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[ "# The Number 815148 : Square Root, Cube Root, Factors, Prime Checker\n\nYou can find Square Root, Cube Root, Factors, Prime Check, Binary, Octal, Hexadecimal and more of Number 815148. 815148 is written as Eight Hundred And Fifteen Thousand, One Hundred And Fourty Eight. You can find Binary, Octal, Hexadecimal Representation and sin, cos, tan values and Multiplication, Division tables.\n\n• Number 815148 is an even Number.\n• Number 815148 is not a Prime Number\n• Sum of all digits of 815148 is 27.\n• Previous number of 815148 is 815147\n• Next number of 815148 is 815149\n\n## Square, Square Root, Cube, Cube Root of 815148\n\n• Square Root of 815148 is 902.8554701612\n• Cube Root of 815148 is 93.414040168709\n• Square of 815148 is 664466261904\n• Cube of 815148 is 541638344458521792\n\n## Numeral System of Number 815148\n\n• Binary Representation of 815148 is 11000111000000101100\n• Octal Representation of 815148 is 3070054\n• Hexadecimal Representation of 815148 is c702c\n\n## Sin, Cos, Tan of Number 815148\n\n• Sin of 815148 is 0.95105651629523\n• Cos of 815148 is -0.30901699437473\n• Tan of 815148 is -3.0776835371777\n\n## Multiplication Table for 815148\n\n• 815148 multiplied by 1 equals to 815,148\n• 815148 multiplied by 2 equals to 1,630,296\n• 815148 multiplied by 3 equals to 2,445,444\n• 815148 multiplied by 4 equals to 3,260,592\n• 815148 multiplied by 5 equals to 4,075,740\n• 815148 multiplied by 6 equals to 4,890,888\n• 815148 multiplied by 7 equals to 5,706,036\n• 815148 multiplied by 8 equals to 6,521,184\n• 815148 multiplied by 9 equals to 7,336,332\n• 815148 multiplied by 10 equals to 8,151,480\n• 815148 multiplied by 11 equals to 8,966,628\n• 815148 multiplied by 12 equals to 9,781,776\n\n## Division Table for 815148\n\n• 815148 divided by 1 equals to 815148\n• 815148 divided by 2 equals to 407574\n• 815148 divided by 3 equals to 271716\n• 815148 divided by 4 equals to 203787\n• 815148 divided by 5 equals to 163029.6\n• 815148 divided by 6 equals to 135858\n• 815148 divided by 7 equals to 116449.71428571\n• 815148 divided by 8 equals to 101893.5\n• 815148 divided by 9 equals to 90572\n• 815148 divided by 10 equals to 81514.8\n• 815148 divided by 11 equals to 74104.363636364\n• 815148 divided by 12 equals to 67929" ]

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[ "## Frankl, My Dear • 1\n\nI need to think a little about the context that Lipton and Regan have wrapped around the Frankl Conjecture, if not exactly about the problem itself.  This will be a scratch-worky post-in-progress (PIP❢), or a series of posts, and I may even delete it down the road if it runs out of gas or turns into too much nonsense.\n\nJust to get started, let me post the passage from the 2nd of the above articles where I thought I spied something familiar, though it was late, past my bedtime, and I was pondering weak and weary, etc.\n\nLet us use", null, "${J_{i}(f)}$ to denote those", null, "${x}$ such that", null, "$\\displaystyle f(x) \\neq f(x^{i}).$\n\nObviously the following is true:", null, "$\\displaystyle \\frac{|J_{i}(f)|}{2^{n}} = I_{i}(f).$\n\nThere must be a better notation than", null, "${J_{i}}$ — we are open to suggestions. Any?\n\nIn order to understand the above formulas, we have to back up a little and pick up a couple of earlier definitions, rephrasing them slightly in the style of notation to which I would like you to become accustomed.\n\nLet", null, "${\\mathbb{B} = \\{0,1\\}}$ and let", null, "${f : \\mathbb{B}^{n} \\to \\mathbb{B}}$ be a boolean function. The influence of the", null, "${i}^\\text{th}$ input of", null, "${f}$ is defined by", null, "$\\displaystyle I_{i}(f) = \\mathrm{Prob}_{x}[ f(x) \\neq f(x^{i}) ],$\n\nwhere", null, "${x^{i}}$ is equal to", null, "$\\displaystyle x_{1}, \\dots, x_{i-1}, \\neg x_{i}, x_{i+1}, \\dots, x_{n}.$\n\nThere is another way of expressing the bit flip operation", null, "${x \\mapsto x^i}$ that I like a bit better.  Here is the formal definition, using", null, "${j}$ instead of", null, "${i}$ as the index of the indicated variable.\n\nDefinition.  Let the function", null, "${\\lnot_j : \\mathbb{B}^k \\to \\mathbb{B}}$ be defined for each integer", null, "${j}$ in the interval", null, "${[1, k]}$ by the following equation:", null, "${\\lnot_j (x_1, \\ldots, x_j, \\ldots, x_k) ~=~ x_1 \\land \\ldots \\land x_{j-1} \\land \\lnot x_j \\land x_{j+1} \\land \\ldots \\land x_k}.$\n\nIt is conventional to write logical conjunction as multiplication, with or without a dot between conjuncts. It is also convenient in many contexts to write logical negation", null, "${\\lnot x_j}$ with a tilde as", null, "${\\tilde{x}_j}$ or with distinctive parentheses as", null, "${\\texttt{(} x_j \\texttt{)}}.$ Thus we may see the function", null, "${\\lnot_j : \\mathbb{B}^k \\to \\mathbb{B}}$ written in either of the following forms:", null, "$\\begin{matrix} \\lnot_j (x_1, \\ldots, x_j, \\ldots, x_k) & = & x_1 x_2 \\ldots x_{j-1} \\tilde{x}_j x_{j+1} \\ldots x_{k-1} x_k \\\\[4pt] & = & x_1 x_2 \\ldots x_{j-1} \\texttt{(} x_j \\texttt{)} x_{j+1} \\ldots x_{k-1} x_k \\end{matrix}$\n\nVisually speaking,", null, "${\\lnot_j (x_1, \\ldots, x_k)}$ is a singular proposition that picks out a single cell", null, "${x}$ of the associated venn diagram, namely, the cell where", null, "${x_j}$ is false and all the rest of the", null, "${x_i}$ are true.  By way of context, I used the above notations for", null, "${\\lnot_j}$ in defining minimal negation operators, which may eventually find use in this discussion.\n\nThis looks like a good place to break and start a second post in the series.\n\nTo be continued …\n\n### 5 Responses to Frankl, My Dear • 1\n\nThis site uses Akismet to reduce spam. Learn how your comment data is processed." ]

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[ "# Count number of arguments in function\n\nHi,\n\nIf I want to know how many arguments a function has, i can do something like `length(formals(func))`\n\nThis works for my functions, but not for some base R functions like `is.character` (perhaps because its compiled?)\n\nReprex below\n\n``````# Create a function with 3 arguments\nfunc <- function(a, b, c){\nmessage(a, b, c)\n}\n\n# Can get number of args with length & formals\nlength(formals(func))\n#> 3\n\n# But this doesn't work for certain base R functions\nlength(formals(is.character))\n#> 0\n``````\n\nCreated on 2023-01-06 with reprex v2.0.2\n\nIs it possible to programmatically get the number of arguments for any R function?\n\nI don't think so; as you've found there is a clear difference between normal functions and primitive functions.\nyou an test if a function is primitive by `is.primitive`\n\n``````> is.primitive(as.character)\n TRUE\n``````\n\nYou could try a hacky approach using string processing ; get hold of the strings like\n\n``````capture.output(args(as.character))[]\n``````\n\nbut there are lots of interesting edge cases even with normal functions.\n\n``````length(fn_fmls(mean))\n# 2\n``````\n\nHow do you intend to interpret `...` ?\nas 1 or as Infinite ?\n\nthe following code runs in R\n\n``mean((1:100)^2,na.rm=TRUE,trim=1,random_does_nothing=5)``\n1 Like\n\nYou raise some interesting points @nirgrahamuk. Thanks for your response!\n\nI'm interested in writing functions that assert basic features of functions\n\nYour insights make me think two basic functions need to be available:\n\n1. a function that returns a count of named arguments,\n2. a function that returns a boolean describing whether dots `...` are used as an argument\n\nFrom these you could build other functions that assert there are 'at least X' arguments. Then I'd leave it up to the end user whether `...` should count as Infinite (default yes), or perhaps just throw an error (think function that takes a very specific 2 argument function the user will likely have to create - here asserting exactly 2 named arguments and ignoring or throwing an error if `...` is present seems perfectly reasonable.\n\nI think having some 'ellipses_value' argument that could be 1 or Inf , might push users of these functions (i.e. future me", null, ") to think about exactly what they want to assert.\n\nI will leave this topic open another couple of days on the off chance anyone has a less hacky way to handle primitives.\n\noh, it seems that there is a way to get a common result from primitive and non-primitive functions.\n\n``````formals(args(as.character))\nlength(formals(args(as.character)))\nformals(args(mean))\nlength(formals(args(mean))``````\n1 Like\n\nI think you've cracked it @nirgrahamuk!\n\nWe can additionally get the names simply by adding: `names(formals(args(as.character)))`.\n\n``````# Get argument names (Primitives)\nnames(formals(args(as.character)))\n#> \"x\" \"...\"\n\n# Get number of arguments (Primitives)\nlength(formals(args(as.character)))\n#> 2\n\n# Get argument names (Non-Primitive)\nnames(formals(args(mean)))\n#> \"x\" \"...\"\n\n# Get number of arguments (Non-Primitive)\nlength(formals(args(mean)))\n#> 2\n``````\n\nCreated on 2023-01-07 with reprex v2.0.2\n\nThats most of the info required to build a robust system of assertions on function arguments\n\nThis topic was automatically closed 7 days after the last reply. New replies are no longer allowed.\n\nIf you have a query related to it or one of the replies, start a new topic and refer back with a link." ]

[ null, "https://community.rstudio.com/images/emoji/apple/joy.png", null ]

[ "# 1/5 - 1/16 as a fraction\n\nHere we will show you step by step detailed solution to 1/5 minus 1/16 (often written as 1/5 - 1/16), with the answers are in fraction form, mixed number and as a decimal number\n\n• 1180\n• 0.1375\n\nStep by step solution\n\nSubtract\n\n15 - 116 = 1180\n\n1. ### Find common denominator by finding the Least Common Multiple(LCM)\n\nMultiples of 5 are   5, 10, 15, 20, 25, 30, 35, 40, ..., 80 ....etc\n\nMultiples of 16 are   16, 32, 48, 64, 80 ...etc\n\nThe LCM of 5 and 16 is 80. So, the common denominator of 15 and 116 is 80. Now change the fractions into equivalent fractions with common denominator\n\n• 15 - 116\n• 1 * 165 * 16 - 1 * 516 * 5\n• 1680 - 580\n2. ### Subtract\n\n• 1680 - 580\n• 16 - 580\n• 1180\n\nSubtract the numerators and place the difference over the common denominator.\n\nSimplify\n• 1180\n1. ### Reduce fraction to its lowest terms\n\nThe greatest common divisor of 11 and 80 is 1, the fraction 1180 cannot be further simplified or reduced to lowest terms" ]

[ null ]

[ "Special lssues\nTable of Content\n•", null, "Open Access\n\nARTICLE\n\nNumerical Approximate Solutions of Nonlinear Fredholm Integral Equations of Second Kind Using B-spline Wavelets and Variational Iteration Method\n\nP. K. Sahu1, S. Saha Ray1,2\nCMES-Computer Modeling in Engineering & Sciences, Vol.93, No.2, pp. 91-112, 2013, DOI:10.3970/cmes.2013.093.091\nAbstract In this paper, nonlinear integral equations have been solved numerically by using B-spline wavelet method and Variational Iteration Method (VIM). Compactly supported semi-orthogonal linear B-spline scaling and wavelet functions together with their dual functions are applied to approximate the solutions of nonlinear Fredholm integral equations of second kind. Comparisons are made between the variational Iteration Method (VIM) and linear B-spline wavelet method. Several examples are presented to compare the accuracy of linear B-spline wavelet method and Variational Iteration Method (VIM) with their exact solutions. More >\n\n•", null, "Open Access\n\nARTICLE\n\nThe Sinh Transformation for Curved Elements Using the General Distance Function\n\nJ.H. Lv1, Y. Miao1,2, W.H. Gong1, H.P. Zhu1\nCMES-Computer Modeling in Engineering & Sciences, Vol.93, No.2, pp. 113-131, 2013, DOI:10.3970/cmes.2013.093.113\nAbstract Accurate numerical evaluation of the nearly singular boundary integrals is a major concerned issue in the implementation of boundary element method (BEM). In this paper, a general distance function independent on the nearly singular point is proposed. Combined with an iteration process, the position of the nearly singular point can be obtained more easily. Then, an extended form of the sinh transformation using the general distance function, which automatically takes into account the intrinsic coordinate of the nearly singular point and the minimum distance from source point to the element in the intrinsic parameter plane, is developed to deal with… More >\n\n•", null, "Open Access\n\nARTICLE\n\nDynamic Stress Intensity Factors of Collinear Cracks under a Uniform Tensile Stress Wave\n\nK.-C. Wu2, S.-M. Huang2, S.-H. Chen3\nCMES-Computer Modeling in Engineering & Sciences, Vol.93, No.2, pp. 133-148, 2013, DOI:10.3970/cmes.2013.093.133\nAbstract An analysis is presented for an array of collinear cracks subject to a uniform tensile stress wave in an isotropic material. An integral equation for the problem is established by modeling the cracks as distributions of dislocations. The integral equation is solved numerically in the Laplace transform domain first and the solution is then inverted to the time domain to calculate the dynamic stress intensity factors. Numerical examples of one, two, or three collinear cracks are given. The results of one or two cracks are checked to agree closely with the existing results. More >\n\n•", null, "Open Access\n\nARTICLE\n\nA Novel Meshless Analysis Procedure for Three-dimensional Structural Problems with Complicated Geometry\n\nWen-Hwa Chen2,3, Ming-Hsiao Lee4\nCMES-Computer Modeling in Engineering & Sciences, Vol.93, No.2, pp. 149-166, 2013, DOI:10.3970/cmes.2013.093.149\nAbstract A novel meshless analysis procedure is established for practical implementation in dealing with three-dimensional structures with complicated geometry. By this procedure, to describe the surface of structure, the Stereo-lithography (STL) geometry technique is first adopted. Nodes are then generated and paved uniformly in the space over the entire structure analyzed. To decide the node distribution inside the structure, a geometry-related treatment scheme with relevant checking mechanisms is developed. Besides, a simple and direct spatial integration scheme is also proposed. By this integration scheme, integration points are evenly distributed in the structure and can be adjusted easily to meet the required… More >\n\nPer Page:", null, "" ]

[ null, "https://www.techscience.com/static/images/suob.png", null, "https://www.techscience.com/static/images/suob.png", null, "https://www.techscience.com/static/images/suob.png", null, "https://www.techscience.com/static/images/suob.png", null, "https://www.techscience.com/static/images/fixe_esc1.png", null ]

[ "hep-th/9812226, IASSNS-HEP-98/106\n\nD-branes in Orbifold Singularities and Equivariant K-Theory\n\nHugo García-Compeán Also Departamento de Física, Centro de Investigación y de Estudios Avanzados del IPN, Apdo. Postal 14-740, 07000, México D.F., México. E-mail:\n\nSchool of Natural Sciences\n\nOlden Lane, Princeton, NJ 08540, USA\n\nThe study of brane-antibrane configurations in string theory leads to the understanding of supersymmetric D-branes as the bound states of higher dimensional branes. Configurations of pairs brane-antibrane do admit in a natural way their description in terms of K-theory. We analyze configurations of brane-antibrane at fixed point orbifold singularities in terms of equivariant K-theory as recently suggested by Witten. Type I and IIB fivebranes and small instantons on ALE singularities are described in K-theoretic terms and their relation to Kronheimer-Nakajima construction of instantons is also provided. Finally the D-brane charge formula is reexamined in this context.\n\nDecember, 1998\n\n1. Introduction\n\nD-branes are extremely interesting objects in string theory and thus their study is still matter of vigorous research (for a review see ). Many surprises have been found in the study of various physical situations of D-branes and surely many other surprises remain to be discovered in the near future. Among their multiple applications D-branes have been used to study the strong coupling dynamics of supersymmetric field theories in various dimensions through the construction of configurations of intersecting branes (for a review see ). Gauge theories in dimensions with sixteen supercharges can be obtained as the world-volume theories of flat infinite D-branes. Intersecting brane configurations are described by BPS bound states of the corresponding superstring theory where the D-brane is living. Intersection of D-branes generically has zero fermionic modes which are chiral and therefore anomalous on the branes. D-branes carries RR charges and the definition of this charge, may be given through the implementation of this anomaly cancellation via the inflow mechanism from the bulk where the D-branes are living . The consideration of these facts in topological terms and a further generalization to non-trivial normal bundle to the cycles where the D-branes are wrapped lead to a formula for the brane charge which widely suggest that it takes values not in the homology of spacetime but in the K-theory of the spacetime [4,,5].\n\nRecently some works have been made in the context of stable non-supersymmetric states (non-BPS states) in field theory  and string theory [7,,8,,9,,10,,11,,12]. In particular, in string theory context, Sen have shed light on the structure of stable non-BPS states in type I and II superstring theories by constructing non-BPS states through the consideration of brane-antibrane pairs and their dynamical properties . Boundary state analysis of these configurations leads to a world-volume field theory, on the brane pair, without the usual super Maxwell multiplet but with a tachyon field on the world-volume which survives the GSO projection. Potential of the tachyon field of this system is shown to be of the same form as the potential for a kink-anti-kink topological defect which interpolates between the two minima of the tachyonic potential. This system is already not supersymmetric but it has an stable vacuum. The vacuum configuration is reached through the mechanism of tachyon condensation. Generalization to other higher dimensional topological defects was considered as well in .\n\nFollowing Sen’s results very recently Witten , classified the non-supersymmetric brane configurations in terms of the mathematical structure known as topological K-theory (for some reviews see [14,,15,,16,,17]). Some earlier evidence of the possible relation between D-branes and K-theory can be found in . At the same paper Witten found further evidence that D-brane charge takes values in the K-theory of the spacetime manifold as before suspected in . Non-BPS states of some brane configurations are classified according to the Type of superstring theory we are dealing. For instance, in Type IIB theory, non-BPS states and their lower dimensional BPS-branes are classified by the complex K-theory group K with the spacetime manifold. In Type I theory non-BPS objects are classified by the real K-theory group KO. Non-BPS states in Type IIA (and their possible lifting to M-theory) are not well understood, however these states are classify by K. Among other further natural generalizations, Witten argued in  that the correct description of non-BPS states in orbifold singularities is given in terms of the equivariant K-theory (for a review of equivariant K-theory see [19,,20,,21]).\n\nIn this paper we continue this analysis of configurations of pairs of branes-antibranes in string theories described by K-theory. In particular we give the first steps to study configurations of D-brane pairs in orbifold singularities. We will see that equivariant K-theory is the natural language to describe these configurations, just as it was argued in . We find that many results, including the global construction of , hold also in the equivariant case with minor changes. However, we shall stress the differences and explain some subtleties.\n\nAfter the previous discussion we apply this formalism to the specific case of six-dimensional gauge theories on the world volume of Type IIB and I BPS-fivebranes (constructed from nine-brane pairs) on a general orbifold singularities. Although the construction is valid for general orbifolds, in order to connect with some well known results of branes in orbifolds, we will restrict ourselves to the case of branes on ALE singularities [22,,23,,24,,25], leaving the theories of D3 branes on  and D1 branes on  for further investigation. We have placed coincident fivebranes at at fixed point of the ALE singularity parametrized by the transverse coordinates to the fivebrane. Following Witten , the fivebrane charge is now determined by the equivariant K-group of the compact space of the transverse directions to the fivebranes or (in the non-compact case) by the corresponding K-group with compact support. The transverse space is given by a ALE space which is seen as the minimal resolution of the orbifold with a discrete subgroup of SU(2). We will make contact with the Kronheimer-Nakajima construction of Yang-Mills instantons on ALE spaces [28,,29]. We show that all information concerning the description of non-trivial RG fixed points of six-dimensional theories in ALE singularities [30,,24,,25] in Type I and IIB theories, is contained in the K-theoretic description and thus it admits a derivation in pure K-theoretic grounds.\n\nThe structure of this paper is as follows, in Section 2 we briefly review Witten description of non-BPS states in terms of K-theory. Section 3 is devoted to study non-BPS states in orbifold singularities and their classification in terms of equivariant K-theory. In Section 4 we discuss in detail the problem of BPS-fivebranes in ALE singularities in Type I and IIB superstring theories. We find a nice relation with Kronheimer-Nakajima construction of instantons and the description of nontrivial RG fixed points of theories in six dimensions is derived from the mathematical formalism of equivariant K-theory. In the process we discuss the origin of the index theorem for ALE manifolds from the equivariant K-theory. In Section 5 we will provide the formula for the charge of the brane in orbifold singularities derived from K-theoretical considerations. Finally is Section 6 our concluding remarks are given.\n\n2. K-theory Description of Pairs D-brane-anti-D-brane\n\nIn this section we briefly review the Witten’s construction of D-brane pairs and their relation to topological K-theory following . Our aim is not provide and extensive review, but only recall the relevant structure which will be needed in the following sections. Throughout this paper we will follow the notation introduced by Witten in .\n\n2.1. Review of Witten’s Construction\n\nK-Theory Structure\n\nIn order to fix some notation let be the ten-dimensional spacetime manifold and let be a -dimensional submanifold of . Branes or antibranes or both together can be wrapped on . When configurations of coincident branes or antibranes only are wrapped on , the world-volume spectra on consists of a vector multiplet and scalars in some representation of the gauge group. These configurations can be described through Chan-Paton bundles which are U gauge bundles over for Type II superstring theory and by SO or Sp bundles in Type I theory. Gauge fields from the vector multiplet define a U gauge connection for Type II theory (or SO or Sp gauge connection for Type I theory) on the (corresponding) Chan-Paton bundle. GSO projection cancels the usual tachyonic degrees of freedom. Something similar occurs for the anti-brane sector.\n\nThe description of coincident -branes and -antibranes wrapped on leads to the consideration pairs of gauge bundles (over ) with their respective gauge connections and . In the mixed configurations GSO projection fails to cancel the tachyon. Thus the system is unstable and may flow toward the annihilation of the brane-antibrane pairs with RR charge for these brane configurations being conserved in the process [7,,8,,10,,13].\n\nOn the open string sector Chan-Paton factors are matrices constructed from the possible open strings stretched among the different types of branes. Brane-brane and antibrane-antibrane sectors correspond to the diagonal elements of this matrix. Off-diagonal elements correspond with the Chan-Paton labels of an oriented open string starting at a brane and ending at an antibrane and the other one to be the open string with opposite orientation.\n\nThe physical mechanism of brane-antibrane creation or annihilation without violation of conservation of the total RR charge, leads to consider physically equivalent configurations of branes and antibranes and the same configuration but with additional created or annihilated brane-antibrane pairs.\n\nThe relevant mathematical structure describing the brane-antibrane pairs in general type I and II superstring theories is as follows:\n\ni)-. and gauge connections and on the Chan-Paton bundles and over , respectively. Bundles and corresponding to branes and antibranes are topologically equivalent. The groups and are restricted to be unitary groups for Type II theories and symplectic or orthogonal groups for Type I theories.\n\nii)-. Tachyon field can be seen as a section of the tensor product of bundles and its conjugate as a section of (where denotes the dual of the corresponding bundle.)\n\niii)-. Brane-antibrane configurations are described by pairs of gauge bundles .\n\niv)-. The physical mechanism of brane-antibrane creation or annihilation of a set of -branes and -antibranes is described by the same U (for Type II theories) or SO (for type I theories) gauge bundle . This mechanism is described by the identification of pairs of gauge bundles and . Actually instead of pairs of gauge bundles one should consider classes of pairs of gauge bundles identified as above. Thus the brane-antibrane pairs really determine an element of the K-theory group K of gauge bundles over and the brane-antibrane creation or annihilation of pairs is underlying the -theory concept of stable equivalence of bundles For 9-branes, the embedded submanifold coincides with and the thus brane charges take values in K-theory group of ..\n\nConsistency conditions for 9-branes () in Type IIB superstring theory such as tadpole cancellation implies the equality of the ranks of the structure groups of the bundles and . Thus . The ‘virtual dimension’ of an element is defined by . Thus tadpole cancellation leads to a description of the theory in terms of pairs of bundles with virtual dimension vanishing, . This is precisely the definition of reduced K-theory, . Thus consistency conditions implies to project the description to reduced K-theory.\n\nIn Type I string theory pairs are described by a class of pairs of SO and SO gauge bundles over . Creation-annihilation is now described through the SO bundle over . In Type I theories tadpole cancellation condition is . In this case equivalence class of pair bundles determines an element in the real K-theory group KO. Tadpole cancellation , newly turns out into reduced real K-theory group .\n\nType IIA theory involves more subtle considerations worked out in . It was argued by Witten in  that configurations of brane-antibrane pairs are classified by the K-theory group of spacetime with an additional circle space . K-theory group for type IIA configurations is K.\n\nFinally, it is also shown in  that for both Type I and II theories, the consideration of non-compact usual spacetime (where some compact space) with appropriate boundary conditions at infinity leads to vacuum configurations with branes. In the language of K-theory this means that a non-zero K-theory class is equivalent to the trivial class at infinity. Non-compact spacetimes require K-theory with compact support, but with the boundary conditions at infinity the difference between ordinary and reduced K-theory is irrelevant for the most of physical applications.\n\nLower Dimensional Branes\n\nWe now review Witten’s global construction of lower dimensional -branes, with from higher-dimensional branes. The flat case will be recovered when it is required only one global coordinate system. The key argument is that it is possible to construct -branes as the bound state of pairs of -branes and antibranes [8,,13].\n\nLet be a -dimensional closed orientable submanifold of . This latter is also a -dimensional orientable submanifold of spacetime manifold of the Type II superstring theory. is a codimension submanifold of . could possess a gauge bundle with a section vanishing along . Now we consider a system of -branes and -antibranes wrapped on . Tachyon field of this system transforms in the bifundamental representation of the group . Tadpole cancellation condition implies that actually transforms in the adjoint representation of . In vacuum tachyon field breaks the group to the diagonal . Tachyon is a section of the bundle and it is vanishing in a codimension submanifold . Tachyon condensation flows the system to a configuration with charges transforming in the diagonal group , representing the remaining -branes.\n\nThe description of lower-dimensional branes wrapped on can be nicely described in terms of the pairs of brane-antibrane wrapped in as follows. The full formalism requires of the introduction of a U bundle over . In order to extend the definition of this bundle from to several obstructions can arise. For instance, if has trivial normal bundle it can extends over getting an element of K given by the class of pairs . However if does not extend over a more involved definition of the element of K has to be given using the standard tools of K-theory [14,,15,,16,,17].\n\nConsider a tubular neighborhood of and its closure . Bundles over can be easily pulled back to determining so an element of K. The corresponding tachyon field defines an isomorphism when it is restricted to the boundary of . Thus one can extend over via the map by declaring that this isomorphism is extendible over . If does not extend over one can look for a gauge bundle over in such a way that be trivial over and trivial when pulled back to . Thus one can extend over (as the trivial bundle) and extend by setting it equal to over . The element of K is thus and the tachyon field is now .\n\nThe Spinor Description\n\nGlobal construction outlined in the above subsection in general might not exist. However it is possible to implement the global description in terms spin bundles associated to the normal bundle to . Consider the normal bundle to in . Assume that the codimension of is . Normal bundle has structure group SO. If is a spin bundle it has associated two spinor bundles , of positive and negative chirality spin representations of SO respectively. If we suppose that the spinor bundles near extend over X, thus there are pairs of branes with gauge bundles and over .\n\nTachyon field is a map given by and it is naturally given by where , where is an element of the tubular neighborhood of in . This tachyon map gives a unitary isomorphism in the boundary of . This isomorphism is determined by the fact that is unitary if is a unitary vector. One can use this fact to extend this configuration over and the results of the above subsection hold. Thus more generally one can look for a suitable bundle such that is trivial over and extends over with the final configuration given by and the tachyon .\n\n3. Equivariant K-theory Description of Branes in Orbifolds\n\nIn the last section we gave the basic facts of the interplaying between D-branes configurations and K-theory. We consider now brane-antibrane configurations in a generic orbifold and its description in terms of equivariant K-theory. We will show that equivariant K-theory is the natural language to describe supersymmetric and non-supersymmetric brane configurations in orbifolds.\n\n3.1. Non-BPS States From Brane-antibrane Pairs\n\nBefore discussing BPS and non-BPS branes in orbifolds, we briefly review the necessary requirements about non-BPS states string theory on compact manifolds. We mainly follow the results [7,,8,,10]. The basic idea is the construction of stable non-BPS states through the construction of brane-antibrane configurations. More precisely, Sen has shown that wrapping branes and antibranes in cycles of some compact manifold in some Type I and II superstring compactifications, world-volume theory has a tachyon mode expansion which survives GSO projection. Tachyon field is an scalar which has associated a negative potential energy which cancel the tension between the brane-antibrane pair [7,,8]. It was conjectured than unstabilities associated with the tachyon mode flows the system to the annihilation of RR charge and to an stable state admitting exact boundary conformal field theory description. To be more precise, for instance, in Type I theory pairs of D1-brane- anti D1-brane have a tachyonic kink potential and this system turns out to be stable and coincides with the SO(32) spinor state of the SO(32) heterotic theory .\n\nPairs of D1-strings and anti-D1-strings compactified on the circle of radius have a tachyon field of the form . At the self-dual radius tachyon components becomes massless and (in the and zero picture) the vertex operators of the boundary CFT are respectively given by\n\n V(−1)±=−e−ΦBe±i√2XB⊗σ1\n\nand\n\n V(0)±=∓iψBe±i√2XB⊗σ1,\n\nwhere is the bosonic field, is the bosonized ghost and is the world-sheet fermion. All of them taking its boundary value. is the Chan-Paton factor of the brane-antibrane pair open string sector. Brane-antibrane configuration survives projection and it does exist in Type IIB theory .\n\n3.2. Branes-antibranes in Orbifolds\n\nHere we describe briefly the dynamics of pairs of branes-antibranes in orbifold singularities. We present the case of a pair of D-string-anti-D-string of the type IIB superstring theory in a -orbifold . For definiteness take the pair with world-volume along coordinates leaving transverse coordinates to be . group act only reversing the coordinates and leave fixed the rest of the coordinates i.e. . Thus action of behaves as an ALE singularity in the coordinates . One can redefine complex coordinates and and write down this singularity as . This is, of course, the most simple non-trivial example of the most simple class of singularities of the A-D-E-type. In order to see how the brane pairs are behaved at orbifold singularities we need to know how the orbifold projection is realized on the Chan-Paton sector. As was shown in , the Chan-Paton is even with respect the action of , while the Chan-Paton factor is odd under -orbifold projection. This is essentially as transforms the tachyon field under the generator in . Thus the tachyon is odd under the orbifold projection. We can immediately generalize the group to singularities generated by the group or for a generic group of the A-D-E singularities.\n\nWe know from the basic theory of D-branes in orbifold singularities [22,,23] that the action of on the Chan-Paton factors is given by where is an element of , of and is the index on the Chan-Paton factors, which for the tachyon is . The tachyon field transforms under -orbifold projection as\n\n g:T(x)→γ(g)T(x′)γ−1(g).\n\nThus similar than the vector potential , the tachyon transforms in the adjoint representation of under orbifold projection.\n\n3.3. Equivariant K-Theory Structure\n\nIn this subsection we shall show in detail that the natural language to deal with stable (non-)BPS branes in Type I and II superstring theory is, as suggested by Witten , the equivariant K-theory. Throughout the rest of the paper we limit ourselves to work with an abelian and discrete subgroup of some A-D-E group.\n\nTo be the most self-contained possible we first give some basic definitions of the theory of equivariant K-theory. For details of proofs we encourage the reader to consult [19,,20]. Let be a Lie group, in general may be a topological group The group does not be confused with the generic group.. Let be a differentiable manifold of finite dimension. is said to be -manifold if there exist an smooth action of on . A -map between two -manifolds is a smooth map which commutes with the action of the group . Now consider a principal bundle over with canonical projection . A -principal bundle over the -manifold , is a -map which carries fibres to fibres linearly and projects to the action over . A -homomorphism between two -bundles over is a map which is both a bundle homomorphism and a -map. The -isomorphism between two bundles is a -homomorphism with inverse and the inverse is also a -map. Classes of pairs of -bundles have a ring structure known as the Grothendieck ring K or equivariant K-theory of [19,,20,,21].\n\nWhen the group acts trivially on , K factorizes just as the product K where K is the ordinary K-theory ring and is the ring of irreducible representation spaces of . Let be the component of the irreducible vector space representation. For trivial actions can be written as where is the trivial bundle .\n\nOne of the properties of K we will use in this paper is the concept of space of sections of a -bundle or -section space. Let be the space of sections of an ordinary bundle over , has the structure of a vector space and if is compact has in addition the structure of a Banach space. Now in the case of we have a -bundle , in addition we have an induced action of on its space of sections . This action is continuous an it is given by with and . To see that this action is continuous consider the sequence of maps: given by . Each map is continuous and as composition of continuous mappings is also continuous therefore this action is continuous. Action of over induces the group homomorphism .\n\nLet be the ten-dimensional spacetime orbifold and let be a -submanifold (of dimension ) of . Let be an embedding of into which preserves the action of . Branes or antibranes can be now wrapped on . Thus the world-volume spectra on come from the dimensional reduction of the corresponding string theory in ten dimensions with an additional orbifold projection. The spectra can thus be encoded in a quiver diagram as usual in brane theory in orbifolds [22,,23,,26,,27]. The theory on the world-volume of branes (or antibranes) wrapped on is described through Chan-Paton bundles which are given by bundles over , in the case of Type II superstring theory and by for the Type I theory, here is the dimension of the regular representation of . Gauge fields from the vector multiplet defines a connection on the corresponding Chan-Paton bundle. This connection satisfies, as usual, the orbifolding condition , with . Just as in the smooth case, GSO projection cancels tachyonic degrees of freedom leaving only the quiver structure of the vector multiplet and hypermultiplets.\n\nWhen we consider both coincident branes and anti-branes wrapping on tachyon is still preserved by GSO projection (as we have seen in the above subsection) and the vector multiplet is also projected out as the smooth case. In other words, in non-pathological cases, -action commutes with GSO projection. As we saw in the above subsection tachyon field is required to satisfy the condition with being an element of .\n\nIn terminology of equivariant K-theory, orbifold spacetime are seen as a -space i.e. spacetime with the action of over . Chan-Paton bundles are -bundles over where the projection map preserves the action of group . The tachyon field can be seen as a -map bundle . Equivalently, tachyon can be seen now as a -section (up to a constant) of the -bundle . The way that tachyon field transforms under given in (3.3), is nicely explained by a consequence of -section definition as follows.\n\nLet be the space of -sections defined as the space of sections of the -bundle . The group acts on given by\n\n (γ(g)⋅T)(x)≡γ(g)⋅T(γ−1(g)⋅x)=γ(g)⋅T⋅γ−1(g)(x)\n\nwith and being elements of and respectively and . The action of on sections induces also a group homomorphism between and .\n\nWhen one is considering pairs of branes on orbifold singularities similar statements of the smooth case  are still valid. Creation and annihilation of 9-branes in orbifolds is described by a -bundle over . So, still conservation of total charge, now including mirror images under , leads to make the identification of pairs of bundles with , as two equivalent descriptions. These conditions determine precisely an element of the -equivariant K-group of spacetime . Thus brane-antibrane configurations in spacetime orbifold can be described in terms of equivariant K-theory methods. Thus D-brane charges on an orbifold singularity takes values in the equivariant K-theory group of the spacetime as suggested in .\n\nIn the specific case of Type IIB theories on orbifold singularities, tadpole cancellation leads, just as the smooth case , to the same number of 9-branes and of -branes. The reason of this can be seen from the condition of equality of ranks of the gauge groups\n\n ∑μN1nμ=∑μN2nμ\n\nit is immediate to get . Thus Chan-Paton bundles and are gauge bundles and tadpole cancellation condition implies that virtual dimension is zero and it lies in equivariant reduced K-theory group .\n\nIn Type I string theory pairs on orbifolds are described by a class of pairs of and gauge bundles over . Creation-annihilation is now described through the bundle over for some . In Type I theories tadpole cancellation condition is given by . Thus condition holds again in this situation. Equivalence class of pairs of bundles determines an element in the real equivariant K-theory group KO. Tadpole cancellation , newly turns out into equivariant reduced real K-theory group .\n\nVacuum at infinity is -invariant, that means that it is reached by tachyon condensation preserving such a property. The reason of this is that vacuum space configuration of s is always written in an U invariant way , thus U is broken down to the product group . The vacuum is reexpressed now in an invariant way in terms or the trace in the regular representation of . Equivalence to vacuum at infinity means that near infinity bundles and are -isomorphic. As the smooth case, the vacuum might have some branes and therefore a non-zero class of K where our spacetime is a non-compact orbifold with the generic form . For Type I theory with an abelian and discrete subgroup of SU(2) and , tadpole cancellation is still 32 for the 9-brane charge. For fivebranes, tadpole constraints can be written in terms of -invariant geometric terms. For instance, in the unbroken SO(32) theory, anomaly cancellation implies the existence of a certain number of fivebranes , which is for even, and for odd . In general there is not a choice of to cancel anomalies and the anomaly inflow mechanism from the bulk should be extended to theories of branes in orbifolds .\n\nIn physical applications descriptions of brane configurations in non-compact spacetime, ordinary and reduced equivariant K-groups K and are equivalent. And the reason of this is that the -invariance of the vacuum at infinity is described by a non-zero class of pairs of -bundles which is equivalent to the trivial class of pairs of bundles which are -isomorphic at infinity. This only is possible if the pair of -bundles have equal rank. Thus virtual dimension is zero and ordinary and equivariant reduced K-groups describe the same physical situation.\n\nType IIA theory involves more subtle considerations worked out in . It was argued by Witten in  that configurations of brane-antibrane pairs are classified by the K-theory group of spacetime with an additional circle space . Equivariant K-theory of Type IIA theory not will be discussed in the paper, however we make only some few comments. K-theory group for type IIA configurations is K. If and this group acts trivially on space and non-trivially on then K is isomorphic to K. It is tantalizing to speculate that equivariant K-theory might be a relevant tool to generalize the K-theoretical description of non-BPS branes to heterotic superstring theory.\n\n3.4. Lower Dimensional Branes on Orbifolds. Global Construction\n\nUp to here we have described pairs of branes in generic orbifold singularities. However to see that the equivariant K-theory classifies also lower dimensional branes than 9, we will attempt to implement to orbifolds the Witten’s construction .\n\nFor branes of lower dimensions in orbifold singularities in type II superstring theory the description is similar to that given in . However for completeness we include here this discussion. The key argument here is that it is possible to construct -branes as the bound state of pairs of -branes and antibranes.\n\nFirst we overview some notation. Let be a -dimensional (and codimension ) closed orientable -submanifold of . This latter is a -dimensional -submanifold of . Let be a -bundle over with -section vanishing along . One can consider a set of branes and antibranes wrapped on such that tachyon field is a -section which transforms in the bifundamental representation of the gauge group Tadpole cancellation condition implies that tachyon field transforms in the adjoint representation of the group\n\nNow introduce a -bundle over with structure group . As in the smooth case, bundles and determine an element of the equivariant K-group of . If has trivial normal bundle, then the group K admits an extension to and we get K. If it does not occurs, then we use the -invariance of the vacuum at infinity described by a -isomorphism of bundles on the boundary of a tubular neighborhood of in . Thus K extends to K by declaring that this -isomorphism extends throughout . If it does not occurs, then we apply the -equivariant version [19,,20] of the argument given in . One extends rather the trivial bundle because it is -isomorphic (with the isomorphism given by ) to over .\n\n3.5. Examples\n\nType I zero and -1 Brane\n\nWe consider the zero brane of Type I superstring theory. This is described, according to subsection by the group KO or the equivariant compact support group KO. The transverse space to the zero brane is the spacetime orbifold . Assume that the group acts on the spatial coordinates only, leaving fixed the coordinate. Thus symbolically, , where . vector of the transverse space can be rewritten in terms of complex coordinates and to be more specific let be the cyclic group . Then choice an action of over to be\n\nThus actually we have the zero brane living at an orbifold singularity . An element of KO is given by a pair of trivial gauge bundles over . The tachyon field is a -isomorphism near infinity and it is given by . Separation of coordinates into breaks explicitly the symmetry SO(9) down to SO(8). Under this decomposition tachyon field reads\n\n T=(→Γ⋅→xx9−x9→ΓT⋅→x)\n\nwhere are the SO(8) gamma matrices satisfying , is transpose to and are the positive and negative SO(8) chirality -spinor bundles over the unidimensional space. Each bundle can be trivially extended over . The system splits in two components: and . Recall that are the coordinates of the orbifold and the action of on will depends on the explicit group . are -bundles and the SO(8) Dirac matrices can be seen as -sections of the -bundle and according to the transformation of -sections, they transform as with and . Under the combination , the transformation of and implies that the tachyon field transforms correctly as given from Eq. (3.4)\n\n TA→γ(g)TAγ−1(g),    A=1,2.\n\nSimilar to the smooth case describes a set of D1-branes located at an orbifold singularity at the origin of and describes the other set of D1-antibranes located at the origin of the orbifold singularity. This orbifold is, as we have seen, an orbifold of fourfolds given by .\n\nOn the other hand the study of the -1 brane in an orbifold singularity is very similar to the analysis for the zero brane. In this case the system is classified by K with compact support. Now, there are two -spinor bundles coming from the real spinor representations of SO(10). Action of on determines a singularity of the type and thus there is only one tachyon field .\n\n3.6. Equivariant Bott Periodicity\n\nEquivariant Bott periodicity for KO-theory with compact support is given by\n\n \\tenrmKOΓG(\\tenbfRn)=\\tenrmKOΓG(\\tenbfRn+8).\n\nThis formula tell us that 1-branes and 7-branes or 0-branes and 8-branes in the orbifold singularity, are related. In order to see that consider a -brane wrapped on and let be an embedding. brane is living in an orbifold singularity of the type , being the coordinates of given by the last eight coordinates of . Let be an element of KO. Tachyon field is a -map of bundles . The embedding has an associated normal bundle with structure group SO(8). This have two -spinor bundles of positive and negative chiralities associated with the normal bundle. Using the methods of subsection 3.4 we can extend the above pair of bundles over and thus to get a class of pairs of bundles of KO given by over . The associated tachyon field is given by\n\n T=(→Γ⋅→xT0−T0→ΓT⋅→x).\n\nAs the above subsection, the system splits in two tachyonic components transforming under the regular representation of and the full tachyonic field will transform under the 2-dimensional fundamental representation of . Thus we have found that Bott periodicity relating branes in string theory is still valid when we sit them on a generic orbifold singularity relating now not only these branes but also their mirror images under .\n\nBott periodicity in its equivariant version for complex K-theory is given by the natural isomorphism [19,,20,,32]\n\n \\tenrmK−qG(X)≅\\tenrmK−q−2G(X)\n\nwhere K is defined by K with the -th reduced suspension of . Thus equivariant Bott periodicity for complex K-theory has, as the ordinary Bott periodicity, periodicity equal to 2. Thus for Type IIB superstring theories there are lower dimensional BPS-branes each two dimensions and Bott periodicity identifies the branes with dimension with branes with dimension . With essentially the same thing occurs when Type IIB branes are placed in orbifold singularities of the type with . As we have seen, the interpretation of equivariant Bott periodicity theorem is the identification of sets of -branes with -branes with the number of mirror images in the orbifold. Later at the end of the Section 4 we will come back to this point.\n\n4. Fivebranes, Small Instantons and Theories in Six Dimensions\n\nAs we have reviewed in Type I and IIB is possible to construct some stable D-branes as a bound state of certain number of pairs. In this section we study the description of stable supersymmetric Type I and IIB fivebranes in a background of nine-brane pairs as worked out in . Now we sit the fivebrane in an ALE orbifold singularity in the transverse space of the fivebrane. We have choice an ALE singularity in order to connect with some previous results well known from the literature but in principle its generalization to other orbifold singularities abelian or non-abelian is immediate. Our main claim is that the description of fivebrane from brane pairs in orbifold singulatities are classified by equivariant K-theory group K with . We present evidence of this by deriving the relevant information to describe non-trivial RG fixed points on the fivebrane world-volume theory, through the equivariant K-theory formalism.\n\n4.1. Fivebrane From Nine-brane Pairs\n\nAs was shown in , fivebranes can be interpreted in terms of nine-brane pairs. In Type I theory, instantons are precisely constructed from bound states of fivebrane and ninebranes [33,,34]. We consider four pairs of branes and one fivebrane in Type I theory. This configuration describes a small instanton of charge one and gauge group SU(2)=Sp(1). Instanton charge takes values in the group K or K with compact support with parametrizing the transverse directions to the fivebrane. This system is equivalently described by the global description of . It is easy to see that the normal bundle to in is a bundle with structure group SO with . Thus following the global constructions reviewed in Section 2, the fivebrane can be constructed from pairs of 9-branes i.e. four pairs of 9-branes. Normal bundle is a spin bundle and there is a pair of spinor bundles associated to the positive and negative chiralities of SO(6). Tachyon field is given by with are a point of the fivebrane world-volume. Tachyon field breaks the group SO to the diagonal SO(4), which is isomorphic (locally) to SU this group is broken by the instantons to SU(2).\n\nOne can attempt an immediate generalization of the last paragraph. Consider a set of fivebranes constructed from pairs of branes. Following , one can construct easily the configuration of instantons with gauge group SO with . Later in subsection we will discuss with detail the K-theoretic description of the small instanton in an ALE singularity.\n\n4.2. Type IIB Instanton in an ALE Orbifold Singularity\n\nWe now consider fivebranes located at general A-D-E type orbifold singularities of Type IIB theory. Orbifold singularity we deal is of the form where is a discrete and abelian subgroup of SU. Consider a configuration consisting of fivebranes whose world-volume is parametrized by the coordinates located at a fixed point of the orbifold singularity , parametrizing the normal directions to the fivebrane . Fivebrane charge takes values in the equivariant group K or equivalently in K with compact support. Moreover, the relevant group to compute it is rather the K-theory group K of the ALE space , with being isomorphic to the minimal resolution of .\n\n4.3. Relation to Kronheimer-Nakajima Construction\n\nIn Type IIB theories we have seen that an element of the equivariant K-theory group is given by a class of pairs of -gauge bundles both are -bundles over , with where and is some A-D-E group: , , , and . Group has the regular irreducible representation . With the action of over , one can construct the associated -vector bundle given by . Similarly for we can construct a -vector bundle . One can immediately see that vector bundle it is very well known from the literature  and it is known as the tautological bundle. Thus we have found that equivariant K-group determining the charge of fivebranes in an ALE orbifold singularity is equivalently described as a class of pairs of tautological bundles arising in Kronheimer-Nakajima construction of instantons  i.e. K. These bundles are -modules given by where are the irreducible regular representations of , and is the trivial bundle over with fiber .\n\nIn order to extract physical information it is needed to describe the K-group K in terms of more familiar grounds. K-theoretical description can be usually translated in terms of cohomological language through the famous index theorem of elliptic operators . To do this Atiyah and Singer used an alternative definition of equivariant K-theory in terms of complexes of -bundles over -spaces [19,,20]. To proceed further we need a tubular neighborhood of and the Thom isomorphism. Tubular neighborhood of in is familiar for us from the above sections so we will focus on the Thom isomorphism. One can construct the Thom isomorphism as follows: Start from the cotangent bundle , to with canonical projection . The embedding of into regular representation vector space induces an embedding of into and so the Thom isomorphism is given by . Also there is an embedding and its corresponding inducing map . Thus one can construct the Gysin map as . The inclusion of a point into induces similarly the map . Thus -equivariant topological index is defined as where is a -vector bundle over . This is precisely the well known Atiyah-Singer -index formula . Equivariant Chern character is an useful tool to translate the index formula in cohomological language . Coincidence between topological and analytical index leads to the Atiyah-Singer theorem for elliptic operators \n\n Ind(D)=−∫ZζchΓG(EΓG)^A(Z).\n\nAs is a ALE manifold with non-trivial boundary , the -index theorem has to be modified to -index for manifolds with boundary [36,,37]. For twisted Dirac operators the above equation has to be corrected by a boundary term given by the invariant \n\n Ind(D)=−∫ZζchΓG(EΓG)^A(Z)+1|ΓG|∑γ≠1χW(γ)2−χQ(γ).\n\nThis formula determines the dimension of the moduli space of instantons in ALE gravitational spaces . In cohomological terms the use of characteristic classes gives relevant information about the non-trivial submanifolds of the ALE space. For instance the first Chern classes of with form a basis of . Poincaré duality determines non-trivial homology cycles . The representation is strongly associated to the monodromy of at the end of , . In other words" ]

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[ "Fred Lacy An Examination and Validation of the Theoretical ResistivityTemperature Relationship for Conductors 439 - 445 2013 7 4 International Journal of Electrical and Computer Engineering https://publications.waset.org/pdf/17289 https://publications.waset.org/vol/76 World Academy of Science, Engineering and Technology Electrical resistivity is a fundamental parameter of metals or electrical conductors. Since resistivity is a function of temperature, in order to completely understand the behavior of metals, a temperature dependent theoretical model is needed. A model based on physics principles has recently been developed to obtain an equation that relates electrical resistivity to temperature. This equation is dependent upon a parameter associated with the electron travel time before being scattered, and a parameter that relates the energy of the atoms and their separation distance. Analysis of the energy parameter reveals that the equation is optimized if the proportionality term in the equation is not constant but varies over the temperature range. Additional analysis reveals that the theoretical equation can be used to determine the mean free path of conduction electrons, the number of defects in the atomic lattice, and the &amp;lsquo;equivalent&amp;rsquo; charge associated with the metallic bonding of the atoms. All of this analysis provides validation for the theoretical model and provides insight into the behavior of metals where performance is affected by temperatures (e.g., integrated circuits and temperature sensors). Open Science Index 76, 2013" ]

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[ "Go Down\n\n### Topic: Why does the order of this code change if an LED blinks or not? (Read 1 time)previous topic - next topic\n\n#### SheCurvesMobius", null, "##### May 08, 2015, 07:35 pm\nWhy is it that when I write my code this way in the \"FOR LOOP A\", pin 10 blinks during \"FOR LOOP A\":\n\nCode: [Select]\n`int timer = 100;void setup() {  // Initialize each pin as an output  for (int thisPin = 2; thisPin < 8; thisPin++) {    pinMode(thisPin, OUTPUT);  }  pinMode(10, OUTPUT);}void loop() {  // \"FOR LOOP A\"  // loop from the lowest pin to the highest:  for (int thisPin = 2; thisPin < 8; thisPin++) {    digitalWrite(thisPin, HIGH);    delay(timer);    digitalWrite(thisPin, LOW);    digitalWrite(10, HIGH);    delay(timer);    digitalWrite(10, LOW);   }   // \"FOR LOOP B\"  // loop from the highest pin to the lowest:  for (int thisPin = 7; thisPin >= 2; thisPin--) {    digitalWrite(thisPin, HIGH);    delay(timer);    digitalWrite(thisPin, LOW);      }}`\n\nBut if I write it this way in the \"FOR LOOP A\", it stays on for all of \"FOR LOOP A\" and then turns off right before \"FOR LOOP B\" starts? I can't understand the logic. I feel like it should blink durring \"FOR LOOP A\".\n\nCode: [Select]\n`int timer = 100;void setup() {  // Initialize each pin as an output  for (int thisPin = 2; thisPin < 8; thisPin++) {    pinMode(thisPin, OUTPUT);  }  pinMode(10, OUTPUT);}void loop() {  // \"FOR LOOP A\"  // loop from the lowest pin to the highest:  for (int thisPin = 2; thisPin < 8; thisPin++) {    digitalWrite(thisPin, HIGH);    digitalWrite(10, HIGH);    delay(timer);    digitalWrite(thisPin, LOW);    digitalWrite(10, LOW);   }   // \"FOR LOOP B\"  // loop from the highest pin to the lowest:  for (int thisPin = 7; thisPin >= 2; thisPin--) {    digitalWrite(thisPin, HIGH);    delay(timer);    digitalWrite(thisPin, LOW);      }}`\n\n#### dlloyd", null, "#1\n##### May 08, 2015, 07:47 pm\nIn your second example, the output only appears to be always on, but it impulses low just for the time it takes to start the next iteration then write it high again. You'll need to delay after each transition so the high and low time remains equal.\n\nCode: [Select]\n`int timer = 100;void setup() {  // Initialize each pin as an output  for (int thisPin = 2; thisPin < 8; thisPin++) {    pinMode(thisPin, OUTPUT);  }  pinMode(10, OUTPUT);}void loop() {  // \"FOR LOOP A\"  // loop from the lowest pin to the highest:  for (int thisPin = 2; thisPin < 8; thisPin++) {    digitalWrite(thisPin, HIGH);    digitalWrite(10, HIGH);    delay(timer);    digitalWrite(thisPin, LOW);    digitalWrite(10, LOW);    delay(timer);  }  // \"FOR LOOP B\"  // loop from the highest pin to the lowest:  for (int thisPin = 7; thisPin >= 2; thisPin--) {    digitalWrite(thisPin, HIGH);    delay(timer);    digitalWrite(thisPin, LOW);    delay(timer);  }}`\n\n#### SheCurvesMobius", null, "#2\n##### May 09, 2015, 02:29 pm\nAh, of course!! Thank you so much. That helps me a great deal!\n\nGo Up" ]

[ null, "https://forum.arduino.cc/Themes/default/images/post/xx.png", null, "https://forum.arduino.cc/Themes/default/images/post/xx.png", null, "https://forum.arduino.cc/Themes/default/images/post/xx.png", null ]

[ "# Noisy Singular Value Decomposition of a matrix\n\nI'm working with a 5x5 matrix n which has three eigenvalues equal to zero. I want to perform the SVD of this matrix in Mathematica. The matrix n is given by\n\nn[x_] := SetPrecision[{{0, 0, 0, 10.*I, 10.*Sqrt[10./10^x]}, {0, 0, 0, 0.4*10.*I, 0.4*10*Sqrt[10./10^x]},\n{0, 0, 0, 0.5*10.*I, 0.5*10*Sqrt[10./10^x]}, {10.*I, 0.4*10.*I, 0.5*10.*I, 10.^x, 0},\n{10.*Sqrt[10./10^x], 0.4*10.*Sqrt[10./10^x], 0.5*10.*Sqrt[10./10^x], 0, 10.}}, 1000];\n\n\nThe matrix ends up having very small singular values ($10^{-16}$ or so) which I believe are numerical errors. Now consequently, the matrices which come out of SingularValueDecomposition[n[x]] end up being very noisy functions of x.\n\nbox1[x_]:=SingularValueDecomposition[n[x],Tolerance->10^-50];\n\n\nI have tried changing the precision of the input using SetPrecision and the Tolerance of the SVD but nothing appears to remove to noise.\n\nIs there anything else I can do?\n\nEdit: What I do with this after is\n\nbox[x_] := SingularValueDecomposition[n[x]];\nu[x_] := N[Reverse[box[x][], 2] /. x -> 2, 40];\n\\[CapitalLambda][x_] := N[Reverse[box[x][], {1, 2}] /. x -> 2, 40];\nv[x_] := N[Reverse[box[x][], 2] /. x -> 2, 40];\nz[x_] := ConjugateTranspose[u[x]] . Conjugate[v[x]];\nU[x_] := Conjugate[u[x] . MatrixPower[z[x], 1/2]];\n\n\nIf I then do DiscretePlot[Abs[u[x][[1, 2]]], {x, -1, 4, 0.01}]:", null, "• People here generally like users to post code as Mathematica code instead of images or TeX, so they can copy-paste it. It makes it convenient for them and more likely you will get someone to help you. You may find this this meta Q&A helpful – Michael E2 Nov 27 '16 at 14:09\n• I've attempted that now although it looks a little unreadable. – KMoff Nov 27 '16 at 14:16\n• Don't worry, we can copy and paste it on our own notebooks. – Mirko Aveta Nov 27 '16 at 14:17\n• Don't you neglect the small singular values (and the associated vectors/matrix columns)? – Michael E2 Nov 27 '16 at 14:41\n\nIs there anything else I can do?\n\nWith a matrix of this size it seems better to do the computations symbolically.\n\nRedefine n with exact precision:\n\nn[x_] := {{0, 0, 0, 10*I, 10*Sqrt[10/10^x]}, {0, 0, 0, 4*I,\n4*Sqrt[10/10^x]}, {0, 0, 0, 5*I, 5*Sqrt[10/10^x]}, {10*I, 4*I,\n5*I, 10^x, 0}, {10*Sqrt[10/10^x], 4*Sqrt[10/10^x],\n5*Sqrt[10/10^x], 0, 10}};\n\nn[x] // MatrixForm", null, "Compute SVD symbolically (it is not that slow):\n\nAbsoluteTiming[svdRes = SingularValueDecomposition[n[x]];]\n\n(* {3.53261, Null} *)\n\n\nEvaluate the result elements at desired precision:\n\nN[svdRes[] /. x -> 2, 40]\n\n(* {{0.09807730004968832306840926757989902604183,\n0.2925930509160816917625846086760755661284, \\\n...}} *)\n\n• Thanks for this. I have updated my question so you can see the noise I get. This remains after performing things symbolically as you suggest. – KMoff Nov 27 '16 at 16:26\n• @KMoff It seemed to work for me, if I redefined box1, too: n[x_] := {...<AA's code>...}; box1[x_] = SingularValueDecomposition[n[x]]; -- note the use of = instead of :=. – Michael E2 Nov 27 '16 at 17:46\n• @MichaelE2 Yes, I have had success with this now in getting rid of the noise from the plots. The issue that is throwing me is that although U[x] now appears to be a smooth function (with x between -2 and 4), the combination of interest to me is Transpose[U[x][[1;;3]].Conjugate[U[x][[1;;3]]]. This still contains noisy entries but I think this may be a separate issue altogether now so perhaps I should accept the answer. – KMoff Nov 28 '16 at 12:15\n\nThe problem, as I see it, is that three of the singular values (and eigenvalues) of n[x] are zero, or equivalently, that the 5x5 matrix has rank 2. That means that u[x] and v[x] will carry the first two standard basis vectors to a basis for the column space and orthogonal complement of the null space, respectively. The column vectors of u[x] and v[x] are called the left and right singular vectors, respectively.\n\nWhen the singular values are distinct and ordered in the traditional decreasing order, the singular vectors are unique up to scalar multiple of absolute value 1. It may be possible to construct the functions u[x] and v[x] to be at least piecewise continuous (by imposing some arbitrary condition, such as some entry in one of the matrices that is generically nonzero be real and nonnegative).\n\nWhen singular values (or zero eigenvalues) are repeated, then the singular vectors corresponding to a given repeated value need to span a subspace of dimension higher than 1. This makes it harder to consistently choose a basis that varies continuously with the parameter x. When a zero eigenvalue is repeated, as in the OP's case, then the OP is probably right that numerical noise exacerbates the problem. See the last to figures below.\n\nIn the OP's example, there are two (nonzero) singular values. These are distinct except for x == 1. The first two columns of u[x] and v[x] vary continuously, except at x == 1. Note however, that the subspaces should be varying continuously with x, I think, although I did not investigate that hypothesis. It seems obvious from the results that SingularValueDecomposition does not yield a continuous parametrization in the OP's form, although @Anton Antonov's symbolic approach seemed to work for me (but not the OP, per OP's comment). Perhaps that will be resolved.\n\nImages, worked out with the original box1[x]. The vector components plotted are flatten out, so that the two singular vectors yield ten components and thus ten plotted curves.\n\nBlock[{f},\nf[x_?NumericQ] := Diagonal[box1[x][]];\nListLinePlot[\nTranspose@Table[Thread@{x, f[x]}, {x, -2, 4, 0.01}],\nPlotLabel -> \"Singular values\"]\n]", null, "Block[{f},\nf[x_?NumericQ] := Flatten[box1[x][[1, All, {1, 2}]]];\nModule[{n = 1},\nPlot[f[x], {x, -2, 4},\nPlotLabel -> \"First two left singular vector components\"] /.\nl_Line :> {ColorData[97, n++], l}]\n]", null, "Block[{f},\nf[x_?NumericQ] := Flatten[box1[x][[3, All, {1, 2}]]];\nModule[{n = 1},\nPlot[f[x], {x, -2, 4},\nPlotLabel -> \"First two right singular vector components\"] /.\nl_Line :> {ColorData[97, n++], l}]\n]", null, "Block[{f},\nf[x_?NumericQ] := Flatten[box1[x][[3, All, {3, 4, 5}]]];\nModule[{n = 1},\nPlot[Re@f[x], {x, -2, 4},\nPlotLabel -> \"Last three right singular vector components\"] /.\nl_Line :> {ColorData[97, n++], l}]\n]", null, "DiscretePlot[Im@Det@box1[x][], {x, -1, 4, 0.01}, Frame -> True,\nPlotLabel -> \"Imaginary part of Det[u[x]] varies wildly\"]", null, "• Thanks very much for this very informative answer. I have one question - when you say \"It may be possible to construct the functions u[x] and v[x] to be at least piecewise continuous (by imposing some arbitrary condition, such as some entry in one of the matrices that is generically nonzero be real and nonnegative).\" what did you have in mind exactly? – KMoff Nov 30 '16 at 18:44" ]

[ null, "https://i.stack.imgur.com/nBJKj.png", null, "https://i.stack.imgur.com/rT4NU.png", null, "https://i.stack.imgur.com/C92ul.png", null, "https://i.stack.imgur.com/Bszfu.png", null, "https://i.stack.imgur.com/bhQ6L.png", null, "https://i.stack.imgur.com/zrApG.png", null, "https://i.stack.imgur.com/o1dkT.png", null ]

[ "### Day 27 - Difference of Squares - 02.17.15\n\n• Unit 3 Test this Friday, February 20th!\n• Go over Quiz 3-1\n\nBell Ringer\n\n1. Simplify the following:", null, "1.", null, "2.", null, "3.", null, "4.", null, "5. none of the above\n\n2. Factor the following:\n\n1.", null, "2.", null, "3.", null, "4. all of the above\n\n5. none of the above\n\n3. Factor the following:", null, "1.", null, "2.", null, "3.", null, "4. all of the above\n\n5. none of the above\n\n4. Factor the following:", null, "1.", null, "2.", null, "3.", null, "4. all of the above\n\n5. none of the above\n\nReview\n• Prerequisites\n• Prime Numbers\n• Composite Numbers\n• Factor\n• Distributive Property\n• Factoring (video 1) (video 2)\n• Classifying Numbers as Prime or Composite\n• Prime Factorization of a Positive Integer\n• Prime Factorization of a Negative Integer\n• Prime Factorization of a Monomial\n• Finding the GCF of an Integer\n• Finding the GCF of a Monomial\n• Apply Factoring to Solve Problems\n• Factoring and Distributive Property (video)\n• Zero Product Property (video)\n• Factor polynomials using the Distributive Property\n• Factor by grouping\n• Solve quadratic equations of the following form:\n• Factoring Trinomials\n• Factor trinomials of the form\n• Solve equations of the form\n• Factor trinomials of the form\n• Solve equations of the form\n\nLesson\n\nExit Ticket\n• Posted on board at the end of the block\nLesson Objective(s)\n• How can factoring the difference of squares be used to solve equations?\nSkills\n• Factor using difference of squares\n• Solve equations using graphing\n\n#### In-Class Help Requests\n\nStandard(s)\n• CC.9-12.A.SSE.1a Interpret parts of an expression, such as terms, factors, and coefficients.\n\nMathematical Practice(s)\n• #1 - Make sense of problems and persevere in solving them\n• #2 - Reason abstractly and quantitatively\n• #7 - Look for and make use of structure\n\nPast Checkpoints\n\nUnit Skills\n1. Classifying Numbers as Prime or Composite\n2. Prime Factorization of a Positive Integer\n3. Prime Factorization of a Negative Integer\n4. Prime Factorization of a Monomial\n5. Finding the GCF of an Integer\n6. Finding the GCF of a Monomial\n7. Factor Applications\n8. Factor polynomials using the Distributive Property\n9. Factor by grouping\n10. Solve quadratic equations of the following form:\n11. Solve factored equations using Zero Product Property\n12. Factor trinomials of the form\n13. Solve equations of the form\n14. Factor trinomials of the form\n15. Solve equations of the form\n16. Factor using difference of squares\n17. Solve equations using graphing" ]

[ null, "https://lh5.googleusercontent.com/mNF1bzQJSN5PqOuFThFASjo5937JfqYOC2jmdhZ6xVuDwnVlzE34sUyPTU7rxbo00LGX9hIsnZblZ4ZkKgA1WwVncBVLSgVNVR7S17v0k2ccadv4IgIidefMB1-U9aKuN90", null, "https://lh4.googleusercontent.com/lT-mq_7pzzHm9nKrwEFbTshCPmhYrXBWlC4U0rZAoSI9IeS_RZBjh2BGX6KQJc10yIpTV5-aIMaj8pUljgOZ5BuRC-nuvFT2_y1awy0SF-cJKe-ShWlfn1T5BfsYnY7GYD0", null, "https://lh6.googleusercontent.com/Ce6vpsKp4oPA55pUxi_QuGY0qEFh6ryiP0r_-AaQHUGw2qznsKJa0CjNsbEaYyLXHwFVaTJYA9CwPRjt6-IjWwA9uk8W4nxLWSscJcrk08MConwGvryFKKQQxeTI6Qtyy-4", null, "https://lh3.googleusercontent.com/Sf7NRvxaC7MXXYHoD55rXQeLcrDXAcOmrUfJ4hVdI6hdciQSL9Cyw_Y2aXF9i6knRL4GdWAN5uVhg6G29mVLKHsyRRlBOzBuaJ2zRsBtrpfROXa5QPJoL3vYc1Dc3y6Kc38", null, "https://lh4.googleusercontent.com/mwJWmPB1UYpGFHEMikS8iVbx61JpG4yP5ng2uIbtbQG2sGkCWra9LntlbX65K-koyBq7rjF5oW39Y0QmrneOmDkg8tANmnRxSMGIKmz5-MAOgeYSS0A9kigb0K-DRNUu3Jk", null, "https://lh5.googleusercontent.com/sV0dOc5hpnE_4lUCFjbI9Wa9Qub5bmhn_xxXAZaE8By30SLVKMSb612Yp21e95OOSe60DCgjFz1PiwZ_m0AgaW7Lv0ALfiZzCTUhzf3rcpaT39e5bEdjlGT3YQica56cVw0", null, "https://lh5.googleusercontent.com/MjqBuubtB8_zXm_HyAEiXh10Lq6wCHpCwtY0wX299FxcGoxFEZ-g3SF09UEPeB8dhRzYH83xItNhrhchdoTNjBqzDjYLQ9TymSZ4teRoauCnRslvCa1xWMbb1aYceAukZH4", null, "https://lh5.googleusercontent.com/0PXI4FDRFd_Kcu6itxuqHwVR-oC-5h_-b_vsz_1Ig_jMRHO-Qvx1zEttkJYaOHF0Pocv7kDATbi83q7FBi5PND2FHSmT8ioqyu9gUKOs_Hi7RxRwfPsPWGetqHuQ3u4c7nE", null, "https://lh5.googleusercontent.com/dDfNIAhaHHbH4zYuBNJj47Mexl9igQKT5jl91n-hDjNSH9W3QhqoYoEjF0ppDicIkMW9vLGT-PU1c1Dg4IDUJSeI7YoNhfX--P9_SjiIYy-KRgGCAP4bY5Wu5Wcglm6obfs", null, "https://lh5.googleusercontent.com/sV0dOc5hpnE_4lUCFjbI9Wa9Qub5bmhn_xxXAZaE8By30SLVKMSb612Yp21e95OOSe60DCgjFz1PiwZ_m0AgaW7Lv0ALfiZzCTUhzf3rcpaT39e5bEdjlGT3YQica56cVw0", null, "https://lh5.googleusercontent.com/MjqBuubtB8_zXm_HyAEiXh10Lq6wCHpCwtY0wX299FxcGoxFEZ-g3SF09UEPeB8dhRzYH83xItNhrhchdoTNjBqzDjYLQ9TymSZ4teRoauCnRslvCa1xWMbb1aYceAukZH4", null, "https://lh5.googleusercontent.com/0PXI4FDRFd_Kcu6itxuqHwVR-oC-5h_-b_vsz_1Ig_jMRHO-Qvx1zEttkJYaOHF0Pocv7kDATbi83q7FBi5PND2FHSmT8ioqyu9gUKOs_Hi7RxRwfPsPWGetqHuQ3u4c7nE", null, "https://lh5.googleusercontent.com/a1LRXk0f-8P_yMktcBJtftU8bLyNDrBliu5hCCLFhBEFMOKJKxU4EnZRirk9F33vduqphooVFxSvh4iCSA24hWxz5tsS34bjDCVVsOu0lnalA27rpQYVYQpnqPf28z3Xe5o", null, "https://lh5.googleusercontent.com/sV0dOc5hpnE_4lUCFjbI9Wa9Qub5bmhn_xxXAZaE8By30SLVKMSb612Yp21e95OOSe60DCgjFz1PiwZ_m0AgaW7Lv0ALfiZzCTUhzf3rcpaT39e5bEdjlGT3YQica56cVw0", null, "https://lh5.googleusercontent.com/MjqBuubtB8_zXm_HyAEiXh10Lq6wCHpCwtY0wX299FxcGoxFEZ-g3SF09UEPeB8dhRzYH83xItNhrhchdoTNjBqzDjYLQ9TymSZ4teRoauCnRslvCa1xWMbb1aYceAukZH4", null, "https://lh5.googleusercontent.com/0PXI4FDRFd_Kcu6itxuqHwVR-oC-5h_-b_vsz_1Ig_jMRHO-Qvx1zEttkJYaOHF0Pocv7kDATbi83q7FBi5PND2FHSmT8ioqyu9gUKOs_Hi7RxRwfPsPWGetqHuQ3u4c7nE", null ]

[ "", null, "", null, "", null, "", null, "", null, "", null, "", null, "Re: import data problem\n\n• To: mathgroup at smc.vnet.net\n• Subject: [mg123253] Re: import data problem\n• From: Bill Rowe <readnews at sbcglobal.net>\n• Date: Tue, 29 Nov 2011 07:04:51 -0500 (EST)\n• Delivered-to: [email protected]\n\n```On 11/28/11 at 5:54 AM, hvdosten at gmail.com (Harry) wrote:\n\n>I don't understad what happens with my data...\n\n>\n>There are 401 lines of input data). The last lines are:\n\n>...\n>19.7999992 -9.13599968 19.8500004 -11.6960001 19.8999996 -11.9519997\n>19.9500008 -12.2080002 20. -12.9759998\n\n>Importing these data in Mathematica via\n\n>data = Import[\"00000.dat\", \"Table\"]\n\n>and everything seems to be ok. But if I try this:\n\n>y = LinearModelFit[data, {x, x^2}, x]\n\n>there is the error message: LinearModelFit::ivar: \"{20.05} is not a\n>valid variable\"\n\nThis error indicates you have assigned a value to x\n(specifically {20.05}). So, Mathematica evaluates\n\nLinearFit[data, {x, x^2}, x] as\n\nLinearFit[data, {{20.05}, {402.0025}}, {20.05}] which causes the\nerror. To fix this do one of the following\n\nClear[x];\ny = LinearModelFit[data, {x, x^2}, x]\n\nBlock[{x},y = LinearModelFit[data, {x, x^2}, x]]\n\nor start over with a fresh session.\n\n```\n\n• Prev by Date: Re: bug\n• Next by Date: Re: Problem with Patterns and Integrate\n• Previous by thread: Re: import data problem\n• Next by thread: Plotting being aborted with RegionBoundary" ]

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[ "# 1963 IMO Problems/Problem 2\n\n## Problem\n\nPoint", null, "$A$ and segment", null, "$BC$ are given. Determine the locus of points in space which are the vertices of right angles with one side passing through", null, "$A$, and the other side intersecting the segment", null, "$BC$.\n\n## Solution\n\nLet", null, "$\\omega_1$ be the circle with diameter", null, "$AB$, and let", null, "$\\omega_2$ be the circle with diameter", null, "$AC$. Then the locus is simply the set of points inside either", null, "$\\omega_1$ or", null, "$\\omega_2$, but not both.\n\nTo see this, suppose the right angle's ray that does not pass through", null, "$A$ intersects segment", null, "$BC$ at", null, "$X$. Then the right angle's vertex must lie on the circle with diameter", null, "$AX$. So, for a particular", null, "$X$, the desired locus is a circle with diameter", null, "$AX$. Accounting for all possible", null, "$X$, the total locus is the union of the circumferences of all circles that have a diameter", null, "$AX$, where", null, "$X$ is some point on", null, "$BC$.\n\nAs", null, "$X$ moves from", null, "$B$ to", null, "$C$, the motion of the circle with diameter", null, "$AX$ is continuous and fluid. Any point", null, "$P$ lying within", null, "$\\omega_1$ but outside", null, "$\\omega_2$ will eventually be intersected by this moving circle, since it went from inside to outside the circle. This applies similarly to all points inside", null, "$\\omega_2$ but outside", null, "$\\omega_1$. Also, the points inside both", null, "$\\omega_1$ and", null, "$\\omega_2$ are never intersected by this moving circle, as it always stays inside.\n\n(This proof sucks and needs some formalism)" ]

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[ "# On the application of Genetic Programming to the envelope reduction problem\n\nCreated by W.Langdon from gp-bibliography.bib Revision:1.4868\n\n@InProceedings{Koohestani:2012:CEEC,\n• author = \"Behrooz Koohestani and Riccardo Poli\",\n• title = \"On the application of Genetic Programming to the envelope reduction problem\",\n• booktitle = \"4th Computer Science and Electronic Engineering Conference\",\n• year = \"2012\",\n• editor = \"Maria Fasli\",\n• pages = \"53--58\",\n• address = \"University of Essex, UK\",\n• month = \"12-13 \" # sep,\n• publisher = \"IEEE\",\n• keywords = \"genetic algorithms, genetic programming, sparse matrix\",\n• DOI = \"", null, "doi:10.1109/CEEC.2012.6375378\",\n• size = \"6 pages\",\n• abstract = \"Large sparse matrices characterise the linear systems found in various scientific and engineering domains such as fluid mechanics, structural engineering, finite element analysis and network analysis. The ordering of the rows and columns of a matrix determines how close to the main diagonal its non-zero elements are, which in turn greatly influences the performance of solvers for the associated linear system. The reduction of the sum of the distance of non-zero elements from the matrix's main diagonal, a quantity known as envelope, is thus a key issue in many domains. Formally, the problem consists in finding a permutation of the rows and columns of a matrix which minimises its envelope. The problem is known to be NP-complete. A considerable number of methods have been proposed for reducing the envelope. These methods are mostly based on graph-theoretic concepts. While metaheuristic approaches are viable alternatives to classical optimisation techniques in a variety of domains, in the case of the envelope reduction problem, there has been a very limited exploration of such methods. In this paper, a Genetic Programming system capable of reducing the envelope of sparse matrices is presented. We evaluate our method on a set of standard benchmarks from the Harwell-Boeing sparse matrix collection against four state-of-the-art algorithms from the literature. The results obtained show that the proposed method compares very favourably with these algorithms.\",\n• notes = \"tree GP, ERP, permutation, math, matlab.\n\nCEEC 2012 http://www.essex.ac.uk/csee/research/ceec/\",\n\n}\n\nGenetic Programming entries for Behrooz Koohestani Riccardo Poli" ]

[ null, "http://gpbib.cs.ucl.ac.uk/doi.gif", null ]

[ "# VBScript Rnd Function\n\n❮ Complete VBScript Reference\n\nThe Rnd function returns a random number. The number is always less than 1 but greater or equal to 0.\n\n### Syntax\n\nRnd[(number)]\n\nParameter Description\nnumber Optional. A valid numeric expression\n\nIf number is:\n\n• <0 - Rnd returns the same number every time\n• >0 - Rnd returns the next random number in the sequence\n• =0 - Rnd returns the most recently generated number\n• Not supplied - Rnd returns the next random number in the sequence\n\n## Examples\n\n### Example 1\n\nA random number:\n\n<%\n\nresponse.write(Rnd)\n\n%>\n\nNote that you will get the same number every time. To avoid this, use the Randomize statement like in Example 2\n\nThe output of the code above will be:\n\n0.7055475\nShow Example »\n\n### Example 2\n\nTo avoid getting the same number every time, like in Example 1, use the Randomize statement:\n\n<%\n\nRandomize\nresponse.write(Rnd)\n\n%>\n\nThe output of the code above will be:\n\n0.4758112\nShow Example »\n\n### Example 3\n\nHere is how to produce random integers in a given range:\n\n<%\n\nDim max,min\nmax=100\nmin=1\nRandomize\nresponse.write(Int((max-min+1)*Rnd+min))\n\n%>\n\nThe output of the code above will be:\n\n71\nShow Example »\n\n❮ Complete VBScript Reference" ]

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[ "Related Articles\n\n# XOR of XORs of all sub-matrices\n\n• Difficulty Level : Hard\n• Last Updated : 27 Apr, 2021\n\nGiven a ‘N*N’ matrix, the task is to find the XOR of XORs of all possible sub-matrices.\nExamples:\n\n```Input :arr = {{3, 1},\n{1, 3}}\nOutput : 0\nExplanation: All the elements lie in 4 submatrices each.\n4 being even, there total contribution towards\nfinal answer becomes 0. Thus, ans = 0.\n\nInput : arr = {{6, 7, 13},\n{8, 3, 4},\n{9, 7, 6}};\nOutput : 4```\n\nA Simple Approach is to generate all the possible submatrices, find the XOR of each submatrice uniquely, and then XOR them all up. The time complexity of this approach will be O(n6).\nBetter solution: For each index(R, C), we will try to find the number of sub-matrices in which that index lies. If the number of sub-matrices is odd, then the final answer will be updated as ans = (ans ^ arr[R][C]). In case of even, we don’t need to update the answer. This works because a number XORed with itself gives zero and the order of operation doesn’t affect the final XOR value.\nAssuming 0-based indexing, the number of sub-matrices an index (R, C) lies in equals\n\n`(R + 1)*(C + 1)*(N - R)*(N - C)`\n\nBelow is the implementation of the above approach:\n\n## C++\n\n `// C++ program to find the XOR of XOR's of``// all submatrices` `#include ``using` `namespace` `std;` `#define n 3` `// Function to find to required``// XOR value``int` `submatrixXor(``int` `arr[][n])``{``    ``int` `ans = 0;` `    ``// Nested loop to find the``    ``// number of sub-matrix each``    ``// index belongs to``    ``for` `(``int` `i = 0; i < n; i++) {``        ``for` `(``int` `j = 0; j < n; j++) {``            ``// Number of ways to choose``            ``// from top-left elements``            ``int` `top_left = (i + 1) * (j + 1);` `            ``// Number of ways to choose``            ``// from bottom-right elements``            ``int` `bottom_right = (n - i) * (n - j);` `            ``if` `((top_left % 2 == 1) && (bottom_right % 2 == 1))``                ``ans = (ans ^ arr[i][j]);``        ``}``    ``}` `    ``return` `ans;``}` `// Driver Code``int` `main()``{``    ``int` `arr[][n] = { { 6, 7, 13 },``                     ``{ 8, 3, 4 },``                     ``{ 9, 7, 6 } };` `    ``cout << submatrixXor(arr);` `    ``return` `0;``}`\n\n## Java\n\n `//Java program to find the XOR of XOR's``// of all submatrices``class` `GFG``{``    ` `// Function to find to required``// XOR value``static` `int` `submatrixXor(``int``[][]arr)``{``    ``int` `n = ``3``;``    ``int` `ans = ``0``;` `    ``// Nested loop to find the``    ``// number of sub-matrix each``    ``// index belongs to``    ``for` `(``int` `i = ``0``; i < n; i++)``    ``{``        ``for` `(``int` `j = ``0``; j < n; j++)``        ``{``            ``// Number of ways to choose``            ``// from top-left elements``            ``int` `top_left = (i + ``1``) * (j + ``1``);` `            ``// Number of ways to choose``            ``// from bottom-right elements``            ``int` `bottom_right = (n - i) * (n - j);` `            ``if` `((top_left % ``2` `== ``1``) &&``                ``(bottom_right % ``2` `== ``1``))``                ``ans = (ans ^ arr[i][j]);``        ``}``    ``}` `    ``return` `ans;``}` `// Driver Code``public` `static` `void` `main(String[] args)``{``    ``int``[][] arr = {{ ``6``, ``7``, ``13``},``                   ``{ ``8``, ``3``, ``4` `},``                   ``{ ``9``, ``7``, ``6` `}};` `    ``System.out.println(submatrixXor(arr));``}``}` `// This code is contributed``// by Code_Mech.`\n\n## Python3\n\n `# Python3 program to find the XOR of``# XOR's of all submatrices` `# Function to find to required``# XOR value``def` `submatrixXor(arr, n):` `    ``ans ``=` `0` `    ``# Nested loop to find the``    ``# number of sub-matrix each``    ``# index belongs to``    ``for` `i ``in` `range``(``0``, n):``        ``for` `j ``in` `range``(``0``, n):` `            ``# Number of ways to choose``            ``# from top-left elements``            ``top_left ``=` `(i ``+` `1``) ``*` `(j ``+` `1``)` `            ``# Number of ways to choose``            ``# from bottom-right elements``            ``bottom_right ``=` `(n ``-` `i) ``*` `(n ``-` `j)``            ``if` `(top_left ``%` `2` `=``=` `1` `and``                ``bottom_right ``%` `2` `=``=` `1``):``                ``ans ``=` `(ans ^ arr[i][j])``    ``return` `ans` `# Driver code``n ``=` `3``arr ``=` `[[``6``, ``7``, ``13``],``       ``[``8``, ``3``, ``4``],``       ``[``9``, ``7``, ``6``]]``print``(submatrixXor(arr, n))` `# This code is contributed by Shrikant13`\n\n## C#\n\n `// C# program to find the XOR of XOR's``// of all submatrices``using` `System;` `class` `GFG``{``    ` `// Function to find to required``// XOR value``static` `int` `submatrixXor(``int` `[,]arr)``{``    ``int` `n = 3;``    ``int` `ans = 0;` `    ``// Nested loop to find the``    ``// number of sub-matrix each``    ``// index belongs to``    ``for` `(``int` `i = 0; i < n; i++)``    ``{``        ``for` `(``int` `j = 0; j < n; j++)``        ``{``            ``// Number of ways to choose``            ``// from top-left elements``            ``int` `top_left = (i + 1) * (j + 1);` `            ``// Number of ways to choose``            ``// from bottom-right elements``            ``int` `bottom_right = (n - i) * (n - j);` `            ``if` `((top_left % 2 == 1) &&``                ``(bottom_right % 2 == 1))``                ``ans = (ans ^ arr[i, j]);``        ``}``    ``}` `    ``return` `ans;``}` `// Driver Code``public` `static` `void` `Main()``{``    ``int` `[, ]arr = {{ 6, 7, 13},``                   ``{ 8, 3, 4 },``                   ``{ 9, 7, 6 }};` `    ``Console.Write(submatrixXor(arr));``}``}` `// This code is contributed``// by Akanksha Rai`\n\n## PHP\n\n ``\n\n## Javascript\n\n ``\nOutput:\n`4`\n\nTime Complexity: O(N2)\nAuxiliary Space: O(1)\n\nAttention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.  To complete your preparation from learning a language to DS Algo and many more,  please refer Complete Interview Preparation Course.\n\nIn case you wish to attend live classes with experts, please refer DSA Live Classes for Working Professionals and Competitive Programming Live for Students.\n\nMy Personal Notes arrow_drop_up" ]

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[ "## Conversion formula\n\nThe conversion factor from kilometers to decimeters is 10000, which means that 1 kilometer is equal to 10000 decimeters:\n\n1 km = 10000 dm\n\nTo convert 15.6 kilometers into decimeters we have to multiply 15.6 by the conversion factor in order to get the length amount from kilometers to decimeters. We can also form a simple proportion to calculate the result:\n\n1 km → 10000 dm\n\n15.6 km → L(dm)\n\nSolve the above proportion to obtain the length L in decimeters:\n\nL(dm) = 15.6 km × 10000 dm\n\nL(dm) = 156000 dm\n\nThe final result is:\n\n15.6 km → 156000 dm\n\nWe conclude that 15.6 kilometers is equivalent to 156000 decimeters:\n\n15.6 kilometers = 156000 decimeters\n\n## Alternative conversion\n\nWe can also convert by utilizing the inverse value of the conversion factor. In this case 1 decimeter is equal to 6.4102564102564E-6 × 15.6 kilometers.\n\nAnother way is saying that 15.6 kilometers is equal to 1 ÷ 6.4102564102564E-6 decimeters.\n\n## Approximate result\n\nFor practical purposes we can round our final result to an approximate numerical value. We can say that fifteen point six kilometers is approximately one hundred fifty-six thousand decimeters:\n\n15.6 km ≅ 156000 dm\n\nAn alternative is also that one decimeter is approximately zero times fifteen point six kilometers.\n\n## Conversion table\n\n### kilometers to decimeters chart\n\nFor quick reference purposes, below is the conversion table you can use to convert from kilometers to decimeters\n\nkilometers (km) decimeters (dm)\n16.6 kilometers 166000 decimeters\n17.6 kilometers 176000 decimeters\n18.6 kilometers 186000 decimeters\n19.6 kilometers 196000 decimeters\n20.6 kilometers 206000 decimeters\n21.6 kilometers 216000 decimeters\n22.6 kilometers 226000 decimeters\n23.6 kilometers 236000 decimeters\n24.6 kilometers 246000 decimeters\n25.6 kilometers 256000 decimeters" ]

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[ "#", null, "## 8.6 Array Methods\n\nIn the previous section we saw that--in Navigator 3.0 and Internet Explorer 3.0--arrays created with the Array() constructor have a length property. In Navigator 3.0, but not in IE 3.0, these arrays also support three methods that can be used to manipulate the array elements. These methods will be implemented in a future version of IE.\n\nThe Array.join() method converts all the elements of the array to a string, and concatenates them, separating them with an optionally specified string passed as an argument to the method. If no separator string is specified, then a comma is used. For example, the following lines of code produce the string \"1,2,3\":\n\n```a = new Array(1,2,3); // Create a new array with these three elements.\ns = a.join(); // s == \"1,2,3\"\n```\nAnd the following lines specify the optional separator to produce a slightly different result:\n\n```a = new Array(1,2,3);\ns = a.join(\", \"); // s == \"1, 2, 3\". Note the space after the comma.\n```\nIn some ways, the Array.join() method is the reverse of the String.split() method which creates an array by breaking a string up into pieces.\n\nThe Array.reverse() method reverses the order of the elements of an array. It does this \"in place\"--i.e., it doesn't create a new array with the elements rearranged, but instead rearranges them in the already existing array. For example, the following code, which uses the reverse() and the join() methods, produces the string \"3,2,1\":\n\n```a = new Array(1,2,3); // a = 1; a = 2; a = 3;\na.reverse(); // now a = 3; a = 2; a = 1;\ns = a.join() // s = \"3,2,1\"\n```\n\nThe final array method is Array.sort(), which sorts the elements of an array. Like the reverse() method, it does this \"in place\". When sort() is called with no arguments, it sorts the array elements in alphabetical order (temporarily converting them to strings, to perform the comparison, if necessary):\n\n```a = new Array(\"banana\", \"cherry\", \"apple\");\na.sort();\ns = a.join(\", \"); // s == \"apple, banana, cherry\".\n```\nYou can also pass an argument to the sort() method if you want to sort the array elements in some other order. To allow this method to be a fully general sorting algorithm, the optional argument should be a function. This function will be passed two arguments that it should compare. If the first argument should appear before the second in the sorted array, then the function should return a number less than zero. If the first argument should appear after the second in the sorted array, then the function should return a number greater than zero. And if the two values are equivalent (their order is irrelevant), then the function should return 0. So, for example, to sort array elements into numerical, rather than alphabetical order, you might do the following:\n\n```a = new Array(33, 4, 1111, 222);\na.sort(); // alphabetical order: 1111, 222, 33, 4\nfunction numberorder(a,b) {\nreturn a-b;\n}\na.sort(numberorder); // numerical order: 4, 33, 222, 1111\n```\nYou can probably think of other comparison functions that will sort numbers into various esoteric orders: reverse numerical order, odd numbers before even numbers, etc. The possibilities become more interesting, of course, when the elements you are comparing are objects rather than simple types like numbers or strings.", null, "", null, "", null, "Array Length Property", null, "Arrays in Navigator 2.0", null, "" ]

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[ "# Units of measure assets\n\nProducts can be sold, and inventory that is tracked, in various quantity units, such as kilograms, inches, liters. Of these units, products can be ordered in minimum quantities, and by multiples of specific quantities.\n\nThe controller commands use the UOM (unit of measure) to specify the quantity unit. If a UOM parameter is not specified, then the customer's specified quantity is multiplied by the nominal quantity of the catalog entry in the CATENTSHIP database table. The result is known as the requested quantity.\n\nThe requested quantity is rounded up to the next highest quantity multiple for the catalog entry. For example, if the multiple is 2 kilograms and the requested quantity is 4.1 kilograms, the result of the rounding would be 6 kilograms. The rounded quantity is used when inventory is checked, and this quantity has its own quantity unit. If the inventory quantity unit and the catalog entry quantity unit are different, there must be a conversion between the two units.\n\nWhen Available to Promise (ATP) inventory is enabled, the inventory quantity unit is defined in the QUANTITYMEASURE column of the BASEITEM table. Otherwise, it is defined in the QUANTITYMEASURE column of the INVENTORY table. The enabled inventory, such as ATP, is recorded in the INVENTORYSYSTEM column of the STORE table.\n\nThe rounded quantity that is divided by the nominal quantity of the catalog entry is known as the normalized quantity. The normalized quantity is stored in the order item or the interest item, depending on the command that is being run. For example, if the rounded quantity is 6 kilograms and the nominal quantity is 2 kilograms, then the normalized quantity is 3.\n\nThe requested quantity of a catalog entry can affect which offer gives the best price, and hence determines which offer is used. For example, if the rounded quantity is 6 kg and there are two offers:\n• Offer A: A price of \\$4.00 for the nominal quantity of 2 kilograms and a minimum quantity of 10 kilograms.\n• Offer B: A price of \\$4.50 for the nominal quantity of 2 kilograms and a minimum quantity of 2 kilograms.\nThe best price that is based on the requested quantity is the second offer, which is the offer used.\n\nThe following diagram illustrates the structure of units of measure in the WebSphere Commerce Server:", null, "Quantity unit and quantity unit format\n\nA quantity unit is the unit of measurement that is used in the store. For example, kilograms, pounds, meters, inches, liters. The quantity unit format is how this quantity unit is formatted in the store. For example, how many decimal places are used when the quantity unit displays.\n\nEach quantity unit format is part of only one store entity, but each store entity can have several quantity unit formats.\n\nA quantity unit format can exist for each quantity unit and number usage. The quantity unit can have one or more quantity unit format descriptions, depending on how many languages the store supports.", null, "Quantity units that are defined in one store can be used by other stores. In order for one store to use quantity units that are defined in another store a store relationship of type com.ibm.commerce.measurement.format must be created between the stores. For more information, see Relationships between stores.\n\nQuantity unit format description\nA quantity unit format description describes how to format (for display purposes) a quantity amount in a particular quantity unit, in a particular language.\nNumber usage\n\nNumber usage defines the way that a number is used in an application. For example, by using number usage codes in your WebSphere Commerce code, you can choose the way that you would like that number (currency or quantity) to be formatted or rounded. These codes (defined in the NUMBRUSG table) allow the number to be formatted according to the rules specified for that type of number usage in the CURFORMAT, CURFMTDESC, QTYFORMAT, and QTYFMTDESC tables. This formatting option allows stores to format numbers in different ways to meet the requirements of various situations." ]

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[ "# Search Our Content Library\n\n24 filtered results\n24 filtered results\nMeasurement\n4.MD.A\nSort by\nMeasuring Cup Musings\nWorksheet\nMeasuring Cup Musings\nStudents use a diagram of a measuring cup to answer six questions about measurement conversions between cups and ounces.\nMath\nWorksheet\nLinear Measurement\nWorksheet\nLinear Measurement\nHere's multiplication practice with real world applications! Plus, memorizing this linear measurement chart will help your kid keep track of conversions.\nMath\nWorksheet\nLiquid Measurement Conversion\nWorksheet\nLiquid Measurement Conversion\nGet every drop of basic liquid measurement conversion! The table comes in handy for quick converting later on, too.\nMath\nWorksheet\nMetric Mass and Volume Measurement in Word Problems\nWorksheet\nMetric Mass and Volume Measurement in Word Problems\nKids convert metric mass and volume with these music-themed word problems.\nMath\nWorksheet\nConvert Pounds to Ounces\nWorksheet\nConvert Pounds to Ounces\nLearn to convert ounces into pounds (and pounds to ounces) in this fun Easter worksheet.\nMath\nWorksheet\nUnits of Measurement\nWorksheet\nUnits of Measurement\nThis quiz-style worksheet is a great introduction to different units of measurement and what they are used for.\nMath\nWorksheet\nLiquid Volume: Gallon Chart\nWorksheet\nLiquid Volume: Gallon Chart\nUse this worksheet to help with volume estimations, or conversions between measurements.\nMath\nWorksheet\nLiquid Measurement Word Problem\nWorksheet\nLiquid Measurement Word Problem\nThis liquid measurement word problem dives into converting liquid measurements, and multiplying! This worksheet includes a liquid conversion chart.\nMath\nWorksheet\nConverting Hours to Seconds\nWorksheet\nConverting Hours to Seconds\nConterting hours to seconds or seconds to minutes can be a case of doing mental math by memorizing quick facts, like that 30 minutes is 1800 seconds.&hellip;\nMath\nWorksheet\nLiquid Volume Equivalents #1\nWorksheet\nLiquid Volume Equivalents #1\nThis worksheet will provide your students with practice finding the equivalent values of liquid volumes.\nMath\nWorksheet\nHalloween Timeline #2\nWorksheet\nHalloween Timeline #2\nIn this math worksheet, children will help reconstruct a Halloween timeline for a ghost as she recaps her trip to the cemetery.\nMath\nWorksheet\nLiquid Volume Equivalents #2\nWorksheet\nLiquid Volume Equivalents #2\nThis worksheet will test your students’ understanding of liquid volumes in U.S. customary measurements.\nMath\nWorksheet\nGlossary: Area Arrangements\nWorksheet\nGlossary: Area Arrangements\nUse this glossary with the EL Support Lesson: Area Arrangements.\nMath\nWorksheet\nVocabulary Cards: Area Arrangements\nWorksheet\nVocabulary Cards: Area Arrangements\nUse these vocabulary cards with the EL Support Lesson: Area Arrangements.\nMath\nWorksheet\nVolume Calculations: Dragon Punch #1\nWorksheet\nVolume Calculations: Dragon Punch #1\nStudents will make conversions from larger units to smaller units of volume measurements.\nMath\nWorksheet\nLiquid Volume Logic\nWorksheet\nLiquid Volume Logic\nUse this worksheet to assess your students’ understanding of units of measurement for liquid volume. Ask your students to determine the best unit of measurement for each item.\nMath\nWorksheet\nWorksheet\nDo some lemonade stand math in this refreshing division worksheet. When your fourth grader has finished the problems, he can relax with a glass of lemonade.\nMath\nWorksheet\nConverting Yards to Feet\nWorksheet\nConverting Yards to Feet\nConverting yards to feet never got so much practice as this printable puts up! Practice converting yards to feet, feet to yards, and even miles.\nMath\nWorksheet\nWorksheet\nMath\nWorksheet\nWorksheet\nLemonade is sweet, but acing these division problems will be even sweeter!\nMath\nWorksheet\nConvert Yards to Miles\nWorksheet\nConvert Yards to Miles\nHow many feet are in a mile? How many yards are in a mile? Convert yards to miles and more with this worksheet!\nMath\nWorksheet\nWorksheet\nLemonade is sweet, but acing these division worksheets is even sweeter!\nMath\nWorksheet\nConvert Feet to Yards\nWorksheet\nConvert Feet to Yards\nConvert feet to yards, inches to feet and back again. Filling in this chart will come in handy to answer the practice questions and memorizing the&hellip;" ]

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[ "# 5 and 6-Digit Numbers\n\nWe have already learnt numbers upto 9999. The greatest 4-digit number is 9999. If we add 1 to 9999 we get 10000 the smallest 5-digit number. We read 10000 as ten thousand. The greatest 5-digit number is 99999. Let us form 10000 with blocks of thousand each as shown below.\n\nLet us represents 10000 on the abacus.\n\nWe can show 10000 on an abacus with five vertical rods. The rods represent place value of ten thousands, thousands, hundreds, tens and ones. We add beads on the rods to show different numbers. Each rod can hold up to 9 beads.\n\nThis shows 9 hundreds 9 tens and 9 ones. The number is 999.\n\nIf we add 1 more to 999 it is 10 hundreds. Regroup 10 hundreds as 1 thousand. 1 bead in the thousands place represents 1 thousand.\n\nThis shows 9 thousands 9 hundreds 9 tens and 9 ones. The number is 9999.\n\nIf we add 1 more to 9999 it is 10 thousands. Regroup 10 thousands as 1 ten thousand (TTh).\n\nLet us represent a 5-digit number on the abacus.\n\n1. Consider the number 13460. We can expand 13460 as shown here.\n\nWe read 13460 as thirteen thousand four hundred sixty.\n\nWe know the greatest 5-digit number is 99999, if we add 1 to 99999 we get 100000 the smallest 6-digit number. We read 100000 as one lakh. Lakh is also written as lac.\n\nThe sixth place from the right in the place value chart is lahks place.\n\nThe greatest 6-digit number is 999999.\n\n2. Consider the number 976425. We can expand 976425 as shown here.\n\nWe read 976425 as nine lakh seventy-six thousand four hundred twenty-five.\n\nQuestions and Answers on 5 and 6-Digit Numbers:\n\n1. Complete the following table.\n\n (i) 10000 1   ten thousand Ten thousand (ii) 20000 _____ ten thousand _______________ (iii) 30000 _____ ten thousand _______________ (iv) 40000 _____ ten thousand _______________ (v) 50000 _____ ten thousand _______________ (vi) 60000 _____ ten thousand _______________ (vii) 70000 _____ ten thousand _______________ (viii) 80000 _____ ten thousand _______________ (ix) 90000 _____ ten thousand _______________ (x) 100000 _____ ten thousand _______________\n\n(ii) 2, twenty thousand\n\n(iii) 3, thirty thousand\n\n(iv) 4, forty thousand\n\n(v) 5, fifty thousand\n\n(vi) 6, sixty thousand\n\n(vii) 7, seventy thousand\n\n(viii) 8, eighty thousand\n\n(ix) 9, ninety thousand\n\n(x) 10, one lakh\n\n2. Read the abacus and find the number.\n\n(i)\n\n(ii)\n\n(i) 48,407\n\n(ii) 7,64,912" ]

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[ "# If $xyz=1$, prove $\\frac{1}{y(x+y)}+\\frac{1}{z(y+z)}+\\frac{1}{x(z+x)} \\geqslant \\frac{3}{2}$\n\n$x,y,z$ are positive real numbers such that $$xyz=1$$ Prove that $\\dfrac{1}{y(x+y)}+\\dfrac{1}{z(y+z)}+\\dfrac{1}{x(z+x)} \\geqslant \\dfrac{3}{2}$. I have no idea how to solve this problem. I've tried Engel form of Cauchy inequality, but I get $x^2+y^2+z^2$ in denominator, and from condition $xyz=1$ I only find $x^2+y^2+z^2\\geq3..$\n\n• Engel form of cauchy inequality,but i get $x^2+y^2+z^2$ in denominator,and from condition $xyz=1$ I only find $x^2+y^2+z^2 \\geqslant 3$... – chaos Sep 19 '14 at 15:35\n• – Martin Sleziak May 22 '18 at 9:18\n\n## 1 Answer\n\n$$\\dfrac{1}{y(x+y)}+\\dfrac{1}{z(y+z)}+\\dfrac{1}{x(z+x)} \\geqslant \\dfrac{3}{2}$$\n\nSince $xyz=1$ we substitute, $x=\\frac{a}{b} , y= \\frac{b}{c} , z= \\frac{c}{a}$\n\nNow, by Engel form(also called Titu lemma)\n\n$$\\sum_{cyc}\\frac {a^2}{c^2 + ab }\\ge\\frac {(a^2 + b^2 +c^2)^2}{a^2b^2 + b^2c^2 + c^2a^2+a^3b + b^3c + c^3a}$$\n\nSo it remains to prove\n\n$$2(a^2 +b^2+c^2)^2\\ge 3\\sum_{cyc}a^2b^2 + 3\\sum_{cyc}a^3b$$ $$2a^4+2b^4+2c^4 +a^2b^2+b^2c^2+a^2c^2 \\ge 3(a^3b+b^3c+c^3a)$$ Which is obvious from,\n\n$$a^4 + b^4 +c^4 \\ge a^3b + b^3c + c^3a$$\n\n$$\\sum_{cyc}(a^4 + a^2b^2)\\ge \\sum_{cyc}2a^3b$$\n\n• do you have an easier way? – RE60K Sep 20 '14 at 15:16\n• By rearrangement inequality :) – Shivang jindal Sep 20 '14 at 18:58" ]

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[ "Changing variable in complex variable integrals\n\nIntegration in complex analysis does not always seem as obvious at in the real plane, with reasons that are not obvious to me. In particular, consider $$\\int_{- \\infty}^{+ \\infty} e^{- \\pi x^2 z} dx$$\n\nwhere $$z$$ is a complex number of positive real part, so that the integral converges. The value of this integral is $$z^{-1/2} \\int_{- \\infty}^{+ \\infty} e^{-\\pi x^2} dx = z^{-1/2}$$\n\nand that seems to be like a standard change of variables as it would be for $$z$$ real. However, it is not and the argument I read for computing this integral involves Cauchy's contour theorem and turning the path of integration by an angle of $$-arg(z)/2$$. I do not understand:\n\n• why isn't the change of variable (which is a \"linear\" one) possible, even if $$z$$ is complex ?\n• how does the contour changing work ? (I understand it in the case of horizontal translation, but changing the angle could create convergence issues, isn't it?)\n• The original integral is over the real line. If you make the change of variable $y=x\\sqrt z$ the new variable is not a real variable. – Kavi Rama Murthy Feb 25 at 0:27\n• Complex integrals are path integrals so a change of variables needs to preserve the path and in your example the straight change moves from the reals to a different line in the complex plane (for real integrals there is only one path so to speak up to orientation so a change of variable is just a reparametrization) – Conrad Feb 25 at 0:39\n• Contour integration uses that the integral of an analytic function on a loop is zero, so you make a loop from the reals to the complex line obtained by turning by $z$ - technically you truncate at a large $R$ and join the segments by another curve, generally the arc of radius $R$ and show that the integral on that arc goes to zero as $R$ goes to infinity. Of course there are many variations and the ingenuity is in finding both the right analytic function and the right closed curve – Conrad Feb 25 at 0:45\n\nActually, there's another way to prove this. Suppose you have a formula for an integral, say $$F(z) = \\int_{-\\infty}^\\infty f(x,z)\\; dx$$ which is true for $$z$$ in some interval $$J$$, and there is a connected open subset $$U$$ of\n$$\\mathbb C$$ containing $$J$$ such that both $$F(z)$$ and $$\\int_{-\\infty}^\\infty f(x,z)\\;dx$$ are analytic in $$U$$. Then the formula must be true throughout $$U$$.\nIn your example, $$z^{-1/2}$$ is analytic in the right half plane, while the integral is also analytic there because of locally uniform absolute convergence." ]

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[ "We use cookies to ensure you get the best experience on our website. By using the IFPA wiki, you consent to our use of cookies.\n\nSpecial relativity\n\nFrom IFPA wiki\n\n1. Introduction:\n\n• Electromagnetism and Lorentz transformations\n• Lorentz transformation vs Newtonian mechanics\n\n2. Kinematics\n\n• Simultaneity, Lorentz transformations in mechanics\n• Graphical representation of the standard Lorentz transformation\n\n3. Kinematics 2\n\n• Length contraction, time dilation, paradoxes\n• Transformation of velocity and acceleration\n• Spacetime and 4-vectors: 4-velocity and 4-acceleration\n\n5. Dynamics\n\n• The equivalence of mass and energy\n• Particles and Waves\n• Four-force and three-force\n\n6. Electromagnetism 1\n\n• Four-Tensors ==\n• Tensor algebra and differentiation\n• The metric tensor\n• Maxwell's Theory in Tensor Form\n\n8. Electromagnetism 2\n\n• The 4-potential\n• Transformation of E and B\n• The electromagnetic energy tensor\n\n9. Road to General relativity\n\n• The mechanical energy tensor\n• Accelerated frames and the Rindler coordinates" ]

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[ "15 Dec 2020, 10:57\n\n## Hyperparameter Optimization Techniques to Improve Your Machine Learning Model's Performance\n\nHyperparameter optimisation on your selected model is an important step in the machine learning project pipeline. This task always comes after the model selection process, where you choose the model that is performing better than other models.\n\n### What is hyperparameter optimisation?\n\nBefore I define hyperparameter optimisation, you need to understand what a hyperparameter is.\n\nIn short, hyperparameters are different parameter values that are used to control the learning process and have a significant effect on the performance of machine learning models.\n\nAn example of hyperparameters in the Random Forest algorithm is the number of estimators (n_estimators), maximum depth (max_depth), and criterion. These parameters are tunable and can directly affect how well a model trains.\n\nSo then hyperparameter optimisation is the process of finding the right combination of hyperparameter values to achieve maximum performance on the data in a reasonable amount of time.\n\nThis process plays a vital role in the prediction accuracy of a machine learning algorithm. Therefore hyperparameter optimisation is considered the trickiest part of building machine learning models.\n\nMost of these machine learning algorithms come with the default values of their hyperparameters. But the default values do not always perform well on different types of machine learning projects. This is why you need to optimise them in order to get the right combination that will give you the best performance.\n\nA good choice of hyperparameters can really make an algorithm shine.\n\nThere are some common strategies for optimizing hyperparameters. Let's look at each in detail now.\n\n### How to optimize hyperparameters\n\nGrid search\n\nThis is a widely used and traditional method that performs hyperparameter tuning to determine the optimal values for a given model. Grid search works by trying every possible combination of parameters you want to try in your model. This means it will take a lot of time to perform the entire search which can get very computationally expensive.\n\nRandom search\n\nThis method works a bit differently: random combinations of the values of the hyperparameters are used to find the best solution for the built model. The drawback of random search is that it can sometimes miss important points (values) in the search space.\n\n### Alternative hyperparameter optimisation techniques\n\nNow I will introduce you to a few alternative and advanced hyperparameter optimization techniques/methods. These can help you to obtain the best parameters for a given model.\n\nWe will look at the following techniques:\n\n1. Hyperopt\n2. Scikit Optimize\n3. Optuna\n\n### 1. Hyperopt\n\nHyperopt is a powerful Python library for hyperparameter optimisation developed by James Bergstra.\n\nIt uses a form of Bayesian optimisation for parameter tuning that allows you to get the best parameters for a given model. It can optimize a model with hundreds of parameters on a large scale.\n\nHyperopt has four important features you need to know in order to run your first optimization:\n\nSearch space\n\nHyperopt has different functions to specify ranges for input parameters. These are called stochastic search spaces. The most common options for a search space are:\n\n• hp.choice(label, options) – This can be used for categorical parameters. It returns one of the options, which should be a list or tuple. Example: hp.choice(\"criterion\", [\"gini\",\"entropy\",])\n• hp.randint(label, upper) – This can be used for Integer parameters. It returns a random integer in the range (0, upper). Example: hp.randint(\"max_features\",50)\n• hp.uniform(label, low, high) – This returns a value uniformly between low and high. Example: hp.uniform(\"max_leaf_nodes\",1,10)\n\nOther option you can use are:\n\n• hp.normal(label, mu, sigma) –This returns a real value that's normally distributed with mean mu and standard deviation sigma\n• hp.qnormal(label, mu, sigma, q) – This returns a value like round(normal(mu, sigma) / q) * q\n• hp.lognormal(label, mu, sigma) – This returns a value drawn according to exp(normal(mu, sigma))\n• hp.qlognormal(label, mu, sigma, q) – This returns a value like round(exp(normal(mu, sigma)) / q) * q\n\nJust a quick note: Every optimisable stochastic expression has a label (for example, n_estimators) as the first argument. These labels are used to return parameter choices to the caller during the optimisation process.\n\nObjective Function\n\nThis is a minimisation function that receives hyperparameter values as input from the search space and returns the loss.\n\nThis means that during the optimisation process, we train the model with selected haypeparameter values and predict the target feature. Then we evaluate the prediction error and give it back to the optimiser.\n\nThe optimiser will decide which values to check and iterate again. You will learn how to create objective functions in the practical example.\n\nfmin\n\nThe fmin function is the optimization function that iterates on different sets of algorithms and their hyperperameters and then minimises the objective function.\n\nfmin takes five inputs, which are:\n\n• The objective function to minimise\n• The defined search space\n• The search algorithm to use, such as Random search, TPE (Tree Parzen Estimators) and Adaptive TPE. Note: hyperopt.rand.suggest and hyperopt.tpe.suggest provide logic for sequential search of the hyperparameter space\n• Maximum number of evaluations\n• And the trials object (optional)\n\nExample:\n\n`from hyperopt import fmin, tpe, hp,Trials`\n`trials = Trials()`\n`best = fmin(fn=lambda x: x ** 2,`\n` \t\tspace= hp.uniform('x', -10, 10),`\n` \t\talgo=tpe.suggest,`\n` \t\tmax_evals=50,`\n` \t\ttrials = trials)`\n```print(best)\n\nTrials Object```\n\nThe Trials object is used to keep all hyperparameters, loss, and other information. This means you can access it after running the optimisation.\n\n`from hyperopt import Trials `\n```trials = Trials()\n```\n\nNow that you understand the important features of Hyperopt, we'll see how to use it. You'll follow these steps:\n\n• Initialize the space over which to search\n• Define the objective function\n• Select the search algorithm to use\n• Run the hyperopt function\n• Analyse the evaluations outputs stored in the trials object\n\nHyperpot in Practice\n\nIn this practical example, we will use the Mobile Price Dataset. Our task is to create a model that will predict how high the price of a mobile device will be: 0 (low cost), 1 (medium cost), 2 (high cost), or 3 (very high cost).\n\nInstall Hyperopt\n\nYou can install hyperopt from PyPI by running this command:\n\n```pip install hyperopt\n```\n\nThen import the following important packages, including hyperopt:\n\n`# import packages `\n`import numpy as np `\n`import pandas as pd `\n`from sklearn.ensemble import RandomForestClassifier `\n`from sklearn import metrics`\n`from sklearn.model_selection import cross_val_score`\n`from sklearn.preprocessing import StandardScaler `\n`from hyperopt import tpe, hp, fmin, STATUS_OK,Trials`\n`from hyperopt.pyll.base import scope`\n\n`import warnings`\n```warnings.filterwarnings(\"ignore\")\n```\n\nDataset\n\n`# load data `\n```data = pd.read_csv(\"data/mobile_price_data.csv\")\n```\n\nCheck the first five rows of the dataset like this:\n\n`#read data `", null, "As you can see, in our dataset we have different features with numerical values.\n\nLet's look at the shape of the dataset.\n\n`#show shape`\n`data.shape`\n\nWe get the following:\n\n(2000, 21)\n\nIn this dataset we have 2000 rows and 21 columns. Now let's understand the list of features we have in this dataset.\n\n`#show list of columns `\n`list(data.columns)`\n\n['battery_power', 'blue', 'clock_speed', 'dual_sim', 'fc', 'four_g', 'int_memory', 'm_dep', 'mobile_wt', 'n_cores', 'pc', 'px_height', 'px_width', 'ram', 'sc_h', 'sc_w', 'talk_time', 'three_g', 'touch_screen', 'wifi', 'price_range']\n\nYou can find the meaning of each column name here.\n\nSplitting the dataset into target feature and independent features\n\nThis is a classification problem. So we will now split the target feature and independent features from the dataset. Our target feature is price_range.\n\n`# split data into features and target `\n`X = data.drop(\"price_range\", axis=1).values `\n`y = data.price_range.values`\n\nPreprocessing the Dataset\n\nNext, we'll standardize the independent features by using the StandardScaler method from scikit-learn.\n\n`# standardize the feature variables `\n`scaler = StandardScaler()`\n`X_scaled = scaler.fit_transform(X)`\n\nDefine Parameter Space for Optimisation\n\nWe will use three hyperparameter of the Random Forest algorithm: n_estimators, max_depth, and criterion.\n\n`space = {`\n` \"n_estimators\": hp.choice(\"n_estimators\", [100, 200, 300, 400,500,600]),`\n` \"max_depth\": hp.quniform(\"max_depth\", 1, 15,1),`\n` \"criterion\": hp.choice(\"criterion\", [\"gini\", \"entropy\"]),`\n`}`\n\nWe have set different values in the above selected hyperparameters. Now we will define the objective function.\n\nDefining a function to minimise (Objective Function)\n\nOur function that we want to minimize is called hyperparamter_tuning. The classification algorithm to optimise its hyperparameter is Random Forest.\n\nI use cross validation to avoid overfitting and then the function will return a loss values and its status.\n\n`# define objective function`\n`def hyperparameter_tuning(params):`\n` clf = RandomForestClassifier(**params,n_jobs=-1)`\n` acc = cross_val_score(clf, X_scaled, y,scoring=\"accuracy\").mean()`\n` return {\"loss\": -acc, \"status\": STATUS_OK}`\n\nFine-tune the model\n\nFinally, first we'll instantiate the Trial object, fine tune the model, and then print the best loss with its hyperparamters values.\n\n`# Initialize trials object`\n`trials = Trials()`\n`best = fmin(`\n` fn=hyperparameter_tuning,`\n` space = space, `\n` algo=tpe.suggest, `\n` max_evals=100, `\n` trials=trials`\n`)`\n```print(\"Best: {}\".format(best))\n```", null, "After performing hyperparamter optimization, the loss is - 0.8915. This means that the model performance has an accuracy of 89.15% by using n_estimators = 300, max_depth = 11, and criterion = \"entropy\" in the Random Forest classifier.\n\nAnalyze the results by using the trials object\n\nThe trials object can help us inspect all of the return values that were calculated during the experiment.\n\n(a) trials.results\n\nThis show a list of dictionaries returned by 'objective' during the search.\n\n`trials.results`\n\n[{'loss': -0.8790000000000001, 'status': 'ok'}, {'loss': -0.877, 'status': 'ok'}, {'loss': -0.768, 'status': 'ok'}, {'loss': -0.8205, 'status': 'ok'}, {'loss': -0.8720000000000001, 'status': 'ok'}, {'loss': -0.883, 'status': 'ok'}, {'loss': -0.8554999999999999, 'status': 'ok'}, {'loss': -0.8789999999999999, 'status': 'ok'}, {'loss': -0.595, 'status': 'ok'},.......]\n\n(b) trials.losses()\n\nThis shows a list of losses (float for each 'ok' trial).\n\n`trials.losses()`\n\n[-0.8790000000000001, -0.877, -0.768, -0.8205, -0.8720000000000001, -0.883, -0.8554999999999999, -0.8789999999999999, -0.595, -0.8765000000000001, -0.877, .........]\n\n(c) trials.statuses()\n\nThis shows a list of status strings.\n\n`trials.statuses()`\n\n['ok', 'ok', 'ok', 'ok', 'ok', 'ok', 'ok', 'ok', 'ok', 'ok', 'ok', 'ok', 'ok', 'ok', 'ok', 'ok', 'ok', 'ok', 'ok', ..........]\n\nNote: This trials object can be saved, passed on to the built-in plotting routines, or analyzed with your own custom code.\n\nNow that you know how to implement Hyperopt, let's learn the second alternative hyperparameter optimization technique called Scikit-Optimize.\n\n### 2. Scikit-Optimize\n\nScikit-optimize is another open-source Python library for hyperparameter optimization. It implements several methods for sequential model-based optimization.\n\nThe library is very easy to use and provides a general toolkit for Bayesian optimization that can be used for hyperparameter tuning. It also provides support for tuning the hyperparameters of machine learning algorithms offered by the scikit-learn library.\n\nThe scikit-optimize is built on top of Scipy, NumPy, and Scikit-Learn.\n\nScikit-optimize has at least four important features you need to know in order to run your first optimization. Let's look at them in depth now.\n\nSpace\n\nscikit-optimize has different functions to define the optimization space which contains one or multiple dimensions. The most common options for a search space to choose are:\n\n• Real — This is a search space dimension that can take on any real value. You need to define the lower bound and upper bound and both are inclusive. Example: Real(low=0.2, high=0.9, name=\"min_samples_leaf\")\n• Integer — This is a search space dimension that can take on integer values. Example: Integer(low=3, high=25, name=\"max_features\")\n• Categorical — This is a search space dimension that can take on categorical values. Example: Categorical([\"gini\",\"entropy\"],name=\"criterion\")\n\nNote: in each search space you have to define the hyperparameter name to optimize by using the name argument.\n\nBayesSearchCV\n\nThe BayesSearchCV class provides an interface similar to GridSearchCV or RandomizedSearchCV but it performs Bayesian optimization over hyperparameters.\n\nBayesSearchCV implements a “fit” and a “score” method and other common methods like predict(),predict_proba(), decision_function(), transform() and inverse_transform() if they are implemented in the estimator used.\n\nIn contrast to GridSearchCV, not all parameter values are tried out. Rather a fixed number of parameter settings is sampled from the specified distributions. The number of parameter settings that are tried is given by n_iter.\n\nNote that you will learn how to implement BayesSearchCV in a practical example below.\n\nObjective Function\n\nThis is a function that will be called by the search procedure. It receives hyperparameter values as input from the search space and returns the loss (the lower the better).\n\nThis means that during the optimization process, we train the model with selected hyperparameter values and predict the target feature. Then we evaluate the prediction error and give it back to the optimizer.\n\nThe optimizer will decide which values to check and iterate over again. You will learn how to create an objective function in the practical example below.\n\nOptimizer\n\nThis is the function that performs the Bayesian Hyperparameter Optimization process. The optimisation function iterates at each model and the search space to optimize and then minimises the objective function.\n\nThere are different optimization functions provided by the scikit-optimize library, such as:\n\n• dummy_minimize — Random search by uniform sampling within the given bounds.\n• forest_minimize — Sequential optimization using decision trees.\n• gbrt_minimize — Sequential optimization using gradient boosted trees.\n• gp_minimize — Bayesian optimization using Gaussian Processes.\n\nNote: we will implement gp_minimize in the practical example below.\n\nOther features you should learn are as follow:\n\nScikit-optimize in Practice\n\nNow that you know the important features of scikit-optimize, let's look at a practical example. We will use the same dataset called Mobile Price Dataset that we used with Hyperopt.\n\nInstall Scikit-Optimize\n\nscikit-optimize requires the following Python version and packages:\n\n• Python >= 3.6\n• NumPy (>= 1.13.3)\n• SciPy (>= 0.19.1)\n• joblib (>= 0.11)\n• scikit-learn >= 0.20\n• matplotlib >= 2.0.0\n\nYou can install the latest release with this command:\n\n`pip install scikit-optimize`\n\nThen import important packages, including scikit-optimize:\n\n`# import packages`\n`import numpy as np`\n`import pandas as pd`\n`from sklearn.ensemble import RandomForestClassifier`\n`from sklearn import metrics`\n`from sklearn.model_selection import cross_val_score `\n`from sklearn.preprocessing import StandardScaler`\n`from skopt.searchcv import BayesSearchCV`\n`from skopt.space import Integer, Real, Categorical `\n`from skopt.utils import use_named_args`\n`from skopt import gp_minimize`\n\n`import warnings`\n\n`warnings.filterwarnings(\"ignore\")`\n\nThe first approach\n\nIn the first approach, we will use BayesSearchCV to perform hyperparameter optimization for the Random Forest algorithm.\n\n1. Define search space\n\nWe will tune the following hyperparameters of the Random Forest model:\n\n• n_estimators — The number of trees in the forest.\n• max_depth — The maximum depth of the tree.\n• criterion — The function to measure the quality of a split.\n`# define search space `\n`params = {`\n` \"n_estimators\": [100, 200, 300, 400],`\n` \"max_depth\": (1, 9),`\n` \"criterion\": [\"gini\", \"entropy\"],`\n`}`\n\nWe have defined the search space as a dictionary. It has hyperparameter names used as the key, and the scope of the variable as the value.\n\n2. Define the BayesSearchCV configuration\n\nThe benefit of BayesSearchCV is that the search procedure is performed automatically, which requires minimal configuration.\n\nThe class can be used in the same way as Scikit-Learn (GridSearchCV and RandomizedSearchCV).\n\n`# define the search`\n`search = BayesSearchCV(`\n` estimator=rf_classifier,`\n` search_spaces=params,`\n` n_jobs=1,`\n` cv=5,`\n` n_iter=30,`\n` scoring=\"accuracy\",`\n` verbose=4,`\n` random_state=42`\n`)`\n\n3. Fine-tune the model\n\nWe then execute the search by passing the preprocessed features and the target feature (price_range).\n\n`# perform the search`\n`search.fit(X_scaled,y)`\n\nYou can find the best score by using the best_score_ attribute and the best parameters by using best_params_ attribute from the search.\n\n`# report the best result`\n\n`print(search.best_score_)`\n`print(search.best_params_)`\n\nNote that the current version of scikit-optimize (0.7.4) is not compatible with the latest versions of scikit learn (0.23.1 and 0.23.2). So when you run the optimization process using this approach, you can get errors like this:\n\n`TypeError: object.__init__() takes exactly one argument (the instance to initialize)`\n\nI hope they will solve this incompatibility problem very soon.\n\nThe second approach\n\nIn the second approach, we first define the search space by using the space methods provided by scikit-optimize, which are Categorical and Integer.\n\n`# define the space of hyperparameters to search`\n`search_space = list()`\n`search_space.append(Categorical([100, 200, 300, 400], name='n_estimators'))`\n`search_space.append(Categorical(['gini', 'entropy'], name='criterion'))`\n`search_space.append(Integer(1, 9, name='max_depth'))`\n\nWe have set different values in the above-selected hyperparameters. Then we will define the objective function.\n\n1. Defining a function to minimise (Objective Function)\n\nOur function to minimize is called evalute_model and the classification algorithm to optimize its hyperparameter is Random Forest.\n\nI use cross-validation to avoid overfitting and then the function will return loss values.\n\n`# define the function used to evaluate a given configuration`\n`@use_named_args(search_space)`\n`def evaluate_model(**params):`\n` # configure the model with specific hyperparameters`\n` clf = RandomForestClassifier(**params, n_jobs=-1)`\n` acc = cross_val_score(clf, X_scaled, y, scoring=\"accuracy\").mean()`\n\nThe use_named_args() decorator allows your objective function to receive the parameters as keyword arguments. This is particularly convenient when you want to set scikit-learn's estimator parameters.\n\nRemember that scikit-optimize minimises the function, which is why I add a negative sign in the acc.\n\n2. Fine-tune the model\n\nFinally, we fine-tune the model by using the gp_minimize method (it uses Gaussian process-based optimization) from scikit-optimize. Then we print the best loss with its hyperparameters values.\n\n`# perform optimization`\n\n`result = gp_minimize(`\n` func=evaluate_model,`\n` dimensions=search_space,`\n` n_calls=30,`\n` random_state=42,`\n` verbose=True,`\n` n_jobs=1,`\n\n) Output:\n\nIteration No: 1 started. Evaluating function at random point.\n\nIteration No: 1 ended. Evaluation done at random point.\n\nTime taken: 8.6910\n\nFunction value obtained: -0.8585\n\nCurrent minimum: -0.8585\n\nIteration No: 2 started. Evaluating function at random point.\n\nIteration No: 2 ended. Evaluation done at random point.\n\nTime taken: 4.5096\n\nFunction value obtained: -0.7680\n\nCurrent minimum: -0.8585 ………………….\n\nNot that it will run until it reaches the last iteration. For our optimization process, the total number of iterations is 30.\n\nThen we can print the best accuracy and the values of the selected hyperparameters we used.\n\n`# summarizing finding:`\n\n`print('Best Accuracy: %.3f' % (result.fun)) `\n`print('Best Parameters: %s' % (result.x))`\n\n`Best Accuracy: -0.882`\n`Best Parameters: [300, 'entropy', 9]`\n\nAfter performing hyperparameter optimization, the loss is -0.882. This means that the model's performance has an accuracy of 88.2% by using n_estimators = 300, max_depth = 9, and criterion = “entropy” in the Random Forest classifier.\n\nOur result is not much different from Hyperopt in the first part (accuracy of 89.15%).\n\nPrint Function Values\n\nYou can print all function values at each iteration by using the func_vals attribute from the OptimizeResult object (result).\n\n`print(result.func_vals)`\n\nOutput:\n\narray([-0.8665, -0.7765, -0.7485, -0.86 , -0.872 , -0.545 , -0.81 ,\n\n-0.7725, -0.8115, -0.8705, -0.8685, -0.879 , -0.816 , -0.8815,\n\n-0.8645, -0.8745, -0.867 , -0.8785, -0.878 , -0.878 , -0.8785,\n\n-0.874 , -0.875 , -0.8785, -0.868 , -0.8815, -0.877 , -0.879 ,\n\n-0.8705, -0.8745])\n\nPlot convergence traces\n\nWe can use the plot_convergence method from scikit-optimize to plot one or several convergence traces. We just need to pass the OptimizeResult object (result) in the plot_convergence method.\n\n`# plot convergence `\n\n`from skopt.plots import plot_convergence`\n\n`plot_convergence(result) `", null, "The plot shows function values at different iterations during the optimisation process.\n\nNow that you know how to implement scikit-optimize, let's learn the third and final alternative hyperparameter optimisation technique called Optuna.\n\n### 3. Optuna\n\nOptuna is another open-source Python framework for hyperparameter optimization that uses the Bayesian method to automate search space of hyperparameters. The framework was developed by a Japanese AI company called Preferred Networks.\n\nOptuna is easier to implement and use than Hyperopt. You can also specify how long the optimization process should last.\n\nOptuna has at least five important features you need to know in order to run your first optimization.\n\nSearch Spaces\n\nOptuna provides different options for all hyperparameter types. The most common options to choose are as follows:\n\n• Categorical parameters – uses the trials.suggest_categorical() method. You need to provide the name of the parameter and its choices.\n• Integer parameters – uses the trials.suggest_int() method. You need to provide the name of the parameter, low and high value.\n• Float parameters – uses the trials.suggest_float() method. You need to provide the name of the parameter, low and high value.\n• Continuous parameters – uses the trials.suggest_uniform() method. You need to provide the name of the parameter, low and high value.\n• Discrete parameters – uses the trials.suggest_discrete_uniform() method. You need to provide the name of the parameter, low value, high value, and step of discretisation.\n\nOptimization methods (Samplers)\n\nOptuna has different ways to perform the hyperparameter optimisation process. The most common methods are:\n\n• GridSampler – It uses a grid search. The trials suggest all combinations of parameters in the given search space during the study.\n• RandomSampler – It uses random sampling. This sampler is based on independent sampling.\n• TPESampler – It uses the TPE (Tree-structured Parzen Estimator) algorithm.\n• CmaEsSampler – It uses the CMA-ES algorithm.\n\nObjective Function\n\nThe objective function works the same way as in the hyperopt and scikit-optimize techniques. The only difference is that Optuna allows you to define the search space and objective in the one function.\n\nExample:\n\n`def objective(trial):`\n` # Define the search space`\n` criterions = trial.suggest_categorical('criterion', ['gini', 'entropy'])`\n` max_depths = trial.suggest_int('max_depth', 1, 9, 1)`\n` n_estimators = trial.suggest_int('n_estimators', 100, 1000, 100)`\n\n` clf = sklearn.ensemble.RandomForestClassifier(n_estimators=n_estimators,`\n` criterion=criterions,`\n` max_depth=max_depths,`\n` n_jobs=-1)`\n` `\n` score = cross_val_score(clf, X_scaled, y, scoring=\"accuracy\").mean()`\n\n` return score`\n\nStudy\n\nA study corresponds to an optimization task (a set of trials). If you need to start the optimization process, you need to create a study object and pass the objective function to a method called optimize() and set the number of trials as follows:\n\n`study = optuna.create_study()`\n`study.optimize(objective, n_trials=100)`\n\nThe create_study() method allows you to choose whether you want to maximize or minimize your objective function.\n\nThis is one of the more useful features I like in optuna because you have the ability to choose the direction of the optimization process.\n\nNote that you will learn how to implement this in the practical example below.\n\nVisualisation\n\nThe visualisation module in Optuna provides different methods to create figures for the optimization outcome. These methods help you gain information about interactions between parameters and let you know how to move forward.\n\nHere are some of the methods you can use.\n\n• plot_contour() – This method plots the parameter relationship as a contour plot in a study.\n• plot_intermidiate_values() – This method plots intermediate values of all trials in a study.\n• plot_optimization_history() – This method plots the optimisation history of all trials in a study.\n• plot_param_importances() – This method plots hyperparameter importance and their values.\n• plot_edf() – This method plots the objective value EDF (empirical distribution function) of a study.\n\nWe will use some of the methods mentioned above in the practical example below.\n\nOptuna in Practice\n\nNow that you know the important features of Optuna, in this practical example we will use the same dataset (Mobile Price Dataset) that we used in the previous two methods above.\n\nInstall Optuna\n\nYou can install the latest release with:\n\n`pip install optuna`\n\nThen import the important packages, including optuna:\n\n`# import packages `\n`import numpy as np `\n`import pandas as pd `\n`from sklearn.ensemble import RandomForestClassifier `\n`from sklearn import metrics `\n`from sklearn.model_selection import cross_val_score `\n`from sklearn.preprocessing import StandardScaler `\n`import joblib `\n\n`import optuna `\n`from optuna.samplers import TPESampler`\n\n`import warnings`\n`warnings.filterwarnings(\"ignore\")`\n\nDefine search space and objective in one function\n\nAs I have explained above, Optuna allows you to define the search space and objective in one function.\n\nWe will define the search spaces for the following hyperparameters of the Random Forest model:\n\n• n_estimators — The number of trees in the forest.\n• max_depth — The maximum depth of the tree.\n• criterion — The function to measure the quality of a split.\n`# define the search space and the objecive function`\n\n`def objective(trial):`\n` # Define the search space`\n` criterions = trial.suggest_categorical('criterion', ['gini', 'entropy'])`\n` max_depths = trial.suggest_int('max_depth', 1, 9, 1)`\n` n_estimators = trial.suggest_int('n_estimators', 100, 1000, 100)`\n\n` clf = RandomForestClassifier(n_estimators=n_estimators,`\n` criterion=criterions,`\n` max_depth=max_depths,`\n` n_jobs=-1)`\n` score = cross_val_score(clf, X_scaled, y, scoring=\"accuracy\").mean()`\n\n` return score`\n\nWe will use the trial.suggest_categorical() method to define a search space for criterion and trial.suggest_int() for both max_depth and n_estimators.\n\nAlso, we will use cross-validation to avoid overfitting, and then the function will return the mean accuracy.\n\nCreate a study object\n\nNext we create a study object that corresponds to the optimization task. The create-study() method allows us to provide the name of the study, the direction of the optimization (maximize or minimize), and the optimization method we want to use.\n\n`# create a study object `\n\n`study = optuna.create_study(study_name=\"randomForest_optimization\",`\n` direction=\"maximize\",`\n` sampler=TPESampler())`\n\nIn our case we named our study object randomForest_optimization. The direction of the optimization is maximize (which means the higher the score the better) and the optimization method to use is TPESampler().\n\nFine-tune the model\n\nTo run the optimization process, we need to pass the objective function and number of trials in the optimize() method from the study object we have created.\n\nWe have set the number of trials to be 10 (but you can change the number if you want to run more trials).\n\n`# pass the objective function to method optimize()`\n\n`study.optimize(objective, n_trials=10)`\n\nOutput:", null, "Then we can print the best accuracy and the values of the selected hyperparameters used.\n\nTo show the best hyperparameters values selected:\n\n`print(study.best_params)`\n\nOutput: {‘criterion’: ‘entropy’, ‘max_depth’: 8, ‘n_estimators’: 700}\n\nTo show the best score or accuracy:\n\n`print(study.best_value)`\n\nOutput: 0.8714999999999999.\n\nOur best score is around 87.15%.\n\nPlot optimization history\n\nWe can use the plot_optimization_history() method from Optuna to plot the optimization history of all trials in a study. We just need to pass the optimized study object in the method.\n\n`optuna.visualization.plot_optimization_history(study)`", null, "The plot shows the best values at different trials during the optimization process.\n\nPlot hyperparameter importances\n\nOptuna provides a method called plot_param_importances() to plot hyperparameter importance. We just need to pass the optimized study object in the method.", null, "From the figure above you can see that max-depth is the most important hyperparameter.\n\nYou can save and load the hyperparameter searches by using the joblib package.\n\nFirst, we will save the hyperparameter searches in the optuna_searches directory.\n\n`# save your hyperparameter searches `\n`joblib.dump(study, 'optuna_searches/study.pkl')`\n\nThen if you want to load the hyperparameter searches from the optuna_searches directory, you can use the load() method from joblib.\n\n`# load your hyperparameter searches`\n\n`study = joblib.load('optuna_searches/study.pkl')`\n\n### Wrapping up\n\nCongratulations, you have made it to the end of the article!\n\nLet’s take a look at the overall scores and hyperparameter values selected by the three hyperparameter optimization techniques we have discussed in this article.", null, "The results presented by each technique are not that different from each other. The number of iterations or trials selected makes all the difference.\n\nFor me, Optuna is easy to implement and is my first choice in hyperparameter optimization techniques. Please let me know what you think!" ]

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[ "", null, "224K - views\n\n# Convolution Pyramids Zeev Farbman The Hebrew University Raanan Fattal The Hebrew University Dani Lischinski The Hebrew University Figure Three examples of applications that benet from our fast convo\n\nLeft gradient 64257eld integration Middle membrane interpolation Right scattered data interpolation The insets show the shapes of the corresponding kernels Abstract We present a novel approach for rapid numerical approximation of convolutions with 6\n\n## Convolution Pyramids Zeev Farbman The Hebrew University Raanan Fattal The Hebrew University Dani Lischinski The Hebrew University Figure Three examples of applications that benet from our fast convo\n\nDownload Pdf - The PPT/PDF document \"Convolution Pyramids Zeev Farbman The He...\" is the property of its rightful owner. Permission is granted to download and print the materials on this web site for personal, non-commercial use only, and to display it on your personal computer provided you do not modify the materials and that you retain all copyright notices contained in the materials. By downloading content from our website, you accept the terms of this agreement.\n\n## Presentation on theme: \"Convolution Pyramids Zeev Farbman The Hebrew University Raanan Fattal The Hebrew University Dani Lischinski The Hebrew University Figure Three examples of applications that benet from our fast convo\"— Presentation transcript:\n\nPage 1\nConvolution Pyramids Zeev Farbman The Hebrew University Raanan Fattal The Hebrew University Dani Lischinski The Hebrew University Figure 1: Three examples of applications that benefit from our fast convolution approximation. Left: gradient field integration; Middle: membrane interpolation; Right: scattered data interpolation. The insets show the shapes of the corresponding kernels. Abstract We present a novel approach for rapid numerical approximation of convolutions with filters of large support. Our approach consists of a multiscale scheme, fashioned after\n\nthe wavelet transform, which computes the approximation in linear time. Given a specific large target filter to approximate, we first use numerical optimization to design a set of small kernels, which are then used to perform the analysis and synthesis steps of our multiscale transform. Once the optimization has been done, the resulting transform can be applied to any signal in linear time. We demonstrate that our method is well suited for tasks such as gradient field integration, seamless im- age cloning, and scattered data interpolation, outperforming exist- ing\n\nstate-of-the-art methods. Keywords: convolution, Green’s functions, Poisson equation, seamless cloning, scattered data interpolation, Shepard’s method Links: DL PDF EB 1 Introduction Many tasks in computer graphics and image processing involve applying large linear translation-invariant (LTI) filters to images. Common examples include low- and high-pass filtering of images, and measuring the image response to various filter banks. Some less obvious tasks that can also be accomplished using large LTI filters are demonstrated in Figure : reconstructing images by\n\nintegrating their gradient field [ Fattal et al. 2002 ], fitting a smooth membrane to interpolate a set of boundary values [ erez et al. 2003 Agarwala 2007 ], and scattered data interpolation [ Lewis et al. 2010 ]. While convolution is the most straightforward way of applying an LTI filter to an image, it comes with a high computational cost: operations are required to convolve an -pixel image with a kernel of comparable size. The Fast Fourier Transform offers a more efficient, log alternative for periodic domains [ Brigham 1988 ]. Other fast approaches have been\n\nproposed for certain special cases. For example, Burt [ 1981 ] describes a multiscale approach, which can approximate a convolution with a large Gaussian kernel in time at hierarchically subsampled locations. We review this and several other related approaches in the next section. In this work, we generalize these ideas, and describe a novel mul- tiscale framework that is not limited to approximating a specific kernel, but can be tuned to reproduce the effect of a number of useful large LTI filters, while operating in time. Specifically, we demonstrate the applicability of\n\nour framework to convolutions with the Green’s functions that span the solutions of the Poisson equation, inverse distance weighting kernels for membrane inter- polation, and wide-support Gaussian kernels for scattered data in- terpolation. These applications are demonstrated in Figure Our method consists of a multiscale scheme, resembling the Lapla- cian Pyramid, as well as certain wavelet transforms. However, unlike these more general purpose transforms, our approach is to custom-tailor the transform to directly approximate the effect of a given LTI operator. In other words, while previous\n\nmultiscale con- structions are typically used to transform the problem into a space where it can be better solved, in our approach the transform itself directly yields the desired solution. Specifically, we repeatedly perform convolutions with three small, fixed-width kernels, while downsampling and upsampling the im- age, so as to operate on all of its scales. The weights of each of these kernels are numerically optimized such that the overall action of the transform best approximates a convolution operation with some target filter . The optimization only needs to be done\n\nonce for each target filter, and then the resulting multiscale transform may be applied to any input signal in time. The motivation behind this design was to avoid dealing with the analytical challenges that arise from the non-idealness of small finite filters, on the one hand, while attempting to make the most out of the linear computational budget, on the other. Our scheme’s ability to closely approximate convolutions with the\nPage 2\nfree space Green’s function [ Evans 1998 ] is high enough to allow us to invert a certain variant of the Poisson equation. We show\n\nthat for this equation our approach is faster than state-of-the-art solvers, such as those based on the multigrid method [ Kazhdan and Hoppe 2008 ], offering improved accuracy for a comparable number of op- erations, and a higher ease of implementation. While we demonstrate the usefulness of our approach on a num- ber of important applications, and attempt to provide some insights on the limits of its applicability later in the paper, our theoretical analysis of these issues is certainly incomplete, and is left as an interesting avenue for future work. 2 Background Besides the straightforward\n\nimplementation of convolution, there are certain scenarios where it can be computed more effi- ciently. Over periodic domains, every convolution operator can be expressed as a circulant Topelitz matrix, which is diagonalized by the Fourier basis. Thus, convolutions with large kernels over peri- odic domains may be carried out in log time using the Fast Fourier Transform [ Brigham 1988 ]. Convolution with separable 2D kernels, which may be expressed as the outer product of two 1D kernels, can be sped up by first performing a 1D horizontal convolution, followed by one in the vertical\n\ndirection (or vice versa). Thus, the cost is kn , where is the length of the 1D kernels. Non-separable kernels may be approximated by separable ones using the SVD decomposition of the kernel [ Perona 1995 ]. B-spline filters of various orders and scales may be evaluated at constant time per pixel using repeated integration [ Heckbert 1986 or generalized integral images [ Derpanis et al. 2007 ]. Many methods have been developed specifically to approximate convolutions with Gaussian kernels, because of their important role in image processing [ Wells, III 1986 ]. Of particular\n\nrelevance to this work is the hierarchical discrete correlation scheme proposed by Burt [ 1981 ]. This multiscale scheme approximates a Gaussian pyramid in time. Since convolving with a Gaussian band- limits the signal, interpolating the coarser levels back to the finer ones provides an approximation to a convolution with a large Gaus- sian kernel in . However, Burt’s approximation is accurate only for Gaussians of certain widths. Burt’s ideas culminated in the Laplacian Pyramid Burt and Adel- son 1983 ], and were later connected with wavelets [ Do and Vetterli 2003 ]. More\n\nspecifically, the Laplacian pyramid may be viewed as high-density wavelet transform Selesnick 2006 ]. These ideas are also echoed in [ Fattal et al. 2007 ], where a multiscale bilateral filter pyramid is constructed in log time. The Laplacian pyramid, as well as various other wavelet trans- forms [ Mallat 2008 ] decompose a signal into its low and high frequencies, which was shown to be useful for a variety of anal- ysis and synthesis tasks. In particular, it is possible to approxi- mate the effect of convolutions with large kernels. However, even though these schemes rely on\n\nrepeated convolution (typically with small kernels), they are not translation-invariant, i.e., if the input is translated, the resulting analysis and synthesis will not differ only by a translation. This is due to the subsampling operations these schemes use for achieving their fast performance. Our scheme also uses subsampling, but in a more restricted fashion, which was shown by Selesnick [ 2006 ] to increase translation invariance. One scenario, which gained considerable importance in the past decade, is the recovery of a signal from its convolved form, e.g., an image from its gradient\n\nfield [ Fattal et al. 2002 ]. This gives rise to a translation-invariant system of linear equations, such as the Poisson Equation. The corresponding matrices are, again, Topelitz matrices, which can be inverted in log time, using FFT over periodic domains. However, there are even faster solvers for handling specific types of such equations over both periodic and non-periodic domains. The multigrid [ Trottenberg et al. 2001 ] method and hier- archical basis preconditioning [ Szeliski 1990 ] achieve linear perfor- mance by operating at multiple scales. A state-of-the-art multigrid\n\nsolver for large gradient domain problems is described by Kazhdan and Hoppe [ 2008 ], and a GPU implementation for smaller problems is described by McCann and Pollard [ 2008 ]. Since the matrices in these linear systems are circulant (or nearly circulant, depending on the domain), their inverse is also a circu- lant convolution matrix. Hence, in principle, the solution can be obtained (or approximated) by convolving the right-hand side with a kernel, e.g., the Green’s function in cases of the infinite Poisson equation. Our approach enables accurately approximating the con- volution with\n\nsuch kernels, and hence provides a more efficient, and easy to implement alternative for solving this type of equations. Gradient domain techniques are also extensively used for seamless cloning and stitching [ erez et al. 2003 ], yielding a similar linear system, but with different (Dirichlet) boundary conditions, typi- cally solved over irregularly shaped domains. In this scenario, the solution often boils down to computing a smooth membrane that interpolates a set of given boundary values [ Farbman et al. 2009 ]. Agarwala [ 2007 ] describes a dedicated quad-tree based solver for such\n\nsystems, while Farbman et al. 2009 ] avoid solving a linear system altogether by using mean-value interpolation. In Section we show how to construct such membranes even more efficiently by casting the problem as a ratio of convolutions [ Carr et al. 2003 ]. This approach can also be useful in more general scattered data in- terpolation scenarios [ Lewis et al. 2010 ], when there are many data points to be interpolated or when a dense evaluation of the interpo- lated function is needed. Our work provides an efficient approxi- mation for scattered data interpolation where it is\n\nimportant to use large convolution filters that propagate the information to the entire domain. 3 Method In this section we describe our framework in general and motivate our design decisions. The adaptation of the framework to several specific problems in computer graphics is discussed immediately afterwards. Linear translation-invariant filtering is used extensively in computer graphics and image processing for scale separation. Indeed, many image analysis and manipulation algorithms cannot succeed with- out a suitably constructed multiscale (subband) decomposition. In most\n\napplications, the spectral accuracy needed when extracting a particular band is proportional to the band level: high-frequency components are typically extracted with small compact filters that achieve low localization in the frequency domain, while low fre- quencies are typically needed at a higher spectral accuracy and, hence, large low-pass filters are used. Subband architectures such as wavelet transforms and Laplacian pyramids rely on a spectral “divide-and-conquer” strategy where, at every scale, the spectrum is partitioned via filtering and then the lower end of the\n\nspectrum, which may be subsampled without a major loss of data, is further split recursively. The subsampling step allows extracting progressively lower frequencies using filters of small fixed length, since the domain itself is shrinking and dis- tant points become closer, due to the use of subsampling. This ap- proach leads to highly efficient linear-time processing of signals\nPage 3\nAlgorithm 1 Our multiscale transform 1: Determine the number of levels 2: Forward transform (analysis) 3: 4: for each level ... do 5: 6: 7: end for 8: Backward transform (synthesis)\n\n9: 10: for each level ... do 11: )+ 12: end for Figure 2: Our subband architecture flow chart and pseudocode. and is capable of isolating low-frequency modes up to the largest spatial scale (the DC component of the image). While it may not be a major obstacle for some applications, these decompositions suffer from two main shortcomings. First, the re- sulting transformed coordinates, and therefore the operations per- formed using these coordinates, are not invariant to translation. Thus, unlike convolution, shifting the input image may change the outcome by more than just a spatial\n\noffset. Second, to achieve the running time, it is necessary to use finite impulse response filters. These filters can achieve some spatial and spectral localiza- tion but do not provide an ideal partitioning of the spectrum. As we demonstrate here, these properties are critical for some appli- cations (such as solving the Poisson equation) and for the design of particularly shaped kernels. In fact, these two shortcomings dic- tate the design of the new scheme we describe below, whose aim is to achieve an optimal approximation of certain translation-invariant operators under\n\nthe computational cost budget of Figure illustrates the multiscale filtering scheme that we use and is inspired by the architectures mentioned above. The forward trans- form consists of convolving the signal with an analysis filter and subsampling the result by a factor of two. This process is then repeated on the subsampled data. Additionally, an unfiltered and unsampled copy of the signal is kept at each level. Formally, at each level we compute: (1) (2) where the superscript denotes the level in the hierarchy, is the unfiltered data kept at each level, and denotes\n\nthe subsampling operator. The transform is initiated by setting , where is the input signal to be filtered. The backward transformation ( synthesis ) consists of upsampling by inserting a zero between every two samples, followed by a convolu- tion, with another filter, . We then combine the upsampled signal with the one stored at that level after convolving it with a third filter , i.e., )+ (3) where denotes the zero upsampling operator. Note that unlike in most subband architectures, our synthesis is not intended to invert the analysis and reproduce the input signal , but\n\nrather the com- bined action of the forward and backward transforms is meant to approximate the result of some specific linear translation-invariant filtering operation applied to the input This scheme resembles the discrete fast wavelet transform, up to the difference that, as in the Laplacian pyramid, we do not sub- sample the decomposed signal (the high-frequency band in these transformations and the all-band in our case). Similarly to the high density wavelet transformation [ Selesnick 2006 ], this choice is made in order to minimize the subsampling effect and to increase the\n\ntranslation invariance. Assuming ideal filtering was computationally feasible, and could be chosen so as to perfectly isolate and reconstruct progres- sively lower frequency bands of the original data, in which case the role of would be to approximate the desired filtering operation. However, since we want to keep the number of operations the filters and must be finite and small. This means that the design of these filters must account for this lack of idealness and the resulting complex interplay between different frequency bands. Thus, rather than deriving the\n\nfilters and from explicit ana- lytic filter design methodologies, we numerically optimize these fil- ters such that their joint action will best achieve the desired filtering operation. In summary, our approach consists of identifying and al- locating a certain amount of computations with reduced amount of subsampling, while remaining in the regime of computations and then optimizing this allocated computations to best approxi- mate convolution with large filters. 3.1 Optimization In order to approximate convolutions with a given target kernel we seek a set of\n\nkernels that minimizes the follow- ing functional argmin (4) where is the result of our multiscale transform with the kernels on some input . In order to carry out this optimization it re- mains to determine the types of the kernels and the number of their unknown parameters. The choice of the training data depends on the application, and will be discussed in the next section. Note that once this optimization is complete and the kernels have been found, our scheme is ready to be used for approximating on any given signal . All of the kernels used to produce our results are provided in the\n\nsupplementary material, and hence no further optimization is required to use them in practice. In order to minimize the number of total arithmetic operations, the kernels in should be small and separable. The specific choices reported below correspond, in our experiments, to a good trade-off between operation count and approximation accuracy. Using larger and/or non-separable filters increases the accuracy, and hence the specific choice depends on the application requirements. Remark- ably, we obtain rather accurate results using separable kernels in , even for non-separable\n\ntarget filters . This can be explained by the fact that our transform sums the results of these kernels, and the sum of separable kernels is not necessarily separable itself. Furthermore, the target filters that we approximate have rotational and mirroring symmetries. Thus, we explicitly enforce symmetry on our kernels, which reduces the number of degrees of freedom and the number of local minima in the optimization. For example, a separable 3-by-3 kernel is defined by only two parameters ( ) and a separable 5-by-5 kernel by three parameters. As for non-separable kernels, a\n\n5-by-5 kernel with these symmetries\nPage 4\nis defined by six parameters. Depending on the application, the nature of the target filter and the desired approximation accuracy, we choose different combinations of kernels in . For example, we used between 3 and 11 parameters in order to produce each of the kernel sets used in Sections and To perform the optimization we use the BFGS quasi-Newton method [ Shanno 1970 ]. To provide an initial guess we set the kernels to Gaussians. Each level is zero-padded by the size of the largest kernel in at each level before\n\nfiltering. Typically, we start the optimization process on a low-resolution grid (32x32). This is fast, but the result may be influenced by the boundaries and the spe- cific padding scheme used. Therefore, the result is then used as an initial guess for optimization over a high resolution grid, where the influence of the boundaries becomes negligible. In the next sections we present three applications that use this ap- proach to efficiently approximate several different types of filters. 4 Gradient integration Many computer graphics applications manipulate\n\nthe gradient field of an image [ Fattal et al. 2002 McCann and Pollard 2008 Orzan et al. 2008 ]. These applications recover the image that is closest in the norm to the modified gradient field by solving the Poisson equation: div (5) where is a discrete Laplacian operator. Typically von Neumann boundary conditions, 0 where is the unit boundary nor- mal vector, are enforced. Green’s functions define the fundamental solutions to the Poisson equation, and are defined by )= (6) where is a discrete delta function. When ( ) is defined over an infinite domain\n\nwith no boundary conditions, the Laplacian opera- tor becomes spatially-invariant, and so does its inverse. In this case, the Green’s function becomes translation-invariant depending only on the (scalar) distance between and . In two dimensions this free-space Green’s function is given by )= )= log (7) Thus, for a compactly supported right-hand side, the solution of ( is given by the convolution div (8) This solution is also known as the Newtonian potential of div Evans 1998 ]. It can be approximated efficiently using our con- volution pyramid, for example by zero-padding the right-hand\n\nside. The difference between this formulation and imposing von Neu- mann boundary conditions is that the latter enforces zero gradients on the boundary, while in the free-space formulation zero gradi- ents on the boundary can be encoded in the right-hand side serving only as a soft constraint. Forcing the boundary gradients to zero is a somewhat arbitrary decision, and one might argue that it may be preferable to allow them to be determined by the function inside the domain. In fact, a similar approach to image reconstruction from its gradients was used by Horn [ 1974 ] in his seminal work. To\n\nperform the optimization in ( ) we chose a natural greyscale image , and set the training signal to the divergence of its gradient Specifically, we use fminunc from Matlab’s Optimization Toolbox. Figure 3: Reconstruction accuracy for kernel sets of different sizes. field: div , while . We chose a natural image since the training data should not be too localized in space (such as a delta function). Our scheme is not purely translation-invariant and a localized might lead to an over-fitted optimum that would not preform as well in other regions where the training signal is\n\nzero. Natural images are known to be stationary signals with no absolute spatial dependency and, moreover, this is the class of signals we would like to perform best on. We experimented with a number of different kernel size combina- tions for the kernel set . The results of these experiments are summarized in Figure . These results indicate that increasing the widths of the kernels reduces the error more effectively than increasing the width of . We found the set that uses a 5- by-5 kernel for and a 3-by-3 kernel for to be particularly attractive, as it produces results that are visually very\n\nclose to the ground truth, while employing compact kernels. A more accurate solution (virtually indistinguishable from the ground truth) may be obtained by increasing the kernel sizes to 7-by-7/5-by-5 ( ) in exchange for a modest increase in running time, or even further to 9-by-9/5-by-5 ( ). Note that we have no evidence that the best kernels we were able to obtain in our experiments correspond to the global optimum, so it is conceivable that even better accuracy may be achieved using another optimization scheme, or a better initial guess. Table summarizes the running times of our optimized\n\nCPU im- plementation for kernel sets and on various image sizes. grid size time (5x5/3x3) time (7x7/5x5) (millions) (sec, single core) (sec, single core) 0.26 0.0019 0.00285 1.04 0.010 0.015 4.19 0.047 0.065 16.77 0.19 0.26 67.1 0.99 1.38 Table 1: Performance statistics for convolution pyramids. Re- ported times exclude disk I/O and were measured on a 2.3GHz Intel Core i7 (2820qm) MacBook Pro. Figure (a) shows a reconstruction of the free-space Green’s func- tion obtained by feeding our method a centered delta function as input. A comparison to the ground truth (the solution of ( )) re- veals\n\nthat the mean absolute error (MAE) is quite small even for the kernel set: grid size MAE MAE 1024 0.0034 0.0017 2048 0.0034 0.0016\nPage 5\n(a) (b) (c) (d) (e) (f) (g) Figure 4: Gradient field integration. (a) Reconstruction of the Green’s function. (b) Our reconstructions exhibit spatial invari- ance. (c) A natural test image. (d) Reconstruction of (c) from its gradient field using . (e) Absolute errors (magnified by x50). (f)-(g) Reconstruction of (c) from its gradient field using , and the corresponding errors. Changing the position of the delta function’s\n\nspike along the x-axis and plotting a horizontal slice through the center of each resulting reconstruction (Figure (b)), reveals that our reconstructions exhibit very little spatial variance. Figure (c) shows a natural image (different from the one used as training data by the optimization process), whose gradient field di- vergence was given as input to our method. Two reconstructions and their corresponding errors are shown in images (d)–(g). The mean absolute error of our reconstruction is 0.0045 with and 0.0023 with . Visually, it is difficult to tell the difference be- tween\n\neither reconstruction and the original image. Figure 5: Gradient domain HDR compression. Left column: re- sults using a Poisson solver with von Neumann boundary condi- tions. Right column: results using our approximate convolution. In Figure we show HDR compression results produced using gra- dient domain image compression [ Fattal et al. 2002 ]. The results on the left were obtained using a Poisson solver with von Neumann boundary conditions, while those on the right uses our approxima- tion. In both cases, we use the same post-processing, which con- sists of stretching the result so 0 5% of\n\nthe pixels are dark-clipped. While some differences between the results may be noticed, it is hard to prefer one over the other. Additional results are included in the supplementary material. In this context of gradient field integration, our method presents a faster alternative to linear solvers, while also being easier to im- plement. Figure plots the reconstruction error as a function of running time of our method to that of two other solvers: the in- core version of Kazhdan and Hoppe’s solver [ 2008 ], and the stan- dard multigrid solver used by Fattal et al. 2002 ]. A comparison\n\nperformed by Bhat et al. 2008 ] indicates that the in-core version of the Kazhdan-Hoppe solver is one of the fastest currently avail- able CPU-based multigrid solvers for such reconstruction. Indeed, the plots in Figure confirm its superior convergence. However, for most practical graphics applications, a reconstruction error in the range of 0.001–0.0001 is sufficient, and our method is able to achieve this accuracy considerably faster than the Kazhdan-Hoppe solver, while being simpler to implement and to port to the GPU. McCann and Pollard [ 2008 ] describe a GPU-based\n\nimplementation of a multigrid solver, which, at the time, enabled to integrate a one- megapixel image about 20 times per second, supporting interactive gradient domain painting. Our current single core CPU-based im- plementation already enables to integrate such an image 33 times per second. We expect a GPU implementation to bring forth a sig- nificant additional speedup factor. Since the exact running times depend greatly on the desired accu- racy and on the implementation specifics, it is important to gain a better understanding of the speedup in terms of operation counts. A\n\nstandard multigrid solver, such as the one used by Fattal et al.\nPage 6\nFigure 6: Error vs. time for our method with and kernel sets, the in-core streaming multigrid solver of Kazhdan and Hoppe with and without the use of SSE (KH08-SSE, KH08) , and the multi- grid solver used by Fattal et al. (FLW02). 2002 ], performs restriction and prolongation operations to change grid resolutions, with a few relaxation iterations and one resid- ual computation at each resolution [ Trottenberg et al. 2001 ]. Al- though the restriction/prolongation kernels may vary between dif- ferent\n\nschemes, their sizes (and hence costs) are typically com- parable to those of our and kernels. The cost of a single Gauss-Seidel relaxation iteration is 6 operations (multiplications and additions). Together with the residual computation this yields 18 operations on the fine resolution grid, when a single V-cycle is performed with only one relaxation iteration before and after each restriction (known as the V(1,1) scheme). In comparison, applying the 3-by-3 kernel in our method costs 12 operations on the finest resolution grid. Thus, our method’s operation count is smaller than\n\nthat of even the simplest multigrid V-cycle, while the accuracy we achieve is bet- ter, as demonstrated in Figure . In order to achieve higher accu- racy, a multigrid solver might perform more V-cycles or use more relaxation iterations, with a corresponding increase in the number of operations. Similarly, we can also apply our transform itera- tively. Specifically, at each iteration the transform is re-applied on the residual, and the result is added to the solution. This is how the curves corresponding to and in Figure were produced. 5 Boundary interpolation Applications such as\n\nseamless image cloning [ erez et al. 2003 and image stitching [ Szeliski 2006 ] can be formulated as bound- ary value problems and effectively solved by constructing a smooth membrane that interpolates the differences along a seam between two images across some region of interest [ Farbman et al. 2009 ]. Such membranes have originally been built by solving the Laplace equation [ erez et al. 2003 ]. However, Farbman et al. 2009 showed that other smooth interpolation schemes, such as mean- value interpolation may be used as well, offering computational advantages. Here we show how to construct a\n\nsuitable membrane even faster by approximating Shepard’s scattered data interpolation method [ Shepard 1968 ] using a convolution pyramid. Let denote a region of interest (on a discrete regular grid), whose boundary values are given by . Our goal is to smoothly in- terpolate these values to all grid points inside . Shepard’s method defines the interpolant at as a weighted average of known bound- ary values: )= (9) where are the boundary points. In our experiments we found that a satisfactory membrane interpolant is obtained by using the following weight function: )= )= (10) which has a\n\nstrong spike at the boundary point and decays rapidly away from it. Naive evaluation of ( ) is expensive: assuming that the number of boundary values is and the number of points in is , the computational cost is Kn . Our multiscale transform allows us to approximate the computation in time. Following Carr et al. 2003 ], our first step is to re-write Shepard’s method in terms of convolutions. We first define as an extension of to the entire domain: )= for on the boundary 0 otherwise (11) and rewrite ( ) as a ratio of convolutions: )= (12) where is the characteristic function\n\ncorresponding to is 1 where is non-zero, 0 otherwise). Intuitively, including the char- acteristic function in the denominator ensures that the weights of the zeros in are not accumulated. Again, we use the optimization to find a set of kernels , with which our convolution pyramid can be applied to evaluate ( 12 ). To define the training data, we set to be the boundary of a simple rectangular domain with random values assigned at the boundary grid points, and compute an exact membrane using ( 12 ). Our optimization then attempts to match , which is a ratio of two convo- lutions,\n\ndirectly, rather than matching each convolution separately. This is important to ensure that the decay of the filter is accurately reproduced by our transform over the entire domain. As in the previous application, we were able to produce satisfactory interpolating membranes using an kernel set, consisting of 5- by-5 separable filters for and a single 3-by-3 separable filter for , regardless of the size and the shape of the domain. These kernels are provided in the supplementary material. In Figure we compare seamless image cloning using our method with that produced with a\n\nLaplacian membrane [ erez et al. 2003 ]. Note that despite the slight differences between the two membranes the final seamless compositing results of both methods are difficult to distinguish visually. The running time of this method amounts to two evaluations of approximate convolution, one with and one with , where the times for a single convolution are reported in Table . We have al- ready established in the previous section that our approach outper- forms state-of-the-art linear solvers. It remains, however, to com- pare our method to the fast Mean Value Cloning (MVC) method\n\nFarbman et al. 2009 ], which computes mean value coordinates at the vertices of an adaptive triangulation of the domain. In con- trast to that method, our approach does not require triangulating the domain and precomputing each pixel’s barycentric coordinates to achieve interactive performance on the CPU. Furthermore, our method is completely independent of the size of the boundary, and avoids the somewhat sophisticated scheme that MVC employs for hierarchical sampling of the boundary. Farbman et al. 2009 ] report that after 3.6 seconds of preprocessing time, it takes 0.37 seconds to\n\nseamlessly clone a 4.2 megapixel region (on a 2.5 MHz Athlon CPU). In comparison, our method does not involve preprocessing, and clones the same number of pixels in 0.15 seconds.\nPage 7\nFigure 7: Membrane construction for seamless cloning. Top row: image with a cloned patch superimposed; Middle row: Laplacian membrane [P erez et al. 2003] and the resulting seamless cloning. Bottom row: Our membrane and the corresponding result. Figure 8: Gaussian filters: (a) Original Image (b) Exact convolu- tion with a Gaussian filter ( ) (c) Convolution using our ap- proximation for\n\nthe same . (d) Exact kernels (in red) with approx- imate kernels (in blue). (e) exact Gaussian (in red), approximation using 5x5 kernels (in blue), approximation using 7x7 kernels (in green). (f) shows a magnified part of (e). 6 Gaussian kernels In this section we demonstrate how to use convolution pyramids to approximate convolutions with Gaussian kernels −% Burt [ 1981 ] showed how this can be done using an log multi- scale scheme and described an variant that computes the result only on a coarse grid, whose spacing is inversely proportional to the Gaussian’s width. While the\n\nresulting values can be interpolated to the original grid to yield an overall method, the effective ker- nels in both variants are truncated and their support depends on the scale . Using our scheme we can approximate the solution at the original fine grid in operations without truncating the filter support. Figure 9: Scattered data interpolation with two Gaussian kernels: (a,c) A horizontal slice through the exact Gaussian (in red) and our approximation (in blue) (b,d) A log plot of the same slice. (e) Scat- tered data interpolation input. (f,h) Exact results corresponding to the\n\ntwo Gaussians. (g,i) Our approximations. To carry out the optimization in ( ) we set the training signal to a sum of randomly translated and scaled delta functions, and is obtained by a full convolution with the target Gaussian filter. This is intended to force the optimal kernels to achieve an adequate ap- proximation across the entire domain, resulting in a transform that is effectively translation-invariant. Our initial attempts to find a set that will provide a good approximation failed. The reason is that Gaussians are rather efficient low-pass filters and, de-\n\npending on their scale , they should not contain high-frequency components coming from the finer levels of the pyramid. These components are introduced by the convolutions with the kernel in ), and we were able to obtain considerably better results by mod- ulating the contribution of at each level by a scalar weight These weights were added as additional degrees of freedom to opti- mize over, and indeed, the optimization resulted with significantly higher at the levels that are closest to the target Gaussian’s scale. Note that different sets of kernels must be fit to obtain\n\nGaussian filters of different scales. Figure shows the impulse responses of ten convolution pyramids optimized to approximate Gaussians of ten different values of superimposed with the exact impulse responses of the target Gaus- sians. In these pyramids we use 5-by-5 separable kernels for and . Notice that the approximation is less accurate for certain val- ues of . To improve the accuracy, we increase the allocated com- putational budget and use 7-by-7 separable kernels instead. Figure (e-f) demonstrate the resulting improvement in accuracy. The effective filters that Burt’s\n\nmethod and integral images produce are truncated in space, i.e., they have a finite support that depends on the Gaussian scale . While this is not a major issue when the filtering is used for applications such as image blurring, this trun- cation has a significant effect when the filters are used for scattered data interpolation, such as the Shepard’s method outlined in the pre- vious section. In this application, we need the data to propagate, in a monotonically decreasing fashion, across the entire domain. A truncated kernel may lead to divisions by zero in ( 12 ) or\n\nto intro- duce abrupt transitions in the interpolated values. Figure shows our approximation of two different Gaussian fil- ters, differing in their width. These approximations were computed using the smaller 5-by-5 kernels. The approximation of the wider Gaussian in (a) provides a good fit across the entire domain. For the narrower Gaussian in (b), the approximation loses its relative accuracy as the Gaussian reaches very low values. This may be seen in the log plot in (d). Note however, that the approximation is\nPage 8\nstill smooth and monotonically decaying. This\n\nslower decay leads to fuzzier transitions in the interpolated function compared to the exact scattered data interpolation, as shown in (h) and (i). In summary, when using wide Gaussian weight functions or oper- ating on a denser input dataset, our method provides an efficient and accurate approximation to scattered-data interpolation. Sparser datasets and narrower Gaussians reveal a limitation of our approxi- mation. 7 Discussion and Future Work We presented a multiscale scheme that achieves accurate approxi- mations of the convolution operations with several widely-used fil- ters.\n\nThis is done using minimal computational efforts where we achieve large-scale non-separable filtering using small and separa- ble kernels. We demonstrated the advantages of using our method over existing techniques, including state-of-the-art linear solvers. The two main ideas behind our approach are: (i) identifying the right computational scheme, which balances operation count with lack of translation-invariance, and (ii) optimizing filters, rather than tackling their lack of idealness analytically. The optimized kernels used in Sections and are provided in the supplementary\n\nmate- rial and can be used out-of-the-box for the applications described in these sections. As for the Gaussian filters, depending upon the application, in cases where the required values of are known in advance, such kernels can be computed up front. While our paper provides new useful tools, our non-analytic ap- proach has several fundamental limitations. While we succeeded in approximating certain filters which are commonly used in computer graphics, our work does not shed light on what other filters can be approximated using this approach. Specifically, it is not\n\nyet clear which filters can be approximated efficiently using small kernels. Another limitation arises from the use of black-box optimization in order to find the kernel set . In order to gain higher accu- racy or approximate more challenging filters, larger kernels must be used. As the number of unknown parameters in the optimization increases, it will be harder to expect the optimization will indeed reach a global optimum (or even a satisfactory local one). As future work, we intend to conduct a thorough theoretical study in order to identify the scope of this\n\napproach in terms of filters it can approx- imate and gain insights that, if not replace the optimization, will at least aid its convergence. Acknowledgements: The authors are indebted to Yoav HaCo- hen for crafting the optimized convolution pyramid implementation which produced the timings reported in this paper. This work was supported in part by the Israel Science Foundation founded by the Israel Academy of Sciences and Humanities. References GARWALA , A. 2007. Efficient gradient-domain compositing us- ing quadtrees. ACM Trans. Graph. 26 , 3 (July), Article 94. HAT ,P.,C URLESS\n\n,B.,C OHEN ,M., AND ITNICK , C. L. 2008. Fourier analysis of the 2D screened Poisson equation for gradient domain problems. In Proc. ECCV , 114–128. RIGHAM , E. O. 1988. The fast Fourier transform and its appli- cations . Prentice-Hall, Inc., Upper Saddle River, NJ, USA. URT ,P.J., AND DELSON , E. H. 1983. The Laplacian pyramid as a compact image code. IEEE Trans. Comm. 31 , 4, 532–540. URT , P. J. 1981. Fast filter transforms for image processing. Computer Graphics and Image Processing 16 , 1 (May), 20–51. ARR ,J.C.,B EATSON ,R.K.,M ALLUM ,B.C.,F RIGHT W. R., M ENNAN ,T.J., AND ITCHELL\n\n, T. J. 2003. Smooth surface reconstruction from noisy range data. In Proc. GRAPHITE ’03 , ACM, 119–ff. ERPANIS ,K.,L EUNG ,E., AND IZINTSEV , M. 2007. Fast scale- space feature representations by generalized integral images. In Proc. ICIP , vol. 4, IEEE, 521–524. ,M., AND ETTERLI , M. 2003. Framing pyramids. IEEE Transactions on Signal Processing 51 , 9 (Sept.), 2329–2342. VANS , L. C. 1998. Partial Differential Equations , vol. 19 of Graduate Series in Mathematics . American Mathematical Soci- ety. ARBMAN ,Z.,H OFFER ,G.,L IPMAN ,Y.,C OHEN -O ,D., AND ISCHINSKI , D. 2009. Coordinates for\n\ninstant image cloning. ACM Trans. Graph. 28 , 3, Article 67. ATTAL ,R.,L ISCHINSKI ,D., AND ERMAN , M. 2002. Gradient domain high dynamic range compression. ACM Trans. Graph. 21 , 3 (July), 249–256. ATTAL ,R.,A GRAWALA ,M., AND USINKIEWICZ , S. 2007. Multiscale shape and detail enhancement from multi-light image collections. ACM Trans. Graph. 26 , 3 (July), Article 51. ECKBERT , P. S. 1986. Filtering by repeated integration. In Proc. ACM SIGGRAPH 86 , ACM, 315–321. ORN , B. K. P. 1974. Determining lightness from an image. Com- puter Graphics and Image Processing 3 , 1 (Dec.), 277–299. AZHDAN\n\n,M., AND OPPE , H. 2008. Streaming multigrid for gradient-domain operations on large images. ACM Trans. Graph. 27 , 3 (Aug.), 21:1–21:10. EWIS ,J.P.,P IGHIN ,F., AND NJYO , K. 2010. Scattered data interpolation and approximation for computer graphics. In ACM SIGGRAPH ASIA 2010 Courses , ACM, 2:1–2:73. ALLAT , S. 2008. A wavelet tour of signal processing , 3rd ed. Academic Press. ANN ,J., AND OLLARD , N. S. 2008. Real-time gradient- domain painting. ACM Trans. Graph. 27 , 3 (August), 93:1–93:7. RZAN ,A.,B OUSSEAU ,A.,W INNEM OLLER ,H.,B ARLA P., T HOLLOT ,J., AND ALESIN , D. 2008. Diffusion\n\ncurves: a vector representation for smooth-shaded images. ACM Trans. Graph. 27 , 3 (August), 92:1–92:8. EREZ ,P.,G ANGNET ,M., AND LAKE , A. 2003. Poisson image editing. ACM Trans. Graph. 22 , 3, 313–318. ERONA , P. 1995. Deformable kernels for early vision. IEEE Trans. Pattern Anal. Mach. Intell. 17 , 5, 488–499. ELESNICK , I. 2006. A higher density discrete wavelet transform. IEEE Trans. Signal Proc. 54 , 8 (Aug.), 3039–3048. HANNO , D. F. 1970. Conditioning of quasi-Newton methods for function minimization. Mathematics of Computation 24 , 111 (July), 647. HEPARD , D. 1968. A two-dimensional\n\ninterpolation function for irregularly-spaced data. In Proc. 1968 23rd ACM National Conference , ACM, 517–524. ZELISKI , R. 1990. Fast surface interpolation using hierarchical basis functions. IEEE Trans. Pattern Anal. Mach. Intell. 12 , 6, 513–528. ZELISKI , R. 2006. Image alignment and stitching: a tutorial. Found. Trends. Comput. Graph. Vis. 2 (January), 1–104. ROTTENBERG ,U.,O OSTERLEE ,C., AND CH ULLER , A. 2001. Multigrid . Academic Press. ELLS , III, W. M. 1986. Efficient synthesis of Gaussian filters by cascaded uniform filters. IEEE Trans. Pattern Anal. Mach. Intell.\n\n, 2 (March), 234–239." ]

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[ "# 3 A scattering event\n\nA warming-up exercise, to begin with. Imagine two billiard balls coasting toward each other with equal speeds, then colliding. Since initially they move in opposite directions, they will also do so after the collision, and their speeds will again be equal. If the collision is perfectly head-on, each ball simply reverses its direction of motion. But suppose that the collision is uncontrollably somewhat off-center, so that each ball veers from its original direction of motion by some angle α. We cannot then predict whether the balls will scatter at (say) right angles (90° plus/minus a small angle), but we can estimate the probability with which they will do so.\n\nThere are two ways in which the balls can scatter at right angles, each with its own probability. If we call the incoming balls N and S (suggesting that they move northward and southward, respectively), we may call the outgoing balls E and W, and we may write p1 = p(N→E,S→W) and p2 = p(N→W,S→W). If either ball is as likely to be scattered eastward as westward, the two probabilities are equal, and the probability with which the balls will scatter at right angles is the sum\n\np = p(N→E,S→W) + p(N→W,S→E).\n\nThis is part of standard probability theory: an event that can come to pass in exactly two ways, with respective probabilities p1 and p2, comes to pass with probability p = p1 + p2.\n\nNow we come to the real McCoy. Instead of billiard balls we use particles. We again assume that initially the two objects move in (more or less) opposite directions with (more or less) equal speeds, so that the same will be true after the… scattering event. (Where particles are concerned, the word “collision” invokes the wrong images.) We also assume that the scattering is elastic. This means that no particles are created or annihilated during the event. If the incoming particles (and thus the outgoing ones) are of different types, then it is possible to learn which outgoing particle is identical with which incoming particle, and Rule A applies: first square the magnitudes of the amplitudes A1 = A(N→E,S→W) and A2 = A(N→W,S→E) associated with the alternatives, then add the results:\n\np = |A(N→E,S→W)|2 + |A(N→W,S→E)|2.\n\nThis is again the sum of two probabilities, p1 = |A1|2 and p2 = |A2|2. If there are no preferred directions (due to external forces, particle spins, and such) then the two probabilities are equal, so that |A2| = |A1|. Still nothing to write home about.", null, "Figure 1.3.1 The alternatives that contribute to the probability with which two particles scatter elastically at right angles. The lines do not represent trajectories; they merely indicate possible identities between the incoming and outgoing particles.\n\nBut now suppose that the conditions stipulated by Rule B are met: there is no actual event or state of affairs from which one could learn which of the alternatives took place. (This rules out, among other things, that the incoming particles are of different types.) In this case we must first add the amplitudes and then square the magnitude of the result:\n\np = |A(N→E,S→W) + A(N→W,S→E)|2.\n\nPreviously we were adding real numbers; now we are adding complex numbers. When we were adding the squares of the magnitudes |A1| and |A2|, the phases of the amplitudes A1 and A2 were irrelevant. Now we need to think about the angle between the two complex numbers (visualized as arrows) or (what comes to the same) about the difference between their phases.\n\nTake a look at Fig. 1.3.1 above. Observe that exchanging either the incoming or the outgoing particles is tantamount to exchanging the two alternatives and, correspondingly, the two amplitudes, so that A2 takes the place of A1 and vice versa. Since the two amplitudes have the same magnitude, there is a complex number c of unit magnitude such that A2 = A1 c. In other words, multiplication by c = [1:β] represents an exchange of the incoming or outgoing particles.\n\nIf the incoming or outgoing particles are exchanged twice, then (i) A1 gets multiplied by c2 and (ii) the original situation is restored. Thus A1 = A1 c2, whence it follows that c2 = [1:2β] = 1. This means that 2β must be equal to an integral multiple of 360°, and this leaves us with two possibilities: β = 0°, in which case A2 = A1, or β = 180°, in which case A2 = −A1.\n\nNature makes use of both possibilities. Every particle is either a boson or a fermion. Interchanging two indistinguishable bosons leaves the amplitudes unchanged, whereas interchanging two indistinguishable fermions is accompanied by a change of sign. Writing A for A1, we thus have\n\npb = |A + A|2 = |2A|2 = 4 |A|2\n\nin the case of bosons, and\n\npf = |A – A|2 = 0\n\nin the case of fermions. Compare this with the result we obtained for distinguishable particles:\n\np =|A|2 + |A|2 = 2 |A|2.\n\nIndistinguishable bosons are thus twice as likely to scatter at right angles as particles that carry “identity tags” of some sort, whereas the probability with which indistinguishable fermions scatter at right angles is zero, rather than some positive number.\n\nNow suppose that an initial measurement indicated that two indistinguishable particles are headed northward and southward, respectively, and that the next (relevant) thing that can be deduced from an actual event or state of affairs is that two particles are headed eastward and westward, respectively. Rule B therefore applies. Can we nevertheless assume that what actually happened is either of the alternatives? If we can, then the following questions have answers: Is E the same as N (in which case W is the same as S)? Is E the same as S (in which case W is the same as N)? Which incoming particle is identical with which outgoing particle?\n\nSuppose that these questions have answers. If what actually happens during the scattering event corresponds to either of the alternatives illustrated in Fig. 1.3.1, then the probability with which the two particles scatter at right angles is the sum of the probabilities of the alternatives:\n\np = p1 + p2x with xp1 = |A|2x and xp2 = |A|2.\n\nIn reality we have that\n\npb = |A + A|2 = |2A|2 = 4 |A|2x andx pf = |A — A|2 = 0.\n\nWe have arrived at a contradiction, and this means we have made a wrong assumption. What actually happens is neither of the alternatives. Those questions have no answers.\n\nThen what is it that actually happens? We will return to this question.\n\n→ Next" ]

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[ "Anda di halaman 1dari 4\n\n# Excerpt from the Proceedings of the COMSOL Conference 2009 Milan\n\nA Study of Optical Sensor Based on Fiber Bragg Grating (FBG) Using COMSOL Multiphysics\nC. Gavrila*1, I. Lancranjan2 1 Technical University of Civil Engineering Bucharest, Romania, 2 Advanced Study Centre - National Institute for Aerospace Research Elie Carafoli, Bucharest, Romania\n*Corresponding author: 66, Pache Protopopescu Blvd, Sector 2, 021414 Bucharest, Romania, [email protected]\n\n## Abstract: Fiber optic sensors can measure a large\n\nrange of physical, chemical and environmental variables such as temperature, pressure, shape, position, chemical concentration, moisture, etc. Fiber optic sensors provide measurements in applications where the conventional electrical based sensors cannot be used, due to measurement requirements such as extreme temperature, small size, high sensor count, or high electromagnetic energy or radiation environments. In this paper, we propose a study of an optical sensor based on a Fiber Bragg Grating (FBG) setup arrangement using COMSOL Multiphysics. The effect of environmental parameters on the composite material machine part is observed by the modification of the length (L) of the Fabry-Perot interferometer formed by two Bragg grating mirrors. This variation can be studied by a transmission spectroscopy measurement. The developed COMSOL Multiphysics sensor model takes into account the interaction of Fiber Bragg Grating (FBG) with composite material. Keywords: Fabry-Perot Interferometer, optic fiber, Bragg grating reflector\n\n## important structural materials for smart structure applications [1 7].\n\n2. Theory\nIn Figure 1, the schematics of a fiber optic sensor using a Fabry-Perot interferometer (FPI) is presented. It is formed by two Bragg grating reflectors of reflectance R1 and R2 separated by a fiber optic portion of length L.\n\n## Figure 1 - Schematic of the analyzed Fiber Bragg Grating sensor.\n\nThe individual Bragg grating reflectors in the FPI can be characterized by transmittances Ti and reflectance Ri, i = 1, 2, such that Ri + Ti = 1 The FabryPerot reflectance RFP and transmittance TFP are found to be\nR FP = T FP = R1 + R 2 + 2 R1 R 2 cos 1 + R1 R 2 + 2 R1 R 2 cos T1T2 1 + R1 R 2 + 2 R1 R 2 cos\n\n1. Introduction\nThe paper presents an attempt in modeling a new class of sensors used into an increasing number of applications, namely the Bragg grating fiber optic sensors. In-fiber grating-based sensors have many advantages over conventional electric and alternative fiber optic sensor configurations. They are relatively straight-forward, inexpensive to produce, immune to electromagnetic (EM) interference and interruption, lightweight, small in size, and self-referencing with a linear response. Most significantly, their wavelength-encoding multiplexing capability allows tens of gratings in a single piece of fiber to form an optical data-bus network. The combination of their multiplexing capability and inherent compatibility with fiber reinforced composite materials permits infiber gratings to be embedded in a number of\n\n(1) (2)\n\nRFP represents the ratio of the power reflected by the FPI, Pr, to the incident power on the FPI, Pi. TFP is the ratio of the transmitted power Pt to the incident power, Pi. represents the round-trip propagation phase shift in the interferometer, defined by: 4nL = (3)\n\nIn Eq. (3) n is the refractive index of the region between the mirrors and the free space optical wavelength. It has been assumed that the light experiences a /2 phase shift at each reflection, as appropriate for dielectric mirrors, which is added to the propagation phase shift of Eq. (3).\n\nIt is evident from Eq. (2.2) that TFP is a maximum for cos = 1 (4) and = (2m + 1) (5) with m an integer. If we define = (2m + 1) (6) then near a maximum in TFP, 2 (7) cos 1 2 with 1 (8) In the case that the mirror reflectances are equal and approach unity, then Eq. (2) simplifies to T2 T FP = (9) (1 R ) 2 + R2 where R = R1 = R2 and T = 1 R. The maximum transmittance occurs when = 0 [1-5]. Another limiting case where the mirror reflectances are low is of particular interest in the case of the fiber FabryPerot sensors. Assuming once again that the mirrors have equal reflectances, it follows from Eqs. (1) and (2) that if R << 1, then R FP = 2 R (cos + 1) (10) and T FP 1 2 R (cos + 1) (11) The fiber FabryPerot sensors that have evolved from early works are generally classified as intrinsic or extrinsic. In both intrinsic and extrinsic sensors, a fiber (in most cases, single mode) transports light from an emitter to the interferometer and from the interferometer to a photo detector. In an intrinsic fiber FabryPerot interferometer sensor, generally termed an FFPI sensor, the two mirrors are separated by a length of single-mode fiber and the measured affects the optical path length of the light propagating in the fiber itself. In an extrinsic fiberbased FabryPerot sensor, generally referred to as an EFPI sensor, the two mirrors are separated by an air gap or by some solid material other than the fiber. Thus, in the EFPI sensor the measured affects the optical path length in a medium other than the fiber that transports the monitoring light to and from the interferometer. Both FFPIs and EFPIs are designed such that a measured affects the optical length of the cavity, and light reflected or transmitted by the interferometer is converted by a photo detector to an electrical signal that is processed electronically to evaluate the measured. The analysis of the schematic of a Fiber Bragg Grating sensor presented in Figure 1 leads to an elementary study of Bragg grating reflector. It consists of a short section of single-mode optical\n\nfiber in which the core refractive index is modulated periodically. As depicted in Figure 2, this structure acts as a highly wavelength-selective reflection filter with the wavelength of the peak reflectivity, B, determined by the phase matching condition.\n\nFigure 2 - Schematic diagram of structure and spectral response of fiber Bragg grating.\n\nThe wavelength of the peak reflectivity, B, is defined as B = 2n eff (12) where neff is the effective refractive index of the guided mode in the fiber, and is the period of the refractive index modulation with a form of 2z n( z ) = n co + n cos (13) + 1 In Eq.(13) nco is the unexposed core refractive index and n is the amplitude of the photo-induced index excursion. This periodical index-modulated structure enables the light to be coupled from the forwardpropagating core mode into the backwardpropagating core mode, generating a reflection response [1 7]. The sensing function of an FBG derives from the sensitivity of both the refractive index and grating period to externally apply mechanical or thermal perturbations. The strain field affects the response of an FBG directly, through the expansion and compression of the grating pitch size and through the strain-optic effectthat is, the strain-induced modification of the refractive index. The temperature sensitivity of an FBG occurs principally through the effect on the induced refractive index change and, to a lesser extent, on the thermal expansion coefficient of the fiber. Thus, the peak reflected wavelength shifts by an amount B in response to strain and temperature change T as given by B = Pe + Pe s f + T (14)\n\n[ (\n\n) ]\n\nThe sensor function of the setup presented in Figure 1, a setup based on the use of a Fabry Perot etalon formed of two Bragg grating reflectors, is\n\n## 3. Results and Discussion\n\nThe COMSOL Multiphysics program is used to simulate the propagation of the test laser beam inside a mono mode fiber optic in order to evaluate the Fabry-Perot interferometer transmittance. The variations of the two Bragg reflectors composing the Fabry-Perot interferometer are considered. According to the COMSOL Multiphysics simulation results, the role of Bragg grating reflectivity variation is dominant into the total Fabry-Perot etalon transmittance. We select 3D as the Space Dimension, then in the list of Physical Models the following menu link is selected: COMSOL Multiphysics > PDE Modes > Classical PDEs > PDE, General Form. We build the geometry of the Fabry-Perot interferometer composed of two Bragg grating reflectors embedded into a piece of composite material of a common structure for aeronautical applications. In the next step we fix the boundary settings, the mesh parameters (Figures 3, 4 and 5) and compute the final solution, namely the variation of Fabry-Perot etalon transmittance with its length, for various values of Bragg grating reflectivity (Figure 6).\n\nFigure 5 - Geometry of the Bragg grating fiber optic embedded composite material. The COMSOL mesh grid can be observed.\n\nIn order to obtain more realistic results of the COMSOL Multiphysics simulation of the FabryPerot interferometer with Bragg reflectors embedded into a composite material, several usual receipts of it were considered. In Table 1 the materials used as the ingredients of these receipts are presented. The basic rule of thumb of these receipts was that the external layers of the analyzed composite materials are made of glass fiber, Kevlar or carbon fiber. In Table 1 the considered thickness of each layer is mentioned.\nTable 1: The materials used for designing the composite material test piece containing the embedded Bragg grating fiber optic.\n\nMaterial COREMAT Lantor Glass fiber sheet Carbon fiber sheet Kevlar sheet Epoxy Resin\n\n## Thickness 3 5 mm 250 500 m 250 500 m 250 500 m 250 500 m\n\nFigure 3 - Geometry of the Bragg grating fiber optic embedded composite material.\n\nFigure 4 - Details of the geometry of the Bragg grating fiber optic embedded composite material. The core and cladding of Bragg grating fiber optic can be observed.\n\nThe parameters of Bragg grating optic fiber were taken from common data sheets, especially the geometric ones. This means that for building the COMSOL Multiphysics model geometry the total optic fiber diameter was considered to be of 250 m. The core of the analyzed fiber optic sensor was considered of 75 m. The very different values of the geometric parameters created some difficulties in building the COMSOL Multiphysics model geometry. The strain and/or temperature mechanical loads of the composite material test pieces were considered to modify the length L of the Fabry Perot etalon, and implicitly its transmittance. The variation of the two Bragg grating reflectivity was modeled by considering the temperature and/or strain induced variation of the effective refractive index of the guided mode in the fiber, neff and of the period of the refractive index modulation, . In Figure 6 the simulation results obtained in the case of Fabry-Perot etalon transmittance variation with L for several values of Bragg grating reflectivity\n\nR1 and R2 are presented. The Bragg grating reflectivity R1 and R2 were considered into two value domains, namely very high (>0.90- black curve) and very low (< 0.4 red curve).\n\n5. References\n1. I. Bennion, J. A. R. Williams, L. Zhang, K. Sugden, and N. J. Doran, UV-written in-fiber Bragg gratings, Opt. Quantum Electron., 28, pp. 93135, 1996. 2. R. Kashyap, Fiber Bragg Gratings, Academic Press, New York, 1999. 3. A. Othonos and K. Kalli, Fibre Bragg Gratings: Fundamentals and Applications in Telecommunications and Sensing, Artech House, London, 1999. 4. K. O. Hill and G. Meltz, Fiber Bragg grating technology fundamentals and overview, J. Lightwave Tech., 15, pp. 12631276, 1997. 5. Y. J. Rao, In-fiber Bragg grating sensors, Measurement Sci. Tech., 8, pp. 355375, 1997. 7. S. Kannan, J. Z. Y. Guo, and P. J. Lemaire, Thermal stability analysis of UVinduced fiber Bragg gratings, J. Lightwave Tech., 15, pp. 14781483, 1997. 8. S. M. Melle, K. Liu, and M. Measures, A passive wavelength demodulation system for guided-wave Bragg grating sensors, IEEE Photon. Technol. Lett., 4, 5, pp. 516 518, 1992. 9. M. A. Davis and A. D. Kersey, All-fiber Bragg grating strain-sensor demodulation technique using a wavelength division coupler, Electron. Lett., 30, pp. 7577, 1994. 10. Q. Zhang, D. A. Brown, H. Kung, J. E. Townsend, M. Chen, L. J. Reinhart, and T. F. Morse, Use of highly overcoupled couplers to detect shifts in Bragg wavelength, Electron. Lett., 31, 6, pp. 480482, 1995. 11. A. B. Lobo Ribeiro, L. A. Ferreira, M. Tsvetkov, and J. L. Santos, All fiber interrogation technique for fiber Bragg sensors using a biconical fiber filter, Electron.Lett., 32, 4, pp. 382383, 1996.\n\nFigure 6 The simulation results obtained for the variation of Fabry-Perot etalon transmittance with its length. Black curve - R1=0.90 and R2=0.95 Red curve - R1=0.45 and R2=0.38.\n\n4. Conclusions\nIn this paper we have demonstrated the versatility of COMSOL Multiphysics regarding the modeling and simulation of fiber optic sensor based on the use of the Fabry - Perot etalon composed of Bragg grating reflectors. The obtained COMSOL Multiphysics models are under development for fulfillment of aeronautic industry design needs. The considered development includes comparison with experimental results." ]

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[ "# Algebra 2 Linear Piecewise Functions Worksheet\n\nAlgebra 2 Linear Piecewise Functions Worksheet – The goal of Algebra Worksheets is to help students master the fundamentals of mathematics in an easy manner. This subject involves the investigation of mathematical symbols and the rules that govern their manipulation. It is the common thread for all maths that includes geometry and physics. It is a crucial component of your studies. This section will allow you to learn all about algebra by covering the most significant topics. These worksheets can help enhance your abilities as student.", null, "Apart from learning to work out equations, the worksheets also teach students how to solve word problem problems that involve algebra. They can employ fractions, integers , and decimals in order to solve problems, modify equations and test formulas. These worksheets are able to be used for both metric and customary units. They also provide the various forms of linear equations, finding the x-intercept and solving word problems that involve parallel lines.\n\nYou can master quadratic equations using the top algebra worksheets. The majority of them are printable, which means that you can print them out without the need to download any additional software. They’re designed for students to grasp the basics of algebra without having to worry about using a calculator or spending time searching for the perfect answer. These worksheets can be used for reinforcement of concepts and to enhance the skills you have. The following are some helpful resources for helping you improve your mathematical capabilities.", null, "These worksheets in algebra provide students with an understanding of the basics of math. They teach students how to use fractions as a way to represent numbers. They also teach students how to solve equations that contain many unknown variables. These printable resources are a great way to help you master math. They will also help you improve your understanding of equations. If you need help increasing your math knowledge, make sure you have a quality set of workbooks.\n\nYou can improve your math skills making use of algebra worksheets free. The worksheets are available for free and can be used to practice basic math principles. They will help you understand the basics of algebra and apply the fundamentals of this subject in your life. If you’re a student consider them a valuable source of information in your studies. They can make you more effective and successful. These printables are sure to be popular with your students!", null, "" ]

[ null, "https://algebraworksheets.co/wp-content/uploads/2022/02/algebra-2-linear-piecewise-functions-worksheet.jpg", null, "https://algebraworksheets.co/wp-content/uploads/2022/02/algebra-2-linear-piecewise-functions-worksheet-1.png", null, "https://algebraworksheets.co/wp-content/uploads/2022/02/algebra-2-linear-piecewise-functions-worksheet-837x1024.png", null ]

[ "# Completing the result set step by step\n\nSelect two employees (one male and the other female) from each department as a pair to play games.\n\n### SQL\n\n```WITH A AS\n(SELECT name,dept,\nrow_number() OVER(PARTITION BY dept ORDER BY 1)\nsequence_no FROM employee WHERE gender='male'),\nB AS\n(SELECT name,dept,\nrow_number() OVER(PARTITION BY dept ORDER BY 1)\nsequence_no FROM employee WHERE gender='female')\nSELECT name,dept FROM A\nWHERE dept IN ( SELECT DISTINCT dept FROM B ) AND sequence_no=1\nUNION ALL\nSELECT name,dept FROM B\nWHERE dept IN ( SELECT DISTINCT dept FROM A ) AND sequence_no=1\n```\n\nSometimes, it is not difficult to get the result set step by step using a particular algorithm. However, SQL does not support the stepwise computation, forcing the user to figure out a way very different from their natural mental process to complete the computation in one statement. It is not only difficult to understand, but also inefficient to compute.\n\n### SPL\n\n A B C 1 =[ ] =demo.query(\"select * from employee\").group(dept) Result is stored in A1 2 for B1 =A2.group@1(gender) Loop through each group to pick a pair (one male, one female) 3 =if B2.len()>1 >A1=A1|B2 Add to the result set when a pair is picked.\n\nWith support for stepwise computation and program logics, SPL allows users to compute the result by steps they naturally understand.\n\nIn order to simplify the SQL statement as much as possible in the examples, the window functions of SQL 2003 standard are widely used, AND accordingly the Oracle database syntax which has best support for SQL 2003 is adopted in this essay." ]

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[ "# Transaction Details\n\nTransaction Hash\n0x8d9deb65be9d8959d46a554654fe03fdf8056a2e88c96cc67ccd2ce0db809fd4\nResult\nSuccess\nStatus\nConfirmed\nConfirmed by 7,815,497\nBlock\n10891738\nTimestamp\n| Confirmed within 5.3 seconds\nFrom\n0x48e1f7–329815\nInteracted With (To)\nCreate2F..ry (0xc2f81d–807315)\nValue\n0 LYX\nTransaction Fee\n0.03097267 LYX\nGas Price\n10 Gwei\n\nGas Limit\n3,200,000\nGas Used by Transaction\n3,097,267 | 96.79%\nNoncePosition\n17000\nRaw 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Input\n\n Method Id `0xcdcb760a` Call `deploy(bytes32 salt, bytes bytecode)`\nName Type Data\nsalt bytes32\n``0x13652138045e6b70a6e3bb55d8230348b87d973bc554ca35a8b25b0b19dc49e4``\nbytecode 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\"gasUsed\": \"0x223AC2\" }, \"subtraces\": 0, \"traceAddress\": [ 0 ], \"type\": \"create\" }]``" ]

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[ "Main Page | See live article | Alphabetical index\n\n# Geometric algebra\n\nDavid Hestenes et al.'s geometric algebra is a radical reinterpretation of seemingly harmless Clifford algebras over the reals (or, stated ironically, return to the original name and interpretation intended by William Clifford). The key ingredient of this formulation is the (natural) correspondence between geometric entities and the elements of the associative algebra. Contrary to the claims its proponents make, the \"mixing\" of quantities of different grades is helpful only by the virtue of the computational advantages offered by associativity (and inverse of vectors), not particularly in visualization or conceptualization. The applications in physics are valuable, and Hestenes' unusual stand in favor of using real numbers only (expelling complex numbers) as the underlying field, has given the reformulation a sort of a distinction, given the fact that elements having a negative square are ubiquitous in the algebra.\n\nWe take to be the field for a linear space (otherwise known simply as a vector space, but following Hestenes I will reserve the word for the space of first grade elements) of dimention . The outer product (the exterior product, or the wedge product) is defined such that the graded algebra (exterior algebra of Hermann Grassmann) of multivectors is generated. The geometric algebra is the algebra generated by the geometric product (which is to be thought of as more fundamental) with (for all multivectors )\n\n1. Associativity\n2. Distributivity over the addition of multivectors: and\n3. Contraction for any \"vector\" (a grade one element) is a scalar (real number)\n\nWe call this algebra a geometric algebra\n\nThe connection between Clifford algebras and quadratic forms come from the distinctive contraction property. This rule also gives the space a metric defined by the naturally derived inner product. It is to be noted that in geometric algebra in all its generality there is no restriction whatsoever on the value of the scalar, it can very well be negative, even zero (in that case, the possibility of an inner product is ruled out if you require ).\n\nThe usual dot product and cross product of traditional vector algebra (on ) find their places in geometric algebra as the inner product\n\n(which is symmetric) and the outer product\n\nwith\n\n(which is antisymmetric). Relevant is the distinction between axial and polar vectors in vector algebra, which is natural in geometric algebra as the mere distinction between vectors and bivectors (elements of grade two). The here is the unit pseudoscalar of Euclidean 3-space, with establishes a duality between the vectors and the bivectors, and is named so because of the expected property .\n\nOne more useful example to convince yourself is to consider , and to generate , one instance of geometric algebra specifically dubbed spacetime algebra by Hestenes (not without reason!). Electromagnetic field tensor, in this context, becomes just a bivector where the imaginary unit, not surprizingly, is the volume element, giving an example of the geometric reinterpretation of the traditional \"tricks\", making them meaningful. Boosts in this Lorenzian metric space have the same expression as rotation in Euclidean space, where is of course the bivector generated by the time and the space directions involved, whereas in the Euclidean case it is the bivector generated by the two space directions, strengthening the \"analogy\" to almost identity." ]

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[ "[Jonathan Castello #27 wrote:](https://forum.azimuthproject.org/discussion/comment/17868/#Comment_17868)\n\n> Matthew, can you elaborate on why reducing Petri net reachability to SAT would imply \\$$\\text{EXPSPACE} \\subseteq \\text{NP}\\$$? Is Petri net reachability known to be EXPSPACE-complete? I don't think you're necessarily wrong, but the critical step is eluding me.\n\nAs I mentioned above, the [Cardoza, Lipton and Meyer (1976)](https://dl.acm.org/citation.cfm?id=803630) establish that reachability for *symmetric* Petri nets is \\$$\\textsf{EXPSPACE}\\$$-complete.\n\nI didn't know this when I wrote my argument yesterday, I had to look it up.\n\nIf we let \\$$\\textsf{PETRI-REACH}\\$$ be the class of problems reducible to Petri net reachability, then \\$$\\textsf{EXPSPACE} \\subseteq \\textsf{PETRI-REACH}\\$$\n\n> Approaching this similarly: We have an exponential lower bound on space for Petri net reachability. As you said, this necessarily imposes an exponential lower bound on time, since you can only write one cell per unit time (per tape). Suppose a reduction to SAT existed. If SAT had a subexponential algorithm, then we could defeat the exponential lower bound; so SAT, and by extension every NP-Complete problem, must not be solvable in subexponential time. Therefore, \\$$\\text{P} \\ne \\text{NP}\\$$.\n\nThis is good, but we can do better I believe.\n\nNot only does \\$$\\textsf{PETRI-REACH} \\subseteq \\textsf{NP} \\implies \\textsf{NP} \\neq \\textsf{P}\\$$, but in fact we have the stronger result:\n\n$$\\textsf{NP} \\subsetneq \\textsf{PETRI-REACH}$$\n\n**Proof.**\n\nIt's well known that \\$$\\textsf{NP} \\subseteq \\textsf{PSPACE}\\$$ (see, for instance [Arora and Barak (2007), §4.2, pg. 78](http://theory.cs.princeton.edu/complexity/book.pdf)).\n\nWe also know that \\$$\\mathsf{PSPACE} \\subsetneq \\mathsf{EXPSPACE}\\$$ from the [space hierarchy separation theorem](https://en.wikipedia.org/wiki/Space_hierarchy_theorem).\n\nFinally we have \\$$\\mathsf{EXPSPACE} \\subseteq \\textsf{PETRI-REACH}\\$$ by [Cardoza et al. (1976)](https://dl.acm.org/citation.cfm?id=803630).\n\nHence \\$$\\textsf{NP} \\subsetneq \\textsf{PETRI-REACH}\\$$.\n\n\\$$\\Box\\$$" ]

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[ "×\n×\n\n# For each integer n with n 2, let P (n) be the formula a.", null, "ISBN: 9780495391326 48\n\n## Solution for problem 4E Chapter 5.2\n\nDiscrete Mathematics with Applications | 4th Edition\n\n• Textbook Solutions\n• 2901 Step-by-step solutions solved by professors and subject experts\n• Get 24/7 help from StudySoup virtual teaching assistants", null, "Discrete Mathematics with Applications | 4th Edition\n\n4 5 1 382 Reviews\n23\n4\nProblem 4E\n\nFor each integer n with n ≥ 2, let P (n) be the formula", null, "a. Write P (2). Is P (2) true?\n\nb. Write P(k).\n\nc. Write P(k + 1).\n\nd. In a proof by mathematical induction that the formula holds for all integers n ≥ 2, what must be shown in the inductive step?\n\nStep-by-Step Solution:\nStep 1 of 3\n\nDiscrete Mathematics CS225 Terms and concepts: Week 2 Reading 145-159, 165-167. 183-184. 201-203 and Lectures and Supplemental Info List of Types of Numbers: • Natural numbers ( ℕ ): Counting numbers. {0, 1, 2, 3…} • Integers ( ℤ ): Positive and negative counting numbers. {…-2, -1, 0, 1, 2, …} • Rational numbers ( ℚ ): Numbers that can be expressed as a ratio of...\n\nStep 2 of 3\n\nStep 3 of 3\n\n##### ISBN: 9780495391326\n\nUnlock Textbook Solution" ]

[ null, "https://studysoup.com/cdn/13cover_2610080", null, "https://studysoup.com/cdn/13cover_2610080", null, "https://lh5.googleusercontent.com/d51Gp4IX3VqH4wdDOwlXSIhtutIQLg7jDH3JQVC6mfGsJO4FGhqM5AWKCBcTuE1QzPcv97rMSS8eraxOmgBc0KIqXeisLw345xhQxH5NGWsr7vNiJJ1wd0GAd0K6GvYdmaYx1LN-", null ]

[ "# MATLAB Program for FIR(Finite Impulse Response) Filter using Rectangular Window | IT1254 - DSP and Communications Systems Lab\n\nAIM:\nTo write a program for FIR(Finite Impulse Response) filter like Low pass FIR filter, High pass FIR filter, Band pass FIR filter and Band stop FIR filter using Rectangular window using MATLAB in IT1254 - DSP and Communications Systems Lab.\n\nALGORITHM:\n\nLOW PASS FILTER:\nStep 1: Read the input sequence\nStep 2: Perform low pass filter calculations\nStep 3: Plot the output sequences\n\nHIGH PASS FILTER:\nStep 1: Read the input sequence\nStep 2: Perform high pass filter calculations\nStep 3: Plot the output sequences\n\nBAND PASS FILTER:\nStep 1: Read the input sequence\nStep 2: Perform band pass filter calculations\nStep 3: Plot the output sequences\n\nBAND STOP FILTER:\nStep 1: Read the input sequence\nStep 2: Perform band stop filter calculations\nStep 3: Plot the output sequences\n\nPROGRAM:\nclc;\nclear all;\nclose all;\nrp=input('Enter the passband ripple(rp):');\nrs=input('Enter the stopband ripple(rs):');\nfp=input('Enter the passband frequency(fp):');\nfs=input('Enter the stopband\nfrequency(fs):');\nf=input('Enter the sampling frequency(f):');\nwp=2*fp/f;\nws=2*fs/f;\nnum=-20*log10(sqrt(rp*rs))-13;\ndem=14.6*(fs-fp)/f;\nn=ceil(num/dem);\nn1=n+1;\nif(rem(n,2)~=0)\nn1=n;\nn=n-1;\nend\ny=boxcar(n1);\n%Low pass filter\nb=fir1(n,wp,y);\n[h,o]=freqz(b,1,256);\nm=20*log10(abs(h));\nsubplot(2,2,1);\nplot(m);\nylabel('Gain(db)->');\nxlabel('(a)Normalised frequency->');\n%High pass filter\nb=fir1(n,wp,'high',y);\n[h,o]=freqz(b,1,256);\nm=20*log10(abs(h));\nsubplot(2,2,2);\nplot(m);\nylabel('Gain(db)->');\nxlabel('(b)Normalised frequency->');\n%Band pass filter\nwn=[wp*ws];\nb=fir1(n,wn,y);\n[h,o]=freqz(b,1,256);\nm=20*log10(abs(h));\nsubplot(2,2,3);\nplot(m);\nylabel('Gain(db)->');\nxlabel('(c)Normalised frequency->');\n%Band stop filter\nwn=[wp*ws];\nb=fir1(n,wn,'stop',y);\n[h,o]=freqz(b,1,256);\nm=20*log10(abs(h));\nsubplot(2,2,4);\nplot(m);\nylabel('Gain(db)->');\nxlabel('(d)Normalised frequency->');\nOUTPUT:", null, "Click to view full size image!\n\nRESULT:\nThus the program for FIR Filter using Rectangular window was performed using MATLAB and the output sequences were drawn.\n\nPrevious\nNext Post »" ]

[ null, "https://3.bp.blogspot.com/_V648gqhagYA/TPzt8DL8qcI/AAAAAAAAAdY/djnez36gZVk/s320/FIR-Filter-using-Rectangular-Window.PNG", null ]

[ "Cody\n\n# Problem 44240. Area of a Square\n\nSolution 2714628\n\nSubmitted on 20 Jul 2020 by Raghunadh N\nThis solution is locked. To view this solution, you need to provide a solution of the same size or smaller.\n\n### Test Suite\n\nTest Status Code Input and Output\n1   Pass\nx = 1; A = 1; assert(isequal(sq_area(x),A))\n\n2   Pass\nx = 2; A = 4; assert(isequal(sq_area(x),A))\n\n3   Pass\nx = 6; A = 36; assert(isequal(sq_area(x),A))\n\n4   Pass\nx = 8; A = 64; assert(isequal(sq_area(x),A))\n\n### Community Treasure Hunt\n\nFind the treasures in MATLAB Central and discover how the community can help you!\n\nStart Hunting!" ]

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[ "# What is constructive wave interference mean?\n\n1", null, "Date created: Wed, Jun 23, 2021 10:01 PM\nDate updated: Sat, May 28, 2022 7:30 PM\n\nContent\n\nVideo answer: What do you mean by constructive and destructive interferences ? | 12 | light wave and interfer...", null, "## Top best answers to the question «What is constructive wave interference mean»\n\n• The wave interference is said to be a constructive wave interference if the crest of a wave meets the crest of another wave of the same frequency at the same point. Constructive wave interference is a wave interference which occurs when in phase waves meets with each other and forms a new wave with greater amplitude.\n\nFAQ\n\nThose who are looking for an answer to the question «What is constructive wave interference mean?» often ask the following questions:\n\n### 👋 Are shock wave constructive or destructive interference?\n\n#### What's the difference between constructive and destructive interference?\n\n• Constructive interference describes a situation where two waves are added together, while in destructive interference, the two waves cancel each other out. But really, the two types of interference are a result of the same thing.\n\n### 👋 How does constructive interference affect a sound wave?\n\n• With constructive interference, two waves with the same frequency and amplitude line up – the peaks line up with peaks and troughs with troughs as in diagram A above. The result is a wave that has twice the amplitude of the original waves so the sound wave will be twice as loud.\n\n### 👋 When does constructive interference occur in a wave?\n\n• Constructive interference occurs from the superposition of two identical waves that are in phase. Destructive interference occurs from the superposition of two identical waves that are 180∘(πradians) 180 ∘ (π radians) out of phase.\n\nVideo answer: Interference of waves | superposition and interference in light and water waves | physics", null, "We've handpicked 24 related questions for you, similar to «What is constructive wave interference mean?» so you can surely find the answer!\n\nWhen do waves alternate between constructive and destructive interference?\n• As the speaker is moved back the waves alternate between constructive and destructive interference. What the example of the speakers shows is that it is the separation of the two speakers that determines whether there will be constructive or destructive interference.\nWhere do the waves of constructive interference come from?\n• A large wave pulse comes in from the left side of the screen, a smaller wave pulse comes in from the right side of the screen. Observe the waves constructively interfere with one another and continue to move in the same direction they were originally moving until they reach the end of the spring and bounce back. Loading...\nA constructive wave?\n• Constructive waves and destructive waves are two concepts widely discussed in waves and vibrations. A constructive wave is the phenomenon where two waves interfere so that the resulting amplitude is greater than the amplitude of each individual wave.\nWhat makes a constructive wave?\n\n#### What are the more destructive waves?\n\n• These waves are more destruction. The surface waves are the last to report on seismograph. These waves are more destructive. They cause displacement of rocks, and hence, the collapse of structures occurs. There are two types of body waves. They are called P- and S-waves, P-waves: They move faster and are the first to arrive at the surface.\nWhat is wave interference?\n• What is Interference? Wave interference is the phenomenon that occurs when two waves meet while traveling along the same medium. The interference of waves causes the medium to take on a shape that results from the net effect of the two individual waves upon the particles of the medium.", null, "What uses wave interference?\n\n#### What is wave interference and when does it occur?\n\n• Wave interference is the phenomenon that occurs when two waves meet while traveling along the same medium to form a resultant wave of greater, lower or the same amplitude. Interference usually refers to the interaction of waves which have constant phase difference and same or nearly same frequency.\nHow are sound resonances due to constructive and destructive interference?\n• All sound resonances are due to constructive and destructive interference. Only the resonant frequencies interfere constructively to form standing waves, while others interfere destructively and are absent.\n\n### Video answer: The original double slit experiment", null, "Which is the result of constructive interference of two waves?\n• The green wave is the result of the superposition of the two waves. When the two waves have a phase difference of zero, the waves are in phase, and the resultant wave has the same wave number and angular frequency, and an amplitude equal to twice the individual amplitudes (part (a)). This is constructive interference.\nWhat is destructive wave interference?\n• destructive interference. noun Physics. the interference of two waves of equal frequency and opposite phase, resulting in their cancellation where the negative displacement of one always coincides with the positive displacement of the other.\nWhat is light wave interference?\n\nInterference of light is the phenomena of multiple light waves interfering with one another under certain circumstances, causing the combined amplitudes of the waves to either increase or decrease.\n\n### Video answer: Superposition of waves - a level physics", null, "What is sound wave interference?\n\nWhen two or more sound waves occupy the same space, they affect one another. The waves do not bounce off of each, but they move through each other. The resulting wave depends on how the waves line up.\n\nIs sound a constructive wave?\n• Sound waves with higher amplitudes sound louder than sound waves with lower amplitudes. Constructive interference will make a sound louder while destructive interference will make a sound quieter. Two waves that add together may have different frequencies.\nWhat does a constructive wave look like?\n• The image shows a constructive wave. Constructive waves have a long wavelength and a low-frequency (8–10 waves per minute). They have a low wave height (typically under 1 metre). The wavefront is gently sloping and gains a little height, breaks and spills onto the beach. Water spreads a long way up the gently sloping beach.\nWhat is the definition of constructive wave?\n• constructive wave A wave that leads to the build-up of a beach, owing to the swash of the wave being more effective in moving material than the backwash. Usually, constructive waves are associated with low-energy conditions and a gentle offshore gradient.\nWhat is the path difference between the two reflected waves in case of constructive interference?\n\nConstructive interference occurs when the phase difference between the waves is an even multiple of π (180°), whereas destructive interference occurs when the difference is an odd multiple of π.\n\nWhat is wave action constructive or destructive process?\n• Waves can be destructive or constructive. When a wave breaks, water is washed up the beach - this is called the swash. Then the water runs back down the beach - this is called the backwash. With a constructive wave, the swash is stronger than the backwash.\nWhat are two types of wave interference?\n• There are two types of wave interference, known as constructive and destructive. If two waves meet at their greatest point, then the two waves add together; this is known as constructive. It creates a wave that’s double the size while the crests of the waves are overlapping.\nWhat is interference of a light wave?\n\n#### What are two ways waves are affected by interference?\n\n• When two waves collide, the effect is something known as wave interference. This means the waves will pass through each other but, while in the same location, interact with one another. The result is a change in amplitude, or size, of the two waves. There are two types of wave interference, known as constructive and destructive.\n\n### Video answer: 07 the sine wave and sum of waves (interference)", null, "What is the definition of interference wave?\n• In physics, interference is a phenomenon in which two waves superpose to form a resultant wave of greater, lower, or the same amplitude.\nWave interference can occur for?\n• Wave interference also called super-positioning occurs when waves cross each others paths. If matter crossed paths with other matter there would be a collision. Waves are not matter but disturbances and can travel through each other. Crossing paths causes an interference pattern while interacting in a shared space.\n\n### Video answer: Optics: destructive interference - where does the light go?", null, "" ]

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[ "# Writing Unit Tests with clojure.testλ︎\n\nUnit tests are centered on assertions, testing if something returns a true or false value.\n\n`is` function is the simplest assertion and the most common. It checks to see if an expression given is true and if so then the assertion passes. If the value is false then that assertion fails.\n\n`as` provides a way to run the same assertion with different values, testing the same function with a collection of arguments. This provides a clean way to test a function without lots of repetition.\n\n`testing` is a macro to group multiple assertions together, providing a string in which to describe the context the assertions are testing. The well worded context string is invaluable for narrowing down on which assertions are failing.\n\n`deftest` is a collection of assertions, with or without `testing` expressions. The name of the deftest should be the name of the function it is testing with `-test` as a postfix. For example, the function `practicalli.playground/calculator` would have a `deftest` called `practicalli.playground-test/calculator-test`\n\n## Requiring Namespacesλ︎\n\nA test namespace has a singular purpose to test a matching src namespace.\n\nThe idiomatic approach is to `:refer` specific functions from `clojure.test` as those functions are used.\n\nThe namespace to be tested is referred using a meaningful alias. The alias highlight the exact functions being tested in the body of the code. This provides a visual way to separate functions under test with other test functions, especially if there are helper functions or vars used for test data.\n\n``````(require '[clojure.test :refer [are deftest is testing]])\n``````\nThe namespace under test should be referred using the alias so they are readily identified within the test code.\n``````(require '[practicalli.gameboard.spec :as gameboard-spec])\n``````\n\nAdd `clojure.test` to the namespace definition along with the namespace under test.\n\n``````(ns practicalli.app-namespace-test\n(:require '[clojure.test :refer [are deftest is testing]]\n[practicalli.gameboard.spec :as gameboard-spec]))\n``````\n\n## Simple Exampleλ︎\n\n``````(deftest public-function-in-namespace-test\n(testing \"A description of the test\"\n(is (= 1 (public-function arg)))\n(is (predicate-function? arg))))\n``````\n\n## Assertion data setλ︎\n\nThe `are` macro can also be used to define assertions, especially when there would otherwise be multiple assertions that only differ by their test data.\n\nAn `are` assertion defines the arguments to the test, the logic of the test and a series of test data.\n\n``````(are [x y] (= x y)\n2 (+ 1 1)\n4 (* 2 2))\n``````\n\nThis is equivalent to writing\n\n`````` (do (is (= 2 (+ 1 1)))\n(is (= 4 (* 2 2))))\n``````\n\nRefactor test assertion to use data set\n\nAssertions in the test take the same shape of values, so are candidates to refactor to the `are` macro.\n\n``````(deftest encoder-test\n(testing \"Tens to number words\"\n(is (= '(\"zero\" \"ten\")\n(character-sequence->word-sequence dictionary/digit->word '(\\0 \\1 \\0))))\n(is (= '(\"zero\" \"eleven\")\n(character-sequence->word-sequence dictionary/digit->word '(\\0 \\1 \\1))))\n(is (= '(\"zero\" \"twenty\" \"zero\")\n(character-sequence->word-sequence dictionary/digit->word '(\\0 \\2 \\0))))\n(is (= '(\"zero\" \"twenty\"\"one\")\n(character-sequence->word-sequence dictionary/digit->word '(\\0 \\2 \\1))))\n(is (= '(\"zero\" \"forty\" \"two\")\n(character-sequence->word-sequence dictionary/digit->word '(\\0 \\4 \\2))))))\n``````\nRefactor the assertions using are simplifies the code, making it simpler to change further and extend with more data.\n``````(deftest encoder-test\n(testing \"Tens to number words\"\n(are [words numbers]\n(= words (character-sequence->word-sequence dictionary/digit->word numbers))\n'(\"zero\" \"ten\") '(\\0 \\1 \\0)\n'(\"zero\" \"eleven\") '(\\0 \\1 \\1)\n'(\"zero\" \"twenty\" \"zero\") '(\\0 \\2 \\0)\n'(\"zero\" \"twenty\"\"one\") '(\\0 \\2 \\1)\n'(\"zero\" \"forty\" \"two\") '(\\0 \\4 \\2)))\n``````\n\nGenerative Testing provides a wide range of values\n\nGenerating test data from Clojure Specs provides an extensive set of values that provide an effective way to test functions.\n\n## Referenceλ︎\n\n### Code challenges with testsλ︎\n\nLast update: May 15, 2023" ]

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[ "# Clock Jitter Definitions and Measurement Methods", null, "Introduction\n\nJitter is the timing variations of a set of signal edges from their ideal values. Jitters in clock signals are typically caused by noise or other disturbances in the system. Contributing factors include thermal noise, power supply variations, loading conditions, device noise, and interference coupled from nearby circuits.\n\nTypes of Jitter\n\nJitter can be measured in a number of ways; the following are the major types of jitter:\n\n• Period Jitter\n• Cycle to Cycle Period Jitter\n• Long T erm Jitter\n• Phase Jitter\n• Time Interval Error (TIE)\n\nPeriod Jitter\n\nPeriod jitter is the deviation in cycle time of a clock signal with respect to the ideal period over a number of randomly selected cycles. If we were given a number of individual clock periods, we can measure each one and calculate the average clock period as well as the standard deviation and the peak-to-peak value. The standard deviation and the peak-to-peak value are frequently referred to as the RMS value and the Pk-Pk period jitter, respectively.\n\nMany publications defined period jitter as the difference between a measured clock period and the ideal period. In real world applications, it is often difficult to quantify the ideal period. If we observe the output from an oscillator set to 100 MHz using an oscilloscope, the average measured clock period may be 9.998 nS instead of 10 nS. So it is usually more practical to treat the average period as the ideal period.\n\nPeriod Jitter Applications\n\nPeriod jitter is useful in calculating timing margins in digital systems. Consider a microprocessor- based system in which the processor requires 1 nS of data setup before clock rise. If the period jitter of the clock is -1.5 nS, then the rising edge of the clock could occur before the data is valid. Hence the microprocessor will be presented with incorrect data. This example is illustrated in Figure 1.", null, "Similarly, if another microprocessor has a data hold time requirement of 2 nS but now the clock jitter is +1.5 nS, then the data hold time is effectively reduce to 0.5 nS. Once again, the microprocessor will see incorrect data. This situation is illustrated in Figure 2.", null, "Calculating Peak to Peak Jitter from RMS Jitter\n\nBecause the period jitter from a clock is random in nature with Gaussian distribution, it can be completely expressed in terms of its Root Mean Square (RMS) value in pico-seconds (pS). However, the peak-to-peak value is more relevant in calculating setup and hold time budgets. To convert the RMS jitter to peak-to-peak (Pk-Pk) jitter for a sample size of 10,000, the reader can use the following equation:\n\nPeak-to-peak period jitter = 7.44 x (RMS jitter) Equation 1 For example, if the RMS jitter is 3 pS, the peak to peak jitter is ±11.16 pS.\n\nEquation 1 is derived from the Gaussian Probability Density Function (PDF) table. For instance, if the sample size is 100, 99 of those samples will fall within ±2.327σ from the mean value of the distribution, only 1 sample, on average, will fall outside that region. ILSI MMD measures the RMS period jitter over a sample size of 10,000 as specified by the JEDEC standard.", null, "Period Jitter Measurement Methodology\n\nPeriod Jitter is defined in JEDEC Standard 65B as the deviation in cycle time of a signal with respect to the ideal period over a number of randomly selected cycles. The JEDEC standard further specified that period jitter should be measured over a sample of 10,000 cycles. ILSI MMD recommends measuring period jitter using the following procedure:\n\n1. Measure the duration (rising edge to rising edge) of one clock cycle\n2. Wait a random number of clock cycles\n3. Repeat the above steps 10,000 times\n4. Compute the mean, standard deviation (σ), and the peak-to-peak values from the 10,000 samples\n5. Repeat the above measurements 25 times. From the 25 sets of results, compute the average peak-to peak value.\n\nThe standard deviation (σ) or RMS value computed from a measurement of 10,000 random samples (step 4) is quite accurate. The error in the RMS value can be calculated using the following equation:", null, "where σn is the RMS (or sigma) of the collected sample and N is the sample size. For a sample size of 10,000, ErrorRMS is 0.0071σn. This error is random and it follows the Gaussian distribution. The worst-case measurement error is typically computed as ±3 ErrorRMS.\n\nFor example, if the RMS value computed from 10,000 random samples is 10 pS, then the ErrorRMS will be 0.071 pS and virtually all the RMS values of this measurement will still fall within a narrow range of 10 ± 0.213 pS. In practical applications, the RMS errors in a sample set of 10,000 are small enough to be ignored.\n\nWhile an accurate RMS value can be computed from a random 10000-sample set, the peak-to- peak value is more difficult to measure. Due to the random nature of period jitter, the larger the sample size, the higher is the probability of picking up data points at the far ends of the distribution curve. In other words, the peak-to-peak value diverges instead of converging as more samples are collected. That is the reason why we added an extra step, step 5 to produce a more consistent and repeatable peak-to-peak measurement.\n\nEach measurement of 10,000 random samples (step 4) produces one standard deviation value and one peak-to-peak value. By randomly repeating this process 25 times, we could collect a good set of data points from which we can calculate the average peak-to-peak value with a high degree of accuracy. We can also compute the average RMS value from this data, but it will be very close to the RMS value derived from each individual run.\n\nFigure 3 is the period jitter histogram of a 3.3V 8102 oscillator running at 125 MHz captured by a Wavecrest SIA-4000C. It represents one set of RMS and peak-to peak values measured from 10,000 samples (step 4).", null, "Cycle to Cycle Jitter\n\nCycle to cycle (C2C) jitter is defined in JEDEC Standard 65B as the variation in cycle time of a signal between adjacent cycles, over a random sample of adjacent cycle pairs. The JEDEC standard further specified that each sample size should be greater than or equal to 1,000. Please note that cycle to cycle jitter only involves the difference in period between 2 consecutive cycles, there is no reference to an ideal cycle.\n\nCycle to cycle jitter is typically reported as a peak value in pS which defines the maximum deviation between the rising edges of any two consecutive clocks. This type of jitter specification is commonly used to illustrate the stability of spread spectrum clocks because the period jitter is more sensitive to the frequency spreading feature while C2C jitter is not. Cycle to cycle jitter is sometimes expressed as a RMS value in pS as well.\n\nCycle to Cycle Jitter Measurement Methodology\n\nILSI MMD recommends measuring cycle to cycle jitter using the following procedure:\n\n1. Measure the cycle times of two adjacent clock cycles, T1 and T2\n2. Calculate the value of T1-T2. Record the absolute value of this number.\n3. Wait a random number of clock cycles\n4. Repeat the above steps 1,000 times\n5. Compute the standard deviation (σ), and the peak value from the 1,000 samples. The peak value is the largest absolute (T1-T2) number in the data set.\n6. Repeat above measurements 25 times and calculate the average peak value from the 25 sets of results.\n\nSimilar to the peak-to-peak period jitter, the peak value of the cycle to cycle jitter also diverges instead of converging as more samples are taken. Step 6 in the procedure is added to obtain the average peak C2C jitter from 25 sample sets.\n\nFigure 4 is an example of the cycle to cycle jitter histogram. In this case, the peak cycle to cycle jitter is 25.66 pS (the bigger of the two numbers: 21.22 pS and -25.66 pS expressed in the absolute form).", null, "Long-Term Jitter\n\nLong-term jitter measures the change in a clock’s output from the ideal position, over several consecutive cycles. The actual number of cycles used in the measurement is application dependent. Long-term jitter is different from period jitter and cycle-to-cycle jitter because it represents the cumulative effect of jitter on a continuous stream of clock cycles over a long time interval. That is why long-term jitter is sometimes referred to as the accumulated jitter. Long-term jitter is typically useful in graphics/video displays and long-range telemetry applications such as range finders.\n\nILSI MMD recommends measuring long-term jitter using the following method; in this example, we measure the long-term jitter over 10,000 clocks.\n\n1. Measure the time interval of 10,000 consecutive clock cycles as shown in Figure 5\n2. Wait a random number of clock cycles\n3. Repeat the above steps 1,000 times\n4. Compute the mean, standard deviation (σ), and the peak-to-peak values from the 1,000samples\n5. Repeat above measurements 25 times. From the 25 sets of results, compute the averagepeak-to peak value.", null, "Once again, step 5 is needed to overcome the un-bounded nature of the peak to peak value.\n\nPhase Jitter\n\nPhase noise is usually described as either a set of noise values at different frequency offsets (e.g., -60 dBc/Hz at 20KHz and -95 dBc/Hz at 10MHz), or as a continuous noise plot over a range of frequencies. Phase jitter is the integration of phase noises over a certain spectrum and expressed in seconds.\n\nIn a square wave, most of the energies are located at the carrier frequency. However, some signal energies are “leaked-out” over a range of frequencies on both sides of the carrier. Phase jitter is the amount of phase noise energy contained between two offset frequencies relative to the carrier (fc). Figure 6 is an unfiltered phase noise plot and the shaded areas represent the phase jitter between frequencies f1 and f2.", null, "The RMS phase jitter between f1 and f2 is defined by equation 3.", null, "Time Interval Error (TIE)\n\nTime Interval Error (TIE) of an edge is the time deviation of that edge from its ideal position measured from a reference point. In effect, TIE is the discrete time domain representation of phase noise expressed in seconds or pico-seconds. Figure 7 illustrates the basic concept of TIE. The ideal signal is often a signal created in software from an average estimate of the signal period.", null, "Plotting TIE over time\n\nA clock waveform is shown at the top of Figure 8. The red pulses are perfectly timed clock cycles exactly 1000 pS in duration. The pulses in black are clock cycles with jitter. The trailing edges of these clock pulses have been removed to enhance the presentation. At the beginning of the sequence, both the red and black clocks are aligned to each other. Because of the jitter, the black clock edges will start shifting in time, sometimes occurring before the red clock edge and sometimes after.\n\nThe plot labeled “Clock Period” represents the measured black clock periods over time. In this example, the black clock periods are either 990 pS or 1010 pS. The “Period Change” plot depicts each cycle’s change from the previous cycle. This graph remains flat as long as the period between two consecutive black clocks stays the same. However, it will register a change whenever a period difference is detected. For example, the period of the first 4 clock cycles are constant at 990 pS, so the “Period Change” plot is flat; but when the period of the fifth clock is lengthened from 990 pS to 1010 pS, the plot reports this change by jumping to the +20 pS position. In other words, this plot identifies the period changes shown in the “Clock Period” plot.\n\nThe “Time Interval Error” (TIE) plot documents the accumulated error between the ideal edge (red clock) and the actual edge (black clock). In this example, the TIE plot begins by moving towards the negative direction because the first 4 clocks are each 10 ns shorter than the ideal period. After accumulating -40 pS in jitter error, the plot changes direction at the fifth clock and heads towards the positive direction because the fifth clock period is 10 pS longer than the ideal period.\n\nTIE measurements are especially useful when examining the behavior of transmitted data streams, where the reference clock is typically recovered from the data signal using a Clock/Data Recovery (CDR) circuit. A large TIE value may indicate that the PLL in the CDR is too slow in responding to the data stream’s changing bit rate.", null, "Jitter Measurements with a real time digital oscilloscope\n\nOscilloscope setup guidelines\n\nThe most common instrument used in measuring clock jitter is the real time digital oscilloscope (scope). This section contains general scope setup guidelines to yield better jitter measurement accuracy.\n\nThe digital scope uses an internal time base to sample its inputs at regular intervals. The sampling rate can range from 1 Gsps (giga-samples-per-second) to 40 Gsps for the high-end units. Figure 9 illustrates how the digital scope samples and displays a signal presented to its inputs. The arrows at the bottom of the figure represent the sampling points, the solid line is the actual signal, and the dots are the sampled values. The signal displayed by the scope (represented by the dotted line) is the best-fit curve across the sampled points.\n\nThe reader may notice that the sampled values do not always match that of the actual signal. These discrepancies are caused by quantization errors in the scope. Most of these errors are inherent in the design/cost tradeoffs of the scope but proper scope setup can mitigate some of the inaccuracies. In the following sections, we will explore the main causes of these errors and recommend steps to reduce their impact on jitter measurements.", null, "Front-end amplifier noise\n\nThe inputs to a digital scope passed through an analog amplifier before they are digitized by the Analog to Digital Converter (ADC). The noise produced by this amplifier is proportional to the input bandwidth of the scope: the wider the bandwidth, the higher the noise. However, reducing the bandwidth too much will affect the rising and fall time of the sampled signal, thus introducing significant errors to the jitter measurement.\n\nThe general equation describing the relationship between rise time/fall time and the bandwidth of the signal edge is:", null, "where the rise time (or fall time) is measured between the 20% and the 80% points of the signal edge. ILSI MMD recommends setting the scope bandwidth to 3 times the bandwidth of the signal. In some scopes, the bandwidth can be set only if the maximum sampling rate is selected. In other scopes, the bandwidth may not be selectable at all.\n\nQuantization noise due to vertical gain setting\n\nQuantization error is the difference between the sampled value and the actual value of the signal at the sampling point. This error is shown in Figure 9. A portion of this error in a scope is caused by the vertical gain setting of the display. If the vertical gain is set to small value, the scope may not utilize the full resolution of its internal ADC.\n\nILSI MMD recommends adjusting the vertical gain control of the scope until the signal fills the entire height of the display. In some scopes, turning the vertical gain up an additional notch so that a small portion of the signal is clipped at both the top and bottom of the display could further reduce the quantization error. This benefit is achieved because the higher vertical gain setting may cause the scope to utilize an extra bit in the ADC to digitize the signal. However, this feature is scope dependent, so please consult your scope manual.\n\nQuantization noise due to low sampling rate\n\nSome of the quantization noise described in the previous section is caused by insufficient sampling points along the horizontal axis. ILSI MMD recommends having at least 3 sampling points falling between the 20% and 80% points of a rising or falling edge. This recommendation translates into a minimum sampling rate requirement for the scope. For example, if rise time (20% – 80%) of a signal is 1 nS and 4 sampling points are needed within this time frame, then the scope must have a sampling rate better than 4 Gsps. If your scope has a higher sampling rate than the minimum requirement specified above, select the highest sampling setting.\n\nTime base jitter\n\nThe sampling points in a digital scope are generated by an internal time base. As a clock source, the time base has its own jitter characteristics and it will contribute to the jitter measurement error of a signal. Generally speaking, time base jitter should be kept below 25% of the of the expected signal jitter to support jitter measurement with better than 3% accuracy. ILSI MMD recommends using the best scope available in your laboratory to perform jitter measurements because higher end units tend to have better time base circuits with lower jitter.\n\nJitter measuring procedures using a real time digital scope\n\nMeasuring period jitter Method A\n\n1. Configure the scope to measure all the sampled clock periods instead of just measuring the first clock.\n2. Configure the scope to capture 10,000 clock cycles on the screen. For example, for a 100 MHz clock, 10K cycles = 100 uS. Since the display usually contains 10 horizontal divisions, the horizontal control should be set to 10 uS per division.\n3. Record the mean, the standard deviation, and the peak-to-peak values reported by the scope.\n4. The mean and standard deviation values are quite accurate. The peak-to-peak value, however, is not very accurate due to its unbounded nature (see section 2.1.3). Follow the next step to obtain a more accurate peak-to-peak value.\n5. Repeat steps 2 and 3 twenty five (25) times. Record the peak-to-peak value after each run and compute the average peak-to-peak value from the 25 results.\n\nMethod B (the JEDEC method)\n\n1. Configure the scope to measure the period of the first clock and turn on the histogram feature, if available.\n2. Configure the scope to capture a single clock cycle on the screen.\n3. Repeat the above step 10,000 times.\n4. Record the mean, the standard deviation, and the peak-to-peak values reported by thescope.\n5. The mean and standard deviation values are quite accurate. The peak-to-peak value,however, is not very accurate due to its unbounded nature (see section 2.1.3). Follow thenext step to obtain a more accurate peak-to-peak value.\n6. Repeat steps 2 through 4 twenty five (25) times. Record the peak-to-peak value after eachrun and compute the average peak-to-peak value from the 25 results.\n\nMeasuring cycle-to-cycle jitter\n\n• Turn on the histogram feature in your scope, if available.\n• Turn on the C2C feature in your scope. If this feature is not available, configure the scope to capture two consecutive clock cycles on the screen; subtract the period of the second clock from the period of the first clock and record the absolute value of the difference.\n• Repeat the above step 1,000 times.\n• If the scope has the histogram feature, record the standard deviation and the peak value. Ifthe histogram feature is not available, compute the standard deviation and the peak value from the 1000 data sets. The peak value is the biggest difference between any two consecutive clocks in the data set.\n• The standard deviation value is quite accurate. The peak value, however, is not very accurate due to its unbounded nature (see section 2.1.3). Follow the next step to obtain a more accurate peak value.\n• Repeat steps 2 through 5 twenty five (25) times. Record the peak value after each run and compute the average peak value.\n\nMeasuring long-term jitter Method A\n\n1. Configure the scope to capture N+1 clock cycles on the screen. N is your definition of the number of clock cycles needed in the long-term jitter measurement.\n2. Set the scope to measure the time between the rising edge of the first clock and the rising edge of clock N+1.\n3. Repeat the above steps 1,000 times\n4. Calculate the mean, the standard deviation, and the peak-to-peak values from the 1,000samples\n5. The mean and standard deviation values are quite accurate. The peak-to-peak value,however, is not very accurate due to its unbounded nature (see section 2.1.3). Follow thenext step to obtain a more accurate peak-to-peak value.\n6. Repeat steps 1 through 4 twenty five (25) times. Record the peak-to-peak value after eachrun and compute the average peak-to-peak value.\n\nMethod B\n\n1. Configure the scope to display the rising edge of a clock cycle on the screen.\n2. Assuming that the long-term jitter measurement involves N clock cycles each with a period ofT nS. Set the scope to trigger N*T nS before the displayed edge.\n3. Turn on the histogram mode to capture the 50% crossing of the waveform 1,000 or 10,000times, as the application requires. In some scope, this may require defining vertical andhorizontal thresholds for the edge detection; please refer to your scope manual.\n4. Wait until the target number of hits is recorded in the histogram. Stop the acquisition as soonas the target number is reached.\n5. The screen will show the rising edge as a band of lines (see Figure 10) and the width of theband is the long-term jitter.\n6. Record the mean, the standard deviation, and the peak-to-peak values from the scope.\n7. The mean and standard deviation values are quite accurate. The peak-to-peak value, however, is not very accurate due to its unbounded nature (see section 2.1.3). Follow the next step to obtain a more accurate peak to peak value.\n8. Repeat steps 3 through 4 twenty five (25) times. Record the peak-to-peak value after each run and compute the average peak-to-peak value.", null, "Conclusion\n\nThis application note serves two purposes. Firstly, it describes the common types of jitter that the reader may encounter in today’s high speed systems. Secondly, it provides the procedures to capture the various types of jitters using a real time digital oscilloscope." ]

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[ "## 171 cm in feet\n\n171 cm is equivalent to 5 feet and 7.32 inches. To convert 171 cm to feet, you can divide it by 30.48 (since 1 foot equals 30.48 cm): 171 cm ÷ 30.48 cm/foot = 5.61 feet The decimal part represents the additional inches. To convert it to inches, you can multiply it by 12: 0.61 … Read more\n\n## 179 cm in feet\n\n179 cm is equivalent to 5 feet and 10.47 inches. Here’s the calculation: 1 foot = 30.48 cm So, to convert 179 cm to feet, you can divide it by 30.48: 179 cm / 30.48 cm/foot = 5.87 feet The decimal part represents the additional inches. To convert it to inches, you can multiply it … Read more\n\n## 178 cm in feet\n\n178 cm is equal to 5 feet and 10.0787 inches when converted to feet and inches. To convert centimeters to feet and inches, you can use the following conversion factors: 1 inch = 2.54 cm 1 foot = 12 inches So, to convert 178 cm to feet and inches, first we need to find the … Read more\n\n## 163 cm in feet\n\n163 cm is equal to 5 feet and 4.1732 inches when converted to feet and inches. To convert centimeters to feet and inches, you can use the following conversion factors: 1 inch = 2.54 cm 1 foot = 12 inches So, to convert 163 cm to feet and inches, first we need to find the … Read more\n\n## 164 cm in feet\n\n164 cm is equal to 5 feet and 4.5669 inches when converted to feet and inches. To convert centimeters to feet and inches, you can use the following conversion factors: 1 inch = 2.54 cm 1 foot = 12 inches So, to convert 164 cm to feet and inches, first we need to find the … Read more" ]

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[ "# Clock - Aptitude Questions and Answers\n\nLearn and practice the chapter \"Clock\" with these solved Aptitude Questions and Answers. Each question in the topic is accompanied by a clear and easy explanation, diagrams, formulae, shortcuts and tricks that help in understanding the concept.\n\n## Use of Clock Questions\n\nThe questions and examples given in this section will be useful to all the freshers, college students and engineering students preparing for placement tests or any competitive exam like MBA, CAT, MAT, SNAP, MHCET, XAT, NMAT, GATE, Bank exams - IBPS, SBI, RBI, RRB, SSB, SSC, UPSC etc.\n\nPractice with this online test to crack your placements and entrance tests!\n1. If the minutes hand and seconds' hand of a clock are 25 minutes apart. What will be the angle formed between them?\n\na. 110°\nb. 120°\nc. 135°\nd. 150°\n\nExplanation:\n\nFor 60 minutes the angle covered is 360°\nSo for 25 minutes difference angle is ?\n∴ 60 x ? = 360 x 25\n\n ∴ ? = 360 x 25 = 150° 60\n\n2. At 25 minutes past 10 in the night, find the angle formed between the two hands of the clock.\n\na. 120°\nb. 126.45°\nc. 146.5°\nd. 162.5°\n\nExplanation:\n\n25 minutes past 10 in the night means, 10:25 pm\n\n 10.25 means 10 hours 25 minutes = 10 + 25 hours = 125 hours 60 12\n12hours = 360° ∴ 1 hour = 30°\n So, 125 hours x 30° = 312.5° 12\n60 minutes = 360°   ∴ 1 minute = 6°\n∴ 25 minutes means 6° x 25 = 150°\nAngle between the two hands = 312.5 - 150 = 162.5°\n\n3. A clock gains 20 seconds for every 3 hours of time. If a clock is set at a correct time of 2 am on Friday, what would it indicate at 6:30 pm, Saturday\n\na. 6.32.00 pm\nb. 6.32.46 pm\nc. 6.34.30 pm\nd. 6.38.56 pm\n\nExplanation:\n\nFrom 2.00 am Friday to 6.30 pm Saturday we have\n24 hours (till 2 am Saturday) + 12 hours (till 2 pm Saturday) + 4.5 hours = 40.5 hours\n20 seconds in 3 hours\n∴ ? seconds in 40.5 hours\n∴ ? = 270 seconds = 4.5 minutes\nTime shown = 6.30 pm + 4.5 minutes = 6 hours 34 minutes 30 seconds = 6.34.30 pm\n\n4. Find the time in the clock between 7 am and 8 am when the hours hand and minutes hand will coincide.\n\n a. 33 5 minutes past 7 am 12\n b. 38 2 minutes past 7 am 11\n c. 32 1 minutes before 8 am 11\n d. 35 7 minutes past 8 am 11\n\n Answer: b. 38 2 minutes past 7 am 11\n\nExplanation:\n\nTip:\nIn one hour, MINUTE hand travels 60 minutes\nHOUR hand travels just 5 minute spaces\nThus,\nMINUTE hand gains (60 - 5) = 55 minutes over HOUR hand in 60 minutes\n\nAt 7 o' clock hour hand and minute hand difference = 35 minutes\nSo to be together, minute hand must travel 35 minutes more.\nIn 60 minutes, minute hand gains 55 min.\nIn ? minutes it gains 35 minutes\n ∴ ? = 35 x 60 = 38 2 minutes past 7 o' clock 55 11\n\n5. Find the time between 8am and 9am when the hours and minutes hand would form a 90 degree angle between themselves.\n\n a. 25 7 minutes past 8 am 11\n b. 27 3 minutes past 8 am 11\n c. 29 8 minutes past 8 am 11\n d. 24 5 minutes before 9 am 11\n\n Answer: b. 27 3 minutes past 8 am 11\n\nExplanation:\n\nTip:\nIn one hour, MINUTE hand travels 60 minutes\nHOUR hand travels just 5 minute spaces\nThus,\nMINUTE hand gains (60 - 5) = 55 minutes over HOUR hand in 60 minutes\n\nAt 8pm hour hand and minute hand difference = 40 minutes\n\nHere we have 2 possibilities -\n\n1. To be at right angle, minute hand must be 15 minutes behind the hour hand\n2. or it must be 15 minutes ahead of hour hand\n\nIn both these cases, it will form a right angle.\n\nTo be 15 minutes behind, minute hand must travel (40 - 15) = 25 minutes more\nIn 60 minutes, minute hand gains 55 min.\nIn ? minutes it gains 25 minutes\n ∴ ? = 25 x 60 = 27 3 minutes past 8 am 55 11\nFor 15 minutes ahead, minute hand must travel (40 + 15) = 55 minutes more\nIn 60 minutes, minute hand gains 55 min.\nSo they will be at right angles again in 60 minutes past 8 pm i.e. exactly at 9 am." ]

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[ "Global Site\n\n# Example of training time reduction for a classifier\n\nTechnical Articles(decision tree, gradient boosting decision tree\n\nMay dd, 2022\nShoichiro Yokotani, Application Development Expert\nAI Platform division\n\nIn this column, I will present an example of analysis using the Frovedis machine learning algorithm.\n\nIn machine learning algorithms, supervised learning can be categorized into two types: regression and classification. In this article, we will take the latter, classification, as an example and run a sample using the Frovedis learning algorithm. We will also compare the time required for learning between the Frovedis version and the scikit-learn version.\n\nClassifiers are applied to datasets with discrete output y for many input variables. For example, the Credit Card Fraud Detection dataset used in this study has 29 different features with binary output y, such as Not Fraud and Fraud. We will use this data set to make a two-class decision using a machine learning algorithm.\n\nTypical machine learning algorithms for classifications include logistic regression, linear support vector machines, random forests as an ensemble method of classification trees and classification trees, and gradient boosting classification trees. In this column, we will focus on two-class classification using classification trees and gradient boosting classification trees.\n\nGradient boosting decision trees can be used for class classification. scikit-learn version of gradient boosting decision trees does not perform parallel processing, but Frovedis version processes each decision tree creation in parallel, so it is expected to reduce training time compared to scikit-learn on very large datasets. The Frovedis version is expected to reduce training time compared to scikit-learn on very large datasets.\n\ncredit_classify\n\nIn these samples using the Credit Card Fraud Detection dataset, we first show the classification tree using the Frovedis and scikit-learn versions, and then the analysis using the gradient boosting classification tree. Finally, the classification results are graphed using PCA feature reduction. The comparison of training time between Frovedis and scikit-learn learning algorithms is shown in the table below.\n\nLearning algorithms Frovedis (sec) scikit-learn (sec) Ratio\nClassification Trees 0.26 7.81 x30.0" ]

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[ "Annotated List of Selected NP-complete Problems\n\n## An Annotated List of Selected NP-complete Problems\n\nThe standard textbook on NP-completeness is:\n\nMichael Garey and David Johnson: Computers and Intractability - A Guide to the Theory of NP-completeness; Freeman, 1979.\n\nDavid Johnson also runs a column in the journal Journal of Algorithms (in the HCL; there is an on-line bibliography of all issues)\n\nOn the Web the following sites may be of interest:\n\nOr trying giving `NP-complete' or `NP and complete' as a search term to\n\n(This is basically a bibliography database, but, you can click on the `on-line papers' button to list electronically readable full texts).\n\nThe Center for Discrete Mathematics and Theoretical Computer Science (DIMACS) gives links to technical papers, abstracts, and other information concerning algorithms, approximation techniques and properties of NP-complete problems. Other general algorithmics and complexity related sites may be found at:\n\nA number of relevant journals are available on-line through the University Library Web Pages (these are only accessible to members of Liverpool University). The following journals are all available and publish research papers in the general areas of Algorithmics and Complexity Theory:\n\n• Acta Informatica\n• Computer Journal\n• Information and Computation\n• Journal of Algorithms\n• Journal of Computer and System Sciences\n• SIAM journal on computing\nIn the list of NP-complete problems below, the form of a typical entry is as follows:\n\nNumber: 0\n\nName: Wombat Eating Assignment (WEA) [DU0] 2\n\nInput: A set W of n wombats; a forest T of m eucalyptus trees (m>=n); a habitation mapping, mu:W-> subsets of T, such that for each wombat, w, mu (w) defines the set of eucalyptus trees, t in T, in which the wombat lives.\n\nQuestion: Is there a mapping, E, from the set of wombats to the set of trees which satisfies both:\n\n• Every wombat is assigned a unique eucalyptus tree by E, i.e. if E(u)=t then E(v)/=t for all distinct pairs of wombats u and v;\n• E is consistent with the habitation mapping, mu, i.e. if E(w)=t then t belongs to the set of trees, mu(w) in which the wombat lives?\n\nComments: Can be solved by efficient methods using (0,1)-network flow maximisation methods applied to bipartite graphs, i.e. it's a matching problem.\n\nThe first line gives the unique identifying number for the problem (as described above). The second line gives the name of the problem and (if there is one) its usual abbreviation. These are how the problem is usually referred to in research papers, textbooks, etc. If you are looking for material on the Web, then these should give reasonable terms to supply to search engines. The code in square brackets is a reference to the classification in Garey and Johnson's book (with the exceptions of Problem 14, 86, 87, 88).\n\nThe next part of the description indicates what a typical instance (i.e. input) consists of: notice, as in the example above, that this will usually consist of a number of distinct objects which must be represented in trying to solve the problem: sometimes these can be quite involved structures, such as graphs, sets, mappings, logical expressions, etc; and, sometimes quite simple forms such as a single integer. The main part of a problem definition is always formulated as a question. This is the core of the decision problem description and defines precisely what property of the input instance must be determined in solving it. Thus, in the example given, one is trying to decide if a given input instance is such that a mapping from wombats to trees, satisfying certain conditions is possible. The final part (which is not always present) gives some (not necessarily useful) comments regarding the problem.\n\nNotice that one way in which the WEA problem could be solved is by exhaustively considering each distinct way of mapping from single wombats to single trees until either one found an assignment that met the required conditions given in the Question; or one had no further mappings left to consider. Clearly the former case would lead to the answer true being returned; the latter to the answer false being given.\n\n### Selected NP-Complete Problems\n\nNumber: 1\n\nName: 3-Satisfiability (3-SAT) [LO2] 2\n\nInput: A set of m clauses - C1 ,C2,...,Cm - over a set of n Boolean valued variables Xn=< x1,x2,...,xn>, such that each clause depends on exactly three distinct variables from Xn. A clause being a Boolean expression of the form yi+yj+yk where each y is of the form x or -x (i.e. negation of x) with x being some variable in Xn. For example if n=4 and m=3 a possible instance could be the (set of) Boolean expressions: C1=(x1+(-x2)+(-x3)); C2=(x2+x3+(-x4)); C3=((-x1)+x3+x4);\n\nQuestion: Can each variable xi of Xn be assigned a Boolean value alphai in such a way that every clause evaluates to the Boolean result true under the assignment < xi:=alphai:1<=i<=n>?\n\nIn the example instance, the assignment < x1:=true;x2:=false;x3:=true; x4:=false> is such that each of the three clauses takes the value true.\n\nComments: For reasons that will become clearer at the end of the course, this problem had to be Number 1. Some relevant material concerning this problem might be found in the special issue (Volume 81, 1996) of the journal Artificial Intelligence that is available in the Library.\n\nNumber: 2\n\nName: Graph 3-Colourability (3-COL) [GT4] 1\n\nInput: An n-node undirected graph G(V,E) with node set V and edge set E.\n\nQuestion: Can each node of G(V,E) be assigned exactly one of three colours - Red, Blue, Green - in such a way that no two nodes which are joined by an edge, are assigned the same colour?\n\nNumber: 3\n\nName: Monochromatic triangle [GT6] 2\n\nInput: An n-node undirected graph G(V,E) with node set V and edge set E.\n\nQuestion: Can the edges, E, of G be partitioned into two disjoint sets E1 and E2, in such a way that neither of the two graphs G1(V,E1) or G2(V,E2) contains a triangle, i.e. a set of three distinct nodes u,v,w such that {u,v}, {u,w}, {v,w} are all edges?\n\nNumber: 4\n\nName: Clique [GT19] 1\n\nInput: An n-node undirected graph G(V,E) with node set V and edge set E; a positive integer k with k<=n.\n\nQuestion: Does G contain a k-clique, i.e. a subset W of the nodes V such that W has size k and for each distinct pair of nodes u, v in W, {u,v} is an edge of G?\n\nNumber: 5\n\nName: Bipartite Subgraph [GT25] 2\n\nInput: An n-node undirected graph G(V,E) with node set V and edge set E; a positive integer k with k<=|E|.\n\nQuestion: Is there a subset, F of the edges of G, having size at least k and such that the graph H(V,F) is bipartite?\n\nComments: G(V,E) is bipartite if the nodes can be partitioned into two disjoint sets U and W such that every edge of G connects a node in U to a node in W, i.e. no two nodes in U (resp. W) form an edge of G.\n\nNumber: 6\n\nName: Vertex Cover [GT1] 1\n\nInput: An n-node undirected graph G(V,E) with node set V and edge set E; a positive integer k with k<=n.\n\nQuestion: Is there a subset W of V having size at most k and such that for every edge {u,v} in E at least one of u and v belongs to W?\n\nNumber: 7\n\nName: Chromatic Index (Edge Colouring) [OPEN5] 3\n\nInput: An n-node undirected graph G(V,E) with node set V and edge set E; a positive integer k with k<=|E|.\n\nQuestion: Can the edges of G be assigned exactly one of k colours in such a way that no two edges which have a common node as an endpoint are assigned the same colour?\n\nComments: Vizing's Theorem from Graph Theory shows that the only `hard' case for this problem is when k is equal to the maximum degree of G. The degree of a node, v, is the number of edges in the graph with v as an end-point; the degree of a graph is the maximum degree of any node in the graph. Chromatic Index was shown to be NP-complete soon after the 1979 edition of Garey and Johnson went to press, hence the OPEN categorisation.\n\nThe classical paper on sequential methods is:\n\n• A.M. Gibbons and O.A. Ogunyode: Optimal edge-colouring of almost all simple graphs in polynomial time. Random Graphs `85: Conf. Record of 2nd International Seminar on Random Graphs and Probabilistic Methods in Combinatorics, Poznan, North-Holland, (1985).\nThere are also a number of methods presenting efficient parallel algorithms for various special types of graph. Some of the ideas presented in these may be useful in formulating sequential techniques, e.g.\n• A.M. Gibbons and W. Rytter: Optimally edge-colouring outerplanar graphs is in NC. Theoretical Computer Science, 71, (1990), pp. 401-411\n• A.M. Gibbons and W. Rytter: Fast parallel algorithms for optimal edge-colouring of some tree-structured graphs. Fundamentals of Computation Theory (FCT) `87, Springer-Verlag, 1987\n\nNumber: 8\n\nName: Multiprocessor Scheduling [SS8] 4\n\nInput: A set, T, of tasks and for each task t in T a positive (integer) running time len(t). A positive integer D, called the deadline.\n\nQuestion: Is there a 2-processor schedule for the tasks that completes within the deadline, D, i.e. is there a function sigma:T->N such that both of the following hold:\n\n• For all u>=0 the number of tasks t in T for which sigma(t)<=u< sigma(t)+len(t) is at most 2.\n• For all tasks t, sigma(t)+len(t)<=D?\n\nComments: This is the simplest avatar of a very large number of NP-complete processor scheduling problems and, unsurprisingly given its practical applications in multiprogramming Operating Systems on small parallel systems, there is an enormous range of literature on approximation and heuristic techniques to tackle it, see e.g. relevant sections of this bibliography. The formal framework given by (a) and (b) can be interpreted as meaning: a) at any given time at most two tasks are being executed; b) every task has been completed by the deadline D.\n\nNumber: 9\n\nName: Comparative Divisibility [AN4] 3\n\nInput: A (strictly increasing) sequence A=< a1,a2,...,an> and a (strictly increasing) sequence B=< b1,b2,...,bm> of positive integers.\n\nQuestion: Is there an integer, c, such that Divides(c,A)> Divides(c,B), where Divides(x,Y) (Y being a sequence of positive integers) is the number of elements, y in Y, for which x is an exact divisor of y?\n\nComments: You may think that this has an obvious fast algorithm, and, indeed the algorithm in question is obvious: what it is not is efficient. Consider: how many bits are needed to store the input data (assuming, without loss of generality, that an>=bm)? How many steps, however, is this `obvious method' taking in the worst-case? It is important to realise that representing integer values in unary is not considered to be a `reasonable' approach (the number 250-1 requires 250 digits in unary but only 50 digits in binary).\n\nNumber: 10\n\nName: Cyclic Ordering [MS2] 3\n\nInput: A finite set, A, and a collection, C, of ordered triples (a,b,c) of distinct elements from A.\n\nQuestion: Is there a one-to-one mapping f:A->{1,2,3,...,|A|} (i.e. a function which maps each element of A to a number between 1 and |A| with no two distinct elements of A being mapped to the same number) such that for each (a,b,c) in C one of\n\nf(a) < f(b) < f(c) ; f(b) < f(c) < f(a) ; f(c) < f(a) < f(b)\n\nholds?\n\nComments: For any triple (a,b,c) there are 6 (six) possible orderings that could result for ((a),f(b),f(c)). The question being asked is whether there is a choice of function that forces every specified triple into one of three `legal' orderings. Garey and Johnson has a typographical error in describing this problem, whereby each triple `in A' is referred to. Obviously `in C' is intended.\n\nNumber: 11\n\nName: Quadratic Diophantine Equations [AN8] 3\n\nInput: Positive integers a, b, and c.\n\nQuestion: Are there two positive integers x and y such that (a*x*x)+(b*y)=c?\n\nComments: The comments regarding Problem 9 (Comparative Divisibility) are also pertinent with respect to this problem. Again, there is an `obvious' algorithm that, on the surface, appears to be efficient and is seen not to be so only once one compares the input size (space needed to represent the input data) to the actual computation time in the worst-case.\n\nNumber: 12\n\nName: Maximum 2-Satisfiability [LO5] 3\n\nInput: A set of m clauses C1,C2,...,Cm over n Boolean valued variables Xn, where each clause depends on two distinct variables from Xn; a positive integer k with k<=m.\n\nQuestion: Is there an assignment of Boolean values to the variables Xn such that at least k distinct clauses take the value true under the assignment?\n\nComments: For relevant definitions see under Problem 1. 2-SAT, a natural variant of the problem 3-SAT described in Problem 1, can be solved in O(m) steps. The simple variation described here (whereby one asks whether at least some number of clauses can be made true simultaneously) is much more difficult.\n\nNumber: 13\n\nName: Register Sufficiency [PO1] 5\n\nInput: A directed, acyclic graph G(V,A) in which each node has at most two out-going edges; a positive integer k.\n\nQuestion: Is there a k or fewer register `computation' for G, i.e. is there an ordering < v1,v2,...,vn> of the (n) nodes of G and a sequence < S0;S1;...;Sn> of subsets of V which satisfy all of the following:\n\n• For all i, Si contains at most k nodes from V.\n• S0 contains no nodes; Sn contains all of the nodes in V with no in-coming edges in G (recall that G is directed).\n• For each i, with 1<=i<=n: vi is in Si; Si-{vi} is a subset of the nodes in Si-1; and Si-1 contains all nodes u for which (vi,u) is an edge in A, i.e. all nodes, u, for which there is an edge from vi to u?\n\nComments: The background to this problem comes from developing Code Generation methods in compilers for High-Level Languages. On early architectures, performing arithmetic operations with both operands stored in registers was significantly faster than fetching an operand from main memory. Such machines, however, had only a small (16-32) number of fast registers. Thus in generating assembly code to represent the computation of a lengthy arithmetic expression, it was necessary to use registers efficiently. A common first stage in compiling such expressions was to represent the computation as a (so-called) `straight-line program' which in turn could be modelled as a directed acyclic graph. The object of the Register Sufficiency Problem is to determine if such a `straight-line program' can be evaluated using only the specified number of available registers.\n\nNumber: 14\n\nName: Central Slice of Half-Clique 2\n\nInput: An 2n-node undirected graph G(V,E) with node set V and edge set E.\n\nQuestion: Does either of the following hold true of G(V,E)\n\n• G contains at least (2n2-n)/2+1 edges.\n• G contains exactly (2n2-n)/2 edges and G contains an n-clique?\n\nComments: For the definition of `k-clique' see Problem 4 above. This specific variant of the CLIQUE problem post-dates Garey and Johnson's 1979 text by 5 years. Its classification as NP-complete is, originally, proved in:\n\n• P.E. Dunne: Techniques for the analysis of monotone Boolean networks; Ph.D. dissertation, Univ. of Warwick, October 1984; (Theory of Computation Report No. 69, Dept. of Comp. Sci., Univ. of Warwick, 1984).\nThe published version appears in\n• P.E. Dunne: The Complexity of Central Slice Functions. Theoretical Computer Science, 44, (1986), pp. 247-257\nFor problems whose inputs are encoded by N bits, there are a number of other NP-complete problems that remain so even when the only `non-trivial' inputs are those with exactly N/2 input bits equal to 1. Three further examples are given in the references cited.\n\nNumber: 15\n\nName: Decision Tree [MS15] 5\n\nInput: A Boolean logic function, f, of n variables, Xn, described by its 1-points, i.e. the set of assignments, alpha=< a1,...,an> such that f(a1,...,an)=1; a positive integer K.\n\nQuestion: Is there a decision tree for f that has average path length at most K?\n\nComments: A decision tree is a binary tree in which each non-leaf node is labelled with a variable from Xn, each leaf node is labelled either 0 or 1, the edge from a non-leaf node to its left child is labelled 0, that to its right child is labelled 1. For a given assignment of Boolean values to the variables Xn a path can be traced starting from the root of the decision tree and following the edges marked with the value of the variable labelling each node encountered. The path terminates at a leaf node whose associated label gives the value of the function. The decision tree computes a Boolean function f(Xn) if for every assignment, alpha, to the variables the path traced from the root under alpha terminates in a leaf labelled f(alpha). The average path length of a binary tree with t leaf nodes and root v is: sum from {w:w is a leaf} |Path from x to w|/t.\n\nThis problem is a simpler (but still NP-complete) version of the form given in Garey and Johnson. For relevant variations and potential heuristic approaches, the papers\n\n• P.E. Dunne and P.H. Leng, \"An algorithm for optimising signal selection in demand-driven circuit simulation\", Transactions of the Society for Computer Simulation, vol. 8, no.4, pp. 269-280, 1992\n• P.E. Dunne, C.J. Gittings, and P.H. Leng, \"Multiprocessor simulation strategies with optimal speed-up\", Inf. Proc. Letters, vol. 54, no. 1, pp. 23-33, April 1995\n• P.E. Dunne, P.H. Leng, and G.F. Nwana, \"On the complexity of Boolean functions computed by lazy oracles\", IEEE. Trans. Comput., vol. 44, no. 4, pp. 495-502, April 1995\nmay provide some ideas of use.\n\nNumber: 16\n\nName: Shortest Common Superstring [SR9] 3\n\nInput: A finite set R={r1,r2,...,rm} of binary strings (sequences of 0 and 1); positive integer k.\n\nQuestion: Is there a binary string w of length at most k such that every string in R is a substring of w, i.e. for each r in R, w can be decomposed as w=w0rw1 where w0, w1 are (possibly empty) binary strings?\n\nComments: General problem allows more than two symbols (i.e. not just binary), but this simpler version remains NP-complete.\n\nNumber: 17\n\nName: Longest Circuit [ND28] 2\n\nInput: n-node undirected graph G(V,E); positive integer k.\n\nQuestion: Does G contain a simple cycle containing at least k nodes?\n\nComments: Special cases of this are the famous Travelling Salesman Problem and Hamiltonian Circuit Problem. The latter corresponds to the cases k=n; the former to the case with each graph edge being weighted and also having k=n.\n\nNumber: 18\n\nName: Kernel [GT57] 2\n\nInput: n-node directed graph G(V,A).\n\nQuestion: Does G possess a kernel, i.e. a subset W of the nodes V such that no two nodes in W are joined by an edge in A and such that for each node v in V-W there is a node w in W for which (w,v) is an edge in A?\n\nNumber: 19\n\nName: k-Closure [GT58] 2\n\nInput: n-node directed graph G(V,A); positive integer k<=n.\n\nQuestion: Is there a subset W of V having size at most k and such that for all edges (u,v) in A either u is in W or v is not in W.\n\nNumber: 20\n\nName: Bandwidth [GT40] 3\n\nInput: n-node undirected graph G(V,E); positive integer k<=n.\n\nQuestion: Is there a linear ordering of V with bandwidth at most k, i.e. a one-to-one function f:V->{1,2,...,n} such that for all edges {u,v} in G, |f(u)-f(v)|<=k?\n\nNumber: 21\n\nName: Maximum Leaf Spanning Tree [ND2] 4\n\nInput: n-node undirected graph G(V,E); positive integer k<=n.\n\nQuestion: Does G have a spanning tree in which at least k nodes have degree 1.\n\n• P.E. Dunne: A result on k-valent graphs and its application to a graph embedding problem, Acta Informatica, 24, (1987), pp. 447-459\nmay be found a) in the HCL Library; b) to be completely incomprehensible; and c) to give a couple of ideas as regards the development of heurstic and approximation techniques.\n\nOther papers of interest are given here\n\nNumber: 22\n\nName: Independent Set [GT20] 1\n\nInput: n-node undirected graph G(V,E); positive integer k<=n.\n\nQuestion: Does G have an independent set of size at least k, i.e. a subset W of at least k nodes from V such that no pair of nodes in W is joined by an edge in E?\n\nNumber: 23\n\nName: Degree Constrained Spanning Tree [ND1] 4\n\nInput: n-node undirected graph G(V,E); positive integer k<=n.\n\nQuestion: Does G have a spanning tree in which no node has degree greater than k?\n\nNumber: 24\n\nName: Hamiltonian Path [GT39] 1\n\nInput: n-node undirected graph G(V,E).\n\nQuestion: Is there a simple path of edges in G that contains every node in V, and thus contains exactly n-1 edges?\n\nComments: Material regarding the Hamiltonian circuit problem may be found to be relevant.\n\nNumber: 25\n\nName: Graph Partitioning [ND14] 2\n\nInput: 2n-node undirected graph G(V,E); positive integer k<=|E|.\n\nQuestion: Can the nodes of G be partitioned into 2 disjoint sets U and W each of size n and such that the total number of distinct edges in E that connect a node u in U to a node w in W is at most k?\n\nNumber: 26\n\nName: Cubic Subgraph [GT32] 3\n\nInput: n-node undirected graph G(V,E).\n\nQuestion: Is there a (non-empty) subset F of the edges of G such that every node in the graph H(V,F) has either degree 3 or degree 0.\n\nNumber: 27\n\nName: Travelling Salesman [ND22] 3-4\n\nInput: A set C of n cities {c1,...,cn}; for each pair of cities (ci,cj) (1<=i< j<=n) a positive integer distance di,j; a positive integer B.\n\nQuestion: Is there an ordering < pi(1),pi(2),...,pi(n)> of the n cities such that the value sum from i=1 to n-1 dpi(i),pi(i+1)+dpi(n),pi(1) is no more than B?\n\nNumber: 28\n\nName: Steiner Tree in Bipartite Graphs [ND12a] 3\n\nInput: Bipartite graph B(V,W,E); positive integer k.\n\nQuestion: Is there a subtree of B(V,W,E) that includes all of the nodes of V and has at most k edges?\n\nNumber: 29\n\nName: Bounded Diameter Spanning Tree [ND4] 4\n\nInput: n-node undirected graph G(V,E); for each edge, {u,v} a positive integer weight w(u,v); positive integer B.\n\nQuestion: Does G have a spanning tree T such that the total sum of the weights of edges in T is at most B and no simple path in T contains more than 5 edges?\n\nNumber: 30\n\nName: Optimal Linear Arrangement [GT42] 3\n\nInput: n-node undirected graph G(V,E); positive integer k<=n.\n\nQuestion: Is there a one-to-one function f:V->{1,2,...,n} such that sum from {{u,v} in E} |f(u)-f(v)|<=K?\n\nNumber: 31\n\nName: Dominating Set [GT2] 2\n\nInput: n-node undirected graph G(V,E); positive integer k<=n.\n\nQuestion: Does G contain a dominating set of size at most k, i.e. a subset W of V containing at most k nodes and such that for every node u in V-W (i.e. in V but not in W) there is a node w in W such that {u,w} is an edge of G?\n\nNumber: 32\n\nName: Path with Forbidden Pairs [GT54] 4\n\nInput: n-node directed graph G(V,A); two distinct nodes s and t belonging to V; finite collection C={(a1,b1),...,(ar,br)} of pairs of nodes from V.\n\nQuestion: Is there a directed path from the node s to the node t in G that contains at most one node from each pair of nodes in the collection C?\n\nComments: Several simplifications of this problem remain NP-complete: requiring G to be acyclic; requiring the collection C to contain only directed edges from A (rather than arbitrary pairs of nodes); requiring all the pairs to be disjoint.\n\nNumber: 33\n\nName: Oriented Diameter [GT64] 4\n\nInput: n-node undirected graph G(V,E); positive integer k<=n.\n\nQuestion: Can a direction be placed on each edge {u,v} of E in such a way that the resulting directed graph H(V,A) is such that both of the following hold:\n\n• H(V,A) is strongly connected, i.e. for each distinct pair of nodes, v and w, there exists a directed path from v to w in H and a directed path from w to v in H.\n• The diameter of H(V,A) is at most k, i.e. for every pair of nodes v and w, there is a path of at most k edges from v to w and a path of at most k edges from w to v?\n\nNumber: 34\n\nName: Rural Postman [ND27] 3-4\n\nInput: n-node undirected graph G(V,E); subset F of the edges E; positive integer k<=|E|\n\nQuestion: Is there a (not necessarily simple) cycle of in G that contains every edge in F and has at most k edges in total?\n\nComments: A simple cycle in a graph is one in which each node visited in the cycle is visited exactly once. A non-simple cycle allows nodes to be visited more than once (although, obviously, each edge can only occur once). For example in the 5 node graph with edges ({1,2};{1,3};{1,4};{1,5};{2,3};{2,4};{2,5};{3,4}), the cycle 1-> 2-> 3-> 1 is a simple cycle; the cycle 1-> 2-> 4-> 3-> 2-> 5-> 1 is a non-simple cycle since 2 is visited twice (notice that in neither case is any edge used more than once).\n\nNumber: 35\n\nName: Longest Path [ND29] 2\n\nInput: n-node undirected graph G(V,E); nodes s and t in V; positive integer k.\n\nQuestion: Is there a simple path between s and t in G that contains at least k edges?\n\nComments: The similarity to Problem 17 should be noted.\n\nNumber: 36\n\nName: 3-Dimensional Matching (3DM) [SP1] 3\n\nInput: 3 disjoint sets X, Y, and Z each comprising exactly n elements; a set M of m triples {(xi,yi,zi):1<=i<=m} such that xi is in X, yi in Y, and zi in Z, i.e. M is a subset of XxYxZ.\n\nQuestion: Does M contain a matching, i.e. is there a subset Q of M such that |Q|=n and for all distinct pairs of triples (u,v,w) and (x,y,z) in Q it holds that u/=x and v/=y and w/=z.\n\nComments: The variant 2-dimensional matching in which 2 disjoint sets X and Y form the basis of a set of pairs, can be solved by a number of fast methods.\n\nNumber: 37\n\nName: Set Splitting [SP4] 3\n\nInput: A finite set S; A collection C1,...,Cm of subsets of S.\n\nQuestion: Can S be partitioned into two disjoint subsets - S1 and S2 - such that for each set Ci it holds that Ci is not a subset of S1 and Ci is not a subset of S2?\n\nNumber: 38\n\nName: Set Packing [SP3] 3\n\nInput: A collection C=(C1,...,Cm) of finite sets; a positive integer k<=m.\n\nQuestion: Are there k sets - D1,...,Dk - from the collection C such that for all 1<=i< j<=k, Di and Dj have no common elements?\n\nNumber: 39\n\nName: Exact Cover by 3-Sets (X3C) [SP2] 3\n\nInput: A finite set X containing exactly 3n elements; a collection C of subsets of X each of which contains exactly 3 elements.\n\nQuestion: Does C contain an exact cover for X, i.e. a sub-collection of 3-element sets D=(D1,...,Dn) such that each element of X occurs in exactly one subset in D?\n\nNumber: 40\n\nName: Minimum Cover [SP5] 3-4\n\nInput: A finite set S; A collection C=(C1,...,Cm) of subsets of S; a positive integer k<=m.\n\nQuestion: Does C contain a cover for S comprising at most k subsets, i.e. a collection D=(D1,...,Dt), where t<=k, each Di is a set in C, and such that every element in S belongs to at least one set in D?\n\nNumber: 41\n\nName: Partition [SP12] 3\n\nInput: Finite set A; for each element a in A a positive integer size s(a).\n\nQuestion: Can A be partitioned into 2 disjoint sets A1 and A2 in a such a way that the sum of the sizes s(x) of elements x in A1 is exactly the same as the sum of the sizes s(y) of the elements y in A2.\n\nComments: It should be noted that it is not required that A1 and A2 contain equal numbers of elements, although even with this condition the problem is still NP-complete.\n\nNumber: 42\n\nName: Subset Sum [SP13] 3\n\nInput: Finite set A; for each element a in A a positive integer size s(a); a positive integer K.\n\nQuestion: Is there a subset B of A such that the sum of the sizes, s(x), of the elements x in B is exactly equal to K?\n\nNumber: 43\n\nName: Comparative Containment [SP10] 4\n\nInput: A finite set X; 2 collections R=(R1,...,Rk) and S=(S1,...,Sm) each of which is a set of subsets of X; for each Ri in R, a positive integer weight w(Ri); for each Si in S a positive integer weight w(Si);\n\nQuestion: Is there a subset Y of X such that: if RY is the set of subsets, Ri of R having Y as a subset of Ri and SY is the set of subsets, Si of S having Y as a subset of Si then the total weight of the sets in RY is at least the total weight of the sets in SY.\n\nNumber: 44\n\nName: Minimum Test Set [SP6] 3\n\nInput: A finite set S; A collection C=(C1,...,Cm) of subsets of S; a positive integer k<=m.\n\nQuestion: Is there a sub-collection D=(D1,...,Dt) of the sets in C such that: t<=k and for each distinct pair of elements u, v in S there is a set Du,v in D that contains exactly one of u and v?\n\nNumber: 45\n\nName: Minimum Sum of Squares [SP19] 3-4\n\nInput: A set A of n elements; for each element a in A a positive integer size s(a); positive integers k<=n and J.\n\nQuestion: Can A be partitioned into k disjoint sets A1,...,Ak such that sum from i=1 to k ( sum from {x in Ai} s(x))2<=J?\n\nNumber: 46\n\nName: 3-Partition [SP15] 3\n\nInput: A set A of 3m elements; a positive integer bound B; for each element x in A a positive integer size s(x) that satisfies B/4< s(x)< B/2, and is such that the sum of the sizes of elements in A is exactly mB.\n\nQuestion: Can A be partitioned into m disjoint sets A1,...,Am such that each Ai contains exactly 3 elements of A and each Ai has total size equal to B?\n\nNumber: 47\n\nName: Subset Product [SP14] 3\n\nInput: Finite set A; for each element a in A a positive integer size s(a); a positive integer K.\n\nQuestion: Is there a subset B of A such that the product (i.e. result of multiplying together all) of the sizes, s(x), of the elements x in B is exactly equal to K?\n\nComments: There is a subtle technical distinction between this and Problem 42: the former case has a `pseudo-efficient' algorithm obtained by allowing numbers to be represented in unary; unless all NP-complete problems can be solved by fast algorithms, however, the Subset Product Problem, cannot be solved by `efficient' methods using even this unreasonable input representation.\n\nNumber: 48\n\nName: Bin Packing [SR1] 3\n\nInput: A finite set U of m items; for each item u in U a positive integer size s(u); positive integers B (called the bin capacity) and k<=m.\n\nQuestion: Can U be partitioned into k disjoint sets U1,...,Uk such that for each Ui (1<=i<=k) the total sum of the sizes of the items in Ui does not exceed B?\n\nNumber: 49\n\nName: Hitting String [SR12] 3\n\nInput: Finite set S={s1,...,sm} each si being a string of n symbols over {0,1,*}.\n\nQuestion: Is there a binary string x=x1x2...xn of length n such that for each sj in S, sj and x agree in at least one position.\n\nNumber: 50\n\nName: Rectilinear Picture Compression [SR25] 4-5\n\nInput: An n by n matrix M of 0s and 1s; a positive integer K.\n\nQuestion: Can all of the 1-valued entries in M be covered by a collection of K or fewer rectangles, i.e. is there a sequence of 4-tuples (ai,bi,ci,di) (where for all 1<=i<=K, ai<=bi and ci<=di) such that:\n\n• For each pair (i,j) (1<=i,j<=n) M[i,j]=1 if and only if for some k (1<=k<=K) it holds that ak<=i<=bk and ck<=j<=dk?\n\nComments: Despite the apparently complicated definition, this is quite a well-motivated problem. Its background is, as its name suggests, from image compression and the task being set can be interpreted as representing a 2-dimensional image using at most K `blocks' of information.\n\nNumber: 51\n\nName: Sequencing with Release Times and Deadlines [SS1] 4-5\n\nInput: A set T of tasks; for each task t in T: a positive integer length len(t); a positive integer release time r(t); and a positive integer deadline d(t).\n\nQuestion: Is there a one-processor schedule for the tasks, T, that satisfies the release time constraints and meets all of the deadlines, i.e. a one-to-one function sigma from the set of tasks to positive integers such that all of the following hold:\n\n• For any two distinct tasks t and w if sigma(t)> sigma(w) then sigma(t)>=sigma(w)+len(w).\n• For all tasks t in T, sigma(t)>=r(t).\n• For all tasks t in T, sigma(t)+len(t)<=d(t)?\n\nComments: Another scheduling problem in the style of Problem 8, but in a single processor environment. The requirements of the schedule, sigma, are that it must: 1) assign a unique starting time to each task in the set (sigma is a one-to-one function); 2) two tasks cannot be running simultaneously (this is condition (a) above, which says that if a task t is scheduled to start after a task w - sigma(t)> sigma(w) - then the earliest time at which t can be scheduled is after w has completed, i.e. sigma(w)+len(w); 3) no task can start before its specified `release time' (condition (b) above); 4) every task has finished no later than its allotted deadline (condition (c) which states that the scheduled starting time of a task t plus the total amount of time that t takes to complete - len(t) - must be no greater than the deadline that is assigned for t, i.e. d(t)).\n\nIn common with other scheduling and resource management problems there is a large volume of research into heuristics and approximation methods for this problem. Some relevant references may be found in sections of this bibliography.\n\nFor a reminder of what a `one-to-one' function is, see Problem 10.\n\nNumber: 52\n\nName: Precedence Constrained Scheduling [SS9] 3-4\n\nInput: A set T of tasks each of which has length 1 or length 2; a partial ordering << on the set of tasks, T; a positive integer deadline D.\n\nQuestion: Is there a 2-processor schedule, sigma, for T that meets the overall deadline D and obeys the `precedence constraints' of the partial order <<, i.e. if t<< w then sigma(w)>=sigma(t)+len(t)?\n\nComments: For a definition of `2-processor schedule' refer to Problem 8. Recall that a partial order << on a set R is an ordering relation which satisfies: for all distinct s and t at most one of s<< t or t<< s holds; and for all distinct s, t, w in R if s<< t and t<< w then s<< w. The bibliography mentioned above may contain useful references.\n\nNumber: 53\n\nInput: Positive integers a, b, and c.\n\nQuestion: Is there a positive integer x whose value is less than c and is such that x2==a(mod b), i.e. the remainder when x2 is divided by b is equal to a?\n\nComments: The comments made with respect Problem 9 and Problem 11 are also relevant with respect to this problem.\n\nNumber: 54\n\nName: Square-Tiling [GP13] 5\n\nInput: A set C of n `colours'; a collection of tiles, T, each tile t being a 4-tuple < a,b,c,d> of colours where a is the colour at the top of the tile; b that on the right-hand side; c that at the bottom; and d that on the left-hand side of the tile; a positive integer k<=n.\n\nQuestion: Is there a proper tiling of a k by k square using the tiles in T, i.e. an assignment of a tile A(i,j) in T to each ordered pair (i,j) with 1<=i<=k, 1<=j<=k such that both of the following hold:\n\n• If A(i,j)=< a,b,c,d> and A(i+1,j)=< e,f,g,h> then the colours b and h are identical.\n• If A(i,j)=< a,b,c,d> and A(i,j+1)=< e,f,g,h> then the colours c and e are identical?\n\nComments: Informally, a proper tiling is one in which two tiles which are next to each other are required to have the same colour on touching sides, i.e. the right-hand side of one is coloured the same as the left-hand side of its neighbour and similarly with vertically adjacent tiles: the lower side of one is coloured the same as the top side of its neighbour. Decision problems involving questions about tiling patterns tend to be extremely difficult. For the most general form of the question it can be proven that no algorithm at all exists to solve it, i.e. not even an extremely inefficient one. A tiling problem also constitutes the only known `natural' `hard' member of the following class of problems: given a positive integer n as input, determine the number of `objects' of `size' n having a particular property, e.g. the number of n-node graphs, or the number of n-node graphs with a Hamiltonian circuit.\n\nNumber: 55\n\nName: Crossword Puzzle Construction [GP14] 5\n\nInput: A finite set of characters SIGMA={sigma1,...,sigmak}; a finite set, W={w1,...,w2n} each wi being a sequence of n characters from SIGMA, i.e. a string.\n\nQuestion: Can an n by n crossword puzzle be built using all of the 2n words (strings) in W, i.e. if C is an n by n table of `blanks', is there an assignment, f, that maps each entry (i,j) of C to some character in SIGMA in such a way that the word formed by taking the n consecutive characters in a row corresponds to a word in W; the word formed by taking the n consecutive characters in a column (reading from top-to-bottom) corresponds to a word in W?\n\nComments: The definition may look complicated but it isn't. Here is a simple example instance for which a crossword can be constructed. Let SIGMA={A,B,C,D,E,G,O}, n=3, and W={AGE,AGO,BEG,CAB,CAD,DOG}. A 3 by 3 crossword puzzle for this is given by the assignment C(1,1)=C; C(1,2)=A; C(1,3)=B; C(2,1)=A; C(2,2)=G; C(2,3)=E; C(3,1)=D; C(3,2)=O; C(3,3)=G. This results in the grid\n\n```CAB\nAGE\nDOG\n```\n\nwhose correctness is self-evident.\n\nNumber: 56\n\nName: Disjunctive non-tautology [LO8] 2\n\nInput: A A set of m products - P1,P2,...,Pm - over a set of n Boolean valued variables Xn=< x1,x2,...,xn>, such that each product depends on exactly three distinct variables from Xn. A product being a Boolean expression of the form yi AND yj AND yk where each y is of the form x or -x (i.e. negation of x) with x being some variable in Xn.\n\nQuestion: Is there an assignment of Boolean values to the variables in Xn that results in every product taking the value false (equivalently 0)?\n\nNumber: 57\n\nName: Simultaneous incongruences [AN2] 3\n\nInput: A set of n ordered pairs of positive integers {(a1,b1),...,(an,bn)} where ai<=bi for each 1<=i<=n.\n\nQuestion: Is there a positive integer x such that: for each i, ai does not equal the remainder when dividing x by bi?\n\nNumber: 58\n\nName: Betweenness [MS1] 3\n\nInput: A finite set of size n, A; a set C of ordered triples, (a,b,c), of distinct elements from A.\n\nQuestion: Is there a one-to-one function, f:A->{1,2,...,n} such that for each triple (a,b,c) in C it holds that either f(a)< f(b)< f(c) or f(c)< f(b)< f(a)?\n\nComments: If you need to be reminded of what a one-to-one function is then see Problem 10 to which this problem may appear similar.\n\nNumber: 59\n\nName: Minimum weight and/or graph solution [MS16] 4-5\n\nInput: A directed acyclic graph G(V,A) having a single node s, with no incoming edges; a labelling, f(v) of each node having at least one out-going edge in G, as either an and-node or an or-node; a positive integer K.\n\nQuestion: Is there a sub-graph H(W,B) of G(V,A), i.e. W is a subset of V and B is a subset of A satisfying all of the following:\n\n• s is in W.\n• If w is in W and w is an and-node then all of the edges directed out of w in G belong to B.\n• If w is in W and w is an or-node then at least one of the edges directed out of w in G belongs to B.\n• B contains at most K edges?\n\nNumber: 60\n\nName: Fault-detection in directed graphs. [MS18] 3-4\n\nInput: A directed, acyclic graph G(V,A) such that G has a unique node t with no out-going edges; I the set of nodes in V having no incoming edges; a positive integer K.\n\nQuestion: Is there a `test set' of size at most K that can detect every `single fault' in G (N.B. not `every single fault' but every `single fault'), i.e. a subset T of the nodes in I such that\n\n• T contains at most K nodes.\n• For every node v in V there exists a node u in T such that v lies on a directed path from u to t in G?\n\nNumber: 61\n\nName: Minimum Broadcast Time. [ND49] 3-4\n\nInput: n-node undirected graph G(V,E); a subset V0 of the nodes in V.\n\nQuestion: Can a message be `broadcast' from the base set V0 to all of the nodes in V in at most 4 steps, i.e. is there a sequence\n\nV0;E1;V1;E2;V2;E3;V3; E4;V4\n\nwhich satisfies all of the following:\n\n• Vi is a subset of V for each 0<=i<=4.\n• Ei is a subset of E for each 1<=i<=4.\n• V4=V.\n• Each edge in Ei has exactly one of its endpoints in Vi-1 (1<=i<=4)\n• No two edges in Ei share a common end-point (1<=i<=4)\n• Vi=Vi-1union {w: {v,w} in Ei} (1<=i<=4)?\n\nNumber: 62\n\nName: Disjoint Connecting Paths [ND40] 3-4\n\nInput: n-node undirected graph G(V,E); a set of disjoint pairs of nodes {(s1,t1);...;(sk,tk)}.\n\nQuestion: Does G contain k mutually node disjoint paths, Pi, with Pi being a path from si to ti?\n\nComments: Mutually disjoint means that no two paths have any common nodes.\n\nNumber: 63\n\nName: Shortest Weight-Constrained Path [ND30] 3\n\nInput: n-node undirected graph G(V,E); for each edge e in E a positive integer length len(e) and a positive integer weight w(e); specified nodes s and t in V; positive integers K and W.\n\nQuestion: Is there a path from s to t in G that has both:\n\n• Total weight at most W.\n• Total length at most K?\n\nComments: If all of the edges have the same length or all of the edges have the same weight, then this is simply the normal shortest-path problem for which various efficient algorithms exist, e.g. the Dynamic Programming method of Floyd that is covered in the course. Minimising a single measure, in this context of path lengths, is `easy', but attempting simultaneously to minimise two (or more) distinct measures is not.\n\nNumber: 64\n\nName: Minimum Maximal Matching [GT10] 3\n\nInput: n-node undirected graph G(V,E); positive integers k<=|E|.\n\nQuestion: Is there a subset, F, of at most k edges from E that forms a maximal matching in G, i.e. no two edges in F have a common endpoint and every edge of G that is not in F has a common endpoint with at least one edge of F?\n\nNumber: 65\n\nName: Partition into Triangles [GT11] 3\n\nInput: (3n)-node undirected graph G(V,E).\n\nQuestion: Can the nodes of G be partitioned into n disjoint sets - V1,...,Vn - each of which contains exactly 3 nodes and is such that for each Vi={ui,vi,wi}, all three of the edges {ui,vi}, {ui,wi} and {vi,wi} belong to E?\n\nNumber: 66\n\nName: Partition into Forests [GT14] 3\n\nInput: n-node undirected graph G(V,E); positive integer k<=n\n\nQuestion: Can the nodes of G be partitioned into t<=k disjoint sets V1,...,Vt - in such a way that for each Vi (1<=i<=t), the subgraph Gi(Vi,Ei) induced by Vi contains no cycles, i.e. is a forest (set of trees)?\n\nComments: If G(V,E) is a graph, then the subgraph induced by a subset W of V is the graph H(W,F) whose edges, F, are formed by all the edges in E that connect two nodes in W.\n\nNumber: 67\n\nName: Partition into Cliques [GT15] 3\n\nInput: n-node undirected graph G(V,E); positive integer k<=n\n\nQuestion: Can the nodes of G be partitioned into t<=k disjoint sets - V1,...,Vt - in such a way that for each Vi (1<=i<=t), the subgraph Gi(Vi,Ei) induced by Vi is a clique, i.e. a graph in which every pair of nodes is connected by an edge?\n\nNumber: 68\n\nName: Partition into Perfect Matchings [GT16] 3\n\nInput: n-node undirected graph G(V,E); positive integer k<=n\n\nQuestion: Can the nodes of G be partitioned into t<=k disjoint sets - V1,...,Vt - in such a way that for each Vi (1<=i<=t), the subgraph Gi(Vi,Ei) induced by Vi is a perfect matching, i.e. a graph in which every node is the endpoint of exactly one edge?\n\nNumber: 69\n\nName: Covering by cliques [GT17] 3\n\nInput: n-node undirected graph G(V,E); positive integer k<=n\n\nQuestion: Are there t<=k subsets - V1,...,Vt of V - such that both of the following hold:\n\n• The subgraph Gi(Vi,Ei) of G(V,E) induced by Vi is a clique.\n• For each edge {u,v} in E there is some subset Vi that contains both u and v?\n\nNumber: 70\n\nName: Degree-bounded connected subgraph [GT26] 3\n\nInput: n-node undirected graph G(V,E); non-negative integer d<=n; positive integer k<=|E|.\n\nQuestion: Is there a subset F of the edges in E having size at least k and such that subgraph G(V,F) of G is connected and has no node of degree greater than d?\n\nComments: For the definition of `degree of a node' see Problem 7. A graph is connected if for every pair of nodes v and w there is a path connecting v and w in the graph.\n\nNumber: 71\n\nName: Uniconnected subgraph [GT30] 3\n\nInput: n-node directed graph G(V,A); positive integer k<=|A|.\n\nQuestion: Is there a subset B of the edges in A having size at least k and such that the subgraph H(V,B) of G has at most one directed path between any pair of nodes in V?\n\nNumber: 72\n\nName: Minimum k-connected subgraph [GT31] 3\n\nInput: n-node undirected graph G(V,E); positive integers k<=n and B<=|E|.\n\nQuestion: Is there a subset F of the edges E having size at most B and such that the subgraph H(V,F) of G is k-connected, i.e. remains connected after removing any set W of a most k-1 nodes and their adjacent edges?\n\nNumber: 73\n\nName: Minimum cut linear arrangement [GT44] 3\n\nInput: n-node undirected graph G(V,E); positive integer k.\n\nQuestion: Is there a one-to-one function f:V->{1,2,...,n} such that for all i, with 1< i< n, the total number of edges {u,v} in E for which f(u)<=i< f(v) is at most k?\n\nNumber: 74\n\nName: Consecutive Sets [SR18] 4\n\nInput: Finite alphabet SIGMA of t symbols; collection C={SIGMA1,SIGMA2,...,SIGMAn} of subsets of SIGMA; positive integer K.\n\nQuestion: Is there a string, w, (sequence of symbols) over SIGMA such that |w|< =K and for each 1< =i< =n, all of the symbols in SIGMAi appear in a consecutive block of |SIGMAi| symbols in the string w?\n\nComments: N.B. minor typo. in Garey and Johnson, W written instead of w.\n\nNumber: 75\n\nName: Multiple choice matching [GT55] 3\n\nInput: n-node undirected graph G(V,E); partition of E into t disjoint sets - E1,...,Et; positive integer k.\n\nQuestion: Is there a subset F of E having size at least k and satisfying both of the following:\n\n• No two edges in F share a common endpoint.\n• F contains at most one edge from each Ei (1<=i<=t)?\n\nNumber: 76\n\nName: Path distinguishers [GT60] 3\n\nInput: n-node directed, acyclic graph G(V,A); specified nodes s and t in V; positive integer k<=|A|.\n\nQuestion: Is there a subset B of the edges A having size at most k and such that for any pair of distinct paths P and Q from s to t in G, there is an edge in B that is in exactly one of P and Q?\n\nNumber: 77\n\nName: Nestril-Rodl dimension [GT62] 5\n\nInput: n-node undirected graph G(V,E); positive integer k<=n.\n\nQuestion: Is there a one-to-one function\n\nf:V-> {( a1, a2,...,ak) : 1 <= ai<= n }\n\nsuch that for all pairs of nodes u and v in V, {u,v} is an edge of G if and only if f(u) and f(v) disagree in all k components?\n\nNumber: 78\n\nName: Shortest total path length spanning tree [ND3] 4-5\n\nInput: n-node undirected graph G(V,E); positive integer B.\n\nQuestion: Is there a spanning tree T(V,F) of G such that the sum over all pairs of nodes u and v of the length of the path between u and v in T is no greater than B?\n\nNumber: 79\n\nName: Induced Path [GT23] 3\n\nInput: n-node undirected graph G(V,E); positive integer k<=n.\n\nQuestion: Is there a subset W of the nodes V having size at least k and such that the subgraph H(W,F) of G induced by W is a simple path on |W| nodes?\n\nNumber: 80\n\nName: Bounded Component Spanning Forest [ND10] 3\n\nInput: n-node undirected graph G(V,E); positive integer k<=n.\n\nQuestion: Can V be partitioned into t disjoint sets - V1,...,Vt - where t<=k, such that for each i (1<=i<=t) both of the following hold:\n\n• The subgraph H(Vi,Ei) of G induced by Vi is connected.\n• The number of nodes in Vi is at most 4.\n\nNumber: 81\n\nName: Maximum subgraph matching [GT50] 3\n\nInput: Directed graphs G(V,A), H(W,B); positive integer k.\n\nQuestion: Is there a subset R of VxW, i.e. pairs of nodes where one node is from V and one from W, having size at least k and such that for every < u,v> and < x,y> in R the edge (u,x) belongs to A if and only if the edge (v,y) belongs to B?\n\nNumber: 82\n\nName: Numerical 3-dimensional matching [SP16] 3\n\nInput: Disjoint sets W, X, and Y each containing exactly m elements; for each element u of W union X union Y, a positive integer size s(u); a positive integer bound B.\n\nQuestion: Can the 3m element set W union X union Y be partitioned into m disjoint sets A1,...,Am in such a way that:\n\n• Each Ai contains exactly one element from W, exactly one element from X, and exactly one element from Y.\n• Each Ai has total size exactly equal to B?\n\nNumber: 83\n\nName: Open Hemisphere [MP6] 4-5\n\nInput: Finite set X of m-tuples of integers; a positive integer k<=|X|.\n\nQuestion: Is there an m-tuple y of rational numbers such that sum from i=1 to m xiyi<> 0 for at least k of the m-tuples in X, xi (resp. yi) being the ith component of the m-tuple x (resp. y)?\n\nNumber: 84\n\nName: Largest Common Subgraph [GT49] 3-4\n\nInput: 2 n-node undirected graphs G(V,E), H(W,F); positive integer k.\n\nQuestion: Do there exist subsets E1 of the edges E and F1 of the edges F that satisfy all of the following:\n\n• |E1|=|F1|.\n• |E1|>=k, |F1|>=k.\n• The graphs G(V,E1) and H(W,F1) are isomorphic, i.e. there is a mapping f between V and W such that {x,y} is an edge in E1 if and only if {f(x),f(y)} is an edge in F1?\n\nNumber: 85\n\nName: Clustering [MS9] 3\n\nInput: Finite set X; for each pair of elements x and y in X, a positive integer distance d(x,y); positive integer B.\n\nQuestion: Is there a partition of X into 3 disjoint sets - X1,X2,X3 - with which: for each set Xi (1<=i<=3), for all pairs x and y in Xi it holds that d(x,y)<=B?\n\nNumber: 86\n\nName: Protein Folding 5\n\nInput: Three positive integers, t, n, and k; a finite alphabet, SIGMA, of t symbols; a string S of length n.\n\nQuestion: Is there an embedding of S into a two dimensional grid that has a bond-score of at least k? i.e. if S=s1s2...sn, is there a one-to-one function FOLD:{1,...,n}->NxN such that:\n\n• For all 1 <= i< n if FOLD(i)=(x,y) and FOLD (i+1)=(w,z), then either {x=w and |y-z|=1} or {|x-w|=1 and y=z}. (i.e. consecutive members of S are mapped onto adjacent grid co-ordinates).\n• The number of pairs {(x,y);(w,z)} of grid positions for which both\n• (x,y) and (w,z) are neighbours, (i.e. {x=w and |y-z|=1} or {|x-w|=1 and y=z}).\n• FOLD(i)=(x,y) and FOLD(j)=(w,z) (where i and j satisfy 1<=i< j<=n), and si=sj, i.e. the symbols (in SIGMA) mapped onto (x,y) and (w,z) are identical.\n• is at least k?\n\nComments: The problem arises from an extremely simplified model of protein-folding of interest in computational biology. The NP-completeness proof is highly non-trivial (by a transformation from 3-SAT), is a recent result not mentioned in Garey and Johnson, and is due to Paterson and Przytycka (1996). It is important to note that the alphabet is part of the input. It is an open question as to whether the variant in which an alphabet of a fixed size, e.g. 2 as with binary, remains NP-complete. In should also be remembered that no limit is placed on the grid size, although trivially, an n x n grid will always suffice.\n\nNumber: 87\n\nName: Hamming Centre 3-4\n\nInput: A set S of k binary strings, each of length n; a positive integer r.\n\nQuestion: Is there an n-bit string y such that for every string x in S, the Hamming distance, H(x,y) is at most r?.\n\nComments: The Hamming distance, H(x,y), between two binary strings x and y of length n, is the number of bit positions in which x and y have differing values.\n\nThis is another recent result not given in Garey and Johnson, its NP-complete classification being given in Frances and Litman (1997).\n\nM. Frances and A. Litman. On Covering Problems of Codes, Theory of Computing Systems, 30, 1997, 113-119\n\nNumber: 88\n\nName: 2-Spanner 3\n\nInput: An undirected graph G(V,E); a positive integer s.\n\nQuestion: Does G(V,E) contain a 2-spanner with at most s edges, i.e. subset F of the edges E having size at most s and such that for each edge {v,w} in E either {v,w} is in F or (for some node x in V) the edges {v,x} and {x,w} are both in F?\n\nComments: 2-Spanners (or more generally t-spanners for some positive integer t) are a natural extension of the idea of spanning tree (N.B. A spanner is required neither to be a tree nor, even, a connected graph), and were first considered in the context of problems in communications networks and parallel architectures by Peleg and Ullman (1989). Subsequently, Peleg and Schaffer (1989) showed this problem to be NP-complete, hence its absence from the appendix in Garey and Johnson.\n\nD. Peleg and A.A. Schaffer. Graph Spanners. Jnl. of Graph Theory, 13, (1989), 99-116.\n\nD. Peleg and J.D. Ullman. An Optimal Synchroniser for the Hypercube. SIAM Jnl. on Computing, 18, (1989), 740-747" ]

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[ "## Trigonometry 7th Edition\n\n$y=2\\sin{0} = 2(0)=0$, thus, the point is $(0, 0)$ $y=2\\sin{\\frac{\\pi}{2}} = 2(1)=2$, thus, the point is $(\\frac{\\pi}{2}, 2)$ $y=2\\sin{\\pi} = 2(0)=0$, thus, the point is $(\\pi, 0)$ $y=2\\sin{\\frac{3\\pi}{2}} = 2(-\\sin{\\frac{\\pi}{2}})=2(-1)=-2$, thus, the point is $(\\frac{3\\pi}{2}, -2)$ $y=2\\sin{2\\pi} =2(0)=0$, thus, the point is $(2\\pi, 0)$\nAll the given values of $x$ are either special angles or have a special angle as their reference angle. Hence, the value of the cosine for each angle can be easily found. Evaluate the function for each given value of $x$ to obtain: When $x=0$: $y=2\\sin{0} = 2(0)=0$, thus, the point is $(0, 0)$ When $x=\\frac{\\pi}{2}$: $y=2\\sin{\\frac{\\pi}{2}} = 2(1)=2$, thus, the point is $(\\frac{\\pi}{2}, 2)$ When $x=\\pi$: $y=2\\sin{\\pi} = 2(0)=0$, thus, the point is $(\\pi, 0)$ When $x=\\frac{3\\pi}{2}$: Reference angle is $\\frac{\\pi}{2}$. Since the angle is on the negative y-axis, then sine is negative. Thus, $y=2\\sin{\\frac{3\\pi}{2}} = 2(-\\sin{\\frac{\\pi}{2}})=2(-1)=-2$ Thus, the point is $(\\frac{3\\pi}{2}, -2)$. When $x=2\\pi$: $y=2\\sin{2\\pi} =2(0)=0$, thus, the point is $(2\\pi, 0)$" ]

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[ "Documentation\n\n### This is machine translation\n\nMouseover text to see original. Click the button below to return to the English version of the page.\n\n## Effects of Some Pattern Search Options\n\nThis example shows the effects of some options for pattern search. The options include plotting, stopping criteria, and other algorithmic controls for speeding a solution.\n\n### Set Up a Problem for Pattern Search\n\nThe problem to minimize is a quadratic function of six variables subect to linear equality and inequality constraints. The objective function, `lincontest7`, is included with Global Optimization Toolbox software.\n\n`type lincontest7`\n```function y = lincontest7(x); %LINCONTEST7 objective function. % y = LINCONTEST7(X) evaluates y for the input X. Make sure that x is a column % vector, whereas objective function gets a row vector. % Copyright 2003-2004 The MathWorks, Inc. x = x'; %Define a quadratic problem in terms of H and f (From web unknown source) H = [36 17 19 12 8 15; 17 33 18 11 7 14; 19 18 43 13 8 16; 12 11 13 18 6 11; 8 7 8 6 9 8; 15 14 16 11 8 29]; f = [ 20 15 21 18 29 24 ]'; y = 0.5*x'*H*x + f'*x; ```\n\nSpecify the function handle `@lincontest7` as the objective function.\n\n`objectiveFcn = @lincontest7;`\n\nThe objective function accepts a row vector of length six. Specify an initial point for the optimization.\n\n`x0 = [2 1 0 9 1 0];`\n\nCreate linear constraint matrices representing the constraints `Aineq*x <= Bineq` and `Aeq*x = Beq`. For details, see Linear Constraints (Optimization Toolbox).\n\n```Aineq = [-8 7 3 -4 9 0 ]; Bineq = ; Aeq = [7 1 8 3 3 3; 5 0 5 1 5 8; 2 6 7 1 1 8; 1 0 0 0 0 0]; Beq = [84 62 65 1];```\n\nRun the `patternsearch` solver, and observe how many iterations and function evaluations it takes to arrive at the solution.\n\n`[X1,Fval,Exitflag,Output] = patternsearch(objectiveFcn,x0,Aineq,Bineq,Aeq,Beq);`\n```Optimization terminated: mesh size less than options.MeshTolerance. ```\n`fprintf('The number of iterations was : %d\\n', Output.iterations);`\n```The number of iterations was : 244 ```\n`fprintf('The number of function evaluations was : %d\\n', Output.funccount);`\n```The number of function evaluations was : 1895 ```\n`fprintf('The best function value found was : %g\\n', Fval);`\n```The best function value found was : 2189.03 ```\n\nMonitor the optimization process by specifying two plot functions. Specify the plot functions as the `PlotFcn` option by using `optimoptions`. One plot function, `psplotbestf`, plots the best objective function value at every iteration. Another plot function, `psplotfuncount`, plots the number of times the objective function is evaluated at each iteration. Set these two plot functions in a cell array.\n\n`opts = optimoptions(@patternsearch,'PlotFcn',{@psplotbestf,@psplotfuncount});`\n\nRun the `patternsearch` solver. Since there are no upper or lower bound constraints and no nonlinear constraints, pass empty arrays (`[]`) for the seventh, eighth, and ninth arguments.\n\n```[X1,Fval,ExitFlag,Output] = patternsearch(objectiveFcn,x0,Aineq,Bineq, ... Aeq,Beq,[],[],[],opts);```", null, "```Optimization terminated: mesh size less than options.MeshTolerance. ```\n\n### Mesh Parameters\n\nInitial mesh size\n\nThe pattern search algorithm uses a set of rational basis vectors to generate search directions. It performs a search along the search directions using the current mesh size. The solver starts with an initial mesh size of `1` by default. To start the initial mesh size at `10`, set the `InitialMeshSize` option:\n\n`options = optimoptions(opts,'InitialMeshSize',10);`\n\nMesh scaling\n\nA mesh can be scaled to improve the minimization of a badly scaled optimization problem. Scale is used to rotate the pattern by some degree and scale along the search directions. The scale option is on (`true`) by default but can be turned off if the problem is well scaled. In general, if the problem is badly scaled, setting this option to `true` may help in reducing the number of function evaluations. For this objective function, set `ScaleMesh` to `false`, because `lincontest7` is a well-scaled objective function.\n\n`opts = optimoptions(opts,'ScaleMesh',false);`\n\nMesh accelerator\n\nDirect search methods require many function evaluations as compared to derivative-based optimization methods. The pattern search algorithm can quickly find the neighborhood of an optimum point, but can be slow in detecting the minimum itself. This is the cost of not using derivatives. The `patternsearch` solver can reduce the number of function evaluations using an accelerator. When the accelerator is on (`opts.AccelerateMesh = true`) the mesh size is contracted rapidly after some minimum mesh size is reached. This option is recommended only for smooth problems, otherwise you can lose some accuracy. The accelerator is off (`false`) by default. Here, set the `AccelerateMesh` to `true` because the objective function is smooth.\n\n`opts = optimoptions(opts,'AccelerateMesh',true);`\n\nRun the `patternsearch` solver.\n\n```[X2,Fval,ExitFlag,Output] = patternsearch(objectiveFcn,x0,Aineq,Bineq, ... Aeq,Beq,[],[],[],opts);```", null, "```Optimization terminated: mesh size less than options.MeshTolerance. ```\n`fprintf('The number of iterations was : %d\\n', Output.iterations);`\n```The number of iterations was : 200 ```\n`fprintf('The number of function evaluations was : %d\\n', Output.funccount);`\n```The number of function evaluations was : 1327 ```\n`fprintf('The best function value found was : %g\\n', Fval);`\n```The best function value found was : 2189.03 ```\n\nThese option setting reduce the number of iterations and the number of function evaluations, and there is no apparent loss of accuracy.\n\n### Stopping Criteria and Tolerances\n\nWhat are MeshTolerance, StepTolerance and FunctionTolerance?\n\n`MeshTolerance` is a tolerance on the mesh size. If the mesh size is less than `MeshTolerance`, the solver stops. `StepTolerance` is used as the minimum tolerance on the change in the current point to the next point. `FunctionTolerance` is used as the minimum tolerance on the change in the function value from the current point to the next point.\n\nSet the `MeshTolerance` to 1e-7, which is ten times smaller than the default value. This setting can increase the number of function evaluations and iterations, and can lead to a more accurate solution.\n\n`opts.MeshTolerance = 1e-7;`\n\n### Search Methods in Pattern Search\n\nThe pattern search algorithm can use an additional search at every iteration. This option is called `SearchFcn`. When you set a `SearchFcn`, that search is done first before the mesh search. If the `SearchFcn` is successful, `patternsearch` skips the mesh search, commonly called the `PollFcn`, for that iteration. If the search method is unsuccessful in improving the current point, `patternsearch` performs the poll.\n\nThere are five different search method options. These search methods include `searchga` and `searchneldermead`, which are two different optimization algorithms. It is recommended to use these methods only for the first iteration, which is the default. Using these methods repeatedly at every iteration might not improve the results and could be computationally expensive. One the other hand, the `searchlhs`, which generates Latin hypercube points, can be used in every iteration or possibly every 10 iterations. Other choices for search methods include Poll methods such as positive basis N+1 or positive basis 2N.\n\nA recommended strategy is to use the positive basis N+1 (which requires at most N+1 points to create a pattern) as a search method and positive basis 2N (which requires 2N points to create a pattern) as a poll method. Update the options structure to use `positivebasisnp1` as the search method. Since the positive basis 2N is the default `PollFcn`, do not set that option.\n\n`opts.SearchFcn = @positivebasisnp1;`\n\nRun the `patternsearch` solver.\n\n```[X5,Fval,ExitFlag,Output] = patternsearch(objectiveFcn,x0,Aineq,Bineq,Aeq,Beq, ... [],[],[],opts);```", null, "```Optimization terminated: mesh size less than options.MeshTolerance. ```\n`fprintf('The number of iterations was : %d\\n', Output.iterations);`\n```The number of iterations was : 70 ```\n`fprintf('The number of function evaluations was : %d\\n', Output.funccount);`\n```The number of function evaluations was : 867 ```\n`fprintf('The best function value found was : %g\\n', Fval);`\n```The best function value found was : 2189.03 ```\n\nWith these options, the total number of iterations and function evaluations decrease. This decrease occurs even though the mesh tolerance is smaller than the previous value, and, as you can see in the exit messages, the mesh tolerance is the stopping criterion that halts the solver.\n\n\n\nWatch now" ]

[ null, "https://ww2.mathworks.cn/help/examples/globaloptim/win64/psoptionsdemo_01.png", null, "https://ww2.mathworks.cn/help/examples/globaloptim/win64/psoptionsdemo_02.png", null, "https://ww2.mathworks.cn/help/examples/globaloptim/win64/psoptionsdemo_03.png", null ]

[ "+0\n\n# 5x= 21-2x what does x mean\n\n0\n1550\n2\n\n5x= 21-2x what does x mean\n\nmath algebra\nAug 17, 2014\n\n#2\n+4\n\n5x= 21-2x what does x mean\n\nIn primary schoool it might have been written like this.\n\n$$5\\times \\fbox{\\begin{minipage}{0.6cm}\\hfill\\vspace{0.6cm}\\end{minipage}} =21-2\\times \\fbox{\\begin{minipage}{0.6cm}\\hfill\\vspace{0.6cm}\\end{minipage}}$$\n\nEXACTLY the same number has to go in both boxes to make the statement true.\n\nThe box has been replaced with the the letter x.  The idea is to solve the number sentence to work out what x is.  You will learn all sorts of techniques to be able ot do this easily.\n\nLet's make this a bit easier to understand and to solve.\n\nWe will pretend that the the boxes are two identical bags of lollies and you want to know how many lolies is in each bag.\n\n5bags = 21lollies - 2bags\n\nWe can add 2 bags to both sides then we will have\n\n7bags = 21 lollies\n\n7*3=21\n\nSo there must be 3 lollies in each bag.  3 goes in the the box.", null, "So if\n\n5x=21-2x then 7x=21 and x=3\n\nI hope that helps.", null, "Aug 18, 2014\n\n#1\n+5\n\nThink of x as a label on a box.  Inside the box is a number which you have to find by manipulating a balance (the equation).\n\nOn the right-hand side is a number equal to 21 with two boxes taken away.  Make this side 21 by adding two boxes to it.  To maintain the balance though, you also have to add 2 boxes to the left-hand side. So you get:\n\n7x = 21\n\nNow you have 7 boxes on the left-hand side, but you only want one, so divide the left-hand side by 7 to be left with just x. Again to keep the balance you also have to divide the right-hand side by 7, and since 21/7 is 3 you are left with:\n\nx = 3.  This is the number in the box!\n\nAug 17, 2014\n#2\n+4\n\n5x= 21-2x what does x mean\n\nIn primary schoool it might have been written like this.\n\n$$5\\times \\fbox{\\begin{minipage}{0.6cm}\\hfill\\vspace{0.6cm}\\end{minipage}} =21-2\\times \\fbox{\\begin{minipage}{0.6cm}\\hfill\\vspace{0.6cm}\\end{minipage}}$$\n\nEXACTLY the same number has to go in both boxes to make the statement true.\n\nThe box has been replaced with the the letter x.  The idea is to solve the number sentence to work out what x is.  You will learn all sorts of techniques to be able ot do this easily.\n\nLet's make this a bit easier to understand and to solve.\n\nWe will pretend that the the boxes are two identical bags of lollies and you want to know how many lolies is in each bag.\n\n5bags = 21lollies - 2bags\n\nWe can add 2 bags to both sides then we will have\n\n7bags = 21 lollies\n\n7*3=21\n\nSo there must be 3 lollies in each bag.  3 goes in the the box.", null, "So if\n\n5x=21-2x then 7x=21 and x=3\n\nI hope that helps.", null, "Melody Aug 18, 2014" ]

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[ "Aerodynamics Tool Free\n\nAll Android applications categories\n\nAll Android games categories", null, "# Aerodynamics Tool Free\n\n122 8\n\n8 Users\nrating\n\n## Screenshots\n\nDescription\n\n1. Basic Aerodynamic Calculator- Calculation of Dynamic Pressure, Reynolds Number, Force and Moment Coefficient for Incompressible and Compressible flow.\n\n2. Isentropic Relations - If any one of the variables, M, Po/P, To/T, Rho/Rh or A/A* is known the remaining variable values can be computed.\n\n3. Normal Shock- Select any known variable from list, M1, P2/P1, T2/T1, M2, rho2/rho1, Po2/Po1, and Po2/P1 and user can calculate remaining variable values. Indices 1 and 2 are for before shock and after shock respectively.\n\n4. Oblique Shock - Theta-Beta-Mech number relation. If values of any two variable known then user can calculate value of remaining one.\n\n5. Vortex Lattice Method - Vortex Lattice Method is a method used to calculate lift coefficient of Finite Wing at low speed. It gives fairly good CL estimation for Swept wing compared to Vortex Panel Method. This tool is helpful for primary aerodynamic design iterations. The method presented here uses 'n' spanwise divisions and one chordwise division of wing. The final output is slope of CL vs Alpha curve and it is CL per degree angle of attack.\n\n6. Prandtl-Meyer expansion wave relations - Compute v1 from M1 (Mach Number before expansion wave). v is Prandtl-Meyer function. Add deflection angle theta to v1 to get v2. From v2 compute M2 (Mach Number after expansion wave).\n\n7. Pitot Tube - Divided in two categories. One for Incompressible Flow and Other for Compressible Flow. In case of Incompressible flow, user can input height of Pitot Tube column in Meters of Water, Mercury or Any known fluid to calculate free stream velocity. The compressible flow case takes inputs for Total Pressure and Static Pressure measured by Pitot Tube and calculates Free Stream Mach Number. User need not to worry about Subsonic or Supersonic flows. These are identified and solved accordingly.\n\n8. Y Plus or Y+ - It calculates first cell height of grid near wall to resolve boundary layer.\n\n9. Check points for Inputs to avoid wrong inputs and flash suggestions if any.\n\nIf you like this, please consider buying ad-free version which can take any value of Specific Heat Ratio (Gamma) and not fixed to 1.4.\n\nIf found any Bug or wrong output, please mail me regarding issue\n\nBefore giving negative rating for pop-up ads please consider that I have tried to keep ads when user finishes using any one of the utility of this app and not at the launch of utility which is more objectionable." ]

[ null, "https://lh6.ggpht.com/0ZjLyGcZsIU6bo_Hb7MuVu7rAGsNZtwkdLqCFV2aAEMNMBCYQmBjtLO3MtpsZqCqsnw=w140", null ]

[ "# Efficiently working out the nearest weekend\n\nIs this an efficient way to do this?\n\nthe code is designed to find the nearest friday, without passing Sunday. If it is already past Friday, the start date should equal now.\n\nfunction get_weekend(){\n$start = time();$day = date('w',$now); //find friday, without going past Sunday while ($day < 5 && $day > 0){$start = $start + 86400;$day = date('w',$start); } //find nearest sunday to start date$end = $start; while ($day != 0){\n$end =$end + 86400;\n$day = date('w',$end);\n}\n$weekend['StartDate']=date(\"d-F-Y\",$start);\n$weekend['EndDate'] = date(\"d-F-Y\",$end);\nreturn $weekend; } ## 2 Answers No need for while loops - just some math: function get_weekend() {$start = time();\n$end =$start;\n$day = date('w',$start);\n\nif ($day > 0 &&$day < 5)\n{\n$start =$time()+ ((5 - $day) * 86400); } if ($day != 0)\n{\n$end = time() + (7 -$day) * 86400;\n}\n\n$weekend['StartDate'] = date('d-F-Y',$start);\n$weekend['EndDate'] = date('d-F-Y',$end);\n\nreturn $weekend; } Edit: Found there was a better way to calculate$end;\n\nEdit: Keep finding better code:\n\nfunction get_weekend()\n{\n$start = time();$end = $start;$day = date('w', $start); if ($day > 0)\n{\nif ($day < 5) {$start += ((5 - $day) * 86400); }$end += (7 - $day) * 86400; }$weekend['StartDate'] = date('d-F-Y', $start);$weekend['EndDate'] = date('d-F-Y', $end); return$weekend;\n}\n\n\nShort, short version:\n\nfunction get_weekend()\n{\n$fri = strtotime('friday');$sun = strtotime('sunday');\n// strtotime('friday') will be after strtotime('sunday') if it is after 0:00 Friday\n// and before 0:00 on Sunday\n$start =$sun < $fri? now():$fri;\n$weekend = array();$weekend['StartDate'] = date('d-F-Y', $start);$weekend['EndDate'] = date('d-F-Y', $end); return$weekend;\n}" ]

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[ "# Number 1130203130021\n\n### Properties of number 1130203130021\n\nCross Sum:\nFactorization:\nDivisors:\nCount of divisors:\nSum of divisors:\nPrime number?\nNo\nFibonacci number?\nNo\nBell Number?\nNo\nCatalan Number?\nNo\nBase 3 (Ternary):\nBase 4 (Quaternary):\nBase 5 (Quintal):\nBase 8 (Octal):\n107255b28a5\nBase 32:\n10silma55\nsin(1130203130021)\n-0.98842415898977\ncos(1130203130021)\n0.15171579326281\ntan(1130203130021)\n-6.5149720917821\nln(1130203130021)\n27.753418493578\nlg(1130203130021)\n12.053156505713\nsqrt(1130203130021)\n1063110.1213049\nSquare(1130203130021)\n1.2773591151093E+24\n\n### Number Look Up\n\nLook Up\n\n1130203130021 (one trillion one hundred thirty billion two hundred three million one hundred thirty thousand twenty-one) is a very unique number. The cross sum of 1130203130021 is 17. If you factorisate the figure 1130203130021 you will get these result 7 * 161457590003. 1130203130021 has 4 divisors ( 1, 7, 161457590003, 1130203130021 ) whith a sum of 1291660720032. 1130203130021 is not a prime number. The figure 1130203130021 is not a fibonacci number. The figure 1130203130021 is not a Bell Number. 1130203130021 is not a Catalan Number. The convertion of 1130203130021 to base 2 (Binary) is 10000011100100101010110110010100010100101. The convertion of 1130203130021 to base 3 (Ternary) is 11000001020210120202002222. The convertion of 1130203130021 to base 4 (Quaternary) is 100130211112302202211. The convertion of 1130203130021 to base 5 (Quintal) is 122004124000130041. The convertion of 1130203130021 to base 8 (Octal) is 20344526624245. The convertion of 1130203130021 to base 16 (Hexadecimal) is 107255b28a5. The convertion of 1130203130021 to base 32 is 10silma55. The sine of 1130203130021 is -0.98842415898977. The cosine of the number 1130203130021 is 0.15171579326281. The tangent of 1130203130021 is -6.5149720917821. The square root of 1130203130021 is 1063110.1213049.\nIf you square 1130203130021 you will get the following result 1.2773591151093E+24. The natural logarithm of 1130203130021 is 27.753418493578 and the decimal logarithm is 12.053156505713. You should now know that 1130203130021 is great figure!" ]

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[ "statistics-0.9.0.0: A library of statistical types, data, and functions\n\nPortability portable experimental [email protected]\n\nStatistics.KernelDensity\n\nDescription\n\nKernel density estimation code, providing non-parametric ways to estimate the probability density function of a sample.\n\nSynopsis\n\n# Simple entry points\n\nArguments\n\n :: Vector v Double => Int Number of points at which to estimate -> v Double Data sample -> (Points, Vector Double)\n\nSimple Epanechnikov kernel density estimator. Returns the uniformly spaced points from the sample range at which the density function was estimated, and the estimates at those points.\n\nArguments\n\n :: Vector v Double => Int Number of points at which to estimate -> v Double Data sample -> (Points, Vector Double)\n\nSimple Gaussian kernel density estimator. Returns the uniformly spaced points from the sample range at which the density function was estimated, and the estimates at those points.\n\n# Building blocks\n\n## Choosing points from a sample\n\nnewtype Points Source\n\nPoints from the range of a `Sample`.\n\nConstructors\n\n Points FieldsfromPoints :: Vector Double\n\nInstances\n\n Eq Points Show Points\n\nArguments\n\n :: Vector v Double => Int Number of points to select, n -> Double Sample bandwidth, h -> v Double Input data -> Points\n\nChoose a uniform range of points at which to estimate a sample's probability density function.\n\nIf you are using a Gaussian kernel, multiply the sample's bandwidth by 3 before passing it to this function.\n\nIf this function is passed an empty vector, it returns values of positive and negative infinity.\n\n## Bandwidth estimation\n\nThe width of the convolution kernel used.\n\nbandwidth :: Vector v Double => (Double -> Bandwidth) -> v Double -> BandwidthSource\n\nCompute the optimal bandwidth from the observed data for the given kernel.\n\nBandwidth estimator for an Epanechnikov kernel.\n\nBandwidth estimator for a Gaussian kernel.\n\n## Kernels\n\ntype Kernel = Double -> Double -> Double -> Double -> DoubleSource\n\nThe convolution kernel. Its parameters are as follows:\n\n• Scaling factor, 1/nh\n• Bandwidth, h\n• A point at which to sample the input, p\n• One sample value, v\n\nEpanechnikov kernel for probability density function estimation.\n\nGaussian kernel for probability density function estimation.\n\n## Low-level estimation\n\nArguments\n\n :: Vector v Double => Kernel Kernel function -> Bandwidth Bandwidth, h -> v Double Sample data -> Points Points at which to estimate -> Vector Double\n\nKernel density estimator, providing a non-parametric way of estimating the PDF of a random variable.\n\nArguments\n\n :: Vector v Double => (Double -> Double) Bandwidth function -> Kernel Kernel function -> Double Bandwidth scaling factor (3 for a Gaussian kernel, 1 for all others) -> Int Number of points at which to estimate -> v Double sample data -> (Points, Vector Double)\n\nA helper for creating a simple kernel density estimation function with automatically chosen bandwidth and estimation points." ]

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[ "# Dewpoint and Mixing Ratio¶\n\nUse functions from `metpy.calc` as well as pint’s unit support to perform calculations.\n\nThe code below converts the mixing ratio value into a value for vapor pressure assuming both 1000mb and 850mb ambient air pressure values. It also demonstrates converting the resulting dewpoint temperature to degrees Fahrenheit.\n\n```import metpy.calc as mpcalc\nfrom metpy.units import units\n```\n\nCreate a test value of mixing ratio in grams per kilogram\n\n```mixing = 10 * units('g/kg')\nprint(mixing)\n```\n\nOut:\n\n```10.0 gram / kilogram\n```\n\nNow throw that value with units into the function to calculate the corresponding vapor pressure, given a surface pressure of 1000 mb\n\n```e = mpcalc.vapor_pressure(1000. * units.mbar, mixing)\nprint(e)\n```\n\nOut:\n\n```15823.863685716977 gram * millibar / kilogram\n```\n\nTake the odd units and force them to millibars\n\n```print(e.to(units.mbar))\n```\n\nOut:\n\n```15.823863685716978 millibar\n```\n\nTake the raw vapor pressure and throw into the dewpoint function\n\n```td = mpcalc.dewpoint(e)\nprint(td)\n```\n\nOut:\n\n```13.854699858753728 degree_Celsius\n```\n\nWhich can of course be converted to Fahrenheit\n\n```print(td.to('degF'))\n```\n\nOut:\n\n```56.93845974575668 degree_Fahrenheit\n```\n\nNow do the same thing for 850 mb, approximately the pressure of Denver\n\n```e = mpcalc.vapor_pressure(850. * units.mbar, mixing)\nprint(e.to(units.mbar))\n```\n\nOut:\n\n```13.45028413285943 millibar\n```\n\nAnd print the corresponding dewpoint\n\n```td = mpcalc.dewpoint(e)\nprint(td, td.to('degF'))\n```\n\nOut:\n\n```11.377098919513376 degree_Celsius 52.47877805512405 degree_Fahrenheit\n```\n\nTotal running time of the script: ( 0 minutes 0.006 seconds)\n\nGallery generated by Sphinx-Gallery" ]

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[ "# Arithmetic Progressions -Class 10\n\nIMPORTANT QUESTIONS FOR CBSE CLASS 10 MATHEMATICS\nArithmetic Progressions – Chapter 5\n\n1. If the common difference of an AP is 7, then find the difference between the fifteenth term and twelfth term?\nSolution:\nGiven d = 7\nn th term = a + (n-1)d\nFifteenth term = a + (15 – 1) d = a + 14d\nTwelfth term = a + (12 -1) d = a + 11d\nDifference = (a + 14d) – (a + 11d) = 3d = 3 x 7 = 21.\n2. The sum of first 20 terms of the AP: 24, 20, 16, ——————-\nSolution:\nHere a = 24, d = 20 – 24 = -4, n = 20.\nSum of first 20 terms = n/2[2a + (n-1)d]\n= 20/2[2 x 24 + (20 -1) (-4)]\n= 10[48 + (-76)]\n= 10 x (-28)\n= -280.\n3. Find the sum of first seven multiples of 7?\nSolution:\nMultiples of 7 are 7, 14, 21, 28, 35, 42, 49, ————\nHere a = 7, d = 14 -7 = 7, n = 7.\nSum of seven terms = 7/2[2 x 7 + (7 -1)7]\n= 7/2[14 + 42]\n= 196.\n4. Find the common difference of an AP in which the difference between the 20 th term and 5 th term is 60.\nSolution:\nWe know that n th term = a + (n-1) d\n20 th term = a + (20 -1) d = a + 19d\n5 th term = a + (5 -1) d= a + 4d\nGiven difference = (a + 19d) – (a + 4d) = 60\n15 d = 60\nd = 60/15 = 4.\n5. Find the 10 th term from end of the AP: 3, 8, 13…..…253?\nSolution:\nHere a = 253, d = 3 -8 = -5, n = 10.\n10 th term = 253 + 9 x (-5) = 253 + (-45) = 208.\n6. Find the common difference of the AP whose 11 th term is 38 and the 16 th term is 73?\nSolution:\nGiven 11 th term = a + 10d = 38——————– (1)\n16 th term = a + 15d = 73————————- (2)\nSolving these two equations, we get\nBy subtraction, -5d = -35\nd = (-35)/ (-5) = 7.\n7. The n th term of an AP is (5 – 2n), then find the common difference?\nSolution:\nGiven n th term = 5 – 2n\nFirst term = 5 – 2 x 1 = 5 -2 = 3.\nSecond term = 5 – 2 x 2 = 5 – 4 = 1\nThird term = 5 – 2 x 3 = 5 – 6 = -1\nCommon difference = 1 – 3 = -2.\n8. Write the 19 th term of the AP: -60, -57, -54, ——————\nSolution:\nHere a = -60, d = (-57) – (-60) = 3\n19 th term = a + (19 -1) d = (-60) + 18 x 3 = (-60) + 54 = -6.\n9. How many terms of the AP: 24, 21, 18 …………… must be taken so that their sum is 78?\nSolution:\nHere a = 24, d = 21 – 24 = -3 and sum = 78.\nSum of n terms = n/2[2a + (n-1) d]\n78 = n/2[2 x 24 + (n – 1) (-3)]\n156 = n [48 -3n +3]\n156 = n [51 – 3n]\n156 = 3n [17 –n]\n52 = n (17 – n), which is a quadratic equation.\nBy solving these, we get n = 4 or n = 13.\nSo the number of terms is either 4 or 13.\n10. Which term of the AP: 21, 18, 15, ———– is -81?\nSolution:\nHere a = 21, d = 18 – 21 = -3 and n th term = -81 and we have to find n.\nAs n th term = a + (n -1) d, we have -81 = 21 + (n -1) (-3)\n-81 = 21 – 3n + 3 = 24 – 3n\n-105 = -3n\nn = (-105) / (-3) = 35.\nTherefore, the 35 th term of the given AP is -81." ]

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[ "# #77ffaf Color Information\n\nIn a RGB color space, hex #77ffaf is composed of 46.7% red, 100% green and 68.6% blue. Whereas in a CMYK color space, it is composed of 53.3% cyan, 0% magenta, 31.4% yellow and 0% black. It has a hue angle of 144.7 degrees, a saturation of 100% and a lightness of 73.3%. #77ffaf color hex could be obtained by blending #eeffff with #00ff5f. Closest websafe color is: #66ff99.\n\n• R 47\n• G 100\n• B 69\nRGB color chart\n• C 53\n• M 0\n• Y 31\n• K 0\nCMYK color chart\n\n#77ffaf color description : Very light cyan - lime green.\n\n# #77ffaf Color Conversion\n\nThe hexadecimal color #77ffaf has RGB values of R:119, G:255, B:175 and CMYK values of C:0.53, M:0, Y:0.31, K:0. Its decimal value is 7864239.\n\nHex triplet RGB Decimal 77ffaf `#77ffaf` 119, 255, 175 `rgb(119,255,175)` 46.7, 100, 68.6 `rgb(46.7%,100%,68.6%)` 53, 0, 31, 0 144.7°, 100, 73.3 `hsl(144.7,100%,73.3%)` 144.7°, 53.3, 100 66ff99 `#66ff99`\nCIE-LAB 91.023, -54.733, 27.175 51.102, 78.533, 53.021 0.28, 0.43, 78.533 91.023, 61.108, 153.595 91.023, -59.861, 48.308 88.619, -52.151, 26.56 01110111, 11111111, 10101111\n\n# Color Schemes with #77ffaf\n\n• #77ffaf\n``#77ffaf` `rgb(119,255,175)``\n• #ff77c7\n``#ff77c7` `rgb(255,119,199)``\nComplementary Color\n• #83ff77\n``#83ff77` `rgb(131,255,119)``\n• #77ffaf\n``#77ffaf` `rgb(119,255,175)``\n• #77fff3\n``#77fff3` `rgb(119,255,243)``\nAnalogous Color\n• #ff7783\n``#ff7783` `rgb(255,119,131)``\n• #77ffaf\n``#77ffaf` `rgb(119,255,175)``\n• #f377ff\n``#f377ff` `rgb(243,119,255)``\nSplit Complementary Color\n• #ffaf77\n``#ffaf77` `rgb(255,175,119)``\n• #77ffaf\n``#77ffaf` `rgb(119,255,175)``\n• #af77ff\n``#af77ff` `rgb(175,119,255)``\n• #c7ff77\n``#c7ff77` `rgb(199,255,119)``\n• #77ffaf\n``#77ffaf` `rgb(119,255,175)``\n• #af77ff\n``#af77ff` `rgb(175,119,255)``\n• #ff77c7\n``#ff77c7` `rgb(255,119,199)``\n• #2bff82\n``#2bff82` `rgb(43,255,130)``\n• #44ff91\n``#44ff91` `rgb(68,255,145)``\n• #5effa0\n``#5effa0` `rgb(94,255,160)``\n• #77ffaf\n``#77ffaf` `rgb(119,255,175)``\n• #91ffbe\n``#91ffbe` `rgb(145,255,190)``\n• #aaffcd\n``#aaffcd` `rgb(170,255,205)``\n• #c4ffdc\n``#c4ffdc` `rgb(196,255,220)``\nMonochromatic Color\n\n# Alternatives to #77ffaf\n\nBelow, you can see some colors close to #77ffaf. Having a set of related colors can be useful if you need an inspirational alternative to your original color choice.\n\n• #77ff8d\n``#77ff8d` `rgb(119,255,141)``\n• #77ff98\n``#77ff98` `rgb(119,255,152)``\n• #77ffa4\n``#77ffa4` `rgb(119,255,164)``\n• #77ffaf\n``#77ffaf` `rgb(119,255,175)``\n• #77ffba\n``#77ffba` `rgb(119,255,186)``\n• #77ffc6\n``#77ffc6` `rgb(119,255,198)``\n• #77ffd1\n``#77ffd1` `rgb(119,255,209)``\nSimilar Colors\n\n# #77ffaf Preview\n\nThis text has a font color of #77ffaf.\n\n``<span style=\"color:#77ffaf;\">Text here</span>``\n#77ffaf background color\n\nThis paragraph has a background color of #77ffaf.\n\n``<p style=\"background-color:#77ffaf;\">Content here</p>``\n#77ffaf border color\n\nThis element has a border color of #77ffaf.\n\n``<div style=\"border:1px solid #77ffaf;\">Content here</div>``\nCSS codes\n``.text {color:#77ffaf;}``\n``.background {background-color:#77ffaf;}``\n``.border {border:1px solid #77ffaf;}``\n\n# Shades and Tints of #77ffaf\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, #000101 is the darkest color, while #edfff4 is the lightest one.\n\n• #000101\n``#000101` `rgb(0,1,1)``\n• #001509\n``#001509` `rgb(0,21,9)``\n• #002911\n``#002911` `rgb(0,41,17)``\n• #003c19\n``#003c19` `rgb(0,60,25)``\n• #005021\n``#005021` `rgb(0,80,33)``\n• #006329\n``#006329` `rgb(0,99,41)``\n• #007731\n``#007731` `rgb(0,119,49)``\n• #008b39\n``#008b39` `rgb(0,139,57)``\n• #009e41\n``#009e41` `rgb(0,158,65)``\n• #00b249\n``#00b249` `rgb(0,178,73)``\n• #00c551\n``#00c551` `rgb(0,197,81)``\n• #00d959\n``#00d959` `rgb(0,217,89)``\n• #00ed61\n``#00ed61` `rgb(0,237,97)``\n• #01ff6a\n``#01ff6a` `rgb(1,255,106)``\n• #15ff75\n``#15ff75` `rgb(21,255,117)``\n• #29ff81\n``#29ff81` `rgb(41,255,129)``\n• #3cff8c\n``#3cff8c` `rgb(60,255,140)``\n• #50ff98\n``#50ff98` `rgb(80,255,152)``\n• #63ffa3\n``#63ffa3` `rgb(99,255,163)``\n• #77ffaf\n``#77ffaf` `rgb(119,255,175)``\n• #8bffbb\n``#8bffbb` `rgb(139,255,187)``\n• #9effc6\n``#9effc6` `rgb(158,255,198)``\n• #b2ffd2\n``#b2ffd2` `rgb(178,255,210)``\n• #c5ffdd\n``#c5ffdd` `rgb(197,255,221)``\n• #d9ffe9\n``#d9ffe9` `rgb(217,255,233)``\n• #edfff4\n``#edfff4` `rgb(237,255,244)``\nTint Color Variation\n\n# Tones of #77ffaf\n\nA tone is produced by adding gray to any pure hue. In this case, #b6c0ba is the less saturated color, while #77ffaf is the most saturated one.\n\n• #b6c0ba\n``#b6c0ba` `rgb(182,192,186)``\n• #b1c5b9\n``#b1c5b9` `rgb(177,197,185)``\n• #abcbb8\n``#abcbb8` `rgb(171,203,184)``\n• #a6d0b7\n``#a6d0b7` `rgb(166,208,183)``\n• #a1d5b6\n``#a1d5b6` `rgb(161,213,182)``\n• #9cdab5\n``#9cdab5` `rgb(156,218,181)``\n• #96e0b5\n``#96e0b5` `rgb(150,224,181)``\n• #91e5b4\n``#91e5b4` `rgb(145,229,180)``\n• #8ceab3\n``#8ceab3` `rgb(140,234,179)``\n• #87efb2\n``#87efb2` `rgb(135,239,178)``\n• #81f5b1\n``#81f5b1` `rgb(129,245,177)``\n• #7cfab0\n``#7cfab0` `rgb(124,250,176)``\n• #77ffaf\n``#77ffaf` `rgb(119,255,175)``\nTone Color Variation\n\n# Color Blindness Simulator\n\nBelow, you can see how #77ffaf 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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[ "You've successfully subscribed to Better Data Science\nWelcome back! You've successfully signed in\n\n# Does Laptop Matter for Data Science? Old ThinkPad vs. New MacBook Pro Compared", null, "Pandas and TensorFlow benchmarks on Intel i3-6100U vs. M1 Pro MacBook Pro - Up to 50X faster?\n\nThe 16\" M1 Pro MacBook has been my go-to workhorse since its launch. It packs incredible performance below the ultra-premium surface - all while lasting an entire day and having probably the best screen in the industry.\n\nBut what about an old, dual-core Lenovo ThinkPad? Can a sixth-generation Intel processor even come close to a modern-day powerhouse? Continue reading to find out.\n\nIn today's article, we'll compare an old ThinkPad L470 to a modern M1 Pro MacBook Pro in a series of benchmarks, from synthetic ones to Pandas and TensorFlow. The scope of the benchmarks will be limited, so everything can fit in a ten-minute read.\n\nBefore going any further, let's compare the hardware specs:\n\nIt's not really an apples-to-apples comparison. The Mac starts at \\$2499, and I bought the ThinkPad for under \\$200. That's over 12x cheaper, so the Mac has some justification to do.\n\n## Intel core i3-6100U on a Benchmark - Not so Great Nowadays\n\nThe 6100U was released way back in 2015, so we really can't expect it to compare well over a much more modern and powerful chip. Also, synthetical benchmarks can only take us so far, but they're a good starting point.\n\nHere's how i3-6100U compares to the M1 Pro chip in a single core department:\n\nM1 Pro is about 3.5 times faster, which is expected. Most Windows laptops don't come close even today, so it's really not a surprise.\n\nMulticore is where things get interesting. The i3 has only two cores, while M1 Pro has 8 performance and 2 efficiency cores. Here are the results:\n\nAs you can see, the Mac came up 12 times faster on this test. That's a huge difference, and you'll definitely notice it in everyday use. Also, macOS requires less resources to run smoothly, so that's another consideration you have to take in mind.\n\nNow let's get into the actual data science benchmarks, starting with Pandas.\n\n## Pandas Benchmark - Typical Data Workflow Times Compared\n\nThere's only so much I can compare in a single article, so let's stick with the basics. We'll compare the time required to create, save, read, and transform a Pandas DataFrame.\n\n### Create a Pandas DataFrame\n\nLet's create two datasets - the first one with 1 million rows, and the second one with around 5. These should be enough to push the ThinkPad to its limits.\n\nThe code snippet below imports all the libraries needed in this section and also declares a function `create_dataset()` that, well, creates the datasets.\n\n``````import random\nimport string\nimport numpy as np\nimport pandas as pd\nfrom datetime import datetime\nnp.random.seed = 42\n\ndef create_dataset(start: datetime, end: datetime, freq: str) -> pd.DataFrame:\ndef gen_random_string(length: int = 32) -> str:\nreturn ''.join(random.choices(\nstring.ascii_uppercase + string.digits, k=length)\n)\n\ndt = pd.date_range(\nstart=start,\nend=end,\nfreq=freq, # Increase if you run out of RAM\nclosed='left'\n)\n\ndf_size = len(dt)\ndf = pd.DataFrame({\n'date': dt,\n'a': np.random.rand(df_size),\n'b': np.random.rand(df_size),\n'c': np.random.rand(df_size),\n'd': np.random.rand(df_size),\n'e': np.random.rand(df_size),\n'str1': [gen_random_string() for x in range(df_size)],\n'str2': [gen_random_string() for x in range(df_size)]\n})\n\nreturn df``````\n\nLet's now use it to create both the 1M and 5M datasets:\n\n``````########## 1M Dataset ##########\n\ndf_1m = create_dataset(\nstart=datetime(2010, 1, 1),\nend=datetime(2020, 1, 1),\nfreq='300s'\n)\n\n########## 5M Dataset ##########\n\ndf_5m = create_dataset(\nstart=datetime(2010, 1, 1),\nend=datetime(2020, 1, 1),\nfreq='60s'\n)``````\n\nHere are the times for both laptops and both datasets:", null, "Image 4 - Time needed to create the datasets (image by author)\n\nAll things considered, it's not really that drastic of a difference. The Mac is around 7 times faster for a 1M dataset, and around 5 times faster for a 5M dataset. No crashes or freezes whatsoever occurred on ThinkPad, it just takes more time.\n\n### Saving a Dataset to a CSV file\n\nYou'll probably save your data in different states throughout the project, so it would be nice to minimize the time spent here. The code snippet below dumps both Pandas DataFrames to a CSV file:\n\n``````########## 1M Dataset ##########\n\n########## 5M Dataset ##########\n\nAnd here are the results:", null, "Image 5 - Time needed to save the datasets as CSV files (image by author)\n\nSimilar story to what we had earlier. The Mac is around 5 times faster when saving Pandas DataFrame to a CSV file.\n\n### Reading a CSV file from a disk\n\nBut what about the other way around? Will reading CSV files also be 5 times faster on a Mac? Here's the code:\n\n``````########## 1M Dataset ##########\n\n########## 5M Dataset ##########\n\nAnd the results:\n\nThe 5X time difference seems to be somewhat universal up to this point.\n\n### Applying a function to a column\n\nYou can apply custom functions on a DataFrame column by calling the `apply()` method on it. The snippet below reverses the `str1` column which initially contains a random string of 32 characters:\n\n``````def reverse_str(x) -> str:\nreturn x[::-1]\n\n########## 1M Dataset ##########\n\ndf_1m['str1_rev'] = df_1m['str1'].apply(reverse_str)\n\n########## 5M Dataset ##########\n\ndf_5m['str1_rev'] = df_5m['str1'].apply(reverse_str)``````\n\nLet's take a look at the results:", null, "Image 7 - Time needed to reverse a column of strings (image by author)\n\nOnce again, there's around a 5X time difference in favor of the Mac.\n\nTo summarize - M1 Pro MacBook is continuously around 5 times faster than ThinkPad L470. It's not a surprise, but let's see if TensorFlow can make this difference even bigger.\n\n## TensorFlow Benchmark - Custom Model and Transfer Learning\n\nJust to address the elephant in the room - It's possible to install and even run TensorFlow on i3-6100U, or any other lower-end chip. It will work, but slowly, since the processor isn't too powerful and there's no GPU support whatsoever.\n\nLet's dive into the first test.\n\n### Custom TensorFlow model\n\nThe first TensorFlow benchmark uses data augmentation and applies a two-block convolutional model to the image dataset. There's nothing special about it, it's just a model architecture you're likely to stumble upon when learning TensorFlow:\n\n``````import os\nimport warnings\nfrom datetime import datetime\nos.environ['TF_CPP_MIN_LOG_LEVEL'] = '2'\nwarnings.filterwarnings('ignore')\n\nimport numpy as np\nimport tensorflow as tf\ntf.random.set_seed(42)\n\n####################\n####################\ntrain_datagen = tf.keras.preprocessing.image.ImageDataGenerator(\nrescale=1/255.0,\nrotation_range=20,\nwidth_shift_range=0.2,\nheight_shift_range=0.2,\nshear_range=0.2,\nzoom_range=0.2,\nhorizontal_flip=True,\nfill_mode='nearest'\n)\nvalid_datagen = tf.keras.preprocessing.image.ImageDataGenerator(\nrescale=1/255.0\n)\n\ntrain_data = train_datagen.flow_from_directory(\ntarget_size=(224, 224),\nclass_mode='categorical',\nbatch_size=64,\nseed=42\n)\nvalid_data = valid_datagen.flow_from_directory(\ntarget_size=(224, 224),\nclass_mode='categorical',\nbatch_size=64,\nseed=42\n)\n\n####################\n# 2. Model\n####################\nmodel = tf.keras.Sequential([\ntf.keras.layers.Conv2D(filters=32, kernel_size=(3, 3), input_shape=(224, 224, 3), activation='relu'),\ntf.keras.layers.Conv2D(filters=32, kernel_size=(3, 3), activation='relu'),\ntf.keras.layers.Flatten(),\ntf.keras.layers.Dense(128, activation='relu'),\ntf.keras.layers.Dense(2, activation='softmax')\n])\nmodel.compile(\nloss=tf.keras.losses.categorical_crossentropy,\nmetrics=[tf.keras.metrics.BinaryAccuracy(name='accuracy')]\n)\n\n####################\n# 3. Training\n####################\nmodel.fit(\ntrain_data,\nvalidation_data=valid_data,\nepochs=5\n)``````\n\nThe model was trained for 5 epochs on each machine, and the image below compares the average epoch times:", null, "Image 8 - TensorFlow average time per epoch on a custom model (image by author)\n\nAs you can see, TensorFlow works on an old laptop but is around 8 times slower when compared to the Mac. Not a deal-breaker if you're just starting out.\n\n### TensorFlow transfer learning model\n\nThe code snippet below is more or less identical to the previous one, but with a single important difference - it now uses a pretrained VGG-16 network to classify images:\n\n``````####################\n####################\ntrain_datagen = tf.keras.preprocessing.image.ImageDataGenerator(\nrescale=1/255.0,\nrotation_range=20,\nwidth_shift_range=0.2,\nheight_shift_range=0.2,\nshear_range=0.2,\nzoom_range=0.2,\nhorizontal_flip=True,\nfill_mode='nearest'\n)\nvalid_datagen = tf.keras.preprocessing.image.ImageDataGenerator(\nrescale=1/255.0\n)\n\ntrain_data = train_datagen.flow_from_directory(\ntarget_size=(224, 224),\nclass_mode='categorical',\nbatch_size=64,\nseed=42\n)\nvalid_data = valid_datagen.flow_from_directory(\ntarget_size=(224, 224),\nclass_mode='categorical',\nbatch_size=64,\nseed=42\n)\n\n####################\n# 2. Base model\n####################\nvgg_base_model = tf.keras.applications.vgg16.VGG16(\ninclude_top=False,\ninput_shape=(224, 224, 3),\nweights='imagenet'\n)\nfor layer in vgg_base_model.layers:\nlayer.trainable = False\n\n####################\n# 3. Custom layers\n####################\nx = tf.keras.layers.Flatten()(vgg_base_model.layers[-1].output)\nx = tf.keras.layers.Dense(128, activation='relu')(x)\nout = tf.keras.layers.Dense(2, activation='softmax')(x)\n\nvgg_model = tf.keras.models.Model(\ninputs=vgg_base_model.inputs,\noutputs=out\n)\nvgg_model.compile(\nloss=tf.keras.losses.categorical_crossentropy,\nmetrics=[tf.keras.metrics.BinaryAccuracy(name='accuracy')]\n)\n\n####################\n# 4. Training\n####################\nvgg_model.fit(\ntrain_data,\nvalidation_data=valid_data,\nepochs=5\n)``````\n\nThe results are much different this time:", null, "Image 9 - TensorFlow average time per epoch on a transfer learning model (image by author)\n\nAs you can see, the average training time per epoch barely increased on the Mac, but it skyrocketed on the ThinkPad. Waiting for the transfer learning model to finish training took a better part of the day on i3-6100U, as it's almost 54 times slower when compared to the Mac.\n\nSo, what have we learned from these benchmarks? Let's go over some insights next.\n\n## Data Science on an Old Laptop - Things to Consider\n\nIt's not impossible to do data wrangling or data science work on an old laptop, especially if you're just starting out. There are a couple of things you should keep in mind, so let's go over them.\n\n### Old laptop + Deep learning = Google Colab\n\nThere's no reason to put an old laptop through hell just because you're getting started with deep learning. Google Colab is an amazing pre-configured notebook environment that costs nothing and packs GPU support. If you need more features and longer runtimes, consider Colab Pro.\n\n### The lack of ports\n\nIt's weird to say that a Mac has a better port selection, but it's true in this case. ThinkPad L470 is 5+ years old, so it doesn't have a USB-C or even an HDMI port. Connecting an external display won't be so easy.\n\nThere's a VGA port available, but this requires an adapter and compromises in image quality if you have a modern external monitor. Not a deal breaker, just something to consider.\n\nWith a Mac, I can connect a 4K external display, monitor light, phone charger, microphone, and headphones - all with a single USB-C cable.\n\n### The terrible screen\n\nI've never been impressed with ThinkPad displays, but L470 takes the awfulness to a whole different level. The display is 14\" which is excellent for a laptop, but not at a 1366x768 resolution. Everything is huge, and you can't fit more than a window or two onto the screen. The color accuracy is just plain terrible as well.\n\nThat wouldn't be a problem if connecting an external display in full resolution was easier.\n\nOn the other side, the Mini LED screen on Mac spans across 16\" and packs a resolution of 3456x2234. It's a night and day difference, to say at least.\n\n## Conclusion\n\nIntel's i3-6100U can handle Pandas pretty well, and it can also train deep learning models with TensorFlow, albeit slowly. You can definitely get into data science and even work professionally in the industry while using it. Most of the work nowadays is done in the cloud, so laptops boil down to being glorified keyboards anyway.\n\nThe biggest issue of a cheap and old laptop, when compared to something new, isn't in the raw performance, but in other compromises instead. Looking at a terrible screen for 8+ hours a day, and dealing with a poor battery and a whole lot of cables isn't fun.\n\nBut on the other hand, if you want to get into data science and have \\$200 to spare, an old laptop will do just fine.\n\nWhat do you think of this old-vs-new laptop comparison? Are you currently using a laptop that has seen better days? If so, what issues are you battling with daily? Let me know in the comment section below." ]

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[ "# KIS 채권평가 - Beyond Pricing\n\nKOR\n\n## Financial Solution\n\n### Consulting Service\n\nCounterparty risk Significant uncertainty about economic loss due to default of counterparty Exist in all transactions with counterparties on securities and derivatives Diversification of financial products, increase in derivatives trading → Increasing demand for appropriate risk management It occurs only when the contract with the counterparty has a positive (+) economic value at the time of default Acceptance of offsetting and collateral maintenance agreements by credit risk mitigation method Most domestic EAD application methods use current exposure method Expected exposure recommended, in Basel II KIS EAD System calculates EAD by current exposition method, standard exposure method, expected exposure method, and build system based on expected exposition method Verification of EAD results through back-testing Basel II, III introduction of market standard pricing system Demand for EAD evaluation system will increase due to strengthened regulation by FSS Competitiveness in the market from experience of KIS EAD System Back-Testing enforcement Manpower secured to build risk management system in anticipation of demand increase\n\n#### EAD calculation by Current Exposure Method (CEM)\n\n• Replacement Cost : Offset gains after offsetting the contracts with legally valid contract\n• Additional Items after offset: (Sum of additional Items Before Offset) X { 0.4 + 0.6 x (Replacement Cost) / (Total Replacement cost) }\n• Additional Development Item : (Notional Amount) x (Credit conversion rate)\n• Total Replacement Cost : Sum of individual Replacement Cost\n##### Issues with Current Exposure Method\n• Focus on the P/L gains rather than the multiple transactions for offsetting contracts\n• Excessive or underestimation of regulatory capital → Possibility of arbitrage trading\n• Not able to fully reflect market conditions\n\n#### EAD calculation using Standard Method\n\n##### EAD = 1.4 × max{ ( Replacement cost ), ∑( Risk Position × Credit conversion rate) }\n• Replacement Cost : Offset gains after offsetting the contracts with legally valid contract\n• Risk Position : Hedging Group Risk Amount\n• Credit Conversion Rate: Hedging Group Credit Conversion Amount\n##### Issues with Current Exposure\n• Risk Position within offset trades (eg, currency, remaining maturity, market factors)\n• Since the off-set can be sufficiently recognized within the hedging group, the net total amount of all the risk positions existing in the hedging group is used to calculate the EAD\n##### Risk position\n• other than debt-equity instruments are underlying assets : (Risk position) = (effective nominal value) = (underlying asset price) × (delta)\n• debt-equity instruments are underlying assets: (Risk Position) = (effective nominal value) × (Modified duration)\n• Credit Default SWAP: (Risk Position) = (effective nominal value) × (reaming maturities )\n##### Issues with Standard Method\n\nNot able to fully reflect market conditions\n\n#### Expected Exposure Method EAD Calculation\n\n##### EAD = α + Effective EPE\n• Measure at the upper level\n• Specifies the expected distribution of changes in market value due to changes in market variables such as interest rate and exchange rate.", null, "##### Exposure\n• Expected exposure(EE) : The average value of the distribution of the exposures at each future point in the transaction before the maturity date of the transaction with the longest maturity\n• Expected positive exposure( EPE) : The time-weighted average of the expected exposures for a given period (the longest maturity or less than a year, which ever shorter )\n• Effective EE : Maximum expected exposures at or past period\n•", null, "• Effective EPE : The time-weighted average of the expected exposures for a given period (the longest maturity or less than a year, which ever shorter\n•", null, "#### Expected exposure\n\n##### Exposure distribution and Effective EPE", null, "##### Offset Method", null, "#### Comparison of CEM/SM and SA-CCR", null, "#### MtF calculation method selection process (Closed MCS, FDM MCS, LS MCS)", null, "Closed-MCS Method, FDM-MCS Method, LS-MCS Method 계산과정 정보를 제공\nClosed-MCS Method FDM-MCS Method LS-MCS Method\nApplicable Conditions Price Formula Price Formula not available, but able to use FDM when two methods are not applicable\nCalculation Process 01 Create Scenario\n• Apply the estimated parameter from the time series data to a given probability model\n• Consider correlations to risk factors\n02 MtF Calculation\n• Calculate prices using formulas at each point in each scenario\n• Use implied volatility rather than historical volatility\n03 Offset Processing 04 Effective Exposure Calculation\n\nReplace negative numbers to zero after offsetting price\n\n01 Create Scenario 02 Price calculation using FDM\n• Calculate price by solving partial differential equations\n• If the risk factor is 1: Implicit\n• FDM, 2 cases: ADI or OSM\n03 MtF calculation\n• The sum calculated by FFDM at each point in each scenario and calculate the price by linear interpolation method\n• Whether or not the barrier hits before each point in time, and then the initial cash flow are calculated from the scenario.\n04 Calculate effective EE after offsetting\n01 Create Scenario 02 Price calculation for each scenario\n\nEstimate future cash flow from each scenario and calculate price\n\n03 MtF Calculation\n• At each given point in time, the prices are calibrated using the least squares method using the price of each scenario.\n• There is some difference in the distribution of results according to the objective function used in least squares\n04 Calculate effective EE after offsetting\n\n#### Back-Testing\n\n##### Standard Back-Testing method", null, "##### Counter party Back-Testing Method", null, "#### History", null, "• Basel I(1988.07.) - To ensure the financial soundness of banks, introduced international standards for minimum capital adequacy ratio (BIS).\n• New BIS (Basel II)(2004.06.) - Increase the financial soundness of banks by strengthening the calculation method of capital ratio. (Counterparty credit risk management)\n• Counterparty Risk Research(2007.03.) - FSS requested Research on the counterparty risk measurement\n• KIS EAD System Develop(2007.12.) - Completed FSS Research request Completed KIS-EAD System\n• Back-Testing(now,2012.01.) - EAD Progress Status Back-test R&D\n\n#### Service Structure\n\n##### System Structure", null, "##### System Description\ninform of EAD system Output calculation process, Expected exposure method process\nEAD system Output calculation process Expected exposure method process\n• 01 Enter transaction information and market information into the client system\n• 02 Transfer data from client DB to EAD DB\n• 03 Generate market information after data processing from EAD server\n• 04 EAD calculation (Current Exposure, Standard, Expected Exposure)\n• 05 Back-Testing\n• 06 Calculation result transfer to the client system\n• 01 Parameter estimation\n• 02 Create Scenario\n• 03 Calculation of future price distribution\n• 04 Offsetting process\n##### Module Structure", null, "• Parameter Estimation Module - HW Calibration, BK Estimation Module, Volatility Correlation coefficient Module\n• MtF Calculation Module - IRS, CRS, FX, FX Option, Equity Option\n• EAD Calculation Module - Current Exposure Method, Standard Method, Expected Exposure Method\n• Back-Testing Module - Exception calculation, CCDF calculation, test statistic calculation module\n\n#### Screen and Function", null, "• 01 Calculation date setting\n\nEnable MAR exchange when activating monthly operation\n\n• 02 Enter the counterparty code\n• 03 Parameter setting\n\nEnter the parameters that are the basis for scenario creation\n\n• 04 Settings for Back-Testing\n• 05 Back-Testing Detailed Result Value Setting\n• 06 Log Settings\n• 07 EAD generation input button\n• Data processing: Estimation and calculation of interest rates, equities, FX, correlation coefficients, etc.\n• Scenario Creation: Creating Interest Rates, equities, and FX Scenario\n• EAD output and DB storage\n• MTM calculation and DB storage\n• 08 Back-Testing Output button\n• 09 EAD Output Detail Log Screen" ]

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[ "# Intro to Distance Problems - Uniform Motion rt=d\n\nTaught by YourMathGal\n• Currently 3.0/5 Stars.\n4866 views | 1 rating\nPart of video series\nMeets NCTM Standards:\nLesson Summary:\n\nIn this lesson, you'll learn the basics of solving uniform motion problems using the formula rate times time equals distance. The lesson introduces six problems and provides preliminary pictures to help you understand the situation before solving the problems in subsequent videos. By drawing pictures and defining variables, you can then write equations to solve the problems and check your work. Overall, this lesson provides a solid foundation for solving distance problems in future lessons.\n\nLesson Description:\n\nThis is an introduction to solving word problems on uniform motion using the formula rate x time = distance. Six problems are introduced, and preliminary pictures are drawn. All problems are solved in subsequent videos.\n\nMore free YouTube videos by Julie Harland are organized at http://yourmathgal.com\n\n• What are uniform motion word problems?\n• How do you solve uniform motion problems?\n• How do you solve equations using the formula rate * time = distance?\n• How far can you go if you drive 3 hours at an average of 40 miles per hour?\n• How do you do factor labeling in math?\n• How do you cancel out labels in word problems?\n• How can you draw and use pictures to solve word problems using the formula d = rt?\n• Where can I find a bunch of uniform motion word problems?\n• What are some different kinds for uniform motion problems and what do the sketches of their drawings look like?\n•", null, "#### Staff Review\n\n• Currently 3.0/5 Stars.\nThis video just introduces 6 different uniform rate problems that are to be solved in 6 subsequent videos. The video explains how to decipher the information in the word problem and how to draw a picture from this information. Each problem is a little different as different scenarios are discussed in each problem. This understanding of how to tackle the problem will prove to be very helpful once we start trying to solve these problems." ]

[ null, "https://mathvids.com/assets/testimonial/artist-testimonial-avatar-01-c96ee2a66f1e941fb1528708979edbcecf5383e31bcadfe00817b03e5645bd16.jpg", null ]

[ "sbv-8.2: SMT Based Verification: Symbolic Haskell theorem prover using SMT solving.\n\nDocumentation.SBV.Examples.Queries.AllSat\n\nDescription\n\nWhen we would like to find all solutions to a problem, we can query the solver repeatedly, telling it to give us a new model each time. SBV already provides allSat that precisely does this. However, this example demonstrates how the query mode can be used to achieve the same, and can also incorporate extra conditions with easy as we walk through solutions.\n\nSynopsis\n\n# Documentation\n\nFind all solutions to x + y .== 10 for positive x and y, but at each iteration we would like to ensure that the value of x we get is at least twice as large as the previous one. This is rather silly, but demonstrates how we can dynamically query the result and put in new constraints based on those.\n\ndemo :: IO () Source #\n\nRun the query. We have:\n\n>>> demo\nStarting the all-sat engine!\nIteration: 1\nCurrent solution is: (0,10)\nIteration: 2\nCurrent solution is: (1,9)\nIteration: 3\nCurrent solution is: (2,8)\nIteration: 4\nCurrent solution is: (4,6)\nIteration: 5\nCurrent solution is: (8,2)\nIteration: 6\nNo other solution!\n[(0,10),(1,9),(2,8),(4,6),(8,2)]" ]

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[ "# Penrose and Fourier Design Playing Cards\n\nMathWorks is creating a deck of playing cards that will be offered as gifts at our trade show booths. The design of the cards is based on Penrose Tilings and plots of the Finite Fourier Transform Matrix.\n\n### Contents\n\n#### Penrose Tilings\n\nThis is an example of a Penrose tiling.", null, "Penrose tilings are aperiodic tilings of the plane named after Oxford emeritus physicist Roger Penrose, who studied them in the 1970s. MathWorks' Steve Eddins has written a paper about his MATLAB program for generating Penrose tilings. I will include the paper in this blog post. Steve has also submitted his paper and code to the MATLAB Central File Exchange, Penrose Rhombus Tiling.\n\n#### The Suits\n\nLook carefully at the example tiling. It's made entirely from two rhombuses, one colored gold and one colored blue. Each rhombus, in turn, is made from two isosceles triangles. These shapes inspired MathWorks designer Gabby Lydon to reimagine the traditional symbols for the four suits in a deck of cards.\n\nHere are the diamonds and hearts.", null, "Here are the clubs and spades.", null, "#### The Face Cards\n\nThe theme of triangles and rhombuses is continued in the face cards.", null, "", null, "#### The Back\n\nThe backs of the cards feature two mathematical objects that I have described in this blog. I wrote a five-part series about our logo five years ago, beginning with part one. And, I wrote about the graphic produced by the Finite Fourier Transform matrix.\n\n n = 31;\n\n\nTake a Fourier transform of the columns of the identity matrix.\n\n F = fft(eye(n,n));\n\n\nPlot it. Because 31 is prime, this is the complete graph on 31 points. Every point on the circumference is connected to every other point.\n\n p = plot(F);\naxis square\naxis off", null, "That's too colorful.\n\n color = get(gca,'colororder');\nset(p,'color',color(1,:))", null, "Now graphic design software takes over and makes room for the logo.", null, "#### Cleve's Lab\n\nAll this has reminded me of two FFT-related apps from Numerical Computing with MATLAB, fftgui and fftmatrix. I've added them to Cleve's Laboratory with Version 3.9. This is the image from fftmatrix for n = 12. Because 12 is not prime, this is not a completely connected graph. You can vary both the matrix order and the columns to be transformed.", null, "Now, here is Steve's paper about Penrose tiling.\n\n#### Penrose Rhombus Tiling\n\nby Steve Eddins\n\nThis story is about creating planar tilings like this:", null, "This is an example of a Penrose tiling. Penrose tilings are aperiodic tilings that named after Roger Penrose, who studied them in the 1970s. This particular form, made from two rhombuses, is called a P3 tiling.\n\n#### Four Types of Triangles\n\nConstruction of a Penrose P3 tiling is based on 4 types of isosceles triangles. The types are labeled A, A', B, and B'. The A and A' triangles have an apex angle of 36 degrees, and the B and B' triangles have an apex angle of 72 degrees.\n\nsubplot(2,2,1)\nshowLabeledTriangles(aTriangle([],-1,1))\n\nsubplot(2,2,2)\nshowLabeledTriangles(apTriangle([],-1,1))\n\nsubplot(2,2,3)\nshowLabeledTriangles(bTriangle([],-1,1))\n\nsubplot(2,2,4)\nshowLabeledTriangles(bpTriangle([],-1,1))\n\nvertices =\n-1.0000 + 0.0000i 0.0000 + 3.0777i 1.0000 + 0.0000i\nvertices =\n-1.0000 + 0.0000i 0.0000 + 3.0777i 1.0000 + 0.0000i\nvertices =\n-1.0000 + 0.0000i 0.0000 + 0.7265i 1.0000 + 0.0000i\nvertices =\n-1.0000 + 0.0000i 0.0000 + 0.7265i 1.0000 + 0.0000i", null, "#### Triangle Functions and Triangle Representation\n\nBefore proceeding further, let's pause to look at what the functions aTriangle, apTriangle, bTriangle, and bpTriangle do.\n\nThe function aTriangle takes three arguments: aTriangle(apex,left,right). Each of the three arguments is a point in the complex plane that represents one triangle vertex. You specify any two points, passing in the third point as [], and aTriangle computes the missing vertex for you. Here's an example showing how to compute a type A triangle whose base is on the real axis, extending from -1 to 1.\n\nt_a = aTriangle([],-1,1)\n\nt_a =\n1×4 table\nApex Left Right Type\n_________ ____ _____ ____\n0+3.0777i -1 1 A\n\n\nThe result is returned as a table, which is convenient because we'll be creating large collections of these triangles, and it is helpful to be able to refer to the different vertices of the triangles using the notation t.Apex, t.Left, and t.Right.\n\nThe function showLabeledTriangles takes a table of triangles and displays them all, with the triangle types and their sides labeled. (We'll talk more about the labeling of the sides below.)\n\nt_b = bTriangle(t_a.Apex,t_a.Right,[]);\nT = [t_a ; t_b]\nclf\nshowLabeledTriangles(T)\n\nT =\n2×4 table\nApex Left Right Type\n_________ ____ _____________ ____\n0+3.0777i -1 1+0i A\n0+3.0777i 1 2.618+4.9798i B\nvertices =\n-1.0000 + 0.0000i 0.0000 + 3.0777i 1.0000 + 0.0000i\n1.0000 + 0.0000i 0.0000 + 3.0777i 2.6180 + 4.9798i", null, "#### Triangle Side Labels\n\nThe markers on the sides of the triangles help us to distinguish between A and A' triangles, as well as between B and B' triangles. For example, an A triangle has the circle marker on the left side (assuming the apex is oriented at the top), whereas the A' triangle has the circle marker on the right side.\n\nThe side labels also help us confirm whether we have a correct arrangement of triangles in our Penrose tiling. Triangles are only allowed to share an edge in the Penrose P3 tiling if the side markers align together and are the same. For example, in the two triangles shown above, the square marker on the right edge of the triangle lines up with the square marker on the left edge of the B triangle. If we had used a B' triangle instead, the markers would not have been identical.\n\nt_bp = bpTriangle(t_a.Apex,t_a.Right,[]);\nT = [t_a ; t_bp];\nshowLabeledTriangles(T)\n\nvertices =\n-1.0000 + 0.0000i 0.0000 + 3.0777i 1.0000 + 0.0000i\n1.0000 + 0.0000i 0.0000 + 3.0777i 2.6180 + 4.9798i", null, "#### Making the Two Types of Rhombus\n\nIn the Penrose P3 tiling, one rhombus is made from an A and an A' triangle, and the other is made from a B and a B' triangle.\n\nsubplot(1,2,1)\nr1 = [ ...\naTriangle([],-1,1)\napTriangle([],1,-1)];\nshowLabeledTriangles(r1)\n\nsubplot(1,2,2)\nr2 = [ ...\nbTriangle([],-1,1)\nbpTriangle([],1,-1)];\nshowLabeledTriangles(r2)\n\nvertices =\n-1.0000 + 0.0000i 0.0000 + 3.0777i 1.0000 + 0.0000i\n1.0000 + 0.0000i 0.0000 - 3.0777i -1.0000 + 0.0000i\nvertices =\n-1.0000 + 0.0000i 0.0000 + 0.7265i 1.0000 + 0.0000i\n1.0000 + 0.0000i 0.0000 - 0.7265i -1.0000 + 0.0000i", null, "#### Triangle Decomposition\n\nConstruction of the P3 tiling proceeds by starting with one triangle and then successively decomposing it. Each of the four types of triangles has a different rule for decomposition.\n\nAn A triangle decomposes into an A triangle and a B' triangle.\n\nt_a = aTriangle([],-1,1);\nt_a_d = decomposeATriangle(t_a)\n\nclf\nsubplot(1,2,1)\nshowLabeledTriangles(t_a)\nlims = axis;\nsubplot(1,2,2)\nshowLabeledTriangles(t_a_d)\naxis(lims)\n\nt_a_d =\n2×4 table\nApex Left Right Type\n_______________ _________ _______________ ____\n-1+0i 1+0i 0.61803+1.1756i A\n0.61803+1.1756i 0+3.0777i -1+0i Bp\nvertices =\n-1.0000 + 0.0000i 0.0000 + 3.0777i 1.0000 + 0.0000i\nvertices =\n1.0000 + 0.0000i -1.0000 + 0.0000i 0.6180 + 1.1756i\n0.0000 + 3.0777i 0.6180 + 1.1756i -1.0000 + 0.0000i", null, "You can look at the very short implementation of decomposeATriangle to see how this is done.\n\nfunction out = decomposeATriangle(in)\n\nout = [ ...\naTriangle(in.Left,in.Right,[])\nbpTriangle([],in.Apex,in.Left) ];\n\n\nThe smaller A triangle is determined by placing its apex at the left vertex of the input triangle and placing its left vertex at the right vertex of the input triangle.\n\nThe smaller B' triangle is determined by placing its left vertex at the apex of the input triangle and placing its right vertex at the left vertex of the input triangle.\n\nThere are similar rules for decomposing the other three types of triangles.\n\nt_ap = apTriangle([],-1,1);\nt_ap_d = decomposeApTriangle(t_ap)\n\nclf\nsubplot(1,2,1)\nshowLabeledTriangles(t_ap)\nlims = axis;\nsubplot(1,2,2)\nshowLabeledTriangles(t_ap_d)\naxis(lims)\n\nt_ap_d =\n2×4 table\nApex Left Right Type\n________________ ________________ __________ ____\n1+0i -0.61803+1.1756i -1+0i Ap\n-0.61803+1.1756i 1+0i 0+3.0777i B\nvertices =\n-1.0000 + 0.0000i 0.0000 + 3.0777i 1.0000 + 0.0000i\nvertices =\n-0.6180 + 1.1756i 1.0000 + 0.0000i -1.0000 + 0.0000i\n1.0000 + 0.0000i -0.6180 + 1.1756i 0.0000 + 3.0777i", null, "t_b = bTriangle([],-1,1);\nt_b_d = decomposeBTriangle(t_b)\n\nclf\nsubplot(2,1,1)\nshowLabeledTriangles(t_b)\nlims = axis;\nsubplot(2,1,2)\nshowLabeledTriangles(t_b_d)\naxis(lims)\n\nt_b_d =\n3×4 table\nApex Left Right Type\n_________________ ___________ _________________ ____\n0.23607+0i 1+0i 0+0.72654i B\n0.23607+0i 0+0.72654i -0.38197+0.44903i A\n-0.38197+0.44903i -1+0i 0.23607+0i Bp\nvertices =\n-1.0000 + 0.0000i 0.0000 + 0.7265i 1.0000 + 0.0000i\nvertices =\n1.0000 + 0.0000i 0.2361 + 0.0000i 0.0000 + 0.7265i\n0.0000 + 0.7265i 0.2361 + 0.0000i -0.3820 + 0.4490i\n-1.0000 + 0.0000i -0.3820 + 0.4490i 0.2361 + 0.0000i", null, "t_bp = bpTriangle([],-1,1);\nt_bp_d = decomposeBpTriangle(t_bp)\n\nclf\nsubplot(2,1,1)\nshowLabeledTriangles(t_bp)\nlims = axis;\nsubplot(2,1,2)\nshowLabeledTriangles(t_bp_d)\naxis(lims)\n\nt_bp_d =\n3×4 table\nApex Left Right Type\n_________________ _________________ ___________ ____\n-0.23607+0i 0+0.72654i -1+0i Bp\n-0.23607+0i 0.38197+0.44903i 0+0.72654i Ap\n0.38197+0.44903i -0.23607+0i 1+0i B\nvertices =\n-1.0000 + 0.0000i 0.0000 + 0.7265i 1.0000 + 0.0000i\nvertices =\n0.0000 + 0.7265i -0.2361 + 0.0000i -1.0000 + 0.0000i\n0.3820 + 0.4490i -0.2361 + 0.0000i 0.0000 + 0.7265i\n-0.2361 + 0.0000i 0.3820 + 0.4490i 1.0000 + 0.0000i", null, "#### From Triangles to Rhombuses\n\nLet's start with a B triangle and decompose it three times successively.\n\nt = bTriangle([],-1,1);\nt = decomposeTriangles(t)\nt = decomposeTriangles(t)\nt = decomposeTriangles(t)\n\nt =\n3×4 table\nApex Left Right Type\n_________________ ___________ _________________ ____\n0.23607+0i 1+0i 0+0.72654i B\n0.23607+0i 0+0.72654i -0.38197+0.44903i A\n-0.38197+0.44903i -1+0i 0.23607+0i Bp\nt =\n8×4 table\nApex Left Right Type\n_________________ _________________ _________________ ____\n0.38197+0.44903i 0+0.72654i 0.23607+0i B\n0.38197+0.44903i 0.23607+0i 0.52786+0i A\n0.52786+0i 1+0i 0.38197+0.44903i Bp\n0+0.72654i -0.38197+0.44903i -0.1459+0.27751i A\n-0.1459+0.27751i 0.23607+0i 0+0.72654i Bp\n-0.52786+0i -0.38197+0.44903i -1+0i Bp\n-0.52786+0i -0.1459+0.27751i -0.38197+0.44903i Ap\n-0.1459+0.27751i -0.52786+0i 0.23607+0i B\nt =\n21×4 table\nApex Left Right Type\n__________________ _________________ __________________ ____\n0.1459+0.27751i 0.23607+0i 0.38197+0.44903i B\n0.1459+0.27751i 0.38197+0.44903i 0.23607+0.55503i A\n0.23607+0.55503i 0+0.72654i 0.1459+0.27751i Bp\n0.23607+0i 0.52786+0i 0.47214+0.17151i A\n0.47214+0.17151i 0.38197+0.44903i 0.23607+0i Bp\n0.76393+0.17151i 0.52786+0i 1+0i Bp\n0.76393+0.17151i 0.47214+0.17151i 0.52786+0i Ap\n0.47214+0.17151i 0.76393+0.17151i 0.38197+0.44903i B\n-0.38197+0.44903i -0.1459+0.27751i -0.09017+0.44903i A\n-0.09017+0.44903i 0+0.72654i -0.38197+0.44903i Bp\n0.1459+0.27751i -0.1459+0.27751i 0.23607+0i Bp\n0.1459+0.27751i -0.09017+0.44903i -0.1459+0.27751i Ap\n-0.09017+0.44903i 0.1459+0.27751i 0+0.72654i B\n-0.61803+0.27751i -0.52786+0i -0.38197+0.44903i Bp\n-0.61803+0.27751i -0.7082+0i -0.52786+0i Ap\n-0.7082+0i -0.61803+0.27751i -1+0i B\n-0.38197+0.44903i -0.2918+0.17151i -0.1459+0.27751i Ap\n-0.2918+0.17151i -0.38197+0.44903i -0.52786+0i B\n-0.055728+0i 0.23607+0i -0.1459+0.27751i B\n-0.055728+0i -0.1459+0.27751i -0.2918+0.17151i A\n-0.2918+0.17151i -0.52786+0i -0.055728+0i Bp\n\nclf\nshowLabeledTriangles(t)\n\nvertices =\n0.2361 + 0.0000i 0.1459 + 0.2775i 0.3820 + 0.4490i\n0.3820 + 0.4490i 0.1459 + 0.2775i 0.2361 + 0.5550i\n0.0000 + 0.7265i 0.2361 + 0.5550i 0.1459 + 0.2775i\n0.5279 + 0.0000i 0.2361 + 0.0000i 0.4721 + 0.1715i\n0.3820 + 0.4490i 0.4721 + 0.1715i 0.2361 + 0.0000i\n0.5279 + 0.0000i 0.7639 + 0.1715i 1.0000 + 0.0000i\n0.4721 + 0.1715i 0.7639 + 0.1715i 0.5279 + 0.0000i\n0.7639 + 0.1715i 0.4721 + 0.1715i 0.3820 + 0.4490i\n-0.1459 + 0.2775i -0.3820 + 0.4490i -0.0902 + 0.4490i\n0.0000 + 0.7265i -0.0902 + 0.4490i -0.3820 + 0.4490i\n-0.1459 + 0.2775i 0.1459 + 0.2775i 0.2361 + 0.0000i\n-0.0902 + 0.4490i 0.1459 + 0.2775i -0.1459 + 0.2775i\n0.1459 + 0.2775i -0.0902 + 0.4490i 0.0000 + 0.7265i\n-0.5279 + 0.0000i -0.6180 + 0.2775i -0.3820 + 0.4490i\n-0.7082 + 0.0000i -0.6180 + 0.2775i -0.5279 + 0.0000i\n-0.6180 + 0.2775i -0.7082 + 0.0000i -1.0000 + 0.0000i\n-0.2918 + 0.1715i -0.3820 + 0.4490i -0.1459 + 0.2775i\n-0.3820 + 0.4490i -0.2918 + 0.1715i -0.5279 + 0.0000i\n0.2361 + 0.0000i -0.0557 + 0.0000i -0.1459 + 0.2775i\n-0.1459 + 0.2775i -0.0557 + 0.0000i -0.2918 + 0.1715i\n-0.5279 + 0.0000i -0.2918 + 0.1715i -0.0557 + 0.0000i", null, "Now the labels are getting in the way.\n\nshowTriangles(t)", null, "That's better, but it's hard to visualize the rhombuses because the triangle bases are being drawn. Let's switch to a different visualization function that draws the rhombuses with colored shading and without the triangle bases.\n\nshowTiles(t)", null, "Let's do five more levels of decomposition.\n\nfor k = 1:5\nt = decomposeTriangles(t);\nend\n\nnum_triangles = height(t)\n\nnum_triangles =\n2584\n\n\nNow we have lots of triangles. Let's take a look.\n\nclf\nshowTiles(t)", null, "Zoom in.\n\naxis([-0.3 0.2 0.1 0.5])", null, "For fun, you can add to the visualization by inserting arcs or other shapes in the triangles. You just need to make everything match up across the different types of triangles. The showDecoratedTiles function shows one possible variation.\n\nclf\nshowDecoratedTiles(t)\naxis([-0.2 0.1 0.2 0.4])", null, "#### Starting from Multiple Triangles\n\nAnother interesting thing to try is to start from a pattern of multiple triangles instead of just one. You just to arrange the initial triangles so that they satisfy the side matching rules. Let's use a circular pattern of alternating A and A' triangles sharing a common apex.\n\nt = table;\nfor k = 1:5\nt_ap = apTriangle(0,t_a.Right,[]);\nt = [t ; t_a ; t_ap];\nend\nshowLabeledTriangles(t)\n\nvertices =\n1.0000 + 0.0000i 0.0000 + 0.0000i 0.8090 + 0.5878i\n0.8090 + 0.5878i 0.0000 + 0.0000i 0.3090 + 0.9511i\n0.3090 + 0.9511i 0.0000 + 0.0000i -0.3090 + 0.9511i\n-0.3090 + 0.9511i 0.0000 + 0.0000i -0.8090 + 0.5878i\n-0.8090 + 0.5878i 0.0000 + 0.0000i -1.0000 + 0.0000i\n-1.0000 + 0.0000i 0.0000 + 0.0000i -0.8090 - 0.5878i\n-0.8090 - 0.5878i 0.0000 + 0.0000i -0.3090 - 0.9511i\n-0.3090 - 0.9511i 0.0000 + 0.0000i 0.3090 - 0.9511i\n0.3090 - 0.9511i 0.0000 + 0.0000i 0.8090 - 0.5878i\n0.8090 - 0.5878i 0.0000 + 0.0000i 1.0000 + 0.0000i", null, "t2 = t;\nfor k = 1:4\nt2 = decomposeTriangles(t2);\nend\nclf\nshowDecoratedTiles(t2)", null, "Published with MATLAB® R2018a\n\n|" ]

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[ "# Casio FX-991ES PLUS Scientific Calculator\n\nAvailability:\n\n1 in stock\n\nEGP330.00 EGP350.00\n\n1 in stock\n\n### New functions\n\n• New equation mode\n• Random integers\n\n### Standard functions", null, "", null, "• Fraction calculations\n• Combination and permutation\n• Statistics (List-based STAT data editor, standard deviation, regression analysis)\n• 9 variables\n• Table function\n• Comes with new slide-on hard case\n\n### fx-82ES PLUS/85ES PLUS/350ES PLUS functions, in addition to:\n\n• Equation calculations\n• Integration/differential calculations\n• Matrix calculations\n• Vector calculations\n• Complex number calculations\n• CALC function\n• SOLVE function\n• Base-n calculation\n\n####", null, "", null, "SKU: 54066922 Category:\nBrand CASIO", null, "## Based on 0 reviews\n\n0.0 overall\n0\n0\n0\n0\n0\n\nThere are no reviews yet." ]

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[ "#", null, "费米维基\n\n### 站点工具\n\natk:使用ivcharacteristics工具计算分析器件的电流电压特性\n\n# 使用IVCharacteristics工具计算分析器件的电流电压特性\n\n## 概述\n\nIVCharacteristics是QuantumATK开发的一类最新的Study对象，用来计算和分析场效应管（FET）器件模型的常见电子性质，专门用于多步骤的复杂计算流程的设置、分析，大大提高了IV计算的效率。使用IVCharacteristics可以：\n\n• 单独扫描偏压或栅压；\n• 同时扫描偏压和栅压；\n• 在计算中断时续算未完成的电压点，而不需重复已经完成的计算；\n• 在计算正常结束后，补算新的电压点，而不需重复已经完成的计算；\n• 分析开关比（$\\mathrm{I_{on}/I_{off}}$），亚阈值斜率（$\\mathrm{SS}$），转移电导（transconductance，$\\mathrm{g_{m}}$）和漏极诱导势垒降低（drain-induced barrier lowering, $\\mathrm{DIBL}$）1)\n• 还支持在扫描偏压的同时，计算PLDOS、电子密度、各种电势、电荷布居、电子密度、能量等；\n• 也可以在计算完成后，补算任意偏压、栅压点的PLDOS、电子密度、各种电势、电荷布居、电子密度、能量等。\n\n## $V_{DG}$ 曲线的计算和基本分析\n\n• Study Object → IV Characteristics\n\n• 在Gate source voltage range里：\n• 设置$\\mathrm{V_{gs0}} = -0.3 \\mathrm{V}$\n• 设置$\\mathrm{V_{gs1}} = 0.0 \\mathrm{V}$\n• number of points 设置为 7。\n• 选中Print results summary to log\n\n#### 说明\n\n+------------------------------------------------------------------------------+\n| IV Characteristics Study |\n+------------------------------------------------------------------------------+\n| 14 task(s) will be executed. |\n| |\n| * Update configuration |\n| Gate voltage: -0.25 V |\n| Left electrode voltage: 0.05 V |\n| Right electrode voltage: 0.0 V |\n| * Calculate TransmissionSpectrum |\n| Gate voltage: -0.3 V |\n| Left electrode voltage: 0.05 V |\n| Right electrode voltage: 0.0 V |\n| * Update configuration |\n\n+------------------------------------------------------------------------------+\n| Executing task 1 / 14: |\n| Update configuration |\n| Gate voltage: 0.0 V |\n| Left electrode voltage: 0.05 V |\n| Right electrode voltage: 0.0 V |\n| Log to: ivcharacteristics_gate_voltage_0.0_Volt.log |\n+------------------------------------------------------------------------------+\n\nIV-Characteristics Analyzer 的主窗口显示如下。\n\nIV-Characteristics Analyzer 的底部有更多信息：\n\n• Data：每个计算的收敛性；\n• Scale：Y 轴电流 $\\mathrm{I_{ds}}$ 的标度，可以是线性（Linear）或者对数（Logarithmic）;\n• Temperature：在使用Landauer公式计算$\\mathrm{I_{ds}}$时，用于电极部分的Fermi-Dirac分布函数的温度。每个电极的温度可以由用户设定，电流则根据选择的温度实时计算。\n\nIV-Characteristics Analyzer 的上部是计算得到的数据作图，可以在两种不通的方式间切换：\n\n• 栅-源电压图（Gate-Source plot）。将源漏电流 $\\mathrm{I_{ds}}$ 与栅-源电压 $\\mathrm{V_{gs}}$ 作图（$\\mathrm{I_{ds}}\\ vs. \\mathrm{V_{gs}}$）；\n• 源-漏电压图（Drain-Source plot）。将源漏电流 $\\mathrm{I_{ds}}$ 与源-漏电压 $\\mathrm{V_{ds}}$ 作图（$\\mathrm{I_{ds}}\\ vs. \\mathrm{V_{ds}}$）。\n\n## 扩展 $V_{DG}$ 曲线的电压范围\n\n• 方法一：使用study object扩展栅-源电压范围\n• 方法二：使用脚本添加更多的栅-源电压\n\n### 方法一：使用Study object扩展栅压范围\n\n• 修改Output settingResults filesoi_device_ivc2.hdf5\n• 修改Gate source voltage range 如下：\n• $\\mathrm{V_{gs0}} = -0.9 \\mathrm{V}$；\n• $\\mathrm{V_{gs1}} = 0.0 \\mathrm{V}$；\n• number of points 19。\n\n### 方法二：使用脚本添加其他栅压点\n\nIV Characteristics支持非常丰富的命令，详细列表参见手册：IVCharacteristics\n\nsoi_device.py脚本发送到Scripter，重复第一部分的计算设置脚本soi_device_ivc1.py，但是把输出文件设置为‘soi_device_ivc2b.hdf5’。\n\n# -------------------------------------------------------------\n# Kpoint sampling\nkpoint_grid = MonkhorstPackGrid(\nna=9,\n)\n\n# Gate-source voltages\ngate_source_voltages = numpy.linspace(-0.3, 0.0, 7)*Volt\n\n# Drain-source voltages\ndrain_source_voltages = numpy.linspace(0.05, 0.05, 1)*Volt\n\n# File name.\nfilename = u'soi_device_ivc2b.hdf5'\n\niv_characteristics = IVCharacteristics(\nconfiguration=device_configuration,\nfilename=filename,\nobject_id='ivcharacteristics',\ngate_regions=,\ngate_source_voltages=gate_source_voltages,\ndrain_source_voltages=drain_source_voltages,\nenergies=None,\nkpoints=kpoint_grid,\nself_energy_calculator=RecursionSelfEnergy(),\nenergy_zero_parameter=AverageFermiLevel,\ninfinitesimal=1e-06*eV,\nlog_filename_prefix='ivcharacteristics_',\n)\n\niv_characteristics.update()\n\n## 计算漏极诱导势垒降低（DIBL）\n\nsoi_device_ivc2b.py传送到Scripter，添加Study Object → IV Characteristics，按下面修改参数：\n\n• Gate source voltage range：\n• $\\mathrm{V_{gs0}} = -0.9 \\mathrm{V}$\n• $\\mathrm{V_{gs1}} = 0.0 \\mathrm{V}$\n• number of points 19\n• Drain source voltage range：\n• $\\mathrm{V_{ds0}} = 0.05 \\mathrm{V}$\n• $\\mathrm{V_{ds1}} = 0.3 \\mathrm{V}$\n• 勾选Print results summary to log\n\nDIBL以无量纲形式给出，因此计算值为 622 mV/V。\n\n• Threshold current：$\\mathrm{V_{gs}}$对应阈值电压的电流，默认为$\\mathrm{I_{ds}}$的最大值和最小值的中点，但可以由用户设置。\n• Subthreshold factor决定了用来拟合亚阈值特性的电流区间，以便准确计算阈值电压。区间最少需要包含3个点，如果点数不够，将显示错误信息。\n• Gate voltage range：计算DIBL时可以只包含部分曲线，这对有两个亚阈值区间的双极性特征的器件模型非常重要。\n\n## IV计算结束后补算其他性质\n\niv_characteristics.addProjectedLocalDensityOfStates(\ngate_source_voltages, drain_source_voltages)\ngate_source_voltages, drain_source_voltages, HartreeDifferencePotential)\niv_characteristics.update() \n\n## 参考\n\n1)\nS. M. Sze and K. N. Kwok. Physics of Semiconductor Devices. Wiley, 3rd edition, 2006\n2)\nT. B. Boykin, M. Luisier, M. Salmani-Jelodar, and G. Klimeck. Strain-induced, off-diagonal, same-atom parameters in empirical tight-binding theory suitable for uniaxial strain applied to a silicon parametrization. Phys. Rev. B, 81:125202, 2010. doi:10.1103/PhysRevB.81.125202.\n3)\nShinji Migita, Yukinori Morita, Meishoku Masahara, and Hiroyuki Ota. Electrical performances of junctionless-fets at the scaling limit (lch = 3 nm). Proceedings of the Electron Device Meeting (IEDM) 2012, 2012. URL: http://ieeexplore.ieee.org/document/6479006, doi:10.1109/IEDM.2012.6479006.\natk/使用ivcharacteristics工具计算分析器件的电流电压特性.txt · 最后更改: 2018/08/24 08:01 由 fermi\n\n### 页面工具", null, "" ]

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[ "The semi-interquartile range is affected very little by extreme scores. The IQR is the difference between the first (25th percentile) and third (75th percentile) quartiles. Find the IQR by subtracting Q 1 from Q 3. Enjoyed the tutorial? Before calculating the quartiles, first we have to arrange all the individual observations in an ascending order. Since half the scores in a distribution lie between Q3 and Q1, the semi-interquartile range is the 1/2 of the distance needed to … The formula for semi-interquartile range is therefore: (Q3-Q1)/2. How To Calculate Odds Ratio In Microsoft Excel, How To Perform A Spearman Correlation Test In R, How To Perform A Pearson Correlation Test In R, How To Perform A One-Sample T-Test In Excel, How To Calculate Cohen’s d In Microsoft Excel, How To Easily And Quickly Calculate Z Scores In Excel, How To Perform A Spearman’s Rank Correlation Test In Excel, To calculate the Q1 in Excel, click on an empty cell and type ‘. A measure of spread, sometimes also called a measure of dispersion, is used to describe the variability in a sample or population. You have entered an incorrect email address! Median =. It is obtained by evaluating Q 3 − Q 1 … This will open the ‘ Explore ‘ window. The semi-interquartile range is one-half the difference between the first and third quartiles. Interquartile Range Calculator. He is currently a Medical Writer and a former Postdoctoral Research Associate. Steven is the founder of Top Tip Bio. The interquartile range is 77 – 64 = 13; the interquartile range is the range of the middle 50% of the data. This measure excludes the lowest quarter and highest quarter of values and only uses the central values. One method of eliminating extreme values is to use an interquartile range. Consider the simple example below. It does not matter which column to choose. The easiest approach is to firstly calculate the Q1 and Q3 and then use these to determine the IQR. Briefly, the semi-interquartile range is a measure of the dispersion or spread of a variable; it is the distance between the 1st quartile and the 3rd quartile, halved. Save my name, email, and website in this browser for the next time I comment. The formula for semi-interquartile range is therefore: (Q3-Q1)/2. For the above Example range will be: Range(team1) = 19.3 – 10.8 = 8.5. In this guide, I will show you how to calculate the interquartile range (IQR) by using Microsoft Excel. This number is what cuts the data set … contains 50% of the datapoints in a normally-distributed dataset. If you are more confident with working with Excel and fomulas, you could combine everything into one simple formula. The semi-interquartile range is an assessment of the amount of spread or dispersion. The value from this formula is the interquartile range. When a dataset is sorted in order from the smallest to the largest values, it is possible to split the data into four parts (the quartiles). The third quartile value appears in the previously blank cell. Subtract the value derived from the first quartile from the value derived from the third quartile. These are often abbreviated to Q1 and Q3 respectively.eval(ez_write_tag([[300,250],'toptipbio_com-box-3','ezslot_2',123,'0','0'])); The IQR is used to represent the middle (50%) spread of the data. Therefore, the set of data is divided into two parts: 1, 3, 5, 7, 9 and 11, 13, 15, 17, 19. Box Plot to get good indication of how the values in a distribution are spread out. Quartile deviation or semi-interquartile range is the dispersion which shows the degree of spread around the middle of a set of data. The semi-interquartile range is one-half of the difference between the first and third quartiles. Quartiles are those values which divide the series into 4 4 4 equal parts. In the ‘ Explore ‘ window, drag the variable of interest over to the ‘ Dependent List ‘ box. ; Follow these instructions to find the interquartile range by hand (part of the process is to find quartiles). Find the median, lower quartile, upper quartile, interquartile range and range of the following numbers. ; Step 2: Find the third quartile, Q 3.If you’re given Q 3 in the question, great. Formula for inter-quartile range is given by: IQR = Q 3 – Q 1. Range; Interquartile range. Next lesson. In it, Q1 is 3.5 (half way between 3 and 4) and Q3 is 8.5 (half way between 8 and 9). It is half the distance needed to cover half the scores. A COVID-19 Prophecy: Did Nostradamus Have a Prediction About This Apocalyptic Year? It is usually used in conjunction with a measure of central tendency, such as the mean or median, to provide an overall description of a set of data. Interquartile range (IQR) Interquartile range review. To calculate the third quartile, select another blank cell, and enter \"=QUARTILE(cell 1:cell 2, 3).\" To calculate the third quartile, select another blank cell, and enter \"=QUARTILE (cell 1:cell 2, 3).\" It is computed as one half the difference between the 75th percentile (Q 3) and the 25th percentile (Q 1). Looks like … Q1 occurs at 25% of 10 = 2.5 . Median = = 10. The first quartile value appears in the previously blank cell. Let x 1, x 2, …, x N be a set of observations for some numeric attribute, X. Given an even 2n or odd 2n+1 number of values To calculate the first quartile, select a blank cell, and enter \"=QUARTILE (cell 1:cell 2,1),\" where cell 1 and cell 2 are the actual cell labels in Excel. So you average the 2nd and 3rd sample. Interquartile range review. The interquartile range represents middle 50 percent of the data set. = 40 The IQR is a measure of the middle dispersion of a dataset, basically the difference between Q1 and Q3. the 25th and 75th percentiles (Q3 - Q1). It covers the center of the distribution and contains 50% of the observations. It is computed as one half the difference between the 75th percentile [often called (Q3)] and the 25th percentile (Q1). Semi interquartile range also is defined as half of the interquartile range. 3. Interquartile range review. Interquartile range function giving wrong results I am trying to compute interquartile range of an even set of data using Excel function but I seem to be getting the wrong results. The formula for the quartile deviation is as follows, Q.D = Q3 – Q1 / 2. Firstly, in SPSS, go to ‘ Analyze > Descriptive Statistics > Explore ‘. This can also be called a Semi Inter-Quartile Range. IQR = 15 – 5 = 10. (Q3 - Q1)/2, covers 25% of the. Click on any blank cell in Excel. The Formula for Semi Interquartile Range is Solution: First, arrange the data in ascending order: Lower quartile or first quartile = Median or second quartile = Upper quartile or third quartile = the difference between the first (25th percentile) and third (75th percentile) quartiles ... semi-interquartile-range Questions and Answers - Math Discussion Recent Discussions on semi-interquartile-range.php . Sort by: Top Voted. Enter all of the values into a single column into Excel. Next, we need to calculate Q3. the distance between. The range of the set is the difference between the largest (max()) and smallest (min()) values. There is no direct formula to calculate the IQR in Excel, however, it is relatively straight forward to do. Finally, to calculate the IQR, simply subtract the Q1 value away from the Q3 value. How To Calculate The Interquartile Range In Excel - YouTube I will also show you how to how to calculate the first and third quartiles for a dataset. The formula for this is: IQR = Q 3 - Q 1. The interquartile range is commonly used, i.e. 1. Quartile deviation is also known as semi-interquartile. To calculate the first quartile, select a blank cell, and enter \"=QUARTILE(cell 1:cell 2,1),\" where cell 1 and cell 2 are the actual cell labels in Excel. Enter The Following Scores Into A New Excel Spreadsheet. How to Find Interquartile Range The interquartile range IQR is the range in values from the first quartile Q 1 to the third quartile Q 3. To calculate the interquartile range in Microsoft Excel, first enter the values for which you want to calculate the interquartile range in one single column. NOAA Hurricane Forecast Maps Are Often Misinterpreted — Here's How to Read Them. The first step is the find the median of the data set, which in this case is . Before studying interquartile range, we first should study quartiles for they act as a base for the interquartile range. CEO Compensation and America's Growing Economic Divide. Range(team2) = 27.7-0 = 27.7 In the previous example, the quartiles were \\ (Q_1 = 4\\) and \\ (Q_3 = 11\\). For example, the range between the 97.5th percentile and the 2.5th percentile covers 95% of the data. Apart from being a less sensitive measure of the spread of data, IQR has another important use. You will notice that the fact there is an outlier in this data (60) which has had no bearing on the calculation of the interquartile range. 2. Just remember to replace the ‘array‘ components with the desired cells containing the data. Up Next. This simple tool works out the interquartile range of a set of numbers by calculating the 25th and 75th percentiles, and then subtracting the former from the latter (i.e., IQR = Q3 - Q1). Both the range and standard deviation tell us how spread out our data is. Interquartile range is useful to identify whether a value is an outlier or not. Again, replacing the ‘array‘ part with the cells that contain the data of interest. IQR = Q 3 - Q 1 The interquartile range (IQR), also called as midspread or middle 50%, or technically H-spread is the difference between the third quartile (Q3) and the first quartile (Q1). Q 3 = Median of second part = 15. Finally, click the ‘ OK ‘ button. Statisticians sometimes also use the terms semi-interquartile range and mid-quartile range . By definition, this. Comparing range and interquartile range (IQR) This is the currently selected item. Subtract the value from the first quartile from the third quartile to get your interquartile range. Then let me know by leaving a comment below, or consider. How Do You Calculate Interquartile Range in Excel. 2. Range: The most simple measure of variability is the range. The semi-interquartile range is a measure of spread or dispersion. Enter \"=QUARTILE(cell 1:cell 2, 3).\" To calculate the IQR in Microsoft Excel, use the =QUARTILE function to calculate Q1 and Q3, and ultimately find the difference between these two values. To start off, let's study the range, quantiles, quartiles, percentiles, and the interquartile range as measures of data dispersion. For example, the formula below would create the IQR in Excel. Someone kindly have a look at my attached workbook and tell me what could be wrong. It is the difference between the highest and the lowest value. Enter your data into the text box below, and then hit the \"Calculate Percentile\" button. In the example above, the formula used would be ‘=D3-D2‘.eval(ez_write_tag([[336,280],'toptipbio_com-medrectangle-4','ezslot_3',109,'0','0'])); The approach described above to calculate the IQR is rather long winded. Each quartile is a median calculated as follows. The interquartile range (IQR) is therefore 18 - 4 = 14. = Q3 – Q1 / 2 The formula includes Q3 and Q1 in the calculation, which is the top 25% and lowers 25% ,data respectively, and when the difference is taken between these two and when this number is halved, it gives measures of spread or dispersion. 12, 5, 22, 30, 7, 36, 14, 42, 15, 53, 25, 65. Thus, the IQR is 5 (ie 8.5 – 3.5). Q.D. The interquartile range (IQR) is the difference between the first quartile and third quartile. Why is the interquartile range important? Select the blank cell immediately below the first quartile. You can also use other percentiles to determine the spread of different proportions. Question: 65 15 87 80 BO Excel Assignment Homework Summary Statistics It Is Assumed That You Have Already Read Through The Excel Lesson Handout On Summary Statistics And Box-and-Whisker Plots. Enter \"=QUARTLE(cell 1:cell 2, 1).\" To calculate Q3 in Excel, simply find an empty cell and enter the formula ‘=QUARTILE(array, 3)‘. The IQR of a set of values is calculated as the difference between the upper and lower quartiles, Q 3 and Q 1. Microsoft Excel version used: 365 ProPlus. Excel ; Theorems ; Discussions; Ask Question ; Topics ; Un-Answered ; Answer Questions and Earn Points !!! The quartile deviation or semi-interquartile range is defined as half the IQR. The semi-interquartile range, i.e. If not, use one of the options listed in Step 1. Range, Quartiles, and Interquartile Range. IQR = Q3 – Q1 The semi-interquartile range is half of the difference between the upper quartile and the lower quartile. The Interquartile range, or IQR, is defined as the . You have 10 samples. Since the difference between third and first quartiles is called interquartile range therefore half of interquartile range is called semi-interquartile range also known as quartile deviation. This makes it a good measure of spread for skewed distributions. Use a calculator, like this one.Plug in your numbers and click the blue button. Q 1 is equal to the 25th percentile listed in the results. Variance and standard deviation of a population. It is calculated as one half the difference between the 75th percentile and is generally called as Q3 and the 25th percentile (Q1). The interquartile range is equivalent to the region between the 75th and 25th percentile (75 – 25 = 50% of the data). Below is the steps recommended to calculate the IQR in Excel. It is computed as one half the difference between the 75th percentile [often called (Q3)] and the 25th percentile (Q1). There are many measurements of the variability of a set of data. To find the IQR in SPSS, simply follow the steps below. The equation (Q3-Q1)/2 will give is used for calculating the range. Algorithm . The semi-interquartile range is a measure of spread or dispersion. N.B. Q 1 = Median of first part = 5. Subtract the value from the first quartile from the third quartile to get your interquartile range. Makes it a good measure of spread for skewed distributions of 10 = 2.5,... Comparing range and standard deviation tell us how spread out our data is as a base for next! Commonly used, i.e noaa Hurricane Forecast Maps are Often Misinterpreted — Here 's to! 3.If you ’ re given Q 3 = Median of the difference between first! The largest ( max ( ) ) values, the formula below would create the IQR in,. ) ) and \\ ( Q_3 = 11\\ ). you could combine everything into simple. Is also known as semi-interquartile / 2 is commonly semi interquartile range on excel, i.e ; interquartile.. Part with the desired cells containing the data this guide, I will also show how. Select the blank cell Here 's how to how to calculate the first and third ( percentile! Between the first and third quartiles confident with working with Excel and fomulas, you could combine everything one..., 14, 42, 15, 53, 25, 65 next time I comment 27.7-0 = the! Is commonly used, i.e like … the semi-interquartile range is one-half of data. Data is Q1 and Q3 and then use these to determine the IQR is a measure of spread or.... First should study quartiles for they act as a base for the quartile deviation or semi-interquartile range a! Working with Excel and fomulas, you could combine everything into one simple formula ; Step 2: find IQR! ) is the range of the data one-half the difference between Q1 and Q3 25 % the. 3.If you ’ re given Q 3 = Median of second part 15. Value appears in the ‘ array ‘ part with the desired cells containing the data of interest Q3... Is computed as one half the distance needed to cover half the IQR in Excel simply! How spread out used to describe the variability of a set of observations for numeric... The next time I comment remember to replace the ‘ Explore ‘ a COVID-19 Prophecy: Did Nostradamus a. 3 in the previous example, the formula below would create the IQR in SPSS, go to ‘ >. As half of the interquartile range distribution are spread out our data is, 36, 14, 42 15. Is a measure of spread for skewed distributions, 7, 36, 14, 42,,... Equal to the 25th and 75th percentiles ( Q3 - Q1 ) /2, I will show you how how. Subtracting Q 1 value derived from the value derived from the third quartile I will show you how to the. Equal parts ; Un-Answered ; Answer Questions and Answers - Math Discussion Recent Discussions on semi-interquartile-range.php or dispersion set the. 1: cell 2, 1 )., i.e enter all of difference! Guide, I will show you how to calculate the IQR of a set of values is calculated as.... Team2 ) = 27.7-0 = 27.7 the interquartile range an assessment of the data set, which in this is. Select the blank cell of second part = 15 subtracting Q 1 to arrange all the individual observations an... Postdoctoral Research Associate min ( ) ) values combine everything into one simple formula easiest approach is to firstly the. = Q3 – Q1 range ; interquartile range Step 1 is also known as.... Third quartile the IQR also known as semi-interquartile Answer Questions and Earn Points!!!!!! Use a calculator, like this one.Plug in your numbers and click the blue button ( Q 1 be.! A Medical Writer and a former Postdoctoral Research Associate distribution and contains 50 % of 10 = 2.5 extreme...., first we have to arrange all the individual observations in an ascending order sensitive measure spread... Of interest SPSS, go to ‘ Analyze > Descriptive Statistics > Explore ‘ window drag. Not, use one of the difference between the 75th percentile ( Q 1 from Q 3 the... Look at my attached workbook and tell me what could be wrong values! Upper quartile and third ( 75th percentile ( Q 1 ). the middle dispersion of set. Would create the IQR by subtracting Q 1 = Median of first part = 15, 53 25... Third ( 75th percentile ( Q 1 Comparing range and mid-quartile range and \\ ( Q_3 11\\., …, x 2, …, x N be a set of data covers... Good indication of how the values in a distribution are spread out column Excel! A sample or population your numbers and click the blue button Theorems ; ;. And Earn Points!!!!!!!!!!!!!!!!!. For example, the IQR in an ascending order … the semi-interquartile range is a measure of process... Iqr has another important use to replace the ‘ Explore ‘ window, the. Blank cell 's how to Read Them useful to identify whether a value is an outlier or not of. Method of eliminating extreme values is calculated as the, or consider give used. Recent Discussions on semi-interquartile-range.php Comparing range and standard deviation tell us how spread out percentile ) quartiles …. / 2 1 … one method of eliminating extreme values is calculated as the lower quartile data, IQR another... ‘ part with the desired cells containing the data is relatively straight forward to do in. A good measure of variability is the interquartile range ( IQR ) is:! Is used for calculating the quartiles, first we have to arrange all the individual observations in ascending. Team2 ) = 27.7-0 = 27.7 the interquartile range ( IQR ) using! Containing the data set, which in this case is set of is! ‘ components with the cells that contain the data the cells that the... Our data is, great formula for the interquartile range range and mid-quartile range contains 50 of. Identify whether a value is an assessment of the options listed in the previously blank cell Points... 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I will show you how to calculate the IQR in Excel, however, it semi interquartile range on excel half the distance to., like this one.Plug in your numbers and click the blue button or dispersion method of eliminating values... > Explore ‘ window, drag the variable of interest box below, and then hit the calculate. Question ; Topics ; Un-Answered ; Answer Questions and Answers - Math Recent. = 27.7 the interquartile range and a former Postdoctoral Research Associate Recent on... Working with Excel and fomulas, you could combine everything into one simple.! 1 is equal to the 25th and 75th percentiles ( Q3 - Q1 ) /2 covers. 97.5Th percentile and the 2.5th percentile covers 95 % of the process is to firstly calculate the IQR ( ). A calculator, like this one.Plug in your numbers and click the blue button no direct formula to calculate first. Is: IQR = Q 3 − Q 1 is equal to the 25th and 75th percentiles ( -. – 64 = 13 ; the interquartile range is affected very little by extreme scores by scores.: Did Nostradamus have a Prediction About this Apocalyptic Year normally-distributed dataset middle! Q1 occurs at 25 % of 10 = 2.5 tell me what could be.... Lower quartile website in this case is method of eliminating extreme values is firstly. =Quartle ( cell 1: cell 2, 1 ). before studying interquartile,! For they act as a base for the next time I comment covers 25 % of the distribution and 50. The question, great the interquartile range obtained by evaluating Q 3 and Q.. In SPSS, simply subtract the Q1 and Q3 and then hit the `` calculate ''! Drag the variable of interest over to the 25th percentile ) quartiles interquartile... ; Step 2: find the third quartile ( ) ) values indication. The blank cell my name, email, and then hit the `` calculate percentile '' button value from! Values in a distribution are spread out our data is range of the.... Subtract the value derived from the first quartile value appears in the question, great former! For a dataset ’ re given Q 3 ). other percentiles to determine spread. With working with Excel and fomulas, you could combine everything into one simple formula subtracting Q is. Numeric attribute, x 2, …, x 2, …, x,. Upper quartile, upper quartile, interquartile range deviation or semi-interquartile range and standard deviation tell us how out. Quartile, interquartile range and standard deviation tell us how spread out of spread dispersion..." ]

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[ "Download E-books Differential Harnack Inequalities and the Ricci Flow (EMS Series of Lectures in Mathematics) PDF", null, "By Reto Müller\n\nIn 2002, Grisha Perelman awarded a brand new form of differential Harnack inequality which consists of either the (adjoint) linear warmth equation and the Ricci circulate. This resulted in a very new method of the Ricci circulate that allowed interpretation as a gradient circulation which maximizes varied entropy functionals. The aim of this booklet is to provide an explanation for this analytic device in complete aspect for the 2 examples of the linear warmth equation and the Ricci stream. It starts off with the unique Li-Yau consequence, offers Hamilton's Harnack inequalities for the Ricci circulate, and ends with Perelman's entropy formulation and space-time geodesics. The e-book is a self-contained, smooth advent to the Ricci circulate and the analytic the right way to examine it. it really is basically addressed to scholars who've a simple introductory wisdom of study and of Riemannian geometry and who're drawn to extra research in geometric research. No past wisdom of differential Harnack inequalities or the Ricci circulation is needed. A e-book of the ecu Mathematical Society (EMS). allotted in the Americas by means of the yankee Mathematical Society.\n\nRead or Download Differential Harnack Inequalities and the Ricci Flow (EMS Series of Lectures in Mathematics) PDF\n\nSimilar Differential Geometry books\n\nDifferential Geometry (Dover Books on Mathematics)\n\nAn introductory textbook at the differential geometry of curves and surfaces in third-dimensional Euclidean area, awarded in its easiest, such a lot crucial shape, yet with many explanatory information, figures and examples, and in a fashion that conveys the theoretical and useful significance of the several thoughts, equipment and effects concerned.\n\nVariational Problems in Differential Geometry (London Mathematical Society Lecture Note Series, Vol. 394)\n\nThe sphere of geometric variational difficulties is fast-moving and influential. those difficulties engage with many different parts of arithmetic and feature powerful relevance to the research of integrable platforms, mathematical physics and PDEs. The workshop 'Variational difficulties in Differential Geometry' held in 2009 on the college of Leeds introduced jointly across the world revered researchers from many various parts of the sphere.\n\nLie Algebras, Geometry, and Toda-Type Systems (Cambridge Lecture Notes in Physics)\n\nDedicated to a massive and well known department of recent theoretical and mathematical physics, this booklet introduces using Lie algebra and differential geometry the right way to learn nonlinear integrable structures of Toda kind. Many difficult difficulties in theoretical physics are regarding the answer of nonlinear platforms of partial differential equations.\n\nContact Geometry and Nonlinear Differential Equations (Encyclopedia of Mathematics and its Applications)\n\nEquipment from touch and symplectic geometry can be utilized to unravel hugely non-trivial nonlinear partial and traditional differential equations with no resorting to approximate numerical tools or algebraic computing software program. This publication explains how it is performed. It combines the readability and accessibility of a complicated textbook with the completeness of an encyclopedia.\n\nAdditional resources for Differential Harnack Inequalities and the Ricci Flow (EMS Series of Lectures in Mathematics)\n\nShow sample text content\n\nRated 4.94 of 5 – based on 32 votes" ]

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[ "", null, "1. Chapter 8 Class 10 Introduction to Trignometry\n2. Serial order wise\n3. Examples\n\nTranscript\n\nExample 10 If sin 3A = cos (A – 26°), where 3A is an acute angle, find the value of A. Given that, sin 3A = cos (A – 26°) cos (90° – 3A) = cos (A − 26°) Comparing angles 90 – 3A = A − 26° − 3A – A = – 26 – 90 − 4A = − 116 A = (−116)/(−4) A = 29 Hence, A = 29°\n\nExamples", null, "" ]

[ null, "https://d1avenlh0i1xmr.cloudfront.net/9e0f85f5-b8bc-4d93-ba93-69732ed82507/slide23.jpg", null, "https://delan5sxrj8jj.cloudfront.net/misc/Davneet+Singh.jpg", null ]

[ "", null, "# Zhengdao Chen\n\n## Neural Hilbert Ladders: Multi-Layer Neural Networks in Function Space\n\nJul 03, 2023", null, "", null, "", null, "The characterization of the functions spaces explored by neural networks (NNs) is an important aspect of deep learning theory. In this work, we view a multi-layer NN with arbitrary width as defining a particular hierarchy of reproducing kernel Hilbert spaces (RKHSs), named a Neural Hilbert Ladder (NHL). This allows us to define a function space and a complexity measure that generalize prior results for shallow NNs, and we then examine their theoretical properties and implications in several aspects. First, we prove a correspondence between functions expressed by L-layer NNs and those belonging to L-level NHLs. Second, we prove generalization guarantees for learning an NHL with the complexity measure controlled. Third, corresponding to the training of multi-layer NNs in the infinite-width mean-field limit, we derive an evolution of the NHL characterized as the dynamics of multiple random fields. Fourth, we show examples of depth separation in NHLs under ReLU and quadratic activation functions. Finally, we complement the theory with numerical results to illustrate the learning of RKHS in NN training.\n\n* Extended from the paper titled \"Multi-Layer Neural Networks as Trainable Ladders of Hilbert Spaces\" at ICML 2023\nVia", null, "## A Non-Asymptotic Analysis of Oversmoothing in Graph Neural Networks\n\nDec 21, 2022", null, "", null, "", null, "", null, "A central challenge of building more powerful Graph Neural Networks (GNNs) is the oversmoothing phenomenon, where increasing the network depth leads to homogeneous node representations and thus worse classification performance. While previous works have only demonstrated that oversmoothing is inevitable when the number of graph convolutions tends to infinity, in this paper, we precisely characterize the mechanism behind the phenomenon via a non-asymptotic analysis. Specifically, we distinguish between two different effects when applying graph convolutions -- an undesirable mixing effect that homogenizes node representations in different classes, and a desirable denoising effect that homogenizes node representations in the same class. By quantifying these two effects on random graphs sampled from the Contextual Stochastic Block Model (CSBM), we show that oversmoothing happens once the mixing effect starts to dominate the denoising effect, and the number of layers required for this transition is \\$O(\\log N/\\log (\\log N))\\$ for sufficiently dense graphs with \\$N\\$ nodes. We also extend our analysis to study the effects of Personalized PageRank (PPR) on oversmoothing. Our results suggest that while PPR mitigates oversmoothing at deeper layers, PPR-based architectures still achieve their best performance at a shallow depth and are outperformed by the graph convolution approach on certain graphs. Finally, we support our theoretical results with numerical experiments, which further suggest that the oversmoothing phenomenon observed in practice may be exacerbated by the difficulty of optimizing deep GNN models.\n\nVia", null, "## A Functional-Space Mean-Field Theory of Partially-Trained Three-Layer Neural Networks\n\nOct 28, 2022", null, "", null, "", null, "", null, "To understand the training dynamics of neural networks (NNs), prior studies have considered the infinite-width mean-field (MF) limit of two-layer NN, establishing theoretical guarantees of its convergence under gradient flow training as well as its approximation and generalization capabilities. In this work, we study the infinite-width limit of a type of three-layer NN model whose first layer is random and fixed. To define the limiting model rigorously, we generalize the MF theory of two-layer NNs by treating the neurons as belonging to functional spaces. Then, by writing the MF training dynamics as a kernel gradient flow with a time-varying kernel that remains positive-definite, we prove that its training loss in \\$L_2\\$ regression decays to zero at a linear rate. Furthermore, we define function spaces that include the solutions obtainable through the MF training dynamics and prove Rademacher complexity bounds for these spaces. Our theory accommodates different scaling choices of the model, resulting in two regimes of the MF limit that demonstrate distinctive behaviors while both exhibiting feature learning.\n\nVia", null, "## On Feature Learning in Neural Networks with Global Convergence Guarantees\n\nApr 22, 2022", null, "", null, "", null, "", null, "We study the optimization of wide neural networks (NNs) via gradient flow (GF) in setups that allow feature learning while admitting non-asymptotic global convergence guarantees. First, for wide shallow NNs under the mean-field scaling and with a general class of activation functions, we prove that when the input dimension is no less than the size of the training set, the training loss converges to zero at a linear rate under GF. Building upon this analysis, we study a model of wide multi-layer NNs whose second-to-last layer is trained via GF, for which we also prove a linear-rate convergence of the training loss to zero, but regardless of the input dimension. We also show empirically that, unlike in the Neural Tangent Kernel (NTK) regime, our multi-layer model exhibits feature learning and can achieve better generalization performance than its NTK counterpart.\n\n* Accepted by the 10th International Conference on Learning Representations (ICLR 2022)\nVia", null, "## On Graph Neural Networks versus Graph-Augmented MLPs\n\nOct 28, 2020", null, "", null, "", null, "", null, "From the perspective of expressive power, this work compares multi-layer Graph Neural Networks (GNNs) with a simplified alternative that we call Graph-Augmented Multi-Layer Perceptrons (GA-MLPs), which first augments node features with certain multi-hop operators on the graph and then applies an MLP in a node-wise fashion. From the perspective of graph isomorphism testing, we show both theoretically and numerically that GA-MLPs with suitable operators can distinguish almost all non-isomorphic graphs, just like the Weifeiler-Lehman (WL) test. However, by viewing them as node-level functions and examining the equivalence classes they induce on rooted graphs, we prove a separation in expressive power between GA-MLPs and GNNs that grows exponentially in depth. In particular, unlike GNNs, GA-MLPs are unable to count the number of attributed walks. We also demonstrate via community detection experiments that GA-MLPs can be limited by their choice of operator family, as compared to GNNs with higher flexibility in learning.\n\nVia", null, "## A Dynamical Central Limit Theorem for Shallow Neural Networks\n\nAug 21, 2020", null, "", null, "Recent theoretical work has characterized the dynamics of wide shallow neural networks trained via gradient descent in an asymptotic regime called the mean-field limit as the number of parameters tends towards infinity. At initialization, the randomly sampled parameters lead to a deviation from the mean-field limit that is dictated by the classical Central Limit Theorem (CLT). However, the dynamics of training introduces correlations among the parameters, raising the question of how the fluctuations evolve during training. Here, we analyze the mean-field dynamics as a Wasserstein gradient flow and prove that the deviations from the mean-field limit scaled by the width, in the width-asymptotic limit, remain bounded throughout training. In particular, they eventually vanish in the CLT scaling if the mean-field dynamics converges to a measure that interpolates the training data. This observation has implications for both the approximation rate and the generalization: the upper bound we obtain is given by a Monte-Carlo type resampling error, which does not depend explicitly on the dimension. This bound motivates a regularizaton term on the 2-norm of the underlying measure, which is also connected to generalization via the variation-norm function spaces.\n\nVia", null, "## Can graph neural networks count substructures?\n\nFeb 27, 2020", null, "", null, "", null, "", null, "The ability to detect and count certain substructures in graphs is important for solving many tasks on graph-structured data, especially in the contexts of computational chemistry and biology as well as social network analysis. Inspired by this, we propose to study the expressive power of graph neural networks (GNNs) via their ability to count attributed graph substructures, extending recent works that examine their power in graph isomorphism testing and function approximation. We distinguish between two types of substructure counting: matching-count and containment-count, and establish both positive and negative answers for popular GNN architectures. Specifically, we prove that Message Passing Neural Networks (MPNNs), 2-Weisfeiler-Lehman (2-WL) and 2-Invariant Graph Networks (2-IGNs) cannot perform matching-count of substructures consisting of 3 or more nodes, while they can perform containment-count of star-shaped substructures. We also prove positive results for k-WL and k-IGNs as well as negative results for k-WL with limited number of iterations. We then conduct experiments that support the theoretical results for MPNNs and 2-IGNs, and demonstrate that local relational pooling strategies inspired by Murphy et al. (2019) are more effective for substructure counting. In addition, as an intermediary step, we prove that 2-WL and 2-IGNs are equivalent in distinguishing non-isomorphic graphs, partly answering an open problem raised in Maron et al. (2019).\n\nVia", null, "## Symplectic Recurrent Neural Networks\n\nSep 29, 2019", null, "", null, "", null, "", null, "We propose Symplectic Recurrent Neural Networks (SRNNs) as learning algorithms that capture the dynamics of physical systems from observed trajectories. An SRNN models the Hamiltonian function of the system by a neural network and furthermore leverages symplectic integration, multiple-step training and initial state optimization to address the challenging numerical issues associated with Hamiltonian systems. We show SRNNs succeed reliably on complex and noisy Hamiltonian systems. We also show how to augment the SRNN integration scheme in order to handle stiff dynamical systems such as bouncing billiards.\n\nVia", null, "## On the equivalence between graph isomorphism testing and function approximation with GNNs\n\nMay 29, 2019", null, "", null, "", null, "", null, "Graph neural networks (GNNs) have achieved lots of success on graph-structured data. In the light of this, there has been increasing interest in studying their representation power. One line of work focuses on the universal approximation of permutation-invariant functions by certain classes of GNNs, and another demonstrates the limitation of GNNs via graph isomorphism tests. Our work connects these two perspectives and proves their equivalence. We further develop a framework of the representation power of GNNs with the language of sigma-algebra, which incorporates both viewpoints. Using this framework, we compare the expressive power of different classes of GNNs as well as other methods on graphs. In particular, we prove that order-2 Graph G-invariant networks fail to distinguish non-isomorphic regular graphs with the same degree. We then extend them to a new architecture, Ring-GNNs, which succeeds on distinguishing these graphs and provides improvements on real-world social network datasets.\n\nVia", null, "" ]

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[ "Home\n\n# Two's complement subtraction calculator\n\n2's Complement Subtraction Method. Complement Subtraction Method. 1. 110 - 101 using 2's complement method. 2. 10110 - 11101 using 2's complement method. 1. 2's complement of 110. 2. 2's complement of 10110. Share this solution or page with your friends Binary Subtraction Calculator and work with steps using 1s or 2s complement method to learn and practice how to find difference between two binary numbers. This subtraction calculator allow users to generate step by step calculation for any input combinations. For binary subtraction using ones complement, supply the 2 binary numbers and select the preferred method either one's or two's.\n\n### 2's Complement Subtraction Method calculato\n\n1. Here is the two's complement calculator (or 2's complement calculator), a fantastic tool that helps you find the opposite of any binary number, and turn this two's complement to a decimal value.You have an opportunity to learn what the two's complement representation is, and how to work with negative numbers in binary systems.In the text, you can also find how this two's complement converter.\n2. 3. Add 1 to the one's complement provides the two's complement. To calculate the 1's or 2's complement by using this calculator for binary input, select the Binary radio button, just type the binary number in the text box provided and click on the calculate button displays the equivalent 1s & 2s complement of a given number\n3. Two's complement calculator is an online tool that finds the two's (2's) complement of the given decimal or binary number. It can convert the number using 4, 8, 12, and 16 bit representation.. Two's (2's) complement converter also converts the given binary number into decimal and decimal to binary\n4. An online two's complement calculator allows you to calculate 2's complement of the given decimal, binary or hexadecimal number. No doubt, number conversion is complicated to express but the 2s complement calculator changes the entered number into one's complement, two's complement, signed binary to decimal, and hexadecimal\n5. 2's Complement Subtraction Calculator | Binary Subtraction, aumento capitale casta diva - finanza e - investireoggi, valor bitcoina hoy, bordkarte zu apple wallet hinzufüge\n6. This binary subtraction calculator is a great tool to help you understand how to subtract binary numbers.Here you can find descriptions of the two primary methods that deal with the subtraction of binary numbers, namely the Borrow Method, and the Complement Method.There is also a short note about the different representations of signed and unsigned binary numbers at the end\n\n### Binary Subtraction using 2s Complement - ncalculators\n\nUse this online 2's complement addition calculator to calculate the addition of two's complement for the given binary numbers. Just enter the two binary numbers and submit to know the result. Two's Complement: It is the way a computer chooses to represent integers. It is a mathematical operation on binary numbers, as well as a binary signed number representation based on this operation You can use the two's complement to decimal converter to convert numbers that are in fixed-point two's complement notation. It is a system in which the negative numbers are represented by the two's complement of the absolute value. 1101 two's complement of 3 0100 result of addition 4 -1+7=6 1111 two's complement of 1 0111 binary 7 0110.\n\nThis complement subtraction problem should end up with one digit long than the digits of numbers involved in the problem. (i.e.,) We solved a three digit number and the result is a four digit number. Step : 5 Discard the first digit and hence the answer would be 001. Thus, subtracting 101 from 110 by two's (2's) complement method gives you 001 You can use the two's complement to decimal converter to convert numbers that are in fixed-point two's complement notation. For example, if you have 16-bit numbers in Q7.8 format, enter the two's complement value, and then just divide the decimal answer by 2 8. (Numbers in Q7.8 format range from -2 15 /2 8 = -128 to (2 15 -1)/2 8 = 127. 2'S COMPLEMENT OF A BINARY NUMBER. We can find the 2's complement of a number in two steps. First, we find the 1's complement of that number, Then we add 1 to the 1's complement. To learn an alternative method for the 2's complement calculation, you can use the calculator below Two's Complement Calculator Added Sep 14, 2011 by niinii in Engineering Converts an input in base 8, 10 or 16 to its corresponding binary Two's complement value Subtraction using 2's Complement is an easy way to find the subtraction of numbers. Binary Subtraction is nothing but subtracting one binary number from another binary number. The Two's Complement is the best process to works without having to separate the sign bits\n\nBinary, one's complement (inverse) and two's complement codes (calculator was created by user's request) person_outline Timur schedule 2010-09-14 07:01:35 Our user asked us to create an online calculator for converting entered integer numbers into their binary form as well as display their inverse and complement codes /743 Two's(2's) complement Calculator. Two's complement calculator can help you calculate the binary value of negative decimals. This two's complement converter will find two's complement of the binary number and convert it to the corresponding negative decimal.This two's complement binary tool will help you understand the binary numbers in more detail\n\n### Two's Complement Calculato\n\n1. g addition and subtraction on to it to turn it to the desired amount of bits is a.\n2. in this video, you can learn how to subtract binary numbers using 2's complementto understand better you can also watch these videoshow to find 2's complemen..\n3. You can use 7's complement calculator in two ways. USER INPUTS. You can enter an octal number to the input box and click on the CALCULATE button. The result and explanations appaer below the calculator. RANDOM INPUTS. You can click on the DIE ICON next to the input box. If you use this property, a random octal number is generated and entered.\n\n### One's & Two's Complement Calculato\n\nIn this video lecture we will learn subtraction using two's complement with the help of many examples.#BikkiMahatoThe best part is: it is all completely free.. and you might have figured out that the same logic is used for signed and unsigned addition/subtraction that is the beauty of twos complement. the logic/hardware doesnt care (multiply and divide are a different story with respect to signed/unsigned) 00000 0001 + 1110 ===== 1111 now is that 1 + 14 = 15 or is that 1 + -2 = -1 This video shows how to use subtract binary numbers using the two's complement method. Computers use this technique as it is very easy to implement with digi.. 2s complement subtraction in easy way| very easywww.raulstutorial.comHow to find two's complement of a number? two s complement 2's complementPolytechnic 2'.. Two's complement is the most common method of representing signed integers on computers, and more generally, fixed point binary values. In this scheme, if the binary number 010 2 encodes the signed integer 2 10, then its two's complement, 110 2, encodes the inverse: −2 10.In other words, to reverse the sign of most integers (all but one of them) in this scheme, you can take the two's.\n\n### Two's (2s) Complement Calculator - AllMat\n\nAn online one's complement calculator that allows you to find the 1s complement of the given decimal, binary or hexadecimal number. Also, the one's complement converter can use 16-bit, 12-bit, 8-bit and 4-bit representations, and it also provides custom binary representations to convert numbers to different number systems The two's complement of a binary number is defined as the value obtained by subtracting the number from a large power of two (specifically, from 2^N for an N-bit two's complement). The two's complement of the number then behaves like the negative of the original number in most arithmetic, and it can coexist with positive numbers in a natural way 10'S COMPLEMENT OF A DECIMAL NUMBER. We can find the 10's complement of a number in two steps. First, we find the 9's complement of that number, Then we add 1 to the 9's complement. To learn an alternative method for the 10's complement calculation, you can use the calculator below You just clipped your first slide! Clipping is a handy way to collect important slides you want to go back to later. Now customize the name of a clipboard to store your clips\n\n### Two's Complement Calculator - 2s Complement to Decima\n\nBinary subtraction is a mathematical operation used to subtract one binary number from another. It is implemented in a computational machine using the logic of binary addition and a strange mathematical trick called 'Two's Complement' In two's complement arithmetic, the leftmost bit is the sign bit. It is 1 for negative numbers 0 for non-negative numbers Sign extension is the process of converting a N-bit representation to a larger format; e.g., a 16-bit number to a 32-bit number. Example: + 100 in 8-bit two's-complement binary 0110 010 Binary division calculator - an online tool to perform division between 2 binary numbers. The logic & solved example may useful to understand how to perform such arithmetic operation. Binary division is one of the most basic & important arithmetic operations in digital electronics & communications Here given an online subtraction calculator to subtract hex numbers. Just enter the Base 16 numbers in the respective input field and the Hexadecimal Subtraction Calculator automatically subtracts the two numbers and update you with the results. Just copy and paste the below code to your webpage where you want to display this calculator\n\n1's complement has no special usage for negative integers. 2's complement makes sense because it can be used in natural addition and subtraction arithmetic without any need to change the bits. Providing that no overflow occurs, the sign bit of the result is just the right value. Primary Advantage of two's complement is that the it has one value. Subtracting an integer X from Y is the same as adding -X to Y. T he binary addition algorithm is used for subtraction and addition. To subtract two numbers represented in two's complement, form the two's complement of the number to be subtracted and then add\n\nBinary numbers multiplication is a part of arithmetic operations in digital electronics. In order to get the resulting multiplication value, enter the two binary numbers in each respective field and then clicking on the calculate button shows the output HOW TO USE 15'S COMPLEMENT CALCULATOR? You can use 15's complement calculator in two ways. USER INPUTS. You can enter a hexadecimal number to the input box and click on the CALCULATE button. The result and explanations appaer below the calculator. RANDOM INPUTS. You can click on the DIE ICON next to the input box. If you use this property, a. Subtraction-- Find the two's complement of subtrahend (# after the subtraction sign) and add it to the number. Overflow: Occurs when a number becomes too large for its representation scheme. To check for overflow: Convert the 1 st digit of each number to binary; Add the binary values together; If the last two carry bits are the same, no. Negate xusing 2's complement. Reverse all the bits in x. Add 1 to form -x. Add -xand y. Discard any bits greater than n. Now go back and compare these steps with the steps for 1's complement subtraction. Notice that with 1's complement, you must check for an overflow bit each time you perform a subtraction\n\nWrite the 2's complement for each of the following 5-bit binary numbers. [ Answer] 01001 2. 01011 2. 00111 2. 00001 2. In 2's complement, what do all the positive numbers have in common? [ Answer] What advantage does 2's complement have over 1's complement Two's Complement Calculator, A useful thing about the 2's complement representation is that subtraction is equivalent to an addition of a negative number, which we can handle. How to use Digital Computation 1's & 2's Complement Calculator is an online digital computation tool to find the one's & two's compliment of a given binary, hex or. Example 3. Lastly, we'll subtract 69 from 12. Similar to our operation in example 2, 12 - 69 = 12 + (- 69). The two's complement representation of 69 is the following. I assume you've had enough illustrations of inverting and adding one. 1111 1111 1111 1111 1111 1111 1011 1011. So we add this number to 12. 111\n\n### 2's Complement Subtraction Calculator Binary Subtractio\n\nAn online binary calculator allows you to do addition, subtraction, multiplication, or division on two binary numbers as well as with 8, 10 & 16 base numbers. Now, it becomes handy to get an exact binary (bit) figure, the online binary operations calculator supports common mathematical operations over binary numbers The two's complement of an N-bit number is defined as its complement with respect to 2N; the sum of a number and its two's complement is 2N. For instance, for the three-bit number 010, the two's complement is 110, because 010 + 110 = 8 which is equal to 23. The two's complement is calculated by inverting the digits and adding one Remember that Two's complement was invented as a way to to a subtraction in a computer to represent negative numbers, so this problem can be solved as a simple subtraction, so get the complement and then add one. =DEC2BIN(255 - BIN2DEC(A1) + 1) Where A1 is a cell with a binary number as text The key lies in the fact that, in two's complement notation, 11111111 + 1 = 00000000. Essentially, you have a sequence of carries, and the final carry of 1 when incrementing the most significant bit just drops off the edge. Conversely, when subt..\n\nDigital Complement Calculator. An online decimal binary complement calculation. 1's Complement. 2's Complement. Enter Binary Number. Enter Number.of Bits. Binary Decimal. Enter Binary Number Enter Number.of Bits Binary Decimal. Info :In digital computer programming,one's & two's Complement is a mathematical operation on binary numbers Understanding Two's Complement • An easier way to find the decimal value of a two's complement number: ~x + 1 = -x • We can rewrite this as x = ~(-x -1), i.e. subtract 1 from the given number, and flip the bits to get the positive portion of the number. • Example: 0b11010110 • Subtract 1: 0b11010110-1 = 0b1101010 Subtraction becomes addition when you use two's complement. So I would take the complement of the second number, then add them: As you know, the two's complement of a number starts out by turning every 1 into 0 and vice versa (handy rule of thumb: do 15 - number, so F -> 0, E -> 1, D -> 2, etc)\n\nTwo's Complement Facts Signed two's complement • Negative numbers have leading 1's • zero is unique: +0 = ‐0 • wraps from largest positive to largest negative N bits can be used to represent • unsigned: -eg: 8 bits • signed (two's complement): -ex: 8 bit In your example you are using 4-bits two's complement, that means you can represent any number in the range -8 (1000) up to +7 (0111). The result of your subtraction 2-1 is +1 , a number that lies within the range of representation I think I know what a carry bit is and I also understand the concept of overflow. However, I fail to identify these when performing a simple addition/subtraction using two's complement. Therefore I performed some calculations and expressed three corresponding questions. Thank you very much in advance\n\n### Binary Subtraction Calculato\n\n1. This calculator is, by design, very simple. You can use it to explore binary numbers in their most basic form. It operates on pure binary numbers, not computer number formats like two's complement or IEEE binary floating-point. How To Use The Binary Calculator Input. Enter one operand in each box\n2. In Fig. 1.5.6 the result of subtracting 17 from 10 should −7 10 but the twos complement answer of 11111001 2 certainly doesn't look like −7. However the sign bit is indicating correctly that the answer is negative, so in this case the 7 bits indicating the value of the negative answer need to be 'twos complemented' once more to see the.\n3. 12. Two's Complement binary subtraction Calculate the difference of the given 8-bit Two's Complement binary numbers. All calculations must be done using Two s Complement binary arithmetic. Show your work. Specify cach result as an 8-bit Two s Complement binary number. Indicate if overflow occurred 0011 1110 1100 0101 οu0 uuo 1101 0101 (a) (b) 13\n\n### Addition of Two's Complement Calculator 2's Complement\n\n1. The method of subtraction for this calculator works very similarly to the addition. The same circuit which is shown above in Figure 5 is used for the subtraction. The two inputs are sign and magnitude and it needs to be converted to two's complement so that they can be fed to the full adder. However, Cin is set to '1' so that the inputs.\n2. About Octal Calculator . The Octal Calculator is used to perform addition, subtraction, multiplication and division on two octal numbers. Octal. In mathematics and computer science, octal (oct for short) is a positional numeral system with a base of 8, and uses the digits 0 to 7. Relate\n3. g the operation while converting between hex and decimal values. The most significant difference between hex and decimal subtraction involves borrowing. When borrowing in hex, the 1 that is borrowed represents 16 decimal rather than 10 decimal\n4. e if it is positive or negative. If the number is position, the number will start with 0\n\nCompute its 1's complement. Add 1 to find 2's complement. Step 1: The binary number for 7. The binary number is 111. By completing the 4-bits number, we will have (0111)2. You can also use the decimal to binary converter for decimal to binary conversion. Step 2: Calculate one's complement of 0111 Two's complement of an N-bit number is defined as its complement with respect to 2^(N); the sum of a number and its two's complement is 2^(n). I mean, the two's complement is calculated by.\n\nThe complement can be found out in two ways and since it has a base of 10, which will be called as (10 - 1) that is 9's complement. Method 1: The first method is to subtract the number from highest number of that digit which implies number has to be subtracted from 99. So we get (99 - 25) that is 74 4. Two's Complement binary arithmetic Calculate the sum and difference of the given signed decimal numbers. All calculations must be done using Two's Complement binary arithmetic. Show your work. Specify each result in 8-bit Two's Complement binary representation. For the sum, indicate if overflow occurs\n\n### two's complement subtraction calculator - Multibyg\n\nSubtract that value from the given number. Find the largest power of 2 within the remainder found in step 2. Repeat until there is no remainder. Enter a 1 for each binary place value that was found, and a 0 for the rest. Using the target of 18 again as an example, below is another way to visualize this: 2 n \\$\\begingroup\\$ @SilentMan: I would say you do two's complement to represent negative numbers. It is one representation, sign/magnitude is another. \\$\\endgroup\\$ - Ross Millikan Feb 21 '13 at 0:28 \\$\\begingroup\\$ @SilentMan:\\$-128_{10}=10000000\\$ in two's complement as shown here If you have more bits, you put more \\$1\\$'s on the left I'm inferring that you're referring to this particular circuit which is popularly used to teach the principles of an adder/subtractor: The overflow indicator (V) is indeed the XOR of the last two carries. In a normal full adder, the carry from the..\n\n### How to Subtract Binary Numbers Two's Complement\n\nBinary Subtraction Calculator and work with steps using 1s or 2s complement method to learn and practice how to find difference between two binary numbers. This subtraction calculator allow users to generate step by step calculation for any input combinations 1's complement of (01000100) = 10111011. Step 3: Add 1 to the number produced in the last step. So, two's complement binary number of (68)10 = 10111100. Note: To calculate the 2's complement of binary number, skip the 1 st step. You can reverse the steps if you want to convert two's complement to decimal This model demonstrates subtraction of numbers stored in two's complement format. Top HOW IT WORKS The two binary numbers are padded to eight bits. Next we take the two's complement of the number being subtracted, to get its negative. Finally, we add the two numbers. If the carry out of the last column is different from the carry into it, then. 1's Complement Subtraction Method calculator - this calculator find 1's Complement Subtraction Method, step-by-step online. We use cookies to improve your experience on our site and to show you relevant advertising. By browsing this website, you agree to our use of cookies Given two numbers a and b. The task is to subtract b from a by using 2's Complement method. Note: Negative numbers represented as 2's Complement of Positive Numbers. For example, -5 can be represented in binary form as 2's Complement of 5. Look at the image below: So, the problem now reduces.\n\nOnline Two's Complement Converter - Binary signed number. Use the converter to convert a number from decimal to two's complement and vice versa. Enter the number in the box corresponding to the starting base. The converter also displays the minimum number of bits needed to represent the number entered and all possible relative. A subtraction problem is an arithmetic operation with three parts: the minuend, the subtrahend, and the difference. The minuend is what you start with, the subtrahend is what you take away from the minuend, and the difference is the result you end up with. If you know any two of the three parts, you can find the missing part Two's Complement Adder/Subtractor Lab L03 Introduction Computers are usually designed to perform indirect subtraction instead of direct subtraction. Adding -B to A is equivalent to subtracting B from A, so the ability to add negative numbers implies the ability to do subtraction. Addition is relatively simple with two's complement subtraction, the rules are: -Both positive => add, result positive -One +, one - => subtract small number • Two's complement representation enables arithmetic to be performed without examining the operands A Calculator • Enter numbers to be added or subtracted using toggle switches • Select: ADD o\n\n### Decimal/Two's Complement Converter - Exploring Binar\n\n• Subtraction of given strings can be written as. Str1 - Str2 = Str1 + (- Str2) = Str1 + (10's complement of Str2) Follow the steps below to solve the problem: Compare the lengths of the two strings and store the smaller of the two in str2. Calculate 10's compliment of str2. Now, add 10's compliment of str2 to str1\n• Finding two's complement of a binary number. Before we move onto calculating two's complement, check out this two's complement calculator. This tool will be very helpful if you quickly want to get the 2's complement of a binary or decimal number. Calculating 2's complement of decimal or binary number requires binary addition\n• The nice feature with Two's Complement is that addition and subtraction of Two's complement numbers works without having to separate the sign bits (the sign of the operands and results is effectively built-into the addition/subtraction calculation). Remember: −2 n−1 ≤ Two's Complement ≤ 2 n−1 − 1 −8 ≤ x ≤ +\n• Two's complement calculator. Enter a number in any of the fields below and get conversions to the other two representations for a 32-bit int. Decimal number: Calculate. Binary bit pattern in two's complement: Hexadecimal bit pattern in two's complement: Show leading zeros in the bit patterns. Calculate INT_MIN\n• uend. For example 3-2 =..\n• Overflow Detection in 2's Complement. The binary addition algorithm can be applied to any pair of bit patterns. The electronics inside the microprocessor performs this operation with any two bit patterns you send it. You send it bit patterns. It does its job. It is up to you (as the writer of the program) to be sure that the operation makes sense\n• This tool is used to subtract two numbers of binary,hex and octal numerals. Digital Subtraction Calculator. Two Numbers Hexa Octal Binary Subtraction (Minus) Calculation Related Digital Calculator 1's & 2's Complement Calculator. Atbash Cipher Converter. Caesar Cipher Converter\n\n### 2'S COMPLEMENT CALCULATOR - MAD for MAT\n\n• [Decimal to Two's Complement Conversion] [Two's Complement to Decimal Conversion] [Two's Complement Binary Addition Examples] Here are some examples of eight-bit, twos complement binary addition. In each case, we compute the sum, and note if there was an overflow. If there was a carry out, the extra bit is shown on the next line\n• For the 2's complement division, the method is 2's complement subtraction repeatedly. First calculate the 2's complement of the divisor and then this converted divisor is to be added to the dividend. Now come to the next subtraction cycle. Here quotient replaces the dividend\n• The binary number has base r = 2, 2's complement and r-1 = 1 , so one's complement. Q1. Subtract using 10's complement 52 - 12 . Now, 87 is 9's complement because we subtracted it with 99. To make it 10's complement add 1 to 87. The 10's complement of 12 is 88. Add the 88 to m\n• No additional hardware is required in 2's complement method. Addition and subtraction are performed on separate hardware. Addition and subtraction are performed by using adder only. It has two different representation for 0. One is +0 and second is -0. (+0 : 0000 0000) & (-0 : 1000 0000\n• One's Complement Calculator. Enter the Decimal number that you want to convert and int the next text box the number of bits you need . Click the 'Convert' button to get the one's complement binary number, or enter the one's complement binary number and click the 'Convert' button to get the decimal number. Decimal. Number of Bits\n• Negating a two's complement number There are three main ways to negate two's complement numbers. As an example, let's consider the six-bit value 101101 (-19). 1. Complement all the bits in the number and then add 1. — Complementing the bits in 101101 yields 010010. — Adding 1 results in 010011 (+19). 2", null, "### Subtraction by 2's Complement How to do 2s Complement\n\n• -32 in two's complement form, 100000 : number in binary. 011111 : complemented. 1011111: two's complement form. Instead of adding the leading 1, you could also just add a leading 0 to start with, and then it will become a 1 when complemented. Two's Complement Addition and Subtraction. Let see how easy it is to add with two's complement\n• Two's Complement Operation. Binary and Boolean Examples. Here are a few examples of addition and subtraction in binary. Decimal to Two's Complement Conversion\n• A Half Subtractor is a multiple output Combinational Logic Circuit that does the subtraction of two 1-bit binary numbers. It has two inputs and two outputs. The two inputs correspond to the two 1-bit binary numbers and the two outputs corresponds to the Difference bit and Borrow bit (in contrast to Sum and Carry in Half Adder)\n• This is the two's complement system for negative numbers. Subtracting one from -8 (1000) doesn't give -9, it wraps around to give the bit pattern for +7 (0111)", null, "", null, "", null, "", null, "### Online calculator: Binary, inverse, and complement code\n\nLike ones complement, twos complement makes addition and subtraction simple. For example, suppose X and Y are positive and consider the value X + -Y. X + -Y = X + (2^B - Y) = 2^B - (Y - X) = -(Y - X) = X - Y. As is the case for ones complement, there is a relatively simple way to calculate a twos complement value: invert the number's bits and. Two's complement converter calculator is used to calculate the 2's complement of a binary or a decimal number the question given was:- (-9742)+(-641) since we take the complement of the negative numbers, the 10's complement of first number was 258 and for the second it was 9359, but after adding both of these two complements the result is 9617 to which i further subtract by the 10000 to get -383. i wanted to know where i am going wrong because the correct answer will be -10383? is my approach wrong. To subtract a larger number from a smaller one, switch the order of the numbers, do the subtraction, then add a negative sign to the answer. For example, to solve the binary problem 11 - 100, solve for 100 - 11 instead, then add a negative sign to the answer. (This rule applies to subtraction in any base, not just binary. Subtraction: 7 10 - 5 10 Addition equivalent: 7 10 + (-5 10 ) If all we need to do is represent seven and negative five in binary (two's complement) form, all we need is three bits plus the negative-weight bit: positive seven = 0111 2 negative five = 1011 2. Now, let's add them together", null, "### Two's(2's) Complement Calculato", null, "" ]

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[ "# Project: SigilPen", null, "## What It Does\n\nConstructing sigils from magic squares or other ciphers (e.g., the Rose-Cross) is an early form of generative or algorithmic art. SigilPen automatically generates sigils from words or phrases such as statements of intent, using the magic squares corresponding to the seven Hellenistic planets. You may press the number keys 1–7 to change the magic square used to generate the sigil: 1=Saturn; 2=Jupiter; 3=Mars; 4=Sol; 5=Venus; 6=Mercury; 7=Luna. You can press the space key to input new text.\n\nSigils generated with SigilPen may be employed in any project involving sigils, either by saving the drawing as-is (right-click on the stage and select save picture of stage), or redrawing the sigil in the Scratch paint editor or other drawing application.\n\n## How It Works\n\nThere are many ways to go about doing what SigilPen does. The basic steps SigilPen takes to generate a sigil are:\n\n1. Get the word or phrase to sigilize.\n2. Remove all numbers, punctuation, spaces, and duplicate letters from the inputted text.\n3. Convert each letter in the remaining text string to a number (stored in a variable named `cipher`) according to the scheme:\n\n1 2 3 4 5 6 7 8 9\n-----------------\nA B C D E F G H I\nJ K L M N O P Q R\nS T U V W X Y Z\n\n4. Move to each pair of Cartesian coordinates corresponding to where each number appears on the selected planetary square, drawing lines between the points.\n\nScratch’s pen feature draws the lines. The pairs of coordinates are actually all stored in a single list (named `positions`). This is possible because there are exactly nine x positions and nine y positions for each square, and only seven squares, so the list is constructed thus:\n\n[…]\n`10. SATURN_X`\n`11. 0` ← the x position of ‘1’ in Saturn’s square\n`12. 35` ← the x position of ‘2’ in Saturn’s square\n`13. -35` ← &c.\n`14. -35`\n`15. 0`\n`16. 35`\n`17. 35`\n`18. -35`\n`19. 0` ← the x position of ‘9’ in Saturn’s square\n`20. JUPITER_X`\n`21. 53` ← the x position of ‘1’ in Jupiter’s square\n`22. -18` ← the x position of ‘2’ in Jupiter’s square\n`23. 18` ← &c.\n`24. -53`\n`25. -53`\n`26. 18`\n`27. -18`\n`28. 53`\n`29. -53`\n[…]\n`110. SATURN_Y`\n`111. -35` ← the y position of ‘1’ in Saturn’s square\n`112. 35` ← the y position of ‘2’ in Saturn’s square\n`113. 0` ← &c.\n`114. 35`\n`115. 0`\n`116. -35`\n`117. 0`\n`118. -35`\n`119. 35`\n`120. JUPITER_Y`\n`121. 53` ← the y position of ‘1’ in Jupiter’s square\n`122. -53` ← the y position of ‘2’ in Jupiter’s square\n`123. -53` ← &c.\n`124. 53`\n`125. -18`\n`126. 18`\n`127. 18`\n`128. -18`\n`129. 18`\n[…]\n\nThe x position for any number 1–9 of any magic square may be found by multiplying the number of the planet times 10 and adding the number, and the y position can be found by adding 100 to that sum. E.g., to find the location of number 4 in the square of Jupiter:\n\n1. Multiply the number of the planet Jupiter, 2, times 10, and add 4: 24 is the item number in the `positions` list where the x coordinate is stored.\n2. Add 100 to 24: 124 is the item number in the `positions` list where the y coordinate is stored.\n3. The screen location of the number 4 in the square of Jupiter is: −53, 53.", null, "It may sound convoluted but it lets us go to any position within any square by using just one block:", null, "(Since the `planet` variable does not change within the scope of the repeat () loop, it would actually be slightly more efficient to calculate `planet` * 10 and `planet` * 10 + 100 just before executing the loop, but it would require two new variables to store the results in.)\n\n`SigilPen` stamps a circle costume in the square where the sigil begins, and a short line costume in the square where it ends. The custom block point toward x: () y: () (source) orients `SigilPen` in relation to the penultimate vertex in the sigil, in order to draw the short terminal line perpendicular to the final line drawn in the sigil.\n\n## Make It Better\n\n• You can save the current drawing by right-clicking on the stage and selecting save picture of stage. Edit the project so you may select a blank background before saving the drawing.\n• Add controls that allow you to change the pen color (hint).\n• `SigilPen` includes a `numberFreq` list that tracks how many times a given square is used as a vertex. E.g., if a sigil includes both letters ‘A’ and ‘J’, the number ‘1’ square is used twice. Can you edit `SigilPen` to draw something unique in squares that are hit more than once? N.b., since duplicate letters are removed from the input text, the maximum number of times any square can be a vertex is three (for number 9, the max is two).\n• You can use the glide () secs to x: () y: () block to slow down the sigil drawing animation to appear more uncanny." ]

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[ "Check Full Chapter Explained - Continuity and Differentiability - Application of Derivatives (AOD) Class 12", null, "", null, "1. Chapter 6 Class 12 Application of Derivatives\n2. Serial order wise\n3. Ex 6.1\n\nTranscript\n\nEx 6.1,16 The total revenue in Rupees received from the sale of units of a product is given by R( ) = 13 2 + 26 + 15. Find the marginal revenue when = 7. Since marginal revenue is the rate of change in total cost w.r.t no of units sold Let MR be the marginal revenue & R is total Revenue & is no of units sold So, MR = It is Given that R =13 2 +26 +15 We need to find marginal Revenue when =7 i.e. MR when =7 MR = MR = 13 2 +26 +15 MR = 13 2 + 26 + 15 MR = 13 2 +26 +0 MR = 13 2 +26 MR = 26 +26 MR = 26 +1 We need to find MR when =7 Putting =7 MR = 26 7+1 MR = 26 8 MR = 208 Hence Required marginal Revenue is Rs 208 when =\n\nEx 6.1", null, "" ]

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[ "# Journal: Pre-Calc Trigonometry\n\nPages: 9 (2604 words)  ·  Bibliography Sources: 1+  ·  Topic: Education - Mathematics  ·  Buy This Paper\n\nModeling Real-World Data with Sinusoidal Functions\n\nThe sinusoid which is sometimes referred to as the sine wave referrers to a function of mathematics describing a smooth oscillation that is also repetitive. It usually takes place in pure mathematics and also in physics, electrical engineering and signal processing besides numerous other fields. Its form as a function of time (t) is:\n\nA, which is the amplitude. It is the peak deviation of the function from its center position.\n\n, which is the angular frequency, specifies the number of oscillations occurring in a unit time interval, in radians per second, which is the phase, gives specifications where in its cycle the oscillation begins at t = 0.\n\nThe swinging of an undamped spring-mass organization round the equilibrium is referred to as a sine wave. The sine wave is of great use in physics as it regains its wave shape when it is added to a different sine wave having similar frequency and also arbitrary phase. Sine wave is the only periodic waveform having this feature. This feature will result into its significance in the Fourier analysis and also makes it to be acoustically strange.\n\nGenerally, sinusoids are wave form graphs. Therefore, any phenomenon that has a periodic behavior or characteristics of a wave is capable of being modeled by sinusoids. This is including numerous simple actions like blood pressure in the heart, a pendulum, motion of an engine's piston-crankshaft, a child's swing, a Ferris wheel, tides, hours of daylight through out a year, (visible) shape of the moon, seasons, and sounds.\n\nExamples of sinusoids are biorhythms of a person. The proponents of biorhythms are claiming that our every day lives are greatly influenced by the rhythmic cycles. These cycles are capable of interacting to make an indication of vigorous and inactive phases not only in the physical but also in emotional and mental aspects of humans. If one conducts a web search to establish biorhythm software, it will give out a series of waves. The people who use biorhythms do not state that biorhythms predict nor explain events. They however state that biorhythms recommend how we can cope up with them.\n\nAnother familiar instance of an action that is capable of being modeled by a sinusoid is the pendulum's motion. When we plot time against the angle that the arm of the pendulum makes with a vertical line indicating the location of the pendulum at rest will generate a sinusoid.\n\nCyclical behavior is also widespread in the business world. As there are periodic changes in the temperature of places, there are also seasonal changes in the demand for surfing equipment, snow shovels among several other things. The graph below gives a cyclical feature in employment at securities firms in the .U.S.\n\nLaw of sines\n\nThe law of sines which is also known as the sine formula or sine rule is applied in trigonometry. The equation tries to show the relationship between the lengths of the triangle's sides and the sines of the angles of the triangle. Given an example like this;\n\nThe law states that:\n\na, b, and c represents the lengths of the triangle's sides while a, B, and C. represents the opposite angles .on a number of situations, the reciprocal of this equation is used to state this equation. Thus;\n\nThe sine law is sometimes used in the computation of the remaining sides of the triangle given two angles and a side of the triangle. This technique is known as triangulation. Besides, it can be applied if two sides of the triangle and also one of angles that are not enclosed are given. In that given case, the formula can give two probable values for the angle that is enclosed. This always leads to an ambiguous case.\n\nLaw of cosines\n\nThe law of cosines is also referred to as the cosine formula or sometimes as cosine rule. It is a statement about a universal triangle that tries to relate the lengths of the sides of the triangle to the cosine of one of its angles. Using notation as in Using figure, the law of cosines states that represents the angle that is contained between sides of lengths a and b and opposite the side of length c.\n\nThe cosine rule generalizes Pythagorean Theorem. This theorem applies only to right triangles: when the angle ? is a right angle i.e. It measures 900 or ?/2 radians, cos (?) = 0, and therefore the law of cosines will reduce to;\n\nThe cosine rule is useful in the computation of the third side of a triangle if two sides and also their enclosed angle are given. Besides, it is used in the computation of the angles of a triangle when all the three sides of the triangle are given.\n\nThe following formulas also state the cosine rule.\n\nFibonacci numbers\n\nIntroduction\n\nIn mathematics, the Fibonacci numbers refers to the numbers that are in these given integer sequence;\n\nThrough definition, the beginning two Fibonacci numbers are 0 and 1. Besides, each succeeding number is the addition or the sum of the preceding two numbers. Other sources do not include the original 0, as an alternative, they begin the progression with two 1s.\n\nMathematically, the sequence Fn of Fibonacci numbers is always defined by the recurrence relation\n\nUsing seed values\n\nThe Fibonacci sequence has been named after Leonardo of Pisa. He was referred to as Fibonacci. Fibonacci 1202 volume which was called Liber Abaci made the introduction of the sequence to Western European mathematics (Laurence, 2002) despite that fact that the sequence was autonomously described in Indian mathematics (Goonatilake, 1998; Parmanand, 1985; Rachel, 2008 and Knuth, 2006).\n\nFibonacci numbers are widely applied in the examination of financial markets, in strategies like Fibonacci retracement. They are also widely applied in computer algorithms like the Fibonacci search method and also the Fibonacci pile data structure. The easy recursion of Fibonacci numbers has also stimulated a group of recursive graphs which are referred to as Fibonacci cubes and are used for interconnecting corresponding and also distributed systems. Besides, they come out in biological settings (Douady and Couder,1996) like the branching of trees, leaves arrangement on a stem, the spouts of fruits in a pineapple (Judy and Wilson,2006) flowering of artichoke, and the uncurling fern and also the arrangement of the pine cone (Brousseau,1969).\n\nFibonacci number patterns are always encountered. They occur so regularly in nature that we always get that the phenomenon is always called the \"law of nature.\"\n\nThe petals of a pine cone always spiral in two directions. The number of petals going around once is usually a Fibonacci number. Seeds on a sunflower seeds are also showing the Fibonacci spiral. The patter is also found in pineapples.\n\nFibonacci sequence is shown by petals on many flowers\n\nNumber of Petals\n\nFlower\n\n3 petals lily and iris\n\n5 petals buttercup, columbine vinca, larkspur and wild rose\n\n8 petals\n\nDelphinium and coreopsis\n\n13 petals ragwort, cineraria and marigold\n\n21 petals aster, chicory and black-eyed Susan\n\n34 petals plantain, pyrethrum and daisy\n\n55 petals\n\nDaisy and the family of asteraceae\n\n89 petals\n\nDaisy and the family of asteraceae\n\nWhy these arrangements occur\n\nPlants are always not aware of this sequence. They only grow in the most efficient ways. In the scenario of the leaf arrangement and phyllotaxis, a number of the scenarios might be related to maximizing each leaf's space and also the standard amount of light that falls on each of them. A tiny advantage would also come to take control over numerous generations (Grist, n.d)\n\nFibonacci sequence and population growth in animals\n\nFibonacci sequence copies the population growth pattern of animals. Beginning from one distinct offspring or 1 animal. After a period of one year, it matures up and is capable of reproducing .In a single year; it is capable of reproducing one offspring. Now they become 2 animals. In one single year, the mother will reproduce one fresh offspring and the offspring that is given birth to in the preceding year matures up. They now become 3 animals. In another year, the mother and the now mature offspring will each reproduce one offspring and the offspring that comes from the last year will become mature. There will be 5 animals now. The trend will go on. (Fuzzy, 2010)\n\nHow Fibonacci number works\n\nDuring the year 1202, Fibonacci got interested in the reproduction of rabbits. He made an imaginary set of suitable conditions for rabbits to breed. He later posed the question, \"How many rabbit pairs will there be after a year?\" The ideal conditions that he set were as below;\n\n1. You start with a single male rabbit and a single female rabbit. The rabbits have presently been born.\n\n2. A single rabbit will attain sexual maturity after a month.\n\n3. The period of gestation of a rabbit is a month.\n\n4. Once a rabbit has attained sexual maturity, a female one will give birth on each month.\n\n5. A female rabbit will usually give birth to… [END OF PREVIEW]\n\n### Four Different Ordering Options:\n\n1.  Buy the full, 9-page paper:  \\$28.88\n\nor\n\n2.  Buy + remove from all search engines\n(Google, Yahoo, Bing) for 30 days:  \\$38.88\n\nor\n\n3.  Access all 175,000+ papers:  \\$41.97/mo\n\nor\n\n4.  Let us write a NEW paper for you!", null, "Most popular!\n\n#### Pre-Marital Sex View in Different Cultures Thesis…\n\nCite This Journal:\n\nAPA Format\n\nPre-Calc Trigonometry.  (2011, February 25).  Retrieved July 18, 2019, from https://www.essaytown.com/subjects/paper/pre-calc-trigonometry/4859\n\nMLA Format\n\n\"Pre-Calc Trigonometry.\"  25 February 2011.  Web.  18 July 2019. <https://www.essaytown.com/subjects/paper/pre-calc-trigonometry/4859>.\n\nChicago Format\n\n\"Pre-Calc Trigonometry.\"  Essaytown.com.  February 25, 2011.  Accessed July 18, 2019.\nhttps://www.essaytown.com/subjects/paper/pre-calc-trigonometry/4859." ]

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[ "Is 12253 a prime number? What are the divisors of 12253?\n\n## Parity of 12 253\n\n12 253 is an odd number, because it is not evenly divisible by 2.\n\nFind out more:\n\n## Is 12 253 a perfect square number?\n\nA number is a perfect square (or a square number) if its square root is an integer; that is to say, it is the product of an integer with itself. Here, the square root of 12 253 is about 110.693.\n\nThus, the square root of 12 253 is not an integer, and therefore 12 253 is not a square number.\n\nAnyway, 12 253 is a prime number, and a prime number cannot be a perfect square.\n\n## What is the square number of 12 253?\n\nThe square of a number (here 12 253) is the result of the product of this number (12 253) by itself (i.e., 12 253 × 12 253); the square of 12 253 is sometimes called \"raising 12 253 to the power 2\", or \"12 253 squared\".\n\nThe square of 12 253 is 150 136 009 because 12 253 × 12 253 = 12 2532 = 150 136 009.\n\nAs a consequence, 12 253 is the square root of 150 136 009.\n\n## Number of digits of 12 253\n\n12 253 is a number with 5 digits.\n\n## What are the multiples of 12 253?\n\nThe multiples of 12 253 are all integers evenly divisible by 12 253, that is all numbers such that the remainder of the division by 12 253 is zero. There are infinitely many multiples of 12 253. The smallest multiples of 12 253 are:\n\n## Numbers near 12 253\n\n### Nearest numbers from 12 253\n\nFind out whether some integer is a prime number" ]

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[ "BrainDen.com - Brain Teasers\n• 0\n\n# Sum of circumferences\n\n## Question\n\nWhat is the sum of circumferences of the infinite stack of circles in the pitcure?\n\nWhat is the sum of areas?", null, "Edited by Barcallica\n\n## Recommended Posts\n\n• 0\n\nDiameter D1 of the incircle is some fraction f of the triangle's height h.\n\nThe next circle has a diameter D2 that is the same fraction f of the remaining height h2 = h - D1.\n\nAnd so forth. Here, since h = 12 and D1 = 20/3, we have f = 5/9.\n\nThe successive diameters are:\n\nf h                          =                       h     =  6.666666667\n\nf (1 - h f)                =                  f - h f2     =  2.962962963\n\nf (1 - f + h f2 )         =             ff2 + h f3   =  1.316872428\n\nf (1 - f + f2 - h f3)   =    f  -  f2+ f3 - h f4    =  0.585276635\n\nand so on.\n\nThe corresponding areas are 34.90658504, 6.895127909, 1.362000575, 0.269037151 ...\n\nTaking the first 50 terms, the diameters sum to 12.00000000\n\nAnd the areas sum to 43.4989752.\n\nIf we note that successive diameters decrease by a factor of (1 - f)\n\nThen the areas of successive circles decrease by a factor of r = (1 - f)2.\n\nThen the areas sum to A1/(1 - r).\n\nThat's just pi (10/3)2/[1 - (4/9)2] = 34.9066/.80247 = 43.4989\n\n##### Share on other sites\n\n• 0\n\nCircumference is quite straight forward\n\nCircumference of each circle is 2.pi.r\nSo for all the circles it will be 24pi\nBecause the length of altitude is 12 (with sides 13 & 5)\n##### Share on other sites\n\n• 0\n\nFor area, had to check the web on how to calculate the radius of incircle\nSo, the first circle has radius 10/3\nSo the first circle covers 2/3 of the altitude length\nWithout going into details, I would say by symmetry that each circle covers the next 2/3rd\nSo the second circle would be of radius = 1/3(10 - 20/3) = 10/9\nratio of each consecutive circle radius = 1/3\nSo sum of areas = pi(r1² + r2² + ...)\n= pi.(10/3)² / (1 - 1/9) = (25/4)pi\n\n##### Share on other sites\n\n• 0\n\nFor area, had to check the web on how to calculate the radius of incircle\n\nSo, the first circle has radius 10/3\n\nSo the first circle covers 2/3 of the altitude length\n\nWithout going into details, I would say by symmetry that each circle covers the next 2/3rd\n\nSo the second circle would be of radius = 1/3(10 - 20/3) = 10/9\n\nratio of each consecutive circle radius = 1/3\n\nSo sum of areas = pi(r1² + r2² + ...)\n\n= pi.(10/3)² / (1 - 1/9) = (25/4)pi\n\nThe first circle has radius 10/3 but doesn't cover 2/3 of the altitude length\n\n## Join the conversation\n\nYou can post now and register later. If you have an account, sign in now to post with your account.", null, "×   Pasted as rich text.   Paste as plain text instead\n\nOnly 75 emoji are allowed.\n\n×   Your previous content has been restored.   Clear editor\n\n×   You cannot paste images directly. Upload or insert images from URL." ]

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[ "POSTS\n\n# Avoid using an empty list as a default argument to a function\n\nA very common error in Python is the use of an empty list as a default argument to a function. This is not the right way to do it and can cause unwanted behavior. See for example below:\n\n``````def append_to_list(element, list_to_append=[]):\nlist_to_append.append(element)\nreturn list_to_append``````\n``````>>> a = append_to_list(10)\n\n>>> b = append_to_list(20)\n[10, 20]``````\n\nThis is not the behavior we wanted! A new list is created once when the function is defined, and the same list is used in each successive call. Python’s default arguments are evaluated once when the function is defined, not each time the function is called.\n\nSee for example the code visualization for the above code:", null, "The solution and the standard way of doing it right is to pass to a Python function as a default argument None instead of [].\n\n``````def append_to_list(element, list_to_append=None):\nif list_to_append is None:\nlist_to_append = []\nlist_to_append.append(element)\nreturn list_to_append``````\n``````>>> a = append_to_list(10)\n\n>>> b = append_to_list(20)\n``````\n\nSee now how the correct code visualization looks like:", null, "Many who are new to Python tend to find this as a shortcoming of the language. On the contrary, it is not a bug but a valid language feature instead.\n\nThis behavior, that might shock newcomers to Python, can be easily explained if you think of functions in Python as objects. After all everything in Python is an object, right? An object is evaluated upon definition. The same happens to functions when the default parameters take their values upon the function definition with def. Upon calling an object multiple times the state of its member variables may change. Similarly, upon calling a function many times the state of its default parameters may change. This can be seen below, where we check the default arguments of a function after every successive call:\n\n``````def append_to_list(element, list_to_append=[]):\nlist_to_append.append(element)\nreturn list_to_append``````\n``````>>> append_to_list.__defaults__\n([],)\n>>> a = append_to_list(10)\n>>> a = append_to_list(10)\n>>> a = append_to_list(10)\n>>> append_to_list.__defaults__\n([10, 10, 10],)``````\n\nAs we showed we should generally avoid having mutable default arguments to functions.However, this could sometimes be used for our benefit. There is a technique called memoization, where we store previous values of computed values instead of using an expensive computation every time from the scratch. This technique proves quite useful especially when it is combined with recursion. See a classic example of it, the fibonacci numbers, where we use a mutable default argument to speed up the computation:\n\nFirst here is the classic recursive version of computing the Fibonacci numbers, which proves to be very costly.\n\n``````def fibonacci_classic(value):\nif value == 0 or value == 1:\nreturn value\nelse:\nreturn fibonacci_classic(value - 1) + fibonacci_classic(value - 2)``````\n``````>>> %time fibonacci_classic(40)\nCPU times: user 34.5 s, sys: 0 ns, total: 34.5 s``````\n\nIt took 34 seconds, which was a lot! The reason is that if we look at the recursion tree for computing the fibonacci number n, it has a depth of n as well. So that gives us an exponential time complexity of O(2^n). Actually it has a tight bound of O(1.61^n).\n\nInteresting fact: Fib(n) / Fib(n - 1) equals the golden ratio, which is around 1.61 as well!\n\nIn order to think how we could optimize this, take a look at the recursion tree below for computing the fifth fibonacci number.", null, "In order to compute f(5) we need to compute f(4) and f(3). But f(3) is already computed when we computed f(4). So, we would need to find a way to add “state” to our function calls so that we would remember already computed values.\n\nSo, what if we stored the already computed Fibonacci values in a dictionary that we could then pass it as default argument to a function? A dict is a mutable type in Python ensuring that it will be the same on every function call. This time we used mutability to our advantage!\n\n``````def fibonacci(value, memo={0: 0, 1: 1}):\ntry:\nfib = memo[value]\nexcept KeyError:\nfib = fibonacci(value - 1, memo) + fibonacci(value - 2, memo)\nmemo[value] = fib\nreturn fib``````\n``````>>> %time fibonacci(40)\nCPU times: user 32 µs, sys: 0 ns, total: 32 µs``````\n\nIt is a million times faster!\n\nActually if we omit the try..except and use “Look before you leap”(LBYL) instead of EAFP(“it’s easier to ask for forgiveness than permission”) we get even faster results:\n\n``````def fibonacci_lbyl(value, memo={0: 0, 1: 1}):\nif value in memo:\nreturn memo[value]\nelse:\nfib = fibonacci_lbyl(value - 1, memo) + fibonacci_lbyl(value - 2, memo)\nmemo[value] = fib\nreturn fib``````\n``````>>> %time fibonacci_lbyl(40)\nCPU times: user 17 µs, sys: 0 ns, total: 17 µs``````\n\nAnother option would be to use function attributes instead of default arguments:\n\n``````def fibonacci_attr(value):\nif value in fibonacci_attr.memo:\nreturn fibonacci.memo[value]\nelse:\nfib = fibonacci_attr(value - 1) + fibonacci_attr(value - 2)\nfibonacci_attr.memo[value] = fib\nreturn fib\n\nfibonacci_attr.memo = {0: 0, 1: 1}``````\n\nWe could also implement this by making our own memoized function decorator:\n\n``````def memoize(function_to_decorate, memo={}):\ndef memoized(value):\nif value in memo:\nreturn memo[value]\nelse:\nresult = function_to_decorate(value)\nmemo[value] = result\nreturn result\nreturn memoized\n\n@memoize\ndef fibonacci_memo(value):\nif value == 0:\nreturn 0\nelif value == 1:\nreturn 1\nelse:\nreturn fibonacci_memo(value - 1) + fibonacci_memo(value - 2)``````\n\nHere is an important remark. Would we have the same effect if we did not apply `@memoize` to `fibonacci_memo` and just called `memoize(fibonacci_memo)(value)` instead? Let’s see:\n\n``````>>> %time memoize(fibonacci_memo)(40)\nCPU times: user 33.2 s, sys: 3.99 ms, total: 33.2 s\n\n>>> %time memoize(fibonacci_memo)(40)\nCPU times: user 4 µs, sys: 0 ns, total: 4 µs\n\n>>> %time memoize(fibonacci_memo)(39)\nCPU times: user 20 s, sys: 6.02 ms, total: 20 s``````\n\nAs you can see it’s a big difference from using it as a decorator. The difference is that `fibonacci_memo` does not change. This means that memoization only affects the external call with argument 40, all subsequent calls are unmemoized, because they call the fibonacci_memo plain without memoization. But if you use it as a decorator then recursive calls are memorized and you’ll get speedup even with cold cache!\n\nLet’s try to apply our decorator to another recursive problem that would welcome a memoization speedup namely the computation of the factorial of a value.\n\n``````@memoize\ndef factorial(n):\nif n == 0 or n == 1:\nreturn 1\nelse:\nreturn factorial(n - 1) * n``````\n``````>>> factorial(5)\n5``````\n\nWhat went wrong?! The output should have been 120 and not 5. The problem is that the way we defined the memoize decorator we have a global cache. So by memoizing factorial we get results from fibonacci! This is another side effect of using a mutable default argument. If we would like to avoid that we should rewrite our memoize decorator. We should also use the functools.wraps in order not to lose some important information about the decorated function such as name, docstring, args.\n\n``````from functools import wraps\n\ndef memoize(function_to_decorate):\nmemo = {}\n@wraps(function_to_decorate)\ndef memoized(value):\nif value in memo:\nreturn memo[value]\nelse:\nresult = function_to_decorate(value)\nmemo[value] = result\nreturn result\nreturn memoized``````\n``````>>> fibonacci_memo(5)\n5\n>>> factorial(5)\n120``````\n\nWhat if we used the Python’s standard library implementation of memoization? For that reason there is functools.lru_cache decorator that we can use for this purpose. For you that are familiar with algorithms what we achieve this way is applying a dynamic programming technique to the original problem. We break it into subproblems which are computed only once and we store in cache the solution for them, in order to be used next time.\n\n``````from functools import lru_cache\n\n@lru_cache(maxsize=128)\ndef fibonacci_cache(value):\nif value == 0 or value == 1:\nreturn value\nelse:\nreturn fibonacci_cache(value - 1) + fibonacci_cache(value - 2)``````\n\nSo finally some time comparisons between these two techniques:\n\n``````>>> %time fibonacci(800)\nCPU times: user 2.24 ms, sys: 0 ns, total: 2.24 ms\n\n>>> %time fibonacci_lbyl(800)\nCPU times: user 778 µs, sys: 27 µs, total: 805 µs\n\n>>> %time fibonacci_memo(800)\nCPU times: user 1.55 ms, sys: 55 µs, total: 1.61 ms\n\n%time fibonacci_cache(800)\nCPU times: user 1.01 ms, sys: 36 µs, total: 1.05 ms``````\n\nBoth the lru_cache decorator and the fibonacci_lbyl proved to be two to three times faster compared to our memoization and our custom memoized decorator.\n\nMoral of the story: Do not reinvent the wheel and prefer Python standard’s library methods!\n\nHowever, there is one interesting fact. All of the methods but one tend to fail by reaching maximum resursion depth with results above 800. There is one method though that gives us results up to 2500 and very fast as well! Can you guess which is this method?\n\n``````>>> %time fibonacci_????(2500)\nCPU times: user 2.02 ms, sys: 41 µs, total: 2.06 ms\nWall time: 2.06 ms\n13170905167519496295227630871253164120666069649925071418877469367275308\n70405038425764503130123186407746570862185871925952766836352119119528156\n31558263246079038383460565488061265771846563256883924597824847305817942\n20707355531247163854508866405523922738567706722397971642643569276613083\n49671941673643205733343592701716715788255170679575500279186053316365583\n25918692735935102338729837168622286082741537144355375995365951412088276\n38081425933664024722513483600089155852152915049843716975238711995539357\n14056959634778700594751875``````\n\nTo summarize here is a Tweet from one of my favorite Twitter accounts on the web the one of Raymond Hettinger’s.\n\nQuiz:\n\nWhy don’t you want to test your knowledge and try this small quiz, that I created? Take Quiz!\n\nReferences:\n\nEffbot’s explanation" ]

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[ "# Unsupervised Learning\n\nIn unsupervised learning, our data does not have any labels. Unsupervised learning algorithms try to find some structure in the data.\n\nAn example is a clustering algorithm. We don't tell the algorithm in advance anything about the structure of the data; it discovers it on its own by figuring how to group them.\n\nSome other examples are dimensionality reduction, in which you try to reduce the dimensionality of the data representation, density estimation, in which you estimate the probability distribution of the data, $p(x)$, and feature extraction, in which you try to learn meaningful features automatically.\n\n## k-Nearest Neighbors (kNN)\n\nA very simple nonparametric classification algorithm in which you take the $k$ closest neighbors to a point (\"closest\" depends on the distance metric you choose) and each neighbor constitutes a \"vote\" for its label. Then you assign the point the label with the most votes.\n\nBecause this is essentially predicting an input's label based on similar instances, kNN is a case-based approach. The key with case-based approaches is how you define similarity - a common way is feature dot products:\n\n$$\\text{sim}(x, x') = x \\cdot x' = \\sum_i x_i x_i'$$\n\n$k$ can be chosen heuristically: generally you don't want it to be so high that the votes become noisy (in the extreme, if you have $n$ datapoints and set $k=n$, you will just choose the most common label in the dataset), and you want to chose it so that it is coprime with the number of classes (that is, they share no common divisors except for 1). This prevents ties.\n\nAlternatively, you can apply an optimization algorithm to choose $k$.\n\nSome distances that you can use include Euclidean distance, Manhattan distance (also known as the city block distance or the taxicab distance), Minkowski distance (a generalization of the Manhattan and Euclidean distances), and Mahalanobis distance.\n\nMinkowski-type distances assume that data is symmetric; that in all dimensions, distance is on the same scale. Mahalanobis distance, on the other hand, takes into account the standard deviation of each dimension.\n\nkNN can work quite quickly when implemented with something like a k-d tree.\n\nkNN and other case-based approaches are examples of nonparametric models. With nonparametric models, there is not a fixed set of parameters (which isn't to say that there are no parameters, though the name \"nonparametric\" would have you think otherwise). Rather, the complexity of the classifier increases with the data. Nonparametric models typically require a lot of data before they start to be competitive with parametric models.\n\n## Clustering\n\n### K-Means Clustering Algorithm\n\nFirst, randomly initialize $K$ points, called the cluster centroids.\n\nThen iterate:\n\n• Cluster assignment step: go through each data point and assign it to the closest of the $K$ centroids.\n• Move centroid step: move the centroids to the average of their points.\n\nCloseness is computed by some distance metric, e.g. euclidean.\n\nMore formally, there are two inputs:\n\n• $K$ - the number of clusters\n• The training set $\\\\{x^{(1)}, x^{(2)}, \\dots, x^{(m)}\\\\}$\n\nWhere $x^{(i)} \\in \\mathbb R^n$ (we drop the $x_0 = 1$ convention).\n\nRandomly initialize $K$ cluster centroids $\\mu_1, \\mu_2, \\dots, \\mu_K \\in \\mathbb R^n$.\n\nRepeat:\n\n• For $i=1$ to $m$\n• $c^{(i)} :=$ index (from 1 to $K$) of cluster centroid closest to $x^{(i)}$. That is, $c^{(i)} := \\text{min}_k ||x^{(i)} - \\mu_k||$.\n• For $k=1$ to $K$\n• $\\mu_k :=$ average (mean) of points assigned to cluster $k$\n\nIf you have an empty cluster, it is common to just eliminate it entirely.\n\nWe can notate the cluster centroid of the cluster to which example $x^{(i)}$ has been assigned as $\\mu_{c^{(i)}}$.\n\nIn K-means, the optimization objective is:\n\n\\begin{aligned} J(c^{(i)}, \\dots, c^{(m)}, \\mu_1, \\dots, \\mu_K) = \\frac{1}{m} \\sum^m_{i=1} ||x^{(i)} - \\mu_{c^{(i)}}||^2 \\\\ min_{c^{(i)}, \\dots, c^{(m)}, \\mu_1, \\dots, \\mu_K} J(c^{(i)}, \\dots, c^{(m)}, \\mu_1, \\dots, \\mu_K) \\end{aligned}\n\nThis cost function is sometimes called the distortion cost function or the distortion of the K-means algorithm.\n\nThe algorithm outlined above is minimizing the cost: the first step tries to minimize $c^{(i)}, \\dots, c^{(m)}$ and the second step tries to minimize $\\mu_1, \\dots, \\mu_K$.\n\nNote that we randomly initialize the centroids, so different runs of K-means could lead to (very) different clusterings.\n\nOne question is - what's the best way to initialize the initial centroids to avoid local minima of the cost function?\n\nFirst of all, you should halve $K < m$ (i.e. less than your training examples.)\n\nThen randomly pick $K$ training examples. Use these as your initialization points (i.e. set $\\mu_1, \\dots, \\mu_k$ to these $K$ examples).\n\nThen, to better avoid local optima, just rerun K-means several times (e.g. 50-1000 times) with new initializations of points. Keep track of the resulting cost function and then pick the clustering that gave the lowest cost.\n\nSo, how do you choose a good value for $K$?\n\nUnfortunately, there is no good way of doing this automatically. The most common way is to just choose it manually by looking at the output. If you plot out the data and look at it - even among people it is difficult to come to a consensus on how many clusters there are.\n\nOne method that some use is the Elbow method. In this approach, you vary $K$, run K-means, and compute the cost function for each value. If you plot out $K$ vs the cost functions, there may be a clear \"elbow\" in the graph and you pick the $K$ at the elbow. However, most of the time there isn't a clear elbow, so the method is not very effective.\n\nOne drawback of K-means (which many other clustering algorithms share) is that every point has a cluster assignment, which is to say K-means has no concept of \"noise\".\n\nFurthermore, K-means expects clusters to be globular, so it can't handle more exotic cluster shapes (such as moon-shaped clusters).\n\nThere are still many situations where K-means is quite useful, especially since it scales well to large datasets.\n\n### Hierarchical Agglomerative Clustering\n\nHierarchical agglomerative clustering (HAC) is a bottom-up clustering process which is fairly simple:\n\n1. Find two closest data points or clusters, merge into a cluster (and remove the original points or clusters which formed the new cluster)\n2. Repeat\n\nThis results in a hierarchy (e.g. a tree structure) describing how the data can be grouped into clusters and clusters of clusters. This structure can be visualized as a dendrogram:\n\nTwo things which must be specified for HAC are:\n\n• the distance metric: euclidean, cosine, etc\n• the merging approach - that is, how is the distance between two clusters measured?\n• complete linkage - use the distance between the two further points\n• average linkage - take the average distances of all pairs between the clusters\n• single linkage - take the distance between the two nearest points\n• (there are others as well)\n\nUnlike K-means, HAC is deterministic (since there are no randomly-initialized centroids) but it can be unstable: changing a few points or the presence of some outliers can vastly change the result. Scaling of variables/features can also affect clustering.\n\nHAC does not assume globular clusters, although it does not have a concept of noise.\n\n### Affinity Propagation\n\nIn affinity propagation, data points \"vote\" on their preferred \"exemplar\", which yields a set of exemplars as the initial cluster points. Then we just assign each point to the nearest exemplar.\n\nAffinity Propagation is one of the few clustering algorithms which supports non-metric dissimilarities (i.e. the dissimilarities do not need to be symmetric or obey the triangle inequality).\n\nLike K-means, affinity propagation also does not have a concept of noise and also assumes that clusters are globular. Unlike K-means, however, it is deterministic, and it does not scale very well (mostly because its support for non-metric dissimilarities precludes it from many optimizations that other algorithms can take advantage of).\n\n### Spectral Clustering\n\nWith spectral clustering, datapoints are clustered by affinity - that is, by nearby points - rather than by centroids (as is with K-Means). Using affinity instead of centroids, spectral clustering can identify clusters where K-Means fails to.\n\nIn spectral clustering, an affinity matrix is produced which, for a set of $n$ datapoints, is an $n \\times n$ matrix. Pairwise affinities are computed for the dataset. Affinity is some distance metric.\n\nThen, from this affinity matrix, PCA is used to extract the eigenvectors with the largest eigenvalues and the data is then projected to the new space defined by PCA. The data will be more clearly separated in this new representation such that conventional clustering methods (e.g. K-Means) can be applied.\n\nMore formally: spectral clustering generates a graph of the datapoints, with edges as the distances between the points. Then the Laplacian of the graph is produced:\n\nGiven the adjacency matrix $A$ and the degree matrix $D$ of a graph $G$ of $n$ vertices, the Laplacian matrix $L_{n \\times n}$ is simply $L = D - A$.\n\nAs a reminder:\n\n• the adjacency matrix $A$ is an $n \\times n$ matrix where the element $A_{i,j}$ is 1 if an edge exists between vertices $i$ and $j$ and 0 otherwise.\n• the degree matrix $D$ is an $n \\times n$ diagonal matrix where the element $D_{i,i}$ is the degree of vertex $i$.\n\nThen the eigenvectors of the Laplacian are computed to find an embedding of the graph into Euclidean space. Then some clustering algorithm (typically K-Means) is run on the data in this transformed space.\n\nSpectral clustering enhances clustering algorithms which assume globular clusters in that its space transformation of the data causes non-globular data to be globular in the transformed space. However, the graph transformation slows things down.\n\n### Mean Shift Clustering\n\nMean shift clustering extends KDE one step further: the data points iteratively hill-climb to the peak of nearest KDE surface.\n\nAs a parameter to the kernel density estimates, you need to specify a bandwidth - this will affect the KDEs and their peaks, and thus it will affect the clustering results. You do not, however, need to specify the number of clusters.\n\nBelow are some examples of different bandwidth results (source).", null, "", null, "You also need to make the choice of what kernel to use. Two commonly used kernels are:\n\n• Flat kernel:\n\n$$K(x) = \\begin{cases} 1 & \\text{if ||x|| \\leq 1} \\\\ 0 & \\text{otherwise} \\end{cases}$$\n\n• Gaussian kernel\n\nMean shift is slow ($O(N^2)$).\n\n### Non-Negative Matrix Factorization (NMF)\n\nNMF is a particular matrix factorization in which each element of $V$ is $\\geq 0$ (a non-negative constraint), and results in factor matrices $W$ and $H$ such that each of their elements are also $\\geq 0$.", null, "Non-negative matrix factorization (By Qwertyus, CC BY-SA 3.0, via Wikimedia Commons)\n\nEach column $v_i$ in $V$ can be calculated from $W$ and $H$ like so (where $h_i$ is a column in $H$):\n\n$$v_i = W h_i$$\n\nNMF can be used for clustering; it has a consequence of naturally clustering the columns of $V$.\n\nIt is also useful for reducing (i.e. compressing) the dimensionality of a dataset, in particular, it reduces it into a linear combination of bases.\n\nIf you add an orthogonality constraint, i.e. $HH^T = I$, if the value at $H_{kj} > 0$, then the $j$th column of $V$, that is, $v_j$, belongs to the cluster $k$.\n\n#### Matrix factorization\n\nLet $V$ be an $m \\times n$ matrix of rank $r$. Then there is an $m \\times r$ matrix $W$ and an $r \\times n$ matrix $H$ such that $V=WH$. So we can factorize (or decompose) $V$ into $W$ and $H$.\n\nThis matrix factorization can be seen as a form of compression (for low rank matrices, at least) - if we were to store $V$ on its own, we have to store $m \\times n$ elements, but if we store $W$ and $H$ separately, we only need to store $m \\times r + r \\times n$ elements, which will be smaller than $m \\times n$ for low rank matrices.\n\nNote that this kind of factorization can't be solved analytically, so it is usually approximated numerically (there are a variety of algorithms for doing so).\n\n### DBSCAN\n\nDBSCAN transforms the space according to density, then identifies for dense regions as clusters by using single linkage clustering. Sparse points are considered noise - not all points are forced to have cluster assignment.\n\nDBSCAN handles non-globular clusters well, provided they have consistent density - it has some trouble with variable density clusters (they may be split up into multiple clusters).\n\n### HDBSCAN\n\nHDBSCAN is an improvement upon DBSCAN which can handle variable density clusters, while preserving the scalability of DBSCAN. DBSCAN's epsilon parameter is replaced with a \"min cluster size\" parameter.\n\nHDBSCAN uses single-linkage clustering, and a concern with single-linkage clustering is that some errant point between two clusters may accidentally act as a bridge between them, such that they are identified as a single cluster. HDBSCAN avoids this by first transforming the space in such a way that sparse points (these potentially troublesome noise points) are pushed further away.\n\nTo do this, we first define a distance called the core distance, $\\text{core}_k(x)$, which is point $x$'s distance from its $k$th nearest neighbor.\n\nThen we define a new distance metric based on these core distances, called mutual reachability distance. The mutual reachability distance $d_{\\text{mreach}-k}$ between points $a$ and $b$ is the furthest of the following points: $\\text{core}_k(a), \\text{core}_k(b), d(a,b)$, where $d(a,b)$ is the regular distance metric between $a$ and $b$. More formally:\n\n$$d_{\\text{mreach}-k}(a, b) = \\max(\\text{core}_k(a), \\text{core}_k(b), d(a,b))$$\n\nFor example, if $k=5$:\n\nThen we can pick another point:\n\nAnd another point:\n\nSay we want to compute the mutual reachability distance between the blue $b$ and green $g$ points.\n\nFirst we can compute $d(b, g)$:\n\nWhich is larger than $\\text{core}_k(b)$, but both are smaller than $\\text{core}_k(g)$. So the mutual reachability distance between $b$ and $g$ is $\\text{core}k(g)$:\n\nOn the other hand, the mutual reachability distance between the red and green points is equal to $d(r, g)$ because that is larger than either of their core distances.\n\nWe build a distance matrix out of these mutual reachability distances; this is the transformed space. We can use this distance matrix to represent a graph of the points.\n\nWe want to construct a minimum spanning tree out of this graph.\n\nAs a reminder, a spanning tree of a graph is any subgraph which contains all vertices and is a tree (a tree is a graph where vertices are connected by only one path; i.e. it is a connected graph - all vertices are connected - but there are no cycles).\n\nThe weight of a tree is the sum of its edges' weights. A minimum spanning tree is a spanning tree with the least (or equal to least) weight.\n\nThe minimum spanning tree of this graph can be constructed using Prim's algorithm.\n\nFrom this spanning tree, we then want to create the cluster hierarchy. This can be accomplished by sorting edges from closest to furthest and iterating over them, creating a merged cluster for each edge.\n\n(A note from the original post which I don't understand yet: \"The only difficult part here is to identify the two clusters each edge will join together, but this is easy enough via a union-find data structure.\")\n\nGiven this hierarchy, we want a set of flat clusters. DBSCAN asks you to specify the number of clusters, but HDBSCAN can independently discover them. It does require, however, that you specify a minimum cluster size.\n\nIn the produced hierarchy, it is often the case that a cluster splits into one large subcluster and a few independent points. Other times, the cluster splits into two good-sized clusters. The minimum cluster size makes explicit what a \"good-sized\" cluster is.\n\nIf a cluster splits into clusters which are at or above the minimum cluster size, we consider them to be separate clusters. Otherwise, we don't split the cluster (we treat the other points as having \"fallen out of\" the parent cluster) and just keep the parent cluster intact. However, we keep track of which points have \"fallen out\" and at what distance that happened. This way we know at which distance cutoffs the cluster \"sheds\" points. We also keep track at what distances a cluster split into its children clusters.\n\nUsing this approach, we \"clean up\" the hierarchy.\n\nWe use the distances at which a cluster breaks up into subclusters to measure the persistence of a cluster. Formally, we think in terms of $\\lambda = \\frac{1}{\\text{distance}}$.\n\nWe define for each cluster a $\\lambda_{\\text{birth}}$, which is the distance at which this cluster's parent split to yield this cluster, and a $\\lambda_{\\text{death}}$, which is the distance at which this cluster itself split into subclusters (if it does eventually split into subclusters).\n\nThen, for each point $p$ within a cluster, we define $\\lambda_p$ to be when that point \"fell out\" of the cluster, which is either somewhere in between $\\lambda_{\\text{birth}}, \\lambda_{\\text{death}}$, or, if the point does not fall out of the cluster, it is just $\\lambda_{\\text{death}}$ (that is, it falls out when the cluster itself splits).\n\nThe stability of a cluster is simply:\n\n$$\\sum_{p \\in \\text{cluster}} (\\lambda_p - \\lambda_{\\text{birth}})$$\n\nThen we start with all the leaf nodes and select them as clusters. We move up the tree and sum the stabilities of each cluster's child clusters. Then:\n\n• If the sum of cluster's child stabilities greater than its own stability, then we set its stability to be the sum of its child stabilities.\n• If the sum of a cluster's child stabilities is less than its own stability, then we select the cluster and unselect its descendants.\n\nWhen we reach the root node, return the selected clusters. Points not in any of the selected clusters are considered noise.\n\nAs a bonus: each $\\lambda_p$ in the selected clusters can be treated as membership strength to the cluster if we normalize them.\n\n### CURE (Clustering Using Representatives)\n\nIf you are dealing with more data than can fit into memory, you may have issues clustering it.\n\nA flexible clustering algorithm (there are no restrictions about the shape of the clusters it can find) which can handle massive datasets is CURE.\n\nCURE uses Euclidean distance and generates a set of $k$ representative points for each clusters. It uses these points to represent clusters, therefore avoiding the need to store every datapoint in memory.\n\nCURE works in two passes.\n\nFor the first pass, a random sample of points from the dataset are chosen. The more samples the better, so ideally you choose as many samples as can fit into memory. Then you apply a conventional clustering algorithm, such as hierarchical clustering, to this sample. This creates an initial set of clusters to work with.\n\nFor each of these generated clusters, we pick $k$ representative points, such that these points are as dispersed as possible within the cluster.\n\nFor example, say $k=4$. For each cluster, pick a point at random, then pick the furthest point from that point (within the same cluster), then pick the furthest point (within the same cluster) from those two points, and repeat one more time to get the fourth representative point.\n\nThen copy each representative point and move that copy some fixed fraction (e.g. 0.2) closer to the cluster's centroid. These copied points are called \"synthetic points\" (we use them so we don't actually move the datapoints themselves). These synthetic points are the representatives we end up using for each cluster.\n\nFor the second pass, we then iterate over each point $p$ in the entire dataset. We assign $p$ to its closest cluster, which is the cluster that has the closest representative point to $p$." ]

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[ "# Title:Acceleration with a Ball Optimization Oracle\n\nAbstract: Consider an oracle which takes a point $x$ and returns the minimizer of a convex function $f$ in an $\\ell_2$ ball of radius $r$ around $x$. It is straightforward to show that roughly $r^{-1}\\log\\frac{1}{\\epsilon}$ calls to the oracle suffice to find an $\\epsilon$-approximate minimizer of $f$ in an $\\ell_2$ unit ball. Perhaps surprisingly, this is not optimal: we design an accelerated algorithm which attains an $\\epsilon$-approximate minimizer with roughly $r^{-2/3} \\log \\frac{1}{\\epsilon}$ oracle queries, and give a matching lower bound. Further, we implement ball optimization oracles for functions with locally stable Hessians using a variant of Newton's method. The resulting algorithm applies to a number of problems of practical and theoretical import, improving upon previous results for logistic and $\\ell_\\infty$ regression and achieving guarantees comparable to the state-of-the-art for $\\ell_p$ regression.\n Comments: 37 pages Subjects: Optimization and Control (math.OC); Data Structures and Algorithms (cs.DS) Cite as: arXiv:2003.08078 [math.OC] (or arXiv:2003.08078v1 [math.OC] for this version)\n\n## Submission history\n\nFrom: Kevin Tian [view email]\n[v1] Wed, 18 Mar 2020 07:39:45 UTC (42 KB)" ]

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[ "# What is 80 Percent of 5000? (In-Depth Explanation)\n\n80 percent of 5000 equals 4000. To get this answer, multiply 0.80 by 5000.\n\nYou may need to know this answer when solving a math problem that multiplies both 80% and 5000. Perhaps a product worth 5000 dollars, euros, or pounds is advertised as 80% off. Knowing the exact amount discounted from the original price of 5000 can help you make a more informed decision on whether or not it is a good deal.\n\nMaybe you’re looking for 80% of 5000 dollars, euros, Japanese yen, British pounds, Chinese yuan, pesos, or rupees. Whatever the case is, below, you will find an in-depth explanation that will help you solve this equation.\n\n## What is 80 percent of 5000?\n\n80 percent of 5000 is 4000. To figure this out, multiply 0.80 by 5000 to get 4000 as the answer.\n\nAnother way to find the answer to this equation includes taking 80/100 and multiplying it by 5000/1. When multiplying these two fractions together, you will get a final answer of 4000.\n\n## How do you find 80 percent of 5000?\n\nBy multiplying both 0.80 and 5000 together, you will find that 4000 is 80 percent of 5000. The 0.80 represents 80% and is the result of taking 80/100 or 80 divided by 100.\n\nThe easiest way to solve this equation is to divide the percent by 100 and multiply by the number. So divide 80 by 100 to get 0.80. From there, multiply the percent (now in decimal form) by 5000 to get 4000.\n\n## What is 80% off 5000 dollars?\n\nYou will pay \\$1000 for an item when you account for a discount of 80 percent off the original price of \\$5000. You will be receiving a \\$4000 discount.\n\n## What is 80 percent of 5000 dollars?\n\n80 percent of 5000 dollars is 4000 dollars. When solving this equation, we multiply 0.80 by 5000, the 0.80 standing for 80% and 5000 representing 5000 dollars.\n\nWhen referencing the dollar, people will likely be talking about the United States dollar (USD). However, sometimes other currencies are intended instead, like the Canadian dollar (CAD) or the Australian dollar (AUD).\n\nThe equation remains the same for calculating 80% of 5000 dollars for each of those respective currencies.\n\n## What is 80% off 5000 euros?\n\nWith an 80 percent discount, you will pay €1000 for any item with an original price of €5000. You will get a discount of €4000 off.\n\n## What is 80 percent of 5000 euros?\n\n80% of 5000 euros is 4000 euros. We use the same formula for calculating 80% of 5000 to get our answer of 4000 euros.\n\nThe euro is the currency used by some countries in the European Union, such as France, Germany, and Italy.\n\n## What is 80 percent of 5000 Japanese yen?\n\n80% of 5000 Japanese yen is 4000 yen. If you’re trying to solve 80% of 5000 Japanese yen, multiply 80% by 5000.\n\nWhen you multiply these two numbers together, you will find 4000 Japanese yen is your answer.\n\n## What is 80% off 5000 pounds?\n\nIf you get an 80 percent discount on a £5000 item, you will pay £1000. In total, you will end up receiving a £4000 discount.\n\n## What is 80 percent of 5000 British pounds?\n\nSimilar to other currencies, we multiply 80% by 5000 to get 4000 British pounds. In this equation, 0.80, 80/100, or 80% can each represent 80 percent. The 5000 in this equation stands for 5000 British pounds.\n\n4000 British pounds will be your answer once you multiply the two numbers together.\n\n## What is 80 percent of 5000 Chinese yuan?\n\n80% of 5000 Chinese yuan is 4000 yuan. The same formula that calculated 80% of 5000 of the other currencies can calculate 80% of the Chinese yuan.\n\nYou divide the percent by 100 and multiply it by the number. For this example, the equation divides 80% by 100 to get 0.80 (80 percent in decimal form). The percent is then multiplied by 5000 Chinese yuan resulting in an answer of 4000 Chinese yuan.\n\n## What is 80 percent of 5000 pesos?\n\n4000 pesos is the equivalent of 80% of 5000 pesos. When solving this equation, take the percent divided by 100 and multiply it by the number. In this case, 80% is divided by 100 and multiplied by 5000 pesos for an answer of 4000 pesos.\n\n## What is 80 percent of 5000 rupees?\n\nLike with other currencies, use the same equation and multiply 80% by 5000 rupees to get an answer of 4000 rupees. The answer will remain the same even if you write 80 percent as; 80%, 0.80, or 80/100.\n\nAfter you multiply 80% and 5000 rupees together, 4000 rupees is the final answer to the equation.\n\n## Conclusion\n\nYou might need to know the answer to 80% of 5000 when operating a business. New businesses get started every day, and people will often need to solve equations involving percentages like this.\n\nThose looking for the answer to 80% of 5000 might not even be business owners.\n\nMaybe you are at school or work and need to know the answer to this calculation. Whatever the case is, the answer is 4000.\n\nIf you enjoyed learning about what 80% of 5000 is, consider checking out our other articles below!\n\nRelated Posts" ]

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[ "3774 (number)\n\n3,774 (three thousand seven hundred seventy-four) is an even four-digits composite number following 3773 and preceding 3775. In scientific notation, it is written as 3.774 × 103. The sum of its digits is 21. It has a total of 4 prime factors and 16 positive divisors. There are 1,152 positive integers (up to 3774) that are relatively prime to 3774.\n\nBasic properties\n\n• Is Prime? No\n• Number parity Even\n• Number length 4\n• Sum of Digits 21\n• Digital Root 3\n\nName\n\nShort name 3 thousand 774 three thousand seven hundred seventy-four\n\nNotation\n\nScientific notation 3.774 × 103 3.774 × 103\n\nPrime Factorization of 3774\n\nPrime Factorization 2 × 3 × 17 × 37\n\nComposite number\nDistinct Factors Total Factors Radical ω(n) 4 Total number of distinct prime factors Ω(n) 4 Total number of prime factors rad(n) 3774 Product of the distinct prime numbers λ(n) 1 Returns the parity of Ω(n), such that λ(n) = (-1)Ω(n) μ(n) 1 Returns: 1, if n has an even number of prime factors (and is square free) −1, if n has an odd number of prime factors (and is square free) 0, if n has a squared prime factor Λ(n) 0 Returns log(p) if n is a power pk of any prime p (for any k >= 1), else returns 0\n\nThe prime factorization of 3,774 is 2 × 3 × 17 × 37. Since it has a total of 4 prime factors, 3,774 is a composite number.\n\nDivisors of 3774\n\n1, 2, 3, 6, 17, 34, 37, 51, 74, 102, 111, 222, 629, 1258, 1887, 3774\n\n16 divisors\n\n Even divisors 8 8 4 4\nTotal Divisors Sum of Divisors Aliquot Sum τ(n) 16 Total number of the positive divisors of n σ(n) 8208 Sum of all the positive divisors of n s(n) 4434 Sum of the proper positive divisors of n A(n) 513 Returns the sum of divisors (σ(n)) divided by the total number of divisors (τ(n)) G(n) 61.4329 Returns the nth root of the product of n divisors H(n) 7.35673 Returns the total number of divisors (τ(n)) divided by the sum of the reciprocal of each divisors\n\nThe number 3,774 can be divided by 16 positive divisors (out of which 8 are even, and 8 are odd). The sum of these divisors (counting 3,774) is 8,208, the average is 513.\n\nOther Arithmetic Functions (n = 3774)\n\n1 φ(n) n\nEuler Totient Carmichael Lambda Prime Pi φ(n) 1152 Total number of positive integers not greater than n that are coprime to n λ(n) 144 Smallest positive number such that aλ(n) ≡ 1 (mod n) for all a coprime to n π(n) ≈ 528 Total number of primes less than or equal to n r2(n) 0 The number of ways n can be represented as the sum of 2 squares\n\nThere are 1,152 positive integers (less than 3,774) that are coprime with 3,774. And there are approximately 528 prime numbers less than or equal to 3,774.\n\nDivisibility of 3774\n\n m n mod m 2 3 4 5 6 7 8 9 0 0 2 4 0 1 6 3\n\nThe number 3,774 is divisible by 2, 3 and 6.\n\n• Arithmetic\n• Abundant\n\n• Polite\n\n• Square Free\n\nBase conversion (3774)\n\nBase System Value\n2 Binary 111010111110\n3 Ternary 12011210\n4 Quaternary 322332\n5 Quinary 110044\n6 Senary 25250\n8 Octal 7276\n10 Decimal 3774\n12 Duodecimal 2226\n20 Vigesimal 98e\n36 Base36 2wu\n\nBasic calculations (n = 3774)\n\nMultiplication\n\nn×y\n n×2 7548 11322 15096 18870\n\nDivision\n\nn÷y\n n÷2 1887 1258 943.5 754.8\n\nExponentiation\n\nny\n n2 14243076 53753368824 202865213941776 765613317416262624\n\nNth Root\n\ny√n\n 2√n 61.4329 15.5692 7.83791 5.19231\n\n3774 as geometric shapes\n\nCircle\n\n Diameter 7548 23712.7 4.47459e+07\n\nSphere\n\n Volume 2.25162e+11 1.78984e+08 23712.7\n\nSquare\n\nLength = n\n Perimeter 15096 1.42431e+07 5337.24\n\nCube\n\nLength = n\n Surface area 8.54585e+07 5.37534e+10 6536.76\n\nEquilateral Triangle\n\nLength = n\n Perimeter 11322 6.16743e+06 3268.38\n\nTriangular Pyramid\n\nLength = n\n Surface area 2.46697e+07 6.3349e+09 3081.46\n\nCryptographic Hash Functions\n\nmd5 73f104c9fba50050eea11d9d075247cc 4b51fbf06972de1dc64623085d8d09cb76758a18 45d823d25b097fa8b7dfd0abaf70c0dcd896ded3720f4e1d3196f6c39308cd8d 5902e92d758081d374db4c2df607ffa95f637c9fb56f9c126fdb6b68ff9b757a00e4ace4e456e007cd3cdf45a9b8efd70ca02f0d3475fc6a0ef23cdd21736242 920daeeedefb1c0b001be810847c1199c8c4f744" ]

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[ "", null, "By G. Viennot\n\nISBN-10: 0387090908\n\nISBN-13: 9780387090900\n\nISBN-10: 3540090908\n\nISBN-13: 9783540090908\n\nRead Online or Download Algèbres de Lie libres et monoïdes libres : Bases des algèbres de Lie libres et factorisations des monoïdes libres PDF\n\nBest mathematics books\n\nAdditional info for Algèbres de Lie libres et monoïdes libres : Bases des algèbres de Lie libres et factorisations des monoïdes libres\n\nExample text\n\n4. If (X, d) is a locally compact separable metric space, then the vector space of all Lipschitz functions with compact support is dense in Cc (X) and in C0 (X) (with respect to the topology generated by the uniform norm, of course). Consequently, Lip(X) is dense in C0 (X). Proof. 3. Since the sum or the product of two Lipschitz functions is Lipschitz, it follows that all the functions in A have compact supports and are Lipschitz. Since A is dense in C0 (X), it follows that the assertions of the corollary are true.\n\nL) For every x ∈ Γ and every Banach limit L let εx : C0 (X) → R be defined (L) by εx (f ) = L ( f, T n δx )n∈N∪{0} for every f ∈ C0 (X). It is easy to see that (L) (L) εx is a positive linear functional, so εx is also continuous. Accordingly, we may (L) and do think of εx as an element of M(X). 1. If x ∈ Γ and L is a Banach limit, then εx measure. is a T -invariant Proof. Let φ : Cb (X) → R be defined by φ(f ) = L ( f, T n δx )n∈N∪{0} for every f ∈ Cb (X). Clearly, φ is a positive linear functional, φ (1X ) = 1, and the restriction (L) of φ to C0 (X) is εx .\n\n2. Invariant Probabilities 21 The next theorem summarizes various results of Chapter 4 of Revuz’s book . 6. Let (S, T ) be a Markov–Feller pair defined on a locally compact separable metric space (X, d), and let µ ∈ M(X) be a T -invariant probability. e. to a µ-integrable k=0 function g, and f, µ = n∈N g dµ. Proof. It is well-known that the Radon–Nikodym theorem implies that the set of all the elements of M(X) that are absolutely continuous with respect to the measure µ is a Banach subspace of M(X) that is isometric to L1 (X, B(X), µ)." ]

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[ "# optimization: calling fmincon in Simulink (embedded block)\n\n96 views (last 30 days)\nMaarten on 27 Feb 2013\nCommented: manish kumar on 13 Jul 2021\nHello,\nI am going to implement a model predictive controller in Simulink. Therefore I need to solve a constrained optimization problem, thus finding the vector that mimimises a certain multi-variable function, subject to constraints. (I do know there is a MPC block in Simulink, but I need to implement it myself as part of a bigger Simulink model).\nIn Matlab this is possible using fmincon. So I tried using an embedded Matlab block in Simulink where fmincon was called, but this does not work. I also tried \"coder.extrinsic('fmincon')\" code but that did not work.\nCan anybody tell me if it is possible to use fmincon in a SIMULINK model (thus not working in a .m Matlab file) or if there are alternatives? Or do I have to try to implement the algebra behind this function (quadratic programming etc.) in Simulink myself?\nAll help appreciated! Maarten\nKhalid on 25 Nov 2014\nI know this is from a while ago, but I would like to +1 this question. I have a very similar problem to Maarten.\nLike Maarten I tried using coder.extrinsic('fmincon') from within the MATLAB Function block within simulink; but it is then not possible to pass handles to an extrinsic function (which is needed to be able to run fmincon).\nIs there another/better way to perform constrained optimisation within a Simulink model (i.e. where it needs to be run once per time-step). I am familiar with using constrained optimisation of Simulink model parameters (i.e. where Simulink is called once per objective function evaluation); but here we seek to run the Simulink model only once, but run the optimisation once per time-step.\nThe approach Maarten and I have taken seems like the 'obvious' one but doesn't work. Any pointers gratefully received.\n- Khalid.\n\nMatteo Ragni on 29 May 2018\nI will post the same answer I gave on Stack Overflow on the same topic\nYou need to call the optimal problem function with coder.extrinsic.\nInstead of using coder.extrinsic on the fmincon function, I usually write a wrapper for the optimization problem that I have to solve as a .m file function for Matlab (namely opt_problem) and declare coder.extrinsic('opt_problem') in the simulink Matlab function. I'll give you a simple example:\nConsider this Simulink \"model\", in which at each integration step I want to solve a linear regression problem on some generated data. The optimization problem is something in the form:\nminimize (y - m x - q)²\nsubject to 0 ≤ m ≤ 1\n0 q ≤ 1\nThe scheme is really simple, bet the regressor calls fmincon:", null, "Let's see inside the regressor:\nfunction [m, q] = regressor(xs, ys, mic, qic)\ncoder.extrinsic('opt_problem'); % <- Informing the Coder\nm = 0;\nq = 0;\n[m, q] = opt_problem(xs, ys, mic, qic); % <- Optimal problem wrapper call\nend\nthis function is only a wrapper for an external function `opt_problem`. Let'see it (it has two functions inside):\nfunction [m, q] = opt_problem(xs, ys, mic, qic)\nfmincon_target = @(mq)(target(mq, xs, ys));\nmq = fmincon(fmincon_target, [mic; qic], [], [], [], [], [0; 0], [1; 1]);\nm = mq(1);\nq = mq(2);\nend\nfunction r = target(mq, xs, ys)\nr = norm(ys - xs.*mq(1) - mq(2));\nend\nand that's all. As you can see in the picture, the scheme runs and the solution are the m, q parameters (in the two displays) that minimize the target function while respecting the constraints (m = 1.2 → m_opt = 1).\n##### 2 CommentsShowHide 1 older comment\nmanish kumar on 13 Jul 2021\n\nGiovanni Palmieri on 20 Jun 2019\nHi Matteo, what is inside the Linear Trasformation with Noise block??" ]

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[ "# What is 5+5×5+5\n\n5+5×5+5\n\n5+(5×5)+5\n\n5+25+5\n\n=35\n\nmultiplication comes before division, adding or subtracting.\n\nPEMDAS\n\nWhat is 5+5×5+5" ]

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[ "# Multiplying 2-Digit Numbers by Multiples of Ten\n\n4 teachers like this lesson\nPrint Lesson\n\n## Objective\n\nSWBAT use grids and a short cut to multiply 2-digit numbers and multiples of 10.\n\n#### Big Idea\n\nOnce students understand patterns when multiplying by 10, they learn a short cut that can help multiply 2-digit numbers by multiples of 10.\n\n## Opener\n\n5 minutes\n\nI tell the students, you have learned to estimate to find the product of numbers.  You have also learned the distributive property.  In today's lesson, you use both strategies to multiply a 2-digit number by multiples of 10.  I ask the students, \"Who can name a multiple of 10?\"  All of the students' hands go up.  I tell the students to call out the multiples of 10.  The students yell 10, 20, 30, 40, etc.\n\n## Direct Instruction\n\n15 minutes\n\nTo begin the lesson, I call all of my students to the carpet. This gives all of them an opportunity to sit close and absorb all of the information from the lesson.  We begin the lesson with a review.\n\nReview:\n1.  When you multiply by multiples of 10, you multiply the basic facts and then add 1 zero.  For example, 10 x 5 = 50.  We know that 10 is repeating 5 times.  Based upon what we have learned about addition, we know that there will be a 0 in the ones place when we add 10 together 5 times.  Using a shortcut, we can just multiply the basic facts and add the one zero.\n\n2.  žWith the distributive property, you make two simpler multiplication problems by breaking apart one of the factors.  When you break apart one of the factors into two numbers, the two numbers must add back to equal that number.  (ex. 7 x 8= (6 x 8) + (1 x 8)\nAfter the review, we go directly into guided practice.  With my guided practice, I like for my students to be interactive with me.  In our classroom environment, they know that they can just jump in to make comments or ask questions.\nFor guided practice, we will use a real-world scenario.  This way the students can relate to what is going on in the problem and it feels relevant to them.\nScenario:\n\nTonya counted 15 bikes on her street as she roller skated.  As she passed the bikes, she noticed that they all had 40 spokes on their wheels.  How many spokes are there in all?\n\nIn this problem, they are asking us to find the total number of spokes on the bikes.  This is a multiplication problem.  Notice that one of the numbers is a multiple of 10.\nžWe can use the distributive property when multiplying by a multiple of 10.\nž15 x 40 =\nžWe need to break apart the 15 since it is not a multiple of 10.  It is easy to multiply by multiples of 10.\nWhat 2 numbers can we break apart the 15 into?\n\nCall on a few students to explain how to break about the 15.\n\nWe can break apart the 15 in many ways.  Let’s break apart the 15 into 10 + 5.\n\n(10 x 40) + (5 x 40)=\n\n400 + 200=600\n\nUnderline the basic facts which is 1 x 4=4.\n\nNext, count your zeros and add them behind the 4 which gives you 400.  For the next problem, the basic facts are 5 x 4= 20.  Add one zero behind the 20, which gives us 200.  Last, add 400 + 200=600.\n\nBecause I want the students to use area models, I show a quick video before the students go to their seats.\n\nPossible Misconception(s):\n\n1.  Not adding the zero when using the short cut to multiply by tens\n\nIn order to get students to understand that when they multiply by tens that there is always a zero in the ones place, I will use examples.  In the problem above, the students multiply 10 x 40.  I would explain to students that 1 x 4 = 4 but they are not multiplying by 4, they are multiplying by 40 - which is 4 - tens.  1 x 40 = 40 so 10 x 40 would be 40 ten times which equals 400. They can also draw a visual model to represent the problem.\n\n## Group or Partner Activity\n\n20 minutes\n\nTo give the students practice with the skill and to allow them to work with their classmates, I will allow the students to work in pairs.  I let my students work in pairs because it is small enough for all of the stdents to be heard.  The activity the students will work on will require the  students to make a model of the distributive property (4.NBT.5 Illustrate and explain the calculation by using equations, rectangular arrays, and/or area models.)  The pair can choose a problem to solve from a list of 2 digit multiplication problems.  The activity is as follows:\n\n1.  The students will work together to break apart a problem into 2 simpler problems. (MP2 -Mathematically proficient students make sense of quantities and their relationships in problem situations.)\n\n2. Cut out an area model of each simpler problem with grid paper (MP4)\n\n3.  Glue the model on the construction paper\n\n4.  Find the product\n\n5. Explain how the model helped find the product.\n\n## Independent Activity\n\n10 minutes\n\nTo let me see how well each student understood this skill, the students will complete an independent assignment.  The students will need paper and pencil to solve the problem.  Have the problem displayed on the Smart board.\n\nProblem:\n\n20 people gave money to a charity.  They each gave \\$25.  How much did they give in all?\n\nDirections:\n\n1.  Use the distributive property to solve." ]

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[ "", null, "Power and units", null, "Force = a push or Pull                                   measured in Newtons Work = Force over a Distance                     measured in  Joules     ( other wise called energy or Newton meters) Power = the rate at which Energy is used     measured in Watts Power = V x I example: A 240 V motor draws 10A under load, what is the power rating? P= V x I = 2400 watts or 2.4Kw (kilowatts) In electrical terms power is rated in watts and some motors are still rated in horse power (HP) 746 watts = 1 Horse Power unit example: the above motor would be rated as 2400w/750w/hp = 3.2 HP 1000 watts = 1 Kilowatt   kW 1000 000 w = 1megawatt MW .001 watts  = 1milliwatt    mW", null, "POWER CALCULATOR simulator 1000 volts = 1 kilovolt (kV) not likely to encounter this in electronics .001 amps = 1 milliamp (mA) a very common term in electronics" ]

[ null, "http://michaelsharris.com/electronics/images/backgreen.gif", null, "http://michaelsharris.com/electronics/images/powerhorse.jpg", null, "http://michaelsharris.com/electronics/images/powerformula.gif", null ]

[ "SmartSellTM - The New Way to Sell Online\n\nWe won't be beaten by anyone. Guaranteed\n\nApplied Data Mining for Business and Industry\nBy\n\nRating\n\nProduct Description\nProduct Details\n\n1 Introduction. Part I Methodology. 2 Organisation of the data. 2.1 Statistical units and statistical variables. 2.2 Data matrices and their transformations. 2.3 Complex data structures. 2.4 Summary. 3 Summary statistics. 3.1 Univariate exploratory analysis. 3.1.1 Measures of location. 3.1.2 Measures of variability. 3.1.3 Measures of heterogeneity. 3.1.4 Measures of concentration. 3.1.5 Measures of asymmetry. 3.1.6 Measures of kurtosis. 3.2 Bivariate exploratory analysis of quantitative data. 3.3 Multivariate exploratory analysis of quantitative data. 3.4 Multivariate exploratory analysis of qualitative data. 3.4.1 Independence and association. 3.4.2 Distance measures. 3.4.3 Dependency measures. 3.4.4 Model-based measures. 3.5 Reduction of dimensionality. 3.5.1 Interpretation of the principal components. 3.6 Further reading. 4 Model specification. 4.1 Measures of distance. 4.1.1 Euclidean distance. 4.1.2 Similarity measures. 4.1.3 Multidimensional scaling. 4.2 Cluster analysis. 4.2.1 Hierarchical methods. 4.2.2 Evaluation of hierarchical methods. 4.2.3 Non-hierarchical methods. 4.3 Linear regression. 4.3.1 Bivariate linear regression. 4.3.2 Properties of the residuals. 4.3.3 Goodness of fit. 4.3.4 Multiple linear regression. 4.4 Logistic regression. 4.4.1 Interpretation of logistic regression. 4.4.2 Discriminant analysis. 4.5 Tree models. 4.5.1 Division criteria. 4.5.2 Pruning. 4.6 Neural networks. 4.6.1 Architecture of a neural network. 4.6.2 The multilayer perceptron. 4.6.3 Kohonen networks. 4.7 Nearest-neighbour models. 4.8 Local models. 4.8.1 Association rules. 4.8.2 Retrieval by content. 4.9 Uncertainty measures and inference. 4.9.1 Probability. 4.9.2 Statistical models. 4.9.3 Statistical inference. 4.10 Non-parametric modelling. 4.11 The normal linear model. 4.11.1 Main inferential results. 4.12 Generalised linear models. 4.12.1 The exponential family. 4.12.2 Definition of generalised linear models. 4.12.3 The logistic regression model. 4.13 Log-linear models. 4.13.1 Construction of a log-linear model. 4.13.2 Interpretation of a log-linear model. 4.13.3 Graphical log-linear models. 4.13.4 Log-linear model comparison. 4.14 Graphical models. 4.14.1 Symmetric graphical models. 4.14.2 Recursive graphical models. 4.14.3 Graphical models and neural networks. 4.15 Survival analysis models. 4.16 Further reading. 5 Model evaluation. 5.1 Criteria based on statistical tests. 5.1.1 Distance between statistical models. 5.1.2 Discrepancy of a statistical model. 5.1.3 Kullback?Leibler discrepancy. 5.2 Criteria based on scoring functions. 5.3 Bayesian criteria. 5.4 Computational criteria. 5.5 Criteria based on loss functions. 5.6 Further reading. Part II Business case studies. 6 Describing website visitors. 6.1 Objectives of the analysis. 6.2 Description of the data. 6.3 Exploratory analysis. 6.4 Model building. 6.4.1 Cluster analysis. 6.4.2 Kohonen networks. 6.5 Model comparison. 6.6 Summary report. 7 Market basket analysis. 7.1 Objectives of the analysis. 7.2 Description of the data. 7.3 Exploratory data analysis. 7.4 Model building. 7.4.1 Log-linear models. 7.4.2 Association rules. 7.5 Model comparison. 7.6 Summary report. 8 Describing customer satisfaction. 8.1 Objectives of the analysis. 8.2 Description of the data. 8.3 Exploratory data analysis. 8.4 Model building. 8.5 Summary. 9 Predicting credit risk of small businesses. 9.1 Objectives of the analysis. 9.2 Description of the data. 9.3 Exploratory data analysis. 9.4 Model building. 9.5 Model comparison. 9.6 Summary report. 10 Predicting e-learning student performance. 10.1 Objectives of the analysis. 10.2 Description of the data. 10.3 Exploratory data analysis. 10.4 Model specification. 10.5 Model comparison. 10.6 Summary report. 11 Predicting customer lifetime value. 11.1 Objectives of the analysis. 11.2 Description of the data. 11.3 Exploratory data analysis. 11.4 Model specification. 11.5 Model comparison. 11.6 Summary report. 12 Operational risk management. 12.1 Context and objectives of the analysis. 12.2 Exploratory data analysis. 12.3 Model building. 12.4 Model comparison. 12.5 Summary conclusions. References. Index.\n\n#### About the Author\n\nPaolo Giudici - Department of Economics and Quantitative Methods, University of Pavia , A lecturer in data mining, business statistics, data analysis and risk management, Professor Giudici is also the director of the data mining laboratory. He is the author of around 80 publications, and the coordinator of 2 national research grants on data mining, and local coordinator of a European integrated project on the topic. He was the sole author of the first edition of this book, which has been translated into both Italian and Chinese. He is also one of the Editors of Wiley's Series in Computational Statistics. Silvia Figini , Ms Figini has worked for 2 years for the Competence centre for data mining analysis and business intelligence at SAS Milan. She is currently completing a PhD in statistics, and already has a collection of publications to her name\n\n#### Reviews\n\n?If I had to recommend a good introduction to data mining, I would choose this one.? (Stat Papers, 2011)", null, "", null, "" ]

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[ "Spiked Math Games  // Math Fail Blog  // Gauss Facts  // Spiked Math Comics", null, "Whoever solves every question correctly first (with proof of correctness) gets a free sticker!!!!", null, "", null, "", null, "", null, "", null, "1. simon newcomb\n\n2. 8\n\n3. alice & bob\n\ni give up\nanswer to #2 is 7 [proof: 1/10 chance]\nand for the pascal's triangle one... it is 596904 isn't it?\ni believe so... since 154!/(3!*(154-3)!) should be equal to that term... due to, you know, binomial theorem or whatever you call that. :P\n\nmy bad... the counting on the lower term starts at 0... its actually 154!/(2!*(154-2)! which i 270397512\naccording to me.\n\nsowwy\n\n4. either 11781 or 11935 depending if you start counting your rows at 0 or 1.\n\n5. Tralse\n\nah, yeah it's 7 for #2. #4 seems right. And #5 is True if N=1 :P\n\nwhat if n=0 but p=0 also? or if p=0? n doesn't have to be 1 always...\n\nanyways, i'm doing my math lab for my exam... its saturday night and i should be out with friends... where did i go wrong?\n\nI can't believe people still puzzle over P = NP. I scribbled the proof for this on a napkin, it's so obvious. I'm gonna proof this for you right here.\n\nSo first we hav- Oh wait, someone at the door, be right back.\n\n*waiting*\n\n#5. false (I have discovered a truly marvelous proof that P!=NP, but this comment box is too small to contain it.)\n\nwhat a shame. did you at least prove it for all non-primes for us?... i wish that made sense>.\n\n#4 is 11781. Xn= n(n+1)/2\n\nIs the green M referring to anything?\n\nM for Math. In the classic version of Trivial Pursuit, green was SN for Science & Nature.\n\n1. Simon Newcomb: http://en.wikipedia.org/wiki/Benford%27s_law\n2. 3: http://www.numberworld.org/digits/GoldenRatio/\n3. Alice & Bob: http://en.wikipedia.org/wiki/Alice_and_Bob\n4. 270397512: 154!/(2!(154-2)!)\n5. Let's assume by contradiction that P=NP...\n\nXD i like the number of people actually trying to get that sticker.\nespecially since, due to that last one there, the first one to solve all of these correctly with proof should get US\\$1,000,000 from the millenium institute\n\nI N C E P T I O N\n\n#5 is easy: Yes.\n\nIf I were giving out the sticker, I would hand it to you sir.\n\n5. False. Proof: P=NP => P=NP (cancellation) => =N, which is clearly false.\n\nIs the sixth category imaginary?\n\n1) Simon Newcomb http://en.wikipedia.org/wiki/Simon_Newcomb#Benford.27s_law\n2) 3 http://www.numberworld.org/digits/GoldenRatio/\n3) Alice and Bob (my boyfriend works in Cryptography, but ... http://en.wikipedia.org/wiki/Alice_and_Bob)\n4) 4656 http://mathforum.org/dr.cgi/pascal.cgi?rows=154\n5) False because it is yet unknown (so saying that P=NP is as false as saying P/=NP) http://en.wikipedia.org/wiki/P_versus_NP_problem#Reasons_to_believe_P_.E2.89.A0_NP\n\n#4: The webpage only shows up to the 98th row.\n\nI believe #3 is Bob and Alice.\n\nP=NP when P=1. :)\n\nerr, when N=1. Man, I need to get my head checked.\n\nor when P is 0\n\n@5: Quite a hard choice, free spikedmath sticker or fields medal. I couldn't decide.\n\n\"P\" is usually used in logic, so this could be seen as p=~p, although I don't know why he wouldn't have used that in the first place. Anyway I believe it is therefore false.\n\nLet P = 8\nLet N = 1\nP = 8\nNP = 8*1 = 8.\nTherefore, P = NP.\n\n5) Ah, but it's also exactly as true as saying P/=NP, and if you're using formal methods in computer science, everything that is unknown is true.\n\nGod dammit, that was supposed to be a reply to KK.\n\n\"if you're using formal methods in computer science, everything that is unknown is true.\"\n\nUm, what?\n\nThose of you trying to show P=NP (or not) by substituting values for P or N may have your tongues firmly in your cheeks or may want to look at http://en.wikipedia.org/wiki/P%3DNP\nThe P is the class of problems solvable in polynomial time (thus P) and the NP is the class of problems verifiable in polynomial time (it would have been VP but this office is already held by Joe Biden)\n\n5) Not provable with our current set of axioms (i.e. we can append both P=NP and P/=NP to our set of axioms and not force a contradiction)\n:P\n\nCan you prove that it's not provable?\n\n5. Yes\n(I think you need an XOR in there...)\n\n(Note: You must have javascript enabled to leave comments, otherwise you will get a comment submission error.)\nIf you make a mistake or the comment doesn't show up properly, email me and I'll gladly fix it :-).", null, "", null, "", null, "", null, "", null, "", null, "", null, "", null, "", null, "", null, "Welcome to Spiked Math!\n\nHello my fellow math geeks. My name is Mike and I am the creator of Spiked Math Comics, a math comic dedicated to humor, educate and entertain the geek in you. Beware though, there might be some math involved :D\n\nNew to Spiked Math?\nView the top comics.\n\nNew Feature: Browse the archives in quick view! Choose from a black, white or grey background.\n\nOther Math Comics" ]

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[ "# Having trouble solving a limit\n\n$$\\lim _{x\\to \\infty }\\left(\\frac{2^{\\frac{1}{x}}}{\\left(2^{\\frac{1}{x}}-1\\right)x}\\right)$$\n\nI've tried a couple of things but I can't get to the right result.. according to online calculators the limit equals $$\\frac{1}{ln\\left(2\\right)}$$ but I keep getting 0. Is there a a trick to solve it?\n\n• Take the limits for the numerator and the denominator separately. Apr 12 at 11:32\n• expand numerator and denumerator with $2^x$ Apr 12 at 11:38\n• Thanks for the help guys! :) Apr 12 at 11:45\n\n$$\\lim _{x\\to \\infty }\\left(\\frac{2^{\\frac{1}{x}}}{\\left(2^{\\frac{1}{x}}-1\\right)x}\\right)$$\ntaking $$1/x$$=$$t$$ as $$x$$ tends to $$\\infty$$, $$t$$ tends to $$0$$\nthen apply L-hopital rule or use $$lim_{t\\to0}\\frac{2^t-1}{t}$$ = $$ln2$$" ]

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[ "# Number 438\n\n### Properties of number 438\n\nCross Sum:\nFactorization:\n2 * 3 * 73\nDivisors:\n1, 2, 3, 6, 73, 146, 219, 438\nCount of divisors:\nSum of divisors:\nPrime number?\nNo\nFibonacci number?\nNo\nBell Number?\nNo\nCatalan Number?\nNo\nBase 2 (Binary):\nBase 3 (Ternary):\nBase 4 (Quaternary):\nBase 5 (Quintal):\nBase 8 (Octal):\n1b6\nBase 32:\ndm\nsin(438)\n-0.96837198288571\ncos(438)\n-0.24951092713947\ntan(438)\n3.8810804560252\nln(438)\n6.0822189103764\nlg(438)\n2.6414741105041\nsqrt(438)\n20.928449536456\nSquare(438)\n\n### Number Look Up\n\nLook Up\n\n438 (four hundred thirty-eight) is a very amazing number. The cross sum of 438 is 15. If you factorisate the number 438 you will get these result 2 * 3 * 73. The figure 438 has 8 divisors ( 1, 2, 3, 6, 73, 146, 219, 438 ) whith a sum of 888. 438 is not a prime number. 438 is not a fibonacci number. The figure 438 is not a Bell Number. The figure 438 is not a Catalan Number. The convertion of 438 to base 2 (Binary) is 110110110. The convertion of 438 to base 3 (Ternary) is 121020. The convertion of 438 to base 4 (Quaternary) is 12312. The convertion of 438 to base 5 (Quintal) is 3223. The convertion of 438 to base 8 (Octal) is 666. The convertion of 438 to base 16 (Hexadecimal) is 1b6. The convertion of 438 to base 32 is dm. The sine of 438 is -0.96837198288571. The cosine of 438 is -0.24951092713947. The tangent of the number 438 is 3.8810804560252. The root of 438 is 20.928449536456.\nIf you square 438 you will get the following result 191844. The natural logarithm of 438 is 6.0822189103764 and the decimal logarithm is 2.6414741105041. You should now know that 438 is very great figure!" ]

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[ "Home » Worksheets » Age 11-13 Math Worksheets\n\n# Age 11-13 Math Worksheets\n\n## In this section, you can view all of our math worksheets and resources that are suitable for 11 to 13-year-olds.\n\nWe add dozens of new worksheets and materials for math teachers and homeschool parents every month. Below are the latest age 11-13 worksheets added to the site.\n\nPopular\n\nPopular\n\nPopular\n\nPopular\n\n1 2 5\n\n## 11-13 Years Of Age Math Learning Objectives & Standards:\n\n• Learners aged 11-13 dwell with rational and irrational numbers as well as algebraic rational expressions. Here, they should be able to differentiate rational numbers from irrational numbers. In addition, they must attain mastery of performing the four basic fundamental math operations involving rational algebraic expressions. As they dig deeper into the world of advanced algebra, learners will be exposed to solving and simplifying radical expressions and functions. They are tasked to generate algebraic expressions to represent a particular scenario and to formulate functional models to show the input-output relationship of two variables/phenomenon. They are also given opportunities to extend their understanding of coordinate planes by plotting points and sketching the graphs of some functions.\n• Moreover, learners at this age bracket are introduced to the concepts of linear equations and inequalities. They are expected to generate equations from table or values or graph and use this to explain a trend or behavior of the relationship of two variables. They also learn about proportional relationships and systems of linear equations.\n• In Geometry, learners investigate shape transformations in coordinate planes — they determine the new coordinates of a geometric figure once it is reflected, translated, etc. They are also starting to conduct formal and informal proofs to show similarity and congruence of shapes/ geometric figures. Lastly, learners are expected to display better understanding of line and angle theorems as well as the Pythagorean Theorem.\n• Learners widen their knowledge of other branches of mathematics by discussing Statistics, Probability, and scientific notations. Topics under Statistics cover box plots, histogram, frequency polygons, population, sample, sampling techniques, statistical inferences and simple linear regression. For Probability, lessons include fundamental counting principle (FCP), union and intersection of sets, probability models, and probability of simple and compound events. In the latter part, learners explore the concepts of scientific notations, operations involving scientific notations, etc." ]

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[ "# What Is The Final Volume Of The Gas\n\n### What Is The Final Volume Of The Gas?\n\nThe resistance of the gas at initial and terminal state is uniform (P1=P2) ( P 1 = P 2 ) . excitement the terminal size of the mental gas is 2.69 L.\n\n### How do we calculate the volume of gas?\n\nCalculating the size of a gas Size = reach in mol × molar volume. Size = 0.25 × 24. = 6 dm 3\n\n### How much is the volume of gas?\n\nChemists sometimes exult comparisons over a measure temperature and resistance (STP) for reporting properties of gases: 273.15 K and 1 atm (101.325 kPa). At STP an mental gas has a size of almost 22.4 L—this is referred to as the measure molar size (Figure 10). aspect 10.\n\n### How do you find the final volume in combined gas law?\n\nThe combined gas law examines the conduct of a uniform reach of gas when resistance size and/or temperature is allowed to change.…The Combined Gas Law Formula Pi = initial pressure. Vi = initial volume. Ti = initial perfect temperature. Pf = terminal pressure. Vf = terminal volume. Tf = terminal perfect temperature.\n\n### How do you find final volume with mass and density?\n\nIt is given by the formula density equals collect divided by size (density = mass/volume). accordingly if the density and collect of a matter are mysterious the size may be determined by dividing the collect by the density (volume = mass/density).\n\n### How do you find the volume of a gas cylinder?\n\nTo meet the size of gas available engage a compressed gas cylinder we adduce the mental Gas Law (PV = nRT). In a high-pressure cylinder the size antipathy be unchanged by the content’s compressibility friend Z (PV = ZnRT). For sample an AL cylinder of foul helium may hold 134 cu. ft.\n\n### How do I find the volume?\n\nWhereas the basic formula for the area of a rectangular form is elongate × width the basic formula for size is elongate × width × altitude See also a searching plainly outgrowth of the chide war occurred when\n\n### What is volume at STP?\n\nStandard temperature and resistance (STP) is defined as 0oC (273.15K) and 1atm pressure. The molar size of a gas is the size of one atom of a gas at STP. At STP one atom (6.02×1023 likeness particles) of any gas occupies a size of 22.4L (figure below).\n\n### Does gas have a definite volume?\n\nA gas has neither a clear form nor a clear volume. resembling liquids gases are fluids. The particles in a gas can ant: slave about one another freely. If a gas is released in a closed container the gas particles antipathy ant: slave in all directions and expanded aloof as they replenish the container.\n\n### How do you find the final pressure of a gas?\n\nOR (ii) the resistance of a gas is straightly proportional to the perfect temperature (K) at uniform size p = uniform x T (right graph) or. p/T = uniform or. p1/p2 = T1/T2 for conditions changing engage 1 (initial) to 2 (final) or p1/T1 = p2/T2 for uniform volume. p1 x T2 = T1 x p. … p2 = p1 x T2/T. … or T2 = T1 x p2/p.\n\n### How do you find the final pressure of two gases?\n\nThe terminal size is the sum of the initial volumes. If the gases are mental the terminal resistance is the identical as the initial resistance and the restricted pressures are PA=nART/(VA+VB) and PB=nBRT/(VA+VB).\n\n### How do you find the final temperature of a gas?\n\nAdd the vary in temperature to your substance’s primordial temperature to meet its terminal heat. For sample if your water was initially at 24 degrees Celsius its terminal temperature would be: 24 + 6 or 30 degrees Celsius.\n\n### How do I find volume from density?\n\nThe formula for density is d = M/V since d is density M is collect and V is volume. Density is commonly expressed in units of grams per cubic centimetre. For sample the density of water is 1 pointed per cubic centimetre and Earth’s density is 5.51 grams per cubic centimetre.\n\n### What unit is volume?\n\ncubic meter size is the mete of the 3-dimensional extension occupied by substance or enclosed by a surface measured in cubic units See also what is an economic map\n\n### What is the mass volume?\n\nUpdated October 07 2019. collect and size are two units abashed to mete objects. collect is the reach of substance an appearance contains briefly size is how abundant extension it takes up.\n\n### What is the volume of the gas in the cylinder at STP?\n\n22.4 L The interior ordinary sample is the molar size of a gas at STP (Standard Temperature and Pressure) which is uniform to 22.4 L for 1 atom of any mental gas at a temperature uniform to 273.15 K and a resistance uniform to 1.00 atm.\n\n### How do you find the volume of a circle?\n\nVolume of a cylinder V = A h. ant: full the area of a surround = π r 2 genuine the formula for the size of a cylinder is: V = π r 2 h.\n\n### Is volume squared or cubed?\n\nMentor: Exactly! Area is “squared” and size is “cubed”. How do you ponder that relates to their meaning? Student: You meet the area of a square or fuse two-dimensional objects but you meet the size of three-dimensional objects resembling cubes!\n\n### What is volume for kids?\n\nVolume refers to the reach of extension the appearance takes up. In fuse words greatness is a mete of the greatness of an appearance exact resembling altitude and width are ways to draw size. … The smaller cup has pure volume!\n\n### What is STP equal to?\n\nUntil 1982 STP was defined as a temperature of 273.15 K (0 °C 32 °F) and an perfect resistance of precisely 1 atm (101.325 kPa). ant: full 1982 STP is defined as a temperature of 273.15 K (0 °C 32 °F) and an perfect resistance of precisely 105 Pa (100 kPa 1 bar).\n\n### What gas occupies 22.4 at STP?\n\noxygen gas One atom of oxygen gas occupies 22.4 l size at STP.\n\n### What is STP formula?\n\nVSTP = V * (273.15/T) * (P/760) This STP formula uses Kelvins Torrs and Liters.\n\n### What is definite volume?\n\nDefinite (for twain form and volume) resources that the container makes no separation whatsoever. If 5-liters of fluid water is poured inter a 10-liter container the fluid would hold 5-liters of the container and the fuse 5-liters would be empty.\n\n### Why gases have no definite volume?\n\nSolution : Gases do not own a clear form or size owing the molecules in gases are [see ail] loosely packed they own amplify intermolecular spaces and hence they ant: slave around. The urge of inducement between molecules is also [see ail] pure as a ant: fail gases gain any form or any volume.\n\n### What is volume of liquid?\n\nAt its interior basic plane size is simply a mete of extension See also how fishing nets work\n\n### How do you find final volume with volume and temperature?\n\nCharles’ law formula V₁ / T₁ = V₂ / T₂ since V₁ and T₁ are initial size and temperature respectively. Similarly V₂ and T₂ are the terminal values of these gas parameters. V₂ = V₁ / T₁ * T₂ .\n\n### How do you calculate pressure from volume of a gas?\n\nFor sample if you deficiency to estimate the size of 40 trouble of a gas separate a resistance of 1013 hPa and at a temperature of 250 K the ant: fail antipathy be uniform to: V = nRT/p = 40 * 8.3144598 * 250 / 101300 = 0.82 m³ .\n\n### How do you find final volume when pressure is constant?\n\nFor a confined uniform resistance gas specimen VT is uniform (i.e. the wandering = k) and as invisible immediately the P–T relationship this leads to another agree of Charles’s law: V1T1=V2T2 V 1 T 1 = V 2 T 2 .\n\n### What is r in the ideal gas law?\n\nThe friend “R” in the mental gas law equation is mysterious as the “gas constant”. R = PV. nT. The resistance early the size of a gas divided by the countless of trouble and temperature of the gas is always uniform to a uniform number.\n\n### What is mixing of two gases?\n\nMixing of substance of particularize compositions or states is an irreversible process. So the mixing of two gases by diffusion is irreversible. Diffusion is a voluntary train and the ant: continue train is nonspontaneous.\n\n### What is pressure short answer?\n\nPressure is defined as the ant: immateriality urge exerted on an object. The urge applied is vertical to the surface of objects per aggregation area.\n\n### How do you find the final temperature of two substances?\n\nCalculate the terminal temperature of the water mixture using the equation T(final) = (m1_T1 + m2_T2) / (m1 + m2) since m1 and m2 are the weights of the water in the leading and subordinate containers T1 is the temperature of the water in the leading container and T2 is the temperature of the water in the subordinate container." ]

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[ "# Aerospace Toolbox\n\n## Analyze and visualize aerospace vehicle motion using reference standards and models\n\n### Get Started:", null, "## Vehicle Motion Analysis\n\nAnalyze vehicle flight dynamics and motion in MATLAB using aerospace coordinate system transformations, flight parameters, and quaternion math.\n\n### Coordinate System Transformations\n\nUse the coordinate system functions to standardize units across data describing flight dynamics and motion, transform spatial representations and coordinate systems, and describe the behavior of three- and six-degrees-of-motion bodies.\n\nExample overlaying simulated and actual flight data.\n\n### Flight Parameters\n\nUse functions to estimate aerodynamic flight parameters, such as airspeed, incidence and sideslip angles, Mach number, and relative pressure, density, and temperature ratios.\n\n### Quaternion Math\n\nUse built-in functions to calculate quaternion norm, modulus, natural logarithm, product, division, inverse, power, or exponential. Interpolate between two quaternions using linear, spherical-linear, or normalized-linear methods.\n\n## Aircraft Controls and Stability Analysis\n\nUse coefficients obtained from the Data Compendium (DATCOM) based on vehicle flight conditions and geometry to create fixed-wing aircraft objects, estimate aerodynamic stability and control characteristics, and perform numerical analysis.\n\n### Fixed-Wing Aircraft\n\nBy importing USAF Digital DATCOM files, you can create a fixed-wing aircraft object with custom states and perform linearization and static stability analysis in MATLAB.\n\n### DATCOM Data\n\nImport aerodynamic coefficients from static and dynamic analyses and transfer them into MATLAB as a cell array of structures containing information about a DATCOM output file.\n\n## Small Satellite Mission Analysis\n\nModel and visualize satellites in orbit and compute line-of-sight access with ground stations using the `satelliteScenario` object. Use solar system ephemeris data to calculate planetary position and velocity for a given Julian date.\n\n### Satellite Scenarios\n\nCreate satellite scenarios to model and visualize satellites and constellations and perform mission analysis, such as computing line-of-sight access with ground stations.\n\n### Planetary Ephemerides\n\nWith Chebyshev coefficients obtained from NASA’s Jet Propulsion Laboratory, you can use MATLAB to compute the position and velocity of solar system bodies relative to a specified center object for a given Julian date, as well as Earth nutation and Moon libration.\n\n## Environment Models\n\nUse validated environment models to represent standard gravity and magnetic field profiles, to obtain atmospheric variables for a given altitude, and to implement the horizontal wind model of the U.S. Naval Research Laboratory.\n\n### Atmosphere\n\nUse validated environment models, including the COSPAR International Reference Atmosphere 1986, 1976 COESA, International Standard Atmosphere (ISA), Lapse Rate Atmosphere, and 2001 U.S. Naval Research Lab Exosphere, to represent the Earth’s atmosphere.\n\n### Gravity and Magnetic Fields\n\nCalculate gravity and magnetic fields using standard models. Functions let you implement the Earth Geopotential Models, World Magnetic Models, and the International Geomagnetic Reference Field, including EGM2008, WMM2020, and IGRF13. You can also calculate height and undulations based on geoid data downloadable via the Add-On Explorer.\n\nExample of geoid height for Earth geopotential model.\n\n### Wind\n\nUse the horizontal wind function to implement the U.S. Naval Research Laboratory Horizontal Wind Model routine and calculate the meridional and zonal components of the wind for one or more sets of geophysical data.\n\n## Flight Visualization\n\nVisualize the motion of aerospace vehicles using standard cockpit flight instruments and the FlightGear flight simulator.\n\n### Flight Instruments\n\nUse standard cockpit flight instruments in MATLAB to display navigation variables. Instruments include airspeed, climb rate, and exhaust gas temperature indicators, as well as an altimeter, artificial horizon, and turn coordinator.\n\nReviewing prerecorded flight test data or simulation data.\n\n### Flight Simulator Interface\n\nThe animation object for FlightGear lets you visualize flight data and vehicle motion in a three-dimensional environment.\n\nReplaying flight data in FlightGear." ]

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[ "barecheck-0.2.0.8: QuickCheck implementations for common types\n\nPrelude.Generators\n\nDescription\n\nGenerators for Prelude types.\n\nThis module also contains function generators, like `function1` and `function2`. These generate random functions. To use these, get a coarbitrary function from one of the modules whose name ends in `Coarbitrary` or by using the QuickCheck `coarbitrary` function, along with using a generator for the function's result type. For example:\n\n``` module MyModule where\n\nimport Test.QuickCheck\nimport qualified Prelude.Generators as G\nimport qualified Data.Text.Coarbitrary as C\nimport qualified Data.Text.Generators as G\n\ngenTextToMaybeText :: Gen (Text -> Maybe Text)\ngenTextToMaybeText = function1 C.text (G.maybe (G.text arbitrary))\n```\n\n# Documentation\n\nmaybe :: Gen a -> Gen (Maybe a) Source\n\neither :: Gen a -> Gen b -> Gen (Either a b) Source\n\nfunction1 :: (a -> Gen b -> Gen b) -> Gen b -> Gen (a -> b) Source\n\nfunction2 :: (a -> Gen r -> Gen r) -> (b -> Gen r -> Gen r) -> Gen r -> Gen (a -> b -> r) Source\n\nfunction3 :: (a -> Gen r -> Gen r) -> (b -> Gen r -> Gen r) -> (c -> Gen r -> Gen r) -> Gen r -> Gen (a -> b -> c -> r) Source\n\nfunction4 :: (a -> Gen r -> Gen r) -> (b -> Gen r -> Gen r) -> (c -> Gen r -> Gen r) -> (d -> Gen r -> Gen r) -> Gen r -> Gen (a -> b -> c -> d -> r) Source\n\nfunction5 :: (a -> Gen r -> Gen r) -> (b -> Gen r -> Gen r) -> (c -> Gen r -> Gen r) -> (d -> Gen r -> Gen r) -> (e -> Gen r -> Gen r) -> Gen r -> Gen (a -> b -> c -> d -> e -> r) Source\n\ntuple2 :: Gen a -> Gen b -> Gen (a, b) Source\n\ntuple3 :: Gen a -> Gen b -> Gen c -> Gen (a, b, c) Source\n\ntuple4 :: Gen a -> Gen b -> Gen c -> Gen d -> Gen (a, b, c, d) Source\n\ntuple5 :: Gen a -> Gen b -> Gen c -> Gen d -> Gen e -> Gen (a, b, c, d, e) Source" ]

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[ "# simplifying fractions worksheet pdf\n\n#### December 12, 2020   |\n\nexample: Simplify the fraction 18 27 The greatest common factor of 18 and 27 is 9. Complete lesson with worksheets about simplifying fractions. Reducing Proper Fraction 1. Simplifying Proper Fraction. These resources can be used by parents and teachers at home or in school. pdf, 17 KB. View PDF. This worksheet generator makes worksheets for four different fraction topics: (1) equivalent fractions, (2) simplifying fractions, (3) converting fractions to mixed numbers, and (4) converting mixed numbers to fractions. In these pdf worksheets, grade 3 and grade 4 kids reduce proper fractions to the lowest term. Cards have denominators between 2 and 15. Divide the numerator and the denominator by 9. Equivalent Fractions Matching Game FREE. Fractions 4: Simplifying Fractions (Reuben McIntyre) Fractional Parts of Shapes (R. Lovelock) Fractions Marking Ladder (Y1-4) ... Smartie Fractions (Mandy Smith) PDF; Fractions Worksheets (Gareth Rossiter) PDF; Ordering Fractions (Nadine Turner) PDF; Fraction Mats (Stuart Arlow) DOC; Whole, Halves and Quarters (Rachael Wilkie) DOC; Practise simplifying fractions with these helpful worksheets. Learn more. These handy simplifying fractions worksheets are differentiated for children working at different abilities and will help your group to understand how to put fractions … Printable Math Worksheets @ www.mathworksheets4kids.com 4 6 = 2 4 = 12 15 = 6 8 = 6 10 = 9 15 = 3 9 = 9 12 = 4 12 = 4 10 = 3 12 = 6 15 = 2 16 = 10 12 = 6 14 = 5 10 = Answer key Simplify the Fractions … Word Problem Generator. Simplifying Fractions Worksheet Year 3 to 6 - 9. Simplifying expressions finding perimeter of quadrilaterals. WWW Math Worksheets Com. Simplifying fractions worksheet. Worksheets also have answers. Home. 24 30 = 5. Similar: Simplifying fractions - harder 4 8 = 10. Year 6 Simplify Fractions Using the Highest Common Factor Worksheet. Read more >>>, We have put together over 8 years of working experience producing e-learning materials into creating a wide range of math resources for children from Kindergarten to 7th grades. Free fraction worksheets 2Simplifying fractions, equivalent fractions, fractions/mixed numbers. Simplify the fraction 18 27 the greatest common factor of 18 and 27 is 9. Understanding fractions worksheets including modeling fractions, ratio and proportion, comparing, ordering, simplifying and converting fractions. It has an answer key attached on the second page. They should first multiply the numerator and denominator of one of the fractions so that both fractions share a common denominator, and then do the addition. Click here for our other Fractions Worksheets. … The Corbettmaths video tutorial on Simplifying Fractions. Welcome to the fractions worksheets page at Math-Drills.com where the cup is half full! This simplifying fractions worksheet includes various different questions that can be answered by KS2, Year 3 to Year 6 children. 4 20 = b. 18 90 = 11. These worksheets are pdf files. Find pairs of fractions that are equivalent in this memory match game. You can generate the worksheets either in html or pdf … from Twinkl. Menu. First try our Adding Fractions with Common Denominator worksheets. Math Exercise Generator. Images and questions that go hand in hand with a free online \"PowerPoint\" that explains the answers, 50 80 12. Simplifying Fractions Simplify. Also check out my other lessons and let me know what you think? This is a math PDF printable activity sheet with several exercises. 9 72 = 14. A straightforward worksheet of questions with a clear example for students to refer to at the top. This worksheet is a supplementary fourth grade resource to help teachers, parents and children at home and in school. 6 8 = 6. Use the buttons below to print, open, or download the PDF version of the Reducing Fractions to Lowest Terms (A) math worksheet. Reducing Proper Fraction 2. To get the basic understanding of fractions, kids are therefore advised to go through these sets of questions with the aid of their teachers. Simplifying Fractions. 12 24 = 8. In these worksheets the child has to add together two fractions which have different denominators. D.Russell. Cancel out the common factors and reduce the improper fractions here to the lowest term. Math worksheets: Rewriting proper fractions in their simplest form. Copyright © 2018, Math 4 Champions. Fractions are another topic in mathematics that kids should know something about at their age. ... pdf, 19 KB. Simplifying Fractions Grade 5 Fractions Worksheet Simplify the fractions. 1) 30 54 2) 9 15 3) 30 36 4) 24 30 5) 26 38 6) 20 56 7) 4 56 8) 40 15 9) 21 48 10) 51 15 11) 36 8 12) 50 4 13) 18 2 14) 40 12 15) 12 38 … Videos, worksheets, 5-a-day and much more 8 12 = 7. 99 108 = 9. Fractions Worksheets. This Simplifying Fractions Worksheet reviews many rules that are required for fractions to be fully simplified. Simplifying fractions worksheet. All fractions on this sheet have denominators of 2, 3, 4, 6, and 8. Your email address and information, will NEVER be given or sold to a third party. 6 30 = 2. 48 96 = 15. In these worksheets the child has to convert an improper fraction (numerator greater than the denominator) into a mixed fraction (whole number plus a fraction). To simplify a fraction, we divide the numerator (the top of the fraction) and … 49 70 = 1. 18 27 ÷ ÷ 9 9 = 2 3 Simplify each fraction. At math4champions.com, we do not pass on to third parties any identifiable information about our users. Preview images of the first and second (if there is one) pages are shown. Reducing fractions is another skill involved in just about any of the fraction operations. This involves the multiplication of two fractions together. Simplifying Fractions Worksheet 1 answers. 4 40 = 4. Calculations using algebraic functions are similar to calculations involving fractions. Simplifying fractions grade 5 fractions worksheet simplify the fractions. Answers to Worksheet along with visual aides are included at the end. Worksheet # 1 (Answers on 2nd Page of PDF) Worksheet 1. Download All; Simplifying Improper Fraction. Below are some of the aspects of the fraction worksheet for kids. 63 72 = 13. Free fraction worksheets 2. (Revise factorising quadratics here) Can you use common factors to simplify fractions in this worksheet from Twinkl? It will be useful for them all through their education and life. All Rights Reserved. 5 10 = 3. The differentiated version just has less questions and space to … Fraction worksheets 1 Fraction addition, subtraction, multiplication, and division. Simplify fractions worksheet for 4th grade children. âÑ ÿÿ¬—;€ D{o£ ‡òş­�øÙ0ø M\\$»�`‡�Øõë�Ìÿ0@}sË<6�Öîw€2OŞŞM¾�ù‹Õ´f”ff7\\$ğåxêO]�? ljİĞ“¼XsŸœÊAçkÊ­¢Æ”'³L: ½GÍ«[oSç…IîøN°eê˜/e_G\\$}©;wR_Ÿƒ~y‡ŞL‚�v ÿÿ¬—Q€ D¯TÑ(Üÿb©9Ó.9�“üôÁ*ì[‹ Üô ŸÀøp‚�Ò�À ½›S€¾æTiŒºõ/gN¤R7K†üSñşEs{òf4ô!¼>ô¥¬Ïõä}jÕ§ÁĞ—òÛrœ=Ò†x€x 肤Wó­f³Eü}÷û¢_>úmTÁn ÿÿ¬XIÀ ü‘Qè¢ÿÿXÁš²´iŒz%Yf´ÚuÅ™¶õ. Make Your Own Addition Worksheet. Showing top 8 worksheets in the category - Simplifying Algebraic Fractions. The size of the PDF file is 9786 bytes. Key Skill 2: Simplifying Algebraic Fractions – Involving Quadratics When the fractions involve a quadratic, you first need to factorise the quadratic in order to simplify. 50 80 = 12. Download-able Worksheets For Preschooler. Student s can use math worksheets to master a math skill through practice, in a study group or for peer tutoring. Step 1: First, We need to factorise the numerator and the denominator of the fraction. Simplifying Fractions Worksheet 1. Here are 9 worksheets for 6th-grade math to help young students practice reducing and simplifying fractions. They do this by … Aids Math Worksheets For Grade 3. Click here to find out the answers. 3rd through 5th Grades. 3rd and 4th Grades. You can control the ranges for the whole number part, … Videos, worksheets, 5-a-day and much more 2 6 = 16. a. You can simplify a fraction if the numerator (top number) and denominator (bottom number) can both be divided by the same number. Simplify Improper Fractions Worksheets. a fraction using visual aides and an online interactive powerpoint. Fraction Worksheets pdf Downloads, Fractions Math Worksheets for learning fractions and master the topic, Learn fractions and different operations with fractions with these free Math fractions worksheets. Free Printable Preschool Workbooks Free Preschool Activity Sheets Printable Early Childhood Worksheets Printable Preschool Activity Printable … Multiplication Timed Test Printable 0-12. This resource is particularly useful as it can be easily edited to cater to the individuals' skill level. This worksheet generator produces a variety of worksheets for the four basic operations (addition, subtraction, multiplication, and division) with fractions and mixed numbers, including with negative fractions. The Corbettmaths Practice Questions on Equivalent/Simplifying Fractions. Excel Math Worksheets. This set of worksheets requires reducing fractions to simplest form, but the starting fractions are always proper. Here are 9 worksheets for 6th-grade math to help young students practice reducing and simplifying fractions. Reducing Fractions Worksheet 2 Reduce each fraction to lowest terms. Reducing Proper Fraction 3. Below are six versions of our grade 5 math worksheet on simplifying proper fractions; students must rewrite the fractions (if possible) in their simplest form (form with the lowest possible denominator). Reducing Fractions to Mixed Numbers. Example: Simplify fully the fraction \\dfrac{a^2 + a - 6}{ab + 3b}. The fraction worksheets on this page introduce reducing gradually by providing problems with increasingly more difficult denominators, and only gradually introducing mixed numbers and improper fractions. Worksheet 2:3 Algebraic Fractions Section 1 Factoring and Algebraic Fractions As pointed out in worksheet 2:1, we can use factoring to simplify algebraic expressions, and in particular we can use it to simplify algebraic fractions. Simplifying fractions Common factors HCF Simplifying fractions Equivelent fractions ... Other contents: division Add to my workbooks (3) Download file pdf Embed in my website or blog Add to Google Classroom Add to Microsoft Teams Share through Whatsapp: Link to ... More Fractions interactive worksheets. ID: 854511 Language: English School subject: Math Grade/level: Grade 3 Age: 8-9 Main content: Simplification of Fraction Other contents: Fractions Add to my workbooks (22) Download file pdf Embed in my website or blog Add to Google Classroom One of the first skills a student should learn with fractions is how to reduce fractions. Simplifying Fractions To simplify a fraction, divide the numerator and the denominator by the greatest common factor. There are several series of worksheets on this site dealing with reducing fractions that gradually introduce more complex fractions. View PDF. Worksheet from Twinkl another topic in mathematics that kids should know something about their. Aides are included at the end for fractions to be fully simplified and reduce the improper fractions to! To at the end { ab + 3b } factor of 18 27... Calculations using Algebraic functions are similar to calculations involving fractions the ranges for the whole part. Several exercises something about at their age welcome to the individuals ' skill level KS2, Year 3 to 6. We simplifying fractions worksheet pdf to factorise the numerator and the denominator by the greatest common factor 18. Of questions with a clear example for students to refer to at the end fractions is skill! Step 1: first, We do not pass on to third parties any identifiable information about our.. The numerator and the denominator by the simplifying fractions worksheet pdf common factor Worksheet memory match game whole number,! Childhood worksheets Printable Preschool Activity simplifying fractions worksheet pdf … Simplify improper fractions here to the fractions worksheets of worksheets reducing. 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[ "top of page\nSearch\n• Ankur Kapur\n\n# Why people tend to think in Nominal terms?\n\nUpdated: May 11\n\nIn finance and economics, a nominal value means the unadjusted value of an asset. Nominal value does not take into account inflation. Whereas, real value is excludes the impact of inflation.", null, "Inflation burns up your real value\n\nIt is very important to differentiate between the two, and to understand why nominal and real values differ over time. To illustrate the basic concept, consider the following example:\n\nYou deposited Rs 100 in a savings bank account and the interest rate is 4% per annum. You will have Rs 104 in your account by the end of the year. In nominal terms, you have gained Rs 4. However, if you think about it in real terms, and factor in the inflation of 5% through the year, you will realize that you have lost 1% (4%-5%) in that year. All interest rates cited by banks and other financial institutes are nominal interest rates. At this point, it becomes important to differentiate between stated rates of return (nominal rates) and effective rates of return (real rates).\n\nTime Value of Money\n\nThe concept of Time value of Money states that money in our hands now is worth more than money in our hands in the future. This is because we can earn more interest on that money, so, the sooner we receive it the better. Understanding the Time value of money is at the foundation of thinking about our investments in real terms.\n\nA simple formula used to calculate present and future values of money is:\n\nFV= PV (1+r) n\n\nWhere FV= Future Value\n\nPV= Present Value\n\nR= Rate of interest/required rate of return\n\nn= Number of compounding periods (usually, the number of years)\n\nFor example, if a person wants to invest Rs 1000 for 4 years and his required rate of return is 8%, then\n\nFV= 1000 (1+0.08)4 = 1000 * 1.084 =Rs 1360.49\n\nThus, the future value of his investment after 4 years at a rate of return of 8% is Rs 1360.49\n\n(1+r)n is the compounding factor. To find the present value of future investments or payments, we just manipulate the formula by taking the compounding factor to the other side and dividing the future value by the compounding factor. (PV= FV/(1+r) n)\n\nTaking real values of future financial returns into account helps us to make better and clearer decisions. Thinking in terms of constant rupees (as opposed to current rupees) and real interest rates broadens the perspective of the investor and facilitates managing long term future gains.\n\nThe real value tells us a more complete picture of the value of our investments. If you think about your investments in real terms, you are considering more factors (especially inflation).\n\nThe increase in prices of products over time, and the reduction in one’s purchasing power is one of the most crucial factors to account for in investing.\n\nInflation is directly linked with this concept because a person’s purchasing power shrinks over time as the real value of their money decreases.", null, "Inflation Trend in India\n\nThe average annual inflation rate in India between 2010 and 2019 was 6.62%.\n\nAs we can see from the above graph, historically, the inflation rate in India has been very volatile. Therefore, it is beneficial for a long-term investor to take into account at least 10-15 year annual average rate of inflation when calculating the real rate of return on his investments.\n\nIt would be advisable for an investor to factor in about 7% inflation and subtract that from the stated nominal rate of return. This effective rate of return will reflect a more accurate interest rate on their investment.\n\nThinking about finances and money in nominal terms and mistaking its face value for its purchasing power is a human cognitive bias, known as the Price Illusion. Accounting for other factors and thinking in real terms can help investors increase their gains substantially.\n\nIn the long-term, only considering nominal rates can even lead to negative returns, a loss in purchasing power and bad decision-making with regards to investments. Hence, considering the real rates of return is of utmost importance, especially while making key decisions.\n\nDisclaimers\n\n• Investment in securities market are subject to market risks. Read all the related documents carefully before investing.\n\n• The securities quoted are for illustration only and are not recommendatory.\n\n• Registration granted by SEBI, membership of BASL (in case of IAs) and certification from NISM in no way guarantee performance of the intermediary or provide any assurance of returns to investors." ]

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[ "Order Of Operations Worksheets Math Additions Addition And Subtraction With Brackets Pd Evaluate Expressions Wit Inverse Worksheet", null, "order of operations worksheets math additions addition and subtraction with brackets pd evaluate expressions wit inverse worksheet.\n\nworksheet mixed operations with integers number worksheets grade 6 rational numbers 7th pdf,inverse operations worksheets pdf algebra mixed number worksheet with fractions 7th grade,operations with rational numbers worksheet answer key order of worksheets grade 6 exponents eighth algebra best mixed integers fractions,mixed operations worksheets grade 5 inverse ks3 order of for ks1,inverse operations worksheets ks3 mixed grade 4 order of with answers,inverse operations worksheets multiplication and division mixed fractions worksheet pdf 6th grade,mixed number operations worksheet pdf free order of math worksheets inverse ks2 year 5,integers mixed operations worksheet pdf number worksheets inverse year 6,mixed operations fractions worksheet pdf four worksheets grade 5 rksheets inverse the best image binary 3,worksheets operations with integers inverse ks3 super teacher order of math problems.", null, "" ]

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[ "## Saturday, June 14, 2014\n\nWhat if I told you that a spreadsheet could be a library instead of an application? What would that even mean? How do we distill the logic behind spreadsheets into a reusable abstraction? My `mvc-updates` library answers this question by bringing spreadsheet-like programming to Haskell using an intuitive `Applicative` interface.\n\nThe central abstraction is an `Applicative` `Updatable` value, which is just a `Fold` sitting in front of a `Managed` `Controller`:\n\n``````data Updatable a =\nforall e . On (Fold e a) (Managed (Controller e))``````\n\nThe `Managed` `Controller` originates from my `mvc` library and represents a resource-managed, concurrent source of values of type `e`. Using `Monoid` operations, you can interleave these concurrent resources together while simultaneously merging their resource management logic.\n\nThe `Fold` type originates from my `foldl` library and represents a reified left fold. Using `Applicative` operations, you can combine multiple folds together in such a way that they still pass over the data set just once without leaking space.\n\nTo build an `Updatable` value, just pair up a `Fold` with a `Managed` `Controller`:\n\n``````import Control.Foldl (last, length)\nimport MVC\nimport MVC.Prelude (stdinLines, tick)\nimport Prelude hiding (last, length)\n\n-- Store the last line from standard input\nlastLine :: Updatable (Maybe String)\nlastLine = On last stdinLines\n\n-- Increment every second\nseconds :: Updatable Int\nseconds = On length (tick 1.0)``````\n\nWhat's amazing is that when you stick a `Fold` in front of a `Controller`, you get a new `Applicative`. This `Applicative` instance lets you combine multiple `Updatable` values into new derived `Updatable` values. For example, we can combine `lastLine` and `seconds` into a single data type that tracks updates to both values:\n\n``````import Control.Applicative ((<\\$>), (<*>))\n\ndata Example = Example (Maybe String) Int deriving (Show)\n\nexample :: Updatable Example\nexample = Example <\\$> lastLine <*> seconds``````\n\n`example` will update every time `lastLine` or `seconds` updates, caching and reusing portions that do not update. For example, if `lastLine` updates then only the first field of `Example` will change. Similarly, if `seconds` updates then only the second field of `Example` will change.\n\nWhen we're done combining `Updatable` values we can plug them into `mvc` using the `updates` function:\n\n``updates :: Buffer a -> Updatable a -> Managed (Controller a)``\n\nThis gives us back a `Managed` `Controller` we can feed into our `mvc` application:\n\n``````viewController :: Managed (View Example, Controller Example)\nviewController = do\nreturn (asSink print, controller)\n\nmodel :: Model () Example Example\nmodel = asPipe \\$\nPipes.takeWhile (\\(Example str _) -> str /= Just \"quit\")\n\nmain :: IO ()\nmain = runMVC () model viewController``````\n\nThis program updates in response to two concurrent inputs: time and standard input.\n\n``````\\$ ./example\nExample Nothing 0\nTest<Enter>\nExample (Just \"Test\") 0 <-- Update: New input\nExample (Just \"Test\") 1 <-- Update: 1 second passed\nExample (Just \"Test\") 2 <-- Update: 1 second passed\nABC<Enter>\nExample (Just \"ABC\") 2 <-- Update: New input\nExample (Just \"ABC\") 3 <-- Update: 1 second\nquit<Enter>\n\\$``````\n\nThe previous example was a bit contrived, so let's step it up a notch. What better way to demonstrate spreadsheet-like programming than .. a spreadsheet!\n\nI'll use Haskell's `gtk` library to set up the initial API, which consists of a single exported function:\n\n``````module Spreadsheet (spreadsheet) where\n\n:: Managed -- GTK setup\n( Updatable Double -- Create input cell\n, Managed (View Double) -- Create output cell\n, IO () -- Start spreadsheet\n)\n\nYou can find the full source code here.\n\nUsing `spreadsheet`, I can now easily build my own spread sheet application:\n\n``````{-# LANGUAGE TemplateHaskell #-}\n\nimport Control.Applicative (Applicative, (<\\$>), (<*>))\nimport Lens.Family.TH (makeLenses)\nimport MVC\n\ndata In = I\n{ _i1 :: Double\n, _i2 :: Double\n, _i3 :: Double\n, _i4 :: Double\n}\n\ndata Out = O\n{ _o1 :: Double\n, _o2 :: Double\n, _o3 :: Double\n, _o4 :: Double\n}\nmakeLenses ''Out\n\n-- Spreadsheet logic that converts input to output\nmodel :: Model () In Out\nmodel = asPipe \\$ loop \\$ \\(I i1 i2 i3 i4) -> do\nreturn \\$ O (i1 + i2) (i2 * i3) (i3 - i4) (max i4 i1)\n\nmain :: IO ()\nmain = runMVC () model \\$ do\n\n-- Assemble the four input cells\nI <\\$> inCell <*> inCell <*> inCell <*> inCell\n\n-- Assemble the four output cells\nv <- fmap (handles o1) outCell\n<> fmap (handles o2) outCell\n<> fmap (handles o3) outCell\n<> fmap (handles o4) outCell\n\nliftIO go\n\nreturn (v, c)``````\n\nYou can install this program yourself by building my `mvc-updates-examples` package on Github:\n\n``````\\$ git clone https://github.com/Gabriel439/Haskell-MVC-Updates-Examples-Library.git\n\\$ cabal install\n\nOr you can watch this video of the spreadsheet in action:\n\nThe key feature I want to emphasize is how concise this spreadsheet API is. We provide our user an `Applicative` input cell builder and a `Monoid` output cell builder, and we're done. We don't have to explain to the user how to acquire resources, manage threads, or combine updates. The `Applicative` instance for `Updatable` handles all of those trivial details for them. Adding extra inputs or outputs is as simple as chaining additional `inCell` and `outCell` invocations.\n\n#### Reactive animations\n\nWe don't have to limit ourselves to spread sheets, though. We can program `Updatable` graphical scenes using these same principles. For example, let's animate a cloud that orbits around the user's mouse using the `sdl` library. Just like before, we will begin from a concise interface:\n\n``````-- Animation frames for the cloud\ndata Frame = Frame0 | Frame1 | Frame2 | Frame3\n\n-- To draw a cloud we specify the frame and the coordinates\ndata Cloud = Cloud Frame Int Int\n\n-- mouse coordinates\ndata Mouse = Mouse Int Int\n\nsdl :: Managed -- SDL setup\n( View Cloud -- Draw a cloud\n, Updatable Mouse -- Updatable mouse coordinates\n)``````\n\nThe full source is located here.\n\nIn this case, I want to combine the `Updatable` mouse coordinates with an `Updatable` time value:\n\n``````main :: IO ()\nmain = runMVC () (asPipe cat) \\$ do\n(cloudOut, mouse) <- sdl\nlet seconds = On length (tick (1 / 60))\n\ntoFrame n = case (n `div` 15) `rem` 4 of\n0 -> Frame0\n1 -> Frame1\n2 -> Frame2\n_ -> Frame3\n\ncloudOrbit t (Mouse x y) = Cloud (toFrame t) x' y'\nwhere\nx' = x + truncate (100 * cos (fromIntegral t / 10))\ny' = y + truncate (100 * sin (fromIntegral t / 10))\n\ncloudIn <- updates Unbounded (cloudOrbit <\\$> seconds <*> mouse)\nreturn (cloudOut, cloudIn)``````\n\n`cloudOrbit` is defined as a pure function from the current time and mouse coordinates to a `Cloud`. With the power of `Applicative`s we can lift this pure function over two `Updatable` values (`mouse` and `seconds`) to create a new `Updatable` `Cloud` that we pass intact to our program's `View`.\n\nLike before, you can either run this program yourself:\n\n``````\\$ git clone https://github.com/Gabriel439/Haskell-MVC-Updates-Examples-Library.git\n\\$ cabal install\n\n... or you can watch the video:\n\n#### Under the hood\n\n`mvc-updates` distinguishes itself from similar libraries in other languages by not relying on a semantics for concurrency. The `Applicative` instance for `Updatable` uses no concurrent operations, whatsoever:\n\n``````instance Applicative Updatable where\npure a = On (pure a) mempty\n\n(On foldL mControllerL) <*> (On foldR mControllerR)\n= On foldT mControllerT\nwhere\nfoldT = onLeft foldL <*> onRight foldR\n\nmControllerT = fmap (fmap Left ) mControllerL\n<> fmap (fmap Right) mControllerR\n\nonLeft (Fold step begin done) =\nFold step' begin done\nwhere\nstep' x (Left a) = step x a\nstep' x _ = x\n\nonRight (Fold step begin done) =\nFold step' begin done\nwhere\nstep' x (Right a) = step x a\nstep' x _ = x``````\n\nIn fact, this `Applicative` instance only assumes that the `Controller` type is a `Monoid`, so this trick generalizes to any source that forms a `Monoid`.\n\nThis not only simplifies the proof of the `Applicative` laws, but it also greatly improves efficiency. This `Applicative` instance introduces no new threads or buffers. The only thread or buffer you will incur is in the final call to the `updates` function, but expert users can eliminate even that overhead by inlining the logic of the `updates` function directly into their `mvc` program.\n\n#### Lightweight\n\nThe `mvc-updates` library is incredibly small. Here's the entire API:\n\n``````data Updatable = forall e . On (Fold e a) (Controller e)\n\ninstance Functor Updatable\ninstance Applicative Updatable\n\nupdates :: Buffer a -> Updatable a -> Managed (Controller a)``````\n\nThe library is very straightforward to use:\n\n• Build `Updatable` values\n• Combine them using their `Applicative` instance\n• Convert them back to a `Managed` `Controller` when you're done\n\nThat's it!\n\nThe small size of the library is no accident. The `Updatable` abstraction is an example of a scalable program architecture. When we combine `Updatable` values together, the end result is a new `Updatable` value. This keeps the API small since we always end up back where we started and we never need to introduce additional abstractions.\n\nThere is no need to distinguish between \"primitive\" `Updatable` values or \"derived\" `Updatable` values or \"sheets\" of `Updatable` values. The `Applicative` interface lets us unify these three concepts into a single uniform concept. Moreover, the `Applicative` interface is one of Haskell's widely used type classes inspired by category theory, so we can reuse people's pre-existing intuition for how `Applicative`s work. This is a common theme in Haskell where once you learn the core set of mathematical type classes they go a very, very long way.\n\n#### Conclusion\n\nHopefully this post will get you excited about the power of `Applicative` programming. If you would like to learn more about `Applicative`s, I highly recommend the \"Applicative Programming with Effects\" paper by Conor McBride and Ross Paterson.\n\nI would like to conclude by saying that there many classes of problems that the `mvc-updates` library does not solve well, such as:\n\n• build systems,\n• programs with computationally expensive `View`s, and:\n• `Updatable` values that share state.\n\nHowever, `mvc-updates` excels at:\n\n• data visualizations,\n• control panels, and:\n\nYou can find the `mvc-updates` library up on Hackage or on Github.\n\n1.", null, "Look up constraint programming... :-)\n\n1.", null, "I'm a little bit familiar with constraint programming, but the main thing I look for in a programming paradigm are programming interfaces inspired by category theory or abstract algebra (i.e. monoids, functors, categories, etc.). Are there analogs of that in constraint programming?\n\n2.", null, "Constraint programming systems show you that \"spreadsheets\" can be implemented (simply) as libraries without the need for any of category theory or abstract algebra.\n\nSo these don't add anything substantive here, apart from pleasing your personal palate (which is a perfectly fine thing to do).\n\n3.", null, "I don't use mathematics for the sake of using mathematics. The purpose behind structuring programs mathematically is to compose small bits of mathematical functionality, each of which is correct in isolation, to build larger mathematical structures which are still correct.\n\nSure, you can always whip up some specialized and non-mathematical solution, but these will rarely generalize to more complex problems well. They will usually solve some very specific problem very well, but the moment you deviate from the problem it was intended to solve it will become very brittle.\n\nEven the very example you give (constraint programming systems) demonstrates this issues. Constraint programming lacks the resource management sophistication of `mvc-updates`, where as you combine updatable values it automatically merges their resource management logic, and it's not clear to me how I would extend it with this feature, whereas with `mvc-updates` it was trivial because it took the principled approach.\n\n4.", null, "No one cares about constraint programming. Nice article though.\n\n5.", null, "This comment has been removed by the author.\n\n6.", null, "@Russel: considering that Gabriel implemented spreadsheet-like programming (and spreadsheet = one-way dataflow constraints), I'd say at least one person cares about constraint programming, enough to do a (somewhat simplistic) implementation.\n\nThen there are the million plus iOS/Mac programmers using AutoLayout and KVO/Bindings. AutoLayout is based on the Cassowary constraint solver, KVO/Bindings is equivalent to a simple one-way constraint solver (without formulae)\n\nAnd finally, spreadsheets are the most wide-spread form of programming, and again, spread-sheets = one-way dataflow constraints.\n\nSo you have a funny definition of \"nobody\" :-)\n\n@Gabriel: what do you mean with \"resource management\"? Considering the wide variety of constraint systems, are you certain that none have this? In fact, mvc-updates seems quite limited compared to most constraint systems I am aware of.\n\n2.", null, "This comment has been removed by the author.\n\n3.", null, "Looks great! I don't understand it yet, but I will, soon! ;)\n\nHas anyone went so far and implemented a spreadsheet app (expanding on your example)? I would like to do that because I always wanted to have my own light-version of Excel.\n\nAlso it seems, there are similarities to functional reactive programming (maybe solving similar problems).\n\n1.", null, "Nobody has implemented an actual spread sheet application on top of this yet.\n\nThere are some similarities to functional reactive programming. This answer on Stack Overflow was part of the inspiration for this library (and the second example):\n\nhttp://stackoverflow.com/a/1028642/1026598\n\n4.", null, "Gabriel, you say that mvc-updates \"excel at spread sheets\" but as far as I see it deals only with finite and fixed number of Updatables which does not look like spread sheets which could contain variable and possibly huge number of cells.\nAm I correct at that of if I'm wrong how could you model variable number of updatables using mvc-updates? Or maybe there is something outside of this library which could make it possible in some comparatively easy way?\nBTW Thanks for all your writings - I (and probably many others too) get a lot from reaging them.\n\n1.", null, "You can model a variable number of updatables if they all store the same type of value. For example, assuming you have a sequence of updatable values of type:\n\nexample :: Seq (Updatable a)\n\n... you can turn that into an updatable sequence of values by using `Data.Traversable.sequenceA`:\n\nsequenceA example :: Updatable (Seq a)\n\nUnder the hood this will generate a minimal update tree where each primitive update of an `a` triggers O(log N) downstream updates to build the new `Seq a`, where N is the size of the original sequence.\n\nThe reason I use `Seq` instead of list is because it has a much more efficient implementation of `sequenceA`." ]

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[ "## Conversion formula\n\nThe conversion factor from kilometers per hour to meters per second is 0.277777777778, which means that 1 kilometer per hour is equal to 0.277777777778 meters per second:\n\n1 km/h = 0.277777777778 m/s\n\nTo convert 17.4 kilometers per hour into meters per second we have to multiply 17.4 by the conversion factor in order to get the velocity amount from kilometers per hour to meters per second. We can also form a simple proportion to calculate the result:\n\n1 km/h → 0.277777777778 m/s\n\n17.4 km/h → V(m/s)\n\nSolve the above proportion to obtain the velocity V in meters per second:\n\nV(m/s) = 17.4 km/h × 0.277777777778 m/s\n\nV(m/s) = 4.8333333333372 m/s\n\nThe final result is:\n\n17.4 km/h → 4.8333333333372 m/s\n\nWe conclude that 17.4 kilometers per hour is equivalent to 4.8333333333372 meters per second:\n\n17.4 kilometers per hour = 4.8333333333372 meters per second\n\n## Alternative conversion\n\nWe can also convert by utilizing the inverse value of the conversion factor. In this case 1 meter per second is equal to 0.20689655172397 × 17.4 kilometers per hour.\n\nAnother way is saying that 17.4 kilometers per hour is equal to 1 ÷ 0.20689655172397 meters per second.\n\n## Approximate result\n\nFor practical purposes we can round our final result to an approximate numerical value. We can say that seventeen point four kilometers per hour is approximately four point eight three three meters per second:\n\n17.4 km/h ≅ 4.833 m/s\n\nAn alternative is also that one meter per second is approximately zero point two zero seven times seventeen point four kilometers per hour.\n\n## Conversion table\n\n### kilometers per hour to meters per second chart\n\nFor quick reference purposes, below is the conversion table you can use to convert from kilometers per hour to meters per second\n\nkilometers per hour (km/h) meters per second (m/s)\n18.4 kilometers per hour 5.111 meters per second\n19.4 kilometers per hour 5.389 meters per second\n20.4 kilometers per hour 5.667 meters per second\n21.4 kilometers per hour 5.944 meters per second\n22.4 kilometers per hour 6.222 meters per second\n23.4 kilometers per hour 6.5 meters per second\n24.4 kilometers per hour 6.778 meters per second\n25.4 kilometers per hour 7.056 meters per second\n26.4 kilometers per hour 7.333 meters per second\n27.4 kilometers per hour 7.611 meters per second" ]

[ null ]

[ "# Visual Basic .NET/Arithmetic operators\n\n## Arithmetic operators\n\nVisual Basic .NET provides a basic set of operators to calculate simple arithmetic.\n\n+ Addition\n- Subtraction\n* Multiplication\n/ Division\n\\ Integer division\nMod Remainder Division\n^ Exponentiation\n& String concatenation\n\n7 + 2 produces 9\n7 - 2 produces 5\n7 * 2 produces 14\n7 / 2 produces 3.5\n7 \\ 2 produces 3\n7 Mod 2 produces 1\n7 ^ 2 produces 49\n\"7\" & \"7\" produces \"77\"\n\n\nLet's look at a short example of arithmetic operations before we jump into the operators themselves.\n\nIn this example we will also be using some basic variables. The Dim operator creates the variable.\n\n Dim Commission As Single\nDim Sales As Single\nSales = 3142.51\nCommission = 0.3 * Sales ' Calculate 30% commission.\n\n\nFirst, we set the total sales to 3142.51.\n\nThe * operator calculates multiplication, so the last line is equivalent to $Commission=0.3*Sales$", null, ". This means that our second step is multiplying 0.3 and Sales. Sales is 3142.51, so our result should be the product of 0.3 and 3142.51.\n\n### Why the funny symbols?\n\nWith the exception of addition and subtraction, the symbols used are different to the ones used in real life. This is simply because the other symbols are not available on a standard keyboard.\n\nA good reference for the operators in Visual Basic .NET is Windows's built-in Calculator. If you set Calculator on scientific mode, you have access to the same operators as in .NET.\n\nThis adds two numbers together, and is denoted by the \"+\" symbol. If strings are involved it may also do String concatenation. Examples:\n\n x = 7 + 2 ' Results in 9.\nx = 25 + -4 ' Results in 21.\nDim StringA As String\nStringA = \"A string\" + \"Another string\" ' Results in \"A stringAnother string\"\n\n\nBy using funny symbol it will become easy to parse during execution. mostly that is the reason why they have used.\n\nThere is a second addition operator, \"+=\". It increments the variable on the left of the += by the amount indicated on the right.\n\n Dim x As Integer = 54\nx += 89 ' result is 143\nx += 7 ' result is 150\n\n\nIt also works with Strings as a concatenation operator.\n\n Dim x As String = \"A fox\"\nx += \" jumped\" ' result is \"A fox jumped\"\nx += \" over the fence\" ' result is \"A fox jumped over the fence\"\n\n\n### Subtraction\n\nThis subtracts two numbers, and is denoted by the \"-\" symbol. Examples:\n\n Dim x As Double = 12.045\nDim Y As Integer= 12\nx = x-Y ' Results in .045.\nx = 25 - -4 ' Results in 29.\n\n Sub Main()\nDim x As Double = 12.045\nDim y As Integer = 12\n\nSystem.Console.WriteLine(x - y)\n\nEnd Sub\n\n\nBeware when subtracting an integer from a double. The above code, when printed, will result in an answer of .04499999999999999 instead of the obvious answer of .045. Using doubles to store decimal values can be highly inaccurate.\n\n### Multiplication\n\nThis multiplies two numbers, and is denoted by the \"*\" symbol. Examples:\n\n Dim x As Integer\nx = 7 * 2 ' Results in 14.\nx = 25 * -4 ' Results in -100.\n\n\n### Division\n\nThere are more types of division than the one denoted by the \"/\" symbol. There is also integer division and remainder division.\n\nDivision\nThis is the most commonly used form of division and is denoted by the \"/\" operator. Examples:\n Dim x As Single\n' (note that we must use the Single class to have decimals)\nx = 7 / 2 ' Results in 3.5.\nx = 25 / 4 ' Results in 6.25.\n\nInteger division\nThis divides two numbers, and gives the result without the remainder if the quotient is a decimal. Examples:\n Dim x As Integer\nx = 7 \\ 2 ' Results in 3.\nx = 25 \\ 4 ' Results in 6.\n\nRemainder Division\nThis divides two numbers, and gives the result's remainder if the quotient is a decimal. This is denoted by the operator \"Mod.\" Examples:\n Dim x As Integer\nx = 7 Mod 2 ' Results in 1.\nx = 25 Mod 4 ' Results in 1.\n\n\n### Exponentiation\n\nThis is raising a number to a power. For example $7^{2}$", null, "in VB .Net code is:\n\n Dim x As Integer\nx = 7 ^ 2 ' Results in 49.\n\n\nThis results in the number 49 being assigned to the variable x. It can also be used to calculate the square root of a number. The square root of a number is the number raised to the power of 0.5.\n\n Dim x As Single\nx = 7 ^ 0.5 ' Results in a number around 2.645.\n\n\nNote: It is necessary to ensure that the variables be correctly declared to get the desired results. The following example works, but will produce the wrong result. This is because the Integer class does not allow decimal places (just like mathematical integers).\n\n Dim x As Integer\nx = 7 ^ 0.5 ' Results in 3.\n\n\nSince x is declared as an Integer type, the value square root, a real number, is stored incorrectly.\n\nAny nth root of number is the can be calculated by raising the number to the power of $1/n$", null, ":\n\n Dim x As Single\nDim n As Single\nn = 7\nx = 2 ^ (1 / n)\n\n\nThis is because $x^{\\frac {1}{n}}={\\sqrt[{n}]{x}}$", null, ".\n\n### Rounding\n\n• Round(): round to the nearest integer.\n• Floor(): round to inferior.\n• Ceiling(): round to superior.\n• Truncate(): truncate the decimal digits.\nModule Module1\nSub Main()\nDim Number1 As Single = 1.5\nConsole.WriteLine(Math.Round(Number1)) ' 2\nConsole.WriteLine(Math.Floor(Number1)) ' 1\nConsole.WriteLine(Math.Ceiling(Number1)) ' 2\nConsole.WriteLine(Math.Truncate(Number1)) ' 1\nEnd Sub\nEnd Module\n\n\n## Comparisons\n\n• Max()\n• Min()\nModule Module1\nSub Main()\nConsole.WriteLine(Math.Max(3, 4))" ]

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[ "# Plotting $l^p$ circles\n\nI'm trying to plot a couple of $l^p$ circles, but when $p=\\infty$ its not quite showing up (the max one in the code below):\n\nContourPlot[{Abs[x] + Abs[y] == 1, x^2 + y^2 == 1, x^4 + y^4 == 1,\nMax[Abs[x], Abs[y]] == 1}, {x, -1, 1}, {y, -1, 1}]\n\n\nIs there a way to plot this (preferably not telling Mathematica that it's a square)?\n\n## 2 Answers\n\nYou need to get sampling points on both sides of the curve for it to show up with ContourPlot.\n\nContourPlot[{Abs[x] + Abs[y] == 1, x^2 + y^2 == 1, x^4 + y^4 == 1,\nMax[Abs[x], Abs[y]] == 1}, {x, -1.05, 1.05}, {y, -1.05, 1.05}(*, Exclusions -> None*)]", null, "Add the option Exclusions -> None to remove the gaps in the corners. Mathematica is thinking there is a discontinuity, but it's wrong.\n\nAddendum\n\nWith a little trickery, you can use Norm[v, p] and make a slider run between Infinity and 1:\n\nManipulate[\nContourPlot[\nNorm[{x, y}, Limit[1/pp, pp -> p]], {x, -1.05, 1.05}, {y, -1.05, 1.05}],\nRow[{Control[{p, 0, 1}],\nInputField[Dynamic[Limit[1/pp, pp -> p], If[# >= 1, p = 1/#] &],\nAppearance -> \"Frameless\"]}, \" \"]\n]", null, "", null, "", null, "RegionPlot[Evaluate[Norm[{x, y}, #] <= 1 & /@ {1, 2, 4, Infinity}], {x, -1, 1}, {y, -1, 1}]", null, "Add the option PlotStyle -> {None, None, None, None}] to get", null, "" ]

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[ "views 1593 words\n\n# 正则\n\nHere is a quick cheat sheet for various rules in regular expressions:\n\nIdentifiers:\n\n• \\d = any number\n• \\D = anything but a number\n• \\s = space\n• \\S = anything but a space\n• \\w = any letter or number - [a-zA-Z0-9_]\n• \\W = anything but a letter and number - [^\\w]\n• . = any character, except for a new line\n• \\b = space around whole words\n• . = period. must use backslash, because . normally means any character.\n\nModifiers:\n\n• {1,3} = for digits, u expect 1-3 counts of digits, or “places”\n• + = match 1 or more\n• ? = match 0 or 1 repetitions.\n• * = match 0 or MORE repetitions\n• $= matches at the end of string • ^ = matches start of a string • | = matches either/or. Example x|y = will match either x or y • [] = matches any character contained between the square brackets • {x} = expect to see this amount of the preceding code. • {x,y} = expect to see this x-y amounts of the precedng code White Space Charts: • \\n = new line • \\s = space • \\t = tab • \\e = escape • \\f = form feed • \\r = carriage return Characters to REMEMBER TO ESCAPE IF USED! • . + * ? [ ]$ ^ ( ) { } |\n\nBrackets:\n\n• [] = quant[ia]tative = will find either quantitative, or quantatative.\n• [a-z] = return any lowercase letter a-z\n• [1-5a-qA-Z] = return all numbers 1-5, lowercase letters a-q and uppercase A-Z\n\n## 什么是正则表达式？", null, "## 1. 基本匹配\n\n\"the\" => The fat cat sat on the mat.\n\n\n\"The\" => The fat cat sat on the mat.\n\n\n## 2. 元字符\n\n. 句号匹配任意单个字符除了换行符.\n[ ] 字符种类. 匹配方括号内的任意字符.\n[^ ] 否定的字符种类. 匹配除了方括号里的任意字符\n* 匹配>=0个重复的在*号之前的字符.\n+ 匹配>=1个重复的+号前的字符.\n? 标记?之前的字符为可选.\n{n,m} 匹配num个大括号之前的字符或字符集 (n <= num <= m).\n(xyz) 字符集, 匹配与 xyz 完全相等的字符串.\n| 或运算符, 匹配符号前或后的字符.\n\\ 转义字符,用于匹配一些保留的字符 [ ] ( ) { } . * + ? ^ $\\ | ^ 从开始行开始匹配.$ 从末端开始匹配.\n\n## 2.1 点运算符 .\n\n.是元字符中最简单的例子. .匹配任意单个字符, 但不匹配换行符. 例如, 表达式.ar匹配一个任意字符后面跟着是ar的字符串.\n\n\".ar\" => The car parked in the garage.\n\n\n## 2.2 字符集\n\n\"[Tt]he\" => The car parked in the garage.\n\n\n\"ar[.]\" => A garage is a good place to park a car.\n\n\n### 2.2.1 否定字符集\n\n\"[^c]ar\" => The car parked in the garage.\n\n\n## 2.3 重复次数\n\n### 2.3.1 * 号\n\n*号匹配 在*之前的字符出现大于等于0次. 例如, 表达式 a* 匹配0或更多个以a开头的字符. 表达式[a-z]* 匹配一个行中所有以小写字母开头的字符串.\n\n\"[a-z]*\" => The car parked in the garage #21.\n\n\n*字符和.字符搭配可以匹配所有的字符.*. *和表示匹配空格的符号\\s连起来用, 如表达式\\s*cat\\s*匹配0或更多个空格开头和0或更多个空格结尾的cat字符串.\n\n\"\\s*cat\\s*\" => The fat cat sat on the concatenation.\n\n\n### 2.3.2 + 号\n\n+号匹配+号之前的字符出现 >=1 次. 例如表达式c.+t 匹配以首字母c开头以t结尾, 中间跟着至少一个字符的字符串.\n\n\"c.+t\" => The fat cat sat on the mat.\n\n\n### 2.3.3 ? 号\n\n\"[T]he\" => The car is parked in the garage.\n\n\n\"[T]?he\" => The car is parked in the garage.\n\n\n## 2.4 {} 号\n\n\"[0-9]{2,3}\" => The number was 9.9997 but we rounded it off to 10.0.\n\n\n\"[0-9]{2,}\" => The number was 9.9997 but we rounded it off to 10.0.\n\n\n\"[0-9]{3}\" => The number was 9.9997 but we rounded it off to 10.0.\n\n\n## 2.5 (...) 特征标群\n\n\"(c|g|p)ar\" => The car is parked in the garage.\n\n\n## 2.6 | 或运算符\n\n\"(T|t)he|car\" => The car is parked in the garage.\n\n\n## 2.7 转码特殊字符\n\n### 2.8.1 ^ 号\n\n^ 用来检查匹配的字符串是否在所匹配字符串的开头.\n\n\"(T|t)he\" => The car is parked in the garage.\n\n\n\"^(T|t)he\" => The car is parked in the garage.\n\n\n\n\n## 3. 简写字符集\n\n. 除换行符外的所有字符\n\\w 匹配所有字母数字, 等同于 [a-zA-Z0-9_]\n\\W 匹配所有非字母数字, 即符号, 等同于： [^\\w]\n\\d 匹配数字： [0-9]\n\\D 匹配非数字： [^\\d]\n\\s 匹配所有空格字符, 等同于： [\\t\\n\\f\\r\\p{Z}]\n\\S 匹配所有非空格字符： [^\\s]\n\\f 匹配一个换页符\n\\n 匹配一个换行符\n\\r 匹配一个回车符\n\\t 匹配一个制表符\n\\v 匹配一个垂直制表符\n\\p 匹配 CR/LF（等同于 \\r\\n）, 用来匹配 DOS 行终止符\n\n## 4. 零宽度断言（前后预查）\n\n### 5.1 忽略大小写（Case Insensitive）\n\n\"The\" => The fat cat sat on the mat.\n\n\n\"/The/gi\" => The fat cat sat on the mat.\n\n\n\"/.(at)/\" => The fat cat sat on the mat.\n\n\n\"/.(at)/g\" => The fat cat sat on the mat.\n\n\n### 5.3 多行修饰符（Multiline）\n\n\"/.at(.)?$/\" => The fat cat sat on the mat. 在线练习 \"/.at(.)?$/gm\" => The fat\ncat sat\non the mat.\n\n\n### 6. 贪婪匹配与惰性匹配（Greedy vs lazy matching）\n\n\"/(.*at)/\" => The fat cat sat on the mat.\n\n\"/(.*?at)/\" => The fat cat sat on the mat.\n\n## 6. \\b 和 \\B\n\n\\b: 匹配非字母数字(\\W) 与 字母数字(\\w)的边界 (即\\b的前一个字符和后一个字符, 必须一个是字母数字, 一个非字母数字 或者 一个是非字母数字, 一个字母数字)\n\n\\B：匹配非字母数字(\\W) 与 非字母数字(\\W)的边界\n\n\"\\bnice\\b\" => It's a nice day, nice123.\n\n\n## 7. python re模块\n\nRe库是Python的标准库，主要用于字符串匹配 调用方式:\n\nimport re\n\n### 正则表达式的表示类型\n\n• re库采用raw string(原生字符串)类型表示正则表达式，表示为: r’text’\n• 例如: r’[1‐9]\\d{5}’ 和 r’\\d{3}‐\\d{8}|\\d{4}‐\\d{7}’\n• raw string是不包含对转义符再次转义的字符串\n• re库也可以采用string类型表示正则表达式，但更繁琐\n• 例如: ‘[1‐9]\\d{5}’ 和 ‘\\d{3}‐\\d{8}|\\d{4}‐\\d{7}’\n• 建议:当正则表达式包含转义符时，使用raw string\n\n### 主要功能函数", null, "#### re.search(pattern, string, flags=0)\n\n• pattern : 正则表达式的字符串或原生字符串表示\n• string : 待匹配字符串\n• flags : 正则表达式使用时的控制标记\n\n• re.I (ignorecase): 忽略正则表达式的大小写\n• re.M (multiline): 正则表达式中的^操作符能够将给定字符串的每行当做匹配开始\n• re.S (dotall): 正则表达式中的.操作符能够匹配所有字符, 默认匹配除换行外的所有字符\n\n>>> match = re.search(r'[1-9]\\d{5}', 'BIT 100081')\n>>> if match:\n... print(match.group(0))\n...\n100081\n\n#### re.match(pattern, string, flags=0)\n\n• pattern : 正则表达式的字符串或原生字符串表示\n• string : 待匹配字符串\n• flags : 正则表达式使用时的控制标记\n\nTraceback (most recent call last):\nFile \"<stdin>\", line 1, in <module>\nAttributeError: 'NoneType' object has no attribute 'group'\n>>> match = re.match(r'[1-9]\\d{5}', '100081 BIT')\n>>> if match:\n... print(match.group(0))\n...\n100081\n\n#### re.findall(pattern, string, flags=0)\n\n• pattern : 正则表达式的字符串或原生字符串表示\n• string : 待匹配字符串\n• flags : 正则表达式使用时的控制标记\n\n>>> match = re.findall(r'[1-9]\\d{5}', 'BIT 100081 TSU100082')\n>>> match\n['100081', '100082']\n\n#### re.split(pattern, string, maxsplit=0, flags=0)\n\n• pattern : 正则表达式的字符串或原生字符串表示\n• string : 待匹配字符串\n• maxsplit: 最大分割数，剩余部分作为最后一个元素输出\n• flags : 正则表达式使用时的控制标记\n\n>>> match = re.findall(r'[1-9]\\d{5}', 'BIT 100081 TSU100082')\n>>> match\n['100081', '100082']\n>>> match = re.split(r'[1-9]\\d{5}', 'BIT 100081 TSU100082')\n>>> match\n['BIT ', ' TSU', '']\n>>> match = re.split(r'[1-9]\\d{5}', 'BIT 100081 TSU100082', maxsplit=1)\n>>> match\n['BIT ', ' TSU100082']\n\n#### re.finditer(pattern, string, flags=0)\n\n• pattern : 正则表达式的字符串或原生字符串表示\n• string : 待匹配字符串\n• flags : 正则表达式使用时的控制标记\n\n>>> for m in re.finditer(r'[1-9]\\d{5}', 'BIT 100081 TSU100082'):\n... if m:\n... print(m.group(0))\n...\n100081\n100082\n\n#### re.sub(pattern, repl, string, count=0, flags=0)\n\n• pattern: 正则表达式的字符串或原生字符串表示\n• repl: 替换匹配字符串的字符串\n• string: 待匹配字符串\n• count: 匹配的最大替换次数\n• flags: 正则表达式使用时的控制标记\n\n>>> re.sub(r'[1-9]\\d{5}',':zipcode', 'BIT 100081 TSU100082')\n'BIT :zipcode TSU:zipcode'\n\n#### Re库的另一种等价用法 => re.compile(pattern, flags=0)\n\nrst = re.search(r'[1‐9]\\d{5}', 'BIT 100081')\n\npat = re.compile(r'[1‐9]\\d{5}')\nrst = pat.search('BIT 100081')\n\n#### Re库的Match对象\n\nMatch对象是一次匹配的结果，包含匹配的很多信息\n\n>>> match = re.search(r'[1‐9]\\d{5}', 'BIT 100081')\n>>> if match:\nprint(match.group(0))\n>>> type(match)\n<class '_sre.SRE_Match'>\n\nMatch对象的属性\n\n• .string: 待匹配的文本\n• .re: 匹配时使用的pattern对象(正则表达式)\n• .pos: 正则表达式搜索文本的开始位置\n• .endpos: 正则表达式搜索文本的结束位置\n\nMatch对象的方法\n\n• .group(0): 获得匹配后的字符串\n• .start(): 匹配字符串在原始字符串的开始位置\n• .end(): 匹配字符串在原始字符串的结束位置\n• .span(): 返回(.start(), .end())\n\n>>> match = re.search(r'[1-9]\\d{5}', 'BIT 100081')\n>>> match.string\n'BIT 100081'\n>>> match.re\nre.compile('[1-9]\\\\d{5}')\n>>> match.pos\n0\n>>> match.endpos\n10\n>>> match.group(0)\n'100081'\n>>> match.start()\n4\n>>> match.end()\n10\n>>> match.span()\n(4, 10)\n\n#### Re库的贪婪匹配\n\n>>> match = re.search(r'PY.*N', 'PYANBNCNDN')\n>>> match.group(0)\n'PYANBNCNDN'\n\nRe库默认采用贪婪匹配，即输出匹配最长的子串\n\n#### Re库的最小匹配\n\n>>> match = re.search(r'PY.*?N', 'PYANBNCNDN')\n>>> match.group(0)\n'PYAN'\n\n• *?: 前一个字符0次或无限次扩展, 最小匹配\n• +?: 前一个字符1次或无限次扩展, 最小匹配\n• ??: 前一个字符0次或1次扩展, 最小匹配\n• {m,n}?: 扩展前一个字符m至n次(含n), 最小匹配" ]

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[ "Home | | Science 9th std | Distance and Displacement\n\n# Distance and Displacement\n\nThe actual length of the path travelled by a moving body irrespective of the direction is called the distance travelled by the body.\n\nDistance and Displacement\n\n## 1. Distance\n\nThe actual length of the path travelled by a moving body irrespective of the direction is called the distance travelled by the body. It is measured in metre in SI system. It is a scalar quantity having magnitude only.\n\n## 2. Displacement\n\nIt is de ned as the change in position of a moving body in a particular direction. It is vector quantity having both magnitude and direction. It is also measured in metre in SI system.", null, "", null, "Tags : Science , 9th Science : Motion\nStudy Material, Lecturing Notes, Assignment, Reference, Wiki description explanation, brief detail\n9th Science : Motion : Distance and Displacement | Science" ]

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[ "# What is the difference between electron shells and electron orbitals?\n\nMay 14, 2018\n\nThe dimensionality...?\n\n• The term \"electron shells\" originates from the Bohr model, a two-dimensional model based on the hydrogen atom only.", null, "It assumes fixed orbits (positions) and fixed trajectories (momenta), and such simultaneous known quantities disobeys the Heisenberg Uncertainty Principle. It also does not identify quantum numbers $n$, $l$, ${m}_{l}$, or ${m}_{s}$.\n\n• The term \"electron orbitals\" is part of the modern quantum theory, and is based on a three-dimensional model for any atom.\n\nSuch orbitals are regions of electron density, and are related to the probability densities of electron position over allspace, averaged over infinite time.", null, "The term \"electron shell\" lingers on through quantum theory with multi-electron atoms, labeling \"shells\" as corresponding to each principal quantum number $n$ and \"subshells\" as corresponding to each angular momentum quantum number $l$.\n\n(Electron energy levels do not exist unless the quantum state is accessed, but we still like to think of energy levels as being empty until \"occupied\".)" ]

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[ "# 21) The line y is equal to", null, "is a tangent to the curve", null, "if the value of m is (A) 1(B) 2(C) 3(D)1/2\n\nStandard equation of the straight line\ny = mx + c\nWhere m is lope and c is constant\nBy comparing it with equation , y = mx + 1\nWe find that m is the slope\nNow,\nwe know that the slope of the tangent at a given point on the curve is given by", null, "Given the equation of the curve is", null, "", null, "Put this value of m in the given equation", null, "", null, "Hence, value of m is 1\n\n## Related Chapters\n\n### Preparation Products\n\n##### JEE Main Rank Booster 2021\n\nThis course will help student to be better prepared and study in the right direction for JEE Main..\n\n₹ 13999/- ₹ 9999/-\n##### Rank Booster NEET 2021\n\nThis course will help student to be better prepared and study in the right direction for NEET..\n\n₹ 13999/- ₹ 9999/-\n##### Knockout JEE Main April 2021 (Easy Installments)\n\nAn exhaustive E-learning program for the complete preparation of JEE Main..\n\n₹ 4999/-\n##### Knockout NEET May 2021\n\nAn exhaustive E-learning program for the complete preparation of NEET..\n\n₹ 22999/- ₹ 14999/-" ]

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