecutive sequence of zeroes is followed by a consecutive sequence of ones in the said string:\nTrue\n\nOriginal sequence: 00\nCheck if every consecutive sequence of zeroes is followed by a consecutive sequence of ones in the said string:\nFalse\n\nOriginal sequence: 000111000111\nCheck if every consecutive sequence of zeroes is followed by a consecutive sequence of ones in the said string:\nTrue\n\nOriginal sequence: 00011100011\nCheck if every consecutive sequence of zeroes is followed by a consecutive sequence of ones in the said string:\nFalse\n```\n\nFlowchart:", null, "## Visualize Python code execution:\n\nThe following tool visualize what the computer is doing step-by-step as it executes the said program:\n\nSample Solution-2:\n\nPython Code:\n\n``````def test(str1):\ntemp=[]\nfor x in str1:\nif (x=='0'):\ntemp.append('0')\nelse:\ntemp.pop()\nreturn not temp\n\nstr1 = \"01010101\"\nprint(\"Original sequence:\",str1)\nprint(\"Check if every consecutive sequence of zeroes is followed by a consecutive sequence of ones in the said string:\")\nprint(test(str1))\nstr1 = \"00\"\nprint(\"\\nOriginal sequence:\",str1)\nprint(\"Check if every consecutive sequence of zeroes is followed by a consecutive sequence of ones in the said string:\")\nprint(test(str1))\nstr1 = \"000111000111\"\nprint(\"\\nOriginal sequence:\",str1)\nprint(\"Check if every consecutive sequence of zeroes is followed by a consecutive sequence of ones in the said string:\")\nprint(test(str1))\nstr1 = \"00011100011\"\nprint(\"\\nOriginal sequence:\",str1)\nprint(\"Check if every consecutive sequence of zeroes is followed by a consecutive sequence of ones in the said string:\")\nprint(test(str1))\n``````\n\nSample Output:\n\n```Original sequence: 01010101\nCheck if every consecutive sequence of zeroes is followed by a consecutive sequence of ones in the said string:\nTrue\n\nOriginal sequence: 00\nCheck if every consecutive sequence of zeroes is followed by a consecutive sequence of ones in the said string:\nFalse\n\nOriginal sequence: 000111000111\nCheck if every consecutive sequence of zeroes is followed by a consecutive sequence of ones in the said string:\nTrue\n\nOriginal sequence: 00011100011\nCheck if every consecutive sequence of zeroes is followed by a consecutive sequence of ones in the said string:\nFalse\n```\n\nFlowchart:", null, "## Visualize Python code execution:\n\nThe following tool visualize what the computer is doing step-by-step as it executes the said program:\n\nPython Code Editor:\n\nHave another way to solve this solution? Contribute your code (and comments) through Disqus.\n\nWhat is the difficulty level of this exercise?\n\nTest your Programming skills with w3resource's quiz.\n\n\n\n## Python: Tips of the Day\n\nHow to make a flat list out of list of lists?\n\nGiven a list of lists l\n\n```flat_list = [item for sublist in l for item in sublist]\n```\n\nwhich means:\n\n```flat_list = []\nfor sublist in l:\nfor item in sublist:\nflat_list.append(item)\n```\n\nis faster than the shortcuts posted so far. (l is the list to flatten.) Here is the corresponding function:\n\nflatten = lambda l: [item for sublist in l for item in sublist]\n\nAs evidence, you can use the timeit module in the standard library:\n\n```\\$ python -mtimeit -s'l=[[1,2,3],[4,5,6], , [8,9]]*99' '[item for sublist in l for item in sublist]'\n10000 loops, best of 3: 143 usec per loop\n\\$ python -mtimeit -s'l=[[1,2,3],[4,5,6], , [8,9]]*99' 'sum(l, [])'\n1000 loops, best of 3: 969 usec per loop\n\\$ python -mtimeit -s'l=[[1,2,3],[4,5,6], , [8,9]]*99' 'reduce(lambda x,y: x+y,l)'\n1000 loops, best of 3: 1.1 msec per loop\n```\n\nExplanation: the shortcuts based on + (including the implied use in sum) are, of necessity, O(L**2) when there are L sublists -- as the intermediate result list keeps getting longer, at each step a new intermediate result list object gets allocated, and all the items in the previous intermediate result must be copied over (as well as a few new ones added at the end). So, for simplicity and without actual loss of generality, say you have L sublists of I items each: the first I items are copied back and forth L-1 times, the second I items L-2 times, and so on; total number of copies is I times the sum of x for x from 1 to L excluded, i.e., I * (L**2)/2.\n\nThe list comprehension just generates one list, once, and copies each item over (from its original place of residence to the result list) also exactly once.\n\nRef: https://bit.ly/3dKsNTR" ]

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[ "Enthalpy Definition in Chemistry and Physics\n\nEnthalpy is a thermodynamic property of a system. It is the sum of the internal energy added to the product of the pressure and volume of the system. It reflects the capacity to do non-mechanical work and the capacity to release heat.\n\nEnthalpy is denoted as H; specific enthalpy denoted as h. Common units used to express enthalpy are the joule, calorie, or BTU (British Thermal Unit.) Enthalpy in a throttling process is constant.\n\nChange in enthalpy is calculated rather than enthalpy, in part because total enthalpy of a system cannot be measured. However, it is possible to measure the difference in enthalpy between one state and another. Enthalpy change may be calculated under conditions of constant pressure.\n\nEnthalpy Formulas\n\nH = E + PV\n\nwhere H is enthalpy, E is internal energy of the system, P is pressure, and V is volume\n\nd H = T d S + P d V\n\nWhat Is the Importance of Enthalpy?\n\n• Measuring the change in enthalpy allows us to determine whether a reaction was endothermic (absorbed heat, positive change in enthalpy) or exothermic (released heat, negative change in enthalpy.)\n• It is used to calculate the heat of reaction of a chemical process.\n• Change in enthalpy is used to measure heat flow in calorimetry.\n• It is measured to evaluate a throttling process or Joule-Thomson expansion.\n• Enthalpy is used to calculate minimum power for a compressor.\n• Enthalpy change occurs during a change in the state of matter.\n• There are many other applications of enthalpy in thermal engineering.\n\nExample Change in Enthalpy Calculation\n\nYou can use the heat of fusion of ice and heat of vaporization of water to calculate the enthalpy change when ice melts into a liquid and the liquid turns to a vapor.\n\nThe heat of fusion of ice is 333 J/g (meaning 333 J is absorbed when 1 gram of ice melts.) The heat of vaporization of liquid water at 100°C is 2257 J/g.\n\nPart A: Calculate the change in enthalpy, ΔH, for these two processes.\n\nH2O(s) → H2O(l); ΔH = ?\nH2O(l) → H2O(g); ΔH = ?\nPart B: Using the values you calculated, find the number of grams of ice you can melt using 0.800 kJ of heat.\n\nSolution\nA. The heats of fusion and vaporization are in joules, so the first thing to do is convert to kilojoules. Using the periodic table, we know that 1 mole of water (H2O) is 18.02 g. Therefore:\nfusion ΔH = 18.02 g x 333 J / 1 g\nfusion ΔH = 6.00 x 103 J\nfusion ΔH = 6.00 kJ\nvaporization ΔH = 18.02 g x 2257 J / 1 g\nvaporization ΔH = 4.07 x 104 J\nvaporization ΔH = 40.7 kJ\nSo the completed thermochemical reactions are:\nH2O(s) → H2O(l); ΔH = +6.00 kJ\nH2O(l) → H2O(g); ΔH = +40.7 kJ\nB. Now we know that:\n1 mol H2O(s) = 18.02 g H2O(s) ~ 6.00 kJ\nUsing this conversion factor:\n0.800 kJ x 18.02 g ice / 6.00 kJ = 2.40 g ice melted" ]

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[ "Select Page\n\n# Difference between == and is operator in Python\n\nby | Aug 7, 2021 | Python\n\nHome » Python » Difference between == and is operator in Python\n\n## Introduction\n\nThe equals (==) operator and is operator of python are the check operators. The equals (==) operators checks whether the values of objects are same or not, whereas the is operators checks whether the object points to the object of the same memory location.\n\nSample code to get the id of objects:\n\n```sample_list1 = []\nsample_list2 = []\n\nprint(id(sample_list1))\nprint(id(sample_list2))```\n\n140214539034112\n140214538406144\n\nThe id() function returns the unique id of the object that is assigned when the object is created. From the above code, we can see that sample_list1 and sample_list2 refer to two different objects.\n\n## Syntax\n\nEquals: If ( var1 == var2 )\n\nIs: If (var1 is var2)\n\n## Program\n\n```sample_list1 = []\nsample_list2 = []\nsample_list3 = sample_list1\nsample_list4 = sample_list1 + sample_list2\n\nif (sample_list1 == sample_list2):\nprint(\"True\")\nelse:\nprint(\"False\")\n\nif (sample_list1 is sample_list2):\nprint(\"True\")\nelse:\nprint(\"False\")\n\nif (sample_list3 is sample_list1):\nprint(\"True\")\nelse:\nprint(\"False\")\n\nif (sample_list3 is sample_list4):\nprint(\"True\")\nelse:\nprint(\"False\")```\n\nTrue\nFalse\nTrue\nFalsex\n\n## Explanation\n\nIn the first case, we are checking the variables whether they are equal or not. The smaple_list1 and sample_list2 are two empty lists, and thus is satisfies the equal’s condition. Therefore, the output is True.\n\nIn the second case, we are checking the two variables whether they point to the same object or not. The variable smaple_list1 and sample_list2 are two different objects pointing to different memory locations. Therefore, the output is False.\n\nIn the third case, we have assigned the value of sample_list1 to variable smaple_list3 and checked whether they point to the object at the same memory location. The output is True as both the variables share the same memory location.\n\nIn the fourth case, we have checked whether the variable sample_list3 and sample_list4 points to the same memory location or not. The two objects are different and therefore the output is False.\n\n## Author\n\n•", null, "A Full Stack Developer with 10+ years of experience in technical content creation.\n\n•", null, "" ]

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[ "# nLab quantum group\n\nContents\n\n### Context\n\n#### Algebra\n\nhigher algebra\n\nuniversal algebra\n\n# Contents\n\n## Idea\n\nThe notion of quantum group refers to various objects which are deformations of (algebras of functions on) groups, but still have very similar properties to (algebras of functions on) groups, and in particular to semisimple Lie groups. Most important are the Hopf algebras deforming the function algebras on semisimple Lie groups or to the enveloping algebras of Kac-Moody Lie algebras.\n\n## Overview\n\nIt is a common experience in representation theory that a number of mathematical structures behaves very similarly to algebraic or Lie groups. After the impetus of the theory of quantum integrable systems, mainly the work of Leningrad’s school of mathematical physics around 1980, several mathematicians (including Drinfeld, Manin, Woronowicz, Jimbo, Faddeev–Reshetikhin–Takhtajan) found, in different formalisms, major series of examples which are mostly noncommutative noncocommutative Hopf algebras and which deform enveloping algebras of (semisimple) Lie algebras, or algebras of functions on the corresponding algebraic groups. These deformations $G_q$ depend on a parameter $q$ (sometimes one prefers a formal parameter $h$ with $q = e^{h}$), which may be taken as belonging to the ground field, but also being formal (transcendental over the ground field). A peculiar case is when the parameter $q$ of the deformation is an $l$-th root of unity; the remaining cases are usually called generic $q$.\n\nThe representation theory for these ‘quantum’ examples is highly developed; in fact many phenomena in the representation theory of semisimple Lie algebras (e.g. canonical bases) were discovered first as a limiting case of constructions in the quantum case, which become degenerate in the classical case (the principle that quantization removes degeneracy). While representations for generic $q$ parallel classical ones, the theory at roots of unity is peculiar and related to the representation theory of affine Lie algebras; the quantum groups at roots of unity as algebras have big centers.\n\nNowadays, both the class of examples and the class of formalisms has been extended a lot, hence the term ‘quantum group’ is not a fixed notion but rather a collective term for a rather author-dependent class of group-like objects, most often subclasses or extensions of the concept of Hopf algebras which are sometimes required to belong to families of deformations of their classical counterparts. One of the common features is that if we forget the group-like features, the examples belong to the class of noncommutative spaces (see noncommutative geometry).\n\nMathematically better defined are notions (sometimes equated by various authors with the class of quantum groups) like quasitriangular Hopf algebras, quantum matrix groups? (quantum linear groups, more general FRT-algebras and Majid’s $A(R)$ where $R$ is a quantum Yang-Baxter equation), quantized enveloping algebras, quantum function algebras, compact matrix pseudogroups, Kac algebras, Yangians etc. The representations of quasitriangular Hopf algebras form braided monoidal categories, which are in main examples related to the mathematics of Iwahori–Hecke algebras, braid groups, knot theory, finite group Chern–Simons theory and Wess–Zumino–Novikov–Witten theory of CFT. One should note that in the classical limit quantum function algebras give not simply (functions on) algebraic (or Lie) groups but also a compatible (= multiplicative) Poisson structure giving rise to Poisson–Lie or Poisson algebraic groups.\n\nThere is an extensive geometric theory of homogeneous spaces for quantum groups and fiber bundles whose structure groups are quantum groups.\n\n## Properties\n\n### Tannaka duality\n\nTannaka duality for categories of modules over monoids/associative algebras\n\nmonoid/associative algebracategory of modules\n$A$$Mod_A$\n$R$-algebra$Mod_R$-2-module\nsesquialgebra2-ring = monoidal presentable category with colimit-preserving tensor product\nbialgebrastrict 2-ring: monoidal category with fiber functor\nHopf algebrarigid monoidal category with fiber functor\nhopfish algebra (correct version)rigid monoidal category (without fiber functor)\nweak Hopf algebrafusion category with generalized fiber functor\nquasitriangular bialgebrabraided monoidal category with fiber functor\ntriangular bialgebrasymmetric monoidal category with fiber functor\nquasitriangular Hopf algebra (quantum group)rigid braided monoidal category with fiber functor\ntriangular Hopf algebrarigid symmetric monoidal category with fiber functor\nsupercommutative Hopf algebra (supergroup)rigid symmetric monoidal category with fiber functor and Schur smallness\nform Drinfeld doubleform Drinfeld center\ntrialgebraHopf monoidal category\n\n2-Tannaka duality for module categories over monoidal categories\n\nmonoidal category2-category of module categories\n$A$$Mod_A$\n$R$-2-algebra$Mod_R$-3-module\nHopf monoidal categorymonoidal 2-category (with some duality and strictness structure)\n\n3-Tannaka duality for module 2-categories over monoidal 2-categories\n\nmonoidal 2-category3-category of module 2-categories\n$A$$Mod_A$\n$R$-3-algebra$Mod_R$-4-module\n• V. G. Drinfel'd, Quantum groups, Proceedings of the International Congress of Mathematicians 986, Vol. 1, 798–820, AMS 1987, djvu:1.3M, pdf:2.5M\n\n• Shahn Majid, Foundations of quantum group theory, Cambridge University Press 1995, 2000.\n\n• Yu. I. Manin, Quantum groups and non-commutative geometry, CRM, Montreal 1988.\n\n• B. Parshall, J.Wang, Quantum linear groups, Mem. Amer. Math. Soc. 89(1991), No. 439, vi+157 pp.\n\n• N. Yu. Reshetikhin, L. A. Takhtajan, L. D. Faddeev, Quantization of Lie groups and Lie algebras, Algebra i analiz 1, 178 (1989) (Russian), English translation in Leningrad Math. J. 1.\n\n• Arun Ram, A survey of quantum groups: background, motivation, and results, in: Geometric analysis and Lie theory in mathematics and physics, A. Carey and M. Murray eds., Australian Math. Soc. Lecture Notes Series 11, Cambridge Univ. Press 1997, pp. 20-104. pdf\n\n• P. Etingof, O. Schiffmann, Lectures on Quantum Groups, Lectures in Math. Phys., International Press (1998).\n\n• P.Etingof, I. Frenkel, Lectures on representation theory and Knizhnik-Zamolodchikov equations\n\n• A. U. Klymik, K. Schmuedgen, Quantum groups and their representations, Springer 1997.\n\n• A. Joseph, Quantum groups and their primitive ideals, Springer 1995.\n\n• Ross Street, Quantum groups : a path to current algebra, Cambridge Univ. Press 2007\n\n• L. I. Korogodski, Ya. S. Soibelman, Algebras of functions on quantum groups I, Math. Surveys and Monographs 56, AMS 1998.\n\n• A. Varchenko, Hypergeometric functions and representation theory of Lie algebras and quantum groups, Advanced Series in Mathematical Physics, Vol. 21, World Scientific (1995)\n\n• George Lusztig, Introduction to quantum groups\n\n• V. Chari, A. Pressley, A guide to quantum groups, Camb. Univ. Press 1994\n\n• C. Kassel, Quantum groups, Graduate Texts in Mathematics 155, Springer 1995 (also errata\n\n• Bangming Deng, Jie Du, Brian Parshall, Jianpan Wang, Finite dimensional algebras and quantum groups, Mathematical Surveys and Monographs 150, Amer. Math. Soc. 2008. xxvi+759 pp. MR2009i:17023)\n\n• Tom Bridgeland, Quantum groups via Hall algebras of complexes, Annals of Mathematics 177:2 (2013) 739-759 (21 pages)" ]

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[ "# Sinusoidal Function In MatLab®(Illustrated Expression)\n\nby,\n\nIn mathematical calculations, sinusoidal values are frequently used. You can define the sinusoidal values in Matlab®. In this article we will show you;\n\n• What kind of sinusoidal values can be calculated in mathematical relations in Matlab®?\n• How to calculate arc values of sinusoidal values in Matlab®?\n\n## How To Use Sinusoidal Functions In Matlab®?\n\n``````>> a = sin(pi/2)\nb = cosd(90)\nc = atan(0.5)\nd = acosd(0.3)\n\na =\n\n1\n\nb =\n\n0\n\nc =\n\n0.4636\n\nd =\n\n72.5424\n\n>> ``````\n\nThe sinusoidal functions are defined as;\n\n• Directly typing a radian value inside the sinusoidal function as shown above,\n• Typing an angle value inside the sinusoidal functions such as; ‘cosd, sind tand etc…’ The ‘d’ at the end of sinuzoidal function defines the degree value that you need to enter inside the sinusoidal function.\n• You can calculate the angle or radian value with reverse sinusoidal functions in Matlab®. For example, the variable c as shown above is the reverse sinusoidal calculation of ‘tan’. The ‘a’ at the ‘tan’ means ‘arc’ in another word ‘arctan’ in Matlab®. The result will be shown as radians.\n• Also, you can take the result as angle value from reverse sinusoidal function calculations in Matlab® as shown above. You just type ‘d’ at the end of ‘atan’ which is ‘atand’ to calculate as degrees as shown in Matlab®.\n``````>> a = sin(pi/2)\nb = cosd(90)\nc = atan(0.5)\nd = acosd(0.3)\na*b/c\n\na =\n\n1\n\nb =\n\n0\n\nc =\n\n0.4636\n\nd =\n\n72.5424\n\nans =\n\n0\n\n>> ``````\n\nMathematical calculations with sinusoidal functions in Matlab®.\nAlso, you can do direct mathematical calculations such as multiplication or dividings with sinusoidal functions as shown in Matlab®.\n\nYOU CAN LEARN MatLab® IN MECHANICAL BASE; Click And Start To Learn MatLab®!\n\nThe use and the features of sinusoidal functions are explained with very basic examples in this article. If you wish, we can add much more complex examples and real engineering examples about sinusoidal functions in Mechanicalland." ]

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[ "## Program 18: Big Bang simulation\n\n### Introduction and Purpose\n\nThe purpose of the program QBasic BIGBANG.BAS is to demonstrate:\n1. The Big Bang in a simple linear world model.\nThe model assumes that the speed of light is not influenced by the expanding space.\n2. That the expansion speed of space is a function of distance.\n3. That the observed age of a galaxy is a function of distance. How further away how younger.\n4. That the youngest observed galaxy is a function of expansion rate (multiple times c)\nFor a copy and a description of \"the same\" program in Visual Basic 5.0 or Visual Studio 2010 select:\nVisual Basic Big Bang simulation program \"VB BigBang.exe\" and \"VB2010 BigBang.exe\nThis program also simulates Milne's Universe.\nFor a general introduction to world models select this: 3 World models\n\nTo get a copy select: BIGBANG.BAS\nTo see the listing of the program select: BIGBANG.HTM\nExecution file select: BIGBANG.EXE and: brun45.exe\n\n### Program Operation\n\nThe program demonstrates the Big Bang over a period of 15 b year subdivided in periods of 1 b year by means of 150 galaxies or clocks. The 150 galaxies are randomly scatered over the Universe. The Observer is at the centre. Each BillionYear is identified by a different colour.\n\nInput for the program is the expansion factor of Space.\n• First enter 1\n• In the second run enter 3\n• In the third run enter 0.3\n• Finally enter 0 to end the program\nThe expansion rate of the Universe is equal to the expansion factor times c (speed of light).\n• An expansion factor of 1 means that the Universe expands with a speed of c. At the end of the demonstration, the simulation shows that the youngest galaxies you can see now are roughly 7.5 billion year old and the oldest 15 b year\n• At the end of the simulation the total size of the universe is 15 b lightyears.\n• An expansion factor of 2 means that the Universe expands with a speed of 2c. At the end of the demonstration, the simulation shows that the youngest galaxies you can see now are roughly 5 b year old. That means you can see further back in time\n• At the end of the simulation the total size of the universe is 30 b lightyears.\n• With an expansion factor of 10, at the end of the simulation you will be able to see the state of the Universe very close at the moment of the Big Bang.\n• With an expansion factor of 0.3, at the end of the demonstration, you will only see the galaxies between 12 and 15 b years.\n\n### Program Description\n\n ``` D /| / | / | / | / | / | / | / | / | / | / | / | / | / | / | / | / C / . | e / . | . / . | . / . | d . | /| . . | / | . . | / | c | / | . | . A / | . | . . | / . | a | / . | |. | . | / . | . | | . | / . . | | | . | / . | | | . | BB...............................O 0 5 8 11 16 ``` The sketch left demonstrates how the program operates. Suppose that expansion factor is 2c. The line BB-O is the time axis t The line OACD is the x axis. The line BB-D represents is the line x=2t. This is the radius of the world model. The line BB-C represents the the path of a galaxy at a factor p from the radius with p between 0 and 1. The line BB-C is represented by the equation x=2*p*t The line oacde is the linecone through O. This line is represented in this case by x=16-t. In general x=age-t The time of point d is calculated as 2t = 16 - t or 3t = 16 or t = age/3 The time of point c is calculated as 2pt = 16 - t or 2pt + t = age or t = age /(2p+1) The distance is for point c is equal to 2p*age*c / (2p+1)\n\n### Feedback\n\nNone\n\nCreated:29 December 2001\nModified: 15 July 2004\nModified: 5 Februari 2014\n\nBack to my home page: Contents of This Document" ]

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