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https://vedicmathsindia.org/blog/polynomial-division-2/ | [
"",
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"# Polynomial Division 2\n\nThis post is, once again, a summary of a longer post authored by me on my own blog. My blog covers a lot of areas, including Vedic Mathematics. If you are interested in reading my thoughts on other topics, please feel free to visit my blog and post comments on the other articles you find there also! Thank you!\n\nIn the previous lesson, we introduced the concept of polynomial division and how it relates to arithmetic division. We also saw how we can use the Madhyamadhyena Adhyamanthyena sutra to rewrite the dividend so that we can factor out the divisor and get a quotient and remainder. In this lesson, we will extend the method we developed, to apply it to divisors that are not linear.\n\nHow do we use the method developed in the previous lesson to perform division when the divisor contains more than 2 terms, and the unknown term is raised to powers higher than 1? One obvious way to do this would be to factorize the divisor into as many linear factors as necessary and then divide by each sequentially. This method requires no extension of the methodology developed in the previous lesson.\n\nHowever, this may not be possible for several reasons. First of all, it may not be possible to factorize the divisor into linear factors easily. Moreover, even if that were possible or easy, it becomes messy to deal with multiple remainders from multiple steps of the operation. So, how do we perform the division directly in one step? To answer this question, it is better to work out an example rather than trying to explain the procedure in abstract mathematical terms.\n\nTherefore, let us try the division of (3x^4 + 4x^3 + 7x^2 + 5x + 4) by (x^2 + x + 1). Notice that the divisor is a quadratic expression (polynomial of degree 2). Notice also that it has three terms, but in general quadratic expressions need not have 3 terms. But a full quadratic expression will have 3 terms. The three is important because it tells us that the dividend has to be arranged in sets of three terms for the division. Lastly, notice that the divisor really has no linear factors that use only real numbers (the quadratic expression itself has only complex roots).\n\nNotice that the coefficients of the three terms in the divisor are in the ratio 1:1:1. This tells us that the sets we derive for division of the dividend have to be in that ratio just as in linear division. But, what do we mean by sets of three terms? What we mean is that the dividend has to be arranged as below:\n\n3x^4 + ax^3 + bx^2 + cx^3 + dx^2 + ex + fx^2 + gx + h + jx + k\n\nOur first set of three terms starts with the x^4 term, and contains the next two lower powers of x (x^3 and x^2). Our next set of 3 terms starts with an x^3 term, and similarly contains the next two lower powers of x (x^2 and x). Our next set of 3 terms starts with an x^2 term, and contains an x term and a constant. We stop making sets when the last set contains a constant term. We will notice that this last set will also start with a term that is of the same degree as the divisor (in this case, our divisor is of degree 2, and our last set starts with an x^2 term). The sets are shown in separate colors above.\n\nAfter the last set, we write our reminder terms. Note that in general, the remainder will always be a polynomial of degree 1 lower than the divisor. Thus, in this case, jx + k is the remainder and it is of degree 1 whereas our divisor is of degree 2. One can consider the remainder term to be an incomplete set of 3 terms starting with an x term (the set contains only 2 terms, and hence is incomplete).\n\nNotice that this method is what was used in the case of division by linear divisors also in the previous lesson. In that case, each set contained 2 terms, with the second term of each set being the same degree as the first term of the next set. This positioning of terms of similar degree next to each other masked the fact that we were creating sets of 2 terms, just as we are creating sets of 3 terms here.\n\nNow that we have decided what the sets should be, what should the coefficients a, b, etc., be? This is where the ratio of coefficients in the divisor comes into play. The coefficients in each set should be in the same ratio as they are in the divisor. Thus, in the expression we wrote above, we can actually impose the following conditions:\n\n3/a = 1/1\na/b = 1/1\nc/d = 1/1\nd/e = 1/1\nf/g = 1/1\ng/h = 1/1\n\nThese establish the ratios in the two sets of terms we created. In addition, we also have to make sure that the like terms can be added to give us the original dividend. So, we write the following equations also:\n\na + c = 4\nb + d + f = 7\ne + g + j = 5\nh + k = 4\n\nWorking from the very first ratio, we can mechanically satisfy the ratios and sums by rewriting the dividend as below:\n\n3x^4 + 3x^3 + 3x^2 + x^3 + x^2 + x + 3x^2 + 3x + 3 + x + 1\n\nWe can now factorize it as below:\n\n3x^2(x^2 + x + 1) + x(x^2 + x + 1) + 3(x^2 + x + 1) + x + 1\n\nThis then tells us that our quotient is 3x^2 + x + 1, and our remainder is 3x + 3. One can indeed verify that (3x^2 + x + 3)(x^2 + x + 1) + x + 1 = 3x^4 + 4x^3 + 7x^2 + 5x + 4. Thus, our division is correct.\n\nJust as in the case of division by a linear polynomial, note that it is possible for any of the coefficients of the quotient or remainder to be fractional or zero. In addition, some of the coefficients, including the remainder terms may be negative too. We will encounter some such cases in the next couple of examples.\n\nNext let us try 5x^5 + 4x^3 + 3x^2 + 6x + 3 divided by 2x^2 + x + 2. Here, we note that the dividend is missing the x^4 term. This does not mean that we leave it out of the factorization entirely. It just means that the sum of all x^4 terms in the rewritten dividend expression should be 0.\n\nOnce again, we notice that the divisor is of degree 2, and has 3 terms. So, we need to make sets of 3, and the remainder will contain an x term and a constant. The dividend is rewritten as below:\n\n5x^5 + ax^4 + bx^3 + cx^4 + dx^3 + ex^2 + fx^3 + gx^2 + hx + jx^2 + kx + m + nx + p\n\nThe different sets of 3 terms have been colored differently to make them easier to identify. Since the terms in the divisor are in the ration 1:2:1, we need to make the terms in each set the same ratio too. In addition, we impose the following conditions:\n\na + c = 0\nb + d + f = 4\ne + g + j = 3\nh + k + n = 6\nm + p = 3\n\nThis then gives us the following rewritten equation and its factorization:\n\n5x^5 + 10x^4 + 5x^3 -10x^4 – 20x^3 – 10x^2 + 19x^3 + 38x^2 + 19x – 25x^2 – 50x – 25 + 37x + 28, which can be factorized as\n5x^3(x^2 + 2x + 1) – 10x^2(x^2 + 2x + 1) + 19x(x^2 + 2x + 1) – 25(x^2 + 2x + 1) + 37x + 28\n\nThus, the quotient is (5x^3 – 10x^2 + 19x – 25), and the remainder is 37x + 28. It is easy to verify that (5x^3 – 10x^2 + 19x – 25)(x^2 + 2x + 1) + 37x + 28 = 5x^5 + 4x^3 + 3x^2 + 6x + 3. Notice that the quotient is of degree 3, which is what we expect given that the dividend is of degree 5 and the divisor is of degree 2 (5 – 2 = 3).\n\nNow let us divide 4x^3 – 7x^2 + 15x – 12 by 2x^2 – 3. In this case, notice that the divisor is of degree 2, but has only 2 terms. It is missing the x term (therefore the coefficient of the x term in the divisor is zero). Note that if the divisor were missing the constant, then x would be a common factor in the divisor and we could divide by x separately before dividing by the rest of the divisor if we chose. But it is not necessary to factor out the divisor in any case. Simply replace the missing term by 0, and proceed as usual.\n\nThus, the above division is actually the division of 4x^3 – 7x^2 + 15x – 12 by 2x^2 + 0x – 3. Thus, we see that we have to create sets of 3 in the dividend, and the coefficients have to be in the ratio 2:0:-3. This is shown below:\n\n4x^3 + 0x^2 – 6x – 7x^2 – 0x + 10.5 + 21x – 22.5, which can be factorized as\n2x(2x^2 – 3) – 3.5(2x^2 – 3) + 24x – 22.5\n\nThus, the quotient is (2x – 3.5), and the remainder is 21x – 22.5. We see that both the quotient and remainder can have fractional terms, as mentioned before.\n\nNote that if the dividend expression is of lower degree than the divisor, the quotient is zero and the dividend becomes the remainder. For instance, if we attempt division of 3x^2 + 4x – 8 by x^3 + 4x^2 – 6x + 4, then the quotient is zero, and the remainder would be the dividend itself, which is 3x^2 + 4x – 8.\n\nNote also that the same rules as we derived in the previous lesson apply regarding divisions in which the divisor has a constant common factor. Consider, for example, the division of 5x^4 + 7x^3 – 6x^2 + 3x – 5 by 2x^2 – 4x + 2. In this case, the common factor in the divisor is 2. There are two alternative approaches to performing this division. We can divide each term in the dividend by the common factor, perform the division as usual, and then multiply the remainder by the common factor at the end. Or, we can divide first, then divide the quotient by 2 while leaving the remainder as is. Let us consider each of these methods in turn below.\n\nUsing the first method, we find the ratio of terms in the divisor is 1:-2:1. We change the dividend as below, by dividing by the common factor in the divisor, 2:\n\n2.5x^4 + 3.5x^3 – 3x^2 + 1.5x – 2.5\n\nWe then create sets of 3 terms as below for factorization:\n\n2.5x^4 – 5x^3 + 2.5x^2 + 8.5x^3 – 17x^2 + 8.5x + 11.5x^2 – 23x + 11.5 + 16x -14, which can be factorized as\n2.5x^2(x^2 – 2x + 1) + 8.5x(x^2 – 2x + 1) + 11.5(x^2 -2x + 1) + 16x – 14\n\nThis tells us that the quotient is 2.5x^2 + 8.5x + 11.5. The remainder is twice the remainder in the above line since we have to multiply it by the common factor in the divisor. So, we get a remainder of 32x – 28.\n\nNow, let us perform the division the other way. We retain the dividend as it is, but rewrite it for factorization as below:\n\n5x^4 – 10x^3 + 5x^2 + 17x^3 – 34x^2 + 17x + 23x^2 – 46x + 23 + 32x – 28 which can be factorized as\n5x^2(x^2 – 2x + 1) + 17x(x^2 – 2x + 1) + 23(x^2 – 2x + 1) + 32x – 28\n\nWe can then conclude that the quotient is one half of 5x^2 + 17x + 23, which is 2.5x^2 + 8.5x + 11.5, and the remainder is 32x – 28. Notice that the two solutions are identical in the end.\n\nFor an example of dividing by a divisor of higher degree than 2, please read the full lesson here. The methodology is the same, but there are obviously more terms involved, so it is important to be organized about it when doing it to avoid errors.\n\nHopefully, these examples have provided a good insight into how these problems need to be approached and tackled. Obviously, things are relatively easy when the ratios in the divisors are easier to deal with rather than being a wild mix of numbers. The principle still does not change, but the difficulty of performing the arithmetic operations required to implement the method will obviously increase as the ratios become more and more unwieldy. Unfortunately, that is something no system of mathematics can alleviate!\n\nWe have not dealt with the extension of this methodology to deal with arithmetic division in this lesson. We will do this in the next lesson. In the meantime, hope you will find the time to practice the kinds of divisions illustrated here so that its application to arithmetic division will be much easier. Good luck, and happy computing!"
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https://brainmass.com/economics/elasticity/calculating-price-and-advertising-elasticity-values-525459 | [
"Explore BrainMass\nShare\n\n# Calculating price and advertising elasticity values\n\nThis content was COPIED from BrainMass.com - View the original, and get the already-completed solution here!\n\nGeneral Cereals is using a regression model to estimate the demand for Tweetie Sweeties, a whistle-shaped, sugar-coated breakfast cereal for children. The following (multiplicative exponential) demand function is being used:\n\nQD = 6,280P-2.15A1.05N3.70\nwhere\nQP=quantity demanded in 10 oz. boxes\nP =price per box in dollars\nA =advertising expenditures on daytime television, in dollars\nN =proportion of the population under 12 years old\n\na. Determine the point price elasticity of demand for Tweetie Sweeties.\nb. Determine the advertising elasticity of demand.\nc. What interpretation would you give to the exponent of N?\n\n#### Solution Preview\n\na. Determine the point price elasticity of demand for Tweetie Sweeties.\n\nQD= 6,280*P^(-2.15)*A^1.05*N^3.70\n\nDifferentiate with respect to P, we get\ndQD/dP=-2.25*6280*P^(-2.15-1)*A^1.05*N^3.70\n=-2.25*6280*P^(-2.15) ...\n\n#### Solution Summary\n\nSolution depicts the steps to calculate the price elasticity and advertising elasticity of demand in the given case.\n\n\\$2.19"
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https://www.wikihow.com/Calculate-Force-of-Gravity | [
"# How to Calculate Force of Gravity",
null,
"Download Article",
null,
"Download Article\n\nGravity is one of the fundamental forces of physics. The most important aspect of gravity is that it is universal: all objects have a gravitational force that attracts other objects to them. The force of gravity acting on any object is dependent upon the masses of both objects and the distance between them.\n\n### Part 1 Part 1 of 2:Calculating the Force of Gravity Between Two Objects\n\n1. 1\nDefine the equation for the force of gravity that attracts an object, Fgrav = (Gm1m2)/d2. In order to properly calculate the gravitational force on an object, this equation takes into account the masses of both objects and how far apart the objects are from each other. The variables are defined below.\n• Fgrav is the force due to gravity\n• G is the universal gravitation constant 6.673 x 10-11 Nm2/kg2\n• m1 is the mass of the first object\n• m2 is the mass of the second object\n• d is the distance between the centers of two objects\n• Sometimes you will see the letter r instead of the letter d. Both symbols represent the distance between the two objects.\n2. 2\nUse the proper metric units. For this particular equation, you must use metric units. The masses of objects need to be in kilograms (kg) and the distance needs to be in meters (m). You must convert to these units before continuing with the calculation.\n3. 3\nDetermine the mass of the object in question. For smaller objects, you can weigh them on a scale or balance to determine their weight in grams. For larger objects, you will have to look-up the approximate mass in a table or online. In physics problems, the mass of the object will generally be provided to you.\n4. 4\nMeasure the distance between the two objects. If you are trying to calculate the force of gravity between an object and the earth, you need to determine how far away the object is from the center of the earth.\n• The distance from the surface of the earth to the center is approximately 6.38 x 106 m.\n• You can find tables and other resources online that will provide you with approximate distances of the center of the earth to objects at various elevations on the surface.\n5. 5\nSolve the equation. Once you have defined the variables of your equation, you can plug them in and solve. Be sure that all of your units are in metric and on the right scale. Mass should be in kilograms and distance in meters. Solve the equation using the proper order of operations.\n• For example: Determine the force of gravity on a 68 kg person on the surface of the earth. The mass of the earth is 5.98 x 1024 kg.\n• Make sure all your variables have the proper units. m1 = 5.98 x 1024 kg, m2 = 68 kg, G = 6.673 x 10-11 Nm2/kg2, and d = 6.38 x 106 m\n• Write your equation: Fgrav = (Gm1m2)/d2 = [(6.67 x 10-11) x 68 x (5.98 x 1024)]/(6.38 x 106)2\n• Multiply the masses of the two objects together. 68 x (5.98 x 1024) = 4.06 x 1026\n• Multiply the product of m1 and m2 by the gravitational constant G. (4.06 x 1026) x (6.67 x 10-11) = 2.708 x 1016\n• Square the distance between the two objects. (6.38 x 106)2 = 4.07 x 1013\n• Divide the product of G x m1 x m2 by the distance squared to find the force of gravity in Newtons (N). 2.708 x 1016/4.07 x 1013 = 665 N\n• The force of gravity is 665 N.\n\n### Part 2 Part 2 of 2:Calculating the Force of Gravity on Earth\n\n1. 1\nUnderstand Newton’s Second Law of Motion, F = ma. Newton’s second law of motion states that any object will accelerate when acted upon by a net or unbalanced force. In other words, if a force is acting upon an object that is greater than the forces acting in the opposite direction, the object will accelerate in the direction of the larger force.\n• This law can be summed up with the equation F = ma, where F is the force, m is the mass of the object, and a is acceleration.\n• Using this law, we can calculate the force of gravity of any object on the surface of the earth, using the known acceleration due to gravity.\n2. 2\nKnow the acceleration due to gravity on earth. On earth, the force of gravity causes objects to accelerate at a rate of 9.8 m/s2. On the earth’s surface, we can use the simplified equation Fgrav = mg to calculate the force of gravity.\n• If you want a more exact approximation of force, you can still use the above equation, Fgrav = (GMearthm)/d2 to determine force of gravity.\n3. 3\nUse the proper metric units. For this particular equation, you must use metric units. The mass of the object needs to be in kilograms (kg) and the acceleration needs to be in meters per second squared (m/s2). You must convert to these units before continuing with the calculation.\n4. 4\nDetermine the mass of the object in question. For smaller objects, you can weigh them on a scale or balance to determine its weight in kilograms (kg). For larger objects, you will have to look-up the approximate mass in a table or online. In physics problems, the mass of the object will generally be provided to you.\n5. 5\nSolve the equation. Once you have defined the variables of your equation, you can plug them in and solve. Be sure that all of your units are in metric and on the right scale. Mass should be in kilograms and distance in meters. Solve the equation using the proper order of operations.\n• Let’s use the same equation from above and see how close the approximation is. Determine the force of gravity on a 68 kg person on the surface of the earth.\n• Make sure all your variables have the proper units: m = 68 kg, g = 9.8 m/s2.\n• Write your equation. Fgrav = mg = 68*9.8 = 666 N.\n• With F = mg the force of gravity is 666 N, while using the more exact equation yields a force of 665 N. As you can see, these values are almost identical.\n\n## Community Q&A\n\nSearch\n• Question\nHow do I find the mass of the moon?",
null,
"Check out same steps as mentioned below. But remember gravity on moon is 1/6th of gravity on earth.\n• Question\nA mass of 25 kg weighs 123 Newtons on another planet. What is the gravity on the planet?",
null,
"The \"gravity\" on the surface of a planet is it's acceleration (the rate of increase in speed as an object falls). Fg (the force of gravity) is m x g (acceleration of gravity), in m/(s squared), so g is Fg / m = 123 N / 25 kg ~= 4.92 m/(s squared).\n• Question\nHow do I find the value of acceleration due to a gravity at a height of 2R from the surface of the earth?",
null,
"If you want to know what the gravity would be when you are 3 earth-radii away from the center of earth, then the gravity would be 1/9th normal gravity. You're multiplying by 3 on the bottom, so 1/3, but then it's squared. Acceleration would then be 1.09 meters per second squared.\n• Question\nCan we use the diameter of a planet for \"R\" in the gravity formula?",
null,
"\"R\" is the radius. The diameter is twice the radius. So just divide the diameter by two to get the radius.\n• Question\nIs the gravitational constant the same on other planets?",
null,
"No, because different planets have different masses.\n• Question\nHow do I calculate gravity if I know the N of gravity and the weight of the item?",
null,
"If you know the force in newtons, 'N' and the mass of the item in kilograms, then you can apply the formula F = m * a. By rearranging this equation, the acceleration of an object (the gravity) is: a = F / m where 'F' is newtons, 'm' is mass in kilograms and 'a' is acceleration (due to gravity in this case) in meters per second per second.\n• Question\nIf I have two identical billy carts placed on top of a hill, one with 150mm wheels the other with 400 mm wheels, which one will get to the bottom of the hill first?",
null,
"Having smaller wheels means the mass is closer to the centre, therefore the gravitational force is higher than with larger wheels.\n• Question\nMr. Thurman has a mass of 67 kg. What is the force of gravity pressing down on him?",
null,
"Startrekker\nW=mg = 67.10 kg m/s2 = 670 N Weight of Mr. Thurman is 670N Fun note: Weight machines actually show the normal reaction, and then divide it by acceleration due to gravity (g). If the weighing machine is in a horizontal position, it will measure the normal reaction as 670N. The computer then divides it by 10 and shows the reading as 67 kg.\n200 characters left\n\n## Tips\n\n• You may round off 9.8m/s2 to 10m/s2, to make calculations easier.\nThanks!\n• These two formulas should give the same result, but the shorter formula is simpler to use when discussing objects on a planet's surface.\nThanks!\n• Use the first formula if you don’t know the acceleration due to gravity on a planet or if you’re determining the force of gravity between two very large objects such as a moon and a planet.\nThanks!\nSubmit a Tip\nAll tip submissions are carefully reviewed before being published\nThanks for submitting a tip for review!",
null,
"Co-authored by:\nwikiHow Staff Writer\nThis article was co-authored by wikiHow Staff. Our trained team of editors and researchers validate articles for accuracy and comprehensiveness. wikiHow's Content Management Team carefully monitors the work from our editorial staff to ensure that each article is backed by trusted research and meets our high quality standards. This article has been viewed 560,742 times.\nCo-authors: 22\nUpdated: June 24, 2022\nViews: 560,742\nCategories: Physics\nArticle SummaryX\n\nTo calculate the force of gravity of an object, use the formula: force of gravity = mg, where m is the mass of the object and g is the acceleration of the object due to gravity. Since g is always 9.8 m/s^2, just multiply the object's mass by 9.8 and you'll get its force of gravity! If you want to learn how to calculate the force of gravity between 2 objects, keep reading the article!"
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https://apcocoa.uni-passau.de/wiki/index.php?title=ApCoCoA-1:LinBox.CharPoly&diff=prev&oldid=8775 | [
"# Difference between revisions of \"ApCoCoA-1:LinBox.CharPoly\"\n\n## LinBox.CharPoly\n\nCompute the characteristic polynomial of a matrix.\n\n### Syntax\n\n```LinBox.CharPoly(M:MAT, X:POLY):LIST\n```\n\n### Description\n\nX is an indeterminate, and M is a square matrix whose entries do not involve X.\n\nThis function returns the characteristic polynomial of M in the indeterminate X computed by the ApCoCoAServer using LinBox functions.\n\n• @param M A matrix with whose components do not contain the indeterminate X.\n\n• @param X An indeterminate.\n\n• @return The characteristic polynomial of M in the indeterminate X.\n\n#### Example\n\n```Use R ::= Z/(19)[x];\nLinBox.CharPoly(BringIn(Mat([[1,2,3],[4,5,6],[7,8,9]])), x);\n-- CoCoAServer: computing Cpu Time = 0\n-------------------------------\nx^3 + 4x^2 + x\n-------------------------------\n\nUse R ::= Z[x];\nLinBox.CharPoly(Mat([[1,2,3],[4,5,6],[7,8,9]]), x);\n-- WARNING: Coeffs are not in a field\n-- GBasis-related computations could fail to terminate or be wrong\n\n-------------------------------\n-- WARNING: Coeffs are not in a field\n-- GBasis-related computations could fail to terminate or be wrong\n-- CoCoAServer: computing Cpu Time = 0\n-------------------------------\nx^3 - 15x^2 - 18x\n-------------------------------\n```"
]
| [
null
]
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https://stats.stackexchange.com/questions/265307/assessing-proportional-hazards-assumption-of-a-cox-model-with-caseweights | [
"# Assessing Proportional Hazards Assumption of a Cox Model with Caseweights\n\nI have data that I have used the R package twang with multinomial propensity scoring, to assign caseweights based on an ATT model. Once these caseweights have been assigned, they are typically used with a survey procedure. In R, there is the survey package which contains svycoxph, but this can also be done using the survival package with + cluster(id) in the model, to compute robust standard errors (gives same result for my caseweights).\n\nI am have satisfactorily fit the non-propensity scored data as such, after reviewing cox.zph test and the Schoenfeld residual plots:\n\n> coxph1 <- coxph(Surv(time,status) ~ x1 + x2 + age + tt(age),\ndata=data, tt=function(x,t,...) x * log(t))\n\n> cox.zph(coxph1)\nrho chisq p\nx1 -0.0306 1.54 0.215\nx2 -0.0367 2.21 0.137\nage -0.5752 37284.88 0.000\ntt(age) 0.5789 48667.57 0.000\nGLOBAL NA 48676.43 0.000\n\n\nx1 and x2 are categorical (1=yes, 0=no) predictors.\n\nI have removed some variables for simplicity of explanation, disregard the age variable.\n\nWhen I fit a model using the weights and robust standard error measurement, I am concerned that the use of the weights are obscuring the cox.zph and Schoenfeld plots:\n\n> coxph2 <- coxph(Surv(time,status)~ x1 + x2 + age + tt(age) + cluster(id), weights=w, data=data, tt=function(x,t,...) x * log(t))\n> cox.zph(coxph2)\nrho chisq p\nx1 0.182 60.1 0.00000000000000888\nx2 -0.128 42.5 0.00000000006934375\nage -0.581 54338.6 0.00000000000000000\ntt(age) 0.586 70624.8 0.00000000000000000\nGLOBAL NA 70734.7 0.00000000000000000\n\n\nThe Schoenfeld plots look accordingly bad.",
null,
"This gets even worse as more predictors are added (in terms of how it looks in comparison to the unweighted version).\n\nSide question: x1 has a maximum follow up time of 60, whereas x2 and the reference category have follow up to 100 months. Is there any chance this is having an impact on the caseweights/their relation to the residual?\n\nThis issue persists if using the survey package and svycox.\n\nI have two questions: 1) Is there in fact an issue with using caseweights and or robust SE estimation in how the Schoenfeld residual is constructed, and should these be avoided? 2) If 1 is true, is there another preferred method for testing the PH assumption? So far, I have been adding a time transformation tt() for each and seeing if that is significant, but that is far from an ideal approach. The only other alternative I see is to treat those that violate the assumption in the unweighted model the same way in the weighted model.\n\nUpdate:\n\nI think I have found a solution to the problem. I edited the original cox.zph function to included weighted Schoenfeld residuals:\n\nHere was the original:\n\ncox.zph <- function(fit, transform='km', global=TRUE) {\ncall <- match.call()\nif (!inherits(fit, 'coxph')) stop (\"Argument must be the result of coxph\")\nif (inherits(fit, 'coxph.null'))\nstop(\"The are no score residuals for a Null model\")\n\nsresid <- resid(fit, 'schoenfeld')\nvarnames <- names(fit$coefficients) nvar <- length(varnames) ndead<- length(sresid)/nvar if (nvar==1) times <- as.numeric(names(sresid)) else times <- as.numeric(dimnames(sresid)[]) And the change: cox.zph <- function(fit, transform='km', global=TRUE) { call <- match.call() if (!inherits(fit, 'coxph')) stop (\"Argument must be the result of coxph\") if (inherits(fit, 'coxph.null')) stop(\"The are no score residuals for a Null model\") sresid <- resid(fit, 'schoenfeld', **weighted=TRUE**) varnames <- names(fit$coefficients)\nnvar <- length(varnames)\n\n• You were focusing on weights but your model has a tt component and the code of cox.zph says: \"stop(\"function not defined for models with tt() terms\")\". And as I understand the situation, the express purpose of tt terms is to allow management of certain kinds on non-proportionality.I think you would get further information in Therneau & Grampsch, ch 6 sect 5."
]
| [
null,
"https://i.stack.imgur.com/1gm5G.png",
null
]
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https://edurev.in/studytube/Combination-of-ResistorsSeries-and-Parallel-Curren/98f50fbc-5b8f-468c-b36b-db494efeb536_t | [
"NEET > Resistors in Series & Parallel Combinations\n\n# Resistors in Series & Parallel Combinations | Physics Class 12 - NEET\n\n 1 Crore+ students have signed up on EduRev. Have you?\n\nCombination of Resistance\n\nA number of resistance can be connected in a circuit and any complicated combination can be, in general, reduced essentially to two different types, namely series and parallel combinations.\n\nResistance in Series",
null,
"• In this combination the resistance are joined end to end. The second end of each resistance is joined to first end of the next resistance and so on. A cell is connected between the first end of first resistance and second end of last resistance. Figure shows three resistances R1, R2 and R3 connected in this way. Let V1, V2 and V3 are the potential differences across these resistances.\n• In this combination current flowing through each resistance will be same and will be equal to current supplied by the battery.\n• As resistances are different and current flowing through them is same, hence potential differences across them will be different. Applied potential difference will be distributed among three resistances directly in their ratio.\n• As i is constant, hence V is directly proportional to R\ni.e., V1 = iR1, V2 = iR2, v3 = iR3\n• If the potential difference between the points A and D is V, then\nV = V1 + V+ V= I(R1 + R+ R3)\n• If the combination of resistances between two points is replaced by a single resistance R such that there is no change in the current of the circuit in the potential difference between those two points, then the single resistance R will be equivalent to combination and V = i R i.e.,\nIR = I(R+ R+ R3)\nR = R+ R2 + R3\n• Thus in series combination of resistances, important conclusion are:\n(i) Equivalent Resistance > highest individual resistance\n(ii) Current supplied by source = Current in each resistance\nor",
null,
"=",
null,
"=",
null,
"=",
null,
"(iii) The total potential difference V between points A and B is shared among the three resistances directly in their ratio\nV1 : V2 : V3 = R1 : R2 : R3\n\nResistance in Parallel",
null,
"• When two or more resistance are combined in such a way that their first ends are connected to one terminal of the battery while other ends are connected to other terminal, then they are said to be connected in parallel. Figure shows three resistances R1, R2 and R3 joined in parallel between two points A and B. Suppose the current flowing from the battery is i. This current gets divided into three parts at the junction A. Let the currents in three resistance R, R2 and R3, are i1, i2, irespectively.\n• Suppose potential difference between points A and B is V. Because each resistance is connected between same two points A and B, hence potential difference across each resistance will be same and will be equal to applied potential difference V.\n• Since potential difference across each resistance is same, hence current approaching the junction A is divided among three resistances reciprocally in their ratio.\nAs V is constant, hence i µ (1/R) i.e.,",
null,
"",
null,
"• Because i the main current which is divided into three parts i1, i2 and i3 at the junction A.\nhence,",
null,
"• If the equivalent resistance between the points A and B is R, then i =",
null,
"Thus,",
null,
"or",
null,
"• Thus in parallel combination of resistance important conclusion are :\n(i) Equivalent resistance < lowest individual resistance\n(ii) Applied potential difference = Potential difference across each resistance.\nor iR = i1R1 = i2R2 = i3R3\n(iii) Current approaching the junction A = Current leaving the junction B and current is shared among the three resistances in the inverse ratio of resistances\ni1 : i2 : i3 =",
null,
"Note: (i) If two or more resistance are joined in parallel then i1R = i1R2 = i3R3............\n\ni.e., iR = constant i.e., a low resistance joined in parallel always draws a higher current.\n\n(ii) When two resistance R1 and R2 are joined in parallel, then",
null,
"or",
null,
"or",
null,
"or",
null,
"i.e., heat produced will be maximum in the lowest resistance.\n\nExample1. Find current which is passing through battery.",
null,
"Sol. Here potential difference across each resistor is not 30 V\n\nQ battery has internal resistance here the concept of combination of resistors is useful.\n\nReq = 1 1 = 2?",
null,
"Example 2. Find equivalent Resistance",
null,
"Sol.",
null,
"Here all the Resistance are connected between the terminals A and B. So, Modified circuit is\n\nSo Req=",
null,
"",
null,
"Example 3. Find the current in Resistance P if voltage supply between A and B is V volts",
null,
"Sol. Req =",
null,
"",
null,
"I =",
null,
"Modified circuit",
null,
"Current in",
null,
"=",
null,
"",
null,
"Example 4. Find the current in 2 W resistance.",
null,
"Sol. 2Ω, 1Ω in series = 3Ω\n\n3Ω, 6Ω in parallel =",
null,
"",
null,
"2Ω, 4Ω in series = 6Ω\n\n6Ω, 3Ω is parallel = 2Ω\n\nReq = 4 + 4 + 2 = 10 Ω",
null,
"So current in 2 ? Resistance =",
null,
"",
null,
"Special Problems",
null,
"We wish to determine equivalent resistance between A and B. In figure shown points (1,2) (3, 4, 5) and (6, 7) are at same potential Equivalent circuit can be redrawn as in figure shown.\n\nThe equivalent resistance of this series combination is",
null,
"",
null,
"",
null,
"In the figure shown, the resistances specified are in ohms. We wish to determine the equivalent resistance between point A and D. Point B and C, E and F are the the same potential so the circuit can be redrawn as in figure shown.\n\nThus the equivalent resistance is 1 ?.",
null,
"",
null,
"",
null,
"In the network shown in figure shown all the resistances are equal, we wish to determine equivalent resistance between A and E. Point B and D have same potential, similarly F and H have same potential. The equivalent circuit is shown in figure shown. The equivalent resistance of network is 7R/2.",
null,
"",
null,
"The document Resistors in Series & Parallel Combinations | Physics Class 12 - NEET is a part of the NEET Course Physics Class 12.\nAll you need of NEET at this link: NEET\n\n## Physics Class 12\n\n157 videos|452 docs|213 tests\n\n## Physics Class 12\n\n157 videos|452 docs|213 tests\n\n### How to Prepare for NEET\n\nRead our guide to prepare for NEET which is created by Toppers & the best Teachers\n\nTrack your progress, build streaks, highlight & save important lessons and more!",
null,
"(Scan QR code)\n\n,\n\n,\n\n,\n\n,\n\n,\n\n,\n\n,\n\n,\n\n,\n\n,\n\n,\n\n,\n\n,\n\n,\n\n,\n\n,\n\n,\n\n,\n\n,\n\n,\n\n,\n\n;"
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null
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| {"ft_lang_label":"__label__en","ft_lang_prob":0.93906003,"math_prob":0.9967197,"size":5283,"snap":"2023-14-2023-23","text_gpt3_token_len":1256,"char_repetition_ratio":0.19909073,"word_repetition_ratio":0.06583072,"special_character_ratio":0.240583,"punctuation_ratio":0.1208577,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9989391,"pos_list":[0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26,27,28,29,30,31,32,33,34,35,36,37,38,39,40,41,42,43,44,45,46,47,48,49,50,51,52,53,54,55,56,57,58,59,60,61,62,63,64,65,66,67,68,69,70,71,72,73,74,75,76,77,78,79,80,81,82,83,84,85,86,87,88,89,90,91,92,93,94],"im_url_duplicate_count":[null,2,null,2,null,2,null,2,null,2,null,2,null,2,null,2,null,2,null,2,null,2,null,2,null,2,null,2,null,2,null,2,null,2,null,2,null,2,null,2,null,2,null,2,null,2,null,2,null,2,null,2,null,2,null,2,null,2,null,2,null,2,null,2,null,2,null,2,null,2,null,2,null,2,null,2,null,2,null,2,null,2,null,2,null,2,null,2,null,2,null,2,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-06-05T13:40:32Z\",\"WARC-Record-ID\":\"<urn:uuid:ddbe660f-bc70-4fde-aefb-044de8e45f37>\",\"Content-Length\":\"437299\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:b0f917e0-ff50-47a5-9972-f4f7c651e3a5>\",\"WARC-Concurrent-To\":\"<urn:uuid:bde9e436-b2b2-4466-b17e-408c9ab77043>\",\"WARC-IP-Address\":\"20.198.201.88\",\"WARC-Target-URI\":\"https://edurev.in/studytube/Combination-of-ResistorsSeries-and-Parallel-Curren/98f50fbc-5b8f-468c-b36b-db494efeb536_t\",\"WARC-Payload-Digest\":\"sha1:HEI4NDM3IZZNAB3YZ6FWU6WQU6H5N2TX\",\"WARC-Block-Digest\":\"sha1:ZZPUPI6J7YN7V524T5GQMMOMBPG356KJ\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-23/CC-MAIN-2023-23_segments_1685224652116.60_warc_CC-MAIN-20230605121635-20230605151635-00329.warc.gz\"}"} |
https://scicomp.stackexchange.com/questions/14369/what-are-some-reasons-that-conjugate-gradient-iteration-does-not-converge | [
"# What are some reasons that Conjugate Gradient iteration does not converge?\n\nI would greatly appreciate it if you could share some reasons the Conjugate Gradient iteration for Ax = b does not converge? My matrix A is symmetric positive definite.\n\nThank you so much!\n\nEdit with more information: My matrix is the reduced Hessian in the optimization algorithms for problems with simple constraints. The matrix is 50% sparse and I am using matlab.\n\n• Can you provide any extra information? What is the structure of your matrix, is it dense or sparse? Where the matrix come from, is from FEM, FDM? Which programming language are you using? – nicoguaro Aug 12 '14 at 23:06\n• Thanks. I edited my question. Let me know if I need to provide more information. – coding Aug 12 '14 at 23:25\n• What's the condition number of your matrix? – Brian Borchers Aug 13 '14 at 2:24\n• And how do you know the matrix is in fact spd (as opposed to \"should be spd\")? Is it explicitly given, or are you using a matrix-free CG (i.e., you have a procedure to compute $Ax$ for given $x$)? And finally, what do you mean by \"does not converge\"? Are you sure it's not just converging (very) slowly? – Christian Clason Aug 13 '14 at 7:27\n• Could you elaborate a little more about the reduced Hessian? Matrices that arise from the Karush-Kuhn-Tucker conditions can be symmetric indefinite, for instance. How much do you know about the spectrum of your matrix? – Geoff Oxberry Aug 13 '14 at 18:51\n\nIf your matrix is symmetric, positive definite, the CG method may converge slowly, but it converges for $n\\to\\infty$. The only reason it does not converge on a computer are round-off errors, in particular if the condition number of the matrix, the quotient of largest and smallest eigenvalue is large.\n\nExperience is, that with double precision arithmetic, even with moderate condition numbers convergence stalls at around $10^{-12}$ to $10^{-14}$ of the original error.\n\n• In the case of symmetric positive definite matrix, the CG method converges (in theory) in $k$ iterations, where $k$ is the size of $b$. Like you said, it will stall much earlier than that in practice with finite precision. – LKlevin Aug 15 '14 at 15:37\n• To add to LKlevin, if CG executes enough iterations, the produced conjugate vectors can lose orthogonality numerically, which can also lead to nonconvergence. CG may be \"restarted\" to offset this. – Jesse Chan Aug 15 '14 at 15:39\n\nProbabily the CG fails to converge because your problem is ill conditioned, the condition number of your matrix is too large. For SPD matrix A you can get the condition number calculating the eigenvalues\n\ncodition_number = largest_eigenvalue/smallest_eigenvalue\n\nIll conditioned means that small changes on some entries of the matrix could give large changes in the result when solving the system of equations (in a well conditioned matrix a small change in any entry of the matrix gives a small change in the result).\n\nTo reduce the condition number of the problem you can use a preconditioned CG, there are several preconditioners. The Jacobi preconditioner (also known as diagonal preconditioner) is easy to implement but not robust."
]
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| {"ft_lang_label":"__label__en","ft_lang_prob":0.92534435,"math_prob":0.96952575,"size":365,"snap":"2019-35-2019-39","text_gpt3_token_len":73,"char_repetition_ratio":0.11634349,"word_repetition_ratio":0.0,"special_character_ratio":0.19452055,"punctuation_ratio":0.08955224,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9939089,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-09-16T10:28:20Z\",\"WARC-Record-ID\":\"<urn:uuid:e521791b-1b0f-4d49-8e67-35b300a50471>\",\"Content-Length\":\"148982\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:270ce8e1-a9f5-421f-884a-e50b547915c2>\",\"WARC-Concurrent-To\":\"<urn:uuid:de7a25cd-d800-42e7-b6bf-4280273acbbc>\",\"WARC-IP-Address\":\"151.101.193.69\",\"WARC-Target-URI\":\"https://scicomp.stackexchange.com/questions/14369/what-are-some-reasons-that-conjugate-gradient-iteration-does-not-converge\",\"WARC-Payload-Digest\":\"sha1:OGEBUNENIEVHU262KEIJ4DX5WEVWDQ2P\",\"WARC-Block-Digest\":\"sha1:Q5ZYU4GV6FADLY7EDC32JAZN2MUJZGZ3\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-39/CC-MAIN-2019-39_segments_1568514572517.50_warc_CC-MAIN-20190916100041-20190916122041-00369.warc.gz\"}"} |
https://apps.dtic.mil/sti/citations/AD0768034 | [
"# Abstract:\n\nThe purpose of the report is to develop a means of estimating, in real time, the average frequency of a random process, given only a time-limited sample of that process. The average frequency of a random process is defined in terms of the power spectral density. Previous estimators of average frequency were mathematically based on estimating the power spectral density. In this work, the average frequency is expressed in terms of the autocorrelation and cross-correlation functions, and a new estimator is obtained which is based on estimates of these correlation functions. Two estimators presented in the literature and the proposed estimator are compared, and expressions are obtained for the power spectral density and autocorrelation function at each point in the systems, exclusive of the outputs. Approximate expressions are obtained for the bias and variance of the proposed estimator and one of the referenced estimators. A lab model of the proposed device was constructed and tested under several sets of operating conditions. It is found that, under typical operating conditions, the proposed estimator performs almost as well as those proposed earlier, and is simpler in form. For very short samples of the random process, however, the difference in variance becomes more noticeable.\n\n# Subject Categories:\n\n• Meteorology\n• Statistics and Probability\n• Cybernetics"
]
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| {"ft_lang_label":"__label__en","ft_lang_prob":0.9449427,"math_prob":0.96060663,"size":1661,"snap":"2020-45-2020-50","text_gpt3_token_len":330,"char_repetition_ratio":0.14121906,"word_repetition_ratio":0.01632653,"special_character_ratio":0.18482842,"punctuation_ratio":0.112676054,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9868369,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-10-31T08:57:30Z\",\"WARC-Record-ID\":\"<urn:uuid:2db0cc1a-dd9a-43bd-907b-645c5135abcf>\",\"Content-Length\":\"60759\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:192f7fce-05d5-4119-a019-8565702c0672>\",\"WARC-Concurrent-To\":\"<urn:uuid:8c7f1201-a40d-4c8f-8ff8-7570aef13a5e>\",\"WARC-IP-Address\":\"131.84.180.30\",\"WARC-Target-URI\":\"https://apps.dtic.mil/sti/citations/AD0768034\",\"WARC-Payload-Digest\":\"sha1:QPIJN3S3PVW3MBVYUO4ZMEVBWDOMN6JQ\",\"WARC-Block-Digest\":\"sha1:3ASR3DD4XJWJATPML2J4INM5OSAR45Q3\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-45/CC-MAIN-2020-45_segments_1603107916776.80_warc_CC-MAIN-20201031062721-20201031092721-00628.warc.gz\"}"} |
http://reference.iucr.org/dictionary/Point_group | [
"# Point group\n\nJump to: navigation, search\n\nزمرة نقطية (Ar). Groupe ponctuel (Fr). Punktgruppe (Ge). Gruppo punto (It). 点群 (Ja). Точечная группа симметрии (Ru). Grupo puntual (Sp).\n\n## Definition\n\nA point group is a group of symmetry operations all of which leave at least one point unmoved. A crystallographic point group is a point group that maps a point lattice onto itself: in three dimensions the symmetry operations of these groups are restricted to 1, 2, 3, 4, 6 and",
null,
"$\\bar 1$,",
null,
"$\\bar 2$ (= m),",
null,
"$\\bar 3$,",
null,
"$\\bar 4$,",
null,
"$\\bar 6$ respectively.\n\n## Occurrence\n\nCrystallographic point groups occur:\n\n• in vector space, as symmetries of the external shapes of crystals (morphological symmetry), as well as symmetry of the physical properties of the crystal ('vector point group');\n• in point space, as site-symmetry groups of points in lattices or in crystal structures, and as symmetries of atomic groups and coordination polyhedra ('point point group').\n\n## Controversy on the nomenclature\n\nThe matrix representation of a symmetry operation consists of a linear part, which represents the rotation or rotoinversion component of the operation, and a vector part, which gives the shift to be added once the linear part of the operation has been applied. The vector part is divided into two components: the intrinsic component, which represents the screw and glide component of the operation, and the location component, which is non-zero when the symmetry element does not pass through the origin. The set of the linear parts of the matrices representing the symmetry operations of a space group is a representation of the point group of the crystal. On the other hand, the set of matrix-vector pairs representing the symmetry operations of a site symmetry group form a group which is isomorphic to a crystallographic point group. The vector part being in general non-zero, some authors reject the term point group for the site-symmetry groups. On the other hand, all the symmetry operations of a site symmetry group do leave invariant at least one point, albeit not necessarily the origin, satisfying the above definition of point group.\n\n## See also\n\n• Chapter 3.2.1 of International Tables for Crystallography, Volume A, 6th edition"
]
| [
null,
"http://reference.iucr.org/mediawiki/images/math/8/1/0/810e8427631fcd88eb979030fdeed2fa.png ",
null,
"http://reference.iucr.org/mediawiki/images/math/7/a/c/7ac549c16e7439d73482ebc8c35750c7.png ",
null,
"http://reference.iucr.org/mediawiki/images/math/3/6/2/36248c80ba63a551f47e4c03d7aced52.png ",
null,
"http://reference.iucr.org/mediawiki/images/math/5/f/a/5fa6e8caaed8dbf6572dfdc9b029d1c1.png ",
null,
"http://reference.iucr.org/mediawiki/images/math/e/8/1/e81825d7900430a10046d735b6479b4b.png ",
null
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| {"ft_lang_label":"__label__en","ft_lang_prob":0.84564483,"math_prob":0.9729746,"size":2188,"snap":"2019-13-2019-22","text_gpt3_token_len":491,"char_repetition_ratio":0.1923077,"word_repetition_ratio":0.037463978,"special_character_ratio":0.20155393,"punctuation_ratio":0.11809045,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9913233,"pos_list":[0,1,2,3,4,5,6,7,8,9,10],"im_url_duplicate_count":[null,null,null,4,null,6,null,4,null,4,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-05-21T09:03:00Z\",\"WARC-Record-ID\":\"<urn:uuid:40c9e277-c63b-41f6-a827-a1a7fe491b87>\",\"Content-Length\":\"16301\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:74fcd8e6-5cba-4017-9b23-0f35a1714f91>\",\"WARC-Concurrent-To\":\"<urn:uuid:4872fc71-b26b-4540-b0b5-191748119ec4>\",\"WARC-IP-Address\":\"192.70.242.91\",\"WARC-Target-URI\":\"http://reference.iucr.org/dictionary/Point_group\",\"WARC-Payload-Digest\":\"sha1:KLQLNN5TBDD3QFAYSWRTSDXUQJURQDKA\",\"WARC-Block-Digest\":\"sha1:QX3TIT7LTND2G54BG72TFO27Q72QOKVA\",\"WARC-Identified-Payload-Type\":\"application/xhtml+xml\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-22/CC-MAIN-2019-22_segments_1558232256314.25_warc_CC-MAIN-20190521082340-20190521104340-00433.warc.gz\"}"} |
https://www.crazy-numbers.com/en/6503 | [
"Discover a lot of information on the number 6503: properties, mathematical operations, how to write it, symbolism, numerology, representations and many other interesting things!\n\n## Mathematical properties of 6503\n\nIs 6503 a prime number? No\nIs 6503 a perfect number? No\nNumber of divisors 4\nList of dividers 1, 7, 929, 6503\nSum of divisors 7440\nPrime factorization 7 x 929\nPrime factors 7, 929\n\n## How to write / spell 6503 in letters?\n\nIn letters, the number 6503 is written as: Six thousand five hundred and three. And in other languages? how does it spell?\n\n6503 in other languages\nWrite 6503 in english Six thousand five hundred and three\nWrite 6503 in french Six mille cinq cent trois\nWrite 6503 in spanish Seis mil quinientos tres\nWrite 6503 in portuguese Seis mil quinhentos três\n\n## Decomposition of the number 6503\n\nThe number 6503 is composed of:\n\n1 iteration of the number 6 : The number 6 (six) is the symbol of harmony. It represents balance, understanding, happiness.... Find out more about the number 6\n\n1 iteration of the number 5 : The number 5 (five) is the symbol of freedom. It represents change, evolution, mobility.... Find out more about the number 5\n\n1 iteration of the number 0 : ... Find out more about the number 0\n\n1 iteration of the number 3 : The number 3 (three) is the symbol of the trinity. He also represents the union.... Find out more about the number 3\n\nOther ways to write 6503\nIn letter Six thousand five hundred and three\nIn roman numeral MMMMMMDIII\nIn binary 1100101100111\nIn octal 14547\nIn US dollars USD 6,503.00 (\\$)\nIn euros 6 503,00 EUR (€)\nSome related numbers\nPrevious number 6502\nNext number 6504\nNext prime number 6521\n\n## Mathematical operations\n\nOperations and solutions\n6503*2 = 13006 The double of 6503 is 13006\n6503*3 = 19509 The triple of 6503 is 19509\n6503/2 = 3251.5 The half of 6503 is 3251.500000\n6503/3 = 2167.6666666667 The third of 6503 is 2167.666667\n65032 = 42289009 The square of 6503 is 42289009.000000\n65033 = 275005425527 The cube of 6503 is 275005425527.000000\n√6503 = 80.641180546914 The square root of 6503 is 80.641181\nlog(6503) = 8.7800188878692 The natural (Neperian) logarithm of 6503 is 8.780019\nlog10(6503) = 3.8131137540079 The decimal logarithm (base 10) of 6503 is 3.813114\nsin(6503) = -0.096641861568776 The sine of 6503 is -0.096642\ncos(6503) = 0.99531922044765 The cosine of 6503 is 0.995319\ntan(6503) = -0.097096348169898 The tangent of 6503 is -0.097096"
]
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| {"ft_lang_label":"__label__en","ft_lang_prob":0.71964115,"math_prob":0.99115765,"size":2117,"snap":"2023-40-2023-50","text_gpt3_token_len":728,"char_repetition_ratio":0.1736867,"word_repetition_ratio":0.046783626,"special_character_ratio":0.45299953,"punctuation_ratio":0.1438849,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99586344,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-09-26T12:50:05Z\",\"WARC-Record-ID\":\"<urn:uuid:f88d8465-7967-4ece-88ec-56d231c3f753>\",\"Content-Length\":\"33556\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:41d7ebc0-b229-47fd-8aee-5e295b093934>\",\"WARC-Concurrent-To\":\"<urn:uuid:685fe292-d9f6-42e3-9038-edf6611f68d6>\",\"WARC-IP-Address\":\"128.65.195.174\",\"WARC-Target-URI\":\"https://www.crazy-numbers.com/en/6503\",\"WARC-Payload-Digest\":\"sha1:BHVD3BITDZPMITWZR6OFTYR3IENTBOKZ\",\"WARC-Block-Digest\":\"sha1:4P5ZVOPUQ2LVZY2V4WHFLNXHMNNBS5YF\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-40/CC-MAIN-2023-40_segments_1695233510208.72_warc_CC-MAIN-20230926111439-20230926141439-00715.warc.gz\"}"} |
https://appdividend.com/2020/02/22/python-numpy-delete-function-numpy-delete/ | [
"AppDividend\nLatest Code Tutorials\n\n# Python Numpy delete() Function | Numpy.delete()\n\nPython numpy delete() is an inbuilt numpy function that is used to delete any subarray from an array along with the mentioned axis. The numpy delete() function returns the new array after performing the deletion operation. For a 1D array, it just deletes the object which we want to delete.\n\n## Python Numpy delete()\n\nPython numpy.delete(array, object, axis = None) method returns the new array with the deletion of sub-arrays along with the mentioned axis.\n\n### Syntax\n\n`numpy.delete(array, object, axis = None)`\n\nThe numpy delete() function takes 3 parameters:\n\n1. array: This is the input array.\n2. object: This can be any single number or a subarray.\n3. axis: This indicates the axis which is to be deleted from the array.\n\n### Return Value\n\nThe numpy delete() function returns the array by deleting the subarray, which was mentioned during the function call.\n\n### Programming Example\n\n#### Deleting elements from a 1D array\n\nSee the following code.\n\n```# Importing numpy\nimport numpy as np\n\n# We will create an 1D array\n\n# this will create an array with values 0 to 5\narr1 = np.arange(6)\n# Printing the array\nprint(\"The array is: \", arr1)\n\n# Now we will call delete() function\n# To delete the element 3\nobject = 3 # 3 is to be deleted\n\n# here arr1 is the main array\n# object is the number which is to be deleted\narr = np.delete(arr1, object)\n\n# Printing new array\nprint(\"After deleting \", object, \" new array is: \")\nprint(arr)\n```\n\n#### Output\n\n```The array is: [0 1 2 3 4 5]\nAfter deleting 3 new array is:\n[0 1 2 4 5]\n```\n\n#### Explanation\n\nIn this program, we have first created a 1D array using the numpy arange() function. Then we have printed the original array.\n\nThen we have initialized the value, which is to be deleted from the 1D array in the object variable. Then, we have called the delete() function, bypassing the object in the function.\n\nPlease note that we had not mentioned axis when we called delete() function because in the 1D array, there is only one axis, so by default axis should be None.\n\nHowever, in the object, we gave the value 3, so it deleted 3 from the original array, and then it returned the new array, which we have printed at last.\n\n## Python Numpy: Delete elements from a 2D array\n\nUsing the NumPy method np.delete(), you can delete any row and column from the NumPy array ndarray. We can also remove elements from a 2D array using numpy delete() function. See the following code.\n\n```# Importing numpy\nimport numpy as np\n\n# We will create a 2D array\n# Of shape 4x3\narr1 = np.array([(1, 2, 3), (4, 5, 6), (7, 8, 9), (50, 51, 52)])\n# Printing the array\nprint(\"The array is: \")\nprint(arr1)\n\n# Now we will call delete() function\n# To delete the subarray present in array\n\n# This indicates this will delete 3rd column\n# Of the array\nobj = 2 # 3rd column\naxis = 1 # column wise\n\n# here arr1 is the main array\n# object is the number which is to be deleted\narr = np.delete(arr1, obj, axis)\n\n# Printing new array\nprint(\"After deleting column wise new array is: \")\nprint(arr)\n\n# Now we will delete 2nd row of the array\narr = np.delete(arr, 1, 0)\n\n# Printing new array\nprint(\"After deleting row wise new array is: \")\nprint(arr)\n```\n\n#### Output\n\n```The array is:\n[[ 1 2 3]\n[ 4 5 6]\n[ 7 8 9]\n[50 51 52]]\nAfter deleting column wise new array is:\n[[ 1 2]\n[ 4 5]\n[ 7 8]\n[50 51]]\nAfter deleting row wise new array is:\n[[ 1 2]\n[ 7 8]\n[50 51]]\n```\n\n#### Explanation\n\nIn this program, we have declared a 2D numpy array of size 4×3, as we can see in the output.\n\nAt first, we wanted to delete the 3rd column of the array. For that, we have passed value 2 to obj (obj=2) as an array index starts from 0 and given axis =1, which indicates it will delete the column. Means, it will delete the 3rd column. If we give axis=0, then it will delete the 3rd row.\n\nHowever, we have deleted the 3rd column and then printed the new array.\n\nAfter that, we wanted to delete the 2nd row of the new array, that’s why we have passed 1 as object value and axis=0, because axis=0 indices the row, and object indicates which row to be deleted.\n\nFinally, we have printed the final array.\n\n## Python Numpy: Delete row\n\nUsing numpy.delete(), and we can remove an entire row from an array. See the following code.\n\n```import numpy as np\n\na = np.arange(12).reshape(3, 4)\nprint(a, '\\n')\n\na_del = np.delete(a, 1, 0)\nprint('After removing second row')\nprint(a_del, '\\n')\n\nprint('Original Array')\nprint(a)\n```\n\n#### Output\n\n``` python3 app.py\n[[ 0 1 2 3]\n[ 4 5 6 7]\n[ 8 9 10 11]]\n\nAfter removing second row\n[[ 0 1 2 3]\n[ 8 9 10 11]]\n\nOriginal Array\n[[ 0 1 2 3]\n[ 4 5 6 7]\n[ 8 9 10 11]]```\n\nFrom the output, we can see that 2nd row(obj=1, axis=0) is removed from the array.\n\nThe original array is not changed, and the new copy of ndarray is returned.\n\n## Python Numpy: Specify the index for row: obj\n\nWe can specify the index (row number or column number) to be deleted in the second parameter obj. The index starts at 0. Specifying a nonexistent index raises an error.\n\n```import numpy as np\n\na = np.arange(12).reshape(3, 4)\nprint(a, '\\n')\n\na_del = np.delete(a, 0, 0)\nprint('After removing first row')\nprint(a_del, '\\n')\n\na_del2 = np.delete(a, 2, 0)\nprint('After removing third row')\nprint(a_del2, '\\n')\n\nprint('Original Array')\nprint(a)\n```\n\n#### Output\n\n```pyt python3 app.py\n[[ 0 1 2 3]\n[ 4 5 6 7]\n[ 8 9 10 11]]\n\nAfter removing first row\n[[ 4 5 6 7]\n[ 8 9 10 11]]\n\nAfter removing third row\n[[0 1 2 3]\n[4 5 6 7]]\n\nOriginal Array\n[[ 0 1 2 3]\n[ 4 5 6 7]\n[ 8 9 10 11]]```\n\n#### Explanation\n\nIn the above example, we have removed the first and third rows of the array. The index starts at 0. So 0 means first row, and 2 means the third row.\n\nLet’s pass the index which does not exist in the array and see the output.\n\n```import numpy as np\n\na = np.arange(12).reshape(3, 4)\nprint(a, '\\n')\n\na_del = np.delete(a, 3, 0)\nprint('After removing fourth row')\nprint(a_del, '\\n')\n\nprint('Original Array')\nprint(a)\n```\n\n#### Output\n\n```python3 app.py\n[[ 0 1 2 3]\n[ 4 5 6 7]\n[ 8 9 10 11]]\n\nTraceback (most recent call last):\nFile \"app.py\", line 6, in <module>\na_del = np.delete(a, 3, 0)\nFile \"/Library/Frameworks/Python.framework/Versions/3.6/lib/python3.6/site-packages/numpy/lib/function_base.py\", line 4371, in delete\n\"size %i\" % (obj, axis, N))\nIndexError: index 3 is out of bounds for axis 0 with size 3```\n\n#### Explanation\n\nWe have specified a nonexistent index; that is why it raises an error.\n\n## Python Numpy: Specify the axis (dimension): axis\n\nSpecify an axis (dimension) to be deleted in a third parameter axis.\n\nAxis number starts from 0. In the case of the two-dimensional array, the row is the first dimension (axis=0), and the column is a second dimension (axis=1). Specifying the nonexistent dimension raises the error.\n\nSee the following code.\n\n```import numpy as np\n\na = np.arange(12).reshape(3, 4)\nprint(a, '\\n')\n\na_del = np.delete(a, 1, 0)\nprint('After removing second row')\nprint(a_del, '\\n')\n\na_del2 = np.delete(a, 1, 1)\nprint('After removing second column')\nprint(a_del2, '\\n')\n\nprint('Original Array')\nprint(a)\n```\n\n#### Output\n\n```python3 app.py\n[[ 0 1 2 3]\n[ 4 5 6 7]\n[ 8 9 10 11]]\n\nAfter removing second row\n[[ 0 1 2 3]\n[ 8 9 10 11]]\n\nAfter removing second column\n[[ 0 2 3]\n[ 4 6 7]\n[ 8 10 11]]\n\nOriginal Array\n[[ 0 1 2 3]\n[ 4 5 6 7]\n[ 8 9 10 11]]```\n\nLet’s define the axis which does not exist in the array and see the output.\n\n```import numpy as np\n\na = np.arange(12).reshape(3, 4)\nprint(a, '\\n')\n\na_del = np.delete(a, 1, 2)\nprint('After removing second row')\nprint(a_del, '\\n')\n\nprint('Original Array')\nprint(a)\n```\n\n#### Output\n\n```python3 app.py\n[[ 0 1 2 3]\n[ 4 5 6 7]\n[ 8 9 10 11]]\n\nTraceback (most recent call last):\nFile \"app.py\", line 6, in <module>\na_del = np.delete(a, 1, 2)\nFile \"/Library/Frameworks/Python.framework/Versions/3.6/lib/python3.6/site-packages/numpy/lib/function_base.py\", line 4300, in delete\naxis = normalize_axis_index(axis, ndim)\nnumpy.AxisError: axis 2 is out of bounds for array of dimension 2```\n\nSo, Specifying a nonexistent dimension raises an error.\n\n## Python Numpy: Delete multiple rows and columns\n\nMultiple rows and columns can be removed at once by specifying the list or a slice in the second parameter obj. We can delete multiple rows and columns at once using the following way.\n\n1. Use a Python list\n2. Use a Python slice\n3. Use a np.s_[]\n\n### Use a Python list\n\nSpecify the row numbers and column numbers in the form of a list to be deleted in an array.\n\nSee the following code.\n\n```import numpy as np\n\na = np.arange(12).reshape(3, 4)\nprint(a, '\\n')\n\na_del = np.delete(a, [0, 3], 1)\nprint('After removing first and fourth columns')\nprint(a_del, '\\n')\n\nprint('Original Array')\nprint(a)\n```\n\n#### Output\n\n```python3 app.py\n[[ 0 1 2 3]\n[ 4 5 6 7]\n[ 8 9 10 11]]\n\nAfter removing first and fourth columns\n[[ 1 2]\n[ 5 6]\n[ 9 10]]\n\nOriginal Array\n[[ 0 1 2 3]\n[ 4 5 6 7]\n[ 8 9 10 11]]```\n\nIf we want to remove the column, then we have to pass 1 in np.delete(a, [0, 3], 1) function, and we need to remove the first and fourth column from the array. So we have written np.delete(a, [0, 3], 1) code.\n\nWe can also remove multiple rows at once. See the following code.\n\n```import numpy as np\n\na = np.arange(12).reshape(3, 4)\nprint(a, '\\n')\n\na_del = np.delete(a, [1, 2], 0)\nprint('After removing second and third rows')\nprint(a_del, '\\n')\n\nprint('Original Array')\nprint(a)\n```\n\n#### Output\n\n```python3 app.py\n[[ 0 1 2 3]\n[ 4 5 6 7]\n[ 8 9 10 11]]\n\nAfter removing second and third rows\n[[0 1 2 3]]\n\nOriginal Array\n[[ 0 1 2 3]\n[ 4 5 6 7]\n[ 8 9 10 11]]```\n\nIn the above example, we have removed the second and third rows from the array.\n\n`np.delete(a, [1, 2], 0)`\n\nIn the above code, we have passed 0, which means we need to remove row or x-axis.\n\nThen we have specified the second and third row to be removed. The array index starts from 0.\n\n### Use a Python slice\n\nIt is also possible to specify the multiple rows and columns by using the slice specifying a range with [start: stop: step].\n\nCreate a slice object with a slice() and specify it as a second parameter obj.\n\nIt is equivalent to [: stop] if there is single argument, [start: stop] if there are two arguments, and [start: stop: step] if there are three arguments. If you want to omit, specify None explicitly.\n\nSee the following code.\n\n```import numpy as np\n\na = np.arange(12).reshape(3, 4)\nprint(a, '\\n')\n\nprint('After removing first and second columns')\na_del = np.delete(a, slice(2), 1)\nprint(a_del, '\\n')\n\nprint('Original Array')\nprint(a)\n```\n\n#### Output\n\n```python3 app.py\n[[ 0 1 2 3]\n[ 4 5 6 7]\n[ 8 9 10 11]]\n\nAfter removing first and second columns\n[[ 2 3]\n[ 6 7]\n[10 11]]\n\nOriginal Array\n[[ 0 1 2 3]\n[ 4 5 6 7]\n[ 8 9 10 11]]```\n\n### Use a np.s_[]\n\nUse numpy.s_[] if you want to write in a form [start:stop:step].\n\nLet’s remove the first and second columns using np.s_[] function. Remember array index starts from 0.\n\nSee the following code.\n\n```import numpy as np\n\na = np.arange(12).reshape(3, 4)\nprint(a, '\\n')\n\nprint('After removing first and second columns')\na_del = np.delete(a, np.s_[:2], 1)\nprint(a_del, '\\n')\n\nprint('Original Array')\nprint(a)\n```\n\n#### Output\n\n```python3 app.py\n[[ 0 1 2 3]\n[ 4 5 6 7]\n[ 8 9 10 11]]\n\nAfter removing first and second columns\n[[ 2 3]\n[ 6 7]\n[10 11]]\n\nOriginal Array\n[[ 0 1 2 3]\n[ 4 5 6 7]\n[ 8 9 10 11]]```\n\n## Conclusion\n\nPython numpy delete() function is used to remove a single row or column, multiple rows, or columns in the array. We can also use Python slice and list to remove multiple elements(rows and columns). We have seen many variations of it in this tutorial. I hope you learn many things from this brief example.\n\nThis site uses Akismet to reduce spam. Learn how your comment data is processed."
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https://numbermatics.com/n/94220/ | [
"# 94220\n\n## 94,220 is an even composite number composed of four prime numbers multiplied together.\n\nWhat does the number 94220 look like?\n\nThis visualization shows the relationship between its 4 prime factors (large circles) and 24 divisors.\n\n94220 is an even composite number. It is composed of four distinct prime numbers multiplied together. It has a total of twenty-four divisors.\n\n## Prime factorization of 94220:\n\n### 22 × 5 × 7 × 673\n\n(2 × 2 × 5 × 7 × 673)\n\nSee below for interesting mathematical facts about the number 94220 from the Numbermatics database.\n\n### Names of 94220\n\n• Cardinal: 94220 can be written as Ninety-four thousand, two hundred twenty.\n\n### Scientific notation\n\n• Scientific notation: 9.422 × 104\n\n### Factors of 94220\n\n• Number of distinct prime factors ω(n): 4\n• Total number of prime factors Ω(n): 5\n• Sum of prime factors: 687\n\n### Divisors of 94220\n\n• Number of divisors d(n): 24\n• Complete list of divisors:\n• Sum of all divisors σ(n): 226464\n• Sum of proper divisors (its aliquot sum) s(n): 132244\n• 94220 is an abundant number, because the sum of its proper divisors (132244) is greater than itself. Its abundance is 38024\n\n### Bases of 94220\n\n• Binary: 101110000000011002\n• Base-36: 20P8\n\n### Squares and roots of 94220\n\n• 94220 squared (942202) is 8877408400\n• 94220 cubed (942203) is 836429419448000\n• The square root of 94220 is 306.9527650959\n• The cube root of 94220 is 45.5038035705\n\n### Scales and comparisons\n\nHow big is 94220?\n• 94,220 seconds is equal to 1 day, 2 hours, 10 minutes, 20 seconds.\n• To count from 1 to 94,220 would take you about two hours.\n\nThis is a very rough estimate, based on a speaking rate of half a second every third order of magnitude. If you speak quickly, you could probably say any randomly-chosen number between one and a thousand in around half a second. Very big numbers obviously take longer to say, so we add half a second for every extra x1000. (We do not count involuntary pauses, bathroom breaks or the necessity of sleep in our calculation!)\n\n• A cube with a volume of 94220 cubic inches would be around 3.8 feet tall.\n\n### Recreational maths with 94220\n\n• 94220 backwards is 02249\n• The number of decimal digits it has is: 5\n• The sum of 94220's digits is 17\n• More coming soon!\n\nMLA style:\n\"Number 94220 - Facts about the integer\". Numbermatics.com. 2021. Web. 1 December 2021.\n\nAPA style:\nNumbermatics. (2021). Number 94220 - Facts about the integer. Retrieved 1 December 2021, from https://numbermatics.com/n/94220/\n\nChicago style:\nNumbermatics. 2021. \"Number 94220 - Facts about the integer\". https://numbermatics.com/n/94220/\n\nThe information we have on file for 94220 includes mathematical data and numerical statistics calculated using standard algorithms and methods. We are adding more all the time. If there are any features you would like to see, please contact us. Information provided for educational use, intellectual curiosity and fun!\n\nKeywords: Divisors of 94220, math, Factors of 94220, curriculum, school, college, exams, university, Prime factorization of 94220, STEM, science, technology, engineering, physics, economics, calculator, ninety-four thousand, two hundred twenty.\n\nOh no. Javascript is switched off in your browser.\nSome bits of this website may not work unless you switch it on."
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https://www.programmingwithbasics.com/2017/04/hacker-rank-solution-for-diagonal.html | [
"# Diagonal Difference Hackerrank Solution in C | C++ Languages\n\nIn this article, we are providing Diagonal Difference Hackerrank Solution in C, C++, and Java programming Languages. This programming problem belongs to hackerrank 30 days of code, and we are going to find the Hackerrank Diagonal Difference Solution in C and C++ language. In this programming challenge. We have to find the Diagonal Difference of an N*N matrix or a square matrix. So why we are waiting, let's do it firstly, we have to find the right diagonal sum of a matrix and left a diagonal sum of magic, then we will find out the Diagonal Difference of a matrix. In this matrix, we already know that our matrix is an N*N shape.\n\n## Diagonal Difference Hackerrank- Table of Content\n\n1. Function description\n2. Input Format\n3. Constraints\n4. Output Format\n5. Sample Input\n6. Sample Output\n7. Explanation Diagonal Difference Hackerrank Solution\n8. The logic of Diagonal Difference\n9. Diagonal Difference Hackerrank Solution in C\n10. Diagonal Difference Hackerrank Solution in C++\n11. And more...\n\nGiven a square matrix, calculate the absolute difference between the sums of its diagonals.\n\nFor example, the square matrix arr is shown below:\n\n1 2 3\n4 5 6\n9 8 9\n\nThe left-to-right diagonal = 1 + 5 + 9 = 15 . The right to left diagonal = 3 + 5 + 9 = 17 . Their absolute difference is |15-17| = 2.\n\n### Function description\n\nComplete the Diagonaldifference function in the editor below. It must return an integer representing the absolute diagonal difference.\n\ndiagonalDifference takes the following parameter:\n\narr: an array of integers.\n\n### Input Format\n\nThe first line contains a single integer, n, the number of rows and columns in the matrix arr. Each of the next n lines describes a row, arr[i], and consists of n space-separated integers arr[i][j].\n\n### Constraints\n\n-100<=arr[i][j]<=100\n\n### Output Format\n\nPrint the absolute difference between the sums of the matrix's two diagonals as a single integer.\n\n3\n11 2 4\n4 5 6\n10 8 -12\n\n15\n\n### Explanation Diagonal Difference Hackerrank Solution Sample\n\nThe primary diagonal is:\n\nSum across the primary diagonal: 11 + 5 - 12 = 4\n\nThe secondary diagonal is\n\nSum across the secondary diagonal: 4 + 5 + 10 = 19\n\nDifference: |4 - 19| = 15\n\nNow the last step is to find the difference between the sum of diagonal, so add first diagonal and second diagonal after that mod the difference so | 4 - 19| = 15. Hence we got our solution.\n\nNote: |x| is the absolute value of x\n\nCheck This- Hackerrank 30 days of code Solution.\n\n## Diagonal Difference Hackerrank Solution Logic\n\nSo the logic is straightforward in this diagonal difference hackerrank problem. We have to find the diagonal sum of the matrix, and after seeing the total amount. We have to find out the difference between both diagonal sums. If the difference of both diagonal matrices is negative, then find the Mod or, in the end, print the output. Get the Hackerrank Diagonal Difference Solution in C language See the above logic solution with an example in the explanation.\n\n## Diagonal Difference Hackerrank Solution in C\n\n`#include <math.h>#include <stdio.h>#include <string.h>#include <stdlib.h>#include <assert.h>#include <limits.h>#include <stdbool.h>int main(){ int n, j; int i=0,RightDiagonalSum=0,LeftDiagonalSum=0, firstarray, secondarray; scanf(\"%d\",&n); int a[n][n]; for(int firstarray = 0; firstarray < n; firstarray++) { for(int secondarray = 0; secondarray < n; secondarray++) { scanf(\"%d\",&a[firstarray][secondarray]); } } while(i<n) { RightDiagonalSum=RightDiagonalSum+a[i][i]; i++; } j=n-1,i=0; while(i<n) { LeftDiagonalSum=LeftDiagonalSum+a[i][j]; i++; j--; } printf(\"%d\",abs(RightDiagonalSum-LeftDiagonalSum)); return 0;}`\n\nSo as we already know, what is the logic,, so without wasting time, let's just implement the above logic in our diagonal difference hackerrank problem with an example? Take a MATRIX with the same number of rows and columns. Below is the Hackerrank Diagonal Difference Solution in C++ Programming language.\n\n## Diagonal Difference Hackerrank Solution in C++\n\n`#include <iostream>using namespace std;int main() { int N, LeftDiagonalSum = 0, RightDiagonal = 0; cin >> N; int a[N][N]; for (int i = 0; i < N; i++) { for (int j = 0; j < N; j++) { cin >> a[i][j]; if (i == j) { LeftDiagonalSum = LeftDiagonalSum + a[i][j]; } } } for (int i = 0; i < N; i++) { for (int j = N - 1 - i; j >= 0;) { RightDiagonal = RightDiagonal + a[i][j]; break; } } cout << abs(LeftDiagonalSum - RightDiagonal) << endl; return 0;}`\n\n### Diagonal Difference Hackerrank Solution Output",
null,
"### Similar to Hackerrank Diagonal Difference Solution\n\nPrevious Post\nNext Post\n\n#### post written by: Ghanendra Yadav\n\nHi, I’m Ghanendra Yadav, SEO Expert, Professional Blogger, Programmer, and UI Developer. Get a Solution of More Than 500+ Programming Problems, and Practice All Programs in C, C++, and Java Languages. Get a Competitive Website Solution also Ie. Hackerrank Solutions and Geeksforgeeks Solutions. If You Are Interested to Learn a C Programming Language and You Don't Have Experience in Any Programming, You Should Start with a C Programming Language, Read: List of Format Specifiers in C."
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https://quant.stackexchange.com/questions/43596/what-is-forward-moneyness-and-how-to-calculate-it | [
"# What is forward moneyness and how to calculate it?\n\nI'm now studying the concept \"implied volatility\", and my teacher gave us a figure about the implied volatility with respect to the moneyness which is expressed by $$\\frac{ln(\\frac{F}{K})}{\\sigma\\sqrt{T}}$$ , where $$F$$ should be the forward price at maturity of the underlying I think?\n\nBased on my knowledge, the moneyness should be $$\\frac{S}{K}$$\n\nCould anyone tell me the meaning of the upper expression and the differences between these two kinds of moneyness?\n\n• How to find $F$ ? If the stock pays no dividend then $F=S e^{r T}$. – Alex C Jan 21 at 0:06\n• There might be a typo. It Should be probably have been $ln(F/K)$. Then the upper expression is known as the standardized momeyness – Sanjay Jan 21 at 0:11\n• Thanks for point out the typo, I've already edited it! – Francis Gong Jan 21 at 10:23\n\n• the simple moneyness is $$\\frac{S}{K}$$ (in some cases you will see $$\\frac{K}{S}$$)\n• the log moneyness is $$\\ln \\frac{S}{K}$$\n• the standardized log moneyness$$\\frac{\\ln(S/K)}{\\sigma\\sqrt T}$$\nIf the forward price $$F$$ is used in place of the underlying price $$S$$ you have (three definitions of) the forward moneyness. The forward moneyness is useful because it is more consistent with the way the Black Scholes formula works, it is more natural.\nHow to find $$F$$ ? If the stock pays no dividend then $$F=S e^{r T}$$. You can also find $$F$$ by comparing the prices of puts and calls."
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| {"ft_lang_label":"__label__en","ft_lang_prob":0.9325034,"math_prob":0.999395,"size":459,"snap":"2019-43-2019-47","text_gpt3_token_len":114,"char_repetition_ratio":0.12087912,"word_repetition_ratio":0.0,"special_character_ratio":0.22875817,"punctuation_ratio":0.05952381,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99988115,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-11-14T01:57:30Z\",\"WARC-Record-ID\":\"<urn:uuid:c6f84131-fdc5-4fbc-9976-40ff5221c902>\",\"Content-Length\":\"135227\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:82a2ead5-de44-4887-ab18-91946888c355>\",\"WARC-Concurrent-To\":\"<urn:uuid:eb924213-b85c-436e-a672-0e0e8e889589>\",\"WARC-IP-Address\":\"151.101.193.69\",\"WARC-Target-URI\":\"https://quant.stackexchange.com/questions/43596/what-is-forward-moneyness-and-how-to-calculate-it\",\"WARC-Payload-Digest\":\"sha1:2OKN2H4LQBRSQW4GA26IA3N5IQ55JXDP\",\"WARC-Block-Digest\":\"sha1:HPZ46WCCKBSPVFEZUXGOIBJ62FXRP54V\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-47/CC-MAIN-2019-47_segments_1573496667767.6_warc_CC-MAIN-20191114002636-20191114030636-00079.warc.gz\"}"} |
https://feet-to-meters.appspot.com/pl/8480-stopa-na-metr.html | [
"Feet To Meters\n\n# 8480 ft to m8480 Foot to Meters\n\nft\n=\nm\n\n## How to convert 8480 foot to meters?\n\n 8480 ft * 0.3048 m = 2584.704 m 1 ft\nA common question is How many foot in 8480 meter? And the answer is 27821.5223097 ft in 8480 m. Likewise the question how many meter in 8480 foot has the answer of 2584.704 m in 8480 ft.\n\n## How much are 8480 feet in meters?\n\n8480 feet equal 2584.704 meters (8480ft = 2584.704m). Converting 8480 ft to m is easy. Simply use our calculator above, or apply the formula to change the length 8480 ft to m.\n\n## Convert 8480 ft to common lengths\n\nUnitLength\nNanometer2.584704e+12 nm\nMicrometer2584704000.0 µm\nMillimeter2584704.0 mm\nCentimeter258470.4 cm\nInch101760.0 in\nFoot8480.0 ft\nYard2826.66666666 yd\nMeter2584.704 m\nKilometer2.584704 km\nMile1.6060606061 mi\nNautical mile1.3956285097 nmi\n\n## What is 8480 feet in m?\n\nTo convert 8480 ft to m multiply the length in feet by 0.3048. The 8480 ft in m formula is [m] = 8480 * 0.3048. Thus, for 8480 feet in meter we get 2584.704 m.\n\n## 8480 Foot Conversion Table",
null,
"## Alternative spelling\n\n8480 ft to Meter, 8480 ft in Meter, 8480 Feet in Meter, 8480 ft to m, 8480 Feet to m, 8480 Feet in m, 8480 Foot to Meter,"
]
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null,
"https://feet-to-meters.appspot.com/image/8480.png",
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| {"ft_lang_label":"__label__en","ft_lang_prob":0.84127474,"math_prob":0.83641636,"size":616,"snap":"2022-40-2023-06","text_gpt3_token_len":207,"char_repetition_ratio":0.20588236,"word_repetition_ratio":0.0,"special_character_ratio":0.43993506,"punctuation_ratio":0.14864865,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.98428434,"pos_list":[0,1,2],"im_url_duplicate_count":[null,4,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-09-24T23:38:52Z\",\"WARC-Record-ID\":\"<urn:uuid:d9ed58b7-01cb-4201-8919-92fd1ca4b273>\",\"Content-Length\":\"28142\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:f78090ae-c985-4226-90da-b6f9895fd9c8>\",\"WARC-Concurrent-To\":\"<urn:uuid:40d345f6-facc-45d6-aa68-90c63268e404>\",\"WARC-IP-Address\":\"142.251.163.153\",\"WARC-Target-URI\":\"https://feet-to-meters.appspot.com/pl/8480-stopa-na-metr.html\",\"WARC-Payload-Digest\":\"sha1:JRV25WOY6ZM75YK4VLYI3DTBQEJS7JR5\",\"WARC-Block-Digest\":\"sha1:DLBYXZ3W6OIRZXI5FFBZC7HPHJ6GEV3E\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-40/CC-MAIN-2022-40_segments_1664030333541.98_warc_CC-MAIN-20220924213650-20220925003650-00567.warc.gz\"}"} |
https://www.geeksforgeeks.org/puzzle-38-tic-tac-toe-puzzle/ | [
"# Puzzle 38 | (Tic Tac Toe Puzzle)\n\nStatement:\nThe game of Tic-Tac-Toe is being played between two players and it is in below state after six moves.",
null,
"1. Who will win the game, O or X?\n2. Which was the sixth mark and at which position?\n\nAssume that both the players are intelligent enough.\n\nSolution:\nO will win the game. The sixth mark was X in square 9.\n\nExplanation:\nThe 7th mark must be placed in square 5 which is the win situation for both X and O. Hence, the 6th mark must be placed in a line already containing two of the opponents marks. There are two such possibilities – the 6th mark would have been either O in square 7 or X in square 9.\n\nAs we know both the players are intelligent enough, the 6th mark could not be O in square 7. Instead, he would have placed O in square 5 and would have won.\n\nHence, the sixth mark must be X placed in square 9. And the seventh mark will be O. Thus O will win the game."
]
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"https://media.geeksforgeeks.org/wp-content/cdn-uploads/tic_tac_toe.jpg",
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| {"ft_lang_label":"__label__en","ft_lang_prob":0.9448432,"math_prob":0.9277169,"size":1860,"snap":"2019-51-2020-05","text_gpt3_token_len":495,"char_repetition_ratio":0.15463363,"word_repetition_ratio":0.016666668,"special_character_ratio":0.2747312,"punctuation_ratio":0.077562325,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9694938,"pos_list":[0,1,2],"im_url_duplicate_count":[null,4,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-01-22T06:12:20Z\",\"WARC-Record-ID\":\"<urn:uuid:3a39b23c-b2f2-4d0b-850f-8cd737686cc6>\",\"Content-Length\":\"119934\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:c68bdbd0-350b-48db-a8d9-e97df7a049e1>\",\"WARC-Concurrent-To\":\"<urn:uuid:c8427bfe-38e9-4978-a222-53d51ef92505>\",\"WARC-IP-Address\":\"23.199.63.170\",\"WARC-Target-URI\":\"https://www.geeksforgeeks.org/puzzle-38-tic-tac-toe-puzzle/\",\"WARC-Payload-Digest\":\"sha1:VPGXHDMAMYY2OOTNUF7GB46YM5MQFXFQ\",\"WARC-Block-Digest\":\"sha1:U7HNI7PBUHVHVHXU7GHLLFHIBAZMBMN7\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-05/CC-MAIN-2020-05_segments_1579250606696.26_warc_CC-MAIN-20200122042145-20200122071145-00132.warc.gz\"}"} |
https://www.matlabcoding.com/2019/10/numerical-methods-for-engineers-and.html | [
"## Search This Blog\n\nThis book provides a pragmatic, methodical and easy-to-follow presentation of numerical methods and their effective implementation using MATLAB, which is introduced at the outset. The author introduces techniques for solving equations of a single variable and systems of equations, followed by curve fitting and interpolation of data. The book also provides detailed coverage of numerical differentiation and integration, as well as numerical solutions of initial-value and boundary-value problems. The author then presents the numerical solution of the matrix eigenvalue problem, which entails approximation of a few or all eigenvalues of a matrix. The last chapter is devoted to numerical solutions of partial differential equations that arise in engineering and science. Each method is accompanied by at least one fully worked-out example showing essential details involved in preliminary hand calculations, as well as computations in MATLAB. This thoroughly-researched resource:\nBuy: Numerical Methods for Engineers and Scientists Using MATLAB® (Tayl01) Hardcover – 4 May 2017 by Ramin S. Esfandiari (Author)\nPDF Download: Numerical Methods for Engineers and Scientists Using MATLAB® by Randy Moore in Computer Science on October 21, 2019\n\nMATLAB",
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"",
null,
""
]
| [
null,
"https://ir-in.amazon-adsystem.com/e/ir",
null,
"https://www.awin1.com/cshow.php",
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| {"ft_lang_label":"__label__en","ft_lang_prob":0.90796,"math_prob":0.8200605,"size":1597,"snap":"2022-40-2023-06","text_gpt3_token_len":315,"char_repetition_ratio":0.13622096,"word_repetition_ratio":0.32894737,"special_character_ratio":0.18722604,"punctuation_ratio":0.08527132,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9952967,"pos_list":[0,1,2,3,4],"im_url_duplicate_count":[null,null,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-02-05T10:24:15Z\",\"WARC-Record-ID\":\"<urn:uuid:21ff7162-cb08-4d33-b010-f954e7c130df>\",\"Content-Length\":\"331297\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:d97a52cc-592d-410e-8958-80c7f4ca1360>\",\"WARC-Concurrent-To\":\"<urn:uuid:5175deda-5137-4dd2-88dc-6cb315ecfe8f>\",\"WARC-IP-Address\":\"172.253.62.121\",\"WARC-Target-URI\":\"https://www.matlabcoding.com/2019/10/numerical-methods-for-engineers-and.html\",\"WARC-Payload-Digest\":\"sha1:HAJFI6XOL65EU46OKXN4YPP5IXS7U53I\",\"WARC-Block-Digest\":\"sha1:7GOGZZTRVP4KEA4ADO235MLVZGHICFCM\",\"WARC-Identified-Payload-Type\":\"application/xhtml+xml\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-06/CC-MAIN-2023-06_segments_1674764500251.38_warc_CC-MAIN-20230205094841-20230205124841-00244.warc.gz\"}"} |
https://www.csdn.net/tags/MtTaEg0sMTg4ODQtYmxvZwO0O0OO0O0O.html | [
"",
null,
"Highest Common Factor(HCF)\n\nGreatest Common Divisor(GCD)\n\n• 问题：请从键盘上输入两个数值 x，y，请用C语言求出这两个数值的最大公约数与最小公倍数。 首先，我们要想解决这道问题，就要了解什么是最大公约数与最小公倍数。 最大公因数；也称最大公约数、最大公因子，指两个或...\nC语言三种算法求解最大公约数与最小公倍数\n\n最大公约数与最小公倍数的求解是很多初学C的人所面临的一道问题。当然这道问题并不难解答，也有很多人已经写过相关的博客，我在此书写此篇博客，一是为了让自己能够夯实基础，另外就是希望能够帮到和我一样的初学者。\n\n当然，在写这篇博客之前，我已经做过相关资料的调查，可能读者会发现此篇博客会与其他人的博客有所重复，但是，我保证绝未抄袭。好了，进入正题！\n\n问题：请从键盘上输入两个数值 x，y，请用C语言求出这两个数值的最大公约数与最小公倍数。\n\n首先，我们要想解决这道问题，就要了解什么是最大公约数与最小公倍数。\n\n最大公因数；也称最大公约数、最大公因子，指两个或多个整数共有约数中最大的一个。----来源百度百科\n\n最小公倍数：两个或多个整数公有的倍数叫做它们的公倍数。----来源百度百科\n\n了解了其含义，接下来就是构思算法，通常而言，求解最大公约数有三种算法，而最小公倍数的求解，我们可以很容易的推断出，最小公倍数等于两个数值的乘积除以这两个数值的最大公约数。那么接下来的算法我将在此一一进行列举和解释。\n\n1.辗转相除法：\n\n又名欧几里德算法（Euclidean algorithm），它是已知最古老的算法， 其可追溯至公元前300年前。 ----来源百度百科\n\n辗转：望文生义，就是翻来覆去。相除就很好理解了，就是进行除法运算。\n\n辗转相除法的核心就是不断的让两个数做除法运算。其原理基于两个整数的最大公约数等于其中较小的数和两数的相除余数的最大公约数。\n\n假设两数为 x，y。\n\n先令 z = x % y ;\n\n之后 y 赋给 x 即令 x = y ;\n\n再将 z 赋给 y 即令 y = z；\n\n辗转相减，其终止条件为：y 等于0时。\n\n代码如下：\n\n#include<stdio.h>\nint main()\n{\nint x, y, z, m, n;\nprintf(\"请输入两个数：\");\nscanf_s(\"%d%d\", &x, &y);\nm = x, n = y;\nwhile (y != 0)\n{\nz = x%y;\nx = y;\ny = z;\n}\nprintf(\"最大公约数是: %d\\n\", x);\nprintf(\"最小公倍数是: %d\\n\", m*n / x);\nsystem(\"pause\");\nreturn 0;\n}\n\n2.辗转相减法：\n\n即尼考曼彻斯法，其特色是做一系列减法，从而求得最大公约数。----来源百度百科\n\n辗转相减法即通过对两数的不断减法运算。\n\n假设两数为 x, y。\n\n当 x > y 时，令 x = x - y;\n\n反之，则令 y = y - x;\n\n之后一直辗转相减，直至 x = y 时，终止。\n\n代码如下：\n\n#include<stdio.h>\nint main()\n{\nint x, y, m, n;\nprintf(\"请输入两个数：\");\nscanf_s(\"%d%d\", &x, &y);\nm = x, n = y;\nwhile (x!=y)\n{\nif (x>y)\nx = x-y;\nelse\ny = y-x;\n}\nprintf(\"最大公约数是: %d\\n\", x);\nprintf(\"最小公倍数是: %d\\n\", m*n / x);\nsystem(\"pause\");\nreturn 0;\n}\n\n3.穷举法：\n\n穷举法的基本思想是根据题目的部分条件确定答案的大致范围，并在此范围内对所有可能的情况逐一验证，直到全部情况验证完毕。----来源百度百科\n\n穷举法又称枚举法，通过对数值范围内的所有数字进行检验，得出其结果。\n\n代码如下：\n\n#include<stdio.h>\nint main()\n{\nint x, y, i, m, n;\nprintf(\"请输入两个数：\");\nscanf_s(\"%d%d\", &x, &y);\nm = x, n = y;\nfor (i = x; i > 0; i--)\n{\nif (x%i == 0 && y%i == 0)\nbreak;\n}\nprintf(\"最大公约数是: %d\\n\", i);\nprintf(\"最小公倍数是: %d\\n\", m*n / i);\nsystem(\"pause\");\nreturn 0;\n}\n\n以上即为求解最大公约数与最小公倍数的三种算法，如有纰漏，还请各位不吝赐教。\n\n\n展开全文",
null,
"• 运行最大公约数的常用算法，并进行程序的调式与测试，要求程序设计风格良好，并添加异常处理模块（如输入非法等）。 二、算法 辗转相除法 辗转相除法（又名欧几里德法）C语言中用于计算两个正整数a,b的最大公约数和...\n一、题目\n运行最大公约数的常用算法，并进行程序的调式与测试，要求程序设计风格良好，并添加异常处理模块（如输入非法等）。\n二、算法\n辗转相除法\n\n辗转相除法（又名欧几里德法）C语言中用于计算两个正整数a,b的最大公约数和最小公倍数，实质它依赖于下面的定理：\n在这里插入图片描述\n根据这一定理可以采用函数嵌套调用和递归调用形式进行求两个数的最大公约数和最小公倍数,现分别叙述如下\n\n三、源代码\n#include<stdio.h>\n\nint main()\n{\nint x = 0;\nint y = 0;\nint z = 0;\nscanf_s(\"%d%d\", &x, &y);\nz = x * y;\n\nif (x < y)\n{\nint temp = x;\nx = y;\ny = temp;\n}\nint k = x % y;\nwhile (k)\n{\nx = y;\ny = k;\nk = x % y;\n}\nprintf(\"最大公约数为：%d\\n\", y);\nprintf(\"最大公约数为：%d\\n\", z/y);\nreturn 0;\n}\n\n四、运行结果",
null,
"展开全文",
null,
"• ## 求最大公约数（辗转相除法）\n\n万次阅读 多人点赞 2019-06-03 16:20:50\n也称最大公因数、最大公因子，a， b的最大公约数记为（a，b），同样的，a，b，c的最大公约数记为（a，b，c），多个 整数的最大公约数也有同样的记号。求最大公约数有多种 方法，常见的有质因数分解法、短除法、辗转...\n\n最大公约数（Greatest Common Divisor）指两个或多个整数共有约数中最大的一个。\n\n也称最大公因数、最大公因子，a， b的最大公约数记为（a，b），同样的，a，b，c的最大 公约数记为（a，b，c），多个 整数的最大公约数也有同样的记号。求最大公约数有多种 方法，常见的有 质因数分解法、 短除法、 辗转相除法、 更相减损法。与最大公约数相对应的概念是 最小公倍数，a，b的 最小公倍数记为[a，b]。\n\n再来介绍一下辗转相除法：\n\n辗转相除法又叫欧几里得算法,是欧几里得最先提出来的.辗转相除法的实现,是基于下面的原理(在这里用(a,b)表示a和b的最大公因数)：\n(a,b)=(a,ka+b),其中a、b、k都为自然数.………………①\n也就是说,两个数的最大公约数,将其中一个数加到另一个数上,得到的新数,其公约数不变,比如(4,6)=(4+6,6)=(4,6+2×4)=2.要证明这个原理很容易：如果p是a和ka+b的公约数,p整除a,也能整除ka+b.那么就必定要整除b,所以p又是a和b的公约数,从而证明他们的最大公约数也是相等的.\n基于上面的原理,就能实现我们的迭代相减法：\n(78,14)=(64,14)=(50,14)=(36,14)=(22,14)=(8,14)=(8,6)=(2,6)=(2,4)=(2,2)=(0,2)=2\n基本上思路就是大数减去小数,一直减到能算出来为止,在作为练习的时候,往往进行到某一步就已经可以看出得值.迭代相减法简单,不过步数比较多,实际上我们可以看到,在上面的过程中,由(78,14)到(8,14)完全可以一步到位,因为(78,14)=(14×5+8,14)=(8,14),由此就诞生出我们的辗转相除法.\n用辗转相除法求(a,b).设r0=b,r1=a,反复运用除法算式,得到一系列整数qi,ri和下面的方程：\n相当于每一步都运用原理①把数字进行缩小,上面右边就是每一步对应的缩小结果,可以看出,最后的余数rn就是a和b的公约数.迭代相减法和辗转相除法在本质上是一样的,相对来说,减法比较简单（需要10步）,但是除法步数少（仅需4步）.\n\n因此可以通过这个原理来求出最大公约数：\n\n#include<iostream>\n#include<cstdio>\nusing namespace std;\n\nint fun(int m,int n){\nint rem;\t\t\t//余数，当余数为0的时候，最后的m即为最大公约数\n//先用较小的数对较大的数取余，再用余数对较小的数求余，直到余数为零\nwhile(n > 0){\nrem = m % n;\nm = n;\nn = rem;\n}\nreturn m;\t\t\t//将结果返回\n}\nint main(){\nint n,m;\ncin>>m>>n;\ncout<<\"m和n的最大公约数为：\"<<fun(m,n)<<endl;\nreturn 0;\n} \n\n因为到余数为零结束，所以还可以将程序简化一下用递归来求：\n\nint fun(int m,int n){\nif(n==0) return m;\nreturn fun(n,m%n);\n}\n\n再简化一下，用一行代码来求：\n\nint gcd(int m, int n) {\nreturn n ? gcd(n, m % n) : m;\n}\n\n\n展开全文",
null,
"",
null,
"辗转相除法\n• 分享一个大牛的人工智能教程。... * 求两个自然数的最大公约数 - C++ - by Chimomo * * Answer：辗转相除法 */ #include <iostream> #include <cassert> #include &...\n分享一个大牛的人工智能教程。零基础！通俗易懂！风趣幽默！希望你也加入到人工智能的队伍中来！请点击http://www.captainbed.net\n\n/*\n* 求两个自然数的最大公约数 - C++ - by Chimomo\n*\n* Answer：辗转相除法\n*/\n\n#include <iostream>\n#include <cassert>\n#include <stack>\n#include <math.h>\n\nusing namespace std;\n\nint GreatestCommonDivisor(int a, int b) {\nint t;\n\nif (a < b) {\n// 交换两个数，使大数放在a的位置上。\nt = a;\na = b;\nb = t;\n}\n\nwhile (b != 0) {\n// 利用辗转相除法，直到b为0为止。\nt = a % b;\na = b;\nb = t;\n}\n\nreturn a;\n}\n\nint main() {\ncout << GreatestCommonDivisor(318, 87632) << endl;\nreturn 0;\n}\n\n// Output:\n/*\n2\n\n*/\n\n\n\n展开全文",
null,
"",
null,
"算法 C++\n• 先求 最大公约数 （辗转相除法．）： 以小数除大数，如果能整除，那么小数就是所求的最大公约数．否则就用余数来除刚才的除数；再用这新除法的余数去除刚才的余数．依此类推，直到一个除法能够整除，这时作为除数的...",
null,
"算法",
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https://everything.explained.today/Ideal_class_group/ | [
"# Ideal class group explained\n\nIn number theory, the ideal class group (or class group) of an algebraic number field is the quotient group where is the group of fractional ideals of the ring of integers of, and is its subgroup of principal ideals. The class group is a measure of the extent to which unique factorization fails in the ring of integers of . The order of the group, which is finite, is called the class number of .\n\nThe theory extends to Dedekind domains and their field of fractions, for which the multiplicative properties are intimately tied to the structure of the class group. For example, the class group of a Dedekind domain is trivial if and only if the ring is a unique factorization domain.\n\n## History and origin of the ideal class group\n\nIdeal class groups (or, rather, what were effectively ideal class groups) were studied some time before the idea of an ideal was formulated. These groups appeared in the theory of quadratic forms: in the case of binary integral quadratic forms, as put into something like a final form by Carl Friedrich Gauss, a composition law was defined on certain equivalence classes of forms. This gave a finite abelian group, as was recognised at the time.\n\nLater Ernst Kummer was working towards a theory of cyclotomic fields. It had been realised (probably by several people) that failure to complete proofs in the general case of Fermat's Last Theorem by factorisation using the roots of unity was for a very good reason: a failure of unique factorization, i.e., the fundamental theorem of arithmetic, to hold in the rings generated by those roots of unity was a major obstacle. Out of Kummer's work for the first time came a study of the obstruction to the factorisation. We now recognise this as part of the ideal class group: in fact Kummer had isolated the p-torsion in that group for the field of p-roots of unity, for any prime number p, as the reason for the failure of the standard method of attack on the Fermat problem (see regular prime).\n\nSomewhat later again Richard Dedekind formulated the concept of ideal, Kummer having worked in a different way. At this point the existing examples could be unified. It was shown that while rings of algebraic integers do not always have unique factorization into primes (because they need not be principal ideal domains), they do have the property that every proper ideal admits a unique factorization as a product of prime ideals (that is, every ring of algebraic integers is a Dedekind domain). The size of the ideal class group can be considered as a measure for the deviation of a ring from being a principal ideal domain; a ring is a principal domain if and only if it has a trivial ideal class group.\n\n## Definition\n\nIf R is an integral domain, define a relation ~ on nonzero fractional ideals of R by I ~ J whenever there exist nonzero elements a and b of R such that (a)I = (b)J. (Here the notation (a) means the principal ideal of R consisting of all the multiples of a.) It is easily shown that this is an equivalence relation. The equivalence classes are called the ideal classes of R.Ideal classes can be multiplied: if [''I''] denotes the equivalence class of the ideal I, then the multiplication [''I''][''J''] = [''IJ''] is well-defined and commutative. The principal ideals form the ideal class [''R''] which serves as an identity element for this multiplication. Thus a class [''I''] has an inverse [''J''] if and only if there is an ideal J such that IJ is a principal ideal. In general, such a J may not exist and consequently the set of ideal classes of R may only be a monoid.\n\nHowever, if R is the ring of algebraic integers in an algebraic number field, or more generally a Dedekind domain, the multiplication defined above turns the set of fractional ideal classes into an abelian group, the ideal class group of R. The group property of existence of inverse elements follows easily from the fact that, in a Dedekind domain, every non-zero ideal (except R) is a product of prime ideals.\n\n## Properties\n\nThe ideal class group is trivial (i.e. has only one element) if and only if all ideals of R are principal. In this sense, the ideal class group measures how far R is from being a principal ideal domain, and hence from satisfying unique prime factorization (Dedekind domains are unique factorization domains if and only if they are principal ideal domains).\n\nThe number of ideal classes (the of R) may be infinite in general. In fact, every abelian group is isomorphic to the ideal class group of some Dedekind domain. But if R is in fact a ring of algebraic integers, then the class number is always finite. This is one of the main results of classical algebraic number theory.\n\nComputation of the class group is hard, in general; it can be done by hand for the ring of integers in an algebraic number field of small discriminant, using Minkowski's bound. This result gives a bound, depending on the ring, such that every ideal class contains an ideal norm less than the bound. In general the bound is not sharp enough to make the calculation practical for fields with large discriminant, but computers are well suited to the task.\n\nThe mapping from rings of integers R to their corresponding class groups is functorial, and the class group can be subsumed under the heading of algebraic K-theory, with K0(R) being the functor assigning to R its ideal class group; more precisely, K0(R) = Z×C(R), where C(R) is the class group. Higher K groups can also be employed and interpreted arithmetically in connection to rings of integers.\n\n## Relation with the group of units\n\nIt was remarked above that the ideal class group provides part of the answer to the question of how much ideals in a Dedekind domain behave like elements. The other part of the answer is provided by the multiplicative group of units of the Dedekind domain, since passage from principal idealsto their generators requires the use of units (and this is the rest of the reason for introducing the concept of fractional ideal, as well):\n\nDefine a map from R× to the set of all nonzero fractional ideals of R by sending every element to the principal (fractional) ideal it generates. This is a group homomorphism; its kernel is the group of units of R, and its cokernel is the ideal class group of R. The failure of these groups to be trivial is a measure of the failure of the map to be an isomorphism: that is the failure of ideals to act like ring elements, that is to say, like numbers.\n\n## Examples of ideal class groups\n\n• The rings Z, Z[ω], and Z[''i''], where ω is a cube root of 1 and i is a fourth root of 1 (i.e. a square root of -1), are all principal ideal domains (and in fact are all Euclidean domains), and so have class number 1: that is, they have trivial ideal class groups.\n• If k is a field, then the polynomial ring k[''X''<sub>1</sub>, ''X''<sub>2</sub>, ''X''<sub>3</sub>, ...] is an integral domain. It has a countably infinite set of ideal classes.\n\n### Class numbers of quadratic fields\n\nIf d is a square-free integer (a product of distinct primes) other than 1, then Q is a quadratic extension of Q. If d < 0, then the class number of the ring R of algebraic integers of Q is equal to 1 for precisely the following values of d: d = -1, -2, -3, -7, -11, -19, -43, -67, and -163. This result was first conjectured by Gauss and proven by Kurt Heegner, although Heegner's proof was not believed until Harold Stark gave a later proof in 1967. (See Stark–Heegner theorem.) This is a special case of the famous class number problem.\n\nIf, on the other hand, d > 0, then it is unknown whether there are infinitely many fields Q with class number 1. Computational results indicate that there are a great many such fields. However, it is not even known if there are infinitely many number fields with class number 1.\n\nFor d < 0, the ideal class group of Q is isomorphic to the class group of integral binary quadratic forms of discriminant equal to the discriminant of Q. For d > 0, the ideal class group may be half the size since the class group of integral binary quadratic forms is isomorphic to the narrow class group of Q.\n\nFor real quadratic integer rings, the class number is given in OEIS A003649; for the imaginary case, they are given in OEIS A000924.\n\n#### Example of a non-trivial class group\n\nThe quadratic integer ring R = Z[{{radic|−5}}] is the ring of integers of Q. It does not possess unique factorization; in fact the class group of R is cyclic of order 2. Indeed, the ideal\n\nJ = (2, 1 +)is not principal, which can be proved by contradiction as follows.\n\nR\n\nhas a norm function\n\nN(a+b\\sqrt{-5})=a2+5b2\n\n, which satisfies\n\nN(uv)=N(u)N(v)\n\n, and\n\nN(u)=1\n\nif and only if\n\nu\n\nis a unit in\n\nR\n\n. First of all,\n\nJ\\neR\n\n, because the quotient ring of\n\nR\n\nmodulo the ideal\n\n(1+\\sqrt{-5})\n\nis isomorphic to\n\nZ/6Z\n\n, so that the quotient ring of\n\nR\n\nmodulo\n\nJ\n\nis isomorphic to\n\nZ/2Z\n\n. If J were generated by an element x of R, then x would divide both 2 and 1 + . Then the norm\n\nN(x)\n\nwould divide both\n\nN(2)=4\n\nand\n\nN(1+\\sqrt{-5})=6\n\n, so N(x) would divide 2. If\n\nN(x)=1\n\n, then\n\nx\n\nis a unit, and\n\nJ=R\n\nN(x)\n\ncannot be 2 either, because R has no elements of norm 2, because the Diophantine equation\n\nb2+5c2=2\n\nhas no solutions in integers, as it has no solutions modulo 5.\n\nOne also computes that J2 = (2), which is principal, so the class of J in the ideal class group has order two. Showing that there aren't any other ideal classes requires more effort.\n\nThe fact that this J is not principal is also related to the fact that the element 6 has two distinct factorisations into irreducibles:\n\n6 = 2 × 3 = (1 +) × (1 -).\n\n## Connections to class field theory\n\nClass field theory is a branch of algebraic number theory which seeks to classify all the abelian extensions of a given algebraic number field, meaning Galois extensions with abelian Galois group. A particularly beautiful example is found in the Hilbert class field of a number field, which can be defined as the maximal unramified abelian extension of such a field. The Hilbert class field L of a number field K is unique and has the following properties:\n\n• Every ideal of the ring of integers of K becomes principal in L, i.e., if I is an integral ideal of K then the image of I is a principal ideal in L.\n• L is a Galois extension of K with Galois group isomorphic to the ideal class group of K.\n\nNeither property is particularly easy to prove."
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https://www.varsitytutors.com/ap_computer_science_a-help/strings | [
"# AP Computer Science A : Strings\n\n## Example Questions\n\n### Example Question #1 : Strings\n\nConsider the following code:\n\nint[] vals = {5,4,2};\n\nString s = \"Hervaeus\";\n\nString s2 = \"\";\n\nfor(int i = 0; i < s.length(); i++) {\n\nfor(int j = 0; j < vals[i % vals.length]; j++) {\n\ns2 += s.charAt(i);\n\n}\n\n}\n\nSystem.out.println(s2);\n\nWhat is the output for the code above?\n\nHHHHHeeeeerrrrrrvvvvvaaaaaeeeeeuuuuusssss\n\nHHHHHeeeerrvaeus\n\nHHHHHeeeerrvvvvvaaaaeeuuuuussss\n\nHervaeeeeeuuuuss\n\nThe code will note execute.\n\nHHHHHeeeerrvvvvvaaaaeeuuuuussss\n\nExplanation:\n\nThe main thing to look at for this question is the main loop for the code. This loop first goes through each of the characters in the String s:\n\nint i = 0; i < s.length(); i++\n\nNext, notice the condition on the inner loop:\n\nvals[i % vals.length]\n\nThe modulus on i will yield values that are between 0 and 2 (given the length of vals). This means that you will loop in the inner loop the following sequence of times:\n\n5,4,2,5,4,2,5,4\n\nThis will replicate the given letter at the index (i) in the initial string, using 5,4,2,5, etc as the replication count. Thus, you will replicate \"H\" 5 times, \"e\" 4, etc. This gives you an output:\n\nHHHHHeeeerrvvvvvaaaaeeuuuuussss\n\n### Example Question #1 : Strings\n\nConsider the following code:\n\nchar[] values = {'I',' ','l','o','v','e',' ','y','o','u','!','!'};\n\nString s = \"\";\n\nfor(int i = 0; i < values.length / 2; i++) {\n\nchar temp = values[i];\n\nvalues[i] = values[values.length - i-1];\n\nvalues[values.length - i-1] = temp;\n\n}\n\nfor(int i = 0; i < values.length; i++) {\n\ns += values[i];\n\n}\n\nSystem.out.println(s);\n\nWhat is the output for the code above?\n\nNone of the others\n\n!uyeolI!o vl\n\n!!uoy evol I\n\n!you I love!\n\n!!you love I\n\n!!uoy evol I\n\nExplanation:\n\nIt is easiest to think of the values array as a String: \"I love you!!\".\n\nNow, the loop is going to run for",
null,
"or",
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"times. Notice what it does on each iteration. It swaps the values at i and values.length - i -1. Thus, it will do the following swaps:\n\n0 and 11\n\n1 and 10\n\netc...\n\nThis sequence of swaps will eventually reverse the array. Thus, your output is:\n\n!!uoy evol I\n\n### Example Question #1 : Common Data Structures\n\nWhich of the following blocks of code makes every other character in the string s to be upper case, starting with the second character?\n\nString s = \"This is a great string!\";\n\nString s2 = \"\";\n\nfor(int i = 0; i < s.length(); i++) {\n\nchar c = Character.toLowerCase(s.charAt(i));\n\ns2+=c;\n\nif(i % 2 == 0) {\n\nc = Character.toUpperCase(c);\n\n}\n\n}\n\nString s = \"This is a great string!\";\n\nString s2 = \"\";\n\nfor(int i = 0; i < s.length(); i++) {\n\nchar c = Character.toLowerCase(s.charAt(i));\n\nif(i % 2 == 0) {\n\nc = Character.toUpperCase(c);\n\n}\n\ns2+=c;\n\n}\n\nString s = \"This is a great string!\";\n\nString s2 = \"\";\n\nfor(int i = 0; i < s.length(); i++) {\n\nchar c = Character.toLowerCase(s.charAt(i));\n\nif(i % 2 == 3) {\n\nc = Character.toUpperCase(c);\n\n}\n\ns2+=c;\n\n}\n\nString s = \"This is a great string!\";\n\nString s2 = \"\";\n\nfor(int i = 0; i < s.length(); i++) {\n\nchar c = s.charAt(i);\n\nif(i % 2 == 1) {\n\nc = Character.toUpperCase(c);\n\n}\n\ns2+=c;\n\n}\n\nString s = \"This is a great string!\";\n\nString s2 = \"\";\n\nfor(int i = 0; i < s.length(); i++) {\n\nchar c = Character.toLowerCase(s.charAt(i));\n\nif(i % 2 == 1) {\n\nc = Character.toUpperCase(c);\n\n}\n\ns2+=c;\n\n}\n\nString s = \"This is a great string!\";\n\nString s2 = \"\";\n\nfor(int i = 0; i < s.length(); i++) {\n\nchar c = Character.toLowerCase(s.charAt(i));\n\nif(i % 2 == 1) {\n\nc = Character.toUpperCase(c);\n\n}\n\ns2+=c;\n\n}\n\nExplanation:\n\nGiven that strings cannot be internally modified, you will have to store your result in a new string, namely s2. Now, for each character in s, you will have to make that charater lower case to begin with:\n\nchar c = Character.toLowerCase(s.charAt(i));\n\nNext, for the odd values of i, you will need to make your value to be upper case. The modulus operator is great for this! You can use % 2 to find the odd values. When the remainder of a division by 2 is equal to 1, you know you have an odd value. Hence, you have the condition:\n\nif(i % 2 == 1) {...\n\nThen, once you appropriately capitalize, you can place your character on s2.\n\n### Example Question #1 : Strings\n\nWhich of the following blocks of code converts an array of characters into a string?\n\nprivate static void string() {\n\nchar[] vals = {'A','t', ' ', '6',' ','a','m','!'};\n\nString s = \"\";\n\nfor(int i = 0; i < vals.length; i++) {\n\ns = vals[i];\n\n}\n\n}\n\nprivate static void string() {\n\nchar[] vals = {'A','t', ' ', '6',' ','a','m','!'};\n\nString s = \"\";\n\nfor(int i = 0; i < vals.length; i++) {\n\nvals[i] = s;\n\n}\n\n}\n\nprivate static void string() {\n\nchar[] vals = {'A','t', ' ', '6',' ','a','m','!'};\n\nString s;\n\nfor(int i = 0; i < vals.length; i++) {\n\ns += vals[i];\n\n}\n\n}\n\nprivate static void string() {\n\nString s = \"At 6 am!\";\n\nchar[] vals = new char[s.length()];\n\nfor(int i = 0; i < s.length(); i++) {\n\nvals[i] = s.charAt(i);\n\n}\n\n}\n\nprivate static void string() {\n\nchar[] vals = {'A','t', ' ', '6',' ','a','m','!'};\n\nString s = \"\";\n\nfor(int i = 0; i < vals.length; i++) {\n\ns += vals[i];\n\n}\n\n}\n\nprivate static void string() {\n\nchar[] vals = {'A','t', ' ', '6',' ','a','m','!'};\n\nString s = \"\";\n\nfor(int i = 0; i < vals.length; i++) {\n\ns += vals[i];\n\n}\n\n}\n\nExplanation:\n\nThe easiest way to consider this is by commenting on the correct answer. You must begin by defining the character array:\n\nchar[] vals = {'A','t', ' ', '6',' ','a','m','!'};\n\nNext, you must initialize the string value s to be an empty string. This is critical. Otherwise, you can't build your string!\n\nString s = \"\";\n\nNext, you have the loop. This goes through the characters and concatenates the values to the variable s. The operation to concatenate the characters is the \"+=\". This will give you the string value of the array of characters.\n\n### Example Question #1 : Strings\n\nSee code below:\n\nString[] books = {\n\n\"De Secundis Intentionibus\",\n\n\"Leviathan\",\n\n\"Averrois Commentaria Magna in Aristotelem De celo et mundo\",\n\n\"Logica Docens for Idiots\",\n\n\"Logica Utens for Tuba Players\"\n\n};\n\nString userInput;\n\n// In code excised from here, a person inputs the value \"Logica Utens for Tuba Players\" ...\n\nfor(int i = 0; i < books.length; i++) {\n\nif(books[i] == userInput) {\n\nSystem.out.println(\"This is a wonderful book!!\");\n\n}\n\n}\n\nWhat is the error in the code above?\n\nThe string comparison.\n\nIt overruns an array.\n\nThe use of the [] operator on the array.\n\nThe way the array is declared.\n\nThe use of if on an array element.\n\nThe string comparison.\n\nExplanation:\n\nThe only major issue with this code is the use of the == operator to compare the two strings. Since the user has input this value, will not have an equality on this test. You must use the method .equals in order to check whether two strings are equal. The code should be:\n\nif(books[i].equals(userInput)){\n\n...\n\n(There are some cases in which == will work for string comparison, namely when literals are involved. However, you should avoid relying on this and always use .equals().)\n\n### Example Question #1 : Strings\n\nString[] books = {\n\n\"Logica sive Ars Rationalis\",\n\n\"Kritik der reinen Vernunft\",\n\n\"Cursus Philosophicus Thomisticus\",\n\n\"Happy words for happy people\",\n\n\"Insane words for an insane world\"\n\n};\n\nString str = \"Kittens in a cart\";\n\nArrayList<String> vals = new ArrayList<String>();\n\nfor(int i = 0; i < books.length; i++) {\n\nif(books[i].compareTo(str) > 0) {\n\n}\n\n}\n\nSystem.out.println(vals);\n\nWhat is the output for this method?\n\n[Happy words for happy people, Insane words for an insane world]\n\n[Cursus Philosophicus Thomisticus, Happy words for happy people, Insane words for an insane world]\n\n[Kritik der reinen Vernunft, Logica sive Ars Rationalis]\n\n[Logica sive Ars Rationalis, Kritik der reinen Vernunft]\n\n[]\n\n[Logica sive Ars Rationalis, Kritik der reinen Vernunft]\n\nExplanation:\n\nThe compareTo method for strings compares to string objects and returns:\n\n• 0 when they are equal\n• Something positive when the given string is alphabetically (really, lexicographically) later than the argument to the method.\n• Something negative when the given string is alphabetically (really, lexicographically) before the argument to the method.\n\nOur strings could be put in this order:\n\n\"Cursus Philosophicus Thomisticus\"\n\n\"Happy words for happy people\"\n\n\"Insane words for an insane world\"\n\n\"Kritik der reinen Vernunft\"\n\n\"Logica sive Ars Rationalis\"\n\nFor each of these, we are asking, \"Is it later in order than 'Kittens in a cart'?\" This is true for \"Kritik der reinen Vernunft\" and \"Logica sive Ars Rationalis\". Thus, our output is:\n\n[Logica sive Ars Rationalis, Kritik der reinen Vernunft]\n\nNotice the order—this is due to the order in the original array.\n\n### Example Question #1 : Strings\n\nString greet = \"Hello \";\nString sub;\nint len = greet.length();\nsub = greet.substring(0, (len/2));\nSystem.out.println (sub);\n\nWhat is printed?\n\nHell\n\nHel\n\nHello\n\nHe\n\nError\n\nHel\n\nExplanation:\n\nThe length of greet is 6 characters including the space at the end.\n\ngreet.substring(0, (len/2)) is equal to greet.substring(0, 3)\n\nThe substring of greet from the zeroth position to second position, not to the third position.\n\n### All AP Computer Science A Resources",
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https://brilliant.org/problems/a-problem-by-efren-medallo/ | [
"# A problem by Efren Medallo\n\nGeometry Level 2",
null,
"In the figure, $\\Delta ABC \\cong \\Delta EDF$ with $\\overline{AC} = \\overline{EF} = 8$.\n\nIf $\\Delta DEF$ is positioned in such a way that $D$ lies along $\\overline{AB}$ and the area of polygon $ADEC$ is $28,$ what is the length of $\\overline{FC}?$\n\n×"
]
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null,
"https://ds055uzetaobb.cloudfront.net/brioche/uploads/g6wWB2bX3I-50071.svg",
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| {"ft_lang_label":"__label__en","ft_lang_prob":0.88259286,"math_prob":1.0000004,"size":281,"snap":"2020-45-2020-50","text_gpt3_token_len":67,"char_repetition_ratio":0.15162455,"word_repetition_ratio":0.25,"special_character_ratio":0.24199288,"punctuation_ratio":0.25,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99991417,"pos_list":[0,1,2],"im_url_duplicate_count":[null,3,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-11-29T11:23:32Z\",\"WARC-Record-ID\":\"<urn:uuid:b9682cd6-50ce-4a02-838c-6dabf7ba71e3>\",\"Content-Length\":\"44108\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:83f1c3f6-ac41-48fe-900d-c95d7185669a>\",\"WARC-Concurrent-To\":\"<urn:uuid:e15358a0-0662-4e1d-8e6d-9cdfc1edea3d>\",\"WARC-IP-Address\":\"104.20.34.242\",\"WARC-Target-URI\":\"https://brilliant.org/problems/a-problem-by-efren-medallo/\",\"WARC-Payload-Digest\":\"sha1:DWYHKINNO5DCNX3OJ35Y5EYV7SAJ27WR\",\"WARC-Block-Digest\":\"sha1:YGPW7EKR75Q6DQVPS3HWCTKRP4JJWLWM\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-50/CC-MAIN-2020-50_segments_1606141197593.33_warc_CC-MAIN-20201129093434-20201129123434-00363.warc.gz\"}"} |
https://www.cliffsnotes.com/cliffsnotes/subjects/math/what-s-the-formula-to-convert-square-feet-into-square-meters | [
"## What's the formula to convert square feet into square meters?\n\nOne square foot equals 0.0929 square meters. So to convert from square feet to square meters, just multiply by 0.0929.\n\nSo let's say you're in the United States, and you've purchased a home that's 2,030 square feet. You're trying to tell your friend in London about the home, and how big it is.\n\n2030 x 0.0929 = 188.587\n\nSo your 2,030-square-foot house is equal to 188.587 square meters. (If you say it's 188 square meters, he'll get the idea.)\n\nTo convert from square meters to square feet, simply divide the number of square meters by 0.0929 to get the number of square feet."
]
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| {"ft_lang_label":"__label__en","ft_lang_prob":0.9263353,"math_prob":0.9969508,"size":635,"snap":"2019-51-2020-05","text_gpt3_token_len":168,"char_repetition_ratio":0.22345483,"word_repetition_ratio":0.0,"special_character_ratio":0.30393702,"punctuation_ratio":0.14685315,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9984419,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-01-20T15:20:13Z\",\"WARC-Record-ID\":\"<urn:uuid:eaca9370-799d-4fc7-b25c-2d91e0dd78a6>\",\"Content-Length\":\"1049331\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:0bd258db-e86a-49ba-9677-0e3f8a09e04a>\",\"WARC-Concurrent-To\":\"<urn:uuid:82257288-0067-4f60-b20e-4841a8feedc7>\",\"WARC-IP-Address\":\"23.47.18.181\",\"WARC-Target-URI\":\"https://www.cliffsnotes.com/cliffsnotes/subjects/math/what-s-the-formula-to-convert-square-feet-into-square-meters\",\"WARC-Payload-Digest\":\"sha1:FZ2KINUS3ESK4O75GHQBMCRAYNFLD3HO\",\"WARC-Block-Digest\":\"sha1:BIM7LO2V3PHK4ANLJHLUKNKYLU2AVZQH\",\"WARC-Truncated\":\"length\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-05/CC-MAIN-2020-05_segments_1579250598800.30_warc_CC-MAIN-20200120135447-20200120164447-00119.warc.gz\"}"} |
https://math.berkeley.edu/~jbrere/HADESfall.html | [
"Berkeley Harmonic Analysis and Differential equations student seminar\nFall\n2017\n\nHere is a list of links to material and conferences/workshops on PDE’s and Harmonic Analysis.\n\n September 12th 3:40 - 5:00 891 Evans Hall Kevin O'Neill (UCB) A Sharp Schrodinger Maximal Estimate in R^2. In this talk, I will present a recent paper of Xiumin Du, Larry Guth, and Xiaochun Li, which proves almost everywhere convergence of solutions to the Schrodinger equation in R^2 for initial data in H^s (s>1/3). I will give an extended introduction to the method of polynomial partitioning, which is used in the proof of their main theorem. A new result which arises during the proof is a bilinear local refinement of the Strichartz inequality which is made possible by the l^2-decoupling theorem of Bourgain and Demeter. September 19th 3:40 - 5:00 891 Evans Hall Cristian Gavurs (UC Berkeley) A wave packet parametrix for wave equations. The aim of this talk is to present the construction of a parametrix for the wave equation with variable coefficients due to Hart Smith. The idea is to write approximate solutions as linear combinations of wave packets by decomposing the initial data using a frame of functions (concentrated in space and frequency) which are then transported across the bicharacteristic flow. Even though this construction and proof are fairly elementary, it works under minimal assumptions on the regularity of the coefficients (two bounded derivatives). The talk is based on the paper https://eudml.org/doc/75304 . September 26th 3:40 - 5:00 891 Evans Hall A. Martina Neuman (UC Berkeley) Decoupling for surfaces in R^4. We present a sharp decoupling result of surfaces in R. The techniques brought forth will demonstrate that the natural decoupling scale to work with non-degenerate surfaces or hypersurfaces in any dimensions is N^(-1/2). October 3rd 3:40 - 5:00 891 Evans Hall No HADES October 10th 3:40 - 5:00 891 Evans Hall Marina Iliopoulou (UC Berkeley) Algebraic structure underlying Kakeya-type problems. Kakeya-type questions ask how much tubes that point to different directions overlap. Such problems are central in harmonic analysis, due to their connection with restriction theory, geometric measure theory, PDE and number theory. Over the years there was some indication that Kakeya-type problems have an underlying algebraic structure, but it was only a decade ago that a tool was introduced in the area to reveal such structure, leading to important advances in the field. This tool is the polynomial method. During this talk we will explain the method and see applications to Kakeya-type problems. October 17th 3:40 - 5:00 891 Evans Hall Albert Ai (UC Berkeley) Strichartz estimates for the gravity water waves. In this talk we will introduce the gravity water waves equations, which describe the motion of a fluid influenced by gravity, under a free interface with a vacuum. We will discuss various formulations of the problem, and in particular a paradifferential reduction due to Alazard, Burq, and Zuily. From this formulation we can exhibit the dispersive properties of the water waves system by establishing Strichartz estimates. October 24th 3:40 - 5:00 891 Evans Hall Grace Liu (UC Berkeley) Scattering and modified scattering for the NLS equation. The NLS with nonlinear term |u|^p u has the property that when 1 < p < 2/n, there is no low energy scattering, when 2/n < p < 4/n, there is low energy scattering. When p=2/n, we expect there will be modified scattering. The talk will be based on the work of Hayashi and Naumkin, they proved that the NLS with critical nonlinearity has low energy scattering when n=1,2,3. October 31st 3:40 - 5:00 891 Evans Hall Peter Hintz (UC Berkeley) Resonances and wave decay on Euclidean and hyperbolic spaces. I will introduce the notion of resonances in potential and obstacle scattering in Euclidean and hyperbolic spaces and explain their relation to the local energy decay of solutions of the wave equation. This talk is based on joint work with Maciej Zworski. November 7th 3:40 - 5:00 891 Evans Hall Justin Brereton (UC Berkeley) Stokes' theorem and construction of an invariant measure on an L^2 ball. An invariant measure is useful in proving almost sure well-posedness of a PDE on a compact set. Thomann and Tzvetkov constructed an invariant measure for the derivative nonlinear Schrodinger (DNLS) equation on the torus. In this talk we construct a DNLS-invariant measure on the set of functions with L^2(T) norm equal to a fixed value m, by an argument that is similar to Stokes' theorem, but in infinitely many dimensions. November 14th 3:40 - 5:00 891 Evans Hall Casey Jao (UC Berkeley) Global quasilinear waves in 1+3 dimensions. I will present a classical result, obtained independently by Christodoulou and Klainerman, on the global existence of solutions to certain second order quasilinear wave equations with small data. November 21st 3:40 - 5:00 891 Evans Hall NO HADES November 28th 3:40 - 5:00 891 Evans Hall Mohandas Pillai (UC Berkeley) Singularity formation for 1-equivariant wave maps into S^2 I will talk about part of a work of Raphael and Rodnianski which constructed singularity forming solutions to the 1-equivariant wave maps equation into S^2 (the work in fact considers several equations at once, but I will only talk about the 1-equivariant wave maps problem)."
]
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https://de.mathworks.com/matlabcentral/answers/1946353-ode-simulation-issue-with-the-size-of-arrays | [
"# ODE simulation: issue with the size of arrays\n\n4 Ansichten (letzte 30 Tage)\nAlessandro Bellocchi am 13 Apr. 2023\nBearbeitet: Davide Masiello am 13 Apr. 2023\nHello, everyone. I am trying to simulate the evolution of the following model. However, I get an error with the size of the arrays (Unable to perform assignment because the size of the left side is 1-by-2 and the size of the right side is 1-by-330201). Can anyone help me solve it?\n% Define the parameters\nm = 0.25;\nbeta = 1.25;\nrho = 0.04;\ndelta = 0.03;\nalpha = 0.2;\npphi = 0.5;\nmu = 0.2;\ngamma = 1;\n% Define the initial condition\nphi0 = 10;\ntheta0 = 1;\n% Define the values of sigma for the two situations\nsigma_values = [0.1, 0.5];\n% Define the time span for the simulation\ntspan = [0 500];\n% Define the Wiener process\ndW = makedist('Normal', 'mu', 0, 'sigma', sqrt(diff(tspan)));\n% Define the number of simulations\nnum_simulations = 5;\n% Initialize arrays to store the phi values for each simulation\nphi_values = zeros(length(sigma_values), length(tspan), num_simulations);\naverage_phi_values = zeros(length(sigma_values), length(tspan));\n% Perform multiple simulations and store the phi values\nfor sim = 1:num_simulations\nfor i = 1:length(sigma_values)\n% Define the current value of sigma\nsigma = sigma_values(i);\n% Define the function to be solved\nfinance = @(t, y) [\n(1/gamma)*((m*y(2)^beta)/(rho+delta-(alpha*(beta*sigma)^2)/2)-(pphi-mu)-delta*y(1));\nsigma*y(2)*dW.random()];\n% Solve the differential equation numerically\n[t, y] = ode45(finance, tspan, [phi0, theta0]);\n% Store the phi values for the current simulation and sigma value\nphi_values(i, :, sim) = y(:, 1)';\nend\nend\n% Calculate the average phi value over all simulations for each time point and sigma value\nfor i = 1:length(sigma_values)\nfor j = 1:length(tspan)\naverage_phi_values(i, j) = mean(phi_values(i, j, :));\nend\nend\n% Plot the average phi values against time for each sigma value\nhold on\nfor i = 1:length(sigma_values)\nplot(tspan, average_phi_values(i, :), 'LineWidth', 2);\nend\n% Add legend, labels, and title\nlegend(['\\sigma = ', num2str(sigma_values(1))], ['\\sigma = ', num2str(sigma_values(2))]);\nxlabel('Time');\nylabel('Average \\phi_t');\ntitle('Evolution of Finance over Time for Different Values of \\sigma and Wiener Process');\n##### 1 KommentarKeine anzeigenKeine ausblenden\nDavide Masiello am 13 Apr. 2023\nUnfortunately, you do not know the size of the t vector before the ODE integration itself.\nHence, you cannot initialize phi_values the way you did.\nEven if you didn't initialize it, you'd probably end up with a size error, because each simulation will require a different number of steps to complete the integration.\nThis is due to the fact that MatLab ode solvers are adaptive solvers, with variable step size.\n\nMelden Sie sich an, um zu kommentieren.\n\n### Antworten (1)\n\nDavide Masiello am 13 Apr. 2023\nBearbeitet: Davide Masiello am 13 Apr. 2023\nIn addition to my comment above, here's a solution to your problem.\nIt's based on the fact that you can specify tspan as a vector of points at which the solution must be reported.\nThis provides a homogeneous output regardless of the choice of the integration time steps.\nclear,clc\n% Define the parameters\nm = 0.25;\nbeta = 1.25;\nrho = 0.04;\ndelta = 0.03;\nalpha = 0.2;\npphi = 0.5;\nmu = 0.2;\ngamma = 1;\n% Define the initial condition\nphi0 = 10;\ntheta0 = 1;\n% Define the values of sigma for the two situations\nsigma_values = [0.1, 0.5];\n% Define the time span for the simulation\ntspan = linspace(0,500,1000);\n% Define the Wiener process\ndW = makedist('Normal', 'mu', 0, 'sigma', sqrt(tspan(end)-tspan(1)));\n% Define the number of simulations\nnum_simulations = 5;\n% Initialize arrays to store the phi values for each simulation\nphi_values = zeros(length(sigma_values), length(tspan), num_simulations);\naverage_phi_values = zeros(length(sigma_values), length(tspan));\n% Perform multiple simulations and store the phi values\nfor sim = 1:num_simulations\nfor i = 1:length(sigma_values)\n% Define the current value of sigma\nsigma = sigma_values(i);\n% Define the function to be solved\nfinance = @(t, y) [\n(1/gamma)*((m*y(2)^beta)/(rho+delta-(alpha*(beta*sigma)^2)/2)-(pphi-mu)-delta*y(1));\nsigma*y(2)*dW.random()];\n% Solve the differential equation numerically\n[t, y] = ode45(finance, tspan, [phi0, theta0]);\n% Store the phi values for the current simulation and sigma value\nphi_values(i, :, sim) = y(:, 1)';\nend\nend\n% Calculate the average phi value over all simulations for each time point and sigma value\nfor i = 1:length(sigma_values)\nfor j = 1:length(tspan)\naverage_phi_values(i, j) = mean(phi_values(i, j, :));\nend\nend\n% Plot the average phi values against time for each sigma value\nplot(tspan, average_phi_values, 'LineWidth', 2);\nWarning: Imaginary parts of complex X and/or Y arguments ignored.\n% Add legend, labels, and title\nlegend(['\\sigma = ', num2str(sigma_values(1))], ['\\sigma = ', num2str(sigma_values(2))]);\nxlabel('Time');\nylabel('Average \\phi_t');\ntitle('Evolution of Finance over Time for Different Values of \\sigma and Wiener Process')",
null,
"##### 0 Kommentare-1 ältere Kommentare anzeigen-1 ältere Kommentare ausblenden\n\nMelden Sie sich an, um zu kommentieren.\n\n### Kategorien\n\nMehr zu Stochastic Differential Equation (SDE) Models finden Sie in Help Center und File Exchange\n\nR2022b\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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https://patents.google.com/patent/EP3335321A1/en | [
"# EP3335321A1 - Rate-compatible polar codes - Google Patents\n\nRate-compatible polar codes\n\n## Info\n\nPublication number\nEP3335321A1\nEP3335321A1 EP16742000.9A EP16742000A EP3335321A1 EP 3335321 A1 EP3335321 A1 EP 3335321A1 EP 16742000 A EP16742000 A EP 16742000A EP 3335321 A1 EP3335321 A1 EP 3335321A1\nAuthority\nEP\nEuropean Patent Office\nPrior art keywords\npolar\nplurality\ncode\ncoded bits\nbits\nPrior art date\nLegal status (The legal status is an assumption and is not a legal conclusion. Google has not performed a legal analysis and makes no representation as to the accuracy of the status listed.)\nPending\nApplication number\nEP16742000.9A\nOther languages\nGerman (de)\nFrench (fr)\nInventor\nSongnam HONG\nDennis Hui\nIvana Maric\nCurrent Assignee (The listed assignees may be inaccurate. Google has not performed a legal analysis and makes no representation or warranty as to the accuracy of the list.)\nTelefonaktiebolaget LM Ericsson AB\nOriginal Assignee\nTelefonaktiebolaget LM Ericsson AB\nPriority date (The priority date is an assumption and is not a legal conclusion. Google has not performed a legal analysis and makes no representation as to the accuracy of the date listed.)\nFiling date\nPublication date\nPriority to US14/824,694 priority Critical patent/US10461779B2/en\nApplication filed by Telefonaktiebolaget LM Ericsson AB filed Critical Telefonaktiebolaget LM Ericsson AB\nPriority to PCT/IB2016/053941 priority patent/WO2017025823A1/en\nPublication of EP3335321A1 publication Critical patent/EP3335321A1/en\nPending legal-status Critical Current\n\n• 230000001702 transmitter Effects 0.000 claims abstract description 16\n• 230000000875 corresponding Effects 0.000 claims description 17\n• 238000000034 methods Methods 0.000 description 16\n\n## Classifications\n\n• HELECTRICITY\n• H03BASIC ELECTRONIC CIRCUITRY\n• H03MCODING; DECODING; CODE CONVERSION IN GENERAL\n• H03M13/00Coding, decoding or code conversion, for error detection or error correction; Coding theory basic assumptions; Coding bounds; Error probability evaluation methods; Channel models; Simulation or testing of codes\n• H03M13/29Coding, decoding or code conversion, for error detection or error correction; Coding theory basic assumptions; Coding bounds; Error probability evaluation methods; Channel models; Simulation or testing of codes combining two or more codes or code structures, e.g. product codes, generalised product codes, concatenated codes, inner and outer codes\n• H03M13/2906Coding, decoding or code conversion, for error detection or error correction; Coding theory basic assumptions; Coding bounds; Error probability evaluation methods; Channel models; Simulation or testing of codes combining two or more codes or code structures, e.g. product codes, generalised product codes, concatenated codes, inner and outer codes using block codes\n• HELECTRICITY\n• H03BASIC ELECTRONIC CIRCUITRY\n• H03MCODING; DECODING; CODE CONVERSION IN GENERAL\n• H03M13/00Coding, decoding or code conversion, for error detection or error correction; Coding theory basic assumptions; Coding bounds; Error probability evaluation methods; Channel models; Simulation or testing of codes\n• H03M13/03Error detection or forward error correction by redundancy in data representation, i.e. code words containing more digits than the source words\n• H03M13/05Error detection or forward error correction by redundancy in data representation, i.e. code words containing more digits than the source words using block codes, i.e. a predetermined number of check bits joined to a predetermined number of information bits\n• H03M13/13Linear codes\n• HELECTRICITY\n• H03BASIC ELECTRONIC CIRCUITRY\n• H03MCODING; DECODING; CODE CONVERSION IN GENERAL\n• H03M13/00Coding, decoding or code conversion, for error detection or error correction; Coding theory basic assumptions; Coding bounds; Error probability evaluation methods; Channel models; Simulation or testing of codes\n• H03M13/63Joint error correction and other techniques\n• H03M13/635Error control coding in combination with rate matching\n• H03M13/6362Error control coding in combination with rate matching by puncturing\n• H03M13/6368Error control coding in combination with rate matching by puncturing using rate compatible puncturing or complementary puncturing\n• HELECTRICITY\n• H04ELECTRIC COMMUNICATION TECHNIQUE\n• H04LTRANSMISSION OF DIGITAL INFORMATION, e.g. TELEGRAPHIC COMMUNICATION\n• H04L1/00Arrangements for detecting or preventing errors in the information received\n• H04L1/004Arrangements for detecting or preventing errors in the information received by using forward error control\n• H04L1/0041Arrangements at the transmitter end\n• HELECTRICITY\n• H04ELECTRIC COMMUNICATION TECHNIQUE\n• H04LTRANSMISSION OF DIGITAL INFORMATION, e.g. TELEGRAPHIC COMMUNICATION\n• H04L1/00Arrangements for detecting or preventing errors in the information received\n• H04L1/004Arrangements for detecting or preventing errors in the information received by using forward error control\n• H04L1/0056Systems characterized by the type of code used\n• H04L1/0064Concatenated codes\n• HELECTRICITY\n• H04ELECTRIC COMMUNICATION TECHNIQUE\n• H04LTRANSMISSION OF DIGITAL INFORMATION, e.g. TELEGRAPHIC COMMUNICATION\n• H04L1/00Arrangements for detecting or preventing errors in the information received\n• H04L1/004Arrangements for detecting or preventing errors in the information received by using forward error control\n• H04L1/0056Systems characterized by the type of code used\n• H04L1/0067Rate matching\n\n## Abstract\n\nSystems and methods are disclosed herein relating to rate-compatible polar codes and the use thereof in a wireless communications system. In some embodiments, a transmit node operable for use in a wireless communications system comprises a rate-compatible polar encoder operable to encode information bits to provide coded bits utilizing parallel concatenated polar codes. The transmit node further comprises a transmitter operable to transmit the plurality of coded bits. In this manner, the transmit node may, in some embodiments, use polar codes having different coding rates to adapt to time- varying channel conditions.\n\n## Description\n\nRATE-COMPATIBLE POLAR CODES\n\nRelated Applications\n\n This application claims the benefit of provisional patent application serial number 14/824,694, filed August 12, 2015, the disclosure of which is hereby incorporated herein by reference in its entirety.\n\nTechnical Field\n\n The present disclosure relates to polar codes and, in particular, rate- compatible polar codes and the use thereof for error correction coding in a wireless communications system.\n\nBackground\n\n Polar codes, proposed by E. Arikan, \"Channel Polarization: A Method for Constructing Capacity-Achieving Codes for Symmetric Binary-Input\n\nMemoryless Channels,\" IEEE Transactions on Information Theory, Vol. 55, Issue 7, pages 3051 -3073, July 2009, are the first class of constructive coding schemes that are provable to achieve symmetric capacity of binary-input discrete memoryless channels under a low-complexity Successive Cancellation (SC) decoder. However, the finite-length performance of polar codes under SC is not competitive over other modern channel coding schemes, such as Low-Density Parity-Check (LDPC) codes and turbo codes. Later, a SC List (SCL) decoder was proposed in I. Tal et al., \"List Decoding of Polar Codes,\" 201 1 IEEE\n\nInternational Symposium on Information Theory Proceedings, pages 1 -5, July 31 - August 5, 201 1. The proposed SCL decoder can approach the performance of an optimal Maximum-Likelihood (ML) decoder. By concatenating a simple Cyclic Redundancy Check (CRC) coding, it was shown that the performance of a concatenated polar code is better than the performance of well-optimized LDPC and turbo codes. This result represents the breakthrough of polar codes in future wireless communications systems (i.e., 5th Generation (5G)). Wireless broadband systems require flexible and adaptive\n\ntransmission techniques since they operate in the presence of time-varying channels. For such systems, Hybrid Automatic Repeat Request (HARQ) based on Incremental Redundancy (HARQ-IR) schemes are often used, where parity bits are sent in an incremental fashion depending on the quality of the time- varying channel. IR systems require the use of rate-compatible punctured codes. According to the rate requirement, an appropriate number of parity bits are sent by the transmitter. Here, the set of parity bits of a higher rate code should be a subset of the set of parity bits of a lower rate code. Therefore, in a HARQ-IR system, if the receiver fails to decode at a particular rate, the receiver only needs to request additional parity bits from the transmitter. For this reason, there has been extensive research on the construction of rate-compatible turbo codes and LDPC codes.\n\n Although polar codes can achieve the capacity of a binary input symmetric output channel, existing polar codes are not rate-compatible and, as such, are not suitable for use in a future wireless communications system, particularly one that utilizes HARQ-IR or some similar IR transmission scheme.\n\nSummary\n\n Systems and methods are disclosed herein relating to rate-compatible polar codes and the use thereof in a wireless communications system. In some embodiments, a transmit node operable for use in a wireless communications system comprises a rate-compatible polar encoder operable to encode\n\ninformation bits to provide coded bits utilizing parallel concatenated polar codes. The transmit node further comprises a transmitter operable to transmit the plurality of coded bits. In this manner, the transmit node may, in some\n\nembodiments, use polar codes having different coding rates to adapt to time- varying channel conditions.\n\n In some embodiments, the rate-compatible polar encoder comprises multiple polar encoders operable to encode the information bits. The polar encoders comprise a first polar encoder operable to encode the information bits at a first code rate n to provide a number ni of coded bits, where n-ι = k / n and k is the number of information bits; and a second polar encoder operable to encode a subset of the information bits at a second code rate r2 to provide a number n2 of coded bits, where n > r2 and the number n2 of coded bits is a number of coded bits that, when concatenated with the number ni of coded bits, transforms a resulting code word from the first code rate Pi to the second code rate r2. The coded bits are provided at a desired code rate rdeSired and are a concatenation of the coded bits output by polar encoders of the plurality of polar encoders for rates Pi through rdeSired- Further, in some embodiments, the rate- compatible polar encoder further comprises a concatenator operable to concatenate the coded bits output by the polar encoders for rates Pi through rdesired to provide the plurality of coded bits at the desired code rate rdeSired- In some embodiments, the rate-compatible polar encoder comprises multiple polar encoders operable to encode the information bits based on corresponding polar code generator matrices for rates η and lengths n,, where:\n\n• i = {1 , ... , T}, where T is a positive integer greater than or equal to\n\n2;\n\n• ni is a length for code rate Pi and, for all other values of i, n, is a number of additional coded bits to transform a code word for code rate η.ι into a code word for code rate η, i.e.,\n\nwhere k is the number of information bits i;\n\n• ∑[=1 nt = n, where n is a maximum code word length of the coded bits (i.e., n = k/nT); and\n\nThe coded bits are provided at a desired code rate rdesired and are a\n\nconcatenation of the coded bits output by polar encoders for code rates Pi through rdeSired- Further, in some embodiments, the rate-compatible polar encoder further comprises a concatenator operable to concatenate outputs of the polar encoders to provide the coded bits at the desired code rate rdesired such that, for the desired code rate rdesired, the coded bits are a concatenation of the outputs of the polar encoders for code rates Pi , ... , rdesired- Further, in some\n\nembodiments, the polar code generator matrices for the polar encoders are submatrices ί¾ ([1: rjnj) of row-permuted generator matrices of i-th polar codes consisting of rows 1 through of G^ .\n\n In some embodiments, the polar encoders comprise a first polar encoder for code rate Pi operable to encode the information bits at the code rate n to provide ni coded bits, and one or more additional polar encoders, each additional polar encoder operable to encode a subset of the information bits consisting of a number ηη, of the information bits at the code rate η to provide n, coded bits. In some embodiments, for each of the one or more additional polar encoders, the subset of the information bits encoded by the additional polar encoder is different than the subsets encoded by the other additional polar encoders. In other embodiments, for each of the one or more additional polar encoders, the subset of the information bits encoded by the additional polar encoder comprises a predefined number of the information bits that are most unreliable with respect to transmission of the number ni of coded bits from the first polar encoder. In some embodiments, for at least one of the one or more additional polar encoders, an ordering of the subset of the information bits encoded by the at least one of the one or more additional polar encoders is different than an ordering of those same information bits when encoded by the first polar encoder.\n\n In some embodiments, the transmit node further comprises at least one processor, memory containing instructions executable by the at least one processor whereby the transmit node is further operable to: select an initial code rate; perform, via the rate-compatible polar encoder, polar encoding of the information bits at the initial code rate; transmit, via the transmitter, the coded bits, having been encoded at the initial code rate; select a new code rate for an incremental redundancy retransmission upon receiving a negative acknowledgement from a receive node for the transmission of the coded bits; perform, via the rate-compatible polar encoder, polar encoding of some of the information bits to provide additional coded bits for the new code rate; and transmit, via the transmitter, the additional coded bits. Further, in some embodiments, the rate-compatible polar encoder comprises multiple polar encoders operable to encode the information bits based on corresponding polar code generator matrices for code rates η and lengths n,, where:\n\n• i = {1 , ... , T}, where T is a positive integer greater than or equal to 2;\n\n· ni is a length for code rate Pi and, for all other values of i, n, is a number of additional coded bits to transform a code word for code rate η.ι into a code word for code rate η, i.e.,\n\nwhere k is a number of information bits i;\n\nn-i = n, where n is a maximum code word length of the coded bits (i.e., n = k/nT); and\n\n• η > ri+1 for i = 1 , T-1.\n\nThe coded bits are provided at the initial code rate rinitiai and are a concatenation of the coded bits output by polar encoders for code rates n through rinitiai- Further, in some embodiments, the rate-compatible polar encoder further comprises a concatenator operable to selectively concatenate outputs of the polar encoders to provide the coded bits at an initial code rate rm a\\ such that, for the initial code rate nnmai, the coded bits are a concatenation of the outputs of the polar encoders for code rates r-i , rinitiai -\n\n[001 1] In some embodiments, the additional coded bits for a new code rate rnew are a concatenation of the outputs of the polar encoders for code rates\n\nHnitial+I , Hiew, Where Tnew Hnitial+I - Embodiments of a method of operation of a transmit node in a wireless communications system are also disclosed.\n\n In some embodiments, a receive node operable for use in a wireless communications system comprises a receiver operable to receive coded bits and a rate-compatible polar decoder operable to decode the coded bits to provide information bits.\n\n In some embodiments, in order to decode the coded bits, the rate- compatible polar decoder is operable to: determine a code rate rd of the coded bits, the code rate rd being one of a predefined set of code rates {η}ί=1 T where η > η+1 for i = 1 , ... ,T-1 and T is a positive integer greater than or equal to 2; and successively perform polar decoding of the last n, coded bits of the coded bits for code rates rd to Pi to provide sets of information bits for the code rates rd to r-i , respectively, where:\n\n· k is a number of information bits encoded into the coded bits; and\n\n• n > n+i for i = 1 , ... ,T, where rT is a lowest code rate.\n\nThe rate-compatible polar decoder is further operable to concatenate the sets of information bits for the code rates rd to Pi to provide the information bits.\n\n In some embodiments, in order to successively perform polar decoding of the last n, coded bits of the coded bits for code rates rd to n to provide the sets of information bits for the code rates rd to n, respectively, the rate-compatible polar decoder is further operable to: perform polar decoding of the last nd coded bits for code rate rd to provide the set of information bits for code rate rd; perform polar decoding of the next nd-i coded bits using the set of information bits for code rate rd as frozen bits to provide the set of information bits for code rate rd-i; and perform polar decoding of the next nd-2 coded bits using the union of the sets of information bits for code rates rd and rd-i as frozen bits to provide the set of information bits for code rate rd-2. Those skilled in the art will appreciate the scope of the present disclosure and realize additional aspects thereof after reading the following detailed description of the embodiments in association with the accompanying drawing figures.\n\nBrief Description of the Drawings\n\n The accompanying drawing figures incorporated in and forming a part of this specification illustrate several aspects of the disclosure, and together with the description serve to explain the principles of the disclosure.\n\n Figure 1 illustrates a wireless communications system in which rate- compatible polar codes are utilized according to some embodiments of the present disclosure;\n\n Figure 2 is a flow chart that illustrates the operation of the transmit node of Figure 1 according to some embodiments of the present disclosure;\n\n Figures 3 and 4 illustrate embodiments of the rate-compatible polar encoder of the transmit node of Figure 1 according to some embodiments of the present disclosure;\n\n Figure 5 is a flow chart that illustrates the operation of the receive node of Figure 1 according to some embodiments of the present disclosure;\n\n Figures 6A through 6C illustrate an embodiment of the rate-compatible polar decoder of the receive node of Figure 1 according to some embodiments of the present disclosure;\n\n Figure 7 illustrates a cellular communications network in which rate- compatible polar codes are utilized according to some embodiments of the present disclosure;\n\n Figure 8 illustrates a Hybrid Automatic Repeat Request (HARQ) based on Incremental Redundancy (HARQ-IR) process implemented in the cellular communications network of Figure 7 that utilizes rate-compatible polar codes according to some embodiments of the present disclosure; and\n\n Figure 9 is a block diagram of a communication node according to some embodiments of the present disclosure. Detailed Description\n\n The embodiments set forth below represent information to enable those skilled in the art to practice the embodiments and illustrate the best mode of practicing the embodiments. Upon reading the following description in light of the accompanying drawing figures, those skilled in the art will understand the concepts of the disclosure and will recognize applications of these concepts not particularly addressed herein. It should be understood that these concepts and applications fall within the scope of the disclosure and the accompanying claims.\n\n Systems and methods are disclosed herein relating to rate-compatible polar codes and the use thereof in a wireless communications system. In this regard, Figure 1 illustrates a wireless communications system 10 including a transmit (TX) node 12 that includes a rate-compatible polar encoder 14 and a transmitter 16, and a receive (RX) node 18 that includes a rate-compatible polar decoder 20 and a receiver 22. The wireless communications system 10 may be any type of wireless communications system such as, but not limited to, a cellular communications network in which the transmit node 12 is, e.g., a radio access node (e.g., a base station) and the receive node 18 is a wireless device or terminal (e.g., a User Equipment device (UE)), or vice versa.\n\n Before describing the operation of the wireless communications system 10 and, in particular, the operation of the rate-compatible polar encoder 14 and the rate-compatible polar decoder 20, a discussion of a family of rate- compatible polar codes is first provided. The family of rate-compatible polar codes is used to encode k information bits with various code rates fa: i = 1, T}, where r > ■■■ > rT and T≥ 2. The disclosed rate-compatible polar code satisfies the condition that the set of parity bits of a higher rate code should be a subset of the set of parity bits of a lower rate code, which can be used for Hybrid Automatic Repeat Request (HARQ) based on Incremental Redundancy (HARQ-IR) systems.\n\n Let Gmother denote a k x n polar code generator matrix of the mother code with lowest code rate rT, where n = k/rT. For given information bits ( 1( ... , uk), the corresponding codeword for the lowest code rate rT is generated as:\n\n^-k)^ mother (^l'■■■> %n) -\n\nThe codeword of a higher-rate code with code rate r( is simply obtained by taking the first n- = k/r^ bits such as ... , xn;), equivalently puncturing the last n - n- bits from the mother code's codeword ... , xn).\n\n In the present disclosure, a polar code generator matrix Gmother\n\n(oftentimes referred to herein simply as a generator matrix Gmother) is\n\nconstructed of a rate-compatible polar code where each punctured polar code of a higher rate is also capacity achieving under a low-complexity Successive Cancellation (SC) decoding when code length n goes to infinity. As discussed below, the rate-compatible polar encoder 14 operates to encode k information bits according to a desired code rate according to the generator matrix Gmother.\n\n In the following discussion, the generator matrix Gmother is described with respect to an example of a rate-compatible polar code that can be used to encode k information bits and supports three code rates {ίι, Γ2, Γ3}, where rx > r2 > r3. This is then extended into a more general case with various rates {7 : i = 1, ... , T}, where rx > ··· > rT.\n\n Let GN denote a generator matrix (which is also referred to herein as a polar code generator matrix) of polar code with code length N. GN is a\n\ndeterministic matrix, as defined in E. Arikan, \"Channel Polarization: A Method for Constructing Capacity-Achieving Codes for Symmetric Binary-Input Memoryless Channels,\" IEEE Transactions on Information Theory, Vol. 55, Issue 7, pages 3051 -3073, July 2009 (hereinafter \"Arikan\"). Specifically, let\n\nG = \\ °l.\n\nLl lJ\n\nThen, GN can be defined as G®n, where ®n denotes the nth Kronecker power and N = 2n. A polar code with length N and rate r is completely defined by GN and an information set A with |A| = Nr. Here, the information set A determines the locations (i.e., rows of the generator matrix GN) to be used for the information bits in the input of polar encoding. The rows of the generator matrix GN that are not to be sued for the information bits are referred to as locations of \"frozen bits.\" Conversely, the rows of the generator matrix GN that are to be used for the information bits are referred to as locations of \"unfrozen bits\" or the locations of the information bits. Following the conventional notation of coding theory, a k x N generator matrix of a polar code is also defined by simply taking the rows of GN corresponding to A, where all frozen bits are assumed to be zero. It is assumed that that the first k' indices of A is again an information set of polar code of rate k'/n, which can be easily obtained by ordering indices of A (see, for example, S.B. Korada, \"Polar Codes for Channel and Source Coding,\" PhD Thesis, Ecole Polytechnique Federale de Lausanne, 2009 (hereinafter \"Korada\")). For a given information set A, Gfi is defined as a row-permuted matrix of GN with the same size such that all rows of GN whose indices belong to A are located in the first |A| rows of Gfi, i.e., the i-th row of Gfi is equal to the A(i)-th row of GN, where A(i) denotes the i-th element of A. Notice that Gfi is also a generator matrix of a polar code of length N. For an index subset D c [1: N] , Gfi(D) denotes a submatrix of Gfi consisting of rows whose indices belong to D.\n\n Letting k = r3n , one example of the k x n generator matrix Gmother has the form of: wmother\n\nwhere\n\nA± denotes the information set of a polar code of code rate length nx = - = {rz/r n ,\n\nA2 denotes the information set of a polar code of code rate r2 and length n2 = - nx = r1 - r^n^jr^ • A3 denotes the information set of a polar code of code rate r3 and length n3 = - nx - n2 = - ^ = (r2 - r3)n2/r3,\n\nr3 r3 r2\n\nand where n(3;1) = - n1; n(3;2) = ^n2 - n2, and n(3;12) = r3n2.\n\n It can be confirmed that Gmother is indeed the k x n generator matrix since\n\nr1n1 = k and n1 + n2 + n3 = n.\n\n For the convenience of notation, the above generator matrix Gmo ther can be represented by\n\nHere, ( is a k x n( matrix with n( = -∑~J n;, and\n\nlGn3(ir n\n\nNotice that the information sets A1, A2, and A3 can be obtained from an efficient algorithm in I. Tal et al., \"How to Construct Polar Codes,\" IEEE Transactions on Information Theory, Vol. 59, Issue 10, July 10, 2013, pages 6562-6582\n\n(hereinafter \"Tal\").\n\n In addition, a specific generator matrix can be defined for each punctured polar code (i.e., for each rate η) as follows:\n\n• rate rs Gri = [5X];\n\n• rate r2: Gr2 = [S1 52]; and\n\n• rate r3: Gr2 = [S1 S2 53].\n\nTherefore, encoding and decoding for each of the code rates can be performed according to the above shortened matrices derived from the generator matrix\n\n^mother Notably, in the example above, X can be used to encode all k information bits to provide nx coded bits, S2 can be used to encode r2n2 of the k information bits (specifically, in this example, the last r2n2 bits of the k information bits) to provide n2 coded bits, and S3 can be used to encode r3n3 of the k information bits (specifically, in this example, the last r3n(3 2) bits of the k information bits and an additional subset of r3n(3 1) bits of the k information bits) to provide n3 coded bits. The coded bits resulting from S± can be used to provide a code word for code rate rl t the concatenation of the coded bits resulting from S± and S2 can be used to provide a code word for code rate r2, and the concatenation of the coded bits resulting from Sl t S2, and S3 can be used to provide a code word for code rate r3.\n\n Importantly, S2 and S3 of the example generator matrix Gmother above are only examples. In this example, S2 includes the polar code matrix\n\nG^([l: r2n2]) for code rate r2, which is used to encode the last r2n2 bits of the k information bits since, in this example, the rows of G^ll-. r^]) are sorted according to channel quality and, therefore, the reliability of successful reception of the corresponding information bits). Therefore, if the goal of S2 is to retransmit the r2n2 most unreliable information bits, then, in this example, S2 is designed to retransmit the last r2n2 bits of the k information bits. However, in other implementations, it may be desirable to transmit some other set of r2n2 bits of the k information bits for code rate r2, in which case G^([l: r2n2]) can be properly positioned within S2 so as to retransmit the desired set of r2n2 bits of the k information bits for code rate r2. Similarly, in the example above, S3 is designed to retransmit the last r3n(3 2) bits of the k information bits and additionally retransmit some subset of r3n(3 ;1) bits of the k information bits that were not retransmitted for code rate r2. However, again, S3 may be designed to retransmit any r3n3 bits of the k information bits.\n\n In the example above, the generator matrix Gmother is described with respect to an example having three code rates. However, more generally, given a sequence of rates {τ0 i = 1, 7} sorted in a descending order, i.e., r > r2 > · · · > rT, and k information bits to be communicated, the transmitter can transmit a sequence of polar codes to achieve each of these rates. Here, T is greater than or equal to 2. Following the expression in (1 ), the generator matrix Gmother is represented by\n\n^mother = [\\$l \\$2 ST]\n\nwhere ( is a k x ^ri; matrix which can be obtained from the row-permuted generator matrix of i-th polar code denoted by G^. The i-th polar code in the sequence has a block length\n\ni-l\n\n7 = 1\n\nfor i = 1, T, such that the total block length is given by\n\nThe number of unfrozen rows (bits) in the i-th polar code is given by\n\ni-l i-l i-l i-l\n\nijTj = / — r j ^ rij = ^ rij— rt ^ rij = ^ q *\n\n; = 1 ; = 1 ; = 1 ; = 1\n\nwhere\n\n(0 r \\ denotes the share of these unfrozen bits that is transmitted to convert the y'-th polar code constructed previously (for 1≤ j < i) from code rate ri_1 to rt . Notice that in the previous process, the rate of the y'-th polar code has been already converted into the code rate ri_1. We define the locations of these unfrozen bits in Gmother recursively as follows. Assume that we are given {G^J. }T j=1 with\n\nA '\n\nthe rows of each Gn J sorted according to the mutual informations of the corresponding bit channels of the y'th polar code. Suppose that we have the submatrices {5;·}}=1 of Gmother up to the th polar code for some i≥ l , and the corresponding index mapping {1,2,■■■, k} that specifies the\n\nA '\n\nindices the non-zero rows of S; containing the unfrozen rows of Gn J (l : n;-r;-), for all j = 1,2,···, i, with simply defined as i(1)(m) = mform = 1,2,•••,k. With q^l+1) 0, define the interval\n\n- {Q + 1, + 2, ...,Qp? + qf+»]\n\nwhere\n\nsuch that {/ l+1)}}=1 forms a partition of the integer set\n\n7 = 1\n\nThe index mapping can now be defined as ι(ί+1): {1,2,••·,ηί+1Γί+1} → {1,2, ···,&} for Si+1 in a piece-wise fashion as _(i+1)(m\n\nform = 1,2,··· ,ni+1ri+1, where _/' m denotes the index of the interval that contains the integer m.\n\n An optional permutation mapping 7r(i+1): {1,2,■■■ ,ni+1ri+1}→\n\n{1,2,■■■ ,ni+1ri+1} may also be combined with i(i+1)(m) to obtain (i+1)(m) (i+1)(m)' (i+1)(m) in order to allow different mappings of the unfrozen rows (or bits) of to the previously constructed polar codes. Examples of 7r i+1)(-) includes the identity mapping 7r(i+1)(m) = m, and the reverse mapping 7r(i+1)(m) = ni+1ri+1 -m + 1 form = 1,2,■■■ ,ni+iri+i. Now that the rate-compatible polar code generator matrix Gmother has been described, Figure 2 illustrates the operation of the transmit node 12 of Figure 1 according to some embodiments of the present disclosure. As illustrated, the transmit node 12, and in particular the rate-compatible polar encoder 14, encodes the k information bits for a desired code rate rt to provide nr. coded bits (step 100). The rate-compatible polar encoder 14 performs rate- compatible polar encoding using parallel concatenated polar codes. In particular, using code rate r2 as an example, the rate-compatible polar encoder 14 performs polar encoding of the k information bits according to:\n\n(u ... , ufc)[5i S2] = (Xl, ... , xni+n2).\n\nIn other words, the k information bits are encoded according to Sx to provide a first polar code (i.e., nx coded bits in the form of a first polar code). The k information bits are also encoded according to S2 to provide a second polar code (i.e., n2 coded bits in the form of a second polar code). These first and second polar codes are concatenated to provide a polar code of length n + n2. This polar code is the n + n2 coded bits (i.e., the codeword) for code rate r2. The transmit node 12, and in particular the transmitter 16, transmits the coded bits (step 102).\n\n Figures 3 and 4 illustrate two example embodiments of the rate- compatible polar encoder 14 of Figure 1. In these examples, the rate-compatible polar encoder 14 operates to provide rate-compatible polar encoding of the k information bits (uv ... , uk) based on the example generator matrix Gmother described above for code rates rl t r2, and r3. However, one of ordinary skill in the art will readily appreciate that the embodiments of Figures 3 and 4 can readily be extended for the general case of code rates rl t ... , rT.\n\n As illustrated in Figure 3, the rate-compatible polar encoder 14\n\nincludes a number of polar encoders 24-1 through 24-3 (generally referred to herein collectively as polar encoders 24 and individually as polar encoder 24) that operate in parallel to encode the k information bits (uv ... , uk) to provide corresponding sets of coded bits for code rates rl t r2, and r3, respectively. Specifically, the polar encoder 24-1 encodes the k information bits (ulf ... , uk) for code rate rx to provide n coded bits [xlt ... , xni) according to:\n\nThe polar encoder 24-2 encodes the k information bits (uv ... , uk) for code rate r2 to provide n2 coded bits (xni+1, --- , xni+n2) according to: Notably, as discussed above, due to the design of S2, only r2n2 of the k information bits (uv ... , uk) are actually encoded by the polar encoder 24-2.\n\nSimilarly, the polar encoder 24-3 encodes the k information bits (uv ... , uk) for code rate r3 to provide n3 coded bits (xni+Tl2+1, --- , xni+n2 +n3) according to:\n\n( l' - - - ' uk ^2 = (^+πζ + Ι' ■■■ · χηί2 +η^) ·\n\nNotably, as discussed above, due to the design of S3, only r3n3 of the k information bits (uv ... , uk) are actually encoded by the polar encoder 24-3.\n\n In this example, the rate-compatible polar encoder 14 also includes a concatenator 26 that operates to concatenate the coded bits, or polar codes, output by the polar encoders 24-1 through 24-3 to provide the final coded bits, or codeword, output by the rate-compatible polar encoder 14. Note that, depending on the particular implantation, a desired code rate rt can be achieved in different manners. For example, in one example embodiment, only the polar encoders 24 needed for the desired code rate rt are active. For instance, if the desired code rate is r2, only the polar encoders 24-1 and 24-2 are active and, as a result, the output of the concatenator 26 consists of the coded bits ..., ½1+Tl2), which is a polar code of length n1 + n2. In another example embodiment, all of the polar encoders 24 are active, but the concatenator 26 is controlled to output only those bits need for the desired code rate. Thus, for example, if the desired code rate is r2, the polar encoders 24-1 , 24-2, and 24-3 are active, but the concatenator 26 is controlled such that the output of the concatenator 26 consists of the coded bits (xv ... , ½1 +n2 ) . As a final example, if all of the polar encoders 24 are active, the concatenator 26 outputs the codeword for code rate r3 (which is a polar code, or mother polar code, of length n = n1 + n2 + n3), and subsequent processing at the transmit node 12 is used to puncture this mother polar code to provide a codeword for the desired code rate.\n\n Figure 4 illustrates an embodiment of the rate-compatible polar encoder 14 that includes a number of polar encoders 28-1 through 28-3\n\n(generally referred to herein collectively as polar encoders 28 and individually as polar encoder 28) that operate in parallel to encode the appropriate subsets of the k information uv ...,uk) to provide corresponding sets of coded bits for code rates rlt r2, and r3, respectively. Specifically, the polar encoder 28-1 encodes all of the k information bits (u ...,uk) for code rate rx to provide nx coded bits (xv ...,xni) according to:\n\nOi, ...,Mfe)G7i 1([l:r1n1]) = (xlt ...,xn .\n\nThe polar encoder 24-2 encodes a subset of r2n2 of the k information bits\n\n. for code rate r2 to provide n2 coded bits - ,xni+n2) according to:\n\n(u , ...,Mfe)Gn2 2([l: r2n2J) = ...,xni+ri2).\n\nSimilarly, the polar encoder 24-3 encodes a subset of r3n3 bits of the k\n\ninformation bits (uv ...,uk) for code rate r3 to provide n3 coded bits\n\n(xni+n2+i, -,Xn1+n2+n3) according to:\n\n(ui>■■■> uk)Gn^([l: r3n3J) = (xni+n2+±,■■■ <xni+n2+n^)-\n\nNotably, 3([l:r3 ]) = Gi([1:r3\"(3,i)D\n\nGi([r3n(3,l) + 1: r3«(3,l) + r3«(3,2)]).\n\n In this example, the rate-compatible polar encoder 14 also includes a divider 30 that operates to divide the k information bits (uv ...,uk) into\n\nappropriate subsets to be encoded by the polar encoders 28. In particular, in this example, the divider 30 provides all of the k information bits (uv ...,uk) to the polar encoder 28-1, which then encodes the k information bits (ι , ...,uk) at the code rate rx to provide the n coded bits (xv ...,½J. In this example, the divider 30 provides the last r2n2 bits of the k information bits to the polar encoder 28-2, which then encodes those r2n2 information bits at the code rate r2 to provide the n2 coded bits ... , xni+ri2). Further, in this example, the divider 30 provides r3n3 bits of the k information bits to the polar encoder 28-3, which then encodes those r3n3 information bits at the code rate r3 to provide the n3 coded bits In this particular example, the r3n3 information bits provided to the polar encoder 28-3 include r3n(3 1 information bits starting at ur3ni+i and tne |ast r 3 (3,2) information bits. Notably, while not illustrated, in some embodiments, the divider 30 may apply any desired mapping (e.g., identity mapping or reverse mapping) of the bits in the appropriate subsets of the k information bits to the input bit locations of the polar encoders 28.\n\n As discussed above with respect to the example of Figure 3, the rate- compatible polar encoder 14 of Figure 4 also includes the concatenator 26 that operates to concatenate the coded bits output by the polar encoders 28-1 through 28-3 to provide the final coded bits, or codeword, output by the rate- compatible polar encoder 14. As discussed above, depending on the particular implementation, a desired code rate rt can be achieved in different manners. For example, in one example embodiment, only the polar encoders 28 needed for the desired code rate rt are active. For instance, if the desired code rate is r2, only the polar encoders 28-1 and 28-2 are active and, as a result, the output of the concatenator 26 consists of the coded bits ... , ½1+Tl2). In another example embodiment, all of the polar encoders 28 are active, but the concatenator 26 is controlled to output only those bits need for the desired code rate. Thus, for example, if the desired code rate is r2, the polar encoders 28-1 , 28-2, and 28-3 are active, but the concatenator 26 is controlled such that the output of the concatenator 26 consists of the coded bits ... , ½1+Tl2). As a final example, if all of the polar encoders 28 are active, the concatenator 26 outputs the codeword for code rate r3, and subsequent processing at the transmit node 12 is used to puncture the codeword for code rate r3 to provide a codeword for the desired code rate.\n\n Thus far, the discussion has focused on the operation of the transmit node 12 and, in particular, the rate-compatible polar encoder 14. Figure 5 is a flow chart that illustrates the operation of the receive node 18 to receive the coded bits from the transmit node 12 and to decode the coded bits to recover the information bits according to some embodiments of the present disclosure. As illustrated, the receive node 18 receives the coded bits from the transmit node 12 (step 200). As discussed above, these coded bits are in the form of a polar code generated from a rate-compatible polar code generator matrix, as described above. The receive node 18, and in particular the rate-compatible polar decoder 20, sequentially decodes the coded bits to provide the information bits (step 202). More specifically, the rate-compatible polar decoder 20 determines the code rate rd of the received coded bits (step 300). The code rate rd may be determined in any suitable manner. For example, the code rate rd may be communicated to the receive node 18 from the transmit node 12 or determined from the number of coded bits. Here, rd is one of the code rates η. Then, for each code rate η in the range of and including rd to n, starting with the determined code rate rd, the rate- compatible polar decoder 20 successively performs polar decoding of the corresponding n, coded bits based on the corresponding polar code generator matrices ί¾ ([1: nj) to provide corresponds sets of decoded information bits for the code rates rd to n (step 302). For each iteration of the polar decoding after the first iteration (i.e. , for code rates rd-i to n), at least some of the previously decoded information bits can be utilized as frozen bits for polar decoding to thereby decrease the effective code rate for decoding, as described in the examples below. The sets of decoded information bits for code rates Pi through rd are then concatenated to provide the decoded information bits (step 304).\n\n A description of the decoding process for a specific case for T=3 is as follows. The extension of this process to the general case is straightforward. For the case where T=3, there are three polar codes whose generator matrices are G^ , respectively. First, the highest code rate (i.e. , the polar code for code rate n) is by itself a polar code with generator matrix G^ . Thus, the conventional polar decoder such as SC and SC List (SCL) described in I. Tal et al. , \"List Decoding of Polar Codes,\" 2011 IEEE International Symposium on Information Theory Proceedings, pages 1 -5, July 31 -August 5, 201 1 can be used. The polar code for code rate r2 can be considered as parallel concatenation of the two polar codes for code rates rx and r2 , with generator matrices for i = 1,2, respectively, as discussed above. One efficient decoding procedure for code rate r2 is as follows:\n\n• First, the r2n2 = (rx - r2)n1 information bits encoded for code rate r2 (as described above) are decoded using the second polar code with generator matrix G^.\n\n• Using the decoded information bits as frozen bits to the first polar code with the generator matrix G^, the resulting code is a polar code of length nx and code rate r2, where the information set of this polar code is obtained by taking first r2n1 elements. That is, inserting frozen bits produces a lower-rate polar code. Then, the r2n1 information bits are decoded using this lower-rate polar code. Using this procedure, two polar decoding iterations result in successful decoding of (rx - r2)n1 + r2n1 = k information bits. Notice that in the two decoding iterations, two polar codes of code rate r2 are used. Since both achieve the capacity as n goes to infinity, the proposed code can also achieve the capacity.\n\n In the above, a sequential decoding process is described where information bits are decoded by two polar codes sequentially. This approach is optimal by achieving the capacity, when code length goes to infinity. Yet, for a finite code length, other decoding algorithms can be considered such as Belief Propagation (BP) decoding over whole graph induced by two parallel\n\nconcatenated polar codes.\n\n Next, the decoding procedure is described for code rate r3. In a manner similar to that described above, three sequential polar decoding operations are used to decode the information bits, where, in this example, all polar codes (used in decoding) are of code rate r3. • First, the r3n3 information bits encoded for code rate r3 (as described above) are decoded using the polar code of code rate r3 and with generator matrix\n\n• Using the some decoded bits as frozen bits to the polar code with generator matrix G^ , a polar code of code rate r3 is produced.\n\nThen, the r3n2 information bits (i.e., the remaining information bits of the r2n2 information bits encoded for code rate r2 after excluding the frozen bits) are decoded using this polar code of code rate r3.\n\n• Using the all decoded bits (i.e., the union of all decoded bits from the previous decoding operations) as frozen bits to the polar code with the generator matrix G^, the resulting code is a polar code of code rate r3 and length nx. Using this polar code, the remaining r3n! information bits are decoded. At this point, all information bits are decoded.\n\n In other words, the successive decoding procedure can be explained as follows. Assuming that the sequential decoding procedure begins with code rate rT (but may alternatively start at any desired code rate rd), the subset of information bits carried by the nT coded bits at the code rate rT (e.g., the last retransmission in the case of HARQ-IR) is first decoded. The resulting decoded information bits are used as frozen bits to decode the subset of the information bits carried by the nT-i coded bits. Notably, this stage of polar coding/decoding is originally designed for code rate rT-i , but due to the use of the previously decoded information bits as frozen bits, the code rate of this polar decoding is effectively reduced to rT as well. Then, the union of the two sets of information bits resulting from the two previous polar decodes is used as frozen bits for the next stage of polar decoding, where the subset of information bits carried by the nT-2 coded bits are decoded. This stage of polar decoding is originally designed for code rate rT-2, but due to the use of the previously decoded information bits as frozen bits, the code rate of this polar decoding is effectively reduced to rT as well. The process continues in this manner until all of the information bits have been decoded.\n\n Figures 6A through 6C illustrate an example embodiment of the rate- compatible polar decoder 20 operating according to the process of Figure 5. In particular, Figure 6A illustrates the operation of the rate-compatible polar decoder 20 when decoding for code rate r-i , Figure 6B illustrates the operation of the rate- compatible polar decoder 20 when decoding for code rate r2, and Figure 6C illustrates the operation of the rate-compatible polar decoder 20 when decoding for code rate r3. As illustrated in Figure 6A, the coded bits are encoded at a code rate Pi and have a length n-i . As such, a polar decoder 32-1 for code rate Pi and length ni is used to perform polar decoding using the polar code with generator matrix \"11 of code rate Pi .\n\n As illustrated in Figure 6B, the coded bits are encoded at a code rate r2 and have a length n-i+n2. In this case, a polar decoder 32-2 for code rate r2 and length n2 decodes the r2n2 information bits encoded at code rate r2 (as discussed above). The decoded information bits are provided to the polar decoder 32-1 as frozen bits. The polar decoder 32-1 uses the frozen bits for the first polar code with generator matrix to provide a resulting polar code of length n and code rate r2. Using this resulting lower-rate polar code, the polar decoder 32-1 decodes the remaining r2n-i information bits. A concatenator 34 concatenates the decoded sets of information bits output by the polar decoders 32-1 and 32-2 to provide the final set of k information bits n coded bits (uv ... , uk).\n\n In Figure 6C, the coded bits received by the receive node 18 are encoded at a code rate r3 and have a length n-i+n2+n3. In this case, a polar decoder 32-3 for code rate r3 and length n3 decodes the r3n3 information bits encoded at code rate r3 (as discussed above). Some of the decoded information bits (i.e., those information bits that were also encoded at code rate r2) are provided to the polar decoder 32-2 as frozen bits. The polar decoder 32-2 uses the frozen bits for the second polar code with generator matrix to provide a resulting polar code of length n2 and code rate r3. Using this resulting lower-rate polar code, the polar decoder 32-2 decodes the remaining r3n2 information bits (of the r2n2 information bits encoded at code rate r2). Lastly, in this example, all of the previously decoded information bits are provided to the polar decoder 32-1 as frozen bits. The polar decoder 32-1 uses the frozen bits for the first polar code with generator matrix to provide a resulting polar code of length n and code rate r3. Using this resulting lower-rate polar code, the polar decoder 32-1 decodes the remaining r3ni information bits. At this point, all k information bits are decoded. The concatenator 34 concatenates the decoded sets of information bits output by the polar decoders 32-1 through 32-3 to provide the final set of k information bits n coded bits (uv ... , uk) .\n\n As one example, encoding and decoding can be performed as follows. This is only an example and is not to be construed as limiting the scope of the present disclosure. Consider a rate-compatible polar code with 3 2m (for some positive integer m) information bits (i.e., k = 3 2m) and code rates (3/4,1/2,1/4) (i.e., {r i = 1, ... ,3} = {3/4, 1/2 , 1/4}). For the encoding, the following polar codes may be employed by the polar encoders 24, 28:\n\n• Encoder 1 (polar encoder 24-1 , 28-1 ): polar code of code rate 3/4 and length 2m+2 ;\n\n• Encoder 2 (polar encoder 24-2, 28-2): polar code of code rate 1/2 and length 2m+1 ; and\n\n• Encoder 3 (polar encoder 24-3, 28-3): polar code of code rate 1/4 and length 2m+2.\n\nFor the decoding, the following polar codes can be employed by the polar decoders 32:\n\n· code rate 3/4: polar decoder (polar decoder 32-1 ) of code rate 3/4 and length 2m+2 ;\n\n• code rate 1/2: using two polar codes:\n\n- Decoder 1 : polar decoder (polar decoder 32-2) of code rate 1/2 and length 2m+1 ; and - Decoder 2: polar decoder (polar decoder 32-1 ) of code rate 1/2 and length 2m+2 ; and\n\n• code rate 1/4: using three polar codes in Figure 3:\n\n- Decoder 1 : polar decoder (polar decoder 32-3) of code rate 1/4 and length 2m+2 ;\n\n- Decoder 2: polar decoder (polar decoder 32-2) of code rate 1/4 and length 2m+1 ; and\n\n- Decoder 3: polar decoder (polar decoder 32-3) of code rate 1/4 and length 2m+2.\n\n The transmit node 12 and the receive node 18 described above may be implemented in any suitable type of wireless communications system 10. In this regard, Figure 7 illustrates a cellular communications network 36 including wireless nodes (e.g. , Radio Access Network (RAN) nodes and/or wireless devices or terminals) that implement the rate-compatible polar encoder 14 and/or the rate-compatible polar decoder 20 according to some embodiments of the present disclosure. Further, as described below, the rate-compatible generator matrix Gmother may, in some embodiments, be used for HARQ-IR\n\ntransmissions/retransmissions.\n\n In this example, the cellular communications network 36 is a Long Term Evolution (LTE) network and, as such, LTE terminology is sometimes used. However, the cellular communications network 36 is not limited to LTE. As illustrated, the cellular communications network 36 includes a Evolved Universal Terrestrial Radio Access Network (EUTRAN) 38 including enhanced or evolved Node Bs (eNBs) 40 (which may more generally be referred to herein as base stations) serving corresponding cells 42. User Equipment devices (UEs) 44 (which may more generally be referred to herein as wireless devices) transmit signals to and receive signals from the eNBs 40. The eNBs 40 communicate with one another via an X2 interface. Further, the eNBs 40 are connected to a Evolved Packet Core (EPC) 46 via S1 interfaces. As will be understood by one of ordinary skill in the art, the EPC 46 includes various types of core network nodes such as, e.g., Mobility Management Entities (MMEs) 48, Serving\n\nGateways (S-GWs) 50, and Packet Data Network Gateways (P-GWs) 52.\n\n In some embodiments, the transmit node 12 is the eNB 40 and the receive node 18 is the UE 44. In other embodiments, the transmit node 12 is the UE 44 and the receive node 18 is the eNB 40. In other words, the eNBs 40 may include the rate-compatible polar encoder 14 and the transmitter 16 of the transmit node 12 of Figure 1 and/or the rate-compatible polar decoder 20 and the receiver 22 of the receive node 18 of Figure 1 . Likewise, the UEs 44 may include the rate-compatible polar encoder 14 and the transmitter 16 of the transmit node 12 of Figure 1 and/or the rate-compatible polar decoder 20 and the receiver 22 of the receive node 18 of Figure 1 .\n\n As mentioned above, the generator matrix Gmother described above can be utilized for HARQ-IR. In this regard, Figure 8 illustrates the operation of one of the eNBs 40 to transmit k information bits to the UE 44 using HARQ-IR according to some embodiments of the present disclosure. A similar process may be used to transmit k information bits from the UE 44 to the eNB 40 using a HARQ-IR procedure. As illustrated, the eNB 40 selects an initial code rate for downlink data transmission to the UE 44 (i.e., an initial code rate for transmission of k information bits) (step 400). The initial code rate is one of the code rates η. In this example, it is assumed that the initial code rate is the highest code rate r-i , but the initial code rate is not limited thereto. The eNB 40, and in particular the rate-compatible polar encoder 14 of the eNB 40, performs polar encoding of the downlink data (i.e., the k information bits) for the initial code rate based on the appropriate submatrices of the generator matrix Gmother, as described above (step 402). Assuming that the initial code rate is r-i , then polar encoding of the k information bits is performed according to the generator matrix G^ ([l: r^ ). The coded bits are then transmitted to the UE 44 (step 404). The UE 44 attempts to decode the received coded bits (step 406). In this example, decoding is unsuccessful and, as a result, the UE 44 transmits a HARQ Negative Acknowledgement (NACK) to the eNB 40 (step 408). Upon receiving the HARQ NACK, the eNB 40 selects a new code rate for a HARQ, or more specifically HARQ-IR, retransmission (step 410) and performs polar encoding of the downlink data (i.e., the k information bits) to provide the additional coded bits to transform the coded bits for the initial code rate into coded bits for the new code rate (step 412). For example, if the initial code rate is Pi and the new code rate is r2, then the polar encoding of step 412 encodes the r2n2 information bits using the generator matrix i¾ 2([l: r2n2]) (or equivalently encodes the k information bits using S2) to provide the n2 coded bits needed to transform the ni coded bits for code rate Pi into n-i+n2 coded bits for code rate r2. The n2 coded bits are transmitted to the UE 44 (step 414). As another example, if the initial code rate is Pi and the new code rate is r3, then the polar encoding of step 414 encodes the r2n2 information bits using the generator matrix r2n2]) (or equivalently encodes the k information bits using S2) to provide the n2 coded bits and encodes the r3n3 information bits using the generator matrix i¾ 3([l: r3n3]) (or equivalently encodes the k information bits using S3) to provide the n3 coded bits. The n2+n3 coded bits are then transmitted in step 414. Notably, while the polar encoding of step 412 is illustrated as being separate from that of step 402, the polar encoding of steps 402 and 412 may be performed in a single polar encoding process that generates the mother polar code from the mother generator matrix. This mother polar code can then be punctured to provide the different subsets of the encoded bits xi, ... ,xn for transmission in steps 404 and 414 for the initial code rate and new code rate, respectively.\n\n In this example, the UE 44 attempts decoding using the additional coded bits received in step 414 (step 416). In particular, the UE 44 performs sequential polar decoding of the ηη, information bits starting at the new code rate, as described above with respect to Figures 5 and 6A-6C. In this example, decoding is successful and, as such, the UE 44 transmits a HARQ\n\nAcknowledgement (ACK) to the eNB 40 (step 418). Figure 9 illustrates a communication node 54 that includes the rate- compatible polar encoder 14 and/or the rate-compatible polar decoder 20 according to some embodiments of the present disclosure. The communication node 54 may be, for example, the transmit node 12, the receive node 18, the eNB 40, or the UE 44 described above. As illustrated, the communication node 54 includes one or more processors 56 (e.g., one or more Central Processing Units (CPUs), one or more Application Specific Integrated Circuits (ASICs), one or more Field Programmable Gate Arrays (FPGAs), or the like). In some embodiments, the rate-compatible polar encoder 14 and/or the rate-compatible polar decoder 20 are implemented in hardware or a combination of hardware and software within the one or more processors 56 (e.g., implemented as software executed by the one or more processors 56). The communication node 54 also includes memory 58, which may be used to, e.g., store software instructions executed by the processor(s) 56 and/or store other data or information. The communication node 54 also include a transceiver 60, which includes a transmitter(s) 62 and receiver(s) 64 coupled to one or more antennas 66. The transmitter(s) 62 and the receiver(s) 64 include components such as, for example, amplifiers, filters, mixers, etc.\n\n In some embodiments, a computer program including instructions which, when executed by at least one processor, causes the at least one processor to carry out at least some of the functionality of the rate-compatible polar encoder 14 and/or the rate-compatible polar decoder 20 according to any of the embodiments described herein is provided. In some embodiments, a carrier containing the aforementioned computer program product is provided. The carrier is one of an electronic signal, an optical signal, a radio signal, or a computer readable storage medium (e.g., a non-transitory computer readable medium such as the memory 58).\n\n The following acronyms are used throughout this disclosure.\n\n• 5G 5th Generation\n\n· ACK Acknowledgement\n\n• ASIC Application Specific Integrated Circuit BP Belief Propagation\n\nCPU Central Processing Unit\n\nCRC Cyclic Redundancy Check\n\neNB Enhanced or Evolved Node B\n\nEPC Evolved Packet Core\n\nEUTRAN Evolved Universal Terrestrial Radio Access Network\n\nFPGA Field Programmable Gate Array\n\nHARQ Hybrid Automatic Repeat Request\n\nHARQ-IR Hybrid Automatic Repeat Request based on\n\nIncremental Redundancy\n\nIR Incremental Redundancy\n\nLDPC Low-Density Parity-Check\n\nLTE Long Term Evolution\n\nML Maximum-Likelihood\n\n· MME Mobility Management Entity\n\nNACK Negative Acknowledgement\n\nP-GW Packet Data Network Gateway\n\n· SC Successive Cancellation\n\nSCL Successive Cancellation List\n\nS-GW Serving Gateway\n\nTX Transmit\n\nUE User Equipment\n\n Those skilled in the art will recognize improvements and modifications to the embodiments of the present disclosure. All such improvements and modifications are considered within the scope of the concepts disclosed herein and the claims that follow.\n\n## Claims\n\nClaims What is claimed is:\n1. A transmit node (12) operable for use in a wireless communications system (10), comprising:\na rate-compatible polar encoder (14) operable to encode a plurality of information bits to provide a plurality of coded bits utilizing a plurality of parallel concatenated polar codes; and\na transmitter (16) operable to transmit the plurality of coded bits.\n2. The transmit node (12) of claim 1 wherein the rate-compatible polar encoder (14) comprises:\na plurality of polar encoders (24, 28) operable to encode the plurality of information bits, the plurality of polar encoders (24, 28) comprising:\na first polar encoder (24-1 , 28-1 ) operable to encode the plurality of information bits at a first code rate Pi to provide a number ni of coded bits, where n-ι = k / n and k is the number of information bits in the plurality of information bits; and\na second polar encoder (24-2, 28-2) operable to encode a subset of the plurality of information bits at a second code rate r2 to provide a number n2 of coded bits, where Pi > r2 and the number n2 of coded bits is a number of coded bits that, when concatenated with the number ni of coded bits, transforms a resulting code word from the first code rate Pi to the second code rate r2;\nwherein the plurality of coded bits are provided at a desired code rate rdesired and are a concatenation of coded bits output by polar encoders (24, 28) of the plurality of polar encoders (24, 28) for the code rates Pi through rdesired-\n3. The transmit node (12) of claim 2 wherein the rate-compatible polar encoder (14) further comprises a concatenator (26) operable to concatenate the coded bits output by the polar encoders (24, 28) for the code rates Pi through rdesired to provide the plurality of coded bits at the desired code rate rdesired-\n4. The transmit node (12) of claim 1 wherein the rate-compatible polar encoder (14) comprises:\na plurality of polar encoders (24, 28) operable to encode the plurality of information bits based on corresponding polar code generator matrices for code rates η and lengths n,, where:\n• i = {1 , ... , T}, where T is a positive integer greater than or equal to 2;\n• ni is a length for code rate Pi and, for all other values of i, n, is a number of additional coded bits to transform a code word for code rate η.ι into a code word for code rate η, i.e.,\nwhere k is the number of information bits in the plurality of information bits;\n• ∑[=1 nt = n, where n is a maximum code word length of the\nplurality of coded bits (i.e., n = k/nT); and\n• n > ri+1 for i = 1 , T-1 ;\nwherein the plurality of coded bits are provided at a desired code rate rdesired and are a concatenation of coded bits output by polar encoders (24, 28) of the plurality of polar encoders (24, 28) for the code rates r-ι through rdeSired-\n5. The transmit node (12) of claim 4 wherein the rate-compatible polar encoder (14) further comprises a concatenator (26) operable to concatenate outputs of the plurality of polar encoders (24, 28) to provide the plurality of coded bits at the desired code rate rdesired such that, for the desired code rate rdesired, the plurality of coded bits is a concatenation of the outputs of the plurality of polar encoders (24, 28) for the code rates r-i , rdesired-\n6. The transmit node (12) of claim 4 wherein the polar code generator matrices for the plurality of polar encoders (24, 28) are submatrices ί¾ ([1: ^]) of row-permuted generator matrices of i-th polar codes consisting of rows 1 through of G^.\n7. The transmit node (12) of claim 6 wherein the plurality of polar encoders (24, 28) comprises:\na first polar encoder (24-1 , 28-1 ) for code rate Pi operable to encode the plurality of information bits at the code rate Pi to provide ni coded bits; and\none or more additional polar encoders (24, 28), each additional polar encoder (24, 28) operable to encode a subset of the plurality of information bits consisting of a number ηη, of the plurality of information bits at the code rate η to provide n, coded bits.\n8. The transmit node (12) of claim 7 wherein, for each of the one or more additional polar encoders (24, 28), the subset of the plurality of information bits encoded by the additional polar encoder (24, 28) is different than the subsets encoded by the other additional polar encoders (24, 28).\n9. The transmit node (12) of claim 7 wherein, for each of the one or more additional polar encoders (24, 28), the subset of the plurality of information bits encoded by the additional polar encoder (24, 28) comprises a predefined number of the plurality of information bits that are most unreliable with respect to transmission of the number ni of coded bits from the first polar encoder (24-1 , 28-1 ).\n10. The transmit node (12) of claim 7 wherein, for at least one of the one or more additional polar encoders (24, 28), an ordering of the subset of the plurality of information bits encoded by the at least one of the one or more additional polar encoders (24, 28) is different than an ordering of those same information bits when encoded by the first polar encoder (24-1 , 28-1 ).\n1 1. The transmit node (12) of claim 1 further comprising:\nat least one processor (56); and\nmemory (58) containing instructions executable by the at least one processor (56) whereby the transmit node (12) is further operable to:\nselect an initial code rate;\nperform, via the rate-compatible polar encoder (14), polar encoding of the plurality of information bits at the initial code rate;\ntransmit, via the transmitter (16), the plurality of coded bits, having been encoded at the initial code rate;\nupon receiving a negative acknowledgement from a receive node (18) for the transmission of the plurality of coded bits, select a new code rate for an incremental redundancy retransmission;\nperform, via the rate-compatible polar encoder (14), polar encoding of some of the plurality of information bits to provide additional coded bits for the new code rate; and\ntransmit, via the transmitter (16), the additional coded bits.\n12. The transmit node (12) of claim 1 1 wherein the rate-compatible polar encoder (14) comprises:\na plurality of polar encoders (24, 28) operable to encode the plurality of information bits based on corresponding polar code generator matrices for code rates η and lengths n,, where:\n• i = {1 , ... , T}, where T is a positive integer greater than or equal to 2; Πι is a length for code rate Pi and, for all other values of i, n, is a number of additional coded bits to transform a code word for rate η. 1 into a code word for code rate η, i.e.,\nwhere k is a number of information bits in the plurality of\ninformation bits;\nn-i = n, where n is a maximum code word length of the plurality of coded bits (i.e., n = k/nT); and\n• η > ri+1 for i = 1 , T-1 ;\nwherein the plurality of coded bits are provided at an initial code rate rinitiai and are a concatenation of coded bits output by polar encoders (24, 28) of the plurality of polar encoders (24, 28) for the code rates Pi through nnmai-\n13. The transmit node (12) of claim 12 wherein the rate-compatible polar encoder (14) further comprises a concatenator (26) operable to selectively concatenate outputs of the plurality of polar encoders (24, 28) to provide the plurality of coded bits at the initial code rate rinitiai such that, for the initial code rate rinitiai, the plurality of coded bits is a concatenation of the outputs of the plurality of polar encoders (24, 28) for the code rates r-i , . . . , nnmai-\n14. The transmit node (12) of claim 12 wherein the additional coded bits for a new code rate rnew are a concatenation of the outputs of the plurality of polar encoders (24, 28) for the code rates rinitiai+i , rnew, where rnew < Hnitiai+i -\n15. A method of operation of a transmit node (12) in a wireless\ncommunications system (10), comprising:\nencoding (100) a plurality of information bits to provide a plurality of bits utilizing a plurality of parallel concatenated polar codes; and transmitting (102) the plurality of coded bits.\n16. The method of claim 15 wherein encoding (100) the plurality of information bits comprises:\nencoding the plurality of information bits via a first polar code generator matrix at a first code rate Pi to provide a number ni of coded bits, where n-ι = k / n and k is a number of information bits in the plurality of information bits;\nencoding a subset of the plurality of information bits via a second polar code generator matrix at a second code rate r2 to provide a number n2 of coded bits, where n > r2 and the number n2 of coded bits is a number of coded bits that, when concatenated with the number ni of coded bits, transforms a resulting code word from the first code rate n to the second code rate r2; and\nselectively concatenating the number ni of coded bits and the number n2 of coded bits to provide the plurality of coded bits at a desired code rate.\n17. The method of claim 15 wherein encoding (100) the plurality of information bits comprises:\nencoding the plurality of information bits based on polar code generator matrices for code rates η and lengths n,, where:\n• i = {1 , ... , T}, where T is a positive integer greater than or equal to\n2;\n• ni is a length for code rate Pi and, for all other values of i, n, is a number of additional coded bits to transform a code word for code rate η.ι into a code word for code rate η, i.e.,\nwhere k is a number of information bits in the plurality of information bits; • n-i = n, where n is a maximum code word length of the plurality of coded bits (i.e., n = k/nT); and\n• η > ri+i for i = 1 , ... , T-1 ; and\nselectively concatenating sets of coded bits resulting from encoding the plurality of information bits based on the polar code generator matrices for the code rates η and the lengths n, to provide the plurality of coded bits at a desired code rate rdesired such that, for the desired code rate rdesired, the plurality of coded bits is a concatenation of the sets of coded bits resulting from encoding the plurality of information bits based on the polar code generator matrices for the code rates n, rdesired.\n18. The method of claim 17 wherein the polar code generator matrices are submatrices ί¾ ([1: rjnj) of row-permuted generator matrices of i-th polar codes consisting of rows 1 through of G^.\n19. The method of claim 15 further comprising:\nselecting (400) an initial code rate; and\nperforming (402) the encoding (100) of the plurality of information bits at the initial code rate;\nwherein transmitting (102) the plurality of coded bits comprises\ntransmitting (404) the plurality of coded bits, having been encoded at the initial code rate.\n20. The method of claim 19 further comprising:\nupon receiving a negative acknowledgement from a receive node (18) for the transmission of the plurality of coded bits, selecting (410) a new code rate for an incremental redundancy retransmission;\nperforming (412) polar encoding of some of the plurality of coded bits to provide additional coded bits for the new code rate; and\ntransmitting (414) the additional coded bits.\n21. The method of claim 20 wherein performing (402) the encoding (100) of the plurality of information bits at the initial code rate comprises:\nencoding the plurality of information bits based on corresponding polar code generator matrices for code rates η and lengths n,, where:\n• i = {1 , INITIAIL}, where INITIAL is an index of the initial code rate and is a positive integer greater than or equal to 1 ;\n• ni is a length for code rate Pi and, for all other values of i, n, is a number of additional coded bits to transform a code word for code rate η.ι into a code word for code rate η, i.e.,\nwhere k is the number of information bits in the plurality of information bits;\nconcatenating sets of coded bits resulting from encoding the plurality of information bits based on the corresponding polar code generator matrices for the code rates η and the lengths n, for i = {1 , INITIAL} to provide the plurality of coded bits at the initial code rate.\n22. The method of claim 21 wherein the additional coded bits for a new code rate rnew are a concatenation of one or more sets of coded bits resulting from encoding at least some of the plurality of information bits based on corresponding polar code generator matrices for code rates η and lengths n, for i = {INITIAL+1 , NEW}, where NEW is an index of the new code rate rnew-\n23. A receive node (18) operable for use in a wireless communications system (10), comprising:\na receiver (22) operable to receive a plurality of coded bits; and a rate-compatible polar decoder (20) operable to decode the plurality of coded bits to provide a plurality of information bits.\n24. The receive node (18) of claim 23 wherein, in order to decode the plurality of coded bits, the rate-compatible polar decoder (20) is operable to:\ndetermine (300) a code rate rd of the plurality of coded bits, the code rate rd being one of a predefined set of code rates {π}ί= T where η > ri+ for i =\n1 , ... ,T-1 and T is a positive integer greater than or equal to 2;\nsuccessively perform (302) polar decoding of the last n, coded bits of the plurality of coded bits for code rates rd to n to provide sets of information bits for the code rates rd to n, respectively, where:\n• k is a number of information bits encoded into the plurality of coded bits; and\n• n > n+i for i = 1 , ... ,T, where rT is a lowest code rate; and concatenate (304) the sets of information bits for the code rates rd to n to provide the plurality of information bits.\n25. The receive node (18) of claim 24 wherein, in order to successively perform (302) polar decoding of the last n, coded bits of the plurality of coded bits for the code rates rd to n to provide the sets of information bits for the code rates rd to r-i, respectively, the rate-compatible polar decoder (20) is further operable to: perform polar decoding of the last nd coded bits of the plurality of coded bits for the code rate rd to provide the set of information bits for the code rate rd; perform polar decoding of the next nd-i coded bits of the plurality of coded bits using the set of information bits for the code rate rd as frozen bits to provide the set of information bits for code rate rd-i; and perform polar decoding of the next nd-2 coded bits of the plurality of coded bits using a union of the sets of information bits for code rates rd and rd-i as frozen bits to provide the set of information bits for code rate rd-2.\nEP16742000.9A 2015-08-12 2016-06-30 Rate-compatible polar codes Pending EP3335321A1 (en)\n\n## Priority Applications (2)\n\nApplication Number Priority Date Filing Date Title\nUS14/824,694 US10461779B2 (en) 2015-08-12 2015-08-12 Rate-compatible polar codes\nPCT/IB2016/053941 WO2017025823A1 (en) 2015-08-12 2016-06-30 Rate-compatible polar codes\n\n## Publications (1)\n\nPublication Number Publication Date\nEP3335321A1 true EP3335321A1 (en) 2018-06-20\n\n# Family\n\n## Family Applications (1)\n\nApplication Number Title Priority Date Filing Date\nEP16742000.9A Pending EP3335321A1 (en) 2015-08-12 2016-06-30 Rate-compatible polar codes\n\n## Country Status (3)\n\nUS (1) US10461779B2 (en)\nEP (1) EP3335321A1 (en)\nWO (1) WO2017025823A1 (en)\n\n## Cited By (1)\n\n* Cited by examiner, † Cited by third party\nPublication number Priority date Publication date Assignee Title\nEP3504818A4 (en) * 2016-09-12 2020-04-15 MediaTek Inc. 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Low complexity rate matching design for polar codes\nCA3026317A1 (en) 2016-07-27 2018-02-01 Qualcomm Incorporated Design of hybrid automatic repeat request (harq) feedback bits for polar codes\nCN108599900B (en) * 2016-08-11 2019-06-07 华为技术有限公司 Method, apparatus and equipment for Polarization Coding\nCN107819545B (en) * 2016-09-12 2020-02-14 华为技术有限公司 Retransmission method and device of polarization code\nEP3535889A1 (en) 2016-11-03 2019-09-11 Telefonaktiebolaget LM Ericsson (PUBL) Systems and methods for rate-compatible polar codes for general channels\nUS10554223B2 (en) * 2016-12-23 2020-02-04 Huawei Technologies Co., Ltd. Apparatus and methods for polar code construction\nUS20180199350A1 (en) * 2017-01-11 2018-07-12 Qualcomm Incorporated Control channel code rate selection\nCN108365850A (en) * 2017-01-26 2018-08-03 华为技术有限公司 Coding method, code device and communication device\nEP3577769A1 (en) 2017-02-06 2019-12-11 MediaTek Inc. Polar code interleaving and bit selection\nWO2018148963A1 (en) * 2017-02-20 2018-08-23 Qualcomm Incorporated Polarization weight calculation for punctured polar code\nWO2018161274A1 (en) * 2017-03-08 2018-09-13 Qualcomm Incorporated Polar coding design for performance and latency\nCN108599891B (en) * 2017-03-17 2020-05-08 华为技术有限公司 Encoding method, encoding device, and communication device\nBR112019019282A2 (en) * 2017-03-23 2020-04-14 Qualcomm Inc assignment of parity bit channel for polar coding\nCN109412608B (en) * 2017-03-24 2019-11-05 华为技术有限公司 Polar coding method and code device, interpretation method and code translator\nCN108768586B (en) * 2017-03-25 2019-07-12 华为技术有限公司 A kind of method and apparatus of rate-matched\nCN108667464A (en) * 2017-04-01 2018-10-16 华为技术有限公司 Polarization code encodes and method, sending device and the receiving device of decoding\nCN108964834A (en) * 2017-05-23 2018-12-07 华为技术有限公司 Data transmission method, chip, transceiver and computer readable storage medium\nUS10312939B2 (en) 2017-06-10 2019-06-04 Qualcomm Incorporated Communication techniques involving pairwise orthogonality of adjacent rows in LPDC code\nCN110061745B (en) * 2017-06-16 2020-04-28 华为技术有限公司 Method and device for rate matching and rate de-matching\nKR20200020782A (en) 2017-07-07 2020-02-26 퀄컴 인코포레이티드 Communication techniques applying low-density parity-check code base graph selection\nEP3652865A1 (en) * 2017-07-12 2020-05-20 Telefonaktiebolaget LM Ericsson (PUBL) Enhanced information sequences for polar codes\nEP3656059A1 (en) * 2017-07-19 2020-05-27 Telefonaktiebolaget LM Ericsson (PUBL) Enhanced information sequences for polar codes\nEP3659259A1 (en) * 2017-07-25 2020-06-03 Telefonaktiebolaget LM Ericsson (PUBL) Enhanced information sequences for polar codes\nWO2019021225A1 (en) * 2017-07-26 2019-01-31 Telefonaktiebolaget Lm Ericsson (Publ) Enhanced information sequences for polar codes\nEP3673581A1 (en) * 2017-08-21 2020-07-01 Telefonaktiebolaget LM Ericsson (PUBL) Upo compliant information sequences for polar codes\nCN109428608A (en) * 2017-08-25 2019-03-05 华为技术有限公司 The interpretation method and decoder of polarization code\nCN108365921B (en) 2017-09-30 2019-03-26 华为技术有限公司 Polar coding method and code device, interpretation method and code translator\nUS10312948B1 (en) * 2018-04-30 2019-06-04 Polaran Yazilim Bilisim Danismanlik Ithalat Ihracat Sanayi Ticaret Limited Sirketi Method and system for retransmitting data using systematic polar coding\n\n## Family Cites Families (5)\n\n* Cited by examiner, † Cited by third party\nPublication number Priority date Publication date Assignee Title\nKR20050046471A (en) 2003-11-14 2005-05-18 삼성전자주식회사 Apparatus for encoding/decoding using parallel concatenation low density parity check code and the method thereof\nUS8347186B1 (en) * 2012-04-19 2013-01-01 Polaran Yazilim Bilisim Danismanlik Ithalat Ihracat Sanayi Ticaret Limited Sirketi Method and system for error correction in transmitting data using low complexity systematic encoder\nUS9362956B2 (en) * 2013-01-23 2016-06-07 Samsung Electronics Co., Ltd. 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A practical construction method for polar codes in AWGN channels\nUS8782499B2 (en) Apparatus and method for transmitting and receiving data in communication/broadcasting system\nUS9118353B2 (en) System and method for communicating with low density parity check codes\nKR101283724B1 (en) Multi-layer cyclic redundancy check code in wireless communication system\nCN106817195B (en) Method and apparatus for rate matching of polar codes\nJP6468526B2 (en) Polar code rate matching method and apparatus, and wireless communication device\nEP2540024B1 (en) System and method for transmitting and receiving acknowledgement information\nUS20180323800A1 (en) Reception apparatus and associated method of receiving encoded data\nRU2310274C1 (en) Device and method for encoding/decoding a channel with usage of parallel cascade even parity check code with low density\nJP4350750B2 (en) Retransmission control method and communication apparatus\nCN101217337B (en) A low density parity code encoding device and method supporting incremental redundancy hybrid automatic repeat\nUS8234551B2 (en) Single CRC polynomial for both turbo code block CRC and transport block CRC\nKR101339120B1 (en) Ldpc encoding and decoding of packets of variable sizes\nUS9362956B2 (en) Method and system for encoding and decoding data using concatenated polar codes\nCN103684680B (en) Decoding a coded data block\nWO2015139316A1 (en) Polar code retransmission method and device\nJP4602418B2 (en) Parity check matrix generation method, encoding method, decoding method, communication apparatus, encoder, and decoder\nUS20050154958A1 (en) Method and apparatus for varying lengths of low density party check codewords\nEP1656737B1 (en) Method and apparatus for varying lengths of low density parity check codewords\nJP5955261B2 (en) Method and system for using reconfigurable rateless code and decoding in-process status feedback to communicate with multimedia\nUS20090006906A1 (en) Method and system for encoding data using rate-compatible irregular LDPC codes based on edge growth and parity splitting\nUS8341490B2 (en) Virtual limited buffer modification for rate matching\nWO2018141277A1 (en) Offset lifting method\n\n## Legal Events\n\nDate Code Title Description\nAX Request for extension of the european patent\n\nExtension state: BA ME\n\n17P Request for examination filed\n\nEffective date: 20180214\n\nAK Designated contracting states\n\nKind code of ref document: A1\n\nDesignated state(s): AL AT BE BG CH CY CZ DE DK EE ES FI FR GB GR HR HU IE IS IT LI LT LU LV MC MK MT NL NO PL PT RO RS SE SI SK SM TR\n\nAV Request for validation of the european patent\n\nExtension state: MA MD\n\nDAV Request for validation of the european patent (deleted)\nDAX Request for extension of the european patent (deleted)\n17Q First examination report despatched\n\nEffective date: 20200424"
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| {"ft_lang_label":"__label__en","ft_lang_prob":0.81047857,"math_prob":0.9304115,"size":68299,"snap":"2020-34-2020-40","text_gpt3_token_len":17345,"char_repetition_ratio":0.23240355,"word_repetition_ratio":0.28183603,"special_character_ratio":0.2665339,"punctuation_ratio":0.10013258,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9566249,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-08-04T00:58:12Z\",\"WARC-Record-ID\":\"<urn:uuid:149e4403-593f-4817-a520-ac858b491fdb>\",\"Content-Length\":\"176342\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:f1b7d4a9-4d34-4c6d-8b83-c1c1364e940f>\",\"WARC-Concurrent-To\":\"<urn:uuid:a367c9c5-cc1d-4f34-bba4-75023bedeb5b>\",\"WARC-IP-Address\":\"172.217.13.238\",\"WARC-Target-URI\":\"https://patents.google.com/patent/EP3335321A1/en\",\"WARC-Payload-Digest\":\"sha1:YF2RC75NILJ2DK4SYCXRL4EF2KO4LVF5\",\"WARC-Block-Digest\":\"sha1:Y6UKOANRS4PAWPANVNTKT55QQL4YBU6R\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-34/CC-MAIN-2020-34_segments_1596439735836.89_warc_CC-MAIN-20200803224907-20200804014907-00098.warc.gz\"}"} |
https://www.intlpress.com/site/pub/pages/journals/items/mrl/content/vols/0022/0005/a012/index.php | [
"# Mathematical Research Letters\n\n## Volume 22 (2015)\n\n### A new Fourier transform\n\nPages: 1541 – 1562\n\nDOI: https://dx.doi.org/10.4310/MRL.2015.v22.n5.a12\n\n#### Author\n\nJonathan Wang (Department of Mathematics, University of Chicago, U.S.A.)\n\n#### Abstract\n\nIn order to define a geometric Fourier transform, one usually works with either $\\ell$-adic sheaves in characteristic $p \\gt 0$ or with $\\mathcal{D}$-modules in characteristic $0$. If one considers $\\ell$-adic sheaves on the stack quotient of a vector bundle $V$ by the homothety action of $\\mathbb{G}_m$, however, Laumon provides a uniform geometric construction of the Fourier transform in any characteristic. The category of sheaves on $[V / \\mathbb{G}_m]$ is closely related to the category of (unipotently) monodromic sheaves on $V$. In this article, we introduce a new functor, which is defined on all sheaves on $V$ in any characteristic, and we show that it restricts to an equivalence on monodromic sheaves. We also discuss the relation between this new functor and Laumon’s homogeneous transform, the Fourier–Deligne transform, and the usual Fourier transform on $\\mathcal{D}$-modules (when the latter are defined).\n\nAccepted 2 April 2015\n\nPublished 13 April 2016"
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https://goprep.co/ex-30.3-q16-log-1-root-x-5x-a-3a-x-cube-root-x-2-6-root-4-x-i-1nl77n | [
"# Differentiate the\n\nGiven,\n\nf(x) =",
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null,
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null,
"Using algebra of derivatives –\n\nf’(x) =",
null,
"Use the formula:",
null,
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null,
"f’(x) =",
null,
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"Rate this question :\n\nHow useful is this solution?\nWe strive to provide quality solutions. Please rate us to serve you better.\nTry our Mini CourseMaster Important Topics in 7 DaysLearn from IITians, NITians, Doctors & Academic Experts\nDedicated counsellor for each student\n24X7 Doubt Resolution\nDaily Report Card\nDetailed Performance Evaluation",
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]
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"https://gradeup-question-images.grdp.co/liveData/PROJ26135/1549538917783789.png",
null,
"https://grdp.co/cdn-cgi/image/height=128,quality=80,f=auto/https://gs-post-images.grdp.co/2020/8/group-7-3x-img1597928525711-15.png-rs-high-webp.png",
null,
"https://gs-post-images.grdp.co/2020/8/group-img1597139979159-33.png-rs-high-webp.png",
null
]
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https://quantumcomputing.stackexchange.com/questions/6655/how-can-a-cnot-gate-change-the-control-qbit-e-g-in-the-deutsch-oracle-problem | [
"# How can a CNOT gate change the control qbit (e.g. in the Deutsch Oracle problem)?\n\nIt is very confusing how the CNOT gate can change the properties of the controlled qubit (possibly because I am still stuck in the classical programming mindset):",
null,
"Many books have explained the phenomenon using a $$4\\times 4$$ matrix multiplication and tensor products to show how this happens. However, the math doesn't seem to agree because if I don't do the tensor product of the two qubits and instead just do the math separately by applying $$2\\times 2$$ matrices to the top and bottom qubit based on what is happening and then tensor the two together in the end, the two methods don't agree.\n\nHere is (what is in my mind) a simple example to show the mathematical error:",
null,
"As we can see from the bottom qubit. The $$X$$ in the very beginning obviously transforms the $$|0\\rangle$$ to a $$|1\\rangle$$. The controlled dot doesn't do anything: it merely observes the bottom qubit in order to decide whether to apply the $$X$$ gate onto the top qubit. Thus, it does nothing to the bottom qubit. As everyone knows, all functions of quantum computing are inverses of each other. Hence, the 2 H gates cancel out. It thus makes no sense to me how the bottom qubit can be 100% off all the time when it should logically be \"On\" (and if we do out the matrices it will work out as such).\n\nThank you so much for your help and I apologize if this question is basic.\n\n• Hi Normen! What do you mean by \"all functions of quantum computing are inverses of each other?\" Quantum gates such as $H$ are reversible, but that does not mean that applying $H$ to one qubit and another qubit is meant to cancel anything out. Jul 1, 2019 at 21:04\n\nThe controlled dot doesn't do anything: it merely observes the bottom qubit in order to decide whether to apply the 𝑋 gate onto the top qubit.\n\nIn the answer below, the qubit that appears first is the control qubit, it maps to the bottom qubit in your diagram, and the second one is the target qubit.\n\n• The CNOT gate works on two qubits at once. You can't think of it's operation separately on the first and second qubit, and you can't do the math using two separate $$2*2$$ matrices.\n• The definition of what CNOT does to a basis is this: $$CNOT(|00\\rangle) = |00\\rangle$$ $$CNOT(|01\\rangle) = |01\\rangle$$ $$CNOT(|10\\rangle) = |11\\rangle$$ $$CNOT(|11\\rangle) = |10\\rangle$$\n\nSure, when you look at the definition of the basis, it seems like CNOT doesn't do anything to the first qubit, and only observes the first and flips the second if necessary. But consider applying CNOT on a non-basis state:\n\n$$CNOT(|+\\rangle |0\\rangle)$$ $$=CNOT \\left(\\frac{|0\\rangle + |1\\rangle}{\\sqrt 2}|0\\rangle \\right)$$ $$=CNOT \\left(\\frac{|00\\rangle}{\\sqrt 2} + \\frac{|10\\rangle}{\\sqrt 2}\\right)$$ Now, exploiting the linear property of quantum gates, we have $$= \\frac{|00\\rangle + |11\\rangle}{\\sqrt 2}$$\n\nNotice that the first qubit is not in the $$|+\\rangle$$ state anymore, even though it was the qubit just being observed. It in fact does not have any pure representation. It has gotten entangled with the second qubit.\n\nSo CNOT does infact effect the control qubit in the sense that it ties it up with outcomes of the target qubit.\n\n• Now, in your first example\n\nAfter applying $$X$$ on both qubits, you have $$|11\\rangle$$.\n\nAfter applying $$H$$ on both qubits you have $$|--\\rangle$$, or $$\\frac{|00\\rangle - |01\\rangle - |10\\rangle + |11\\rangle}{\\sqrt 2}$$.\n\nAfter applying the CNOT on this qubit set, you have $$\\frac{|00\\rangle - |01\\rangle - |11\\rangle + |10\\rangle}{\\sqrt 2}$$ which you can factorise as $$|+-\\rangle$$\n\nAnd now applying the two $$H$$'s, you have $$|01\\rangle$$. Hence the bottom qubit is 'off' and the top qubit is 'on'.\n\nAs everyone knows, all functions of quantum computing are inverses of each other. Hence, the 2 H gates cancel out.\n\nQuantum gates all have inverses, but the inverse of a gate is not necessarily the same gate, though Hadamard gates, which are the ones being most considered here, are their own inverse. I'm not sure if by \"2 H gates\" you mean the two gates before the CNOT in the smaller diagram, which are on different qubits, or the two $$H^{\\otimes 2}$$ gates flanking the CNOT in the larger diagram. In the former case, there's no cancellation since they are on different qubits. In the other case, they don't cancel because there is the CNOT between them and matrix multiplication is non-commutative: since $$AB \\neq BA$$ is possible, you can't conclude $$A^{-1}BA = A^{-1}AB = B$$, and thus the two sets of Hadamards don't cancel out.\n\nNow that we know that they don't just cancel out, you can derive that a CNOT with $$H^{\\otimes 2}$$ flankings actually together make another CNOT with swapped control and target qubits. First, here is the conceptually simple but really tedious matrix multiplication of that:\n\n$$\\frac{1}{4}\\begin{bmatrix}1 & 1 & 1 & 1 \\\\ 1 & -1 & 1 & -1 \\\\ 1 & 1 & -1 & -1 \\\\ 1 & -1 & -1 & 1 \\end{bmatrix} \\begin{bmatrix}1 & 0 & 0 & 0 \\\\ 0 & 1 & 0 & 0 \\\\ 0 & 0 & 0 & 1 \\\\ 0 & 0 & 1 & 0\\end{bmatrix} \\begin{bmatrix}1 & 1 & 1 & 1 \\\\ 1 & -1 & 1 & -1 \\\\ 1 & 1 & -1 & -1 \\\\ 1 & -1 & -1 & 1 \\end{bmatrix} = \\frac{1}{4}\\begin{bmatrix}1 & 1 & 1 & 1 \\\\ 1 & -1 & 1 & -1 \\\\ 1 & 1 & -1 & -1 \\\\ 1 & -1 & -1 & 1 \\end{bmatrix} \\begin{bmatrix}1 & 1 & 1 & 1 \\\\ 1 & -1 & 1 & -1 \\\\ 1 & -1 & -1 & 1 \\\\ 1 & 1 & -1 & -1\\end{bmatrix} = \\begin{bmatrix}1 & 0 & 0 & 0 \\\\ 0 & 0 & 0 & 1 \\\\ 0 & 0 & 1 & 0 \\\\ 0 & 1 & 0 & 0\\end{bmatrix}$$.\n\nWith $$a\\left|00\\right> + b\\left|01\\right> + c\\left|10\\right> + d\\left|11\\right> = \\begin{bmatrix} a \\\\ b \\\\ c \\\\ d \\end{bmatrix}$$, we see the original in the flank swaps the probability amplitudes of $$\\left|10\\right>$$ and $$\\left|11\\right>$$ (left qubit is control), while the new one swaps the amplitudes of $$\\left|01\\right>$$ and $$\\left|11\\right>$$ (right qubit is control).\n\nA harder but interesting and less tedious way to recognize this is through the symmetry of the controlled-$$Z$$ gate, as described in page 5 of https://arxiv.org/pdf/1110.2998.pdf. A controlled-$$Z$$ gate's control and target qubits are indistinguishable since the only probability amplitude in computational basis that changes is the sign flip of the probability amplitude of $$\\left|11\\right>$$. However, if one of the qubits in the controlled-$$Z$$ has Hadamards flanking it, they together are a CNOT with the Hadamard-flanked qubit being the target. Add the $$H^{\\otimes 2}$$ on both sides to this CNOT and you have the \"control\" qubit of the controlled-$$Z$$ gate with a Hadamard behind and in front of it, and the \"target\" qubit with two on each side. The two Hadamards on each side for that qubit cancel out since there is nothing between them, so it ends up with a controlled-$$Z$$ with Hadamards behind and in front of what we previously noted as the control, but, since a controlled-$$Z$$ doesn't distinguish control and target, that is instead the target of a CNOT due to the Hadamards, and we have the switch.\n\nI am not sure what the second question is about, but in the first one you are probably inquiring about the concept of \"phase kickback\":\n\nWhy does the \"Phase Kickback\" mechanism work in the Quantum phase estimation algorithm?"
]
| [
null,
"https://i.stack.imgur.com/XlpCU.png",
null,
"https://i.stack.imgur.com/ruauW.png",
null
]
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http://multistrand.org/tutorials/Tutorial%2012%20-%20NUPACK%20calls.html | [
"# Tutorial 12 - NUPACK calls¶\n\nMultistrand is equipped with Python wrapper functions for many of Nupack's executables, including pfunc, energy, defect, prob, sample, mfe, and pairs. Here we provide usage examples for all provided wrapper functions. See the Nupack User Manual for more information about the arguments to each executable. Many of these examples are adapted from examples provided with Nupack. Note that these functions assume you have Nupack version 3.0.2 or higher. Certain functions, such as sample(), will not work with earlier versions of Nupack.\n\nIn :\n# In general, the following arguments are required:\n# sequences - a list of DNA/RNA sequences\n# structure - a dot-paren structure (required by energy, prob, and defect)\n# energy_gap - the max allowed energy gap from the minimum free energy (required by subopt)\n# samples - an integer specifying how many samples to draw (required by sample)\n#\n# The following optional arguments may also be included in all function calls:\n# ordering - a list of integer indices indicating the ordering of the given strand sequences\n# in the complex, or None if each strand is distinguishable, occurs exactly once,\n# and is ordered as in the sequences argument. Indices may be repeated to indicate\n# identical, indistinguishable strands. This argument is ignored if multi == False.\n# (default: None)\n# material - a string representing the material parameters to use (e.g. 'rna' or 'dna')\n# The Nupack User Manual has a complete list of possible values.\n# (default: 'rna')\n# multi - whether or not to allow multistrand structures with identical strands\n# only works with DNA (default: True)\n# pseudo - whether or not to allow pseudoknotted structures\n# only works with RNA (default: False)\n# dangles - 'none', 'all', or 'some' to specify what types of dangle corrections to use (default: 'some')\n# T - temperature of the system in degrees C (default: 37)\n# sodium - concentration of Na+ in molar (default: 1.0)\n# magnesium - concentration of Mg++ in molar (default: 0.0)\n#\n# The following optional arguments are allowed by only some functions:\n# cutoff - (for use with pairs) probabilities less than cutoff are excluded (default: 0.001)\n# degenerate - (for use with mfe) degenerate structures are all returned (default: False)\n# mfe - (for use with defect) returns mfe defect rather than ensemble defect (default: False)\n\n\nImport the NUPACK wrapper:\n\nIn :\nfrom nupack import *\n\nPython interface to NUPACK 3.0 (Pierce lab, Caltech, www.nupack.org) loaded.\n\nIn :\n## Sequences used throughout this file:\nrna_seq = ['GGGCUGUUUUUCUCGCUGACUUUCAGCCCCAAACAAAAAAUGUCAGCA'];\ndna_seqs = ['AGTCTAGGATTCGGCGTGGGTTAA',\n'TTAACCCACGCCGAATCCTAGACTCAAAGTAGTCTAGGATTCGGCGTG',\n'AGTCTAGGATTCGGCGTGGGTTAACACGCCGAATCCTAGACTACTTTG']\ndna_struct = '((((((((((((((((((((((((+))))))))))))))))))))))))((((((((((((.(((((((((((+........................))))))))))).))))))))))))'\n\n\n### Examples with pfunc()¶\n\nFind the partition function of this set of three DNA strands.\n\nIn :\npfunc(dna_seqs, material = 'dna')\n\nOut:\n-62.66453454292297\n\nFind the partition function of this single RNA strand, including pseudoknotted structures.\n\nIn :\npfunc(rna_seq, material = 'rna', multi = False, pseudo = True)\n\nOut:\n-17.2786076\n\n### Examples with energy()¶\n\nFind the energy (in kcal/mol) of this DNA structure.\n\nIn :\nenergy(dna_seqs, dna_struct, material = 'dna')\n\nOut:\n-61.79270283815892\n\n### Examples with defect()¶\n\nFind the ensemble defect of this DNA structure.\n\nIn :\ndefect(dna_seqs, dna_struct, material = 'dna')\n\nOut:\n8.297\n\nFind the MFE defect of this DNA structure.\n\nIn :\ndefect(dna_seqs, dna_struct, material = 'dna', mfe = True)\n\nOut:\n2.0\n\n### Examples with prob()¶\n\nFind the probability of this DNA structure.\n\nIn :\nprob(dna_seqs, dna_struct, material = 'dna')\n\nOut:\n7.992e-05\n\n### Examples with sample()¶\n\nBoltzmann sample 3 structures with these DNA strands.\n\nIn :\nsample(dna_seqs, 3, material = 'dna')\n\nOut:\n['.(((((((((((((((((((((..+..)))))))))))))))))))))..(((((((((((((((((((((((+........................))))))))))))))))))))))).',\n'((((((((((((((((((((((..+..)))))))))))))))))))))).(((((((((((((((((((((((+..(.....)...............))))))))))))))))))))))).',\n'((((((((((((((((((((((((+))))))))))))))))))))))))(((((((((((((((((((((((.+......(.......)..........)))))))))))))))))))))))']\n\n### Examples with mfe()¶\n\nFind the MFE structure of this ordering of DNA sequences (the energy is also provided).\n\nIn :\nmfe(dna_seqs, material = 'dna')\n\nOut:\n[('((((((((((((((((((((((((+))))))))))))))))))))))))((((((((((((((((((((((((+........................))))))))))))))))))))))))',\n'-65.913')]\n\n### Examples with pairs()¶\n\nFind the pair probabilities of the following unpseudoknotted DNA strands. The output is a list of tuples of the form (i, j, p) where p is the probability that the ith and jth bases are bound.\n\nIn :\npairs(['ACGT','GCTT'], material = 'dna')\n\nOut:\n[(1, 7, 0.031271),\n(1, 8, 0.049495),\n(2, 5, 0.2163),\n(3, 6, 0.73677),\n(1, 9, 0.91923),\n(2, 9, 0.7837),\n(3, 9, 0.26323),\n(4, 9, 1.0),\n(5, 9, 0.7837),\n(6, 9, 0.26323),\n(7, 9, 0.96873),\n(8, 9, 0.95051)]\nIn [ ]:"
]
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https://www.mathdoubts.com/parallel-lines/ | [
"# Parallel lines\n\nTwo or more straight lines which do not intersect each other at any point on the plane are called parallel lines.",
null,
"Parallel lines are straight lines basically and they appear on a plane side by side. They never intersect each other anywhere on a plane because of having same angle. Therefore, if two or more straight lines have same angle, then they do not meet each other.\n\nSimilarly, the vertical distance between opposite points of them is always same due to the parallelism of the parallel lines.\n\n## Example",
null,
"$\\overleftrightarrow{AB}$ and $\\overleftrightarrow{CD}$ are two straight lines and they are extended infinitely but they do not intersect each other even though they are appearing side by side.\n\nThey maintain equal distance between them at all their opposite points. The reason for this is the angles of both straight lines are same.\n\nThe straight lines $\\overleftrightarrow{AB}$ and $\\overleftrightarrow{CD}$ are called parallel lines.\n\n### Representation\n\nA double vertical line ($\\|$) is used to represent parallelism of the lines mathematically.\n\nThe parallel symbol is displayed between straight lines to express that they both are parallel lines.\n\n$\\overleftrightarrow{AB} \\,\\, \\| \\,\\, \\overleftrightarrow{CD}$\n\nIt is read as $\\overleftrightarrow{AB}$ parallel to the $\\overleftrightarrow{CD}$.\n\nSimilarly, it can also be displayed in other way.\n\n$\\overleftrightarrow{CD} \\,\\, \\| \\,\\, \\overleftrightarrow{AB}$\n\nIt is read as $\\overleftrightarrow{CD}$ parallel to the $\\overleftrightarrow{AB}$.\n\nLatest Math Topics\nApr 18, 2022\nApr 14, 2022\n\nA best free mathematics education website for students, teachers and researchers.\n\n###### Maths Topics\n\nLearn each topic of the mathematics easily with understandable proofs and visual animation graphics.\n\n###### Maths Problems\n\nLearn how to solve the maths problems in different methods with understandable steps.\n\nLearn solutions\n\n###### Subscribe us\n\nYou can get the latest updates from us by following to our official page of Math Doubts in one of your favourite social media sites."
]
| [
null,
"https://www.mathdoubts.com/cimgs/line/parallel-lines/graphical.png",
null,
"https://www.mathdoubts.com/cimgs/line/parallel-lines/animation.gif",
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https://open.kattis.com/problems/collatzconjecture | [
"# Collatz Conjecture",
null,
"Picture by mscolly via Flickr.\nIn 1978 AD the great Sir Isaac Newton, whilst proving that $\\mathcal{P}$ is a strict superset of $\\mathcal{NP}$, defined the Beta Alpha Pi Zeta function $f$ as follows over any sequence of positive integers $a_1, \\dots , a_ n$. Given integers $1\\leq i\\leq j\\leq n$, we define $f(i, j)$ as $\\gcd (a_ i, a_{i+1}, \\dots , a_{j-1}, a_ j)$.\n\nAbout a century later Lothar Collatz applied this function to the sequence $1, 1, 1, \\dots , 1$, and observed that $f$ always equalled $1$. Based on this, he conjectured that $f$ is always a constant function, no matter what the sequence $a_ i$ is. This conjecture, now widely known as the Collatz Conjecture, is one of the major open problems in botanical studies. (The Strong Collatz Conjecture claims that however many values $f$ takes on, the real part is always $\\frac{1}{2}$.)\n\nYou, a budding young cultural anthropologist, have decided to disprove this conjecture. Given a sequence $a_ i$, calculate how many different values $f$ takes on.\n\n## Input\n\nThe input consists of two lines.\n\n• A single integer $1 \\leq n \\leq 5 \\cdot 10^5$, the length of the sequence.\n\n• The sequence of integers $a_1, a_2, \\dots , a_ n$. It is given that $1 \\leq a_ i \\leq 10^{18}$.\n\n## Output\n\nOutput a single line containing a single integer, the number of distinct values $f$ takes on over the given sequence.\n\nSample Input 1 Sample Output 1\n4\n9 6 2 4\n\n6\n\nSample Input 2 Sample Output 2\n4\n9 6 3 4\n\n5"
]
| [
null,
"https://open.kattis.com/problems/collatzconjecture/file/statement/en/img-0001.jpg",
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https://mathematica.stackexchange.com/questions/103000/replacement-producing-unexpected-result | [
"# Replacement producing unexpected result\n\nIf I use a replacement rule like\n\nTimes[2, Plus[Times[AuRr, BuRb], Times[AuRb, BuRr]], h10] /.\n{Times[r_, Plus[Times[x_, y_], Times[v_, w_]], h10] :>\nr*TensorProduct[(TensorProduct[x, y] + TensorProduct[v, w]), h10]}\n\n\nI get correctly\n\n(* 2 (AuRb \\[TensorProduct] BuRr + AuRr \\[TensorProduct] BuRb) \\[TensorProduct]h10 *)\n\n\nNevertheless, if I try to do the same thing more general, i.e. allowing for a general last coefficient, not specifically h10:\n\nTimes[2, Plus[Times[AuRr, BuRb], Times[AuRb, BuRr]], h10] /.\n{Times[r_, Plus[Times[x_, y_], Times[v_, w_]], h_] :>\nr*TensorProduct[(TensorProduct[x, y] + TensorProduct[v, w]), h]}\n\n\nI get the wrong result\n\n (* 2 h10 (AuRb \\[TensorProduct] BuRr + AuRr \\[TensorProduct] BuRb) *)\n\n\nAny ideas about what is going wrong here, would be much appreciated!\n\n• Please don't add bugs tag until the community has conformed it. BTW it's almost impossible to find a bug in core language of Mathematica. – xzczd Dec 29 '15 at 9:27\n\nMathematica automatically factors out constants from TensorProducts:\n\nTensorProduct[a, 2 b] (* returns 2 a\\[TensorProduct]b *)\n\n\nIf you consider\n\nTimes[2, Plus[Times[AuRr, BuRb], Times[AuRb, BuRr]], h10] /.\n{Times[r_, Plus[Times[x_, y_], Times[v_, w_]], h_]\n:>\nEcho[r]*TensorProduct[(TensorProduct[x, y] + TensorProduct[v, w]), Echo@h]\n}\n\n\nyou will see that the pattern-matcher sets r = h10 and h = 2, which accounts for why the constants end up outside the expression. Then it ends up with a TensorProduct of an expression with 1, and that gets rid of the outermost TensorProduct.\n\n• Thanks a lot for the explanation! I now simply add the condition /; Element[r, Reals] and everything works perfectly – jak Dec 29 '15 at 9:13"
]
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https://www.onlinemathlearning.com/divide-multi-digit-6ns2.html | [
"",
null,
"# Multi-Digit Division (Grade 6)\n\nRelated Topics:\nCommon Core for Grade 6\nMore Lessons for Grade 6\n\nExamples, solutions, videos, and lessons to help Grade 6 students learn to fluently divide multi-digit numbers using the standard algorithm.\n\nCommon Core: 6.NS.2\n\n### Suggested Learning Targets\n\n• I can divide multi-digit numbers.\n• I can tell if my quotient is reasonable.\nI can divide multi-digit numbers using the standard algorithm with speed and accuracy, without any math tools (i.e., calculator, multiplication chart).\nComponent Skills from Previous Grades\n 5.NBT.6 Add, subtract, multiply, and divide decimals to hundredths, using concrete models or drawings and strategies based on place value, properties of operations, and/or the relationship between addition and subtraction; relate the strategy to a written method and explain the reasoning used.\n6NS2 Dividing Multi-digit Numbers with Standard Algorithm - Part 1\nExample:\n7452 ÷ 54\n8525 ÷ 20 6NS2 Dividing Multi-digit Numbers with Standard Algorithm part 2\nRepeating Decimals in the quotient.\nExample:\n6432 ÷ 396\n\nWhy the Standard Algorithm for Long Division Works?\nDividing by a two digit number\nExample:\n7182 ÷ 42 2 × 3 Division: Learn how to solve 2 digit by 3 digit long division problems\nExample:\n528 divide by 43\n228 divide by 43 Divide Multi Digit Numbers\n3 Steps\n1) Divide\n2) Multiply\n3) Subtract\n4) Bring Down\n5) Repeat Dividing Multi Digit Numbers\nExamples:\n1) 192 ÷ 8\n2) 278 ÷ 18\n3) 5,272 ÷ 64\n4) An elephant will eat about 6,820 pounds of vegetation in a 31-day month. What is the average amount an elephant will eat every day?\n\nTry the free Mathway calculator and problem solver below to practice various math topics. Try the given examples, or type in your own problem and check your answer with the step-by-step explanations.",
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"We welcome your feedback, comments and questions about this site or page. Please submit your feedback or enquiries via our Feedback page."
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https://ww2.mathworks.cn/help/matlab/matlab_prog/vectorization.html | [
"## 向量化\n\n### 向量化的应用\n\nMATLAB® 针对涉及矩阵和向量的运算进行了优化。修正基于循环且面向标量的代码以使用 MATLAB 矩阵和向量运算的过程称为向量化。由于多种原因，代码的向量化是可取的：\n\n• 外观：向量化的数学代码在外观上更像教科书中的数学表达式，从而使代码更便于理解。\n\n• 不容易出错：由于不带循环，向量化的代码通常更短。代码行的减少意味着引入编程错误的机会也减少了。\n\n• 性能：向量化代码的运行速度通常比相应的、包含循环的代码更快。\n\n#### 实现代码向量化以进行常规计算\n\n```i = 0; for t = 0:.01:10 i = i + 1; y(i) = sin(t); end```\n\n```t = 0:.01:10; y = sin(t);```\n\n#### 针对特定任务实行代码向量化\n\n```x = 1:10000; ylength = (length(x) - mod(length(x),5))/5; y(1:ylength) = 0; for n= 5:5:length(x) y(n/5) = sum(x(1:n)); end ```\n\n```x = 1:10000; xsums = cumsum(x); y = xsums(5:5:length(x)); ```\n\n### 数组运算\n\n`V = 1/12*pi*(D^2)*H;`\n\n```for n = 1:10000 V(n) = 1/12*pi*(D(n)^2)*H(n); end```\n\n```% Vectorized Calculation V = 1/12*pi*(D.^2).*H;```\n\n```A = [97 89 84; 95 82 92; 64 80 99;76 77 67;... 88 59 74; 78 66 87; 55 93 85]; mA = mean(A); B = zeros(size(A)); for n = 1:size(A,2) B(:,n) = A(:,n) - mA(n); end```\n\n`devA = A - mean(A)`\n```devA = 18 11 0 16 4 8 -15 2 15 -3 -1 -17 9 -19 -10 -1 -12 3 -24 15 1```\n\n`F(x,y) = x*exp(-x2 - y2)`\n\n```x = (-2:0.2:2)'; % 21-by-1 y = -1.5:0.2:1.5; % 1-by-16 F = x.*exp(-x.^2-y.^2); % 21-by-16```\n\n### 逻辑数组运算\n\n```D = [-0.2 1.0 1.5 3.0 -1.0 4.2 3.14]; D >= 0```\n```ans = 0 1 1 1 0 1 1```\n\n`Vgood = V(D >= 0);`\n\nMATLAB 允许您分别使用 `all``any` 函数对整个向量的元素执行逻辑 AND 或 OR 运算。如果 `D` 的所有值都小于零，则会引发以下警告：\n\n```if all(D < 0) warning('All values of diameter are negative.') return end```\n\nMATLAB 也可以比较两个大小兼容的向量，并允许您施加更多限制。此代码查找 V 为非负且 `D` 大于 `H` 的所有值：\n\n`V((V >= 0) & (D > H))`\n\n```x = [2 -1 0 3 NaN 2 NaN 11 4 Inf]; xvalid = x(~isnan(x))```\n```xvalid = 2 -1 0 3 2 11 4 Inf```\n\n`Inf == Inf` 返回 true；但 `NaN == NaN` 始终返回 false。\n\n### 矩阵运算\n\n`A = ones(5,5)*10;`\n\n```v = 1:5; A = repmat(v,3,1)```\n```A = 1 2 3 4 5 1 2 3 4 5 1 2 3 4 5```\n\n```A = repmat(1:3,5,2) B = repmat([1 2; 3 4],2,2)```\n```A = 1 2 3 1 2 3 1 2 3 1 2 3 1 2 3 1 2 3 1 2 3 1 2 3 1 2 3 1 2 3 B = 1 2 1 2 3 4 3 4 1 2 1 2 3 4 3 4```\n\n### 排序、设置和计数运算\n\n#### 消除冗余元素\n\n```x = [2 1 2 2 3 1 3 2 1 3]; x = sort(x); difference = diff([x,NaN]); y = x(difference~=0)```\n```y = 1 2 3```\n\n`y=unique(x);`\n\n#### 计算向量中的元素数\n\n```x = [2 1 2 2 3 1 3 2 1 3]; x = sort(x); difference = diff([x,max(x)+1]); count = diff(find([1,difference])) y = x(find(difference))```\n```count = 3 4 3 y = 1 2 3 ```\n`find` 函数不返回 `NaN` 元素的索引。您可以使用 `isnan``isinf` 函数计算 `NaN``Inf` 值的数目。\n\n```count_nans = sum(isnan(x(:))); count_infs = sum(isinf(x(:)));```\n\n### 向量化的常用函数\n\n`all`\n\n`any`\n\n`cumsum`\n\n`diff`\n\n`find`\n\n`ind2sub`\n\n`ipermute`\n\nN 维数组的逆置换维度\n\n`logical`\n\n`meshgrid`\n\n`ndgrid`\n\nN 维空间中的矩形网格\n\n`permute`\n\n`prod`\n\n`repmat`\n\n`reshape`\n\n`shiftdim`\n\n`sort`\n\n`squeeze`\n\n`sub2ind`\n\n`sum`"
]
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https://insideaiml.com/blog/Python---Deque-444 | [
"",
null,
"",
null,
"Download our e-book of Attention Mechanism\n\n#### Top Courses",
null,
"#### Machine Learning with Python & Statistics",
null,
"4 (4,001 Ratings)",
null,
"220 Learners\n\n#### How To Land a Job in Data Science",
null,
"Jun 24th (7:00 PM) 240 Registered\nMore webinars\n##### Python - Deque",
null,
"Sanjeev Shah\n\n3 days ago\n\nIn this article, we will try to learn about deque with some examples.",
null,
"Python - Deque | Insideaiml\n\n## Table Of Content\n\n• Python - Deque\n• What is Deque?\n• Deque Methods.\n• Deque - extend() Method\n• Deque - extendleft()\n• Method to Add Numbers To Left End\n• Deque - rotate() Method To Rotate The Deque\n• Deque - reverse() Method To Reverse The Deque\n• Conclusion\n\n## What is Deque?\n\nDeque is a double-ended queue. It has two ends, a front and a rear end. In Python, we can implement deque using a collection module. Deque provides a fast update, insert and delete operations than a list data structure. For appending and pop operation, deque provides time complexity O(1).\nWe can add or remove the item from deque either ends. The new item can be added from the front or rear end. The characteristics of the deque are the same as stack and queue, but it does not need to follow the Last In First Out (LIFO) or First In Last Out (FILO) order.\n``````# importing \"collections\" module or deque operations\nimport collections\n\ndq = collections.deque([\"Mon\",\"Tue\",\"Wed\"])\n\ndq.append(\"Thu\")\n\nprint (\"Appended at right - \")\nprint (dq)\n\ndq.appendleft(\"Sun\")\n\nprint (\"Appended at right at left is - \")\nprint (dq)\n\ndq.pop()\n\nprint (\"Deleting from right - \")\nprint (dq)\n\nDoubleEnded.popleft()\n\nprint (\"Deleting from left - \")\nprint (dq)\n``````\nOutput:\n``````Appended at right -\ndeque(['Mon', 'Tue', 'Wed', 'Thu'])\nAppended at right at left is -\ndeque(['Sun', 'Mon', 'Tue', 'Wed', 'Thu'])\nDeleting from right -\ndeque(['Sun', 'Mon', 'Tue', 'Wed'])\nDeleting from left -\ndeque(['Mon', 'Tue', 'Wed'])``````\n\nRecommended blog for you : Introduction to Data Structures\n\n## Deque methods :\n\n• extend()\n• extendleft()\n• rotate()\n• reverse()\n\n### Deque - extend() Method\n\n``````# importing \"collections\" module for deque operations\nimport collections\n\n# initializing deque\ndq = collections.deque([10, 20, 30,])\n\n# add numbers to right end using extend() method\n# adds 40,50,60 to right end\ndq.extend([40,50,60])\n\n# extended Deque\nprint (\"extended deque at the end is: \")\nprint (dq)``````\nOutput:\n``````The deque after extending deque at end is :\ndeque([10, 20, 30, 40, 50, 60])``````\n\n### Deque - extendleft() Method to Add Numbers To Left End\n\n``````# add numbers to left end using extendleft() method\n# adds 70,80,90 to right end\ndq.extendleft([70,80,90])\n\n# Ater extendleft deque operation\nprint (\"extended deque is : \")\nprint (dq) ``````\nOutput:\n``````The deque after extending deque at beginning is :\ndeque([90, 80, 70, 10, 20, 30, 40, 50, 60])``````\n\n### Deque - rotate() Method To Rotate The Deque\n\n``````# to rotate the deque use rotate() method\n# rotates by 3 to left\ndq.rotate(-3)\n\n# printing rorated deque\nprint (\"rotated deque is : \")\nprint (dq) ``````\nOutput:\n``````The deque after rotating deque is :\ndeque([10, 20, 30, 40, 50, 60, 90, 80, 70])``````\n\n### Deque - reverse() Method To Reverse The Deque\n\n``````# to reverse the deque using reverse() method\ndq.reverse()\n\n# printing reverse deque\nprint (\" reverse deque is: \")\nprint (dq)``````\nOutput:\n``````The deque after reversing deque is :\ndeque([70, 80, 90, 60, 50, 40, 30, 20, 10])``````\n\n## Conclusion\n\nI hope you enjoyed reading this article and finally, you came to know about Python - Deque and its operations.\nTo know more about python programming language follow the insideaiml youtube channel.\nFor more such blogs/courses on data science, machine learning, artificial intelligence and emerging new technologies do visit us at InsideAIML.\nThanks for reading…\nHappy Learning…\n\nRecommended Course for you\n\nRecommended blog for you"
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https://stackoverflow.com/questions/4740748/when-to-use-common-table-expression-cte | [
"# When to use Common Table Expression (CTE)\n\nI have begun reading about Common Table Expression and cannot think of a use case where I would need to use them. They would seem to be redundant as the same can be done with derived tables. Is there something I am missing or not understanding well? Can someone give me a simple example of limitations with regular select, derived or temp table queries to make the case of CTE? Any simple examples would be highly appreciated.\n\nOne example, if you need to reference/join the same data set multiple times you can do so by defining a CTE. Therefore, it can be a form of code re-use.\n\nAn example of self referencing is recursion: Recursive Queries Using CTE\n\nFor exciting Microsoft definitions Taken from Books Online:\n\nA CTE can be used to:\n\n• Create a recursive query. For more information, see Recursive Queries Using Common Table Expressions.\n\n• Substitute for a view when the general use of a view is not required; that is, you do not have to store the definition in metadata.\n\n• Enable grouping by a column that is derived from a scalar subselect, or a function that is either not deterministic or has external access.\n\n• Reference the resulting table multiple times in the same statement.\n\n• Yep. You can't self join a derived table. Worth making the point that a self join on a CTE will still leave you with 2 separate invocations of it though. Jan 19 '11 at 21:11\n• @Martin - I am surprised. Can you back up that statement? Jan 19 '11 at 21:25\n• @John Thanks, I am finding 4guysfromrolla.com/webtech/071906-1.shtml quite useful too\n– imak\nJan 19 '11 at 21:25\n• @cyberkiwi - Which bit? That a self join will lead to 2 different invocations? See the example in this answer stackoverflow.com/questions/3362043/… Jan 19 '11 at 21:28\n• Interesting fact about CTE. I always wondered why NEWID() in the CTE changes when the CTE is referenced more than once. `select top 100 * into #tmp from master..spt_values order by 1,2,3,4 select A.number, COUNT(*) from #tmp A inner join #tmp B ON A.number = B.number+1 group by A.number` vs `with CTE AS (select top 100 * from master..spt_values order by 1,2,3,4) select A.number, COUNT(*) from CTE A inner join CTE B ON A.number = B.number+1 group by A.number` Jan 19 '11 at 21:37\n\nI use them to break up complex queries, especially complex joins and sub-queries. I find I'm using them more and more as 'pseudo-views' to help me get my head around the intent of the query.\n\nMy only complaint about them is they cannot be re-used. For example, I may have a stored proc with two update statements that could use the same CTE. But the 'scope' of the CTE is the first query only.\n\nTrouble is, 'simple examples' probably don't really need CTE's!\n\nStill, very handy.\n\n• ok. Can you make case with some relatively complex example that can help my head around this concept?\n– imak\nJan 19 '11 at 21:10\n• \"My only complaint about them is they cannot be re-used\" -- a CTE that you want to re-use should be considered a candidate for a `VIEW` :) Jan 20 '11 at 10:38\n• @onedaywhen: Understood, but that implies a global-scope I'm not always comfortable with. Sometimes within the scope of a proc I'd like to define a CTE and use it for selects and updates, or selects of similar data from different tables. Jun 5 '13 at 13:02\n• When I need the same CTE more than once, I feed it into a temporary table and then use the temporary table as much as I want. Jan 27 '16 at 5:32\n\nThere are two reasons I see to use cte's.\n\nTo use a calculated value in the where clause. This seems a little cleaner to me than a derived table.\n\nSuppose there are two tables - Questions and Answers joined together by Questions.ID = Answers.Question_Id (and quiz id)\n\n``````WITH CTE AS\n(\nSelect Question_Text,\nFROM Questions Q\n)\nSELECT * FROM CTE\n``````\n\nHere's another example where I want to get a list of questions and answers. I want the Answers to be grouped with the questions in the results.\n\n``````WITH cte AS\n(\nSELECT [Quiz_ID]\n,[ID] AS Question_Id\n,[Question_Text]\n,1 AS Is_Question\nFROM [Questions]\n\nUNION ALL\n\nSELECT Q.[Quiz_ID]\n,[Question_ID]\n,Q.Question_Text\n,0 AS Is_Question\nFROM [Answers] A INNER JOIN [Questions] Q ON Q.Quiz_ID = A.Quiz_ID AND Q.Id = A.Question_Id\n)\nSELECT\nQuiz_Id,\nQuestion_Id,\nIs_Question,\n(CASE WHEN Answer IS NULL THEN Question_Text ELSE Answer END) as Name\nFROM cte\norder by Quiz_Id, Question_Id, Is_Question Desc, Name\n``````\n• Can't your first example be simplified down to just use a nested query instead of the CTE?\n– Sam\nSep 17 '14 at 6:50\n• Both examples could be. May 12 '16 at 7:44\n• You should've added the first one without the CTE, then it's immediately apparent why is the latter useful.\n– Ufos\nJan 19 '17 at 13:36\n• `HAVING` is another way to do a late-stage filter which can be similar to using a sub-`SELECT` Jan 31 '20 at 18:49\n\nOne of the scenarios I found useful to use CTE is when you want to get DISTINCT rows of data based on one or more columns but return all columns in the table. With a standard query you might first have to dump the distinct values into a temp table and then try to join them back to the original table to retrieve the rest of the columns or you might write an extremely complex partition query that can return the results in one run but in most likelihood, it will be unreadable and cause performance issue.\n\nBut by using CTE (as answered by Tim Schmelter on Select the first instance of a record)\n\n``````WITH CTE AS(\nSELECT myTable.*\n, RN = ROW_NUMBER()OVER(PARTITION BY patientID ORDER BY ID)\nFROM myTable\n)\nSELECT * FROM CTE\nWHERE RN = 1\n``````\n\nAs you can see, this is much easier to read and maintain. And in comparison to other queries, is much better at performance.\n\nPerhaps its more meaningful to think of a CTE as a substitute for a view used for a single query. But doesn't require the overhead, metadata, or persistence of a formal view. Very useful when you need to:\n\n• Create a recursive query.\n• Use the CTE's resultset more than once in your query.\n• Promote clarity in your query by reducing large chunks of identical subqueries.\n• Enable grouping by a column derived in the CTE's resultset\n\nHere's a cut-and-paste example to play with:\n\n``````WITH [cte_example] AS (\nSELECT 1 AS [myNum], 'a num' as [label]\nUNION ALL\nSELECT [myNum]+1,[label]\nFROM [cte_example]\nWHERE [myNum] <= 10\n)\nSELECT * FROM [cte_example]\nUNION\nSELECT SUM([myNum]), 'sum_all' FROM [cte_example]\nUNION\nSELECT SUM([myNum]), 'sum_odd' FROM [cte_example] WHERE [myNum] % 2 = 1\nUNION\nSELECT SUM([myNum]), 'sum_even' FROM [cte_example] WHERE [myNum] % 2 = 0;\n``````\n\nEnjoy\n\nToday we are going to learn about Common table expression that is a new feature which was introduced in SQL server 2005 and available in later versions as well.\n\nCommon table Expression :- Common table expression can be defined as a temporary result set or in other words its a substitute of views in SQL Server. Common table expression is only valid in the batch of statement where it was defined and cannot be used in other sessions.\n\nSyntax of declaring CTE(Common table expression) :-\n\n``````with [Name of CTE]\nas\n(\nBody of common table expression\n)\n``````\n\nLets take an example :-\n\n``````CREATE TABLE Employee([EID] [int] IDENTITY(10,5) NOT NULL,[Name] [varchar](50) NULL)\n\ninsert into Employee(Name) values('Neeraj')\ninsert into Employee(Name) values('dheeraj')\ninsert into Employee(Name) values('shayam')\ninsert into Employee(Name) values('vikas')\ninsert into Employee(Name) values('raj')\n\nCREATE TABLE DEPT(EID INT,DEPTNAME VARCHAR(100))\ninsert into dept values(10,'IT')\ninsert into dept values(15,'Finance')\ninsert into dept values(25,'HR')\ninsert into dept values(10,'Payroll')\n``````\n\nI have created two tables employee and Dept and inserted 5 rows in each table. Now I would like to join these tables and create a temporary result set to use it further.\n\n``````With CTE_Example(EID,Name,DeptName)\nas\n(\nselect Employee.EID,Name,DeptName from Employee\ninner join DEPT on Employee.EID =DEPT.EID\n)\nselect * from CTE_Example\n``````\n\nLets take each line of the statement one by one and understand.\n\nTo define CTE we write \"with\" clause, then we give a name to the table expression, here I have given name as \"CTE_Example\"\n\nThen we write \"As\" and enclose our code in two brackets (---), we can join multiple tables in the enclosed brackets.\n\nIn the last line, I have used \"Select * from CTE_Example\" , we are referring the Common table expression in the last line of code, So we can say that Its like a view, where we are defining and using the view in a single batch and CTE is not stored in the database as an permanent object. But it behaves like a view. we can perform delete and update statement on CTE and that will have direct impact on the referenced table those are being used in CTE. Lets take an example to understand this fact.\n\n``````With CTE_Example(EID,DeptName)\nas\n(\nselect EID,DeptName from DEPT\n)\ndelete from CTE_Example where EID=10 and DeptName ='Payroll'\n``````\n\nIn the above statement we are deleting a row from CTE_Example and it will delete the data from the referenced table \"DEPT\" that is being used in the CTE.\n\n• I still don't get the point. What's the difference between this and just deleting from DEPT with exactly the same condition? It doesn't seem to make anything easier. May 7 '14 at 19:27\n• Please correct me if I'm wrong, but the execution plan may be different, and I think that is Neeraj's point, that there are many ways to achieve the same objective, but some will have advantages over others depending on the situation. For instance, it might be easier to read a CTE over a DELETE FROM statement in some circumstances, also the reverse might be true in others. Performance might improve or worsen. etc. Oct 11 '19 at 11:19\n\nIt is very useful when you want to perform an \"ordered update\".\n\nMS SQL does not allow you to use ORDER BY with UPDATE, but with help of CTE you can do it that way:\n\n``````WITH cte AS\n(\nSELECT TOP(5000) message_compressed, message, exception_compressed, exception\nFROM logs\nWHERE Id >= 5519694\nORDER BY Id\n)\nUPDATE cte\nSET message_compressed = COMPRESS(message), exception_compressed = COMPRESS(exception)\n``````\n\nLook here for more info: How to update and order by using ms sql\n\nOne point not pointed out yet, is the speed. I know it's an old answered question, but I think this deserves direct comment/answer:\n\nThey would seem to be redundant as the same can be done with derived tables\n\nWhen I used CTE the very first time I was absolutely stunned by it's speed. It was a case like from a textbook, very suitable for CTE, but in all ocurences I ever used CTE, there was a significant speed gain. My first query was complex with derived tables, taking long minutes to execute. With CTE it took fractions of seconds and left me shocked, that it is even possible.\n\n`````` ;with cte as\n(\nSelect Department, Max(salary) as MaxSalary\nfrom test\ngroup by department\n)\nselect t.* from test t join cte c on c.department=t.department\nwhere t.salary=c.MaxSalary;\n``````\n\ntry this"
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https://gradeup.co/torsion-of-circular-shafts-i-0bb3f8c4-bea6-11e5-ba24-afe88cbc1fef | [
"# Torsion Study Notes for Mechanical Engineering\n\nUpdated : Aug 21, 2019, 9:05\nBy : Vineet Vijay\n\nTorsion means twisting a structural Member when it is loaded by couple that Produces rotation about longitudinal axis.\n\n• If be the intensity of shear stress, on any layer at a distance r from the centre of shaft, then\n\nSign Convention\n\n• Sign convention of torque can be explained by right hand thumb rule.\n• A positive torque is that in which there is tightening effect of nut on the bolt. From either side of the cross-section. If torque is applied in the direction of right hand fingers than right hand thumbs direction represents movement of the nut.\n\nTMD = Torsion moment diagram\n\nT = Torque\n\n• Rate of twist :\n• Total angle of twist :\n\nWhere, T = Torque,\n\n• J = Polar moment of inertia\n• G = Modulus of rigidity,\n• θ = Angle of twist\n• L = Length of shaft,\n• GJ = Torsional rigidity\n• Torsional stiffness;\n• Torsional flexibility\n• Axial stiffness;\n• Axial flexibility\n\nMoment of Inertia About polar Axis:\n\n• For solid circular shaft,:\n• For hollow circular shaft:\n\nPower Transmitted in the Shaft\n\n• Power transmitted by shaft:\n\nWhere, N = Rotation per minute.\n\nCompound Shaft\n\nAn improved type of compound coupling for connecting in series and parallel are given below\n\n1. Series connection: Series connection of compound shaft as shown in figure. Due to series connection the torque on shaft 1 will be equal to shaft 2 and the total angular deformation will be equal to the sum of deformation of 1st shaft and 2nd shaft.\n\nTherefore,\n\nWhere,\n\nθ1 = Angular deformation of 1st shaft\n\nθ2 = Angular deformation of 2nd shaft\n\n1. Parallel connection: Parallel connection of compound shaft as shown in figure. Due to parallel connection of compound shaft the total torque will be equal to the sum of torque of shaft 1 and torque of shaft 2 and the deflection will be same in both the shafts.\n\nTherefore,\n\n### Strain energy (U) stored in shaft due to torsion:\n\n• G = Shear modulus\n• T = Torque\n• J = Moment of inertia about polar axis\n\n### Effect of Pure Bending on Shaft\n\nThe effect of pure bending on shaft can be defined by the relation for the shaft,\n\nWhere, σ = Principal stress\n\n• D = Diameter of shaft\n• M = Bending moment\n\n### Effect of Pure Torsion on Shaft\n\n• It can be calculated by the formula, which are given below\n\n• Where, τ = Torsion\n• D = Diameter of shaft\n\n### Combined effect of Moment & Torque\n\n• The equivalent bending moment Me may be defined as the bending moment which will produce the same direct stress as produced by the bending moment and torque acting separately.\n• Similarly, the equivalent torque Te, may be defined as the torque which will produce the same maximum shear stress as produced by the bending moment and torque acting separately.\n\nFor a circular shaft of diameter, d = 2r\n\nPrincipal stress\n\nMaximum Shear Stress\n\n• If P = 0\n\nPower transmitted by a shaft\n\n• Torque =\n\nWhere power is measured in watts and n is the speed in rpm\n\nShear Stress Distribution:\n\n• Solid Circulation Section:\n\n• Hollow Circulation Section\n\n• Composite Circular Section\n\n• Thin Tubular section: In view of small thickness-shear stress is assumed to be uniform\n\nThanks\n\nPrep Smart. Score Better. Go Gradeup!\n\nAug 21Mechanical Engg.\n\nPosted by:",
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"Vineet is a Community Manager for GATE and various other Engineering exams at Gradeup. A Mechanical engineer by profession, Vineet has a keen interest in guiding the pursuing/graduated engineers for successful careers and believes in illuminating their path through solving their queries. Accompanying his role in Gradeup, he also aims to develop content beneficial for all possible exams for engineers. His email address is [email protected].\nMember since Apr 2016",
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"GradeStack Learning Pvt. Ltd.Windsor IT Park, Tower - A, 2nd Floor,Sector 125, Noida,Uttar Pradesh [email protected]"
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"https://gs-post-images.grdp.co/2018/12/screenshot-2018-12-21-at-3-img1545387500435-75.png-rs-high-webp.png",
null,
"https://gradeup.co/torsion-of-circular-shafts-i-0bb3f8c4-bea6-11e5-ba24-afe88cbc1fef",
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https://docs.kde.org/trunk5/en/kmplot/kmplot/coords-config.html | [
"## Coordinate System Configuration\n\nTo open this dialog select ViewCoordinate System... from the menubar.",
null,
"### Axes Configuration\n\nHorizontal axis Range\n\nSets the range for the horizontal axis scale. Note that you can use the predefined functions and constants (see the section called “Predefined Function Names and Constants”) as the extremes of the range (e.g., set Min: to `2*pi`). You can even use functions you have defined to set the extremes of the axis range. For example, if you have defined a function `f(x) = x^2`, you could set Min: to `f(3)`, which would make the lower end of the range equal to 9.\n\nVertical axis Range\n\nSets the range for the vertical axis. See Horizontal axis Range above.\n\nHorizontal axis Grid Spacing\n\nThis controls the spacing between grid lines in the horizontal direction. If Automatic is selected, then KmPlot will try to find a grid line spacing of about two centimeters that is also numerically nice. If Custom: is selected, then you can enter the horizontal grid spacing. This value will be used regardless of the zoom. For example, if a value of 0.5 is entered, and the x range is 0 to 8, then 16 grid lines will be shown.\n\nVertical axis Grid Spacing\n\nThis controls the spacing between grid lines in the vertical direction. See Horizontal axis Grid Spacing above."
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"https://docs.kde.org/trunk5/en/kmplot/kmplot/settings-coords.png",
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https://pldb.com/languages/eulisp.html | [
"# EuLisp\n\nEuLisp is a pl created in 1985.\n\n #178on PLDB 37Years Old 90Users 0Books 0Papers\n\nEuLisp is a statically and dynamically scoped Lisp dialect developed by a loose formation of industrial and academic Lisp users and developers from around Europe. The standardizers intended to create a new Lisp \"less encumbered by the past\" (compared to Common Lisp), and not so minimalist as Scheme. Another objective was to integrate the object-oriented programming paradigm well. Read more on Wikipedia...\n\nExample from Wikipedia:\n```(defmodule hanoi (syntax (syntax-0) import (level-0) export (hanoi)) ;;;------------------------------------------------- ;;; Tower definition ;;;------------------------------------------------- (defconstant *max-tower-height* 10) (defclass <tower> () ((id reader: tower-id keyword: id:) (blocks accessor: tower-blocks))) (defun build-tower (x n) (labels ((loop (i res) (if (= i 0) res (loop (- i 1) (cons i res))))) ((setter tower-blocks) x (loop n ())) x)) (defmethod generic-print ((x <tower>) (s <stream>)) (sformat s \"#<tower ~a: ~a>\" (tower-id x) (tower-blocks x))) ;;;------------------------------------------------- ;;; Access to tower blocks ;;;------------------------------------------------- (defgeneric push (x y)) (defmethod push ((x <tower>) (y <fpi>)) (let ((blocks (tower-blocks x))) (if (or (null? blocks) (< y (car blocks))) ((setter tower-blocks) x (cons y blocks)) (error <condition> (fmt \"cannot push block of size ~a on tower ~a\" y x))))) (defgeneric pop (x)) (defmethod pop ((x <tower>)) (let ((blocks (tower-blocks x))) (if blocks (progn ((setter tower-blocks) x (cdr blocks)) (car blocks)) (error <condition> (fmt \"cannot pop block from empty tower ~a\" x))))) ;;;------------------------------------------------- ;;; Move n blocks from tower x1 to tower x2 using x3 as buffer ;;;------------------------------------------------- (defgeneric move (n x1 x2 x3)) (defmethod move ((n <fpi>) (x1 <tower>) (x2 <tower>) (x3 <tower>)) (if (= n 1) (progn (push x2 (pop x1)) (print x1 nl x2 nl x3 nl nl)) (progn (move (- n 1) x1 x3 x2) (move 1 x1 x2 x3) (move (- n 1) x3 x2 x1)))) ;;;------------------------------------------------- ;;; Initialize and run the 'Towers of Hanoi' ;;;------------------------------------------------- (defun hanoi () (let ((x1 (make <tower> id: 0)) (x2 (make <tower> id: 1)) (x3 (make <tower> id: 2))) (build-tower x1 *max-tower-height*) (build-tower x2 0) (build-tower x3 0) (print x1 nl x2 nl x3 nl nl) (move *max-tower-height* x1 x2 x3))) (hanoi) ;;;------------------------------------------------- ) ;; End of module hanoi ;;;-------------------------------------------------```\n\n#### Language features\n\nFeature Supported Example Token\n`; A comment`\n`; A comment`"
]
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| {"ft_lang_label":"__label__en","ft_lang_prob":0.57707745,"math_prob":0.7526293,"size":2770,"snap":"2022-40-2023-06","text_gpt3_token_len":735,"char_repetition_ratio":0.28488794,"word_repetition_ratio":0.032,"special_character_ratio":0.47509027,"punctuation_ratio":0.147806,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.974542,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-10-01T07:54:36Z\",\"WARC-Record-ID\":\"<urn:uuid:0c4f6826-97cf-4866-8f5e-dab95c0e4bde>\",\"Content-Length\":\"11134\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:db2aa523-7043-495d-ac23-ab9adba4c7a2>\",\"WARC-Concurrent-To\":\"<urn:uuid:a2e42c6b-085b-45a1-bc16-809b2f8224fe>\",\"WARC-IP-Address\":\"143.198.107.186\",\"WARC-Target-URI\":\"https://pldb.com/languages/eulisp.html\",\"WARC-Payload-Digest\":\"sha1:OTX4X3AETQ2WXNFNU6VMFY4LTAWVRQ5H\",\"WARC-Block-Digest\":\"sha1:3LZQDGUDEO5L7PJ2VXEAU5PKQBBUPZES\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-40/CC-MAIN-2022-40_segments_1664030335573.50_warc_CC-MAIN-20221001070422-20221001100422-00233.warc.gz\"}"} |
https://physics.stackexchange.com/questions/233013/induced-current-i | [
"induced Current I\n\nI am working with the Problem which I've seen in a book. That Problem was : https://physics.stackexchange.com/questions/227582/direction-of-induced-current\n\nI can't find any meaningful explanation for this problem in any source. So, I am working with this problem. The answer of the first question is No Current is induced in the chord between the loop. And the Answer of the second question is No Current induced in connector. And in all the current is in clockwise direction.\n\nHere, I have many questions regarding Electromagnetic Induction, and I don't think any theory provides its answer. If there is a theory please help me. I am a 12th grade student but cannot find the answer of these questions.\n\nIn my efforts to answer these problem,\n\n• I am confused how to apply lenz's law here?\n\n• Sudden Question arises, why Current or EMF is not generetad in unclosed wire. I know Maxwell's equation is defined on closed integral, but it doesn't means that unclosed wire don't generate current and emf.\n\n• If we are not going to apply Lenz's law, we should take interest in Induced Electric Field. (Maybe Electric field is zero in that connectors. But How?)\n\nI 've read Halliday, Rennik, Walker's Fundamental Physics extended on Google books. But still confused about Induced electric field. How Symmetry implies Direction of Induced Electric field is tangential.\n\nI know I've many questions and should not type it here together. But My main question is to find the direction of induced current in the figure given in the link.\n\n• Another Question is what about induced emf and induced current in a circuit rather than a loop? Are there any answers?\n\nFirst realise that you can induce an emf in any closed loop. Only if the loop is conducting do you get an induced current.\n\nThe direction of the induced current will always be in such a direction of oppose the change producing it.\n\nHere is a relatively simple way of assigning an induced current",
null,
"Diagram 1 There is a loop with a decreasing B-field into the screen.\n\nDiagram 2 Lenz tells you that the induced current must try and counteract that decreasing B-field into the screen by producing a B-field out of the screen.\n\nDiagram 3 Use the right hand grip rule to assign the direction of the current.\n\nNow your problems require a bit more thought eg as to how to deal with the link between the outer and inner ring in the left hand diagram. I suggest you apply Lenz to each of the semicircles and see what you get.\n\nYou can do an great deal with these simple ideas and you do not need to go back all the way to Maxwell.\n\nSo for your first question see what Lenz predicts for these three loops.",
null,
"• Please Explain How I apply Lenz to each diagram. I am just confused whether I am right or wrong. – Hardey Pandya Feb 1 '16 at 12:25\n• Do you mean the diagrams in your problem? – Farcher Feb 1 '16 at 12:56\n• Yeah. Do you mean current in opposite directions will vanish by themselves? – Hardey Pandya Feb 1 '16 at 13:05\n• If you think that the currents and equal in magnitude but opposite in direction then yes the current is zero. – Farcher Feb 1 '16 at 13:08\n• What is actual phenomenon on micro level then ? What happens to electrons flowing in the conductor ? They repel each other ? – Hardey Pandya Feb 1 '16 at 13:11"
]
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"https://i.stack.imgur.com/qoGhe.jpg",
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"https://i.stack.imgur.com/5cwpb.jpg",
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https://www.geeksforgeeks.org/how-to-color-scatter-plot-points-in-r/ | [
"# How to Color Scatter Plot Points in R ?\n\n• Last Updated : 30 May, 2021\n\nA scatter plot is a set of dotted points to represent individual pieces of data in the horizontal and vertical axis. But by default, the color of these points is black and sometimes there might be a need to change the color of these points.\n\nIn this article, we will discuss how to change the color of points in scatterplot in the R Programming Language.\n\n### Method 1: Using plot()\n\nThe simple scatterplot is created using the plot() function.\n\nSyntax: plot(x, y, main, xlab, ylab, xlim, ylim, axes)\n\nLet us first create a scatterplot without any color so that the difference is apparent.\n\nExample:\n\n## R\n\n `df<-``read.csv``(``\"bestsellers.csv\"``)`` ` `plot``(df\\$Reviews,df\\$Price,pch=16)`\n\nOutput:",
null,
"Now to change the colors of a scatterplot using plot(), simply select the column on basis of which different colors should be assigned to various points. Pass the column that will help differentiate between points to “col” attribute.\n\nExample:\n\n## R\n\n `df<-``read.csv``(``\"bestsellers.csv\"``)`` ` `plot``(df\\$Reviews,df\\$Price,pch=16,col=df\\$Genre)`\n\nOutput:",
null,
"### Method 2: Using ggplot\n\nggplot2 module supports geom_point() function that can help plot scatterplot. Let us first see how a scatterplot will appear without providing any mechanism for changing colors.\n\nExample:\n\n## R\n\n `library``(``\"ggplot2\"``)`` ` `df<-``read.csv``(``\"bestsellers.csv\"``)`` ` `ggplot``(df,``aes``(x=Reviews,y=Price))+``geom_point``()`\n\nOutput:",
null,
"By using, ggplot there are various ways of adding colors to a scatterplot. Let us first discuss how colors are changed by default. For this simply pass the differentiating column to col attribute.\n\nExample:\n\n## R\n\n `library``(``\"ggplot2\"``)`` ` `df<-``read.csv``(``\"bestsellers.csv\"``)`` ` `ggplot``(df,``aes``(x=Reviews,y=Price,col=Genre))+``geom_point``()`\n\nOutput:",
null,
"Another way of producing the same result is to provide the grouping column to group attribute and again to color in geom_point()\n\nExample:\n\n## R\n\n `library``(``\"ggplot2\"``)`` ` `df<-``read.csv``(``\"bestsellers.csv\"``)`` ` `ggplot``(df,``aes``(x=Reviews,y=Price,group=Genre))+`` ``geom_point``(``aes``(color=Genre))`\n\nOutput:",
null,
"We can also add custom colors by using scale_color_manual() function with the list of colors to chose from.\n\nExample\n\n## R\n\n `library``(``\"ggplot2\"``)`` ` `df<-``read.csv``(``\"bestsellers.csv\"``)`` ` `ggplot``(df,``aes``(x=Reviews,y=Price,group=Genre))+`` ``geom_point``(``aes``(color=Genre))+`` ``scale_color_manual``(values=``c``(``'Yellow'``,``'Green'``))`\n\nOutput:",
null,
"A scatterplot can also display colors only from grey scale, for this use scale_color_grey() function.\n\nExample:\n\n## R\n\n `library``(``\"ggplot2\"``)`` ` `df<-``read.csv``(``\"bestsellers.csv\"``)`` ` `ggplot``(df,``aes``(x=Reviews,y=Price,group=Genre))+`` ``geom_point``(``aes``(color=Genre))+`` ``scale_color_grey``()`\n\nOutput:",
null,
"scale_color_brewer() function is also a method to add colors to a scatterplot. This function takes the name of the palette to pick colors from.\n\nExample:\n\n## R\n\n `library``(``\"ggplot2\"``)`` ` `df<-``read.csv``(``\"bestsellers.csv\"``)`` ` `ggplot``(df,``aes``(x=Reviews,y=Price,group=Genre))+`` ``geom_point``(``aes``(color=Genre))+`` ``scale_color_brewer``(palette=``\"Accent\"``)`\n\nOutput:",
null,
"My Personal Notes arrow_drop_up"
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https://www.dominaexcel.cl/image-frames/ | [
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"×\n\nCARACTERISTICAS DEL CURSO DE EXCEL BASICO\n\n• Más de 100 ejercicios por nivel aplicados al ámbito empresarial\n• Trabajo con un relator experto\n• Guías de ejercicios y funciones\n• Manual del Curso\n• Plataforma de Capacitación"
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https://mtrading.com/education/articles/forex-basics/the-rule-of-72-in-finance-explained-for-beginners | [
"• Products\n• Partnership\n• Promotions\n• Analysis & Education\n• Market Insights\n• Education\n• Legal\nEn\nEnglish\nLogout\nAre you sure you want to exist?\n\n# The Rule of 72 in Finance Explained for Beginners\n\nThe Rule of 72 in finance is generally associated with a formula that is used to estimate how much time it will take for investment to double its value. The rule of 72 formula is used when earning a fixed rate of return per year. The easiest way to calculate the shortcut is to divide 72 by the interest rate you get annually.",
null,
"In this article, we will show a real rule of 72 example as well as explain how the formula works and how to use it when calculating potential returns.\n\n## What Is the Rule of 72?\n\nOnce again, the formula makes it possible to calculate how many years you may need to double your investments in value when considering a given annual rate of return. Additionally, the calculation can serve as the instrument to estimate the rate of annual compounded return.\n\nOf course, one can benefit from niche-specific software like Microsoft Excel, Google Sheets, and other spreadsheet apps. However, they work more for precise calculation while the rule of 72 is more like a mental tool to benefit from instant results when identifying the approximate value.\n\nDeposit bonus\nup to 200%",
null,
"from 0 pips",
null,
"Awarded Copy Trading platform",
null,
"This is what makes the formula a great tool for beginner traders. It teaches them to both comprehend and calculate in real time without using side applications or tools. What’s more, the rule is used by financial institutions and organizations, for example, the Security and Exchange Commission for maintaining grade-level financial literacy channels.\n\n### The Story behind the Rule of 72\n\nThe rule originated in 1494. Luca Pacioli was the first one to introduce it in his mathematical work known as Summa de Arithmetica. The main issue here is that Pacioli does not provide any explanation of how he worked out the formula. Also, you will not find its explanation in his book. So, some people believe that the mathematician used the formula that pre-dates his work.\n\n#### The Rule of 72 Calculation\n\nThe rule is very simple to use. All you need is to divide 72 by the annual return generated by your investments. In the end, you will know how many years it will take you to double your original income.\n\nThe Rule of 72 example: let’s say, you are planning to invest in a security that promises 8% of the annual return. The calculation goes like this: 72/8 = 9. It means you will need 9 years to double your investment.\n\nThe cool thing about the formula is the ability to leverage it differently. You can use two ways to estimate the period needed to double the required rate of return. Here are these two methods:\n\n1. Time to Double – the first approach supposes determining doubling time. To calculate it, you need to divide 72 by the expected rate of return (as in example shown earlier). The main factor here is that the formula uses only a single average rate throughout the initial investment lifespan.\n2. Expected Rate of Return – another approach is to calculate the expected rate of return rather than timeframes. Here you will need to divide 72 by the years to double the investment. The number of years should not necessarily be a whole number. You can easily complete the calculation using portions or fractions of the year. The good news is that the result you get will cover compounding interest rate over the whole investment holding period.\n\n## The Bottom Line\n\nThe Rule of 72 can be a simple and useful tool to quickly calculate potential time needed to double your investment as well as expected rate of return. The formula is very simple, which makes it a good option for beginners whenever they need to estimate the potential profit in the long run when deciding whether to select an asset or not.\n\nThis material does not contain and should not be construed as containing investment advice, investment recommendations, an offer of or solicitation for any transactions in financial instruments. Before making any investment decisions, you should seek advice from independent financial advisors to ensure you understand the risks."
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| {"ft_lang_label":"__label__en","ft_lang_prob":0.94978124,"math_prob":0.8942739,"size":3052,"snap":"2022-40-2023-06","text_gpt3_token_len":607,"char_repetition_ratio":0.118766405,"word_repetition_ratio":0.007677543,"special_character_ratio":0.19888598,"punctuation_ratio":0.08462867,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.97960836,"pos_list":[0,1,2,3,4,5,6,7,8],"im_url_duplicate_count":[null,2,null,null,null,null,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-02-06T23:34:55Z\",\"WARC-Record-ID\":\"<urn:uuid:46b5a528-d247-4ae2-adef-5a37ba2a2521>\",\"Content-Length\":\"32926\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:23ff340f-60b4-4a95-9259-08410a3d7679>\",\"WARC-Concurrent-To\":\"<urn:uuid:88aaf5d5-12b6-4ef2-931e-aa83364f30bc>\",\"WARC-IP-Address\":\"190.115.31.249\",\"WARC-Target-URI\":\"https://mtrading.com/education/articles/forex-basics/the-rule-of-72-in-finance-explained-for-beginners\",\"WARC-Payload-Digest\":\"sha1:YZJTJCDD4P4RTRVDUJY77CAYDRG6QLEB\",\"WARC-Block-Digest\":\"sha1:GTA3N3LPU7ZML7MJ6FMU3V3P53CPRH3S\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-06/CC-MAIN-2023-06_segments_1674764500365.52_warc_CC-MAIN-20230206212647-20230207002647-00710.warc.gz\"}"} |
https://www.jbwere.com.au/resources/jbwere-cash-account-rates | [
"# JBWere Cash Account Rates\n\n## 4.10% p.a. ^\n\nThe JBWere Cash Account Interest Rate has increased to 4.10% p.a.\n\nThe rate is effective 8 November 2023.\n\n^This interest rate is payable net of management costs (which vary up to 0.70% pa) and may vary depending on a range of factors, including (but not limited to) changes in the cash rate target determined by the Reserve Bank of Australia.\n\n## Historical cash account interest rates\n\n3.85% p.a.7 June 2023\n\nFollowing the increase of the RBA cash rate, the JBWere Cash Account interest rate will increase to 3.85% p.a.\n\nThis increased rate was effective 7 June 2023.\n\n3.60% p.a.3 May 2023\n\nFollowing the increase of the RBA cash rate, the JBWere Cash Account interest rate will increase to 3.60% p.a.\n\nThis increased rate was effective 3 May 2023.\n\n3.35% p.a.8 March 2023\n\nFollowing the increase of the RBA cash rate, the JBWere Cash Account interest rate will increase to 3.35% p.a.\n\nThis increased rate was effective 8 March 2023.\n\n3.10% p.a.8 February 2023\n\nFollowing the increase of the RBA cash rate, the JBWere Cash Account interest rate will increase to 3.10% p.a.\n\nThis increased rate was effective 8 February 2023.\n\n2.85% p.a.7 December 2022\n\nFollowing the increase of the RBA cash rate, the JBWere Cash Account interest rate will increase to 2.85% p.a.\n\nThis increased rate was effective 7 December 2022.\n\n2.65% p.a.2 November 2022\n\nFollowing the increase of the RBA cash rate, the JBWere Cash Account interest rate will increase to 2.65% p.a.\n\nThis increased rate was effective 2 November 2022.\n\n2.40% p.a.5 October 2022\n\nFollowing the increase of the RBA cash rate, the JBWere Cash Account interest rate will increase to 2.40% p.a.\n\nThis increased rate was effective 5 October 2022.\n\n2.15% p.a.7 September 2022\n\nFollowing the increase of the RBA cash rate, the JBWere Cash Account interest rate will increase to 2.15% p.a.\n\nThis increased rate was effective 7 September 2022.\n\n1.65% p.a.3 August 2022\n\nFollowing the increase of the RBA cash rate, the JBWere Cash Account interest rate will increase to 1.65% p.a.\n\nThis increased rate was effective 3 August 2022.\n\n1.25% p.a.6 July 2022\n\nFollowing the increase of the RBA cash rate, the JBWere Cash Account interest rate will increase to 1.25% p.a.\n\nThis increased rate was effective 6 July 2022.\n\n0.80% p.a.8 June 2022\n\nFollowing the increase of the RBA cash rate, the JBWere Cash Account interest rate will increase to 0.80% p.a.\n\nThis increased rate was effective 8 June 2022.\n\n0.30% p.a4 May 2022\n\nFollowing the increase of the RBA cash rate, the JBWere Cash Account interest rate will increase to 0.30% p.a.\n\nThis increased rate was effective 4 May 2022.\n\n0.05% p.a.1 July 2021\n\nFollowing the reduction of the RBA cash rate, the JBWere Cash Account interest rate will decrease to 0.05% p.a.\n\nThis reduced rate was effective 1 July 2021.\n\n0.11% p.a.4 November 2020\n\nFollowing the reduction of the RBA cash rate, the JBWere Cash Account interest rate will decrease to 0.11% p.a.\n\nThis reduced rate was effective 4 November 2020.\n\n0.26% p.a.20 March 2020Cash Account first offered."
]
| [
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| {"ft_lang_label":"__label__en","ft_lang_prob":0.93002605,"math_prob":0.747769,"size":3038,"snap":"2023-40-2023-50","text_gpt3_token_len":883,"char_repetition_ratio":0.22247858,"word_repetition_ratio":0.4448743,"special_character_ratio":0.31468073,"punctuation_ratio":0.17329545,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.96717817,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-12-10T17:22:01Z\",\"WARC-Record-ID\":\"<urn:uuid:f037eb41-aced-41f4-9ef3-2fdb03182c51>\",\"Content-Length\":\"34575\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:a1db217e-a09b-4e78-95e1-e184bf57062c>\",\"WARC-Concurrent-To\":\"<urn:uuid:61e48133-f9b7-47a9-a491-cd3e06f288c3>\",\"WARC-IP-Address\":\"104.96.227.212\",\"WARC-Target-URI\":\"https://www.jbwere.com.au/resources/jbwere-cash-account-rates\",\"WARC-Payload-Digest\":\"sha1:OSF4OH75XK7KL5T3A6MPQIGCKQUHXXOD\",\"WARC-Block-Digest\":\"sha1:QZPJOLZC7HKRG4FLFAPILI2DTVVTNS3D\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-50/CC-MAIN-2023-50_segments_1700679102612.80_warc_CC-MAIN-20231210155147-20231210185147-00707.warc.gz\"}"} |
https://www.w3cschool.cn/coffeescript/pezi1oca.html | [
"# CoffeeScript 策略模式\n\n2022-06-29 17:16 更新\n\n## 解决方案\n\n### 基类\n\n``````StringSorter = (algorithm) ->\nsort: (list) -> algorithm list``````\n\n### 策略\n\n``````bubbleSort = (list) ->\nanySwaps = false\nswapPass = ->\nfor r in [0..list.length-2]\nif list[r] > list[r+1]\nanySwaps = true\n[list[r], list[r+1]] = [list[r+1], list[r]]\n\nswapPass()\nwhile anySwaps\nanySwaps = false\nswapPass()\nlist\n\nreverseBubbleSort = (list) ->\nanySwaps = false\nswapPass = ->\nfor r in [list.length-1..1]\nif list[r] < list[r-1]\nanySwaps = true\n[list[r], list[r-1]] = [list[r-1], list[r]]\n\nswapPass()\nwhile anySwaps\nanySwaps = false\nswapPass()\nlist``````\n\n### 使用策略\n\n``````sorter = new StringSorter bubbleSort\n\nunsortedList = ['e', 'b', 'd', 'c', 'x', 'a']\n\nsorter.sort unsortedList\n\n# => ['a', 'b', 'c', 'd', 'e', 'x']\n\nunsortedList.push 'w'\n\n# => ['a', 'b', 'c', 'd', 'e', 'x', 'w']\n\nsorter.algorithm = reverseBubbleSort\n\nsorter.sort unsortedList\n\n# => ['a', 'b', 'c', 'd', 'e', 'w', 'x']``````\n\n## 讨论\n\n“没有作战计划在第一次接触敌人时便能存活下来。” 用户如是，但是我们可以运用从变化的情况中获得的知识来做出适应改变。在示例末尾，例如，数组中的最新项是乱序排列的，知道了这个细节，我们便可以通过切换算法来加速排序，只要简单地重赋值就可以了。\n\n## 练习\n\n• 将StringSorter扩展为AlwaysSortedArray类来实现规则序列的所有功能，但是要基于插入方法自动分类新的项（例如push对比shift）。\n\nApp下载",
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"",
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]
| [
null,
"https://7n.w3cschool.cn/statics/images/w3c/app-qrcode2.png",
null,
"https://7n.w3cschool.cn/statics/images/w3c/mp-qrcode.png",
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| {"ft_lang_label":"__label__zh","ft_lang_prob":0.51350915,"math_prob":0.6820505,"size":1214,"snap":"2023-40-2023-50","text_gpt3_token_len":628,"char_repetition_ratio":0.1446281,"word_repetition_ratio":0.27083334,"special_character_ratio":0.33031303,"punctuation_ratio":0.18811882,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9641203,"pos_list":[0,1,2,3,4],"im_url_duplicate_count":[null,null,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-10-02T17:39:03Z\",\"WARC-Record-ID\":\"<urn:uuid:59ba02b9-5baf-4bd6-8ded-50145769ccef>\",\"Content-Length\":\"68173\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:d8c78f70-4f36-47df-ab47-20b687f28646>\",\"WARC-Concurrent-To\":\"<urn:uuid:56517c8d-b31e-48ad-ab3d-2c24e445bf97>\",\"WARC-IP-Address\":\"120.79.88.157\",\"WARC-Target-URI\":\"https://www.w3cschool.cn/coffeescript/pezi1oca.html\",\"WARC-Payload-Digest\":\"sha1:GLFHIJS6MZDA4K77KFOAFEAGGUKRT5GI\",\"WARC-Block-Digest\":\"sha1:VCTZLL7WJHE7CGAMOG22CHBA3MNS7YHR\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-40/CC-MAIN-2023-40_segments_1695233511002.91_warc_CC-MAIN-20231002164819-20231002194819-00607.warc.gz\"}"} |
https://www.colorhexa.com/8d9c62 | [
"# #8d9c62 Color Information\n\nIn a RGB color space, hex #8d9c62 is composed of 55.3% red, 61.2% green and 38.4% blue. Whereas in a CMYK color space, it is composed of 9.6% cyan, 0% magenta, 37.2% yellow and 38.8% black. It has a hue angle of 75.5 degrees, a saturation of 22.8% and a lightness of 49.8%. #8d9c62 color hex could be obtained by blending #ffffc4 with #1b3900. Closest websafe color is: #999966.\n\n• R 55\n• G 61\n• B 38\nRGB color chart\n• C 10\n• M 0\n• Y 37\n• K 39\nCMYK color chart\n\n#8d9c62 color description : Mostly desaturated dark green.\n\n# #8d9c62 Color Conversion\n\nThe hexadecimal color #8d9c62 has RGB values of R:141, G:156, B:98 and CMYK values of C:0.1, M:0, Y:0.37, K:0.39. Its decimal value is 9280610.\n\nHex triplet RGB Decimal 8d9c62 `#8d9c62` 141, 156, 98 `rgb(141,156,98)` 55.3, 61.2, 38.4 `rgb(55.3%,61.2%,38.4%)` 10, 0, 37, 39 75.5°, 22.8, 49.8 `hsl(75.5,22.8%,49.8%)` 75.5°, 37.2, 61.2 999966 `#999966`\nCIE-LAB 61.931, -15.219, 28.634 25.077, 30.321, 16.086 0.351, 0.424, 30.321 61.931, 32.428, 117.991 61.931, -6.375, 38.929 55.065, -15.073, 21.225 10001101, 10011100, 01100010\n\n# Color Schemes with #8d9c62\n\n• #8d9c62\n``#8d9c62` `rgb(141,156,98)``\n• #71629c\n``#71629c` `rgb(113,98,156)``\nComplementary Color\n• #9c8e62\n``#9c8e62` `rgb(156,142,98)``\n• #8d9c62\n``#8d9c62` `rgb(141,156,98)``\n• #709c62\n``#709c62` `rgb(112,156,98)``\nAnalogous Color\n• #8e629c\n``#8e629c` `rgb(142,98,156)``\n• #8d9c62\n``#8d9c62` `rgb(141,156,98)``\n• #62709c\n``#62709c` `rgb(98,112,156)``\nSplit Complementary Color\n• #9c628d\n``#9c628d` `rgb(156,98,141)``\n• #8d9c62\n``#8d9c62` `rgb(141,156,98)``\n• #628d9c\n``#628d9c` `rgb(98,141,156)``\n• #9c7162\n``#9c7162` `rgb(156,113,98)``\n• #8d9c62\n``#8d9c62` `rgb(141,156,98)``\n• #628d9c\n``#628d9c` `rgb(98,141,156)``\n• #71629c\n``#71629c` `rgb(113,98,156)``\n• #636d44\n``#636d44` `rgb(99,109,68)``\n• #717d4e\n``#717d4e` `rgb(113,125,78)``\n• #7f8c58\n``#7f8c58` `rgb(127,140,88)``\n• #8d9c62\n``#8d9c62` `rgb(141,156,98)``\n• #98a671\n``#98a671` `rgb(152,166,113)``\n• #a4b081\n``#a4b081` `rgb(164,176,129)``\n• #afba91\n``#afba91` `rgb(175,186,145)``\nMonochromatic Color\n\n# Alternatives to #8d9c62\n\nBelow, you can see some colors close to #8d9c62. Having a set of related colors can be useful if you need an inspirational alternative to your original color choice.\n\n• #9c9c62\n``#9c9c62` `rgb(156,156,98)``\n• #979c62\n``#979c62` `rgb(151,156,98)``\n• #929c62\n``#929c62` `rgb(146,156,98)``\n• #8d9c62\n``#8d9c62` `rgb(141,156,98)``\n• #889c62\n``#889c62` `rgb(136,156,98)``\n• #839c62\n``#839c62` `rgb(131,156,98)``\n• #7f9c62\n``#7f9c62` `rgb(127,156,98)``\nSimilar Colors\n\n# #8d9c62 Preview\n\nThis text has a font color of #8d9c62.\n\n``<span style=\"color:#8d9c62;\">Text here</span>``\n#8d9c62 background color\n\nThis paragraph has a background color of #8d9c62.\n\n``<p style=\"background-color:#8d9c62;\">Content here</p>``\n#8d9c62 border color\n\nThis element has a border color of #8d9c62.\n\n``<div style=\"border:1px solid #8d9c62;\">Content here</div>``\nCSS codes\n``.text {color:#8d9c62;}``\n``.background {background-color:#8d9c62;}``\n``.border {border:1px solid #8d9c62;}``\n\n# Shades and Tints of #8d9c62\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, #0a0b07 is the darkest color, while #fffffe is the lightest one.\n\n• #0a0b07\n``#0a0b07` `rgb(10,11,7)``\n• #15170f\n``#15170f` `rgb(21,23,15)``\n• #202416\n``#202416` `rgb(32,36,22)``\n• #2b301e\n``#2b301e` `rgb(43,48,30)``\n• #363c25\n``#363c25` `rgb(54,60,37)``\n• #41482d\n``#41482d` `rgb(65,72,45)``\n• #4c5435\n``#4c5435` `rgb(76,84,53)``\n• #57603c\n``#57603c` `rgb(87,96,60)``\n• #616c44\n``#616c44` `rgb(97,108,68)``\n• #6c784b\n``#6c784b` `rgb(108,120,75)``\n• #778453\n``#778453` `rgb(119,132,83)``\n• #82905a\n``#82905a` `rgb(130,144,90)``\n• #8d9c62\n``#8d9c62` `rgb(141,156,98)``\n• #96a46e\n``#96a46e` `rgb(150,164,110)``\n• #9fab7a\n``#9fab7a` `rgb(159,171,122)``\n• #a7b386\n``#a7b386` `rgb(167,179,134)``\n• #b0bb92\n``#b0bb92` `rgb(176,187,146)``\n• #b9c29e\n``#b9c29e` `rgb(185,194,158)``\n• #c1caaa\n``#c1caaa` `rgb(193,202,170)``\n``#cad1b6` `rgb(202,209,182)``\n• #d3d9c2\n``#d3d9c2` `rgb(211,217,194)``\n• #dce0ce\n``#dce0ce` `rgb(220,224,206)``\n• #e4e8da\n``#e4e8da` `rgb(228,232,218)``\n• #edefe6\n``#edefe6` `rgb(237,239,230)``\n• #f6f7f2\n``#f6f7f2` `rgb(246,247,242)``\n• #fffffe\n``#fffffe` `rgb(255,255,254)``\nTint Color Variation\n\n# Tones of #8d9c62\n\nA tone is produced by adding gray to any pure hue. In this case, #848876 is the less saturated color, while #bcfe00 is the most saturated one.\n\n• #848876\n``#848876` `rgb(132,136,118)``\n• #88926c\n``#88926c` `rgb(136,146,108)``\n• #8d9c62\n``#8d9c62` `rgb(141,156,98)``\n• #92a658\n``#92a658` `rgb(146,166,88)``\n• #96b04e\n``#96b04e` `rgb(150,176,78)``\n• #9bb945\n``#9bb945` `rgb(155,185,69)``\n• #a0c33b\n``#a0c33b` `rgb(160,195,59)``\n• #a5cd31\n``#a5cd31` `rgb(165,205,49)``\n• #a9d727\n``#a9d727` `rgb(169,215,39)``\n• #aee01e\n``#aee01e` `rgb(174,224,30)``\n• #b3ea14\n``#b3ea14` `rgb(179,234,20)``\n• #b7f40a\n``#b7f40a` `rgb(183,244,10)``\n• #bcfe00\n``#bcfe00` `rgb(188,254,0)``\nTone Color Variation\n\n# Color Blindness Simulator\n\nBelow, you can see how #8d9c62 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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| {"ft_lang_label":"__label__en","ft_lang_prob":0.53504026,"math_prob":0.65277386,"size":3730,"snap":"2019-26-2019-30","text_gpt3_token_len":1695,"char_repetition_ratio":0.12614064,"word_repetition_ratio":0.011070111,"special_character_ratio":0.5522788,"punctuation_ratio":0.23756906,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9768456,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-07-17T22:25:42Z\",\"WARC-Record-ID\":\"<urn:uuid:68ebb1a6-b4de-4eb2-95ea-4269d2a9bef3>\",\"Content-Length\":\"36473\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:84789503-7a91-4d59-b3db-a9de7b7ac2d5>\",\"WARC-Concurrent-To\":\"<urn:uuid:81145a59-1fcf-4896-9d0c-4930d3d0529a>\",\"WARC-IP-Address\":\"178.32.117.56\",\"WARC-Target-URI\":\"https://www.colorhexa.com/8d9c62\",\"WARC-Payload-Digest\":\"sha1:EFFYLBVAY52AFSPTOPVFUNZYYIBAQ4L3\",\"WARC-Block-Digest\":\"sha1:ZBKIBY6JX2VEH5AJHVLAJ2XMVGBCFTT7\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-30/CC-MAIN-2019-30_segments_1563195525414.52_warc_CC-MAIN-20190717221901-20190718003901-00460.warc.gz\"}"} |
https://answers.everydaycalculation.com/multiply-fractions/84-3-times-45-49 | [
"Solutions by everydaycalculation.com\n\n## Multiply 84/3 with 45/49\n\n1st number: 28 0/3, 2nd number: 45/49\n\nThis multiplication involving fractions can also be rephrased as \"What is 84/3 of 45/49?\"\n\n84/3 × 45/49 is 180/7.\n\n#### Steps for multiplying fractions\n\n1. Simply multiply the numerators and denominators separately:\n2. 84/3 × 45/49 = 84 × 45/3 × 49 = 3780/147\n3. After reducing the fraction, the answer is 180/7\n4. In mixed form: 255/7\n\nMathStep (Works offline)",
null,
"Download our mobile app and learn to work with fractions in your own time:\nAndroid and iPhone/ iPad\n\n×"
]
| [
null,
"https://answers.everydaycalculation.com/mathstep-app-icon.png",
null
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| {"ft_lang_label":"__label__en","ft_lang_prob":0.8447545,"math_prob":0.9630183,"size":415,"snap":"2022-40-2023-06","text_gpt3_token_len":151,"char_repetition_ratio":0.13625304,"word_repetition_ratio":0.0,"special_character_ratio":0.4120482,"punctuation_ratio":0.09677419,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.96667707,"pos_list":[0,1,2],"im_url_duplicate_count":[null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-10-02T03:12:49Z\",\"WARC-Record-ID\":\"<urn:uuid:81ca13ec-3909-4b12-b606-62e1ec98ae84>\",\"Content-Length\":\"7001\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:5c2d493c-f14b-427a-b9dd-41a2099c0880>\",\"WARC-Concurrent-To\":\"<urn:uuid:6fbe6c50-5f7d-47e6-bd75-862886a92d6b>\",\"WARC-IP-Address\":\"96.126.107.130\",\"WARC-Target-URI\":\"https://answers.everydaycalculation.com/multiply-fractions/84-3-times-45-49\",\"WARC-Payload-Digest\":\"sha1:P7ZEGCJ6JD4CPHC2QYCQIK27SPB64HW3\",\"WARC-Block-Digest\":\"sha1:4L6MJ35XSJDL6FUAPZ2RVVB2CGQKRVX7\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-40/CC-MAIN-2022-40_segments_1664030337244.17_warc_CC-MAIN-20221002021540-20221002051540-00393.warc.gz\"}"} |
https://blogs.msdn.microsoft.com/garyhope/2009/02/26/sql-clr-base64-conversion-performance/ | [
"# SQL & CLR Base64 Conversion Performance\n\nAs a follow-up to my previous post I thought I would have a look at the performance of this type of conversion using the XML query and T-SQL type conversions versus the equivalent code written in C# and executed by the SQL CLR engine.\n\nThe first thing was to convert the code to the equivalent T-SQL functions:\n\n```CREATE FUNCTION ConvertToBase64SQL(@toEncode varchar(max)) RETURNS varchar(max)\nAS\nBEGIN\nDECLARE @xml xml\nDECLARE @bin varbinary(max)\nSELECT @bin = CAST(@toEncode as varbinary(max))\nSET @xml = (SELECT * FROM (SELECT @bin as data) as row FOR XML AUTO, BINARY BASE64)\nRETURN @xml.value('(/row/@data)', 'varchar(max)')\nEND```\n```CREATE FUNCTION ConvertFromBase64SQL(@toDecode varchar(max)) RETURNS varchar(max)\nAS\nBEGIN\nDECLARE @xml xml\nDECLARE @bin varbinary(max)\nSELECT @xml = CAST( '<row data=\"' + @toDecode + '\" />' as xml )\nSELECT @bin = @xml.value('(/row/@data)', 'varbinary(max)')\nRETURN CAST(@bin as varchar(max))\nEND ```\n\nSecondly to create the equivalent code in C# and deploy this to the SQL instance:\n\n``` [Microsoft.SqlServer.Server.SqlFunction]\npublic static SqlString ConvertToBase64CLR(SqlString toEncode)\n{\nbyte[] toEncodeAsBytes = System.Text.ASCIIEncoding.ASCII.GetBytes(toEncode.ToString());\nSqlString returnValue = System.Convert.ToBase64String(toEncodeAsBytes);\nreturn returnValue;\n}```\n``` [Microsoft.SqlServer.Server.SqlFunction]\npublic static SqlString ConvertFromBase64CLR(SqlString toDecode)\n{\nbyte[] DecodedBytes = System.Convert.FromBase64String(toDecode.ToString());\nSqlString returnValue = System.Text.ASCIIEncoding.ASCII.GetString(DecodedBytes);\nreturn returnValue;\n}```\n\nFinally compare the performance:\n\n```DECLARE @Counter INT = 100000\nDECLARE @Output varchar(max)\nDECLARE @StartTime DateTime = GetDate()\nWHILE @Counter > 0\nBEGIN\nSET @Output = dbo.ConvertFromBase64SQL(dbo.ConvertToBase64SQL('This is a small test message'));\nSET @Counter = @Counter - 1\nEND\nPRINT @Output\nPRINT 1.0*DATEDIFF(MS,@StartTime,GetDate())/1000\n-- 25.440000 seconds```\n\n```DECLARE @Counter INT = 100000\nDECLARE @Output varchar(max)\nDECLARE @StartTime DateTime = GetDate()\nWHILE @Counter > 0\nBEGIN\nSET @Output = dbo.ConvertFromBase64CLR(dbo.ConvertToBase64CLR('This is a small test message'));\nSET @Counter = @Counter - 1\nEND\nPRINT @Output\nPRINT 1.0*DATEDIFF(MS,@StartTime,GetDate())/1000\n-- 2.663000 seconds```\n\nThe results indicate that whilst the C# CLR implementation is significantly faster, the benefit of this performance difference (almost 10 times faster) it would likely only have a real impact if there where a large number of documents being converted.\n\nI have not run this test on any larger documents which may result in a different performance results.\n\nI think the real point here is that today there are a number of ways of doing these types of operations and you need to keep your mind open and evaluate the performance of your chosen implementation.\n\n<Gary>"
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| {"ft_lang_label":"__label__en","ft_lang_prob":0.5698236,"math_prob":0.9137202,"size":2854,"snap":"2019-51-2020-05","text_gpt3_token_len":672,"char_repetition_ratio":0.11754386,"word_repetition_ratio":0.20821919,"special_character_ratio":0.24211633,"punctuation_ratio":0.11157895,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9745991,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-12-07T00:28:31Z\",\"WARC-Record-ID\":\"<urn:uuid:45f8b162-bc64-44ba-8209-6b79d7c214f7>\",\"Content-Length\":\"64248\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:48e7b7ab-9289-4972-9d39-8fb2a1d836fa>\",\"WARC-Concurrent-To\":\"<urn:uuid:46629b4e-abc8-4f2c-b2b9-cc29791c6f5a>\",\"WARC-IP-Address\":\"23.49.183.10\",\"WARC-Target-URI\":\"https://blogs.msdn.microsoft.com/garyhope/2009/02/26/sql-clr-base64-conversion-performance/\",\"WARC-Payload-Digest\":\"sha1:LB5WFKLLP4EEH5WFNHFPE2I7OXCNAZ4A\",\"WARC-Block-Digest\":\"sha1:2PATKFPPSWKLNB362H3TYYFKZAN4KYUL\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-51/CC-MAIN-2019-51_segments_1575540491491.18_warc_CC-MAIN-20191206222837-20191207010837-00414.warc.gz\"}"} |
https://www.easycalculation.com/engineering/electrical/gapped-core-inductance-calculator.php | [
"# Gapped Core Inductance Calculator\n\nGapped cores were used in low power transformers to control the inductance and increase the quality factor (Q) of air core inductors. This online calculator allows you to calculate the air core inductance in any applications.\n\n## Gapped Core Inductance Calculator\n\nmm\nmm\nmm\nuH\nnH\ngauss/A\n\nGapped cores were used in low power transformers to control the inductance and increase the quality factor (Q) of air core inductors. This online calculator allows you to calculate the air core inductance in any applications.\n\nCode to add this calci to your website",
null,
"",
null,
"#### Formula:",
null,
"Al = (4π*Ae) / ((le/μr)+lg) L = Al*N210-3 B/I = (0.4π*N) / ((le/μr)+lg)\nWhere, Al = Inductance per turn2 N = Number of turns Ae = Gap area Le = Core length μr = Relative permeability Lg = Gap length B/I = Flux Density per Amp"
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"https://www.easycalculation.com/images/embed-plus.gif",
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"https://www.easycalculation.com/images/embed-minus.gif",
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"https://www.easycalculation.com/engineering/electrical/GappedCoreInductance.png",
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| {"ft_lang_label":"__label__en","ft_lang_prob":0.88226455,"math_prob":0.99138564,"size":451,"snap":"2021-04-2021-17","text_gpt3_token_len":94,"char_repetition_ratio":0.14317673,"word_repetition_ratio":0.9411765,"special_character_ratio":0.1751663,"punctuation_ratio":0.05263158,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9976582,"pos_list":[0,1,2,3,4,5,6],"im_url_duplicate_count":[null,null,null,null,null,2,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-04-12T20:26:25Z\",\"WARC-Record-ID\":\"<urn:uuid:ef9f59ab-cca4-4b54-9356-45ea1089d86d>\",\"Content-Length\":\"27171\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:334ab261-ac0f-41ee-aa14-12ef0d7f5159>\",\"WARC-Concurrent-To\":\"<urn:uuid:533dba0d-94d8-4bfc-856c-22887be5c1d2>\",\"WARC-IP-Address\":\"50.116.14.108\",\"WARC-Target-URI\":\"https://www.easycalculation.com/engineering/electrical/gapped-core-inductance-calculator.php\",\"WARC-Payload-Digest\":\"sha1:7WT5M46HIAUEGEQBMUKF7G5MUADS6JBG\",\"WARC-Block-Digest\":\"sha1:MV2DZWGQJR2L6FXE5XKQMQAG64RUA3D5\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-17/CC-MAIN-2021-17_segments_1618038069133.25_warc_CC-MAIN-20210412175257-20210412205257-00378.warc.gz\"}"} |
https://calculomates.com/en/divisors/of/979 | [
"# Divisors of 979\n\n## Divisors of 979\n\nThe list of all positive divisors (that is, the list of all integers that divide 22) is as follows :\n\nAccordingly:\n\n979 is multiplo of 1\n\n979 is multiplo of 11\n\n979 is multiplo of 89\n\n979 has 3 positive divisors\n\n## Parity of 979\n\n979is an odd number,as it is not divisible by 2\n\n## The factors for 979\n\nThe factors for 979 are all the numbers between -979 and 979 , which divide 979 without leaving any remainder. Since 979 divided by -979 is an integer, -979 is a factor of 979 .\n\nSince 979 divided by -979 is a whole number, -979 is a factor of 979\n\nSince 979 divided by -89 is a whole number, -89 is a factor of 979\n\nSince 979 divided by -11 is a whole number, -11 is a factor of 979\n\nSince 979 divided by -1 is a whole number, -1 is a factor of 979\n\nSince 979 divided by 1 is a whole number, 1 is a factor of 979\n\nSince 979 divided by 11 is a whole number, 11 is a factor of 979\n\nSince 979 divided by 89 is a whole number, 89 is a factor of 979\n\n## What are the multiples of 979?\n\nMultiples of 979 are all integers divisible by 979 , i.e. the remainder of the full division by 979 is zero. There are infinite multiples of 979. The smallest multiples of 979 are:\n\n0 : in fact, 0 is divisible by any integer, so it is also a multiple of 979 since 0 × 979 = 0\n\n979 : in fact, 979 is a multiple of itself, since 979 is divisible by 979 (it was 979 / 979 = 1, so the rest of this division is zero)\n\n1958: in fact, 1958 = 979 × 2\n\n2937: in fact, 2937 = 979 × 3\n\n3916: in fact, 3916 = 979 × 4\n\n4895: in fact, 4895 = 979 × 5\n\netc.\n\n## Is 979 a prime number?\n\nIt is possible to determine using mathematical techniques whether an integer is prime or not.\n\nfor 979, the answer is: No, 979 is not a prime number.\n\n## How do you determine if a number is prime?\n\nTo know the primality of an integer, we can use several algorithms. The most naive is to try all divisors below the number you want to know if it is prime (in our case 979). We can already eliminate even numbers bigger than 2 (then 4 , 6 , 8 ...). Besides, we can stop at the square root of the number in question (here 31.289 ). Historically, the Eratosthenes screen (which dates back to Antiquity) uses this technique relatively effectively.\n\nMore modern techniques include the Atkin screen, probabilistic tests, or the cyclotomic test."
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| {"ft_lang_label":"__label__en","ft_lang_prob":0.93679607,"math_prob":0.9672718,"size":1943,"snap":"2021-31-2021-39","text_gpt3_token_len":529,"char_repetition_ratio":0.19030428,"word_repetition_ratio":0.09947644,"special_character_ratio":0.32732886,"punctuation_ratio":0.12993039,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9992803,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-07-27T09:20:58Z\",\"WARC-Record-ID\":\"<urn:uuid:8dafd4be-76fb-4ae7-9387-7ce95b1c50fc>\",\"Content-Length\":\"16263\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:d89c0925-37e9-4b85-9b53-2fdd0e1c1abe>\",\"WARC-Concurrent-To\":\"<urn:uuid:f448f0b1-a57e-4db8-a3a7-662f38e73fa4>\",\"WARC-IP-Address\":\"172.67.175.160\",\"WARC-Target-URI\":\"https://calculomates.com/en/divisors/of/979\",\"WARC-Payload-Digest\":\"sha1:OQH75DHZJHALA6TBP7QSRZPMWWNWZEZR\",\"WARC-Block-Digest\":\"sha1:TBILK45X6EQFGCW2I6VOUDSJSDWYODID\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-31/CC-MAIN-2021-31_segments_1627046153223.30_warc_CC-MAIN-20210727072531-20210727102531-00504.warc.gz\"}"} |
https://www.mathworks.com/matlabcentral/answers/49027-how-to-manipulate-plot-legend-to-show-markers-and-colours-seperatly | [
"# How to manipulate Plot Legend to show markers and colours seperatly?\n\n355 views (last 30 days)\nJakob on 25 Sep 2012\nHi there everyone!\nI want to display the results of a numerical computation in which I varyed two discretisation parameters.\nNow I want to show in which way any parameter effects the results. By using the same color I want to show the lines representing calculations with the same number of discrete elements. And by using the same marker I want to show the lines that represent calculations with the same number of time steps.\nI got as far as changing the colors and markers in the plot itself goes, but now I need a legend that clearly displays the correlations of color and marker. My legend contains one entry for every line which is not useful. I want a legend that first shows:\n- this color means that many elements\n- this marker means that many time steps.\nIs it possible to tweak the legend like that?\nYour help will be greatly appreciated! Thank you,\n-Jakob.\n\nDr. Seis on 25 Sep 2012\nEdited: Dr. Seis on 25 Sep 2012\nIt can be done... you will have to do some finagling, but it can be done. Here is an example where we have 2 different datasets representing 2 different \"elements\" and each dataset has 3 different sets of data representing 3 different \"time steps\":\n% Generate random data\nsample_points = 0:0.1:10;\ncalc_data_1 = sin(2*pi*[sample_points;sample_points;sample_points]) + ...\n0.1*randn(3,101);\ncalc_data_2 = cos(2*pi*[sample_points;sample_points;sample_points]) + ...\n0.1*randn(3,101);\n% Generate dummy info for plot handles \"h\"\nh = zeros(5,1);\nh(1) = plot(sample_points,calc_data_1(1,:),'r-'); hold on;\nh(2) = plot(sample_points,calc_data_2(1,:),'b-');\nh(3) = plot(sample_points,calc_data_1(1,:),'ks');\nh(4) = plot(sample_points,calc_data_1(2,:),'kd');\nh(5) = plot(sample_points,calc_data_1(3,:),'ko');\n% Plot things the right way\nplot(sample_points,calc_data_1(1,:),'rs-');\nplot(sample_points,calc_data_1(2,:),'rd-');\nplot(sample_points,calc_data_1(3,:),'ro-');\nplot(sample_points,calc_data_2(1,:),'bs-');\nplot(sample_points,calc_data_2(2,:),'bd-');\nplot(sample_points,calc_data_2(3,:),'bo-'); hold off;\n% Define legend according to handles \"h\"\nlegend(h,'M elements used','N elements used','P time steps','Q time steps','R time steps')",
null,
"Mohammad Saiful Islam on 27 Sep 2016\nThanks for your code. But i am facing some problem during plot. Mathworks shows me the line 22 have some error. They shows the following message: Error using line There is no Line property on the Line class.\nError in diff_color_legend_2 (line 22) l1(i, j) = line(w, z, 'Color', colors(i, :), 'Marker', marks{j}, 'Line', 'none'); How can I solve this problem?"
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"https://www.mathworks.com/matlabcentral/answers/uploaded_files/172061/image.jpeg",
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| {"ft_lang_label":"__label__en","ft_lang_prob":0.74132603,"math_prob":0.9916805,"size":2908,"snap":"2022-05-2022-21","text_gpt3_token_len":787,"char_repetition_ratio":0.1656336,"word_repetition_ratio":0.043269232,"special_character_ratio":0.29470426,"punctuation_ratio":0.20711975,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9939028,"pos_list":[0,1,2],"im_url_duplicate_count":[null,2,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-05-28T08:07:14Z\",\"WARC-Record-ID\":\"<urn:uuid:e8438626-449c-4234-9ef7-67547fc08520>\",\"Content-Length\":\"154175\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:e1997b92-ae63-4503-af25-9298e674a857>\",\"WARC-Concurrent-To\":\"<urn:uuid:92e9d056-88ac-444a-bf2f-678effc9ead6>\",\"WARC-IP-Address\":\"104.68.243.15\",\"WARC-Target-URI\":\"https://www.mathworks.com/matlabcentral/answers/49027-how-to-manipulate-plot-legend-to-show-markers-and-colours-seperatly\",\"WARC-Payload-Digest\":\"sha1:QYIQIWXM3JX26GX5GQGOXFB5BUL22G6W\",\"WARC-Block-Digest\":\"sha1:PV2GTC2N2QDUEONFBZF5H6VRQRW62ZWL\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-21/CC-MAIN-2022-21_segments_1652663013003.96_warc_CC-MAIN-20220528062047-20220528092047-00100.warc.gz\"}"} |
https://www.physicsforums.com/threads/line-integral.402176/ | [
"# Line integral\n\nCan line integral of a vector field ever be zero? If can, what is the interpretation of this value (0) ?\n\nThanks.\n\n## Answers and Replies\n\nThat's a very peculiar question. One of the first things students learn about path integrals is that if a f(x,y)dx+ g(x,y)dy is an \"exact differential\" (There exist a function F(x,y) such that $\\partial F/\\partial x= f(x)$ and $\\partial F/\\partial y= g(x,y)$ then the integral around any closed path is 0. And, since you refer to \"vector fields\", we can think of that as the integral of the vector field $f(x,y)\\vec{i}+ g(x,y)\\vec{j}$ over the path."
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| {"ft_lang_label":"__label__en","ft_lang_prob":0.94548905,"math_prob":0.9986663,"size":902,"snap":"2022-40-2023-06","text_gpt3_token_len":238,"char_repetition_ratio":0.16146994,"word_repetition_ratio":0.0,"special_character_ratio":0.25166297,"punctuation_ratio":0.1010101,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99960595,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-10-05T14:33:43Z\",\"WARC-Record-ID\":\"<urn:uuid:18d3b84b-3a43-42cb-9340-dbb786f9dbc1>\",\"Content-Length\":\"57889\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:66899574-df0b-497b-9cd1-3691332dcf89>\",\"WARC-Concurrent-To\":\"<urn:uuid:cb83ee7e-0849-4e67-89c3-1abbca6e94b6>\",\"WARC-IP-Address\":\"172.67.68.135\",\"WARC-Target-URI\":\"https://www.physicsforums.com/threads/line-integral.402176/\",\"WARC-Payload-Digest\":\"sha1:IL4O6VEIHH3AJVS7MTI3JK2D3X7UQ3D2\",\"WARC-Block-Digest\":\"sha1:G2UHZDCDKBBM7OXLME62WBSIDUUX4XDN\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-40/CC-MAIN-2022-40_segments_1664030337631.84_warc_CC-MAIN-20221005140739-20221005170739-00060.warc.gz\"}"} |
https://www.mathsai.com/second-grade.htm | [
"# 2nd Grade Mathematics Personalized Learning\n\nNumbers in Words\nWriting 1 to 50 Number Names in English Words\nWriting 51 to 60 Number Names in English Words\nWriting 61 to 70 Number Names in English Words\nWriting 71 to 80 Number Names in English Words\nWriting 81 to 90 Number Names in English Words\nWriting 91 to 100 Number Names in English Words\nWrite Number Names (1 to 100) in English Words\nNumber in Words to Number (1 to 100) Writing\nCounting Numbers\nCircle all Numbers (1 to 100) of Count in 2's\nCircle all Numbers (1 to 100) of Count in 3's\nCircle all Numbers (1 to 100) of Count in 5's\nWrite all Numbers (1 to 100) of Count in 2's\nWrite all Numbers (1 to 100) of Count in 3's\nFill in the Missing Values for Skip Counting within 1000 by 10s\nFill in the Missing Values for Skip Counting within 1000 by 100s\nWhat Comes Next? - Count in 3's\nWhat Comes Next? - Count in 4's\nWhat Comes Next? - Count in 5's\nWhat Comes Next? - Backward Count in 2's\nWhat Comes Next? - Backward Count in 3's\nWhat Comes Next? - Backward Count in 5's\nOrdinal & Cardinal Numbers\nFill in the Missing Values for Skip Counting within 1000 by 5s\nNumbers Comparison\nCompare Numbers by Greater (>) or Lesser (<) Symbol\nArrange Numbers in Descending Order\nArrange Numbers in Ascending Order\nMaking 2 Digit Bigger & Smaller Numbers\nMaking Biggest & Smallest 2 Digit Numbers (Repetition)\nGreatest & Smallest 2 Digit Numbers (3 Nos No Repetition)\nOdd or Even Number\nPlace Values\nWriting Tens in Words\nPlace Values to Number Writing\nHow Many Tens or Ones in a Place Value\nPlace Value Beads to Number Writing\nWrite Place Values\nDraw Place Value Beads for Number\n3-digit Number Represents Amounts of 100s, 10s, and 1s\nStandard, Expanded and Word Form of a Number\nCompare Two 3-digit Numbers\n2 by 1 Digit Addition with Regrouping\nTwo 2 Digit Addition without Regrouping\n2 Two Digit Numbers Addition with Regrouping\nAddition (9x9) Fill in the Blanks\nAdd up to Four 2-digit Numbers using Strategies based on Place Value\nSubtraction\n2 by 1 Digit Subtraction with Regrouping\n2 Digit Number Subtraction without Regrouping\n2 Digit Numbers Subtraction with Regrouping\nAdd and Subtract within 30 Contextual Problems\nAdd Subtract 10 or 100 from Number between 100 to 900\nFluently Add or Subtract within 100\nMultiplication\nWrite Addition & Multiplication Facts for Pictures\nMoney\nSubtracting Money without Exchanging\nAdding Money Word Problems without Regrouping\nSubtracting Money Word Problems without Exchanging\nGeometry\n2D Shape Names, Sides & Corners\nColor Similar 2D Shapes\nDraw Shapes for Object Names\nComplete the 2D Shapes\nMatch Objects for given 3D Shapes\nWrite 3D Shape Names\nWrite 3D Shape Names for given Objects\nDraw Two-Dimensional Shapes having Specified Attributes\nPattern\nWrite Missing Numbers to Complete the Pattern\nObserve & Complete the Number Pattern\nColor Odd Object in a Group of Objects\nComplete the Image Pattern\nMatch Same Kind of Objects\nMeasurement & Conversion\nMeasure the Length of an Object using Two Different Units\nEstimate Units of Inches, Feet, Yards, Centimeters, & Meters\nTell and Write Time in Nearest Five Minutes using Analog\nTell and Write Time in Nearest Five Minutes using Digital\nSum and Difference of Whole Numbers on a Number Line\nHow Much Longer One Object is Than Another\nTotal Number of Unit Squares in a Rectangle\nAlgebra\nDetermine Odd or Even by Pairing Group of Objects\nExpress an Even Number as a sum of two equal addends\nWrite an Equation for Repeated Addition as a Sum of Equal Addends\n\nAlign math worksheets, homework, step-by-step solutions, challenges and calculators to your curriculum to learn, practice, assess 2nd grade math skills in more personalized way! The large number of practice problems, online test, quizzes, puzzles and calculators that show work fosters the personalized learning and breaks the dependency on teaching or learning second grade mathematics in best ways."
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| {"ft_lang_label":"__label__en","ft_lang_prob":0.9029282,"math_prob":0.7025266,"size":400,"snap":"2019-51-2020-05","text_gpt3_token_len":74,"char_repetition_ratio":0.1010101,"word_repetition_ratio":0.0,"special_character_ratio":0.1725,"punctuation_ratio":0.14492753,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9549439,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-01-23T07:41:00Z\",\"WARC-Record-ID\":\"<urn:uuid:3460abcd-8497-454a-9e42-e1aa76eb06e9>\",\"Content-Length\":\"91915\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:ac97ac1c-dc6d-400c-9641-5299683b319c>\",\"WARC-Concurrent-To\":\"<urn:uuid:b8f2fe2e-40cf-4a2d-81d1-638f5d57f460>\",\"WARC-IP-Address\":\"35.184.30.27\",\"WARC-Target-URI\":\"https://www.mathsai.com/second-grade.htm\",\"WARC-Payload-Digest\":\"sha1:D4VGCI26CK566WMOHJXHNY6WZZ4SYOXB\",\"WARC-Block-Digest\":\"sha1:ZHGRM6RWXLHRNPZX6TQF45U56JARDMSM\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-05/CC-MAIN-2020-05_segments_1579250609478.50_warc_CC-MAIN-20200123071220-20200123100220-00049.warc.gz\"}"} |
https://whatisconvert.com/110-centimeters-in-kilometers | [
"## Convert 110 Centimeters to Kilometers\n\nTo calculate 110 Centimeters to the corresponding value in Kilometers, multiply the quantity in Centimeters by 1.0E-5 (conversion factor). In this case we should multiply 110 Centimeters by 1.0E-5 to get the equivalent result in Kilometers:\n\n110 Centimeters x 1.0E-5 = 0.0011 Kilometers\n\n110 Centimeters is equivalent to 0.0011 Kilometers.\n\n## How to convert from Centimeters to Kilometers\n\nThe conversion factor from Centimeters to Kilometers is 1.0E-5. To find out how many Centimeters in Kilometers, multiply by the conversion factor or use the Length converter above. One hundred ten Centimeters is equivalent to zero point zero zero one one Kilometers.\n\n## Definition of Centimeter\n\nThe centimeter (symbol: cm) is a unit of length in the metric system. It is also the base unit in the centimeter-gram-second system of units. The centimeter practical unit of length for many everyday measurements. A centimeter is equal to 0.01(or 1E-2) meter.\n\n## Definition of Kilometer\n\nThe kilometer (symbol: km) is a unit of length in the metric system, equal to 1000m (also written as 1E+3m). It is commonly used officially for expressing distances between geographical places on land in most of the world.\n\n## Using the Centimeters to Kilometers converter you can get answers to questions like the following:\n\n• How many Kilometers are in 110 Centimeters?\n• 110 Centimeters is equal to how many Kilometers?\n• How to convert 110 Centimeters to Kilometers?\n• How many is 110 Centimeters in Kilometers?\n• What is 110 Centimeters in Kilometers?\n• How much is 110 Centimeters in Kilometers?\n• How many km are in 110 cm?\n• 110 cm is equal to how many km?\n• How to convert 110 cm to km?\n• How many is 110 cm in km?\n• What is 110 cm in km?\n• How much is 110 cm in km?"
]
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| {"ft_lang_label":"__label__en","ft_lang_prob":0.86794996,"math_prob":0.9899639,"size":1784,"snap":"2022-40-2023-06","text_gpt3_token_len":453,"char_repetition_ratio":0.25,"word_repetition_ratio":0.096666664,"special_character_ratio":0.25504485,"punctuation_ratio":0.10755814,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9988335,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-02-06T19:00:34Z\",\"WARC-Record-ID\":\"<urn:uuid:13fbe66a-6ba1-4dd8-b753-c12e5c89caef>\",\"Content-Length\":\"28594\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:9db3af2f-3f0f-4087-a186-e1cb9e0412f7>\",\"WARC-Concurrent-To\":\"<urn:uuid:fc4e416f-92a9-49f1-9c63-101fc7ac951f>\",\"WARC-IP-Address\":\"172.67.133.29\",\"WARC-Target-URI\":\"https://whatisconvert.com/110-centimeters-in-kilometers\",\"WARC-Payload-Digest\":\"sha1:P5RVHTLGKNZT4P4OSCBXYLAAB6KY4NCW\",\"WARC-Block-Digest\":\"sha1:IIOURAZCNA7L7KKQZSRYYEIRFASJYSP6\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-06/CC-MAIN-2023-06_segments_1674764500357.3_warc_CC-MAIN-20230206181343-20230206211343-00829.warc.gz\"}"} |
https://www.juxhub.com/course_detail/326/data-analytics-using-r | [
"• Have any questions?\n• +91 9606333101/ 102/ 118\n• [email protected]\nData Analytics using R\nCategories:\nIT",
null,
"Description\nR is a programming language software environment for statistical computing and graphics supported by the R Foundation for Statistical Computing. The R language is widely used among statisticians and data miners for developing statistical software and data analysis. It is an approach of data analysis employed for summarizing and visualizing data set, and the focus of the approach is to analyze data’s basic structures and variables to develop a basic understanding of the data set. in this data science certification course, students will learn data exploration, data visualization, predictive analytics and descriptive analytics techniques with the R language.\n\nCurriculum\n\nModule 1: Introduction to R Environment\n\nHistory and development of R Statistical computing programming language, installing R and R studio, getting started with R, creating new working directory, changing existing working directory, understanding the different data types, installing the available packages, calling the installed packages, arithmetic operations, variable definition in R, simple functions, vector definition and logical expressions, matrix calculation and manipulation using matrix data types, workspace management.\n\nModule 2: Data Structures and Control Statements\n\nIntroduction to different data types, vectors, atomic vectors, types and tests, coercion, lists, list indexing, function applying on the lists, adding and deleting the elements of lists, attributes, name and factors, matrices and arrays, matrix indexing, filtering on matrix, generating a covariance matrix, applying function to row and column of the matrix, data frame-creating, coercion, combining data frames, special types in data frames, operations in data frame,applying functions: lapply ( ) and supply ( ) on data frames, control statements, loops, looping over non vector sets, arithmetic and Boolean operators and values, branching with if, looping with for, if-else control structure, looping with while, vector based programming.\n\nModule 3: I/O Operations and String Manipulations\n\nIntroduction to I/O functions in R, accessing I/O devices, using of scan( ), readline ( ) function, comparison and usage of scan and readline function, reading different format files into R: text file, CSV file, Statistical package files, xls and xlsx files, reading data frame files, converting from one format to another using in built function, writing different file format in to the local machine directory, getting file directory information, accessing the internet : overview of TCP/IP, sockets in R, implementation of parallel R, basics of string manipulations – grep ( ), nchar ( ), paste( ), sprintf( ), substr( ), regexpr( ), strsplit( ), testing of file name with given suffix.\n\nModule 4: R for Summary and Parametric Tests\n\nDescriptive statistics – summery statistics for vectors, making contingency tables, creating contingency tables from vectors, converting objects in to tables, complex flat tables, making ‘Flat’ contingency tables, testing tables and flat table objects, cross tables, testing cross tabulation, recreating original data from contingency tables, switching class, mean (arithmetic, geometric and harmonic ) median, mode for raw and grouped data, measure of dispersion-range, standard deviation, variance, coefficient of variation, testing of hypothesis – small sample test, large sample test – for comparing mean, proportion, variance (dependent and independent samples) correlation and regression – significance of correlation and regression coefficients.\n\nModule 5: R for Graphs, Non parametric Tests and ANOVA\n\nIntroduction To Graphs, Box-Whisker Plot, Scatter Plots, Pairs Plots, Line Chart, Pie Chart, Cleveland Dot Charts, Bar Charts, Customization Of Charts, Non Parametric Tests: The Wilcoxon U-Test (Mann-Whitney: One And Two Sample U-Test, Tests For Association: Chi Square Tests.\n\nLearning Outcomes\n\n• Illustrate the concept the importance of multivariate analysis in data analysis\n• Recognise the importance of classification rule mining in data mining business and engineering.\n• Understand the importance of R in customising the analytics in\n• Understand the different options in I/O operations in R programming\n• Understand the basic concepts of statistical functions in R for the analysis.\n\nWho Should Attend?\n\n• Engineering and IT students\n• Graduates with a programming background\n\nJob Prospects\n\n• Market Data Analyst\n• Data Scientist\n• Data Science modelers\n• Manager Data Analytics\n• Associate - Data Engineer\n\nCertification\n\nAfter completing this course and successfully passing the certification examination, the student will be awarded the “Data Analytics using R” certification.\n\nIf a learner chooses not to take up the examination, they will still get a 'Participation Certificate'\n\nFaculty Profiles"
]
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null,
"https://www.juxhub.com/img/courses_video.png",
null
]
| {"ft_lang_label":"__label__en","ft_lang_prob":0.78464496,"math_prob":0.6865682,"size":3529,"snap":"2019-51-2020-05","text_gpt3_token_len":696,"char_repetition_ratio":0.1177305,"word_repetition_ratio":0.0,"special_character_ratio":0.19098894,"punctuation_ratio":0.19344263,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.96652734,"pos_list":[0,1,2],"im_url_duplicate_count":[null,2,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-12-15T03:37:03Z\",\"WARC-Record-ID\":\"<urn:uuid:36da5acf-6848-4733-b750-a7eeefcea16f>\",\"Content-Length\":\"36438\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:d9c687e0-66f1-444b-9f1b-26d19281ef8c>\",\"WARC-Concurrent-To\":\"<urn:uuid:d41518af-5d40-4334-9d6b-f0023838e6b6>\",\"WARC-IP-Address\":\"52.66.196.134\",\"WARC-Target-URI\":\"https://www.juxhub.com/course_detail/326/data-analytics-using-r\",\"WARC-Payload-Digest\":\"sha1:2DJEX4ZAGZAEFY5IX4ZX3GMLIESL2JCB\",\"WARC-Block-Digest\":\"sha1:JKRCR2EDUG6RTC564PRGPKIYST32D3SZ\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-51/CC-MAIN-2019-51_segments_1575541301014.74_warc_CC-MAIN-20191215015215-20191215043215-00414.warc.gz\"}"} |
http://www.numbersaplenty.com/2145254203 | [
"Search a number\nBaseRepresentation\nbin111111111011101…\n…1111101100111011\n312112111200012112101\n41333313133230323\n513343141113303\n6552512124231\n7104106234301\noct17767375473\n95474605471\n102145254203\n11a00a38264\n124ba536677\n132825a40b1\n14164caa271\n15c8505d1d\nhex7fddfb3b\n\n2145254203 has 4 divisors (see below), whose sum is σ = 2145391888. Its totient is φ = 2145116520.\n\nThe previous prime is 2145254197. The next prime is 2145254213. The reversal of 2145254203 is 3024525412.\n\nAdding to 2145254203 its reverse (3024525412), we get a palindrome (5169779615).\n\nIt is a semiprime because it is the product of two primes.\n\nIt is a cyclic number.\n\nIt is not a de Polignac number, because 2145254203 - 229 = 1608383291 is a prime.\n\nIt is a Duffinian number.\n\nIt is not an unprimeable number, because it can be changed into a prime (2145254213) by changing a digit.\n\nIt is a polite number, since it can be written in 3 ways as a sum of consecutive naturals, for example, 41976 + ... + 77797.\n\nIt is an arithmetic number, because the mean of its divisors is an integer number (536347972).\n\nAlmost surely, 22145254203 is an apocalyptic number.\n\n2145254203 is a deficient number, since it is larger than the sum of its proper divisors (137685).\n\n2145254203 is a wasteful number, since it uses less digits than its factorization.\n\n2145254203 is an odious number, because the sum of its binary digits is odd.\n\nThe sum of its prime factors is 137684.\n\nThe product of its (nonzero) digits is 9600, while the sum is 28.\n\nThe square root of 2145254203 is about 46316.8889607236. Note that the first 3 decimals coincide. The cubic root of 2145254203 is about 1289.7125339462.\n\nThe spelling of 2145254203 in words is \"two billion, one hundred forty-five million, two hundred fifty-four thousand, two hundred three\".\n\nDivisors: 1 17911 119773 2145254203"
]
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http://pages.vassar.edu/magnes/advanced-em/angelica/ | [
"# Precession of Mercurian Planets: Project Plan\n\nGoals:\n\nFor my computational physics project I would like to investigate the relationship between the precession and the eccentricity of a planet’s orbit due to general relativity as detailed in chapter 4, section 3 of Computational Physics, by Nicholas J. Giordano and Hisao Nakanishi. The rate of precession examined in this project will only be that which is caused by the general relativistic effect of the model planet’s host star.\n\nWeek 1: Preliminary Results\n\nAs a check to make sure my code will run correctly, I will first model the precession of Mercury’s orbit as caused by the Sun. I must first write a code that will model the movement of a Mercury through it’s orbit according to Kepler’s laws. The force law predicted by General Relativity is given by:",
null,
"where G is the Gravitational constant of the universe, Ms is the mass of the Sun, Mm is the mass of Mercury, and a Constant*α gives the rate of precession for the orbit. For the case of Mercury,",
null,
".\n\nApplying the Euler-Chromer method to the above equation to approximate the x and y velocities and positions of the planet as it travels throughout its orbit yields:",
null,
"",
null,
"",
null,
"",
null,
"Where the initial conditions are (assuming",
null,
"to be at the planet’s aphelion):",
null,
"",
null,
"where a is the semi-major axis length of the planet’s orbit.\n\nIn Computational Physics, Giordano and Nakanishi find the precession of Mercury by first finding the precession rate for a much larger value of",
null,
", because it is difficult to measure the precession for a value this small (it would require computing the motion for hundreds of thousands of years). Using different values of α give different values for the precession rate, and graphing these two parameters against one another will provide the desired constant C. Once C is obtained, I can multiply this number by α to get the true precession of Mercury caused by the Sun.\n\nWeek 2: Results with Data Analysis\n\nOnce this code is working correctly, I can modify the initial conditions for the orbital motion portion of the code to represent new orbits. (Changing the parameters a, e, but maintaining the value of the perihelion as that of Mercury, so that I may compare my results to that for Mercury.)\n\nIf I have extra time I will add in other planets to the original Mercury-Sun system to observe the resulting effects on Mercury’s precession.\n\nWeek 3: Finalize project: make sure the code’s comments are informative and helpful, and review the program/results for any mistakes.\n\nWeek 4: Provide constructive criticism to other students on their projects/Prepare for project presentation.\n\nWeek 5: Give my project presentation to the class.\n\nReferences:\nComputational Physics by Nicholas J. Giordano and Hisao Nakanishi"
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null,
"http://pages.vassar.edu/magnes/wp-content/ql-cache/quicklatex.com-3ac83872ee29128f6ee32ee0b3a114f6_l3.png",
null,
"http://pages.vassar.edu/magnes/wp-content/ql-cache/quicklatex.com-e3079c89aaa30a8b38bc0fc5ea04ecfd_l3.png",
null,
"http://pages.vassar.edu/magnes/wp-content/ql-cache/quicklatex.com-f352f92596048574aa7931be9b7320a5_l3.png",
null,
"http://pages.vassar.edu/magnes/wp-content/ql-cache/quicklatex.com-3372ca4c0050c4a3aa48588004478e49_l3.png",
null,
"http://pages.vassar.edu/magnes/wp-content/ql-cache/quicklatex.com-240947f04aef107a9cad01218288b994_l3.png",
null,
"http://pages.vassar.edu/magnes/wp-content/ql-cache/quicklatex.com-65a1ac1ba616631c4006abd521f14cb2_l3.png",
null,
"http://pages.vassar.edu/magnes/wp-content/ql-cache/quicklatex.com-106820ffee1dfe3a9ae6cc19000dd772_l3.png",
null,
"http://pages.vassar.edu/magnes/wp-content/ql-cache/quicklatex.com-c20943647f6bc8d0830d773f6d2e445d_l3.png",
null,
"http://pages.vassar.edu/magnes/wp-content/ql-cache/quicklatex.com-33923f0ad5e89f0ed8417fb3ef58c002_l3.png",
null,
"http://pages.vassar.edu/magnes/wp-content/ql-cache/quicklatex.com-9db374010925d42a047b3bd26b154b8e_l3.png",
null
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http://wiki.nars2000.org/index.php?title=Combinatorial | [
"# Combinatorial\n\n Z←a‼R returns an array whose values depend upon which Combinatorial Function is chosen by the left operand a.\nR is a non-negative numeric scalar or one- or two-element vector. If R has only one element, it is treated as if the value were duplicated as in 2⍴R. For convenience in the description below, the two elements are referred to as M and N as in (M N)←2⍴R.\na is a non-negative numeric scalar or one- or two- element vector which serves as the Selector for the Twelve Combinatorial Functions.\n\nThe first element (Function Selector) in a is a non-negative integer for each of the Twelve functions (where the number is written here with three digits to emphasize that each digit has a separate meaning):\n\n 000 M Pigeons in N holes 001 Partitions of the number M into no more than N parts 002 Partitions of the number M into N parts 010 M Combinations of N items 011 M Multisets of N items 012 Compositions of the number M into N parts a.k.a. Partitions of the number M into N ordered parts 100 M Pigeons in N holes 101 Partitions of the set {⍳M} into no more than N parts 102 Partitions of the set {⍳M} into N parts 110 M Permutations of N items 111 M Tuples of N items 112 Partitions of the set {⍳M} into N ordered parts\nThe second element (Count/Generate Flag) in a is an optional Integer value where 0 (the default) means Count the number of elements in the Combinatorial Function as applied to R and values greater than 0 mean generate the array of elements. If the Generate Flag is 1, generate the array in an unspecified order. Certain Function Selectors (see the individual Combinatorial Functions) accept values greater than 1. In particular, the value 2 generates the array in Lexicographic order, and 3 generates the array in Gray Code order.\nThe symbol (U+203C) can be entered from the default keyboard layout with Alt-’k’ or Ctrl-’k’, depending upon your choice of keyboard layouts.\n\n## Introduction\n\nCounting and generating items is fundamental in mathematics, but has been sorely lacking in APL (notwithstanding the counting functions !N and M!N); instead we have had to rely upon a patchwork of various library routines.\n\nMoreover, most APL papers on the topic have focused on the implementation of the algorithms rather than their organization and syntax mostly because, at the time, there was no unifying concept nor common syntax.\n\nThe main purpose of this document is to present in APL a unified organizing principle to classify and access various Combinatorial Algorithms.\n\nA secondary purpose is to shed light on the relationships between the various algorithms through a new perspective provided by Gian-Carlo Rota’s clever way to fit them into a single organizational framework.\n\nThe goal of this document is to describe a single APL primitive to both Count and Generate various Combinatorial Arrays: permutations, combinations, compositions, partitions, etc. The unifying (and very APL-like) principle for such a primitive is Gian-Carlo Rota's Twelvefold Way as described in Richard Stanley's \"Enumerative Combinatorics\" and Knuth’s TAoCP, Vol. 4A among other references.\n\n## The Twelvefold Way\n\nThis elegant notion consolidates twelve Combinatorial Algorithms into a single 2×2×3 array based on the simple concept of placing balls into boxes (urns, to you old-timers). The three dimensions of the array can be described as follows:\n\n• The Balls may be labeled (or not) {2 ways},\n• The Boxes may be labeled (or not) {2 ways}, and\n• The Capacity of Balls in a Box may be one of (at most one | unrestricted | at least one) {3 ways}.\n\nAmazingly, these twelve choices spanning three dimensions knit together within a single concept (balls in boxes) all of the following interesting, fundamental, and previously disparate and disorganized Combinatorial Algorithms:\n\n• Permutations\n• Combinations\n• Compositions\n• Multisets\n• Partitions of a set\n• Partitions of a number\n• Tuples\n• Pigeon Holes\n\nAs mentioned above, although the first element of the Function Selector is an integer, it is written here with three digits to emphasize that each digit has a separate meaning. Those meanings are exactly related to the 2×2×3 array mentioned above.\n\n• The first digit represents the Balls as Unlabeled (0) or Labeled (1)\n• The second digit represents the Boxes as Unlabeled (0) or Labeled (1)\n• The third digit represents the Capacity of Balls in a Box as one of At most One (0), Unrestricted (1), or At Least One (2).\n\nFor example:\n\n• A function selector of 010 means unlabeled balls (0), labeled boxes (1), and at most one ball per box (0). If we have 2 unlabeled balls (●●) and 4 labeled boxes (1234) with at most one ball per box, there are 6 (↔ 2!4) ways to meet these criteria:\n ● ● 1 2 3 4\n ● ● 1 2 3 4\n ● ● 1 2 3 4\n ● ● 1 2 3 4\n ● ● 1 2 3 4\n ● ● 1 2 3 4\n ⇐ Box Contents ⇐ Box Labels (blank = Unlabeled)\n\nfrom which it is easy to see that these criteria correspond to M combinations of N items (↔ M!N).\n\n► Notice how we obtain the generated answer as a 6×M matrix. In this example, it is obtained by reading the Box Labels in ascending order (6 M⍴1 2, 1 3, 1 4, 2 3, 2 4, 3 4). Also note that because each individual Combination is always written in ascending order, that explains why the Function Selector uses Unlabeled Balls and Labeled Boxes with At Most One Ball per Box. That is, the use of Unlabeled Balls forces us to read in ascending order the labels on the Labeled Boxes; because there are M Balls and At Most One Ball Per Box we are assured of obtaining exactly M box labels per row.\n• A function selector of 110 means labeled balls (1), labeled boxes (1), and at most one ball in each box (0). If we have 3 labeled balls (❶❷❸) and 3 labeled boxes (123) with at most one ball per box, there are 6 (↔ (!⍠¯3)3 ↔ 3×2×1) ways to meet these criteria:\n ❶ ❷ ❸ 1 2 3\n ❷ ❶ ❸ 1 2 3\n ❷ ❸ ❶ 1 2 3\n ❶ ❸ ❷ 1 2 3\n ❸ ❶ ❷ 1 2 3\n ❸ ❷ ❶ 1 2 3\n\nIf we have 2 labeled balls (❶❷) and 3 labeled boxes (123) with at most one ball per box, there are 6 (↔ (!⍠¯2)3 ↔ 3×2) ways to meet these criteria:\n\n ❶ ❷ 1 2 3\n ❷ ❶ 1 2 3\n ❷ ❶ 1 2 3\n ❶ ❷ 1 2 3\n ❶ ❷ 1 2 3\n ❷ ❶ 1 2 3\n\nfrom which it is easy to see that these criteria correspond to M permutations of N items. When M=N, this is the # permutations of ⍳N, (↔ !N), and when M<N, this is the # M-permutations, also called the Falling Factorial (!⍠(-M) N).\n\n► Notice how we obtain the generated answer as a 6×M matrix differently from the Combinations example. In the two Permutations examples, it is obtained by reading the Box Labels in ascending order of the Ball Labels. For the first Permutations example (with M=3), the generated answer is 6 M⍴1 2 3, 2 1 3, 3 1 2, 1 3 2, 2 3 1, 3 2 1, and for the second one (with M=2), it is 6 M⍴1 2, 2 1, 3 1, 1 3, 2 3, 3 2. It is common to need a different method of generating the answer for many of these Combinatorial Algorithms.\n\nAs a side note, the above examples reveal one of the many insights the Twelvefold Way provides into Combinatorial Algorithms. Previously, you might not have seen any connection between the algorithms for Combinations and Permutations, but, as the above examples show, they are closely related in that they differ only in the use of labeled (1) vs. unlabeled (0) balls; both algorithms use labeled boxes (1) with at most one ball per box (0).\n\n## Labeled vs. Unlabeled\n\n### Boxes\n\nFor most cases, the boxes are the columns of the result. Two or more labeled boxes may hold identical content, but because the boxes are labeled, they are considered distinct. On the other hand, unlabeled boxes with identical content are indistinguishable. For example, the following (partial) configurations of 3 unlabeled balls (●●●) in 3 unlabeled boxes\n\n ●●●\n ●●●\n ●●●\n\nare all considered equivalent and are counted only once because the boxes are unlabeled.\n\nSimilarly, the following (partial) configurations of 3 labeled balls (❶❷❸) in 2 unlabeled boxes\n\n ❶❷ ❸\n ❸ ❶❷\n ❷❶ ❸\n ❸ ❷❶\n\nare also all considered equivalent and are counted only once, again because the boxes are unlabeled.\n\nNote that the order of the (labeled) balls within a box is ignored which means that even if the boxes were labeled, the first and third configurations above are equivalent, as are the second and fourth.\n\n### Balls\n\nIn a similar manner, the counts and generations for combinations (010) and permutations (110) differ by a factor of !M, this time because of the balls: one is unlabeled and the other labeled. That is, the count for M combinations of N items is\n\nM!N ↔ (!N)÷(!N-M)×!M\n\nand the count for M permutations of N items is\n\n!⍠(-M) N ↔ (!N)÷!N-M\n\nOf course, when M=N, the permutation count is the familiar !N.\n\n## The Functions\n\nThe array of functions can be displayed as follows in a table organized by the Function Selector:\n\nFS Table Balls Per Box\n At Most One xx0\n Unrestricted xx1\n At least One xx2\n M unlabeled balls 00x N unlabeled boxes\n M pigeons 000 into N holes\n partitions of M 001 into ≤N parts\n partitions of M 002 into N parts\n M unlabeled balls 01x N labeled boxes\n M-combinations 010 of N items\n M-multisets 011 of N items\n compositions of M 012 into N parts\n M labeled balls 10x N unlabeled boxes\n M pigeons 100 into N holes\n partitions of {⍳M} 101 into ≤N parts\n partitions of {⍳M} 102 into N parts\n M labeled balls 11x N labeled boxes\n M-permutations 110 of N items\n M-tuples 111 of N items\n partitions of {⍳M} 112 into N ordered parts\n\nClick on one of the above colored cells to see more detail on that function.\n\n## Examples\n\n► Almost without exception, the counts for these functions grow quite rapidly as the arguments increase (hence the term Combinatorial Explosion). It's a good idea to check the Count value before trying to Generate the corresponding array.\n\nThe expression 110‼M N produces M Permutations of N items. When M=N, it represents the usual Permutation function.\n\n``` 110‼3 ⍝ Count the !3 Permutations\n6\n110 1‼3 ⍝ Generate the !3 Permutations\n1 2 3\n1 3 2\n3 1 2\n3 2 1\n2 3 1\n2 1 3\n110 2‼3 ⍝ Generate the !3 Permutations in Lexicographic order\n1 2 3\n1 3 2\n2 1 3\n2 3 1\n3 1 2\n3 2 1\n110 1‼2 3 ⍝ Generate the 2 Permutations of 3 items\n1 2\n2 1\n1 3\n3 1\n2 3\n3 2\nperm←{110 1‼⍵} ⍝ Permutation function\nperm 3\n1 2 3\n1 3 2\n3 1 2\n3 2 1\n2 3 1\n2 1 3\n```\n\nThe expression 10‼M N produces M Combinations of N items.\n\n``` comb←{10 1‼⍺ ⍵} ⍝ Combinations function\n3 comb 5\n1 2 3\n1 2 4\n1 3 4\n2 3 4\n1 2 5\n1 3 5\n2 3 5\n1 4 5\n2 4 5\n3 4 5\n10 2‼3 5 ⍝ Combinations in Lexicographic order\n1 2 3\n1 2 4\n1 2 5\n1 3 4\n1 3 5\n1 4 5\n2 3 4\n2 3 5\n2 4 5\n3 4 5\n```\n\nThe expression 1‼M N produces Partitions of M into at most N parts (as a nested array).\n\n``` 1‼7 3\n8\n⍪1 1‼7 3\n7\n6 1\n5 2\n5 1 1\n4 3\n4 2 1\n3 3 1\n3 2 2\n```\n\nIf you have seen the movie “The Man Who Knew Infinity” (2015) (about the life and academic career of the brilliant Indian mathematician Srinivasa Ramanujan), you may recall that at one point it focuses on the problem of calculating p(200) — the number of Partitions of the number 200 into at most 200 parts. This number can be calculated by\n\n``` 1‼200\n3972999029388```\n\nin a few hundred-thousandths of a second.\n\n## Special Cases\n\nThe problem of counting Partitions of M Into At Most M Parts has been well studied. As alluded to above, the brilliantly intuitive Indian mathematician Srinivasa Ramanujan along with G. H. Hardy and Hans Rademacher have provided a convergent series which can produce an exact value for this number. Fortunately, this algorithm (referred to as Hardy-Ramanujan-Rademacher or HRR) has been coded by Frederik Johansson using the open source library MPFR (Multiple Precision Floating Point) and is available in the open source library FLINT (Fast Library for Number Theory). This code is used for the function 1‼M when M is in the range 395 < M < 231 on the 32-bit version of NARS2000, and in the range 395 < M < 263 on the 64-bit version. The code is so efficient that it can calculate 1‼1E12 in under a minute (on a 64-bit machine only!). However, if you try this, be sure to assign the result to a variable as otherwise it will display a number with 1,113,996 digits!\n\nCorrespondingly, because partitions of M into N non-negative parts (001) is the same as partitions of N+M into N positive parts (002), (i.e., 1‼M N ←→ 2‼(M+N) N), the case of 2‼(2×M) M also takes advantage of the fast algorithm.\n\n## Memoization\n\nThis technique is a form of caching used to speed up certain algorithms, particularly recursive ones.\n\nTwo of the Combinatorial Functions (001 and 002) are dependent on the following recurrence relation for Partition Numbers defined on integer n and k:\n\n```P(0,0) = 1\nP(n,k) = 0\t\t\t\tfor n≤0 or k≤0\nP(n,k) = P(n-k,k) + P(n-1,k-1)```\n\nWithin a session of the interpreter, these values are cached internally so that subsequent requests for already calculated Partition Numbers are sped up significantly.\n\nThree other Combinatorial Functions (101, 102, and 112) are dependent on the Stirling numbers of the 2nd kind. They satisfy the following recurrence relation defined on integer n≥0 and k≥0:\n\n```S(0,0) = 1\nS(0,n) = S(n,0) = 0\t\t\tfor n>0\nS(n,k) = k × S(n-1,k) + S(n-1,k-1)```\n\nThese numbers are also cached internally by the interpreter so as to speed up subsequent access.\n\nIn case you need to clear the cache so as to time the internal algorithms starting with an empty cache, use the expression\n\n``` ∘‼1\nCache cleared```\n\nNote that this expression does not clear certain caching internal to the above HRR algorithm in FLINT.\n\n## History\n\nThe idea of consolidating these twelve algorithms into a single primitive is credited to Gian-Carlo Rota through a series of lectures given at the Massachusetts Institute of Technology (MIT). The mathematics behind the Twelvefold Way is described in several places, most notably in Richard Stanley's Enumerative Combinatorics, and Wikipedia. The name was suggested by Joel Spencer.\n\n## Implementation\n\nThis Combinatorial Operator is implemented in the Released version of NARS2000 and may be downloaded from here. For an in-depth look at the Twelvefold Way and its implementation in APL, see Smith's paper.\n\n## Conclusions\n\n• Rota’s amazing Twelvefold Way of consolidating numerous Combinatorial Algorithms through the unifying concept of Balls in Boxes into a single organizational framework is presented and each algorithm is discussed in detail with examples.\n• This organizational framework is ideally suited for implementation in APL for both counting and generation by referencing the individual algorithms using a function selector as the (left) operand to a new monadic primitive operator.\n• Insight into these Combinatorial Algorithms is gained when viewed from the perspective of the Twelvefold Way. To wit:\n• The relationships among the algorithms is made clearer when comparing their APL versions, especially through identities.\n• The algorithms are shown to have considerable dependence amongst themselves as shown through APL identities.\n• Interesting similarities within the function selector table are identified and are worthy of further investigation.\n• Thanks to the work of D. E. Knuth in his TAoCP Vol 4A, each of the twelve ways has a high quality algorithm behind it.\n• Finally, APL programmers need no longer search for the fastest APL program to generate any of several Combinatorial Counts or Generations as the fastest way is now available primitively.\n\n## Acknowledgments\n\nNo paper is written in isolation, and this paper is no exception. I’d like to thank David Liebtag, Roy Sykes, Norman Thomson, Jim Brown, Roger Hui, and Michael Turniansky for their helpful advice and suggestions."
]
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https://blog.ojhaintech.com/javascript-jumpstart-functions | [
"# Javascript Jumpstart - Functions\n\n## a.k.a Methods, we explore core concepts of functions.",
null,
"### What we'll cover\n\n• Exploring functions and their syntax\n\n• Differences between function declaration and function expression\n\n• Reusing parameter names and understanding their properties",
null,
"In JavaScript, functions are blocks of reusable code that perform specific tasks or a set of related tasks. They promote code reusability, modularity, and maintainability. Functions have a name, parameters (or arguments), and a body containing the instructions or statements that define their behaviour.\n\nWhen a function is called in JavaScript, control is transferred to that function, its statements are executed, and then control returns to the calling code. Functions can also return a value to the caller using the `return` statement.\n\n``````// let's see an example before we go further\nvar sum = a + b;\nreturn sum;\n}\n\nconsole.log(result); // Output: 8\n``````\n\nJavaScript functions can be called multiple times with different arguments, enabling code reuse. They can also be organized into modules or libraries, making them accessible to other parts of the program.\n\nNow let's go in-depth...\n\n## Functions\n\n• In javascript, `functions` are first-class citizens.\n\nIn JavaScript, the term \"functions are first-class citizens\" means that functions are treated as values and have the same capabilities as any other data type, such as numbers or strings. This concept enables functions to be assigned to variables, passed as arguments to other functions, returned as values from functions, and stored in data structures.\n\n• A function is an `object`. Remember, almost everything in javascript is an object.\n\n1. Functions can be assigned to variables and properties in objects.\n\n2. We can pass functions as parameters/arguments in other functions.\n\n3. We can return functions as values from other functions.\n\n4. This behaviour of functions allows us to do Functional Programming in JavaScript.\n\n*we might explore functional programming in another series.\n\n• When we create a function using `function declaration` we create a new variable in memory.\n\n`````` // this is called a function declaration\n// it is one way of creating a function\nfunction myFn() {}\n``````\n\nThis will create a variable `myFn` which holds `function` as a value. Let's inspect this function in the developer console in the browser to get a look under the hood.\n\nThe method `console. dir()` displays an interactive list of the properties of the specified JavaScript object.\n\nSee below, we expanded the `[[Prototype]]` property of the function `myFn` and we see, it contains `f ()` and then within that, it contains `Object`. So we verified that a `function` is an object.",
null,
"• In JavaScript, `[[Prototype]]` and `__proto__` are both mechanisms used to establish the prototype chain, which is the mechanism for object inheritance. *there are differences between the two and we will explore the prototypal chain and related topics in depth when we see advanced concepts.\n\n• `name: \"myFn\"` is the name of the function.\n\n• `[[FunctionLocation]]` points to the location of the function.\n\n• `[[Scopes]]: Scopes` describes the scope chain. This is related to the accessibility and visibility of variables and functions in the execution context. For now, it's enough to be aware of this without understanding it.",
null,
"• A function is a set of reusable statements and expressions.\n\n`````` let a = 5;\nlet b = 3;\nlet c;\nc = a + b;\nconsole.log(c); // 8\na = 20;\nb = 10;\nc = a + b;\nconsole.log(c); // 30\n``````\n\nWe are repeating some lines of code in our script\n\n`````` // repeated code\nc = a + b;\nconsole.log(c);\n``````\n\nThis is a good candidate for us to turn into a function, let's create one\n\n`````` let a = 5;\nlet b = 3;\n// function declaration\nfunction sum(a, b) {\nlet c = a + b;\nconsole.log(c);\n}\n// invoking function with arguments\nsum(a, b); // 8\na = 10;\nb = 20;\nsum(a, b); // 30\n``````\n\n### Syntax of a function\n\n• Function Declaration\n\n`````` // this way of creatiing a function\n// syntax is called function declaration\nfunction myFn(a, b) {\nlet c;\na = a + 1;\nc = a + b;\nreturn c;\n}\n``````\n1. Here, the `name` is `myFn` which is variable, (a, b) is `parameters`, everything between `{}` is called `function body`, the `return` keyword is used to return the result of a function.\n\n2. After the closing curly brace, there is no `;` semi-colon. In JavaScript after a block of statements we don't need to add a semi-colon.\n\n3. If we add this function to the existing code, nothing will happen. To use the function we need to call the function\n\n`````` // myFn(10, 3) is calling the function\n// we wrap it in console.log() to see the output\nconsole.log(myFn(10, 3)); // 13\n``````\n4. Calling the function will execute it\n\n• `\"a\"` is assigned value 10 & `\"b\"` is assigned value 3 ( parameters are assigned corresponding values of arguments )\n\n• variable `\"c\"` is declared\n\n• value of `\"a\"` is incremented by 1\n\n• the sum of the values \"a\" and \"b\" is assigned to \"c\"\n\n• value of \"c\" is returned\n\nThis completes the function call, the function stops after the return statement.\n\n• The parameters of the function are variables, we do not need to declare them with `var, let or const`, these variables are automatically declared and assigned at the time of each function call.\n\n• Parameters, a return statement and arguments in a function call are optional.\n\n`````` // This is a valid function declaration\nfunction myFn() {}\nmyFn(); // undefined\nmyFn(10, true); // undefined\n``````\n\nWe can call a function with arguments even if there are no parameters, it will return undefined. *this kind of behaviour is not good, using typescript we can avoid such issues.\n\n• A function can be of the following types.",
null,
"• Reusing parameter names in code.\n\n`````` // global scope\nfunction myFn1(a, b) { /*function scope*/ }\nfunction myFn2(a, b) { /*function scope*/ }\nlet a = 5;\nlet b = 6;\n``````\n1. We can reuse parameter names.\n\n2. Parameters (a, b) are limited to that function only and hence can be reused in different functions.\n\n3. Variables a, and b are declared in the global scope. These are valid declarations.\n\n• Function Declaration vs Function Expression\n\n1. Function Declaration consists of function keyword, function name, parameters which are optional and function body.\n\n2. Function Expression does not have a function name which means function expression is anonymous, it is an anonymous function expression.\n\n`````` // anonymous function expression\nfunction(a, b) {\nlet c;\na = a + 1;\nc = a + b;\nreturn c;\n}\n``````\n• Tabe to outline the usage of `function declaration vs function expression`",
null,
"The most common use case of Function Expression is being passed as an argument in a call to another function i.e. callback function.\n\n• `Function expression` assigned to a variable.\n\n`````` const myFn = function(a, b) {\nlet c;\na = a + 1;\nc = a + b;\nreturn c;\n}\n// invoking/calling the function\nmyFn(5, 2); // 7\n``````\n\n`const` is mostly used to declare variables that hold a function as a value, to prevent reassignment of this variable. *we will revisit this in scopes later\n\n• In case of a function expression passed as an argument in a call to another function we don't need to assign the anonymous function expression to any variable.\n\n`````` // setTimeout global scoped function like we discussed in Javascript Jumpstart - Types & Variables\n\n// accepts a callback function which is a anonymous function expression being passed to the setTimeout function.\n\n// this anonymous function expression is called a callback\nsetTimeout(function() {\nconsole.log('Delayed Message')\n}, 1000);\n``````\n\nThis article explores all the core concepts we need to know to get a good handle on functions in javascript. Write and execute all different code snippets.\n\nNext article, we will explore operators in-depth Javascript Jumpstart - Operators",
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"data:image/gif;base64,R0lGODlhAQABAIAAAAAAAP///yH5BAEAAAAALAAAAAABAAEAAAIBRAA7",
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"https://cdn.hashnode.com/res/hashnode/image/upload/v1685438680826/ec4ecab9-8721-4f75-93bd-1349d7606a3b.webp",
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https://www.5jt.com/finding-primes-with-q | [
"# Finding primes with q\n\n## Part 1: functional solutions",
null,
"Finding prime numbers is a compute-intensive task familiar to computer-science students. It is typically tackled with tightly-iterating algorithms in a language close to the hardware, such as C.\n\nIt might appear an unrewarding topic for a vector-programming language. We explore it here to study efficient vector solutions in a well-understood problem domain. We are interested not so much in the solutions themselves, but in the process of developing them, which we illustrate.\n\nThe q language handles large data tables efficiently, and supports coders familiar with SQL. It is easy to overlook the vector handling on which its power is based, and the direct access to it that the language affords. We hope our account will be of value to q developers who want to deepen their vector-programming skills.\n\nPart 1 of this article examines functional solutions to two familiar questions:\n\n• Is x prime?\n• What are the primes up to and including N?\n\nSubsequent parts will explore prime factorization and stateful solutions.\n\n## Background\n\nIt is sometimes said that in programming languages derived from Iverson’s notation (e.g. APL, J, k, q) “iteration is free”. Operators and built-in functions iterate implicitly through arrays. They are supplemented by higher-order operators that specify more elaborate iteration patterns such as Converge, Map-reduce and Fold. This has three important, but quite different consequences.\n\n• The writer is relieved of the work of writing loops and can specify common patterns of iteration with one or two symbols. With iteration as subordinated detail the remaining parts of an algorithm are easier to study.\n• Vectors correspond to machine architectures. Algorithms expressed in vectors are more tractable for parallelization. Higher levels of abstraction leave implementors more scope for optimization.\n• Condensing an algorithm to a few lines helps to focus on the iteration, implied or specified. Thinking about iteration is easier in an Iversonian language.\n\nAt the British APL Association’s 40th anniversary celebration at the Royal Society in London in 2004, implementor Arthur Whitney spoke of k code volumes being two orders of magnitude smaller than C:\n\nIt is theoretically impossible for a k program to outperform hand-coded C, because every k program compiles to a C program that is exactly as fast. Yet k programs routinely outperform hand-coded C. How is this possible? It’s because it is a lot easier to see your errors in four lines of k than in 400 lines of C.\n\nThe world does not need new prime numbers, nor even code for finding them. We offer this article as a close study of developing efficient vector solutions to problems more usually tackled with less abstract languages. We take the most elementary problems with primes:\n\n• Is X prime?\n• What are the prime numbers up to and including X?\n• What are the prime factors of X?\n\nFunctional Programming was born from the first Iversonian language, APL, so we begin with purely functional solutions.\n\nThe q Reference is a useful companion to this work.\n\n## Is X prime?\n\nA number is prime if it has exactly two positive divisors: itself and 1.\n\nConsider the first twenty integers.\n\n``````q)n:1+til@ / first x natural numbers\nq)show i:n 20\n1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20``````\n\nAbove, `n` is defined as a composition of the unary projection `1+` and the unary keyword `til`, equivalent to `{1+til x}`.\n\nEvery number is divisible by itself and 1. What other positive divisors might it have?\n\n``````q)i mod n each i\n,0\n0 0\n0 1 0\n0 0 1 0\n0 1 2 1 0\n0 0 0 2 1 0\n0 1 1 3 2 1 0\n0 0 2 0 3 2 1 0\n0 1 0 1 4 3 2 1 0\n0 0 1 2 0 4 3 2 1 0\n0 1 2 3 1 5 4 3 2 1 0\n0 0 0 0 2 0 5 4 3 2 1 0\n0 1 1 1 3 1 6 5 4 3 2 1 0\n0 0 2 2 4 2 0 6 5 4 3 2 1 0\n0 1 0 3 0 3 1 7 6 5 4 3 2 1 0\n0 0 1 0 1 4 2 0 7 6 5 4 3 2 1 0\n0 1 2 1 2 5 3 1 8 7 6 5 4 3 2 1 0\n0 0 0 2 3 0 4 2 0 8 7 6 5 4 3 2 1 0\n0 1 1 3 4 1 5 3 1 9 8 7 6 5 4 3 2 1 0\n0 0 2 0 0 2 6 4 2 0 9 8 7 6 5 4 3 2 1 0\n\nq)(i; 2=sum each 0=i mod n each i)\n1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20\n0 1 1 0 1 0 1 0 0 0 1 0 1 0 0 0 1 0 1 0``````\n\nAbove we exploited the iteration implicit in Equal and `mod`, but had to use `each` to iterate `sum` and `n`. Would we do better with a lambda and a single iterator?\n\nThe command `\\ts` lets us compare execution time and memory use of the two expressions. We have ideas about what is likely to be efficient; this is a reality check. Run each a thousand times.\n\n``````q)\\ts:1000 2=sum each 0=i mod n each i\n23 13568\nq)\\ts:1000 {2=sum 0=x mod n x}each i\n32 2432``````\n\nAbove, the iteration implicit in `mod` saves time but uses an order of magnitude more memory. With larger numbers to test, the saving disappears.\n\n``````q)show R / random ints\n5397760 8463309 3233691 3184915 9057464 4191115 6302698 7578959 893383 5312847\nq)\\ts 2=sum each 0=R mod n each R\n522 2583692800\nq)\\ts {2=sum 0=x mod n x}each R\n476 536871952``````\n\nThat gives us our first functional version of Is X prime?.\n\n``````q)ipf0:{2=sum 0=x mod n x}'\nq)ipf0 R\n0000000110b``````\n\nThe above illustrates an important development practice. (Iversonian languages pioneered the REPL.)\n\nDevelop an algorithm by experiments in the REPL not on one but a list of values.\n\nIf we were using formal software-development methods, we would have started by writing a suite of tests. Our informal equivalent is to experiment with a list of values. This is sometimes called exploratory programming.\n\nWe can use the Display operator `0N!` to instrument our function and observe intermediate values.\n\n``````q){2=sum 0=x mod 0N!n x}R 0\n1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29..\n0b``````\n\n### Fewer right arguments to `mod`\n\nThat’s a long right argument to `mod`. We can shorten it in two ways.\n\nFirst, if `X` has any divisors, any larger than its square root must be paired with another smaller than its square root. If we find no divisors in `n floor sqrt X`, then `X` is prime.\n\n``````q)(i; {not 0 in x mod 1 _ n floor sqrt x}each i)\n1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20\n1 1 1 0 1 0 1 0 0 0 1 0 1 0 0 0 1 0 1 0``````\n\n1 is a special case, equal to its square root.\n\n``````q)ipf1:{(x<>1)and not 0 in x mod 1 _ n floor sqrt x}\nq)(i; ipf1 each i)\n1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20\n0 1 1 0 1 0 1 0 0 0 1 0 1 0 0 0 1 0 1 0``````\n\nThat makes a big difference for big numbers.\n\n``````q)R 0\n5397760\nq)\\ts:1000 ipf0 1#R\n46696 268436304\nq)\\ts:1000 ipf1 each 1#R\n12 131360``````\n\nThe other way we can shorten the right argument to `mod` is by including only prime numbers.\n\nThat suggests some circularity: primality is what we are testing! It suggests the possibility of a stateful solution. We shall return to this in later parts of the article.\n\nBut an efficient functional primes-to function might make a difference here. We’ll return to this question after developing one.\n\n### Fewer left arguments to `mod`\n\nWe can also filter the left argument of `mod`. The value tested above is not prime. We can see that without using `mod`, because an even number cannot be prime; nor a number ending in 5. Where `x` is above 10 we need test only values ending in 1, 3, 7, or 9.\n\n``````q)R\n5397760 8463309 3233691 3184915 9057464 4191115 6302698 7578959 893383 5312847``````\n\nOnly four items of `R` require testing by `mod`. We could iterate Cond `\\$`.\n\n``````q)ld:last 10 vs / last digit/s\nq){\\$[ld[x]in 1 3 7 9;ipf1 x;0b]}each R\n0000000110b``````\n\nBut lists will do the job faster.\n\n``````q)@'[;R] ({0b};ipf1)0 1 0 1 0 0 0 1 0 1 ld R\n0000000110b\n\nq)\\ts:1000 {\\$[ld[x]in 1 3 7 9;ipf1 x;0b]}each R\n101 132320\nq)\\ts:1000 @'[;R] ({0b};ipf1)0 1 0 1 0 0 0 1 0 1 ld R\n61 132864``````\n\nNote above how tersely the values in `R` are mapped; by `ld` to their last digits; by the flag vector to 1s and 0s; finally to either `{0b}` or `ipf1`. (The projection `@'[;R]` allows us to avoid writing `...@'R` with a long left argument.)\n\nJust for contrast, consider the same algorithm in pseudo-code.\n\n``````for-each( n in R ){\nswitch( ld(n) ) {\ncaselist 1 3 5 7:\nreturn ipf1(n);\nbreak;\ndefault:\nreturn false;\nbreak;\n}\n}``````\n\nNothing forbidding in that, but the q REPL lets us in one line edit the algorithm and test it on a list of values.\n\nExtend to small values of `x`:\n\n``````q)ipf2:{@[;where x in 2 3 5 7;:;1b] @'[;x] ({0b};ipf1)0 1 0 1 0 0 0 1 0 1 ld x}\nq)\\ts:1000 ipf1 R\n104 131664\nq)\\ts:1000 ipf2 R\n64 133376``````\n\nFiltering suggests Amend At `@` rather than a test in each iteration. But saves us no time.\n\n``````q)\\ts:1000 @[;where R in 2 3 5 7;:;1b] @[count[R]#0b;i;:;ipf1 each R i:where(ld R)in 1 3 7 9]\n68 133248``````\n\nWe have unfinished business. If we had a `pt` (‘primes to’) function that returned primes up to its argument we could replace\n\n``{(x<>1)and not 0 in x mod 1 _ n floor sqrt x}``\n\nwith\n\n``{(x<>1)and not 0 in x mod 1 _ pt floor sqrt x}``\n\nand shorten the right argument to `mod`.\n\nSo we turn now to our second task, the `pt` function: finding the primes up to `x`.\n\n## Primes to X\n\nOur first solution is: select the prime numbers from `n X`. With a test function already written, we can separate primes from composite numbers.\n\n``````q)X:1000\nq)i where ipf2 i:n X\n2 3 5 7 11 13 17 19 23 29 31 37 41 43 47 53 59 61 67 71 73 79 83 89 97 101 10..``````\n\nAnother simple strategy is to generate all the composite numbers and eliminate them.\n\n``````q){x except raze x*/:\\:x}1_i\n2 3 5 7 11 13 17 19 23 29 31 37 41 43 47 53 59 61 67 71 73 79 83 89 97 101 10..``````\n\nThis has the virtue of simplicity. By eliminating all multiples of all numbers below `x` the algorithm is obviously correct. But it generates even more composite numbers above `x` than below, and there is a price for that.\n\n``````q)\\ts i where ipf2 i\n1 131968\nq)\\ts {x except raze x*/:\\:x}1_i\n51 16794496``````\n\nSometimes we know values will be small, and simple code outweighs the cost of overcomputing. Not here.\n\n``````q)ptf0:{x where ipf2 x}n@\nq)ptf0 100\n2 3 5 7 11 13 17 19 23 29 31 37 41 43 47 53 59 61 67 71 73 79 83 89 97``````\n\nThe definition of `ptf0` above uses `@` to compose two unary functions: the lambda and `n`. It follows a pattern in which if `u` and `v` are unaries, `u v@` is equivalent to `{u v x}`.\n\nA quite different strategy is the Sieve of Eratosthenes, which does no arithmetic at all. As we discover primes, we eliminate their multiples from the candidates.\n\nTo find the primes below 100, flag the candidates. Start by eliminating 1 and the even numbers.\n\n``````q)show s:0b,99#01b\n00101010101010101010101010101010101010101010101010101010101010101010101010101..``````\n\nFor convenience, visualize the candidates as a matrix.\n\n``````q)see:{10 cut ?[x;1+til count x;0N]}\nq)see s\n3 5 7 9\n11 13 15 17 19\n21 23 25 27 29\n31 33 35 37 39\n41 43 45 47 49\n51 53 55 57 59\n61 63 65 67 69\n71 73 75 77 79\n81 83 85 87 89\n91 93 95 97 99``````\n\nAbove, the ternary operator `?` is Vector Conditional.\n\nThe first listed candidate is always prime.\n\n``````q)1+s?1b\n3``````\n\nEliminate its multiples. And the next. And so on.\n\n``````q)see s:s and 100#10b where((0N!1+s?1b)-1),1\n3\n5 7\n11 13 17 19\n23 25 29\n31 35 37\n41 43 47 49\n53 55 59\n61 65 67\n71 73 77 79\n83 85 89\n91 95 97\n\nq)see s:s and 100#10b where((0N!1+s?1b)-1),1\n5\n7\n11 13 17 19\n23 29\n31 37\n41 43 47 49\n53 59\n61 67\n71 73 77 79\n83 89\n91 97\n\nq)see s:s and 100#10b where((0N!1+s?1b)-1),1\n7\n\n11 13 17 19\n23 29\n31 37\n41 43 47\n53 59\n61 67\n71 73 79\n83 89\n97``````\n\nWe can stop here. The next prime, 11, exceeds the square root of 100. There are no more numbers to eliminate. We found the primes 2 3 5 7 – and whatever else remains in the sieve.\n\n``````q)2 3 5 7,1+where s\n2 3 5 7 11 13 17 19 23 29 31 37 41 43 47 53 59 61 67 71 73 79 83 89 97``````\n\nThe iteration above fits the While iterator pattern. The Converge, Do, and While iterators apply to unary functions, so we represent state as a list: the unary returns a new list for the next iteration.\n\nOur state is a pair of lists: known primes, and the sieve flagging the candidates. We might start with 2 and the odd numbers as candidates.\n\n``````q)X:100\nq)show is:(2;0b,1_X#10b) / initial state\n2\n00101010101010101010101010101010101010101010101010101010101010101010101010101..``````\n\nThe sieve finds the next prime, appends it to the list, and removes its multiples from the bitmap. It returns the revised pair.\n\n``````q){n:1+y?1b;(x,n;y and count[y]#10b where(n-1),1)}. is\n2 3\n00001010001010001010001010001010001010001010001010001010001010001010001010001..``````\n\nNote the projection of Apply `.`, which lets the lambda refer to the pair items simply as `x` and `y`. Here the Do iterator `\\` applies it three times, returning the results of 0, 1, 2, and 3 iterations.\n\n``````q)sieve:{n:1+y?1b;(x,n;y and count[y]#10b where(n-1),1)}.\nq)3 sieve\\is / the Do iterator\n2 001010101010101010101010101010101010101010101010101010101010101010101..\n2 3 000010100010100010100010100010100010100010100010100010100010100010100..\n2 3 5 000000100010100010100010000010100000100010100010100010000010100000100..\n2 3 5 7 000000000010100010100010000010100000100010100010000010000010100000100..``````\n\nWhen to stop?\n\nWe could stop sieving when we run out of 1s. For that we have the Converge iterator, `\\` or`/`.\n\n``````q).[{\\$[any y;[n:1+y?1b;(x,n;y and count[y]#10b where(n-1),1)];(x;y)]}]\\ [is]\n2 001010101010101..\n2 3 000010100010100..\n2 3 5 000000100010100..\n2 3 5 7 000000000010100..\n2 3 5 7 11 000000000000100..\n2 3 5 7 11 13 000000000000000..\n2 3 5 7 11 13 17 000000000000000..\n2 3 5 7 11 13 17 19 000000000000000..\n2 3 5 7 11 13 17 19 23 000000000000000..\n2 3 5 7 11 13 17 19 23 29 000000000000000..\n2 3 5 7 11 13 17 19 23 29 31 000000000000000..\n2 3 5 7 11 13 17 19 23 29 31 37 000000000000000..\n2 3 5 7 11 13 17 19 23 29 31 37 41 000000000000000..\n2 3 5 7 11 13 17 19 23 29 31 37 41 43 000000000000000..\n2 3 5 7 11 13 17 19 23 29 31 37 41 43 47 000000000000000..\n2 3 5 7 11 13 17 19 23 29 31 37 41 43 47 53 000000000000000..\n2 3 5 7 11 13 17 19 23 29 31 37 41 43 47 53 59 000000000000000..\n2 3 5 7 11 13 17 19 23 29 31 37 41 43 47 53 59 61 000000000000000..\n2 3 5 7 11 13 17 19 23 29 31 37 41 43 47 53 59 61 67 000000000000000..\n2 3 5 7 11 13 17 19 23 29 31 37 41 43 47 53 59 61 67 71 000000000000000..\n2 3 5 7 11 13 17 19 23 29 31 37 41 43 47 53 59 61 67 71 73 000000000000000..\n2 3 5 7 11 13 17 19 23 29 31 37 41 43 47 53 59 61 67 71 73 79 000000000000000..``````\n\nBut we don’t need all these iterations! We can stop when we have found 2, 3, 5, and 7. By then we have eliminated multiples of all the numbers up to the square root of `X`. Any remaining 1s in the list mark primes.\n\nHere we use the While iterator with `{any z#y}[;;10].` as the test function.\n\n``````q)({any z#y}[;;10].)({n:1+y?1b;(x,n;y and count[y]#10b where(n-1),1)}.)\\ is\n2 001010101010101010101010101010101010101010101010101010101010101010101..\n2 3 000010100010100010100010100010100010100010100010100010100010100010100..\n2 3 5 000000100010100010100010000010100000100010100010100010000010100000100..\n2 3 5 7 000000000010100010100010000010100000100010100010000010000010100000100..``````\n\nThat gives us\n\n``````sieve1:{n:1+y?1b;(x,n;y and count[y]#10b where(n-1),1)}.\nes:{[s;N]{x,1+where y}. ({any z#y}[;;floor sqrt N].)s/(2;0b,01b where 1,N-2)}``````\n\nIn `es`, `N` is the number up to which to find primes, and `s` is the sieve function. (We shall compare some alternatives.)\n\nThe test function `{any z#y}[;;floor sqrt N]` checks whether all the candidates up to the square root of `N` have been eliminated. Projecting a ternary lambda on `floor sqrt N` binds the test to the square-root. (The algorithm has just one arithmetic calculation, and evaluates it just once.)\n\nFinally `{x,1+where y}.` combines the pair: the found primes and the bitmask.\n\n``````q)es[sieve1] 100\n2 3 4 5 7 11 13 17 19 22 23 26 29 31 34 37 38 41 43 46 47 53 58 59 61 62 67 71 73 74 79 82 83 86 89 94 97``````\n\nStudying `sieve1`, we see each iteration performs an AND between two bitmasks. This looks like another array-language ‘overcompute’. We actually only need to set certain indexes to false. That suggests a sieve that uses Amend At `@`. Which is faster?\n\n• calculate the indexes and amend at them\n``````q)sieve2:{n:1+y?1b;(x,n;@[y;1_-[;1]n*til 1+count[y]div n;:;0b])}.\nq)X:2000000\nq)es[sieve1;X]~es[sieve2;X]\n1b\nq)\\ts:100 es[sieve1;X]\n4505 18874640\nq)\\ts:100 es[sieve2;X]\n3597 18875088``````\n\nA solid win for Amend At over `and`? Not quite: `sieve1` runs faster for smaller values of `X`, and only falls behind with `X` at a million or more. But `sieve2` works better for the Project Euler challenge.\n\n``ptf1:es[sieve2]``\n\nHow does this compare with testing for primes? Project Euler asks for primes below 2,000,000.\n\n``````q)\\ts ptf0 2000000\n4538 486540304\nq)\\ts ptf1 2000000\n43 8391088``````\n\nEratosthenes wins by two orders of magnitude.\n\n## The test revisited\n\nNow we have an efficient primes-to function, can we use it to improve our test, by using only primes as the right argument of `mod`? Recall when testing for primality we saw the possibility of replacing\n\n``x mod 1_ n floor sqrt x``\n\nwith\n\n``x mod 1_ pt floor sqrt x``\n\nwhere `pt` is the primes-to function. Perhaps it’s faster to find the primes than to calculate the `mod`s with non-primes?\n\n``````ipf1:{(x<>1)and not 0 in x mod 1 _ n floor sqrt x}\nipf2:{@[;where x in 2 3 5 7;:;1b] @'[;x] ({0b};ipf1)0 1 0 1 0 0 0 1 0 1 ld x}\n\nipf3:{(x<>1)and not 0 in x mod 1 _ ptf1 floor sqrt x}\nipf4:{@[;where x in 2 3 5 7;:;1b] @'[;x] ({0b};ipf3)0 1 0 1 0 0 0 1 0 1 ld x}``````\n``````q)\\ts:10000 ipf2 R\n697 131808\nq)\\ts:10000 ipf4 R\n2863 131920``````\n\nNo such luck. But perhaps, if we did not have to compute the result of `pt sqrt N` for each test? If we knew all the primes up to two million? We shall revisit this when we look at stateful solutions.\n\n## Conclusion\n\nHow far have we got? We have functional answers to the questions Is X prime? and What are the primes up to N? For the former we saved time by making tests smarter. We used the smarter tests to identify primes up to N but found Eratosthenes’s sieve faster by two orders of magnitude.\n\nWe achieved all this without a single control-flow construct, and our vector solutions exploit machine architectures well. The resulting code occupies a handful of lines and imports no libraries.\n\nIn the next parts of this article we shall look at prime decomposition, and at how our functional solutions can be improved with state.\n\n## Acknowledgements\n\nOur thanks to Ferenć Bodon, Geo Carncross, and Rian Ó’Cuinneagáin for reviewing earlier versions of this work. Any remaining errors are of course ours."
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| {"ft_lang_label":"__label__en","ft_lang_prob":0.707262,"math_prob":0.9681306,"size":16915,"snap":"2021-43-2021-49","text_gpt3_token_len":6259,"char_repetition_ratio":0.16876589,"word_repetition_ratio":0.21678744,"special_character_ratio":0.45060596,"punctuation_ratio":0.119893,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.96379775,"pos_list":[0,1,2],"im_url_duplicate_count":[null,2,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-12-08T16:31:43Z\",\"WARC-Record-ID\":\"<urn:uuid:45740af8-dbff-4b93-be9c-fd527e76a9eb>\",\"Content-Length\":\"97430\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:b425eb22-544b-4d31-b102-ee5abc688197>\",\"WARC-Concurrent-To\":\"<urn:uuid:24a960c4-b14c-463b-bc4e-eae584f45d1d>\",\"WARC-IP-Address\":\"54.191.179.131\",\"WARC-Target-URI\":\"https://www.5jt.com/finding-primes-with-q\",\"WARC-Payload-Digest\":\"sha1:MG3AL3WWRRQQL7O26ZAKCXC6N247NLSP\",\"WARC-Block-Digest\":\"sha1:2GY6LPQ46NVVBCP4MQQ6XQIWZBHRDZG7\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-49/CC-MAIN-2021-49_segments_1637964363515.28_warc_CC-MAIN-20211208144647-20211208174647-00411.warc.gz\"}"} |
https://axiomsofchoice.org/sheaf | [
"## Sheaf\n\n### Collection\n\n context $\\langle X,\\mathcal T\\rangle$ … topological space definiendum $F$ in it inclusion $F$ … seperated presheaf for all $U\\in \\mathcal T$ for all $C_U$ … open cover$(U)$ for all $S:\\prod_{V\\in C} FV$ postulate $\\left(\\forall(V,W\\in C_U).\\ S(V)|_{V\\cap W}=S(W)|_{V\\cap W}\\right) \\implies \\exists (s\\in FU).\\ \\forall V.\\ S(V)=s|_V$\n\n### Discussion\n\nAs in seperated presheaf, $t|_V$ denotes the image of $t$ under $F(i)$ with $i:V\\to W$.\n\n#### Elaboration\n\nPreviously, we defined the 'Locality axiom' which makes a presheaf into a seperated presheaf. The postulate in this entry is the second axiom which makes a seperated presheaf into a sheaf.\n\nGluing axiom: If $V\\in C_U$ are elements of a cover $C_U$ of an open set $U$, the function $S$ selects one section $S(V)$ from every $FV$. A seperated presheaf is a sheaf if such a collection of selected to-be-partions-of-a-section $S(V)$, which locally agree with each other, indeed come from a globally defined section $s$ which is an element of the big $FU$. So the postulate says that the image of $F$ contains all sections which can arise from gluing together other available sections.\n\n#### Example\n\n“Any continuous map of topological spaces determines a sheaf of sets. Let f : Y → X be a continuous map. We define a sheaf Γ(Y/X) on X by setting Γ(Y/X)(U) equal to the sections U → Y, that is, Γ(Y/X)(U) is the set of all continuous functions s : U → Y such that f ∘ s = id_U. Restriction is given by restriction of functions. This sheaf is called the sheaf of sections of f, and it is especially important when f is the projection of a fiber bundle onto its base space. Notice that if the image of f does not contain U, then Γ(Y/X)(U) is empty. For a concrete example, take X = C \\ {0}, Y = C, and f(z) = exp(z). Γ(Y/X)(U) is the set of branches of the logarithm on U.”\n\nWikipedia: Sheaf\n\n### Parents\n\n#### Subset of",
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https://www.sololearn.com/Discuss/2756150/plz-help-me-to-solve-the-covid-data-analysis-project-in-python-for-data-science | [
"",
null,
"plz help me to solve the covid data analysis project in python for data science | Sololearn: Learn to code for FREE!\n\n0\n\n# plz help me to solve the covid data analysis project in python for data science\n\n+4\n\nok u add ratio. Now u need filter dates with max ratio for this u can use simple equal df[\"ratio\"]==df[\"ratio\"].max() put this equal in df[here] and u got dates with max ratio https://code.sololearn.com/co287aYKt37I/?ref=app\n\n+2\n\nI know what this task is. I want to know what your problem is. you want to get a solution or you want to get knowledge of how to solve.\n\n0\n\nHmm, where your try or problem question?\n\n0\n\nIn sololearn python for data science course data manipulation with pandas chapter project\n\n0\n\nimport pandas as pd df = pd.read_csv(\"/usercode/files/ca-covid.csv\") df.drop('state', axis=1, inplace=True) df.set_index('date', inplace=True) df['ratio']=df['deaths']/df['cases'] maxindx=df['ratio'].idxmax() print(df.iloc[maxindx]) i tried like this.. its not getting perfect\n\n0\n\nI got it.. thnk u so much 😊\n\n0\n\nimport pandas as pd df = pd.read_csv(\"/usercode/files/ca-covid.csv\") df.drop('state', axis=1, inplace=True) df.set_index('date', inplace=True) df[\"ratio\"] = df[\"deaths\"]/df[\"cases\"] max_ratio = df.loc[df[\"ratio\"]==df[\"ratio\"].max()] print(max_ratio)"
]
| [
null,
"https://www.facebook.com/tr",
null
]
| {"ft_lang_label":"__label__en","ft_lang_prob":0.66122246,"math_prob":0.86750877,"size":1198,"snap":"2022-27-2022-33","text_gpt3_token_len":338,"char_repetition_ratio":0.1080402,"word_repetition_ratio":0.10666667,"special_character_ratio":0.28464106,"punctuation_ratio":0.148,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9994253,"pos_list":[0,1,2],"im_url_duplicate_count":[null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-07-01T23:20:59Z\",\"WARC-Record-ID\":\"<urn:uuid:b115dd2e-0878-45bd-a949-b11d28b0386c>\",\"Content-Length\":\"36197\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:692d1616-620f-4b49-8bc6-390c6f59b4e7>\",\"WARC-Concurrent-To\":\"<urn:uuid:f3533ec7-28ec-40ef-a02e-597ff36ac9f5>\",\"WARC-IP-Address\":\"104.18.16.27\",\"WARC-Target-URI\":\"https://www.sololearn.com/Discuss/2756150/plz-help-me-to-solve-the-covid-data-analysis-project-in-python-for-data-science\",\"WARC-Payload-Digest\":\"sha1:OIH5K7475UXE4ZMN25LIGPC75QD742IS\",\"WARC-Block-Digest\":\"sha1:7EA6MC7FAIC4NF5MDDXECGUIVUFCVFMT\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-27/CC-MAIN-2022-27_segments_1656103947269.55_warc_CC-MAIN-20220701220150-20220702010150-00742.warc.gz\"}"} |
https://anvio.org/help/8/artifacts/variability-profile-xml/ | [
"# variability-profile-xml",
null,
"A XML-type anvi’o artifact. This artifact can be generated, used, and/or exported by anvi’o. It can also be provided by the user for anvi’o to import into its databases, process, and/or use.\n\n🔙 To the main page of anvi’o programs and artifacts.\n\n## Provided by\n\nanvi-gen-variability-network\n\n## Required or used by\n\nThere are no anvi’o tools that use or require this artifact directly, which means it is most likely an end-product for the user.\n\n## Description\n\nAn XML formatted network file that can be read and visualized by the program Gephi.\n\nAt the time of preparing this particular artifact document, An example output looked like this:\n\n``````<?xml version=\"1.0\" encoding=\"UTF-8\"?>\n<gexf xmlns:viz=\"http:///www.gexf.net/1.1draft/viz\" xmlns=\"http://www.gexf.net/1.2draft\" version=\"1.2\">\n<meta lastmodifieddate=\"2010-01-01+23:42\">\n<creator>Oligotyping pipeline</creator>\n</meta>\n<graph type=\"static\" defaultedgetype=\"undirected\">\n\n<nodes>\n<node id=\"FIN_0226C_014D\" label=\"FIN_0226C_014D\">\n<viz:size value=\"8\"/>\n</node>\n<node id=\"FIN_0226C_030D\" label=\"FIN_0226C_030D\">\n<viz:size value=\"8\"/>\n</node>\n<node id=\"FIN_0226C_090D\" label=\"FIN_0226C_090D\">\n<viz:size value=\"8\"/>\n</node>\n<node id=\"FIN_0226M_000A\" label=\"FIN_0226M_000A\">\n<viz:size value=\"8\"/>\n</node>\n<node id=\"FIN_0226M_000B\" label=\"FIN_0226M_000B\">\n<viz:size value=\"8\"/>\n</node>\n(...)\n</nodes>\n<edges>\n<edge id=\"0\" source=\"AA_1004_0\" target=\"FIN_0226C_014D\" weight=\"154.000000\" />\n<edge id=\"1\" source=\"AA_1031_0\" target=\"FIN_0226C_014D\" weight=\"144.000000\" />\n<edge id=\"2\" source=\"AA_1050_0\" target=\"FIN_0226C_014D\" weight=\"133.000000\" />\n<edge id=\"3\" source=\"AA_1250_0\" target=\"FIN_0226C_014D\" weight=\"50.000000\" />\n<edge id=\"4\" source=\"AA_1268_0\" target=\"FIN_0226C_014D\" weight=\"41.000000\" />\n<edge id=\"5\" source=\"AA_1280_0\" target=\"FIN_0226C_014D\" weight=\"33.000000\" />\n<edge id=\"6\" source=\"AA_1298_0\" target=\"FIN_0226C_014D\" weight=\"19.000000\" />\n<edge id=\"7\" source=\"AA_1299_0\" target=\"FIN_0226C_014D\" weight=\"19.000000\" />\n<edge id=\"8\" source=\"AA_1304_0\" target=\"FIN_0226C_014D\" weight=\"11.000000\" />\n(...)\n<edge id=\"3709\" source=\"GT_1013_0\" target=\"USA_0129M_000D\" weight=\"3.000000\" />\n<edge id=\"3710\" source=\"GT_1016_0\" target=\"USA_0129M_000D\" weight=\"11.000000\" />\n<edge id=\"3711\" source=\"GT_1120_0\" target=\"USA_0129M_000D\" weight=\"1.000000\" />\n<edge id=\"3712\" source=\"GT_1172_0\" target=\"USA_0129M_000D\" weight=\"3.000000\" />\n<edge id=\"3713\" source=\"GT_389_0\" target=\"USA_0129M_000D\" weight=\"2.000000\" />\n<edge id=\"3714\" source=\"GT_458_0\" target=\"USA_0129M_000D\" weight=\"4.000000\" />\n<edge id=\"3715\" source=\"GT_512_0\" target=\"USA_0129M_000D\" weight=\"1.000000\" />\n<edge id=\"3716\" source=\"GT_518_0\" target=\"USA_0129M_000D\" weight=\"2.000000\" />\n<edge id=\"3717\" source=\"GT_533_0\" target=\"USA_0129M_000D\" weight=\"1.000000\" />\n<edge id=\"3718\" source=\"GT_570_0\" target=\"USA_0129M_000D\" weight=\"1.000000\" />\n<edge id=\"3719\" source=\"GT_581_0\" target=\"USA_0129M_000D\" weight=\"1.000000\" />\n<edge id=\"3720\" source=\"GT_602_0\" target=\"USA_0129M_000D\" weight=\"3.000000\" />\n<edge id=\"3721\" source=\"GT_707_0\" target=\"USA_0129M_000D\" weight=\"3.000000\" />\n<edge id=\"3722\" source=\"GT_762_0\" target=\"USA_0129M_000D\" weight=\"3.000000\" />\n<edge id=\"3723\" source=\"GT_878_0\" target=\"USA_0129M_000D\" weight=\"2.000000\" />\n<edge id=\"3724\" source=\"GT_887_0\" target=\"USA_0129M_000D\" weight=\"5.000000\" />\n<edge id=\"3725\" source=\"GT_899_0\" target=\"USA_0129M_000D\" weight=\"6.000000\" />\n</edges>\n</graph>\n</gexf>\n``````\n\nEdit this file to update this information."
]
| [
null,
"https://anvio.org/help/8/images/icons/XML.png",
null
]
| {"ft_lang_label":"__label__en","ft_lang_prob":0.5040027,"math_prob":0.8776532,"size":3396,"snap":"2023-40-2023-50","text_gpt3_token_len":1318,"char_repetition_ratio":0.27948114,"word_repetition_ratio":0.0,"special_character_ratio":0.50706714,"punctuation_ratio":0.12149533,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99454486,"pos_list":[0,1,2],"im_url_duplicate_count":[null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-12-04T17:07:17Z\",\"WARC-Record-ID\":\"<urn:uuid:c7b25539-2696-417c-9bde-2ad97b4c687c>\",\"Content-Length\":\"20614\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:e554f601-6420-450f-ac21-be425851e131>\",\"WARC-Concurrent-To\":\"<urn:uuid:07437749-db1e-4103-b84e-835f034878e6>\",\"WARC-IP-Address\":\"185.199.108.153\",\"WARC-Target-URI\":\"https://anvio.org/help/8/artifacts/variability-profile-xml/\",\"WARC-Payload-Digest\":\"sha1:MA2T777HTCI4JHGRJYHSTBETG5PHRVHT\",\"WARC-Block-Digest\":\"sha1:E2OPBWSP4XNUDDVZI66Y54DG7LL5SB4C\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-50/CC-MAIN-2023-50_segments_1700679100531.77_warc_CC-MAIN-20231204151108-20231204181108-00142.warc.gz\"}"} |
https://www.makuwang.com/goods/jz2 | [
"## 数量\n\nSPQ：\n\n10+: 100+: 1000+: 3000+:\n\nSPQ：\n\n10+: 100+: 1000+: 3000+:\n\nSPQ：\n\n10+: 100+: 1000+: 3000+:\n\nSPQ：\n\n10+: 100+: 1000+: 3000+:\n\nSPQ：\n\n10+: 100+: 1000+: 3000+:\n\nSPQ：\n\n10+: 14.691746 100+: 11.130111 1000+: 10.450808 3000+: 9.785401 5000+: 9.481978 10000+: 9.341850\n\nSPQ：\n\n10+: 100+: 1000+: 3000+:\n\nSPQ：\n\n10+: 36.123609 100+: 27.366370 1000+: 25.696122 3000+: 24.060039 5000+: 23.313992 10000+: 22.969450\n\nSPQ：\n\n10+: 71.018639 100+: 53.801999 1000+: 50.518309 3000+: 47.301788 5000+: 45.835066 10000+: 45.157700\n\nSPQ：\n\n10+: 104.668028 100+: 79.293960 1000+: 74.454423 3000+: 69.713879 5000+: 67.552209 10000+: 66.553900\n\nSPQ：\n\n10+: 126.782433 100+: 96.047298 1000+: 90.185256 3000+: 84.443124 5000+: 81.824733 10000+: 80.615500\n\nSPQ：\n\n10+: 209.199547 100+: 158.484505 1000+: 148.811742 3000+: 139.336837 5000+: 135.016315 10000+: 133.021000\n\nSPQ：\n\n10+: 298.749239 100+: 226.325181 1000+: 212.511907 3000+: 198.981186 5000+: 192.811227 10000+: 189.961800\n\nSPQ：\n\n10+: 599.972697 100+: 454.524771 1000+: 426.783822 3000+: 399.610320 5000+: 387.219303 10000+: 381.496850\n\nSPQ：\n\n10+: 1522.174126 100+: 1153.162216 1000+: 1082.781424 3000+: 1013.840285 5000+: 982.403377 10000+: 967.885100\n\nSPQ：\n\n10+: 1.706358 100+: 1.292696 1000+: 1.213799 3000+: 1.136516 5000+: 1.101275 10000+: 1.085000\n\nSPQ：\n\n10+: 100+: 1000+: 3000+:\n\nSPQ：\n\n10+: 100+: 1000+: 3000+:\n\nSPQ：\n\n10+: 100+: 1000+: 3000+:\n\nSPQ：\n\n10+: 100+: 1000+: 3000+:\n共有3857个记录 每页显示20条，本页1-20条 1/193页 1 2 3 4 5 下一页\n\n### 在线客服\n\n2355289363\n\n2355289360\n\n2355289361\n\n2355289368",
null,
"0755-28435922"
]
| [
null,
"https://www.makuwang.com/Public/Mark/img/wx1.png",
null
]
| {"ft_lang_label":"__label__zh","ft_lang_prob":0.8047827,"math_prob":0.9949785,"size":55694,"snap":"2022-40-2023-06","text_gpt3_token_len":47693,"char_repetition_ratio":0.35026038,"word_repetition_ratio":0.9680426,"special_character_ratio":0.49705535,"punctuation_ratio":0.19205382,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9948001,"pos_list":[0,1,2],"im_url_duplicate_count":[null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-01-27T07:18:42Z\",\"WARC-Record-ID\":\"<urn:uuid:cfad8fe0-25ce-4834-8716-42a5a22939ce>\",\"Content-Length\":\"1029254\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:363e8ec1-33bf-4696-96ed-1059c21883e5>\",\"WARC-Concurrent-To\":\"<urn:uuid:93408f07-31a5-417f-bcf2-a568891f81ed>\",\"WARC-IP-Address\":\"8.48.85.207\",\"WARC-Target-URI\":\"https://www.makuwang.com/goods/jz2\",\"WARC-Payload-Digest\":\"sha1:PSCYG2RQ45M3XWGK5ZRU7LRTMZFV5XQC\",\"WARC-Block-Digest\":\"sha1:N6W54RPJKBKWAHB6PV7GFYEFSXD5VM7U\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-06/CC-MAIN-2023-06_segments_1674764494974.98_warc_CC-MAIN-20230127065356-20230127095356-00533.warc.gz\"}"} |
https://la.mathworks.com/help/nav/ref/posegraph.addrelativepose.html | [
"Add relative pose to pose graph\n\n## Syntax\n\n``addRelativePose(poseGraph,measurement)``\n``addRelativePose(poseGraph,measurement,infoMat)``\n``addRelativePose(poseGraph,measurement,infoMat,fromNodeID)``\n``addRelativePose(poseGraph,measurement,infoMat,fromNodeID,toNodeID)``\n``[nodePair,edgeID] = addRelativePose(___)``\n\n## Description\n\n````addRelativePose(poseGraph,measurement)` creates a node based on the input `measurement` that connects to the last pose node in the pose graph. To add landmark nodes, see the `addPointLandmark` function.```\n``` `addRelativePose(poseGraph,measurement,infoMat)` also specifies the information matrix as part of the edge constraint, which represents the uncertainty of the pose measurement.```\n``` `addRelativePose(poseGraph,measurement,infoMat,fromNodeID)` creates a new pose node and connects it to the specific node specified by `fromNodeID`.```\n\nexample\n\n``` `addRelativePose(poseGraph,measurement,infoMat,fromNodeID,toNodeID)` creates an edge by specifying a relative pose measurement between existing nodes specified by `fromNodeID` and `toNodeID`. This edge is called a loop closure. If a loop closure already exists, the function appends the new measurement. Calling the `optimizePoseGraph` function combines multiple appended measurements into a single edge. This syntax does not support adding edges to a landmark node.```\n``` `[nodePair,edgeID] = addRelativePose(___)` returns the newly added edge and edge ID using any of the previous syntaxes.```\n\n## Examples\n\ncollapse all\n\nThis example shows how to identify and remove spurious loop closures from pose graph. To do this, you can modify the relative pose of a loop closure edge and try optimizing the pose graph with and without removing the auto spurious loop closure and compare the results.\n\nLoad the Intel Research Lab Dataset that contains a 2-D pose graph. Optimize the pose graph. Plot the pose graph with IDs off. Red lines indicate loop closures identified in the dataset.\n\n```load intel-2d-posegraph.mat pg optimizedPG = optimizePoseGraph(pg); show(optimizedPG,IDs=\"off\"); title(\"Optimized Pose Graph\")```",
null,
"Modify the relative pose of the loop closure edge `1386` to some random values.\n\n```loopclosureId = 1386; nodePair = edgeNodePairs(optimizedPG,loopclosureId); [relPose,infoMat] = edgeConstraints(optimizedPG,loopclosureId); relPose(2) = -5; relPose(3) = 1.5; addRelativePose(optimizedPG,relPose,infoMat,nodePair(1),nodePair(2));```\n\nOptimize the pose graph without auto loop closure trimming. Plot the optimized pose graph to see the poor adjustment of the nodes with loop closures.\n\n```[updatedPG,solutionInfo] = optimizePoseGraph(optimizedPG); show(updatedPG,IDs=\"off\"); title(\"Updated Pose Graph\")```",
null,
"Certain loop closures should be trimmed from the pose graph. Use the `trimLoopClosures` function to trim these bad loop closures. Set the truncation threshold and maximum iterations for the trimmer parameters.\n\n`trimParams = struct(\"TruncationThreshold\",0.5,\"MaxIterations\",100);`\n\nGenerate solver options.\n\n`solverOptions = poseGraphSolverOptions(\"g2o-levenberg-marquardt\");`\n\nUse the `trimLoopClosures` function with the trimmer parameters and solver options. Plot the new pose graph to see the bad loop closures were removed.\n\n```[newPG,trimInfo] = trimLoopClosures(updatedPG,trimParams,solverOptions); show(newPG,IDs=\"off\"); title(\"New Pose Graph\")```",
null,
"## Input Arguments\n\ncollapse all\n\nPose graph, specified as a `poseGraph` or `poseGraph3D` object.\n\nRelative pose between nodes, specified as one of the following:\n\nFor `poseGraph` (2-D), the pose is a `[x y theta]` vector, which defines a xy-position and orientation angle, `theta`.\n\nFor `poseGraph3D`, the pose is a ```[x y z qw qx qy qz]``` vector, which defines by an xyz-position and quaternion orientation, ```[qw qx qy qz]```\n\nNote\n\nMany other sources for 3-D pose graphs, including `.g2o` formats, specify the quaternion orientation in a different order, for example, `[qx qy qz qw]`. Check the source of your pose graph data before adding nodes to your `poseGraph3D` object.\n\nInformation matrices, specified in compact form as a 6-element vector or 21-element vector.\n\nEach row is the upper triangle of the square information matrix. An information matrix represents the uncertainty of the measurement. The matrix is calculated as the inverse of the covariance. If the measurement is an `[x y theta]` vector, the covariance matrix is a 3-by-3 of pairwise covariance calculations. Typically, the uncertainty is determined by the sensor model.\n\nFor `poseGraph` (2-D), each information matrix is a six-element vector. The default is `[1 0 0 1 0 1]`. For landmark nodes, the last three elements are returned as `NaN`.\n\nFor `poseGraph3D`, each information matrix is a 21-element vector. The default is ```[1 0 0 0 0 0 1 0 0 0 0 1 0 0 0 1 0 0 1 0 1]```.\n\nNode to attach from, specified as a positive integer. This integer corresponds to the node ID of a node in `poseGraph`. When specified without `toNodeID`, `addRelativePose` creates a new node and adds an edge between the new node and the `fromNodeID` node.\n\nNode to attach to, specified as a positive integer. This integer corresponds to the node ID of a node in `poseGraph`. `addRelativePose` adds an edge between this node and the `fromNodeID` node.\n\n## Output Arguments\n\ncollapse all\n\nEdge node pairs in pose graph, returned as two-element vector that lists the IDs of the two nodes that each edge connects. Multiple edges may exist between the same pair of nodes.\n\nID of added edge, returned as a positive integer.\n\n## Version History\n\nIntroduced in R2019b"
]
| [
null,
"https://la.mathworks.com/help/examples/nav/win64/IdentifyAndRemoveSpuriousLoopClosuresFromPoseGraphExample_01.png",
null,
"https://la.mathworks.com/help/examples/nav/win64/IdentifyAndRemoveSpuriousLoopClosuresFromPoseGraphExample_02.png",
null,
"https://la.mathworks.com/help/examples/nav/win64/IdentifyAndRemoveSpuriousLoopClosuresFromPoseGraphExample_03.png",
null
]
| {"ft_lang_label":"__label__en","ft_lang_prob":0.69268024,"math_prob":0.93548155,"size":5625,"snap":"2022-40-2023-06","text_gpt3_token_len":1364,"char_repetition_ratio":0.15584415,"word_repetition_ratio":0.04411765,"special_character_ratio":0.20871112,"punctuation_ratio":0.12663317,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.98074627,"pos_list":[0,1,2,3,4,5,6],"im_url_duplicate_count":[null,1,null,1,null,1,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-09-24T22:47:18Z\",\"WARC-Record-ID\":\"<urn:uuid:c1c8e45d-9910-4691-a7ab-3e7aad46160f>\",\"Content-Length\":\"105200\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:6d6357f9-c720-4238-9016-08ddc82cf0c0>\",\"WARC-Concurrent-To\":\"<urn:uuid:38d6b4b2-11ad-4ba4-a32d-1c9c33a717ef>\",\"WARC-IP-Address\":\"104.68.243.15\",\"WARC-Target-URI\":\"https://la.mathworks.com/help/nav/ref/posegraph.addrelativepose.html\",\"WARC-Payload-Digest\":\"sha1:KLPPR5RLAJHWL24BXWL375QXUJS3AGWT\",\"WARC-Block-Digest\":\"sha1:PY52NHRWGDEBONQZ4A6D7SGEHPZFSAUY\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-40/CC-MAIN-2022-40_segments_1664030333541.98_warc_CC-MAIN-20220924213650-20220925003650-00759.warc.gz\"}"} |
https://portalanalyse.com/domain-info/mconverge.in | [
"Detailed Website Analysis\n\n# Mconverge.in summary\n\nName of website MConverge Services Pvt. Ltd. No entry at this time\n\nThe index page of the site has 6 links to other websites.\n\n## Analysis of hosting server\n\nGeographically, the server for mconverge.in is located in Singapore.\n\nGeographical area of the server Singapore 182.50.147.1 ns68.domaincontrol.com173.201.71.44United States, Arizona, Scottsdalens67.domaincontrol.com216.69.185.44United States, Arizona, Scottsdale\n\nContent-Length: 7202 HTTP/1.1 200 OK Content-Type: text/html Vary: Accept-Encoding Accept-Ranges: bytes Date: Sun, 18 Jun 2017 09:34:34 GMT Server: Apache\n\n## Most frequent website mistypes (typos)\n\n1. mconnverge.in\n2. mcconverge.in\n3. mconvarga.in\n4. mconvurgu.in\n5. mconvegre.in\n6. mcounverge.in\n7. mconverrge.in\n8. mcoverge.in\n9. msionverge.in\n10. mconvyrgy.in\n11. mconverg.in\n12. mkonverge.in\n13. mconv3rg3.in\n14. converge.in\n15. mocnverge.in\n16. mconvege.in\n17. mcanverge.in\n18. cmonverge.in\n19. mdonverge.in\n20. mconvere.in\n21. mc0nverge.in\n22. mconvereg.in\n23. mcpnverge.in\n24. mconvrege.in\n25. mconevrge.in\n26. mclnverge.in\n27. mconvirgi.in\n28. mconvergee.in\n29. mcnverge.in\n30. mmconverge.in\n31. mconveerge.in\n32. mcunverge.in\n33. mcenverge.in\n34. mcynverge.in\n35. mconvverge.in\n36. mconvrge.in\n37. mconerge.in\n38. mcnoverge.in\n39. kconverge.in\n40. jconverge.in\n41. mconveargea.in\n42. mxonverge.in\n43. mconwerge.in\n44. nconverge.in\n45. mcinverge.in\n46. mfonverge.in\n47. mvonverge.in\n48. mcovnerge.in\n49. mcknverge.in\n50. monverge.in\n51. mconvorgo.in\n52. msyonverge.in\n53. mcoonverge.in\n54. mconverge.in\n55. mconvergge.in\n56. mconverde.in\n57. mconverte.in\n58. mconvfrge.in\n59. mconvsrge.in\n60. mkconverge.in\n61. mcomverge.in\n62. mconverve.in\n63. mconvergr.in\n64. mconderge.in\n65. mconvdrge.in\n66. mconvrrgr.in\n67. mcohverge.in\n68. mconberge.in\n69. mconvergd.in\n70. nmconverge.in\n71. mconvsrgs.in\n72. mconvedge.in\n73. mconvfrgf.in\n74. mcfonverge.in\n75. mconvwrgw.in\n76. mcongerge.in\n77. mxconverge.in\n78. mcionverge.in\n79. kmconverge.in\n80. mjconverge.in\n81. mcoinverge.in\n82. mconvwrge.in\n83. mconverne.in\n84. mconvergw.in\n85. mconverre.in\n86. mconverhe.in\n87. mconveege.in\n88. mconvegge.in\n89. mconvefge.in\n90. mconverfe.in\n91. mconvdrgd.in\n92. mconvergf.in\n93. mnconverge.in\n94. mcdonverge.in\n95. mdconverge.in\n96. mcojverge.in\n97. mfconverge.in\n98. mconcerge.in\n99. mcxonverge.in\n100. mconvetge.in\n101. mvconverge.in\n102. mcvonverge.in\n103. jmconverge.in\n104. mcponverge.in\n105. mconvergs.in\n106. mconvrrge.in\n107. mconferge.in\n108. mconverye.in\n109. mcobverge.in\n110. mconverbe.in\n111. mconvedrge.in\n112. mconvgerge.in\n113. mconcverge.in\n114. mconjverge.in\n115. mconvervge.in\n116. mckonverge.in\n117. mconvwerge.in\n118. mconverege.in\n119. mcobnverge.in\n120. mcojnverge.in\n121. mconvergte.in\n122. mclonverge.in\n123. mconhverge.in\n124. mconvefrge.in\n125. mconvergye.in\n126. mconverdge.in\n127. mconvferge.in\n128. mconveryge.in\n129. mconverged.in\n130. mconvergre.in\n131. mcohnverge.in\n132. mconvergve.in\n133. mconvergwe.in\n134. mconverghe.in\n135. mconverhge.in\n136. mconvergew.in\n137. mcomnverge.in\n138. mconvrerge.in\n139. mconverfge.in\n140. mcongverge.in\n141. mconvesrge.in\n142. mconvderge.in\n143. mconvcerge.in\n144. mcondverge.in\n145. mconvserge.in\n146. mconvertge.in\n147. mconvetrge.in\n148. mconvergde.in\n149. mconvernge.in\n150. mconvergbe.in\n151. mcolnverge.in\n152. mconvergne.in\n153. mcoknverge.in\n154. mconverbge.in\n155. mconfverge.in\n156. mconvergse.in\n157. mconverges.in\n158. mconvergfe.in\n159. mconverger.in\n160. mconvegrge.in\n161. mconmverge.in\n162. mconbverge.in\n163. mconvberge.in\n164. mcopnverge.in\n165. mconvewrge.in\n166. mconvergef.in\n167. mconnverge.ib\n168. mcconverge.ib\n169. mconvarga.ib\n170. mconvurgu.ib\n171. mconvegre.ib\n172. mcounverge.ib\n173. mconverrge.ib\n174. mcoverge.ib\n175. msionverge.ib\n176. mconvyrgy.ib\n177. mconverg.ib\n178. mkonverge.ib\n179. mconv3rg3.ib\n180. converge.ib\n181. mocnverge.ib\n182. mconvege.ib\n183. mcanverge.ib\n184. cmonverge.ib\n185. mdonverge.ib\n186. mconvere.ib\n187. mc0nverge.ib\n188. mconvereg.ib\n189. mcpnverge.ib\n190. mconvrege.ib\n191. mconevrge.ib\n192. mclnverge.ib\n193. mconvirgi.ib\n194. mconvergee.ib\n195. mcnverge.ib\n196. mmconverge.ib\n197. mconveerge.ib\n198. mcunverge.ib\n199. mcenverge.ib\n200. mcynverge.ib\n201. mconvverge.ib\n202. mconvrge.ib\n203. mconerge.ib\n204. mcnoverge.ib\n205. kconverge.ib\n206. jconverge.ib\n207. mconveargea.ib\n208. mxonverge.ib\n209. mconwerge.ib\n210. nconverge.ib\n211. mcinverge.ib\n212. mfonverge.ib\n213. mvonverge.ib\n214. mcovnerge.ib\n215. mcknverge.ib\n216. monverge.ib\n217. mconvorgo.ib\n218. msyonverge.ib\n219. mcoonverge.ib\n220. mconverge.ib\n221. mconvergge.ib\n222. mconverde.ib\n223. mconverte.ib\n224. mconvfrge.ib\n225. mconvsrge.ib\n226. mkconverge.ib\n227. mcomverge.ib\n228. mconverve.ib\n229. mconvergr.ib\n230. mconderge.ib\n231. mconvdrge.ib\n232. mconvrrgr.ib\n233. mcohverge.ib\n234. mconberge.ib\n235. mconvergd.ib\n236. nmconverge.ib\n237. mconvsrgs.ib\n238. mconvedge.ib\n239. mconvfrgf.ib\n240. mcfonverge.ib\n241. mconvwrgw.ib\n242. mcongerge.ib\n243. mxconverge.ib\n244. mcionverge.ib\n245. kmconverge.ib\n246. mjconverge.ib\n247. mcoinverge.ib\n248. mconvwrge.ib\n249. mconverne.ib\n250. mconvergw.ib\n251. mconverre.ib\n252. mconverhe.ib\n253. mconveege.ib\n254. mconvegge.ib\n255. mconvefge.ib\n256. mconverfe.ib\n257. mconvdrgd.ib\n258. mconvergf.ib\n259. mnconverge.ib\n260. mcdonverge.ib\n261. mdconverge.ib\n262. mcojverge.ib\n263. mfconverge.ib\n264. mconcerge.ib\n265. mcxonverge.ib\n266. mconvetge.ib\n267. mvconverge.ib\n268. mcvonverge.ib\n269. jmconverge.ib\n270. mcponverge.ib\n271. mconvergs.ib\n272. mconvrrge.ib\n273. mconferge.ib\n274. mconverye.ib\n275. mcobverge.ib\n276. mconverbe.ib\n277. mconvedrge.ib\n278. mconvgerge.ib\n279. mconcverge.ib\n280. mconjverge.ib\n281. mconvervge.ib\n282. mckonverge.ib\n283. mconvwerge.ib\n284. mconverege.ib\n285. mcobnverge.ib\n286. mcojnverge.ib\n287. mconvergte.ib\n288. mclonverge.ib\n289. mconhverge.ib\n290. mconvefrge.ib\n291. mconvergye.ib\n292. mconverdge.ib\n293. mconvferge.ib\n294. mconveryge.ib\n295. mconverged.ib\n296. mconvergre.ib\n297. mcohnverge.ib\n298. mconvergve.ib\n299. mconvergwe.ib\n300. mconverghe.ib\n301. mconverhge.ib\n302. mconvergew.ib\n303. mcomnverge.ib\n304. mconvrerge.ib\n305. mconverfge.ib\n306. mcongverge.ib\n307. mconvesrge.ib\n308. mconvderge.ib\n309. mconvcerge.ib\n310. mcondverge.ib\n311. mconvserge.ib\n312. mconvertge.ib\n313. mconvetrge.ib\n314. mconvergde.ib\n315. mconvernge.ib\n316. mconvergbe.ib\n317. mcolnverge.ib\n318. mconvergne.ib\n319. mcoknverge.ib\n320. mconverbge.ib\n321. mconfverge.ib\n322. mconvergse.ib\n323. mconverges.ib\n324. mconvergfe.ib\n325. mconverger.ib\n326. mconvegrge.ib\n327. mconmverge.ib\n328. mconbverge.ib\n329. mconvberge.ib\n330. mcopnverge.ib\n331. mconvewrge.ib\n332. mconvergef.ib\n333. mconnverge.ih\n334. mcconverge.ih\n335. mconvarga.ih\n336. mconvurgu.ih\n337. mconvegre.ih\n338. mcounverge.ih\n339. mconverrge.ih\n340. mcoverge.ih\n341. msionverge.ih\n342. mconvyrgy.ih\n343. mconverg.ih\n344. mkonverge.ih\n345. mconv3rg3.ih\n346. converge.ih\n347. mocnverge.ih\n348. mconvege.ih\n349. mcanverge.ih\n350. cmonverge.ih\n351. mdonverge.ih\n352. mconvere.ih\n353. mc0nverge.ih\n354. mconvereg.ih\n355. mcpnverge.ih\n356. mconvrege.ih\n357. mconevrge.ih\n358. mclnverge.ih\n359. mconvirgi.ih\n360. mconvergee.ih\n361. mcnverge.ih\n362. mmconverge.ih\n363. mconveerge.ih\n364. mcunverge.ih\n365. mcenverge.ih\n366. mcynverge.ih\n367. mconvverge.ih\n368. mconvrge.ih\n369. mconerge.ih\n370. mcnoverge.ih\n371. kconverge.ih\n372. jconverge.ih\n373. mconveargea.ih\n374. mxonverge.ih\n375. mconwerge.ih\n376. nconverge.ih\n377. mcinverge.ih\n378. mfonverge.ih\n379. mvonverge.ih\n380. mcovnerge.ih\n381. mcknverge.ih\n382. monverge.ih\n383. mconvorgo.ih\n384. msyonverge.ih\n385. mcoonverge.ih\n386. mconverge.ih\n387. mconvergge.ih\n388. mconverde.ih\n389. mconverte.ih\n390. mconvfrge.ih\n391. mconvsrge.ih\n392. mkconverge.ih\n393. mcomverge.ih\n394. mconverve.ih\n395. mconvergr.ih\n396. mconderge.ih\n397. mconvdrge.ih\n398. mconvrrgr.ih\n399. mcohverge.ih\n400. mconberge.ih\n401. mconvergd.ih\n402. nmconverge.ih\n403. mconvsrgs.ih\n404. mconvedge.ih\n405. mconvfrgf.ih\n406. mcfonverge.ih\n407. mconvwrgw.ih\n408. mcongerge.ih\n409. mxconverge.ih\n410. mcionverge.ih\n411. kmconverge.ih\n412. mjconverge.ih\n413. mcoinverge.ih\n414. mconvwrge.ih\n415. mconverne.ih\n416. mconvergw.ih\n417. mconverre.ih\n418. mconverhe.ih\n419. mconveege.ih\n420. mconvegge.ih\n421. mconvefge.ih\n422. mconverfe.ih\n423. mconvdrgd.ih\n424. mconvergf.ih\n425. mnconverge.ih\n426. mcdonverge.ih\n427. mdconverge.ih\n428. mcojverge.ih\n429. mfconverge.ih\n430. mconcerge.ih\n431. mcxonverge.ih\n432. mconvetge.ih\n433. mvconverge.ih\n434. mcvonverge.ih\n435. jmconverge.ih\n436. mcponverge.ih\n437. mconvergs.ih\n438. mconvrrge.ih\n439. mconferge.ih\n440. mconverye.ih\n441. mcobverge.ih\n442. mconverbe.ih\n443. mconvedrge.ih\n444. mconvgerge.ih\n445. mconcverge.ih\n446. mconjverge.ih\n447. mconvervge.ih\n448. mckonverge.ih\n449. mconvwerge.ih\n450. mconverege.ih\n451. mcobnverge.ih\n452. mcojnverge.ih\n453. mconvergte.ih\n454. mclonverge.ih\n455. mconhverge.ih\n456. mconvefrge.ih\n457. mconvergye.ih\n458. mconverdge.ih\n459. mconvferge.ih\n460. mconveryge.ih\n461. mconverged.ih\n462. mconvergre.ih\n463. mcohnverge.ih\n464. mconvergve.ih\n465. mconvergwe.ih\n466. mconverghe.ih\n467. mconverhge.ih\n468. mconvergew.ih\n469. mcomnverge.ih\n470. mconvrerge.ih\n471. mconverfge.ih\n472. mcongverge.ih\n473. mconvesrge.ih\n474. mconvderge.ih\n475. mconvcerge.ih\n476. mcondverge.ih\n477. mconvserge.ih\n478. mconvertge.ih\n479. mconvetrge.ih\n480. mconvergde.ih\n481. mconvernge.ih\n482. mconvergbe.ih\n483. mcolnverge.ih\n484. mconvergne.ih\n485. mcoknverge.ih\n486. mconverbge.ih\n487. mconfverge.ih\n488. mconvergse.ih\n489. mconverges.ih\n490. mconvergfe.ih\n491. mconverger.ih\n492. mconvegrge.ih\n493. mconmverge.ih\n494. mconbverge.ih\n495. mconvberge.ih\n496. mcopnverge.ih\n497. mconvewrge.ih\n498. mconvergef.ih\n499. mconnverge.im\n500. mcconverge.im\n501. mconvarga.im\n502. mconvurgu.im\n503. mconvegre.im\n504. mcounverge.im\n505. mconverrge.im\n506. mcoverge.im\n507. msionverge.im\n508. mconvyrgy.im\n509. mconverg.im\n510. mkonverge.im\n511. mconv3rg3.im\n512. converge.im\n513. mocnverge.im\n514. mconvege.im\n515. mcanverge.im\n516. cmonverge.im\n517. mdonverge.im\n518. mconvere.im\n519. mc0nverge.im\n520. mconvereg.im\n521. mcpnverge.im\n522. mconvrege.im\n523. mconevrge.im\n524. mclnverge.im\n525. mconvirgi.im\n526. mconvergee.im\n527. mcnverge.im\n528. mmconverge.im\n529. mconveerge.im\n530. mcunverge.im\n531. mcenverge.im\n532. mcynverge.im\n533. mconvverge.im\n534. mconvrge.im\n535. mconerge.im\n536. mcnoverge.im\n537. kconverge.im\n538. jconverge.im\n539. mconveargea.im\n540. mxonverge.im\n541. mconwerge.im\n542. nconverge.im\n543. mcinverge.im\n544. mfonverge.im\n545. mvonverge.im\n546. mcovnerge.im\n547. mcknverge.im\n548. monverge.im\n549. mconvorgo.im\n550. msyonverge.im\n551. mcoonverge.im\n552. mconverge.im\n553. mconvergge.im\n554. mconverde.im\n555. mconverte.im\n556. mconvfrge.im\n557. mconvsrge.im\n558. mkconverge.im\n559. mcomverge.im\n560. mconverve.im\n561. mconvergr.im\n562. mconderge.im\n563. mconvdrge.im\n564. mconvrrgr.im\n565. mcohverge.im\n566. mconberge.im\n567. mconvergd.im\n568. nmconverge.im\n569. mconvsrgs.im\n570. mconvedge.im\n571. mconvfrgf.im\n572. mcfonverge.im\n573. mconvwrgw.im\n574. mcongerge.im\n575. mxconverge.im\n576. mcionverge.im\n577. kmconverge.im\n578. mjconverge.im\n579. mcoinverge.im\n580. mconvwrge.im\n581. mconverne.im\n582. mconvergw.im\n583. mconverre.im\n584. mconverhe.im\n585. mconveege.im\n586. mconvegge.im\n587. mconvefge.im\n588. mconverfe.im\n589. mconvdrgd.im\n590. mconvergf.im\n591. mnconverge.im\n592. mcdonverge.im\n593. mdconverge.im\n594. mcojverge.im\n595. mfconverge.im\n596. mconcerge.im\n597. mcxonverge.im\n598. mconvetge.im\n599. mvconverge.im\n600. mcvonverge.im\n601. jmconverge.im\n602. mcponverge.im\n603. mconvergs.im\n604. mconvrrge.im\n605. mconferge.im\n606. mconverye.im\n607. mcobverge.im\n608. mconverbe.im\n609. mconvedrge.im\n610. mconvgerge.im\n611. mconcverge.im\n612. mconjverge.im\n613. mconvervge.im\n614. mckonverge.im\n615. mconvwerge.im\n616. mconverege.im\n617. mcobnverge.im\n618. mcojnverge.im\n619. mconvergte.im\n620. mclonverge.im\n621. mconhverge.im\n622. mconvefrge.im\n623. mconvergye.im\n624. mconverdge.im\n625. mconvferge.im\n626. mconveryge.im\n627. mconverged.im\n628. mconvergre.im\n629. mcohnverge.im\n630. mconvergve.im\n631. mconvergwe.im\n632. mconverghe.im\n633. mconverhge.im\n634. mconvergew.im\n635. mcomnverge.im\n636. mconvrerge.im\n637. mconverfge.im\n638. mcongverge.im\n639. mconvesrge.im\n640. mconvderge.im\n641. mconvcerge.im\n642. mcondverge.im\n643. mconvserge.im\n644. mconvertge.im\n645. mconvetrge.im\n646. mconvergde.im\n647. mconvernge.im\n648. mconvergbe.im\n649. mcolnverge.im\n650. mconvergne.im\n651. mcoknverge.im\n652. mconverbge.im\n653. mconfverge.im\n654. mconvergse.im\n655. mconverges.im\n656. mconvergfe.im\n657. mconverger.im\n658. mconvegrge.im\n659. mconmverge.im\n660. mconbverge.im\n661. mconvberge.im\n662. mcopnverge.im\n663. mconvewrge.im\n664. mconvergef.im\n665. mconnverge.jn\n666. mcconverge.jn\n667. mconvarga.jn\n668. mconvurgu.jn\n669. mconvegre.jn\n670. mcounverge.jn\n671. mconverrge.jn\n672. mcoverge.jn\n673. msionverge.jn\n674. mconvyrgy.jn\n675. mconverg.jn\n676. mkonverge.jn\n677. mconv3rg3.jn\n678. converge.jn\n679. mocnverge.jn\n680. mconvege.jn\n681. mcanverge.jn\n682. cmonverge.jn\n683. mdonverge.jn\n684. mconvere.jn\n685. mc0nverge.jn\n686. mconvereg.jn\n687. mcpnverge.jn\n688. mconvrege.jn\n689. mconevrge.jn\n690. mclnverge.jn\n691. mconvirgi.jn\n692. 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mconvergr.iin\n894. mconderge.iin\n895. mconvdrge.iin\n896. mconvrrgr.iin\n897. mcohverge.iin\n898. mconberge.iin\n899. mconvergd.iin\n900. nmconverge.iin\n901. mconvsrgs.iin\n902. mconvedge.iin\n903. mconvfrgf.iin\n904. mcfonverge.iin\n905. mconvwrgw.iin\n906. mcongerge.iin\n907. mxconverge.iin\n908. mcionverge.iin\n909. kmconverge.iin\n910. mjconverge.iin\n911. mcoinverge.iin\n912. mconvwrge.iin\n913. mconverne.iin\n914. mconvergw.iin\n915. mconverre.iin\n916. mconverhe.iin\n917. mconveege.iin\n918. mconvegge.iin\n919. mconvefge.iin\n920. mconverfe.iin\n921. mconvdrgd.iin\n922. mconvergf.iin\n923. mnconverge.iin\n924. mcdonverge.iin\n925. mdconverge.iin\n926. mcojverge.iin\n927. mfconverge.iin\n928. mconcerge.iin\n929. mcxonverge.iin\n930. mconvetge.iin\n931. mvconverge.iin\n932. mcvonverge.iin\n933. jmconverge.iin\n934. mcponverge.iin\n935. mconvergs.iin\n936. mconvrrge.iin\n937. mconferge.iin\n938. mconverye.iin\n939. mcobverge.iin\n940. mconverbe.iin\n941. mconvedrge.iin\n942. 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mconverger.iin\n990. mconvegrge.iin\n991. mconmverge.iin\n992. mconbverge.iin\n993. mconvberge.iin\n994. mcopnverge.iin\n995. mconvewrge.iin\n996. mconvergef.iin\n997. mconnverge.inn\n998. mcconverge.inn\n999. mconvarga.inn\n1000. mconvurgu.inn\n1001. mconvegre.inn\n1002. mcounverge.inn\n1003. mconverrge.inn\n1004. mcoverge.inn\n1005. msionverge.inn\n1006. mconvyrgy.inn\n1007. mconverg.inn\n1008. mkonverge.inn\n1009. mconv3rg3.inn\n1010. converge.inn\n1011. mocnverge.inn\n1012. mconvege.inn\n1013. mcanverge.inn\n1014. cmonverge.inn\n1015. mdonverge.inn\n1016. mconvere.inn\n1017. mc0nverge.inn\n1018. mconvereg.inn\n1019. mcpnverge.inn\n1020. mconvrege.inn\n1021. mconevrge.inn\n1022. mclnverge.inn\n1023. mconvirgi.inn\n1024. mconvergee.inn\n1025. mcnverge.inn\n1026. mmconverge.inn\n1027. mconveerge.inn\n1028. mcunverge.inn\n1029. mcenverge.inn\n1030. mcynverge.inn\n1031. mconvverge.inn\n1032. mconvrge.inn\n1033. mconerge.inn\n1034. mcnoverge.inn\n1035. kconverge.inn\n1036. jconverge.inn\n1037. mconveargea.inn\n1038. mxonverge.inn\n1039. mconwerge.inn\n1040. nconverge.inn\n1041. mcinverge.inn\n1042. mfonverge.inn\n1043. mvonverge.inn\n1044. mcovnerge.inn\n1045. mcknverge.inn\n1046. monverge.inn\n1047. mconvorgo.inn\n1048. msyonverge.inn\n1049. mcoonverge.inn\n1050. mconverge.inn\n1051. mconvergge.inn\n1052. mconverde.inn\n1053. mconverte.inn\n1054. mconvfrge.inn\n1055. mconvsrge.inn\n1056. mkconverge.inn\n1057. mcomverge.inn\n1058. mconverve.inn\n1059. mconvergr.inn\n1060. mconderge.inn\n1061. mconvdrge.inn\n1062. mconvrrgr.inn\n1063. mcohverge.inn\n1064. mconberge.inn\n1065. mconvergd.inn\n1066. nmconverge.inn\n1067. mconvsrgs.inn\n1068. mconvedge.inn\n1069. mconvfrgf.inn\n1070. mcfonverge.inn\n1071. mconvwrgw.inn\n1072. mcongerge.inn\n1073. mxconverge.inn\n1074. mcionverge.inn\n1075. kmconverge.inn\n1076. mjconverge.inn\n1077. mcoinverge.inn\n1078. mconvwrge.inn\n1079. mconverne.inn\n1080. mconvergw.inn\n1081. mconverre.inn\n1082. mconverhe.inn\n1083. mconveege.inn\n1084. mconvegge.inn\n1085. mconvefge.inn\n1086. mconverfe.inn\n1087. mconvdrgd.inn\n1088. mconvergf.inn\n1089. mnconverge.inn\n1090. mcdonverge.inn\n1091. mdconverge.inn\n1092. mcojverge.inn\n1093. mfconverge.inn\n1094. mconcerge.inn\n1095. mcxonverge.inn\n1096. mconvetge.inn\n1097. mvconverge.inn\n1098. mcvonverge.inn\n1099. jmconverge.inn\n1100. mcponverge.inn\n1101. mconvergs.inn\n1102. mconvrrge.inn\n1103. mconferge.inn\n1104. mconverye.inn\n1105. mcobverge.inn\n1106. mconverbe.inn\n1107. mconvedrge.inn\n1108. mconvgerge.inn\n1109. mconcverge.inn\n1110. mconjverge.inn\n1111. mconvervge.inn\n1112. mckonverge.inn\n1113. mconvwerge.inn\n1114. mconverege.inn\n1115. mcobnverge.inn\n1116. mcojnverge.inn\n1117. mconvergte.inn\n1118. mclonverge.inn\n1119. mconhverge.inn\n1120. mconvefrge.inn\n1121. mconvergye.inn\n1122. mconverdge.inn\n1123. mconvferge.inn\n1124. mconveryge.inn\n1125. mconverged.inn\n1126. mconvergre.inn\n1127. mcohnverge.inn\n1128. mconvergve.inn\n1129. mconvergwe.inn\n1130. mconverghe.inn\n1131. mconverhge.inn\n1132. mconvergew.inn\n1133. mcomnverge.inn\n1134. mconvrerge.inn\n1135. mconverfge.inn\n1136. mcongverge.inn\n1137. mconvesrge.inn\n1138. mconvderge.inn\n1139. mconvcerge.inn\n1140. mcondverge.inn\n1141. mconvserge.inn\n1142. mconvertge.inn\n1143. mconvetrge.inn\n1144. mconvergde.inn\n1145. mconvernge.inn\n1146. mconvergbe.inn\n1147. mcolnverge.inn\n1148. mconvergne.inn\n1149. mcoknverge.inn\n1150. mconverbge.inn\n1151. mconfverge.inn\n1152. mconvergse.inn\n1153. mconverges.inn\n1154. mconvergfe.inn\n1155. mconverger.inn\n1156. mconvegrge.inn\n1157. mconmverge.inn\n1158. mconbverge.inn\n1159. mconvberge.inn\n1160. mcopnverge.inn\n1161. mconvewrge.inn\n1162. mconvergef.inn\n1163. mconnverge.kn\n1164. mcconverge.kn\n1165. mconvarga.kn\n1166. mconvurgu.kn\n1167. mconvegre.kn\n1168. mcounverge.kn\n1169. mconverrge.kn\n1170. mcoverge.kn\n1171. msionverge.kn\n1172. mconvyrgy.kn\n1173. mconverg.kn\n1174. mkonverge.kn\n1175. 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mconverve.kn\n1225. mconvergr.kn\n1226. mconderge.kn\n1227. mconvdrge.kn\n1228. mconvrrgr.kn\n1229. mcohverge.kn\n1230. mconberge.kn\n1231. mconvergd.kn\n1232. nmconverge.kn\n1233. mconvsrgs.kn\n1234. mconvedge.kn\n1235. mconvfrgf.kn\n1236. mcfonverge.kn\n1237. mconvwrgw.kn\n1238. mcongerge.kn\n1239. mxconverge.kn\n1240. mcionverge.kn\n1241. kmconverge.kn\n1242. mjconverge.kn\n1243. mcoinverge.kn\n1244. mconvwrge.kn\n1245. mconverne.kn\n1246. mconvergw.kn\n1247. mconverre.kn\n1248. mconverhe.kn\n1249. mconveege.kn\n1250. mconvegge.kn\n1251. mconvefge.kn\n1252. mconverfe.kn\n1253. mconvdrgd.kn\n1254. mconvergf.kn\n1255. mnconverge.kn\n1256. mcdonverge.kn\n1257. mdconverge.kn\n1258. mcojverge.kn\n1259. mfconverge.kn\n1260. mconcerge.kn\n1261. mcxonverge.kn\n1262. mconvetge.kn\n1263. mvconverge.kn\n1264. mcvonverge.kn\n1265. jmconverge.kn\n1266. mcponverge.kn\n1267. mconvergs.kn\n1268. mconvrrge.kn\n1269. mconferge.kn\n1270. mconverye.kn\n1271. mcobverge.kn\n1272. mconverbe.kn\n1273. mconvedrge.kn\n1274. mconvgerge.kn\n1275. mconcverge.kn\n1276. mconjverge.kn\n1277. mconvervge.kn\n1278. mckonverge.kn\n1279. mconvwerge.kn\n1280. mconverege.kn\n1281. mcobnverge.kn\n1282. mcojnverge.kn\n1283. mconvergte.kn\n1284. mclonverge.kn\n1285. mconhverge.kn\n1286. mconvefrge.kn\n1287. mconvergye.kn\n1288. mconverdge.kn\n1289. mconvferge.kn\n1290. mconveryge.kn\n1291. mconverged.kn\n1292. mconvergre.kn\n1293. mcohnverge.kn\n1294. mconvergve.kn\n1295. mconvergwe.kn\n1296. mconverghe.kn\n1297. mconverhge.kn\n1298. mconvergew.kn\n1299. mcomnverge.kn\n1300. mconvrerge.kn\n1301. mconverfge.kn\n1302. mcongverge.kn\n1303. mconvesrge.kn\n1304. mconvderge.kn\n1305. mconvcerge.kn\n1306. mcondverge.kn\n1307. mconvserge.kn\n1308. mconvertge.kn\n1309. mconvetrge.kn\n1310. mconvergde.kn\n1311. mconvernge.kn\n1312. mconvergbe.kn\n1313. mcolnverge.kn\n1314. mconvergne.kn\n1315. mcoknverge.kn\n1316. mconverbge.kn\n1317. mconfverge.kn\n1318. mconvergse.kn\n1319. mconverges.kn\n1320. mconvergfe.kn\n1321. mconverger.kn\n1322. mconvegrge.kn\n1323. mconmverge.kn\n1324. mconbverge.kn\n1325. mconvberge.kn\n1326. mcopnverge.kn\n1327. mconvewrge.kn\n1328. mconvergef.kn\n1329. mconnverge.i\n1330. mcconverge.i\n1331. mconvarga.i\n1332. mconvurgu.i\n1333. mconvegre.i\n1334. mcounverge.i\n1335. mconverrge.i\n1336. mcoverge.i\n1337. msionverge.i\n1338. mconvyrgy.i\n1339. mconverg.i\n1340. mkonverge.i\n1341. mconv3rg3.i\n1342. converge.i\n1343. mocnverge.i\n1344. mconvege.i\n1345. mcanverge.i\n1346. cmonverge.i\n1347. mdonverge.i\n1348. mconvere.i\n1349. mc0nverge.i\n1350. mconvereg.i\n1351. mcpnverge.i\n1352. mconvrege.i\n1353. mconevrge.i\n1354. mclnverge.i\n1355. mconvirgi.i\n1356. mconvergee.i\n1357. mcnverge.i\n1358. mmconverge.i\n1359. mconveerge.i\n1360. mcunverge.i\n1361. mcenverge.i\n1362. mcynverge.i\n1363. mconvverge.i\n1364. mconvrge.i\n1365. mconerge.i\n1366. mcnoverge.i\n1367. kconverge.i\n1368. jconverge.i\n1369. mconveargea.i\n1370. mxonverge.i\n1371. mconwerge.i\n1372. nconverge.i\n1373. mcinverge.i\n1374. mfonverge.i\n1375. mvonverge.i\n1376. mcovnerge.i\n1377. mcknverge.i\n1378. monverge.i\n1379. mconvorgo.i\n1380. msyonverge.i\n1381. mcoonverge.i\n1382. mconverge.i\n1383. mconvergge.i\n1384. mconverde.i\n1385. mconverte.i\n1386. mconvfrge.i\n1387. mconvsrge.i\n1388. mkconverge.i\n1389. mcomverge.i\n1390. mconverve.i\n1391. mconvergr.i\n1392. mconderge.i\n1393. mconvdrge.i\n1394. mconvrrgr.i\n1395. mcohverge.i\n1396. mconberge.i\n1397. mconvergd.i\n1398. nmconverge.i\n1399. mconvsrgs.i\n1400. mconvedge.i\n1401. mconvfrgf.i\n1402. mcfonverge.i\n1403. mconvwrgw.i\n1404. mcongerge.i\n1405. mxconverge.i\n1406. mcionverge.i\n1407. kmconverge.i\n1408. mjconverge.i\n1409. mcoinverge.i\n1410. mconvwrge.i\n1411. mconverne.i\n1412. mconvergw.i\n1413. mconverre.i\n1414. mconverhe.i\n1415. mconveege.i\n1416. mconvegge.i\n1417. mconvefge.i\n1418. mconverfe.i\n1419. mconvdrgd.i\n1420. mconvergf.i\n1421. mnconverge.i\n1422. mcdonverge.i\n1423. mdconverge.i\n1424. mcojverge.i\n1425. mfconverge.i\n1426. mconcerge.i\n1427. mcxonverge.i\n1428. mconvetge.i\n1429. mvconverge.i\n1430. mcvonverge.i\n1431. jmconverge.i\n1432. mcponverge.i\n1433. mconvergs.i\n1434. mconvrrge.i\n1435. mconferge.i\n1436. mconverye.i\n1437. mcobverge.i\n1438. mconverbe.i\n1439. mconvedrge.i\n1440. mconvgerge.i\n1441. mconcverge.i\n1442. mconjverge.i\n1443. mconvervge.i\n1444. mckonverge.i\n1445. mconvwerge.i\n1446. mconverege.i\n1447. mcobnverge.i\n1448. mcojnverge.i\n1449. mconvergte.i\n1450. mclonverge.i\n1451. mconhverge.i\n1452. mconvefrge.i\n1453. mconvergye.i\n1454. mconverdge.i\n1455. mconvferge.i\n1456. mconveryge.i\n1457. mconverged.i\n1458. mconvergre.i\n1459. mcohnverge.i\n1460. mconvergve.i\n1461. mconvergwe.i\n1462. mconverghe.i\n1463. mconverhge.i\n1464. mconvergew.i\n1465. mcomnverge.i\n1466. mconvrerge.i\n1467. mconverfge.i\n1468. mcongverge.i\n1469. mconvesrge.i\n1470. mconvderge.i\n1471. mconvcerge.i\n1472. mcondverge.i\n1473. mconvserge.i\n1474. mconvertge.i\n1475. mconvetrge.i\n1476. mconvergde.i\n1477. mconvernge.i\n1478. mconvergbe.i\n1479. mcolnverge.i\n1480. mconvergne.i\n1481. mcoknverge.i\n1482. mconverbge.i\n1483. mconfverge.i\n1484. mconvergse.i\n1485. mconverges.i\n1486. mconvergfe.i\n1487. mconverger.i\n1488. mconvegrge.i\n1489. mconmverge.i\n1490. mconbverge.i\n1491. mconvberge.i\n1492. mcopnverge.i\n1493. mconvewrge.i\n1494. mconvergef.i\n1495. mconnverge.ij\n1496. mcconverge.ij\n1497. mconvarga.ij\n1498. mconvurgu.ij\n1499. mconvegre.ij\n1500. mcounverge.ij\n1501. mconverrge.ij\n1502. mcoverge.ij\n1503. msionverge.ij\n1504. mconvyrgy.ij\n1505. mconverg.ij\n1506. mkonverge.ij\n1507. mconv3rg3.ij\n1508. converge.ij\n1509. mocnverge.ij\n1510. mconvege.ij\n1511. mcanverge.ij\n1512. cmonverge.ij\n1513. mdonverge.ij\n1514. mconvere.ij\n1515. mc0nverge.ij\n1516. mconvereg.ij\n1517. mcpnverge.ij\n1518. mconvrege.ij\n1519. mconevrge.ij\n1520. mclnverge.ij\n1521. mconvirgi.ij\n1522. mconvergee.ij\n1523. mcnverge.ij\n1524. mmconverge.ij\n1525. mconveerge.ij\n1526. mcunverge.ij\n1527. mcenverge.ij\n1528. mcynverge.ij\n1529. mconvverge.ij\n1530. mconvrge.ij\n1531. mconerge.ij\n1532. mcnoverge.ij\n1533. kconverge.ij\n1534. jconverge.ij\n1535. mconveargea.ij\n1536. mxonverge.ij\n1537. mconwerge.ij\n1538. nconverge.ij\n1539. mcinverge.ij\n1540. mfonverge.ij\n1541. mvonverge.ij\n1542. mcovnerge.ij\n1543. mcknverge.ij\n1544. monverge.ij\n1545. mconvorgo.ij\n1546. msyonverge.ij\n1547. mcoonverge.ij\n1548. mconverge.ij\n1549. mconvergge.ij\n1550. mconverde.ij\n1551. mconverte.ij\n1552. mconvfrge.ij\n1553. mconvsrge.ij\n1554. mkconverge.ij\n1555. mcomverge.ij\n1556. mconverve.ij\n1557. mconvergr.ij\n1558. mconderge.ij\n1559. mconvdrge.ij\n1560. mconvrrgr.ij\n1561. mcohverge.ij\n1562. mconberge.ij\n1563. mconvergd.ij\n1564. nmconverge.ij\n1565. mconvsrgs.ij\n1566. mconvedge.ij\n1567. mconvfrgf.ij\n1568. mcfonverge.ij\n1569. mconvwrgw.ij\n1570. mcongerge.ij\n1571. mxconverge.ij\n1572. mcionverge.ij\n1573. kmconverge.ij\n1574. mjconverge.ij\n1575. mcoinverge.ij\n1576. mconvwrge.ij\n1577. mconverne.ij\n1578. mconvergw.ij\n1579. mconverre.ij\n1580. mconverhe.ij\n1581. mconveege.ij\n1582. mconvegge.ij\n1583. mconvefge.ij\n1584. mconverfe.ij\n1585. mconvdrgd.ij\n1586. mconvergf.ij\n1587. mnconverge.ij\n1588. mcdonverge.ij\n1589. mdconverge.ij\n1590. mcojverge.ij\n1591. mfconverge.ij\n1592. mconcerge.ij\n1593. mcxonverge.ij\n1594. mconvetge.ij\n1595. mvconverge.ij\n1596. mcvonverge.ij\n1597. jmconverge.ij\n1598. mcponverge.ij\n1599. mconvergs.ij\n1600. mconvrrge.ij\n1601. mconferge.ij\n1602. mconverye.ij\n1603. mcobverge.ij\n1604. mconverbe.ij\n1605. mconvedrge.ij\n1606. mconvgerge.ij\n1607. mconcverge.ij\n1608. mconjverge.ij\n1609. mconvervge.ij\n1610. mckonverge.ij\n1611. mconvwerge.ij\n1612. mconverege.ij\n1613. mcobnverge.ij\n1614. mcojnverge.ij\n1615. mconvergte.ij\n1616. mclonverge.ij\n1617. mconhverge.ij\n1618. mconvefrge.ij\n1619. mconvergye.ij\n1620. mconverdge.ij\n1621. mconvferge.ij\n1622. mconveryge.ij\n1623. mconverged.ij\n1624. mconvergre.ij\n1625. mcohnverge.ij\n1626. mconvergve.ij\n1627. mconvergwe.ij\n1628. mconverghe.ij\n1629. mconverhge.ij\n1630. mconvergew.ij\n1631. mcomnverge.ij\n1632. mconvrerge.ij\n1633. mconverfge.ij\n1634. mcongverge.ij\n1635. mconvesrge.ij\n1636. mconvderge.ij\n1637. mconvcerge.ij\n1638. mcondverge.ij\n1639. mconvserge.ij\n1640. mconvertge.ij\n1641. mconvetrge.ij\n1642. mconvergde.ij\n1643. mconvernge.ij\n1644. mconvergbe.ij\n1645. mcolnverge.ij\n1646. mconvergne.ij\n1647. mcoknverge.ij\n1648. mconverbge.ij\n1649. mconfverge.ij\n1650. mconvergse.ij\n1651. mconverges.ij\n1652. mconvergfe.ij\n1653. mconverger.ij\n1654. mconvegrge.ij\n1655. mconmverge.ij\n1656. mconbverge.ij\n1657. mconvberge.ij\n1658. mcopnverge.ij\n1659. mconvewrge.ij\n1660. mconvergef.ij\n1661. mconnverge.un\n1662. mcconverge.un\n1663. mconvarga.un\n1664. mconvurgu.un\n1665. mconvegre.un\n1666. mcounverge.un\n1667. mconverrge.un\n1668. mcoverge.un\n1669. msionverge.un\n1670. mconvyrgy.un\n1671. mconverg.un\n1672. mkonverge.un\n1673. mconv3rg3.un\n1674. converge.un\n1675. mocnverge.un\n1676. mconvege.un\n1677. mcanverge.un\n1678. cmonverge.un\n1679. mdonverge.un\n1680. mconvere.un\n1681. mc0nverge.un\n1682. mconvereg.un\n1683. mcpnverge.un\n1684. mconvrege.un\n1685. mconevrge.un\n1686. mclnverge.un\n1687. mconvirgi.un\n1688. mconvergee.un\n1689. mcnverge.un\n1690. mmconverge.un\n1691. mconveerge.un\n1692. mcunverge.un\n1693. mcenverge.un\n1694. mcynverge.un\n1695. mconvverge.un\n1696. mconvrge.un\n1697. mconerge.un\n1698. mcnoverge.un\n1699. kconverge.un\n1700. jconverge.un\n1701. mconveargea.un\n1702. mxonverge.un\n1703. mconwerge.un\n1704. nconverge.un\n1705. mcinverge.un\n1706. mfonverge.un\n1707. mvonverge.un\n1708. mcovnerge.un\n1709. mcknverge.un\n1710. monverge.un\n1711. mconvorgo.un\n1712. msyonverge.un\n1713. mcoonverge.un\n1714. mconverge.un\n1715. mconvergge.un\n1716. mconverde.un\n1717. mconverte.un\n1718. mconvfrge.un\n1719. mconvsrge.un\n1720. mkconverge.un\n1721. mcomverge.un\n1722. mconverve.un\n1723. mconvergr.un\n1724. mconderge.un\n1725. mconvdrge.un\n1726. mconvrrgr.un\n1727. mcohverge.un\n1728. mconberge.un\n1729. mconvergd.un\n1730. nmconverge.un\n1731. mconvsrgs.un\n1732. mconvedge.un\n1733. mconvfrgf.un\n1734. mcfonverge.un\n1735. mconvwrgw.un\n1736. mcongerge.un\n1737. mxconverge.un\n1738. mcionverge.un\n1739. kmconverge.un\n1740. mjconverge.un\n1741. mcoinverge.un\n1742. mconvwrge.un\n1743. mconverne.un\n1744. mconvergw.un\n1745. mconverre.un\n1746. mconverhe.un\n1747. mconveege.un\n1748. mconvegge.un\n1749. mconvefge.un\n1750. mconverfe.un\n1751. mconvdrgd.un\n1752. mconvergf.un\n1753. mnconverge.un\n1754. mcdonverge.un\n1755. mdconverge.un\n1756. mcojverge.un\n1757. mfconverge.un\n1758. mconcerge.un\n1759. mcxonverge.un\n1760. mconvetge.un\n1761. mvconverge.un\n1762. mcvonverge.un\n1763. jmconverge.un\n1764. mcponverge.un\n1765. mconvergs.un\n1766. mconvrrge.un\n1767. mconferge.un\n1768. mconverye.un\n1769. mcobverge.un\n1770. mconverbe.un\n1771. mconvedrge.un\n1772. mconvgerge.un\n1773. mconcverge.un\n1774. mconjverge.un\n1775. mconvervge.un\n1776. mckonverge.un\n1777. mconvwerge.un\n1778. mconverege.un\n1779. mcobnverge.un\n1780. mcojnverge.un\n1781. mconvergte.un\n1782. mclonverge.un\n1783. mconhverge.un\n1784. mconvefrge.un\n1785. mconvergye.un\n1786. mconverdge.un\n1787. mconvferge.un\n1788. mconveryge.un\n1789. mconverged.un\n1790. mconvergre.un\n1791. mcohnverge.un\n1792. mconvergve.un\n1793. mconvergwe.un\n1794. mconverghe.un\n1795. mconverhge.un\n1796. mconvergew.un\n1797. mcomnverge.un\n1798. mconvrerge.un\n1799. mconverfge.un\n1800. mcongverge.un\n1801. mconvesrge.un\n1802. mconvderge.un\n1803. mconvcerge.un\n1804. mcondverge.un\n1805. mconvserge.un\n1806. mconvertge.un\n1807. mconvetrge.un\n1808. mconvergde.un\n1809. mconvernge.un\n1810. mconvergbe.un\n1811. mcolnverge.un\n1812. mconvergne.un\n1813. mcoknverge.un\n1814. mconverbge.un\n1815. mconfverge.un\n1816. mconvergse.un\n1817. mconverges.un\n1818. mconvergfe.un\n1819. mconverger.un\n1820. mconvegrge.un\n1821. mconmverge.un\n1822. mconbverge.un\n1823. mconvberge.un\n1824. mcopnverge.un\n1825. mconvewrge.un\n1826. mconvergef.un\n1827. mconnverge.on\n1828. mcconverge.on\n1829. mconvarga.on\n1830. mconvurgu.on\n1831. mconvegre.on\n1832. mcounverge.on\n1833. mconverrge.on\n1834. mcoverge.on\n1835. msionverge.on\n1836. mconvyrgy.on\n1837. mconverg.on\n1838. mkonverge.on\n1839. mconv3rg3.on\n1840. converge.on\n1841. mocnverge.on\n1842. mconvege.on\n1843. mcanverge.on\n1844. cmonverge.on\n1845. mdonverge.on\n1846. mconvere.on\n1847. mc0nverge.on\n1848. mconvereg.on\n1849. mcpnverge.on\n1850. mconvrege.on\n1851. mconevrge.on\n1852. mclnverge.on\n1853. mconvirgi.on\n1854. mconvergee.on\n1855. mcnverge.on\n1856. mmconverge.on\n1857. mconveerge.on\n1858. mcunverge.on\n1859. mcenverge.on\n1860. mcynverge.on\n1861. mconvverge.on\n1862. mconvrge.on\n1863. mconerge.on\n1864. mcnoverge.on\n1865. kconverge.on\n1866. jconverge.on\n1867. mconveargea.on\n1868. mxonverge.on\n1869. mconwerge.on\n1870. nconverge.on\n1871. mcinverge.on\n1872. mfonverge.on\n1873. mvonverge.on\n1874. mcovnerge.on\n1875. mcknverge.on\n1876. monverge.on\n1877. mconvorgo.on\n1878. msyonverge.on\n1879. mcoonverge.on\n1880. mconverge.on\n1881. mconvergge.on\n1882. mconverde.on\n1883. mconverte.on\n1884. mconvfrge.on\n1885. mconvsrge.on\n1886. mkconverge.on\n1887. mcomverge.on\n1888. mconverve.on\n1889. mconvergr.on\n1890. mconderge.on\n1891. mconvdrge.on\n1892. mconvrrgr.on\n1893. mcohverge.on\n1894. mconberge.on\n1895. mconvergd.on\n1896. nmconverge.on\n1897. mconvsrgs.on\n1898. mconvedge.on\n1899. mconvfrgf.on\n1900. mcfonverge.on\n1901. mconvwrgw.on\n1902. mcongerge.on\n1903. mxconverge.on\n1904. mcionverge.on\n1905. kmconverge.on\n1906. mjconverge.on\n1907. mcoinverge.on\n1908. mconvwrge.on\n1909. mconverne.on\n1910. mconvergw.on\n1911. mconverre.on\n1912. mconverhe.on\n1913. mconveege.on\n1914. mconvegge.on\n1915. mconvefge.on\n1916. mconverfe.on\n1917. mconvdrgd.on\n1918. mconvergf.on\n1919. mnconverge.on\n1920. mcdonverge.on\n1921. mdconverge.on\n1922. mcojverge.on\n1923. mfconverge.on\n1924. mconcerge.on\n1925. mcxonverge.on\n1926. mconvetge.on\n1927. mvconverge.on\n1928. mcvonverge.on\n1929. jmconverge.on\n1930. mcponverge.on\n1931. mconvergs.on\n1932. mconvrrge.on\n1933. mconferge.on\n1934. mconverye.on\n1935. mcobverge.on\n1936. mconverbe.on\n1937. mconvedrge.on\n1938. mconvgerge.on\n1939. mconcverge.on\n1940. mconjverge.on\n1941. mconvervge.on\n1942. mckonverge.on\n1943. mconvwerge.on\n1944. mconverege.on\n1945. mcobnverge.on\n1946. mcojnverge.on\n1947. mconvergte.on\n1948. mclonverge.on\n1949. mconhverge.on\n1950. mconvefrge.on\n1951. mconvergye.on\n1952. mconverdge.on\n1953. mconvferge.on\n1954. mconveryge.on\n1955. mconverged.on\n1956. mconvergre.on\n1957. mcohnverge.on\n1958. mconvergve.on\n1959. mconvergwe.on\n1960. mconverghe.on\n1961. mconverhge.on\n1962. mconvergew.on\n1963. mcomnverge.on\n1964. mconvrerge.on\n1965. mconverfge.on\n1966. mcongverge.on\n1967. mconvesrge.on\n1968. mconvderge.on\n1969. mconvcerge.on\n1970. mcondverge.on\n1971. mconvserge.on\n1972. mconvertge.on\n1973. mconvetrge.on\n1974. mconvergde.on\n1975. mconvernge.on\n1976. mconvergbe.on\n1977. mcolnverge.on\n1978. mconvergne.on\n1979. mcoknverge.on\n1980. mconverbge.on\n1981. mconfverge.on\n1982. mconvergse.on\n1983. mconverges.on\n1984. mconvergfe.on\n1985. mconverger.on\n1986. mconvegrge.on\n1987. mconmverge.on\n1988. mconbverge.on\n1989. mconvberge.on\n1990. mcopnverge.on\n1991. mconvewrge.on\n1992. mconvergef.on\n1993. mconnverge.ln\n1994. mcconverge.ln\n1995. mconvarga.ln\n1996. mconvurgu.ln\n1997. mconvegre.ln\n1998. mcounverge.ln\n1999. mconverrge.ln\n2000. mcoverge.ln\n2001. msionverge.ln\n2002. mconvyrgy.ln\n2003. mconverg.ln\n2004. mkonverge.ln\n2005. mconv3rg3.ln\n2006. converge.ln\n2007. mocnverge.ln\n2008. mconvege.ln\n2009. mcanverge.ln\n2010. cmonverge.ln\n2011. mdonverge.ln\n2012. mconvere.ln\n2013. mc0nverge.ln\n2014. mconvereg.ln\n2015. mcpnverge.ln\n2016. mconvrege.ln\n2017. mconevrge.ln\n2018. mclnverge.ln\n2019. mconvirgi.ln\n2020. mconvergee.ln\n2021. mcnverge.ln\n2022. mmconverge.ln\n2023. mconveerge.ln\n2024. mcunverge.ln\n2025. mcenverge.ln\n2026. mcynverge.ln\n2027. mconvverge.ln\n2028. mconvrge.ln\n2029. mconerge.ln\n2030. mcnoverge.ln\n2031. kconverge.ln\n2032. jconverge.ln\n2033. mconveargea.ln\n2034. mxonverge.ln\n2035. mconwerge.ln\n2036. nconverge.ln\n2037. mcinverge.ln\n2038. mfonverge.ln\n2039. mvonverge.ln\n2040. mcovnerge.ln\n2041. mcknverge.ln\n2042. monverge.ln\n2043. mconvorgo.ln\n2044. msyonverge.ln\n2045. mcoonverge.ln\n2046. mconverge.ln\n2047. mconvergge.ln\n2048. mconverde.ln\n2049. mconverte.ln\n2050. mconvfrge.ln\n2051. mconvsrge.ln\n2052. mkconverge.ln\n2053. mcomverge.ln\n2054. mconverve.ln\n2055. mconvergr.ln\n2056. mconderge.ln\n2057. mconvdrge.ln\n2058. mconvrrgr.ln\n2059. mcohverge.ln\n2060. mconberge.ln\n2061. mconvergd.ln\n2062. nmconverge.ln\n2063. mconvsrgs.ln\n2064. mconvedge.ln\n2065. mconvfrgf.ln\n2066. mcfonverge.ln\n2067. mconvwrgw.ln\n2068. mcongerge.ln\n2069. mxconverge.ln\n2070. mcionverge.ln\n2071. kmconverge.ln\n2072. mjconverge.ln\n2073. mcoinverge.ln\n2074. mconvwrge.ln\n2075. mconverne.ln\n2076. mconvergw.ln\n2077. mconverre.ln\n2078. mconverhe.ln\n2079. mconveege.ln\n2080. mconvegge.ln\n2081. mconvefge.ln\n2082. mconverfe.ln\n2083. mconvdrgd.ln\n2084. mconvergf.ln\n2085. mnconverge.ln\n2086. mcdonverge.ln\n2087. mdconverge.ln\n2088. mcojverge.ln\n2089. mfconverge.ln\n2090. mconcerge.ln\n2091. mcxonverge.ln\n2092. mconvetge.ln\n2093. mvconverge.ln\n2094. mcvonverge.ln\n2095. jmconverge.ln\n2096. mcponverge.ln\n2097. mconvergs.ln\n2098. mconvrrge.ln\n2099. mconferge.ln\n2100. mconverye.ln\n2101. mcobverge.ln\n2102. mconverbe.ln\n2103. mconvedrge.ln\n2104. mconvgerge.ln\n2105. mconcverge.ln\n2106. mconjverge.ln\n2107. mconvervge.ln\n2108. mckonverge.ln\n2109. mconvwerge.ln\n2110. mconverege.ln\n2111. mcobnverge.ln\n2112. mcojnverge.ln\n2113. mconvergte.ln\n2114. mclonverge.ln\n2115. mconhverge.ln\n2116. mconvefrge.ln\n2117. mconvergye.ln\n2118. mconverdge.ln\n2119. mconvferge.ln\n2120. mconveryge.ln\n2121. mconverged.ln\n2122. mconvergre.ln\n2123. mcohnverge.ln\n2124. mconvergve.ln\n2125. mconvergwe.ln\n2126. mconverghe.ln\n2127. mconverhge.ln\n2128. mconvergew.ln\n2129. mcomnverge.ln\n2130. mconvrerge.ln\n2131. mconverfge.ln\n2132. mcongverge.ln\n2133. mconvesrge.ln\n2134. mconvderge.ln\n2135. mconvcerge.ln\n2136. mcondverge.ln\n2137. mconvserge.ln\n2138. mconvertge.ln\n2139. mconvetrge.ln\n2140. mconvergde.ln\n2141. mconvernge.ln\n2142. mconvergbe.ln\n2143. mcolnverge.ln\n2144. mconvergne.ln\n2145. mcoknverge.ln\n2146. mconverbge.ln\n2147. mconfverge.ln\n2148. mconvergse.ln\n2149. mconverges.ln\n2150. mconvergfe.ln\n2151. mconverger.ln\n2152. mconvegrge.ln\n2153. mconmverge.ln\n2154. mconbverge.ln\n2155. mconvberge.ln\n2156. mcopnverge.ln\n2157. mconvewrge.ln\n2158. mconvergef.ln\n2159. mconnverge.n\n2160. mcconverge.n\n2161. mconvarga.n\n2162. mconvurgu.n\n2163. mconvegre.n\n2164. mcounverge.n\n2165. mconverrge.n\n2166. mcoverge.n\n2167. msionverge.n\n2168. mconvyrgy.n\n2169. mconverg.n\n2170. mkonverge.n\n2171. mconv3rg3.n\n2172. converge.n\n2173. mocnverge.n\n2174. mconvege.n\n2175. mcanverge.n\n2176. cmonverge.n\n2177. mdonverge.n\n2178. mconvere.n\n2179. mc0nverge.n\n2180. mconvereg.n\n2181. mcpnverge.n\n2182. mconvrege.n\n2183. mconevrge.n\n2184. mclnverge.n\n2185. mconvirgi.n\n2186. mconvergee.n\n2187. mcnverge.n\n2188. mmconverge.n\n2189. mconveerge.n\n2190. mcunverge.n\n2191. mcenverge.n\n2192. mcynverge.n\n2193. mconvverge.n\n2194. mconvrge.n\n2195. mconerge.n\n2196. mcnoverge.n\n2197. kconverge.n\n2198. jconverge.n\n2199. mconveargea.n\n2200. mxonverge.n\n2201. mconwerge.n\n2202. nconverge.n\n2203. mcinverge.n\n2204. mfonverge.n\n2205. mvonverge.n\n2206. mcovnerge.n\n2207. mcknverge.n\n2208. monverge.n\n2209. mconvorgo.n\n2210. msyonverge.n\n2211. mcoonverge.n\n2212. mconverge.n\n2213. mconvergge.n\n2214. mconverde.n\n2215. mconverte.n\n2216. mconvfrge.n\n2217. mconvsrge.n\n2218. mkconverge.n\n2219. mcomverge.n\n2220. mconverve.n\n2221. mconvergr.n\n2222. mconderge.n\n2223. mconvdrge.n\n2224. mconvrrgr.n\n2225. mcohverge.n\n2226. mconberge.n\n2227. mconvergd.n\n2228. nmconverge.n\n2229. mconvsrgs.n\n2230. mconvedge.n\n2231. mconvfrgf.n\n2232. mcfonverge.n\n2233. mconvwrgw.n\n2234. mcongerge.n\n2235. mxconverge.n\n2236. mcionverge.n\n2237. kmconverge.n\n2238. mjconverge.n\n2239. mcoinverge.n\n2240. mconvwrge.n\n2241. mconverne.n\n2242. mconvergw.n\n2243. mconverre.n\n2244. mconverhe.n\n2245. mconveege.n\n2246. mconvegge.n\n2247. mconvefge.n\n2248. mconverfe.n\n2249. mconvdrgd.n\n2250. mconvergf.n\n2251. mnconverge.n\n2252. mcdonverge.n\n2253. mdconverge.n\n2254. mcojverge.n\n2255. mfconverge.n\n2256. mconcerge.n\n2257. mcxonverge.n\n2258. mconvetge.n\n2259. mvconverge.n\n2260. mcvonverge.n\n2261. jmconverge.n\n2262. mcponverge.n\n2263. mconvergs.n\n2264. mconvrrge.n\n2265. mconferge.n\n2266. mconverye.n\n2267. mcobverge.n\n2268. mconverbe.n\n2269. mconvedrge.n\n2270. mconvgerge.n\n2271. mconcverge.n\n2272. mconjverge.n\n2273. mconvervge.n\n2274. mckonverge.n\n2275. mconvwerge.n\n2276. mconverege.n\n2277. mcobnverge.n\n2278. mcojnverge.n\n2279. mconvergte.n\n2280. mclonverge.n\n2281. mconhverge.n\n2282. mconvefrge.n\n2283. mconvergye.n\n2284. mconverdge.n\n2285. mconvferge.n\n2286. mconveryge.n\n2287. mconverged.n\n2288. mconvergre.n\n2289. mcohnverge.n\n2290. mconvergve.n\n2291. mconvergwe.n\n2292. mconverghe.n\n2293. mconverhge.n\n2294. mconvergew.n\n2295. mcomnverge.n\n2296. mconvrerge.n\n2297. mconverfge.n\n2298. mcongverge.n\n2299. mconvesrge.n\n2300. mconvderge.n\n2301. mconvcerge.n\n2302. mcondverge.n\n2303. mconvserge.n\n2304. mconvertge.n\n2305. mconvetrge.n\n2306. mconvergde.n\n2307. mconvernge.n\n2308. mconvergbe.n\n2309. mcolnverge.n\n2310. mconvergne.n\n2311. mcoknverge.n\n2312. mconverbge.n\n2313. mconfverge.n\n2314. mconvergse.n\n2315. mconverges.n\n2316. mconvergfe.n\n2317. mconverger.n\n2318. mconvegrge.n\n2319. mconmverge.n\n2320. mconbverge.n\n2321. mconvberge.n\n2322. mcopnverge.n\n2323. mconvewrge.n\n2324. mconvergef.n\n2325. mconnverge.ni\n2326. mcconverge.ni\n2327. mconvarga.ni\n2328. mconvurgu.ni\n2329. mconvegre.ni\n2330. mcounverge.ni\n2331. mconverrge.ni\n2332. mcoverge.ni\n2333. msionverge.ni\n2334. mconvyrgy.ni\n2335. mconverg.ni\n2336. mkonverge.ni\n2337. mconv3rg3.ni\n2338. converge.ni\n2339. mocnverge.ni\n2340. mconvege.ni\n2341. mcanverge.ni\n2342. cmonverge.ni\n2343. mdonverge.ni\n2344. mconvere.ni\n2345. mc0nverge.ni\n2346. mconvereg.ni\n2347. mcpnverge.ni\n2348. mconvrege.ni\n2349. mconevrge.ni\n2350. mclnverge.ni\n2351. mconvirgi.ni\n2352. mconvergee.ni\n2353. mcnverge.ni\n2354. mmconverge.ni\n2355. mconveerge.ni\n2356. mcunverge.ni\n2357. mcenverge.ni\n2358. mcynverge.ni\n2359. mconvverge.ni\n2360. mconvrge.ni\n2361. mconerge.ni\n2362. mcnoverge.ni\n2363. kconverge.ni\n2364. jconverge.ni\n2365. mconveargea.ni\n2366. mxonverge.ni\n2367. mconwerge.ni\n2368. nconverge.ni\n2369. mcinverge.ni\n2370. mfonverge.ni\n2371. mvonverge.ni\n2372. mcovnerge.ni\n2373. mcknverge.ni\n2374. monverge.ni\n2375. mconvorgo.ni\n2376. msyonverge.ni\n2377. mcoonverge.ni\n2378. mconverge.ni\n2379. mconvergge.ni\n2380. mconverde.ni\n2381. mconverte.ni\n2382. mconvfrge.ni\n2383. mconvsrge.ni\n2384. mkconverge.ni\n2385. mcomverge.ni\n2386. mconverve.ni\n2387. mconvergr.ni\n2388. mconderge.ni\n2389. mconvdrge.ni\n2390. mconvrrgr.ni\n2391. mcohverge.ni\n2392. mconberge.ni\n2393. mconvergd.ni\n2394. nmconverge.ni\n2395. mconvsrgs.ni\n2396. mconvedge.ni\n2397. mconvfrgf.ni\n2398. mcfonverge.ni\n2399. mconvwrgw.ni\n2400. mcongerge.ni\n2401. mxconverge.ni\n2402. mcionverge.ni\n2403. kmconverge.ni\n2404. mjconverge.ni\n2405. mcoinverge.ni\n2406. mconvwrge.ni\n2407. mconverne.ni\n2408. mconvergw.ni\n2409. mconverre.ni\n2410. mconverhe.ni\n2411. mconveege.ni\n2412. mconvegge.ni\n2413. mconvefge.ni\n2414. mconverfe.ni\n2415. mconvdrgd.ni\n2416. mconvergf.ni\n2417. mnconverge.ni\n2418. mcdonverge.ni\n2419. mdconverge.ni\n2420. mcojverge.ni\n2421. mfconverge.ni\n2422. mconcerge.ni\n2423. mcxonverge.ni\n2424. mconvetge.ni\n2425. mvconverge.ni\n2426. mcvonverge.ni\n2427. jmconverge.ni\n2428. mcponverge.ni\n2429. mconvergs.ni\n2430. mconvrrge.ni\n2431. mconferge.ni\n2432. mconverye.ni\n2433. mcobverge.ni\n2434. mconverbe.ni\n2435. mconvedrge.ni\n2436. mconvgerge.ni\n2437. mconcverge.ni\n2438. mconjverge.ni\n2439. mconvervge.ni\n2440. mckonverge.ni\n2441. mconvwerge.ni\n2442. mconverege.ni\n2443. mcobnverge.ni\n2444. mcojnverge.ni\n2445. mconvergte.ni\n2446. mclonverge.ni\n2447. mconhverge.ni\n2448. mconvefrge.ni\n2449. mconvergye.ni\n2450. mconverdge.ni\n2451. mconvferge.ni\n2452. mconveryge.ni\n2453. mconverged.ni\n2454. mconvergre.ni\n2455. mcohnverge.ni\n2456. mconvergve.ni\n2457. mconvergwe.ni\n2458. mconverghe.ni\n2459. mconverhge.ni\n2460. mconvergew.ni\n2461. mcomnverge.ni\n2462. mconvrerge.ni\n2463. mconverfge.ni\n2464. mcongverge.ni\n2465. mconvesrge.ni\n2466. mconvderge.ni\n2467. mconvcerge.ni\n2468. mcondverge.ni\n2469. mconvserge.ni\n2470. mconvertge.ni\n2471. mconvetrge.ni\n2472. mconvergde.ni\n2473. mconvernge.ni\n2474. mconvergbe.ni\n2475. mcolnverge.ni\n2476. mconvergne.ni\n2477. mcoknverge.ni\n2478. mconverbge.ni\n2479. mconfverge.ni\n2480. mconvergse.ni\n2481. mconverges.ni\n2482. mconvergfe.ni\n2483. mconverger.ni\n2484. mconvegrge.ni\n2485. mconmverge.ni\n2486. mconbverge.ni\n2487. mconvberge.ni\n2488. mcopnverge.ni\n2489. mconvewrge.ni\n2490. mconvergef.ni\n\n## Associated IDs\n\nAdsense account No entry at this time 112540346323389620576 No entry at this time No entry at this time\n\n## Safety summary\n\nWOT Ranking No entry at this time No entry at this time No entry at this time\n\n## Domain Whois registration info\n\nPotential website owners savinder sarna 11-12-2007 07-06-2015 11-12-2017"
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https://thenewstandardgallery.com/torsional-and-vibrational/ | [
"## Torsional and Vibrational Essay\n\nSTUDENT NAME STUDENT ID COURSE LECTURER SUBMISSION DATE SUBJECT MD ATIQUR RAHMAN FAISAL SCM-012154 BACHELOR IN MECHANICAL ENGINEERING DR. CHIA 19th April 2013 ENGINEERING MECHANICS (EAT 227) Torsion Vibration Experiment Title: Introduction : Torsion Vibration.\n\nTorsion is the twisting of a metallic rod shaped object, when a torque is applied on two sides’ perpendicular to the radius of a uniform cross-sectional bar. To study the response of materials under a torsional force, the torsion test is performed by mounting the specimen onto a torsion testing machine, then applying the twisting moment till failure.The torque and degree of rotation are measured and plotted.\n\nIt can be seen higher torsion force is required at higher degree of rotation. Normally the test specimens used are of a cylindrical rod type since the stress distribution across the section of the rod is the simplest geometry, which is easy for the calculation of the stresses. Both ends of the cylindrical specimen are tightened to hexagonal socket in which one is fitted to a torque shaft and another is fitted to an input shaft. The twisting moment is applied by turning the input hand wheel to produce torque until specimen fails.\n\nWhen the twisting moment is applied, the torque is reacted by a torque shaft, which moves in relation to the deflection arm. The movement of the deflection arm is measured by a linear potentiometer, which is connected to a calibrated TQ digital torque meter to give readout of the torque in unit of Nm or lb. in. The move we turned the input hand wheel clockwise to increase the degree of rotation, the more torque is produced. At the initial stage, the graphical relationship of the torque and degree of rotation measured is linear as demonstrated.The specimen is elastically deformed and the recovery of the specimen to its original shape is possible if the specimen is unloaded. However, if a high degree of rotation is applied passing a proportional limit, the specimen starts to deform plastically and will not return to its original shape when the input hand wheel is turned anti-clockwise. Determining the natural frequency of a system undergoing torsional vibration.\n\nUsing Newton’s second law of torsional system. o Mechanism : Objective Theory ? : : ? ? T ? I ?? …………………. (Equation 1) Where Io = mass moment of inertia of the disk Hence, ? k? ? I o?? …….. ……..\n\n. (Equation 2) Where k = torsional stiffness of the shaft Rearrange Equation 2 ? ? ?? ? ? n ? ? 0 . ………..\n\n……… (Equation 3) 2 Where natural frequency of the system, MD. Atiqur Rahman Faisal Page 2 Torsion Vibration ?n ? k ….\n\n. ……. ….. …….\n\n.. (Equation 4) Io From Simple Theory of Torsion, T ? G? ? ? J R L J = Polar second moment of area R = Radius of shaft Where T = Applied torque ? = Shear stress G = Shear modulus L = Length of shaft As torsional stiffness k ? Apparatus ? ? ? : ? = Angle of twist T ? , it can be determined through k ? GJ ………….. Equation 5) L One solid circular disk with mass = 4. 536kg, diameter = 150mm and thickness = 30mm. One annular circular disk with mass 1. 89kg, outer diameter 150mm, inner diameter = 110mm and thickness = 30mm.\n\nTwo chucks; one steel rod; one stopwatch. MD. Atiqur Rahman Faisal Page 3 Torsion Vibration Procedure : 1. The diameter of the provided rod is measured at three different locations to get the average diameter of the rod. 2. The anchor is chucked tightly to the solid circular disk. 3. The length of the rod or the distance between the two chucks is initially kept 30cm.\n\n4.The disk is displaced slightly, so that the rod can be twisted. 5. The disk is released and the stopwatch is switched on simultaneously. 6. The time taken is recorded according 10, 20, 30, 40 and 50 cycles of the disk.\n\n7. From step 3 to step 6 is repeated by increasing the length between the two chucks from 35 cm to 40 cm. 8. The whole procedure is repeated by attaching the annular circular disk on top of the solid disk. Results : Time, (s) 0. 3m(With Annular disk) Length Number of cycles 10 20 30 40 50 0.\n\n3m 0. 4m 0. 5m 0. 4m(With Annular disk) 0. 5m(With Annular disk) 2.\n\n63 6. 6 9. 25 11. 84 14. 67 3.\n\n30 6. 12 8. 94 13.\n\n31 16. 38 3. 41 6. 50 9. 61 12. 56 16.\n\n56 3. 75 7. 55 10. 93 14. 3 18. 0 3. 95 7.\n\n90 12. 13 15. 90 20.\n\n15 4. 15 8. 52 12. 59 16. 7 21. 15 Given Information Diameter Of the shaft : D1 = 40.\n\n6 mm, D2 = 42 mm, D3 = 45. 8 mm. Davg = 42. 8 mm.\n\nMass of the circular disk = 4. 536kg. Diameter of the circular disk = 150mm. Thickness of the circular disk = 20mm. Mass of the annular disk = 1. 86kg. Outer diameter of the annular disk = 150mm.\n\nInner diameter of the annular disk = 110mm. Thickness of the annular disk = 30mm. G (Shear module) = 80MPa.MD. Atiqur Rahman Faisal Page 4 Torsion Vibration Sample calculation : = = We know, J (polar second moment of area) = We know, We know, (moment of inertia) = mr2 = (4. 54) (0. 075)2 = 0.\n\n013 (circular disk) (moment of inertia) for annular disk = =0. 28728. = We know, K (torsion stiffness of the shaft) = We know, (angular speed) = = = = = 82.\n\n2 rad/s = 87. 84 We know, ? (shear stress) = = 0. 076 Comparing experimental and theoretical value of = 2??? f, where f is collected from graph. = 2??? 3. 3387 = 20. 98rad/s Percentage errors ? 100%, : , = 291. 8% MD. Atiqur Rahman Faisal Page 5Torsion Vibration Plotted Graph : N vs T (without annular disk) L=30cm 60 y = 3.\n\n3387x + 0. 3193 50 40 30 20 10 0 0 5 10 15 20 N vs T (without annular disk) Linear (N vs T (without annular disk)) N vs T (without annular disk) L=35cm 60 y = 2. 9785x + 1. 3763 50 40 30 20 10 0 0 5 10 15 20 N vs T (without annular disk) L=35cm Linear (N vs T (without annular disk) L=35cm) MD. Atiqur Rahman Faisal Page 6 Torsion Vibration N vs T (without annular disk) L=40 cm 60 50 40 30 20 10 0 0 5 10 15 20 10 20 30 40 N vs T (without annular disk) L=40 cm Linear (N vs T (without annular disk) L=40 cm) y = 3.\n\n805x + 0. 0328 50 N vs T (with annular disk) L=30 cm 60 50 40 30 20 10 0 0 5 10 15 20 N vs T (with annular disk) L=30 cm Linear (N vs T (with annular disk) L=30 cm) y = 2. 8355x – 0. 9242 MD. Atiqur Rahman Faisal Page 7 Torsion Vibration N vs T (with annular disk) L=35 cm 60 50 40 30 20 10 0 0 5 10 15 20 25 N vs T (with annular disk) L=35 cm Linear (N vs T (with annular disk) L=35 cm) y = 2. 4746x + 0. 2905 N vs T (with annular disk) L=40cm 60 50 40 30 20 10 0 0 5 10 15 20 25 N vs T (with annular disk) Linear (N vs T (with annular disk)) y = 2. 3702x + 0.\n\n0832 MD.Atiqur Rahman Faisal Page 8 Torsion Vibration Graph Comparison : 60 50 y = 3. 3387x + 0.\n\n3193 40 30 20 10 0 0 5 10 15 20 y = 2. 8355x – 0. 9242 N vs T (With annular disk) L=30cm N vs T (Without annular disk) L=30cm Linear (N vs T (With annular disk) L=30cm) Linear (N vs T (Without annular disk) L=30cm) 60 50 y = 2. 9785x + 1.\n\n3763 40 30 20 10 0 0 5 10 15 20 25 y = 2. 4746x + 0. 2905 N vs T (with Annular disk) L=35 cm N vs T (without Annular disk) L=35cm Linear (N vs T (with Annular disk) L=35 cm) Linear (N vs T (without Annular disk) L=35cm) MD. Atiqur Rahman FaisalPage 9 Torsion Vibration 60 y = 2. 3702x + 0.\n\n0832 50 y = 3. 0805x + 0. 0328 40 N vs T (with annular disk) L=40cm N vs T(without annular disk) L=40cm Linear (N vs T (with annular disk) L=40cm) Linear (N vs T(without annular disk) L=40cm) 30 20 10 0 0 5 10 15 20 25 Discussion : From the graph it can be observed that, the specimen at first deformed elastically and then deformed plastically. The initial stage of the elastic behavior describes a linear relationship between torque and degree of rotation with the slope representing the shear modulus of rigidity, G.Beyond the proportional limit, specimen deformed in a plastic manner. The stresses vary from section to section of the specimen. To get an average value, the diameter of the specimen is measured from 3 different places and then calculated the average value.\n\nConclusion : Comparing results, it is observed that the error percentage is quiet high; this is due to the high sensitive values that have been collected during the experiment. Possible reasons for the errors are, when the disk was rotating, it was quiet difficult to take the exact turn, or rotation of the disk.So there was also some human reflex and visual errors. Even though, when the experiment was performing, there was some unwanted air resistance application running beside the experiment. Some of the measurement was so sensitive, and it was hard to take the measurement. The apparatus tools, and measurement machines was also not so accurate.\n\nThe experiment could be repeated for some more time to get some more result to find it more accurate. Considering all the facts, the experiment can be accepted with errors. MD. Atiqur Rahman Faisal Page 10\n\nx",
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"Hi!\nI'm Tamara!\n\nWould you like to get a custom essay? How about receiving a customized one?\n\nCheck it out"
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https://plovdent.com/sin-2x-x/ | [
"I just want lớn make sure I\"m on the right path with the problem. The problem is as follows:\n\n\\$\\$lim_x oinftyfracsin^2xx^2\\$\\$\n\nI rewrote it as follows:\n\n\\$\\$lim_x oinftyfrac(sin x)^2x^2\\$\\$\n\nNow \\$sin(x)^2\\$ does oscillate as \\$x\\$ approaches infinity và therefore a limit does not exist. However it oscillates between the numbers \\$-1\\$ & \\$1\\$. Since the denominator would increase without bound và the numerator would only move between \\$-1\\$ and \\$1\\$, part of me wants khổng lồ say that the limit is zero.\n\nBạn đang xem: Sin2x = x\n\nHowever a smarter part of me wants to lớn say that the limit does not exist due to the numerator. Could someone shed some light on this problem?\n\ncalculus algebra-precalculus limits trigonometry\nnói qua\nCite\nFollow\nedited Jan 20, năm 2016 at 17:57",
null,
"Martin Sleziak\nasked Jul 6, 2014 at 0:41",
null,
"Irresponsible NewbIrresponsible Newb\n\\$endgroup\\$\n4\nđịa chỉ cửa hàng a comment |\n\nSorted by: Reset to mặc định\nHighest score (default) Date modified (newest first) Date created (oldest first)\n11\n\\$\begingroup\\$\nTo solve the following question, recall the Squeeze Theorem:\n\nSo, as we know: \\$\\$0 le sin^2 x le 1\\$\\$\n\nIf we divide by \\$x^2\\$\n\n\\$\\$0 le fracsin^2 xx^2 le frac1x^2\\$\\$\n\nIf we evaluate the limit from at either ends: \\$\\$lim_x o infty 0 = 0\\$\\$ \\$\\$lim_x o infty frac 1x^2 = 0\\$\\$\n\nTherefore, by the squeeze theorem:\n\n\\$\\$lim_x o infty fracsin^2 xx^2 = 0\\$\\$\n\ngiới thiệu\nCite\nFollow\nedited Dec 23, năm ngoái at 12:15\nanswered Jul 6, năm trước at 1:17",
null,
"Varun IyerVarun Iyer\n\\$endgroup\\$\n4\n\\$\begingroup\\$\nWhat we have is as follows: for all \\$x>0\\$,\\$\\$0 leq fracsin^2 xx^2 leq frac 1x^2\\$\\$Now, chú ý that \\$lim_x o infty 0 = lim_x o infty frac 1x^2 = 0\\$. What theorem can we use here to lớn get the answer?\n\nchia sẻ\nCite\nFollow\nanswered Jul 6, năm trước at 0:44",
null,
"Ben GrossmannBen Grossmann\n\\$endgroup\\$\nshowroom a comment |\n\nThanks for contributing an answer lớn plovdent.comematics Stack Exchange!\n\nBut avoid\n\nAsking for help, clarification, or responding to other answers.Making statements based on opinion; back them up with references or personal experience.\n\nUse plovdent.comJax khổng lồ format equations. plovdent.comJax reference.\n\nXem thêm: Top 29 Trường Khương Đình - Trung Học Phổ Thông Khương Đình\n\nDraft saved\n\nSubmit\n\n### Post as a guest\n\nName\nemail Required, but never shown\n\n### Post as a guest\n\nName\ne-mail\n\nRequired, but never shown\n\n## Not the answer you're looking for? Browse other questions tagged calculus algebra-precalculus limits trigonometry or ask your own question.\n\nFeatured on Meta\n2\nWhat is the value for \\$lim_x oinfty fracsin xx\\$?\n3\nWhat is the limit of \\$limlimits_x→∞fracsin xx\\$\nRelated\n11\nStrange behavior of \\$lim_x o0fracsinleft(xsinleft(frac1x ight) ight)xsinleft(frac1x ight)\\$\n5\nThe behavior of the graph of \\$f(x) = sin (pi/x)\\$ as x approaches 0, Why?\n2\nIs limit of \\$sin x\\$ at infinity finite?\n0\nHorizontal Asymptote- Limits at Infinity\n3\nTrigonometric functions limit to lớn complex infinity\n0\nWhat is the Correct Approach to lớn Solving a Limit that Tends to Negative Infinity?\n4\nL'hopital rule fails with limits to infinity?\n4\nSpivak Calculus Chapter 5-33d: \\$lim_x oinftyfracx^2(1+sin^2(x))(x+sin(x))^2\\$\n1\nFind \\$limlimits_x o -infty (e^-x cosx)\\$\nHot Network Questions more hot questions\n\nQuestion feed",
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| {"ft_lang_label":"__label__en","ft_lang_prob":0.7771906,"math_prob":0.9732629,"size":4162,"snap":"2022-05-2022-21","text_gpt3_token_len":1255,"char_repetition_ratio":0.09042809,"word_repetition_ratio":0.0147929,"special_character_ratio":0.2659779,"punctuation_ratio":0.091593474,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9894565,"pos_list":[0,1,2,3,4,5,6,7,8,9,10],"im_url_duplicate_count":[null,2,null,2,null,2,null,2,null,2,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-05-17T08:52:29Z\",\"WARC-Record-ID\":\"<urn:uuid:73196c74-25cf-4ee9-b20b-1e2f7606bea8>\",\"Content-Length\":\"14053\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:397e376d-1ed2-4567-91c3-ed05fbc48ab8>\",\"WARC-Concurrent-To\":\"<urn:uuid:49c877ef-6878-45b2-8b1d-5586ffd9c0dd>\",\"WARC-IP-Address\":\"172.67.150.44\",\"WARC-Target-URI\":\"https://plovdent.com/sin-2x-x/\",\"WARC-Payload-Digest\":\"sha1:UAA2G3HIV2VOYEUM7NUWQP6LAMYZ7WHR\",\"WARC-Block-Digest\":\"sha1:K6Z3FSCBJUXLOW43UL6X2ED5UOPMHDUA\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-21/CC-MAIN-2022-21_segments_1652662517018.29_warc_CC-MAIN-20220517063528-20220517093528-00231.warc.gz\"}"} |
http://bms.zeugner.eu/doc/htmlhelp/lps.bma.html | [
"lps.bma {BMS} R Documentation\n\n## Log Predictive Score\n\n### Description\n\nComputes the Log Predictive Score to evaluate a forecast based on a bma object\n\n### Usage\n\n``` lps.bma(object, realized.y, newdata = NULL)\n```\n\n### Arguments\n\n `object` an object of class `pred.density`, or class `bma` (cf. `bms`), or class `zlm` `realized.y` a vector with realized values of the dependent variables to be plotted in addition to the predictive density, must have its length conforming to `newdata` `newdata` Needs to be provided if `object` is not of class `pred.density`: a data.frame, matrix or vector containing variables with which to predict.\n\n### Details\n\nThe log predictive score is an indicator for the likelihood of several forecasts.\nIt is definied as minus the arithmethic mean of the logarithms of the point densities for `realized.y` given `newdata`.\nNote that in most cases is more efficient to first compute the predictive density object via a call to `pred.density` and only then pass the result on to `lps.bma`.\n\n### Value\n\nA scalar denoting the log predictive score\n\n### Author(s)\n\nMartin Feldkircher and Stefan Zeugner\n\n`pred.density` for constructing predictive densities, `bms` for creating `bma` objects, `density.bma` for plotting coefficient densities\n\n### Examples\n\n``` data(datafls)\nmm=bms(datafls,user.int=FALSE,nmodel=100)\n\n#LPS for actual values under the used data (static forecast)\nlps.bma(mm, realized.y=datafls[,1] , newdata=datafls[,-1])\n\n#the same result via predicitve.density\npd=pred.density(mm, newdata=datafls[,-1])\nlps.bma(pd,realized.y=datafls[,1])\n\n# similarly for a linear model (not BMA)\nzz = zlm(datafls)\nlps.bma(zz, realized.y=datafls[,1] , newdata=datafls[,-1])\n\n```\n\n[Package BMS version 0.3.1 Index]"
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https://convertoctopus.com/9-grams-to-kilograms | [
"## Conversion formula\n\nThe conversion factor from grams to kilograms is 0.001, which means that 1 gram is equal to 0.001 kilograms:\n\n1 g = 0.001 kg\n\nTo convert 9 grams into kilograms we have to multiply 9 by the conversion factor in order to get the mass amount from grams to kilograms. We can also form a simple proportion to calculate the result:\n\n1 g → 0.001 kg\n\n9 g → M(kg)\n\nSolve the above proportion to obtain the mass M in kilograms:\n\nM(kg) = 9 g × 0.001 kg\n\nM(kg) = 0.009 kg\n\nThe final result is:\n\n9 g → 0.009 kg\n\nWe conclude that 9 grams is equivalent to 0.009 kilograms:\n\n9 grams = 0.009 kilograms\n\n## Alternative conversion\n\nWe can also convert by utilizing the inverse value of the conversion factor. In this case 1 kilogram is equal to 111.11111111111 × 9 grams.\n\nAnother way is saying that 9 grams is equal to 1 ÷ 111.11111111111 kilograms.\n\n## Approximate result\n\nFor practical purposes we can round our final result to an approximate numerical value. We can say that nine grams is approximately zero point zero zero nine kilograms:\n\n9 g ≅ 0.009 kg\n\nAn alternative is also that one kilogram is approximately one hundred eleven point one one one times nine grams.\n\n## Conversion table\n\n### grams to kilograms chart\n\nFor quick reference purposes, below is the conversion table you can use to convert from grams to kilograms\n\ngrams (g) kilograms (kg)\n10 grams 0.01 kilograms\n11 grams 0.011 kilograms\n12 grams 0.012 kilograms\n13 grams 0.013 kilograms\n14 grams 0.014 kilograms\n15 grams 0.015 kilograms\n16 grams 0.016 kilograms\n17 grams 0.017 kilograms\n18 grams 0.018 kilograms\n19 grams 0.019 kilograms"
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| {"ft_lang_label":"__label__en","ft_lang_prob":0.7470011,"math_prob":0.99728423,"size":1609,"snap":"2020-45-2020-50","text_gpt3_token_len":424,"char_repetition_ratio":0.24735202,"word_repetition_ratio":0.0,"special_character_ratio":0.31199503,"punctuation_ratio":0.10810811,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9973378,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-10-30T07:29:30Z\",\"WARC-Record-ID\":\"<urn:uuid:257b0f8f-daa7-419f-8f77-542f4f1f2b43>\",\"Content-Length\":\"27969\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:00e8d64a-4314-4d25-9292-4b99d71f84af>\",\"WARC-Concurrent-To\":\"<urn:uuid:3caa5888-da3e-440d-b6bc-28a734c340ae>\",\"WARC-IP-Address\":\"104.27.143.66\",\"WARC-Target-URI\":\"https://convertoctopus.com/9-grams-to-kilograms\",\"WARC-Payload-Digest\":\"sha1:76RHHEKU5LHNADFYLHRP4JH7YFLUSJJE\",\"WARC-Block-Digest\":\"sha1:LCTLBXXKG436JPOZ3BWW2725IV4CNLXB\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-45/CC-MAIN-2020-45_segments_1603107909746.93_warc_CC-MAIN-20201030063319-20201030093319-00270.warc.gz\"}"} |
https://www.52b2c.com/archives/1169.html | [
"### 陈设参数\n\n• 特色：金贝 TSportage-A八 数码有线引闪器\n影室闪光灯摄影灯雕塑棚触发器遥控器各品牌相机通用\n\nPython: 函数参数法力\n\n2016-12-01\n\n`````` >>> def myfun(x,y):\n>>> return x + y\n``````\n\n`````` >>> s = myfun(10)\n---------------------------------------------------------------------------\nTypeError Traceback (most recent call last)\n<ipython-input-18-135f1c06450c> in <module>()\n----> 1 s = myfun(10)\nTypeError: sum() missing 1 required positional argument: 'y'\n``````\n\n`````` >>> def myfun(x,y,z,h):\n>>> return (x + y)*z\n``````\n\n>>> def myfun(x,y,z = 1,h=0):\n>>> return (x + y)*z + h\n\n1）好处\n\n2）注意点\n\n• 必选参数在前，私下认可参数在后\n• 充实私下认可参数要挂念不要影响旧代码的调用结果，大家能够透过扩展暗中同意参数扩张大家的函数成效\n• 1旦有多少个暗许参数，既能够根据顺序\n如myfun(十,20,3,八),也足以不依据顺序，当不依据顺序传入时索要把参数名称写上如\nmyfun(10,20,h=捌),此时 z未有传到值，默以为1\n• 纪事一点：暗中同意参数必须指向不改变对象\n\n1）定义\n\n1\n\n`````` >>> def myfun (number):\n>>> sum = 0\n>>> for i in number:\n>>> sum = sum + i*i\n>>> return sum\n``````\n\n2\n\nnumber 是多个list\n\n>>> myfun([1,2,3])\n>>> 14\n2）用法\n\n>>> def myfun (number):\n>>> sum = 0\n>>> for i in number:\n>>> sum = sum + i\ni\n>>> return sum\n\n3\n\n>>> myfun(1,3)\n>>> 10\n>>> myfun()\n>>> 0\n\n>>> num = [1,3]\n>>> myfun(*num)\n*nums表示把nums这些list的保有因素作为可变参数传进去。那种写法极度有用，而且很普及\n\n4\n\n** 四.首要字参数 **\n\n5\n\n`````` def myfun (name,email,**kw):\nprint('name:', name, 'email:', email, 'other:', kw)\n``````\n\n6\n\n1. 用法\n在概念关键字参数的时候\n，在函数的最后扩展**kw一言九鼎字即可，函数内部，会把主要字参数转化为dict,大家在扩散的时候也要选用k=v传入\n\nmyfun(‘shuang’,’[email protected]‘,age=18,sex=”女”)\nname: shuang email:\[email protected]\nother: {‘sex’: ‘女’, ‘age’: 18}\n大家也足以像可变参数一个 事先在函数外部定义好2个dict\nextra = {‘age’: ’18’, ‘sex’: ‘女’}\nmyfun(“shuang”,”[email protected]“,**extra)\nname: shuang email:\[email protected]\nother: {‘age’: ’18’, ‘sex’: ‘女’}\n**extra表示把extra那一个dict的有着key-value用关键字参数字传送入到函数的**kw参数，kw将收获3个dict。\n注意：kw获得的dict是extra的一份拷贝，对kw的更动不会潜移默化到函数外的extra\n\n7\n\n5.命名重中之重字参数\n\n8\n\n9\n\n`````` >>> def myfun(name, email, **kw):\n>>> if 'age' in kw:\n>>> # 有age参数\n>>> pass\n>>> if 'sex' in kw:\n>>> # 有sex参数\n>>> pass\n``````\n\n10\n\n11\n\n``````def myfun(name, email, *, age, sex):\nprint(name, email, age, sex)\n``````\n\n12\n\n3）用法\n\n13\n\n>>> def add(x, y): # x, y 是必选参数\n\n`````` >>> myfun('Jack', '[email protected]', age=18, sex='女')\nJack [email protected] 18 女\n``````\n\n… print x + y\n\n4）注意点\n\n• 即便函数定义中1度有了三个可变参数，后边跟着的命名首要字参数就不再须求二个非同小可分隔符*了：\n>>> def myfun(name, email, *args, age, sex):\nprint(name, email, args, age, sex)\n\n• 取名首要字参数必须传入参数名，那和岗位参数分歧。如若未有传来参数名，调用将报错：\n\n`````` >>> myfun('Jack', '[email protected]', '18', '女')\nTraceback (most recent call last):\nFile \"<stdin>\", line 1, in <module>\nTypeError: myfun() takes 2 positional arguments but 4 were given\n``````\n\n>>> add() # 啥都没传，不行\n\nTraceback (most recent call last):\n\n• 缺省值：\n>>> def myfun(name, email, *, age=’18’, sex):\n>>> print(name, email, age, sex)\n鉴于命名第二字参数age具备暗中认可值，调用时，可不传播age参数：\n\n`````` >>> myfun('Jack', '[email protected]', sex='女')\nJack [email protected] 18 女\n``````\n• 1旦没有可变参数，就务须加一个用作特种分隔符。要是贫乏，Python解释器将不可能识别地方参数和命名首要字参数：\n\n`````` def myfun(name, email, age, sex):\n``````\n\n缺少 *，age和sex被视为地点参数\n\nFile “”, line 1, in\n\n6.参数组合\n\nTypeError: add() takes exactly 2 arguments (0 given)\n\n>>> add(1) # 只传了3个，也十分\n\nTraceback (most recent call last):\n\n`````` def f1(a, b, c=0, *args, **kw):\nprint('a =', a, 'b =', b, 'c =', c, 'args =', args, 'kw =', kw)\n\ndef f2(a, b, c=0, *, d, **kw):\nprint('a =', a, 'b =', b, 'c =', c, 'd =', d, 'kw =', kw)\n``````\n\nFile “”, line 1, in\n\nTypeError: add() takes exactly 2 arguments (1 given)\n\n``````>>> f1(1, 2)\na = 1 b = 2 c = 0 args = () kw = {}\n>>> f1(1, 2, c=3)\na = 1 b = 2 c = 3 args = () kw = {}\n>>> f1(1, 2, 3, 'a', 'b')\na = 1 b = 2 c = 3 args = ('a', 'b') kw = {}\n>>> f1(1, 2, 3, 'a', 'b', x=99)\na = 1 b = 2 c = 3 args = ('a', 'b') kw = {'x': 99}\n>>> f2(1, 2, d=99, ext=None)\na = 1 b = 2 c = 0 d = 99 kw = {'ext': None}\n``````\n\n>>> add(1, 2) # 数量一样，通过\n\n3\n\n`````` >>> args = (1, 2, 3, 4)\n>>> kw = {'d': 99, 'x': '#'}\n>>> f1(*args, **kw)\na = 1 b = 2 c = 3 args = (4,) kw = {'d': 99, 'x': '#'}\n>>> args = (1, 2, 3)\n>>> kw = {'d': 88, 'x': '#'}\n>>> f2(*args, **kw)\na = 1 b = 2 c = 3 d = 88 kw = {'x': '#'}\n``````\n\n**kw)的款式调用它，无论它的参数是何许定义的。（个人以为那一个地方即使灵活，可是太灵敏反而有些糊涂，须要多推行技巧深刻明白他的优势）\n\n1\n\n2\n\n3\n\n4\n\n5\n\n6\n\n7\n\n>>> def add(x, y, z=1): # x, y 是必选参数，z\n\n… print x + y + z\n\n>>> add(1, 2, 3) # 1+2+3\n\n6\n\n>>> add(1, 2) # 未有传递 z，自动使用 z=一，即\n1+贰+一\n\n4\n\n1\n\n2\n\n3\n\n4\n\n5\n\n6\n\n7\n\n8\n\n9\n\n10\n\n11\n\n>>> def add(x=1, y, z): # x\n\n… return x + y + z\n\nFile “”, line 1\n\nSyntaxError: non-default argument follows default argument\n\n>>>\n\n>>> def add(x, y=1, z): # y\n\n… return x + y + z\n\nFile “”, line 1\n\nSyntaxError: non-default argument follows default argument\n\n1\n\n2\n\n3\n\n>>> def add_to_list(L=[]):\n\n… L.append(‘END’)\n\n… return L\n\n1\n\n2\n\n3\n\n4\n\n5\n\n6\n\n7\n\n8\n\n9\n\n10\n\n>>> add_to_list([1, 2, 3]) # 没啥难题\n\n[1, 2, 3, ‘END’]\n\n>>> add_to_list([‘a’, ‘b’, ‘c’]) # 没啥难题\n\n[‘a’, ‘b’, ‘c’, ‘END’]\n\n>>> add_to_list() #\n\n[‘END’]\n\n>>> add_to_list() #\n\n[‘END’, ‘END’]\n\n>>> add_to_list() # 糟糕了，三个 ‘END’\n\n[‘END’, ‘END’, ‘END’]\n\nL=[]，经过管理，L 应该只有3个成分，怎么会出现调用函数一回，L\n\n1\n\n2\n\n3\n\n4\n\n5\n\n6\n\n>>> def add(*numbers):\n\n… sum = 0\n\n… for i in numbers:\n\n… sum += i\n\n… print ‘numbers:’, numbers\n\n… return sum\n\n1\n\n2\n\n3\n\n4\n\n5\n\n6\n\n7\n\n8\n\n9\n\n10\n\n11\n\n12\n\n>>> add() # 传递 0 个参数\n\nnumbers: ()\n\n0\n\n>>> add(1) # 传递 1 个参数\n\nnumbers: (1,)\n\n1\n\n>>> add(1, 2) # 传递 2 个参数\n\nnumbers: (1, 2)\n\n3\n\n>>> add(1, 2, 3) # 传递 3 个参数\n\nnumbers: (1, 2, 3)\n\n6\n\n1\n\n2\n\n3\n\n4\n\n5\n\n6\n\n7\n\n8\n\n9\n\n10\n\n11\n\n>>> def add(x, y, z): # 有 三 个必选参数\n\n… return x + y + z\n\n>>> a = [1, 2, 3]\n\n>>> add(a, a, a) # 那样传递参数很麻烦\n\n6\n\n>>> add(*a) # 使用 *a，也就是地方的做法\n\n6\n\n>>> b = (4, 5, 6)\n\n>>> add(*b) # 对元组同样适用\n\n15\n\n1\n\n2\n\n3\n\n4\n\n5\n\n6\n\n7\n\n8\n\n9\n\n10\n\n11\n\n12\n\n>>> def add(*numbers): # 函数参数是可变参数\n\n… sum = 0\n\n… for i in numbers:\n\n… sum += i\n\n… return sum\n\n>>> a = [1, 2]\n\n>>> add(*a) # 使用 *a 给函数字传送递参数\n\n3\n\n>>> a = [1, 2, 3, 4]\n\n>>> add(*a)\n\n10\n\n1\n\n2\n\n3\n\n4\n\n5\n\n6\n\n7\n\n8\n\n>>> def add(**kwargs):\n\nreturn kwargs\n\n>>> add() # 未有参数，kwargs 为空字典\n\n{}\n\n>>> add(x=1) # x=1 => kwargs={‘x’: 1}\n\n{‘x’: 1}\n\n>>> add(x=1, y=2) # x=1, y=2 => kwargs={‘y’: 2, ‘x’: 1}\n\n{‘y’: 2, ‘x’: 1}\n\n1\n\n2\n\n3\n\n4\n\n5\n\n6\n\n7\n\n8\n\n>>> def add(x, y, z):\n\n… return x + y + z\n\n>>> dict1 = {‘z’: 3, ‘x’: 1, ‘y’: 6}\n\n>>> add(dict1[‘x’], dict1[‘y’], dict1[‘z’]) #\n\n10\n\n>>> add(**dict1) # 使用 **dict一来传参，等价于上边包车型地铁做法\n\n10\n\n1\n\n2\n\n3\n\n4\n\n5\n\n6\n\n7\n\n8\n\n9\n\n10\n\n11\n\n12\n\n13\n\n>>> def sum(**kwargs): # 函数参数是首要字参数\n\n… sum = 0\n\n… for k, v in kwargs.items():\n\n… sum += v\n\n… return sum\n\n>>> sum() # 未有参数\n\n0\n\n>>> dict1 = {‘x’: 1}\n\n>>> sum(**dict1) # 相当于 sum(x=1)\n\n1\n\n>>> dict2 = {‘x’: 2, ‘y’: 6}\n\n>>> sum(**dict2) # 相当于 sum(x=2, y=6)\n\n8\n\n1\n\n2\n\n3\n\n4\n\n5\n\n6\n\n>>> def func(x, y, z=0, *args, **kwargs):\n\nprint ‘x =’, x\n\nprint ‘y =’, y\n\nprint ‘z =’, z\n\nprint ‘args =’, args\n\nprint ‘kwargs =’, kwargs\n\n1\n\n2\n\n3\n\n4\n\n5\n\n6\n\n7\n\n8\n\n9\n\n10\n\n11\n\n12\n\n13\n\n14\n\n15\n\n16\n\n17\n\n18\n\n19\n\n20\n\n21\n\n22\n\n23\n\n24\n\n25\n\n26\n\n27\n\n28\n\n29\n\n30\n\n>>> func(1, 2) # 至少提供多少个参数，因为 x,\ny 是必选参数\n\nx = 1\n\ny = 2\n\nz = 0\n\nargs = ()\n\nkwargs = {}\n\n>>> func(1, 2, 3) # x=1, y=2, z=3\n\nx = 1\n\ny = 2\n\nz = 3\n\nargs = ()\n\nkwargs = {}\n\n>>> func(1, 2, 3, 4, 5, 6) # x=1, y=2, z=3, args=(4,\n5, 6), kwargs={}\n\nx = 1\n\ny = 2\n\nz = 3\n\nargs = (4, 5, 6)\n\nkwargs = {}\n\n>>> func(1, 2, 4, u=6, v=7) # args = (), kwargs = {‘u’:\n6, ‘v’: 7}\n\nx = 1\n\ny = 2\n\nz = 4\n\nargs = ()\n\nkwargs = {‘u’: 6, ‘v’: 7}\n\n>>> func(1, 2, 3, 4, 5, u=6, v=7) # args = (4, 5), kwargs =\n{‘u’: 6, ‘v’: 7}\n\nx = 1\n\ny = 2\n\nz = 3\n\nargs = (4, 5)\n\nkwargs = {‘u’: 6, ‘v’: 7}\n\n1\n\n2\n\n3\n\n4\n\n5\n\n6\n\n7\n\n8\n\n>>> a = (1, 2, 3)\n\n>>> b = {‘u’: 6, ‘v’: 7}\n\n>>> func(*a, **b)\n\nx = 1\n\ny = 2\n\nz = 3\n\nargs = ()\n\nkwargs = {‘u’: 6, ‘v’: 7}\n\n*args 表示可变参数，**kwargs 表示关键字参数。\n\n*args 和 **kwargs 是 Python 的惯用写法。\n\nargs 和 *kwargs · Python进阶"
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| {"ft_lang_label":"__label__zh","ft_lang_prob":0.5046592,"math_prob":0.98979056,"size":9351,"snap":"2019-13-2019-22","text_gpt3_token_len":6131,"char_repetition_ratio":0.13448165,"word_repetition_ratio":0.25499335,"special_character_ratio":0.43685168,"punctuation_ratio":0.18785976,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99006134,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-05-22T00:36:16Z\",\"WARC-Record-ID\":\"<urn:uuid:ba5ac649-c6ac-4785-88e3-ab2a04bcb4d2>\",\"Content-Length\":\"36738\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:b2a57e5f-186c-4a48-a2b3-abd284403581>\",\"WARC-Concurrent-To\":\"<urn:uuid:4082409b-15a2-420d-90f8-39caaf8b3876>\",\"WARC-IP-Address\":\"170.130.29.232\",\"WARC-Target-URI\":\"https://www.52b2c.com/archives/1169.html\",\"WARC-Payload-Digest\":\"sha1:YTH4X5VMUKZBC4F2D3RHS7725H4QX5T6\",\"WARC-Block-Digest\":\"sha1:CWAJVVWW5FGC6B4FAWYUEXDZP7GJCZYQ\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-22/CC-MAIN-2019-22_segments_1558232256600.32_warc_CC-MAIN-20190522002845-20190522024845-00325.warc.gz\"}"} |
https://gcc.gnu.org/pipermail/gcc-bugs/2000-December/042032.html | [
"# pa reload problem\n\nAlan Modra [email protected]\nSat Dec 9 05:57:00 GMT 2000\n\n```On Sat, 9 Dec 2000, Alan Modra wrote:\n\n> Ah. What I had to start with was REGNO_REG_CLASS (i) != NO_REGS.\n> Should have stuck with that. Hmm, I see local_alloc uses\n> HARD_REGNO_MODE_OK, so is this better?\n\nGrrr, that allocates r1 first, which loses on 32-bit hppa. Here's yet\nanother try, a little more sophisticated this time.\n\n* except.c (eh_regs): Save results of build_pointer_type to a temp\nas FUNCTION_VALUE macro may evaluate its args multiple times.\nWhen allocating registers, do so according to REG_ALLOC_ORDER and\nchoose GENERAL_REGS.\n\nI was tempted to remove the call_used_regs and fixed_regs tests when\nREG_ALLOC_ORDER is defined, but on checking some targets found i860 has a\nfixed reg as it's no. 1 priority reg. Strange.\n\nAlan Modra\n--\nLinuxcare. Support for the Revolution.\n\n--- gcc/except.c~\tMon Dec 4 14:55:35 2000\n+++ gcc/except.c\tSun Dec 10 00:11:29 2000\n@@ -2999,16 +2999,16 @@ eh_regs (pcontext, psp, pra, outgoing)\nint outgoing ATTRIBUTE_UNUSED;\n{\nrtx rcontext, rsp, rra;\n- unsigned int i;\n+ unsigned int i, r;\n+ tree t;\n\n+ t = build_pointer_type (void_type_node);\n#ifdef FUNCTION_OUTGOING_VALUE\nif (outgoing)\n- rcontext = FUNCTION_OUTGOING_VALUE (build_pointer_type (void_type_node),\n-\t\t\t\t current_function_decl);\n+ rcontext = FUNCTION_OUTGOING_VALUE (t, current_function_decl);\nelse\n#endif\n- rcontext = FUNCTION_VALUE (build_pointer_type (void_type_node),\n-\t\t\t current_function_decl);\n+ rcontext = FUNCTION_VALUE (t, current_function_decl);\n\n#ifdef STATIC_CHAIN_REGNUM\nif (outgoing)\n@@ -3023,22 +3023,39 @@ eh_regs (pcontext, psp, pra, outgoing)\nif (rsp == NULL_RTX)\n{\nfor (i = 0; i < FIRST_PSEUDO_REGISTER; ++i)\n-\tif (call_used_regs[i] && ! fixed_regs[i] && i != REGNO (rcontext))\n-\t break;\n+\t{\n+#ifdef REG_ALLOC_ORDER\n+\t r = reg_alloc_order[i];\n+#else\n+\t r = i;\n+#endif\n+\t if (call_used_regs[r] && ! fixed_regs[r]\n+\t && TEST_HARD_REG_BIT (reg_class_contents[(int) GENERAL_REGS], r)\n+\t && r != REGNO (rcontext))\n+\t break;\n+\t}\nif (i == FIRST_PSEUDO_REGISTER)\nabort();\n\n- rsp = gen_rtx_REG (Pmode, i);\n+ rsp = gen_rtx_REG (Pmode, r);\n}\n\nfor (i = 0; i < FIRST_PSEUDO_REGISTER; ++i)\n- if (call_used_regs[i] && ! fixed_regs[i]\n-\t&& i != REGNO (rcontext) && i != REGNO (rsp))\n- break;\n+ {\n+#ifdef REG_ALLOC_ORDER\n+ r = reg_alloc_order[i];\n+#else\n+ r = i;\n+#endif\n+ if (call_used_regs[r] && ! fixed_regs[r]\n+\t && TEST_HARD_REG_BIT (reg_class_contents[(int) GENERAL_REGS], r)\n+\t && r != REGNO (rcontext) && r != REGNO (rsp))\n+\tbreak;\n+ }\nif (i == FIRST_PSEUDO_REGISTER)\nabort();\n\n- rra = gen_rtx_REG (Pmode, i);\n+ rra = gen_rtx_REG (Pmode, r);\n\n*pcontext = rcontext;\n*psp = rsp;\n\n```"
]
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| {"ft_lang_label":"__label__en","ft_lang_prob":0.557321,"math_prob":0.94303894,"size":2697,"snap":"2022-40-2023-06","text_gpt3_token_len":828,"char_repetition_ratio":0.10063127,"word_repetition_ratio":0.24938875,"special_character_ratio":0.3466815,"punctuation_ratio":0.22197309,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9657808,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-10-05T07:46:01Z\",\"WARC-Record-ID\":\"<urn:uuid:9e5c4050-8620-466c-8cce-5e3c9f711f0e>\",\"Content-Length\":\"5801\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:64a7faba-3b9d-4851-81bc-8c6854723e41>\",\"WARC-Concurrent-To\":\"<urn:uuid:2180d357-b1f0-409a-be52-64ce4997a186>\",\"WARC-IP-Address\":\"8.43.85.97\",\"WARC-Target-URI\":\"https://gcc.gnu.org/pipermail/gcc-bugs/2000-December/042032.html\",\"WARC-Payload-Digest\":\"sha1:MK4FL3B7SCOXVPIKPE2EWUSWJCZOE4JI\",\"WARC-Block-Digest\":\"sha1:WMQQSYHM6OOXECA7A6B4EQ7B4CYRBBBQ\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-40/CC-MAIN-2022-40_segments_1664030337595.1_warc_CC-MAIN-20221005073953-20221005103953-00137.warc.gz\"}"} |
https://tex.stackexchange.com/questions/201122/tikz-draw-a-circle-only-in-selected-nodes | [
"Tikz: draw a circle only in selected nodes\n\nI created a table of nodes using tikz. And now I want to draw a circle inside every even node. I started from just trying to draw a circle in the second node. But my code is not working:\n\n\\begin{center}\n\\begin{tikzpicture}[->,shorten >=1pt,auto]\n\\hashcells{12}\n\\foreach [count=\\i from 0] \\number in {0, 0, 1, 0, 1, 0, 0}{\n\\node at (hashcell\\i) {\\number};\n\\if \\i==2\n\\draw [fill=yellow] (0,0) circle (0.1cm) node (sun) {};\n\\fi\n}\n\\end{tikzpicture}\n\\end{center}\n\nWhereas the code below works fine and draws a circle in the first node. So I suppose there is something wrong with my if statement:\n\n\\begin{center}\n\\begin{tikzpicture}[->,shorten >=1pt,auto]\n\\hashcells{12}\n\\foreach [count=\\i from 0] \\number in {0, 0, 1, 0, 1, 0, 0}{\n\\node at (hashcell\\i) {\\number};\n\\draw [fill=yellow] (0,0) circle (0.1cm) node (sun) {};\n}\n\\end{tikzpicture}\n\n\\end{center}\n\nIf someone can suggest what I am doing wrong I will be grateful.\n\n• Could you please provide a compilable MWE? Sep 14 '14 at 14:18\n• \\ifnum\\i=2? ..\n– cfr\nSep 14 '14 at 14:23\n• Try \\ifodd\\i\\else...\\fi or \\ifnum\\i=2...\\fi.\n– user30471\nSep 14 '14 at 14:24\n• Where is \\hashcells from? It is not in the TiKZ manual.\n– cfr\nSep 14 '14 at 15:27\n• @Ziva The point is that you should post an example which people can compile. If the definition is not needed, that's fine - leave it out of the code you post in your question.\n– cfr\nSep 14 '14 at 15:46\n\nPrevious note: Please, next time provide a COMPLETE minimal working (or not) example: starting from \\documentclass... and finishing with \\end{document} with all libraries and personal functions you use to test your code.\n\nAnswer: I'm not sure about what you want to do but if you just want to add some filled circles to previous defined nodes you can do it outside the foreach loop. Node names are known after foreach so you can make reference to them. There's no need for if inside the loop to add color to a particular node. If you don't know the particular name or want to calculate it forget my answer.\n\nIf you draw a filled circle at a node center you'll cover its contents, but you can use backgrounds library to add the filled circle behind the node. Both examples are shown in next code.\n\nFinal note: I've suposed that \\hashcell was related to your previous question Tikz: define label font size under the node\n\n\\documentclass[tikz,border=2mm]{standalone}\n\n\\usetikzlibrary{backgrounds}\n\\tikzset{\ncell/.style = {draw, minimum width=0.5cm, minimum height=0.5cm, minimum size = 0.5cm}\n}\n\n\\begin{document}\n\n\\begin{tikzpicture}\n\n\\foreach [count=\\i from 0] \\j in {1,...,7} {\n\\node[cell,label=below:{\\scriptsize \\i}] (cell\\i) at (\\i*0.5,0) {};}\n\n\\foreach [count=\\i from 0] \\number in {0, 0, 1, 0, 1, 0, 0}{\n\\node at (cell\\i) {\\number};\n}\n\n\\draw [fill=yellow] (0,0) circle (0.2cm) node (sun) {};\n\n\\begin{scope}[on background layer]\n\\draw [fill=yellow] (cell2.center) circle (0.2cm) node (sun) {};\n\\end{scope}\n\n\\end{tikzpicture}\n\\end{document}",
null,
"I think an easy approach is to use a matrix of nodes to draw the initial table. This automatically asigns a name to each cell, in the form matrix_name-Row-Column so it is easy to later add the circle via a loop.\n\n\\usetikzlibrary{matrix}\n\\begin{tikzpicture}\n\\tikzset{\ncell/.style = {inner sep=2pt, minimum width=2em, minimum height=2em, draw}\n}\n\\matrix[matrix of nodes, every node/.append style=cell] (M) {\n0 & 0 & 1 & 0 & 1 & 0 & 0 \\\\\n};\n\n\\foreach \\cell in {2,4,6}\n\\draw[fill=yellow, fill opacity=0.4] (M-1-\\cell) circle(.8em);\n\\end{tikzpicture}\n\nResult:",
null,
"Here another approach using the pic-environment (available since pgf 3.0). I got two code examples using even/odd or custom positions for the yellow circles. This way the elements are predefined and can easily be reused in the \\foreach statement.\n\nThe code should be self-explanatory, in doubt just ask.\n\nInfo: The both examples just differ in the \\foreach-loop and a \\usepackage{xifthen}. Style definitions are all the same. I just put it in two separate examples for a better understanding.\n\nEven/Odd-Approach\n\n\\documentclass[tikz, border=5mm]{standalone}\n\n\\begin{document}\n\\begin{tikzpicture}\n\\newcommand{\\mysize}{.5cm}\n\\tikzset{\nbase/.style={minimum size=\\mysize, draw, rectangle, label={below:{\\scriptsize \\x}}},\ncell/.pic={ %\n\\node [base] at (0,0) {\\lbl};\n},\nsun/.pic={ %\n\\draw [fill=yellow] (0,0) circle (.2cm);\n\\pic at (0,0) {cell};\n},\n}\n\n\\foreach \\x\\lbl [count=\\i] in {1/,2/0,3/1,4/0,5/1,6/0,7/0} {\n\\newcommand{\\mypos}{\\x*\\mysize, 0}\n\\ifodd\\i % if you need even, just swap the pic-commands\n\\pic at (\\mypos) {sun};\n\\else\n\\pic at (\\mypos) {cell};\n\\fi\n}\n\\end{tikzpicture}\n\\end{document}",
null,
"Custom position approach (needs package xifthen)\n\n\\documentclass[tikz, border=5mm]{standalone}\n\n\\usepackage{xifthen}\n\n\\begin{document}\n\\begin{tikzpicture}\n\\newcommand{\\mysize}{.5cm}\n\\tikzset{\nbase/.style={minimum size=\\mysize, draw, rectangle, label={below:{\\scriptsize \\x}}},\ncell/.pic={ %\n\\node [base] at (0,0) {\\lbl};\n},\nsun/.pic={ %\n\\draw [fill=yellow] (0,0) circle (.2cm);\n\\pic at (0,0) {cell};\n},\n}\n\n\\foreach \\x\\lbl [count=\\i] in {1/,2/0,3/1,4/0,5/1,6/0,7/0} {\n\\newcommand{\\mypos}{\\x*\\mysize,0}\n\\ifthenelse { %\n\\i=1 \\OR \\i=3 \\OR \\i=6 % here go your x-positions\n} { %\n\\pic at (\\mypos) {sun};\n} { %\n\\pic at (\\mypos) {cell};\n}\n}\n\\end{tikzpicture}\n\\end{document}",
null,
""
]
| [
null,
"https://i.stack.imgur.com/s8Hxm.png",
null,
"https://i.stack.imgur.com/qUriB.png",
null,
"https://i.stack.imgur.com/XEgLe.png",
null,
"https://i.stack.imgur.com/P4iqc.png",
null
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https://www.rdocumentation.org/packages/matlib/versions/0.9.2/topics/workers | [
"# workers\n\n0th\n\nPercentile\n\n##### Workers Data\n\nA toy data set comprised of information on workers Income in relation to other variables, used for illustrating linear and multiple regression.\n\nKeywords\ndatasets\n##### Usage\ndata(\"workers\")\n##### Format\n\nA data frame with 10 observations on the following 4 variables.\n\nIncome\n\nincome from the job, a numeric vector\n\nExperience\n\nnumber of years of experience, a numeric vector\n\nSkill\n\nskill level in the job, a numeric vector\n\nGender\n\na factor with levels Female Male\n\n• workers\n##### Examples\n# NOT RUN {\ndata(workers)\nplot(Income ~ Experience, data=workers, main=\"Income ~ Experience\", pch=20, cex=2)\n\n# simple linear regression\nreg1 <- lm(Income ~ Experience, data=workers)\nabline(reg1, col=\"red\", lwd=3)\n\nplot(Income ~ Experience, data=workers, main=\"Income ~ poly(Experience,2)\", pch=20, cex=2)\nreg2 <- lm(Income ~ poly(Experience,2), data=workers)\nfit2 <-predict(reg2)\nabline(reg1, col=\"red\", lwd=1, lty=1)\nlines(workers$Experience, fit2, col=\"blue\", lwd=3) # How does Income depend on a factor? plot(Income ~ Gender, data=workers, main=\"Income ~ Gender\") points(workers$Gender, jitter(workers$Income), cex=2, pch=20) means<-aggregate(workers$Income,list(workers\\$Gender),mean)\npoints(means,col=\"red\", pch=\"+\", cex=2)\nlines(means,col=\"red\", lwd=2)\n\n# }\n\nDocumentation reproduced from package matlib, version 0.9.2, License: GPL (>= 2)\n\n### Community examples\n\nLooks like there are no examples yet."
]
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https://dsp.stackexchange.com/questions/22661/how-do-i-obtain-the-frequencies-of-each-value-in-a-2d-fft-and-what-do-they-mean | [
"# How do I obtain the frequencies of each value in a 2D FFT and what do they mean?\n\nMy goal is to analyze the frequencies in an image representing water simulation data before and after, for example, a Gaussian filter. The direction of these frequencies is not important to me. Ideally, I think I want to plot these frequencies in a 2D graph, where x represents the frequency, ranging from 0 to 0.5, and y represents the amplitude.\n\nI understand how to obtain the 1D frequencies, as is explained here: https://stackoverflow.com/questions/4364823/how-do-i-obtain-the-frequencies-of-each-value-in-a-fft. From this, I also understand how to obtain the frequencies in x- and y- direction.\n\nWhat I do not fully understand is what the direction-independent frequencies of my output values are. Intuitively, I would expect to calculate them like so:\n\n// from output to frequencies to periods\nfx = kx / N; fy = ky / M\npx = 1/fx; py = 1/fy\n\n// direction independent frequency of output value\nperiod = sqrt(px*px + py*py)\nfrequency = 1/p",
null,
"From what I understand, u and v are the x- and y- frequencies respectively, in cycles per unit distance. This would mean that, if I have a signal with x- and y- frequencies of both 0.5, the resulting frequency is 0.707... ? This is over the Nyquist limit, even more so diagonally! I have a feeling I am missing something, but I can't find what.\n\nAdditionally, what do the frequencies with an x- or y- frequency of 0 or 0.5 mean? In 1D, they would be the DC and Nyquist component. How does this translate to 2D, where the other may have a perfectly fine value? More practically, which frequencies can I use for my graph?"
]
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null,
"https://i.stack.imgur.com/xfOG0.png",
null
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http://talkstats.com/threads/proving-regression-with-dummy-variables-gives-same-estimates-as-separate-models.58901/ | [
"# proving regression with dummy variables gives same estimates as separate models\n\n#### statsmeister\n\n##### New Member\nSee attachment for problem description:\n\nTo prove this I started by using: vector of Beta estimates for category k = [Bk], matrix of estimates for category k = [Xk], Y estimates for category = [Yk].\n1. Find betas for single category:\nCalculating [Ba] = (Xa' * Xa)inv * (Xa' *Ya) for the\n\n2. Find betas for matrix of combined categories for the general case.\nIn block diagonal form:\n\n[X] becomes [Xa 0 ... 0]\nbecomes [0 Xb... 0]\nbecomes [0 0... Xk]\n\n[Y] becomes [Ya]\n[Yb] ...\n[Yk]\nand you do the same multiplication B = (X' * X)inverse * (X'*Y).\n\nMultiplying this out I get (X' * X)inverse =\n[1/Xa^2 0 ... 0]\n[0 1/Xb^2... 0]\n[0 0... 1/Xk^2]\nand X' * Y:\n[XaYa ..... 0]\n[0 XbYb.... 0]\n[0 0.. XkYk]\nand eventually I get B = Ya * Xa'.\n\nNow that I'm this far I am not sure how to prove that the singular case for Ba = Ya*Xa' without adding specific numbers. Does anyone know how to proceed? Any help would be appreciated."
]
| [
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]
| {"ft_lang_label":"__label__en","ft_lang_prob":0.80755675,"math_prob":0.9928081,"size":1803,"snap":"2022-27-2022-33","text_gpt3_token_len":607,"char_repetition_ratio":0.1295164,"word_repetition_ratio":0.9655172,"special_character_ratio":0.36272877,"punctuation_ratio":0.22041763,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99102056,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-08-10T19:44:19Z\",\"WARC-Record-ID\":\"<urn:uuid:dc8cc65c-3df2-4ec8-ac35-813d225cf398>\",\"Content-Length\":\"32287\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:71ae66d4-9fb4-48f7-bff7-046994804338>\",\"WARC-Concurrent-To\":\"<urn:uuid:1b4e6703-7cd0-4966-8b34-b89270100756>\",\"WARC-IP-Address\":\"199.167.200.62\",\"WARC-Target-URI\":\"http://talkstats.com/threads/proving-regression-with-dummy-variables-gives-same-estimates-as-separate-models.58901/\",\"WARC-Payload-Digest\":\"sha1:OBST2G74W46KJNJCZ74GZEZTAT4MRJPJ\",\"WARC-Block-Digest\":\"sha1:KIWLMGIMLVEGIIW2DFVE5DGWHODWA5SV\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-33/CC-MAIN-2022-33_segments_1659882571210.98_warc_CC-MAIN-20220810191850-20220810221850-00441.warc.gz\"}"} |
https://www.quesba.com/questions/3-paste-the-data-hand-width-in-the-carr-regression-applet-and-make-a-scatte-41178 | [
"# 3. Paste the data, hand width, in the Carr/Regression applet and make a scatterplot. Describe the...\n\n3. Paste the data, hand width, in the Carr/Regression applet and make a scatterplot. Describe the direction, form, and strength of the scatterplot.\n\n4. Use the applet to determine the least squares regression line for predicting estimated weight based on hand width. What is the value of the slope of the regression line? What does this number imply in terms of hand width and estimated weight?\n\nMar 24 2020| 06:04 AM |",
null,
""
]
| [
null,
"https://www.quesba.com/images/bluer_bg.webp",
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]
| {"ft_lang_label":"__label__en","ft_lang_prob":0.87112796,"math_prob":0.91107845,"size":393,"snap":"2021-31-2021-39","text_gpt3_token_len":89,"char_repetition_ratio":0.13881747,"word_repetition_ratio":0.0,"special_character_ratio":0.20101781,"punctuation_ratio":0.14102565,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9933153,"pos_list":[0,1,2],"im_url_duplicate_count":[null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-07-30T14:51:38Z\",\"WARC-Record-ID\":\"<urn:uuid:7947b900-e46e-4bd9-8b91-6cc2ce1ba585>\",\"Content-Length\":\"121491\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:48011b26-fc56-4717-b3e6-f0e7c5b0537c>\",\"WARC-Concurrent-To\":\"<urn:uuid:7ac7194a-2943-4acc-944e-997ce5d1115f>\",\"WARC-IP-Address\":\"52.71.126.16\",\"WARC-Target-URI\":\"https://www.quesba.com/questions/3-paste-the-data-hand-width-in-the-carr-regression-applet-and-make-a-scatte-41178\",\"WARC-Payload-Digest\":\"sha1:NTA4AORAZVLLYJKNADXHGUPCCNCYQGYV\",\"WARC-Block-Digest\":\"sha1:XJONUIE5ZKDEXREZH2QJPTRUIPU7FSV4\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-31/CC-MAIN-2021-31_segments_1627046153966.60_warc_CC-MAIN-20210730122926-20210730152926-00167.warc.gz\"}"} |
http://artfulsoftware.com/infotree/qrytip.php?id=2063&m= | [
"## List equivalences\n\n### from the Artful Common Queries page\n\n A table `(name, ref, equiv)` list refs and their equivalents. ```drop table if exists t; create table t(name char(1),ref char(2), equivalent char(2) ); insert into t values ('A','A1','B1'),('B','B1','A1'),('C','C1','B1'), ('D','D1','E1'),('E','E1','A1'),('F','F1','R1'); ``` How to retrieve a list of all equivalencies? Equivalence is symmetric---`a≡b` entails `b≡a`---so a complete list of equivalences requires that we take both versions of each equivalence ... ``` select ref,equivalent from t union select equivalent,ref from t ``` Then aggregate with `group_concat()` over `ref`... ```sselect ref,group_concat(equivalent) as equiv from ( select ref,equivalent from t union select equivalent,ref from t ) x group by ref order by ref; +------+-------+ | ref | equiv | +------+-------+ | A1 | B1,E1 | | B1 | A1,C1 | | C1 | B1 | | D1 | E1 | | E1 | A1,D1 | | F1 | R1 | | R1 | F1 | +------+-------+ ```Last updated 26 Apr 2020",
null,
""
]
| [
null,
"http://artfulsoftware.com/ukraine.jpg",
null
]
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https://homework.cpm.org/category/CON_FOUND/textbook/a2c/chapter/12/lesson/12.5.1/problem/12-195 | [
"",
null,
"",
null,
"### Home > A2C > Chapter 12 > Lesson 12.5.1 > Problem12-195\n\n12-195.\n\nUse the Binomial Theorem to write the expansion of each of the following expressions.\n\n1. $( 1 + \\frac { 1 } { n } ) ^ { 3 }$\n\n$1+\\frac{3}{n}+\\frac{3}{n^2}+\\frac{1}{n^3}$\n\n1. $( 1 + \\frac { 1 } { n } ) ^ { 5 }$\n\nUse the same method as in part (a)."
]
| [
null,
"https://homework.cpm.org/dist/7d633b3a30200de4995665c02bdda1b8.png",
null,
"data:image/png;base64,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",
null
]
| {"ft_lang_label":"__label__en","ft_lang_prob":0.79486597,"math_prob":0.9992447,"size":308,"snap":"2022-27-2022-33","text_gpt3_token_len":82,"char_repetition_ratio":0.14473684,"word_repetition_ratio":0.6792453,"special_character_ratio":0.29220778,"punctuation_ratio":0.109375,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9994448,"pos_list":[0,1,2,3,4],"im_url_duplicate_count":[null,null,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-08-15T01:36:22Z\",\"WARC-Record-ID\":\"<urn:uuid:472275b2-4318-431f-a49f-cf247e74cf90>\",\"Content-Length\":\"38383\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:8e818ba0-5883-4e4c-9079-f6cd234ece7e>\",\"WARC-Concurrent-To\":\"<urn:uuid:d3fec71b-cbca-42e1-b662-75dfb7eca04c>\",\"WARC-IP-Address\":\"104.26.6.16\",\"WARC-Target-URI\":\"https://homework.cpm.org/category/CON_FOUND/textbook/a2c/chapter/12/lesson/12.5.1/problem/12-195\",\"WARC-Payload-Digest\":\"sha1:XJH2PMVQ5EBEJL7CS5NOPY6DUUKAKYE6\",\"WARC-Block-Digest\":\"sha1:KYR6TPQW73BKCCQGHPZLBGPFEWO6G5WH\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-33/CC-MAIN-2022-33_segments_1659882572089.53_warc_CC-MAIN-20220814234405-20220815024405-00456.warc.gz\"}"} |
http://www.algorist.com/algowiki/index.php?title=TADM2E_4.24&oldid=63 | [
"It's not clear to me if the first $n - \\sqrt n$ element being sorted means that the remaining $\\sqrt n$ elements are all bigger or not, but let's suppose they are not, since that's more general.\nFirst sort the remaining $\\sqrt n$ elements in $O(\\sqrt n \\log n)$ time. After that we can do a simple merge step to merge the two sorted arrays in $O(n)$ time. Since $\\sqrt n$ dominates $\\log n$ we can come to the conclusion that the total running time of this algorithm is: $O(\\sqrt n \\log n + n) = O(\\sqrt n \\sqrt n + n) = O(n + n) = O(n)$"
]
| [
null
]
| {"ft_lang_label":"__label__en","ft_lang_prob":0.8615508,"math_prob":0.99997044,"size":732,"snap":"2019-51-2020-05","text_gpt3_token_len":215,"char_repetition_ratio":0.17857143,"word_repetition_ratio":0.016528925,"special_character_ratio":0.30464482,"punctuation_ratio":0.06,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":1.0000069,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-01-18T11:21:41Z\",\"WARC-Record-ID\":\"<urn:uuid:eeac4794-70ae-48c6-82ff-00eceac99431>\",\"Content-Length\":\"14734\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:6e52321b-8635-483a-ba58-b210f00ab49e>\",\"WARC-Concurrent-To\":\"<urn:uuid:359ee1dc-6b63-4869-ac35-ea658a3b81c8>\",\"WARC-IP-Address\":\"54.214.216.106\",\"WARC-Target-URI\":\"http://www.algorist.com/algowiki/index.php?title=TADM2E_4.24&oldid=63\",\"WARC-Payload-Digest\":\"sha1:UG7I53GDXKF6BHNOG6Y3BGSNUMT2MCUI\",\"WARC-Block-Digest\":\"sha1:UYUUCYTAAW2F6XSFWPD3FSCZJ4JENGUM\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-05/CC-MAIN-2020-05_segments_1579250592565.2_warc_CC-MAIN-20200118110141-20200118134141-00455.warc.gz\"}"} |
https://stackoverflow.com/questions/36780324/center-of-a-cluster-of-latitude-and-longitude-points-in-google-map-android | [
"# Center of a cluster of latitude and longitude points in Google map android\n\nI am trying to get the center of a shape in android. The shape is draw by hand on the map. I have all of the co-ordinates but the mean values are turning out to be trickier than I thought. I want to plot a marker on the mean latitude and longitude.\n\nI have tried summing up the latitude and longitude separately and then dividing by the number of points. This does not give the correct answer. The marker always seems to be trailing behind the draw figure. I have also tried using the implementation but it gives the same answer, a previous SO question, Calculate the center point of multiple latitude/longitude coordinate pairs\n\nThe code I have been using:\n\n``````private void calculateFreeHandPolygonParameters(){\n\ndouble xValues = 0;\ndouble yValues = 0;\ndouble zValues = 0;\nint count = 0;\n\n// linesForPolygon a list of the lines in the polygon\nfor(Polyline line : linesForPolygon){\nfor (LatLng point : line.getPoints()) {\ncount++;\n}\n}\n\ndouble meanX = xValues/count;\ndouble meanY = yValues/count;\ndouble meanZ = zValues/count;\n\ndouble centralLongitude = Math.atan2(meanY, meanX);\ndouble centralSquareRoot = Math.sqrt(Math.pow(meanX, 2) + Math.pow(meanX, 2) + Math.pow(meanX, 2));\ndouble centralLatitude = Math.atan2(meanZ, centralSquareRoot);\n\ndouble latitude = Math.toDegrees(centralLatitude);\ndouble longitude = Math.toDegrees(centralLongitude);\n\nLog.i(\"MAPS\", \"Freehand Parameters: x mean -> \" + latitude + \" y mean -> \" + longitude);\n\n.position(new LatLng(latitude, longitude))\n.title(\"Polygon center\")\n.snippet(\"lat: \" + latitude + \" long: \" + longitude));\n}\n``````\n• The method you describe should work except perhaps in cases where the longitude is near the international dateline, or if the locations are near the poles. Perhaps you should share your code. – JerryM Apr 21 '16 at 21:03\n\nHere is my centroid code:\n\n``````public class PolygonCentroid {\n\nprivate List<GeoPoint> points;\nprivate int pointsSize;\n\npublic PolygonCentroid(List<GeoPoint> points) {\nthis.points = points;\nthis.pointsSize = points.size();\n}\n\nprotected double polygonArea() {\ndouble area = 0;\nfor (int i = 0, j; i < pointsSize; i++) {\nj = (i + 1) % pointsSize;\narea += points.get(i).getLongitude() * points.get(j).getLatitude();\narea -= points.get(i).getLatitude() * points.get(j).getLongitude();\n}\narea /= 2.0;\nreturn area;\n}\n\npublic GeoPoint centroid() {\ndouble cx = 0, cy = 0;\ndouble factor;\nfor (int i = 0, j; i < pointsSize; i++) {\nj = (i + 1) % pointsSize;\nfactor = (points.get(i).getLongitude() * points.get(j).getLatitude() - points.get(j).getLongitude() * points.get(i).getLatitude());\ncx += (points.get(i).getLongitude() + points.get(j).getLongitude()) * factor;\ncy += (points.get(i).getLatitude() + points.get(j).getLatitude()) * factor;\n}\ndouble A = polygonArea();\nfactor = 1.0 / (6.0 * A);\ncx *= factor;\ncy *= factor;\nreturn new GeoPoint(cy, cx);\n}\n\n}\n``````",
null,
"Also, please pay attention that the centroid can be outside of the polygon:",
null,
"• This method works but only for the first iteration. I haven't figured out why it doesn't work if I draw another polygon. The marker trails behind as in the previous version of my code. – Michael Richardson Apr 22 '16 at 7:44\n• Don't forget to reset the list of co-ordinates like I did. – Michael Richardson Apr 22 '16 at 7:50\n• >\"marker trails behind\" I guess you should pay attention to Z-ordering, if you want to show some of your overlays on top always. – Mikalai Daronin Apr 22 '16 at 8:09\n1. You actually need to calculate what is called CENTROID. It's not a trivial thing, look here: http://www.spatialanalysisonline.com/HTML/index.html?centroids_and_centers.htm\n2. Pay attention to what is called GREAT CIRCLE, recall flight plans on an airplane maps... Thus if the distances are relatively large, you need to bare in mind this phenomena.\n3. This is a classic GIS question, after you understand the pseudo code of the algorithm, it's simple coding task... try gis.stackexchange.com\n• How come the code I have been using does not work? It makes perfect sense to calculate the average like I have done? The first method in the link you gave is what I tried initially but kept trailing behind too. The method which require the area to be calculated will become extremely complicated – Michael Richardson Apr 21 '16 at 21:26\n• Yes, the low level code of this kind of algorithms indeed complicated... Usually one use gis library for this kind of task, or if the locations stored in data base with postgis, so postgis has this kind of query methods too. – michael Apr 21 '16 at 21:58"
]
| [
null,
"https://i.stack.imgur.com/h3I1N.png",
null,
"https://i.stack.imgur.com/1IY9h.png",
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| {"ft_lang_label":"__label__en","ft_lang_prob":0.5763807,"math_prob":0.9458624,"size":1844,"snap":"2019-43-2019-47","text_gpt3_token_len":429,"char_repetition_ratio":0.17663044,"word_repetition_ratio":0.008032128,"special_character_ratio":0.24295011,"punctuation_ratio":0.20474778,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.995683,"pos_list":[0,1,2,3,4],"im_url_duplicate_count":[null,1,null,1,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-11-12T13:00:12Z\",\"WARC-Record-ID\":\"<urn:uuid:b32b63a5-d5a2-4ee5-a5f1-63ec843e6142>\",\"Content-Length\":\"155040\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:a34f414e-645c-491f-9d14-994048aaa51b>\",\"WARC-Concurrent-To\":\"<urn:uuid:2c41c3c6-2299-4267-8972-d62c247fa5e9>\",\"WARC-IP-Address\":\"151.101.129.69\",\"WARC-Target-URI\":\"https://stackoverflow.com/questions/36780324/center-of-a-cluster-of-latitude-and-longitude-points-in-google-map-android\",\"WARC-Payload-Digest\":\"sha1:R3WHYDELN5C5YZSTB72CEQIWBD2BODOU\",\"WARC-Block-Digest\":\"sha1:ZY4MY3JQLMNUU2MGHMNVOCHUMDIAHG6Q\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-47/CC-MAIN-2019-47_segments_1573496665573.50_warc_CC-MAIN-20191112124615-20191112152615-00342.warc.gz\"}"} |
https://www.pharmacy180.com/article/concentration-calculations-2724/ | [
"# Concentration calculations\n\n| Home | | Pharmaceutical Drugs and Dosage | | Pharmaceutical Industrial Management |\n\n## Chapter: Pharmaceutical Drugs and Dosage: Pharmacy math and statistics\n\nA formulation is essentially a multicomponent mixture. The relative amount of a substance in a multicomponent system represents its concentration.\n\nConcentration calculations\n\nA formulation is essentially a multicomponent mixture. The relative amount of a substance in a multicomponent system represents its concentration. It could be the concentration of a dissolved drug in a solution, a suspended drug in a suspension, or a drug powder in a triturate of solid powders. The expression of concentration, its relation to the total amounts, and calculations involving changes to the concentration or total amount are an essential part of pharmacy practice. This section discusses the common ways of expressing concentrations, their basic principles, and the calcula-tions involving drug amounts in such preparations.\n\nPercentage solutions\n\nConcentrations of ingredients in a formula are often represented as a percentage (%). Percentage represents parts of 100 (cent). In liquid prepara-tions, percentage values can represent % weight/weight (% w/w, e.g., 2 g of solid in a 100 g of liquid = 2% w/w), % weight/volume (% w/v, e.g., 2 g of solid in 100 mL of liquid = 2% w/v), or % volume/volume (% v/v, e.g., 2 mL of liquid A in 100 mL of liquid B = 2% v/v of liquid A).\n\nCalculations for the exact amount of an ingredient to be used in a for-mulation when the percentage composition of the formula is known can be made using ratio and proportion. Thus, to dispense 240 mL of a 10% w/v solution of a drug substance, the amount of drug substance needed can be calculated as:",
null,
"Concentrations based on moles and equivalents\n\nMolecular weights or moles of a compound are more useful for calculations when two or more chemical compounds are to be compared for a given attribute. Thus, during drug discovery, relative potencies of different com-pounds are compared on a molar basis.\n\nThe concepts of solution concentrations of compounds are based on their molecular or equivalent weights, which can be defined as follows:\n\n· The molecular weight of a compound represents the weight of one mole (abbreviation: mol) of a compound, in grams. Thus, 1 mole of glucose is 180.16 g of glucose. The molar mass of glucose is thus represented as 180.16 g/mol.\n\n· An equivalent weight of a compound represents its molecular weight divided by the number of valence or ionic charges in solution. It takes into account the chemical activity of an electrolyte. One equivalent (abbreviation: Eq), in grams, of a compound represents 1 mole of compound in grams divided by its valence. Thus, molecular weight of Mg2+ ions is 24.3 g, indicating that 24.3 g of Mg2+ ions represents 1 mole of Mg2+. On the other hand, the equivalent weight of Mg2+ ions is 24.3/2 = 12.15 g, since there are two charges on Mg2+ ions. Thus, when used for charge-neutralization calculations, 1 mole or molecular weight of Mg2+ ions represents two equivalents.\n\nSolutions of electrolytes are often prepared in terms of molarity, molality, and normality.\n\n· Molarity (abbreviation: M) is defined as the moles of solute per liter of solution. Therefore, 1 M of sulfuric acid solution represents 98 g (molecular weight) of H2SO4 dissolved in 1 L of solution.\n\n· Normality (abbreviation: N) represents gram equivalent weight of solute per liter of solution. The difference between molarity and normality is representative of the difference between moles and equiv-alents of a compound. Thus, 1 N of sulfuric acid solution represents 49 g (equivalent weight) of H2SO 4 dissolved in 1 L of solution. The equivalent weight of H2SO4 is half its molecular weight, since H2SO4 is a diprotic acid (i.e., dissociates to release 2 H+ ions in solution).\n\n· Molality (abbreviation: m) is a less frequently used term that represents the number of moles of solute per kilogram of solvent.\n\n· Formality (abbreviation: F) is another less frequently used term that represents the formula weight of a compound in 1 L of solution. It differs from molarity in indicating the amount of solute added to the solution, but it does not consider the nature of the chemical species that actually exist in solution. For example, when 1 mole of sodium carbonate (Na2CO3) or sodium bicarbonate (NaHCO3) is dissolved in a total of 1 L of an acidic solution of hydrochloric acid (HCl), the concentration of Na2CO3 or NaHCO3 may be represented as 1 F (indicates the amount added) but not as 1 M (indicates amount in solution)—even though quantitatively they would be the same—since the compound reacts with acid in the solution and does not remain as the same species that was added.\n\n· The amount of a solute may also be represented as its mole fraction. The mole fraction of a solute is the number of moles of solute as a proportion of the total number of moles (of solute + solvent) in a solu-tion. For example, a mole fraction of 0.2 indicates 2 moles of solute dissolved in 8 moles of solvent. Mole fraction is a dimensionless quan-tity. Mole fraction is frequently used to represent the relative amount of two different solutes in a system.\n\nConcentrations and amounts can be represented in fractions by using the prefixes used in the metric system of measure. Thus, 1 mEq is one milliequivalent of a solute, thus representing 1/1,000th of an equivalent weight of the solute. Similarly, 1 μM would represent 1 micromolar, or 1/1,000,000th of a molar (1 mol/L) concentration of a solute.\n\n### Parts per unit concentrations\n\nParts per unit concentrations are commonly expressed for very low con-centrations of solutes. The commonly used parts per unit concentrations are as follows:\n\n· Parts per million (ppm) represents 1 part of a substance in 1 million (106) parts of the total mixture. Parts per million (ppm) is dimension-less, since the parts of both the substance and the total mixture are represented in the same units. In addition, this measure is applicable to both solutions and solids. Thus, 1 ppm of NaCl in a solid powder may represent 1 μg/g or 1 mg/kg of NaCl. Moreover, 1 ppm of NaCl solution can represent 1 μL/L of NaCl in water.\n\n· Parts per billion (ppb) represents 1 part of a substance in 1 billion (109) parts of the total mixture. Similar to ppm, it is dimensionless and does not represent a state (solid or liquid) of the substance.\n\n· Other less commonly used parts per unit measures are parts per thousand, parts per trillion (ppt, 1 in 1012), and parts per quadrillion (ppq, 1 in 1015).\n\n### Dilution of stock solutions\n\nA stock solution is a concentrated solution of a substance that can be diluted to a lower, desired concentration by adding the solvent immediately before use or dispensing. Stock solutions are frequently used in pharmacy dispensing to increase the efficiency, ease, and accuracy of dispensing, as well as the space and cost advantages with the transportation and storage of lower-volume concentrated solutions.\n\nA common calculation required for the dilution of a concentrated stock solution to a desired concentration is the amount of solvent needed to achieve the desired concentration. This can be derived from the formula for concentration (c) based on the volume (v) of solution and the weight (w) of the substance.",
null,
"Therefore, if the stock solution were designated by the subscript “1” and the final solution to be prepared by the subscript “2,”",
null,
"Since it is the same weight of the solute that would be transferred from the stock solution into the final, diluted solution,\n\nw1 = w2\n\nTherefore,\n\nc1 × v1 = c 2 × v2 (5.3)\n\nThis formula can be used to calculate the volume of solvent required to make a diluted solution. For example, to dilute a 50% w/v stock solution to make 200 mL of a 5% w/v solution, c1 = 50, c 2 = 5, and v2 = 200.\n\nv1 = ( c2 × v2 ) / c1 = ( 5 × 200 ) / 50 = 20mL\n\nHence, the amount of stock solution needed = 20 mL and the amount of solvent needed = 200 − 20 = 180 mL to make a total of 200 mL of the diluted solution.\n\nThe measurements can also be carried out in weight rather than in vol-ume for the stock and the diluted solutions. Thus,\n\nc1 × w1 = c 2 × w2 (5.4)\n\n### Mixing solutions of different concentrations\n\nOften, mixing of two products made of the same solute but having different concentrations is required. A convenient approach to solve these problems is the alligation method. Two kinds of alligation methods are commonly used: alligation medial and alligation alternate.\n\nAlligation medial\n\nFor two ingredients\n\nThis method is based on finding the proportion for a formula. The strength (e.g., % w/w) of an ingredient is multiplied by its amount (e.g., quantity in grams) to obtain the product of each ingredient. The products of all ingre-dients and their quantities in the original formula are added together sepa-rately. Dividing the sum of products by the sum of quantities in the original formula gives a quotient, which represents the strength (e.g., % w/w) of the final mixture.\n\nFor example, to calculate the strength of the final mixture when 12 g of a 10% w/v sucrose solution is mixed with 24 g of a 40% w/v sucrose solution, one would write the alligation medial method as indicated in Table 5.1. Working through this table, the final solution would be of 30.0% w/w strength.\n\nFor more than two ingredients\n\nThis method is also applicable for more than two ingredients. For example, to calculate the strength of the final mixture when 12 g of a 10% w/v sucrose solution is mixed with 24 g of a 40% w/v and 36 g of a 5% w/v sucrose solu-tion, one would write the alligation medial method as indicated in Table 5.2. Working through this table, the final solution would be of 17.5% w/w strength.\n\nTable 5.1 Alligation medial method for two ingredients",
null,
"Table 5.2 Alligation medial method for more than two ingredients",
null,
"Alligation alternate\n\nFor two ingredients\n\nThis method can be used to calculate the amount of a diluent, solute, or different-concentration product that would need to be added to a given concentration product to make a new concentration preparation. The num-ber of parts required for the lower- and higher-concentration preparations to make the target-concentration preparation is obtained by constructing a matrix and doing the calculation as shown in Table 5.3.\n\nTable 5.3 Alligation alternate method for two ingredients",
null,
"Thus, subtracting the target concentration from the lower concentration gives the target amount of the higher-concentration preparation, and subtracting the higher concentration from the target concentration gives the target amount of the lower-concentration preparation.\n\nThus, the total amount of the target-concentration preparation that would be prepared can be obtained by adding together the target amounts of higher- and lower-concentration preparations needed. If the required amount of the target-concentration preparation is dif-ferent than the amount obtained by the formula, the principles of propor-tion, discussed earlier, can be used to calculate the quantities needed for the required total amount of the target-concentration preparation.\n\nFor example, to prepare 200 mL of a 12% w/v sucrose solution using a 40% w/v and another 5% w/v sucrose solution, one would write the alligation matrix as shown in Table 5.4.\n\nThus, combining 7 mL of 40% w/v solution with 28 mL of 5% w/v solution would give 7 + 28 = 35 mL of 12% w/v solution. To make 200 mL of 12% w/v solution, one would use the principles of proportion as follows:\n\nFor the quantity of 40% w/v solution,",
null,
"For the quantity of 5% w/v solution,",
null,
"Alternatively, a conversion factor could be derived for the calculation:",
null,
"The required quantities of low- and high-concentration solutions can then simply be obtained by multiplying their quantities obtained by the alliga-tion formula by this factor. Thus, the quantity of 40% w/v solution required = 7 × 5.714 = 39.998 = 40 mL. Therefore, the quantity of the 5% w/v solution required = 200 − 40 = 160 mL or 28 × 5.714 = 159.992 = 160 mL.",
null,
"Table 5.5 An example of alligation alternate method for more than two ingredients",
null,
"For more than two ingredients\n\nThe alligation alternate method can be used for more than two ingredients by pairing off the values of one higher (than the desired) strength ingredient with two lower (than the desired) strength ingredients, or vice versa. This is illustrated by the following example:\n\nTo prepare a 17.5% w/w solution using a 10% w/v, a 40% w/v, and a 5% w/v sucrose solution, one would write the alligation alternate method, as shown in Table 5.5.\n\nThus, combining 20 mL of 40% w/v solution with 22.5 mL of 10% w/v solution and 22.5 mL of a 5% w/w solution would give 20 + 22.5 + 22.5 = 65 mL of 17.5% w/v solution. The laws of proportion can be used, as described earlier, to calculate specific quantities of starting solutions that would be needed to prepare a desired quantity of the final solution.\n\nThe alligation alternate method for more than two ingredients can use any pairing of higher (than the desired) strength ingredient(s) with lower (than the desired) strength ingredient(s). The pairings can be any number depending on the number of ingredients.\n\nThe alligation methods are applicable to all forms of preparations, including powders. In addition, the alligation method can also be used for calculating the required quantities for dilution of a preparation with the solvent or diluent alone by making the concentration of the lower-concentration preparation zero.\n\n### Tonicity, osmolarity, and preparation of isotonic solutions\n\nOf the two compartments of solution separated by a semipermeable membrane, the solvent tends to flow from the solution with the lower solute concentration to the solution with the higher solute concentration. If uninterrupted flow of solvent is allowed, it would result in the equaliza-tion of concentration across the membrane. This phenomenon is called osmosis. The pressure of solvent involved in this phenomenon is termed osmotic pressure. A solution containing a nonpermeable solute creates a pressure for the inward flow of solvent across the semipermeable mem-brane. Thus, osmotic pressure can also be defined as the pressure that must be applied to a solution to prevent the inward flow of solvent across a semipermeable membrane.\n\nTonicity is the osmotic pressure of two solutions separated by a semiper-meable membrane. Tonicities of solutions are often represented with refer-ence to that of normal body fluids. Thus, solutions that exert lower osmotic pressure than the body fluids are termed hypotonic, while solutions that exert higher osmotic pressure than the body fluids are termed hypertonic. Hypotonic solutions have lower and hypertonic solutions have higher impermeable solute concentration than the body fluids. Two solutions that have the same osmotic pressure are termed isosmotic, while a solution that has the same osmotic pressure as a reference body fluid is termed isotonic.\n\nTo define the osmotic amount and concentration of a solute without referring to another solution, the concepts of osmole, osmolarity, and osmolality are introduced. An osmole is the amount of a substance that represents the number of moles of particles that it forms in a solution. For a nondissociating substance, that is, a nonelectrolyte such as dextrose, 1 osmole = 1 mole. Thus, 1 osmole of dextrose = 186 g (molecular weight) of dextrose.\n\nSimilar to the concept of molarity, osmolarity (abbreviation: Osmol) is defined as the osmoles of solute per liter of solution. Therefore, 1 Osmol of glucose solution represents 186 g (molecular weight) of glucose dissolved in 1 L of solution.Similar to the concept of molality, osmolality is defined as the osmoles of solute per kg of solvent. These quantities can be used with prefixes in the metric system, such as milli and micro. Thus, a commonly used term is milliosmole (abbreviation: mOsmol), which represents 1/1,000th of an Osmol. Moreover, while osmole represents the quantity of solute in grams, Osmol represents the concentration of solute in a solution.\n\nFor a dissociating solute, such as an electrolyte, 1 mole ≠ 1 Osmol and 1 M solution ≠ 1 Osmol solution. The osmoles and osmolarity of such a solute are calculated by multiplying with the number of particles formed on dissociation and the fractional degree of dissociation of a substance in solution. Thus, assuming complete dissociation, NaCl, CaCl2, and FeCl3 form 2, 3, and 4 particles in solution. Thus, 1 mM solution of NaCl, CaCl2, or FeCl3 represents their 2, 3, or 4 mOsmol solution, respectively. Assuming, 80% degree of dissociation for dilute solutions, 2 M of NaCl, CaCl2, and FeCl3 solutions represent:\n\n2 × (1 + 80/100) = 3.6 Osmol of NaCl solution\n\n2 × (1 + 80/100 + 80/100 ) = 5.2 Osmol of CaCl2 solution\n\n2 × (1 + 80/100 + 80/100 + 80/100 ) = 6.8 Osmol of FeCl3 solution\n\nThe normal serum osmolality is in the range of 275–300 mOsmol/kg. osmolality of solutions can be measured in the laboratory by using an osmometer.\n\nTonicity is an important concept in the administration of ophthalmic and parenteral solutions. Hypertonic solutions tend to draw fluids out of body tissues, leading to irritation and dehydration. Hypotonic solu-tions, on the other hand, can provide excess fluid to the body tissues. However, since the volume of the administered solution is much lower than that of body fluids and fluid elimination is a regulated physiological phenomenon, hypotonic solutions are relatively inconsequential. Thus, administration of hypertonic solutions tends to be more tissue-damaging and painful than the administration of hypotonic solutions. Nonetheless, isotonic solutions are better tolerated by patients than either extremes of tonicity.\n\nPreparation of isotonic solutions requires the use of one of the col-ligative properties of solutions. Colligative properties are the solution properties that depend on the number of molecules of solvent in a given volume of solution but are independent of the properties of the solute. These properties include lowering of vapor pressure, elevation of boiling point, osmotic pressure, and depression of freezing point of a solution with increasing solute concentration. Of these, the depression of freezing point is conveniently used to calculate the amount of solute required to prepare an isotonic solution.\n\nFor example, given that the freezing point of blood serum and ophthalmic lachrymal fluid is −0.52°C and that 1 M aqueous solution of a nonelectro-lyte depresses the freezing point of water by 1.86°C, we can calculate the amount of glucose (molecular weight: 180 g/mol) required to prepare an isotonic solution by solving for the amount of glucose that would produce a freezing point depression of 0.52°C. Thus, to make 1 L of isotonic glucose solution, the amount of glucose required (x) can be calculated as:",
null,
"This corresponds to 5% w/v glucose solution. The commonly available dextrose solution for intravenous (IV) administration has this concentra-tion. Similar concentration for an electrolyte, such as sodium chloride, should take into consideration the dissociation constant of the solute and the number of species produced in the solution. Thus, assuming that NaCl in weak solutions is about 80% dissociated, the total number of solutes in solution would be 1.8 times the number of molecules added to the solution. This (1.8) dissociation factor (abbreviation: i) is used in the calculation of isotonic concentrations of electrolytes. Thus, to make a 1 L isotonic NaCl (molecular weight: 58 g/m) solution, the amount of NaCl required (x) can be calculated as:",
null,
"This corresponds to 0.9% w/v NaCl solution, which is commonly available as an isotonic solution for experiments involving living cell and tissues. From these calculations, note that 50 g/L of glucose solution is isotonic to 9 g/L of NaCl solution. Therefore, in quantities of solutes, 50 g of glucose is tonic equivalent to 9 g of NaCl. The tonic equivalence of two substances represents their amounts that would produce the same osmotic pressure. Thus, the quantity of any substance divided by its dissociation factor, i, represents its tonic equivalent quantity to any other substance. This principle is used in the preparation of isotonic solutions by the addi-tion of NaCl to hypotonic drug solutions to increase the tonicity to the physiological equivalent of 0.9% w/v NaCl. Using the above conversion of tonic equivalents, NaCl equivalents (E values) of various substances are known in the literature. The number of grams of all ingredients in a pre-scription is multiplied by their E values and added together to determine the osmotic equivalent of NaCl amount represented by the substances. In addition, the amount of NaCl that would be required to make a 0.9% w/v solution of the same volume as the prescription is determined. Subtracting the former from the latter gives the amount of NaCl needed to make the solution isotonic. Any substance other than NaCl, such as dextrose, can also be used to increase the tonicity of a solution by dividing the amount of NaCl needed by the NaCl equivalent of the other substance.\n\nFor example, to compound 10 mL of an ophthalmic preparation of 3% w/v pilocarpine nitrate, we first determine the amount of drug in 10 mL of solution.\n\nDrug amount = 3/100 ×10 = 0 .3 g\n\nThe NaCl equivalent (E value) of pilocarpine nitrate (molecular weight 271, dissociates into 2 ions, and i value = 1.8) can be read from the literature or calculated as:\n\nE value = (58. 5 / 1.8) / (271 / 1.8) = 0.216\n\nNow, we multiply the E value with the drug amount in solution to get NaCl equivalents represented by the drug amount in the solution:\n\nNaCl equivalent in prescription = 0.3× 0.216 = 0.0648 g\n\nThis is the amount of particles in solution equivalent to NaCl, which must be subtracted from the amount of NaCl that would be needed to make an isotonic solution of the same volume as the prescription (i.e., 10 mL). This is calculated as:\n\nTotal amount of NaCl needed for isotonicity = (0.9/100) × 10 = 0 .09 g\n\nHence, the amount of NaCl that must be added to the prescription to make an isotonic solution = 0.09−0.0648 = 0.0252 g.\n\nIf a prescription contains multiple components, NaCl equivalent for each component is calculated separately and added together to make the total NaCl equivalents in the prescription. This total amount is then subtracted from the total NaCl that would be needed for isotonicity of the volume of pre-scription to obtain the amount of NaCl that must be added to the prescription."
]
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https://postgis.net/docs/manual-2.3/ST_Difference.html | [
"## Name\n\nST_Difference — Returns a geometry that represents that part of geometry A that does not intersect with geometry B.\n\n## Synopsis\n\n`geometry ST_Difference(`geometry geomA, geometry geomB`)`;\n\n## Description\n\nReturns a geometry that represents that part of geometry A that does not intersect with geometry B. One can think of this as GeometryA - ST_Intersection(A,B). If A is completely contained in B then an empty geometry collection is returned.",
null,
"Note - order matters. B - A will always return a portion of B\n\nPerformed by the GEOS module",
null,
"Do not call with a GeometryCollection as an argument",
null,
"This method implements the OpenGIS Simple Features Implementation Specification for SQL 1.1. s2.1.1.3",
null,
"This method implements the SQL/MM specification. SQL-MM 3: 5.1.20",
null,
"This function supports 3d and will not drop the z-index. However it seems to only consider x y when doing the difference and tacks back on the Z-Index\n\n## Examples",
null,
"The original linestrings shown together.",
null,
"The difference of the two linestrings\n```--Safe for 2d. This is same geometries as what is shown for st_symdifference\nSELECT ST_AsText(\nST_Difference(\nST_GeomFromText('LINESTRING(50 100, 50 200)'),\nST_GeomFromText('LINESTRING(50 50, 50 150)')\n)\n);\n\nst_astext\n---------\nLINESTRING(50 150,50 200)\n```\n```\n--When used in 3d doesn't quite do the right thing\nSELECT ST_AsEWKT(ST_Difference(ST_GeomFromEWKT('MULTIPOINT(-118.58 38.38 5,-118.60 38.329 6,-118.614 38.281 7)'), ST_GeomFromEWKT('POINT(-118.614 38.281 5)')));\nst_asewkt\n---------\nMULTIPOINT(-118.6 38.329 6,-118.58 38.38 5)\n```"
]
| [
null,
"https://postgis.net/docs/manual-2.3/images/note.png",
null,
"https://postgis.net/docs/manual-2.3/images/note.png",
null,
"https://postgis.net/docs/manual-2.3/images/check.png",
null,
"https://postgis.net/docs/manual-2.3/images/check.png",
null,
"https://postgis.net/docs/manual-2.3/images/check.png",
null,
"https://postgis.net/docs/manual-2.3/images/st_symdifference01.png",
null,
"https://postgis.net/docs/manual-2.3/images/st_difference01.png",
null
]
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http://at.metamath.org/ileuni/anc2r.html | [
"",
null,
"Intuitionistic Logic Explorer < Previous Next > Nearby theorems Mirrors > Home > ILE Home > Th. List > anc2r GIF version\n\nTheorem anc2r 311\n Description: Conjoin antecedent to right of consequent in nested implication. (Contributed by NM, 15-Aug-1994.)\nAssertion\nRef Expression\nanc2r ((𝜑 → (𝜓𝜒)) → (𝜑 → (𝜓 → (𝜒𝜑))))\n\nProof of Theorem anc2r\nStepHypRef Expression\n1 pm3.21 251 . . 3 (𝜑 → (𝜒 → (𝜒𝜑)))\n21imim2d 48 . 2 (𝜑 → ((𝜓𝜒) → (𝜓 → (𝜒𝜑))))\n32a2i 11 1 ((𝜑 → (𝜓𝜒)) → (𝜑 → (𝜓 → (𝜒𝜑))))\n Colors of variables: wff set class Syntax hints: → wi 4 ∧ wa 97 This theorem was proved from axioms: ax-1 5 ax-2 6 ax-mp 7 ax-ia3 101 This theorem is referenced by: ssorduni 4213\n Copyright terms: Public domain W3C validator"
]
| [
null,
"http://at.metamath.org/ileuni/_icon-il.gif",
null
]
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https://www.npmjs.com/search?q=keywords%3ATest&page=1&perPage=20 | [
"# npm\n\n## 13657 packages found\n\nReturns true if a number or string value is a finite number. Useful for regex matches, parsing, user input, etc.\n\nM\nQ\nP\nM\nQ\nP\n\nReturns `true` if the given string looks like a glob pattern or an extglob pattern. This makes it easy to create code that only uses external modules like node-glob when necessary, resulting in much faster code execution and initialization time, and a bet\n\nM\nQ\nP\nM\nQ\nP\nM\nQ\nP\nM\nQ\nP\nM\nQ\nP\nM\nQ\nP\nM\nQ\nP\nM\nQ\nP\nM\nQ\nP\nM\nQ\nP\nM\nQ\nP\nM\nQ\nP\nM\nQ\nP\nM\nQ\nP\nM\nQ\nP\nM\nQ\nP\nM\nQ\nP\nM\nQ\nP"
]
| [
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| {"ft_lang_label":"__label__en","ft_lang_prob":0.68201184,"math_prob":0.55146956,"size":1405,"snap":"2023-14-2023-23","text_gpt3_token_len":325,"char_repetition_ratio":0.09493219,"word_repetition_ratio":0.0,"special_character_ratio":0.22135231,"punctuation_ratio":0.07172996,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9899132,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-03-27T00:31:06Z\",\"WARC-Record-ID\":\"<urn:uuid:2df4b9ce-9a70-487e-8029-a2dde9db3e1f>\",\"Content-Length\":\"139861\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:ccb941e4-0589-4dc2-b400-ad1ad750ff91>\",\"WARC-Concurrent-To\":\"<urn:uuid:fa481f47-6887-4d11-acd5-fc995dfe42cf>\",\"WARC-IP-Address\":\"104.16.92.83\",\"WARC-Target-URI\":\"https://www.npmjs.com/search?q=keywords%3ATest&page=1&perPage=20\",\"WARC-Payload-Digest\":\"sha1:6ICI6RJNAPP5ASW2ATXDI2AR5KYHDSLX\",\"WARC-Block-Digest\":\"sha1:2X6OIYQXRLVJ5LPJS2KEDGLDPBSOCF6Z\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-14/CC-MAIN-2023-14_segments_1679296946584.94_warc_CC-MAIN-20230326235016-20230327025016-00723.warc.gz\"}"} |
https://id.scribd.com/document/118865830/hypothesis-testing | [
"Anda di halaman 1dari 27\n\n# Econometrics\n\n## Hypothesis Testing (Inference) in Linear Regression Models Yuyi LI\n\nUniversity of Manchester\n\n2012\n\n## Inference in Linear Regression Models\n\n2012\n\n1 / 27\n\nOutline\nFinite-sample inference in linear models: the normality assumption Sampling distributions of the OLS estimators Hypothesis tests: a single restriction The t test Hypothesis tests: multiple linear restrictions The F test Reading: Wooldridge Chapter 4, Appendix C6 Mathematics and Statistics: standard normal, t and F distributions, statistical hypotheses, test statistics, rejection rules, signicance level, critical values, p-value, degrees of freedom. . .\nYuyi LI (University of Manchester) Inference in Linear Regression Models 2012 2 / 27\n\n## Assumption MLR.6: Normality\n\nRecall the MLR model y = 0 + 1 x1 + 2 x2 + . . . + k xk + u Can test hypotheses about j based on the sample distribution of its estimator j . Need an extra assumption on u in nite samples: Assumption MLR.6: (Normality) Error term u is independent of the regressors x1 , x2 , . . . , xk and is normally distributed: u N (0, 2 ). MLR.6 can be justied by the Central Limit Theorem as sample size tends to innity. MLR.1-6 are the Classical Linear Model (CLM) assumptions. Inference in Linear Regression Models Yuyi LI (University of Manchester) 2012 3 / 27\n\n## Normality of OLS Estimators\n\nNormality of u translates to OLS estimators. Theorem (Normality of OLS Estimators) Under CLM assumptions (MLR.1-MLR.6),\n\nj N (j , Var (j )),\n\nj = 0, 1, 2, . . . , k\n\nwhere Var (j ) = 2 (X X)1 and (X X)1 denotes the (j + 1)-th jj jj leading diagonal element of (X X)1 .\nThen, an infeasible distribution follows\n\nj j Var (j )\nYuyi LI (University of Manchester)\n\nN (0, 1)\n\n2012\n\n4 / 27\n\n## The t Distribution of OLS Estimators\n\nEstimate Var (j ) : Var (j ) = 2 (X X)1 = [se (j )]2 jj where 2 = SSR /(n 1 k )\nTheorem (t Distribution of OLS Estimators) Under the CLM assumptions (MLR.1-MLR.6),\n\nj j tn1k , for j = 0, 1, . . . , k se (j )\nwhere tdf is a t distribution with df degrees of freedom tdf approximates N (0, 1) when df is large (> 120) This can be used to test hypothesis on j\nYuyi LI (University of Manchester) Inference in Linear Regression Models 2012 5 / 27\n\n## Hypothesis Testing (Inference)\n\nHypotheses\nConsider MLR model y = 0 + 1 x1 + 2 x2 + . . . + k xk + u The null and alternative hypotheses (H0 and H1 ) Examples of hypotheses:\n(i) H0 : 1 = 0 (a restriction on one parameter) x1 has no impact on y (ii) H0 : 1 2 = 0 (a linear restriction on parameters) x1 and x2 have equivalent effects on y (iii) H0 : 1 = 2 = k = 0 (multiple linear restrictions) x1 , x2 and xk have no joint explanatory power on y\n\n## What test to use?\n\nt : (i) a restriction on one parameter or (ii) a linear restriction on parameters F : (iii) multiple linear restrictions (and (i) & (ii))\nYuyi LI (University of Manchester) Inference in Linear Regression Models 2012 6 / 27\n\n## Classical Testing Procedure\n\nThe Classical procedure involves steps to\nSelect signicance level (): probability of rejecting H0 when it is true. Common choices: 1%, 5%, 10% Choose H1 : may affect the rejection rule Calculate the test statistic Reject H0 if the calculated test statistic is in the rejection region; Otherwise, do not reject H0\n\nNote, selection of is somewhat arbitrary Note, rejection decision is made by comparing test statistic and corresponding critical value Note, critical values depend on , H1 , and statistic distribution under H0\nYuyi LI (University of Manchester) Inference in Linear Regression Models 2012 7 / 27\n\n## Hypothesis Testing using the p-Value\n\nChoice of (e.g. 5%) is somewhat arbitrary\n\nAn alternative: Given the test statistic, what is the smallest signicance level at which H0 would be rejected? This is called the p-value of the test. The p-value (p) is the probability of observing a test statistic (T ) as extreme as we did if H0 is true:\nE.g. t test for H0 : j = 0 against H1 : j 0 = p = Pr(|T | > |t |), where t is the observed t test statistic\n\n## Small p-values provides evidence against H0\n\nE.g. H0 : 1 = 0, H1 : 1 0, t t930 , observed value of t = 2.137642 and p = 0.0328. Then H0 is rejected at 5% level, but not at 1% level. Precisely, H0 is rejected at all signicance levels 3.28%\nYuyi LI (University of Manchester) Inference in Linear Regression Models 2012 8 / 27\n\nThe t Test\n\nA Single Parameter\n\n## t Test: A Restriction on One Parameter\n\nThe null hypothesis (j = 0, 1, 2, . . . , k ): H0 : j = c , The t statistic t= where c is a constant\n\nj j j c = tn1k under H0 se (j ) se (j )\n\n(1)\n\nExample: if j = 0.8 and se (j ) = 0.2, the test statistic for H0 : j = 1 is t = (0.8 1)/0.2 = 1 Next, we introduce a special case when c = 0\nYuyi LI (University of Manchester) Inference in Linear Regression Models 2012 9 / 27\n\nThe t Test\n\nA Single Parameter\n\n## Signicant Regressor & Rejection Rule\n\nIf c = 0, then H0 : j = 0. H0 : once x1 , x2 , . . . , xj 1 , xj +1 , . . . , xk have been controlled for, xj has no effect on the expected value of y (i.e. xj is statistically insignicant). The t statistic is\n\nj j j 0 j t= = = tn1k under H0 se (j ) se (j ) se (j )\nThese t statistics are routinely reported in EViews Alternatives and Rejection Rules:\nAlternative Rejection Rule\nYuyi LI (University of Manchester)\n\n## H1 : j < 0 t < tcv\n\nH1 : j 0 |t | > tcv\n2012 10 / 27\n\n## Inference in Linear Regression Models\n\nThe t Test\n\nA Single Parameter\n\nExample\nTest H0 : 1 = 1 against H1 : 1 < 1 with = 0.01 y = 0 + 1 x1 + 2 x2 + u y = 0.2 + 0.8 x1 1.2 x2\n(0.03) (0.2) (0.05)\n\n## 1 1 tn12 = t350 under H0 se (1 ) 0.8 1 = = 1 0 .2\n\nCritical value tcv = 2.326 (from t350 , 1-tailed H1 , = 0.01) Reject H0 if t < tcv : Fail to reject H0 , as t > tcv here\nYuyi LI (University of Manchester) Inference in Linear Regression Models 2012 11 / 27\n\nThe t Test\n\n## A Linear Restriction on Parameters\n\nH0 involves several parameters, but only one restriction: e.g. H0 : 1 = 2 or H0 : 1 + 2 = 3 Example: Cobb-Douglas production function Yi = ALi 1 Ki 2 Ui where Y =production, A =technology, L =labour, K =capital and U=unobservables. Taking logs: log(Yi ) = log(A ) + 1 log(Li ) + 2 log(Ki ) + ui Constant returns to scale: H0 : 1 + 2 = 1. Note, t tests can be used to test H0 , but statistics are not easily computed due to the complexity of corresponding standard errors\nYuyi LI (University of Manchester) Inference in Linear Regression Models 2012 12 / 27\n\nThe t Test\n\n## A Single Linear Combination of Parameters\n\nEquivalence of Educations\nConsider the population model log(wage ) = 0 + 1 x1 + 2 x2 + 3 x3 + u\nx1 =the number of years in junior college x2 =the number of years in university x3 =the number of months in workforce\n\n(2)\n\nH0 : 1 = 2 . What does it mean? H1 : 1 < 2 . What does this imply? The hypotheses can be rewritten as H0 : 1 2 = 0, H1 : 1 2 < 0.\nThe following slides are based on Wooldridge section 4.4, which mainly details two methods of obtaining the standard error of a linear combination of paramter estimators. Wooldrige section 4.4 (page 140- )\nYuyi LI (University of Manchester) Inference in Linear Regression Models 2012 13 / 27\n\nThe t Test\n\n## A t Statistic and A Problem\n\nA t statistic can be constructed to test H0 : t=\n\n1 2 se (1 2 )\n\n(3)\n\n## where estimates 1 and 2 are easily obtained. Problem: How to get se (1 2 )?\n\nEstimate the variance of (1 2 ): Var (1 2 ) = Var (1 ) + Var (2 ) 2Cov (1 , 2 ) se (1 2 ) =\n=\n\n## Var (1 ) + Var (2 ) 2Cov (1 , 2 )\n\n[se (1 )]2 + [se (2 )]2 2s12 , where a hat indicates an unbiased estimator and s12 = Cov (1 , 2 ).\nYuyi LI (University of Manchester) Inference in Linear Regression Models 2012 14 / 27\n\nThe t Test\n\n## The Problem and Solutions\n\nNote, se (1 ) and se (2 ) are computed by any software, but s12 is not = se (1 2 ) on the previous page is unknown = Statistic (3) is not computable = Testing H0 is not possible? Solutions: Two methods: Method 1: Compute s12 by estimating the covariance matrix of Method 2: Rewrite model (2) by introducing a new parameter These two methods are explained sequentially. . .\nYuyi LI (University of Manchester) Inference in Linear Regression Models 2012 15 / 27\n\nThe t Test\n\n## Method One (1/2)\n\nWrite model (2) in matrix form (Revise notes L3) y = X + u\nwhere y = [y1 , y2 , , yn ] , y log(wage ), X = [x1 , x2 , , xn ] , xi = [1, xi1 , xi2 , xi3 ], i = 1, 2, . . . , n, = [0 , 1 , 2 , 3 ] and u = [u1 , u2 , , un ]\n\nOLS estimator is = (X X)1 X y. Covariance matrix is Var (|X) = 2 (X X)1 , and an unbiased estimator is SSR (X X)1 Var (|X) = 2 (X X)1 = n1k What is the structure of Var (|X)?\nYuyi LI (University of Manchester) Inference in Linear Regression Models 2012 16 / 27\n\nThe t Test\n\n## Method One (2/2)\n\nVar (|X) = [se (0 )]2 s01 s02 2 s10 [se (1 )] s12 s20 s21 [se (2 )]2\ns30 s31 s32 s03 s13 s23 [se (3 )]2\n\nIf we compute Var (|X), it is straightforward to obtain se (1 ), se (2 ) and s12 = se (1 2 ) can be obtained = Statistic (3) can be computed = Testing H0 is possible! is 4 1 here. How to generalise to (k + 1) 1?\nYuyi LI (University of Manchester) Inference in Linear Regression Models 2012 17 / 27\n\nThe t Test\n\n## Method Two (1/2)\n\nDene = 1 2 = (i) H0 : = 0 is the same as H0 : 1 2 = 0 = (ii) 1 = + 2 = (iii) statistic (3) becomes t = /se () Note: Computing (iii) to test (i) is now the solution! How to compute (iii)? Rewrite model (2): y = 0 + 1 x1 + 2 x2 + 3 x3 + u = 0 + ( + 2 )x1 + 2 x2 + 3 x3 + u = 0 + x1 + 2 x1 + 2 x2 + 3 x3 + u = 0 + x1 + 2 (x1 + x2 ) + 3 x3 + u Note, models (2) and (4) are equivalent\nYuyi LI (University of Manchester) Inference in Linear Regression Models 2012 18 / 27\n\n(4)\n\nThe t Test\n\n## Method Two (2/2)\n\nModel (4) can be estimated by regressing y on an intercept, x1 , (x1 + x2 ) and x3 = This yields and se () directly = Statistic t = /se () in (iii) can be computed = Testing H0 : = 0 in (i) is possible! Either method enables a t test on hypothesis involving a linear restriction on several parameters Note, t test is used to test a single linear restriction with one or more parameters Note, if there are multiple linear restrictions, F test can be used\nYuyi LI (University of Manchester) Inference in Linear Regression Models 2012 19 / 27\n\nThe F Test\n\n## Multiple Linear Restrictions\n\nConsider a model y = 0 + 1 x1 + 2 x2 + 3 x3 + 4 x4 + u Example: q linear restrictions hypotheses\nH0 : 1 = 3 = 0 H0 : 1 = 0, 2 = 3 H0 : 1 = 2 = 3 = 4 = 0 exclusion restrictions two restrictions reg. overall signicance\n\n(5)\n\nWhat is q under each of the above H0 ? Alternative hypotheses: H1 : H0 does not hold. Note, H0 is violated as long as one restriction fails To construct the F test, we need to identify the restricted model and unrestricted model. . .\nYuyi LI (University of Manchester) Inference in Linear Regression Models 2012 20 / 27\n\nThe F Test\n\n## Restricted and Unrestricted Models\n\nRestricted model: (5) with q restrictions under H0 Example, restricted models are\nH0 : 1 = 3 = 0 (5) = y = 0 + 2 x2 + 4 x4 + u H0 : 1 = 0, 2 = 3 (5) = y = 0 + 2 (x2 + x3 ) + 4 x4 + u H0 : 1 = 2 = 3 = 4 = 0 (5) = y = 0 + u\n\nUnrestricted model: (5) without any restrictions Estimating restricted model yields SSRr (Sum of Squared Residuals in the restricted model) Estimating unrestricted model yields SSRur (Sum of Squared Residuals in the unrestricted model) Always, SSRur SSRr .\nYuyi LI (University of Manchester) Inference in Linear Regression Models 2012 21 / 27\n\nThe F Test\n\n## The F Statistic and Rejection Rule\n\nThe F statistic F=\n\n## (SSRr SSRur )/q Fq,n1k under H0 SSRur /(n 1 k )\n\nwhere q = number of restrictions under H0 (numerator df ), n 1 k = unrestricted model df (denominator df ), Fq,n1k = an F distribution with q and (n 1 k ) df df = degrees of freedom\n\nF 0. Why? Reject H0 if F > Fcv (critical value from Fq,n1k ). Fail to reject H0 otherwise.\nYuyi LI (University of Manchester) Inference in Linear Regression Models 2012 22 / 27\n\nThe F Test\n\n## An Example for F Test 1/2\n\nConsider the wage equation model lw = 0 + 1 ed + 2 ex + 3 te + 4 hr + u (6)\n\nwhere lw = log (wage ) (monthly), ed = years in education, ex = years of work experience, te = years in the current employment, hr = average weekly working hours, and u denotes the random error term.\n\nte and hr are irrelevant to wage (dropped in (6))? H0 : 3 = 4 = 0 against H1 : 3 0 or/and 4 0. OLS regression of (unrestricted) model (6): SSRur Restricted model (under H0 ; 2 restrictions, q = 2) lw = 0 + 1 ed + 2 ex + u: SSRr .\nYuyi LI (University of Manchester) Inference in Linear Regression Models 2012 23 / 27\n\nThe F Test\n\n## An Example for F Test 2/2\n\nIn our case, n = 935, k = 4, q = 2, SSRur = 139.28, SSRr = 143.98. The F statistic is\nF=\n\n## (SSRr SSRur )/q (143.98 139.28)/2 = 15.70 SSRur /(n 1 k ) 139.28/(935 1 4)\n\nCritical value Fcv = 4.61 (from F2,930 , = 0.01) Since F 15.70 > Fcv , H0 is rejected at 1% signicance level Statistically, years in the current employment (te) and average weekly working hours (hr) have some joint effect on wage.\nYuyi LI (University of Manchester) Inference in Linear Regression Models 2012 24 / 27\n\nThe F Test\n\n## The R-Squared Form of the F Statistic\n\nThe F statistic can also be computed based on the R 2 values in the unrestricted and restricted models\n2 (Rur Rr2 )/q F= 2 (1 Rur )/(n 1 k ) 2 where Rur is the R 2 from the unrestricted model and Rr2 is from the restricted model. Note, this formula works only if unrestricted and restricted models have same dependent variables For example, R 2 form F statistic is invalid to test general linear hypothesis H0 : 0 = 0, 1 2 = 1\nYuyi LI (University of Manchester) Inference in Linear Regression Models 2012 25 / 27\n\nThe F Test\n\n## Overall Signicance of the Model\n\nF test for the joint signicance of all the regressors (excluding the constant) Consider (unrestricted) model: y = 0 + 1 x1 + 2 x2 + . . . + k xk + u. H0 : 1 = 2 = . . . = k = 0 against H1 : any j 0, j = 1, 2, . . . , k Restricted model: y = 0 + e; Note, Rr2 = 0. why? R 2 form of F statistic 2 2 (Rur Rr2 )/q Rur /q F= = 2 2 (1 Rur )/(n 1 k ) (1 Rur )/(n 1 k ) This statistic is routinely reported in EViews and most statistical softwares. Can you nd it?\nYuyi LI (University of Manchester) Inference in Linear Regression Models 2012 26 / 27\n\nAppendix\n\n## Deriving R-Squared Form of the F Statistic\n\nR-Squared Form of the F Statistic. By denition, R2 = SSE SSR =1 = SSR = SST (1 R 2 ) SST SST\n\n2 This means, SSRur = SST (1 Rur ), SSRr = SST (1 Rr2 ) in the unrestricted and restricted models. Then F statistic becomes 2 [SST (1 Rr2 ) SST (1 Rur )]/q (SSRr SSRur )/q = 2 SSRur /(n 1 k ) [SST (1 Rur )]/(n 1 k ) 2 2 [(1 Rr2 ) (1 Rur )]/q (Rur Rr2 )/q = = 2 2 (1 Rur )/(n 1 k ) (1 Rur )/(n 1 k )\n\nF =\n\n2012\n\n27 / 27"
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https://dev.goldbook.iupac.org/terms/view/P04540/plain | [
"## phenomenological equation\n\nhttps://doi.org/10.1351/goldbook.P04540\nIn the following only media which are isotropic with respect to mass transport (i.e. the transport coefficients are independent of direction) are being considered. In the linear range (not too far from equilibrium), for uniform temperature and neglecting external fields such as the earth's gravitational field, the flux density of species B is related to the gradients of the electrochemical potentials of all species by the phenomenological equation: $N_{\\text{B}}- c_{\\text{B}}\\ v_{\\text{A}}=- \\sum _{\\begin{array}{c} i\\\\ i\\neq A \\end{array}}L_{\\text{B}i}^{\\text{A}}\\ \\nabla \\mu _{i}$ with $i=\\text{B},\\text{C},...$ where ∇µi is the gradient of the electrochemical potential of species i. The proportionality factors LBiA are called phenomenological coefficients. Their values depend on the frame of reference. The latter is taken here to move with the velocity v A of species A, and hence: $L_{\\text{A}i}^{\\text{A}}=0$\nSource:\nPAC, 1981, 53, 1827. 'Nomenclature for transport phenomena in electrolytic systems' on page 1830 (https://doi.org/10.1351/pac198153101827)"
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https://www.nileshblog.tech/tag/leetcode-array/page/2/ | [
"# Learn and Grow\n\n### Leet Code: Longest Substring Without Repeating Characters Java | JavaScript | CPP | Python Solution\n\nFind the length of the longest substring without repeated characters given a string s. Example 1: Input: s = “abcabcbb” Output: 3 Explanation: The answer is “abc”, with the length of 3. Example 2: Input: s = “bbbbb” Output: 1\n\n### Leet Code : Add Two List – Java | CPP | Javascript | Python\n\nYou are given two non-empty linked lists that each represent a non-negative integer. The digits are kept in reverse order, with each node containing only one digit. Add the two numbers together and return the total as a linked list.\n\n### Leet Code : Median of two sorted array -CPP || java || Python || Java Script\n\nProvide the median of the two sorted arrays given two sorted arrays of sizes m and n, respectively. The entire complexity of the run time should be O(log (m+n)). Eg 1: Numbers1 = [1,3], Numbers2 = Result: 2.00000 Explanation:\n\n### leet code : Two sum Problem solution – Java || CPP || JavaScript || Python\n\nIn this article, we will be discussing the two sum problem and how it can be solved using the programming languages C++, Java, and Python Problem Statement two sum java This is another article in the series leetcode problem solutions\n\n### Binary Tree Level Order Traversal\n\nGiven the root of a binary tree, return the level order traversal of its nodes’ values. (i.e., from left to right, level by level). Example 1: Input: root = [3,9,20,null,null,15,7] Output: [,[9,20],[15,7]] Example 2: Input: root = Output: [] Example 3:"
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https://studysoup.com/tsg/10028/physics-principles-with-applications-6-edition-chapter-4-problem-38p | [
"×\n×\n\n# (I) Suppose you are standing on a train accelerating at",
null,
"ISBN: 9780130606204 3\n\n## Solution for problem 38P Chapter 4\n\nPhysics: Principles with Applications | 6th Edition\n\n• Textbook Solutions\n• 2901 Step-by-step solutions solved by professors and subject experts\n• Get 24/7 help from StudySoup virtual teaching assistants",
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"Physics: Principles with Applications | 6th Edition\n\n4 5 1 339 Reviews\n26\n3\nProblem 38P\n\nProblem\n\n(I) Suppose you are standing on a train accelerating at 0.20 g. What minimum coefficient of static friction must exist between your feet and the floor if you are not to slide?\n\nStep-by-Step Solution:\n\nSolution 38P: We have to find the minimum value of coefficient of static friction, to prevent the body from sliding as the train accelerates. Step 1 of 3 Concept: Newton’s second law: The net force F acting on an object of mass m produces an acceleration a in that object. Mathematically, F= ma. Frictional force also acts on the surface of the bodies in contact whenever their is relative motion between between the two bodies. Free body diagram of a body gives the diagrammatical representation of all the forces acting on the body in terms of magnitude and diagram, under the given situation. Step 2 of 3 Figure below shows the free body diagram of the person standing on an accelerating train. a Acceleration of the train = 0.20g F NNormal reaction on the person in upward direction mg Weight of the person acting vertically downward direction F frrictional force acting at the contact surfaces Coefficient of static friction\n\nStep 3 of 3\n\n##### ISBN: 9780130606204\n\nThis full solution covers the following key subjects: accelerating, coefficient, exist, feet, floor. This expansive textbook survival guide covers 35 chapters, and 3914 solutions. Physics: Principles with Applications was written by and is associated to the ISBN: 9780130606204. Since the solution to 38P from 4 chapter was answered, more than 826 students have viewed the full step-by-step answer. The full step-by-step solution to problem: 38P from chapter: 4 was answered by , our top Physics solution expert on 03/03/17, 03:53PM. The answer to “(I) Suppose you are standing on a train accelerating at 0.20 g. What minimum coefficient of static friction must exist between your feet and the floor if you are not to slide?” is broken down into a number of easy to follow steps, and 32 words. This textbook survival guide was created for the textbook: Physics: Principles with Applications, edition: 6.\n\nUnlock Textbook Solution"
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"https://studysoup.com/cdn/62cover_2418946",
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"https://studysoup.com/cdn/62cover_2418946",
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| {"ft_lang_label":"__label__en","ft_lang_prob":0.90327656,"math_prob":0.7891378,"size":1113,"snap":"2020-45-2020-50","text_gpt3_token_len":236,"char_repetition_ratio":0.12804328,"word_repetition_ratio":0.0,"special_character_ratio":0.20125785,"punctuation_ratio":0.07619048,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.98585206,"pos_list":[0,1,2,3,4],"im_url_duplicate_count":[null,null,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-12-01T08:10:57Z\",\"WARC-Record-ID\":\"<urn:uuid:82331b57-218b-4bda-a05e-05070c4e4cb6>\",\"Content-Length\":\"81256\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:c7371209-6925-4813-b576-2b833751cd2b>\",\"WARC-Concurrent-To\":\"<urn:uuid:f18391a2-e08f-4905-9917-8583d7c6e8d9>\",\"WARC-IP-Address\":\"54.189.254.180\",\"WARC-Target-URI\":\"https://studysoup.com/tsg/10028/physics-principles-with-applications-6-edition-chapter-4-problem-38p\",\"WARC-Payload-Digest\":\"sha1:PV236HETW7VSA5XJ4HILPXBBJEYCV5PB\",\"WARC-Block-Digest\":\"sha1:UPNPVNG2RBHQYL6ZEC5MNTDLTJXNDYAS\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-50/CC-MAIN-2020-50_segments_1606141672314.55_warc_CC-MAIN-20201201074047-20201201104047-00518.warc.gz\"}"} |
https://ask.sagemath.org/users/1119/bpeacock/?sort=recent | [
"2018-09-28 10:51:31 -0500 received badge ● Notable Question (source) 2017-08-09 12:02:20 -0500 received badge ● Famous Question (source) 2017-08-09 12:02:20 -0500 received badge ● Notable Question (source) 2017-08-09 12:02:20 -0500 received badge ● Popular Question (source) 2016-12-10 06:42:14 -0500 received badge ● Favorite Question (source) 2016-12-10 06:41:43 -0500 received badge ● Popular Question (source) 2015-03-10 10:37:57 -0500 received badge ● Student (source) 2013-07-10 21:58:42 -0500 received badge ● Taxonomist 2013-01-08 19:45:27 -0500 commented answer latex Function Prevent Simplification Thats a shame but it makes sense. I guess that would have to be added to the sage preparser. Thanks! 2013-01-08 19:41:30 -0500 marked best answer latex Function Prevent Simplification Since you can do x.mul(1/x, hold=True), you can also do latex(x.mul(1/x, hold=True)). I suppose you could also write the expression using LaTeX: sage: LatexExpr(r\"\\frac{%s}{%s}\" % (latex(x), latex(x))) \\frac{x}{x} sage: LatexExpr(r\"\\frac{%s}{%s}\" % (latex(factor(x^2+3*x + 2)), latex(x))) \\frac{{\\left(x + 1\\right)} {\\left(x + 2\\right)}}{x} But in general, latex(foo) first computes foo, which means that typically it will get simplified, and then it computes its LaTeX representation. So I don't think there is a simple way to do what you want. 2013-01-08 17:42:24 -0500 asked a question latex Function Prevent Simplification Hi, I am trying to use the latex function without the subject simplifying before it converting to LaTeX. This is similar to the hold=True for algebra functions. For example, I want latex(x/x) to return \\frac{x}{x} rather than 1. Thanks so much! 2012-09-03 15:26:09 -0500 received badge ● Scholar (source) 2012-09-03 15:26:09 -0500 marked best answer Difference Between var(), QQ() and PolynomialRing() var constructs symbolic ring (SR) variables: sage: var('x') sage: sin(x) sin(x) sage: x in SR True sage: x.parent() Symbolic Ring Polynomial rings are much better at working with polynomials, but nothing else: sage: R. = QQ[] sage: x in PolynomialRing(QQ,1,'x') True sage: x.parent() Univariate Polynomial Ring in x over Rational Field Using a polynomial variable in a non-polynomial manner automatically converts it to the symbolic ring: sage: type(x) sage: type(sin(x)) 2012-09-02 20:21:54 -0500 asked a question Difference Between var(), QQ() and PolynomialRing() I am rather new to Sage and am trying to understand the internals of Sage better. I encountered some confusion when reading through the reference manual as to the difference between the different ring constructs used in sage. The var() function is of course used to declare a variable for symbolic manipulation but when should one use QQ[] or PolynomialRing()? I ran into this issue with the convolution() function which requires variables within functions to be declared using QQ[] or Polynomial ring and will not work with var(). Why is this? Is QQ the default namespace? How do these namespaces relate to the symbolic ring used with var? Thank you for your help!"
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https://www.dummies.com/programming/r/how-to-look-at-the-structure-of-vectors-in-r/ | [
"",
null,
"How to Look at the Structure of Vectors in R - dummies\n\n# How to Look at the Structure of Vectors in R\n\nA vector is a one-dimensional set of values, all the same type. It’s the smallest unit you can work with in R. A single value is technically a vector as well — a vector with only one element. In mathematics vectors are almost always used with numerical values, but in R they also can include other types of data, like character strings.\n\nR gives you an easy way to look at the structure of any object. This method comes in handy whenever you doubt the form of the result of a function or a script you wrote. To take a peek inside R objects, use the str() function.\n\nThe str() function gives you the type and structure of the object.\n\nTake a look at the vector baskets.of.Granny:\n\n```> str(baskets.of.Granny)\nnum [1:6] 12 4 5 6 9 3```\n\nR tells you a few things here:\n\n• First, it tells you that this is a num (numeric) type of vector.\n\n• Next to the vector type, R gives you the dimensions of the vector. This example has only one dimension, and that dimension has indices ranging from 1 to 6.\n\n• Finally, R gives you the first few values of the vector. In this example, the vector has only six values, so you see all of them.\n\nIf you want to know only how long a vector is, you can simply use the length() function, as follows:\n\n```> length(baskets.of.Granny)\n 6```\n\nVectors in R can have other types as well. If you look at the vector authors, for example, you see a small difference:\n\n```> authors <- c(\"Andrie\", \"Joris\")\n> str(authors)\nchr [1:2] \"Andrie\" \"Joris\"```\n\nAgain, you get the dimensions, the range of the indices, and the values. But this time, R tells you the type of vector is chr, or character.\n\nThe following types are common types of vectors:\n\n• Numeric vectors, containing all kind of numbers.\n\n• Integer vectors, containing integer values. (An integer vector is a special kind of numeric vector.)\n\n• Logical vectors, containing logical values (TRUE and/or FALSE)\n\n• Character vectors, containing text\n\n• Datetime vectors, containing dates and times in different formats\n\n• Factors, a special type of vector to work with categories.\n\nR makes clear distinctions among these types of vectors, partly for reasons of logic. Multiplying two words, for example, doesn’t make sense."
]
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"https://www.facebook.com/tr",
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https://discourse.julialang.org/t/pyplot-error-plot-3d-line/5668 | [
"# PyPlot error, plot 3D line\n\nThe error in the following code.\nax.plot(xs, ys, zs) # no method found\nCould anyone give me some help? Thanks!\n\n``````using PyPlot; const plt = PyPlot\nxs = Float64[]\nys = Float64[]\nzs = Float64[]\n# initialize xs, ys, zs\n...\n\n# show 3D line\nfig = plt.figure()\nax.plot(xs, ys, zs) # error, \"type PyObject has no field plot \"\n``````\n\nThe aforementioned code is from the following python example. Can anyone convert the following code from Python to Julia? Thanks\n\n``````import matplotlib as mpl\nfrom mpl_toolkits.mplot3d import Axes3D\nimport numpy as np\nimport matplotlib.pyplot as plt\n\nmpl.rcParams['legend.fontsize'] = 10\n\nfig = plt.figure()\nax = fig.gca(projection='3d')\ntheta = np.linspace(-4 * np.pi, 4 * np.pi, 100)\nz = np.linspace(-2, 2, 100)\nr = z**2 + 1\nx = r * np.sin(theta)\ny = r * np.cos(theta)\nax.plot(x, y, z, label='parametric curve')\nax.legend()\n\nplt.show()\n``````\n\nShouldn’t that be plot3d instead of plot?\n\nYou need ax[:plot]\n\nIt is solved. The code is as following.\ncan use keys( obj ) to show the attributes and member functions of the PyObject.\n\n``````using PyPlot; const plt = PyPlot\n\ntheta = collect( linspace( -4π, 4π, 100 ) )\nz = collect( linspace(-2, 2, 100) )\nr = z.^2 + 1;\nx = r .* sin(theta)\ny = r .* cos(theta)\n\nfig = plt.figure()\nax[:plot](x, y, z) # PyObject, use obj[:function_name] to call the member function\n``````",
null,
"``````**# Code updated to Julia version:=0.7.0 for Matplotlib/PyPlot.version:=2.1.2 running on Python version:=3.6.4**\n\n**# Updated Julia code by Marc Cox / MSC**\n\n**# file: 3d-Matplotlib-PyPlot-example.jl**\n\n**using PyPlot**\n\n# Marc Cox / MSC add logging some basic debugging info\n\nprintln("## START Running Marc Cox 3d-Matplotlib-PyPlot-example.jl")\n\nprintln("## Julia VERSION:=", VERSION)\n\nprintln("## matplotlib.__version__():=", matplotlib[:__version__] )\n\nprintln("## Matplotlib/PyPlot.version:=", PyPlot.version)\n\n# Original code shows deprecation warnings\n\n# which become hard stop errors in Julia version 1.0+\n\n# Warning: `linspace(start, stop, length::Integer)` is deprecated, use `range(start, stop=stop, length=length)` instead.\n\n# theta = collect(Float16, linspace( -4π, 4π, 100 ) )\n\n# println("## typeof(theta) :=", typeof(theta))\n\n# z = collect(Float16, linspace(-2, 2, 100) )\n\n# println("## typeof(z) :=", typeof(z))\n\n#\n\n# range(start; length, stop, step=1)\n\ntheta = collect(Float16, range( -4π, length=100, stop=4π) )\n\nprintln("## typeof(theta) :=", typeof(theta))\n\nz = collect(Float16, range( -2, length=100, stop=2) )\n\nprintln("## typeof(z) :=", typeof(z))\n\n#\n\n# MSC reconsider #man-dot-operators-1 ,\n\n# Source: https://docs.julialang.org/en/v1/manual/mathematical-operations/index.html#man-dot-operators-1\n\n# More specifically MUST UNDERSTAND\n\n# Dot Syntax for Vectorizing Functions #man-vectorized-1\n\n# Source: https://docs.julialang.org/en/v0.6.1/manual/functions/#man-vectorized-1\n\n# Therefore, any Julia function f can be applied elementwise to any array (or other collection)\n\n# with the syntax f.(A).\n\n# For example sin can be applied to all elements in the vector A, like so:\n\n# julia> sin.(A)\n\n# Warning: `a::AbstractArray + b::Number` is deprecated, use `a .+ b` instead.\n\nr = z.^2 .+1;\n\nprintln("## typeof(r), typeof(theta) :=", typeof(r), ", " ,typeof(theta))\n\n# ORIGINAL WAS -- x = r .* sin(theta)\n\nx = r.*sin.(theta)\n\n# ORIGINAL WAS -- y = r .* cos(theta)\n\ny = r.*cos.(theta)\n\n# NOGO in Julia but Original OLD Python was -- fig = plt.figure()\n\n# I feel like this should be more prominently explained in the documentation,\n\n# but if you scroll down to the bottom of the Readme for PyCall (which PyPlot uses) it says:\n\n# Important: The biggest difference from Python is that object attributes/members\n\n# are accessed with o[:attribute] rather than o.attribute,\n\n# so that o.method(...) in Python is replaced by o[:method](...)\n\nfig = plt[:figure]()\n\n#\n\n# Source: https://github.com/JuliaPy/PyPlot.jl\n\n# you must first call the using3D() function to ensure that mplot3d is loaded\n\nusing3D() # << WORKS AOK per shows 3D axes below ..."
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https://bytes.com/topic/python/answers/162678-interesting-python-problem | [
"440,550 Members | 1,161 Online",
null,
"Need help? Post your question and get tips & solutions from a community of 440,550 IT Pros & Developers. It's quick & easy.\n\n# An interesting python problem\n\n P: n/a Hi, Look at the follow command in python command line, See what's interesting?:) class A: i = 0 a = A() b = A() a.i = 1 print a.i, b.i 1 0 --------------------------------------- class A: arr = [] a = A() b = A() a <__main__.A instance at 0x00C96698> b <__main__.A instance at 0x00CA0760> A a.arr.append(\"haha\") print a.arr , b.arr ['haha'] ['haha'] a.arr = [\"xixi\"] print a.arr , b.arr ['xixi'] ['haha'] A.arr ['haha'] A.arr.append(\"xx\") A.arr ['haha', 'xx'] a.arr ['xixi'] b.arr ['haha', 'xx'] b.arr.pop() 'xx' b.arr ['haha'] A.arr ['haha'] ------------------------------------- class X: def __init__(self): self.arr = [] m = X() n = X() m.arr.append(\"haha\") print m.arr, n.arr ['haha'] [] Sep 14 '05 #1\n3 Replies\n\n P: n/a Johnny Lee wrote: Hi, Look at the follow command in python command line, See what's interesting?:)class A: i = 0a = A()b = A()a.i = 1print a.i, b.i 1 0 Quite what I would expect. First you declare i as being a *class* attribute of A, with value 0. Then you create 2 instances a and b of A. Then you add to a an *instance* variable named i (that then shadows the class variable of the same name), with value 1. Then you print a.i, wihci is the instance variable i of a, and b.i, which is the class variable i of A, with value 0. ---------------------------------------class A: arr = []a = A()b = A()a <__main__.A instance at 0x00C96698>b <__main__.A instance at 0x00CA0760>A a.arr.append(\"haha\")print a.arr , b.arr ['haha'] ['haha'] Now you create a class A with a *class* variable arr which is an empty list, and 2 instances a and b of A. Then you append to a.arr - which is A.arr, so when you print a.arr and b.arr, you in fact print A.arr a.arr = [\"xixi\"] Then you add an instance variable arr to a, shadowing A.arr print a.arr , b.arr ['xixi'] ['haha'] So now you print a.arr and A.arr (accessed thru b) (snip) class X: def __init__(self): self.arr = []m = X()n = X()m.arr.append(\"haha\")print m.arr, n.arr ['haha'] [] Here you define a class X with an *instance* variable arr, and two instances m and n of X, then append to m.arr, which of course has no impact on n. I dont see anything interesting nor problematic here. If you understand the difference between class attributes and instance attributes, the difference between mutating an object and rebinding a name, and the attribute lookup rules in Python, you'll find that all this is the normal and expected behavior. Or did I miss something ? -- bruno desthuilliers - is Python much more readable than Perl ??? python -c \"print '@'.join(['.'.join([w[::-1] for w in p.split('.')]) for p in 'o****@xiludom.gro'.split('@')])\" Sep 14 '05 #2\n\n P: n/a bruno modulix wrote: I dont see anything interesting nor problematic here. If you understand the difference between class attributes and instance attributes, the difference between mutating an object and rebinding a name, and the attribute lookup rules in Python, you'll find that all this is the normal and expected behavior. Or did I miss something ? No, you didn't miss anything as I can see. Thanks for your help:) Sep 14 '05 #3\n\n P: n/a Johnny Lee wrote: bruno modulix wrote:I dont see anything interesting nor problematic here. If you understandthe difference between class attributes and instance attributes, thedifference between mutating an object and rebinding a name, and theattribute lookup rules in Python, you'll find that all this is thenormal and expected behavior.Or did I miss something ? No, you didn't miss anything as I can see. Thanks for your help:) You're welcome !-) Ok, I guess all this is not that intuitive, specially when comes from less dynamic languages. Python's object model is quite powerful, but one need to have a good understanding of it to understand *why* it works that way. There's an interesting slideshow about metaclasses and descriptors that may help (if your brain is robust enough !-) : http://www.python.org/pycon/dc2004/p...sses-pycon.pdf HTH -- bruno desthuilliers python -c \"print '@'.join(['.'.join([w[::-1] for w in p.split('.')]) for p in 'o****@xiludom.gro'.split('@')])\" Sep 14 '05 #4\n\n### This discussion thread is closed\n\nReplies have been disabled for this discussion.",
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https://www.slideserve.com/cynthiafrancis/chapter-9-powerpoint-ppt-presentation | [
"",
null,
"Download",
null,
"Download Presentation",
null,
"Chapter 9\n\n# Chapter 9\n\nDownload Presentation",
null,
"## Chapter 9\n\n- - - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - - -\n##### Presentation Transcript\n\n1. Chapter 9 Debt Instruments Quantitative Issues\n\n2. Pricing a Bond where P0 = price of bond today T = maturity of the bond Y = appropriate discount rate PAR = par or face value of the bond\n\n3. Bond Prices with Semiannual Payments • Divide coupon payment by two • Multiply maturity of bond by two. • Divide discount rate by two\n\n4. Bond Yields & Rates • Coupon rate (nominal yield) • Current yield (coupon / price) • Yield to maturity (YTM = IRR) • Realized compound yield to maturity (RCYTM) • Yield to First (earliest) Call • Realized return\n\n5. ABC Example • Coupon: \\$40 per year • Par Value: \\$1,000 • Maturity: 6 years • Callable: in 3 years @ \\$1040 • Price: \\$950\n\n6. Coupon Rate • Stated dollar return of fixed-income investment • Equals annual interest payments divided by par value\n\n7. Current Yield • Bond’s coupon rate divided by current market price OR • Stock’s indicated dividend rate divided by per-share price\n\n8. Yield to Maturity • Measure of bond yield that takes into account capital gain or loss, as well as coupon payments • Discount rate that would make present value of bond’s cash flows (payments plus face value at maturity) equal purchase price of bond where C= the coupon payment\n\n9. Yield Relationships\n\n10. Yield to Call where Tc = time to earliest call Yc = yield to first call • Almost identical to YTM, except • Call price replaces par value • Time to call replaces term to maturity\n\n11. Realized Rate (Yield) • Ex post rate of return or yield from investment (internal rate of return) where TH = holding period YH = realized rate of return\n\n12. Bond Price Volatility • Bond prices and interest rates inversely related • Maturity effect: longer a bond’s term to maturity, greater percentage change in price for given change in interest rates • Coupon effect:lower a bond’s coupon rate, greater percentage change in price for given change in interest rates • Yield-to-maturity effect: For given change in interest rates, bonds with lower YTMs have greater percentage price changes than bonds with higher YTMs – all other things equal\n\n13. Which Bond’s Price Is Most Volatile? • Bond X: 25 years to maturity, 10% coupon rate, and a 6% YTM • Bond Y: 10 years to maturity, 2% coupon rate, and a 6% YTM • Bond Z: 17.5 years to maturity, 6% coupon rate, and a 4% YTM\n\n14. Answer • Based on maturity effect, it would be X • Based on coupon effect, it would be Y • Based on yield-to-maturity effect, it would be Z\n\n15. Duration • Weighted average amount of time until present value of bond’s purchase price repaid to the investor • Based on time-weighted present value of bond’s principal and interest payments divided by the bond’s price • Used as measure of bond’s sensitivity to interest rate changes\n\n16. Formula for Duration Where P0 = price of the bond today Y = yield to maturity Ct = cash flow in period t (coupon, principal or both) T = term to maturity\n\n17. Equation 9-6 • Insert Equation • Where Y = yield to maturity C = coupon rate T = term to maturity\n\n18. Uses of Duration • Price volatility index • Larger duration statistic, more volatile price of bond • Immunization • Interest rate risk minimized on bond portfolio by maintaining portfolio with duration equal to investor’s planning horizon\n\n19. Major Characteristics of Duration • Duration of zero-coupon bond equal to term to maturity • Duration of coupon bond always less than term to maturity • Inverse relationship between coupon rate and duration (continued)\n\n20. Major Characteristics of Duration (continued) • Inverse relationship between yield to maturity and duration • Direct relationship between maturity and duration\n\n21. Modified Duration • Adjusted measure of duration used to estimate a bond’s interest rate sensitivityD* = D (1 + YTM)% Chg in price of bond = –D x % Chg in YTM% Chg in price of bond = – D* x [Chg in YTM]\n\n22. Convexity\n\n23. Portfolio Duration • Market value weighted average of durations of individual securities in the portfolio\n\n24. Components of InterestRate Risk • Price Risk • Reinvestment Rate Risk\n\n25. Price Risk • Risk of existing bond’s price changing in response to unknown future interest rate changes • If rates increase, bond’s price decreases • If rates decrease, bond’s price increases\n\n26. Reinvestment Rate Risk • Risk associated with reinvesting coupon payments at unknown future interest rates • If rates increase, coupons are reinvested at higher rates than previously expected • If rates decrease, coupons are reinvested at lower rates than previously expected\n\n27. Immunizing a Portfolio • If a single time horizon goal, purchasing zero-coupon bond whose maturity corresponds with planning horizon • If multiple goals, purchasing series of zero-coupon bonds whose maturities correspond with multiple planning horizons (continued)\n\n28. Immunizing a Portfolio (continued) • Assembling and managing bond portfolio whose duration is kept equal to planning horizon Note: this strategy involves regular adjustment of portfolio because duration of portfolio will change at SLOWER rate than will time itself\n\n29. Bond Swaps • Technique for managing bond portfolio by selling some bonds and buying others • Possible benefits achieved: • tax treatment • yields • maturity structure • trading profits\n\n30. Types of Swaps • Substitution swap • Tax swap • Intermarket spread swap • Pure-yield pick-up swap • Rate anticipation swap\n\n31. Strategies for Managing a Bond Portfolio • Bullet Portfolio – Entire portfolio is placed in one maturity • Bond ladders • Equally distributed dollar allocations over time • Barbells • Majority of dollar allocations in shortest-term and longest-term holdings\n\n32. Yield Curve or Term Structure • Vertical axis: yield to maturity • Horizontal axis: term to maturity • Bonds of like quality • Always based on Treasuries\n\n33. Shapes of Yield Curve • Rising: Most common (used to be only one observed) • Falling: Next most common • Humped • Flat: Rare\n\n34. Types of Yield Curves\n\n35. Theories of the Yield Curve • Unbiased expectations • Long-term rates reflect market’s expectation of current and future short-term rates. • Preferred habitat • Significantly more attractive rates can induce investors and borrowers out of their preferred maturity structures (continued)\n\n36. Theories of Yield Curve(continued) • Market Segmentation: • Yields reflect supply and demand for each maturity class. • Liquidity Preference: • Borrowers are risk averse and demand premium for buying long-term securities • Yield curves tend to be upward sloping. (continued)\n\n37. Theories of the Yield Curve (continued) • Preferred habitat • Significantly more attractive rates can induce investors and borrowers out of their preferred maturity structures • Unbiased expectations • Long-term rates reflect market’s expectation of current and future short-term rates.\n\n38. Factors Affecting Bond Yields • General credit conditions: Credit conditions affect all yields to one degree or another. • Default risk: Riskier issues require higher promised yields. • Term structure: Yields vary with maturity • Duration: Weighted average amount of time until present value of purchase price is recouped. • Coupon effect: Low-coupon issues offer yields that are partially taxed as capital gains. (continued)\n\n39. Factors Affecting Bond Yields (continued) • Seasonings: Newly issued bonds may sell at slight discount to otherwise-equivalent established issues. • Marketability: Actively traded issues tend to be worth more than similar issues less actively traded. • Call protection: Protection from early call tends to enhance bond’s value. • Sinking fund provisions: Sinking funds reduce probability of default, thereby tending to enhance bond’s value. • Me-first rules: Bonds protected from diluting effect of additional borrowings are generally worth more than otherwise-equivalent unprotected issues."
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| {"ft_lang_label":"__label__en","ft_lang_prob":0.8572278,"math_prob":0.900544,"size":7601,"snap":"2021-21-2021-25","text_gpt3_token_len":1583,"char_repetition_ratio":0.14242464,"word_repetition_ratio":0.07909604,"special_character_ratio":0.2085252,"punctuation_ratio":0.057142857,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.974583,"pos_list":[0,1,2,3,4,5,6,7,8],"im_url_duplicate_count":[null,null,null,null,null,1,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-06-14T12:52:56Z\",\"WARC-Record-ID\":\"<urn:uuid:c87837f5-776c-47aa-9a0e-f5e99dfa8eae>\",\"Content-Length\":\"95239\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:6407b069-9ce5-47e8-a1ac-42082ad4da2a>\",\"WARC-Concurrent-To\":\"<urn:uuid:48d29043-06b5-454a-8c8e-aca204c9711e>\",\"WARC-IP-Address\":\"52.37.133.166\",\"WARC-Target-URI\":\"https://www.slideserve.com/cynthiafrancis/chapter-9-powerpoint-ppt-presentation\",\"WARC-Payload-Digest\":\"sha1:57G2Z667MGIOJ2JYX5XTVH6CRYMWLQA7\",\"WARC-Block-Digest\":\"sha1:BQ2QZ7ZPIL6Q66XV63QZHJMG77V6PRYI\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-25/CC-MAIN-2021-25_segments_1623487612154.24_warc_CC-MAIN-20210614105241-20210614135241-00089.warc.gz\"}"} |
https://tutorialsinhand.com/Articles/program-to-swap-two-numbers-using-python-assignment.aspx | [
"Articles\n\n# Python program to swap 2 variables using assignment operator\n\nIn this python programming guide, we are going to learn:\n\n• python program to swap 2 variables assignment operation\n\n### Python program to swap 2 variables using assignment operator\n\nThe python program to swap two numbers using assignment operator is as follows:\n\n``````# Owner : TutorialsInhand Author : Devjeet Roy\n\na = int(input(\"Enter a: \"))\nb = int(input(\"Enter b: \"))\nc = 0\n\nprint(\"Before swapping:\")\nprint(\"a = {}, b = {}\".format(a,b))\n\na, b = b, a\n\nprint(\"After swapping:\")\nprint(\"a = {}, b = {}\".format(a,b))``````\n\nThe output of python program to swap two numbers using assignment operator is as follows:\n\n``````PS C:\\Users\\DEVJEET\\Desktop\\tutorialsInHand> python code.py\nEnter a: 10\nEnter b: 20\nBefore swapping:\na = 10, b = 20\nAfter swapping:\na = 20, b = 10``````\n\n1. Python provides this special double assignment functionality which helps us to directly swap two numbers at ease.\n\n2. The double assginment thing is : a,b = b,a. This interchanges the value between a and b.",
null,
"Basic Python Programs\n\nWould you like to see your article here on tutorialsinhand. Join Write4Us program by tutorialsinhand.com",
null,
""
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| [
null,
"https://tutorialsinhand.com/readwritedata/Articles/1081/swap-two-numbers-with-double-assignment.png",
null,
"https://tutorialsinhand.com/readwritedata/profileimages/profileimage-1081.png",
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http://petrofaq.org/wiki/Unit_Converter | [
"Blog | Tools | Glossary | Search Share: | feedback | Join us",
null,
"# Unit Converter\n\n• Length\n• Weight\n• Volume\n• Temperature\n• Area\n• Pressure\n• Energy\n• Power\n• Force\n• Time\n• Velocity\n• Angle\n• Density\n• Thermal conductivity\n• Specific heat capacity\n• Heat density\n• Heat flux density\n• Heat transfer coefficient\n• Flow\n• Flow – mass\n• Mass flux density\n• Concentration – solution\n• Viscosity – dynamic\n• Viscosity – kinematic\n• Surface tension\n\n From: To: meter [m]kilometer [km]decimeter [dm]centimeter [cm]millimeter [mm]micrometer [µm]micron [µ]nanometer [nm]mile [mi, mi(Int)]yard [yd]foot [ft]inch [in] meter [m]kilometer [km]decimeter [dm]centimeter [cm]millimeter [mm]micrometer [µm]micron [µ]nanometer [nm]mile [mi, mi(Int)]yard [yd]foot [ft]inch [in] Result:\n\n From: To: kilogram [kg]gram [g]milligram [mg]microgram [µg]nanogram [ng]pound [lb]ounce [oz]ton (metric) [t] kilogram [kg]gram [g]milligram [mg]microgram [µg]nanogram [ng]pound [lb]ounce [oz]ton (metric) [t] Result:\n\n From: To: cubic meter [m^3]cubic kilometer [km^3]cubic decimeter [dm^3]cubic centimeter [cm^3, cc]cubic millimeter [mm^3]liter [L, l]milliliter [mL]barrel (oil) [bbl (oil)]barrel (US) [bbl (US)]barrel (UK) [bbl (UK)]gallon (US) [gal (US)]gallon (UK) [gal (UK)]cubic mile [mi^3]cubic yard [yd^3]cubic foot [ft^3]cubic inch [in^3]ccfhundred-cubic footacre-foot [ac*ft]acre-inch [ac*in] cubic meter [m^3]cubic kilometer [km^3]cubic decimeter [dm^3]cubic centimeter [cm^3, cc]cubic millimeter [mm^3]liter [L, l]milliliter [mL]barrel (oil) [bbl (oil)]barrel (US) [bbl (US)]barrel (UK) [bbl (UK)]gallon (US) [gal (US)]gallon (UK) [gal (UK)]cubic mile [mi^3]cubic yard [yd^3]cubic foot [ft^3]cubic inch [in^3]ccfhundred-cubic footacre-foot [ac*ft]acre-inch [ac*in] Result:\n\n From: To: kelvin [K]degree Celsius [°C]degree Fahrenheit [°F]degree Rankine [°R]degree Reaumur [°r]Triple point of water kelvin [K]degree Celsius [°C]degree Fahrenheit [°F]degree Rankine [°R]degree Reaumur [°r]Triple point of water Result:\n\n From: To: square meter [m^2]square kilometer [km^2]square decimeter [dm^2]square centimeter [cm^2]square millimeter [mm^2]square micrometer [µm^2]square nanometer [nm^2]are [a]square mile [mi^2]square yard [yd^2]square foot [ft^2]square inch [in^2]acre [ac]square mil [mil^2] square meter [m^2]square kilometer [km^2]square decimeter [dm^2]square centimeter [cm^2]square millimeter [mm^2]square micrometer [µm^2]square nanometer [nm^2]are [a]square mile [mi^2]square yard [yd^2]square foot [ft^2]square inch [in^2]acre [ac]square mil [mil^2] Result:\n\n From: To: pascal [Pa]megapascal [MPa]kilopascal [kPa]psi [psi]baratmosphere technical [at]Standard atmosphere [atm]millipascal [mPa]newton/square meternewton/square centimeternewton/square millimeterkilonewton/square metermillibar [mbar]microbar [µbar]dyne/square centimeterkilogram-force/square meterkilogram-force/sq. centimeterkilogram-force/sq. millimetergram-force/sq. centimeterton-force (short)/square footton-force (short)/square inchton-force (long)/square footton-force (long)/square inchpound-force/square footpound-force/square inch pascal [Pa]megapascal [MPa]kilopascal [kPa]psi [psi]baratmosphere technical [at]Standard atmosphere [atm]millipascal [mPa]newton/square meternewton/square centimeternewton/square millimeterkilonewton/square metermillibar [mbar]microbar [µbar]dyne/square centimeterkilogram-force/square meterkilogram-force/sq. centimeterkilogram-force/sq. millimetergram-force/sq. centimeterton-force (short)/square footton-force (short)/square inchton-force (long)/square footton-force (long)/square inchpound-force/square footpound-force/square inch Result:\n\n From: To: joule [J]gigajoule [GJ]megajoule [MJ]kilojoule [kJ]millijoule [mJ]microjoule [µJ]nanojoule [nJ]gigawatt-hour [GW*h]megawatt-hour [MW*h]kilowatt-hour [kW*h]kilowatt-second [kW*s]watt-hour [W*h]watt-second [W*s]newton meter [N*m]horsepower hour [hp*h]horsepower (metric) hourBtu (th) [Btu (th)]mega Btu (IT) [MBtu (IT)]fuel oil equivalent @kiloliter joule [J]gigajoule [GJ]megajoule [MJ]kilojoule [kJ]millijoule [mJ]microjoule [µJ]nanojoule [nJ]gigawatt-hour [GW*h]megawatt-hour [MW*h]kilowatt-hour [kW*h]kilowatt-second [kW*s]watt-hour [W*h]watt-second [W*s]newton meter [N*m]horsepower hour [hp*h]horsepower (metric) hourBtu (th) [Btu (th)]mega Btu (IT) [MBtu (IT)]fuel oil equivalent @kiloliter Result:\n\n From: To: watt [W]gigawatt [GW]megawatt [MW]kilowatt [kW]horsepower [hp, hp (UK)]horsepower (550 ft*lbf/s) [hp]horsepower (metric)Btu (th)/hour [Btu (th)/h]Btu (th)/minute [Btu (th)/min]Btu (th)/second [Btu (th)/s]newton meter/second [N*m/s]joule/second [J/s] watt [W]gigawatt [GW]megawatt [MW]kilowatt [kW]horsepower [hp, hp (UK)]horsepower (550 ft*lbf/s) [hp]horsepower (metric)Btu (th)/hour [Btu (th)/h]Btu (th)/minute [Btu (th)/min]Btu (th)/second [Btu (th)/s]newton meter/second [N*m/s]joule/second [J/s] Result:\n\n From: To: newton [N]dyne [dyn]kilogram-force [kgf]ton-force (metric) [tf]pound-force [lbf] newton [N]dyne [dyn]kilogram-force [kgf]ton-force (metric) [tf]pound-force [lbf] Result:\n\n From: To: second [s]millisecond [ms]microsecond [µs]nanosecond [ns]minute [min]hour [h]day [d]weekmonthyear [y] second [s]millisecond [ms]microsecond [µs]nanosecond [ns]minute [min]hour [h]day [d]weekmonthyear [y] Result:\n\n From: To: meter/second [m/s]meter/hour [m/h]meter/minute [m/min]kilometer/hour [km/h]kilometer/minute [km/min]kilometer/second [km/s]centimeter/hour [cm/h]centimeter/minute [cm/min]centimeter/second [cm/s]millimeter/hour [mm/h]millimeter/minute [mm/min]millimeter/second [mm/s]foot/hour [ft/h]foot/minute [ft/min]foot/second [ft/s]yard/hour [yd/h]yard/minute [yd/min]yard/second [yd/s]mile/hour [mi/h]mile/minute [mi/min]mile/second [mi/s] meter/second [m/s]meter/hour [m/h]meter/minute [m/min]kilometer/hour [km/h]kilometer/minute [km/min]kilometer/second [km/s]centimeter/hour [cm/h]centimeter/minute [cm/min]centimeter/second [cm/s]millimeter/hour [mm/h]millimeter/minute [mm/min]millimeter/second [mm/s]foot/hour [ft/h]foot/minute [ft/min]foot/second [ft/s]yard/hour [yd/h]yard/minute [yd/min]yard/second [yd/s]mile/hour [mi/h]mile/minute [mi/min]mile/second [mi/s] Result:\n\n From: To: kilogram/cubic meterkilogram/cubic centimetergram/cubic meter [g/m^3]gram/cubic centimetergram/cubic millimetermilligram/cubic metermilligram/cubic centimetermilligram/cubic millimeterpound/cubic inch [lb/in^3]pound/cubic foot [lb/ft^3]pound/cubic yard [lb/yd^3]psi/1000 feet kilogram/cubic meterkilogram/cubic centimetergram/cubic meter [g/m^3]gram/cubic centimetergram/cubic millimetermilligram/cubic metermilligram/cubic centimetermilligram/cubic millimeterpound/cubic inch [lb/in^3]pound/cubic foot [lb/ft^3]pound/cubic yard [lb/yd^3]psi/1000 feet Result:\n\n From: To: watt/meter/K [W/(m*K)]watt/centimeter/°Ckilowatt/meter/K [kW/(m*K)]Btu (th) foot/hour/sq. foot/°FBtu (th) inch/hour/sq. foot/°F watt/meter/K [W/(m*K)]watt/centimeter/°Ckilowatt/meter/K [kW/(m*K)]Btu (th) foot/hour/sq. foot/°FBtu (th) inch/hour/sq. foot/°F Result:\n\n From: To: joule/kilogram/K [J/(kg*K)]joule/kilogram/°C [J/(kg*°C)]joule/gram/°C [J/(g*°C)]kilojoule/kilogram/Kkilojoule/kilogram/°Cpound-force foot/pound/°RBtu (th)/pound/°FBtu (th)/pound/°RCHU/pound/°C [CHU/(lb*°C)] joule/kilogram/K [J/(kg*K)]joule/kilogram/°C [J/(kg*°C)]joule/gram/°C [J/(g*°C)]kilojoule/kilogram/Kkilojoule/kilogram/°Cpound-force foot/pound/°RBtu (th)/pound/°FBtu (th)/pound/°RCHU/pound/°C [CHU/(lb*°C)] Result:\n\n From: To: joule/square meter [J/m^2]langleyBtu (IT)/square footBtu (th)/square foot joule/square meter [J/m^2]langleyBtu (IT)/square footBtu (th)/square foot Result:\n\n From: To: watt/square meter [W/m^2]kilowatt/square meterwatt/square centimeterwatt/square inch [W/in^2]joule/second/square meterdyne/hour/centimetererg/hour/square millimeterfoot pound/minute/sq. foothorsepower/square foothorsepower (metric)/sq. footBtu (th)/second/square inchBtu (th)/second/square footBtu (th)/minute/square footBtu (th)/hour/square footCHU/hour/square foot watt/square meter [W/m^2]kilowatt/square meterwatt/square centimeterwatt/square inch [W/in^2]joule/second/square meterdyne/hour/centimetererg/hour/square millimeterfoot pound/minute/sq. foothorsepower/square foothorsepower (metric)/sq. footBtu (th)/second/square inchBtu (th)/second/square footBtu (th)/minute/square footBtu (th)/hour/square footCHU/hour/square foot Result:\n\n From: To: watt/square meter/Kwatt/square meter/°Cjoule/second/square meter/KBtu (th)/second/square foot/°FBtu (th)/hour/square foot/°FCHU/hour/square foot/°C watt/square meter/Kwatt/square meter/°Cjoule/second/square meter/KBtu (th)/second/square foot/°FBtu (th)/hour/square foot/°FCHU/hour/square foot/°C Result:\n\n From: To: cubic meter/second [m^3/s]cubic meter/day [m^3/d]cubic meter/hour [m^3/h]cubic meter/minutecubic centimeter/daycubic centimeter/hourcubic centimeter/minutecubic centimeter/secondliter/day [L/d]liter/hour [L/h]liter/minute [L/min]liter/second [L/s]milliliter/day [mL/d]milliliter/hour [mL/h]milliliter/minute [mL/min]milliliter/second [mL/s]gallon (US)/day [gal (US)/d]gallon (US)/hour [gal (US)/h]gallon (US)/minutegallon (US)/secondgallon (UK)/day [gal (UK)/d]gallon (UK)/hour [gal (UK)/h]gallon (UK)/minutegallon (UK)/secondkilobarrel (US)/daybarrel (US)/day [bbl (US)/d]barrel (US)/hour [bbl (US)/h]barrel (US)/minutebarrel (US)/secondacre-foot/year [ac*ft/y]acre-foot/day [ac*ft/d]acre-foot/hour [ac*ft/h]hundred-cubic foot/dayhundred-cubic foot/hourhundred-cubic foot/minuteounce/hour [oz/h]ounce/minute [oz/min]ounce/second [oz/s]ounce (UK)/hour [oz (UK)/h]ounce (UK)/minuteounce (UK)/secondcubic yard/hour [yd^3/h]cubic yard/minutecubic yard/second [yd^3/s]cubic foot/hour [ft^3/h]cubic foot/minute [ft^3/min]cubic foot/second [ft^3/s]cubic inch/hour [in^3/h]cubic inch/minute [in^3/min]cubic inch/second [in^3/s] cubic meter/second [m^3/s]cubic meter/day [m^3/d]cubic meter/hour [m^3/h]cubic meter/minutecubic centimeter/daycubic centimeter/hourcubic centimeter/minutecubic centimeter/secondliter/day [L/d]liter/hour [L/h]liter/minute [L/min]liter/second [L/s]milliliter/day [mL/d]milliliter/hour [mL/h]milliliter/minute [mL/min]milliliter/second [mL/s]gallon (US)/day [gal (US)/d]gallon (US)/hour [gal (US)/h]gallon (US)/minutegallon (US)/secondgallon (UK)/day [gal (UK)/d]gallon (UK)/hour [gal (UK)/h]gallon (UK)/minutegallon (UK)/secondkilobarrel (US)/daybarrel (US)/day [bbl (US)/d]barrel (US)/hour [bbl (US)/h]barrel (US)/minutebarrel (US)/secondacre-foot/year [ac*ft/y]acre-foot/day [ac*ft/d]acre-foot/hour [ac*ft/h]hundred-cubic foot/dayhundred-cubic foot/hourhundred-cubic foot/minuteounce/hour [oz/h]ounce/minute [oz/min]ounce/second [oz/s]ounce (UK)/hour [oz (UK)/h]ounce (UK)/minuteounce (UK)/secondcubic yard/hour [yd^3/h]cubic yard/minutecubic yard/second [yd^3/s]cubic foot/hour [ft^3/h]cubic foot/minute [ft^3/min]cubic foot/second [ft^3/s]cubic inch/hour [in^3/h]cubic inch/minute [in^3/min]cubic inch/second [in^3/s] Result:\n\n From: To: kilogram/second [kg/s]gram/second [g/s]gram/minute [g/min]gram/hour [g/h]gram/day [g/d]milligram/minute [mg/min]milligram/hour [mg/h]milligram/day [mg/d]kilogram/minute [kg/min]kilogram/hour [kg/h]kilogram/day [kg/d]exagram/second [Eg/s]petagram/second [Pg/s]teragram/second [Tg/s]gigagram/second [Gg/s]megagram/second [Mg/s]hectogram/second [hg/s]dekagram/second [dag/s]decigram/second [dg/s]centigram/second [cg/s]milligram/second [mg/s]microgram/second [µg/s]ton (metric)/second [t/s]ton (metric)/minute [t/min]ton (metric)/hour [t/h]ton (metric)/day [t/d]ton (short)/hour [ton (US)/h]pound/second [lb/s]pound/minute [lb/min]pound/hour [lb/h]pound/day [lb/d] kilogram/second [kg/s]gram/second [g/s]gram/minute [g/min]gram/hour [g/h]gram/day [g/d]milligram/minute [mg/min]milligram/hour [mg/h]milligram/day [mg/d]kilogram/minute [kg/min]kilogram/hour [kg/h]kilogram/day [kg/d]exagram/second [Eg/s]petagram/second [Pg/s]teragram/second [Tg/s]gigagram/second [Gg/s]megagram/second [Mg/s]hectogram/second [hg/s]dekagram/second [dag/s]decigram/second [dg/s]centigram/second [cg/s]milligram/second [mg/s]microgram/second [µg/s]ton (metric)/second [t/s]ton (metric)/minute [t/min]ton (metric)/hour [t/h]ton (metric)/day [t/d]ton (short)/hour [ton (US)/h]pound/second [lb/s]pound/minute [lb/min]pound/hour [lb/h]pound/day [lb/d] Result:\n\n From: To: gram/second/square meterkilogram/hour/square meterkilogram/hour/square footkilogram/second/square metergram/second/sq. centimeterpound/hour/square footpound/second/square foot gram/second/square meterkilogram/hour/square meterkilogram/hour/square footkilogram/second/square metergram/second/sq. centimeterpound/hour/square footpound/second/square foot Result:\n\n From: To: kilogram/liter [kg/L]gram/liter [g/L]milligram/liter [mg/L]part/million (ppm)grain/gallon (US) [gr/gal (US)]grain/gallon (UK) [gr/gal (UK)]pound/gallon (US)pound/gallon (UK)pound/million gallon (US)pound/million gallon (UK)pound/cubic foot [lb/ft^3] kilogram/liter [kg/L]gram/liter [g/L]milligram/liter [mg/L]part/million (ppm)grain/gallon (US) [gr/gal (US)]grain/gallon (UK) [gr/gal (UK)]pound/gallon (US)pound/gallon (UK)pound/million gallon (US)pound/million gallon (UK)pound/cubic foot [lb/ft^3] Result:\n\n From: To: pascal second [Pa*s]newton second/square metermillinewton second/sq. meterdyne second/sq. centimeterpoise [P]megapoise [MP]kilopoise [kP]centipoise [cP]millipoise [mP]pound-force second/sq. inchpound-force second/sq. footpoundal second/square footgram/centimeter/secondpound/foot/second [lb/(ft*s)]pound/foot/hour [lb/(ft*h)] pascal second [Pa*s]newton second/square metermillinewton second/sq. meterdyne second/sq. centimeterpoise [P]megapoise [MP]kilopoise [kP]centipoise [cP]millipoise [mP]pound-force second/sq. inchpound-force second/sq. footpoundal second/square footgram/centimeter/secondpound/foot/second [lb/(ft*s)]pound/foot/hour [lb/(ft*h)] Result:\n\n From: To: square meter/second [m^2/s]square meter/hour [m^2/h]square centimeter/secondsquare millimeter/secondsquare foot/second [ft^2/s]square foot/hour [ft^2/h]square inch/second [in^2/s]stokes [St]centistokes [cSt] square meter/second [m^2/s]square meter/hour [m^2/h]square centimeter/secondsquare millimeter/secondsquare foot/second [ft^2/s]square foot/hour [ft^2/h]square inch/second [in^2/s]stokes [St]centistokes [cSt] Result:\n\n From: To: newton/meter [N/m]millinewnon/meter [mN/m]gram-force/centimeter [gf/cm]dyne/centimeter [dyn/cm]erg/square centimetererg/square millimeterpoundal/inch [pdl/in]pound-force/inch [lbf/in] newton/meter [N/m]millinewnon/meter [mN/m]gram-force/centimeter [gf/cm]dyne/centimeter [dyn/cm]erg/square centimetererg/square millimeterpoundal/inch [pdl/in]pound-force/inch [lbf/in] Result:"
]
| [
null,
"https://mc.yandex.ru/watch/11846299",
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https://cookup.ai/u/solve-codechef-problems-ry2olupf/ | [
"Solve Codechef problems Credits: taranjeetio\n\n# Input:\n\nProblem Two friends Chef and Chefina are currently on floors A and B respectively. They hear an announcement that prizes are being distributed on the ground floor and so decide to reach the ground floor as soon as possible. Chef can climb down X floors per minute while Chefina can climb down Y floors per minute. Determine who will reach the ground floor first. In case both reach the ground floor together, print Both. Input Format • The first line of input will contain a single integer T. denoting the number of test cases. The first line of each test case contains four space-separated integers A, B, X, and Y - the current floor of Chef, the current floor of Chefina, speed of Chef and speed of Chefina in floors per minute respectively. Output Format For each test case, output on a new line: • Chef if Chef reaches the ground floor first. • Chefina if she reaches the ground floor first. ⚫Both if both reach the ground floor at the same time. You may print each character of the string in uppercase or lowercase. For example, the strings CHEF, chef, Chef, and chEF are all considered the same.\n\n# Output:\n\nTo determine which of Chef and Chefina will reach the ground floor first, you can compare the time it will take each of them to reach the ground floor. You can do this by dividing the distance they each need to travel (the difference between their current floor and the ground floor) by their speed in floors per minute. The one with the shorter time is the one who will reach the ground floor first. If the times are equal, then both will reach the ground floor at the same time. Here is some example code in Python that demonstrates how you might solve this problem: Copy code `# Read the number of test cases t = int(input()) # Loop over the test cases for _ in range(t): # Read the current floors and speeds a, b, x, y = map(int, input().split()) # Calculate the time it will take each of them to reach the ground floor chef_time = (a - 0) / x chefina_time = (b - 0) / y # Determine which one reaches the ground floor first if chef_time < chefina_time: print(\"Chef\") elif chef_time > chefina_time: print(\"Chefina\") else: print(\"Both\")` This code reads the input for the problem, calculates the time it will take each of Chef and Chefina to reach the ground floor, and then prints the appropriate output based on who reaches the ground floor first. I hope this helps! Let me know if you have any questions."
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https://www.jiskha.com/similar?question=1%29Find+the+sixth+term+of+the+geometric+sequence+for+which+a1%3D5+and+r%3D3+A%291215+B%293645+C%299375+D%2923+I+chose+A+2%29Write+an+equation+for+the+nth+term+of+the+geometric+sequence+-12%2C4%2C-4%2F3...+A%29aN%3D-12%281%2F3%29n-1+B%29aN%3D12%28-1%2F3%29n-1+C%29aN%3D-12%28-1%2F3%29-n%2B1+D%29aN%3D-12%28-1%2F3%29n-1+I+chose+C+3%29Find&page=642 | [
"# 1)Find the sixth term of the geometric sequence for which a1=5 and r=3 A)1215 B)3645 C)9375 D)23 I chose A 2)Write an equation for the nth term of the geometric sequence -12,4,-4/3... A)aN=-12(1/3)n-1 B)aN=12(-1/3)n-1 C)aN=-12(-1/3)-n+1 D)aN=-12(-1/3)n-1 I chose C 3)Find\n\n92,916 questions, page 642\n1. ## chem lab\n\n0.8100g MgO(s)(40.3g/mol) was reacted with 52.3 ml 1.00M HCl(density= 1.00g/ml) in the coffee cup calorimeter, and the observed temperature change was +10.4 degree Celsius. write the equation for this reaction, determine the limiting reactant and find\n\nasked by mann on February 12, 2014\n2. ## physics\n\nA block of mass 1.4kg is placed on a rough surface. the coefficient of friction between the surfaces is 0.58. 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No need for the answers unless you want to give it to me. :D Let triangle ABC be a right triangle with angleC=90degrees. Given the tangent of one of the complementary angles of\n\nasked by Anonymous on March 2, 2017\n22. ## foundation of mechanics\n\nA particle is released from rest at a point P on an inclined plane. The inclination of the plane is tan\"1 (4/3) to the horizontal. If the coefficient of friction between the particle and the plane is 1/3, find i) the speed of the particle when it passes Q,\n\nasked by khairul on December 27, 2011\n23. ## Chemistry\n\n1. What does 'the time taken to reach completion decreases if 20cm^3 of hcl containing 4mol/dm^3' means? How to work out to find the concentration? 2(a)What does concentrated hcl means? (b)What does dilute hcl means? Which will react faster or slower / a\n\nasked by StuartKess on October 30, 2011\n24. ## Algebra 2\n\nFind the values of the inverse function in radians. sin^-1(0.65) a. 0.71+2pi n and -0.71+2pi n b. 0.71+2pi n and -3.85+2pi n c. 0.86+2pi n and -0.86 +2pi n d. -0.61+2pi n and 2.54+2pi n 2. tan^-1(0.09) a.-0.09+2pi n b. no such angle exists c.-1.48+ pi n d.\n\nasked by Fading on May 21, 2017\n25. ## Maths\n\nin an examination 50% of the studentspassedin mathematics and 70% of studentspassed in science while 10 % students failed in both subjects .300 students passed in at leastone subject .find total number of students who appeared in the examination if they\n\nasked by Set launguage on June 25, 2018\n26. ## business and society\n\nI need to find 3 business in my comunity: one that is formed as a sole proprietory, one that is formed as a general partnership, and one that is formed as a corportion. Identify the business and explain what characteristic of the business form led the\n\nasked by omar on October 7, 2007\n27. ## maths\n\nA water tank can be filled up by two pipes together in 30 minutes. One pipe fills up the tank in 32 min sooner than the other pipe when they are used separately. Find the time for each pipe to fill up the tank. Can you explain how you set up the equation\n\nasked by kim jung hee on September 2, 2013\n28. ## calculus2\n\nA tank is full of water. Find the work required to pump the water out of the spout. (Use 9.8 m/s^2 for g. Use 1000 kg/m^3 as the density of water. Assume r = 9 m and h = 3 m.) The tank is a spherical shape with r as the radius(9m). On top of the tank is a\n\nasked by A on June 28, 2018\n29. ## Physics - Tension Question\n\nA uniform bar of mass 12 kg and length 1.4 m is held by a hinge on a vertical wall. A string is attached to the end so that the bar is in mechanical equilibrium. The angle between the bar and the string is 25o. Find the tension in the string.\n\nasked by Anon on November 11, 2016\n30. ## Finance\n\nI calculated the following cash flows for the 5 year project. Can somone help me find the payback period?? year 0 = -80000 year 1= -75000 year2= 105000 year3= 115000 year4=115000 year5=115000\n\nasked by Maya on July 18, 2019\n31. ## Trigonometry\n\nSomeone please help me!!! I think I've found the correct formula for this but I don't understand exactly why this is a sine function and how to find the days of the year when Portland has 11 and 15 hours of daylight. PLEASE HELP!! 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Plz tell me how to find ionization potential of any element???????\n\nasked by Hk on October 20, 2018\n39. ## math\n\nYou find 30 coins in your drawer. 20 % of the coins are nickels. How much money you have in nickels?\n\nasked by Mely on September 5, 2010\n40. ## Algebra 1\n\nFind an equation of the line containing the given pair of points. (1/4,-1/2) and (3/4,5). What is the equation of the line? y= Thank you for all your help.\n\nasked by Esther on January 26, 2011\n41. ## Math 116a\n\nFind the slope and the y intercept f(x)=-2x-10 I have the slope as -2 and the y intercept as (0,-10) I hope this is correct?\n\nasked by Katie A on September 28, 2011\n42. ## algebra\n\nslope of 4 and y-intercept of (0,3) find slope-intercept equation that has same characteristics?\n\nasked by lisa on September 10, 2008\n43. ## math\n\nFind the equation of the line with the following slope and y intercept. 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I need a little help with absolute values on complex numbers.\n\nasked by Aditya on August 16, 2014\n45. ## Precalculus\n\nFind the exact value of: Cos(inverse tan 4/3 + inverse cos 5/13)\n\nasked by Thomas on April 21, 2014\n46. ## maths\n\ni have got to find the perimeter of a triangle that has one side of 8x-7w another that says 3x+9y and then another that says 3y+5w whats the perimeter of the triangle ??? please help\n\nasked by sophie on January 21, 2011\n47. ## Math - Sum of a Series\n\nFind the sum of the series 23 + 18 + 13 + . . .+ (-82) A) -590 B) 590 C) -649 D) 649\n\nasked by Robo123 on December 11, 2010\n48. ## geometry\n\ngiven H is the midpoint of GI. the length of GH is represented by 4x-5 and the length of HI is represented by 3x+20. find the value of x and the measures of GI, Gh, and HI.\n\nasked by Anonymous on September 7, 2010\n49. ## precalc\n\nFind the indicated partial sum of the series. infinity E 3(1/2)^i i=1 third partial sum\n\nasked by bbbb on November 11, 2010\n50. ## Geometry\n\nthe measures of two complementary angles are 16z-9 and 4z+3. 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What is the limit as x goes to 5+(from the right side?\n\nasked by Anonymous on January 25, 2013\n65. ## Math\n\nThe shopkeeper has 75 3/4 kg of sugar and sold 1/4 of it, find the amount of sugar sold.\n\nasked by Deepak on April 27, 2018\n66. ## Maths\n\nThe differencein the measures of 2 complementary angles is 18.Find the measures of angles.\n\nasked by anonymus on December 2, 2011\n67. ## Calculus\n\nLet f(x)=-x^2+3x on the interval [1,3]. Find the absolute maximum and absolute minimum of f(x) on this interval.\n\nasked by Anonymous on August 5, 2013\n68. ## geometry\n\nY is a point on HR with HO=(19x) miles and OR=48 miles. If HR=(27x) miles, find HO (in miles)\n\nasked by sam on September 16, 2012\n69. ## 8th grade\n\nFind the equation of the line With a slope of (-1/3) that goes through the point (2, -4) With a slope of 3 that goes through the point (-5, -1)\n\nasked by kate on January 14, 2009\n70. ## algebra\n\nA number is double of another number.If the difference of the number is twenty, find the number.\n\nasked by pranjita on November 23, 2016\n71. ## Math\n\nThe Perimeter of a right angled triangle is 40 CM. Find the area of the right angled triangle.\n\nasked by Debasish on December 22, 2015\n72. ## Calc\n\nLet ∫-7to-1 f(x)dx=9,∫-7to-5f(x)dx=9,∫-3to-1f9x)dx=10. Find ∫-5to-3 f(x)dx=? and ∫-3to-5 (9f(x)-9)dx=?\n\nasked by Anon. on June 20, 2013\n73. ## math homework\n\nfind a unit vector in the direction of u u = (-5, -12, 9) a. (-5, -12, 9) b. 1 / squareroot 26 (-5, -12, 9) c. squareroot 26 (-5, -12, 9 d. 1/ 5 squareroot 10 (-5, -12, 9) e. 1/ 5 squareroot 10 (5, 12, -9)\n\nasked by lottie on August 30, 2016\n74. ## maths\n\nThe sum of two numbers Is 31.2/3 of one number is eaual to 5/8 of the other number.find the numbers\n\nasked by mma on September 4, 2016\n75. ## Pre-Calculus\n\nGiven cos A = -3/5, π\n\nasked by Anonymous on October 24, 2018\n76. ## Geometry again\n\nFind the measure of angle T if the measure of angle T is 20 more than 4 times its supplement.\n\nasked by Monique N. on September 27, 2009\n77. ## trig\n\nplease help me. use trig. identities to find the exact value. tan 25° + tan 5° / 1- tan 25° tan 5°\n\nasked by ann on April 30, 2010\n78. ## Math/Algebra\n\nFind each sum or difference (3x^2 + 7x- 2)+ (2x^2 - 3x + 4) Answer: 5x^2 + 4x + 2 (4x + 8y) + (-7y + 3x) Answer: 7x+1y (x^2 + 4x - 1)+ (9x^2 - 8x + 9) Answer: 10x^2 - 4x + 8 (8x + 6) - (2x - 1) Answer: 6x+ 11\n\nasked by Cindy on March 24, 2009\n79. ## mathematics\n\nThe HCF of 2 no. Is 1/28 of their LCM.The sum of the HCF and the LCM is 116.If one of the no. Is 16 , find the other.\n\nasked by nitya on July 30, 2014\n80. ## Algebra\n\nI have the equation 5x + 3y = -2 and I am asked to find the rate of change. Is the rate of change -5/3?\n\nasked by Anonymous on January 2, 2014\n81. ## math\n\nHow do you find the sum of the mode from each set of data? Set A: ( 4,9,7,2,4 ) Set B: ( 11,6,4,6,2 ) Set C: ( -3,2,9,-3,-3)\n\nasked by Allison on April 23, 2014\n82. ## Algebra\n\nUse the rational zero theorem to find all rational zeros, and use those zeros to write f(x) = 2x^4-x^3-9x^2+4x+4\n\nasked by Arneta on October 10, 2015\n83. ## Algebra I\n\nDid I do these probelms correctly? Find the intercepts 1. y^2 = x^2 - 4x y^2 = x(x - 4) Y intercepts y^2 = 0(0 - 4) (0,0) x intercepts 0 = x(x - 4) 0 = x - 4 4 = x (4,0) 2. y = (x (x + 3))/(3x + 1)^2 y intercetps y = (0(0 + 3))/(3(0) +1)^2 y = 3/1 (0,3) x\n\nasked by Kate on July 21, 2010\n84. ## algebra II\n\nFind the equation of the line With a slope of (-1/3) that goes through the point (2, -4) With a slope of 3 that goes through the point (-5, -1)\n\nasked by kate on January 14, 2009\n85. ## Math\n\nHow to find the decimal equivalent of the mixed number and the mixed number is 4 2/5\n\nasked by Suzie on April 26, 2008\n86. ## mathematics\n\nfind the equatn of a circle with (a).centre (1,2) and radius 3.(b) centre (1,-3) and radius 2\n\nasked by aminat on May 16, 2010\n87. ## Algebra\n\nFind an equation of the line having the the given slope and containing the given point. M=4/5,(7,-4) The equation of the line is y=?\n\nasked by Pam on February 13, 2014\n88. ## math\n\nFind the probability which odds in favor are given in problem. Odds in favor are 7 to 2. 1 0\n\nasked by Anonymous on August 21, 2012\n89. ## geometry\n\nOne endpoint of a line segment is (8, −1). The point (5, −2) is one-third of the way from that endpoint to the other endpoint. Find the other endpoint.\n\nasked by Mark on May 31, 2015\n90. ## math\n\nFind the exact value of the expression. cos(π/16)cos(3π/16) − sin(π/16) sin(3π/16)\n\nasked by Stella on June 10, 2018\n91. ## Math\n\nFind the LCM of 11(s-3) and 33(s-3). LCM of 11 is 11*1 LCM of 33 is 3*11 LCM is 11*3*1=33 What do I do from here? I put down 33(s-3) was the answer but I don't think that is right. Thanks for your help.\n\nasked by B.B. on May 16, 2010\n92. ## Re\n\ni need to find eveidence of 3 things that jesus did and 3 things jesus didn`t do. thanks!\n\nasked by david on September 22, 2008\n93. ## geometry\n\nin a translation, how can you use the relationship between some points and their images to find the images of other points?\n\nasked by susan on January 28, 2009\n94. ## competition\n\nif means '¡Á' ,'q' means'-' , 'r' means ' ' 's' means '¡' then find the values of the following expression. 16s64q2p5r10\n\nasked by sagar on November 28, 2014\n95. ## G.p.\n\nThe sum of three numbers in g.p. is 21 and the sum of their squares is 189. Find the numbers.\n\nasked by Vijay on October 19, 2016\n96. ## algebra\n\nI am having trouble with the following : find the slope and the y intercept of the line 7x=6y+12. I have the slope as 3 and the y intercept (0,3)\n\nasked by Marla on September 12, 2010\n97. ## geometry\n\nthe #'s given are the coordinates of two points on a # line. find the distance between the two points. -3 and 5 what is the distance between these\n\nasked by taz manian on September 22, 2008\n98. ## Algebra\n\nFind the equation of the line passing through (-5, -4) and parallel to the line passing through (-3,2) and (6,8)\n\nasked by Pat on November 9, 2009\n99. ## Math\n\nfind all the other zeroes of the polynomial x^4+4x^3-4x^2-32x-32 if two of its zeroes are 2 root 2 and -2 root 2\n\nasked by Prithvi on April 9, 2016\n100. ## Algebra\n\nFind an equation of the line having the given slope and containing the given point. m=7/8, (5,-6) The equation of the line is y=?\n\nasked by Ann on May 24, 2011"
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https://whatisconvert.com/339-grams-in-ounces | [
"## Convert 339 Grams to Ounces\n\nTo calculate 339 Grams to the corresponding value in Ounces, multiply the quantity in Grams by 0.03527396194958 (conversion factor). In this case we should multiply 339 Grams by 0.03527396194958 to get the equivalent result in Ounces:\n\n339 Grams x 0.03527396194958 = 11.957873100908 Ounces\n\n339 Grams is equivalent to 11.957873100908 Ounces.\n\n## How to convert from Grams to Ounces\n\nThe conversion factor from Grams to Ounces is 0.03527396194958. To find out how many Grams in Ounces, multiply by the conversion factor or use the Mass converter above. Three hundred thirty-nine Grams is equivalent to eleven point nine five eight Ounces.\n\n## Definition of Gram\n\nThe gram (alternative spelling: gramme; SI unit symbol: g) is a metric system unit of mass. A gram is defined as one one-thousandth of the SI base unit, the kilogram, or 1×10−3 kg, which itself is now defined, not in terms of grams, but as being equal to the mass of a physical prototype of a specific alloy kept locked up and preserved by the International Bureau of Weights and Measures.\n\n## Definition of Ounce\n\nThe ounce (abbreviation: oz) is a unit of mass with several definitions, the most popularly used being equal to approximately 28 grams. The size of an ounce varies between systems. Today, the most commonly used ounces are the international avoirdupois ounce (equal to 28.3495231 grams) and the international troy ounce (equal to 31.1034768 grams).\n\n## Using the Grams to Ounces converter you can get answers to questions like the following:\n\n• How many Ounces are in 339 Grams?\n• 339 Grams is equal to how many Ounces?\n• How to convert 339 Grams to Ounces?\n• How many is 339 Grams in Ounces?\n• What is 339 Grams in Ounces?\n• How much is 339 Grams in Ounces?\n• How many oz are in 339 g?\n• 339 g is equal to how many oz?\n• How to convert 339 g to oz?\n• How many is 339 g in oz?\n• What is 339 g in oz?\n• How much is 339 g in oz?"
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