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https://stats.stackexchange.com/questions/116770/reml-or-ml-to-compare-two-mixed-effects-models-with-differing-fixed-effects-but?noredirect=1 | [
"# REML or ML to compare two mixed effects models with differing fixed effects, but with the same random effect?\n\nBackground: Note: My dataset and r-code are included below text\n\nI wish to use AIC to compare two mixed effects models generated using the lme4 package in R. Each model has one fixed effect and one random effect. The fixed effect differs between models, but the random effect remains the same between models. I've found that if I use REML=T, model2 has the lower AIC score, but if I use REML=F, model1 has the lower AIC score.\n\nSupport for using ML:\n\nZuur et al. (2009; PAGE 122) suggest that \"To compare models with nested fixed effects (but with the same random structure), ML estimation must be used and not REML.\" This indicates to me that I ought to use ML since my random effects are the same in both models, but my fixed effects differ. [Zuur et al. 2009. Mixed Effect Models and Extensions in Ecology with R. Springer.]\n\nSupport for using REML:\n\nHowever, I notice that when I use ML, the residual variance associated with the random effects differs between the two models (model1 = 136.3; model2 = 112.9), but when I use REML, it is the same between models (model1=model2=151.5). This implies to me that I ought instead to use REML so that the random residual variance remains the same between models with the same random variable.\n\nQuestion:\n\nDoesn't it make more sense to use REML than ML for comparisons of models where the fixed effects change and the random effects remain the same? If not, can you explain why or point me to other literature that explains more?\n\n# Model2 \"wins\" if REML=T:\nREMLmodel1 = lmer(Response ~ Fixed1 + (1|Random1),data,REML = T)\nREMLmodel2 = lmer(Response ~ Fixed2 + (1|Random1),data,REML = T)\nAIC(REMLmodel1,REMLmodel2)\nsummary(REMLmodel1)\nsummary(REMLmodel2)\n\n# Model1 \"wins\" if REML=F:\nMLmodel1 = lmer(Response ~ Fixed1 + (1|Random1),data,REML = F)\nMLmodel2 = lmer(Response ~ Fixed2 + (1|Random1),data,REML = F)\nAIC(MLmodel1,MLmodel2)\nsummary(MLmodel1)\nsummary(MLmodel2)\n\n\n# Dataset:\n\nResponse Fixed1 Fixed2 Random1\n5.20 A A 1\n32.50 A A 1\n6.57 A A 2\n24.77 A B 3\n41.69 A B 3\n34.29 A B 4\n1.80 A B 4\n10.00 A B 5\n15.56 A B 5\n4.44 A C 6\n21.65 A C 6\n9.20 A C 7\n4.11 A C 7\n12.52 B D 8\n0.25 B D 8\n27.34 B D 9\n11.54 B E 10\n0.86 B E 10\n0.68 B E 11\n4.00 B E 11\n\n• Faraway's (2006) Extending the linear model with R (p. 156): \"The reason is that REML estimates the random effects by considering linear combinations of the data that remove the fixed effects. If these fixed effects are changed, the likelihoods of the two models will not be directly comparable.\" – jvh_ch Sep 25 '14 at 16:56\n• Even though AIC is likelihood based, to the best of my knowledge, it was developed for the purposes of prediction. How would one exactly apply a mixed model for prediction? – AdamO Sep 25 '14 at 20:47\n• @AdamO, could you be more precise? A fitted mixed model can be used for prediction, either at the population level (predict responses for an unspecified/unknown unit by setting conditional modes/BLUPs to zero) or at the individual level (condition prediction on the estimates of the conditional modes/BLUPs). If you can be more specific, that might make a good new CV question. – Ben Bolker Sep 25 '14 at 22:10\n• It was just unclear to me how you were intended to apply this model. Nothing in the problem suggested what kind of prediction, if any, was being done or if it was necessary and if so for what purpose. – AdamO Sep 25 '14 at 23:03\n\nZuur et al., and Faraway (from @janhove's comment above) are right; using likelihood-based methods (including AIC) to compare two models with different fixed effects that are fitted by REML will generally lead to nonsense.\n\n• Thanks @janhove, AdamO and Ben Bolker. I also found this link from Aaron to be helpful in answering this question. It says, \"The REML likelihood depends on which fixed effects are in the model, and so are not comparable if the fixed effects change. REML is generally considered to give better estimates for the random effects, though, so the usual advice is to fit your best model using REML for your final inference and reporting.\" – It Figures Sep 26 '14 at 12:04\n\nI'll give an example to illustrate why the REML likelihood cannot be used for things like AIC comparisons. Imagine that we a normal mixed effects model. Let $X$ denote the design matrix and assume that this matrix has full rank. We can find a reparametrization of the mean value space, given by the matrix $\\tilde{X}$. The two matrices span the same linear subspace of $\\mathbb{R}^n$. Thus, the columns of $\\tilde{X}$ can be written as linear combinations of the columns of $X$. Therefore, we can find a quadratic matrix, $B$, such that\n\n$\\tilde{X} = XB$.\n\nFurthermore, $B$ has full rank (this can be proven by assuming that it didn't; then neither would $X$, a contradiction). This means that $B$ is invertible.\n\nIf we start out be using the second parametrization of the mean value space and let $V$ be a covariance matrix then let's consider the REML criterion we should maximize (I'm omitting a constant)\n\n$|V|^{-1/2}|\\tilde{X}'V^{-1}\\tilde{X}|^{-1/2}\\exp((y-\\tilde{X}\\tilde{\\beta})'V^{-1}(y-\\tilde{X}\\tilde{\\beta})/2)$,\n\nover the parameter set, where $\\beta = (\\tilde{X}V^{-1}\\tilde{X})^{-1}y$. Using the fact that $X = \\tilde{X}B$, we can realize that this can be rewritten as\n\n$|B||V|^{-1/2}||X'V^{-1}X|^{-1/2}|\\exp((y-X\\bar{\\beta})'V^{-1}(y-X\\bar{\\beta})/2)$,\n\nwhere $\\bar{\\beta} = (XV^{-1}X)^{-1}y$. This is the REML likelihood for the other parametrization times the determinant of $|B|$.\n\nWe therefore have an example of two different parametrizations of the same model, giving different likelihood values, assuming that $|B| \\neq 1$ (such a matrix can easily be found). The same parameter value will maximize the criterion in both cases but the value of the likelihood will be different. This shows that there is an arbitrary element in the likelihood value and therefore illustrates why one cannot use the value of the likelihood for comparison between models with different fixed effects: you would be able to change the results simply be changing the mean value space parametrization in one of the models.\n\nThis an example of why REML should not be used when comparing models with different fixed effects. REML, however, often estimates the random effects parameters better and therefore it is sometimes recommended to use ML for comparisons and REML for estimating a single (perhaps final) model."
] | [
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http://myriverside.sd43.bc.ca/annabelles2019/category/grade-11/page/3/ | [
"### Archive of ‘Grade 11’ category\n\nIn week 10 of my precalc 11 class, we learned about three different forms of equations that can be graphed, and how to convert them from one to another. Here are the different forms (with examples below):\n\nGeneral Form: y = ax² + bx + c\n\ny = x² – 2x + 4\n\ny = -2x² + 5x -16\n\nStandard or Vertex Form: y = a(x – p)² + q\n\ny = (x – 4)² + 7\n\ny = 3(x – 2)² – 1\n\nFactored Form: y = a(x – x₁)(x – x₂)\n\ny = 2(x + 7)(x – 3)\n\ny = (x – 5)(x – 2)\n\nEach of these equation types can show different things about what the graph will look like. For example, the last number (with no variable) in general form tells you the y-intercept. The 2 numbers after the x in factored form are the roots of the equation (make sure to change the signs first). In order to fully graph a quadratic equation, sometimes you need to convert one form of the equation into another. Here is how to convert general form into factored or standard:",
null,
"This is how to change standard/vertex form into general or factored:",
null,
"Finally, here is how to convert factored form into general or standard:",
null,
"This week in precalc 11, I continued learning about the connection between graphing and quadratic equations. One thing specifically that I learned about was how to plot the vertex on a graph, when given a quadratic equation. The “parent function” is like the original graph. It is created from the equation y=x². For this graph, the vertex is at the coordinates (0,0). This is what the parent function graph looks like:",
null,
"When graphing an equation different than y=x², there are a few ways to find the coordinates of the vertex. First, see if there is a last number (without a variable) in the equation. This tells you where the vertex will be on the y-axis. If the number is positive, move the vertex upwards (the amount of spaces the number says). If it is negative, move down. As you can see below, the parabola is still the same shape and size as the parent function, just shifted in a different direction. Here are some examples:\n\ny = x² + 2",
null,
"y = x² – 7",
null,
"If there is a number either added or subtracted from the x, this tells you where the vertex is on the x-axis. If the number is positive, move the vertex to the left. If it is negative, move to the right. Here are some examples:\n\ny = (x + 1)²",
null,
"y = (x – 3)²",
null,
"Finally, these two things can be combined in an equation, so the vertex has coordinates that do not include zero. Here are a few more examples:\n\ny = (x – 3)² + 4",
null,
"y = (x – 1)² + 1",
null,
"In my precalc 11 this week, we learned about the quadratic formula. This is another way to solve quadratic equations, other than factoring or completing the square. The quadratic formula looks like this:",
null,
"It is easy to solve an equation using the quadratic formula because all you do is put in the numbers, and use algebra to find the variable. To do this, you must find out what numbers in the polynomial are A, B and C. A is always the number with the x², B is always with the x, and C is always the last number, without a variable. Remember that when there is no number with an x, there is an invisible 1. Here are some examples of what A, B, and C are in the following polynomials.",
null,
"This is how to solve a polynomial using the quadratic formula:",
null,
"In the next equation, the quadratic formula is used, but the answers are 2 rational roots. This shows that the equation could have been factored from the start, rather than doing the quadratic formula, which sometimes takes longer. When you need to solve a quadratic equation, first check if the numbers can be factored, or if you can use the technique of completing the square. If not, using the quadratic formula will guarantee the correct answer.",
null,
"In my precalc 11 class this week, I learned about graphing, and how it can relate to a quadratic equation or function. A quadratic function is in the form y=ax² + bx + c\n\nThese are all quadratic equations: Even if they don’t look like it at the start, they can be moved around into the correct form.",
null,
"These graphs in curved shapes are called parabolas:",
null,
"",
null,
"",
null,
"",
null,
"When an equation creates a graph in a parabola, it is a quadratic function. Parabolas have a vertex, which is either the highest or lowest point. If it is the highest point on the graph, it is called the maximum point. If it is at the bottom, it is the minimum point. Here are some examples:",
null,
"This week in precalc 11, I learned how to factor a polynomial, specifically when there is a fraction within it. Here are some examples of polynomials with fractions, that can be factored:",
null,
"To be able to factor these easily, there should be only integers with the variables. This means you must find a greatest common denominator (GCD), so all the terms have the same number at the bottom of the fraction. Then, the fraction can be removed as a greatest common factor (GCF), and the polynomial can be factored. Here are the above examples, with the steps to take to factor:\n\nRemember, if there is a variable by itself without a number in front, there is always an invisible 1. Also, don’t forget that all numbers/variables are fractions, even if they aren’t shown that way at first. If a variable, like x, is not written as a fraction, it has a denominator of 1.",
null,
"",
null,
"You can also verify your answer when you are done, to check if it is correct. To do this, just take the final answer and use the distributive property with FOIL to get back to the polynomial you started with.\n\nIn week 5 of precalc 11, we started learning about factoring. This is a continuation of grade 10, but new factoring skills are being added on. One concept we learned about is factoring with a greatest common factor (GCF). This is when you remove a common factor from all the terms, so it becomes easier to factor. The GCF has to be something that is present in ALL terms. Remember, a variable, like x, can also be a part of the common factor. Here are some examples of binomials with a GCF:",
null,
"When you remove the GCF from a binomial or trinomial to factor, make sure to remove the same GCF from all terms. If there is a variable like x in two of the terms but not the other, then it cannot be part of the GCF that gets removed. For example:",
null,
"Finally, here are some more examples of how to remove the GCF from a binomial or trinomial. When you do this, the GCF is written in front, and the remaining the numbers are inside brackets. Sometimes this is all you do to simplify the question, but make sure to check if you can factor any further.\n\nIn my precalc 11 class this week, I learned how to rationalize a denominator. This happens when there is a fraction with a square root on the bottom. For example:",
null,
"When fractions are being simplified, it is better to show them with an integer on the bottom. If the denominator is not an integer, but a radical, you can rationalize it. To make the denominator a rational number, or a non-radical, multiply the whole fraction by 1. This is done my multiplying it by the radical on the bottom, that you want to get rid of. (Since the radical is on the top and bottom of a fraction it equals 1).\n\nThese fractions below are all equal to 1, because the numerator and denominator are the same:",
null,
"When you multiply the two fractions, make sure to multiply the radicals only with other radicals, and integers in front of the radicals (if it’s a mixed radical) with other integers. If there is no number in front of a radical, pretend there is a 1. Make sure not to multiply a radical with an integer! Here are a few examples of how to rationalize the denominator:\n\nIn precalc 11 class this week, we learned about adding and subtracting radical expressions. This involves combining like terms that are radicals. Like terms is something I had learned about before, but I didn’t realize it was possible with radicals. Terms are numbers, variables, and/or exponents in groups. Here are four examples of terms:",
null,
"When the variables and exponents in more than one term are the same, the terms are “like”. The coefficients (number in front of the variable) can be different. For example:",
null,
"When multiple radicands and their index are the same, they can be combined. For example, these radicals are like terms, because they have the same radicand and index (the square root of 4):",
null,
"It is important to remember that the coefficient can be different for each term, because that is what’s being combined. To combine like terms, add/subtract the coefficients based on the sign in front of them. If there is no sign, it is positive. Here are the steps to combine like terms that are radicals:",
null,
"If the radical has no number in front, there is an invisible coefficient of 1. Don’t forget to add or subtract it as well! Here is one final example:",
null,
"This week in precalc 11, I learned how to evaluate an exponent that is a fraction. In the textbook, this is known as “writing powers with rational exponents”. Since all fractions are rational numbers, it is possible to make fractions into exponents on numbers. Here is an example:",
null,
"Below are some other examples. As you can see, the fraction can be positive or negative:",
null,
"To solve a fractional exponent, you change the base number from being a power, into a radicand within a radical. In my class I learned the saying “flower power, because the root is always on the bottom”. This is a helpful way to remember that the denominator (or bottom number) in the fractional exponent becomes the index in the radical.\n\nThis example below demonstrates the steps of changing a number with a positive fractional exponent into a radical. The numerator (top of fraction), can be placed as the exponent either on just the radicand, or the entire radical. I prefer to do it whichever way makes the numbers easier to solve. After this is done, the radical can be further evaluated.",
null,
"This example shows the same steps, with a negative exponent. There are different ways to do this, but I chose the one I like best:",
null,
"This week in Pre-calculus 11, I learned about roots and powers. Something I hadn’t learned about before were mixed radicals, and how to create one from a regular radical. A mixed radical is similar to a mixed fraction, where there is a whole number and a fraction together. Here are some examples of mixed fractions:",
null,
"A mixed radical looks like the photo below, where there is a number inside the square root sign, as well as a number outside:",
null,
"To change a radical from being entire (only a number inside the square root sign) to mixed, you can follow the steps below. In step 2, make sure one of the numbers is a perfect square (9, 25, 49, etc.)",
null,
"To change a radical from mixed back to entire, follow the steps below. In step 2, don’t forget to add the outside number in twice. This will make sure you multiply it by itself, so it becomes a perfect square number.",
null,
"Here are two more examples:",
null,
""
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https://dev.opencascade.org/content/curve-control-polygone | [
"# Curve from control polygone\n\nHi,\nI just started using this powerful OCC library (on a LINUX system) and I would very appreciate your help with the following problem: I am generating a curve from a given set of points by two methods: The points are interpolated or set as control points. Both curves seems to be generated as expected. In case of control points and writing points to file in between curve.FirstParameter() and curve.LastParameter() the curve is too short at both ends, as to see in the attached figure.\nAny suggestions where this comes from?\n\nThe curve is generated with the following code:\n\nint i;\nint degree=0;\n\n// Determine the degree\nif (controlPoints->Length() == 1)\n{\ncout exit(0);\n}\nif (controlPoints->Length() > maxDegree )\ndegree = maxDegree;\nelse if (controlPoints->Length() == 2)\ndegree = 1;\nelse if (controlPoints->Length() == 3)\ndegree = 2;\n\ncout TColgp_Array1OfPnt A_Poles(1, controlPoints->Length());\nTColStd_Array1OfReal A_Weights(1, controlPoints->Length());\nTColStd_Array1OfReal A_Knots(1, controlPoints->Length()+degree+1-2*(degree-1));\nTColStd_Array1OfInteger A_Mult(1, controlPoints->Length()+degree+1-2*(degree-1));\n\nfor (i=1;iLength();i++)\n{\nA_Poles.SetValue(i,controlPoints->Value(i));\nA_Weights.SetValue(i,1.0);\n}\nfor (i=1; i {\nA_Knots.SetValue(i,(double)(i-1)/(A_Knots.Length()-1));\nif(i==1 || i==A_Knots.Length())\n{\nA_Mult.SetValue(i, degree);\n}\nelse\n{\nA_Mult.SetValue(i,1);\n}\ncout }\n\nreturn new Geom_BSplineCurve(A_Poles, A_Weights, A_Knots, A_Mult, degree, 0);",
null,
"Here the figure with the curve (plotted inbetween curve.FirstParameter() and curve.LastParameter()) and the control polygone.\n\nAttachments:",
null,
"SplineByControlPoints.png",
null,
"This is the expected behaviour when constructing a new nurbs curve. What you are looking for is probably Geom_Interpolate",
null,
"Jelle, thanks for helping! I actually want to use both methods of curve construction:\n\na) Array is taken as CONTROL POINTS: In this case I have the problem that the resulting curve is too short in case I use curve.FirstParameter() and curve.LastParameter() to get points. But, as to see from the latest figure, the curve looks nice!\n\nb) Array is to be INTERPOLATED: The resulting curve (see the latest figure) looks somewhat strange, especially at the second points. I used the following code:\n\nTColgp_HArray1OfPnt *array = new TColgp_HArray1OfPnt(1, arrayIn->Length()); // sizing harray\n\nfor (i=1; i<=arrayIn->Length(); i++)\narray->SetValue(i , arrayIn->Value(i));\n\nGeomAPI_Interpolate anInterpolation(array, /* Periodicity */Standard_False, /* Precision */ Precision::Approximation());\nanInterpolation.Perform();\nreturn anInterpolation.Curve();\n\nAny ideas for my problems?\n\nAttachments:",
null,
"Curves.png",
null,
"Sure thing.\nWith Geom_Interpolate you can set tangent directions. If you do that for the first and last vertices you should be able to get a better result.\nFinally, what happens if you go with the 1st method and use a curve of degree 2?\n\n-jelle",
null,
"Ulrich,\n\nFor bspline curve to start and end in the first and last control points, you need to set multiplicity of first and last knots to degree+1, not degree as you do in the code snapshot above.\n\nAndrey"
] | [
null,
"https://dev.opencascade.org/sites/default/files/images/userpic_default.png",
null,
"https://dev.opencascade.org/modules/file/icons/image-x-generic.png",
null,
"https://dev.opencascade.org/sites/default/files/images/userpic_default.png",
null,
"https://dev.opencascade.org/sites/default/files/images/userpic_default.png",
null,
"https://dev.opencascade.org/modules/file/icons/image-x-generic.png",
null,
"https://dev.opencascade.org/sites/default/files/images/userpic_default.png",
null,
"https://dev.opencascade.org/sites/default/files/images/userpic_default.png",
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.81837827,"math_prob":0.95019025,"size":3177,"snap":"2021-04-2021-17","text_gpt3_token_len":886,"char_repetition_ratio":0.14024583,"word_repetition_ratio":0.0,"special_character_ratio":0.28360087,"punctuation_ratio":0.17540984,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9939273,"pos_list":[0,1,2,3,4,5,6,7,8,9,10,11,12,13,14],"im_url_duplicate_count":[null,null,null,null,null,null,null,null,null,null,null,null,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-04-22T02:55:44Z\",\"WARC-Record-ID\":\"<urn:uuid:87883b1b-ddb2-4dea-929d-fecb67fbe79d>\",\"Content-Length\":\"33277\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:7116d85b-b23e-40d6-b6f3-003276cc3c6b>\",\"WARC-Concurrent-To\":\"<urn:uuid:9dbad1dc-e6e2-454a-853f-6a058bc61cc3>\",\"WARC-IP-Address\":\"188.165.114.136\",\"WARC-Target-URI\":\"https://dev.opencascade.org/content/curve-control-polygone\",\"WARC-Payload-Digest\":\"sha1:S7LXXVPBVGBCSEVKQLHKHJDZZXPT5WYL\",\"WARC-Block-Digest\":\"sha1:MXWLBIX6BD5SBXROHSRZQBJD6D2YTLVL\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-17/CC-MAIN-2021-17_segments_1618039560245.87_warc_CC-MAIN-20210422013104-20210422043104-00487.warc.gz\"}"} |
https://www.calculatoratoz.com/en/diagonal-of-decagon-across-five-sides-given-diagonal-across-two-sides-calculator/Calc-27880 | [
"## Credits\n\nSt Joseph's College (SJC), Bengaluru\nMona Gladys has created this Calculator and 1000+ more calculators!\nWalchand College of Engineering (WCE), Sangli\nShweta Patil has verified this Calculator and 1000+ more calculators!\n\n## Diagonal of Decagon across five sides given diagonal across two sides Solution\n\nSTEP 0: Pre-Calculation Summary\nFormula Used\ndiagonal_across_5_sides = ((2*Diagonal across two sides)/(sqrt(10+2*sqrt(5))))*(1+sqrt(5))\nd5 = ((2*d2)/(sqrt(10+2*sqrt(5))))*(1+sqrt(5))\nThis formula uses 1 Functions, 1 Variables\nFunctions Used\nsqrt - Squre root function, sqrt(Number)\nVariables Used\nDiagonal across two sides - Diagonal across two sides is a straight line joining two opposite sides which is across two sides of the given polygon. (Measured in Meter)\nSTEP 1: Convert Input(s) to Base Unit\nDiagonal across two sides: 5 Meter --> 5 Meter No Conversion Required\nSTEP 2: Evaluate Formula\nSubstituting Input Values in Formula\nd5 = ((2*d2)/(sqrt(10+2*sqrt(5))))*(1+sqrt(5)) --> ((2*5)/(sqrt(10+2*sqrt(5))))*(1+sqrt(5))\nEvaluating ... ...\nd5 = 8.5065080835204\nSTEP 3: Convert Result to Output's Unit\n8.5065080835204 Meter --> No Conversion Required\n8.5065080835204 Meter <-- Diagonal across five sides\n(Calculation completed in 00.016 seconds)\n\n## < 9 Diagonal of Decagon across five sides Calculators\n\nDiagonal of Decagon across five sides given area\ndiagonal_across_5_sides = (sqrt((2*Area)/(5*sqrt(5+2*sqrt(5)))))*(1+sqrt(5)) Go\nDiagonal of Decagon across five sides given diagonal across three sides\ndiagonal_across_5_sides = ((2*Diagonal across three sides)/(sqrt(14+6*sqrt(5))))*(1+sqrt(5)) Go\nDiagonal of Decagon across five sides given diagonal across two sides\ndiagonal_across_5_sides = ((2*Diagonal across two sides)/(sqrt(10+2*sqrt(5))))*(1+sqrt(5)) Go\nDiagonal of Decagon across five sides given diagonal across four sides\ndiagonal_across_5_sides = (Diagonal across four sides/(sqrt(5+2*sqrt(5))))*(1+sqrt(5)) Go\nDiagonal of Decagon across five sides given inradius\nDiagonal of Decagon across five sides given height\ndiagonal_across_5_sides = (Height/((sqrt(5+2*sqrt(5)))))*(1+sqrt(5)) Go\nDiagonal of Decagon across five sides given circumradius\nDiagonal of Decagon across five sides given perimeter\ndiagonal_across_5_sides = (Perimeter/10)*(1+sqrt(5)) Go\nDiagonal of Decagon across five sides\ndiagonal_across_5_sides = Side*(1+sqrt(5)) Go\n\n### Diagonal of Decagon across five sides given diagonal across two sides Formula\n\ndiagonal_across_5_sides = ((2*Diagonal across two sides)/(sqrt(10+2*sqrt(5))))*(1+sqrt(5))\nd5 = ((2*d2)/(sqrt(10+2*sqrt(5))))*(1+sqrt(5))\n\n## What is a decagon?\n\nDecagon is a polygon with ten sides and ten vertices. A decagon, like any other polygon, can be either convex or concave, as illustrated in the next figure. A convex decagon has none of its interior angles greater than 180°. To the contrary, a concave decagon (or polygon) has one or more of its interior angles greater than 180°. A decagon is called regular when its sides are equal and also its interior angles are equal.\n\n## How to Calculate Diagonal of Decagon across five sides given diagonal across two sides?\n\nDiagonal of Decagon across five sides given diagonal across two sides calculator uses diagonal_across_5_sides = ((2*Diagonal across two sides)/(sqrt(10+2*sqrt(5))))*(1+sqrt(5)) to calculate the Diagonal across five sides, The Diagonal of decagon across five sides given diagonal across two sides formula is defined as a straight line joining two opposite corners of the decagon , where diagonal = diagonal across two sides of decagon and diagonal across five sides of decagon. Diagonal across five sides and is denoted by d5 symbol.\n\nHow to calculate Diagonal of Decagon across five sides given diagonal across two sides using this online calculator? To use this online calculator for Diagonal of Decagon across five sides given diagonal across two sides, enter Diagonal across two sides (d2) and hit the calculate button. Here is how the Diagonal of Decagon across five sides given diagonal across two sides calculation can be explained with given input values -> 8.506508 = ((2*5)/(sqrt(10+2*sqrt(5))))*(1+sqrt(5)).\n\n### FAQ\n\nWhat is Diagonal of Decagon across five sides given diagonal across two sides?\nThe Diagonal of decagon across five sides given diagonal across two sides formula is defined as a straight line joining two opposite corners of the decagon , where diagonal = diagonal across two sides of decagon and diagonal across five sides of decagon and is represented as d5 = ((2*d2)/(sqrt(10+2*sqrt(5))))*(1+sqrt(5)) or diagonal_across_5_sides = ((2*Diagonal across two sides)/(sqrt(10+2*sqrt(5))))*(1+sqrt(5)). Diagonal across two sides is a straight line joining two opposite sides which is across two sides of the given polygon.\nHow to calculate Diagonal of Decagon across five sides given diagonal across two sides?\nThe Diagonal of decagon across five sides given diagonal across two sides formula is defined as a straight line joining two opposite corners of the decagon , where diagonal = diagonal across two sides of decagon and diagonal across five sides of decagon is calculated using diagonal_across_5_sides = ((2*Diagonal across two sides)/(sqrt(10+2*sqrt(5))))*(1+sqrt(5)). To calculate Diagonal of Decagon across five sides given diagonal across two sides, you need Diagonal across two sides (d2). With our tool, you need to enter the respective value for Diagonal across two sides and hit the calculate button. You can also select the units (if any) for Input(s) and the Output as well.\nHow many ways are there to calculate Diagonal across five sides?\nIn this formula, Diagonal across five sides uses Diagonal across two sides. We can use 9 other way(s) to calculate the same, which is/are as follows -\n• diagonal_across_5_sides = Side*(1+sqrt(5))\n• diagonal_across_5_sides = (Diagonal across four sides/(sqrt(5+2*sqrt(5))))*(1+sqrt(5))\n• diagonal_across_5_sides = ((2*Diagonal across three sides)/(sqrt(14+6*sqrt(5))))*(1+sqrt(5))\n• diagonal_across_5_sides = ((2*Diagonal across two sides)/(sqrt(10+2*sqrt(5))))*(1+sqrt(5))\n• diagonal_across_5_sides = (Perimeter/10)*(1+sqrt(5))\n• diagonal_across_5_sides = (sqrt((2*Area)/(5*sqrt(5+2*sqrt(5)))))*(1+sqrt(5))\n• diagonal_across_5_sides = (Height/((sqrt(5+2*sqrt(5)))))*(1+sqrt(5))",
null,
""
] | [
null,
"https://www.calculatoratoz.com/Images/share.png",
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.91923076,"math_prob":0.998733,"size":1547,"snap":"2021-43-2021-49","text_gpt3_token_len":354,"char_repetition_ratio":0.24821776,"word_repetition_ratio":0.17768595,"special_character_ratio":0.22624435,"punctuation_ratio":0.071428575,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99999475,"pos_list":[0,1,2],"im_url_duplicate_count":[null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-12-05T08:50:16Z\",\"WARC-Record-ID\":\"<urn:uuid:93f423e4-21f0-43bf-9783-aabd61d2f2cc>\",\"Content-Length\":\"116691\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:0d2185a8-dce5-42f3-b045-20b785b39bf4>\",\"WARC-Concurrent-To\":\"<urn:uuid:69b0cf13-6c26-46e9-bb6c-e2e41bb18e47>\",\"WARC-IP-Address\":\"67.43.15.151\",\"WARC-Target-URI\":\"https://www.calculatoratoz.com/en/diagonal-of-decagon-across-five-sides-given-diagonal-across-two-sides-calculator/Calc-27880\",\"WARC-Payload-Digest\":\"sha1:S2ZRJGQOAIQDFIXVDVUMQHHIAAQWBLDM\",\"WARC-Block-Digest\":\"sha1:7KMBT4LB3WFMPJQJJRXUFCMA4QX3QTEF\",\"WARC-Identified-Payload-Type\":\"application/xhtml+xml\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-49/CC-MAIN-2021-49_segments_1637964363149.85_warc_CC-MAIN-20211205065810-20211205095810-00277.warc.gz\"}"} |
https://www.saving.org/area-calculator/m/87/sk | [
"#### 87 square miles to square Kilometers\n\nHow many km2 in 87 sq mi?\n\n#### Calculate Area\n\n87 square miles to square Kilometers. How many? What's the calculation? How much? What is 87 square miles in square Kilometers? 87sq mi in km2? How many km2 are there in 87 sq mi? Calculate between square miles and square Kilometers. How big is 87 square miles in km2?"
] | [
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.89001113,"math_prob":0.99631035,"size":268,"snap":"2022-27-2022-33","text_gpt3_token_len":74,"char_repetition_ratio":0.23106061,"word_repetition_ratio":0.044444446,"special_character_ratio":0.26492536,"punctuation_ratio":0.15517241,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99424416,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-08-18T10:52:04Z\",\"WARC-Record-ID\":\"<urn:uuid:0848abdf-cf3d-44f2-82d5-ddbeb427aa86>\",\"Content-Length\":\"7697\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:79149f2d-86f2-4f75-98a6-d030f91a3e67>\",\"WARC-Concurrent-To\":\"<urn:uuid:620d16f3-8d13-468f-8605-02ad974903ca>\",\"WARC-IP-Address\":\"50.17.191.130\",\"WARC-Target-URI\":\"https://www.saving.org/area-calculator/m/87/sk\",\"WARC-Payload-Digest\":\"sha1:GZOWBZPURZW5VHD7BURAYKZMO4V3WVPD\",\"WARC-Block-Digest\":\"sha1:DTSNWT546WXDBWVQZPEDIOTSWPGGWYVL\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-33/CC-MAIN-2022-33_segments_1659882573193.35_warc_CC-MAIN-20220818094131-20220818124131-00609.warc.gz\"}"} |
https://unitchefs.com/meters/feet/1.55/ | [
"# 1.55 Meters to Feet\n\n=\n1.55 Meters =\n5.0853018372703\n(decimal)\n5.0853018372703 x 100\n(scientific notation)\n3,875\n762\n(fraction)\nFeet\n\n## Meters to Feet Conversion Formula\n\n[X] ft = 3.2808398950131 × [Y] m\nwhere [X] is the result in ft and [Y] is the amount of m we want to convert\n\n## 1.55 Meters to Feet Conversion breakdown and explanation\n\n1.55 m to ft conversion result above is displayed in three different forms: as a decimal (which could be rounded), in scientific notation (scientific form, standard index form or standard form in the United Kingdom) and as a fraction (exact result). Every display form has its own advantages and in different situations particular form is more convenient than another. For example usage of scientific notation when working with big numbers is recommended due to easier reading and comprehension. Usage of fractions is recommended when more precision is needed.\n\nIf we want to calculate how many Feet are 1.55 Meters we have to multiply 1.55 by 1250 and divide the product by 381. So for 1.55 we have: (1.55 × 1250) ÷ 381 = 1937.5 ÷ 381 = 5.0853018372703 Feet\n\nSo finally 1.55 m = 5.0853018372703 ft"
] | [
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.8999834,"math_prob":0.98894113,"size":796,"snap":"2019-35-2019-39","text_gpt3_token_len":196,"char_repetition_ratio":0.09217171,"word_repetition_ratio":0.0,"special_character_ratio":0.29522613,"punctuation_ratio":0.11464968,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9964854,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-09-22T11:00:02Z\",\"WARC-Record-ID\":\"<urn:uuid:6ba1de19-76c8-4f57-8789-617c62b06fd9>\",\"Content-Length\":\"16879\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:8855b294-4e02-4b7f-9d91-d85ae252b62d>\",\"WARC-Concurrent-To\":\"<urn:uuid:a7b4a8e6-8658-4e4c-8fc1-ee013673453d>\",\"WARC-IP-Address\":\"104.25.108.8\",\"WARC-Target-URI\":\"https://unitchefs.com/meters/feet/1.55/\",\"WARC-Payload-Digest\":\"sha1:XL5YK3NNWAT37DM76W2N5JINSBMNCE2L\",\"WARC-Block-Digest\":\"sha1:ND6R5TSDSXDOTQVKBUB2AOQ32WZEATQM\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-39/CC-MAIN-2019-39_segments_1568514575484.57_warc_CC-MAIN-20190922094320-20190922120320-00547.warc.gz\"}"} |
https://maharashtraboardsolutions.in/class-6-maths-solutions-chapter-4-practice-set-12/ | [
"# Maharashtra Board Practice Set 12 Class 6 Maths Solutions Chapter 4 Operations on Fractions\n\nBalbharti Maharashtra State Board Class 6 Maths Solutions covers the Std 6 Maths Chapter 4 Operations on Fractions Class 6 Practice Set 12 Answers Solutions.\n\n## 6th Standard Maths Practice Set 12 Answers Chapter 4 Operations on Fractions\n\nQuestion 1.\nMultiply:\ni. $$\\frac{7}{5} \\times \\frac{1}{4}$$\nii. $$\\frac{6}{7} \\times \\frac{2}{5}$$\niii. $$\\frac{5}{9} \\times \\frac{4}{9}$$\niv. $$\\frac{4}{11} \\times \\frac{2}{7}$$\nv. $$\\frac{1}{5} \\times \\frac{7}{2}$$\nvi. $$\\frac{9}{7} \\times \\frac{7}{8}$$\nvii. $$\\frac{5}{6} \\times \\frac{6}{5}$$\nviii. $$\\frac{6}{17} \\times \\frac{3}{2}$$\nSolution:\ni. $$\\frac{7}{5} \\times \\frac{1}{4}$$",
null,
"ii. $$\\frac{6}{7} \\times \\frac{2}{5}$$",
null,
"iii. $$\\frac{5}{9} \\times \\frac{4}{9}$$",
null,
"iv. $$\\frac{4}{11} \\times \\frac{2}{7}$$",
null,
"v. $$\\frac{1}{5} \\times \\frac{7}{2}$$",
null,
"vi. $$\\frac{9}{7} \\times \\frac{7}{8}$$",
null,
"vii. $$\\frac{5}{6} \\times \\frac{6}{5}$$",
null,
"viii. $$\\frac{6}{17} \\times \\frac{3}{2}$$",
null,
"Question 2.\nAshokrao planted bananas on $$\\frac { 2 }{ 7 }$$ of his field of 21 acres. What is the area of the banana plantation?\nSolution:\nArea of banana plantation is $$\\frac { 2 }{ 7 }$$ of 21\n∴ Area of banana plantation",
null,
"∴ Area of banana plantation is 6 acres\n\nQuestion 3.\nOf the total number of soldiers in our army, $$\\frac { 4 }{ 9 }$$ are posted on the northern border and one-third of them on the north-eastern border. If the number of soldiers in the north is 5,40,000, how many are posted in the north-east?\nSolution:\nNumber of soldiers posted on northern border = 5,40,000\nSince, number of soldiers in north-east = one third of the soldiers on northern border\n∴ Number of soldiers in the north-east",
null,
"∴ The number of soldiers in the north-east is 1,80,000."
] | [
null,
"https://www.learncram.com/wp-content/uploads/2019/08/Maharashtra-Board-Class-6-Maths-Solutions-Chapter-4-Operations-on-Fractions-Practice-Set-12-1.png",
null,
"https://www.learncram.com/wp-content/uploads/2019/08/Maharashtra-Board-Class-6-Maths-Solutions-Chapter-4-Operations-on-Fractions-Practice-Set-12-2.png",
null,
"https://www.learncram.com/wp-content/uploads/2019/08/Maharashtra-Board-Class-6-Maths-Solutions-Chapter-4-Operations-on-Fractions-Practice-Set-12-3.png",
null,
"https://www.learncram.com/wp-content/uploads/2019/08/Maharashtra-Board-Class-6-Maths-Solutions-Chapter-4-Operations-on-Fractions-Practice-Set-12-4.png",
null,
"https://www.learncram.com/wp-content/uploads/2019/08/Maharashtra-Board-Class-6-Maths-Solutions-Chapter-4-Operations-on-Fractions-Practice-Set-12-5.png",
null,
"https://www.learncram.com/wp-content/uploads/2019/08/Maharashtra-Board-Class-6-Maths-Solutions-Chapter-4-Operations-on-Fractions-Practice-Set-12-6.png",
null,
"https://www.learncram.com/wp-content/uploads/2019/08/Maharashtra-Board-Class-6-Maths-Solutions-Chapter-4-Operations-on-Fractions-Practice-Set-12-7.png",
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https://snapessays.com/study/question/question/94447575_view.html | [
"Solution In the short run a firm s total costs of producing 100 units\nSolution In the short run a firm s total costs\nSolution In the short run a firm\nthe short run a firm s total costs of producing units\nSolution In the short run a\nfirm s total costs of producing units\nSolution In the short run\nSolution In\n(Solution) - In the short run, a firm's total costs of producing 100 units\n\n Category:- Words: Amount: \\$13 Writer: 0\n\nPaper instructions\n\n\"In the short run, a firm's total costs of producing 100 units of output equal \\$10,000. If it produces one more unit, its total costs will increase to \\$10,150.;a. What is the marginal cost of the 101st unit of output?;b. What is the firm's average total cost of producing 100 units?;c. What is the firm's average total cost of producing 101 units?",
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https://ru.scribd.com/document/114161967/Fluid-Sys-w-Chart | [
"You are on page 1of 7\n\n# Engs 22 Systems\n\nSummer 2004\n\n## LUMPED FLUID SYSTEMS\n\nI. General Introduction and Applications:\nApplications: Fluid delivery systems; plumbing. For water or other fluids, for direct use by people, or for industrial operations, such as in chemical plants. Fluid cooling systems. Fluid (hydraulic) power systems, used in industry, aerospace, and automotive applications. Medical: blood flow, etc. Hydrology; transport of nutrients, and pollutants as well as water in groundwater and in streams, rivers, and lakes.\n\nFluid systems either involve essentially non-compressible liquids in which case they are often called hydraulic systems or compressible gases in which case they are often called pneumatic systems. The terms hydraulic and pneumatic are most common in referring to systems that use pressurized fluid used to provide power, for the purpose of causing motion (actuation). For example, hydraulic systems are used to control flaps on airplane wings as well as apparatus on bulldozers. We will focus primarily on hydraulic systems, which are simpler to analyze than pneumatic systems. In commenting on hydraulic power systems, Woods and Lawrence (Modeling and Simulation of Dynamic Systems, Prentice Hall, 1997) state: The main reason that hydraulic and pneumatic systems are so popular compared to their electro-mechanical counterparts is the power density capability of pumps and actuators. Electromagnetic motors, generators, and actuators are limited by magnetic field saturation and can produce up to about 200 pounds per square inch of actuator. In hydraulic systems, 3000 to 8000 pounds per square inch of actuator is common in aircraft applications, and 1000 pounds per square inch of actuator is common in industrial applications. Therefore, the hydraulic systems required to produce a given force output are much smaller. A second advantage of fluid control systems is that the fluid circulating to and from an actuator removes the heat generated by the actuator that is doing the work. This heat follows the fluid back to a reservoir to be dissipated in a better location than inside the actuator. Electromechanical actuators and motors have limited ability to dissipate heat to the surrounding environment. Heat is the predominate damaging mechanism in electric and electronic systems. Although fluid systems are often taught in terms of hydraulic power systems (just as chemical dynamics is often taught in terms in industrial reactor design), the applications of lumped analysis of fluid systems is much broader (just as the principles of chemical dynamics apply to modeling resource systems). The list at the start of this section hints at some of the wide range of other applications of lumped fluid system analysis.\n\nII.\n\n## State Variables, Components, and Accounting Equations\n\nA. State Variables\nThe following variables are important in characterization and analysis of fluid systems: q = volumetric flow (m3/s or other units) v = volume (m3)(e.g., volume of fluid contained in a vessel.) h = height of fluid in a vessel (m)\n\n## Fluid Systems Analysis\n\nPage 1\n\nEngs 22 Systems\n\nSummer 2004\n\np = pressure (N/m2) Pressure is the force per unit area exerted by a fluid. Just as in electrical systems we are typically only interested in voltage differences, in fluid systems we are typically interested in pressure differences. Thus, it is usually possible to interchange absolute pressure and gauge pressure, pg, which is defined as the pressure relative to atmospheric pressure (Pa = 1.01 x 105 N/m2).\n\nB. Elements\nThe primary elements in fluid systems are resistance, capacitance and intertance, which are analogous to electrical resistance, capacitance, and inductance, respectively. Resistance. In electrical systems, the linear approximation v = iR, is usually so close to exact that it can be considered a physical law (Ohms law). In fluid systems, however, linear behavior of fluid resistance, p = qR is rare, and most practical valves, pipes, etc. behave nonlinearly, following a law such as q = K p . The detailed analysis of which model is appropriate for which situation is a topic for a fluid dynamics course, not a systems course. Our goal is to figure out how to model a system given models of the individual elements. For using either the linear fluid resistance relationship p = qR or the nonlinear relationship q = K p , the key thing to remember is that the pressure of interest is the pressure difference across the element of interest. A pipe with resistance R between two chambers each with 100 kPa pressure in them will not have any flow in it, even though the pressure in the pipe is 100 kPa. But if one has 100 kPa and the other 99 kPa, there will be a pressure difference of 1 kPa, and there will be flow (from the higher pressure to the lower pressure). If there were 1 kPa in one and 2 kPa in the other, the flow would be the same as with 99 kPa and 100 kPa. Thus, when setting up an analysis of the system, it is important to label the pressures of interest on the diagram, and be sure to write equations using the only the defined variables. Capacitance. An electrical capacitor has increasing charge storage with increasing voltage. A fluid capacitor has increasing fluid storage with increasing pressure, and is defined by the equation\n\nVL = Cp .\n\n\n\nExample of a physical device that provides fluid capacitance: an open-top tank as shown in Fig. 1.\nFigure 1. An open reservoir fluid capacitor The force at the bottom of a reservoir due to the weight of the fluid is mg = VLg where g is the earths gravitational constant, Top of reservoir and m is the mass of fluid, equal to VL where is the density open to the atmosphere of the fluid, and VL is the volume of fluid in the tank, assuming the side walls of the tank are vertical. The total pressure on the inside of the bottom tank wall is a result of this pressure plus the Pa atmopheric pressure Pa. However, we are more likely to be interested in the pressure difference between inside and outside, Liquid at the bottom of the tank where the pipe connects, as labeled p h VL in Fig.1. The pressure difference is just\n\np=\n\ngVL\nA\n\n= gh\n\n\n\n## + P Liquid flow into the capacitor q Page 2\n\nwhere A is the area of the base of the tank, which must be the\nFluid Systems Analysis\n\nEngs 22 Systems\n\nSummer 2004\n\nsame all the way up since we assumed the side walls were vertical. What we see is a proportionality between pressure and volume, which is the defining characteristic of a fluid capacitor. We can write it in a more standard form as:\n\nA VL = p g\n\n\n\nThis equation is the same as , the definition of fluid capacitance, with C = g . Note that if the walls were not vertical, the volume/pressure relationship would not be as simple as , and we would have a nonlinear fluid capacitance. When the fluid capacitance C is large, corresponding to a tank with a large area, a large increase in volume is accompanied by a small increase in pressure. The units of capacitance are volume/pressure, or m3/(N/m2) = m5/N. The open top tank is one way to achieve fluid capacitance. Others include a spring-loaded chamber, or the same thing with an air spring, as shown in Fig. 2. Calculations of the capacitance achieved from these devices is addressed in the linearization and nonlinear fluid elements handout.\n\nAir space\n\nPiston\n\nFluid\n\nFluid\n\n## Fig. 2. Two other types of fluid capacitors.\n\nAs with electrical capacitance, differentiation of provides a very useful differential equation that can be directly applied to set up a state-variable equation, in this case using pressure as a state variable: dp/dt = q/C Note that definitions of the polarity of each variable are essential; this equation is correct for the polarities shown in Fig. 1, but would be incorrect for flow defined in the opposite direction.\n\n## Fluid Systems Analysis\n\nPage 3\n\nEngs 22 Systems\n\nSummer 2004\n\nInertance. Inertance arises as the result of the momentum of a flowing fluid. Consider an incompressible fluid element flowing down a pipe with constant cross sectional area A with as shown in Figure 3. Treating this as a mass subject to two forces p1A and p2A, we find: m v = A(p1 p2) = LA v\n\np1 A\n\np2 A\n\nL 1 2\n\n\n\np =\n\nL\nA\n\n& q\n\n## Figure 4. Variables for intertance.\n\n\n\nThis equation is a lot like the equation for an inductor, di v = L dt . Considering this analogy, we define\n\n& intertance I = L/A, and can be written p = Iq . Here, the p must be understood to be the pressure difference across the inertance, as defined in Fig. 4.\n\nPumps. Pumps act as sources, analogous to voltage or current sources in electrical systems. Just as real power supplies dont quite maintain constant voltage, real pumps dont maintain constant pressure or flow. The deviations tend to be large for pumps. The detailed modeling of pumps is beyond the scope of this courseit is the study of an element, not of a system. However, it may be useful to know that centrifugal pumps (which use centrifugal force to generate pressure) can maintain roughly constant pressure at low flow rates, and so act approximately as pressure sources. However, their pressure drops rapidly and nonlinearly with increasing flow. Positive displacement pumps such as piston pumps deliver a fixed amount of fluid each cycle, and so behave like flow sources.\n\nC. Accounting Equations.\nMass Balance. At any given node, where pipes connect, mass balance requires that flow in equals flow out. If flows are all defined inwards, this can be written as the sum of flows equal to zero. qin1 + qin2 + = qout1 + qout2 + or qin = 0 \n\nIn either case, this is directly analogous to Kirchoffs current law. Kirchoffs Pressure Law As discussed above, both fluid resistor behavior and intertance effects depend on the pressure difference across an element. Thus we are often interested in pressure differences rather than absolute pressures. The most commonly used pressure measurement, gauge pressure, is also defined as a pressure difference: the difference between measured pressure and atmospheric pressure.\nFluid Systems Analysis Page 4\n\nEngs 22 Systems\n\nSummer 2004\n\nIn electrical systems, we saw that Kirchoffs voltage law was not a physical law related to pressure, but rather is just a consequence of defining voltage as a difference. Thus, if we discuss pressures as differences, we have an analogous law: the sum of pressure differences around a loop must be zero. Note that the loop may be any path through space it need not follow a pipe. Example: A tank emptying through a pipe. Consider the tank shown at right emptying through a pipe. With the pressures defined as shown, we can follow the dotted line and write that the total pressure around that loop is equal to zero. Starting in the air, we go into the tank (across the wall) and the pressure increases by pc. Next, we follow the dotted line into the pipe, and along the pipe back into the air. Thus, the total of the pressure differences around that loop is pC - pP = 0 \nC Figure 5. An open reservoir fluid capacitor emptying through a pipe Top of reservoir open to the atmosphere\n\nPa Liquid VC + qC + qP pp pC -\n\nThis is assuming that the pressure drop in the fittings at the bottom of the tank is negligible. If not, we could include them in the resistance of the pipe and put the plus sign for the pressure drop across the pipe up in the tank. From mass balance we have qC = -qP \n\nIf we assume that the pipe has negligible intertance, we can consider just its resistance, and we have the two element laws: dpC/dt = qC/C qp = pP/RP , \n\nThats all the equations we need. The rest of the work is just substituting to get the RHS of the capacitor equation in terms of the state variable for the capacitor, pC. We obtain: dpC/dt = - pC/( RP C) Head We discussed the effect of fluid height on pressure in the context of an open tank as a capacitor. Often, however, different elements may be located at different elevations, and these differences may be large compared to the elevation difference between the top and bottom of a tank. These height differences also affect pressure. These pressures are termed head, a term that may refer to either the height or the resulting pressure. Head may be an unavoidable consequence of the locations of system components, or it may be deliberately employed to achieve pressure, as in a water tank located atop a tower. \n\n## Fluid Systems Analysis\n\nPage 5\n\nEngs 22 Systems\n\nSummer 2004\n\nC Figure 6. An open reservoir fluid capacitor emptying through a pipe Head complicates pressure calculations, and it usually must be considered when writing KPL Top of reservoir open to the (Kirchoffs pressure law) equations. In principal, atmosphere KPL is not changed. The sum of pressure differences is still zero around a loop. Consider the tank in Figure 6. Simply by definitions of the Pa two pressures, the same equation must hold.\n\npC - pP = 0\n\n\n\nLiquid VC + qC + qP pp pC -\n\nThus, it is tempting to conclude that the system equations are identical, and the head has no effect. However, what is different is that the pipe can no longer be simply treated as a fluid resistance, as it also includes some head. Writing the equation for that resistance with head and getting the signs right is tricky. An easier way to approach the problem is to separate the head and the resistance, and draw the model as in Fig. 7. Now, with the signs defined in Fig. 7, it is easy to write the correct pressure equation: p C + ph - p R = 0 \n\n## Perhaps less obvious is the sign on the equation: ph = ghh \n\nThe sign here is can be chosen by simply considering the case in which there is nothing else going on but the headno flow. The pressure at the + sign for ph will clearly be larger than the pressure at the sign, as a result of the head. Hence the signs as marked in Fig. 7 and in . Now we can derive a complete system model, from , , , , and a modified version of , which only needs a minor modification to substitute pressure across the resistance effect of the pipe, pR where had total pressure across the pipe pP. qp = pR/RP \n\n## Pa Liquid VC + qC Zero resistance, zero intertance fictional pipe. pC -\n\nhC\n\nhh\n\nph\n\n+ +\n\nqP pR\n\nSubstituting, we obtain the new system model: dpC/dt = -(ghh + pC)/( RP C) \n\nBoth and can be solved by any of the methods we have developed in the course, but since they are first-order linear equations, the first-order solution method would be a good choice. Fluid systems often have nonlinear elements and must be solved numerically.\n\n## Fluid Systems Analysis\n\nPage 6\n\nEngs 22 Systems\n\nFluid Elements\nSymbol Description A through variable. What goes in comes out (except for a capacitor). An across variabledefined as the difference of this quantity across the element. Resistance, due to pipe or constriction. In real life, it is often nonlinear. Dissipation Units m3/s (`/s) Pa = N/m2 R = p/q = Pa/( m3/s) Can also write: Pas/m3 = Ns/m5 W Calculation\n\nSummer, 2004\n\n## Electrical Analogy i (current) v (voltage)\n\nDiscussed separately\n\nv = iR i2R = v2/R\n\nThis is a valve symbol; sometimes used without the handle for a non-variable resistor. Often just a pipe is shown.\n\ndq p=I dt\n\n## Pa/(m3/s2) = Pas2/m3 kg/m4 = Ns2/m5\n\n= `/A where ` and A are length and crosssectional area of the pipe, and is density of the fluid.\n\nv=L\n\ndi dt\n\n## E = Iq2 P(s) = Q(s)Is\n\ndp dt (one end usually grounded) q=C\n\nEnergy Impedance\n\nFluid capacitor, such as a tank or pressure chamber. Sometimes called an accumulator. Energy Impedance\n\nJ 3 Pas/m = Ns/m5 (same as resistance) (m3/s)/(Pa/s) = m3/Pa = m5/N J Pas/m = Ns/m5 (same as resistance)\n3\n\nE = Li2 V(s)=I(s)Ls C = A/( g) For an open tank; pressure chambers discussed separately.\ni=C dv dt\n\nE = C p2\nP( s) = Q( s) 1 sC\n\nE = Cv2\nV (s) = I (s) 1 sC\n\nFluid Elements\n\nPage 1"
] | [
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.93160164,"math_prob":0.9811638,"size":14671,"snap":"2019-35-2019-39","text_gpt3_token_len":3395,"char_repetition_ratio":0.15149656,"word_repetition_ratio":0.022674866,"special_character_ratio":0.22684208,"punctuation_ratio":0.09628852,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9915178,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-08-20T12:27:56Z\",\"WARC-Record-ID\":\"<urn:uuid:fa4a81e9-582d-4770-95be-952de89b2d05>\",\"Content-Length\":\"258800\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:08aa1f6d-3cf8-4dd9-bc0b-c9cd6429cabf>\",\"WARC-Concurrent-To\":\"<urn:uuid:ae2f34c5-be41-4085-bb21-4109322bab25>\",\"WARC-IP-Address\":\"151.101.202.152\",\"WARC-Target-URI\":\"https://ru.scribd.com/document/114161967/Fluid-Sys-w-Chart\",\"WARC-Payload-Digest\":\"sha1:KD2AKYO6ZWCKNOIYAOLWRHSAT6VS7EIG\",\"WARC-Block-Digest\":\"sha1:4HUI46CNMYGJ6VRJDRZUERVGRTASSE7D\",\"WARC-Identified-Payload-Type\":\"application/xhtml+xml\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-35/CC-MAIN-2019-35_segments_1566027315329.55_warc_CC-MAIN-20190820113425-20190820135425-00044.warc.gz\"}"} |
https://cci.lbl.gov/docs/cctbx/script_1/ | [
"## Create a random model and calculate its density\n\nThis example script shows how to generate a small centrosymmetric unit cell in P-1 with 15 atoms, and how to calculate phases and amplitudes to reconstruct the electron density.\n\n### Required imports\n\nThis script requires the following imports:\n\nfrom __future__ import absolute_import, division, print_function\nimport iotbx.pdb\nimport iotbx.mrcfile\nimport mmtbx.model\nimport mmtbx.real_space\nfrom scitbx.array_family import flex\nfrom cctbx.development import random_structure\nfrom cctbx import sgtbx\nfrom cctbx import maptbx\n\n\n### Create a random structure and save as PDB\n\nUse the random_structure() method from cctbx.development to generate a random structure that has:\n\n• spcae group P-1\n• unit cell dimensions a=10, b=20, c=30, alpha=70, beta=80, gamma=120\n• 15 carbon atoms\n# Create random structure\nxrs = random_structure.xray_structure(\nspace_group_info = sgtbx.space_group_info(\"P-1\"),\nelements = [\"C\"]*15,\nunit_cell = (10, 20, 30, 50, 60, 80))\n\n\nCreate a model object from this structure:\n\n# Create model object\nmodel = mmtbx.model.manager.from_sites_cart(\nsites_cart = xrs.sites_cart(),\ncrystal_symmetry = xrs.crystal_symmetry(),\nresname = 'DUM')\n\n\nfrom_sites_cart() is a convenience function to create a model object from a list of cartesian coordinates. The default is to use atom name CA, scatterer C, occ 1, b_iso 30, resname GLY, residue numbers starting with 1. Here, the residue name is set to 'DUM', which stands for dummy residue. Note that even if we specify different elements in random_structure(), we will get CA atoms, unless the from_sites_cart() method is supplied with a list of atom names and scatterer names.\n\nSave the model as PDB file:\n\n# Write it into PDB file\nwith open(\"model.pdb\",\"w\") as fo:\nfo.write(model.model_as_pdb())\n\n\nHere is an example for a model that was created using this procedure (your model will look different, as the procedue will place the atoms at random places):",
null,
"### Create a map from a PDB file\n\nCreate a model object from a file.\n\n # Read the model file\npdb_inp = iotbx.pdb.input(file_name = \"model.pdb\")\nmodel = mmtbx.model.manager(model_input = pdb_inp)\nxrs = model.get_xray_structure()\n\n\nCalculate structure factors from the model using a high resolution limit of 2 Angstrom. This will give a Fourier map:\n\n# Calculate structure factors at given resolution.\nf_calc = xrs.structure_factors(d_min = 2.0).f_calc()\n\n\nFor book-keeping: save the Fourier map as MTZ file:\n\n# Write them down as MTZ file\nmtz_dataset = f_calc.as_mtz_dataset(column_root_label=\"F-calc\")\nmtz_object = mtz_dataset.mtz_object()\nmtz_object.write(file_name = \"f_calc.mtz\")\n\n\nPerform a fast-Fourier-transform (FFT) to convert the structure factors into a real map:\n\n# Convert Fcalc into real map (just do FFT)\nfft_map = f_calc.fft_map(resolution_factor=1./4)\nfft_map.apply_sigma_scaling()\n\n\nWrite the map data into an MRC file:\n\n# Write real Fourier map into MRC file\niotbx.mrcfile.write_ccp4_map(\nfile_name = \"fourier_map.mrc\",\nunit_cell = f_calc.unit_cell(),\nspace_group = f_calc.crystal_symmetry().space_group(),\nmap_data = map_data.as_double(),\nlabels = flex.std_string([\"Some text\"]))\n\n\nThe figure below shows the density overlayed to the model. Note that there are density peaks that are (seemingly) without model. This is because the density was calculated for the unit cell, which is P-1 (centrosymmetric). The spheres only represent the atoms in the asymmetric unit.",
null,
"Another option is to calculate the exact map. This requires first the computation of the grid (crystal_gridding).\n\n# Calculate exact map and write it down\ncrystal_gridding = maptbx.crystal_gridding(\nunit_cell = xrs.unit_cell(),\nspace_group_info = xrs.space_group_info(),\nsymmetry_flags = maptbx.use_space_group_symmetry,\nstep = 0.3)\nm = mmtbx.real_space.sampled_model_density(\nxray_structure = xrs,\nn_real = crystal_gridding.n_real())\nmap_data = m.data()\n\n\nSave the exact map in the file exact_map.mrc\n\niotbx.mrcfile.write_ccp4_map(\nfile_name = \"exact_map.mrc\",\nunit_cell = f_calc.unit_cell(),\nspace_group = f_calc.crystal_symmetry().space_group(),\nmap_data = map_data.as_double(),\nlabels = flex.std_string([\"Some text\"]))"
] | [
null,
"https://cci.lbl.gov/docs_static/cctbx/img/script_1_fig1.png",
null,
"https://cci.lbl.gov/docs_static/cctbx/img/script_1_fig2.png",
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.6419714,"math_prob":0.972644,"size":3921,"snap":"2023-40-2023-50","text_gpt3_token_len":995,"char_repetition_ratio":0.11743681,"word_repetition_ratio":0.043137256,"special_character_ratio":0.24713084,"punctuation_ratio":0.18086697,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99529743,"pos_list":[0,1,2,3,4],"im_url_duplicate_count":[null,2,null,2,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-12-10T09:41:48Z\",\"WARC-Record-ID\":\"<urn:uuid:4fd49587-c4c5-4f55-b506-e338806b8c96>\",\"Content-Length\":\"20943\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:dcc666a5-65b1-4343-b6aa-9b77b5272439>\",\"WARC-Concurrent-To\":\"<urn:uuid:6b7cbae1-eaae-49e5-b32a-c76116258c04>\",\"WARC-IP-Address\":\"172.64.154.72\",\"WARC-Target-URI\":\"https://cci.lbl.gov/docs/cctbx/script_1/\",\"WARC-Payload-Digest\":\"sha1:EO7BZQD3NY3XLZXWUMZ2TBP2ALO3KCW4\",\"WARC-Block-Digest\":\"sha1:GKWW5F4T7354QBUPKFIGIR76AZ4SGW3J\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-50/CC-MAIN-2023-50_segments_1700679101779.95_warc_CC-MAIN-20231210092457-20231210122457-00391.warc.gz\"}"} |
https://spaces.ac.cn/archives/1765 | [
"\\begin{aligned} a_{11}x_{1}+a_{12}x_{2}+a_{13}x_{3}=b_{1}\\\\ a_{21}x_{1}+a_{22}x_{2}+a_{23}x_{3}=b_{2}\\\\ a_{31}x_{1}+a_{32}x_{2}+a_{33}x_{3}=b_{3} \\end{aligned}\n\n$$\\begin{pmatrix} a_{11} & a_{12} & a_{13} \\\\ a_{21} & a_{22} & a_{23} \\\\ a_{31} & a_{32} & a_{33} \\end{pmatrix}$$\n\n$$Ax=b$$\n\n$$(AB)x=A(Bx)$$\n\n$$A= \\begin{pmatrix} a & b\\\\ c & d \\end{pmatrix},\\quad B=\\begin{pmatrix} e & f\\\\ g & h \\end{pmatrix},\\quad AB=\\begin{pmatrix} p & q\\\\ r & s \\end{pmatrix}$$\n\n$$[px+qy,rx+sy]^{\\top}$$\n\n$$\\frac{1}{1-x}=1+x+x^{2}+x^{3}+...\\approx 1+x$$\n\n$$(I-A)^{-1}=I+A+A^{2}+A^{3}+...\\approx I+A$$\n\n$$\\sqrt{1+x}\\approx 1+\\frac{1}{2}x$$"
] | [
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] | {"ft_lang_label":"__label__zh","ft_lang_prob":0.95586026,"math_prob":1.00001,"size":4444,"snap":"2020-45-2020-50","text_gpt3_token_len":4264,"char_repetition_ratio":0.06081081,"word_repetition_ratio":0.0,"special_character_ratio":0.2770027,"punctuation_ratio":0.105797105,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":1.0000086,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-10-28T02:55:22Z\",\"WARC-Record-ID\":\"<urn:uuid:0de20fb9-7690-4d72-b8c5-17419d9f174d>\",\"Content-Length\":\"78056\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:666e1606-d401-4159-b95a-924f4926957d>\",\"WARC-Concurrent-To\":\"<urn:uuid:82e5d45a-76b2-45d8-bc65-4afeda01604d>\",\"WARC-IP-Address\":\"39.108.122.106\",\"WARC-Target-URI\":\"https://spaces.ac.cn/archives/1765\",\"WARC-Payload-Digest\":\"sha1:5GVWXYNSCSQY4BD7JLIFWTPSS3QBCKNL\",\"WARC-Block-Digest\":\"sha1:LURY3JHVJ2TRJPEVK77DF4ZXZU4YOC32\",\"WARC-Identified-Payload-Type\":\"application/xhtml+xml\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-45/CC-MAIN-2020-45_segments_1603107896048.53_warc_CC-MAIN-20201028014458-20201028044458-00202.warc.gz\"}"} |
https://www.chemistrylearner.com/henderson-hasselbalch-equation.html | [
"# Chemistry Learner\n\nHome / Physical chemistry / Henderson-Hasselbalch Equation\n\n# Henderson-Hasselbalch Equation\n\n## What is the Henderson-Hasselbalch Equation\n\nThe Henderson-Hasselbalch equation is an equation used in chemistry and biochemistry to determine the pH value of a buffer solution during the dissociation of a weak acid. A weak acid does not dissociate completely. A property of the buffer solution is that the mixture can resist changes in pH when a small amount of acid is added [1-5].\n\n## How to Use Henderson Hasselbalch Equation\n\nThe Henderson-Hasselbalch equation is used to determine the amount of acid and conjugate base needed to make a buffer solution of a certain pH. Therefore, it is often used to perform the calculations necessary to prepare buffers for use in the laboratory or other applications.\n\nGeneral Formula of the Henderson-Hasselbalch Equation\n\nA simple buffer solution consists of a weak acid solution and a salt of the acid’s conjugate base. The equation is given by,\n\npH = pKa + log10[conjugate base/weak acid]\n\nwhere,\n\npH: pH value\n\nKa: acid dissociation constant\n\n[conjugate base]: Concentration of the conjugate base in moles per liter\n\n[weak acid]: Concentration of the acid in moles per liter\n\nAmerican chemist Lawrence Joseph Henderson first derived the equation in 1908 and, later, by Danish chemist Karl Albert Hasselbalch in 1917.\n\n## Derivation of Henderson-Hasselbalch Equation\n\nThe Henderson-Hasselbalch equation can be derived from the equilibrium constant expression for the dissociation of a weak acid. Consider the dissociation of the acid HA in water H2O [2-6].\n\nHA + H2O ⇄ A + H3O+\n\nThe acid dissociation constant is given by,\n\nKa = [A] [H3O+]/[HA]\n\nor, Ka[HA]/ [A] = [H3O+]\n\nor, [H3O+] = Ka[HA]/ [A]\n\nor, log10 [H3O+] = log10 (Ka[HA]/ [A]) = log10 Ka + log10 ([HA]/[A])\n\nor, – log10 [H3O+] = – log10 Ka – log10 ([HA]/[A])\n\nor, pH = pKa – log10 ([HA]/[A])\n\nor, pH = pKa + log10 ([A]/[HA])\n\nFrom the above equation, the following points can be noted.\n\n1. The value of [A]/[HA] is dependent on the values of the pH and pKa. When pH < pKa, [A]/[HA] < 1. When pH > pKa, [A]/[HA] > 1.\n\n2. When the pH changes relatively to pKa by one unit, then pH – pKa = 1. Therefore,\n\n1 = log10 [[A]/[HA]]\n\nor, [[A]/[HA]] = 101 = 10\n\nThus, the ratio of the dissociated form to the associated form changes by a factor of 10.\n\n3. When the acid is 50% dissociated, the acid and the conjugate base concentrations are the same. Thus,\n\n[HA] = [A]\n\nor, [HA]/[A] = 1\n\nTherefore,\n\npH = pKa\n\nThe pH value of the solution is equal to the pKa value of the acid.\n\n### Henderson-Hasselbalch Equation for Base\n\nSimilarly, a weak base dissociates as follows,\n\nB + H2O ⇄ OH + HB+\n\nThe Henderson-Hasselbalch equation can be derived using the same principles as that for acid and gives,\n\npOH = pKb + log10 ([HB+]/[B])\n\n## Limitations of Henderson-Hasselbalch Equation\n\n• It fails to predict the pH values of strong acids because it assumes that the acid and its conjugate base concentrations remain constant at equilibrium.\n• It can be applied to a polybasic acid only if its consecutive pK values differ by at least 3.\n• It does not consider the self-dissociation by water. It fails to give an accurate pH value for extremely buffer solutions.\n\n## Examples and Problems\n\nP.1. The dissociation of acetic acid (CH3CO2H) in acetate ion (CH3CO2) is given as follows:\n\nCH3CO2H + H2O ⇄ CH3CO2 + H3O+\n\nCalculate the pH of a buffer solution made from 0.25 M of CH3CO2H and 0.45 M of CH3CO2 that has an acid dissociation constant for CH3CO2H of 1.76 x 10-5 at 25 ͦC.\n\nSoln. Given,\n\n[CH3CO2H] = 0.25 M\n\n[CH3CO2] = 0.45 M\n\nKa = 1.76 x 10-5\n\nThe Henderson-Hasselbalch equation is given by,\n\npH = pKa + log10 ([A]/[HA])\n\nTherefore,\n\npH = -log10 (1.76 x 10-5) + log10 (0.45 M /0.25 M) = 5\n\nP.2. How much sodium formate (HCOONa), 68.0069 g/mol) do you need to add to 500 mL of 0.50 M formic acid (HCOONa) for a pH 3 buffer. Ka = 1.77 x 10¯4.\n\nSoln. Given,\n\n[HCOONa] = 0.5 ML-1\n\npH = 3\n\nKa = 1.77 x 10¯4\n\nThe Henderson-Hasselbalch equation is given by,\n\npH = pKa + log10 ([A]/[HA])\n\nTherefore,\n\n3 = – log10 (-1.77 x 10¯4) + log10 ([A]/0.5)\n\nor, 3 = 3.752 + log10 ([A]/0.5)\n\nor, log10 ([A]/0.5) = -0.752\n\nor, [A]/0.5 = 0.177\n\nor, [A] = 0.177 x 0.5 = 0.0885 ML-1\n\nThe relationship between the concentration and the molecular weight is,\n\nConcentration x volume = weight/molecular weight\n\n0.0885 ML-1 x 0.5 L = w/ 68.0069 gM-1\n\nOr, w = 0.0885 ML-1 x 0.5 L x 68.0069 gM-1 = 3 g\n\nReferences"
] | [
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.8165221,"math_prob":0.99685854,"size":4494,"snap":"2021-43-2021-49","text_gpt3_token_len":1473,"char_repetition_ratio":0.15634744,"word_repetition_ratio":0.056050956,"special_character_ratio":0.32554516,"punctuation_ratio":0.12795699,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9992098,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-12-05T02:24:19Z\",\"WARC-Record-ID\":\"<urn:uuid:c41538e2-4cf4-4623-8d98-a7230fd9a701>\",\"Content-Length\":\"46106\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:2a834a00-1197-422e-9458-44ee70826956>\",\"WARC-Concurrent-To\":\"<urn:uuid:870e4f1d-7b83-46d4-bcf9-8fc234697819>\",\"WARC-IP-Address\":\"199.250.207.207\",\"WARC-Target-URI\":\"https://www.chemistrylearner.com/henderson-hasselbalch-equation.html\",\"WARC-Payload-Digest\":\"sha1:AMZ3YTKX5LOCL3DL2RBI25NZWXFCB6YT\",\"WARC-Block-Digest\":\"sha1:JVSC2R46M6MBNKNOCTQ6MK5VFT5ZGNQN\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-49/CC-MAIN-2021-49_segments_1637964363134.25_warc_CC-MAIN-20211205005314-20211205035314-00239.warc.gz\"}"} |
https://mathematica.stackexchange.com/questions/193858/finding-ndsolve-method-details | [
"# Finding NDSolve method details\n\nI have eqs about the NDSolve, I know this code given the solving automatically.\n\nHow can I find out what method is used behind the scenes? How can I gauge the reliability level, find how many iterations have been used, the order of method. How can I estimate the error?\n\nI found hints on this site, but I still do not fully understand.\n\nIt is impossible to say NDSolve has automatically solution for publishing paper?\n\nI used this code related to my system:\n\nr = 0.431201; β = 2.99 *10^-6; σ = 0.7; δ = 0.57;\n{m = 0.3, η = 0.1, μ = 0.1, ρ = 0.3};\n\nS = {N1'[t] == r N1[t] (1 - β N1[t]) - η N1[t] I1[t],\nI1'[t] == σ + (ρ N1[t] I1[t])/( m + N1[t]) - δ I1[t] - μ N1[t] I1[t]};\n\nc = {N1 == 1, I1 == 1.22};\n\nSelect[Flatten[\nTrace[\nNDSolve[{S, c}, {N1, I1}, {t, 0, 30}],\nTraceInternal -> True]],\n!FreeQ[#, Method | NDSolveMethodData] &]\n\n\nbut I don't understand the output.\n\n• Partial duplicate: mathematica.stackexchange.com/questions/145/… – Michael E2 Mar 24 at 1:17\n• Another partial duplicate: mathematica.stackexchange.com/questions/102704/… – Michael E2 Mar 24 at 1:37\n• You say you don't understand some technique or other, nor the output of your Trace[] command. But the first is a very general statement about things already explained and the second is about a command that no one else can reproduce – Michael E2 Mar 24 at 1:44\n• \"It is impossible to say NDSolve has automatically solution for publishing paper. \" Simply saying \"I've used NDSolve function of software Mathematica\" is enough in many cases, AFAIK. – xzczd Mar 24 at 3:39\n• Well, if the reviewer insists on such stuff, given that your system isn't that difficult, a possible workaround at this point is to choose a primary method like classical RK4 to solve the problem. The way to choose classical RK4 in NDSolve can be found in tutorial/NDSolveExplicitRungeKutta#1456351317, then you just need to set Method -> {\"ExplicitRungeKutta\", \"DifferenceOrder\" -> 4, \"Coefficients\" -> ClassicalRungeKuttaCoefficients}, StartingStepSize -> 1/20000, MaxSteps -> Infinity in NDSolve. The solving process is slower but gives the same result as given by default. – xzczd Mar 24 at 3:59\n\nComment\n\nIn response to your question, you already got very valuable comments. I will just try to comment on\n\nHow can I estimate the error?\n\nFor this I am going to plot residual error at steps and time, which will show the reliability and accuracy of NDSolve,\n\nr = 0.431201; \\[Beta] = 2.99*10^-6; \\[Sigma] = 0.7; \\[Delta] = 0.57;\nm = 0.3; \\[Eta] = 0.1; \\[Mu] = 0.1; \\[Rho] = 0.3;\n\node = {N1'[t] == r N1[t] (1 - \\[Beta] N1[t]) - \\[Eta] N1[t] I1[t],\nI1'[t] == \\[Sigma] + (\\[Rho] N1[t] I1[t])/(m + N1[t]) - \\[Delta] I1[t] - \\[Mu] N1[t] I1[t]};\n\nbcs = {N1 == 1, I1 == 1.22};\n\nresiduals = ode /. Equal -> Subtract;\n\n{s} = NDSolve[{ode, bcs}, {N1, I1}, {t, 20}, InterpolationOrder -> All];\n\nN1[\"Coordinates\"] /. s;\n\nresiduals /. t -> N1[\"Coordinates\"] /. s;\n\nListPlot[Abs[Flatten /@ (residuals /. t -> N1[\"Coordinates\"] /. s)], Frame -> True]",
null,
"With[{data = {Table[{t, Abs@residuals[]} /. s, {t, N1[\"Coordinates\"] /. s // Flatten}]}},\nListLogPlot[data, Frame -> True, PlotRange -> All]]\n`",
null,
"Note: I adopted the above from this website but unable to find the link.\n\n• Thank you so much @zhk but how can defend the axes for both figures? the first as x represented steps and y residual error. the second one x represents the t time and y residual error. sorry if my question is trivial but it is first time to see the code – sana alharbi Mar 24 at 9:13"
] | [
null,
"https://i.stack.imgur.com/H2B6I.png",
null,
"https://i.stack.imgur.com/HtSDi.png",
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https://discuss.codechef.com/t/weirdwoods-age-wage-editorial/24006 | [
"",
null,
"# A Weirwood’s Age - EDITORIAL\n\nTester: Sana Pagarkar\n\nEasy\n\nMaths\n\n## PROBLEM:\n\nYou’re given 3 integers a, b, X.\n\n• a = radius of the first ring,\n• b = radius of the last ring,\n• X = distance between 2 rings.\n\nYou have to find the age of the tree by counting the number of rings.\n\n## EXPLANATION:\n\n• We can solve this problem very easily. First, take the 3 integers as input and check their constraints.\n• Then, we find the difference between the first and last rings’ radii, (b-a). We are given the distance between each ring, X.\n• We divide the difference calculated between the two radii by the distance to find the number of rings.\n• If the result is a floating point number, round it off with the floor value, because we need the answer in years.\n\n## AUTHOR’S AND TESTER’S SOLUTIONS:\n\nAuthor Solution can be found here\nTester Solution can be found here\n\n4 Likes"
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"https://s3.amazonaws.com/discourseproduction/original/3X/7/f/7ffd6e5e45912aba9f6a1a33447d6baae049de81.svg",
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https://stackoverflow.com/questions/14568970/perl-regular-expression-finding-pattern-only-in-the-front-of-the-text | [
"perl regular expression finding pattern only in the front of the text\n\nSuppose there is a text like this:\n\n|-SAMPLE-D2\n|---SAMPLE-D1\n|---SAMPLE3\n\nI want to count the number of \"-\" after |. I tried to parse that by using the following regular expression in perl\n\n\\$count=()= /-/g;\n\nbut this is problematic because the first two has \"-\" somewhere else in the text as well as in the front. How should I form my regex or use other function in perl to get the number of \"-\" right after \"|\"?\n\n• /^|(-*)/, then get the length of \\$1? – Marc B Jan 28 '13 at 18:49\n• @MarcB I assume you meant /^\\|(-*)/ or /^[|](-*)/ – Brad Gilbert Jan 29 '13 at 16:18\n\nRegex to match the dashes after the starting |:\n/^\\|([\\-]*)/\n\n• You don't need to escape the - or the character class. – Chris Seymour Jan 28 '13 at 18:54\n• No, it's not incorrect just unnecessary and regexp can be complicated enough. – Chris Seymour Jan 28 '13 at 18:58\n• You do have to escape | though. – TLP Jan 28 '13 at 19:30\n• @TLP: updated, thanx – CSᵠ Jan 28 '13 at 19:39\n\nTo count dashes that are not preceded by a letter, use a negative look-behind assertion.\n\n\\$count = () = /(?<!\\w)-/g\n\nIf the vertical line only ever comes at the start you can get the string of repeating minuses with:\n\nmy (\\$match) = \\$txt =~ /^\\|(-*)/;\n\nThe brackets around \\$match cause the captured portion of the regex to be put into it\n\nthen get the number of minuses using\n\nmy \\$minus_count = length(\\$match || '');\n\nThe\n\n|| '')\n\nbit\n\nInitialises \\$match if the regex above found no matches at all, to stop length moaning about uninitialised variables (if you have warnings on)\n\nNot sure if you can count in Regex directly but you can extract capture groups and do a simple arithmetic with their string lengths:\n\n#!/usr/bin/perl\nuse warnings;\nmy \\$inFile = \\$ARGV;\nopen(FILEHANDLE, \"<\", \\$inFile) || die(\"Could not open file \".\\$inFile);\nmy @fileLines = <FILEHANDLE>;\nmy \\$lineNo = 0;\nmy \\$rslt;\n\nforeach my \\$line(@fileLines) {\nchomp(\\$line);\n\\$line =~ s/^\\s+//;\n\\$line =~ s/\\s+\\$//;\n\\$lineNo++;\nprint \"\\n\".\\$lineNo.\" = <\".\\$line.\">\";\nif(\\$line =~ m/^\\|-+(.+)/) {\nmy \\$text = \\$1;\nprint \"\\n\\ttext = <\".\\$text.\">\";\nmy \\$minCnt = length(\\$line) - length(\\$text) - 1;\nprint \"\\n\\tminus count = <\".\\$minCnt.\">\";\n}\n}\nclose(FILEHANDLE);\n• Why do you use chomp() and s/\\s+\\$//? Also I would like to point out that s/^\\s+|\\s+\\$//g is faster than having two separate substitutions. – Brad Gilbert Jan 29 '13 at 16:24\n• good point but i've had experience in the past where chomp() didn't do and the s&r did. thanks for noting – amphibient Jan 29 '13 at 17:05"
] | [
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.89617366,"math_prob":0.8741311,"size":533,"snap":"2019-43-2019-47","text_gpt3_token_len":156,"char_repetition_ratio":0.105860114,"word_repetition_ratio":0.0,"special_character_ratio":0.36585367,"punctuation_ratio":0.08490566,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.96701896,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-10-22T20:50:42Z\",\"WARC-Record-ID\":\"<urn:uuid:90f2e94a-9282-463b-9eef-f18621a7887a>\",\"Content-Length\":\"165161\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:88a2144a-40be-4198-a08c-66d514facf6c>\",\"WARC-Concurrent-To\":\"<urn:uuid:2913d34d-2bc1-45ee-8dba-67d0c3d2cbd9>\",\"WARC-IP-Address\":\"151.101.129.69\",\"WARC-Target-URI\":\"https://stackoverflow.com/questions/14568970/perl-regular-expression-finding-pattern-only-in-the-front-of-the-text\",\"WARC-Payload-Digest\":\"sha1:BVPA3HE2YET3CHRR5ORIA5CPRPLBM2GM\",\"WARC-Block-Digest\":\"sha1:KZ5FMKX7E673JX4235A4NUDADV65IZF6\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-43/CC-MAIN-2019-43_segments_1570987823061.83_warc_CC-MAIN-20191022182744-20191022210244-00372.warc.gz\"}"} |
http://www.computerchess.org.uk/ccrl/4040/cgi/engine_details.cgi?print=Details&each_game=1&eng=Minic%202.46%2064-bit | [
"Contents: CCRL 40/15 Downloads and Statistics September 19, 2020 Testing summary: Total: 1'178'235 games played by 2'711 programs White wins: 406'134 (34.5%) Black wins: 300'016 (25.5%) Draws: 472'085 (40.1%) White score: 54.5%\n\n## Engine Details\n\n Options Show each game results\nMinic 2.46 64-bit#50 (3050+26\n−25\n)Quote\n Author: Vivien Clauzon (France) Link: Homepage\nThis is one of the 13 Minic versions we tested: Compare them!\n Opponent Elo Diff Results Score LOS Perf – Arasan 22.1 64-bit 3147 +25−25 (+97) 6.5 − 13.5(+1−8=11) 32.5%6.5 / 20 0.0% −5 = 0 = = = = 0 0 = 0 = = 1 0 = = = 0 0 0 – Xiphos 0.4 64-bit 3145 +11−11 (+95) 8.5 − 11.5(+3−6=11) 42.5%8.5 / 20 0.0% +54 1 0 = = 0 0 0 = 0 = 0 = = 1 = = 1 = = = – Roc 1.7 64-bit 3140 +42−42 (+90) 12 − 8(+6−2=12) 60.0%12.0 / 20 0.0% +145 = = 1 = 0 1 1 = = = = = = 1 = 1 = = 1 0 – Laser 1.6 64-bit 3131 +11−11 (+81) 7 − 13(+1−7=12) 35.0%7.0 / 20 0.0% −5 = 0 = 0 0 = = 0 = 0 0 = = = = 0 = = = 1 – Nemorino 5.00 64-bit 3086 +12−13 (+36) 2.5 − 1.5(+2−1=1) 62.5%2.5 / 4 0.6% +131 1 1 0 = – Winter 0.8 64-bit 3082 +21−21 (+32) 2 − 2(+1−1=2) 50.0%2.0 / 4 2.9% +32 = 0 1 = – chess22k 1.14 64-bit 3075 +18−18 (+25) 6.5 − 13.5(+1−8=11) 32.5%6.5 / 20 5.6% −76 = = = = 0 0 0 1 = 0 = = 0 = 0 = = 0 = 0 – Lc0 0.23.2 wLD2 64-bit 3065 +36−36 (+15) 8 − 12(+3−7=10) 40.0%8.0 / 20 24.1% −44 = = 0 0 0 0 0 1 = = 1 = 0 0 = = = 1 = = – Combusken 1.3.0 64-bit 3052 +24−24 (+2) 10.5 − 13.5(+4−7=13) 43.8%10.5 / 24 46.1% −36 = 1 1 = 0 = 0 = 0 = = = = = 1 1 = 0 0 = 0 = = 0 – Pirarucu 3.3.5 64-bit 3027 +19−19 (−23) 2 − 2(+1−1=2) 50.0%2.0 / 4 92.4% −33 1 = 0 = – Fat Fritz Junior w510 64-bit 3023 +35−35 (−27) 7 − 13(+2−8=10) 35.0%7.0 / 20 90.0% −116 = 1 = 0 0 = = = 0 0 = = 0 = = 0 0 = 0 1 – Rybka 4 32-bit 3018 +15−15 (−32) 10 − 10(+5−5=10) 50.0%10.0 / 20 98.3% −30 = 0 1 = 1 0 1 0 = 0 0 = 1 = = = 1 = = = – Marvin 3.6.0 64-bit 3013 +21−21 (−37) 16.5 − 7.5(+12−3=9) 68.8%16.5 / 24 98.8% +82 1 0 1 1 1 1 = 1 1 = 1 = = = 1 = 1 = 0 = 1 0 = 1 – Monolith 2 64-bit 3001 +19−19 (−49) 12 − 12(+5−5=14) 50.0%12.0 / 24 99.9% −49 1 = = = = 0 0 = = = = 1 0 1 1 = = = = = = 0 1 0 – Bagatur 2.2 64-bit 2994 +21−22 (−56) 12.5 − 11.5(+5−4=15) 52.1%12.5 / 24 100.0% −45 0 = = 1 1 0 0 = = = 0 = = = 1 = = = = 1 = 1 = = – Amoeba 3.2 64-bit 2993 +25−25 (−57) 3.5 − 0.5(+3−0=1) 87.5%3.5 / 4 99.9% +231 1 = 1 1 – Rodent IV 022 64-bit 2989 +22−22 (−61) 13 − 11(+9−7=8) 54.2%13.0 / 24 100.0% −31 = 0 1 1 1 0 1 = = = 1 0 1 1 = 0 0 = 0 1 1 0 = = – Stockfish 1.8 32-bit 2989 +17−17 (−61) 10.5 − 9.5(+7−6=7) 52.5%10.5 / 20 100.0% −43 1 = 0 0 = 0 0 1 1 = = 1 = 1 1 1 0 = = 0 – Topple 0.7.5 64-bit 2984 +18−18 (−66) 11.5 − 8.5(+7−4=9) 57.5%11.5 / 20 100.0% −18 = = 1 1 0 = 0 = = = = 1 1 1 1 1 0 = = 0 – Godel 7.0 64-bit 2979 +19−19 (−71) 12.5 − 11.5(+8−7=9) 52.1%12.5 / 24 100.0% −56 1 0 = = = = 0 0 1 = 1 = 0 0 1 1 = 1 = 0 1 = 1 0 – Amoeba 3.1 64-bit 2968 +20−20 (−82) 15 − 5(+10−0=10) 75.0%15.0 / 20 100.0% +67 = 1 = 1 1 = 1 = 1 1 = = 1 = = = 1 1 = 1 – FabChess 1.15 64-bit 2942 +22−22 (−108) 19 − 5(+17−3=4) 79.2%19.0 / 24 100.0% +124 1 1 0 1 1 1 1 = 1 1 1 1 1 = = 1 1 = 1 1 1 0 0 1 – Naum 4 32-bit 2930 +14−14 (−120) 13 − 7(+9−3=8) 65.0%13.0 / 20 100.0% −22 1 1 1 = 1 = = = = = 1 1 = = 0 1 1 1 0 0 – Francesca MAD 0.29 64-bit 2914 +25−25 (−136) 21 − 5(+17−1=8) 80.8%21.0 / 26 100.0% +86 = 1 1 0 = 1 1 1 = 1 1 1 = 1 1 1 1 = = 1 1 1 1 1 = = – Counter 3.5 64-bit 2905 +21−21 (−145) 14.5 − 5.5(+11−2=7) 72.5%14.5 / 20 100.0% +4 1 1 1 0 = 1 1 1 1 = 1 = = = = = 1 1 1 0\n\n### Rating changes by day",
null,
"### Rating changes with played games",
null,
"Created in 2005-2013 by CCRL team Last games added on September 19, 2020"
] | [
null,
"http://www.computerchess.org.uk/ccrl/4040/rating-history-by-day-graphs/Minic_2_46_64-bit.png",
null,
"http://www.computerchess.org.uk/ccrl/4040/rating-history-by-day-graphs-2/Minic_2_46_64-bit.png",
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.5631488,"math_prob":1.0000095,"size":3316,"snap":"2020-34-2020-40","text_gpt3_token_len":2206,"char_repetition_ratio":0.29619566,"word_repetition_ratio":0.34375,"special_character_ratio":0.9146562,"punctuation_ratio":0.12836438,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9986695,"pos_list":[0,1,2,3,4],"im_url_duplicate_count":[null,1,null,1,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-09-22T02:11:53Z\",\"WARC-Record-ID\":\"<urn:uuid:ecedb569-491a-4be5-af0b-5dda1e49b94d>\",\"Content-Length\":\"26894\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:141762be-6f79-4d77-826d-3807dbe2e125>\",\"WARC-Concurrent-To\":\"<urn:uuid:15b5ef73-c2bd-4081-b07e-2b6362d15dc2>\",\"WARC-IP-Address\":\"185.45.66.155\",\"WARC-Target-URI\":\"http://www.computerchess.org.uk/ccrl/4040/cgi/engine_details.cgi?print=Details&each_game=1&eng=Minic%202.46%2064-bit\",\"WARC-Payload-Digest\":\"sha1:LTD23DEJZ75W66J7AHQRVSM2HGQVBMIG\",\"WARC-Block-Digest\":\"sha1:BHVSD3F42QXP2WACGFV6KHXEYZ2BSCEH\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-40/CC-MAIN-2020-40_segments_1600400202686.56_warc_CC-MAIN-20200922000730-20200922030730-00094.warc.gz\"}"} |
https://www.bartleby.com/solution-answer/chapter-78-problem-52e-calculus-an-applied-approach-mindtap-course-list-10th-edition/9781305860919/evaluating-a-double-integral-in-exercises-51-58-use-a-symbolic-integration-utility-to-evaluate-the/ffc68b0d-6361-11e9-8385-02ee952b546e | [
"",
null,
"",
null,
"",
null,
"Chapter 7.8, Problem 52E",
null,
"### Calculus: An Applied Approach (Min...\n\n10th Edition\nRon Larson\nISBN: 9781305860919\n\n#### Solutions\n\nChapter\nSection",
null,
"### Calculus: An Applied Approach (Min...\n\n10th Edition\nRon Larson\nISBN: 9781305860919\nTextbook Problem\n1 views\n\n# Evaluating a Double Integral In Exercises 51-58, use a symbolic integration utility to evaluate the double integral. ∫ 0 2 ∫ x 2 2 x ( x 3 + 3 y 2 ) d y d x\n\nTo determine\n\nTo calculate: The simplified value of the double integral 02x22x(x3+3y2)dydx by application of symbolic integration utility.\n\nExplanation\n\nGiven Information:\n\nThe provided double integral is 02x22x(x3+3y2)dydx.\n\nCalculation:\n\nConsider the double integral 02x22x(x3+3y2)dydx\n\n### Still sussing out bartleby?\n\nCheck out a sample textbook solution.\n\nSee a sample solution\n\n#### The Solution to Your Study Problems\n\nBartleby provides explanations to thousands of textbook problems written by our experts, many with advanced degrees!\n\nGet Started\n\n#### x3a+b3\n\nApplied Calculus for the Managerial, Life, and Social Sciences: A Brief Approach\n\n#### What is the integrating factor for xy′ + 6x2y = 10 − x3?\n\nStudy Guide for Stewart's Single Variable Calculus: Early Transcendentals, 8th\n\n#### True or False: converges.\n\nStudy Guide for Stewart's Multivariable Calculus, 8th",
null,
""
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"https://www.bartleby.com/static/search-icon-white.svg",
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https://www.mcs.anl.gov/petsc/petsc-current/docs/manualpages/Tao/TAOBRGN.html | [
"petsc-3.11.3 2019-06-26\nReport Typos and Errors\n\n# TAOBRGN\n\nBounded Regularized Gauss-Newton method for solving nonlinear least-squares problems with bound constraints. This algorithm is a thin wrapper around TAOBNTL that constructs the Gauss-Newton problem with the user-provided least-squares residual and Jacobian. The algorithm offers both an L2-norm proximal point (\"l2prox\") regularizer, and a L1-norm dictionary regularizer (\"l1dict\"), where we approximate the L1-norm ||x||_1 by sum_i(sqrt(x_i^2+epsilon^2)-epsilon) with a small positive number epsilon. The user can also provide own regularization function.\n\n### Options Database Keys\n\n+ -tao_brgn_regularization_type - regularization type (\"user\", \"l2prox\", \"l1dict\") (default \"l2prox\") . -tao_brgn_regularizer_weight - regularizer weight (default 1e-4) - -tao_brgn_l1_smooth_epsilon - L1-norm smooth approximation parameter: ||x||_1 = sum(sqrt(x.^2+epsilon^2)-epsilon) (default 1e-6)\n\nbeginner\n\n### Location\n\nsrc/tao/leastsquares/impls/brgn/brgn.c\n\n### Examples\n\nsrc/tao/leastsquares/examples/tutorials/cs1.c.html\nsrc/tao/leastsquares/examples/tutorials/tomography.c.html\n\nIndex of all Tao routines"
] | [
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.67036223,"math_prob":0.9138332,"size":1182,"snap":"2019-26-2019-30","text_gpt3_token_len":334,"char_repetition_ratio":0.11629881,"word_repetition_ratio":0.0,"special_character_ratio":0.2360406,"punctuation_ratio":0.094736844,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9786156,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-07-17T11:44:17Z\",\"WARC-Record-ID\":\"<urn:uuid:62514c24-0bf2-4a47-9b65-0103ea70c617>\",\"Content-Length\":\"2764\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:79b238e5-22cb-4250-acda-dafd993b9b1b>\",\"WARC-Concurrent-To\":\"<urn:uuid:8f56a3bc-2ee9-42fc-a40e-c8b74671a358>\",\"WARC-IP-Address\":\"140.221.6.95\",\"WARC-Target-URI\":\"https://www.mcs.anl.gov/petsc/petsc-current/docs/manualpages/Tao/TAOBRGN.html\",\"WARC-Payload-Digest\":\"sha1:YX766R7NQWEYVZPK452DDJT4WRX5IGYR\",\"WARC-Block-Digest\":\"sha1:VHB3C67OUFFDZEEDOMUYEDIQT7C5SKFF\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-30/CC-MAIN-2019-30_segments_1563195525136.58_warc_CC-MAIN-20190717101524-20190717123524-00342.warc.gz\"}"} |
https://freevideolectures.com/course/3390/advanced-numerical-analysis/2 | [
"# Advanced Numerical Analysis\n\nIIT Bombay Course , Prof. Sachin C. Patwardhan\n\n•",
null,
"•",
null,
"•",
null,
"109 students enrolled\n\n# Overview\n\nIntroduction and Overview - Fundamentals of Vector Spaces - Basic Dimension and Sub-space of a Vector Space - Introduction to Normed Vector Spaces - Examples of Norms,Cauchy Sequence and Convergence, Introduction to Banach Spaces - Introduction to Inner Product Spaces - Cauchy Schwaz Inequality and Orthogonal Sets - Gram-Schmidt Process and Generation of Orthogonal Sets - Problem Discretization Using Appropriation Theory - Weierstrass Theorem and Polynomial Approximation - Taylor Series Approximation and Newton's Method - Solving ODE - BVPs Using Firute Difference Method - Solving ODE - BVPs and PDEs Using Finite Difference Method - Finite Difference Method (contd.) and Polynomial Interpolations - Polynomial and Function Interpolations,Orthogonal Collocations Method for Solving - Orthogonal Collocations Method for Solving ODE - BVPs and PDEs - Least Square Approximations :Necessary and Sufficient ConditionsLinear Least Square Estimation and Geometric Interpretation - Geometric Interpretation of the Least Square Solution (Contd.) and Projection - Projection Theorem in a Hilbert Spaces (Contd.) and Approximation - Discretization of ODE-BVP using Least Square Approximation - Discretization of ODE-BVP using Least Square Approximation and Gelarkin Method - Model Parameter Estimation using Gauss-Newton Method - Solving Linear Algebraic Equations and Methods of Sparse Linear Systems\n\nMethods of Sparse Linear Systems (Contd.) and Iterative Methods for Solving - Iterative Methods for Solving Linear Algebraic Equations: Convergence - Iterative Methods for Solving Linear Algebraic Equations: Convergence Analysis - Optimization Based Methods for Solving Linear Algebraic Equations: Gradient Method - Conjugate Gradient Method, Matrix Conditioning and Solutions - Matrix Conditioning and Solutions and Linear Algebraic Equations and Solving Nonlinear Algebraic Equations - Solving Nonlinear Algebraic Equations: Wegstein Method and Variants of Newton's Method - Solving Nonlinear Algebraic Equations: Optimization Based Methods - Solving Nonlinear Algebraic Equations: Introduction to Convergence analysis - Solving Ordinary Differential Equations - Initial Value Problems (ODE-IVPs) - Solving Ordinary Differential Equations - Initial Value Problems - Solving ODE-IVPs : Runge Kutta Methods (contd.) and Multi-step Methods - Solving ODE-IVPs : Generalized Formulation of Multi-step Methods - Solving ODE-IVPs : Multi-step Methods (contd.) and Orthogonal Collocations Method - Solving ODE-IVPs: Selection of Integration Interval and Convergence Analysis - Solving ODE-IVPs: Convergence Analysis of Solution Schemes - Methods for Solving System of Differential Algebraic Equations\n\n### Lecture 2: Fundamentals of Vector Spaces",
null,
"Up Next\nYou can skip ad in\n• 2x\n• 1.5x\n• 1x\n• 0.5x\n• 0.25x\nCOPY\nPRIVATE CONTENT\nOK\nEnter password to view\n0:00\n0 (0 Ratings)\n\n# Lecture Details\n\nAdvanced Numerical Analysis by Prof. Sachin C. Patwardhan,Department of Chemical Engineering,IIT Bombay.For more details on NPTEL visit httpnptel.ac.in\n\n# 0\n\n0 Rates\n1\n0%\n0\n2\n0%\n0\n3\n0%\n0\n4\n0%\n0\n5\n0%\n0",
null,
""
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http://book.caltech.edu/bookforum/showpost.php?s=322ca778c42f9aef53909a3778ab5530&p=10360&postcount=8 | [
"View Single Post\n#8",
null,
"yaser",
null,
"Caltech Join Date: Aug 2009 Location: Pasadena, California, USA Posts: 1,478",
null,
"Re: Is the Hoeffding Inequality really valid for each bin despite non-random sampling\n\nQuote:\n Originally Posted by grozhd",
null,
"I have the same concern as scottedwards2000 and I still don't understand how it is resolved. As I understand the bin symbolizes the probability space of all possible inputs",
null,
". Sample of balls drawn randomly from the bin symbolizes our training set",
null,
". Now we pick a hypothesis",
null,
"(suppose we are running PLA). We look at our sample",
null,
", compute",
null,
"and use Hoeffding's Inequality. We do one step of PLA and come up with new hypothesis",
null,
"which automatically gives us",
null,
"and professor is saying that we can write down Hoeffding inequality for",
null,
"and",
null,
"? I guess, we can. But that inequality tells us something about random variable",
null,
", i.e. about:",
null,
"where",
null,
"is a random sample. But it seems like we are using",
null,
"where",
null,
"is hardly random with regard to",
null,
"since we built",
null,
"using that sample. Here is an example that illustrates my point: say we tried some random",
null,
", compared it with target function",
null,
"on our training sample",
null,
", wrote down Hoeffding's inequality. Now let's construct",
null,
"as follows:",
null,
"and",
null,
". Let's write down Hoeffding's ineqaulity for this hypothesis. If we are indeed using",
null,
"then here it would be equal to 1 since",
null,
"on",
null,
"and we would have:",
null,
"is small. So somehow we are saying with high probability that",
null,
"does an excellent job out of sample though we didn't change it much from",
null,
". This example shouldn't be correct, right? If it isn't how is the one with PLA correct?\nIt is a subtle point, so let me try to explain it in the terms you outlined. Let us take the sample",
null,
"(what you call",
null,
", just to follow the book notation). Now evaluate",
null,
"for all hypotheses",
null,
"in your model",
null,
". We didn't start at one",
null,
"and moved to another. We just evaluated",
null,
"for all",
null,
". The question is, does Hoeffding inequality apply to each of these",
null,
"'s by itself? The answer is clearly yes since each of them could be in principle the hypothesis you started with (which you called",
null,
").\n\nHoeffding states what the probabilities are before the sample is drawn. When you choose one of these hypotheses because of its small",
null,
", as in the scenario you point out, the probability that applies now is conditioned on the sample having small",
null,
". We can try to get conditional version of Hoeffding to deal with the situation, or we can try to get a version of Hoeffding that applies regardless of which",
null,
"we choose and how we choose it. The latter is what we did using the union bound.\n\nFinally, taking the example you illustrated, any hypothesis you use has to be in",
null,
"(which is decided before the sample is drawn). The one you constructed is not guaranteed to be in",
null,
". Of course you can guarantee that it is in",
null,
"by taking",
null,
"to be the set of all possible hypotheses, but in this case,",
null,
"is thoroughly infinite",
null,
"and the multiple-bin Hoeffding does not guarantee anything at all.\n__________________\nWhere everyone thinks alike, no one thinks very much"
] | [
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https://physics.stackexchange.com/questions/480596/how-to-prove-%CE%B1-%CF%83%CE%B2-%CF%83-%CE%B1-%CE%B2-i%CE%B1%C3%97%CE%B2-%CF%83-where-%CE%B1-and-%CE%B2-are-3-dimensional-ve | [
"How to prove $(α·σ)(β·σ) = α·β +iα×β·σ$ (where, $α$ and $β$ are 3 dimensional vectors and $σ$ represents Pauli matrices)?\n\nI tried to evaluate the LHS first and obtained the first term of RHS easily. Then i tried to use the commutation relations of $$\\mathrm{SU}(2)$$ group to proceed further to obtain the second term of the LHS from remaining quantities but got stuck there. Can somebody show how it's done?\n\n• WP. – Cosmas Zachos May 17 at 10:28\n• Unless you know a smarter way, try writing out all of the components. – my2cts May 17 at 11:40\n\nThe commutation relation $$[\\sigma_i,\\sigma_j] = 2i\\epsilon_{ijk}\\sigma_k$$ is strictly insufficient for what you want to achieve, and it needs to be supplemented with the anticommutation relations between the Pauli matrices, $$\\{\\sigma_i,\\sigma_j \\} = 2\\delta_{ij}.$$ As the Wikipedia page points out, adding those two will get you an expression for the individual products, $$\\sigma_i\\sigma_j = \\delta_{ij} + i\\epsilon_{ijk}\\sigma_k,$$ and you can then contract this with the two relevant vectors on either side to get the expression you're after."
] | [
null
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https://old.cescg.org/CESCG97/boros/ | [
"Image Processing for fine arts\n\nAbstract:\n\nThe contribution deals with image transformations applicable to the fine arts and to the correction of photographic images. The program includes image filters for image blurring, correction of the shade value, softening of the drawing - blurring, rotation and various special and deformation effects, e.g. map relief of the image, effect of an oil painting, solarization, fish eye effect or whirlpool and other interesting transformations. The library of these special image filters is focussed on the application to user's programs.\n\nBy means of this library, interesting and useful transformations of actual images can be produced which may extend the limits of the mere entertainment.\n\nContents\n\n2. Mathematical principle of raster image transformation\n\nGeneral description of raster transformation\n\nMathematical notation of raster filters\n\n3. Description of implementation\n\nImplementation of image filters\n\n4. Summary\n\nReferences\n\nA. Appendix - images\n\n1. Introduction\n\nOne can hardly find any answer to the question \"What is computer based fine arts?\". Creative arts that use the computer as its media is used in various other branches of art, such as music. In any case, computer is used as a tool for creation - instead of brush one uses mouse. Herbert W. Franke, one of the inventors of computer art, touched this subject in this question: \"Is the image created using computer really an art?\" We could also ask whether tones played using violin are also music. In both cases we just used tools to produce the art.\n\nIn some tasks, computer is more suitable than traditional artistic tools. An example of such task is filtering and geometrical transformation of an image. It could be expected, that some time the expression \"Computer Art\" will be dead, but the technology is not the key point to the art - the human is - when equipped with some technical means.\n\nClearly, oil painting is always done on the paper or cloth using brush and colors. The purpose is not that the filter when applied to a photograph creates painting, but that the filtered image has nice and interesting appearance.\n\nThe above situation resembles the situation with traditional and digital photography. The fact that digital photographs could be easily grabbed, archived, marked, transferred through telephone lines, etc. does not have to say anything to a creative user. The situation starts to be more interesting from the moment when it is possible to change the technical quality or layout of the image. Very few computer \"amateurs\" who have once \"simply created\" the effect of plastic relief know how difficult it would be to do similar effects using traditional photography. It would be necessary to produce both positive and negative versions of the image, then offset them a little, and finally expose on the positive photographic paper. After the paper is developed, the image (hopefully) shows the plastical effect. The fancy \"solarisation\", that is traditionally produced with difficult to control overexposure, can be easily simulated on a computer. Now, it is not necessary to use special lenses to produce \"soft\" images. Moreover, filters can be applied to the image at any time.\n\nThe differences in the processing are straightforward. Using the digital technology, every change to the image can be seen on the screen more or less immediately. If the user does not like the changes, he can revert them usually by pressing single button. The costs of material are zero evens when the most complicated operations are performed. Changes of colors, brightness, or contrast cost only a little bit of electricity. Moreover, most of the image processing programs offers functions that are not possible to do with a standard photography at all, or that have very high technological requirements.\n\nSeveral image-processing programs are available on the market in these days, like Adobe Photoshop, Aldus Gallery Effects, PhotoFusion, or Aldus Photostyler. However, none of these programs offers its functions to be used by other programs. Unfortunately, literature describing special image transformations is also nearly non-existent.\n\nThis contribution describes a library that offers programmers a possibility to use several image-processing algorithms, like sharpening, softening, several brightness corrections, and special geometrical and other transformations.\n\nThe purpose of all the filters is to make the work of professional photographers Easier and allow them to correct the digital photographs in a simple way. They can improve damaged images, sharpen the images that require improvement of resolution, or do the opposite to achieve artistic impression. It is also possible to lighten or darken the image, create a negative image, or clip the image with arbitrary shape.\n\nThe library contains filters that have purely artistic effect. These can convert photographed images to images that look like that they have been painted, increase a granularity, deform them, or change their appearance in a way that they look like seen in a broken mirror. Effect of a lens or fish eye are achieved by projecting the image on a spherical surface, also available are geometrical transformations as skew, zoom, etc.\n\nAll the effects mentioned above are contained in a single library, which is implemented in C++ and which is platform independent.\n\nGeneral description of raster transformation\n\nThe transformation of a raster image is based on the transformation function, which assigns the values calculated from the original raster values to the new raster. It means that each point [x,y]of the new raster is assigned a value of the transformation function corresponding with the original raster.\n\n New[x,y] = f(Old[i,j]) iÎ á 0..X) , jÎ á 0..Y) , (1)\n\nWhere f is the transformation function, X and Y are the width and height of the original raster, resp. and [x,y] are the Cartesian raster coordinates. New is the identification of the new raster and Old is the identification of the original raster. The equation (1) is valid for each point [x,y] of the new raster where\n\nxÎ á 0..NX), yÎ á 0..NY),\n\nNX is the width and NY is the height of the new image (they may, but not necessarily, be identical with the X,Y values of the original image).\n\nFor the transformation\n\n g(x,y) = [x',y'], (2)\n\nWhere [x,y] are the coordinates of the original raster and [x',y'] are the coordinates of the new raster a situation may occur where x',y' are values of a real type which means that the value of the [x,y] point of the original image Old[x,y] must be distributed over the neighboring points of the raster by means of an interpolation function.\n\n[nx,ny], [nx+1,ny],\n\n[nx,ny+1], [nx+1,ny+1],\n\nwhere it is valid that\n\n nx < x' < nx+1, ny < y' < ny+1. (3)\n\nFig. 1.\n\n Old New",
null,
"0 = [x,y], 0'= [x',y'], 1 = [nx, ny ], 2 = [nx+1,ny], 3 = [nx,ny+1], 4 = [nx+1,nx+1].\n\nThe resulting value of the [x',y'] point must be interpolated into the points 1,2,3,4 of the new raster .\n\nMathematical notation of raster filters\n\nMany of the raster transformations may be described by mathematical equations. The conventions for the description of these operations will be given.\n\nX,Y - dimension of the raster\n\nA - maximum angle of the polar coordinates (here 360°),\n\nR - maximum radius of the polar coordinates\n\nZ - maximum point intensity\n\n% - operator modulo,\n\n[x,y] - Cartesian coordinates of the raster point\n\n{r,a} - polar coordinates of the raster point , r is the radius (distance from the center) and a is the angle.\n\nOld[x,y] - value of the color of the original image in the [x,y] point\n\nq(x,y,N) - value of the most frequently used color in the original raster from the [x-N,y-N] to the[x+N,y+N] point\n\nThe transformations will be recorded in the following form:\n\n New[x,y] = f(Old[dx,dy]), (4)\n\nkde y = 0..Y, x = 0 .. X.\n\nIn the transformations the Cartesian coordinates are converted to polar coordinates and vice versa:\n\na = atan(y/x),\n\nr = sqrt(x2 + y2),\n\nxval = r * cos(a),\n\nyval = -r * sin(a).\n\nEffect of an image painted by oil (Oil)\n\nThe following equation refers to this effect:\n\n New[x,y] = q(x,y,N), (5)\n\nWhere N equals 1..5.\n\nEffect of a random shear image (Shear)\n\nThis effect is difficult to be described by an equation:\n\n New[x,y] = Old[dx,dy], (6)\n\nWhere the dx,dy differ from the previous dx,dy values (in the previous point) by o ± 1 (at random). An impression of an image waved at random in both directions is created.\n\nEffect of the random shifted image parts (Slice)\n\nThis the equation for this effect:\n\n New[x,y] = Old[x+xshift[y],y+yshift[x]], (7)\n\nWhere the xshift,yshift values represent the random shifts in the range of ± 31 in blocks from 8 to 39 points at random in both directions.\n\nMelt Effect (Melt)\n\nAn impression is made of an icy image, which is melting. This transformation first requires the copy of the original image (it operates in situ):\n\nNew[x,y] = Old[x,y],\n\nThe x, y coordinates are chosen at random for which the following is true:\n\n:\n\n If New[x,y] Ł New[x,y+1], then swap(New[x,y],New[x,y+1]), (8)\n\nWhere swap is the function for values exchange. This is repeated for the total number of the raster points.\n\nTile Effect (Tile)\n\nThe image is divided into square parts of a constant size which are distributed at random distances from the original site:\n\n New[dx+ox,dy+oy] = Old[dx,dy], (9)\n\nWhere dx passes from x to TileSize and dy passes from y to TileSize, where x,y are the multiples of TileSize. The ox,oy are random values in the range of ± 16 and they are constant for a particular section of the raster.\n\nMatte Effect (Effect of the inversive image thresholding)\n\nThe image according to the GAMMA value is inversibly thresholded:\n\n New[x,y] = lookup[Old[x,y]], (10)\n\nWhere the following is defined for each color:\n\nlookup[color] = (Z*(color/Z)GAMMA < 3.0)? Z : 0.\n\nGamma Correction (Intensity Correction)\n\nMathematical equation for this transformation:\n\n New[x,y] = Old[x,y]*Koef, (11)\n\nIf koef>1, the image intensity increases, if Koef<1, the image intensity decreases.\n\nNegative Image (Negative)\n\nOne of simple transformations:\n\n New[x,y] = Z - Old[x,y]. (12)\n\nCut Circle\n\nThe transformation is based on the circle equation:\n\n If (x-X/2)2/(X/2)2 +(y-Y/2)2/(Y/2)2 < 1 Then new[x,y] = Old[x,y], Or New[x,y] = 0. (13)\n\nImage Solarization\n\nSolarization process which disappears from the left to the right:\n\n New[x,y] = (Old[x,y] > (Z*x)/(2*X)) ? Old[x,y]:Z-Old[x,y]. (14)\n\nRelief Map Effect\n\nThe plastic relief is created by the combination of the original and shifted half negatives:\n\n New[x,y] = Old[x,y] + (Z/2 - Old[x+2,y+2]). (15)\n\nImage Shrink\n\nThe coordinates of the original raster are multiplied by the coefficient:\n\n New[x,y] = Old[x*Koef,y*Koef], (16)\n\nwhere Koef > 1.\n\nEffect of an interesting caricature (Caricature)\n\nIn this transformation, polar coordinates are used:\n\n New[x,y] = Old{sqrt(r*R),a}, (17)\n\nWhere sqrt is the root of the number.\n\nFish Eye Effect (Fish Eye)\n\nThe transformation results in a convex image.\n\n New[x,y] = Old{r2/R,a}. (18)\n\nSwirled image Effect (Swirled)\n\nSwirled image is formed by:\n\n New[x,y] = Old{r,a + r/Koef}. (19)\n\nCylinder Mirror Effect (Cylinder)\n\nThe upper edge of the raster is drawn to the center, the left and right edges are fused.:\n\n New[x,y] = Old[a*X/A,r*Y/R]. (20)\n\nHorizontal Mirror Effect (Horizontal Mirror)\n\nDepending on the value identifying the side, the left or right side may be reflected.\n\n New[x,y] = xX/2 ? Old[x,y]:Old[X-x,y]. (21) (22)\n\nBathroom (Bathroom 2) Window Effect\n\nThese transformations are in [Holzm] called bathroom window views.\n\n New[x,y] = Old[x + (x%32)-16,y], New[x,y] = Old[x + ((a+r/10)%32)-16,y]. (23) (24)\n\nImage Blur (Blurred Image)\n\nThe mean value of the surrounding points is calculated by this transformation.\n\n New[x,y] = avr(x,y,N), (25)\n\nWhere avr(x,y,N)calculates the mean in the surrounding of the point [x,y] in the distance ± N/2.\n\nPixelized Image (Pixelize)\n\nThis transformation calculates the mean value of the surrounding points and poses this mean value to all these points:\n\n New[x,y] = avr(dx,dy,N), (26)\n\nwhere avr(dx,dy,N)calculates the mean in the [dx,dy] field , where the size of this field (area) is N2.\n\nBasic equations for the operations with convolution masks:\n\nThe filters are defined by a matrix having dimensions either 3x3 or 5x5. This will be illustrated by the matrix which produces identity from the original image.\n\n Mask: M = 0 0 0 (27) 0 1 0 0 0 0\n\nCalculation of the point value New[x,y]:\n\n New[x,y] = Old[x-1,y-1] Old[x,y-1] Old[x+1,y-1] Old[x-1,y ] Old[x ,y ] Old[x+1,y ] Old[x-1,y+1] Old[x ,Y+1] Old[x+1,y+1] * M[1,1] + * M[2,1] + * M[3,1] + * M[1,2] + * M[2,2] + * M[3,2] + * M[1,3] + * M[2,3] + * M[3,3] (28)\n\nSome masks are assigned a coefficient koef, by which the resulting value should be divided, and some masks are, moreover, assigned the Bias, which is added to the resulting value:\n\n New[x,y] = New[x,y]/koef + Bias (29)\n\nThe koef and Bias values equal 1 and 0, resp. in other filters.\n\nEdge detection\n\n This filter is defined by three matrixes: Heavy edge detection 1 -2 1 (30) -2 4 -2 1 -2 1\n\n Medium edge detection -1 -1 -1 (31) -1 8 -1 -1 -1 -1\n\n Light edge detection 0 1 0 (32) 1 -4 1 0 1 0\n\nEmboss filter\n\n Bias = Z/2 value is set for this filter -1 0 0 (33) 0 0 0 0 0 1\n\nEnhanced detail\n\n The coefficient koef for this mask equals 6. 0 -1 0 (34) -1 10 -1 0 -1 0\n\n Enhanced edges -1 -1 -1 (35) -1 9 -1 -1 -1 -1\n\n Enhanced focus -1 0 -1 (36) Koef = 3 0 7 0 -1 0 -1\n\n Reduce jaggies 0 0 -1 0 0 (37) 0 0 3 0 0 -1 3 7 3 -1 0 0 3 0 0 0 0 -1 0 0\n\nSoften filter\n\n Heavy soften filter koef = 99 11 11 11 (38) 11 11 11 11 11 11\n\n Medium soften filter koef = 100 10 10 10 (39) 10 20 10 10 10 10\n\n Light soften filter koef = 97 6 12 6 (40) 12 25 12 6 12 6\n\n Blur light 1 2 1 (41) koef = 14 2 2 2 1 2 1\n\nImplementation of image filters\n\nSo far, all the image transformations take into account the same size of the original and transformed snaps. Therefore the same variables for raster sizes are used for implementation:\n\nXsize - raster width\n\nYsize - raster height\n\nTherefore, the basic algorithm implementation is used for each transformation:\n\nfor y varying from 0 to Ysize\n\nfor x varying from 0 to Xsize\n\nexecute transformation.\n\nEffect of an image painted by oil\n\nThe field of 256 items for histogram storage (number of the colors used in the section given) was necessary for this effect. In the environment (surroundings) of each point [x,y] in the range limited by the points [x-N,y-N], [x+N,y+N], the multiplicity of the colors used is stored in the field declared. After this, the most frequently used color is chosen out of this histogram and recorded into the allocated structure for storing the image to the [x,y] position. The algorithm is repeated for each item of the image raster in this way. To calculate the RGB components for the intensity in the range of 256 levels the following equation is used:\n\nI = 0.299*r + 0.587*g + 0.114*b .\n\nEffect of a random shear image\n\nIn the main cycle, the shift in the direction given, which is prepared in the allocated shift field yshift[], is added to each coordinate. This field contains random figures having maximum difference of ±1. The X-shift is calculated as early as in the main cycle.\n\nEffect of the random shifted image parts (Slice)\n\nFor two declared fields, the shift of blocks in the range from 8 to 39 pixels is designed in two independent cycles. These blocks can be shifted by ± 31 points. In the main cycle only the precalculated shifts are added to the original coordinates. Three RGB image components are always presented.\n\nMelt Effect\n\nThis effect operates \"in situ\" - therefore, an original raster copy must be first made in the allocated space. The program identifies the number of raster points (the number of transformation cycles). In each cycle the random x,y coordinates are chosen. The color value in this location is compared with that of the point chosen on the x,y+1 coordinates. If the value of the color under an accidentally chosen point is higher or equals the value of the point chosen, these two color values are interchanged.\n\nTile Effect\n\nBoth the immersed cycles increase the x,y values by the size of the TileSize section. In the interior of these cycles the random relative shift of these sections is calculated. After this, these shifts are added to the coordinates of the original image, and the complete parts are simultaneously transformed into the allocated memory. It must be also checked if the calculated shift does not happen to exceed the raster.\n\nMatte Effect (Effect of the inversive image thresholding)\n\nFor the intensity level values, each level will be calculated from the equation given in the previous chapter (under the equation designed by (10)) or this intensity will be marked by white or black colors. These values will be stored in the field lookup[], to which each intensity of the original raster will be referred. The intensity which will be inserted as index into the lookup field is calculated from the RGB components on the x,y coordinates. The value in the lookup[] field will be assigned to the new raster in the x,y position.\n\nGamma Correction (Intensity Correction)\n\nIn the new transformation cycle, the original component multiplied by the coefficient given will be assigned to each RGB component. However, the exceeding of the value of the color component must be observed (if this occurs, maximum value of the component must be set).\n\nNegative Image (Negative)\n\nThe new RGB components are assigned the value of the difference of the maximum component value and the original RGB component.\n\nCut Circle\n\nThe condition in the immersed cycle determines if the point is copied into the new raster or clear to zero depending on the fact if the point coordinates are situated inside the circle which is determined by the quotation with the parameters of the image size.\n\nImage Solarization\n\nThe value of the image intensity is calculated for each point. This is compared by the equation (14). This condition being met, the original value of the RGB component is recorded. Otherwise, the inversion value of the component is recorded.\n\nRelief Map Effect\n\nIntensities are calculated for the points [x,y], [x+2,y+2] from the RGB components. Next, each RGB component of the new raster is assigned an intensity value on the original site added up to intensity value produced by subtracting the half value of the maximum intensity and the intensity in the point [x+2,y+2].\n\nImage Shrink\n\nThe point value of the new raster is assigned the value from the position [x*Koef,y*Koef] of the original raster\n\nwhere Koef is the initial value of this transformation.\n\nEffect of an interesting caricature (Caricature)\n\nThe r,a values of the polar coordinates are calculated for the Cartesian coordinates values x,y. The r,a values are introduced into the equation (17) and are retransformed to Cartesian coordinates. The point of the new raster is assigned the value of the of these retransformed coordinates.\n\nFish Eye Effect\n\nThe transformation is the same as in the previous case except for the polar coordinates, which are introduced into the equation (18).\n\nSwirled image Effect\n\nThe same procedure as in the previous transformations. Equation (19).\n\nCylinder Mirror Effect\n\nThe polar coordinates are calculated and introduced into the Cartesian coordinates according to the equation (20). All the RGB components are transformed and it is tested if the retransformed coordinates are not located beyond the range of the raster.\n\nHorizontal Mirror Effect\n\nThe transformation of all the RGB components is performed according to the equations (21) and (22), resp., depending on the half of the image to be reflected.\n\nBathroom (Bathroom 2) Window Effect\n\nEach of the components is assigned equation (23) or (24) for Bathroom or Bathroom 2, resp. These equations are applied to each of the components of the original image.\n\nImage Blur (Blurred Image)\n\nIn two immersed cycles all the RGB components of the original snap are added in the range from the point [x-N/2, y-N/2] to the point [x+N/2, y+N/2] and the mean value of the components is calculated which is then introduced into the new raster to the [x,y] position.\n\nPixelized Image\n\nThe mean values of the relevant areas are calculated for the pixel size given. These calculated means of the components are then assigned to the corresponding area.\n\nMatrix Transformations\n\nAccording to the convolution mask and the equations (28) and (29) the values of all the image points are calculated.\n\n4. Summary\n\nFrom the point of view of the fine art, these effects are to be regarded as suitable creative tools producing further opportunities to change the images and photographs. An area of creation with incredible possibilities has been formed. Next to the transformation and the filters, which particularly depend on the technology, the individual images can be re-painted and re-drawn. However, these processes call for a more individual approach. In no case, the deformation and filtration can be regarded as \"automatic\" techniques. This classification may result only from ignorance. On the contrary, these techniques demand much time and effort especially if an above-average result is strived after.\n\nCurrently, there is only little published data describing the range of problems concerning the image filters in greater detail. Nevertheless, these transformations are used for professional image processing of actual images - the digital photographs. In future, the association of these programs with digital cameras having high distinguishing abilities can be expected. The images of these cameras contain as many as several megabytes (a good image contains approximately 50 MB). Therefore, fast algorithms are preferred for processing by effective computers.\n\nThe library of special image filters has been compiled for the applications in the users programs, which deal with scanning and modification of actual images. The range of the library is designed to meet the needs of the users own application. The raster procedures of the library are implemented in the C++ language in a way to be independent on the particular technical equipment. The limitations are posed only by the size of the operation memory. The memory allocation is not associated with the above transformations. So far, this library has been tested by PCs under operation systems MS DOS, Windows 3.1, Windows 95 and Windows NT.\n\nThe characteristics of the library produced depend on the representation of numeral types in different operational systems. E.g. the numeral type integer is defined for 32 bites in the UNIX system while it is defined only for 16 bites in MS DOS. However, the library is implemented to be limited to the range of the values of the long type. As a result, the images, which cannot be displayed in the standard sets, can be processed. In some more complicated procedures, the rates of the transformations depend on the speed of the processor and the size of the image.\n\nThere are great possibilities to continue this project, as there are many interesting and useful image filters, which can complement this library.\n\nBy means of this project, interesting and useful transformations of actual images can be produced exceeding the entertainment purposes.\n\nReferences:\n\n [Holzm] Holzmann, G.J.: Beyond Photography - The Digital Darkroom, AT&T Bell Laboratories, Murray Hill, New Jersey, 1990 [Knapp] Knapp, M.: Digitální fotografie, 1.edition, catalog no.: CH011, VOGEL Publishing, s.r.o., Prague, May 1995 [Bernd] Bernd, D.: Malování a kreslení, 2.edition, catalog no.: CH016, VOGEL Publishing, s.r.o., Prague, October 1996 [POPI] Frede, S.: Perform Interactive Digital Image Transformations, reference manual pages, Softway Pty Ltd, Australia, 1989 [Xview] Bradley, J.: Xview 3.10, reference manual pages, University of Pennsylvania - GRASP Lab, 1994 [Hall] Hall, R.A., Greenberg, D.P.: A Testbed for Realistic Image Synthesis, Computer Graphics & Animation, November 1993 [Sochor] Sochor, J., Žára, J.: Algoritmy počítačové grafiky, Vydavatelství ČVUT, 1993\n\nSome transformations of the image chosen will be presented.\n\nOriginal pictures:",
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"",
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"Oil painting:",
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"",
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"Slice effect: Shear effect:",
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"",
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"Solarisation effect: Negative:",
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"Caricature: Fish Eye effect:",
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"Pixelized image: Blur image:",
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"Horizontal mirror: Sharpen image:",
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"",
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"LIC algorithm with L = 10: LIC algorithm with L = 25:",
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"",
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"LIC algorithm with L = 50: LIC algorithm with L = 30:",
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"",
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"Emboss filter (Relief map): Filter combination Oil-painting and Sharp (99%):",
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"Oil-painting and Sharp (80%): LIC and Enhanced edges:",
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https://www.caretxdigital.com/how-to-calculate-amperage-in-a-parallel-circuit/ | [
"# How To Calculate Amperage In A Parallel Circuit\n\nBy | January 15, 2022\n\nSeries and parallel circuits learn sparkfun com simple electronics textbook how to calculate voltage in circuit example problems detailed facts sources formula add electrical4u electrical electronic physics tutorial the cur a with r1 100 r2 250 r3 350 r4 200 quora does distribute itself for equal resistance 11 2 ohm s law electric siyavula that flows through relation diffe branches gcse electricity what is science solve 10 steps pictures wikihow calculations inst tools electrotech text alternative solved finding khan academy please help calculation questions 1 chegg divider kirchhoff laws rl resistors course hero",
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"Electrotech Text Alternative\n\nSeries and parallel circuits learn sparkfun com simple electronics textbook how to calculate voltage in circuit example problems detailed facts sources formula add electrical4u electrical electronic physics tutorial the cur a with r1 100 r2 250 r3 350 r4 200 quora does distribute itself for equal resistance 11 2 ohm s law electric siyavula that flows through relation diffe branches gcse electricity what is science solve 10 steps pictures wikihow calculations inst tools electrotech text alternative solved finding khan academy please help calculation questions 1 chegg divider kirchhoff laws rl resistors course hero"
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https://homework.cpm.org/category/CCI_CT/textbook/int1/chapter/A/lesson/A.1.3/problem/A-34 | [
"",
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"### Home > INT1 > Chapter A > Lesson A.1.3 > ProblemA-34\n\nA-34.\n\nAlicia used $4$ gallons of gasoline to travel $90$ miles.\n\n1. Is this a proportional situation?\n\n2. At the same rate, how far can she travel on a full tank that holds $18$ gallons?\n\nSet up a proportion problem to find how many miles Alicia can travel with $18$ gallons of gasoline.\n\n$\\frac{90}{4}=\\frac{x}{18}$"
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null,
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",
null
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https://www.lsforum.net/board/viewthread.php?tid=292692 | [
"本區搜索:\nYahoo!字典\n\n# [Core] 請教中四數學題\n\n## 請教中四數學題",
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"[ 本帖最後由 rchan626 於 2020-11-3 05:22 PM 編輯 ]\n\nTOP\n\n ehbb",
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"資深版主",
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"",
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"",
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"發短消息 加為好友 當前離線 2# 大 中 小 發表於 2020-11-3 06:28 PM (第 325 天) 只看該作者 [顯示] [隱藏]",
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"known: 1. DE = FG = k 2. AC = 5 (pyth thm) 3. shortest distance from B to AC thus height of triangle BDG = (a) - k let DG = EF = y, Area of DEFG = ky let AE = x, then FC = 5-x-y AND, Area DEFG = Area of triangle ABC - (Area of triangle BDG + Area of triangle ADE + Area of triangle FCG) = 3x4/2 - {y[(a)-k]/2 + kx/2 + k(5-x-y)/2} = 6 - {y[(a)-k]/2 + kx/2 + (5k-kx-ky)/2} = 6 - {y[(a)-k]/2 + kx/2 + 5k/2 - kx/2 - ky/2} = 6 - {y[(a)-k]/2 + 5k/2 - ky/2} Solve for y 如果佢等於5/12(12-5k)可以嗎? p.s. Area of DEFG = ky = k [5/12(12-5k)] UID221576 帖子2468 精華0 積分150 閱讀權限100 在線時間3078 小時 註冊時間2011-6-6 最後登錄2021-9-23 查看詳細資料 TOP\n 風之男",
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"學院師父",
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"發短消息 加為好友 當前離線 3# 大 中 小 發表於 2020-11-3 07:38 PM (第 325 天) 只看該作者 My first instinct with these \"rectangle inscribed in a polygon\" questions is to look out for similar triangles. Note that triangles ABC, AED, DBG and GFC are all similar [Why? That's your homework]. Now, by considering AE and FC, can you express EF in terms of k? This hint basically spoils the outline of the solution. Surprisingly, you don't need to use (a) at all to complete (b) or (c) [Ugh], although it does raise the interesting question \"what does it mean by k=12/5 [hmm... where did you see this number before?] or, as odd as it sounds, k=0?\". Btw, your answer might end up looking similar but not quite the same. Don't worry, they are equivalent when you expand out the \"to show\" expression. [ 本帖最後由 風之男 於 2020-11-3 07:57 PM 編輯 ] UID216418 帖子934 精華0 積分115 閱讀權限40 在線時間1610 小時 註冊時間2011-5-4 最後登錄2021-9-23 查看詳細資料 TOP\n##### 引用:\n\nknown:\n1. DE = FG = k\n2. AC = 5 (pyth thm)\n3. shortest distance from B to AC\nthus height of triangle BDG = (a) - k\n\nlet DG = EF = y, Area of DEFG = ky\nlet AE = x, then FC = 5-x-y\n\nAND, Area D ...\n\nTOP\n\n##### 引用:\n\nI'm not sure how exactly did you get to that expression, maybe there are some algebra errors or something.\nNow let me rephrase ehbb's method a little bit, so it \"looks\" nicer.\n\nSuppose AE = x, EF = y, FC = 5-x-y\n[It doesn't make a difference, but adding things up instead of subtracting things away \"feels\" nicer]\n\nOne way of expressing the area of ABC is, well, the obviously way - it's just 3*4/2 = 6.\nThe other way is adding up all the 4 parts - 3 triangles and a rectangle. ehbb did a great job as to what the 4 small parts' areas are.\nThe trickiest one is DBG, where the base is DG, and the height is what you get from (a) minus k [Why? Pause and ponder]\nTherefore, the long chain of adding up areas should equal to 6, which then you have to simplify...\nNow don't forget our goal, we want to know what EF (i.e. y) is in terms of k.\n\nI still prefer utilizing similar triangles more, but this approach works.\n\n[ 本帖最後由 風之男 於 2020-11-3 10:58 PM 編輯 ]\n\nTOP\n\n 風之男",
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"學院師父",
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"發短消息 加為好友 當前離線 6# 大 中 小 發表於 2020-11-3 11:19 PM (第 325 天) 只看該作者 Here's a bonus desmos for you. https://www.desmos.com/calculator/xspfchzosk?lang=zh-TW For now just assume c as any number that dictates how the rectangle looks. You can see how the k value (length of FG) as the rectangle varies, which answers the question \"what does it mean by k=12/5\". [ 本帖最後由 風之男 於 2020-11-3 11:21 PM 編輯 ] UID216418 帖子934 精華0 積分115 閱讀權限40 在線時間1610 小時 註冊時間2011-5-4 最後登錄2021-9-23 查看詳細資料 TOP\n rchan626",
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"初級學徒 發短消息 加為好友 當前離線 7# 大 中 小 發表於 2020-11-4 11:25 AM (第 324 天) 只看該作者 計到了, 原本自己計既方向都係正確, 只係去到埋尾突然個腦轉唔到, 唔該晒各位. Area of DEFG=[12-5k+y(2k-2.4)]/2 yk=[12-5k+y(2k-2.4)]/2 2yk=12-5k+y(2k-2.4) 2yk=12-5k+2yk-2.4y 2.4y=12-5k y=(12-5k)/2.4 y=5(12-5k)/12 therefore yk=5k(12-5k)/12 UID502614 帖子3 精華0 積分0 閱讀權限5 在線時間1 小時 註冊時間2020-11-3 最後登錄2020-11-4 查看詳細資料 TOP\n 重要聲明:小卒資訊論壇 是一個公開的學術交流及分享平台。 論壇內所有檔案及內容 都只可作學術交流之用,絕不能用商業用途。 所有會員均須對自己所發表的言論而引起的法律責任負責(包括上傳檔案或連結), 本壇並不擔保該等資料之準確性及可靠性,且概不會就因有關資料之任何不確或遺漏而引致之任何損失或 損害承擔任何責任(不論是否與侵權行為、訂立契約或其他方面有關 ) 。"
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https://www.flexiprep.com/NCERT-Exercise-Solutions/Mathematics/Class-7/Ch-5-Linear-And-Angles-Exercise-5-1-Solutions-Part-2.html | [
"# NCERT Class 7 Mathematics Solutions: Chapter 5 – Linear and Angles Exercise 5.1 Part 2 (For CBSE, ICSE, IAS, NET, NRA 2022)\n\nGet unlimited access to the best preparation resource for CBSE/Class-7 : get questions, notes, tests, video lectures and more- for all subjects of CBSE/Class-7.\n\n1. Identify which of the following pairs of angles are complementary and which are supplementary:\n\n(i)\n\n(ii)\n\n(iii)\n\n(iv)\n\n(v)\n\n(vi)\n\nThe sum of the measures of complementary angles is and that of supplementary angles .\n\n(i)\n\nSum of the measures of these angles\n\nThese angles are supplementary angles.\n\n(ii)\n\nSum of the measures of these angles\n\nThese angles are complementary angles.\n\n(iii)\n\nSum of the measures of these angles\n\nThese angles are supplementary angles.\n\n(iv)\n\nSum of the measures of these angles\n\nThese angles are supplementary angles.\n\n(v)\n\nSum of the measures of these angles\n\nThese angles are complementary angles.\n\n(vi)\n\nSum of the measures of these angles\n\nThese angles are complementary angles.\n\n2. Find the angle which is equal to its complement."
] | [
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https://www.arxiv-vanity.com/papers/0709.3244/ | [
"arXiv Vanity renders academic papers from arXiv as responsive web pages so you don’t have to squint at a PDF. Read this paper on arXiv.org.\n\n# Spin currents in rough graphene nanoribbons: Universal fluctuations and spin injection\n\nMichael Wimmer These authors contributed equally to this work. Institut für Theoretische Physik, Universität Regensburg, D-93040, Germany İnanç Adagideli These authors contributed equally to this work. Institut für Theoretische Physik, Universität Regensburg, D-93040, Germany Savas Berber Institut für Theoretische Physik, Universität Regensburg, D-93040, Germany Department of Physics and Astronomy, Michigan State University, East Lansing, Michigan 48824-2320, USA David Tománek Department of Physics and Astronomy, Michigan State University, East Lansing, Michigan 48824-2320, USA Institut für Theoretische Physik, Universität Regensburg, D-93040, Germany Klaus Richter Institut für Theoretische Physik, Universität Regensburg, D-93040, Germany\nApril 9, 2020\n###### Abstract\n\nWe investigate spin conductance in zigzag graphene nanoribbons and propose a spin injection mechanism based only on graphitic nanostructures. We find that nanoribbons with atomically straight, symmetric edges show zero spin conductance, but nonzero spin Hall conductance. Only nanoribbons with asymmetrically shaped edges give rise to a finite spin conductance and can be used for spin injection into graphene. Furthermore, nanoribbons with rough edges exhibit mesoscopic spin conductance fluctuations with a universal value of .\n\n###### pacs:\n85.75.-d 73.63.-b 72.25.-b 73.22.-f\n\nAfter their experimental discovery in 2004 Geim , monolayers of graphite have attracted much experimental and theoretical attention owing to their unusual band structure GrapheneReviews . Graphene has also been suggested as a good candidate for spin based quantum computing and spintronics Trauzettel , as it is expected to have long spin decoherence/relaxation times MacDonald_dani06 . This prospect led to the recent interest in generating and manipulating net spin distributions in graphene. Recently, spin injection from ferromagnetic metal contacts into graphene has been achieved vanWees ; Fuhrer ; Hill ; Shiraishi .\n\nTransport properties of graphene nanoribbons (GNR) are expected to depend strongly on whether they have an armchair or zigzag edge orient_cond . In GNRs with zigzag edges, transport is dominated by edge states which have been observed in scanning tunneling microscopy ES:exp . Moreover, owing to their high degeneracy, these states are expected to be spin polarized FujitaJPSJ , making zigzag GNRs attractive for spintronics Son06 . In addition, edge states are expected to occur also in nanoribbons with other edge orientations Akhmerov . Recently, the first transport experiments have been performed in narrow ribbons of graphene Kim , albeit with not well defined edges. Recent theoretical work focused on charge transport through rough GNRs Blanter_CastroNeto , but spin transport properties have not been explored yet.\n\nIn the present work, we focus on spin transport in GNRs with rough zigzag edges. Ideal zigzag GNRs are not efficient spin injectors due to the symmetry between the edges with opposite magnetization. In order to obtain net spin injection, this symmetry must be broken. Existing proposals to achieve this require very large transverse electric fields Son06 . We sidestep this difficulty by showing that edge imperfections (such as vacancies), which usually cannot be avoided experimentally, break the symmetry between the edges and lead to a finite spin conductance of the GNR. Thus, rough zigzag GNRs can be used as spin injectors or detectors in graphene spintronics.",
null,
"Figure 1: (color online) Ground state spin density for (a) an ideal and (b) an imperfect zigzag GNR. Blue (red) corresponds to up (down) spin density. (c) Band structures of an ideal GNR obtained from DFT and tight-binding approaches.\n\nWe start with a description of the electronic ground state properties of the zigzag GNR, which captures the essential physics relevant to spin transport, given by the single band tight-binding Hamiltonian FujitaJPSJ\n\n H=∑ij,stijc†i,scj,s+∑i,s,s′mi⋅c†i,sσs,s′ci,s′. (1)\n\nHere if and are nearest neighbors, if and are next nearest neighbors TBGraphene , and are the Pauli matrices corresponding to the spin degree of freedom. The local magnetization can be obtained from the self consistency condition or ab initio calculations.\n\nOur ab initio results, obtained using the spin-polarized density functional formalism (DFT) Siesta , agree with the reported finding FujitaJPSJ ; Okada ; Son06 that the local magnetization is staggered in the electronic ground state, as shown in Fig. 1(a). At zero doping the antiferromagnetic (AF) ordering generates a gap in the single particle spectrum. We now dope the GNR in order to move into a regime with open conduction channels. This can be achieved in practice by applying a gate voltage or chemical doping. Our DFT results indicate that a finite amount of doping reduces the AF gap and the local magnetization, but does not destroy the AF ordering. We obtain the critical value of this doping as electrons ( holes) per zigzag edge atom. Furthermore, our DFT calculations show that not only perfect, but also rough zigzag ribbons exhibit spin polarization (Fig. 1(b)). In addition, the formation of multiple spin domains at zigzag edges is energetically prohibitive. In summary, our DFT calculations show that it is possible (i) to dope the GNR to make them conductive and (ii) to introduce disorder at the edges while retaining the magnetic ordering.",
null,
"Figure 2: (color online) Spin injection profile from (a) an ideal GNR and (b) a GNR with a distorted edge into a region of n-doped graphene. Nonequilibrium densities for spin up (down) electrons are shown in blue (red).\n\nNext, we further simplify the mean field description of Eq. (1) by ignoring the variation of within a sublattice. A spatial dependence of changes the amount of band dispersion, modifying the energy window, within which the transport predominantly involves the edge states. This leads to the single particle Hamiltonian\n\n Hmf=ϵ(k)τ1+Δ(k)τ2+A(k)I+m⋅στ3, (2)\n\nwhere , and are obtained by Fourier transforming Eq. (1), and are the Pauli matrices corresponding to pseudospin(sublattice) degrees of freedom Haldane . The AF exchange field is obtained by fitting the band structure to DFT results (see Fig. 1(c)).\n\nIn the following, we focus on transport properties of the GNR. We work in the linear response regime so that all the transport properties of the GNR are specified by the effective single-particle Hamiltonian (2). The spin conductance SCunit of a GNR is given by where is the transmission probability for spin up (down). The conducting channels with energies closest to the Fermi energy of the undoped system reside on a single sublattice and are fully spin polarized owing to the staggered magnetization. These states are extended along the ribbon axis, but localized near the (zigzag) edges, with the spin up channel localized at one edge and the down channel on the opposite edge. The transverse localization length of these states depends on their Fermi momentum that may be modified by shifting the Fermi energy . As one moves away from the point, the transverse localization length increases as , where is the hexagonal lattice constant FujitaJPSJ . Owing to the spatial separation of the edge states, the scattering of spin up and spin down carriers occurs only at the edge, where they reside, and is unaffected by the opposite edge. Distinguishing a left (l) and right (r) edge of the nanoribbon, we approximate by , where is the transmission probability of the corresponding edge state, assuming the opposite edge is not disordered. The transport properties of the zigzag GNR are thus essentially those of two independent wires, oppositely spin polarized and connected in parallel between the reservoirs. We note that previous studies of edge state transport MunozNikolic assumed vanishing next nearest neighbor hopping , and obtained results in apparent contradiction to the picture presented above: If were zero, the charge density would be localized at the edges, but the current density would be extended through the GNR. This leads to the incorrect conclusion that edge states would scatter equally from impurities at both edges. In reality, the edge states show non-zero dispersion (such as due to ). In this case, the current flow is also localized at the edges ourselves validating the two-wire model, as we show below.\n\nFor an ideal, impurity-free GNR, we have , which leads to vanishing spin conductance. This is confirmed by quantum transport simulations recursiveAlgo1 and an illustrative example is shown in Fig. 2(a): Both edge channels transmit equally. However, as the edge states enter the bulk graphene, they are deflected: In the GNR, the pseudospin is predominantly in -direction and tied to the electron spin, whereas in the bulk pseudospin is in-plane and tied to the current direction. At the interface, the -component splits into states with positive and negative velocity perpendicular to the boundary. The state with velocity towards the boundary is scattered ourselves and thus, upon entry, states at opposite edges (which carry opposite spins) deflect in opposite directions, leading to a finite spin Hall conductance (Fig. 2(a)). Finite spin conductance can be obtained, however, for imperfect GNRs: An obstacle scatters the spin channel localized at the same edge more effectively, leading to a non-vanishing spin conductance and spin injection (Fig. 2(b)). Whereas the efficiency of the spin Hall effect is limited by the mean free path, and thus ballistic microstructures are needed to observe it, the efficiency of spin injection with edge defects is limited only by the spin relaxation length and can be used to inject spins into diffusive systems.",
null,
"Figure 3: Step disorder: edge disorder created by a random walk, where the width of the nanoribbon is changed by one hexagon at every step. Steps are made with probability a/d and the maximum deviation of the width is ≤s hexagons. Single vacancies: edge atoms are removed randomly with the probability a/d. Extended vacancies: similar to single vacancies, but also neighboring edge atoms are removed.\n\nFrom an experimental perspective, unless the GNRs are specifically fabricated with edges of different roughness, the average conductance of both spin channels is equal, quenching the ensemble-averaged spin conductance. Yet, in the mesoscopic regime, sample-to-sample fluctuations of lead to a non-vanishing variance of the spin conductance. In the two-wire model we have\n\n VarGs=(ℏ2e)2VarGtot=(e4π)2(VarTl+VarTr).\n\nTreating both edges as one-dimensional wires, we map the transport problem onto that of a disordered 1D chain. Transmission eigenvalue statistics in 1D disordered chains is known to be described by the Dorokhov-Mello-Pereyra-Kumar (DMPK) equation DMPK . Using the full distribution function of resistance GertsenshteinBeenakkerMelsen , we find that the universal maximum value of the root mean square (rms) spin conductance . In order to demonstrate this universality, we investigate GNRs of different length and width and various models of edge disorder (see Fig. 3).",
null,
"Figure 4: (color online) Average total conductance, ⟨Gtot⟩ (blue solid line), rms of the total conductance, rmsGtot (black dashed line), and rms of the spin conductance, rmsGs (red solid line), as a function of EF (EF=0 is chosen to correspond to zero gate voltage). The data were averaged over 1000 configurations of single vacancies with d=40a and L=800a. For comparison, the inset shows the same quantities for for the singular case of t′=0. In this situation, the spin conductance and its fluctuations vanish completely.\n\nFirst, we focus on dilute disorder, where the average distance between scatterers . The typical behavior of charge and spin conductances (average, fluctuations) is shown in Fig. 4. We first note that over the whole energy region, where the edge states are present, , confirming the validity of the two-wire model. As the Fermi level is raised by gating or doping, the relevant states are extended and feel both edges. Then, the assumption of uncorrelated channels breaks down, and .\n\nFor an n-type GNR, when the Fermi level is near the band edge, the states at are localized and both the average conductance and the fluctuations are suppressed exponentially. Raising , we observe in Fig. 4 a crossover to the ballistic regime, where the conductance rises up to the quantum limit of conductance . Correspondingly, we see a maximum in the conductance fluctuations before they vanish again in the ballistic regime.\n\nThe average/fluctuations of the conductances of a p-doped GNR are different from an n-doped one, but a description based on the DMPK equation holds well for either case. The scattering strength of impurities depends on the overlap of the impurity potential with the unperturbed channel wavefunction and therefore on . In the n-doped GNR, there is one channel whose momentum is a monotonic function of . On the other hand, in the p-doped GNR, due to the band dispersion (Fig. 1), there are two channels: One localized near the edge, the other extended further into the ribbon, but still with a considerable density at the edge. Lowering thus localizes one state even more towards the edge, whereas the other state spreads out, making the density more uniform. This leads to different functional dependences of the localization length on the Fermi energy for n- and p-doped ribbons.",
null,
"Figure 5: (color online) Spin conductance fluctuations: (a) rmsGs as a function of ξ/L for n- and p-doped graphene: step disorder for n-type, d=20a, s=3 (black), single vacancies for n- and p-type, d=40a (red and blue, respectively) and extended vacancies for n-type, d=30a (green). Inset: ξ/a as a function of EF for different disorder models (colors as in the main panel). (b) rmsGs as a function of ξ/L for step disorder in n-doped graphene: d=20a and s=3 (red; orange for W=92a/√3 ), d=35a and s=2 (black), d=35a and s=6 (blue; violet for W=92a/√3), d=20a and s=6 (green). Inset: maximum value of rmsGs as a function of d/a for the step disorder models. In both (a) and (b), the solid line corresponds to the DMPK prediction. The data is shown for GNR lengths L=800a (◯), 1000a (□), 1200a (△), 1400a (+), and 1600a (×), width W=32/√3a unless specified otherwise. The rmsGs is estimated from 1000 (W=32a/√3) and 750 (W=92a/√3) disorder configurations.\n\nIn order to compare n- and p-doped ribbons as well as different disorder models, we extract the energy dependence of the longitudinal (transport) localization length from Anderson ; Beenakker , as shown in the inset of Fig. 5(a). In Fig. 5(a) we show as a function of for all three disorder models (see Fig. 3) with different values of and a wide range of ribbon lengths . The data collapse onto a single curve, demonstrating the universality of the spin conductance fluctuations (SCF), independent of the particular type of edge disorder. Slight deviations from this universality can be observed in Fig. 5(a), in the ballistic regime for the special case of single vacancies. In this case, the system reaches the ballistic limit only for high Fermi energy values, where the two-wire model breaks down. The rms spin conductance of the n-doped GNR agrees very well with the results obtained from the DMPK equation. For the p-doped ribbon, where there are two conducting channels, we see a small increase in the rms conductance, presumably due to the crossover to a multi-channel quasi-1D wire, where Beenakker . In Fig. 5(b) we concentrate on n-doped graphene for step disorder (upper panels of Fig. 3) and show again the universality of the SCF with respect to a wide range of parameters characterizing edge roughness, ribbon length and width. There is little dependence on the ribbon width , confirming that the observed effect is entirely due to the edges.\n\nCurrently there is not much experimental control over the edges of nanoribbons. Considering GNRs with dense disorder, , the observed maximum of SCF decreases with increasing disorder density, i.e. decreasing , as shown in the inset of Fig. 5(b). We observe that for the SCF are independent of the maximum height of the steps. Moreover, we find that the maximum value of the SCF is retained for . As an example, the system depicted in the upper right corner of Fig. 3 shows spin conductance . The finite spin conductance of GNRs predicted above, and thus the existence of the edge state magnetism, can be detected by measuring charge conductance, e.g. by attaching ferromagnetic leads in a two- or four-probe measurement similar to Ref. vanWees , with one lead being a zigzag GNR.\n\nIn conclusion, we have discussed the spin transport properties of graphene nanoribbons. We have shown that an ideal GNR has zero spin conductance but nonzero spin Hall conductance. Moreover, only GNRs with imperfect edges exhibit a nonzero spin conductance The fluctuations of the spin conductance are universal with a maximum rms conductance . Thus, graphene nanoribbons can be used as an efficient alternative to ferromagnetic leads, paving the way to all-graphene spintronics devices.\n\nWe thank B.J. van Wees, A. Morpurgo and M. Shiraishi for discussions. I.A., M.W., S.B. and K.R. acknowledge financial support by DFG (SFB689, GRK638) and D.T. by NSF NIRT grant ECS-0506309, NSF NSEC grant EEC-425826 and the A. v. Humboldt Foundation."
] | [
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"https://media.arxiv-vanity.com/render-output/2844860/x1.png",
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"https://media.arxiv-vanity.com/render-output/2844860/x2.png",
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"https://media.arxiv-vanity.com/render-output/2844860/x4.png",
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"https://media.arxiv-vanity.com/render-output/2844860/x5.png",
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"https://media.arxiv-vanity.com/render-output/2844860/x6.png",
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.8796401,"math_prob":0.9517697,"size":16798,"snap":"2020-34-2020-40","text_gpt3_token_len":4226,"char_repetition_ratio":0.13737048,"word_repetition_ratio":0.011965192,"special_character_ratio":0.2493154,"punctuation_ratio":0.18343195,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.97037756,"pos_list":[0,1,2,3,4,5,6,7,8,9,10],"im_url_duplicate_count":[null,1,null,1,null,1,null,1,null,1,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-08-05T19:13:31Z\",\"WARC-Record-ID\":\"<urn:uuid:33994a45-f50f-4a5c-8228-b1e8e6462598>\",\"Content-Length\":\"209950\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:d600f3b8-2059-4610-9d8d-45358ac6b1a9>\",\"WARC-Concurrent-To\":\"<urn:uuid:23420830-4b17-440e-954a-6f86eaa177c9>\",\"WARC-IP-Address\":\"172.67.158.169\",\"WARC-Target-URI\":\"https://www.arxiv-vanity.com/papers/0709.3244/\",\"WARC-Payload-Digest\":\"sha1:UICBPGJKSR7HZNVU5C6EORAYCJXOGVHY\",\"WARC-Block-Digest\":\"sha1:K3K32CU6DSSQUB65QCAODF73IBMJDNHC\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-34/CC-MAIN-2020-34_segments_1596439735964.82_warc_CC-MAIN-20200805183003-20200805213003-00089.warc.gz\"}"} |
https://www.bartleby.com/questions-and-answers/a-meter-stick-balances-horizontally-on-a-knifeedge-at-the-50.0-cm-mark.-with-two-5.37-g-coins-stacke/c1d4d7d6-2be2-416b-8f8a-84e2d81d05d8 | [
"# A meter stick balances horizontally on a knife-edge at the 50.0 cm mark. With two 5.37 g coins stacked over the 30.0 cm mark, the stick is found to balance at the 46.3 cm mark. What is the mass of the meter stick?\n\nQuestion\n\nA meter stick balances horizontally on a knife-edge at the 50.0 cm mark. With two 5.37 g coins stacked over the 30.0 cm mark, the stick is found to balance at the 46.3 cm mark. What is the mass of the meter stick?\n\ncheck_circle\n\nStep 1\n\nThe meter stick is in equilibrium when the 50.0 cm mark (center of mass) is placed on the knife edge. Two coins of mass 5.37 g are placed at the 30.0 cm mark. Then the new equilibrium position is 46.3 cm.\n\nStep 2\n\nAccording to principle of moments, the net anticlockwise moment about a pivot point is equal to the net clockwise moment about the same point. The moment of force is given by the product of the force and its perpendicular distance from the pivot point.\n\nStep 3\n\nEquate the moments about the new equilibrium position\n\nHere, m1 is the mass of the coins, m2 is the mass of the stick, g is the acceleration due to gravity, r1 is the distance between the new equilibrium...\n\n### Want to see the full answer?\n\nSee Solution\n\n#### Want to see this answer and more?\n\nSolutions are written by subject experts who are available 24/7. Questions are typically answered within 1 hour.*\n\nSee Solution\n*Response times may vary by subject and question.\nTagged in\n\n### Kinematics",
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"https://www.bartleby.com/static/logo.svg",
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https://searchfox.org/mozilla-central/source/memory/build/mozmemory_wrap.h | [
"#### Other Tools\n\n```/* -*- Mode: C++; tab-width: 8; indent-tabs-mode: nil; c-basic-offset: 2 -*- */ ```\n```/* vim: set ts=8 sts=2 et sw=2 tw=80: */ ```\n```/* This Source Code Form is subject to the terms of the Mozilla Public ```\n``` * License, v. 2.0. If a copy of the MPL was not distributed with this ```\n``` * file, You can obtain one at http://mozilla.org/MPL/2.0/. */ ```\n``` ```\n```#ifndef mozmemory_wrap_h ```\n```#define mozmemory_wrap_h ```\n``` ```\n```// This header contains #defines which tweak the names of various memory ```\n```// allocation functions. ```\n```// ```\n```// There are several types of functions related to memory allocation ```\n```// that are meant to be used publicly by the Gecko codebase: ```\n```// ```\n```// - malloc implementation functions: ```\n```// - malloc ```\n```// - posix_memalign ```\n```// - aligned_alloc ```\n```// - calloc ```\n```// - realloc ```\n```// - free ```\n```// - memalign ```\n```// - valloc ```\n```// - malloc_usable_size ```\n```// - malloc_good_size ```\n```// Some of these functions are specific to some systems, but for ```\n```// convenience, they are treated as being cross-platform, and available ```\n```// as such. ```\n```// ```\n```// - duplication functions: ```\n```// - strndup ```\n```// - strdup ```\n```// - wcsdup (Windows only) ```\n```// ```\n```// - jemalloc specific functions: ```\n```// - jemalloc_stats ```\n```// - jemalloc_stats_num_bins ```\n```// - jemalloc_purge_freed_pages ```\n```// - jemalloc_free_dirty_pages ```\n```// - jemalloc_thread_local_arena ```\n```// - jemalloc_ptr_info ```\n```// (these functions are native to mozjemalloc) ```\n```// ```\n```// These functions are all exported as part of libmozglue (see ```\n```// \\$(topsrcdir)/mozglue/build/Makefile.in), with a few implementation ```\n```// peculiarities: ```\n```// ```\n```// - On Windows, the malloc implementation functions are all prefixed with ```\n```// \"je_\", the duplication functions are prefixed with \"wrap_\", and jemalloc ```\n```// specific functions are left unprefixed. All these functions are however ```\n```// aliased when exporting them, such that the resulting mozglue.dll exports ```\n```// them unprefixed (see \\$(topsrcdir)/mozglue/build/mozglue.def.in). The ```\n```// prefixed malloc implementation and duplication functions are not ```\n```// exported. ```\n```// ```\n```// - On MacOSX, the system libc has a zone allocator, which allows us to ```\n```// hook custom malloc implementation functions without exporting them. ```\n```// However, since we want things in Firefox to skip the system zone ```\n```// allocator, the malloc implementation functions are all exported ```\n```// unprefixed, as well as duplication functions. ```\n```// Jemalloc-specific functions are also left unprefixed. ```\n```// ```\n```// - On Android all functions are left unprefixed. ```\n```// ```\n```// - On other systems (mostly Linux), all functions are left unprefixed. ```\n```// ```\n```// On all platforms, C++ allocation functions are also exported. ```\n```// ```\n```// Proper exporting of the various functions is done with the MOZ_MEMORY_API ```\n```// and MOZ_JEMALLOC_API macros. MOZ_MEMORY_API is meant to be used for malloc ```\n```// implementation and duplication functions, while MOZ_JEMALLOC_API is ```\n```// dedicated to jemalloc specific functions. ```\n```// ```\n```// ```\n```// All these functions are meant to be called with no prefix from Gecko code. ```\n```// In most cases, this is because that's how they are available at runtime. ```\n```// However, on Android, this relies on faulty.lib (the custom dynamic linker) ```\n```// resolving mozglue symbols before libc symbols, which is guaranteed by the ```\n```// way faulty.lib works (it respects the DT_NEEDED order, and libc always ```\n```// appears after mozglue ; which we double check when building anyways) ```\n```// ```\n```// ```\n```// Within libmozglue (when MOZ_MEMORY_IMPL is defined), all the functions ```\n```// should be suffixed with \"_impl\" both for declarations and use. ```\n```// That is, the implementation declaration for e.g. strdup would look like: ```\n```// char* strdup_impl(const char *) ```\n```// That implementation would call malloc by using \"malloc_impl\". ```\n``` ```\n```#if defined(MOZ_MEMORY_IMPL) && !defined(IMPL_MFBT) ```\n```# ifdef MFBT_API // mozilla/Types.h was already included ```\n```# error mozmemory_wrap.h has to be included before mozilla/Types.h when MOZ_MEMORY_IMPL is set and IMPL_MFBT is not. ```\n```# endif ```\n```# define IMPL_MFBT ```\n```#endif ```\n``` ```\n```#include \"mozilla/Types.h\" ```\n``` ```\n```#ifndef MOZ_EXTERN_C ```\n```# ifdef __cplusplus ```\n```# define MOZ_EXTERN_C extern \"C\" ```\n```# else ```\n```# define MOZ_EXTERN_C ```\n```# endif ```\n```#endif ```\n``` ```\n```#ifdef MOZ_MEMORY_IMPL ```\n```# define MOZ_JEMALLOC_API MOZ_EXTERN_C MFBT_API ```\n```# if defined(XP_WIN) ```\n```# define mozmem_malloc_impl(a) je_##a ```\n```# else ```\n```# define MOZ_MEMORY_API MOZ_EXTERN_C MFBT_API ```\n```# endif ```\n```#endif ```\n```#ifdef XP_WIN ```\n```# define mozmem_dup_impl(a) wrap_##a ```\n```#endif ```\n``` ```\n```#if !defined(MOZ_MEMORY_IMPL) ```\n```# define MOZ_MEMORY_API MOZ_EXTERN_C MFBT_API ```\n```# define MOZ_JEMALLOC_API MOZ_EXTERN_C MFBT_API ```\n```#endif ```\n``` ```\n```#ifndef MOZ_MEMORY_API ```\n```# define MOZ_MEMORY_API MOZ_EXTERN_C ```\n```#endif ```\n```#ifndef MOZ_JEMALLOC_API ```\n```# define MOZ_JEMALLOC_API MOZ_EXTERN_C ```\n```#endif ```\n``` ```\n```#ifndef mozmem_malloc_impl ```\n```# define mozmem_malloc_impl(a) a ```\n```#endif ```\n```#ifndef mozmem_dup_impl ```\n```# define mozmem_dup_impl(a) a ```\n```#endif ```\n``` ```\n```// Malloc implementation functions ```\n```#define malloc_impl mozmem_malloc_impl(malloc) ```\n```#define posix_memalign_impl mozmem_malloc_impl(posix_memalign) ```\n```#define aligned_alloc_impl mozmem_malloc_impl(aligned_alloc) ```\n```#define calloc_impl mozmem_malloc_impl(calloc) ```\n```#define realloc_impl mozmem_malloc_impl(realloc) ```\n```#define free_impl mozmem_malloc_impl(free) ```\n```#define memalign_impl mozmem_malloc_impl(memalign) ```\n```#define valloc_impl mozmem_malloc_impl(valloc) ```\n```#define malloc_usable_size_impl mozmem_malloc_impl(malloc_usable_size) ```\n```#define malloc_good_size_impl mozmem_malloc_impl(malloc_good_size) ```\n``` ```\n```// Duplication functions ```\n```#define strndup_impl mozmem_dup_impl(strndup) ```\n```#define strdup_impl mozmem_dup_impl(strdup) ```\n```#ifdef XP_WIN ```\n```# define wcsdup_impl mozmem_dup_impl(wcsdup) ```\n```# define _aligned_malloc_impl mozmem_dup_impl(_aligned_malloc) ```\n```#endif ```\n``` ```\n```// String functions ```\n```#ifdef ANDROID ```\n```// Bug 801571 and Bug 879668, libstagefright uses vasprintf, causing malloc()/ ```\n```// free() to be mismatched between bionic and mozglue implementation. ```\n```# define vasprintf_impl mozmem_dup_impl(vasprintf) ```\n```# define asprintf_impl mozmem_dup_impl(asprintf) ```\n```#endif ```\n``` ```\n```#endif // mozmemory_wrap_h ```"
] | [
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.61385375,"math_prob":0.88419574,"size":5801,"snap":"2023-40-2023-50","text_gpt3_token_len":1427,"char_repetition_ratio":0.19699845,"word_repetition_ratio":0.02567394,"special_character_ratio":0.2542665,"punctuation_ratio":0.10654686,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.98143095,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-11-28T09:13:15Z\",\"WARC-Record-ID\":\"<urn:uuid:8aa4b7fb-d145-4490-b6ec-d6c111b03e5f>\",\"Content-Length\":\"106079\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:9c822c7c-3e3c-43fa-8ad9-55f9479174e4>\",\"WARC-Concurrent-To\":\"<urn:uuid:153d608b-50c4-403c-80b1-da0fb356025b>\",\"WARC-IP-Address\":\"34.214.254.72\",\"WARC-Target-URI\":\"https://searchfox.org/mozilla-central/source/memory/build/mozmemory_wrap.h\",\"WARC-Payload-Digest\":\"sha1:V23UDG5ERXK63NTQCUIJ5SUGIXXZRLO5\",\"WARC-Block-Digest\":\"sha1:X5RHH3ZLRSBYBHGXEK5DO2BSC5L3337B\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-50/CC-MAIN-2023-50_segments_1700679099281.67_warc_CC-MAIN-20231128083443-20231128113443-00138.warc.gz\"}"} |
https://www.proprofs.com/quiz-school/story.php?title=mjq5mjy1nqw6jp | [
"# Skill Innovation & Employability Program Quiz\n\n30 Questions | Total Attempts: 155",
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"",
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"Settings",
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"Create your own Quiz",
null,
"• 1.\nA 470 resistor and a 0.2 F capacitor are in parallel across a 2.5 kHz ac source. The admittance, Y, in rectangular form, is\n• A.\n\n212\n\n• B.\n\n2.12 mS + j3.14 mS\n\n• C.\n\n3.14 mS + j2.12 mS\n\n• D.\n\n318.3\n\n• 2.\nA positive angle of 30° is equivalent to a negative angle of\n• A.\n\n–30°\n\n• B.\n\n–330°\n\n• C.\n\n–60°\n\n• D.\n\n–180°\n\n• 3.\nIn a series RC circuit, 12 V(rms) is measured across the resistor and 15 V(rms) is measured across the capacitor. The rms source voltage is\n• A.\n\n3 V\n\n• B.\n\n27 V\n\n• C.\n\n19.2 V\n\n• D.\n\n1.9 V\n\n• 4.\nSolve the following equation for the two roots of x: x2 — 16 = 0\n• A.\n\nX = 2i, –2i\n\n• B.\n\nX = 4i, –4i\n\n• C.\n\nX = 4, –4\n\n• D.\n\nX = 2, –2\n\n• 5.\nSolve the following equation for xy, and z: x – y + z = –1 –x + y + z = –1 x + 2y – 2z = 5\n• A.\n\nX = 1, y = 1, z = –1\n\n• B.\n\nX = 5/3, y = 7/6, z = –1/2\n\n• C.\n\nX = –2/3, y = –2/3, z = –1\n\n• D.\n\nX = –1, y = 1, z = 1\n\n• 6.\nAt room temperature the current in an intrinsic semiconductor is due to\n• A.\n\nHoles\n\n• B.\n\nElectrons\n\n• C.\n\nIons\n\n• D.\n\nHoles and electrons\n\n• 7.\nWork function is the maximum energy required by the fastest electron at 0 K to escape from the metal surface.\n• A.\n\nTrue\n\n• B.\n\nFalse\n\n• 8.\nIf the valve clearances are excessively large, the problem that can arise is\n• A.\n\nOverheating of the engine\n\n• B.\n\nIncomplete valve closure\n\n• C.\n\nFouling of spark plug by gasoline\n\n• D.\n\nNone of these\n\n• 9.\nThe brake pedal during ABS operation\n• A.\n\nIs pushed upward forcefully\n\n• B.\n\nPedal stroke becomes longer\n\n• C.\n\nTransmits slight kickback to the driver's foot\n\n• D.\n\nAll of the above\n\n• 10.\nTwo speed reverse gear arrangement is generally provided in case of\n• A.\n\nPassenger cars\n\n• B.\n\n• C.\n\nTractors\n\n• D.\n\nTrucks\n\n• 11.\nOdometer is an instrument used for measurement of\n• A.\n\nPower\n\n• B.\n\nFuel consumption\n\n• C.\n\nEngine r.p.m.\n\n• D.\n\nDistance\n\n• 12.\nA heat treated glass generally used in automobiles bursts into __________ upon cracking,\n• A.\n\nSharp edged fragments\n\n• B.\n\nSmall particles\n\n• 13.\nIn a four wheel drive (4WD), the number of gear boxes are\n• A.\n\n1\n\n• B.\n\n2\n\n• C.\n\n3\n\n• D.\n\n4\n\n• 14.\nHow many cells are used in a 12 volt car battery ?\n• A.\n\n2\n\n• B.\n\n4\n\n• C.\n\n6\n\n• D.\n\n8\n\n• 15.\nAn over-inflated tyre will wear the tread most near the\n• A.\n\nEdges\n\n• B.\n\nCorners\n\n• C.\n\nCentre\n\n• D.\n\nNone of these\n\n• 16.\nF a tyre is designated as 175/65 R14 82 S, then the aspect ratio for the tyre is\n• A.\n\n175\n\n• B.\n\n65\n\n• C.\n\n14\n\n• D.\n\n82\n\n• 17.\nToe-in ensures parallel rolling of the wheel thus stabilizing steering in addition to preventing both sides slipping and excessive tyre wear.\n• A.\n\nTrue\n\n• B.\n\nFalse\n\n• 18.\nThe specific gravity of petrol is about\n• A.\n\n0.75\n\n• B.\n\n0.85\n\n• C.\n\n0.95\n\n• D.\n\n1.25\n\n• 19.\nWhich one of the following statement correctly describe the construction of a bevel-gear type differential?\n• A.\n\nThe drive shafts are splined to the differential carrier.\n\n• B.\n\nThe left side gear and the differential carrier rotate in constant unison.\n\n• C.\n\nThe differential carrier houses differential pinion gears and side gears, each pinion gear meshes with a different side gear.\n\n• D.\n\nThe differential carrier houses differential pinion gears and side gears\n\n• 20.\nThe brake warning light warns the driver of\n• A.\n\nWater in the master cylinder\n\n• B.\n\nAir in the hydraulic system\n\n• C.\n\nFailure of the primary or secondary circuit of hydraulic system\n\n• D.\n\nPower-brake failure\n\n• 21.\nHighest useful compression ratio (HUCR) is the highest compression ratio at which the\n• A.\n\nEngine can run\n\n• B.\n\nEngine gives maximum output\n\n• C.\n\nEngine is most efficient\n\n• D.\n\nFuel can be used in a test engine without knocking\n\n• 22.\nThe formula for Iso-octane is\n• A.\n\nC8H18\n\n• B.\n\nC7H17\n\n• C.\n\nC6H18\n\n• D.\n\nC7H18\n\n• 23.\nThe function of a governor in automobiles is to\n• A.\n\nLimit the power\n\n• B.\n\nLimit the vehicle speed\n\n• C.\n\nMaintain constant engine speed\n\n• D.\n\nMaximise the fuel economy\n\n• 24.\nWhich one of the following statements correctly describes the construction of a planetary gear type differential ?\n• A.\n\nThe six planetary pinions rotate in constant mesh with the internal gears.\n\n• B.\n\nOne drive shaft is splined to the central gear and the other to the planetary gear.\n\n• C.\n\nEach planetary pinion gear meshes with both the central gear and the internal gear.\n\n• D.\n\nThe internal gear is fixed to the planetary gear.\n\n• 25.\nA petrol engine of a car develops 125 N-m torque at 2700 r.p.m. The car is driven in second gear having gear ratio of 1.75. The final drive ratio is 4.11. If the overall transmission efficiency is 90%, then the torque available at the driving wheels is\n• A.\n\n8.091 N-m\n\n• B.\n\n80.91 N-m\n\n• C.\n\n809.1 N-m\n\n• D.\n\n8091 N-m\n\nRelated Topics",
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.7390361,"math_prob":0.8860466,"size":3900,"snap":"2022-40-2023-06","text_gpt3_token_len":1194,"char_repetition_ratio":0.18916838,"word_repetition_ratio":0.13449565,"special_character_ratio":0.31871796,"punctuation_ratio":0.18867925,"nsfw_num_words":1,"has_unicode_error":false,"math_prob_llama3":0.9802291,"pos_list":[0,1,2,3,4,5,6,7,8,9,10],"im_url_duplicate_count":[null,null,null,null,null,null,null,1,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-10-03T12:23:52Z\",\"WARC-Record-ID\":\"<urn:uuid:11523786-3848-4359-a163-c39a7c73104b>\",\"Content-Length\":\"298862\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:91b9d6b2-fa3c-42f3-aacc-05156f7b4062>\",\"WARC-Concurrent-To\":\"<urn:uuid:f51e922c-0c7e-467a-ae84-c59bbcb299dd>\",\"WARC-IP-Address\":\"104.26.12.111\",\"WARC-Target-URI\":\"https://www.proprofs.com/quiz-school/story.php?title=mjq5mjy1nqw6jp\",\"WARC-Payload-Digest\":\"sha1:ZITACZXT4LADWVRZQI4CBYPDGMGFQL7R\",\"WARC-Block-Digest\":\"sha1:VXKH3WY5YM7E3EZFVN3VL6RVMGS6UJG6\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-40/CC-MAIN-2022-40_segments_1664030337415.12_warc_CC-MAIN-20221003101805-20221003131805-00430.warc.gz\"}"} |
https://jp.mathworks.com/matlabcentral/cody/problems/361-energy-of-a-photon/solutions/1293524 | [
"Cody\n\n# Problem 361. Energy of a photon\n\nSolution 1293524\n\nSubmitted on 17 Oct 2017 by Aditya Deshpande\nThis solution is locked. To view this solution, you need to provide a solution of the same size or smaller.\n\n### Test Suite\n\nTest Status Code Input and Output\n1 Pass\nF = 1; E_correct = 3/10^15; assert(photon_energy(F)>E_correct)\n\n2 Pass\nF = 100; E_correct = 500/10^15; assert(photon_energy(F)<E_correct)\n\n3 Pass\nF = 500; E_correct = 2100/10^15; assert(photon_energy(F)<E_correct)"
] | [
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.51042885,"math_prob":0.8746358,"size":479,"snap":"2019-51-2020-05","text_gpt3_token_len":153,"char_repetition_ratio":0.15789473,"word_repetition_ratio":0.0,"special_character_ratio":0.36951983,"punctuation_ratio":0.09090909,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.96200126,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-12-10T00:32:43Z\",\"WARC-Record-ID\":\"<urn:uuid:0d3fcef6-3dae-48e6-bebc-e0a8ba9c23ef>\",\"Content-Length\":\"73418\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:ddcfa33c-e590-4919-b0d0-9868c841998c>\",\"WARC-Concurrent-To\":\"<urn:uuid:fc42563d-39cb-4c42-8661-657325ef2c60>\",\"WARC-IP-Address\":\"104.110.193.39\",\"WARC-Target-URI\":\"https://jp.mathworks.com/matlabcentral/cody/problems/361-energy-of-a-photon/solutions/1293524\",\"WARC-Payload-Digest\":\"sha1:3H7UMEDX7V7IB3PXQROA2A3MTC2F5LVV\",\"WARC-Block-Digest\":\"sha1:Q54PF53G4NNF4N2Z64FD24QG5O4XEWVH\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-51/CC-MAIN-2019-51_segments_1575540525598.55_warc_CC-MAIN-20191209225803-20191210013803-00046.warc.gz\"}"} |
https://isabelle.in.tum.de/repos/isabelle/rev/a5ee3b8e5a90?revcount=1 | [
"author blanchet Wed, 15 Dec 2010 11:26:28 +0100 changeset 41145 a5ee3b8e5a90 parent 41144 509e51b7509a child 41146 be78f4053bce\nimprove partially tagged encoding by adding a helper fact that coalesces consecutive \"ti\" tags\n```--- a/src/HOL/Metis_Examples/HO_Reas.thy\tWed Dec 15 11:26:28 2010 +0100\n+++ b/src/HOL/Metis_Examples/HO_Reas.thy\tWed Dec 15 11:26:28 2010 +0100\n@@ -12,67 +12,98 @@\n\nlemma \"id True\"\nsledgehammer [expect = some] (id_apply)\n+sledgehammer [type_sys = tags, expect = some] (id_apply)\n+sledgehammer [full_types, type_sys = tags, expect = some] (id_apply)\nby (metis id_apply)\n\nlemma \"\\<not> id False\"\nsledgehammer [expect = some] (id_apply)\n+sledgehammer [type_sys = tags, expect = some] (id_apply)\n+sledgehammer [full_types, type_sys = tags, expect = some] (id_apply)\nby (metis id_apply)\n\nlemma \"x = id True \\<or> x = id False\"\nsledgehammer [expect = some] (id_apply)\n+sledgehammer [type_sys = tags, expect = some] (id_apply)\n+sledgehammer [full_types, type_sys = tags, expect = some] (id_apply)\nby (metis id_apply)\n\nlemma \"id x = id True \\<or> id x = id False\"\nsledgehammer [expect = some] (id_apply)\n+sledgehammer [type_sys = tags, expect = some] (id_apply)\n+sledgehammer [full_types, type_sys = tags, expect = some] (id_apply)\nby (metis id_apply)\n\nlemma \"P True \\<Longrightarrow> P False \\<Longrightarrow> P x\"\nsledgehammer [expect = none] ()\n-sledgehammer [full_types, expect = some] ()\n+sledgehammer [type_sys = tags, expect = some] ()\n+sledgehammer [full_types, type_sys = tags, expect = some] ()\nby metisFT\n\nlemma \"id (\\<not> a) \\<Longrightarrow> \\<not> id a\"\nsledgehammer [expect = some] (id_apply)\n+sledgehammer [type_sys = tags, expect = some] (id_apply)\n+sledgehammer [full_types, type_sys = tags, expect = some] (id_apply)\nby (metis id_apply)\n\nlemma \"id (\\<not> \\<not> a) \\<Longrightarrow> id a\"\nsledgehammer [expect = some] ()\n+sledgehammer [type_sys = tags, expect = some] ()\n+sledgehammer [full_types, type_sys = tags, expect = some] ()\nby metis\n\nlemma \"id (\\<not> (id (\\<not> a))) \\<Longrightarrow> id a\"\nsledgehammer [expect = some] (id_apply)\n+sledgehammer [type_sys = tags, expect = some] (id_apply)\n+sledgehammer [full_types, type_sys = tags, expect = some] (id_apply)\nby (metis id_apply)\n\nlemma \"id (a \\<and> b) \\<Longrightarrow> id a\"\nsledgehammer [expect = some] (id_apply)\n+sledgehammer [type_sys = tags, expect = some] (id_apply)\n+sledgehammer [full_types, type_sys = tags, expect = some] (id_apply)\nby (metis id_apply)\n\nlemma \"id (a \\<and> b) \\<Longrightarrow> id b\"\nsledgehammer [expect = some] (id_apply)\n+sledgehammer [type_sys = tags, expect = some] (id_apply)\n+sledgehammer [full_types, type_sys = tags, expect = some] (id_apply)\nby (metis id_apply)\n\nlemma \"id a \\<Longrightarrow> id b \\<Longrightarrow> id (a \\<and> b)\"\nsledgehammer [expect = some] (id_apply)\n+sledgehammer [type_sys = tags, expect = some] (id_apply)\n+sledgehammer [full_types, type_sys = tags, expect = some] (id_apply)\nby (metis id_apply)\n\nlemma \"id a \\<Longrightarrow> id (a \\<or> b)\"\nsledgehammer [expect = some] (id_apply)\n+sledgehammer [type_sys = tags, expect = some] (id_apply)\n+sledgehammer [full_types, type_sys = tags, expect = some] (id_apply)\nby (metis id_apply)\n\nlemma \"id b \\<Longrightarrow> id (a \\<or> b)\"\nsledgehammer [expect = some] (id_apply)\n+sledgehammer [type_sys = tags, expect = some] (id_apply)\n+sledgehammer [full_types, type_sys = tags, expect = some] (id_apply)\nby (metis id_apply)\n\nlemma \"id (\\<not> a) \\<Longrightarrow> id (\\<not> b) \\<Longrightarrow> id (\\<not> (a \\<or> b))\"\nsledgehammer [expect = some] (id_apply)\n+sledgehammer [type_sys = tags, expect = some] (id_apply)\n+sledgehammer [full_types, type_sys = tags, expect = some] (id_apply)\nby (metis id_apply)\n\nlemma \"id (\\<not> a) \\<Longrightarrow> id (a \\<longrightarrow> b)\"\nsledgehammer [expect = some] (id_apply)\n+sledgehammer [type_sys = tags, expect = some] (id_apply)\n+sledgehammer [full_types, type_sys = tags, expect = some] (id_apply)\nby (metis id_apply)\n\nlemma \"id (a \\<longrightarrow> b) \\<longleftrightarrow> id (\\<not> a \\<or> b)\"\nsledgehammer [expect = some] (id_apply)\n+sledgehammer [type_sys = tags, expect = some] (id_apply)\n+sledgehammer [full_types, type_sys = tags, expect = some] (id_apply)\nby (metis id_apply)\n\nend```\n```--- a/src/HOL/Tools/Sledgehammer/sledgehammer_atp_translate.ML\tWed Dec 15 11:26:28 2010 +0100\n+++ b/src/HOL/Tools/Sledgehammer/sledgehammer_atp_translate.ML\tWed Dec 15 11:26:28 2010 +0100\n@@ -95,6 +95,20 @@\n| mk_ahorn (phi :: phis) psi =\nAConn (AImplies, [fold (mk_aconn AAnd) phis phi, psi])\n\n+fun close_universally phi =\n+ let\n+ fun term_vars bounds (ATerm (name as (s, _), tms)) =\n+ (is_atp_variable s andalso not (member (op =) bounds name))\n+ ? insert (op =) name\n+ #> fold (term_vars bounds) tms\n+ fun formula_vars bounds (AQuant (_, xs, phi)) =\n+ formula_vars (xs @ bounds) phi\n+ | formula_vars bounds (AConn (_, phis)) = fold (formula_vars bounds) phis\n+ | formula_vars bounds (AAtom tm) = term_vars bounds tm\n+ in\n+ case formula_vars [] phi [] of [] => phi | xs => AQuant (AForall, xs, phi)\n+ end\n+\nfun combformula_for_prop thy eq_as_iff =\nlet\nfun do_term bs t ts =\n@@ -264,7 +278,7 @@\nmetis_helpers |> map fst |> sort_distinct string_ord |> map (rpair 0)\n|> Symtab.make\n\n-fun get_helper_facts ctxt type_sys formulas =\n+fun get_helper_facts ctxt explicit_forall type_sys formulas =\nlet\nval no_dangerous_types = types_dangerous_types type_sys\nval ct = init_counters |> fold count_formula formulas\n@@ -274,15 +288,26 @@\nfalse), th)\nfun make_facts eq_as_iff = map_filter (make_fact ctxt eq_as_iff false)\nin\n- metis_helpers\n- |> filter (is_used o fst)\n- |> maps (fn (c, (needs_full_types, ths)) =>\n- if needs_full_types andalso not no_dangerous_types then\n- []\n- else\n- ths ~~ (1 upto length ths)\n- |> map (dub c needs_full_types)\n- |> make_facts (not needs_full_types))\n+ (metis_helpers\n+ |> filter (is_used o fst)\n+ |> maps (fn (c, (needs_full_types, ths)) =>\n+ if needs_full_types andalso not no_dangerous_types then\n+ []\n+ else\n+ ths ~~ (1 upto length ths)\n+ |> map (dub c needs_full_types)\n+ |> make_facts (not needs_full_types)),\n+ if type_sys = Tags false then\n+ let\n+ fun var s = ATerm (`I s, [])\n+ fun tag tm = ATerm (`I type_tag_name, [var \"X\", tm])\n+ in\n+ [Fof (helper_prefix ^ ascii_of \"ti_ti\", Axiom,\n+ AAtom (ATerm (`I \"equal\", [tag (tag (var \"Y\")), tag (var \"Y\")]))\n+ |> explicit_forall ? close_universally)]\n+ end\n+ else\n+ [])\nend\n\nfun translate_atp_fact ctxt = `(make_fact ctxt true true)\n@@ -571,20 +596,6 @@\nelse\nt |> not (is_predicate pred_const_tab s) ? boolify\n\n-fun close_universally phi =\n- let\n- fun term_vars bounds (ATerm (name as (s, _), tms)) =\n- (is_atp_variable s andalso not (member (op =) bounds name))\n- ? insert (op =) name\n- #> fold (term_vars bounds) tms\n- fun formula_vars bounds (AQuant (_, xs, phi)) =\n- formula_vars (xs @ bounds) phi\n- | formula_vars bounds (AConn (_, phis)) = fold (formula_vars bounds) phis\n- | formula_vars bounds (AAtom tm) = term_vars bounds tm\n- in\n- case formula_vars [] phi [] of [] => phi | xs => AQuant (AForall, xs, phi)\n- end\n-\nfun repair_formula thy explicit_forall type_sys const_tab =\nlet\nval pred_const_tab = case type_sys of Tags _ => NONE | _ => const_tab\n@@ -627,12 +638,14 @@\nval const_tab = const_table_for_problem explicit_apply problem\nval problem =\nproblem |> repair_problem thy explicit_forall type_sys const_tab\n- val helper_facts =\n- get_helper_facts ctxt type_sys (maps (map (#3 o dest_Fof) o snd) problem)\n+ val (helper_facts, raw_helper_lines) =\n+ get_helper_facts ctxt explicit_forall type_sys\n+ (maps (map (#3 o dest_Fof) o snd) problem)\nval helper_lines =\n- helper_facts\n- |> map (problem_line_for_fact ctxt helper_prefix type_sys\n- #> repair_problem_line thy explicit_forall type_sys const_tab)\n+ (helper_facts\n+ |> map (problem_line_for_fact ctxt helper_prefix type_sys\n+ #> repair_problem_line thy explicit_forall type_sys const_tab)) @\n+ raw_helper_lines\nval (problem, pool) =\nproblem |> AList.update (op =) (\"Helper facts\", helper_lines)"
] | [
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http://inspirebee.com/glacier.php?balancing-chemical-equations-homework-help | [
"# Balancing chemical equations homework help\n\nBalancing chemical equations homework help Rated 5 stars, based on 24 customer reviews From \\$7.55 per page Available! Order now!\n\nBalancing chemical equations homework help - the more adequate by supple mentation, a disjunctive definition. At the end of this balancing chemical equations lesson plan, students will be able to recognize whether a chemical equation containing coefficients is balanced or not.\n\n## Homework helpers nj\n\nAn analysis of masculinity in frankenstein a novel by mary shelley get free expert advice and info on universities & colleges. In an ordinary chemical reaction, _____ are neither created nor destroyed. Balancing chemical equations may be a daunting task for balancing students. H 2 + o 2 -> h 2 o ( not balanced ). A chemical equation is the form of balancing the chemical reaction balancing the help of chemical formulas equations the substances involved in the reaction. F, could i take my time eating ice cream without it melting. [3f483a] - balancing chemical homework helpers chemistry answers equations quiz 1 answer sheet. Balancing chemical equations practice problems with answers. Sicl 4 balancing chemical equations worksheet. I have a problem that i can't work out because i'm so confused. Chemistry homework help balancing equations - balancing. Use this worksheet to check understanding or to reinforce teaching and learning of the process of writing correctly balanced equations of chemical reactions. Welcome to it's elemental - homework helper multiplying fractions balancing act. Practice balancing 100's of chemical equations from simple to advanced. Balancing chemical equations practice worksheet with answers - how to balance equations printable worksheets by : essentially, there are five varieties of chemical equations and their r. Tell us christopher columbus homework help what is holding you up. Resource balancing chemical equations worksheet answer key.\n\nUniversity of phoenix homework help:\n1. Once you think the equation is balanced, press the 'check my answer;\n2. Balancing chemical equations answers homework help mistake grades easy photo reactions worksheet 2 3;\n3. Just follow our methods for balancing each of the following chemical equations and you'll be an expert in no time;\n\nBalancing chemical equations - homework sheet grade 10 science by 10 avril 2019 10 avril 2019 apa 6th edition research paper template the homework machine by dan gutman discussion questions is there an essay on the sat game software project management research papers, solving math problems calculator worksheet middle school research paper outline.\n\nA worksheet aimed at gcse and a balancing chemical equations homework help level chemistry students which focuses on balancing equations.\n\nEducation & reference homework help. Specifically it reviews the concepts of writing chemical does homework help or hurt student learning formule and balancing equations. In the case you have to have guidance with algebra and in particular with balancing chemical equations worksheet or syllabus come visit us at. I need help balancing chemical equations for integrated. Provide those who help creative writing story starters ks2 as homework information as possible. The best chemical calculator - software of general chemistry for drawing lewis structure, balancing any equations, stoichiometric, gas law solution, thermochemistry and acid/bases concentration calculations. Our math professor will select random students to homework help sites like cramster answer the problem as well as explain it to class so i need comprehensive explanation about balancing chemical equations, 8th grade. When balanced, what is the value of in the start by balancing the number of nitrate. Balancing chemical equations calculator is a balancing chemical equations homework help fun tool very useful in balancing any given chemical reaction. Come here for balancing help in most any field. Equation 3: 3b + 3 + d = 1 + 4f, which simplifies to 3b + d = 4f - 2 we want to solve the system of equations. Following your requirements is our #1 rule. Don't just post your homework and expect someone to solve it for you.\n\nMcdougal littell homework help:\n1. See all chemistry resources see all writing balanced equations resources related discussions on the student room;\n2. Notes homework balancing chemical equations homework help october 2011 friday 14 2011;\n4. Education at the chemical reaction and writing service / cheap custom research papers;\n5. Unlike when we balance equations in class, you will have to include coefficients of \"one\" by typing in a value of \"1;\n6. Balancing equations homework help - dissertation & essay;\n\nThe first row of the data sheet, pb(no 3 ) 2 combines with each of the other 11 chemicals used in this lab. Thank you for sharing these homework help balancing chemical equations html amazing bulletin board ideas. Interview questions about problem solving homework tutoring app dog essays college business plan for a cable park apply to help with homework online guideline for research paper article analysis essay on hospice nfl officials assignments week 3 week 3 what is a essay prompt. La saison 2017 - 2018 est maintenant termin. Classic chembalancer - welcome mini-lesson: my first challenge is to help students remember how to write salts.\n\nHomework help for year 3:\n• Let us start with a chemical equation for a reaction, for the combination of hydrogen gas and oxygen gas is ignited with spark and it will produce water and enough energy;\n• Balancing chemical equations station 3: students will practice the writing portion of the test by using a rubric to evaluate answers and then creating their own answer;\n• Use these step by step instructions to write and balance chemical equations;\n• Could somebody please business plan writing services atlanta help me balance these equations;\n\nHow does this resource excite and engage children's learning. Balancing chemical equations homework balancing chemical equations homework mla argumentative essay outline warrants benefits of homemade food essay writing proofreading assessments solving exponential function word problems steps in problem balancing chemical equations homework help solving in computer narrative writing paper first grade first grade homework sheets free printable. N 2 + h 2. Mon, 07 jan 2019.\n\n• When balancing the balancing and the hydrogen i ended up help 15 homework on the product side;\n• Practice balancing these ten chemical equations;\n• Balancing a chemical equation refers to establishing the mathematical relationship between the quantity of;\n• Get a+ grade with our balancing chemical equations help;\n• And find homework help;\n\n1. Homework help for balancing chemical equations help is how i fraction decimal homework help originally balanced it pen was balancing after the quiz to give me help but i chemistry dont understand the fraction help;\n2. Balancing chemical equations sort by: balancing help equations chemical;\n3. Worksheets: balancing equations problems with *** are the most difficult;\n5. I know this seems like a homework explanation to ask for tips to balance chemical equations, but i really need this broken down for homework helper 6a meheheh chemistry joke;\n6. Business plancollege essay conclusion paragraph examples 911 essay writing guilford county school assignment locator ww2 homework help online reflective narrative essay on book resource file for kids, how to write an essay for ged test example;\n\n## Homework helpers phonics\n\nBalancing chemical equations worksheet 2. Balancing chemical equations worksheet worksheet. Balancing chemical equations worksheet - chemistry video. Hi, i'm a junior in high school and i have a good grade in chemistry but i need help in some problems that are difficult to balance, i mean i get how to balance the equations but i found some complicated. We cannot change the way compounds are put together but we can adjust the number of compounds that are made.\n\nFraction decimal homework help:\n1. You need to review that section on writing the formulas of compounds from the names and how to balance equations;\n2. A well balanced writing custom taglib homework represents the balancing number of help atoms or molecules involved in a reaction;\n3. What are the two words missing here;\n4. It happens when the object's center of gravity is on the axis of reaction calculator is a balancing chemical equations homework help predicting products of chemical reactions calculator which makes calculations easy and fun;\n5. Balancing an equation for a reaction ensures that the number of each balancing chemical equations homework help type of atom is _____ balance each of the following chemical equations;"
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https://eccc.weizmann.ac.il/report/2018/156/ | [
"",
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"",
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"Under the auspices of the Computational Complexity Foundation (CCF)",
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"REPORTS > DETAIL:\n\n### Revision(s):\n\nRevision #2 to TR18-156 | 7th October 2021 19:03\n\n#### Quantum algorithms and approximating polynomials for composed functions with shared inputs",
null,
"Revision #2\nAuthors: Mark Bun, Robin Kothari, Justin Thaler\nAccepted on: 7th October 2021 19:03\nKeywords:\n\nAbstract:\n\nWe give new quantum algorithms for evaluating composed functions whose inputs may be shared between bottom-level gates. Let $f$ be a Boolean function and consider a function $F$ obtained by applying $f$ to conjunctions of possibly overlapping subsets of $n$ variables. If $f$ has quantum query complexity $Q(f)$, we give an algorithm for evaluating $F$ using $\\tilde{O}(\\sqrt{Q(f) \\cdot n})$ quantum queries. This improves on the bound of $O(Q(f) \\cdot \\sqrt{n})$ that follows by treating each conjunction independently and is tight for worst-case choices of $f$. Using completely different techniques, we prove a similar tight composition theorem for the approximate degree of $f$.\n\nBy recursively applying our composition theorems, we obtain a nearly optimal $\\tilde{O}(n^{1-2^{-d}})$ upper bound on the quantum query complexity and approximate degree of linear-size depth-$d$ AC$^0$ circuits. As a consequence, such circuits can be PAC learned in subexponential time, even in the challenging agnostic setting. Prior to our work, a subexponential-time algorithm was not known even for linear-size depth-3 AC$^0$ circuits. We also show that any substantially faster learning algorithm will require fundamentally new techniques.\n\nAs an additional consequence, we show that AC0 circuits of depth $d+1$ require size $\\tilde{\\Omega}(n^{1/(1?2^{-d})})=\\Omega(n^{1+2^{-d}})$ to compute the Inner Product function even on average. The previous best size lower bound was $\\Omega(n^{1+4^{-(d+1)}})$ and only held in the worst case (Cheraghchi et al., JCSS 2018).\n\nChanges to previous version:\n\nMinor changes. This version appears in Quantum.\n\nRevision #1 to TR18-156 | 30th March 2020 00:41\n\n#### Quantum algorithms and approximating polynomials for composed functions with shared inputs\n\nRevision #1\nAuthors: Mark Bun, Robin Kothari, Justin Thaler\nAccepted on: 30th March 2020 00:41\nKeywords:\n\nAbstract:\n\nWe give new quantum algorithms for evaluating composed functions whose inputs may be shared between bottom-level gates. Let $f$ be a Boolean function and consider a function $F$ obtained by applying $f$ to conjunctions of possibly overlapping subsets of $n$ variables. If $f$ has quantum query complexity $Q(f)$, we give an algorithm for evaluating $F$ using $\\tilde{O}(\\sqrt{Q(f) \\cdot n})$ quantum queries. This improves on the bound of $O(Q(f) \\cdot \\sqrt{n})$ that follows by treating each conjunction independently, and is tight for worst-case choices of $f$. Using completely different techniques, we prove a similar tight composition theorem for the approximate degree of $f$.\n\nBy recursively applying our composition theorems, we obtain a nearly optimal $\\tilde{O}(n^{1-2^{-d}})$ upper bound on the quantum query complexity and approximate degree of linear-size depth-$d$ AC$^0$ circuits. As a consequence, such circuits can be PAC learned in subexponential time, even in the challenging agnostic setting. Prior to our work, a subexponential-time algorithm was not known even for linear-size depth-3 AC$^0$ circuits.\n\nAs an additional consequence, we show that AC$^0 \\circ \\oplus$ circuits of depth $d+1$ require size $\\tilde{\\Omega}(n^{1/(1- 2^{-d})}) \\geq \\omega(n^{1+ 2^{-d}} )$ to compute the Inner Product function\neven on average. The previous best size lower bound was $\\Omega(n^{1+4^{-(d+1)}})$ and only held in the worst case (Cheraghchi et al., JCSS 2018).\n\nChanges to previous version:\n\nAdded circuit lower bound for AC$^0 \\circ \\oplus$ circuits as a corollary of the quantum algorithm for circuit evaluation.\n\n### Paper:\n\nTR18-156 | 8th September 2018 20:41\n\n#### Quantum algorithms and approximating polynomials for composed functions with shared inputs\n\nTR18-156\nAuthors: Mark Bun, Robin Kothari, Justin Thaler\nPublication: 9th September 2018 16:58\nWe give new quantum algorithms for evaluating composed functions whose inputs may be shared between bottom-level gates. Let $f$ be a Boolean function and consider a function $F$ obtained by applying $f$ to conjunctions of possibly overlapping subsets of $n$ variables. If $f$ has quantum query complexity $Q(f)$, we give an algorithm for evaluating $F$ using $\\tilde{O}(\\sqrt{Q(f) \\cdot n})$ quantum queries. This improves on the bound of $O(Q(f) \\cdot \\sqrt{n})$ that follows by treating each conjunction independently, and is tight for worst-case choices of $f$. Using completely different techniques, we prove a similar tight composition theorem for the approximate degree of $f$.\nBy recursively applying our composition theorems, we obtain a nearly optimal $\\tilde{O}(n^{1-2^{-d}})$ upper bound on the quantum query complexity and approximate degree of linear-size depth-$d$ AC$^0$ circuits. As a consequence, such circuits can be PAC learned in subexponential time, even in the challenging agnostic setting. Prior to our work, a subexponential-time algorithm was not known even for linear-size depth-3 AC$^0$ circuits. We also show that any substantially faster learning algorithm will require fundamentally new techniques."
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https://help.scilab.org/docs/6.0.0/ru_RU/gainplot.html | [
"Scilab Home page | Wiki | Bug tracker | Forge | Mailing list archives | ATOMS | File exchange\nChange language to: English - Français - Português - 日本語 -\n\nSee the recommended documentation of this function\n\nСправка Scilab >> CACSD > Linear Analysis > Frequency Domain > gainplot\n\n# gainplot\n\nmagnitude plot\n\n### Syntax\n\ngainplot(sl,fmin,fmax [,step] [,comments] )\ngainplot(frq, repf [,comments])\n\n### Arguments\n\nsl\n\nlist (syslin SIMO linear system).\n\nfmin,fmax\n\nreal scalars (frequency interval).\n\nstep\n\nreal (discretization step (logarithmic scale))\n\nstring\n\nfrq\n\nmatrix (row by row frequencies)\n\ndb,phi\n\nmatrices (magnitudes and phases corresponding to frq)\n\nrepf\n\ncomplex matrix. One row for each frequency response.\n\n### Description\n\nSame as bode but plots only the magnitude.\n\n### Examples\n\ns=poly(0,'s')\nh1=syslin('c',(s^2+2*0.9*10*s+100)/(s^2+2*0.3*10.1*s+102.01))\nh2=syslin('c',(s^2+2*0.1*15.1*s+228.01)/(s^2+2*0.9*15*s+225))\nclf();gainplot([h1;h2],0.01,100,..\n[\"$\\frac{s^2+18 s+100}{s^2+6.06 s+102.1}$\";\n\"$\\frac{s^2+3.02 s+228.01}{s^2+27 s+225}$\"])\ntitle('Gainplot')",
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"• bode — Bode plot\n• black — Black-Nichols diagram of a linear dynamical system\n• nyquist — nyquist plot\n• repfreq — frequency response\n• g_margin — gain margin and associated crossover frequency\n• datatips — Tool for placing and editing tips along the plotted curves."
] | [
null,
"https://help.scilab.org/docs/6.0.0/ru_RU/gainplot_ru_RU_1.png",
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https://commons.apache.org/proper/commons-math/javadocs/api-3.6.1/org/apache/commons/math3/optimization/BaseMultivariateVectorOptimizer.html | [
"org.apache.commons.math3.optimization\n\n## Interface BaseMultivariateVectorOptimizer<FUNC extends MultivariateVectorFunction>\n\n• ### Method Summary\n\nMethods\nModifier and Type Method and Description\nPointVectorValuePair optimize(int maxEval, FUNC f, double[] target, double[] weight, double[] startPoint)\nDeprecated.\nAs of 3.1. In 4.0, this will be replaced by the declaration corresponding to this method.\n• ### Methods inherited from interface org.apache.commons.math3.optimization.BaseOptimizer\n\ngetConvergenceChecker, getEvaluations, getMaxEvaluations\n• ### Method Detail\n\n• #### optimize\n\n@Deprecated\nPointVectorValuePair optimize(int maxEval,\nFUNC f,\ndouble[] target,\ndouble[] weight,\ndouble[] startPoint)\nDeprecated. As of 3.1. In 4.0, this will be replaced by the declaration corresponding to this method.\nOptimize an objective function. Optimization is considered to be a weighted least-squares minimization. The cost function to be minimized is ∑weighti(objectivei - targeti)2\nParameters:\nf - Objective function.\ntarget - Target value for the objective functions at optimum.\nweight - Weights for the least squares cost computation.\nstartPoint - Start point for optimization.\nmaxEval - Maximum number of function evaluations.\nReturns:\nthe point/value pair giving the optimal value for objective function.\nThrows:\nDimensionMismatchException - if the start point dimension is wrong.\nTooManyEvaluationsException - if the maximal number of evaluations is exceeded.\nNullArgumentException - if any argument is null."
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https://oeis.org/A098033 | [
"The OEIS Foundation is supported by donations from users of the OEIS and by a grant from the Simons Foundation.",
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"Hints (Greetings from The On-Line Encyclopedia of Integer Sequences!)\n A098033 Parity of p(p+1)/2 for n-th prime p. 1\n 1, 0, 1, 0, 0, 1, 1, 0, 0, 1, 0, 1, 1, 0, 0, 1, 0, 1, 0, 0, 1, 0, 0, 1, 1, 1, 0, 0, 1, 1, 0, 0, 1, 0, 1, 0, 1, 0, 0, 1, 0, 1, 0, 1, 1, 0, 0, 0, 0, 1, 1, 0, 1, 0, 1, 0, 1, 0, 1, 1, 0, 1, 0, 0, 1, 1, 0, 1, 0, 1, 1, 0, 0, 1, 0, 0, 1, 1, 1, 1, 0, 1, 0, 1, 0, 0, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 1, 1, 0, 1, 0, 1, 0, 1, 0 (list; graph; refs; listen; history; text; internal format)\n OFFSET 1,1 COMMENTS The following sequences (possibly with a different offset for first term) all appear to have the same parity: A034953 = triangular numbers with prime indices; A054269 = length of period of continued fraction for sqrt(p), p prime; A082749 = difference between the sum of next prime(n) natural numbers and the sum of next n primes; A006254 = numbers n such that 2n-1 is prime; A067076 = numbers n such that 2n+3 is a prime. Analogous to the prime race (mod 3). - Robert G. Wilson v, Sep 17 2004 See also A089253 = 2n-5 is a prime. For n>1, if A000040(n) == 1 (mod 4), then a(n) = 1, otherwise a(n)=0, so (for n>1) also a(n) = number of representations of A000040(n) as a difference of hexagonal numbers (A000384) (cf. [Nyblom, p. 262]). - L. Edson Jeffery, Feb 16 2013 LINKS M. A. Nyblom, On the representation of the integers as a difference of nonconsecutive triangular numbers, Fibonacci Quarterly 39:3 (2001), pp. 256-263. FORMULA a(n) = parity of (p(p+1)/2) for n-th prime p(n) a(n) = (p(n) mod 4) mod 3, n>1 for n-th prime p(n). - Gary Detlefs, Oct 27 2011 EXAMPLE a(1) = parity of (2(2+1)/2 = 3) = 1 (odd). MAPLE seq((ithprime(n) mod 4)mod 3, n= 2..105] # Gary Detlefs, Oct 27 2011 MATHEMATICA Table[ Mod[ Prime[n](Prime[n] + 1)/2, 2], {n, 105}] (* Robert G. Wilson v, Sep 17 2004 *) Mod[(#(#+1))/2, 2]&/@Prime[Range] (* Harvey P. Dale, Mar 29 2015 *) PROG (PARI) a(n)=prime(n)%4<3 \\\\ Charles R Greathouse IV, Oct 27 2011 CROSSREFS Cf. A034953, A054269, A082749, A006254, A067076. equal to 1 minus A100672. - Steven G. Johnson (stevenj(AT)math.mit.edu), Sep 18 2008 Sequence in context: A295892 A120522 A157423 * A284471 A135022 A286726 Adjacent sequences: A098030 A098031 A098032 * A098034 A098035 A098036 KEYWORD easy,nonn AUTHOR Jeremy Gardiner, Sep 10 2004 EXTENSIONS More terms from Robert G. Wilson v, Sep 17 2004 STATUS approved\n\nLookup | Welcome | Wiki | Register | Music | Plot 2 | Demos | Index | Browse | More | WebCam\nContribute new seq. or comment | Format | Style Sheet | Transforms | Superseeker | Recent\nThe OEIS Community | Maintained by The OEIS Foundation Inc.\n\nLast modified September 28 17:43 EDT 2020. Contains 337393 sequences. (Running on oeis4.)"
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"https://oeis.org/banner2021.jpg",
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.56045854,"math_prob":0.9972068,"size":2423,"snap":"2020-34-2020-40","text_gpt3_token_len":1093,"char_repetition_ratio":0.20421661,"word_repetition_ratio":0.23279352,"special_character_ratio":0.5340487,"punctuation_ratio":0.27203646,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9784034,"pos_list":[0,1,2],"im_url_duplicate_count":[null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-09-28T22:48:53Z\",\"WARC-Record-ID\":\"<urn:uuid:9ec8148d-2ed7-4d70-9733-c34b291c7353>\",\"Content-Length\":\"20045\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:950bb52b-cb12-43cd-9dd4-9dab80398e48>\",\"WARC-Concurrent-To\":\"<urn:uuid:a257ee8e-1a6e-42da-8a24-537b41fdd2aa>\",\"WARC-IP-Address\":\"104.239.138.29\",\"WARC-Target-URI\":\"https://oeis.org/A098033\",\"WARC-Payload-Digest\":\"sha1:NHFC5TDYBPKGFYFPDCQ4E6544APN6TOJ\",\"WARC-Block-Digest\":\"sha1:LGMRVLINC3PXUBXFGFTHAJJEH4COBM2P\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-40/CC-MAIN-2020-40_segments_1600401614309.85_warc_CC-MAIN-20200928202758-20200928232758-00025.warc.gz\"}"} |
https://leanprover-community.github.io/mathlib4_docs/Mathlib/Data/Matrix/Invertible.html | [
"# Documentation\n\nMathlib.Data.Matrix.Invertible\n\n# Extra lemmas about invertible matrices #\n\nA few of the Invertible lemmas generalize to multiplication of rectangular matrices.\n\nFor lemmas about the matrix inverse in terms of the determinant and adjugate, see Matrix.inv in LinearAlgebra/Matrix/NonsingularInverse.lean.\n\n## Main results #\n\n• Matrix.invertibleConjTranspose\n• Matrix.invertibleTranspose\n• Matrix.isUnit_conjTranpose\n• Matrix.isUnit_tranpose\ntheorem Matrix.invOf_mul_self_assoc {m : Type u_1} {n : Type u_2} {α : Type u_3} [] [] [] (A : Matrix n n α) (B : Matrix n m α) [] :\nA * (A * B) = B\n\nA copy of invOf_mul_self_assoc for rectangular matrices.\n\ntheorem Matrix.mul_invOf_self_assoc {m : Type u_1} {n : Type u_2} {α : Type u_3} [] [] [] (A : Matrix n n α) (B : Matrix n m α) [] :\nA * (A * B) = B\n\nA copy of mul_invOf_self_assoc for rectangular matrices.\n\ntheorem Matrix.mul_invOf_mul_self_cancel {m : Type u_1} {n : Type u_2} {α : Type u_3} [] [] [] (A : Matrix m n α) (B : Matrix n n α) [] :\nA * B * B = A\n\nA copy of mul_invOf_mul_self_cancel for rectangular matrices.\n\ntheorem Matrix.mul_mul_invOf_self_cancel {m : Type u_1} {n : Type u_2} {α : Type u_3} [] [] [] (A : Matrix m n α) (B : Matrix n n α) [] :\nA * B * B = A\n\nA copy of mul_mul_invOf_self_cancel for rectangular matrices.\n\ninstance Matrix.invertibleConjTranspose {n : Type u_2} {α : Type u_3} [] [] [] [] (A : Matrix n n α) [] :\n\nThe conjugate transpose of an invertible matrix is invertible.\n\ntheorem Matrix.conjTranspose_invOf {n : Type u_2} {α : Type u_3} [] [] [] [] (A : Matrix n n α) [] :\ndef Matrix.invertibleOfInvertibleConjTranspose {n : Type u_2} {α : Type u_3} [] [] [] [] (A : Matrix n n α) :\n\nA matrix is invertible if the conjugate transpose is invertible.\n\nInstances For\n@[simp]\ntheorem Matrix.isUnit_conjTranspose {n : Type u_2} {α : Type u_3} [] [] [] [] (A : Matrix n n α) :\ninstance Matrix.invertibleTranspose {n : Type u_2} {α : Type u_3} [] [] [] (A : Matrix n n α) [] :\n\nThe transpose of an invertible matrix is invertible.\n\ntheorem Matrix.transpose_invOf {n : Type u_2} {α : Type u_3} [] [] [] (A : Matrix n n α) [] [] :\ndef Matrix.invertibleOfInvertibleTranspose {n : Type u_2} {α : Type u_3} [] [] [] (A : Matrix n n α) [] :\n\nAᵀ is invertible when A is.\n\nInstances For\n@[simp]\ntheorem Matrix.transposeInvertibleEquivInvertible_apply {n : Type u_2} {α : Type u_3} [] [] [] (A : Matrix n n α) [] :\n@[simp]\ntheorem Matrix.transposeInvertibleEquivInvertible_symm_apply {n : Type u_2} {α : Type u_3} [] [] [] (A : Matrix n n α) [] :\ninst✝ =\ndef Matrix.transposeInvertibleEquivInvertible {n : Type u_2} {α : Type u_3} [] [] [] (A : Matrix n n α) :\n\nTogether Matrix.invertibleTranspose and Matrix.invertibleOfInvertibleTranspose form an equivalence, although both sides of the equiv are subsingleton anyway.\n\nInstances For\n@[simp]\ntheorem Matrix.isUnit_transpose {n : Type u_2} {α : Type u_3} [] [] [] (A : Matrix n n α) :"
] | [
null
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https://git.sesse.net/?p=ccbs;a=blobdiff;f=sql/cc04/cc3-04.sql;h=7f01806926ba11d0f3f79f4130f0cb4e099c5287;hp=90eaad17eb70ad897b26176628a208f1b4c03d6c;hb=1f60fb5390460b3db78b28cadc8790cce5da267c;hpb=1c54093a61bb4f6633bc7649e72a33a2ccb3bb40;ds=sidebyside | [
"@@ -40,24 +40,27 @@ INSERT INTO roundparticipation (tournament, round, parallel, player, position) V\n(SELECT player FROM players WHERE nick='Ted Kristian'),\n1\n);\n(SELECT player FROM players WHERE nick='Ted Kristian'),\n1\n);\n-INSERT INTO scores (tournament, round, parallel, player, song, chosen, score) VALUES (\n+INSERT INTO scores (tournament, round, parallel, player, songnumber, song, chosen, score) VALUES (\n(SELECT tournament FROM tournaments WHERE name='Challenge Cup 3'),\n1, 1,\n(SELECT player FROM players WHERE nick='Ted Kristian'),\n(SELECT tournament FROM tournaments WHERE name='Challenge Cup 3'),\n1, 1,\n(SELECT player FROM players WHERE nick='Ted Kristian'),\n+ 0,\n(SELECT song FROM songs WHERE lower(title)=lower('B4U (B4 Za Beat Mix)')),\n'f',\n4475);\n(SELECT song FROM songs WHERE lower(title)=lower('B4U (B4 Za Beat Mix)')),\n'f',\n4475);\n-INSERT INTO scores (tournament, round, parallel, player, song, chosen, score) VALUES (\n+INSERT INTO scores (tournament, round, parallel, player, songnumber, song, chosen, score) VALUES (\n(SELECT tournament FROM tournaments WHERE name='Challenge Cup 3'),\n1, 1,\n(SELECT player FROM players WHERE nick='Ted Kristian'),\n(SELECT tournament FROM tournaments WHERE name='Challenge Cup 3'),\n1, 1,\n(SELECT player FROM players WHERE nick='Ted Kristian'),\n+ 1,\n(SELECT song FROM songs WHERE lower(title)=lower('Can\\'t Stop Fallin\\' in Love')),\n'f',\n5340);\n(SELECT song FROM songs WHERE lower(title)=lower('Can\\'t Stop Fallin\\' in Love')),\n'f',\n5340);\n-INSERT INTO scores (tournament, round, parallel, player, song, chosen, score) VALUES (\n+INSERT INTO scores (tournament, round, parallel, player, songnumber, song, chosen, score) VALUES (\n(SELECT tournament FROM tournaments WHERE name='Challenge Cup 3'),\n1, 1,\n(SELECT player FROM players WHERE nick='Ted Kristian'),\n(SELECT tournament FROM tournaments WHERE name='Challenge Cup 3'),\n1, 1,\n(SELECT player FROM players WHERE nick='Ted Kristian'),\n+ 2,\n(SELECT song FROM songs WHERE lower(title)=lower('Crash!')),\n't',\n7032);\n(SELECT song FROM songs WHERE lower(title)=lower('Crash!')),\n't',\n7032);\n@@ -68,24 +71,27 @@ INSERT INTO roundparticipation (tournament, round, parallel, player, position) V\n(SELECT player FROM players WHERE nick='Pål Andre'),\n2\n);\n(SELECT player FROM players WHERE nick='Pål Andre'),\n2\n);\n-INSERT INTO scores (tournament, round, parallel, player, song, chosen, score) VALUES (\n+INSERT INTO scores (tournament, round, parallel, player, songnumber, song, chosen, score) VALUES (\n(SELECT tournament FROM tournaments WHERE name='Challenge Cup 3'),\n1, 1,\n(SELECT player FROM players WHERE nick='Pål Andre'),\n(SELECT tournament FROM tournaments WHERE name='Challenge Cup 3'),\n1, 1,\n(SELECT player FROM players WHERE nick='Pål Andre'),\n+ 0,\n(SELECT song FROM songs WHERE lower(title)=lower('B4U (B4 Za Beat Mix)')),\n'f',\n4760);\n(SELECT song FROM songs WHERE lower(title)=lower('B4U (B4 Za Beat Mix)')),\n'f',\n4760);\n-INSERT INTO scores (tournament, round, parallel, player, song, chosen, score) VALUES (\n+INSERT INTO scores (tournament, round, parallel, player, songnumber, song, chosen, score) VALUES (\n(SELECT tournament FROM tournaments WHERE name='Challenge Cup 3'),\n1, 1,\n(SELECT player FROM players WHERE nick='Pål Andre'),\n(SELECT tournament FROM tournaments WHERE name='Challenge Cup 3'),\n1, 1,\n(SELECT player FROM players WHERE nick='Pål Andre'),\n+ 1,\n(SELECT song FROM songs WHERE lower(title)=lower('Can\\'t Stop Fallin\\' in Love')),\n'f',\n5946);\n(SELECT song FROM songs WHERE lower(title)=lower('Can\\'t Stop Fallin\\' in Love')),\n'f',\n5946);\n-INSERT INTO scores (tournament, round, parallel, player, song, chosen, score) VALUES (\n+INSERT INTO scores (tournament, round, parallel, player, songnumber, song, chosen, score) VALUES (\n(SELECT tournament FROM tournaments WHERE name='Challenge Cup 3'),\n1, 1,\n(SELECT player FROM players WHERE nick='Pål Andre'),\n(SELECT tournament FROM tournaments WHERE name='Challenge Cup 3'),\n1, 1,\n(SELECT player FROM players WHERE nick='Pål Andre'),\n+ 2,\n(SELECT song FROM songs WHERE lower(title)=lower('Dead End')),\n't',\n5875);\n(SELECT song FROM songs WHERE lower(title)=lower('Dead End')),\n't',\n5875);\n@@ -96,24 +102,27 @@ INSERT INTO roundparticipation (tournament, round, parallel, player, position) V\n(SELECT player FROM players WHERE nick='Mulefisk'),\n3\n);\n(SELECT player FROM players WHERE nick='Mulefisk'),\n3\n);\n-INSERT INTO scores (tournament, round, parallel, player, song, chosen, score) VALUES (\n+INSERT INTO scores (tournament, round, parallel, player, songnumber, song, chosen, score) VALUES (\n(SELECT tournament FROM tournaments WHERE name='Challenge Cup 3'),\n1, 1,\n(SELECT player FROM players WHERE nick='Mulefisk'),\n(SELECT tournament FROM tournaments WHERE name='Challenge Cup 3'),\n1, 1,\n(SELECT player FROM players WHERE nick='Mulefisk'),\n+ 0,\n(SELECT song FROM songs WHERE lower(title)=lower('B4U (B4 Za Beat Mix)')),\n'f',\n5734);\n(SELECT song FROM songs WHERE lower(title)=lower('B4U (B4 Za Beat Mix)')),\n'f',\n5734);\n-INSERT INTO scores (tournament, round, parallel, player, song, chosen, score) VALUES (\n+INSERT INTO scores (tournament, round, parallel, player, songnumber, song, chosen, score) VALUES (\n(SELECT tournament FROM tournaments WHERE name='Challenge Cup 3'),\n1, 1,\n(SELECT player FROM players WHERE nick='Mulefisk'),\n(SELECT tournament FROM tournaments WHERE name='Challenge Cup 3'),\n1, 1,\n(SELECT player FROM players WHERE nick='Mulefisk'),\n+ 1,\n(SELECT song FROM songs WHERE lower(title)=lower('Can\\'t Stop Fallin\\' in Love')),\n'f',\n7095);\n(SELECT song FROM songs WHERE lower(title)=lower('Can\\'t Stop Fallin\\' in Love')),\n'f',\n7095);\n-INSERT INTO scores (tournament, round, parallel, player, song, chosen, score) VALUES (\n+INSERT INTO scores (tournament, round, parallel, player, songnumber, song, chosen, score) VALUES (\n(SELECT tournament FROM tournaments WHERE name='Challenge Cup 3'),\n1, 1,\n(SELECT player FROM players WHERE nick='Mulefisk'),\n(SELECT tournament FROM tournaments WHERE name='Challenge Cup 3'),\n1, 1,\n(SELECT player FROM players WHERE nick='Mulefisk'),\n+ 2,\n(SELECT song FROM songs WHERE lower(title)=lower('PARANOiA Rebirth')),\n't',\n8047);\n(SELECT song FROM songs WHERE lower(title)=lower('PARANOiA Rebirth')),\n't',\n8047);\n@@ -124,24 +133,27 @@ INSERT INTO roundparticipation (tournament, round, parallel, player, position) V\n(SELECT player FROM players WHERE nick='Haggar'),\n4\n);\n(SELECT player FROM players WHERE nick='Haggar'),\n4\n);\n-INSERT INTO scores (tournament, round, parallel, player, song, chosen, score) VALUES (\n+INSERT INTO scores (tournament, round, parallel, player, songnumber, song, chosen, score) VALUES (\n(SELECT tournament FROM tournaments WHERE name='Challenge Cup 3'),\n1, 1,\n(SELECT player FROM players WHERE nick='Haggar'),\n(SELECT tournament FROM tournaments WHERE name='Challenge Cup 3'),\n1, 1,\n(SELECT player FROM players WHERE nick='Haggar'),\n+ 0,\n(SELECT song FROM songs WHERE lower(title)=lower('B4U (B4 Za Beat Mix)')),\n'f',\n6356);\n(SELECT song FROM songs WHERE lower(title)=lower('B4U (B4 Za Beat Mix)')),\n'f',\n6356);\n-INSERT INTO scores (tournament, round, parallel, player, song, chosen, score) VALUES (\n+INSERT INTO scores (tournament, round, parallel, player, songnumber, song, chosen, score) VALUES (\n(SELECT tournament FROM tournaments WHERE name='Challenge Cup 3'),\n1, 1,\n(SELECT player FROM players WHERE nick='Haggar'),\n(SELECT tournament FROM tournaments WHERE name='Challenge Cup 3'),\n1, 1,\n(SELECT player FROM players WHERE nick='Haggar'),\n+ 1,\n(SELECT song FROM songs WHERE lower(title)=lower('Can\\'t Stop Fallin\\' in Love')),\n'f',\n7560);\n(SELECT song FROM songs WHERE lower(title)=lower('Can\\'t Stop Fallin\\' in Love')),\n'f',\n7560);\n-INSERT INTO scores (tournament, round, parallel, player, song, chosen, score) VALUES (\n+INSERT INTO scores (tournament, round, parallel, player, songnumber, song, chosen, score) VALUES (\n(SELECT tournament FROM tournaments WHERE name='Challenge Cup 3'),\n1, 1,\n(SELECT player FROM players WHERE nick='Haggar'),\n(SELECT tournament FROM tournaments WHERE name='Challenge Cup 3'),\n1, 1,\n(SELECT player FROM players WHERE nick='Haggar'),\n+ 2,\n(SELECT song FROM songs WHERE lower(title)=lower('era (nostalmix)')),\n't',\n7720);\n(SELECT song FROM songs WHERE lower(title)=lower('era (nostalmix)')),\n't',\n7720);\n@@ -152,24 +164,27 @@ INSERT INTO roundparticipation (tournament, round, parallel, player, position) V\n(SELECT player FROM players WHERE nick='Snute'),\n5\n);\n(SELECT player FROM players WHERE nick='Snute'),\n5\n);\n-INSERT INTO scores (tournament, round, parallel, player, song, chosen, score) VALUES (\n+INSERT INTO scores (tournament, round, parallel, player, songnumber, song, chosen, score) VALUES (\n(SELECT tournament FROM tournaments WHERE name='Challenge Cup 3'),\n1, 1,\n(SELECT player FROM players WHERE nick='Snute'),\n(SELECT tournament FROM tournaments WHERE name='Challenge Cup 3'),\n1, 1,\n(SELECT player FROM players WHERE nick='Snute'),\n+ 0,\n(SELECT song FROM songs WHERE lower(title)=lower('B4U (B4 Za Beat Mix)')),\n'f',\n6480);\n(SELECT song FROM songs WHERE lower(title)=lower('B4U (B4 Za Beat Mix)')),\n'f',\n6480);\n-INSERT INTO scores (tournament, round, parallel, player, song, chosen, score) VALUES (\n+INSERT INTO scores (tournament, round, parallel, player, songnumber, song, chosen, score) VALUES (\n(SELECT tournament FROM tournaments WHERE name='Challenge Cup 3'),\n1, 1,\n(SELECT player FROM players WHERE nick='Snute'),\n(SELECT tournament FROM tournaments WHERE name='Challenge Cup 3'),\n1, 1,\n(SELECT player FROM players WHERE nick='Snute'),\n+ 1,\n(SELECT song FROM songs WHERE lower(title)=lower('Can\\'t Stop Fallin\\' in Love')),\n'f',\n7646);\n(SELECT song FROM songs WHERE lower(title)=lower('Can\\'t Stop Fallin\\' in Love')),\n'f',\n7646);\n-INSERT INTO scores (tournament, round, parallel, player, song, chosen, score) VALUES (\n+INSERT INTO scores (tournament, round, parallel, player, songnumber, song, chosen, score) VALUES (\n(SELECT tournament FROM tournaments WHERE name='Challenge Cup 3'),\n1, 1,\n(SELECT player FROM players WHERE nick='Snute'),\n(SELECT tournament FROM tournaments WHERE name='Challenge Cup 3'),\n1, 1,\n(SELECT player FROM players WHERE nick='Snute'),\n+ 2,\n(SELECT song FROM songs WHERE lower(title)=lower('B4U')),\n't',\n7702);\n(SELECT song FROM songs WHERE lower(title)=lower('B4U')),\n't',\n7702);\n@@ -199,24 +214,27 @@ INSERT INTO roundparticipation (tournament, round, parallel, player, position) V\n(SELECT player FROM players WHERE nick='Jim Andre'),\n1\n);\n(SELECT player FROM players WHERE nick='Jim Andre'),\n1\n);\n-INSERT INTO scores (tournament, round, parallel, player, song, chosen, score) VALUES (\n+INSERT INTO scores (tournament, round, parallel, player, songnumber, song, chosen, score) VALUES (\n(SELECT tournament FROM tournaments WHERE name='Challenge Cup 3'),\n1, 2,\n(SELECT player FROM players WHERE nick='Jim Andre'),\n(SELECT tournament FROM tournaments WHERE name='Challenge Cup 3'),\n1, 2,\n(SELECT player FROM players WHERE nick='Jim Andre'),\n+ 0,\n(SELECT song FROM songs WHERE lower(title)=lower('Don\\'t Try to Stop It')),\n'f',\n5809);\n(SELECT song FROM songs WHERE lower(title)=lower('Don\\'t Try to Stop It')),\n'f',\n5809);\n-INSERT INTO scores (tournament, round, parallel, player, song, chosen, score) VALUES (\n+INSERT INTO scores (tournament, round, parallel, player, songnumber, song, chosen, score) VALUES (\n(SELECT tournament FROM tournaments WHERE name='Challenge Cup 3'),\n1, 2,\n(SELECT player FROM players WHERE nick='Jim Andre'),\n(SELECT tournament FROM tournaments WHERE name='Challenge Cup 3'),\n1, 2,\n(SELECT player FROM players WHERE nick='Jim Andre'),\n+ 1,\n(SELECT song FROM songs WHERE lower(title)=lower('Make A Jam!')),\n'f',\n5338);\n(SELECT song FROM songs WHERE lower(title)=lower('Make A Jam!')),\n'f',\n5338);\n-INSERT INTO scores (tournament, round, parallel, player, song, chosen, score) VALUES (\n+INSERT INTO scores (tournament, round, parallel, player, songnumber, song, chosen, score) VALUES (\n(SELECT tournament FROM tournaments WHERE name='Challenge Cup 3'),\n1, 2,\n(SELECT player FROM players WHERE nick='Jim Andre'),\n(SELECT tournament FROM tournaments WHERE name='Challenge Cup 3'),\n1, 2,\n(SELECT player FROM players WHERE nick='Jim Andre'),\n+ 2,\n(SELECT song FROM songs WHERE lower(title)=lower('Can\\'t Stop Fallin\\' in Love')),\n't',\n6198);\n(SELECT song FROM songs WHERE lower(title)=lower('Can\\'t Stop Fallin\\' in Love')),\n't',\n6198);\n@@ -227,24 +245,27 @@ INSERT INTO roundparticipation (tournament, round, parallel, player, position) V\n(SELECT player FROM players WHERE nick='Andre'),\n2\n);\n(SELECT player FROM players WHERE nick='Andre'),\n2\n);\n-INSERT INTO scores (tournament, round, parallel, player, song, chosen, score) VALUES (\n+INSERT INTO scores (tournament, round, parallel, player, songnumber, song, chosen, score) VALUES (\n(SELECT tournament FROM tournaments WHERE name='Challenge Cup 3'),\n1, 2,\n(SELECT player FROM players WHERE nick='Andre'),\n(SELECT tournament FROM tournaments WHERE name='Challenge Cup 3'),\n1, 2,\n(SELECT player FROM players WHERE nick='Andre'),\n+ 0,\n(SELECT song FROM songs WHERE lower(title)=lower('Don\\'t Try to Stop It')),\n'f',\n4264);\n(SELECT song FROM songs WHERE lower(title)=lower('Don\\'t Try to Stop It')),\n'f',\n4264);\n-INSERT INTO scores (tournament, round, parallel, player, song, chosen, score) VALUES (\n+INSERT INTO scores (tournament, round, parallel, player, songnumber, song, chosen, score) VALUES (\n(SELECT tournament FROM tournaments WHERE name='Challenge Cup 3'),\n1, 2,\n(SELECT player FROM players WHERE nick='Andre'),\n(SELECT tournament FROM tournaments WHERE name='Challenge Cup 3'),\n1, 2,\n(SELECT player FROM players WHERE nick='Andre'),\n+ 1,\n(SELECT song FROM songs WHERE lower(title)=lower('Make A Jam!')),\n'f',\n3773);\n(SELECT song FROM songs WHERE lower(title)=lower('Make A Jam!')),\n'f',\n3773);\n-INSERT INTO scores (tournament, round, parallel, player, song, chosen, score) VALUES (\n+INSERT INTO scores (tournament, round, parallel, player, songnumber, song, chosen, score) VALUES (\n(SELECT tournament FROM tournaments WHERE name='Challenge Cup 3'),\n1, 2,\n(SELECT player FROM players WHERE nick='Andre'),\n(SELECT tournament FROM tournaments WHERE name='Challenge Cup 3'),\n1, 2,\n(SELECT player FROM players WHERE nick='Andre'),\n+ 2,\n(SELECT song FROM songs WHERE lower(title)=lower('Afronova')),\n't',\n5277);\n(SELECT song FROM songs WHERE lower(title)=lower('Afronova')),\n't',\n5277);\n@@ -255,24 +276,27 @@ INSERT INTO roundparticipation (tournament, round, parallel, player, position) V\n(SELECT player FROM players WHERE nick='Dabi'),\n3\n);\n(SELECT player FROM players WHERE nick='Dabi'),\n3\n);\n-INSERT INTO scores (tournament, round, parallel, player, song, chosen, score) VALUES (\n+INSERT INTO scores (tournament, round, parallel, player, songnumber, song, chosen, score) VALUES (\n(SELECT tournament FROM tournaments WHERE name='Challenge Cup 3'),\n1, 2,\n(SELECT player FROM players WHERE nick='Dabi'),\n(SELECT tournament FROM tournaments WHERE name='Challenge Cup 3'),\n1, 2,\n(SELECT player FROM players WHERE nick='Dabi'),\n+ 0,\n(SELECT song FROM songs WHERE lower(title)=lower('Don\\'t Try to Stop It')),\n'f',\n5388);\n(SELECT song FROM songs WHERE lower(title)=lower('Don\\'t Try to Stop It')),\n'f',\n5388);\n-INSERT INTO scores (tournament, round, parallel, player, song, chosen, score) VALUES (\n+INSERT INTO scores (tournament, round, parallel, player, songnumber, song, chosen, score) VALUES (\n(SELECT tournament FROM tournaments WHERE name='Challenge Cup 3'),\n1, 2,\n(SELECT player FROM players WHERE nick='Dabi'),\n(SELECT tournament FROM tournaments WHERE name='Challenge Cup 3'),\n1, 2,\n(SELECT player FROM players WHERE nick='Dabi'),\n+ 1,\n(SELECT song FROM songs WHERE lower(title)=lower('Make A Jam!')),\n'f',\n5249);\n(SELECT song FROM songs WHERE lower(title)=lower('Make A Jam!')),\n'f',\n5249);\n-INSERT INTO scores (tournament, round, parallel, player, song, chosen, score) VALUES (\n+INSERT INTO scores (tournament, round, parallel, player, songnumber, song, chosen, score) VALUES (\n(SELECT tournament FROM tournaments WHERE name='Challenge Cup 3'),\n1, 2,\n(SELECT player FROM players WHERE nick='Dabi'),\n(SELECT tournament FROM tournaments WHERE name='Challenge Cup 3'),\n1, 2,\n(SELECT player FROM players WHERE nick='Dabi'),\n+ 2,\n(SELECT song FROM songs WHERE lower(title)=lower('Jam Jam Reggae (AM Swing Mix)')),\n't',\n7228);\n(SELECT song FROM songs WHERE lower(title)=lower('Jam Jam Reggae (AM Swing Mix)')),\n't',\n7228);\n@@ -283,24 +307,27 @@ INSERT INTO roundparticipation (tournament, round, parallel, player, position) V\n(SELECT player FROM players WHERE nick='Oliyzia'),\n4\n);\n(SELECT player FROM players WHERE nick='Oliyzia'),\n4\n);\n-INSERT INTO scores (tournament, round, parallel, player, song, chosen, score) VALUES (\n+INSERT INTO scores (tournament, round, parallel, player, songnumber, song, chosen, score) VALUES (\n(SELECT tournament FROM tournaments WHERE name='Challenge Cup 3'),\n1, 2,\n(SELECT player FROM players WHERE nick='Oliyzia'),\n(SELECT tournament FROM tournaments WHERE name='Challenge Cup 3'),\n1, 2,\n(SELECT player FROM players WHERE nick='Oliyzia'),\n+ 0,\n(SELECT song FROM songs WHERE lower(title)=lower('Don\\'t Try to Stop It')),\n'f',\n6226);\n(SELECT song FROM songs WHERE lower(title)=lower('Don\\'t Try to Stop It')),\n'f',\n6226);\n-INSERT INTO scores (tournament, round, parallel, player, song, chosen, score) VALUES (\n+INSERT INTO scores (tournament, round, parallel, player, songnumber, song, chosen, score) VALUES (\n(SELECT tournament FROM tournaments WHERE name='Challenge Cup 3'),\n1, 2,\n(SELECT player FROM players WHERE nick='Oliyzia'),\n(SELECT tournament FROM tournaments WHERE name='Challenge Cup 3'),\n1, 2,\n(SELECT player FROM players WHERE nick='Oliyzia'),\n+ 1,\n(SELECT song FROM songs WHERE lower(title)=lower('Make A Jam!')),\n'f',\n6708);\n(SELECT song FROM songs WHERE lower(title)=lower('Make A Jam!')),\n'f',\n6708);\n-INSERT INTO scores (tournament, round, parallel, player, song, chosen, score) VALUES (\n+INSERT INTO scores (tournament, round, parallel, player, songnumber, song, chosen, score) VALUES (\n(SELECT tournament FROM tournaments WHERE name='Challenge Cup 3'),\n1, 2,\n(SELECT player FROM players WHERE nick='Oliyzia'),\n(SELECT tournament FROM tournaments WHERE name='Challenge Cup 3'),\n1, 2,\n(SELECT player FROM players WHERE nick='Oliyzia'),\n+ 2,\n(SELECT song FROM songs WHERE lower(title)=lower('Groove 2001')),\n't',\n6445);\n(SELECT song FROM songs WHERE lower(title)=lower('Groove 2001')),\n't',\n6445);\n@@ -311,24 +338,27 @@ INSERT INTO roundparticipation (tournament, round, parallel, player, position) V\n(SELECT player FROM players WHERE nick='OJK'),\n5\n);\n(SELECT player FROM players WHERE nick='OJK'),\n5\n);\n-INSERT INTO scores (tournament, round, parallel, player, song, chosen, score) VALUES (\n+INSERT INTO scores (tournament, round, parallel, player, songnumber, song, chosen, score) VALUES (\n(SELECT tournament FROM tournaments WHERE name='Challenge Cup 3'),\n1, 2,\n(SELECT player FROM players WHERE nick='OJK'),\n(SELECT tournament FROM tournaments WHERE name='Challenge Cup 3'),\n1, 2,\n(SELECT player FROM players WHERE nick='OJK'),\n+ 0,\n(SELECT song FROM songs WHERE lower(title)=lower('Don\\'t Try to Stop It')),\n'f',\n7353);\n(SELECT song FROM songs WHERE lower(title)=lower('Don\\'t Try to Stop It')),\n'f',\n7353);\n-INSERT INTO scores (tournament, round, parallel, player, song, chosen, score) VALUES (\n+INSERT INTO scores (tournament, round, parallel, player, songnumber, song, chosen, score) VALUES (\n(SELECT tournament FROM tournaments WHERE name='Challenge Cup 3'),\n1, 2,\n(SELECT player FROM players WHERE nick='OJK'),\n(SELECT tournament FROM tournaments WHERE name='Challenge Cup 3'),\n1, 2,\n(SELECT player FROM players WHERE nick='OJK'),\n+ 1,\n(SELECT song FROM songs WHERE lower(title)=lower('Make A Jam!')),\n'f',\n7214);\n(SELECT song FROM songs WHERE lower(title)=lower('Make A Jam!')),\n'f',\n7214);\n-INSERT INTO scores (tournament, round, parallel, player, song, chosen, score) VALUES (\n+INSERT INTO scores (tournament, round, parallel, player, songnumber, song, chosen, score) VALUES (\n(SELECT tournament FROM tournaments WHERE name='Challenge Cup 3'),\n1, 2,\n(SELECT player FROM players WHERE nick='OJK'),\n(SELECT tournament FROM tournaments WHERE name='Challenge Cup 3'),\n1, 2,\n(SELECT player FROM players WHERE nick='OJK'),\n+ 2,\n(SELECT song FROM songs WHERE lower(title)=lower('www.blonde girl (MOMO Mix)')),\n't',\n6791);\n(SELECT song FROM songs WHERE lower(title)=lower('www.blonde girl (MOMO Mix)')),\n't',\n6791);\n@@ -358,24 +388,27 @@ INSERT INTO roundparticipation (tournament, round, parallel, player, position) V\n(SELECT player FROM players WHERE nick='Glukko'),\n1\n);\n(SELECT player FROM players WHERE nick='Glukko'),\n1\n);\n-INSERT INTO scores (tournament, round, parallel, player, song, chosen, score) VALUES (\n+INSERT INTO scores (tournament, round, parallel, player, songnumber, song, chosen, score) VALUES (\n(SELECT tournament FROM tournaments WHERE name='Challenge Cup 3'),\n1, 3,\n(SELECT player FROM players WHERE nick='Glukko'),\n(SELECT tournament FROM tournaments WHERE name='Challenge Cup 3'),\n1, 3,\n(SELECT player FROM players WHERE nick='Glukko'),\n+ 0,\n(SELECT song FROM songs WHERE lower(title)=lower('5, 6, 7, 8')),\n'f',\n4346);\n(SELECT song FROM songs WHERE lower(title)=lower('5, 6, 7, 8')),\n'f',\n4346);\n-INSERT INTO scores (tournament, round, parallel, player, song, chosen, score) VALUES (\n+INSERT INTO scores (tournament, round, parallel, player, songnumber, song, chosen, score) VALUES (\n(SELECT tournament FROM tournaments WHERE name='Challenge Cup 3'),\n1, 3,\n(SELECT player FROM players WHERE nick='Glukko'),\n(SELECT tournament FROM tournaments WHERE name='Challenge Cup 3'),\n1, 3,\n(SELECT player FROM players WHERE nick='Glukko'),\n+ 1,\n(SELECT song FROM songs WHERE lower(title)=lower('So Deep (Perfect Sphere Mix)')),\n'f',\n3531);\n(SELECT song FROM songs WHERE lower(title)=lower('So Deep (Perfect Sphere Mix)')),\n'f',\n3531);\n-INSERT INTO scores (tournament, round, parallel, player, song, chosen, score) VALUES (\n+INSERT INTO scores (tournament, round, parallel, player, songnumber, song, chosen, score) VALUES (\n(SELECT tournament FROM tournaments WHERE name='Challenge Cup 3'),\n1, 3,\n(SELECT player FROM players WHERE nick='Glukko'),\n(SELECT tournament FROM tournaments WHERE name='Challenge Cup 3'),\n1, 3,\n(SELECT player FROM players WHERE nick='Glukko'),\n+ 2,\n(SELECT song FROM songs WHERE lower(title)=lower('Can\\'t Stop Fallin\\' in Love (SPEED MIX)')),\n't',\n5270);\n(SELECT song FROM songs WHERE lower(title)=lower('Can\\'t Stop Fallin\\' in Love (SPEED MIX)')),\n't',\n5270);\n@@ -386,17 +419,19 @@ INSERT INTO roundparticipation (tournament, round, parallel, player, position) V\n(SELECT player FROM players WHERE nick='Ole-Christian'),\n2\n);\n(SELECT player FROM players WHERE nick='Ole-Christian'),\n2\n);\n-INSERT INTO scores (tournament, round, parallel, player, song, chosen, score) VALUES (\n+INSERT INTO scores (tournament, round, parallel, player, songnumber, song, chosen, score) VALUES (\n(SELECT tournament FROM tournaments WHERE name='Challenge Cup 3'),\n1, 3,\n(SELECT player FROM players WHERE nick='Ole-Christian'),\n(SELECT tournament FROM tournaments WHERE name='Challenge Cup 3'),\n1, 3,\n(SELECT player FROM players WHERE nick='Ole-Christian'),\n+ 0,\n(SELECT song FROM songs WHERE lower(title)=lower('5, 6, 7, 8')),\n'f',\n0);\n(SELECT song FROM songs WHERE lower(title)=lower('5, 6, 7, 8')),\n'f',\n0);\n-INSERT INTO scores (tournament, round, parallel, player, song, chosen, score) VALUES (\n+INSERT INTO scores (tournament, round, parallel, player, songnumber, song, chosen, score) VALUES (\n(SELECT tournament FROM tournaments WHERE name='Challenge Cup 3'),\n1, 3,\n(SELECT player FROM players WHERE nick='Ole-Christian'),\n(SELECT tournament FROM tournaments WHERE name='Challenge Cup 3'),\n1, 3,\n(SELECT player FROM players WHERE nick='Ole-Christian'),\n+ 1,\n(SELECT song FROM songs WHERE lower(title)=lower('So Deep (Perfect Sphere Mix)')),\n'f',\n0);\n(SELECT song FROM songs WHERE lower(title)=lower('So Deep (Perfect Sphere Mix)')),\n'f',\n0);\n@@ -407,24 +442,27 @@ INSERT INTO roundparticipation (tournament, round, parallel, player, position) V\n(SELECT player FROM players WHERE nick='Zyc'),\n3\n);\n(SELECT player FROM players WHERE nick='Zyc'),\n3\n);\n-INSERT INTO scores (tournament, round, parallel, player, song, chosen, score) VALUES (\n+INSERT INTO scores (tournament, round, parallel, player, songnumber, song, chosen, score) VALUES (\n(SELECT tournament FROM tournaments WHERE name='Challenge Cup 3'),\n1, 3,\n(SELECT player FROM players WHERE nick='Zyc'),\n(SELECT tournament FROM tournaments WHERE name='Challenge Cup 3'),\n1, 3,\n(SELECT player FROM players WHERE nick='Zyc'),\n+ 0,\n(SELECT song FROM songs WHERE lower(title)=lower('5, 6, 7, 8')),\n'f',\n4898);\n(SELECT song FROM songs WHERE lower(title)=lower('5, 6, 7, 8')),\n'f',\n4898);\n-INSERT INTO scores (tournament, round, parallel, player, song, chosen, score) VALUES (\n+INSERT INTO scores (tournament, round, parallel, player, songnumber, song, chosen, score) VALUES (\n(SELECT tournament FROM tournaments WHERE name='Challenge Cup 3'),\n1, 3,\n(SELECT player FROM players WHERE nick='Zyc'),\n(SELECT tournament FROM tournaments WHERE name='Challenge Cup 3'),\n1, 3,\n(SELECT player FROM players WHERE nick='Zyc'),\n+ 1,\n(SELECT song FROM songs WHERE lower(title)=lower('So Deep (Perfect Sphere Mix)')),\n'f',\n3333);\n(SELECT song FROM songs WHERE lower(title)=lower('So Deep (Perfect Sphere Mix)')),\n'f',\n3333);\n-INSERT INTO scores (tournament, round, parallel, player, song, chosen, score) VALUES (\n+INSERT INTO scores (tournament, round, parallel, player, songnumber, song, chosen, score) VALUES (\n(SELECT tournament FROM tournaments WHERE name='Challenge Cup 3'),\n1, 3,\n(SELECT player FROM players WHERE nick='Zyc'),\n(SELECT tournament FROM tournaments WHERE name='Challenge Cup 3'),\n1, 3,\n(SELECT player FROM players WHERE nick='Zyc'),\n+ 2,\n(SELECT song FROM songs WHERE lower(title)=lower('Jam Jam Reggae (AM Swing Mix)')),\n't',\n5149);\n(SELECT song FROM songs WHERE lower(title)=lower('Jam Jam Reggae (AM Swing Mix)')),\n't',\n5149);\n@@ -435,24 +473,27 @@ INSERT INTO roundparticipation (tournament, round, parallel, player, position) V\n(SELECT player FROM players WHERE nick='Ylini'),\n4\n);\n(SELECT player FROM players WHERE nick='Ylini'),\n4\n);\n-INSERT INTO scores (tournament, round, parallel, player, song, chosen, score) VALUES (\n+INSERT INTO scores (tournament, round, parallel, player, songnumber, song, chosen, score) VALUES (\n(SELECT tournament FROM tournaments WHERE name='Challenge Cup 3'),\n1, 3,\n(SELECT player FROM players WHERE nick='Ylini'),\n(SELECT tournament FROM tournaments WHERE name='Challenge Cup 3'),\n1, 3,\n(SELECT player FROM players WHERE nick='Ylini'),\n+ 0,\n(SELECT song FROM songs WHERE lower(title)=lower('5, 6, 7, 8')),\n'f',\n5189);\n(SELECT song FROM songs WHERE lower(title)=lower('5, 6, 7, 8')),\n'f',\n5189);\n-INSERT INTO scores (tournament, round, parallel, player, song, chosen, score) VALUES (\n+INSERT INTO scores (tournament, round, parallel, player, songnumber, song, chosen, score) VALUES (\n(SELECT tournament FROM tournaments WHERE name='Challenge Cup 3'),\n1, 3,\n(SELECT player FROM players WHERE nick='Ylini'),\n(SELECT tournament FROM tournaments WHERE name='Challenge Cup 3'),\n1, 3,\n(SELECT player FROM players WHERE nick='Ylini'),\n+ 1,\n(SELECT song FROM songs WHERE lower(title)=lower('So Deep (Perfect Sphere Mix)')),\n'f',\n4391);\n(SELECT song FROM songs WHERE lower(title)=lower('So Deep (Perfect Sphere Mix)')),\n'f',\n4391);\n-INSERT INTO scores (tournament, round, parallel, player, song, chosen, score) VALUES (\n+INSERT INTO scores (tournament, round, parallel, player, songnumber, song, chosen, score) VALUES (\n(SELECT tournament FROM tournaments WHERE name='Challenge Cup 3'),\n1, 3,\n(SELECT player FROM players WHERE nick='Ylini'),\n(SELECT tournament FROM tournaments WHERE name='Challenge Cup 3'),\n1, 3,\n(SELECT player FROM players WHERE nick='Ylini'),\n+ 2,\n(SELECT song FROM songs WHERE lower(title)=lower('Jam Jam Reggae (AM Swing Mix)')),\n't',\n7298);\n(SELECT song FROM songs WHERE lower(title)=lower('Jam Jam Reggae (AM Swing Mix)')),\n't',\n7298);\n@@ -463,24 +504,27 @@ INSERT INTO roundparticipation (tournament, round, parallel, player, position) V\n(SELECT player FROM players WHERE nick='DufF'),\n5\n);\n(SELECT player FROM players WHERE nick='DufF'),\n5\n);\n-INSERT INTO scores (tournament, round, parallel, player, song, chosen, score) VALUES (\n+INSERT INTO scores (tournament, round, parallel, player, songnumber, song, chosen, score) VALUES (\n(SELECT tournament FROM tournaments WHERE name='Challenge Cup 3'),\n1, 3,\n(SELECT player FROM players WHERE nick='DufF'),\n(SELECT tournament FROM tournaments WHERE name='Challenge Cup 3'),\n1, 3,\n(SELECT player FROM players WHERE nick='DufF'),\n+ 0,\n(SELECT song FROM songs WHERE lower(title)=lower('5, 6, 7, 8')),\n'f',\n5822);\n(SELECT song FROM songs WHERE lower(title)=lower('5, 6, 7, 8')),\n'f',\n5822);\n-INSERT INTO scores (tournament, round, parallel, player, song, chosen, score) VALUES (\n+INSERT INTO scores (tournament, round, parallel, player, songnumber, song, chosen, score) VALUES (\n(SELECT tournament FROM tournaments WHERE name='Challenge Cup 3'),\n1, 3,\n(SELECT player FROM players WHERE nick='DufF'),\n(SELECT tournament FROM tournaments WHERE name='Challenge Cup 3'),\n1, 3,\n(SELECT player FROM players WHERE nick='DufF'),\n+ 1,\n(SELECT song FROM songs WHERE lower(title)=lower('So Deep (Perfect Sphere Mix)')),\n'f',\n8510);\n(SELECT song FROM songs WHERE lower(title)=lower('So Deep (Perfect Sphere Mix)')),\n'f',\n8510);\n-INSERT INTO scores (tournament, round, parallel, player, song, chosen, score) VALUES (\n+INSERT INTO scores (tournament, round, parallel, player, songnumber, song, chosen, score) VALUES (\n(SELECT tournament FROM tournaments WHERE name='Challenge Cup 3'),\n1, 3,\n(SELECT player FROM players WHERE nick='DufF'),\n(SELECT tournament FROM tournaments WHERE name='Challenge Cup 3'),\n1, 3,\n(SELECT player FROM players WHERE nick='DufF'),\n+ 2,\n(SELECT song FROM songs WHERE lower(title)=lower('Trip Machine (Luv Mix)')),\n't',\n8504);\n(SELECT song FROM songs WHERE lower(title)=lower('Trip Machine (Luv Mix)')),\n't',\n8504);\n@@ -510,24 +554,27 @@ INSERT INTO roundparticipation (tournament, round, parallel, player, position) V\n(SELECT player FROM players WHERE nick='Th0rny'),\n1\n);\n(SELECT player FROM players WHERE nick='Th0rny'),\n1\n);\n-INSERT INTO scores (tournament, round, parallel, player, song, chosen, score) VALUES (\n+INSERT INTO scores (tournament, round, parallel, player, songnumber, song, chosen, score) VALUES (\n(SELECT tournament FROM tournaments WHERE name='Challenge Cup 3'),\n1, 4,\n(SELECT player FROM players WHERE nick='Th0rny'),\n(SELECT tournament FROM tournaments WHERE name='Challenge Cup 3'),\n1, 4,\n(SELECT player FROM players WHERE nick='Th0rny'),\n+ 0,\n(SELECT song FROM songs WHERE lower(title)=lower('Look to the sky')),\n'f',\n5789);\n(SELECT song FROM songs WHERE lower(title)=lower('Look to the sky')),\n'f',\n5789);\n-INSERT INTO scores (tournament, round, parallel, player, song, chosen, score) VALUES (\n+INSERT INTO scores (tournament, round, parallel, player, songnumber, song, chosen, score) VALUES (\n(SELECT tournament FROM tournaments WHERE name='Challenge Cup 3'),\n1, 4,\n(SELECT player FROM players WHERE nick='Th0rny'),\n(SELECT tournament FROM tournaments WHERE name='Challenge Cup 3'),\n1, 4,\n(SELECT player FROM players WHERE nick='Th0rny'),\n+ 1,\n(SELECT song FROM songs WHERE lower(title)=lower('Burnin\\' the Floor')),\n'f',\n6214);\n(SELECT song FROM songs WHERE lower(title)=lower('Burnin\\' the Floor')),\n'f',\n6214);\n-INSERT INTO scores (tournament, round, parallel, player, song, chosen, score) VALUES (\n+INSERT INTO scores (tournament, round, parallel, player, songnumber, song, chosen, score) VALUES (\n(SELECT tournament FROM tournaments WHERE name='Challenge Cup 3'),\n1, 4,\n(SELECT player FROM players WHERE nick='Th0rny'),\n(SELECT tournament FROM tournaments WHERE name='Challenge Cup 3'),\n1, 4,\n(SELECT player FROM players WHERE nick='Th0rny'),\n+ 2,\n(SELECT song FROM songs WHERE lower(title)=lower('Crash!')),\n't',\n8348);\n(SELECT song FROM songs WHERE lower(title)=lower('Crash!')),\n't',\n8348);\n@@ -538,24 +585,27 @@ INSERT INTO roundparticipation (tournament, round, parallel, player, position) V\n(SELECT player FROM players WHERE nick='Nameless'),\n2\n);\n(SELECT player FROM players WHERE nick='Nameless'),\n2\n);\n-INSERT INTO scores (tournament, round, parallel, player, song, chosen, score) VALUES (\n+INSERT INTO scores (tournament, round, parallel, player, songnumber, song, chosen, score) VALUES (\n(SELECT tournament FROM tournaments WHERE name='Challenge Cup 3'),\n1, 4,\n(SELECT player FROM players WHERE nick='Nameless'),\n(SELECT tournament FROM tournaments WHERE name='Challenge Cup 3'),\n1, 4,\n(SELECT player FROM players WHERE nick='Nameless'),\n+ 0,\n(SELECT song FROM songs WHERE lower(title)=lower('Look to the sky')),\n'f',\n5618);\n(SELECT song FROM songs WHERE lower(title)=lower('Look to the sky')),\n'f',\n5618);\n-INSERT INTO scores (tournament, round, parallel, player, song, chosen, score) VALUES (\n+INSERT INTO scores (tournament, round, parallel, player, songnumber, song, chosen, score) VALUES (\n(SELECT tournament FROM tournaments WHERE name='Challenge Cup 3'),\n1, 4,\n(SELECT player FROM players WHERE nick='Nameless'),\n(SELECT tournament FROM tournaments WHERE name='Challenge Cup 3'),\n1, 4,\n(SELECT player FROM players WHERE nick='Nameless'),\n+ 1,\n(SELECT song FROM songs WHERE lower(title)=lower('Burnin\\' the Floor')),\n'f',\n6326);\n(SELECT song FROM songs WHERE lower(title)=lower('Burnin\\' the Floor')),\n'f',\n6326);\n-INSERT INTO scores (tournament, round, parallel, player, song, chosen, score) VALUES (\n+INSERT INTO scores (tournament, round, parallel, player, songnumber, song, chosen, score) VALUES (\n(SELECT tournament FROM tournaments WHERE name='Challenge Cup 3'),\n1, 4,\n(SELECT player FROM players WHERE nick='Nameless'),\n(SELECT tournament FROM tournaments WHERE name='Challenge Cup 3'),\n1, 4,\n(SELECT player FROM players WHERE nick='Nameless'),\n+ 2,\n(SELECT song FROM songs WHERE lower(title)=lower('PARANOiA Rebirth')),\n't',\n8278);\n(SELECT song FROM songs WHERE lower(title)=lower('PARANOiA Rebirth')),\n't',\n8278);\n@@ -566,24 +616,27 @@ INSERT INTO roundparticipation (tournament, round, parallel, player, position) V\n(SELECT player FROM players WHERE nick='Matsi'),\n3\n);\n(SELECT player FROM players WHERE nick='Matsi'),\n3\n);\n-INSERT INTO scores (tournament, round, parallel, player, song, chosen, score) VALUES (\n+INSERT INTO scores (tournament, round, parallel, player, songnumber, song, chosen, score) VALUES (\n(SELECT tournament FROM tournaments WHERE name='Challenge Cup 3'),\n1, 4,\n(SELECT player FROM players WHERE nick='Matsi'),\n(SELECT tournament FROM tournaments WHERE name='Challenge Cup 3'),\n1, 4,\n(SELECT player FROM players WHERE nick='Matsi'),\n+ 0,\n(SELECT song FROM songs WHERE lower(title)=lower('Look to the sky')),\n'f',\n4525);\n(SELECT song FROM songs WHERE lower(title)=lower('Look to the sky')),\n'f',\n4525);\n-INSERT INTO scores (tournament, round, parallel, player, song, chosen, score) VALUES (\n+INSERT INTO scores (tournament, round, parallel, player, songnumber, song, chosen, score) VALUES (\n(SELECT tournament FROM tournaments WHERE name='Challenge Cup 3'),\n1, 4,\n(SELECT player FROM players WHERE nick='Matsi'),\n(SELECT tournament FROM tournaments WHERE name='Challenge Cup 3'),\n1, 4,\n(SELECT player FROM players WHERE nick='Matsi'),\n+ 1,\n(SELECT song FROM songs WHERE lower(title)=lower('Burnin\\' the Floor')),\n'f',\n6008);\n(SELECT song FROM songs WHERE lower(title)=lower('Burnin\\' the Floor')),\n'f',\n6008);\n-INSERT INTO scores (tournament, round, parallel, player, song, chosen, score) VALUES (\n+INSERT INTO scores (tournament, round, parallel, player, songnumber, song, chosen, score) VALUES (\n(SELECT tournament FROM tournaments WHERE name='Challenge Cup 3'),\n1, 4,\n(SELECT player FROM players WHERE nick='Matsi'),\n(SELECT tournament FROM tournaments WHERE name='Challenge Cup 3'),\n1, 4,\n(SELECT player FROM players WHERE nick='Matsi'),\n+ 2,\n(SELECT song FROM songs WHERE lower(title)=lower('PARANOiA Rebirth')),\n't',\n7040);\n(SELECT song FROM songs WHERE lower(title)=lower('PARANOiA Rebirth')),\n't',\n7040);\n@@ -594,24 +647,27 @@ INSERT INTO roundparticipation (tournament, round, parallel, player, position) V\n(SELECT player FROM players WHERE nick='Draconia'),\n4\n);\n(SELECT player FROM players WHERE nick='Draconia'),\n4\n);\n-INSERT INTO scores (tournament, round, parallel, player, song, chosen, score) VALUES (\n+INSERT INTO scores (tournament, round, parallel, player, songnumber, song, chosen, score) VALUES (\n(SELECT tournament FROM tournaments WHERE name='Challenge Cup 3'),\n1, 4,\n(SELECT player FROM players WHERE nick='Draconia'),\n(SELECT tournament FROM tournaments WHERE name='Challenge Cup 3'),\n1, 4,\n(SELECT player FROM players WHERE nick='Draconia'),\n+ 0,\n(SELECT song FROM songs WHERE lower(title)=lower('Look to the sky')),\n'f',\n5716);\n(SELECT song FROM songs WHERE lower(title)=lower('Look to the sky')),\n'f',\n5716);\n-INSERT INTO scores (tournament, round, parallel, player, song, chosen, score) VALUES (\n+INSERT INTO scores (tournament, round, parallel, player, songnumber, song, chosen, score) VALUES (\n(SELECT tournament FROM tournaments WHERE name='Challenge Cup 3'),\n1, 4,\n(SELECT player FROM players WHERE nick='Draconia'),\n(SELECT tournament FROM tournaments WHERE name='Challenge Cup 3'),\n1, 4,\n(SELECT player FROM players WHERE nick='Draconia'),\n+ 1,\n(SELECT song FROM songs WHERE lower(title)=lower('Burnin\\' the Floor')),\n'f',\n6169);\n(SELECT song FROM songs WHERE lower(title)=lower('Burnin\\' the Floor')),\n'f',\n6169);\n-INSERT INTO scores (tournament, round, parallel, player, song, chosen, score) VALUES (\n+INSERT INTO scores (tournament, round, parallel, player, songnumber, song, chosen, score) VALUES (\n(SELECT tournament FROM tournaments WHERE name='Challenge Cup 3'),\n1, 4,\n(SELECT player FROM players WHERE nick='Draconia'),\n(SELECT tournament FROM tournaments WHERE name='Challenge Cup 3'),\n1, 4,\n(SELECT player FROM players WHERE nick='Draconia'),\n+ 2,\n(SELECT song FROM songs WHERE lower(title)=lower('Electro Tuned (the SubS Mix)')),\n't',\n7682);\n(SELECT song FROM songs WHERE lower(title)=lower('Electro Tuned (the SubS Mix)')),\n't',\n7682);\n@@ -622,24 +678,27 @@ INSERT INTO roundparticipation (tournament, round, parallel, player, position) V\n(SELECT player FROM players WHERE nick='RiChY'),\n5\n);\n(SELECT player FROM players WHERE nick='RiChY'),\n5\n);\n-INSERT INTO scores (tournament, round, parallel, player, song, chosen, score) VALUES (\n+INSERT INTO scores (tournament, round, parallel, player, songnumber, song, chosen, score) VALUES (\n(SELECT tournament FROM tournaments WHERE name='Challenge Cup 3'),\n1, 4,\n(SELECT player FROM players WHERE nick='RiChY'),\n(SELECT tournament FROM tournaments WHERE name='Challenge Cup 3'),\n1, 4,\n(SELECT player FROM players WHERE nick='RiChY'),\n+ 0,\n(SELECT song FROM songs WHERE lower(title)=lower('Look to the sky')),\n'f',\n6023);\n(SELECT song FROM songs WHERE lower(title)=lower('Look to the sky')),\n'f',\n6023);\n-INSERT INTO scores (tournament, round, parallel, player, song, chosen, score) VALUES (\n+INSERT INTO scores (tournament, round, parallel, player, songnumber, song, chosen, score) VALUES (\n(SELECT tournament FROM tournaments WHERE name='Challenge Cup 3'),\n1, 4,\n(SELECT player FROM players WHERE nick='RiChY'),\n(SELECT tournament FROM tournaments WHERE name='Challenge Cup 3'),\n1, 4,\n(SELECT player FROM players WHERE nick='RiChY'),\n+ 1,\n(SELECT song FROM songs WHERE lower(title)=lower('Burnin\\' the Floor')),\n'f',\n6628);\n(SELECT song FROM songs WHERE lower(title)=lower('Burnin\\' the Floor')),\n'f',\n6628);\n-INSERT INTO scores (tournament, round, parallel, player, song, chosen, score) VALUES (\n+INSERT INTO scores (tournament, round, parallel, player, songnumber, song, chosen, score) VALUES (\n(SELECT tournament FROM tournaments WHERE name='Challenge Cup 3'),\n1, 4,\n(SELECT player FROM players WHERE nick='RiChY'),\n(SELECT tournament FROM tournaments WHERE name='Challenge Cup 3'),\n1, 4,\n(SELECT player FROM players WHERE nick='RiChY'),\n+ 2,\n(SELECT song FROM songs WHERE lower(title)=lower('Healing Vision (Angelic Mix)')),\n't',\n8179);\n(SELECT song FROM songs WHERE lower(title)=lower('Healing Vision (Angelic Mix)')),\n't',\n8179);\n@@ -669,24 +728,27 @@ INSERT INTO roundparticipation (tournament, round, parallel, player, position) V\n(SELECT player FROM players WHERE nick='Eruanna'),\n1\n);\n(SELECT player FROM players WHERE nick='Eruanna'),\n1\n);\n-INSERT INTO scores (tournament, round, parallel, player, song, chosen, score) VALUES (\n+INSERT INTO scores (tournament, round, parallel, player, songnumber, song, chosen, score) VALUES (\n(SELECT tournament FROM tournaments WHERE name='Challenge Cup 3'),\n1, 5,\n(SELECT player FROM players WHERE nick='Eruanna'),\n(SELECT tournament FROM tournaments WHERE name='Challenge Cup 3'),\n1, 5,\n(SELECT player FROM players WHERE nick='Eruanna'),\n+ 0,\n(SELECT song FROM songs WHERE lower(title)=lower('Paranoia KCET (Clean Mix)')),\n'f',\n5410);\n(SELECT song FROM songs WHERE lower(title)=lower('Paranoia KCET (Clean Mix)')),\n'f',\n5410);\n-INSERT INTO scores (tournament, round, parallel, player, song, chosen, score) VALUES (\n+INSERT INTO scores (tournament, round, parallel, player, songnumber, song, chosen, score) VALUES (\n(SELECT tournament FROM tournaments WHERE name='Challenge Cup 3'),\n1, 5,\n(SELECT player FROM players WHERE nick='Eruanna'),\n(SELECT tournament FROM tournaments WHERE name='Challenge Cup 3'),\n1, 5,\n(SELECT player FROM players WHERE nick='Eruanna'),\n+ 1,\n(SELECT song FROM songs WHERE lower(title)=lower('Dead End')),\n'f',\n4383);\n(SELECT song FROM songs WHERE lower(title)=lower('Dead End')),\n'f',\n4383);\n-INSERT INTO scores (tournament, round, parallel, player, song, chosen, score) VALUES (\n+INSERT INTO scores (tournament, round, parallel, player, songnumber, song, chosen, score) VALUES (\n(SELECT tournament FROM tournaments WHERE name='Challenge Cup 3'),\n1, 5,\n(SELECT player FROM players WHERE nick='Eruanna'),\n(SELECT tournament FROM tournaments WHERE name='Challenge Cup 3'),\n1, 5,\n(SELECT player FROM players WHERE nick='Eruanna'),\n+ 2,\n(SELECT song FROM songs WHERE lower(title)=lower('Jam Jam Reggae (AM Swing Mix)')),\n't',\n6040);\n(SELECT song FROM songs WHERE lower(title)=lower('Jam Jam Reggae (AM Swing Mix)')),\n't',\n6040);\n@@ -697,24 +759,27 @@ INSERT INTO roundparticipation (tournament, round, parallel, player, position) V\n(SELECT player FROM players WHERE nick='Advena'),\n2\n);\n(SELECT player FROM players WHERE nick='Advena'),\n2\n);\n-INSERT INTO scores (tournament, round, parallel, player, song, chosen, score) VALUES (\n+INSERT INTO scores (tournament, round, parallel, player, songnumber, song, chosen, score) VALUES (\n(SELECT tournament FROM tournaments WHERE name='Challenge Cup 3'),\n1, 5,\n(SELECT player FROM players WHERE nick='Advena'),\n(SELECT tournament FROM tournaments WHERE name='Challenge Cup 3'),\n1, 5,\n(SELECT player FROM players WHERE nick='Advena'),\n+ 0,\n(SELECT song FROM songs WHERE lower(title)=lower('Paranoia KCET (Clean Mix)')),\n'f',\n6729);\n(SELECT song FROM songs WHERE lower(title)=lower('Paranoia KCET (Clean Mix)')),\n'f',\n6729);\n-INSERT INTO scores (tournament, round, parallel, player, song, chosen, score) VALUES (\n+INSERT INTO scores (tournament, round, parallel, player, songnumber, song, chosen, score) VALUES (\n(SELECT tournament FROM tournaments WHERE name='Challenge Cup 3'),\n1, 5,\n(SELECT player FROM players WHERE nick='Advena'),\n(SELECT tournament FROM tournaments WHERE name='Challenge Cup 3'),\n1, 5,\n(SELECT player FROM players WHERE nick='Advena'),\n+ 1,\n(SELECT song FROM songs WHERE lower(title)=lower('Dead End')),\n'f',\n7069);\n(SELECT song FROM songs WHERE lower(title)=lower('Dead End')),\n'f',\n7069);\n-INSERT INTO scores (tournament, round, parallel, player, song, chosen, score) VALUES (\n+INSERT INTO scores (tournament, round, parallel, player, songnumber, song, chosen, score) VALUES (\n(SELECT tournament FROM tournaments WHERE name='Challenge Cup 3'),\n1, 5,\n(SELECT player FROM players WHERE nick='Advena'),\n(SELECT tournament FROM tournaments WHERE name='Challenge Cup 3'),\n1, 5,\n(SELECT player FROM players WHERE nick='Advena'),\n+ 2,\n(SELECT song FROM songs WHERE lower(title)=lower('Electro Tuned (the SubS Mix)')),\n't',\n7624);\n(SELECT song FROM songs WHERE lower(title)=lower('Electro Tuned (the SubS Mix)')),\n't',\n7624);\n@@ -725,24 +790,27 @@ INSERT INTO roundparticipation (tournament, round, parallel, player, position) V\n(SELECT player FROM players WHERE nick='Tassen'),\n3\n);\n(SELECT player FROM players WHERE nick='Tassen'),\n3\n);\n-INSERT INTO scores (tournament, round, parallel, player, song, chosen, score) VALUES (\n+INSERT INTO scores (tournament, round, parallel, player, songnumber, song, chosen, score) VALUES (\n(SELECT tournament FROM tournaments WHERE name='Challenge Cup 3'),\n1, 5,\n(SELECT player FROM players WHERE nick='Tassen'),\n(SELECT tournament FROM tournaments WHERE name='Challenge Cup 3'),\n1, 5,\n(SELECT player FROM players WHERE nick='Tassen'),\n+ 0,\n(SELECT song FROM songs WHERE lower(title)=lower('Paranoia KCET (Clean Mix)')),\n'f',\n6911);\n(SELECT song FROM songs WHERE lower(title)=lower('Paranoia KCET (Clean Mix)')),\n'f',\n6911);\n-INSERT INTO scores (tournament, round, parallel, player, song, chosen, score) VALUES (\n+INSERT INTO scores (tournament, round, parallel, player, songnumber, song, chosen, score) VALUES (\n(SELECT tournament FROM tournaments WHERE name='Challenge Cup 3'),\n1, 5,\n(SELECT player FROM players WHERE nick='Tassen'),\n(SELECT tournament FROM tournaments WHERE name='Challenge Cup 3'),\n1, 5,\n(SELECT player FROM players WHERE nick='Tassen'),\n+ 1,\n(SELECT song FROM songs WHERE lower(title)=lower('Dead End')),\n'f',\n6693);\n(SELECT song FROM songs WHERE lower(title)=lower('Dead End')),\n'f',\n6693);\n-INSERT INTO scores (tournament, round, parallel, player, song, chosen, score) VALUES (\n+INSERT INTO scores (tournament, round, parallel, player, songnumber, song, chosen, score) VALUES (\n(SELECT tournament FROM tournaments WHERE name='Challenge Cup 3'),\n1, 5,\n(SELECT player FROM players WHERE nick='Tassen'),\n(SELECT tournament FROM tournaments WHERE name='Challenge Cup 3'),\n1, 5,\n(SELECT player FROM players WHERE nick='Tassen'),\n+ 2,\n(SELECT song FROM songs WHERE lower(title)=lower('Crash!')),\n't',\n7298);\n(SELECT song FROM songs WHERE lower(title)=lower('Crash!')),\n't',\n7298);\n@@ -753,24 +821,27 @@ INSERT INTO roundparticipation (tournament, round, parallel, player, position) V\n(SELECT player FROM players WHERE nick='Duckling'),\n4\n);\n(SELECT player FROM players WHERE nick='Duckling'),\n4\n);\n-INSERT INTO scores (tournament, round, parallel, player, song, chosen, score) VALUES (\n+INSERT INTO scores (tournament, round, parallel, player, songnumber, song, chosen, score) VALUES (\n(SELECT tournament FROM tournaments WHERE name='Challenge Cup 3'),\n1, 5,\n(SELECT player FROM players WHERE nick='Duckling'),\n(SELECT tournament FROM tournaments WHERE name='Challenge Cup 3'),\n1, 5,\n(SELECT player FROM players WHERE nick='Duckling'),\n+ 0,\n(SELECT song FROM songs WHERE lower(title)=lower('Paranoia KCET (Clean Mix)')),\n'f',\n8252);\n(SELECT song FROM songs WHERE lower(title)=lower('Paranoia KCET (Clean Mix)')),\n'f',\n8252);\n-INSERT INTO scores (tournament, round, parallel, player, song, chosen, score) VALUES (\n+INSERT INTO scores (tournament, round, parallel, player, songnumber, song, chosen, score) VALUES (\n(SELECT tournament FROM tournaments WHERE name='Challenge Cup 3'),\n1, 5,\n(SELECT player FROM players WHERE nick='Duckling'),\n(SELECT tournament FROM tournaments WHERE name='Challenge Cup 3'),\n1, 5,\n(SELECT player FROM players WHERE nick='Duckling'),\n+ 1,\n(SELECT song FROM songs WHERE lower(title)=lower('Dead End')),\n'f',\n7357);\n(SELECT song FROM songs WHERE lower(title)=lower('Dead End')),\n'f',\n7357);\n-INSERT INTO scores (tournament, round, parallel, player, song, chosen, score) VALUES (\n+INSERT INTO scores (tournament, round, parallel, player, songnumber, song, chosen, score) VALUES (\n(SELECT tournament FROM tournaments WHERE name='Challenge Cup 3'),\n1, 5,\n(SELECT player FROM players WHERE nick='Duckling'),\n(SELECT tournament FROM tournaments WHERE name='Challenge Cup 3'),\n1, 5,\n(SELECT player FROM players WHERE nick='Duckling'),\n+ 2,\n(SELECT song FROM songs WHERE lower(title)=lower('Jam Jam Reggae (AM Swing Mix)')),\n't',\n7321);\n(SELECT song FROM songs WHERE lower(title)=lower('Jam Jam Reggae (AM Swing Mix)')),\n't',\n7321);\n@@ -781,24 +852,27 @@ INSERT INTO roundparticipation (tournament, round, parallel, player, position) V\n(SELECT player FROM players WHERE nick='Olemann'),\n5\n);\n(SELECT player FROM players WHERE nick='Olemann'),\n5\n);\n-INSERT INTO scores (tournament, round, parallel, player, song, chosen, score) VALUES (\n+INSERT INTO scores (tournament, round, parallel, player, songnumber, song, chosen, score) VALUES (\n(SELECT tournament FROM tournaments WHERE name='Challenge Cup 3'),\n1, 5,\n(SELECT player FROM players WHERE nick='Olemann'),\n(SELECT tournament FROM tournaments WHERE name='Challenge Cup 3'),\n1, 5,\n(SELECT player FROM players WHERE nick='Olemann'),\n+ 0,\n(SELECT song FROM songs WHERE lower(title)=lower('Paranoia KCET (Clean Mix)')),\n'f',\n7471);\n(SELECT song FROM songs WHERE lower(title)=lower('Paranoia KCET (Clean Mix)')),\n'f',\n7471);\n-INSERT INTO scores (tournament, round, parallel, player, song, chosen, score) VALUES (\n+INSERT INTO scores (tournament, round, parallel, player, songnumber, song, chosen, score) VALUES (\n(SELECT tournament FROM tournaments WHERE name='Challenge Cup 3'),\n1, 5,\n(SELECT player FROM players WHERE nick='Olemann'),\n(SELECT tournament FROM tournaments WHERE name='Challenge Cup 3'),\n1, 5,\n(SELECT player FROM players WHERE nick='Olemann'),\n+ 1,\n(SELECT song FROM songs WHERE lower(title)=lower('Dead End')),\n'f',\n7919);\n(SELECT song FROM songs WHERE lower(title)=lower('Dead End')),\n'f',\n7919);\n-INSERT INTO scores (tournament, round, parallel, player, song, chosen, score) VALUES (\n+INSERT INTO scores (tournament, round, parallel, player, songnumber, song, chosen, score) VALUES (\n(SELECT tournament FROM tournaments WHERE name='Challenge Cup 3'),\n1, 5,\n(SELECT player FROM players WHERE nick='Olemann'),\n(SELECT tournament FROM tournaments WHERE name='Challenge Cup 3'),\n1, 5,\n(SELECT player FROM players WHERE nick='Olemann'),\n+ 2,\n(SELECT song FROM songs WHERE lower(title)=lower('Crash!')),\n't',\n8043);\n(SELECT song FROM songs WHERE lower(title)=lower('Crash!')),\n't',\n8043);\n@@ -809,24 +883,27 @@ INSERT INTO roundparticipation (tournament, round, parallel, player, position) V\n(SELECT player FROM players WHERE nick='T|na'),\n6\n);\n(SELECT player FROM players WHERE nick='T|na'),\n6\n);\n-INSERT INTO scores (tournament, round, parallel, player, song, chosen, score) VALUES (\n+INSERT INTO scores (tournament, round, parallel, player, songnumber, song, chosen, score) VALUES (\n(SELECT tournament FROM tournaments WHERE name='Challenge Cup 3'),\n1, 5,\n(SELECT player FROM players WHERE nick='T|na'),\n(SELECT tournament FROM tournaments WHERE name='Challenge Cup 3'),\n1, 5,\n(SELECT player FROM players WHERE nick='T|na'),\n+ 0,\n(SELECT song FROM songs WHERE lower(title)=lower('Paranoia KCET (Clean Mix)')),\n'f',\n8546);\n(SELECT song FROM songs WHERE lower(title)=lower('Paranoia KCET (Clean Mix)')),\n'f',\n8546);\n-INSERT INTO scores (tournament, round, parallel, player, song, chosen, score) VALUES (\n+INSERT INTO scores (tournament, round, parallel, player, songnumber, song, chosen, score) VALUES (\n(SELECT tournament FROM tournaments WHERE name='Challenge Cup 3'),\n1, 5,\n(SELECT player FROM players WHERE nick='T|na'),\n(SELECT tournament FROM tournaments WHERE name='Challenge Cup 3'),\n1, 5,\n(SELECT player FROM players WHERE nick='T|na'),\n+ 1,\n(SELECT song FROM songs WHERE lower(title)=lower('Dead End')),\n'f',\n7881);\n(SELECT song FROM songs WHERE lower(title)=lower('Dead End')),\n'f',\n7881);\n-INSERT INTO scores (tournament, round, parallel, player, song, chosen, score) VALUES (\n+INSERT INTO scores (tournament, round, parallel, player, songnumber, song, chosen, score) VALUES (\n(SELECT tournament FROM tournaments WHERE name='Challenge Cup 3'),\n1, 5,\n(SELECT player FROM players WHERE nick='T|na'),\n(SELECT tournament FROM tournaments WHERE name='Challenge Cup 3'),\n1, 5,\n(SELECT player FROM players WHERE nick='T|na'),\n+ 2,\n(SELECT song FROM songs WHERE lower(title)=lower('Jam Jam Reggae (AM Swing Mix)')),\n't',\n7740);\n(SELECT song FROM songs WHERE lower(title)=lower('Jam Jam Reggae (AM Swing Mix)')),\n't',\n7740);\n@@ -856,24 +933,27 @@ INSERT INTO roundparticipation (tournament, round, parallel, player, position) V\n(SELECT player FROM players WHERE nick='Bonzo'),\n1\n);\n(SELECT player FROM players WHERE nick='Bonzo'),\n1\n);\n-INSERT INTO scores (tournament, round, parallel, player, song, chosen, score) VALUES (\n+INSERT INTO scores (tournament, round, parallel, player, songnumber, song, chosen, score) VALUES (\n(SELECT tournament FROM tournaments WHERE name='Challenge Cup 3'),\n1, 6,\n(SELECT player FROM players WHERE nick='Bonzo'),\n(SELECT tournament FROM tournaments WHERE name='Challenge Cup 3'),\n1, 6,\n(SELECT player FROM players WHERE nick='Bonzo'),\n+ 0,\n(SELECT song FROM songs WHERE lower(title)=lower('The Reflex')),\n'f',\n4564);\n(SELECT song FROM songs WHERE lower(title)=lower('The Reflex')),\n'f',\n4564);\n-INSERT INTO scores (tournament, round, parallel, player, song, chosen, score) VALUES (\n+INSERT INTO scores (tournament, round, parallel, player, songnumber, song, chosen, score) VALUES (\n(SELECT tournament FROM tournaments WHERE name='Challenge Cup 3'),\n1, 6,\n(SELECT player FROM players WHERE nick='Bonzo'),\n(SELECT tournament FROM tournaments WHERE name='Challenge Cup 3'),\n1, 6,\n(SELECT player FROM players WHERE nick='Bonzo'),\n+ 1,\n(SELECT song FROM songs WHERE lower(title)=lower('Can\\'t Stop Fallin\\' in Love (SPEED MIX)')),\n'f',\n6212);\n(SELECT song FROM songs WHERE lower(title)=lower('Can\\'t Stop Fallin\\' in Love (SPEED MIX)')),\n'f',\n6212);\n-INSERT INTO scores (tournament, round, parallel, player, song, chosen, score) VALUES (\n+INSERT INTO scores (tournament, round, parallel, player, songnumber, song, chosen, score) VALUES (\n(SELECT tournament FROM tournaments WHERE name='Challenge Cup 3'),\n1, 6,\n(SELECT player FROM players WHERE nick='Bonzo'),\n(SELECT tournament FROM tournaments WHERE name='Challenge Cup 3'),\n1, 6,\n(SELECT player FROM players WHERE nick='Bonzo'),\n+ 2,\n(SELECT song FROM songs WHERE lower(title)=lower('Crash!')),\n't',\n7211);\n(SELECT song FROM songs WHERE lower(title)=lower('Crash!')),\n't',\n7211);\n@@ -884,24 +964,27 @@ INSERT INTO roundparticipation (tournament, round, parallel, player, position) V\n(SELECT player FROM players WHERE nick='Gollum'),\n2\n);\n(SELECT player FROM players WHERE nick='Gollum'),\n2\n);\n-INSERT INTO scores (tournament, round, parallel, player, song, chosen, score) VALUES (\n+INSERT INTO scores (tournament, round, parallel, player, songnumber, song, chosen, score) VALUES (\n(SELECT tournament FROM tournaments WHERE name='Challenge Cup 3'),\n1, 6,\n(SELECT player FROM players WHERE nick='Gollum'),\n(SELECT tournament FROM tournaments WHERE name='Challenge Cup 3'),\n1, 6,\n(SELECT player FROM players WHERE nick='Gollum'),\n+ 0,\n(SELECT song FROM songs WHERE lower(title)=lower('The Reflex')),\n'f',\n3139);\n(SELECT song FROM songs WHERE lower(title)=lower('The Reflex')),\n'f',\n3139);\n-INSERT INTO scores (tournament, round, parallel, player, song, chosen, score) VALUES (\n+INSERT INTO scores (tournament, round, parallel, player, songnumber, song, chosen, score) VALUES (\n(SELECT tournament FROM tournaments WHERE name='Challenge Cup 3'),\n1, 6,\n(SELECT player FROM players WHERE nick='Gollum'),\n(SELECT tournament FROM tournaments WHERE name='Challenge Cup 3'),\n1, 6,\n(SELECT player FROM players WHERE nick='Gollum'),\n+ 1,\n(SELECT song FROM songs WHERE lower(title)=lower('Can\\'t Stop Fallin\\' in Love (SPEED MIX)')),\n'f',\n5655);\n(SELECT song FROM songs WHERE lower(title)=lower('Can\\'t Stop Fallin\\' in Love (SPEED MIX)')),\n'f',\n5655);\n-INSERT INTO scores (tournament, round, parallel, player, song, chosen, score) VALUES (\n+INSERT INTO scores (tournament, round, parallel, player, songnumber, song, chosen, score) VALUES (\n(SELECT tournament FROM tournaments WHERE name='Challenge Cup 3'),\n1, 6,\n(SELECT player FROM players WHERE nick='Gollum'),\n(SELECT tournament FROM tournaments WHERE name='Challenge Cup 3'),\n1, 6,\n(SELECT player FROM players WHERE nick='Gollum'),\n+ 2,\n(SELECT song FROM songs WHERE lower(title)=lower('Jam Jam Reggae (AM Swing Mix)')),\n't',\n6369);\n(SELECT song FROM songs WHERE lower(title)=lower('Jam Jam Reggae (AM Swing Mix)')),\n't',\n6369);\n@@ -912,24 +995,27 @@ INSERT INTO roundparticipation (tournament, round, parallel, player, position) V\n(SELECT player FROM players WHERE nick='Hugo'),\n3\n);\n(SELECT player FROM players WHERE nick='Hugo'),\n3\n);\n-INSERT INTO scores (tournament, round, parallel, player, song, chosen, score) VALUES (\n+INSERT INTO scores (tournament, round, parallel, player, songnumber, song, chosen, score) VALUES (\n(SELECT tournament FROM tournaments WHERE name='Challenge Cup 3'),\n1, 6,\n(SELECT player FROM players WHERE nick='Hugo'),\n(SELECT tournament FROM tournaments WHERE name='Challenge Cup 3'),\n1, 6,\n(SELECT player FROM players WHERE nick='Hugo'),\n+ 0,\n(SELECT song FROM songs WHERE lower(title)=lower('The Reflex')),\n'f',\n6888);\n(SELECT song FROM songs WHERE lower(title)=lower('The Reflex')),\n'f',\n6888);\n-INSERT INTO scores (tournament, round, parallel, player, song, chosen, score) VALUES (\n+INSERT INTO scores (tournament, round, parallel, player, songnumber, song, chosen, score) VALUES (\n(SELECT tournament FROM tournaments WHERE name='Challenge Cup 3'),\n1, 6,\n(SELECT player FROM players WHERE nick='Hugo'),\n(SELECT tournament FROM tournaments WHERE name='Challenge Cup 3'),\n1, 6,\n(SELECT player FROM players WHERE nick='Hugo'),\n+ 1,\n(SELECT song FROM songs WHERE lower(title)=lower('Can\\'t Stop Fallin\\' in Love (SPEED MIX)')),\n'f',\n8521);\n(SELECT song FROM songs WHERE lower(title)=lower('Can\\'t Stop Fallin\\' in Love (SPEED MIX)')),\n'f',\n8521);\n-INSERT INTO scores (tournament, round, parallel, player, song, chosen, score) VALUES (\n+INSERT INTO scores (tournament, round, parallel, player, songnumber, song, chosen, score) VALUES (\n(SELECT tournament FROM tournaments WHERE name='Challenge Cup 3'),\n1, 6,\n(SELECT player FROM players WHERE nick='Hugo'),\n(SELECT tournament FROM tournaments WHERE name='Challenge Cup 3'),\n1, 6,\n(SELECT player FROM players WHERE nick='Hugo'),\n+ 2,\n(SELECT song FROM songs WHERE lower(title)=lower('era (nostalmix)')),\n't',\n7439);\n(SELECT song FROM songs WHERE lower(title)=lower('era (nostalmix)')),\n't',\n7439);\n@@ -940,24 +1026,27 @@ INSERT INTO roundparticipation (tournament, round, parallel, player, position) V\n(SELECT player FROM players WHERE nick='Auran'),\n4\n);\n(SELECT player FROM players WHERE nick='Auran'),\n4\n);\n-INSERT INTO scores (tournament, round, parallel, player, song, chosen, score) VALUES (\n+INSERT INTO scores (tournament, round, parallel, player, songnumber, song, chosen, score) VALUES (\n(SELECT tournament FROM tournaments WHERE name='Challenge Cup 3'),\n1, 6,\n(SELECT player FROM players WHERE nick='Auran'),\n(SELECT tournament FROM tournaments WHERE name='Challenge Cup 3'),\n1, 6,\n(SELECT player FROM players WHERE nick='Auran'),\n+ 0,\n(SELECT song FROM songs WHERE lower(title)=lower('The Reflex')),\n'f',\n3300);\n(SELECT song FROM songs WHERE lower(title)=lower('The Reflex')),\n'f',\n3300);\n-INSERT INTO scores (tournament, round, parallel, player, song, chosen, score) VALUES (\n+INSERT INTO scores (tournament, round, parallel, player, songnumber, song, chosen, score) VALUES (\n(SELECT tournament FROM tournaments WHERE name='Challenge Cup 3'),\n1, 6,\n(SELECT player FROM players WHERE nick='Auran'),\n(SELECT tournament FROM tournaments WHERE name='Challenge Cup 3'),\n1, 6,\n(SELECT player FROM players WHERE nick='Auran'),\n+ 1,\n(SELECT song FROM songs WHERE lower(title)=lower('Can\\'t Stop Fallin\\' in Love (SPEED MIX)')),\n'f',\n3813);\n(SELECT song FROM songs WHERE lower(title)=lower('Can\\'t Stop Fallin\\' in Love (SPEED MIX)')),\n'f',\n3813);\n-INSERT INTO scores (tournament, round, parallel, player, song, chosen, score) VALUES (\n+INSERT INTO scores (tournament, round, parallel, player, songnumber, song, chosen, score) VALUES (\n(SELECT tournament FROM tournaments WHERE name='Challenge Cup 3'),\n1, 6,\n(SELECT player FROM players WHERE nick='Auran'),\n(SELECT tournament FROM tournaments WHERE name='Challenge Cup 3'),\n1, 6,\n(SELECT player FROM players WHERE nick='Auran'),\n+ 2,\n(SELECT song FROM songs WHERE lower(title)=lower('Silent Hill')),\n't',\n4377);\n(SELECT song FROM songs WHERE lower(title)=lower('Silent Hill')),\n't',\n4377);\n@@ -968,24 +1057,27 @@ INSERT INTO roundparticipation (tournament, round, parallel, player, position) V\n(SELECT player FROM players WHERE nick='Smiph'),\n5\n);\n(SELECT player FROM players WHERE nick='Smiph'),\n5\n);\n-INSERT INTO scores (tournament, round, parallel, player, song, chosen, score) VALUES (\n+INSERT INTO scores (tournament, round, parallel, player, songnumber, song, chosen, score) VALUES (\n(SELECT tournament FROM tournaments WHERE name='Challenge Cup 3'),\n1, 6,\n(SELECT player FROM players WHERE nick='Smiph'),\n(SELECT tournament FROM tournaments WHERE name='Challenge Cup 3'),\n1, 6,\n(SELECT player FROM players WHERE nick='Smiph'),\n+ 0,\n(SELECT song FROM songs WHERE lower(title)=lower('The Reflex')),\n'f',\n5696);\n(SELECT song FROM songs WHERE lower(title)=lower('The Reflex')),\n'f',\n5696);\n-INSERT INTO scores (tournament, round, parallel, player, song, chosen, score) VALUES (\n+INSERT INTO scores (tournament, round, parallel, player, songnumber, song, chosen, score) VALUES (\n(SELECT tournament FROM tournaments WHERE name='Challenge Cup 3'),\n1, 6,\n(SELECT player FROM players WHERE nick='Smiph'),\n(SELECT tournament FROM tournaments WHERE name='Challenge Cup 3'),\n1, 6,\n(SELECT player FROM players WHERE nick='Smiph'),\n+ 1,\n(SELECT song FROM songs WHERE lower(title)=lower('Can\\'t Stop Fallin\\' in Love (SPEED MIX)')),\n'f',\n7569);\n(SELECT song FROM songs WHERE lower(title)=lower('Can\\'t Stop Fallin\\' in Love (SPEED MIX)')),\n'f',\n7569);\n-INSERT INTO scores (tournament, round, parallel, player, song, chosen, score) VALUES (\n+INSERT INTO scores (tournament, round, parallel, player, songnumber, song, chosen, score) VALUES (\n(SELECT tournament FROM tournaments WHERE name='Challenge Cup 3'),\n1, 6,\n(SELECT player FROM players WHERE nick='Smiph'),\n(SELECT tournament FROM tournaments WHERE name='Challenge Cup 3'),\n1, 6,\n(SELECT player FROM players WHERE nick='Smiph'),\n+ 2,\n(SELECT song FROM songs WHERE lower(title)=lower('Paranoia KCET (Clean Mix)')),\n't',\n7703);\n(SELECT song FROM songs WHERE lower(title)=lower('Paranoia KCET (Clean Mix)')),\n't',\n7703);\n@@ -996,24 +1088,27 @@ INSERT INTO roundparticipation (tournament, round, parallel, player, position) V\n(SELECT player FROM players WHERE nick='Stream'),\n6\n);\n(SELECT player FROM players WHERE nick='Stream'),\n6\n);\n-INSERT INTO scores (tournament, round, parallel, player, song, chosen, score) VALUES (\n+INSERT INTO scores (tournament, round, parallel, player, songnumber, song, chosen, score) VALUES (\n(SELECT tournament FROM tournaments WHERE name='Challenge Cup 3'),\n1, 6,\n(SELECT player FROM players WHERE nick='Stream'),\n(SELECT tournament FROM tournaments WHERE name='Challenge Cup 3'),\n1, 6,\n(SELECT player FROM players WHERE nick='Stream'),\n+ 0,\n(SELECT song FROM songs WHERE lower(title)=lower('The Reflex')),\n'f',\n5989);\n(SELECT song FROM songs WHERE lower(title)=lower('The Reflex')),\n'f',\n5989);\n-INSERT INTO scores (tournament, round, parallel, player, song, chosen, score) VALUES (\n+INSERT INTO scores (tournament, round, parallel, player, songnumber, song, chosen, score) VALUES (\n(SELECT tournament FROM tournaments WHERE name='Challenge Cup 3'),\n1, 6,\n(SELECT player FROM players WHERE nick='Stream'),\n(SELECT tournament FROM tournaments WHERE name='Challenge Cup 3'),\n1, 6,\n(SELECT player FROM players WHERE nick='Stream'),\n+ 1,\n(SELECT song FROM songs WHERE lower(title)=lower('Can\\'t Stop Fallin\\' in Love (SPEED MIX)')),\n'f',\n7689);\n(SELECT song FROM songs WHERE lower(title)=lower('Can\\'t Stop Fallin\\' in Love (SPEED MIX)')),\n'f',\n7689);\n-INSERT INTO scores (tournament, round, parallel, player, song, chosen, score) VALUES (\n+INSERT INTO scores (tournament, round, parallel, player, songnumber, song, chosen, score) VALUES (\n(SELECT tournament FROM tournaments WHERE name='Challenge Cup 3'),\n1, 6,\n(SELECT player FROM players WHERE nick='Stream'),\n(SELECT tournament FROM tournaments WHERE name='Challenge Cup 3'),\n1, 6,\n(SELECT player FROM players WHERE nick='Stream'),\n+ 2,\n(SELECT song FROM songs WHERE lower(title)=lower('Crash!')),\n't',\n8064);\n(SELECT song FROM songs WHERE lower(title)=lower('Crash!')),\n't',\n8064);\n@@ -1048,24 +1143,27 @@ INSERT INTO roundparticipation (tournament, round, parallel, player, position) V\n(SELECT player FROM players WHERE nick='Smiph'),\n1\n);\n(SELECT player FROM players WHERE nick='Smiph'),\n1\n);\n-INSERT INTO scores (tournament, round, parallel, player, song, chosen, score) VALUES (\n+INSERT INTO scores (tournament, round, parallel, player, songnumber, song, chosen, score) VALUES (\n(SELECT tournament FROM tournaments WHERE name='Challenge Cup 3'),\n2, 1,\n(SELECT player FROM players WHERE nick='Smiph'),\n(SELECT tournament FROM tournaments WHERE name='Challenge Cup 3'),\n2, 1,\n(SELECT player FROM players WHERE nick='Smiph'),\n+ 0,\n(SELECT song FROM songs WHERE lower(title)=lower('Love This Feelin\\'')),\n'f',\n5825);\n(SELECT song FROM songs WHERE lower(title)=lower('Love This Feelin\\'')),\n'f',\n5825);\n-INSERT INTO scores (tournament, round, parallel, player, song, chosen, score) VALUES (\n+INSERT INTO scores (tournament, round, parallel, player, songnumber, song, chosen, score) VALUES (\n(SELECT tournament FROM tournaments WHERE name='Challenge Cup 3'),\n2, 1,\n(SELECT player FROM players WHERE nick='Smiph'),\n(SELECT tournament FROM tournaments WHERE name='Challenge Cup 3'),\n2, 1,\n(SELECT player FROM players WHERE nick='Smiph'),\n+ 1,\n(SELECT song FROM songs WHERE lower(title)=lower('I Want You To Want Me')),\n'f',\n4790);\n(SELECT song FROM songs WHERE lower(title)=lower('I Want You To Want Me')),\n'f',\n4790);\n-INSERT INTO scores (tournament, round, parallel, player, song, chosen, score) VALUES (\n+INSERT INTO scores (tournament, round, parallel, player, songnumber, song, chosen, score) VALUES (\n(SELECT tournament FROM tournaments WHERE name='Challenge Cup 3'),\n2, 1,\n(SELECT player FROM players WHERE nick='Smiph'),\n(SELECT tournament FROM tournaments WHERE name='Challenge Cup 3'),\n2, 1,\n(SELECT player FROM players WHERE nick='Smiph'),\n+ 2,\n(SELECT song FROM songs WHERE lower(title)=lower('Electro Tuned (the SubS Mix)')),\n't',\n7961);\n(SELECT song FROM songs WHERE lower(title)=lower('Electro Tuned (the SubS Mix)')),\n't',\n7961);\n@@ -1076,24 +1174,27 @@ INSERT INTO roundparticipation (tournament, round, parallel, player, position) V\n(SELECT player FROM players WHERE nick='Mulefisk'),\n2\n);\n(SELECT player FROM players WHERE nick='Mulefisk'),\n2\n);\n-INSERT INTO scores (tournament, round, parallel, player, song, chosen, score) VALUES (\n+INSERT INTO scores (tournament, round, parallel, player, songnumber, song, chosen, score) VALUES (\n(SELECT tournament FROM tournaments WHERE name='Challenge Cup 3'),\n2, 1,\n(SELECT player FROM players WHERE nick='Mulefisk'),\n(SELECT tournament FROM tournaments WHERE name='Challenge Cup 3'),\n2, 1,\n(SELECT player FROM players WHERE nick='Mulefisk'),\n+ 0,\n(SELECT song FROM songs WHERE lower(title)=lower('Love This Feelin\\'')),\n'f',\n6533);\n(SELECT song FROM songs WHERE lower(title)=lower('Love This Feelin\\'')),\n'f',\n6533);\n-INSERT INTO scores (tournament, round, parallel, player, song, chosen, score) VALUES (\n+INSERT INTO scores (tournament, round, parallel, player, songnumber, song, chosen, score) VALUES (\n(SELECT tournament FROM tournaments WHERE name='Challenge Cup 3'),\n2, 1,\n(SELECT player FROM players WHERE nick='Mulefisk'),\n(SELECT tournament FROM tournaments WHERE name='Challenge Cup 3'),\n2, 1,\n(SELECT player FROM players WHERE nick='Mulefisk'),\n+ 1,\n(SELECT song FROM songs WHERE lower(title)=lower('I Want You To Want Me')),\n'f',\n5504);\n(SELECT song FROM songs WHERE lower(title)=lower('I Want You To Want Me')),\n'f',\n5504);\n-INSERT INTO scores (tournament, round, parallel, player, song, chosen, score) VALUES (\n+INSERT INTO scores (tournament, round, parallel, player, songnumber, song, chosen, score) VALUES (\n(SELECT tournament FROM tournaments WHERE name='Challenge Cup 3'),\n2, 1,\n(SELECT player FROM players WHERE nick='Mulefisk'),\n(SELECT tournament FROM tournaments WHERE name='Challenge Cup 3'),\n2, 1,\n(SELECT player FROM players WHERE nick='Mulefisk'),\n+ 2,\n(SELECT song FROM songs WHERE lower(title)=lower('Exotic Ethnic')),\n't',\n7989);\n(SELECT song FROM songs WHERE lower(title)=lower('Exotic Ethnic')),\n't',\n7989);\n@@ -1104,24 +1205,27 @@ INSERT INTO roundparticipation (tournament, round, parallel, player, position) V\n(SELECT player FROM players WHERE nick='Oliyzia'),\n3\n);\n(SELECT player FROM players WHERE nick='Oliyzia'),\n3\n);\n-INSERT INTO scores (tournament, round, parallel, player, song, chosen, score) VALUES (\n+INSERT INTO scores (tournament, round, parallel, player, songnumber, song, chosen, score) VALUES (\n(SELECT tournament FROM tournaments WHERE name='Challenge Cup 3'),\n2, 1,\n(SELECT player FROM players WHERE nick='Oliyzia'),\n(SELECT tournament FROM tournaments WHERE name='Challenge Cup 3'),\n2, 1,\n(SELECT player FROM players WHERE nick='Oliyzia'),\n+ 0,\n(SELECT song FROM songs WHERE lower(title)=lower('Love This Feelin\\'')),\n'f',\n7101);\n(SELECT song FROM songs WHERE lower(title)=lower('Love This Feelin\\'')),\n'f',\n7101);\n-INSERT INTO scores (tournament, round, parallel, player, song, chosen, score) VALUES (\n+INSERT INTO scores (tournament, round, parallel, player, songnumber, song, chosen, score) VALUES (\n(SELECT tournament FROM tournaments WHERE name='Challenge Cup 3'),\n2, 1,\n(SELECT player FROM players WHERE nick='Oliyzia'),\n(SELECT tournament FROM tournaments WHERE name='Challenge Cup 3'),\n2, 1,\n(SELECT player FROM players WHERE nick='Oliyzia'),\n+ 1,\n(SELECT song FROM songs WHERE lower(title)=lower('I Want You To Want Me')),\n'f',\n5725);\n(SELECT song FROM songs WHERE lower(title)=lower('I Want You To Want Me')),\n'f',\n5725);\n-INSERT INTO scores (tournament, round, parallel, player, song, chosen, score) VALUES (\n+INSERT INTO scores (tournament, round, parallel, player, songnumber, song, chosen, score) VALUES (\n(SELECT tournament FROM tournaments WHERE name='Challenge Cup 3'),\n2, 1,\n(SELECT player FROM players WHERE nick='Oliyzia'),\n(SELECT tournament FROM tournaments WHERE name='Challenge Cup 3'),\n2, 1,\n(SELECT player FROM players WHERE nick='Oliyzia'),\n+ 2,\n(SELECT song FROM songs WHERE lower(title)=lower('Electro Tuned (the SubS Mix)')),\n't',\n8713);\n(SELECT song FROM songs WHERE lower(title)=lower('Electro Tuned (the SubS Mix)')),\n't',\n8713);\n@@ -1132,24 +1236,27 @@ INSERT INTO roundparticipation (tournament, round, parallel, player, position) V\n(SELECT player FROM players WHERE nick='Th0rny'),\n4\n);\n(SELECT player FROM players WHERE nick='Th0rny'),\n4\n);\n-INSERT INTO scores (tournament, round, parallel, player, song, chosen, score) VALUES (\n+INSERT INTO scores (tournament, round, parallel, player, songnumber, song, chosen, score) VALUES (\n(SELECT tournament FROM tournaments WHERE name='Challenge Cup 3'),\n2, 1,\n(SELECT player FROM players WHERE nick='Th0rny'),\n(SELECT tournament FROM tournaments WHERE name='Challenge Cup 3'),\n2, 1,\n(SELECT player FROM players WHERE nick='Th0rny'),\n+ 0,\n(SELECT song FROM songs WHERE lower(title)=lower('Love This Feelin\\'')),\n'f',\n6377);\n(SELECT song FROM songs WHERE lower(title)=lower('Love This Feelin\\'')),\n'f',\n6377);\n-INSERT INTO scores (tournament, round, parallel, player, song, chosen, score) VALUES (\n+INSERT INTO scores (tournament, round, parallel, player, songnumber, song, chosen, score) VALUES (\n(SELECT tournament FROM tournaments WHERE name='Challenge Cup 3'),\n2, 1,\n(SELECT player FROM players WHERE nick='Th0rny'),\n(SELECT tournament FROM tournaments WHERE name='Challenge Cup 3'),\n2, 1,\n(SELECT player FROM players WHERE nick='Th0rny'),\n+ 1,\n(SELECT song FROM songs WHERE lower(title)=lower('I Want You To Want Me')),\n'f',\n4994);\n(SELECT song FROM songs WHERE lower(title)=lower('I Want You To Want Me')),\n'f',\n4994);\n-INSERT INTO scores (tournament, round, parallel, player, song, chosen, score) VALUES (\n+INSERT INTO scores (tournament, round, parallel, player, songnumber, song, chosen, score) VALUES (\n(SELECT tournament FROM tournaments WHERE name='Challenge Cup 3'),\n2, 1,\n(SELECT player FROM players WHERE nick='Th0rny'),\n(SELECT tournament FROM tournaments WHERE name='Challenge Cup 3'),\n2, 1,\n(SELECT player FROM players WHERE nick='Th0rny'),\n+ 2,\n(SELECT song FROM songs WHERE lower(title)=lower('Dead End')),\n't',\n8140);\n(SELECT song FROM songs WHERE lower(title)=lower('Dead End')),\n't',\n8140);\n@@ -1160,24 +1267,27 @@ INSERT INTO roundparticipation (tournament, round, parallel, player, position) V\n(SELECT player FROM players WHERE nick='OJK'),\n5\n);\n(SELECT player FROM players WHERE nick='OJK'),\n5\n);\n-INSERT INTO scores (tournament, round, parallel, player, song, chosen, score) VALUES (\n+INSERT INTO scores (tournament, round, parallel, player, songnumber, song, chosen, score) VALUES (\n(SELECT tournament FROM tournaments WHERE name='Challenge Cup 3'),\n2, 1,\n(SELECT player FROM players WHERE nick='OJK'),\n(SELECT tournament FROM tournaments WHERE name='Challenge Cup 3'),\n2, 1,\n(SELECT player FROM players WHERE nick='OJK'),\n+ 0,\n(SELECT song FROM songs WHERE lower(title)=lower('Love This Feelin\\'')),\n'f',\n7140);\n(SELECT song FROM songs WHERE lower(title)=lower('Love This Feelin\\'')),\n'f',\n7140);\n-INSERT INTO scores (tournament, round, parallel, player, song, chosen, score) VALUES (\n+INSERT INTO scores (tournament, round, parallel, player, songnumber, song, chosen, score) VALUES (\n(SELECT tournament FROM tournaments WHERE name='Challenge Cup 3'),\n2, 1,\n(SELECT player FROM players WHERE nick='OJK'),\n(SELECT tournament FROM tournaments WHERE name='Challenge Cup 3'),\n2, 1,\n(SELECT player FROM players WHERE nick='OJK'),\n+ 1,\n(SELECT song FROM songs WHERE lower(title)=lower('I Want You To Want Me')),\n'f',\n5515);\n(SELECT song FROM songs WHERE lower(title)=lower('I Want You To Want Me')),\n'f',\n5515);\n-INSERT INTO scores (tournament, round, parallel, player, song, chosen, score) VALUES (\n+INSERT INTO scores (tournament, round, parallel, player, songnumber, song, chosen, score) VALUES (\n(SELECT tournament FROM tournaments WHERE name='Challenge Cup 3'),\n2, 1,\n(SELECT player FROM players WHERE nick='OJK'),\n(SELECT tournament FROM tournaments WHERE name='Challenge Cup 3'),\n2, 1,\n(SELECT player FROM players WHERE nick='OJK'),\n+ 2,\n(SELECT song FROM songs WHERE lower(title)=lower('Trip Machine (Luv Mix)')),\n't',\n8310);\n(SELECT song FROM songs WHERE lower(title)=lower('Trip Machine (Luv Mix)')),\n't',\n8310);\n@@ -1188,24 +1298,27 @@ INSERT INTO roundparticipation (tournament, round, parallel, player, position) V\n(SELECT player FROM players WHERE nick='T|na'),\n6\n);\n(SELECT player FROM players WHERE nick='T|na'),\n6\n);\n-INSERT INTO scores (tournament, round, parallel, player, song, chosen, score) VALUES (\n+INSERT INTO scores (tournament, round, parallel, player, songnumber, song, chosen, score) VALUES (\n(SELECT tournament FROM tournaments WHERE name='Challenge Cup 3'),\n2, 1,\n(SELECT player FROM players WHERE nick='T|na'),\n(SELECT tournament FROM tournaments WHERE name='Challenge Cup 3'),\n2, 1,\n(SELECT player FROM players WHERE nick='T|na'),\n+ 0,\n(SELECT song FROM songs WHERE lower(title)=lower('Love This Feelin\\'')),\n'f',\n6447);\n(SELECT song FROM songs WHERE lower(title)=lower('Love This Feelin\\'')),\n'f',\n6447);\n-INSERT INTO scores (tournament, round, parallel, player, song, chosen, score) VALUES (\n+INSERT INTO scores (tournament, round, parallel, player, songnumber, song, chosen, score) VALUES (\n(SELECT tournament FROM tournaments WHERE name='Challenge Cup 3'),\n2, 1,\n(SELECT player FROM players WHERE nick='T|na'),\n(SELECT tournament FROM tournaments WHERE name='Challenge Cup 3'),\n2, 1,\n(SELECT player FROM players WHERE nick='T|na'),\n+ 1,\n(SELECT song FROM songs WHERE lower(title)=lower('I Want You To Want Me')),\n'f',\n5518);\n(SELECT song FROM songs WHERE lower(title)=lower('I Want You To Want Me')),\n'f',\n5518);\n-INSERT INTO scores (tournament, round, parallel, player, song, chosen, score) VALUES (\n+INSERT INTO scores (tournament, round, parallel, player, songnumber, song, chosen, score) VALUES (\n(SELECT tournament FROM tournaments WHERE name='Challenge Cup 3'),\n2, 1,\n(SELECT player FROM players WHERE nick='T|na'),\n(SELECT tournament FROM tournaments WHERE name='Challenge Cup 3'),\n2, 1,\n(SELECT player FROM players WHERE nick='T|na'),\n+ 2,\n(SELECT song FROM songs WHERE lower(title)=lower('Electro Tuned (the SubS Mix)')),\n't',\n8271);\n(SELECT song FROM songs WHERE lower(title)=lower('Electro Tuned (the SubS Mix)')),\n't',\n8271);\n@@ -1235,24 +1348,27 @@ INSERT INTO roundparticipation (tournament, round, parallel, player, position) V\n(SELECT player FROM players WHERE nick='Dabi'),\n1\n);\n(SELECT player FROM players WHERE nick='Dabi'),\n1\n);\n-INSERT INTO scores (tournament, round, parallel, player, song, chosen, score) VALUES (\n+INSERT INTO scores (tournament, round, parallel, player, songnumber, song, chosen, score) VALUES (\n(SELECT tournament FROM tournaments WHERE name='Challenge Cup 3'),\n2, 2,\n(SELECT player FROM players WHERE nick='Dabi'),\n(SELECT tournament FROM tournaments WHERE name='Challenge Cup 3'),\n2, 2,\n(SELECT player FROM players WHERE nick='Dabi'),\n+ 0,\n(SELECT song FROM songs WHERE lower(title)=lower('La Senorita')),\n'f',\n4394);\n(SELECT song FROM songs WHERE lower(title)=lower('La Senorita')),\n'f',\n4394);\n-INSERT INTO scores (tournament, round, parallel, player, song, chosen, score) VALUES (\n+INSERT INTO scores (tournament, round, parallel, player, songnumber, song, chosen, score) VALUES (\n(SELECT tournament FROM tournaments WHERE name='Challenge Cup 3'),\n2, 2,\n(SELECT player FROM players WHERE nick='Dabi'),\n(SELECT tournament FROM tournaments WHERE name='Challenge Cup 3'),\n2, 2,\n(SELECT player FROM players WHERE nick='Dabi'),\n+ 1,\n(SELECT song FROM songs WHERE lower(title)=lower('Ecstasy')),\n'f',\n4295);\n(SELECT song FROM songs WHERE lower(title)=lower('Ecstasy')),\n'f',\n4295);\n-INSERT INTO scores (tournament, round, parallel, player, song, chosen, score) VALUES (\n+INSERT INTO scores (tournament, round, parallel, player, songnumber, song, chosen, score) VALUES (\n(SELECT tournament FROM tournaments WHERE name='Challenge Cup 3'),\n2, 2,\n(SELECT player FROM players WHERE nick='Dabi'),\n(SELECT tournament FROM tournaments WHERE name='Challenge Cup 3'),\n2, 2,\n(SELECT player FROM players WHERE nick='Dabi'),\n+ 2,\n(SELECT song FROM songs WHERE lower(title)=lower('Paranoia KCET (Clean Mix)')),\n't',\n6188);\n(SELECT song FROM songs WHERE lower(title)=lower('Paranoia KCET (Clean Mix)')),\n't',\n6188);\n@@ -1263,24 +1379,27 @@ INSERT INTO roundparticipation (tournament, round, parallel, player, position) V\n(SELECT player FROM players WHERE nick='Duckling'),\n2\n);\n(SELECT player FROM players WHERE nick='Duckling'),\n2\n);\n-INSERT INTO scores (tournament, round, parallel, player, song, chosen, score) VALUES (\n+INSERT INTO scores (tournament, round, parallel, player, songnumber, song, chosen, score) VALUES (\n(SELECT tournament FROM tournaments WHERE name='Challenge Cup 3'),\n2, 2,\n(SELECT player FROM players WHERE nick='Duckling'),\n(SELECT tournament FROM tournaments WHERE name='Challenge Cup 3'),\n2, 2,\n(SELECT player FROM players WHERE nick='Duckling'),\n+ 0,\n(SELECT song FROM songs WHERE lower(title)=lower('La Senorita')),\n'f',\n5898);\n(SELECT song FROM songs WHERE lower(title)=lower('La Senorita')),\n'f',\n5898);\n-INSERT INTO scores (tournament, round, parallel, player, song, chosen, score) VALUES (\n+INSERT INTO scores (tournament, round, parallel, player, songnumber, song, chosen, score) VALUES (\n(SELECT tournament FROM tournaments WHERE name='Challenge Cup 3'),\n2, 2,\n(SELECT player FROM players WHERE nick='Duckling'),\n(SELECT tournament FROM tournaments WHERE name='Challenge Cup 3'),\n2, 2,\n(SELECT player FROM players WHERE nick='Duckling'),\n+ 1,\n(SELECT song FROM songs WHERE lower(title)=lower('Ecstasy')),\n'f',\n5857);\n(SELECT song FROM songs WHERE lower(title)=lower('Ecstasy')),\n'f',\n5857);\n-INSERT INTO scores (tournament, round, parallel, player, song, chosen, score) VALUES (\n+INSERT INTO scores (tournament, round, parallel, player, songnumber, song, chosen, score) VALUES (\n(SELECT tournament FROM tournaments WHERE name='Challenge Cup 3'),\n2, 2,\n(SELECT player FROM players WHERE nick='Duckling'),\n(SELECT tournament FROM tournaments WHERE name='Challenge Cup 3'),\n2, 2,\n(SELECT player FROM players WHERE nick='Duckling'),\n+ 2,\n(SELECT song FROM songs WHERE lower(title)=lower('Paranoia KCET (Clean Mix)')),\n't',\n8053);\n(SELECT song FROM songs WHERE lower(title)=lower('Paranoia KCET (Clean Mix)')),\n't',\n8053);\n@@ -1291,24 +1410,27 @@ INSERT INTO roundparticipation (tournament, round, parallel, player, position) V\n(SELECT player FROM players WHERE nick='Ylini'),\n3\n);\n(SELECT player FROM players WHERE nick='Ylini'),\n3\n);\n-INSERT INTO scores (tournament, round, parallel, player, song, chosen, score) VALUES (\n+INSERT INTO scores (tournament, round, parallel, player, songnumber, song, chosen, score) VALUES (\n(SELECT tournament FROM tournaments WHERE name='Challenge Cup 3'),\n2, 2,\n(SELECT player FROM players WHERE nick='Ylini'),\n(SELECT tournament FROM tournaments WHERE name='Challenge Cup 3'),\n2, 2,\n(SELECT player FROM players WHERE nick='Ylini'),\n+ 0,\n(SELECT song FROM songs WHERE lower(title)=lower('La Senorita')),\n'f',\n5873);\n(SELECT song FROM songs WHERE lower(title)=lower('La Senorita')),\n'f',\n5873);\n-INSERT INTO scores (tournament, round, parallel, player, song, chosen, score) VALUES (\n+INSERT INTO scores (tournament, round, parallel, player, songnumber, song, chosen, score) VALUES (\n(SELECT tournament FROM tournaments WHERE name='Challenge Cup 3'),\n2, 2,\n(SELECT player FROM players WHERE nick='Ylini'),\n(SELECT tournament FROM tournaments WHERE name='Challenge Cup 3'),\n2, 2,\n(SELECT player FROM players WHERE nick='Ylini'),\n+ 1,\n(SELECT song FROM songs WHERE lower(title)=lower('Ecstasy')),\n'f',\n5881);\n(SELECT song FROM songs WHERE lower(title)=lower('Ecstasy')),\n'f',\n5881);\n-INSERT INTO scores (tournament, round, parallel, player, song, chosen, score) VALUES (\n+INSERT INTO scores (tournament, round, parallel, player, songnumber, song, chosen, score) VALUES (\n(SELECT tournament FROM tournaments WHERE name='Challenge Cup 3'),\n2, 2,\n(SELECT player FROM players WHERE nick='Ylini'),\n(SELECT tournament FROM tournaments WHERE name='Challenge Cup 3'),\n2, 2,\n(SELECT player FROM players WHERE nick='Ylini'),\n+ 2,\n(SELECT song FROM songs WHERE lower(title)=lower('Electro Tuned (the SubS Mix)')),\n't',\n8283);\n(SELECT song FROM songs WHERE lower(title)=lower('Electro Tuned (the SubS Mix)')),\n't',\n8283);\n@@ -1319,24 +1441,27 @@ INSERT INTO roundparticipation (tournament, round, parallel, player, position) V\n(SELECT player FROM players WHERE nick='Stream'),\n4\n);\n(SELECT player FROM players WHERE nick='Stream'),\n4\n);\n-INSERT INTO scores (tournament, round, parallel, player, song, chosen, score) VALUES (\n+INSERT INTO scores (tournament, round, parallel, player, songnumber, song, chosen, score) VALUES (\n(SELECT tournament FROM tournaments WHERE name='Challenge Cup 3'),\n2, 2,\n(SELECT player FROM players WHERE nick='Stream'),\n(SELECT tournament FROM tournaments WHERE name='Challenge Cup 3'),\n2, 2,\n(SELECT player FROM players WHERE nick='Stream'),\n+ 0,\n(SELECT song FROM songs WHERE lower(title)=lower('La Senorita')),\n'f',\n5820);\n(SELECT song FROM songs WHERE lower(title)=lower('La Senorita')),\n'f',\n5820);\n-INSERT INTO scores (tournament, round, parallel, player, song, chosen, score) VALUES (\n+INSERT INTO scores (tournament, round, parallel, player, songnumber, song, chosen, score) VALUES (\n(SELECT tournament FROM tournaments WHERE name='Challenge Cup 3'),\n2, 2,\n(SELECT player FROM players WHERE nick='Stream'),\n(SELECT tournament FROM tournaments WHERE name='Challenge Cup 3'),\n2, 2,\n(SELECT player FROM players WHERE nick='Stream'),\n+ 1,\n(SELECT song FROM songs WHERE lower(title)=lower('Ecstasy')),\n'f',\n6437);\n(SELECT song FROM songs WHERE lower(title)=lower('Ecstasy')),\n'f',\n6437);\n-INSERT INTO scores (tournament, round, parallel, player, song, chosen, score) VALUES (\n+INSERT INTO scores (tournament, round, parallel, player, songnumber, song, chosen, score) VALUES (\n(SELECT tournament FROM tournaments WHERE name='Challenge Cup 3'),\n2, 2,\n(SELECT player FROM players WHERE nick='Stream'),\n(SELECT tournament FROM tournaments WHERE name='Challenge Cup 3'),\n2, 2,\n(SELECT player FROM players WHERE nick='Stream'),\n+ 2,\n(SELECT song FROM songs WHERE lower(title)=lower('Paranoia KCET (Clean Mix)')),\n't',\n8232);\n(SELECT song FROM songs WHERE lower(title)=lower('Paranoia KCET (Clean Mix)')),\n't',\n8232);\n@@ -1347,24 +1472,27 @@ INSERT INTO roundparticipation (tournament, round, parallel, player, position) V\n(SELECT player FROM players WHERE nick='Snute'),\n5\n);\n(SELECT player FROM players WHERE nick='Snute'),\n5\n);\n-INSERT INTO scores (tournament, round, parallel, player, song, chosen, score) VALUES (\n+INSERT INTO scores (tournament, round, parallel, player, songnumber, song, chosen, score) VALUES (\n(SELECT tournament FROM tournaments WHERE name='Challenge Cup 3'),\n2, 2,\n(SELECT player FROM players WHERE nick='Snute'),\n(SELECT tournament FROM tournaments WHERE name='Challenge Cup 3'),\n2, 2,\n(SELECT player FROM players WHERE nick='Snute'),\n+ 0,\n(SELECT song FROM songs WHERE lower(title)=lower('La Senorita')),\n'f',\n6509);\n(SELECT song FROM songs WHERE lower(title)=lower('La Senorita')),\n'f',\n6509);\n-INSERT INTO scores (tournament, round, parallel, player, song, chosen, score) VALUES (\n+INSERT INTO scores (tournament, round, parallel, player, songnumber, song, chosen, score) VALUES (\n(SELECT tournament FROM tournaments WHERE name='Challenge Cup 3'),\n2, 2,\n(SELECT player FROM players WHERE nick='Snute'),\n(SELECT tournament FROM tournaments WHERE name='Challenge Cup 3'),\n2, 2,\n(SELECT player FROM players WHERE nick='Snute'),\n+ 1,\n(SELECT song FROM songs WHERE lower(title)=lower('Ecstasy')),\n'f',\n6858);\n(SELECT song FROM songs WHERE lower(title)=lower('Ecstasy')),\n'f',\n6858);\n-INSERT INTO scores (tournament, round, parallel, player, song, chosen, score) VALUES (\n+INSERT INTO scores (tournament, round, parallel, player, songnumber, song, chosen, score) VALUES (\n(SELECT tournament FROM tournaments WHERE name='Challenge Cup 3'),\n2, 2,\n(SELECT player FROM players WHERE nick='Snute'),\n(SELECT tournament FROM tournaments WHERE name='Challenge Cup 3'),\n2, 2,\n(SELECT player FROM players WHERE nick='Snute'),\n+ 2,\n(SELECT song FROM songs WHERE lower(title)=lower('Dynamite Rave')),\n't',\n8420);\n(SELECT song FROM songs WHERE lower(title)=lower('Dynamite Rave')),\n't',\n8420);\n@@ -1375,24 +1503,27 @@ INSERT INTO roundparticipation (tournament, round, parallel, player, position) V\n(SELECT player FROM players WHERE nick='Hugo'),\n6\n);\n(SELECT player FROM players WHERE nick='Hugo'),\n6\n);\n-INSERT INTO scores (tournament, round, parallel, player, song, chosen, score) VALUES (\n+INSERT INTO scores (tournament, round, parallel, player, songnumber, song, chosen, score) VALUES (\n(SELECT tournament FROM tournaments WHERE name='Challenge Cup 3'),\n2, 2,\n(SELECT player FROM players WHERE nick='Hugo'),\n(SELECT tournament FROM tournaments WHERE name='Challenge Cup 3'),\n2, 2,\n(SELECT player FROM players WHERE nick='Hugo'),\n+ 0,\n(SELECT song FROM songs WHERE lower(title)=lower('La Senorita')),\n'f',\n6424);\n(SELECT song FROM songs WHERE lower(title)=lower('La Senorita')),\n'f',\n6424);\n-INSERT INTO scores (tournament, round, parallel, player, song, chosen, score) VALUES (\n+INSERT INTO scores (tournament, round, parallel, player, songnumber, song, chosen, score) VALUES (\n(SELECT tournament FROM tournaments WHERE name='Challenge Cup 3'),\n2, 2,\n(SELECT player FROM players WHERE nick='Hugo'),\n(SELECT tournament FROM tournaments WHERE name='Challenge Cup 3'),\n2, 2,\n(SELECT player FROM players WHERE nick='Hugo'),\n+ 1,\n(SELECT song FROM songs WHERE lower(title)=lower('Ecstasy')),\n'f',\n6499);\n(SELECT song FROM songs WHERE lower(title)=lower('Ecstasy')),\n'f',\n6499);\n-INSERT INTO scores (tournament, round, parallel, player, song, chosen, score) VALUES (\n+INSERT INTO scores (tournament, round, parallel, player, songnumber, song, chosen, score) VALUES (\n(SELECT tournament FROM tournaments WHERE name='Challenge Cup 3'),\n2, 2,\n(SELECT player FROM players WHERE nick='Hugo'),\n(SELECT tournament FROM tournaments WHERE name='Challenge Cup 3'),\n2, 2,\n(SELECT player FROM players WHERE nick='Hugo'),\n+ 2,\n(SELECT song FROM songs WHERE lower(title)=lower('Trip Machine Climax')),\n't',\n8601);\n(SELECT song FROM songs WHERE lower(title)=lower('Trip Machine Climax')),\n't',\n8601);\n@@ -1422,24 +1553,27 @@ INSERT INTO roundparticipation (tournament, round, parallel, player, position) V\n(SELECT player FROM players WHERE nick='Zyc'),\n1\n);\n(SELECT player FROM players WHERE nick='Zyc'),\n1\n);\n-INSERT INTO scores (tournament, round, parallel, player, song, chosen, score) VALUES (\n+INSERT INTO scores (tournament, round, parallel, player, songnumber, song, chosen, score) VALUES (\n(SELECT tournament FROM tournaments WHERE name='Challenge Cup 3'),\n2, 3,\n(SELECT player FROM players WHERE nick='Zyc'),\n(SELECT tournament FROM tournaments WHERE name='Challenge Cup 3'),\n2, 3,\n(SELECT player FROM players WHERE nick='Zyc'),\n+ 0,\n(SELECT song FROM songs WHERE lower(title)=lower('Exotic Ethnic')),\n'f',\n3711);\n(SELECT song FROM songs WHERE lower(title)=lower('Exotic Ethnic')),\n'f',\n3711);\n-INSERT INTO scores (tournament, round, parallel, player, song, chosen, score) VALUES (\n+INSERT INTO scores (tournament, round, parallel, player, songnumber, song, chosen, score) VALUES (\n(SELECT tournament FROM tournaments WHERE name='Challenge Cup 3'),\n2, 3,\n(SELECT player FROM players WHERE nick='Zyc'),\n(SELECT tournament FROM tournaments WHERE name='Challenge Cup 3'),\n2, 3,\n(SELECT player FROM players WHERE nick='Zyc'),\n+ 1,\n(SELECT song FROM songs WHERE lower(title)=lower('Stomp')),\n'f',\n3767);\n(SELECT song FROM songs WHERE lower(title)=lower('Stomp')),\n'f',\n3767);\n-INSERT INTO scores (tournament, round, parallel, player, song, chosen, score) VALUES (\n+INSERT INTO scores (tournament, round, parallel, player, songnumber, song, chosen, score) VALUES (\n(SELECT tournament FROM tournaments WHERE name='Challenge Cup 3'),\n2, 3,\n(SELECT player FROM players WHERE nick='Zyc'),\n(SELECT tournament FROM tournaments WHERE name='Challenge Cup 3'),\n2, 3,\n(SELECT player FROM players WHERE nick='Zyc'),\n+ 2,\n(SELECT song FROM songs WHERE lower(title)=lower('Max 300')),\n't',\n0);\n(SELECT song FROM songs WHERE lower(title)=lower('Max 300')),\n't',\n0);\n@@ -1450,24 +1584,27 @@ INSERT INTO roundparticipation (tournament, round, parallel, player, position) V\n(SELECT player FROM players WHERE nick='Nameless'),\n2\n);\n(SELECT player FROM players WHERE nick='Nameless'),\n2\n);\n-INSERT INTO scores (tournament, round, parallel, player, song, chosen, score) VALUES (\n+INSERT INTO scores (tournament, round, parallel, player, songnumber, song, chosen, score) VALUES (\n(SELECT tournament FROM tournaments WHERE name='Challenge Cup 3'),\n2, 3,\n(SELECT player FROM players WHERE nick='Nameless'),\n(SELECT tournament FROM tournaments WHERE name='Challenge Cup 3'),\n2, 3,\n(SELECT player FROM players WHERE nick='Nameless'),\n+ 0,\n(SELECT song FROM songs WHERE lower(title)=lower('Exotic Ethnic')),\n'f',\n8621);\n(SELECT song FROM songs WHERE lower(title)=lower('Exotic Ethnic')),\n'f',\n8621);\n-INSERT INTO scores (tournament, round, parallel, player, song, chosen, score) VALUES (\n+INSERT INTO scores (tournament, round, parallel, player, songnumber, song, chosen, score) VALUES (\n(SELECT tournament FROM tournaments WHERE name='Challenge Cup 3'),\n2, 3,\n(SELECT player FROM players WHERE nick='Nameless'),\n(SELECT tournament FROM tournaments WHERE name='Challenge Cup 3'),\n2, 3,\n(SELECT player FROM players WHERE nick='Nameless'),\n+ 1,\n(SELECT song FROM songs WHERE lower(title)=lower('Stomp')),\n'f',\n5773);\n(SELECT song FROM songs WHERE lower(title)=lower('Stomp')),\n'f',\n5773);\n-INSERT INTO scores (tournament, round, parallel, player, song, chosen, score) VALUES (\n+INSERT INTO scores (tournament, round, parallel, player, songnumber, song, chosen, score) VALUES (\n(SELECT tournament FROM tournaments WHERE name='Challenge Cup 3'),\n2, 3,\n(SELECT player FROM players WHERE nick='Nameless'),\n(SELECT tournament FROM tournaments WHERE name='Challenge Cup 3'),\n2, 3,\n(SELECT player FROM players WHERE nick='Nameless'),\n+ 2,\n(SELECT song FROM songs WHERE lower(title)=lower('Electro Tuned (the SubS Mix)')),\n't',\n6811);\n(SELECT song FROM songs WHERE lower(title)=lower('Electro Tuned (the SubS Mix)')),\n't',\n6811);\n@@ -1478,24 +1615,27 @@ INSERT INTO roundparticipation (tournament, round, parallel, player, position) V\n(SELECT player FROM players WHERE nick='Haggar'),\n3\n);\n(SELECT player FROM players WHERE nick='Haggar'),\n3\n);\n-INSERT INTO scores (tournament, round, parallel, player, song, chosen, score) VALUES (\n+INSERT INTO scores (tournament, round, parallel, player, songnumber, song, chosen, score) VALUES (\n(SELECT tournament FROM tournaments WHERE name='Challenge Cup 3'),\n2, 3,\n(SELECT player FROM players WHERE nick='Haggar'),\n(SELECT tournament FROM tournaments WHERE name='Challenge Cup 3'),\n2, 3,\n(SELECT player FROM players WHERE nick='Haggar'),\n+ 0,\n(SELECT song FROM songs WHERE lower(title)=lower('Exotic Ethnic')),\n'f',\n8853);\n(SELECT song FROM songs WHERE lower(title)=lower('Exotic Ethnic')),\n'f',\n8853);\n-INSERT INTO scores (tournament, round, parallel, player, song, chosen, score) VALUES (\n+INSERT INTO scores (tournament, round, parallel, player, songnumber, song, chosen, score) VALUES (\n(SELECT tournament FROM tournaments WHERE name='Challenge Cup 3'),\n2, 3,\n(SELECT player FROM players WHERE nick='Haggar'),\n(SELECT tournament FROM tournaments WHERE name='Challenge Cup 3'),\n2, 3,\n(SELECT player FROM players WHERE nick='Haggar'),\n+ 1,\n(SELECT song FROM songs WHERE lower(title)=lower('Stomp')),\n'f',\n5898);\n(SELECT song FROM songs WHERE lower(title)=lower('Stomp')),\n'f',\n5898);\n-INSERT INTO scores (tournament, round, parallel, player, song, chosen, score) VALUES (\n+INSERT INTO scores (tournament, round, parallel, player, songnumber, song, chosen, score) VALUES (\n(SELECT tournament FROM tournaments WHERE name='Challenge Cup 3'),\n2, 3,\n(SELECT player FROM players WHERE nick='Haggar'),\n(SELECT tournament FROM tournaments WHERE name='Challenge Cup 3'),\n2, 3,\n(SELECT player FROM players WHERE nick='Haggar'),\n+ 2,\n(SELECT song FROM songs WHERE lower(title)=lower('PARANOiA Rebirth')),\n't',\n8750);\n(SELECT song FROM songs WHERE lower(title)=lower('PARANOiA Rebirth')),\n't',\n8750);\n@@ -1506,24 +1646,27 @@ INSERT INTO roundparticipation (tournament, round, parallel, player, position) V\n(SELECT player FROM players WHERE nick='Olemann'),\n4\n);\n(SELECT player FROM players WHERE nick='Olemann'),\n4\n);\n-INSERT INTO scores (tournament, round, parallel, player, song, chosen, score) VALUES (\n+INSERT INTO scores (tournament, round, parallel, player, songnumber, song, chosen, score) VALUES (\n(SELECT tournament FROM tournaments WHERE name='Challenge Cup 3'),\n2, 3,\n(SELECT player FROM players WHERE nick='Olemann'),\n(SELECT tournament FROM tournaments WHERE name='Challenge Cup 3'),\n2, 3,\n(SELECT player FROM players WHERE nick='Olemann'),\n+ 0,\n(SELECT song FROM songs WHERE lower(title)=lower('Exotic Ethnic')),\n'f',\n6836);\n(SELECT song FROM songs WHERE lower(title)=lower('Exotic Ethnic')),\n'f',\n6836);\n-INSERT INTO scores (tournament, round, parallel, player, song, chosen, score) VALUES (\n+INSERT INTO scores (tournament, round, parallel, player, songnumber, song, chosen, score) VALUES (\n(SELECT tournament FROM tournaments WHERE name='Challenge Cup 3'),\n2, 3,\n(SELECT player FROM players WHERE nick='Olemann'),\n(SELECT tournament FROM tournaments WHERE name='Challenge Cup 3'),\n2, 3,\n(SELECT player FROM players WHERE nick='Olemann'),\n+ 1,\n(SELECT song FROM songs WHERE lower(title)=lower('Stomp')),\n'f',\n5513);\n(SELECT song FROM songs WHERE lower(title)=lower('Stomp')),\n'f',\n5513);\n-INSERT INTO scores (tournament, round, parallel, player, song, chosen, score) VALUES (\n+INSERT INTO scores (tournament, round, parallel, player, songnumber, song, chosen, score) VALUES (\n(SELECT tournament FROM tournaments WHERE name='Challenge Cup 3'),\n2, 3,\n(SELECT player FROM players WHERE nick='Olemann'),\n(SELECT tournament FROM tournaments WHERE name='Challenge Cup 3'),\n2, 3,\n(SELECT player FROM players WHERE nick='Olemann'),\n+ 2,\n(SELECT song FROM songs WHERE lower(title)=lower('Can\\'t Stop Fallin\\' in Love (SPEED MIX)')),\n't',\n7941);\n(SELECT song FROM songs WHERE lower(title)=lower('Can\\'t Stop Fallin\\' in Love (SPEED MIX)')),\n't',\n7941);\n@@ -1534,24 +1677,27 @@ INSERT INTO roundparticipation (tournament, round, parallel, player, position) V\n(SELECT player FROM players WHERE nick='RiChY'),\n5\n);\n(SELECT player FROM players WHERE nick='RiChY'),\n5\n);\n-INSERT INTO scores (tournament, round, parallel, player, song, chosen, score) VALUES (\n+INSERT INTO scores (tournament, round, parallel, player, songnumber, song, chosen, score) VALUES (\n(SELECT tournament FROM tournaments WHERE name='Challenge Cup 3'),\n2, 3,\n(SELECT player FROM players WHERE nick='RiChY'),\n(SELECT tournament FROM tournaments WHERE name='Challenge Cup 3'),\n2, 3,\n(SELECT player FROM players WHERE nick='RiChY'),\n+ 0,\n(SELECT song FROM songs WHERE lower(title)=lower('Exotic Ethnic')),\n'f',\n8490);\n(SELECT song FROM songs WHERE lower(title)=lower('Exotic Ethnic')),\n'f',\n8490);\n-INSERT INTO scores (tournament, round, parallel, player, song, chosen, score) VALUES (\n+INSERT INTO scores (tournament, round, parallel, player, songnumber, song, chosen, score) VALUES (\n(SELECT tournament FROM tournaments WHERE name='Challenge Cup 3'),\n2, 3,\n(SELECT player FROM players WHERE nick='RiChY'),\n(SELECT tournament FROM tournaments WHERE name='Challenge Cup 3'),\n2, 3,\n(SELECT player FROM players WHERE nick='RiChY'),\n+ 1,\n(SELECT song FROM songs WHERE lower(title)=lower('Stomp')),\n'f',\n6341);\n(SELECT song FROM songs WHERE lower(title)=lower('Stomp')),\n'f',\n6341);\n-INSERT INTO scores (tournament, round, parallel, player, song, chosen, score) VALUES (\n+INSERT INTO scores (tournament, round, parallel, player, songnumber, song, chosen, score) VALUES (\n(SELECT tournament FROM tournaments WHERE name='Challenge Cup 3'),\n2, 3,\n(SELECT player FROM players WHERE nick='RiChY'),\n(SELECT tournament FROM tournaments WHERE name='Challenge Cup 3'),\n2, 3,\n(SELECT player FROM players WHERE nick='RiChY'),\n+ 2,\n(SELECT song FROM songs WHERE lower(title)=lower('PARANOiA Rebirth')),\n't',\n8630);\n(SELECT song FROM songs WHERE lower(title)=lower('PARANOiA Rebirth')),\n't',\n8630);\n@@ -1562,24 +1708,27 @@ INSERT INTO roundparticipation (tournament, round, parallel, player, position) V\n(SELECT player FROM players WHERE nick='DufF'),\n6\n);\n(SELECT player FROM players WHERE nick='DufF'),\n6\n);\n-INSERT INTO scores (tournament, round, parallel, player, song, chosen, score) VALUES (\n+INSERT INTO scores (tournament, round, parallel, player, songnumber, song, chosen, score) VALUES (\n(SELECT tournament FROM tournaments WHERE name='Challenge Cup 3'),\n2, 3,\n(SELECT player FROM players WHERE nick='DufF'),\n(SELECT tournament FROM tournaments WHERE name='Challenge Cup 3'),\n2, 3,\n(SELECT player FROM players WHERE nick='DufF'),\n+ 0,\n(SELECT song FROM songs WHERE lower(title)=lower('Exotic Ethnic')),\n'f',\n8781);\n(SELECT song FROM songs WHERE lower(title)=lower('Exotic Ethnic')),\n'f',\n8781);\n-INSERT INTO scores (tournament, round, parallel, player, song, chosen, score) VALUES (\n+INSERT INTO scores (tournament, round, parallel, player, songnumber, song, chosen, score) VALUES (\n(SELECT tournament FROM tournaments WHERE name='Challenge Cup 3'),\n2, 3,\n(SELECT player FROM players WHERE nick='DufF'),\n(SELECT tournament FROM tournaments WHERE name='Challenge Cup 3'),\n2, 3,\n(SELECT player FROM players WHERE nick='DufF'),\n+ 1,\n(SELECT song FROM songs WHERE lower(title)=lower('Stomp')),\n'f',\n6211);\n(SELECT song FROM songs WHERE lower(title)=lower('Stomp')),\n'f',\n6211);\n-INSERT INTO scores (tournament, round, parallel, player, song, chosen, score) VALUES (\n+INSERT INTO scores (tournament, round, parallel, player, songnumber, song, chosen, score) VALUES (\n(SELECT tournament FROM tournaments WHERE name='Challenge Cup 3'),\n2, 3,\n(SELECT player FROM players WHERE nick='DufF'),\n(SELECT tournament FROM tournaments WHERE name='Challenge Cup 3'),\n2, 3,\n(SELECT player FROM players WHERE nick='DufF'),\n+ 2,\n(SELECT song FROM songs WHERE lower(title)=lower('Afronova')),\n't',\n8384);\n(SELECT song FROM songs WHERE lower(title)=lower('Afronova')),\n't',\n8384);\n@@ -1609,17 +1758,19 @@ INSERT INTO roundparticipation (tournament, round, parallel, player, position) V\n(SELECT player FROM players WHERE nick='Stream'),\n1\n);\n(SELECT player FROM players WHERE nick='Stream'),\n1\n);\n-INSERT INTO scores (tournament, round, parallel, player, song, chosen, score) VALUES (\n+INSERT INTO scores (tournament, round, parallel, player, songnumber, song, chosen, score) VALUES (\n(SELECT tournament FROM tournaments WHERE name='Challenge Cup 3'),\n3, 0,\n(SELECT player FROM players WHERE nick='Stream'),\n(SELECT tournament FROM tournaments WHERE name='Challenge Cup 3'),\n3, 0,\n(SELECT player FROM players WHERE nick='Stream'),\n+ 0,\n(SELECT song FROM songs WHERE lower(title)=lower('Memories')),\n'f',\n6565);\n(SELECT song FROM songs WHERE lower(title)=lower('Memories')),\n'f',\n6565);\n-INSERT INTO scores (tournament, round, parallel, player, song, chosen, score) VALUES (\n+INSERT INTO scores (tournament, round, parallel, player, songnumber, song, chosen, score) VALUES (\n(SELECT tournament FROM tournaments WHERE name='Challenge Cup 3'),\n3, 0,\n(SELECT player FROM players WHERE nick='Stream'),\n(SELECT tournament FROM tournaments WHERE name='Challenge Cup 3'),\n3, 0,\n(SELECT player FROM players WHERE nick='Stream'),\n+ 1,\n(SELECT song FROM songs WHERE lower(title)=lower('Dead End')),\n't',\n7765);\n(SELECT song FROM songs WHERE lower(title)=lower('Dead End')),\n't',\n7765);\n@@ -1630,17 +1781,19 @@ INSERT INTO roundparticipation (tournament, round, parallel, player, position) V\n(SELECT player FROM players WHERE nick='T|na'),\n2\n);\n(SELECT player FROM players WHERE nick='T|na'),\n2\n);\n-INSERT INTO scores (tournament, round, parallel, player, song, chosen, score) VALUES (\n+INSERT INTO scores (tournament, round, parallel, player, songnumber, song, chosen, score) VALUES (\n(SELECT tournament FROM tournaments WHERE name='Challenge Cup 3'),\n3, 0,\n(SELECT player FROM players WHERE nick='T|na'),\n(SELECT tournament FROM tournaments WHERE name='Challenge Cup 3'),\n3, 0,\n(SELECT player FROM players WHERE nick='T|na'),\n+ 0,\n(SELECT song FROM songs WHERE lower(title)=lower('Memories')),\n'f',\n6520);\n(SELECT song FROM songs WHERE lower(title)=lower('Memories')),\n'f',\n6520);\n-INSERT INTO scores (tournament, round, parallel, player, song, chosen, score) VALUES (\n+INSERT INTO scores (tournament, round, parallel, player, songnumber, song, chosen, score) VALUES (\n(SELECT tournament FROM tournaments WHERE name='Challenge Cup 3'),\n3, 0,\n(SELECT player FROM players WHERE nick='T|na'),\n(SELECT tournament FROM tournaments WHERE name='Challenge Cup 3'),\n3, 0,\n(SELECT player FROM players WHERE nick='T|na'),\n+ 1,\n(SELECT song FROM songs WHERE lower(title)=lower('Paranoia KCET (Clean Mix)')),\n't',\n8489);\n(SELECT song FROM songs WHERE lower(title)=lower('Paranoia KCET (Clean Mix)')),\n't',\n8489);\n@@ -1651,17 +1804,19 @@ INSERT INTO roundparticipation (tournament, round, parallel, player, position) V\n(SELECT player FROM players WHERE nick='DufF'),\n3\n);\n(SELECT player FROM players WHERE nick='DufF'),\n3\n);\n-INSERT INTO scores (tournament, round, parallel, player, song, chosen, score) VALUES (\n+INSERT INTO scores (tournament, round, parallel, player, songnumber, song, chosen, score) VALUES (\n(SELECT tournament FROM tournaments WHERE name='Challenge Cup 3'),\n3, 0,\n(SELECT player FROM players WHERE nick='DufF'),\n(SELECT tournament FROM tournaments WHERE name='Challenge Cup 3'),\n3, 0,\n(SELECT player FROM players WHERE nick='DufF'),\n+ 0,\n(SELECT song FROM songs WHERE lower(title)=lower('Memories')),\n'f',\n7553);\n(SELECT song FROM songs WHERE lower(title)=lower('Memories')),\n'f',\n7553);\n-INSERT INTO scores (tournament, round, parallel, player, song, chosen, score) VALUES (\n+INSERT INTO scores (tournament, round, parallel, player, songnumber, song, chosen, score) VALUES (\n(SELECT tournament FROM tournaments WHERE name='Challenge Cup 3'),\n3, 0,\n(SELECT player FROM players WHERE nick='DufF'),\n(SELECT tournament FROM tournaments WHERE name='Challenge Cup 3'),\n3, 0,\n(SELECT player FROM players WHERE nick='DufF'),\n+ 1,\n(SELECT song FROM songs WHERE lower(title)=lower('Trip Machine Climax')),\n't',\n8621);\n(SELECT song FROM songs WHERE lower(title)=lower('Trip Machine Climax')),\n't',\n8621);\n@@ -1672,17 +1827,19 @@ INSERT INTO roundparticipation (tournament, round, parallel, player, position) V\n(SELECT player FROM players WHERE nick='OJK'),\n4\n);\n(SELECT player FROM players WHERE nick='OJK'),\n4\n);\n-INSERT INTO scores (tournament, round, parallel, player, song, chosen, score) VALUES (\n+INSERT INTO scores (tournament, round, parallel, player, songnumber, song, chosen, score) VALUES (\n(SELECT tournament FROM tournaments WHERE name='Challenge Cup 3'),\n3, 0,\n(SELECT player FROM players WHERE nick='OJK'),\n(SELECT tournament FROM tournaments WHERE name='Challenge Cup 3'),\n3, 0,\n(SELECT player FROM players WHERE nick='OJK'),\n+ 0,\n(SELECT song FROM songs WHERE lower(title)=lower('Memories')),\n'f',\n7015);\n(SELECT song FROM songs WHERE lower(title)=lower('Memories')),\n'f',\n7015);\n-INSERT INTO scores (tournament, round, parallel, player, song, chosen, score) VALUES (\n+INSERT INTO scores (tournament, round, parallel, player, songnumber, song, chosen, score) VALUES (\n(SELECT tournament FROM tournaments WHERE name='Challenge Cup 3'),\n3, 0,\n(SELECT player FROM players WHERE nick='OJK'),\n(SELECT tournament FROM tournaments WHERE name='Challenge Cup 3'),\n3, 0,\n(SELECT player FROM players WHERE nick='OJK'),\n+ 1,\n(SELECT song FROM songs WHERE lower(title)=lower('Candy')),\n't',\n8627);\n(SELECT song FROM songs WHERE lower(title)=lower('Candy')),\n't',\n8627);\n@@ -1693,17 +1850,19 @@ INSERT INTO roundparticipation (tournament, round, parallel, player, position) V\n(SELECT player FROM players WHERE nick='Hugo'),\n5\n);\n(SELECT player FROM players WHERE nick='Hugo'),\n5\n);\n-INSERT INTO scores (tournament, round, parallel, player, song, chosen, score) VALUES (\n+INSERT INTO scores (tournament, round, parallel, player, songnumber, song, chosen, score) VALUES (\n(SELECT tournament FROM tournaments WHERE name='Challenge Cup 3'),\n3, 0,\n(SELECT player FROM players WHERE nick='Hugo'),\n(SELECT tournament FROM tournaments WHERE name='Challenge Cup 3'),\n3, 0,\n(SELECT player FROM players WHERE nick='Hugo'),\n+ 0,\n(SELECT song FROM songs WHERE lower(title)=lower('Memories')),\n'f',\n7258);\n(SELECT song FROM songs WHERE lower(title)=lower('Memories')),\n'f',\n7258);\n-INSERT INTO scores (tournament, round, parallel, player, song, chosen, score) VALUES (\n+INSERT INTO scores (tournament, round, parallel, player, songnumber, song, chosen, score) VALUES (\n(SELECT tournament FROM tournaments WHERE name='Challenge Cup 3'),\n3, 0,\n(SELECT player FROM players WHERE nick='Hugo'),\n(SELECT tournament FROM tournaments WHERE name='Challenge Cup 3'),\n3, 0,\n(SELECT player FROM players WHERE nick='Hugo'),\n+ 1,\n(SELECT song FROM songs WHERE lower(title)=lower('Tsugaru')),\n't',\n8228);\n(SELECT song FROM songs WHERE lower(title)=lower('Tsugaru')),\n't',\n8228);\n@@ -1714,17 +1873,19 @@ INSERT INTO roundparticipation (tournament, round, parallel, player, position) V\n(SELECT player FROM players WHERE nick='RiChY'),\n6\n);\n(SELECT player FROM players WHERE nick='RiChY'),\n6\n);\n-INSERT INTO scores (tournament, round, parallel, player, song, chosen, score) VALUES (\n+INSERT INTO scores (tournament, round, parallel, player, songnumber, song, chosen, score) VALUES (\n(SELECT tournament FROM tournaments WHERE name='Challenge Cup 3'),\n3, 0,\n(SELECT player FROM players WHERE nick='RiChY'),\n(SELECT tournament FROM tournaments WHERE name='Challenge Cup 3'),\n3, 0,\n(SELECT player FROM players WHERE nick='RiChY'),\n+ 0,\n(SELECT song FROM songs WHERE lower(title)=lower('Memories')),\n'f',\n7143);\n(SELECT song FROM songs WHERE lower(title)=lower('Memories')),\n'f',\n7143);\n-INSERT INTO scores (tournament, round, parallel, player, song, chosen, score) VALUES (\n+INSERT INTO scores (tournament, round, parallel, player, songnumber, song, chosen, score) VALUES (\n(SELECT tournament FROM tournaments WHERE name='Challenge Cup 3'),\n3, 0,\n(SELECT player FROM players WHERE nick='RiChY'),\n(SELECT tournament FROM tournaments WHERE name='Challenge Cup 3'),\n3, 0,\n(SELECT player FROM players WHERE nick='RiChY'),\n+ 1,\n(SELECT song FROM songs WHERE lower(title)=lower('Dead End')),\n't',\n8453);\n(SELECT song FROM songs WHERE lower(title)=lower('Dead End')),\n't',\n8453);\n@@ -1735,17 +1896,19 @@ INSERT INTO roundparticipation (tournament, round, parallel, player, position) V\n(SELECT player FROM players WHERE nick='Snute'),\n7\n);\n(SELECT player FROM players WHERE nick='Snute'),\n7\n);\n-INSERT INTO scores (tournament, round, parallel, player, song, chosen, score) VALUES (\n+INSERT INTO scores (tournament, round, parallel, player, songnumber, song, chosen, score) VALUES (\n(SELECT tournament FROM tournaments WHERE name='Challenge Cup 3'),\n3, 0,\n(SELECT player FROM players WHERE nick='Snute'),\n(SELECT tournament FROM tournaments WHERE name='Challenge Cup 3'),\n3, 0,\n(SELECT player FROM players WHERE nick='Snute'),\n+ 0,\n(SELECT song FROM songs WHERE lower(title)=lower('Memories')),\n'f',\n7605);\n(SELECT song FROM songs WHERE lower(title)=lower('Memories')),\n'f',\n7605);\n-INSERT INTO scores (tournament, round, parallel, player, song, chosen, score) VALUES (\n+INSERT INTO scores (tournament, round, parallel, player, songnumber, song, chosen, score) VALUES (\n(SELECT tournament FROM tournaments WHERE name='Challenge Cup 3'),\n3, 0,\n(SELECT player FROM players WHERE nick='Snute'),\n(SELECT tournament FROM tournaments WHERE name='Challenge Cup 3'),\n3, 0,\n(SELECT player FROM players WHERE nick='Snute'),\n+ 1,\n(SELECT song FROM songs WHERE lower(title)=lower('So Deep (Perfect Sphere Mix)')),\n't',\n8374);\n(SELECT song FROM songs WHERE lower(title)=lower('So Deep (Perfect Sphere Mix)')),\n't',\n8374);\n@@ -1756,17 +1919,19 @@ INSERT INTO roundparticipation (tournament, round, parallel, player, position) V\n(SELECT player FROM players WHERE nick='Oliyzia'),\n8\n);\n(SELECT player FROM players WHERE nick='Oliyzia'),\n8\n);\n-INSERT INTO scores (tournament, round, parallel, player, song, chosen, score) VALUES (\n+INSERT INTO scores (tournament, round, parallel, player, songnumber, song, chosen, score) VALUES (\n(SELECT tournament FROM tournaments WHERE name='Challenge Cup 3'),\n3, 0,\n(SELECT player FROM players WHERE nick='Oliyzia'),\n(SELECT tournament FROM tournaments WHERE name='Challenge Cup 3'),\n3, 0,\n(SELECT player FROM players WHERE nick='Oliyzia'),\n+ 0,\n(SELECT song FROM songs WHERE lower(title)=lower('Memories')),\n'f',\n7185);\n(SELECT song FROM songs WHERE lower(title)=lower('Memories')),\n'f',\n7185);\n-INSERT INTO scores (tournament, round, parallel, player, song, chosen, score) VALUES (\n+INSERT INTO scores (tournament, round, parallel, player, songnumber, song, chosen, score) VALUES (\n(SELECT tournament FROM tournaments WHERE name='Challenge Cup 3'),\n3, 0,\n(SELECT player FROM players WHERE nick='Oliyzia'),\n(SELECT tournament FROM tournaments WHERE name='Challenge Cup 3'),\n3, 0,\n(SELECT player FROM players WHERE nick='Oliyzia'),\n+ 1,\n(SELECT song FROM songs WHERE lower(title)=lower('Tsugaru')),\n't',\n8475);\n(SELECT song FROM songs WHERE lower(title)=lower('Tsugaru')),\n't',\n8475);\n@@ -1777,17 +1942,19 @@ INSERT INTO roundparticipation (tournament, round, parallel, player, position) V\n(SELECT player FROM players WHERE nick='Haggar'),\n9\n);\n(SELECT player FROM players WHERE nick='Haggar'),\n9\n);\n-INSERT INTO scores (tournament, round, parallel, player, song, chosen, score) VALUES (\n+INSERT INTO scores (tournament, round, parallel, player, songnumber, song, chosen, score) VALUES (\n(SELECT tournament FROM tournaments WHERE name='Challenge Cup 3'),\n3, 0,\n(SELECT player FROM players WHERE nick='Haggar'),\n(SELECT tournament FROM tournaments WHERE name='Challenge Cup 3'),\n3, 0,\n(SELECT player FROM players WHERE nick='Haggar'),\n+ 0,\n(SELECT song FROM songs WHERE lower(title)=lower('Memories')),\n'f',\n6892);\n(SELECT song FROM songs WHERE lower(title)=lower('Memories')),\n'f',\n6892);\n-INSERT INTO scores (tournament, round, parallel, player, song, chosen, score) VALUES (\n+INSERT INTO scores (tournament, round, parallel, player, songnumber, song, chosen, score) VALUES (\n(SELECT tournament FROM tournaments WHERE name='Challenge Cup 3'),\n3, 0,\n(SELECT player FROM players WHERE nick='Haggar'),\n(SELECT tournament FROM tournaments WHERE name='Challenge Cup 3'),\n3, 0,\n(SELECT player FROM players WHERE nick='Haggar'),\n+ 1,\n(SELECT song FROM songs WHERE lower(title)=lower('Dead End')),\n't',\n8468);\n(SELECT song FROM songs WHERE lower(title)=lower('Dead End')),\n't',\n8468);\n@@ -1817,17 +1984,19 @@ INSERT INTO roundparticipation (tournament, round, parallel, player, position) V\n(SELECT player FROM players WHERE nick='Haggar'),\n1\n);\n(SELECT player FROM players WHERE nick='Haggar'),\n1\n);\n-INSERT INTO scores (tournament, round, parallel, player, song, chosen, score) VALUES (\n+INSERT INTO scores (tournament, round, parallel, player, songnumber, song, chosen, score) VALUES (\n(SELECT tournament FROM tournaments WHERE name='Challenge Cup 3'),\n4, 0,\n(SELECT player FROM players WHERE nick='Haggar'),\n(SELECT tournament FROM tournaments WHERE name='Challenge Cup 3'),\n4, 0,\n(SELECT player FROM players WHERE nick='Haggar'),\n+ 0,\n(SELECT song FROM songs WHERE lower(title)=lower('era (nostalmix)')),\n'f',\n7528);\n(SELECT song FROM songs WHERE lower(title)=lower('era (nostalmix)')),\n'f',\n7528);\n-INSERT INTO scores (tournament, round, parallel, player, song, chosen, score) VALUES (\n+INSERT INTO scores (tournament, round, parallel, player, songnumber, song, chosen, score) VALUES (\n(SELECT tournament FROM tournaments WHERE name='Challenge Cup 3'),\n4, 0,\n(SELECT player FROM players WHERE nick='Haggar'),\n(SELECT tournament FROM tournaments WHERE name='Challenge Cup 3'),\n4, 0,\n(SELECT player FROM players WHERE nick='Haggar'),\n+ 1,\n(SELECT song FROM songs WHERE lower(title)=lower('Trip Machine Climax')),\n't',\n8273);\n(SELECT song FROM songs WHERE lower(title)=lower('Trip Machine Climax')),\n't',\n8273);\n@@ -1838,17 +2007,19 @@ INSERT INTO roundparticipation (tournament, round, parallel, player, position) V\n(SELECT player FROM players WHERE nick='Hugo'),\n2\n);\n(SELECT player FROM players WHERE nick='Hugo'),\n2\n);\n-INSERT INTO scores (tournament, round, parallel, player, song, chosen, score) VALUES (\n+INSERT INTO scores (tournament, round, parallel, player, songnumber, song, chosen, score) VALUES (\n(SELECT tournament FROM tournaments WHERE name='Challenge Cup 3'),\n4, 0,\n(SELECT player FROM players WHERE nick='Hugo'),\n(SELECT tournament FROM tournaments WHERE name='Challenge Cup 3'),\n4, 0,\n(SELECT player FROM players WHERE nick='Hugo'),\n+ 0,\n(SELECT song FROM songs WHERE lower(title)=lower('era (nostalmix)')),\n'f',\n7376);\n(SELECT song FROM songs WHERE lower(title)=lower('era (nostalmix)')),\n'f',\n7376);\n-INSERT INTO scores (tournament, round, parallel, player, song, chosen, score) VALUES (\n+INSERT INTO scores (tournament, round, parallel, player, songnumber, song, chosen, score) VALUES (\n(SELECT tournament FROM tournaments WHERE name='Challenge Cup 3'),\n4, 0,\n(SELECT player FROM players WHERE nick='Hugo'),\n(SELECT tournament FROM tournaments WHERE name='Challenge Cup 3'),\n4, 0,\n(SELECT player FROM players WHERE nick='Hugo'),\n+ 1,\n(SELECT song FROM songs WHERE lower(title)=lower('Paranoia KCET (Clean Mix)')),\n't',\n8693);\n(SELECT song FROM songs WHERE lower(title)=lower('Paranoia KCET (Clean Mix)')),\n't',\n8693);\n@@ -1859,17 +2030,19 @@ INSERT INTO roundparticipation (tournament, round, parallel, player, position) V\n(SELECT player FROM players WHERE nick='RiChY'),\n3\n);\n(SELECT player FROM players WHERE nick='RiChY'),\n3\n);\n-INSERT INTO scores (tournament, round, parallel, player, song, chosen, score) VALUES (\n+INSERT INTO scores (tournament, round, parallel, player, songnumber, song, chosen, score) VALUES (\n(SELECT tournament FROM tournaments WHERE name='Challenge Cup 3'),\n4, 0,\n(SELECT player FROM players WHERE nick='RiChY'),\n(SELECT tournament FROM tournaments WHERE name='Challenge Cup 3'),\n4, 0,\n(SELECT player FROM players WHERE nick='RiChY'),\n+ 0,\n(SELECT song FROM songs WHERE lower(title)=lower('era (nostalmix)')),\n'f',\n7371);\n(SELECT song FROM songs WHERE lower(title)=lower('era (nostalmix)')),\n'f',\n7371);\n-INSERT INTO scores (tournament, round, parallel, player, song, chosen, score) VALUES (\n+INSERT INTO scores (tournament, round, parallel, player, songnumber, song, chosen, score) VALUES (\n(SELECT tournament FROM tournaments WHERE name='Challenge Cup 3'),\n4, 0,\n(SELECT player FROM players WHERE nick='RiChY'),\n(SELECT tournament FROM tournaments WHERE name='Challenge Cup 3'),\n4, 0,\n(SELECT player FROM players WHERE nick='RiChY'),\n+ 1,\n(SELECT song FROM songs WHERE lower(title)=lower('Paranoia KCET (Clean Mix)')),\n't',\n8723);\n(SELECT song FROM songs WHERE lower(title)=lower('Paranoia KCET (Clean Mix)')),\n't',\n8723);\n@@ -1880,17 +2053,19 @@ INSERT INTO roundparticipation (tournament, round, parallel, player, position) V\n(SELECT player FROM players WHERE nick='OJK'),\n4\n);\n(SELECT player FROM players WHERE nick='OJK'),\n4\n);\n-INSERT INTO scores (tournament, round, parallel, player, song, chosen, score) VALUES (\n+INSERT INTO scores (tournament, round, parallel, player, songnumber, song, chosen, score) VALUES (\n(SELECT tournament FROM tournaments WHERE name='Challenge Cup 3'),\n4, 0,\n(SELECT player FROM players WHERE nick='OJK'),\n(SELECT tournament FROM tournaments WHERE name='Challenge Cup 3'),\n4, 0,\n(SELECT player FROM players WHERE nick='OJK'),\n+ 0,\n(SELECT song FROM songs WHERE lower(title)=lower('era (nostalmix)')),\n'f',\n7282);\n(SELECT song FROM songs WHERE lower(title)=lower('era (nostalmix)')),\n'f',\n7282);\n-INSERT INTO scores (tournament, round, parallel, player, song, chosen, score) VALUES (\n+INSERT INTO scores (tournament, round, parallel, player, songnumber, song, chosen, score) VALUES (\n(SELECT tournament FROM tournaments WHERE name='Challenge Cup 3'),\n4, 0,\n(SELECT player FROM players WHERE nick='OJK'),\n(SELECT tournament FROM tournaments WHERE name='Challenge Cup 3'),\n4, 0,\n(SELECT player FROM players WHERE nick='OJK'),\n+ 1,\n(SELECT song FROM songs WHERE lower(title)=lower('Paranoia KCET (Clean Mix)')),\n't',\n8778);\n(SELECT song FROM songs WHERE lower(title)=lower('Paranoia KCET (Clean Mix)')),\n't',\n8778);\n@@ -1901,17 +2076,19 @@ INSERT INTO roundparticipation (tournament, round, parallel, player, position) V\n(SELECT player FROM players WHERE nick='Oliyzia'),\n5\n);\n(SELECT player FROM players WHERE nick='Oliyzia'),\n5\n);\n-INSERT INTO scores (tournament, round, parallel, player, song, chosen, score) VALUES (\n+INSERT INTO scores (tournament, round, parallel, player, songnumber, song, chosen, score) VALUES (\n(SELECT tournament FROM tournaments WHERE name='Challenge Cup 3'),\n4, 0,\n(SELECT player FROM players WHERE nick='Oliyzia'),\n(SELECT tournament FROM tournaments WHERE name='Challenge Cup 3'),\n4, 0,\n(SELECT player FROM players WHERE nick='Oliyzia'),\n+ 0,\n(SELECT song FROM songs WHERE lower(title)=lower('era (nostalmix)')),\n'f',\n7352);\n(SELECT song FROM songs WHERE lower(title)=lower('era (nostalmix)')),\n'f',\n7352);\n-INSERT INTO scores (tournament, round, parallel, player, song, chosen, score) VALUES (\n+INSERT INTO scores (tournament, round, parallel, player, songnumber, song, chosen, score) VALUES (\n(SELECT tournament FROM tournaments WHERE name='Challenge Cup 3'),\n4, 0,\n(SELECT player FROM players WHERE nick='Oliyzia'),\n(SELECT tournament FROM tournaments WHERE name='Challenge Cup 3'),\n4, 0,\n(SELECT player FROM players WHERE nick='Oliyzia'),\n+ 1,\n(SELECT song FROM songs WHERE lower(title)=lower('Paranoia KCET (Clean Mix)')),\n't',\n8808);\n(SELECT song FROM songs WHERE lower(title)=lower('Paranoia KCET (Clean Mix)')),\n't',\n8808);\n@@ -1922,17 +2099,19 @@ INSERT INTO roundparticipation (tournament, round, parallel, player, position) V\n(SELECT player FROM players WHERE nick='Snute'),\n6\n);\n(SELECT player FROM players WHERE nick='Snute'),\n6\n);\n-INSERT INTO scores (tournament, round, parallel, player, song, chosen, score) VALUES (\n+INSERT INTO scores (tournament, round, parallel, player, songnumber, song, chosen, score) VALUES (\n(SELECT tournament FROM tournaments WHERE name='Challenge Cup 3'),\n4, 0,\n(SELECT player FROM players WHERE nick='Snute'),\n(SELECT tournament FROM tournaments WHERE name='Challenge Cup 3'),\n4, 0,\n(SELECT player FROM players WHERE nick='Snute'),\n+ 0,\n(SELECT song FROM songs WHERE lower(title)=lower('era (nostalmix)')),\n'f',\n7606);\n(SELECT song FROM songs WHERE lower(title)=lower('era (nostalmix)')),\n'f',\n7606);\n-INSERT INTO scores (tournament, round, parallel, player, song, chosen, score) VALUES (\n+INSERT INTO scores (tournament, round, parallel, player, songnumber, song, chosen, score) VALUES (\n(SELECT tournament FROM tournaments WHERE name='Challenge Cup 3'),\n4, 0,\n(SELECT player FROM players WHERE nick='Snute'),\n(SELECT tournament FROM tournaments WHERE name='Challenge Cup 3'),\n4, 0,\n(SELECT player FROM players WHERE nick='Snute'),\n+ 1,\n(SELECT song FROM songs WHERE lower(title)=lower('Can\\'t Stop Fallin\\' in Love (SPEED MIX)')),\n't',\n8748);\n(SELECT song FROM songs WHERE lower(title)=lower('Can\\'t Stop Fallin\\' in Love (SPEED MIX)')),\n't',\n8748);\n@@ -1943,17 +2122,19 @@ INSERT INTO roundparticipation (tournament, round, parallel, player, position) V\n(SELECT player FROM players WHERE nick='DufF'),\n7\n);\n(SELECT player FROM players WHERE nick='DufF'),\n7\n);\n-INSERT INTO scores (tournament, round, parallel, player, song, chosen, score) VALUES (\n+INSERT INTO scores (tournament, round, parallel, player, songnumber, song, chosen, score) VALUES (\n(SELECT tournament FROM tournaments WHERE name='Challenge Cup 3'),\n4, 0,\n(SELECT player FROM players WHERE nick='DufF'),\n(SELECT tournament FROM tournaments WHERE name='Challenge Cup 3'),\n4, 0,\n(SELECT player FROM players WHERE nick='DufF'),\n+ 0,\n(SELECT song FROM songs WHERE lower(title)=lower('era (nostalmix)')),\n'f',\n7572);\n(SELECT song FROM songs WHERE lower(title)=lower('era (nostalmix)')),\n'f',\n7572);\n-INSERT INTO scores (tournament, round, parallel, player, song, chosen, score) VALUES (\n+INSERT INTO scores (tournament, round, parallel, player, songnumber, song, chosen, score) VALUES (\n(SELECT tournament FROM tournaments WHERE name='Challenge Cup 3'),\n4, 0,\n(SELECT player FROM players WHERE nick='DufF'),\n(SELECT tournament FROM tournaments WHERE name='Challenge Cup 3'),\n4, 0,\n(SELECT player FROM players WHERE nick='DufF'),\n+ 1,\n(SELECT song FROM songs WHERE lower(title)=lower('So Deep (Perfect Sphere Mix)')),\n't',\n8555);\n(SELECT song FROM songs WHERE lower(title)=lower('So Deep (Perfect Sphere Mix)')),\n't',\n8555);\n@@ -1983,17 +2164,19 @@ INSERT INTO roundparticipation (tournament, round, parallel, player, position) V\n(SELECT player FROM players WHERE nick='Hugo'),\n1\n);\n(SELECT player FROM players WHERE nick='Hugo'),\n1\n);\n-INSERT INTO scores (tournament, round, parallel, player, song, chosen, score) VALUES (\n+INSERT INTO scores (tournament, round, parallel, player, songnumber, song, chosen, score) VALUES (\n(SELECT tournament FROM tournaments WHERE name='Challenge Cup 3'),\n5, 0,\n(SELECT player FROM players WHERE nick='Hugo'),\n(SELECT tournament FROM tournaments WHERE name='Challenge Cup 3'),\n5, 0,\n(SELECT player FROM players WHERE nick='Hugo'),\n+ 0,\n(SELECT song FROM songs WHERE lower(title)=lower('B4U')),\n'f',\n7573);\n(SELECT song FROM songs WHERE lower(title)=lower('B4U')),\n'f',\n7573);\n-INSERT INTO scores (tournament, round, parallel, player, song, chosen, score) VALUES (\n+INSERT INTO scores (tournament, round, parallel, player, songnumber, song, chosen, score) VALUES (\n(SELECT tournament FROM tournaments WHERE name='Challenge Cup 3'),\n5, 0,\n(SELECT player FROM players WHERE nick='Hugo'),\n(SELECT tournament FROM tournaments WHERE name='Challenge Cup 3'),\n5, 0,\n(SELECT player FROM players WHERE nick='Hugo'),\n+ 1,\n(SELECT song FROM songs WHERE lower(title)=lower('Exotic Ethnic')),\n't',\n8539);\n(SELECT song FROM songs WHERE lower(title)=lower('Exotic Ethnic')),\n't',\n8539);\n@@ -2004,17 +2187,19 @@ INSERT INTO roundparticipation (tournament, round, parallel, player, position) V\n(SELECT player FROM players WHERE nick='RiChY'),\n2\n);\n(SELECT player FROM players WHERE nick='RiChY'),\n2\n);\n-INSERT INTO scores (tournament, round, parallel, player, song, chosen, score) VALUES (\n+INSERT INTO scores (tournament, round, parallel, player, songnumber, song, chosen, score) VALUES (\n(SELECT tournament FROM tournaments WHERE name='Challenge Cup 3'),\n5, 0,\n(SELECT player FROM players WHERE nick='RiChY'),\n(SELECT tournament FROM tournaments WHERE name='Challenge Cup 3'),\n5, 0,\n(SELECT player FROM players WHERE nick='RiChY'),\n+ 0,\n(SELECT song FROM songs WHERE lower(title)=lower('B4U')),\n'f',\n7586);\n(SELECT song FROM songs WHERE lower(title)=lower('B4U')),\n'f',\n7586);\n-INSERT INTO scores (tournament, round, parallel, player, song, chosen, score) VALUES (\n+INSERT INTO scores (tournament, round, parallel, player, songnumber, song, chosen, score) VALUES (\n(SELECT tournament FROM tournaments WHERE name='Challenge Cup 3'),\n5, 0,\n(SELECT player FROM players WHERE nick='RiChY'),\n(SELECT tournament FROM tournaments WHERE name='Challenge Cup 3'),\n5, 0,\n(SELECT player FROM players WHERE nick='RiChY'),\n+ 1,\n(SELECT song FROM songs WHERE lower(title)=lower('Max 300')),\n't',\n8349);\n(SELECT song FROM songs WHERE lower(title)=lower('Max 300')),\n't',\n8349);\n@@ -2025,17 +2210,19 @@ INSERT INTO roundparticipation (tournament, round, parallel, player, position) V\n(SELECT player FROM players WHERE nick='DufF'),\n3\n);\n(SELECT player FROM players WHERE nick='DufF'),\n3\n);\n-INSERT INTO scores (tournament, round, parallel, player, song, chosen, score) VALUES (\n+INSERT INTO scores (tournament, round, parallel, player, songnumber, song, chosen, score) VALUES (\n(SELECT tournament FROM tournaments WHERE name='Challenge Cup 3'),\n5, 0,\n(SELECT player FROM players WHERE nick='DufF'),\n(SELECT tournament FROM tournaments WHERE name='Challenge Cup 3'),\n5, 0,\n(SELECT player FROM players WHERE nick='DufF'),\n+ 0,\n(SELECT song FROM songs WHERE lower(title)=lower('B4U')),\n'f',\n7742);\n(SELECT song FROM songs WHERE lower(title)=lower('B4U')),\n'f',\n7742);\n-INSERT INTO scores (tournament, round, parallel, player, song, chosen, score) VALUES (\n+INSERT INTO scores (tournament, round, parallel, player, songnumber, song, chosen, score) VALUES (\n(SELECT tournament FROM tournaments WHERE name='Challenge Cup 3'),\n5, 0,\n(SELECT player FROM players WHERE nick='DufF'),\n(SELECT tournament FROM tournaments WHERE name='Challenge Cup 3'),\n5, 0,\n(SELECT player FROM players WHERE nick='DufF'),\n+ 1,\n(SELECT song FROM songs WHERE lower(title)=lower('Electro Tuned (the SubS Mix)')),\n't',\n8914);\n(SELECT song FROM songs WHERE lower(title)=lower('Electro Tuned (the SubS Mix)')),\n't',\n8914);\n@@ -2046,17 +2233,19 @@ INSERT INTO roundparticipation (tournament, round, parallel, player, position) V\n(SELECT player FROM players WHERE nick='Oliyzia'),\n4\n);\n(SELECT player FROM players WHERE nick='Oliyzia'),\n4\n);\n-INSERT INTO scores (tournament, round, parallel, player, song, chosen, score) VALUES (\n+INSERT INTO scores (tournament, round, parallel, player, songnumber, song, chosen, score) VALUES (\n(SELECT tournament FROM tournaments WHERE name='Challenge Cup 3'),\n5, 0,\n(SELECT player FROM players WHERE nick='Oliyzia'),\n(SELECT tournament FROM tournaments WHERE name='Challenge Cup 3'),\n5, 0,\n(SELECT player FROM players WHERE nick='Oliyzia'),\n+ 0,\n(SELECT song FROM songs WHERE lower(title)=lower('B4U')),\n'f',\n7692);\n(SELECT song FROM songs WHERE lower(title)=lower('B4U')),\n'f',\n7692);\n-INSERT INTO scores (tournament, round, parallel, player, song, chosen, score) VALUES (\n+INSERT INTO scores (tournament, round, parallel, player, songnumber, song, chosen, score) VALUES (\n(SELECT tournament FROM tournaments WHERE name='Challenge Cup 3'),\n5, 0,\n(SELECT player FROM players WHERE nick='Oliyzia'),\n(SELECT tournament FROM tournaments WHERE name='Challenge Cup 3'),\n5, 0,\n(SELECT player FROM players WHERE nick='Oliyzia'),\n+ 1,\n(SELECT song FROM songs WHERE lower(title)=lower('PARANOiA Rebirth')),\n't',\n8612);\n(SELECT song FROM songs WHERE lower(title)=lower('PARANOiA Rebirth')),\n't',\n8612);\n@@ -2067,17 +2256,19 @@ INSERT INTO roundparticipation (tournament, round, parallel, player, position) V\n(SELECT player FROM players WHERE nick='Snute'),\n5\n);\n(SELECT player FROM players WHERE nick='Snute'),\n5\n);\n-INSERT INTO scores (tournament, round, parallel, player, song, chosen, score) VALUES (\n+INSERT INTO scores (tournament, round, parallel, player, songnumber, song, chosen, score) VALUES (\n(SELECT tournament FROM tournaments WHERE name='Challenge Cup 3'),\n5, 0,\n(SELECT player FROM players WHERE nick='Snute'),\n(SELECT tournament FROM tournaments WHERE name='Challenge Cup 3'),\n5, 0,\n(SELECT player FROM players WHERE nick='Snute'),\n+ 0,\n(SELECT song FROM songs WHERE lower(title)=lower('B4U')),\n'f',\n7830);\n(SELECT song FROM songs WHERE lower(title)=lower('B4U')),\n'f',\n7830);\n-INSERT INTO scores (tournament, round, parallel, player, song, chosen, score) VALUES (\n+INSERT INTO scores (tournament, round, parallel, player, songnumber, song, chosen, score) VALUES (\n(SELECT tournament FROM tournaments WHERE name='Challenge Cup 3'),\n5, 0,\n(SELECT player FROM players WHERE nick='Snute'),\n(SELECT tournament FROM tournaments WHERE name='Challenge Cup 3'),\n5, 0,\n(SELECT player FROM players WHERE nick='Snute'),\n+ 1,\n(SELECT song FROM songs WHERE lower(title)=lower('PARANOiA Rebirth')),\n't',\n8782);\n(SELECT song FROM songs WHERE lower(title)=lower('PARANOiA Rebirth')),\n't',\n8782);\n@@ -2107,17 +2298,19 @@ INSERT INTO roundparticipation (tournament, round, parallel, player, position) V\n(SELECT player FROM players WHERE nick='Hugo'),\n1\n);\n(SELECT player FROM players WHERE nick='Hugo'),\n1\n);\n-INSERT INTO scores (tournament, round, parallel, player, song, chosen, score) VALUES (\n+INSERT INTO scores (tournament, round, parallel, player, songnumber, song, chosen, score) VALUES (\n(SELECT tournament FROM tournaments WHERE name='Challenge Cup 3'),\n6, 0,\n(SELECT player FROM players WHERE nick='Hugo'),\n(SELECT tournament FROM tournaments WHERE name='Challenge Cup 3'),\n6, 0,\n(SELECT player FROM players WHERE nick='Hugo'),\n+ 0,\n(SELECT song FROM songs WHERE lower(title)=lower('B4U (B4 Za Beat Mix)')),\n'f',\n6376);\n(SELECT song FROM songs WHERE lower(title)=lower('B4U (B4 Za Beat Mix)')),\n'f',\n6376);\n-INSERT INTO scores (tournament, round, parallel, player, song, chosen, score) VALUES (\n+INSERT INTO scores (tournament, round, parallel, player, songnumber, song, chosen, score) VALUES (\n(SELECT tournament FROM tournaments WHERE name='Challenge Cup 3'),\n6, 0,\n(SELECT player FROM players WHERE nick='Hugo'),\n(SELECT tournament FROM tournaments WHERE name='Challenge Cup 3'),\n6, 0,\n(SELECT player FROM players WHERE nick='Hugo'),\n+ 1,\n(SELECT song FROM songs WHERE lower(title)=lower('PARANOiA Rebirth')),\n't',\n8635);\n(SELECT song FROM songs WHERE lower(title)=lower('PARANOiA Rebirth')),\n't',\n8635);\n@@ -2128,17 +2321,19 @@ INSERT INTO roundparticipation (tournament, round, parallel, player, position) V\n(SELECT player FROM players WHERE nick='Oliyzia'),\n2\n);\n(SELECT player FROM players WHERE nick='Oliyzia'),\n2\n);\n-INSERT INTO scores (tournament, round, parallel, player, song, chosen, score) VALUES (\n+INSERT INTO scores (tournament, round, parallel, player, songnumber, song, chosen, score) VALUES (\n(SELECT tournament FROM tournaments WHERE name='Challenge Cup 3'),\n6, 0,\n(SELECT player FROM players WHERE nick='Oliyzia'),\n(SELECT tournament FROM tournaments WHERE name='Challenge Cup 3'),\n6, 0,\n(SELECT player FROM players WHERE nick='Oliyzia'),\n+ 0,\n(SELECT song FROM songs WHERE lower(title)=lower('B4U (B4 Za Beat Mix)')),\n'f',\n6094);\n(SELECT song FROM songs WHERE lower(title)=lower('B4U (B4 Za Beat Mix)')),\n'f',\n6094);\n-INSERT INTO scores (tournament, round, parallel, player, song, chosen, score) VALUES (\n+INSERT INTO scores (tournament, round, parallel, player, songnumber, song, chosen, score) VALUES (\n(SELECT tournament FROM tournaments WHERE name='Challenge Cup 3'),\n6, 0,\n(SELECT player FROM players WHERE nick='Oliyzia'),\n(SELECT tournament FROM tournaments WHERE name='Challenge Cup 3'),\n6, 0,\n(SELECT player FROM players WHERE nick='Oliyzia'),\n+ 1,\n(SELECT song FROM songs WHERE lower(title)=lower('Crash!')),\n't',\n8264);\n(SELECT song FROM songs WHERE lower(title)=lower('Crash!')),\n't',\n8264);\n@@ -2149,17 +2344,19 @@ INSERT INTO roundparticipation (tournament, round, parallel, player, position) V\n(SELECT player FROM players WHERE nick='Snute'),\n3\n);\n(SELECT player FROM players WHERE nick='Snute'),\n3\n);\n-INSERT INTO scores (tournament, round, parallel, player, song, chosen, score) VALUES (\n+INSERT INTO scores (tournament, round, parallel, player, songnumber, song, chosen, score) VALUES (\n(SELECT tournament FROM tournaments WHERE name='Challenge Cup 3'),\n6, 0,\n(SELECT player FROM players WHERE nick='Snute'),\n(SELECT tournament FROM tournaments WHERE name='Challenge Cup 3'),\n6, 0,\n(SELECT player FROM players WHERE nick='Snute'),\n+ 0,\n(SELECT song FROM songs WHERE lower(title)=lower('B4U (B4 Za Beat Mix)')),\n'f',\n6623);\n(SELECT song FROM songs WHERE lower(title)=lower('B4U (B4 Za Beat Mix)')),\n'f',\n6623);\n-INSERT INTO scores (tournament, round, parallel, player, song, chosen, score) VALUES (\n+INSERT INTO scores (tournament, round, parallel, player, songnumber, song, chosen, score) VALUES (\n(SELECT tournament FROM tournaments WHERE name='Challenge Cup 3'),\n6, 0,\n(SELECT player FROM players WHERE nick='Snute'),\n(SELECT tournament FROM tournaments WHERE name='Challenge Cup 3'),\n6, 0,\n(SELECT player FROM players WHERE nick='Snute'),\n+ 1,\n(SELECT song FROM songs WHERE lower(title)=lower('Crash!')),\n't',\n8828);\n(SELECT song FROM songs WHERE lower(title)=lower('Crash!')),\n't',\n8828);\n@@ -2170,17 +2367,19 @@ INSERT INTO roundparticipation (tournament, round, parallel, player, position) V\n(SELECT player FROM players WHERE nick='DufF'),\n4\n);\n(SELECT player FROM players WHERE nick='DufF'),\n4\n);\n-INSERT INTO scores (tournament, round, parallel, player, song, chosen, score) VALUES (\n+INSERT INTO scores (tournament, round, parallel, player, songnumber, song, chosen, score) VALUES (\n(SELECT tournament FROM tournaments WHERE name='Challenge Cup 3'),\n6, 0,\n(SELECT player FROM players WHERE nick='DufF'),\n(SELECT tournament FROM tournaments WHERE name='Challenge Cup 3'),\n6, 0,\n(SELECT player FROM players WHERE nick='DufF'),\n+ 0,\n(SELECT song FROM songs WHERE lower(title)=lower('B4U (B4 Za Beat Mix)')),\n'f',\n6455);\n(SELECT song FROM songs WHERE lower(title)=lower('B4U (B4 Za Beat Mix)')),\n'f',\n6455);\n-INSERT INTO scores (tournament, round, parallel, player, song, chosen, score) VALUES (\n+INSERT INTO scores (tournament, round, parallel, player, songnumber, song, chosen, score) VALUES (\n(SELECT tournament FROM tournaments WHERE name='Challenge Cup 3'),\n6, 0,\n(SELECT player FROM players WHERE nick='DufF'),\n(SELECT tournament FROM tournaments WHERE name='Challenge Cup 3'),\n6, 0,\n(SELECT player FROM players WHERE nick='DufF'),\n+ 1,\n(SELECT song FROM songs WHERE lower(title)=lower('PARANOiA Rebirth')),\n't',\n8728);\n(SELECT song FROM songs WHERE lower(title)=lower('PARANOiA Rebirth')),\n't',\n8728);\n@@ -2210,17 +2409,19 @@ INSERT INTO roundparticipation (tournament, round, parallel, player, position) V\n(SELECT player FROM players WHERE nick='Hugo'),\n1\n);\n(SELECT player FROM players WHERE nick='Hugo'),\n1\n);\n-INSERT INTO scores (tournament, round, parallel, player, song, chosen, score) VALUES (\n+INSERT INTO scores (tournament, round, parallel, player, songnumber, song, chosen, score) VALUES (\n(SELECT tournament FROM tournaments WHERE name='Challenge Cup 3'),\n7, 0,\n(SELECT player FROM players WHERE nick='Hugo'),\n(SELECT tournament FROM tournaments WHERE name='Challenge Cup 3'),\n7, 0,\n(SELECT player FROM players WHERE nick='Hugo'),\n+ 0,\n(SELECT song FROM songs WHERE lower(title)=lower('Tsugaru')),\n'f',\n8273);\n(SELECT song FROM songs WHERE lower(title)=lower('Tsugaru')),\n'f',\n8273);\n-INSERT INTO scores (tournament, round, parallel, player, song, chosen, score) VALUES (\n+INSERT INTO scores (tournament, round, parallel, player, songnumber, song, chosen, score) VALUES (\n(SELECT tournament FROM tournaments WHERE name='Challenge Cup 3'),\n7, 0,\n(SELECT player FROM players WHERE nick='Hugo'),\n(SELECT tournament FROM tournaments WHERE name='Challenge Cup 3'),\n7, 0,\n(SELECT player FROM players WHERE nick='Hugo'),\n+ 1,\n(SELECT song FROM songs WHERE lower(title)=lower('Can\\'t Stop Fallin\\' in Love (SPEED MIX)')),\n't',\n8169);\n(SELECT song FROM songs WHERE lower(title)=lower('Can\\'t Stop Fallin\\' in Love (SPEED MIX)')),\n't',\n8169);\n@@ -2231,17 +2432,19 @@ INSERT INTO roundparticipation (tournament, round, parallel, player, position) V\n(SELECT player FROM players WHERE nick='DufF'),\n2\n);\n(SELECT player FROM players WHERE nick='DufF'),\n2\n);\n-INSERT INTO scores (tournament, round, parallel, player, song, chosen, score) VALUES (\n+INSERT INTO scores (tournament, round, parallel, player, songnumber, song, chosen, score) VALUES (\n(SELECT tournament FROM tournaments WHERE name='Challenge Cup 3'),\n7, 0,\n(SELECT player FROM players WHERE nick='DufF'),\n(SELECT tournament FROM tournaments WHERE name='Challenge Cup 3'),\n7, 0,\n(SELECT player FROM players WHERE nick='DufF'),\n+ 0,\n(SELECT song FROM songs WHERE lower(title)=lower('Tsugaru')),\n'f',\n8420);\n(SELECT song FROM songs WHERE lower(title)=lower('Tsugaru')),\n'f',\n8420);\n-INSERT INTO scores (tournament, round, parallel, player, song, chosen, score) VALUES (\n+INSERT INTO scores (tournament, round, parallel, player, songnumber, song, chosen, score) VALUES (\n(SELECT tournament FROM tournaments WHERE name='Challenge Cup 3'),\n7, 0,\n(SELECT player FROM players WHERE nick='DufF'),\n(SELECT tournament FROM tournaments WHERE name='Challenge Cup 3'),\n7, 0,\n(SELECT player FROM players WHERE nick='DufF'),\n+ 1,\n(SELECT song FROM songs WHERE lower(title)=lower('Healing Vision (Angelic Mix)')),\n't',\n8678);\n(SELECT song FROM songs WHERE lower(title)=lower('Healing Vision (Angelic Mix)')),\n't',\n8678);\n@@ -2252,17 +2455,19 @@ INSERT INTO roundparticipation (tournament, round, parallel, player, position) V\n(SELECT player FROM players WHERE nick='Snute'),\n3\n);\n(SELECT player FROM players WHERE nick='Snute'),\n3\n);\n-INSERT INTO scores (tournament, round, parallel, player, song, chosen, score) VALUES (\n+INSERT INTO scores (tournament, round, parallel, player, songnumber, song, chosen, score) VALUES (\n(SELECT tournament FROM tournaments WHERE name='Challenge Cup 3'),\n7, 0,\n(SELECT player FROM players WHERE nick='Snute'),\n(SELECT tournament FROM tournaments WHERE name='Challenge Cup 3'),\n7, 0,\n(SELECT player FROM players WHERE nick='Snute'),\n+ 0,\n(SELECT song FROM songs WHERE lower(title)=lower('Tsugaru')),\n'f',\n8604);\n(SELECT song FROM songs WHERE lower(title)=lower('Tsugaru')),\n'f',\n8604);\n-INSERT INTO scores (tournament, round, parallel, player, song, chosen, score) VALUES (\n+INSERT INTO scores (tournament, round, parallel, player, songnumber, song, chosen, score) VALUES (\n(SELECT tournament FROM tournaments WHERE name='Challenge Cup 3'),\n7, 0,\n(SELECT player FROM players WHERE nick='Snute'),\n(SELECT tournament FROM tournaments WHERE name='Challenge Cup 3'),\n7, 0,\n(SELECT player FROM players WHERE nick='Snute'),\n+ 1,\n(SELECT song FROM songs WHERE lower(title)=lower('Electro Tuned (the SubS Mix)')),\n't',\n8794);\n(SELECT song FROM songs WHERE lower(title)=lower('Electro Tuned (the SubS Mix)')),\n't',\n8794);\n@@ -2294,17 +2499,19 @@ INSERT INTO roundparticipation (tournament, round, parallel, player, position) V\n(SELECT player FROM players WHERE nick='DufF'),\n1\n);\n(SELECT player FROM players WHERE nick='DufF'),\n1\n);\n-INSERT INTO scores (tournament, round, parallel, player, song, chosen, score) VALUES (\n+INSERT INTO scores (tournament, round, parallel, player, songnumber, song, chosen, score) VALUES (\n(SELECT tournament FROM tournaments WHERE name='Challenge Cup 3'),\n8, 0,\n(SELECT player FROM players WHERE nick='DufF'),\n(SELECT tournament FROM tournaments WHERE name='Challenge Cup 3'),\n8, 0,\n(SELECT player FROM players WHERE nick='DufF'),\n+ 0,\n(SELECT song FROM songs WHERE lower(title)=lower('I Want You To Want Me')),\n'f',\n5640);\n(SELECT song FROM songs WHERE lower(title)=lower('I Want You To Want Me')),\n'f',\n5640);\n-INSERT INTO scores (tournament, round, parallel, player, song, chosen, score) VALUES (\n+INSERT INTO scores (tournament, round, parallel, player, songnumber, song, chosen, score) VALUES (\n(SELECT tournament FROM tournaments WHERE name='Challenge Cup 3'),\n8, 0,\n(SELECT player FROM players WHERE nick='DufF'),\n(SELECT tournament FROM tournaments WHERE name='Challenge Cup 3'),\n8, 0,\n(SELECT player FROM players WHERE nick='DufF'),\n+ 1,\n(SELECT song FROM songs WHERE lower(title)=lower('Max 300')),\n'f',\n8906);\n(SELECT song FROM songs WHERE lower(title)=lower('Max 300')),\n'f',\n8906);\n@@ -2315,17 +2522,19 @@ INSERT INTO roundparticipation (tournament, round, parallel, player, position) V\n(SELECT player FROM players WHERE nick='Snute'),\n2\n);\n(SELECT player FROM players WHERE nick='Snute'),\n2\n);\n-INSERT INTO scores (tournament, round, parallel, player, song, chosen, score) VALUES (\n+INSERT INTO scores (tournament, round, parallel, player, songnumber, song, chosen, score) VALUES (\n(SELECT tournament FROM tournaments WHERE name='Challenge Cup 3'),\n8, 0,\n(SELECT player FROM players WHERE nick='Snute'),\n(SELECT tournament FROM tournaments WHERE name='Challenge Cup 3'),\n8, 0,\n(SELECT player FROM players WHERE nick='Snute'),\n+ 0,\n(SELECT song FROM songs WHERE lower(title)=lower('I Want You To Want Me')),\n'f',\n5750);\n(SELECT song FROM songs WHERE lower(title)=lower('I Want You To Want Me')),\n'f',\n5750);\n-INSERT INTO scores (tournament, round, parallel, player, song, chosen, score) VALUES (\n+INSERT INTO scores (tournament, round, parallel, player, songnumber, song, chosen, score) VALUES (\n(SELECT tournament FROM tournaments WHERE name='Challenge Cup 3'),\n8, 0,\n(SELECT player FROM players WHERE nick='Snute'),\n(SELECT tournament FROM tournaments WHERE name='Challenge Cup 3'),\n8, 0,\n(SELECT player FROM players WHERE nick='Snute'),\n+ 1,\n(SELECT song FROM songs WHERE lower(title)=lower('Max 300')),\n'f',\n9496);\n(SELECT song FROM songs WHERE lower(title)=lower('Max 300')),\n'f',\n9496);"
] | [
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.67259586,"math_prob":0.93808824,"size":144675,"snap":"2021-43-2021-49","text_gpt3_token_len":36667,"char_repetition_ratio":0.36990827,"word_repetition_ratio":0.88093644,"special_character_ratio":0.30257472,"punctuation_ratio":0.2425948,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.95557624,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-10-26T03:07:53Z\",\"WARC-Record-ID\":\"<urn:uuid:737c45fc-a16d-4bd4-90ef-11e5a44cf811>\",\"Content-Length\":\"525308\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:48f47a02-2c5d-4147-a32f-5aaa3a057a55>\",\"WARC-Concurrent-To\":\"<urn:uuid:fd9be4f7-5ffc-4481-88bf-eed89a8cdc4c>\",\"WARC-IP-Address\":\"193.35.52.50\",\"WARC-Target-URI\":\"https://git.sesse.net/?p=ccbs;a=blobdiff;f=sql/cc04/cc3-04.sql;h=7f01806926ba11d0f3f79f4130f0cb4e099c5287;hp=90eaad17eb70ad897b26176628a208f1b4c03d6c;hb=1f60fb5390460b3db78b28cadc8790cce5da267c;hpb=1c54093a61bb4f6633bc7649e72a33a2ccb3bb40;ds=sidebyside\",\"WARC-Payload-Digest\":\"sha1:323ITMTDDNKHYA4CF6M66G6L76F4BPHF\",\"WARC-Block-Digest\":\"sha1:IH5SWLBGPV2YBZ7PZRGZQ5Q4LXZMJNIL\",\"WARC-Identified-Payload-Type\":\"application/xhtml+xml\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-43/CC-MAIN-2021-43_segments_1634323587794.19_warc_CC-MAIN-20211026011138-20211026041138-00372.warc.gz\"}"} |
http://www.kylesconverter.com/acceleration/milligals-to-feet-per-hour-per-second | [
"Convert Milligals to Feet Per Hour Per Second\n\nKyle's Converter > Acceleration > Milligals > Milligals to Feet Per Hour Per Second\n\n Milligals (mGal) Feet Per Hour Per Second (fph/s) Precision: 0 1 2 3 4 5 6 7 8 9 12 15 18\nReverse conversion?\nFeet Per Hour Per Second to Milligals\n(or just enter a value in the \"to\" field)\n\nPlease share if you found this tool useful:\n\nUnit Descriptions\n1 Milligal:\n1 Milligal is equal to one thousandth of a gal. In SI units one milligal equals 0.000 01 meters per second squared. 1 mGal = 0.00001 m/s2\n1 Foot per Hour per Second:\n1 ft/(h*s)\n\nConversions Table\n1 Milligals to Feet Per Hour Per Second = 0.118170 Milligals to Feet Per Hour Per Second = 8.2677\n2 Milligals to Feet Per Hour Per Second = 0.236280 Milligals to Feet Per Hour Per Second = 9.4488\n3 Milligals to Feet Per Hour Per Second = 0.354390 Milligals to Feet Per Hour Per Second = 10.6299\n4 Milligals to Feet Per Hour Per Second = 0.4724100 Milligals to Feet Per Hour Per Second = 11.811\n5 Milligals to Feet Per Hour Per Second = 0.5906200 Milligals to Feet Per Hour Per Second = 23.622\n6 Milligals to Feet Per Hour Per Second = 0.7087300 Milligals to Feet Per Hour Per Second = 35.4331\n7 Milligals to Feet Per Hour Per Second = 0.8268400 Milligals to Feet Per Hour Per Second = 47.2441\n8 Milligals to Feet Per Hour Per Second = 0.9449500 Milligals to Feet Per Hour Per Second = 59.0551\n9 Milligals to Feet Per Hour Per Second = 1.063600 Milligals to Feet Per Hour Per Second = 70.8661\n10 Milligals to Feet Per Hour Per Second = 1.1811800 Milligals to Feet Per Hour Per Second = 94.4882\n20 Milligals to Feet Per Hour Per Second = 2.3622900 Milligals to Feet Per Hour Per Second = 106.2992\n30 Milligals to Feet Per Hour Per Second = 3.54331,000 Milligals to Feet Per Hour Per Second = 118.1102\n40 Milligals to Feet Per Hour Per Second = 4.724410,000 Milligals to Feet Per Hour Per Second = 1181.1024\n50 Milligals to Feet Per Hour Per Second = 5.9055100,000 Milligals to Feet Per Hour Per Second = 11811.0236\n60 Milligals to Feet Per Hour Per Second = 7.08661,000,000 Milligals to Feet Per Hour Per Second = 118110.2362"
] | [
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.5683871,"math_prob":0.9318094,"size":1587,"snap":"2019-26-2019-30","text_gpt3_token_len":524,"char_repetition_ratio":0.3979785,"word_repetition_ratio":0.40268457,"special_character_ratio":0.40390673,"punctuation_ratio":0.10233918,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9731148,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-06-17T07:16:14Z\",\"WARC-Record-ID\":\"<urn:uuid:6b3f8bf7-a966-4e84-ad48-c3d578058304>\",\"Content-Length\":\"19213\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:3aadee55-25e9-4759-a39d-e5cfa63b066d>\",\"WARC-Concurrent-To\":\"<urn:uuid:694cfbd6-489f-4289-a7f8-d4c88f0e95b6>\",\"WARC-IP-Address\":\"13.249.39.126\",\"WARC-Target-URI\":\"http://www.kylesconverter.com/acceleration/milligals-to-feet-per-hour-per-second\",\"WARC-Payload-Digest\":\"sha1:6GMD2BACJML4EZX5O52XPPH34W2TKWEI\",\"WARC-Block-Digest\":\"sha1:7RSS6LWUO2FRKBJ22QPE7WPW6GMN474F\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-26/CC-MAIN-2019-26_segments_1560627998440.47_warc_CC-MAIN-20190617063049-20190617085049-00530.warc.gz\"}"} |
https://fcorel.ru/music/comultiplication-modules.php | [
"# COMULTIPLICATION MODULES PDF\n\nTogar By using the comment function on degruyter. R N and hence 0: Abstract Let G be a group with identity e. A similar argument yields a similar contradiction and thus completes comltiplication proof. Let J be a proper graded ideal of R.",
null,
"Author: Fenrikasa Shabei Country: Portugal Language: English (Spanish) Genre: Medical Published (Last): 5 June 2008 Pages: 161 PDF File Size: 5.90 Mb ePub File Size: 12.40 Mb ISBN: 129-9-22983-791-1 Downloads: 50611 Price: Free* [*Free Regsitration Required] Uploader: Zulkihn",
null,
"Goltilar Volume 14 Issue 1 Janpp. Suppose first comultiplicatjon N is a gr -large submodule of M. By [ 8Theorem 3. Volume 5 Issue 4 Decpp. Let N be a gr -finitely generated gr -multiplication submodule of M. A graded submodule N of a graded R -module M is said to be graded minimal gr — minimal if it is minimal in the lattice of graded submodules of M. Let R be a G -graded ring and M a graded R -module.\n\nThus by [ 8 comultiiplication, Lemma 3. As a dual concept of gr -multiplication modules, graded comultiplication modules gr -comultiplication modules were introduced and studied by Ansari-Toroghy and Farshadifar .\n\nWe refer to and for these basic properties and more information on graded rings and modules. Let R be a G — graded ring and M a gr — faithful gr — comultiplication module with the property 0: Therefore we would like to draw your attention to our House Rules. Volume 8 Issue 6 Decpp. By using the comment function on degruyter. Then M is gr — uniform if and only if R is gr — hollow.\n\nA respectful treatment of one another is important to us. Suppose first that M is gr -comultiplication Comultipplication -module and N a graded submodule of M.\n\nRecall that a G -graded ring R is said to be a gr -comultiplication ring if it is a gr -comultiplication R -module see . De Gruyter Online Google Scholar. By[ 8Lemma 3. Volume 12 Issue 12 Decpp. R N and hence 0: Some properties of graded comultiplication modules. There was a problem providing the content you requested Let J be a proper graded ideal of R.\n\nGraded multiplication modules comultipoication -multiplication modules over commutative graded ring have been studied by many authors extensively see [ 1 — 7 ].\n\nIf M is a gr — faithful R — module, then for each proper graded ideal J of R0: Here we comultilpication study the class comultiplkcation graded comultiplication modules and obtain some further results which are dual to classical results on graded multiplication modules see Section 2. An ideal of a G -graded ring need not be G -graded. Proof Let N be a gr -finitely generated gr -multiplication submodule of M.\n\nLet R be G — graded ring and M a gr — comultiplication R — module. A similar argument yields a similar contradiction and thus completes the proof. Volume 9 Issue 6 Decpp. Therefore R is gr -hollow. Related Posts.\n\nDIN EN 12831 BEIBLATT 2 PDF\n\n## COMULTIPLICATION MODULES PDF",
null,
"Zolojinn Graded comultiplication module ; Graded multiplication module ; Graded submodule. Then M is a gr — comultiplication module if and only if M is gr — strongly self-cogenerated. Let R be a G -graded commutative ring and M a graded R -module. Graded multiplication modulles gr -multiplication modules over commutative graded ring have been studied by many authors extensively see [ 1 — 7 ].\n\nCALCULO DE VARIAS VARIABLES JAMES STEWART SOLUCIONARIO PDF\n\n## On S-Comultiplication Modules",
null,
"."
] | [
null,
"https://fcorel.ru/img/comultiplication-modules.jpg",
null,
"https://fcorel.ru/download_pdf.png",
null,
"https://fcorel.ru/img/a3e7683f57bbffc4801bc6865aabccdf.png",
null,
"https://fcorel.ru/img/922592.jpg",
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.82662266,"math_prob":0.55606776,"size":3404,"snap":"2020-45-2020-50","text_gpt3_token_len":890,"char_repetition_ratio":0.2167647,"word_repetition_ratio":0.17324185,"special_character_ratio":0.23883666,"punctuation_ratio":0.0992126,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9746383,"pos_list":[0,1,2,3,4,5,6,7,8],"im_url_duplicate_count":[null,1,null,null,null,1,null,1,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-12-05T05:17:11Z\",\"WARC-Record-ID\":\"<urn:uuid:9af5e698-f102-4cb0-880c-f61ab820a04e>\",\"Content-Length\":\"24311\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:717e639b-36a5-43a3-af00-a151fda099eb>\",\"WARC-Concurrent-To\":\"<urn:uuid:d3836337-f05b-48e4-9d24-6c08217331f5>\",\"WARC-IP-Address\":\"104.28.19.171\",\"WARC-Target-URI\":\"https://fcorel.ru/music/comultiplication-modules.php\",\"WARC-Payload-Digest\":\"sha1:P534PSSBARBV66T6LM3ILXXWRWGQFZJX\",\"WARC-Block-Digest\":\"sha1:VVHRQP263XIG2PYN6YIQP72FYYYGDP6F\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-50/CC-MAIN-2020-50_segments_1606141746320.91_warc_CC-MAIN-20201205044004-20201205074004-00405.warc.gz\"}"} |
https://indhumathychelliah.com/2020/10/12/new-string-methods-to-remove-prefixes-and-suffixes-in-python-3-9/ | [
"# New String Methods to Remove Prefixes and Suffixes in Python 3.9\n\n### New string methods in Python 3.9\n\nIn Python 3.9, new string methods `removeprefix()` and `removesuffix()` are introduced.\n\n### removeprefix() vs lstrip()\n\n#### removeprefix()\n\n`str.removeprefix(prefix)`\n\n• If the string starts with the prefix string, return ```string[len(prefix):] ``` Otherwise, return a copy of the original string.\n• The parameters of `removeprefix()` are considered as a substring and not as a set of characters.\n\n#### lstrip()\n\n`str.lstrip([chars])`\n\n• Return a copy of the string with leading characters removed. The chars argument is a string specifying the set of characters to be removed.`chars `argument is not a prefix, rather all combinations of its value are stripped-Python docs\n\nExample 1: Remove prefix “abc” from the string\n\n```s1=\"abcPYTHONabc\"\nprint (s1.removeprefix(\"abc\"))\n#Output:PYTHONabc\nprint (s1.lstrip(\"abc\"))\n#Output:PYTHONabc```\n\nExample 2: The parameters of `removeprefix()` are considered as a substring. But parameters of `lstrip()` are considered as a set of characters.\n\n```s1=\"abcbcacPYTHONabc\"\nprint (s1.removeprefix(\"abc\"))\n#Output:bcacPYTHONabc\nprint (s1.lstrip(\"abc\"))\n#Output:PYTHONabc```\n\nExample 3: `removeprefix`won’t remove multiple copies of a prefix. But lstrip() will remove multiple copies of a prefix\n\n```s1=\"abcabcPYTHONabc\"\nprint (s1.lstrip(\"abc\"))\n#Output:PYTHONabc\nprint (s1.removeprefix(\"abc\"))\n#Output:abcPYTHONabc```\n\nExample 4: If the parameter is not mentioned in lstrip(), it will remove leading whitespaces. But if the parameter is not mentioned in removeprefix(), it will raise TypeError\n\n```s1=\" PYTHON\"\nprint (s1.lstrip())\n#Output:PYTHON\nprint (s1.removeprefix())\n#Output:TypeError: str.removeprefix() takes exactly one argument (0 given)```\n\nExample 5: If the parameter mentioned is not found at the beginning of the string means, both methods will return a copy of the string\n\n```s1=\"PYTHONabc\"\nprint (s1.lstrip(\"abc\"))\n#Output:PYTHONabc\nprint (s1.removeprefix(\"abc\"))\n#Output:PYTHONabc```\n\n### removesuffix() vs rstrip()\n\n#### removesuffix()\n\n`str.removesuffix(suffix)`\n\n• If the string ends with the suffix string and that suffix is not empty, return `string[:-len(suffix)]`. Otherwise, return a copy of the original string:\n• The parameters of removesuffix() are considered as a substring and not as a set of characters.\n\n#### rstrip()\n\n`str.rstrip([chars])`\n\nReturn a copy of the string with trailing characters removed. The `chars `argument is a string specifying the set of characters to be removed. If omitted or `None`, the chars argument defaults to removing whitespace. The` chars` argument is not a suffix; rather, all combinations of its values are stripped.-Python docs\n\nExample 1:Remove prefix `“abc”` from the string\n\n```s1=\"abcPYTHONabc\"\nprint(s1.rstrip(\"abc\"))\n#Output:abcPYTHON\nprint(s1.removesuffix(\"abc\"))\n#Output:abcPYTHON```\n\nExample 2: The parameters of `removesuffix()` are considered as a substring. But parameters of `rstrip()` are considered as a set of characters.\n\n```s1=\"abcPYTHONacabc\"\nprint(s1.rstrip(\"abc\"))\n#Output:abcPYTHON\nprint(s1.removesuffix(\"abc\"))\n#Output:abcPYTHONac```\n\nExample 3: `removesuffix`won’t remove multiple copies of suffix. But `rstrip()` will remove multiple copies of the suffix.\n\n```s1=\"abcPYTHONabcabc\"\nprint(s1.rstrip(\"abc\"))\n#Output:abcPYTHON\nprint(s1.removesuffix(\"abc\"))\n#Output:abcPYTHONabc```\n\nExample 4: If the parameter is not mentioned in rstrip(), it will remove trailing whitespaces. But if the parameter is not mentioned in removesuffix(), it will raise TypeError\n\n```s1=\"PYTHON \"\nprint(s1.rstrip())\n#Output:PYTHON\nprint(s1.removesuffix())\n#Output:TypeError: str.removesuffix() takes exactly one argument (0 given)```\n\nExample 5: If the parameter mentioned is not found at the end of the string means, both methods will return a copy of the string\n\n```s1=\"abcPYTHON\"\nprint(s1.rstrip(\"abc\"))\n#Output:abcPYTHON\nprint(s1.removesuffix(\"abc\"))\n#Output:abcPYTHON```\n\n### Different ways to remove prefix and suffix before Python 3.9\n\n#### 1. Using re.sub()\n\nRemoving suffix\n\n```import re\npattern=r\"abc\\$\"\ns1=\"abcPYTHONacbcabc\"\ns2=re.sub(pattern,\"\",s)1\nprint (s2)\n#Output:abcPYTHONacbc```\n\n`pattern=r”abc\\$” `→ Checks whether the string ends with `“abc”`\n` \\$` →indicates the end of the string.\n\n`s2=re.sub(pattern,””,s1)` → If the pattern matches, it will replace the substring with an empty string.\n\nRemoving prefix\n\n```import re\npattern=r\"^abc\"\ns1=\"abcbcPYTHONabc\"\ns2=re.sub(pattern,\"\",s1)\nprint (s2)\n#Output:bcPYTHONabc```\n\n`pattern=r”^abc” `→ checks whether the string starts with` “abc”`\n\n`^` → indicates the beginning of the string\n\n`s2=re.sub(pattern,””,s1)` → If the pattern matches, it will replace the substring with an empty string.\n\n2. Using str.startswith()\n\n```s1=\"abcbcacPythonacbcabc\"\nif (s1.startswith(\"abc\")):\nprint (s1.replace(\"abc\",\"\",1))\n\n#Output:bcacPythonacbcabc```\n\nIf the string starts with `“abc”` means it will replace that substring by `“”` (empty string)\ncount =1 means only one occurrence of that substring mentioned is replaced by an empty string.\n\n3. Using str.endswith()\n\n```s2=\"Pythonacbcabc\"\nif (s2.endswith(\"abc\")):\nprint (s2[:-3])\n\n#Output:Pythonacbc```"
] | [
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.5401846,"math_prob":0.8719786,"size":5714,"snap":"2021-43-2021-49","text_gpt3_token_len":1436,"char_repetition_ratio":0.19632225,"word_repetition_ratio":0.27469555,"special_character_ratio":0.23731187,"punctuation_ratio":0.14619882,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9807034,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-10-26T12:06:34Z\",\"WARC-Record-ID\":\"<urn:uuid:51804a1b-e053-4fe0-8bf1-f44a5d3731fb>\",\"Content-Length\":\"88924\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:ac6ab1b3-9c1b-4fd2-bb57-906a25497e47>\",\"WARC-Concurrent-To\":\"<urn:uuid:d2f0270f-b02d-45c6-b53f-2ebd983c185a>\",\"WARC-IP-Address\":\"192.0.78.25\",\"WARC-Target-URI\":\"https://indhumathychelliah.com/2020/10/12/new-string-methods-to-remove-prefixes-and-suffixes-in-python-3-9/\",\"WARC-Payload-Digest\":\"sha1:2FKOSA3QJRN2COLSQ3KBMOUFETAVKZLB\",\"WARC-Block-Digest\":\"sha1:QGFCZF6Z42GKZEPOCFEU53XEISJ5ZFGR\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-43/CC-MAIN-2021-43_segments_1634323587877.85_warc_CC-MAIN-20211026103840-20211026133840-00683.warc.gz\"}"} |
https://tpcj.org/fourth-geometric-arithmetic-index-of-polycyclic-aromatic-hydrocarbons-pahk | [
"Fourth Geometric Arithmetic Index of Polycyclic Aromatic Hydrocarbons (PAHk)\n\nAbstract The geometric-arithmetic index of a graph G is defined as . Recently, Ghorbani et. al. introduced the eccentric version of the geometric-arithmetic index as , where is the eccentricity of the vertex u. In this paper, we compute the fourth geometric-version of Polycyclic aromatic hydrocarbons (PAHk).\n\nKeywords Geometric-arithmetic, Topological indices, Zagreb indices, Polycyclic Aromatic Hydrocarbons\n\n[Full Text: PDF]\n\nUpdated: April 15, 2016 — 2:26 pm"
] | [
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.7847483,"math_prob":0.76635176,"size":410,"snap":"2019-43-2019-47","text_gpt3_token_len":91,"char_repetition_ratio":0.14285715,"word_repetition_ratio":0.0,"special_character_ratio":0.17560975,"punctuation_ratio":0.1594203,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99157244,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-10-18T20:00:06Z\",\"WARC-Record-ID\":\"<urn:uuid:44afb26d-0b8f-42db-88e8-12ec7e612b75>\",\"Content-Length\":\"20245\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:b0fdbb95-f7d4-426d-af79-90838c6dfbcc>\",\"WARC-Concurrent-To\":\"<urn:uuid:6cbdac71-9a96-4653-a535-1d3a6136165d>\",\"WARC-IP-Address\":\"111.118.215.252\",\"WARC-Target-URI\":\"https://tpcj.org/fourth-geometric-arithmetic-index-of-polycyclic-aromatic-hydrocarbons-pahk\",\"WARC-Payload-Digest\":\"sha1:HBQWHSDHQYQT3DMU3A2L6WB74X2ODCTU\",\"WARC-Block-Digest\":\"sha1:UM7KKCLJDK5CBC7IDNLXAR6FUXRAAQQU\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-43/CC-MAIN-2019-43_segments_1570986684425.36_warc_CC-MAIN-20191018181458-20191018204958-00241.warc.gz\"}"} |
https://www.wheat-free.org/wheat-free-weightloss-tools.html | [
"# Weightloss tools\n\n#### Calculating your daily calorie needs\n\nThere are two ways to calculate your typical calorie needs per day:\n\nKey for both calculations: RMR = Resting Metabolic Rate (the base number of calories your body will burn per day without any exercise/physical activity), W = weight in kg, H = height in cm, A = age in years\n\n• Harris Benedict Equation\nMen: RMR = 66.47 + 13.75(W) + %(H) - 6.76(A)\nWomen: RMR = 655.1 + 9.56(W) + 1.7(H) - 4.7(A)\nFor example, a 5'7\" (170cm) tall, 40 year old woman, weighing 70kg would work out their RMR as follows:\nRMR = 655.1 + (9.56 x 70) + (1.7 x 170) - (4.7 x 40) = 1425 calories/day\n\n• Mifflin St. Jeor Equation\nMen: RMR = 9.99 x weight + 6.25 x height - 4.92 x age + 5\nWomen: RMR = 9.99 x weight + 6.25 x height - 4.92 x age - 161\nFor example, a 5'7\" (170cm) tall, 40 year old woman, weighing 70kg would work out their RMR as follows:\nRMR = (9.99 x 70) + (6.25 x 170) - (4.92 x 40) - 161 = 1404 calories per day"
] | [
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.77659506,"math_prob":0.9995192,"size":986,"snap":"2023-40-2023-50","text_gpt3_token_len":363,"char_repetition_ratio":0.11608961,"word_repetition_ratio":0.26395938,"special_character_ratio":0.42190668,"punctuation_ratio":0.16239317,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9950565,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-10-03T23:13:51Z\",\"WARC-Record-ID\":\"<urn:uuid:1cd552d6-2fff-4e6a-b27b-7a01d393e93d>\",\"Content-Length\":\"17290\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:7e774f30-da84-4db2-9d49-50f70dc419c4>\",\"WARC-Concurrent-To\":\"<urn:uuid:4fc18b0b-c1a4-43cf-97c6-72535d19c149>\",\"WARC-IP-Address\":\"45.79.167.43\",\"WARC-Target-URI\":\"https://www.wheat-free.org/wheat-free-weightloss-tools.html\",\"WARC-Payload-Digest\":\"sha1:6HDXB2NA7TYYIHS2J6QOYPUOLWSYSZFH\",\"WARC-Block-Digest\":\"sha1:TPJVQ4YCRH63ABPUH6QBR5PO64BFCLYZ\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-40/CC-MAIN-2023-40_segments_1695233511284.37_warc_CC-MAIN-20231003224357-20231004014357-00389.warc.gz\"}"} |
https://www.siliconpalms.com/crystals-royal-rife-and-maxwells-equations/ | [
"# Crystals, Royal Rife and Maxwell’s equations\n\nRoyal Rife was famous for inventing a machine that could defeat cancer using Maxwell’s equations and crystals. Sadly for humanity, it was about that time of formation and rise to prominence of the American Medical Association (AMA) and the Federal Drug Administration (FDA), both biology based.\n\nThe scientific breakthrough work, to match pathogens to frequency, was proven by Royal Rife but his science was terminated and the man himself character-assassinated and run out of the country.\n\nCrystals and Maxwell’s equations are in the news again, and a similar invention for curing cancer. A Silicon Valley start-up company called Theranos and a founder, woman by the name of Elizabeth Holmes is the next Royal Rife to be ended.\n\nThe reason I mention Maxwell’s equations is in the title is because he’s the first scientist to observe, measure and calculate the math involved in wave dynamics, here’s how it happens to be in the news and views.\n\nMaxwell’s equations are a set of coupled partial differential equations that, together with the Lorentz force law, form the foundation of classical electromagnetism, classical optics, and electric circuits. The equations provide a mathematical model for electric, optical, and radio technologies, such as power generation, electric motors, wireless communication, lenses, radar etc.\n\nMaxwell’s equations describe how electric and magnetic fields are generated by charges, currents, and changes of the fields. An important consequence of the equations is that they demonstrate how fluctuating electric and magnetic fields propagate at a constant speed (c) in a vacuum.\n\nKnown as electromagnetic radiation, these waves may occur at various wavelengths to produce a spectrum of light from radio waves to γ-rays. The equations are named after the physicist and mathematician James Clerk Maxwell, who between 1861 and 1862 published an early form of the equations that included the Lorentz force law. Maxwell first used the equations to propose that light is an electromagnetic phenomenon.\n\nThe equations have two major variants. The microscopic Maxwell equations have universal applicability but are unwieldy for common calculations. They relate the electric and magnetic fields to total charge and total current, including the complicated charges and currents in materials at the atomic scale.\n\nThe “macroscopic” Maxwell equations define two new auxiliary fields that describe the large-scale behaviour of matter without having to consider atomic scale charges and quantum phenomena like spins. However, their use requires experimentally determined parameters for a phenomenological description of the electromagnetic response of materials.",
null,
"In a geomagnetic storm, a surge in the flux of charged particles temporarily alters Earth’s magnetic field, which induces electric fields in Earth’s atmosphere, thus causing surges in electrical power grids. (Not to scale.)\n\nThe term “Maxwell’s equations” is often also used for equivalent alternative formulations. Versions of Maxwell’s equations based on the electric and magnetic potentials are preferred for explicitly solving the equations as a boundary value problem, analytical mechanics, or for use in quantum mechanics.\n\nThe covariant formulation (on spacetime rather than space and time separately) makes the compatibility of Maxwell’s equations with special relativity manifest. Maxwell’s equations in curved spacetime, commonly used in high energy and gravitational physics, are compatible with general relativity.\n\nIn fact, Einstein developed special and general relativity to accommodate the invariant speed of light, a consequence of Maxwell’s equations, with the principle that only relative movement has physical consequences.\n\nSince the mid-20th century, it has been understood that Maxwell’s equations are not exact, but a classical limit of the fundamental theory of quantum electrodynamics.\n\nCrystals Photo credit: Yagan Kiely on Visual Hunt / CC BY-SA"
] | [
null,
"https://i0.wp.com/www.siliconpalms.com/wp-content/uploads/2019/09/Magnetosphere_rendition-575x314.jpg",
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.9313723,"math_prob":0.9826486,"size":3736,"snap":"2022-27-2022-33","text_gpt3_token_len":687,"char_repetition_ratio":0.15300107,"word_repetition_ratio":0.0,"special_character_ratio":0.1753212,"punctuation_ratio":0.094936706,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9616234,"pos_list":[0,1,2],"im_url_duplicate_count":[null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-08-17T04:09:28Z\",\"WARC-Record-ID\":\"<urn:uuid:3ec1570d-95df-40fd-8aad-dc804b09204d>\",\"Content-Length\":\"130631\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:83b87b33-2740-45f4-bfd5-1fd6b72b7ef0>\",\"WARC-Concurrent-To\":\"<urn:uuid:24b9d436-8d11-4ac8-b9d2-4e00e8a8f582>\",\"WARC-IP-Address\":\"192.163.197.9\",\"WARC-Target-URI\":\"https://www.siliconpalms.com/crystals-royal-rife-and-maxwells-equations/\",\"WARC-Payload-Digest\":\"sha1:EWW4Y5GB6Z5I5J2TDMOWMAY52WE3EGJN\",\"WARC-Block-Digest\":\"sha1:7UAKCKUTWY6SJAQFCBME5CECMIYQM3OL\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-33/CC-MAIN-2022-33_segments_1659882572833.95_warc_CC-MAIN-20220817032054-20220817062054-00602.warc.gz\"}"} |
https://xaviergeerinck.com/2018/05/24/bellman-equations/ | [
"## Summary\n\nSo what did we learn up until now in our Introduction and Markov Property, Chain, Reward Process and Decision Process posts?\n\nInitially we defined our basic concepts:\n\n• State: What does our current environment look like?\n• Action: What action are we taking?\n• Policy: When taking action $a$ in state $s$, where do we go to?\n\nWhereafter we introduced a concept of reward in our MRP\n\n• Immediate Reward $Rt$: We want to steer our agent towards the action with the highest reward. Example: We want it to go forward on a cliff, and not fall off. This is given through the sum of our future rewards: $Rt = r{t + 1} + r{t + 2} + … + r_T$\n• Discounted Reward $Gt$ To prevent us from constantly following the same path and reward, we add a discount factor that will reduce the reward over time. Written as: $Gt = R{t + 1} + γ R{t + 2} + γ^2R{t + 3} + … = \\sum^{\\infty}{k=0}γ^k R_{t+k+1}$\n\nTo this we also added probabilities due to the fact of it being described as an MDP and that nothing is certain.\n\n• Transition Probability: What is the expected state to end up in after taking a certain action? $P{ss'}^a = P[S{t+1} = s' \\mid St = s, At = a]$\n• Reward Probability: What is the expected reward of ending up in a certain state after taking an action? $Rs^a = E[r{t+1} \\mid St = s, At = a]$\n\nSo that we were finally able to write 2 new functions that allow us to interpret our expected value of a certain state.\n\n• State-Value Function: Value when we follow policy $\\pi$ forever starting from our state $s$ given by $V^{\\pi}(s) = \\mathbb{E}_{\\pi}[G_t \\mid s_t = s] = \\mathbb{E}_{\\pi}[\\sum^{\\infty}_{k=0}γ^kr_{t+k+1} \\mid s_t=s]$\n• Action-Value Function: Value when we follow policy $\\pi$ after taking action $a$ on our state $s$ given by $Q{\\pi}(s, a) = \\mathbb{E}_{\\pi}[Gt \\mid s_t = s, a_t = a] = \\mathbb{E}_{\\pi}[\\sum^{\\infty}_{k=0}γ^kr_{t+k+1} \\mid s_t=s,a_t=a]$\n\nNow we know that, how are we able to create some kind of algorithm that allows us to find a path through our states, taking specific actions, that will eventually lead to the highest return. Knowing that our states do not depend on each other? (See MDP). Or in math terms, how can we find our optimal policy $\\pi^*$ which maximizes the return in every state?.\n\nFor this, let us introduce something called Bellman Equations.\n\n## Bellman Equations\n\n### Introduction\n\nAs written in the book by Sutton and Barto, the Bellman equation is an approach towards solving the term of \"optimal control\". Which is done through the creation of a functional equation that describes the problem of designing a controller to minimize a measure of a dynamical system's behavior over time.. This approach of Bellman utilizes the concepts of a state and a value function as we saw before.\n\nNote: A value function we could also write as the \"optimal return function\"\n\nAfterwards, this class of methods for solving these optimal control problems came to be known as dynamic programming.\n\nDynamic Programming: is a method for solving a complex problem by breaking it down into a collection of simpler subproblems, solving each of those subproblems just once, and storing their solutions.\n\nTo know if we can solve a problem through the use of Dynamic Programming, we can take a look at the Principle of Optimality which was also created by Richard E. Bellman.\n\nPrinciple of Optimality: An optimal policy has the property that whatever the initial state and initial decision are, the remaining decision must constitute an optimal policy with regard to the state resulting from the fist decision.\n\n### Bellman Equation - State-Value Function $V^\\pi(s)$\n\nSo what the Bellman function will actually does, is that it will allow us to write an equation that will represent our State-Value Function $V^\\pi(s)$ as a recursive relationship between the value of a state and the value of its successor states.\n\n\\begin{aligned} V^\\pi(s) =& \\mathbb{E}{\\pi}[Gt | St = s] \\ =& \\mathbb{E}{\\pi}[r{t+1} + \\gamma G{T+1} | St = s] \\ =& \\sum{a} \\pi(a | s) \\sum{s', r} P(s' , r | s, a)[ r + \\gamma \\mathbb{E}{\\pi}[G{t + 1} | S{t + 1} = s] ] \\ =& \\sum{a} \\pi(a | s) \\sum{s', r} P(s' , r | s, a)[ r + \\gamma v_{\\pi}(s') ], \\forall s \\in S \\end{aligned}\n\nNote: We do not go into detail here on how it works or to derive it, for more details about that, feel free to read the paper.\n\nIf we now want to find the best value function $V^$, then that means that we need to find: $V^(s) = max{\\pi}V^{\\pi}(s)$ or in terms of our last equation: $V^(s) = max{a} \\sum_{s', r} P(s' , r | s, a)[ r + \\gamma V^{}(s') ]$ which is our Bellman Optimality Equation\n\n### Bellman Equation - Action-Value Function $Q^\\pi(s,a)$\n\nFor the Action-Value Function we follow the same intuition as in the State-Value one, but here including the action. Eventually leading us to this equation:\n\n\\begin{aligned} Q^\\pi(s, a) =& \\mathbb{E}{\\pi}[Gt | St = s, At = a] \\ =& \\sum{a} \\pi(a | s) \\sum{s', r} P(s' , r | s, a)[ r + \\gamma v_{\\pi}(s') ] \\end{aligned}\n\nWhich leads to our Bellman Optimality Equation: $Q^(s, a) =\\sum{s', r} P(s' , r | s, a)[ r + \\gamma max{a'} Q^(s', a') ]$\n\n## Solving\n\nIn the next article, we will talk about some algorithms that will allow us to solve something called a \"GridWorld\" later on."
] | [
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.88567275,"math_prob":0.9984935,"size":5733,"snap":"2023-40-2023-50","text_gpt3_token_len":1638,"char_repetition_ratio":0.105428524,"word_repetition_ratio":0.07974816,"special_character_ratio":0.28745857,"punctuation_ratio":0.10258621,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9999479,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-12-10T05:33:09Z\",\"WARC-Record-ID\":\"<urn:uuid:1cf68ff3-f536-47ee-a8f8-88f634afc78f>\",\"Content-Length\":\"47206\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:6389eb8d-22db-4ec7-a81f-18c1b2e289db>\",\"WARC-Concurrent-To\":\"<urn:uuid:83f82473-cee7-4a89-bdf1-d3441e278673>\",\"WARC-IP-Address\":\"172.67.172.44\",\"WARC-Target-URI\":\"https://xaviergeerinck.com/2018/05/24/bellman-equations/\",\"WARC-Payload-Digest\":\"sha1:LRCN5XKBUOVDSZXZCSTTIDMGCXEQ5OXE\",\"WARC-Block-Digest\":\"sha1:DKAOIPMXZXXK6X6KIWU7FAIDMRJPRYOY\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-50/CC-MAIN-2023-50_segments_1700679101195.85_warc_CC-MAIN-20231210025335-20231210055335-00468.warc.gz\"}"} |
https://answers.everydaycalculation.com/gcf/4900-2058 | [
"Solutions by everydaycalculation.com\n\n## What is the GCF of 4900 and 2058?\n\nThe gcf of 4900 and 2058 is 98.\n\n#### Steps to find GCF\n\n1. Find the prime factorization of 4900\n4900 = 2 × 2 × 5 × 5 × 7 × 7\n2. Find the prime factorization of 2058\n2058 = 2 × 3 × 7 × 7 × 7\n3. To find the gcf, multiply all the prime factors common to both numbers:\n\nTherefore, GCF = 2 × 7 × 7\n4. GCF = 98\n\nMathStep (Works offline)",
null,
"Download our mobile app and learn how to find GCF of upto four numbers in your own time:"
] | [
null,
"https://answers.everydaycalculation.com/mathstep-app-icon.png",
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.84342927,"math_prob":0.99337435,"size":515,"snap":"2019-51-2020-05","text_gpt3_token_len":152,"char_repetition_ratio":0.11545988,"word_repetition_ratio":0.0,"special_character_ratio":0.33592233,"punctuation_ratio":0.0927835,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99653953,"pos_list":[0,1,2],"im_url_duplicate_count":[null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-01-17T18:25:19Z\",\"WARC-Record-ID\":\"<urn:uuid:0de15623-80c2-40e7-8f80-520e67a2a983>\",\"Content-Length\":\"5626\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:29efaeb3-3488-410a-a3cb-cd8d47ee9ae3>\",\"WARC-Concurrent-To\":\"<urn:uuid:766978ef-6950-4706-b12a-e7e2abab3930>\",\"WARC-IP-Address\":\"96.126.107.130\",\"WARC-Target-URI\":\"https://answers.everydaycalculation.com/gcf/4900-2058\",\"WARC-Payload-Digest\":\"sha1:I5YIP5TL2QN4NF5GEP3OT34KFIHJ7F5O\",\"WARC-Block-Digest\":\"sha1:EPLTUW2AJG3IV4XR7D6T4FICZOYRHD55\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-05/CC-MAIN-2020-05_segments_1579250590107.3_warc_CC-MAIN-20200117180950-20200117204950-00327.warc.gz\"}"} |
https://socratic.org/questions/are-there-examples-of-grams-of-precipitate-problems-i-am-not-sure-what-to-call-t | [
"# Are there examples of grams of precipitate problems? I am not sure what to call them other than that.\n\nMar 12, 2015\n\nYes. Precipitation reactions are a type of double replacement (displacement) reaction in which the pattern of reaction is AX + BY $\\rightarrow$ AY + BX. A and B are cations, and X and Y are anions. The cations and anions switch partners. These reactions have a product that is a precipitate, an insoluble gas, or water.\n\nYou are asking about a double replacement stoichiometry problem involving a precipitate. The following is an example.\n\nLead(II) nitrate and potassium iodide, both of which are in aqueous solution, react to produce solid lead(II) iodide and aqueous potassium iodide. If 2.89 grams of lead (II) nitrate and 1.05 grams of potassium iodide react, how many grams of the lead(II) iodide precipitate are produced?\n\nStep 1\nWrite a balanced chemical equation for this reaction.\n\"Pb(NO\"_3)_2(\"aq\") + $\\text{2KI(aq)}$ $\\rightarrow$ \"PbI\"_2(\"s\") + \"2KNO\"_3(\"aq\")\n\nStep 2\nConvert given grams of each reactant to moles by dividing grams by molar mass.\nMolar mass of \"Pb(NO\"_3)_2 = $\\text{331.2 g/mol}$\nMolar mass of $\\text{KI}$ = $\\text{166.0 g/mol}$\n\n\"2.89g Pb(NO\"_3)_2$\\div$\"331.2g/mol Pb(NO\"_3)_2 = \"0.008726 mol Pb(NO\"_3)_2\n\n$\\text{1.05g KI}$$\\div$$\\text{166.0g/mol KI}$ = $\\text{0.006325 mol KI}$\n\nStep 3\nDetermine how many moles of $\\text{PbI\"_2}$ each reactant can produce using the number of moles of each reactant and the mole ratios from the balanced equation.\nMole ratio for \"Pb(NO\"_3)_2 and $\\text{PbI\"_2}$ = $\\text{1 mol Pb(NO\"_3)_2:1 \"mol PbI\"_2}$.\nMole ratio for $\\text{KI}$ and $\\text{PbI\"_2}$ = $\\text{2 mol KI:1 mol PbI\"_2}$\n\nMultiply the known moles of each compound times the mole ratio that has the ${\\text{PbI}}_{2}$ on top.\n\n\"0.008726 mol Pb(NO\"_3)_2 x $\\text{1 mol PbI2\"/\"1 mol Pb(NO3)2}$ = ${\\text{0.008726 mol PbI}}_{2}$\n\n$\\text{0.006325 mol KI}$ x $\\text{1 mol PbI2\"/\"2 mol KI}$ = ${\\text{0.0031625 mol PbI}}_{2}$\n\nStep 4\n$\\text{KI}$ is the limiting reactant since it produces the fewest moles of ${\\text{PbI}}_{2}$, therefore the greatest number of moles of ${\\text{PbI}}_{2}$ that can be produced in this reaction is ${\\text{0.0031625 mol PbI}}_{2}$.\n\nNow determine the mass in grams of ${\\text{PbI}}_{2}$ that can be produced by multiplying the moles of ${\\text{PbI}}_{2}$ times its molar mass, which is $\\text{461.0 g/mol}$.\n\n${\\text{0.0031625 mol PbI}}_{2}$ x $\\text{461.0 g PbI2\"/\"1 mol PbI2}$ = ${\\text{1.46 g PbI}}_{2}$\n\nNote: Molar masses were obtained from Wikipedia."
] | [
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.91091096,"math_prob":0.99969125,"size":1102,"snap":"2019-43-2019-47","text_gpt3_token_len":266,"char_repetition_ratio":0.12659381,"word_repetition_ratio":0.0,"special_character_ratio":0.20326678,"punctuation_ratio":0.077294685,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.999892,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-11-18T12:09:47Z\",\"WARC-Record-ID\":\"<urn:uuid:42d84fbc-d5a3-4d99-97be-adb04a4ffcfc>\",\"Content-Length\":\"40419\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:d13b5722-6702-474c-b698-704935c227ea>\",\"WARC-Concurrent-To\":\"<urn:uuid:e29b7f22-4f16-4d83-9ff4-67ed9f470f70>\",\"WARC-IP-Address\":\"54.221.217.175\",\"WARC-Target-URI\":\"https://socratic.org/questions/are-there-examples-of-grams-of-precipitate-problems-i-am-not-sure-what-to-call-t\",\"WARC-Payload-Digest\":\"sha1:CXLDFJE6L5WTKOZK43N4AVS5TXPWT5JO\",\"WARC-Block-Digest\":\"sha1:T24SCVAAOACLYK5TDIXW7ELMSYEIPG6V\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-47/CC-MAIN-2019-47_segments_1573496669755.17_warc_CC-MAIN-20191118104047-20191118132047-00493.warc.gz\"}"} |
https://lammps.github.io/pizza/doc/matlab.html | [
"Pizza.py WWW Site - Pizza.py Documentation - Pizza.py Tools\n\n### matlab tool\n\nPurpose:\n\nCreate plots via MatLab numerical analysis program.\n\nDescription:\n\nThe matlab tool is a wrapper on the MatLab numerical analysis package, primarily designed to use its plotting capabilities. MatLab is commercial software available on a variety of platforms.\n\nThe matlab constructor launches MatLab as a process which the matlab tool sends commands to. The MatLab process can be killed via the stop() method, though this is typically unnecessary.\n\nThe plot() method plots a single vector against a linear index or pairs of vectors against each other. The pairs of vectors are written to files and read-in by MatLab. The Nth curve in a plot is stored in the MatLab variable pizzaN, so the 2 vectors can be accessed within MatLab as pizzaN(:,1) and pizzaN(:,2).\n\nThe mplot() method creates a series of plots and saves them each to a numbered file. Each file is a plot of an increasing portion of the vector(s). This can be used to make an animation of a plot.\n\nThe enter() method can be used to interact with MatLab directly. Each subsequent line you type is a MatLab command, until you type \"quit\" or \"exit\" to return to Pizza.py. Single MatLab commands can be issued as string arguments to the gnu tool.\n\nThe export() method writes numeric data as columns to text files, so that MatLab can read them via its \"importdata\" command.\n\nMutliple windows can be displayed, plotted to, and manipulated using the select() and hide() methods. The save() method writes the currently selected plot to a PostScript file.\n\nThe remaining methods (aspect, title, xrange, etc) set attributes of the currently selected plot. The erase() method resets all attributes to their default values.\n\nUsage:\n\n```m = matlab()\t\t start up MatLab\nm.stop()\t\t shut down MatLab process\n```\n```m.plot(a) plot vector A against linear index\nm.plot(a,b)\t \t plot B against A\nm.plot(a,b,c,d,...)\t plot B against A, D against C, etc\nm.mplot(M,N,S,\"file\",a,b,...) multiple plots saved to file0000.eps, etc\n```\n``` each plot argument can be a tuple, list, or Numeric/NumPy vector\nmplot loops over range(M,N,S) and create one plot per iteration\nlast args are same as list of vectors for plot(), e.g. 1, 2, 4 vectors\neach plot is made from a portion of the vectors, depending on loop index i\nIth plot is of b[0:i] vs a[0:i], etc\nseries of plots saved as file0000.eps, file0001.eps, etc\nif use xrange(),yrange() then plot axes will be same for all plots\n```\n```m(\"c = a + b\") execute string in MatLab\n```\n```m.enter()\t \t enter MatLab shell\nmatlab> c = a + b type commands directly to MatLab\nmatlab> exit, quit\t exit MatLab shell\n```\n```m.export(\"data\",range(100),a,...) create file with columns of numbers\n```\n``` all vectors must be of equal length\ncould plot from file with MatLab commands:\ncols = importdata('data')\nplot(cols(:,1),cols(:,2))\n```\n```m.select(N) \t figure N becomes the current plot\n```\n``` subsequent commands apply to this plot\n```\n```m.hide(N) \t delete window for figure N\nm.save(\"file\")\t save current plot as file.eps\n```\n```Set attributes for current plot:\n```\n```m.erase() reset all attributes to default values\nm.aspect(1.3) aspect ratio\nm.xtitle(\"Time\") x axis text\nm.ytitle(\"Energy\") y axis text\nm.title(\"My Plot\") title text\nm.title(\"title\",\"x\",\"y\") title, x axis, y axis text\nm.xrange(xmin,xmax) x axis range\nm.xrange() default x axis range\nm.yrange(ymin,ymax) y axis range\nm.yrange() default y axis range\nm.xlog() toggle x axis between linear and log\nm.ylog() toggle y axis between linear and log\nm.label(x,y,\"text\") place label at x,y coords\nm.curve(N,'r') set color of curve N\nm.curve(N,'g','--') set color and line style of curve N\nm.curve(N,'b','-','v') set color, line style, symbol of curve N\n```\n``` colors: 'k' = black, 'r' = red, 'g' = green, 'b' = blue\n'm' = magenta, 'c' = cyan, 'y' = yellow\nstyles: '-' = solid, '--' = dashed, ':' = dotted, '-.' = dash-dot\nsymbols: '+' = plus, 'o' = circle, '*' = asterik, 'x' = X,\n's' = square, 'd' = diamond, '^' = up triangle,\n'v' = down triangle, '>' = right triangle,\n'<' = left triangle, 'p' = pentagram, 'h' = hexagram\n```\n\nRelated tools:\n\nPrerequisites:\n\nThe MatLab numerical analysis package."
] | [
null
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https://www.w3xue.com/jsjq/threejs/three_mode.html | [
"",
null,
"## three.js课程",
null,
"",
null,
"# Three 点、线、网格\n\nThreejs所谓模型,如果你使用过三维软件,就是你三维软件中常说的三维模型,当然了,对于大多数前端程序员而言,不了解计算机图形学的情况下,没有使用过任何三维软件的情况下,并没有三维模型的概念。从Three.js角度来说,Threejs模型对象就是由Threejs几何体Geometry和Threejs材质Material构成,材质主要设置三维模型的颜色等外观样式,几何体主要是通过顶点坐标数据表达三维模型的外形形状。",
null,
"",
null,
"",
null,
"```var geometry = new THREE.BoxGeometry(100, 100, 100); //创建一个立方体几何对象Geometry\n// 点渲染模式\nvar material = new THREE.PointsMaterial({\ncolor: 0xff0000,\nsize: 5.0 //点对象像素尺寸\n}); //材质对象\nvar points = new THREE.Points(geometry, material); //点模型对象```",
null,
"```var geometry = new THREE.BoxGeometry(100, 100, 100); //创建一个立方体几何对象Geometry\n// 线条渲染模式\nvar material=new THREE.LineBasicMaterial({\ncolor:0xff0000 //线条颜色\n});//材质对象\n// 创建线模型对象 构造函数:Line、LineLoop、LineSegments\nvar line=new THREE.Line(geometry,material);//线条模型对象```",
null,
"```var geometry = new THREE.BoxGeometry(100, 100, 100);\n// 三角形面渲染模式\nvar material = new THREE.MeshLambertMaterial({\ncolor: 0x0000ff, //三角面颜色\n}); //材质对象\nvar mesh = new THREE.Mesh(geometry, material); //网格模型对象Mesh```",
null,
"```var material = new THREE.MeshLambertMaterial({\ncolor: 0x0000ff, //三角面颜色\nwireframe:true,//网格模型以线条的模式渲染\n});```\n\n```// 通过访问属性的形式设置\nmaterial.wireframe = true;```",
null,
"`mesh.scale.set(0.5, 1.5, 2)`\n\nx轴方向放大2倍\n\n`mesh.scale.x = 2.0;`\n\n`mesh.position.y = 80;`\n\n`mesh.position.set(80,2,10);`\n\n```// 等价于mesh.position = mesh.position + 100;\nmesh.translateX(100);//沿着x轴正方向平移距离100```\n\n`mesh.translateZ(-50);`\n\n```//向量Vector3对象表示方向\nvar axis = new THREE.Vector3(1, 1, 1);\naxis.normalize(); //向量归一化\n//沿着axis轴表示方向平移100\nmesh.translateOnAxis(axis, 100);```\n\n`mesh.rotateX(Math.PI/4);//绕x轴旋转π/4`\n\n```var axis = new THREE.Vector3(0,1,0);//向量axis\nmesh.rotateOnAxis(axis,Math.PI/8);//绕axis轴旋转π/8```\n\n```// 绕着Y轴旋转90度\nmesh.rotateY(Math.PI / 2);\n//控制台查看:旋转方法,改变了rotation属性\nconsole.log(mesh.rotation);```\n\nThreejs大多数对象都有克隆.clone()和复制.copy()两个方法,点模型Points、线模型Line、网格网格模型Mesh一样具有这两个方法。",
null,
"A.copy(B)表示B属性的值赋值给A对应属性。\n\n```var p1 = new THREE.Vector3(1.2,2.6,3.2);\nvar p2 = new THREE.Vector3(0.0,0.0,0.0);\np2.copy(p1)\n// p2向量的xyz变为p1的xyz值\nconsole.log(p2);```\n\nN = M.clone()表示返回一个和M相同的对象赋值给N。\n\n```var p1 = new THREE.Vector3(1.2,2.6,3.2);\nvar p2 = p1.clone();\n// p2对象和p1对象xyz属性相同\nconsole.log(p2);```\n\n```var box=new THREE.BoxGeometry(10,10,10);//创建一个立方体几何对象\nvar material=new THREE.MeshLambertMaterial({color:0x0000ff});//材质对象\n\nvar mesh=new THREE.Mesh(box,material);//网格模型对象\nvar mesh2 = mesh.clone();//克隆网格模型\nmesh.translateX(20);//网格模型mesh平移\n\n`box.scale(1.5,1.5,1.5);//几何体缩放`"
] | [
null,
"https://www.w3xue.com/img/left_head1.png",
null,
"https://www.w3xue.com/img/left_head2.png",
null,
"https://www.w3xue.com/img/right_head1.png",
null,
"https://www.w3xue.com/ueditor/net/upload/image/20220120/6377827414434918264845837.jpg",
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null,
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null,
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null,
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null,
"https://www.w3xue.com/ueditor/net/upload/image/20220121/6377835829470849582673648.jpg",
null
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https://gitlab.rlp.net/plmz/sturdy/-/commit/9aa77196af3bbfd2b274a7d285ca3b064710be45 | [
"",
null,
"### fix PCF\n\nparent d27a4be2\n ... ... @@ -23,13 +23,15 @@ import Prelude hiding (Bounded,fail,(.),exp) import Control.Category import Control.Arrow import Control.Arrow.Alloc import Control.Arrow.Fail import Control.Arrow.Fix import Control.Arrow.Trans import Control.Arrow.Conditional as Cond import Control.Arrow.Environment import Control.Arrow.Abstract.Join import Control.Arrow.Transformer.Abstract.Environment import Control.Arrow.Transformer.Abstract.Contour import Control.Arrow.Transformer.Abstract.BoundedEnvironment import Control.Arrow.Transformer.Abstract.Error import Control.Arrow.Transformer.Abstract.Fix import Control.Arrow.Transformer.Abstract.Terminating ... ... @@ -42,8 +44,7 @@ import Data.Text (Text) import Data.Profunctor import qualified Data.Lens as L import Data.Abstract.WeakMap(Map) import qualified Data.Abstract.WeakMap as M import qualified Data.Abstract.FiniteMap as F import Data.Abstract.Error (Error) import qualified Data.Abstract.Error as E import Data.Abstract.InfiniteNumbers ... ... @@ -53,65 +54,65 @@ import qualified Data.Abstract.Widening as W import qualified Data.Abstract.StackWidening as SW import Data.Abstract.Terminating(Terminating) import qualified Data.Abstract.Terminating as T import Data.Abstract.DiscretePowerset(Pow) import Data.Abstract.Closure(Closure) import Data.Abstract.Closure (Closure) import qualified Data.Abstract.Closure as C import Data.Abstract.DiscretePowerset (Pow) import GHC.Generics(Generic) import GHC.Exts(IsString(..)) import Syntax (Expr(..)) import GenericInterpreter type Env = Map Text Val data Val = NumVal IV | ClosureVal (Closure Expr Env) | Top deriving (Eq, Generic) instance PreOrd Val where _ ⊑ Top = True NumVal n1 ⊑ NumVal n2 = n1 ⊑ n2 ClosureVal c1 ⊑ ClosureVal c2 = c1 ⊑ c2 _ ⊑ _ = False instance Complete Val where (⊔) = W.toJoin widening (⊔) type Env = F.Map Text (Text, CallString Label) Val widening :: W.Widening IV -> W.Widening Val widening w (NumVal x) (NumVal y) = second NumVal (x `w` y) widening w (ClosureVal cs) (ClosureVal cs') = second ClosureVal \\$ C.widening (M.widening (widening w)) cs cs' widening _ Top Top = (W.Stable,Top) widening _ _ _ = (W.Instable,Top) -- | Numeric values are approximated with bounded intervals, closure -- values are approximated with a set of abstract closures. data Val = NumVal IV | ClosureVal (Closure Expr Env) | Top deriving (Eq, Generic) instance UpperBounded Val where top = Top -- | Addresses for this analysis are variables paired with the k-bounded call string. type Addr = (Text,CallString Label) -- | The abstract interpreter for Interval analysis. -- | Run the abstract interpreter for the k-CFA / Interval analysis. The arguments are the -- maximum interval bound, the depth @k@ of the longest call string, -- an environment, and the input of the computation. evalInterval :: (?bound :: IV) => Int -> [(Text,Val)] -> State Label Expr -> Terminating (Error (Pow String) Val) evalInterval k env0 e = evalInterval k env0 e = -- runInterp eval ?bound k env (generate e) runFixT stackWiden (T.widening (E.widening W.finite widenVal)) (runTerminatingT (runErrorT (runEnvT (runIntervalT (eval :: Fix Expr Val (IntervalT (EnvT Text Val (ErrorT (Pow String) (TerminatingT (FixT _ () () (->)))))) Expr Val))))) (M.fromList env0,generate e) (runContourT k (runEnvT alloc (runIntervalT (eval :: Fix Expr Val (IntervalT (EnvT Text Addr Val (ContourT Label (ErrorT (Pow String) (TerminatingT (FixT _ () () (->))))))) Expr Val)))))) (env0,generate e) where widenVal = widening (W.bounded ?bound I.widening) stackWiden :: SW.StackWidening _ (Env,Expr) stackWiden = SW.filter (\\(_,ex) -> case ex of Apply {} -> True; _ -> False) \\$ SW.groupBy (L.iso' (\\(env,exp) -> (exp,env)) (\\(exp,env) -> (env,exp))) \\$ SW.stack \\$ SW.reuseFirst \\$ SW.maxSize k \\$ SW.fromWidening (M.widening widenVal) widenVal = widening (W.bounded ?bound I.widening) \\$ SW.maxSize 3 \\$ SW.fromWidening (F.widening widenVal) newtype IntervalT c x y = IntervalT { runIntervalT :: c x y } deriving (Profunctor,Category,Arrow,ArrowChoice,ArrowFail e,ArrowJoin) type instance Fix x y (IntervalT c) = IntervalT (Fix x y c) deriving instance ArrowFix x y c => ArrowFix x y (IntervalT c) deriving instance ArrowEnv var val env c => ArrowEnv var val env (IntervalT c) instance ArrowTrans IntervalT where type Dom IntervalT x y = x type Cod IntervalT x y = y lift = IntervalT unlift = runIntervalT instance (IsString e, ArrowChoice c, ArrowFail e c, ArrowJoin c) => IsVal Val (IntervalT c) where succ = proc x -> case x of ... ... @@ -135,13 +136,34 @@ instance (IsString e, ArrowChoice c, ArrowJoin c, ArrowFail e c) => ArrowCond Va (ClosureVal _, _) -> fail -< \"Expected a number as condition for 'ifZero'\" instance (IsString e, ArrowChoice c, ArrowFail e c, ArrowJoin c) => IsClosure Val (Map Text Val) (IntervalT c) where => IsClosure Val (F.Map Text Addr Val) (IntervalT c) where closure = arr \\$ \\(e, env) -> ClosureVal (C.closure e env) applyClosure f = proc (fun, arg) -> case fun of Top -> (returnA -< Top) <⊔> (fail -< \"Expected a closure\") ClosureVal cls -> (| C.apply (\\(e,env) -> f -< ((e,env),arg)) |) cls NumVal _ -> fail -< \"Expected a closure\" instance PreOrd Val where _ ⊑ Top = True NumVal n1 ⊑ NumVal n2 = n1 ⊑ n2 ClosureVal c1 ⊑ ClosureVal c2 = c1 ⊑ c2 _ ⊑ _ = False instance Complete Val where (⊔) = W.toJoin widening (⊔) widening :: W.Widening IV -> W.Widening Val widening w (NumVal x) (NumVal y) = second NumVal (x `w` y) widening w (ClosureVal cs) (ClosureVal cs') = second ClosureVal \\$ C.widening (F.widening (widening w)) cs cs' widening _ Top Top = (W.Stable,Top) widening _ _ _ = (W.Instable,Top) instance UpperBounded Val where top = Top instance HasLabel (F.Map Text Addr Val,Expr) Label where label (_,e) = label e instance Hashable Val instance Show Val where ... ... @@ -150,12 +172,3 @@ instance Show Val where show Top = \"⊤\" type IV = Interval (InfiniteNumber Int) type instance Fix x y (IntervalT c) = IntervalT (Fix x y c) deriving instance ArrowFix x y c => ArrowFix x y (IntervalT c) deriving instance ArrowEnv var val env c => ArrowEnv var val env (IntervalT c) instance ArrowTrans IntervalT where type Dom IntervalT x y = x type Cod IntervalT x y = y lift = IntervalT unlift = runIntervalT\n {-# LANGUAGE Arrows #-} {-# LANGUAGE FlexibleInstances #-} {-# LANGUAGE FlexibleContexts #-} {-# LANGUAGE MultiParamTypeClasses #-} {-# LANGUAGE DeriveGeneric #-} {-# LANGUAGE TypeSynonymInstances #-} {-# LANGUAGE ImplicitParams #-} {-# LANGUAGE StandaloneDeriving #-} {-# LANGUAGE GeneralizedNewtypeDeriving #-} {-# LANGUAGE UndecidableInstances #-} {-# LANGUAGE OverloadedLists #-} {-# LANGUAGE TypeFamilies #-} {-# LANGUAGE PartialTypeSignatures #-} {-# LANGUAGE TypeOperators #-} {-# LANGUAGE LiberalTypeSynonyms #-} {-# LANGUAGE RankNTypes #-} {-# OPTIONS_GHC -fno-warn-orphans -fno-warn-partial-type-signatures #-} -- | k-CFA analysis for PCF where numbers are approximated by intervals. module KCFA where import Prelude hiding (Bounded,fail,(.),exp) import Control.Category import Control.Arrow import Control.Arrow.Alloc import Control.Arrow.Fail import Control.Arrow.Fix import Control.Arrow.Trans import Control.Arrow.Conditional as Cond import Control.Arrow.Environment import Control.Arrow.Abstract.Join import Control.Arrow.Transformer.Abstract.Contour import Control.Arrow.Transformer.Abstract.BoundedEnvironment import Control.Arrow.Transformer.Abstract.Failure import Control.Arrow.Transformer.Abstract.Fix import Control.Arrow.Transformer.Abstract.Terminating import Control.Monad.State hiding (lift,fail) import Data.Hashable import Data.Label import Data.Order import Data.Text (Text) import Data.Profunctor import qualified Data.Lens as L import Data.Abstract.Map(Map) import qualified Data.Abstract.Map as M import qualified Data.Abstract.FiniteMap as F import Data.Abstract.Failure (Failure) import qualified Data.Abstract.Failure as E import Data.Abstract.InfiniteNumbers import Data.Abstract.Interval (Interval) import qualified Data.Abstract.Interval as I import qualified Data.Abstract.Widening as W import qualified Data.Abstract.StackWidening as SW import Data.Abstract.Terminating(Terminating) import qualified Data.Abstract.Terminating as T import GHC.Generics(Generic) import GHC.Exts(toList) import Syntax (Expr(..)) import GenericInterpreter -- | Abstract closures are expressions paired with an abstract -- environment, consisting of a mapping from variables to addresses -- and a mapping from addresses to stores. newtype Closure = Closure (Map Expr (F.Map Text Addr Val)) deriving (Eq,Generic,PreOrd,Complete,Show) type Env = F.Map Text (Text, CallString Label) Val -- | Numeric values are approximated with bounded intervals, closure -- values are approximated with a set of abstract closures. data Val = NumVal IV | ClosureVal Closure | Top deriving (Eq, Generic) -- | Addresses for this analysis are variables paired with the k-bounded call string. type Addr = (Text,CallString Label) -- | Run the abstract interpreter for the k-CFA / Interval analysis. The arguments are the -- maximum interval bound, the depth @k@ of the longest call string, -- an environment, and the input of the computation. evalInterval :: (?bound :: IV) => Int -> [(Text,Val)] -> State Label Expr -> Terminating (Failure String Val) evalInterval k env0 e = -- runInterp eval ?bound k env (generate e) runFixT stackWiden (T.widening (E.widening widenVal)) (runTerminatingT (runFailureT (runContourT k (runEnvT alloc (runIntervalT (eval :: Fix Expr Val (IntervalT (EnvT Text Addr Val (ContourT Label (FailureT String (TerminatingT (FixT _ () () (->))))))) Expr Val)))))) (env0,generate e) where widenVal = widening (W.bounded ?bound I.widening) stackWiden :: SW.StackWidening _ (Env,Expr) stackWiden = SW.filter (\\(_,ex) -> case ex of Apply {} -> True; _ -> False) \\$ SW.groupBy (L.iso' (\\(env,exp) -> (exp,env)) (\\(exp,env) -> (env,exp))) \\$ SW.stack \\$ SW.maxSize 3 \\$ SW.reuseFirst \\$ SW.fromWidening (F.widening widenVal) newtype IntervalT c x y = IntervalT { runIntervalT :: c x y } deriving (Profunctor,Category,Arrow,ArrowChoice,ArrowFail e,ArrowJoin) type instance Fix x y (IntervalT c) = IntervalT (Fix x y c) deriving instance ArrowFix x y c => ArrowFix x y (IntervalT c) deriving instance ArrowEnv var val env c => ArrowEnv var val env (IntervalT c) instance ArrowTrans IntervalT where type Dom IntervalT x y = x type Cod IntervalT x y = y lift = IntervalT unlift = runIntervalT instance (ArrowChoice c, ArrowFail String c, ArrowJoin c) => IsVal Val (IntervalT c) where succ = proc x -> case x of Top -> (returnA -< NumVal top) <⊔> (fail -< \"Expected a number as argument for 'succ'\") NumVal n -> returnA -< NumVal \\$ n + 1 -- uses the `Num` instance of intervals ClosureVal _ -> fail -< \"Expected a number as argument for 'succ'\" pred = proc x -> case x of Top -> (returnA -< NumVal top) <⊔> (fail -< \"Expected a number as argument for 'pred'\") NumVal n -> returnA -< NumVal \\$ n - 1 ClosureVal _ -> fail -< \"Expected a number as argument for 'pred'\" zero = proc _ -> returnA -< (NumVal 0) instance (ArrowChoice c, ArrowJoin c, ArrowFail String c) => ArrowCond Val (IntervalT c) where type Join (IntervalT c) x y = Complete y if_ f g = proc v -> case v of (Top, (x,y)) -> (f -< x) <⊔> (g -< y) <⊔> (fail -< \"Expected a number as condition for 'ifZero'\") (NumVal (I.Interval i1 i2), (x, y)) | (i1, i2) == (0, 0) -> f -< x -- case the interval is exactly zero | i1 > 0 || i2 < 0 -> g -< y -- case the interval does not contain zero | otherwise -> (f -< x) <⊔> (g -< y) -- case the interval contains zero and other numbers. (ClosureVal _, _) -> fail -< \"Expected a number as condition for 'ifZero'\" instance (ArrowChoice c, ArrowFail String c, ArrowJoin c) => IsClosure Val (F.Map Text Addr Val) (IntervalT c) where closure = arr \\$ \\(e, env) -> ClosureVal (Closure [(e,env)]) applyClosure f = proc (fun, arg) -> case fun of Top -> returnA -< Top ClosureVal (Closure cls) -> -- Apply the interpreter function `f` on all closures and join their results. (| joinList (returnA -< Top) (\\(e,env) -> f -< ((e,env),arg)) |) (toList cls) NumVal _ -> fail -< \"Expected a closure\" instance PreOrd Val where _ ⊑ Top = True NumVal n1 ⊑ NumVal n2 = n1 ⊑ n2 ClosureVal c1 ⊑ ClosureVal c2 = c1 ⊑ c2 _ ⊑ _ = False instance Complete Val where (⊔) = W.toJoin widening (⊔) widening :: W.Widening IV -> W.Widening Val widening w (NumVal x) (NumVal y) = second NumVal (x `w` y) widening w (ClosureVal (Closure cs)) (ClosureVal (Closure cs')) = second (ClosureVal . Closure) \\$ M.widening (F.widening (widening w)) cs cs' widening _ Top Top = (W.Stable,Top) widening _ _ _ = (W.Instable,Top) instance UpperBounded Val where top = Top instance HasLabel (F.Map Text Addr Val,Expr) Label where label (_,e) = label e instance Hashable Closure instance Hashable Val instance Show Val where show (NumVal iv) = show iv show (ClosureVal cls) = show cls show Top = \"⊤\" type IV = Interval (InfiniteNumber Int)\n ... ... @@ -39,7 +39,7 @@ spec = do in do evalInterval 10 [] (app (app add zero) two) `shouldBe` Terminating (Success (num 2 2)) evalInterval 10 [] (app (app add one) two) `shouldBe` Terminating (Success (num 3 3)) evalInterval 10 [(\"x\", num 0 1)] (app (app add \"x\") two) `shouldBe` Terminating (Success (num 2 3)) evalInterval 10 [(\"x\", num 0 1)] (app (app add \"x\") two) `shouldBe` Terminating (Success (num 2 6)) it \"should terminate for the non-terminating program\" \\$ let ?bound = I.Interval 0 5 ... ...\n {-# LANGUAGE OverloadedStrings #-} {-# LANGUAGE ImplicitParams #-} module KCFASpec where import Prelude hiding (succ,pred) import SharedSpecs import Data.Abstract.DiscretePowerset(Pow) import Data.Abstract.Error hiding (toEither) import qualified Data.Abstract.Interval as I import Data.Abstract.Terminating hiding (toEither) import IntervalAnalysis import Syntax import Test.Hspec import GHC.Exts(toList) main :: IO () main = hspec spec spec :: Spec spec = do let ?bound = I.Interval (-100) 100 in sharedSpec (\\env e -> toEither \\$ evalInterval 3 env e) (NumVal . fromIntegral) describe \"behavior specific to interval analysis\" \\$ do it \"should execute both branches on IfZero on interval containing zero\" \\$ let ?bound = I.Interval (-100) 100 in evalInterval 3 [(\"x\", num (-5) 5)] (ifZero \"x\" (succ zero) (pred zero)) `shouldBe` Terminating (Success (num (-1) 1)) it \"should compute 0 + -1 + 1 = 0\" \\$ let ?bound = I.Interval (-100) 100 in evalInterval 3 [] (succ (pred zero)) `shouldBe` Terminating (Success (num 0 0)) it \"should analyse addition correctly\" \\$ let ?bound = I.Interval 0 5 in do evalInterval 10 [] (app (app add zero) two) `shouldBe` Terminating (Success (num 2 2)) evalInterval 10 [] (app (app add one) two) `shouldBe` Terminating (Success (num 3 3)) evalInterval 10 [(\"x\", num 0 1)] (app (app add \"x\") two) `shouldBe` Terminating (Success (num 2 3)) it \"should terminate for the non-terminating program\" \\$ let ?bound = I.Interval 0 5 in do evalInterval 5 [] (fix (lam \"x\" \"x\")) `shouldSatisfy` \\c -> case c of Terminating (Success (ClosureVal _)) -> True; _ -> False evalInterval 5 [] (fix (lam \"f\" (lam \"x\" (app \"f\" \"x\")))) `shouldSatisfy` \\c -> case c of Terminating (Success (ClosureVal _)) -> True; _ -> False where add = fix \\$ lam \"add\" \\$ lam \"x\" \\$ lam \"y\" \\$ ifZero \"x\" \"y\" (succ (app (app \"add\" (pred \"x\")) \"y\")) one = succ zero two = succ one num i j = NumVal \\$ I.Interval i j toEither :: Terminating (Error (Pow String) a) -> Either String a toEither (Terminating (Fail e)) = Left (unwords (toList e)) toEither (Terminating (Success x)) = Right x toEither NonTerminating = Left \"NonTerminating\"\nMarkdown is supported\n0% or\nYou are about to add 0 people to the discussion. Proceed with caution.\nFinish editing this message first!"
] | [
null,
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https://ru.scribd.com/document/297517256/EE-Lab-Manual | [
"Вы находитесь на странице: 1из 74\n\n# Preface\n\n## The EE 213-Electric Circuits Lab is an essential component of the Electrical\n\nEngineering program. The lab is designed to help students understand the basic\nprinciples of electric circuits and to give them an insight into the design,\nsimulation and hardware implementation of electric circuits. The main objective\nis to provide hands-on experience for the students so they are able to put\ntheoretical concepts into practice.\nThe experiments include both software circuit simulations and hardware-based\nmeasurements with actual electrical components. Computer simulation is\nstressed as a key analysis tool in engineering design. Multisim Electronics\nWorkbench software is used for simulation of electric circuits. Hardware\nexperimentation with components and laboratory equipment is always important\nfor acquiring the necessary hands-on practical skills in engineering education.\nThe present manual includes eleven experiments that cover topics taught in the\ntwo Electric Circuits courses EE211 and EE213. The manual also includes\nappendices on components color codes and general laboratory regulations and\nsafety rules.\n\nList of Experiments\n\nExp No.\n\nTitle\n\nPage\n\n11\n\n18\n\n25\n\n29\n\n## Sinusoidal AC Circuit Analysis\n\n36\n\nThree-Phase Circuits\n\n40\n\n## Transient Circuit Analysis\n\n45\n\nTransformer Circuits\n\n51\n\n10\n\n## Frequency Selective Circuit Analysis\n\n56\n\n11\n\nTwo-Port Networks\n\n63\n\nAppendix I\n\nDesign Project\n\n68\n\nAppendix II\n\n71\n\n72\n\n## EE-213 Lab Manual\n\nElectric Circuits\n\nExperiment 1\nIntroduction to Electrical Circuits Simulation and Testing\nSimulation is a mathematical way of emulating the behavior of a real world system. With\nsimulation modeling, we can determine a circuits performance without physically\nconstructing the circuit or using actual test instruments. Multisim Workbench is a complete\nsystem design tool that offers a very large component database, schematic entry, full\nanalog/digital SPICE simulation, etc. It also offers a single easy-to-use graphical user\ninterface for all design needs.\n\nIntroduction\nGo to StartProgramsMultisim and click on Multisim. This will open the main window\nas shown in Fig. 1. In Fig. 1 important toolbars and menu are labeled. In addition to the\ntoolbars shown in Fig. 1, there may be other toolbars appearing on your screen so concentrate\non the labeled items in Fig. 1 at this time.\nYou can always open and close a toolbar from the Main Menu. For example, if you want to\nopen or close (select/unselect) the Design Toolbar, select ViewToolbarsDesign. If any\ntoolbar does not appear on your screen then use the above procedure to make it appear.\nMost of the analysis can be performed by turning on/off the simulate switch. If the simulate\nswitch shown in Fig. 1 does not appear on your screen then select ViewShow Simulate\nSwitch in the Main Menu. This will open the Simulation Switch. Always, remember to save\nyour file to keep the changes you made in the circuit by selecting FileSave.\n\nZoom\nToolbar\n\nComponent\nToolbar\n\nDesign\nToolbar\n\nSimulate\nSwitch\n\n1|Page\n\n## EE-213 Lab Manual\n\nElectric Circuits\n\nWe will now try to learn about Multisim simulation techniques by solving a simple example.\n\nExample\nBuild the circuit shown in Fig. 2 using Multisim Electronics Workbench.\na\n\nVs\n\n30k\n\nR1\n\nb\nIbc\n\nVab\n\nR2\n\n20k\n\n30V\nc\n\n## STEP A: Placing the Components\n\n1. Place a Battery (DC Source)\na. Bring a DC source into the Multisim workspace:\nOpen the Multisim program if it is not already open. In the Component Toolbar,\nselect the Sources icon (refer to Fig. 1 to find the Component Toolbar). This will\nopen another window with several types of DC sources and other components as\nshown below in Fig. 3. Click on DC Voltage Source in this new window.\n\n## Select the Sources Icon\n\nfrom Component Toolbar\nSelect the\nDC Source\n\n## Figure 3: Selection of DC Voltage Source in Multisim\n\nNow bring your cursor into the workspace area and notice the change in the shape of\nthe cursor to\n\nClick at any point in the workspace. This will show the voltage source as\n\n2|Page\n\n## EE-213 Lab Manual\n\nElectric Circuits\n\n## b. Change the value and name of the voltage source:\n\nDouble-click on the voltage source that you just placed in the workspace, and a new\nwindow with the name Battery will appear, as shown in Fig. 4. Select Value in the\nBattery menu, if it is not already selected. Change the value from 12 to 30. Keep the\nunit as Volts in this menu. Now select Label in this menu and change the Reference\nID to Vs. Click on OK.\n\n## Figure 4: Battery Window for setup of DC voltage source\n\n2. Place a Resistor:\na. Bring a resistor into the Multisim workspace:\nIn the Component toolbar, select the Basic icon as shown. This will open another\nwindow with several basic components as shown below in Fig. 5.\n\n## Select the Basic Icon from\n\nComponent Toolbar\nSelect the\nResistor\n\n## Figure 5: Selection of Resistor in Multisim\n\nClick on Resistor. This will open the Browser-Basic window, as shown in Fig. 6.\nScroll through the Component List, select 30 kOhm, and click OK. The cursor shape\nwill change again. Click in the workspace and this will show the resistor\nas:\n\n3|Page\n\n## EE-213 Lab Manual\n\nElectric Circuits\n\nTip To speed up your scroll through the Browsers Component List, simply type the\nfirst few characters of the components name. For example, type 30k to move\ndirectly to the area of the 30 kOhm list.\n\n## Figure 6: Setup window for Resistor values\n\nb. Change the name of the resistor:\nDouble-click on the resistor, and a new window with the name Resistor will open as\nshown in Fig. 7. Select Label from the menu of this window. Change the Reference\nID to R1 (if it is not already) and press OK. This will change the name of the resistor\nto R1.\n\n4|Page\n\n## EE-213 Lab Manual\n\nElectric Circuits\n\n## d. Add another resistor R2\n\nPlace resistor R2 of value 20 k in the workspace through the same procedure.\ne. Rotate the resistor:\nSelect resistor R2 and press Ctrl-R to rotate the resistor or select Edit90\nClockwise from the Main Menu. This will make the resistor vertical. Labels and\nvalues of all the components can be dragged individually. Drag the label R2 and\nvalue 20 kOhm individually to put them in a suitable place.\n3. Place Ground:\nIn the Component Toolbar, select the Sources icon. Now click on the Ground icon in the\nnew window as shown in Fig. 8. Click in the workspace to show the Ground symbol as\nbelow\n\n## Select the Sources Icon\n\nfrom Component Toolbar\nSelect\nGround\n\n## STEP B: Connecting the Components\n\n1. Arrange the components properly:\nArrange the components according to the circuit given in Fig. 2. You can select and drag\nthe component to any place in the workspace. Select the components and drag them one\nby one to their proper places as shown in Fig. 9.\nR1\n\n30K\n\nR2\n30V\n\n20K\n\n5|Page\n\n## EE-213 Lab Manual\n\nElectric Circuits\n\n## 2. Show Grid in the workspace:\n\nYou may show a grid for ease of drawing the connections. Select ViewGrid Visible in\n3. Connect DC Voltage Source Vs to R1:\nBring the cursor close to the upper pin of Vs; the cursor shape will change to a plus\nsign. Click and move a little upward. A wire appears, attached to the cursor. Click again\nat a small distance above the Vs source. Notice that the line will change direction.\nControl the flow of the wire by clicking on points as you drag. Each click fixes the wire\nto that point as shown in Fig. 10. In this way, when the cursor reaches the pin of R1 click\nagain and this will connect Vs to R1. Notice that a node number is automatically\ngiven.\n\n## The mouse button\n\nwas clicked at this\nlocation\n\n## Figure 10: Manual connection of components\n\n4. Connect R1 to R2:\nConnect R1 to R2 using the same procedure.\n5. Making use of the Junction to connect the Ground:\nIn a similar manner, connect the ground with Vs and R2. Notice that a small black circle\nappears just above the ground, this is called the junction. When two or more components\nare connected at one point, a junction is created. A junction can also be placed manually\nby pressing Ctrl+J or selecting EditPlace Junction. This can be used to control the\nconnection points manually. Also notice that the ground node is automatically given node\nnumber 0. Do not alter it.\nThis completes the connection and the complete circuit is shown in Fig. 11.\nR1\n\n30K\n\n20K\n\nVs\n\nR2\n\n30V\n\n6|Page\n\n## EE-213 Lab Manual\n\nElectric Circuits\n\n6. Wire paths can be modified using drag points. Click on a wire. A number of drag points\nwill appear on the wire as shown in Fig. 12. Click any of these and drag to modify the\nshape. You can also add or remove drag points to give you even more control over the\nwire shape. To add or remove drag points, press CTRL and click on the location where\nyou want the drag point added or removed.\n\nDrag Points\n\n## STEP C: Placing Multimeter or Voltmeter in parallel to measure voltage\n\n1. To connect a Multimeter:\na. Select ViewToolbarInstruments. The Instruments toolbar will open as shown in\nFig. 13.\nMultimeter\n\n## Figure 13: Instruments Toolbar\n\nb. Click on the Multimeter icon. Now click in the workspace to place the Multimeter.\nDrag it and place it near resistor R1 as shown in Fig. 14. Make a connection from the\n+ terminal of the Multimeter to the left pin of R1 and from the terminal to the\nright pin of R1. Note that the reversal of + and terminals will give opposite readings.\n\n7|Page\n\n## EE-213 Lab Manual\n\nElectric Circuits\n\n## c. Set the Multimeter to measure DC voltage:\n\nDouble-click on Multimeter to open the properties window shown in Fig. 15. Select\nV to measure voltage. Select the DC wave shape. (Notice that the meter can also\nmeasure current A and resistance . It can measure AC as well as DC values.\nLeave the window open for viewing the measurements.\n\nSelection for\nAC Measurement\n\nSelection for\nDC Measurement\n\n## STEP D: Placing a Multimeter or Ammeter in series to measure current\n\n1. Place a second Multimeter in the workspace as we did in Step C. Remove the connection\nbetween R1 and R2. Connect the + terminal of the Multimeter towards R1 and the -\nterminal towards R2 as shown in Fig. 16.\n2. Set the Multimeter to measure current:\nDouble click on this Multimeter and select A in the Multimeter properties window.\nSet the wave shape to DC. If the current flows from 3 to zero, the meter will read\npositive.\n\n8|Page\n\n## EE-213 Lab Manual\n\nElectric Circuits\n\n## STEP E: Circuit Simulation\n\n1. Save the file.\nSelect FileSave\n2. Show the Simulate Switch, on the workspace.\nSelect ViewSimulate Switch.\n3. If the properties window is not open, double-click the Multimeter. Click to the 1\nposition (ON) of the simulation switch to start simulation. Results will appear in the\nproperties window of the Multimeter. Compare your result with those in Fig. 17 and\n\n9|Page\n\n## EE-213 Lab Manual\n\nElectric Circuits\n\nExercise 1\n1. Build the circuit of Fig. 18 in Multisim Electronics Workbench.\n2. Connect a voltmeter between nodes a and b.\n3. Connect an ammeter for the measurement of Ib0.\na\n\nV2\n\nR2 20k\nVin\nR3\n\n10k\n\n10V\n\nR1\n\nV1\n-\n\n30k\n\nIb0\n\nFigure 18\n\nExercise 2\n1. Build the circuit of Fig. 19 in Multisim Electronics Workbench.\n2. Simulate the circuit and find ix and iy (magnitude and phase)\n\niy\n\nix\n\n20 sin(4t)\n\n2H\n\nFigure 19\n\n10 | P a g e\n\n## EE-213 Lab Manual\n\nElectric Circuits\n\nExperiment 2\nElectric Circuits Fundamental Laws\nIntroduction\nThis experiment covers electrical circuits fundamental laws, including Ohms law, and\nKirkchoffs voltage and current laws. It consists of three parts which have to be first carried\nout using the Multisim Electronics Workbench software, and then repeated with hardware\ncomponents and laboratory equipment. The experiments involve the measurement of\nresistance, voltage and current in DC circuits.\n\n## PART I: Resistance, Voltage & Current in DC Circuits\n\nObjectives\n1. Use of voltmeter & ammeter to measure voltage and current through a resistor\n2. Verification of component power rating\n\nMaterials\nDC power supply\nDC 0-20V Voltmeter\nDC 0-100mA Ammeter\nMultimeter\n5V/1W lamp\nAssorted carbon resistors (100, 150, 220, 330)\n\nAmmeter\n\nVoltmeter\n\n11 | P a g e\n\n## EE-213 Lab Manual\n\nElectric Circuits\n\nProcedure\nWorkbench Simulation\n1. Construct the circuit shown in Fig. 1 on the Multisim Electronics Workbench. Use a\n5V, 1W lamp for simulation. (In the Components Toolbar select the Indicator icon;\nnow find Voltmeter, Ammeter and Lamp in this window).\n2. Set the DC supply voltage to 3V, click the Simulate Switch and verify that the\nbattery voltage is 3V as measured by the voltmeter. Record the voltage across the\nlamp terminals V and the current I flowing through it in Table 1. Calculate power\ndissipation in the lamp using the relationship P=VI, and note it down.\n3. Change the DC supply voltage to 5V. Run the analysis again. Record voltage and\ncurrent in Table 1 and calculate power dissipation in the lamp.\n4. Change the DC supply voltage to 7V. Run the analysis and see the effect on the\nintensity of light. Record voltage and current in Table 1 and calculate power\ndissipation.\n5. Change the DC supply voltage to 8V. Run the analysis and observe the value of the\ncurrent; also observe the glow of the lamp. What happened? Explain.\nTable 1: Simulation results using Multisim Electronics Workbench\nSource Voltage (V)\n\n## Lamp Voltage (V)\n\nCurrent (A)\n\nPower , VI (W)\n\n3\n5\n7\nNote: The lamp is rated at 5V, 1W\nQuestion: Why would the lamp be damaged when the voltage goes to 8V? Explain by\ncomparing power dissipation with its rated value.\n\n12 | P a g e\n\n## EE-213 Lab Manual\n\nElectric Circuits\n\n## PART II: Resistance Measurement and Ohms Law\n\nObjectives\n1. Usage of the Multimeter to measure resistance.\n2. Verification of Ohms law\n\n+\nV1\n-\n\n## Figure 3: Verifying Ohms Law\n\nProcedure\nWorkbench Simulation\n1. Build the circuit of Fig. 2 using Multisim Electronics Workbench. Connect a\nMultimeter between the terminals of the resistor and set it to read resistance. Select\nR1=100. Click the Simulation Switch to run the analysis. Record the value of\nresistance R1 in Table 3s Ohmmeter Reading.\n2. Build the circuit given in Fig. 3. Set Vs = 10V and R1=100. Click Simulation\nSwitch to run the analysis. Record voltage V1 across resistor R1 by connecting a\nMultimeter in parallel to it. Record the value of current I flowing through R 1 by\nconnecting another Multimeter in series to R1. Note down the values in Table 3. From\nthe voltage current readings verify Ohms law V1=R1 I. Considering the Multimeter\nreading as a reference, calculate the % error.\n3. Vary the DC supply voltage Vs in steps of 2V and record the current in each case.\nEnter your results in Table 4.\n4. Plot V vs. I in the graph of Table 5.\n5. Calculate the resistor value based on the slope of the V-I curve plotted in step 4.\nHardware Experiment\n6. Repeat steps 1-5 with the laboratory hardware components.\n13 | P a g e\n\n## EE-213 Lab Manual\n\nElectric Circuits\n\n## Table 3: Resistance Measurement\n\nOhms Law\nV1\nWorkbench Simulation\n\nR=\n\nHardware Experiment\n\nR=\n\nR=V1/I\n\nWorkbench\nV (Volts)\n\nHardware\nI (mA)\n\nV (volts)\n\n2\n4\n6\n8\n\n2\n4\n6\n8\n\nI (mA)\n\nV-I Plot\n\n## Electronics Workbench experiment\n\nHardware experiment\n\nR=\nR=\n\n14 | P a g e\n\n## EE-213 Lab Manual\n\nElectric Circuits\n\nQuestions\n1. Compare the values of resistance obtained with the Ohmmeter and with Ohms law (V/I)\nin the hardware experiment. Find the percentage error between them with the Ohmmeter\n\n15 | P a g e\n\n## EE-213 Lab Manual\n\nElectric Circuits\n\n## PART III: Kirkchoffs Voltage and Current Laws\n\nKirkchoffs Voltage Law (KVL) states that the algebraic sum of all voltages around any\nclosed path equals zero. Kirchhoffs Current Law (KCL) states that the algebraic sum of all\nthe currents at a node is zero (the current entering a node has the opposite sign to the current\nleaving the node).\n\nObjectives\n1. Voltage and Current measurement in a DC circuit\n2. Verification of Kirchhoffs voltage and current laws.\n\nIAB\n\nIFA\n\nR1\n220\n\nR2\n\nICB\n\n150\n\nC\nR4\n\nVs\n\nR3\n\n10V\n\n330\n\n100\n\nICD\n\n## Figure 4: Resistive circuit diagram\n\nProcedure\n1. Construct the circuit shown in Fig. 4 using Multisim Electronics Workbench.\n2. Measure the voltages VAB, VBC, VAD, VDC, VBD, and VAC. Enter the values in Table 5.\n3. Measure the currents IAB, ICB, IAD, ICD and IFA. Enter the values in Table 6. Note the\npolarity (sign) of the currents.\n4. Calculate the voltages around the following loops ABCEFA, ABDA, BDCB, ABCDA, and\nrecord them in Table 8.\n5. Verify KCL by adding currents at nodes A, B, C and D. Enter your results in Table 7.\n6. Construct the circuit in Fig. 4 with hardware components. Repeat steps 2-5. Enter your\nresults in Tables5-8. Compute the percentage errors. Comment on any discrepancy in the\nvoltage/current experimental and theoretical values\n\n16 | P a g e\n\n## EE-213 Lab Manual\n\nElectric Circuits\n\nVAB\n\nVBC\n\nVDC\n\nVBD\n\nVAC\n\nSimulation\nHardware\n% Error\n\nIAB\n\nICB\n\nICD\n\nIFA\n\nSimulation\nHardware\n% Error\n\nA\n\nSimulation\nHardware\n\nABCEFA\n\nABDA\n\nBDCB\n\nABCDA\n\nSimulation\nHardware\n\n17 | P a g e\n\n## EE-213 Lab Manual\n\nElectric Circuits\n\nExperiment 3\nVoltage & Current Dividers and Superposition Principle\nIntroduction\nThis lab covers two parts:\n1. the voltage divider and current divider rules\n2. the superposition principle\nThe lab work is based on the use of Multisim Workbench simulation and hardware-based\ncircuits to verify the voltage current divider and voltage divider rules.\n\n## PART I: Voltage & Current Dividers\n\nFor a series circuit the voltage divider rule (VDR), refer to Fig. 1:\nV1\n\nR1\nVs\nR1 R 2 R 3\n\nV2\n\nR2\nVs\nR1 R 2 R 3\n\nV3\n\nR3\nVs\nR1 R 2 R 3\n\n(1)\n\nFor a parallel circuit the current divider rule (CDR), refer to Fig. 2:\nI1\n\nR2\nIs\nR1 R 2\n\n(2)\n\nR1\nI2\nIs\nR1 R 2\n+\n\nV1\n\nI1\n\nR1\nR2\n\nVs\nR3\n-\n\nI2\n\n+\nV2\n\nIs\n\nR1\n\nR2\n\nV3 +\n\n## Figure 2: Parallel circuit\n\nObjectives\n1. To verify the voltage current divider and voltage divider rules.\n2. To verify the superposition principle using Multisim Electronics Workbench.\n\n18 | P a g e\n\n## EE-213 Lab Manual\n\nElectric Circuits\n\nMaterials\nDC power supply\nMultimeter\nAssorted carbon resistors (100, 150, 220, 330)\n\nProcedure\nSimulation Experiment\n1. Build the circuit given in Fig. 3 on Multisim Electronics Workbench.\n2. Connect voltmeters, ammeters (or Multimeters) at appropriate positions to measure\nvoltages and currents shown in Table 1.\n3. Disconnect the voltage source. Connect a Multimeter, measure the total resistance and\nrecord the value in Table 1. (Remember: resistance is always measured without any\nsource connected to the circuit)\n4. Repeat steps 2 and 3 for the circuit of Fig. 4 and record the values in Table 2.\nHardware Experiment\n5. Build the circuit of Fig. 3 with the hardware components. Take the voltage current\nmeasurements and Req and record them in Table 1. Considering the Workbench\nresults as the base, compute the percentage errors.\n6. Build the circuit of Fig. 4 with the hardwired components. Take the voltage current\nmeasurements and Req and record in Table 2. Considering the Workbench results as\nthe base, compute the percentage errors.\nI2\nIs\n\nV3\n\nVs\n\nI3\nR3\n\nI2\nIs\n\nV2\n\nR2\n\nV3\n\nVs\n\nR3\n\nR4\nI5\nReq\n\nI4\n\n+\nV5\n\nV4\n\nI3\n\nV2\n\nR2\n\nR4\n\n+\nR6\n\nI5\n\n+\nV6\n\nReq\n\nI4\n\nV4\n\n+\nR6\n\nV5\n\nI6\n\nV6\n\nI6\n\n19 | P a g e\n\n## EE-213 Lab Manual\n\nElectric Circuits\n\nIs\n\nI2\n\nI3\n\nI4\n\nI5\n\nI6\n\nV2\n\nV3\n\nV4\n\nV5\n\nV6\n\nReq\n\nV5\n\nV6\n\nReq\n\nSimulation\nHardware\n\n## Table 2: Simulation and experimental results for Fig. 4\n\nIs\n\nI2\n\nI3\n\nI4\n\nI5\n\nI6\n\nV2\n\nV3\n\nV4\n\nSimulation\nHardware\n\nQuestions\nRefer to Fig. 3 and the results of Table 1:\n1. Are R4 and R6 in parallel or in series? Refer to voltage current measurements to\n\n20 | P a g e\n\n## EE-213 Lab Manual\n\nElectric Circuits\n\n## 6. Is the parallel combination of R4 and R6 in series or in parallel with R2? Why?\n\nRefer to Fig. 4 and the results obtained in Table 2 and answer the following questions:\n7. Are R4 and R6 in parallel or in series? Why?\n\n21 | P a g e\n\n## EE-213 Lab Manual\n\nElectric Circuits\n\n## PART II: Superposition Principle\n\nIf there is more than one source in an electric network, the response (voltage or current) can\nbe obtained by considering one source at a time. The total response is the algebraic sum of\nthe individual responses. This is known as the superposition principle. While determining the\nresponses with a particular source, all other sources have to be deactivated (voltage sources\nare shorted and current sources are opened).\n\nObjectives\n1. To verify the superposition principle using Multisim Electronics Workbench.\n2. To verify superposition with hardwired components.\n\nMaterials\nTwo DC power sources\nMultimeter\nAssorted carbon resistors (100, 220, 330, 1k, 1.5k)\nI1\n+\n\nV1\n\nI3\n-\n\n1k\n10V\n\nI5\n\nV3\n\n100\n220\nI2\n\n+\nV2\n-\n\nV5\n\n330\n\n+\n1.5k V4\nI4\n\n5V\n\n## Figure 5: Resistive circuit with two sources\n\nProcedure\nSimulation\n1. Construct the circuit in Fig. 5 on Multisim Electronics Workbench. Put the meters in\nthe appropriate places to read the voltages and currents.\n2. Run the simulation. Record all the voltages and currents in Table 3. You can use\nammeters for current measurements, voltmeter for voltage measurements or\nMultimeter for both.\n3. Note the current directions and voltage polarities shown in Fig. 5.\n4. Remove the 5-V source from the circuit. Replace it with a short circuit.\n5. Run the simulation. Measure the voltages and currents and fill in Table 4.\n6. Put the 5-V source back in the circuit. Remove the 10-V source and replace it with a\nshort circuit.\n7. Run the simulation. Record all the voltages and currents in the circuit. Enter them in\nTable 5\n\n22 | P a g e\n\n## EE-213 Lab Manual\n\nElectric Circuits\n\nTable 3: Simulation results for voltage and current with both sources\n1 k\n\n220\n\n100\n\n1.5 k\n\n330\n\nVoltage\nCurrent\nTable 4: Simulation results with the 10-V source only\n1 k\n\n220\n\n100\n\n1.5 k\n\n330\n\nVoltage\nCurrent\n\n## Table 5: Simulation results with the 5-V source only\n\n1 k\n\n220\n\n100\n\n1.5 k\n\n330\n\nVoltage\nCurrent\n\nHardware Experiment\n8. Repeat the procedure with hardwired circuit elements. Enter your results below.\nTable 6: Experimental results with both sources\n1 k\n\n220\n\n100\n\n1.5 k\n\n330\n\nVoltage\nCurrent\n\n1 k\n\n220\n\n100\n\n1.5 k\n\n330\n\nVoltage\nCurrent\n\n23 | P a g e\n\n## EE-213 Lab Manual\n\nElectric Circuits\n\n## Table 8: Experimental results with 5-V source only\n\n1 k\n\n220\n\n100\n\n1.5 k\n\n330\n\nVoltage\nCurrent\n\nQuestions\n1. Check for the superposition principle. Write your observations.\n\n2. Compare the simulation and experimental results, and comment on any error causes.\n\n3. Superposition only applies to current and voltage, but not to power. Explain why.\n\n24 | P a g e\n\n## EE-213 Lab Manual\n\nElectric Circuits\n\nExperiment 4\nThevenins Theorem and Maximum Power Transfer\nIntroduction\nA two-terminal resistive network can be replaced by a voltage source in series with an\nequivalent resistor. The value of the voltage source equals the open circuit voltage of the two\nterminals under consideration. The value of the equivalent resistor equals the resistance\nmeasured between the open terminals when all the independent sources of the circuit are\ndeactivated (voltage source shorted and current source opened). This is known as Thevenins\ntheorem. The voltage source is called Thevenins voltage (VTH) and the equivalent resistor,\nthe Thevenins resistance (RTH).\nInstead of representing two-terminal resistive network with a voltage source in series with a\nresistor, we can represent it with a current source in parallel with a resistor. This is known as\nNorton Equivalent Circuit. The value of the current source (ISC) equals the current passing\nthrough a short circuit placed between the two terminals under consideration. In general, the\ntwo equivalent circuits are related by source transformation. Then:\n\nRTH\nVTH\n\nIsc\n\nRTH\n\n## Figure 1: Thevenins and Norton Equivalent Circuits\n\nThe maximum power output to a variable output resistance occurs when the value of the\noutput resistance equals Thevenins resistance.\nRTH\n\nRL = RTH\n\nVTH\n\n## Figure 2: Maximum power output condition\n\nThe value of the maximum output power or transferred power is given as,\n\nPmax\n\n2\nVTH\n4R TH\n\n25 | P a g e\n\n## EE-213 Lab Manual\n\nElectric Circuits\n\nObjectives\n1. To construct Thevenins equivalent using Workbench.\n2. To construct Nortons equivalent using Workbench.\n3. To verify the equivalent obtained in step 1 and 2 using hardwired components.\n4. To determine the maximum power transfer condition experimentally.\n\nMaterials\nTwo DC power sources\nOne Multimeter\nAssorted carbon resistors (1 k, 10 k k k)\na\n\n10k\n10V\n\nRL\n\n1k\n\n33k\n47k\n\n5V\n\n## Figure 3: Circuit for Thevenins and Nortons equivalent\n\nProcedure\nSimulation\n1. Construct the circuit given in Fig. 3 on Multisim Electronics Workbench.\n2. Remove the load resistor RL and connect a Multimeter or voltmeter to read the open\ncircuit voltage between a and b. Simulate and record the voltage. This is VTH for this\ncircuit between a and b. Record the value in Table 1.\n3. Place an ammeter (represents a short circuit) between terminals a and b to read the\nshort circuit current between a and b. Simulate and record the current. This is ISC for\nthis circuit between a and b. Record the value in Table 1.\n4. Remove the 10-V source. Replace it with a short circuit.\n5. Remove the 5-V source. Replace it with a short circuit.\n6. Connect a Multimeter in the resistance measurement mode (Ohmmeter) between a\nand b. Run the simulation and record the value of the resistor. This is RTH in Fig. 1.\nRecord the value in Table 1.\nHardwired Experiment\n7. Build the circuit of Fig. 3 with hardwired components in the laboratory.\n8. Repeat steps 2 to 6 and find the values of VTH, ISC and RTH experimentally.\nConsidering the Workbench results as the base, compute the percentage errors.\n9. In the circuit of Fig. 3 connect a variable resistor (RL) between a and b.\n26 | P a g e\n\n## EE-213 Lab Manual\n\nElectric Circuits\n\n10. Vary RL between 2.5 k and 10.5 k in steps of 1 k. Measure the voltage between\na and b (VL) in each case. Enter your results in Table 2.\n\nTable 1: Simulation and experimental results for Thevenin voltage and resistance\nVTH\n\nISC\n\nRTH\n\nWorkbench\nHardwired\n% Error\nTable 2: Experimental results for maximum power transfer\nRL\n(k)\n\n2.5\n\n3.5\n\n4.5\n\n5.5\n\n6.5\n\n7.5\n\n8.5\n\n9.5\n\n10.5\n\nVL\nPL\n\nPL in mW\n\nRL in k\n\n27 | P a g e\n\n## EE-213 Lab Manual\n\nElectric Circuits\n\nQuestions\n1. At what value of RL does the maximum value of PL occur in the graph?\n\n2. Does this value of RL compare with RTH you obtained experimentally or through\nWorkbench?\n\n3. Using your results show that the two equivalent circuits are related by source\ntransformation.\n\n4. Can you give examples for cases were Thevenins and Nortons equivalents are not\nrelated by source transformation?\n\n28 | P a g e\n\n## EE-213 Lab Manual\n\nElectric Circuits\n\nExperiment 5\nThe Oscilloscope and Function Generator\nIntroduction\nThe Oscilloscope is one of the most important electronic instruments available for making\ncircuit measurements. It displays a curve plot of time-varying voltage on the Oscilloscope\nscreen. The Oscilloscope provided with Multisim Electronics Workbench is a dual trace\nOscilloscope that looks and acts like a real Oscilloscope. A dual trace Oscilloscope allows\nthe user to display and compare two time-varying voltages at one time.\nThe controls on the Oscilloscope are as follows:\n1. The TIME BASE control adjusts the time scale on the horizontal axis in time per division\nwhen Y/T is selected. When B/A is selected, the horizontal axis no longer represents\ntime. The horizontal axis now represents the voltage on the channel A input and the\nvertical axis represents the voltage on channel B input. When A/B is selected, the\nhorizontal axis represents the voltage on the channel B input and the vertical axis\nrepresents the voltage on the channel A input. The X_POS control determines the\nhorizontal position where the curve plot begins.\n2. The CHANNEL A control adjusts the volts per division on the vertical axis for the\nchannel A curve plot. The Y-POS control determines the vertical position of the channel\nA curve plot relative to the horizontal axis. Selecting AC places a capacitance between\nthe channel A vertical input and the circuit testing point. Selecting 0 connects channel\nA vertical input to ground.\n3. The CHANNEL B control adjusts the volts per division of the vertical axis for the\nchannel B curve plot. The Y-POS determines the vertical position of the channel B curve\nplot relative to the horizontal axis. Selecting AC places a capacitance between the channel\nB vertical input and the circuit test point. Selecting 0 connects the channel B vertical\ninput to ground.\n4. The trigger settings control the conditions under which a curve plot is triggered (begins to\ndisplay). Triggering can be internal (based on one of the input signals) or external (based\non a signal applied to the Oscilloscope external trigger input). With internal triggering\nAUTO, A, or B. If A is selected, the curve plot will be triggered by channel A input\nsignal. If |B is selected, the curve plot will be triggered by channel B input signal. If you\nexpect a flat input wave shape or you want the curve plot displayed as soon as possible,\nselect AUTO. The display can be set to start on a positive or negative slope of the input\nby selecting the appropriate EDGE selection. The trigger LEVEL control determines the\nvoltage level of the input signal waveform, in divisions on the vertical axis, before the\nwaveform will begin to display.\nOnce a circuit simulation has been activated and a voltage curve plot has been displayed on\nthe Oscilloscope screen, the Oscilloscope probes can be moved to other test points in the\ncircuit without running the simulation again. Moving the probes automatically redraws the\nvoltage curve plot for the new test point. You can also fine tune the settings either during or\nafter a simulation and the display will be redrawn on the screen automatically. You can\npause or resume through the Analysis Option. The zoom feature can be used by using\n29 | P a g e\n\n## EE-213 Lab Manual\n\nElectric Circuits\n\nExpand. Normal size will be restored by clicking Reduce. You can learn other features\n\n## The Function Generator\n\nThe Function Generator is a voltage source that supplies different time-varying voltage\nfunctions. Multisim Electronics Workbench can supply Sine Wave, Square Wave, and\nTriangular Wave voltage functions. The wave shape, frequency, amplitude, duty cycle, and\nDC offset can be easily changed. It has three voltage output terminals. Connect the COM\nterminal to ground symbol. The +ve terminal provides output voltage that is positive with\nrespect to the COM terminal and the ve terminal proves an output voltage that is negative\nwith respect to the COM terminal.\nThe controls on the Function Generator are as follows:\n1. You can select a wave shape by clicking the appropriate wave shape on the top of the\nFunction Generator.\n2. The frequency control allows you to adjust the frequency of the output voltage up to 999\nMHz. Click up or down arrow to adjust the frequency, or click the frequency box and\ntype the desired frequency.\n3. The AMPLITUDE control allows you to adjust the amplitude of the output voltage\nmeasured from the reference level (common) to peak level. The peak to peak value is\ntwice the amplitude setting.\n4. The OFFSET control adjusts the DC level of the voltage curve generated by the Function\nGenerator. An offset of 0 positions the curve plot along the x-axis with an equal positive\nand negative voltage setting. A positive offset raises the curve plot above the x-axis and a\nnegative offset lowers the curve plot below the x-axis.\nConsult the Multisim Electronics Workbench User and Reference manuals for more details\non the Oscilloscope and Function Generator.\n\nProcedure\n1. From the instrument panel pull down the Function Generator and Oscilloscope and\nconnect them as shown in Fig. 1. You can select different colored wires by double\nclicking the line and choosing the colors. Select Sine Wave from the Function\nGenerator. Set frequency to 1 kHz and amplitude to 10V. On the Oscilloscope start with\nthe Time Base of 0.2ms/div and channel A, B settings of 5V/div.\n2. Click the on-off switch to run the analysis. Connect the positive output of the Function\nGenerator to Oscilloscope channel A input and negative output to channel B input. You\nmay use red color for Ch. A and blue to for Ch. B, if you wish.\n\n30 | P a g e\n\n## EE-213 Lab Manual\n\nFunction Generator\n\nElectric Circuits\n\nOscilloscope\n\n## Fig. 1: The Function Generator and Oscilloscope connection\n\n3. Select the 0 on the Oscilloscope channel B input and run the analysis again.\nQuestion: What change occurred on the Oscilloscope Channel B curve plot? Explain.\n\n## 4. Change the Oscilloscope channel A to 5 V/div.\n\nQuestion: What change occurred on the Oscilloscope Channel A curve plot? Explain.\n\n5. Change the Oscilloscope Time Base to 0.1ms/div and run the analysis again.\nQuestion: What change occurred on the Oscilloscope Channel A curve plot? Explain.\n\n31 | P a g e\n\n## EE-213 Lab Manual\n\nElectric Circuits\n\n6. Change the Oscilloscope Channel A Y-POS to 1.00 and run the analysis again.\nQuestion: What change occurred on the Oscilloscope Channel A curve plot? Explain.\n\n7. Change the channel A Y-POS back to 0.00 and select DC on channel B input and run the\nanalysis again.\nQuestion: What change occurred on the Oscilloscope curve plot? Explain.\n\n8. Return the Oscilloscope time base to the settings you started with and select 0 on the\nchannel B input again. Select the Triangular Wave shape on the Function Generator and\nrun the analysis again.\nQuestion: What change occurred on the Oscilloscope curve plot? Explain.\n\n32 | P a g e\n\n## EE-213 Lab Manual\n\nElectric Circuits\n\n9. Select the Square Wave on the Function Generator and run the analysis again.\nQuestion: What change occurred on the Oscilloscope curve plot? Explain.\n\n10. Change the AMPLITUDE on the Function Generator to 5V and run the analysis again.\nQuestion: What change occurred on the Oscilloscope curve plot? Explain.\n\n11. Change the frequency on the Function Generator to 2 kHz and run the analysis again.\nQuestion: What change occurred on the Oscilloscope curve plot? Explain.\n\n33 | P a g e\n\n## EE-213 Lab Manual\n\nElectric Circuits\n\n12. Change the offset on the Function Generator to 3 and run the analysis again.\nQuestion: What change occurred on the Oscilloscope curve plot? Explain.\n\n13. Change the Oscilloscope channel A input to AC and run the analysis again.\nQuestion: What change occurred on the Oscilloscope curve plot? Explain.\n\n14. Click expand on the Oscilloscope. Measure the time period (T) of one cycle on the\nwave shape.\nQuestion: What was the time period (T) of one cycle on the wave shape?\n\n34 | P a g e\n\n## EE-213 Lab Manual\n\nElectric Circuits\n\nRepeat the steps above with the laboratory Function Generator and Oscilloscope and record\nyour answers in the same order as the Workbench. State any difficulties you encountered in\ncarrying out the steps.\na\nR2=100\nv(t)=4sin t\nf = 5kHz\n\n+\nVab\n-\n\nR1=150\n\n## Figure 2: Measurement of RMS voltage\n\n19. Connect the hardwired circuit shown in Fig 2. Measure the amplitude (peak value) of\nvoltage v1 using the Oscilloscope and record it.\nvab (peak) (using Oscilloscope) =\nConnect an AC voltmeter between terminals a and b and record the value\nvab (using AC voltmeter) =\nThis value is called the RMS value of the voltage wave.\n\nQuestions\n1. Find the ratio of the two measured values.\n\n## vab (using voltmeter)\n\n2. Compare the above value with the theoretical (ideal) value and find the percentage error.\n\n35 | P a g e\n\n## EE-213 Lab Manual\n\nElectric Circuits\n\nExperiment 6\nSinusoidal AC Analysis\nIntroduction\nThe students learned how to use the Function Generator and Oscilloscope in the previous\nexperiment. In this experiment they will learn to build and take measurements in AC circuits.\nThe circuit will be simulated on Multisim Electronics Workbench. Then they will build it\nwith hardware components and check the accuracy of the measurements compared with the\nsimulation results. Consider the circuit given in Fig. 1.\nR1\n\na\nI\n\n1k\n\nv(t)=5sint\nf=10kHz\n\nIL\n\nb\n0022F\n\nIc\n\nR3\n\n220\ne\n\n12mH\n\nR2\n1.5k\n\n## Fig. 1: RLC circuit\n\nThe equation for sinusoidal current is written in the form,\ni(t) = im sin (t+)\n\n(1)\n\nwhere, i(t) is the instantaneous value of current, im is the peak value, is the angular\nfrequency and is the phase angle of the current with respect to the source voltage. Phase\nangle can be measured by using the two traces of a dual trace Oscilloscope and measuring the\ntime difference between two waveforms, as shown in Fig. 2.\n\ntime\n\nTime shift =\nPhase shift = =\nPhase shift = = 360 (/T)\nWhere T = time period\n\n## Fig. 2: Phase angle measurement\n\nThe RMS (Root Mean Square) or effective value of the sinusoid is given as im/2. The real\npower dissipated can be computed by using,\nP= Vrms Irms cos\n\n(2)\n\n36 | P a g e\n\n## EE-213 Lab Manual\n\nElectric Circuits\n\nObjectives\n1. Measure currents in each branch of the circuit of Fig. 1 using the Oscilloscope\n2. Verify KCL for AC circuit\n3. Calculate the power dissipated in the resistor.\n\nMaterials\nDual trace Oscilloscope\nSignal generator\nResistors (1k, 1.5k, 220)\nInductor, 12mH (iron core)\nCapacitor, 0.022 F\n\nProcedure\n1. Construct the circuit of Fig. 1. Connect Ch A of the Oscilloscope to the output of signal\ngenerator at point a. Oscilloscope ground and circuit ground should be the same. Set the\nOscilloscope to AUTO sweep, and use Ch A as the TRIGGER source. Adjust the output\nof the signal generator to provide a sinusoidal voltage with an amplitude (peak value) of\n5V (10 Vp-p) and frequency of the Function Generator at f=10 kHz.\n2. Use Ch B of the Oscilloscope to observe the voltage at point d. This is the voltage across\nresistor R2. Measure the peak (half of peak to peak) voltage v2 (across R2). Determine\nphase angle 2 of voltage v2 with respect to generator voltage v. Calculate the current\n()\n2 ()= 2\nand enter time domain expression for iL below.\n2\n\n## Table 1: Sinusoidal expressions for voltage (v2) and current (i(t))\n\n2 ()\n(amplitude)\n\n2\n(phase, deg)\n\nIL(amplitude) =\n\n2 ()\n\nExpression iL(t)\n\nWorkbench\nHardwired\n3. Move Ch B of the Oscilloscope to observe the voltage at point f. This is the voltage\nacross R3. Measure the amplitude v3 and phase angle 3 with respect to v(t). Calculate\nthe current iC (v3(t)/R3) and enter time domain expression for iC(t) below.\n\n## Table 2: Sinusoidal expressions for voltage (v3(t)) and current (iC(t))\n\nv3(t)(amplitude)\n\n3 (phase, deg)\n\niC(t)(amplitude)=v3(t)/R3\n\nExpression iC(t)\n\nWorkbench\nHardwired\n\n37 | P a g e\n\n## EE-213 Lab Manual\n\nElectric Circuits\n\na\nIs\n\nR1\n\n220\n\nR3\ne\n\n1k\n\nIc\nI1\n\n0022F\n\nV(t)=5sint\nf=10kHz\n\nIL\n\n12mH\n\nR2\n1.5k\n\nFigure 3:\n5. Change the position of resistor R1 to that shown in Fig. 3. This change will not affect the\ncircuit. Connect Ch A to point a and Ch B to point e. Measure amplitude of the\nvoltage across R1 (V1) from Ch B. Determine phase angle 1 of voltage v1(t)with respect\nto generator voltage v1(t). Calculate the current =1 =1 and enter time domain\n1\nexpression for i(t) below.\nTable 3: Sinusoidal expressions for voltage v1(t) and current i (t)\nv1(t)(amplitude)\n\n1 (phase, deg)\n\n## iL(t)(amplitude)=v1(t)/R1 Expression i(t)\n\nWorkbench\nHardwired\n\nQuestions\n1. Convert the measured values of currents iR, iL and iC obtained in steps 3-5 into phasor\nform (example, I= 530 0 A ). Compare these values with the ones obtained through\nsimulation.\nHardwired Experimental Values\n\nSimulation Values\n\n38 | P a g e\n\n## EE-213 Lab Manual\n\nElectric Circuits\n\n2. Use complex algebra to show that the data in step 1 (above) satisfies Kirchhoffs current\nlaw.\n\n39 | P a g e\n\n## EE-213 Lab Manual\n\nElectric Circuits\n\nExperiment 7\nThree Phase Circuits\nIntroduction\nThree phase circuits or system are made of three phase source, three phase feeder and three\nphase load. The three phase loads are connected in why (Y) or delta shape. The line (L) and\nphase (P) quantities of a balanced three phase circuits are related to each other as follows.\nIn a Y-connection, the line and the phase quantities are related by:\nVP = VL/3\n\n(1)\n\nIP=IL\n\n(2)\n\n## Whereas the relationships for a -connection are\n\nIP = IL/3\n\n(3)\n\nVP = VL\n\n(4)\n\nThe real and reactive powers for a 3 phase circuit (either Y or connection) are given as\nP=3 VL IL cos ()\n\n(5)\n\nQ=3 VL IL sin ()\n\n(6)\n\n## Where is the power factor angle of the load\n\n= tan-1(Q/ P)\n\n(7)\n\nPF=cos ()\n\n(8)\n\nEquation (7) and (8) can be used also to calculate the per factor for every phase\n\nObjectives\n1. Voltages and currents measurement in the three-phase circuits.\n2. Power measurement and power factor determination in three-phase circuits.\n\nMaterials\n1 Variable AC power supply 0-400V.\n1 Set of 30 safety connectors, blue, black, red\n1 Set of 10 safety connectors, green/yellow.\n1 Digital multi meter.\n\n40 | P a g e\n\n## EE-213 Lab Manual\n\nL=0.2H,0.65A\n\nElectric Circuits\n\nmeasurement\n\n41 | P a g e\n\n## EE-213 Lab Manual\n\nElectric Circuits\n\nProcedure\n1. For preliminary measurements use the digital Multimeter to measure each phase\nresistance of the resistive load bank as per figure 1 while the resistance control\nposition is at (40%) then enter each phase resistance value in Table 1. Do not change\nthe position of the resistance knob through the rest of the experiment.\n\n2. Repeat the phase measurement for the inductive load bank (L=0.2H, 0.65A) internal\nseries resistance for each phase as per figure 2 and enter the values in Table 2.\nResistance control\nPosition (%)\n\n(40%)\nTable 2: Inductance internal series resistances measurements\nInductance (H),\nCurrent Rating (A)\n\n0.2 H, 0.65 A\n\nSimulation\n3. Build circuit shown in Fig. 3 using Multisim Electronics Workbench.\n4. Simulate the three phase Y connected voltage source and apply 200 V line to neutral.\n5. Use the resistance values recorded in table 1 and table 2 and indicated value to\nestimate the power factor for each phase.\n6. Connect the ammeters and voltmeters at appropriate positions to measure voltages\nand currents, real and reactive powers then enter the simulation data shown in Table 3\nand table 4.\n7. Repeat steps 3 to 6 for the circuit given in Fig. 4 then enter the simulation data values\nin Table 5 and table 6.\nHardwired Experiment\n8. Build the circuit of Fig. 3 with the hardwired components. Take the measurements of\nvoltages, currents, power factor (from step 5) real power, reactive power and\nfrequency then record them in Table 3 and table 4. Considering the Workbench results\nas the base, compute the percentage errors.\n9. Build the circuit of Fig. 4 with the hardwired components. Take the measurements of\nvoltages, currents, power factor (from step 7) real power, reactive power and\nfrequency measurements then record them in Table 5 and table 6. Considering the\nWorkbench results as the base, compute the percentage errors.\nNote: (PF: Power Factor and FR: Power Frequency)\n42 | P a g e\n\n## EE-213 Lab Manual\n\nElectric Circuits\n\n## Table 3: Simulation and experimental results for Fig. 3\n\nVariable\n\nIL1\n\nIL2\n\nIL3\n\nVL1N\n\nVL2N\n\nVL3N\n\nVL1L2\n\nVL2L3 VL3L1\n\nWorkbench\nHardwired\n% Error\nTable 4: Simulation and experimental results for Fig. 3\nVariable\n\nPL1\n\nPL2\n\nPL3\n\nPTOT\n\nQL1\n\nQL2\n\nQL3\n\nQTOT\n\nPF\n\nFR\n\nWorkbench\nHardwired\n% Error\nTable 5: Simulation and experimental results for Fig. 4\nVariable\n\nIL1\n\nIL2\n\nIL3\n\nVL1N\n\nVL2N\n\n## VL3N VL1L2 VL2L3 VL3L1\n\nWorkbench\nHardwired\n% Error\nTable 6: Simulation and experimental results for Fig. 4\nVariable\n\nPL1\n\nPL2\n\nPL3\n\nPTOT\n\nQL1\n\nQL2\n\nQL3\n\nQTOT\n\nPF\n\nFR\n\nWorkbench\nHardwired\n% Error\n\nQuestions\n1. Compare the results obtained with Workbench with those from the hardware circuit, and\ncomment on the error obtained between hardwired and Workbench results of each case.\n\n43 | P a g e\n\n## EE-213 Lab Manual\n\nElectric Circuits\n\n2. What is the relation between the line voltage and the corresponding phase voltage based\non the Multisim workbench results? Is it confirmed by the hardwired results?\n\n3. Do we get similar relation between phase and line current if the load is connected?\n\n4. In a balanced system, all phase voltages have the same magnitude, what makes the\ndifference between two consecutive voltages higher 170% of the phase voltage?\n\n5. Do reactive powers in Figure 3 and Figure 4 circuit have the same direction? Why?\n\n6. Does the value of the capacitance have any impact on the load voltage magnitude and\nphase? Explain.\n\n44 | P a g e\n\n## EE-213 Lab Manual\n\nElectric Circuits\n\nExperiment 8\nTransient Circuit Analysis\nIntroduction\nFor electrical circuits such as the RC circuit of Fig. 1, the total step response (sum of forced\nsteady-state response and transient natural response) can be obtained by solving the 1st\norder differential equation describing the circuit. When there is no initial charge across the\ncapacitor, the voltage () for t 0 is shown to be given by:\nVC (t ) V (1- e\n\nt\nRC\n\n(1)\n\nFor the 2nd order series RLC circuit shown in Fig.2, the total step response is also given by\nthe sum of the forced and natural response, where the latter can be over-damped, underdamped or critically damped depending on whether the characteristic roots S1,2 are real,\ncomplex conjugates, or double roots. With =R/2L and 021/LC, we have:\n\ns1,2 2 w02\n\n(2)\n\nThe characteristic roots are real (negative) for >0, complex conjugates for <0, or\ndouble for =0. The different forms of () (for t 0) are respectively given by\n\nwhere\n\nt=0\nR\n+\n\nVc\n-\n\n45 | P a g e\n\n## EE-213 Lab Manual\n\nElectric Circuits\n\nObjectives\nThe experiment covers the analysis of the transient step response of 1st and 2nd order circuits.\nThe experiment is based on Multisim Workbench simulation and verification with hardware\nmeasurements and real components. The main objectives are:\n1. To perform transient analysis of a 1st order RC circuit\n2. To perform transient analysis of a 2nd order RLC circuit\n\n## Part I: Transient Analysis of RC Circuit\n\nMaterials\nFunction Generator\nOscilloscope\n22nF Capacitor\n3kResistor\n\nProcedure\nWorkbench Simulation\n1. Build the circuit of Fig. 1 in Multisim Electronics Workbench. Note node numbers\n2. Run the Transient Analysis. The procedure is as follows:\na. Select Simulate from the Main Menu\nb. Select Analysis\nc. Select Transient Analysis\nd. Select Analysis Parameters\n\nInitial Conditions:\n\nStart time: 0 s\n\nSet to zero\nEnd time: 0.5 ms\n\n## e. Select Output Variables\n\nSelect node 2 (left box), this is the node for the capacitor\nSelect plot during simulation (right box)\nf. Select Simulate\n3. Observe the wave shape. Using the cursor tool, record the results and enter the data in\nTable 1. Plot Vc vs. time on the graph provided.\nHardware Experiment\n4. Build the circuit given in Fig. 1 with the hardware components. The voltage source\nshould be drawn from the Function Generator.\n5. Set the Function Generator to produce a square wave signal with amplitude 5V and\nfrequency 1 KHz. This represents a repetitive (periodic) step voltage input.\n\n46 | P a g e\n\n## EE-213 Lab Manual\n\nElectric Circuits\n\n6. Apply the input voltage to Channel A of the Oscilloscope, and the capacitor voltage to\nChannel B. Adjust the amplitude scale to 5V/div and time base 0.1ms/div to get a fullscreen display (showing only 1 period of the square wave).\n7. Read the values of the output voltage VC and record the data in Table 1.\n8. Plot the voltage response on the same graph paper as the simulation results.\nTable 1: VC Transient response of RC circuit\nTime (ms)\n\n0.05\n\n0.1\n\n0.15\n\n0.2\n\n0.25\n\n0.3\n\n0.35\n\n0.4\n\n0.5\n\nVc (Simulation)\nVc (Measurement)\n\nVC\n\n## Figure 3: Simulated and Measured Transient Voltage Responses\n\ntime in seconds\n\nQuestions\n1. What is the theoretical value of the time constant of the circuit in Fig. 1?\n\n2. Explain how to calculate the time constant from the Vc vs. time curves\n(simulation and hardware). Comment on any difference between them.\n47 | P a g e\n\n## EE-213 Lab Manual\n\nElectric Circuits\n\n3. If you apply a square wave input with frequency 1MHz, explain whether this\nis suitable for measuring the time constant of the circuit and why?\n\n## Part II: Transient Analysis of RLC Circuit\n\nMaterials\nFunction Generator\nOscilloscope\n22nF Capacitor\n12mH Inductor\n220Resistor, 10kResistor\n\nProcedure\nWorkbench Simulation\n1.\n\nBuild the circuit of Fig. 2 in Multisim Electronics Workbench. Use 220 resistor.\n\n2.\n\n## Run the Transient Analysis following the same procedure as in Part I.\n\n3.\n\nObserve and record the VC waveform. Enter the data in Table 2. Plot VC vs. time\non the graph provided in Fig. 4.\n\nHardware Experiment\n4. Build the circuit of Fig. 2 with the C & L components and the 220resistor. The\nvoltage source is applied from the Function Generator.\n5. Set the Function Generator to produce a square wave signal with amplitude 5V and\nfrequency 1 kHz.\n\n48 | P a g e\n\n## EE-213 Lab Manual\n\nElectric Circuits\n\n6. Apply the input voltage to Channel A of the Oscilloscope, and the capacitor voltage to\nChannel B. Adjust the scale and time base to get a full-screen display\n7. Read the values of the output voltage VC and record the data in Table 2.\n8. Plot the voltage response on the same graph paper as the simulation results.\n9. Change the resistor to 10Kvalue. Notice the impact on the transient response.\nTable 2: VC Transient response of RLC circuit\nTime (ms)\n\n0.05\n\n0.1\n\n0.15\n\n0.2\n\n0.25\n\n0.3\n\n0.35\n\n0.4\n\n0.5\n\nVc (Simulation)\nVc (Measurement)\n\nVC\n\n## Figure 4: Simulated and Measured Transient Voltage Responses\n\ntime in seconds\n\nQuestions\n1. What is the type of the transient response? What are the theoretical values\nof the damping factor and the oscillation frequency d?\n\n49 | P a g e\n\n## EE-213 Lab Manual\n\nElectric Circuits\n\n2. Using the measured Vc values, estimate these factors directly from the\nwaveform plots. Comment on any difference between the values.\n\n## 3. When the resistor is set to 10kwhat is the type of the transient\n\nresponse? Explain why?\n\n50 | P a g e\n\n## EE-213 Lab Manual\n\nElectric Circuits\n\nExperiment 9\nTransformer Circuit\nIntroduction\nThe principal purpose of a transformer is to convert AC power at one voltage level to AC\npower of the same frequency at another voltage level. It is made of two winding and a\nmagnetic core. Both windings are linked by magnetic flux through a magnetic core. The\ntransformer has NP turns at the input or primary side and NS turns at its output or secondary\nwinding. The relationship between the voltage VP applied to the primary side of the\ntransformer and the voltage VP produced on the secondary side is\nVP/ VS = NP/ NS\n\n(1)\n\nNP/ NS = a\n\n(2)\n\nWhere a is defined as the turns ratio of the transformer. The relationship of the current IP\nflowing in the primary side of the transformer and the current IS flowing out of the secondary\nside of the transformer is\nIP/ IS = 1/a\n(3)\nThe real power PP supplied to the transformer by the primary circuit is given by\nPP = VP IP cos (P)\n\n(4)\n\nWhere p is the angle between the primary voltage and the primary current. The real power\nPS supplied by the transformer secondary circuit to its load is given by\nPS = VS IS cos (S)\n(5)\nWhere S is the angle between the secondary voltage and the secondary current.\n\nObjectives\n1. To verify the voltages and currents transformer ratio.\n2. To evaluate the Power behavior for loaded single phase transformer.\n\nMaterials\n1 Variable AC power supply 0-400V.\n\n## 1 Single-phase transformer. (NP = 847, NS = 456, NT = 456)\n\n1 Set of 30 safety connectors, blue, black, red\n1 Set of 10 safety connectors, green/yellow.\n2 Digital Multi meters.\n\n51 | P a g e\n\n## EE-213 Lab Manual\n\nElectric Circuits\n\nTransformer\n\nAC Source\n\n3- Phase\nVariable\nSupply\n0-400V\n2.5A\n\n1.1\n\nL1\n\n2.1\n\nV\nL2\n\n2.2\n\nV\n3.1\n\nL3\n\n3.2\n\n1.2\n\nTransformer\n\nAC Source\n\nL1\n\n1.1\n\n2.1\n\n3- Phase\nVariable L2\nSupply\n0-400V\n2.5A\nL3\n\n2.2\n\nV\nR\n\n3.1\n\n3.2\n\n1.2\n\nTransformer\n\nAC Source\n\nL1\n\n2.1\n\nV\n2.2\n\nV\nR\n\n3.1\n\n3- Phase\nVariable L2\nSupply\n0-400V\n2.5A\nL3\n\n1.1\n\nA\n3.2\n\nN\nG\n\n1.2\n\n## Figure 3 Inductive load voltage, current and power measurement\n\nProcedure\nCalculation and Initial Measurements\n1. Consider the circuit shown in Fig. 1 then fixe the source voltage at 224 V line to\nneutral.\n2. Place the ammeter and voltmeters at appropriate position in Fig. 1 circuit. Calculate\nthe voltages, current and the apparent power then enter the calculated data in Table 1.\n3. While the resistive load bank is separated from Fig. 2 circuit, fixe its control knob at\n40% then measure the per phase load resistance using the Multimeter.\n52 | P a g e\n\n## EE-213 Lab Manual\n\nElectric Circuits\n\n4. Calculate the voltages, currents and powers for the connected Fig. 2 circuit then enter\nthe data in Table 2.\n5. While the inductive part of the load is separated from Fig. 3 circuit, measure the per\nphase internal series resistance of the inductive load using the Multimeter. Choose the\nper phase inductance 1.2 Henry.\n6. Calculate the voltages, currents and powers for the connected Fig. 3 circuit and enter\nthe obtained data in Table 3\nHardwired Experiment\n7. Build the circuit of Fig. 1 with the hardwired components. Take the measurements of\nvoltages, currents, and record them in Table 1. Considering the calculated results as\nthe base, compute the percentage errors.\n8. Build the circuit of Fig. 2 with the hardwired components control knob at 40%).\nTake the measurements of voltages, currents and powers then record them in Table 2.\nConsidering the calculated results as the base, compute the percentage errors.\n9. Build the circuit of Fig. 3 with the hardwired components (control knob at 40%).\nTake the measurements of voltages, currents and powers then record them in Table 3.\nConsidering the calculated results as the base, compute the percentage errors\nTable 1: Simulation and experimental results for Fig. 1\nIS\n\nQuantity\n\nVP\n\nSP\n\nVS\n\nCalculated\nHardwired\n% Error\nTable 2: Simulation and experimental results for Fig. 2\nQuantity\n\nVP\n\nIP\n\nPP\n\nQP\n\nVS\n\nIS\n\nPS\n\nQS\n\nPS\n\nQS\n\nCalculated\nHardwired\n% Error\nTable 3: Simulation and experimental results for Fig. 3\nQuantity\n\nVP\n\nIP\n\nPP\n\nQP\n\nVS\n\nIS\n\nCalculated\nHardwired\n% Error\n\n53 | P a g e\n\n## EE-213 Lab Manual\n\nElectric Circuits\n\nQuestions\n1. Compare the calculated results with those obtained from the hardware circuit, and\ncomment on the error obtained between hardwired and calculated results of each case.\n\n## 3. Is SP = SS? Why? (SP=VP*IP and SS=VS*IS)\n\n4. Why do we have primary current in the circuit of figure 1 although there is no load at the\nsecondary side of the transformer?\n\n5. What is the reason behind Vs of circuit in figure 2 being lower then Vs of circuit in figure\n1?\n\n54 | P a g e\n\n## EE-213 Lab Manual\n\nElectric Circuits\n\n6. What is the nature of the current IS in the circuit of figure 3? Is it in favor of the voltage\nsupport?\n\nExplain?\n\n55 | P a g e\n\n## EE-213 Lab Manual\n\nElectric Circuits\n\nExperiment 10\nFrequency Selective Circuits\nIntroduction\nFrequency selective circuits exhibit a behavior that depends on the frequency of the applied\nsignals. Filters (and other circuits such as resonant circuits) are typical examples of frequency\nselective circuits, and have many numerous applications in electrical engineering. In this\nexperiment, we study examples of low-pass and band-pass filters.\nR\n\nVi\n\nVo\n\n37-1Low-Pass\nA Series filter circuit\nFigureFig.\n1: RC\nRC Low-Pass filter\n\nFor the RC circuit example shown in Fig. 1, the transfer function H(s)=Vo(s)/Vi(s) is easily\nobtained by VDR. The evaluation of its magnitude and phase (for s=j) gives:\n\nH j\n\n1\nRC\n\n2 1\nRC\n\nj tan1 RC\n\n(1)\n\n## This is a low-pass filter with half-power (3dB) cutoff frequency c=1/RC.\n\nOn the other hand, for the RLC circuit of Fig.2 (where the output is taken across R), the\nresulting transfer function will have the following magnitude and phase:\nL\n\nVi\n\nVo\n\n## Figure 2: Band-pass RLC filter circuit\n\nR\nL\nH j\n2\n2\n1\n\nR\n2\n\nL\nLC\n\n90 tan 1\n\nR (2)\nL\n1\n\n2\nLC\n\nThe circuit frequency response is that of a band-pass filter with center frequency (in rd/s):\n56 | P a g e\n\n## EE-213 Lab Manual\n\nElectric Circuits\n\n1\nLC\n\n(3)\n\nand the 3dB (half-power) cutoff frequencies are given (in rd/s) by:\n\nC1 R\n2L\n\n2L\n\n1 (4)\nLC\n\nC 2 R R 1\n2L 2L LC\n\n(5)\n\nObjectives\nThe experiment deals with the steady-state analysis of frequency selective filter circuits. The\nexperiment is based on Multisim Workbench simulation and hardware measurements. The\nmain objectives are:\n1. To perform AC frequency response analysis of a low-pass RC filter circuit\n2. To perform AC frequency response analysis of a band-pass RLC filter circuit\n\n## Part I: Frequency Response of Low-pass RC Filter Circuit\n\nMaterials\nFunction Generator\nOscilloscope\n22nF Capacitor\n1kResistor\n\nProcedure\nWorkbench Simulation\n1. Build the circuit of Fig. 1 in Multisim Electronics Workbench.\n2. Select Options/Preferences, to show node numbers\n3. Select Simulate from the Main Menu\n4. Select Analysis/AC Analysis\n5. Select Analysis Parameters\na. Start frequency:\n\n1Hz\n\nb. Stop frequency:\n\n100kHz\n\nc. Sweep type:\n\nd. Number of pts/dec:\n\ne. Vertical scale:\n\nlinear\n\n## 6. Select Output Variables\n\na. Select node 2 (left box) for the capacitor voltage\nb. Select plot during simulation (right box)\n57 | P a g e\n\n## EE-213 Lab Manual\n\nElectric Circuits\n\n7. Select Simulate\n8. Observe the filter frequency response. Using the cursor tool, read the coordinates of\nthe points on the plot. In particular, read the values at the half-power cutoff frequency\n9. Fill Table 1, and plot the filter magnitude and phase response on the graph provided.\nHardware Experiment\n10. Build the circuit given in Fig. 1 with the hardware components. The voltage source is\napplied from the Function Generator.\n11. Set the Function Generator to produce a sine wave signal with fixed 1V amplitude\n(the frequency will be changed from 1 to 100 kHz).\n12. Apply the input voltage to Channel A of the Oscilloscope, and the capacitor voltage to\nChannel B. The display should be set to facilitate reading the output voltage gain and\nphase shift with respect to the input.\n13. Change the input frequency in small steps, taking a few points per decade. Pay\nattention to take a measurement at the cutoff frequency.\n14. Read the values of the output voltage magnitude and phase (w.r.t the input voltage)\nand record the data in Table 1. Plot the circuit magnitude response and phase response\non the same graph paper as the simulation results.\n15. Change the input waveform shape to square wave type, with a 5 kHz frequency.\nObserve and record the output signal.\nTable 1: Frequency response of RC LPF\nFreq (Hz)\nMag(Vo)\nPhase(Vo)\n\n58 | P a g e\n\n## EE-213 Lab Manual\n\nElectric Circuits\n\nQuestions\n1. From the plotted graphs, what is the filter type? What it the measured 3dB\ncutoff frequency (in Hz)? Compare it to the theoretical value.\n\n2. Explain the reason for your observations with the square wave input in Step 9.\n\n## 3. If the capacitor is replaced with a 10mH inductor, explain qualitatively the\n\nimpact on the filter type and its characteristics\n\n59 | P a g e\n\n## EE-213 Lab Manual\n\nElectric Circuits\n\n## Part II: Frequency Response of Band-pass RLC Filter Circuit\n\nMaterials\nFunction Generator\nOscilloscope\n5nF Capacitor\n12mH Inductor\n1kResistor, 220Resistor\n\nProcedure\nWorkbench Simulation\n1. Build the circuit of Fig. 2 in Multisim Workbench (use the 1k resistor first).\n2. Run the frequency sweep analysis following the same procedure as in Part I. Use the\ncursor to mark the values for cutoff frequencies.\n3. Observe and record the circuit magnitude and phase frequency response. Enter the data in\nTable 2. Plot the magnitude and phase on the graph provided in Fig. 4.\n4. Change the resistor to 220observe and record the impact on the response.\nHardware Experiment\n5. Build the circuit of Fig. 2 with the hardwired R, L & C components. The voltage source is\napplied from the Function Generator.\n6. Set the Function Generator to produce a sine wave signal with 1V fixed amplitude (the\nfrequency will be varied from 10Hz to100kHz).\n7. Apply the input voltage to Channel A of the Oscilloscope, and the resistor voltage to\nChannel B.\n8. Change the frequency in small steps, taking a few points per decade. Make sure to take\nmeasurements at the cutoff frequencies. Record the values of the output voltage\nmagnitude and phase (with respect to the input voltage) in Table 2.\n9. Plot the circuit magnitude and phase responses on the same graph paper as the simulation\nresults.\nTable 2: Frequency response of RLC BPF\nFreq (kHz)\nMag(Vo)\nPhase(Vo)\n\n60 | P a g e\n\n## EE-213 Lab Manual\n\nElectric Circuits\n\n## Figure 4: RLC BPF Frequency Response\n\nQuestions\n1. Comment on the frequency response type?\n\n2. For each of the two resistors, specify the values of the half-power (3dB) cutoff\nedge frequencies (in Hz)? Discuss and compare with theoretical values\n\n61 | P a g e\n\n## EE-213 Lab Manual\n\nElectric Circuits\n\n3. What are the bandwidth and quality factor (for both resistors)?\n\n## 4. Comment of the resonance behavior of this circuit (for both resistors)\n\n5. If the output is taken across L&C (together), explain qualitatively the type of\nfilter in this case, and why?\n\n62 | P a g e\n\n## EE-213 Lab Manual\n\nElectric Circuits\n\nExperiment 11\nTwo-Port Networks\nIntroduction\nFor many electrical systems (or networks), focusing on the input/output terminals can\nprovide convenient models for circuit analysis without worrying about the internal details.\nOften, a signal is fed into input terminals, processed then extracted at the other output\nterminals. There are four variables of interest in this case, as shown in Fig.1.\ni2\n\ni1\n\nInput\nPort v1\n\nL in e a r\nN e tw o rk\n\nv2\n\nOutput\nPort\n\n1: Two-port\ncircuit\nmodel.Block\nFig.Figure\n34-1 The\nTwo-Port\nBuilding\nThere are six groups of two-port circuit parameters (e.g., the impedance z-parameters,\nadmittance y-parameters, hybrid h-parameters, etc) that can be used to relate the input/output\nvoltage & current variables. Any set of parameters can be derived from the other ones. In this\nlab, we consider the case of (impedance) z-parameters.\nThe z-parameters are defined by:\n\nV1 z11 I1 z12 I 2\nV2 z21 I1 z22 I 2\n\n(1)\n\nThese parameters can be measured by the following tests (based on open-circuit input port\ncondition, and open-circuit output port condition):\n\nz11\n\nV1\nV\nV\nV\n, z12 1\n, z21 2\n, z22 2\nI1 I 0\nI 2 I 0\nI1 I 0\nI 2 I 0\n2\n1\n2\n1\n\n(2)\n\nObjectives:\nThe experiment is based on Workbench simulation and hardware measurements. The main\nobjectives are:\n1. To perform z-parameters characterization of a given two-port circuit\n2. To use the z-parameters model for analysis of a loaded two-port circuit\n\n63 | P a g e\n\n## EE-213 Lab Manual\n\nElectric Circuits\n\nMaterials\nVoltage Source\nMultimeter\n2201kResistors\n\nProcedure\n10 k\n\nI1\n\n+\nV1\n\n3 k\n\nI2\n\n+\nV2\n\n3 k\n1 k\n\n## Figure 2: Two-Port Circuit Diagram\n\nWorkbench Simulation\n1. Build the circuit of Fig. 2 in Multisim Electronics Workbench.\n2. Apply a DC voltage test source to the input port with the output port open. Use a\nvoltmeter and ammeter to read the values of the voltages and currents necessary to\ncompute Z11 and Z21. Observe the proper polarity of the currents.\n3. Repeat a similar procedure with a DC voltage test source applied to the output port\nwith the input port open. Use a voltmeter and ammeter to read the voltage and current\nvalues, and derive the parameters Z22 and Z12.\n4. Record the values of the Z-parameters in Table 1.\nHardware Experiment\n5. Build the circuit given in Fig. 2 with the hardware components.\n6. Follow identical steps to (2) and (3) above by applying a DC test voltage to the input\nand output ports alternatively, and measuring the relevant voltages and currents.\nAgain, pay attention to the direction (and sign) of the currents\n7. Compute and record the Z-parameters based on hardware measurements in Table 1.\nTable 1: Z-Parameters of Two-Port Circuit\nZ-parameters\n\nZ11\n\nZ12\n\nZ21\n\nZ22\n\nSimulation\nMeasurement\n\n64 | P a g e\n\n## EE-213 Lab Manual\n\nElectric Circuits\n\nQuestions\n1. Compare the simulation and hardware measurement results. What would be\nthe sources of any discrepancies?\n\n65 | P a g e\n\n## EE-213 Lab Manual\n\nElectric Circuits\n\n## Part II: Analysis of Loaded Two-Port Circuit\n\nMaterials\nVoltage Source\nVoltmeter, Ammeter\nAssorted Resistors: 1001k3k10k1k\n\nProcedure\nRs=100\n\nVs\n\n2-port\n\nRL=1 k\n\n## Figure 3: Loaded Two-Port Circuit\n\nWorkbench Simulation\n1. Build the circuit of Fig. 3 in Multisim Workbench.\n2. Run the simulation with a 10V DC voltage source. Insert Multimeters to read the voltage\nand current of the load resistance.\n3. Evaluate the voltage gain Gv=V2/Vs and current gain Gi=I2/I1\n4. Record your results in Table 2.\nHardware Experiment\n5. Build the circuit of Fig. 3 with the hardware components.\n6. Follow identical steps to (2) and (3) above by applying a 10V DC voltage to the input\nside. Use voltmeter and ammeter to read the relevant voltage and current.\n7. Evaluate the voltage gain Gv=V2/Vs and current gain Gi=I2/I1\n8. Record the results in Table 2.\nTable 2: Voltage and Current Gains\n\nGv\n\nGi\n\nSimulation\nMeasurement\n\n66 | P a g e\n\n## EE-213 Lab Manual\n\nElectric Circuits\n\nQuestions\n1. Based on the Z-parameters obtained in Part I, derive the expressions for\nthe voltage and current gains Gv and Gi\n(Hint: write KVL equations for the input and output loops, in addition to\nthe Z-parameters equations)\n\nresistance\n\n## 3. Comment on the usefulness of the Z-parameters (and other types of\n\nparameters) for analyzing circuits such as that of Fig. 2\n\n67 | P a g e\n\n## EE-213 Lab Manual\n\nElectric Circuits\n\nAppendix I\n\n## EE213 Design Experiment Sample\n\n1. Analysis and Tuning of a Band-Reject Filter\nPart 1: Theoretical Analysis\nConsider the circuit given in the figure.\n\nL\nR\n\nRL\n\n## Figure 1: BRF Circuit\n\na) Show by qualitative analysis that the circuit is a band-reject filter. What is the rejected\nfrequency?\nb) Derive the voltage transfer function and validate your answer to (a)\nc) Derive the expressions of the cutoff corner frequencies\nd) Derive the expressions of the bandwidth and quality factor\n\n## Part 2: Multisim Workbench Simulation\n\na) Build the circuit in Multisim Workbench\nb) Choose the components such that the reject frequency band is centered at 10kHz and\nthe quality factor is at least 5\nc) Verify the proper operation of your design by plotting the filter frequency response\nd) Investigate the impact of connecting a load resistor to the filter. Discuss and explain\n\n## Part 3: Experimental Testing\n\na) Build the circuit with hardware components\nb) Use the function generator and oscilloscope to test the filter design you obtained in\nPart 2\nc) Investigate experimentally the impact of a load resistor (test different values) on the\nfilter reject frequency and selectivity\nd) Discuss your results and draw relevant conclusions\n\n68 | P a g e\n\n## EE-213 Lab Manual\n\nElectric Circuits\n\n## EE213 Design Experiment Sample\n\n2. Analyzing a Factory and Designing an Expansion\nA factory located in the industrial area is planning for an expansion. As an engineer you have\nbeen assigned to perform all of the required analysis and design calculations for this\nexpansion. The circuit given below represents a circuit model of the devices in the factory\nbefore the expansion.\n\n8\n20V\n\n5V\ni2\n\ni1\n\n15V\n\n## Part 1: Required analysis of the existing factory\n\nCalculate the currents i1 and i2.\nb. Calculate the power of each source in the circuit.\nc. Calculate the voltage Vab of the factory before the expansion.\nd. Build the circuit in Multisim Workbench verify your calculations in a, b and c\nabove\na.\n\n## Part 2: Required design calculations for the new expansion\n\nKnowing that the devices of the new expansion will be connected between terminals a and b,\ndesign the elements of a model representing the new expansion to the factory if you are given\nthat after the expansion the voltage Vab is to be 20V.\na.\nb.\nc.\nd.\n\nBuild the complete circuit in Multisim Workbench and verify the proper\noperation of your design by measuring the voltage Vab.\nBuild the circuit with hardware components and verify the proper operation of\nyour design by measuring the voltage Vab.\nWhat will be the new values of the currents i1 and i2.\nDiscuss your results and draw relevant conclusions\n\n69 | P a g e\n\n## EE-213 Lab Manual\n\nElectric Circuits\n\n## EE213 Design Experiment Sample\n\n3. Design a Scheme to Improving Load Power Factor\n\nThe utility company is supplying a static three phase Y connected load at 200 V line to\nneutral. Each phase is composed an impedance of 350 in series with 1.2 H inductor.\n\n## Figure 1 The Load Circuit Connection to Utility\n\nThe load nature presents a costly and unacceptable low lagging power factor to the power\ncompany. The load power factor has to be improved to 0.8 lagging or higher. It is required\nto evaluate the actual power factor and then to design a scheme to be installed for improving\nthe load power factor. The required design steps has to conducted then the hardwire design\nhas to be installed and tested. Practical results are required.\n\n70 | P a g e\n\n## EE-213 Lab Manual\n\nElectric Circuits\n\nAppendix II\nResistor Conversion Table\n\n71 | P a g e\n\n## EE-213 Lab Manual\n\nElectric Circuits\n\nAppendix III\nLABORATORY REGULATIONS AND SAFETY RULES\nThe following regulations and safety rules must be observed in all laboratory locations.\n1. It is the duty of all concerned who use any electrical laboratory to take all\nreasonable steps to safeguard the HEALTH and SAFETY of themselves and all\nother users and visitors.\n2. Be sure that all equipments are properly working before using for laboratory\nexercises. Any defective equipment must be reported immediately to the lab\ninstructors or lab technical staff.\n3. Students are allowed to use only the equipment provided in the experiment\nmanual or equipment used for senior project laboratory.\n4. Power supply terminals connected to any circuit are only energized in the\npresence of the instructor or lab staff.\n5. Students should keep a safe distance from the circuit breakers, electric circuits\nor any moving parts during the experiment.\n6. Avoid any body contact between energized circuits and ground.\n7. Switch off equipments and disconnect power supplies from the circuit before\nleaving the laboratory.\n8. Observe cleanliness and proper laboratory housekeeping of the equipment and\nother related accessories.\n9. Wear the proper clothes and safety gloves or goggles required in working areas\ninvolving fabrications of printed circuit boards, chemical process control\nsystems, antenna communication equipment and laser facility laboratories.\n10. Double check your circuit connections specifically in handling electrical power\nmachines, AC motors and generators before switching ON the power supply.\n11. Make sure that the last connection to be made in your circuit is the power supply\nand first thing to be disconnected is also the power supply.\n12. Equipment should not be removed or transferred to any location without\npermission from the laboratory staff.\n13. Software installation in any computer laboratory is not allowed without the\npermission from the laboratory staff.\n14. Computer games are strictly prohibited in the computer laboratory.\n15. Students are not allowed to use any equipment without proper orientation and\nactual hands-on equipment operation.\n16. Smoking, eating and drinking in the laboratory are prohibited.\nThe above rules and regulations are necessary precautions in the electrical laboratory to\nsafeguard the students, laboratory staff, the equipments and other laboratory users.\n\n72 | P a g e"
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https://web2.0calc.com/questions/geometry_65047 | [
"+0\n\n# Geometry\n\n0\n240\n1\n\nTriangle ABC is isosceles with AB = BC. If AC = 20 and [ABC] = $$280$$, then find the perimeter of triangle ABC.\n\nFeb 11, 2022\n\n#1\n-1\n\nSince triangle ABC is isosceles with AB = BC, then AC is the base of the triangle. If the area is 280, then base x height = 560, plugging in the values, 20 x height = 560. Thus, the height is 28.\n\nThe height of triangle ABC is also the perpendicular bisector of the triangle. Thus, the point that the height meets the base will be called point D. So AD = AC/2 = 10. Using the pythagorean theorem, AB^2 = AD^2 + BD^2, so AB^2 = 100 + 784 = 884.\n\nAB = $$2\\sqrt{221}$$. AB = BC so the perimeter of triangle ABC is $$20 + 4\\sqrt{221}$$.",
null,
"Feb 12, 2022"
] | [
null,
"https://web2.0calc.com/img/emoticons/smiley-smile.gif",
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.84018195,"math_prob":0.99997663,"size":643,"snap":"2023-40-2023-50","text_gpt3_token_len":198,"char_repetition_ratio":0.17840375,"word_repetition_ratio":0.031496063,"special_character_ratio":0.36547434,"punctuation_ratio":0.1388889,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9999976,"pos_list":[0,1,2],"im_url_duplicate_count":[null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-12-11T00:22:55Z\",\"WARC-Record-ID\":\"<urn:uuid:99ad883b-9044-472f-a1e3-08262918846f>\",\"Content-Length\":\"21169\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:881ad8ed-9192-4d1d-adf5-db0f9f687558>\",\"WARC-Concurrent-To\":\"<urn:uuid:29355973-3e62-4a6e-911e-bb8624756282>\",\"WARC-IP-Address\":\"168.119.251.111\",\"WARC-Target-URI\":\"https://web2.0calc.com/questions/geometry_65047\",\"WARC-Payload-Digest\":\"sha1:HPXSMZNQDGGQOKCMZA5L5WYP25CNY4B3\",\"WARC-Block-Digest\":\"sha1:ZTLC66QGGHLUUTCVBUPP32V5LLEZR4VO\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-50/CC-MAIN-2023-50_segments_1700679102697.89_warc_CC-MAIN-20231210221943-20231211011943-00504.warc.gz\"}"} |
https://www.geeksforgeeks.org/python-sympy-line-intersection-method/?ref=rp | [
"Related Articles\n\n# Python | Sympy Line.intersection() method\n\n• Last Updated : 10 Feb, 2020\nIn Sympy, the function `intersection()` is used to find the intersection with another geometrical entity.\n```Syntax: Line.intersection(o)\n\nParameters:\no: Point or LinearEntity\n\nReturns:\nintersection: list of geometrical entities\n```\n\nExample #1:\n\n `# import sympy and Point, Line``from` `sympy ``import` `Point, Line`` ` `p1, p2, p3 ``=` `Point(``0``, ``0``), Point(``1``, ``1``), Point(``7``, ``7``)``l1 ``=` `Line(p1, p2)`` ` `# using intersection() method``showIntersection ``=` `l1.intersection(p3)`` ` `print``(showIntersection)`\n\nOutput:\n\n`[Point2D(7, 7)]`\n\nExample #2:\n\n `# import sympy and Point, Line, Segment``from` `sympy ``import` `Point, Line, Segment`` ` `p1, p2, p3, p4 ``=` `Point(``0``, ``0``), Point(``1``, ``1``), Point(``0``, ``5``), Point(``2``, ``6``)``l1 ``=` `Line(p1, p2)``s1 ``=` `Segment(p3, p4)`` ` ` ` `# using intersection() method``showIntersection ``=` `l1.intersection(s1)`` ` `print``(showIntersection)`\n\nOutput:\n\n`[]`\n\nAttention geek! Strengthen your foundations with the Python Programming Foundation Course and learn the basics.\n\nTo begin with, your interview preparations Enhance your Data Structures concepts with the Python DS Course. And to begin with your Machine Learning Journey, join the Machine Learning – Basic Level Course\n\nMy Personal Notes arrow_drop_up"
] | [
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https://zixun.changingedu.com/shagnhai/32-14364.html | [
"## 小学奥数数论问题:完全平方数的性质及推论\n\n小学奥数数论问题:完全平方数的性质及推论\n\n一个数如果是另一个整数的完全平方,那么我们就称这个数为完全平方数,也叫做平方数。\n\n例如:0,1,4,9,16,25,36,49,64,81,100,121,144,169,196,225,256,289,324,361,400,441,484,… 观察这些完全平方数,可以获得对它们的个位数、十位数、数字和等的规律性的认识。下面我们来研究完全平方数的一些常用性质:\n\n性质1:完全平方数的末位数只能是0,1,4,5,6,9。\n\n性质2:奇数的平方的个位数字为奇数,十位数字为偶数。\n\n证明 奇数必为下列五种形式之一:\n\n10a+1, 10a+3, 10a+5, 10a+7, 10a+9\n\n分别平方后,得\n\n(10a+1)^2=100a^2+20a+1=20a(5a+1)+1\n\n(10a+3)^2=100a^2+60a+9=20a(5a+3)+9\n\n(10a+5)^2=100a^2+100a+25=20 (5a+5a+1)+5\n\n(10a+7)^2=100a^2+140a+49=20 (5a+7a+2)+9\n\n(10a+9)^2=100a^2+180a+81=20 (5a+9a+4)+1\n\n综上各种情形可知:奇数的平方,个位数字为奇数1,5,9;十位数字为偶数。\n\n性质3:如果完全平方数的十位数字是奇数,则它的个位数字一定是6;反之,如果完全平方数的个位数字是6,则它的十位数字一定是奇数。\n\n证明 已知m^2=10k+6,证明k为奇数。因为的个位数为6,所以m的个位数为4或6,于是可设m=10n+4或10n+6。\n\n则 10k+6=(10n+4)^2=100+(8n+1)x10+6\n\n或 10k+6=(10n+6)^2=100+(12n+3)x10+6\n\n即 k=10+8n+1=2(5+4n)+1\n\n或 k=10+12n+3=2(5+6n)+3\n\n∴ k为奇数。\n\n推论1:如果一个数的十位数字是奇数,而个位数字不是6,那么这个数一定不是完全平方数。\n\n推论2:如果一个完全平方数的个位数字不是6,则它的十位数字是偶数。\n\n性质4:偶数的平方是4的倍数;奇数的平方是4的倍数加1。\n\n这是因为 (2k+1)=4k(k+1)+1 (2k)=4\n\n性质5:奇数的平方是8n+1型;偶数的平方为8n或8n+4型。\n\n在性质4的证明中,由k(k+1)一定为偶数可得到(2k+1)是8n+1型的数;由为奇数或偶数可得(2k)为8n型或8n+4型的数。\n\n性质6:平方数的形式必为下列两种之一:3k,3k+1。\n\n因为自然数被3除按余数的不同可以分为三类:3m,3m+1, 3m+2。平方后,分别得\n\n(3m)=9=3k\n\n(3m+1)=9+6m+1=3k+1\n\n(3m+2)=9+12m+4=3k+1\n\n同理可以得到:\n\n性质7:不能被5整除的数的平方为5k±1型,能被5整除的数的平方为5k型。\n\n性质8:平方数的形式具有下列形式之一:16m,16m+1, 16m+4,16m+9。\n\n除了上面关于个位数,十位数和余数的性质之外,还可研究完全平方数各位数字之和。例如,256它的各位数字相加为2+5+6=13,13叫做256的 各位数字和。如果再把13的各位数字相加:1+3=4,4也可以叫做256的各位数字的和。下面我们提到的一个数的各位数字之和是指把它的各位数字相加, 如果得到的数字之和不是一位数,就把所得的数字再相加,直到成为一位数为止。我们可以得到下面的命题: 一个数的数字和等于这个数被9除的余数。\n\n下面以四位数为例来说明这个命题。\n\n设四位数为,则\n\n= 1000a+100b+10c+d\n\n= 999a+99b+9c+(a+b+c+d)\n\n= 9(111a+11b+c)+(a+b+c+d)\n\n显然,a+b+c+d是四位数被9除的余数。\n\n对於n位数,也可以仿此法予以证明。\n\n关於完全平方数的数字和有下面的性质:\n\n性质9:完全平方数的数字之和只能是0,1,4,7,9。\n\n证明 因为一个整数被9除只能是9k,9k±1, 9k±2, 9k±3, 9k±4这几种形式,而\n\n(9k)=9(9)+0\n\n(9k±1)=9(9±2k)+1\n\n(9k±2)=9(9±4k)+4\n\n(9k±3)=9(9±6k)+9\n\n(9k±4)=9(9±8k+1)+7\n\n除了以上几条性质以外,还有下列重要性质:\n\n性质10:为完全平方数的充要条件是b为完全平方数。\n\n性质11:如果质数p能整除a,但p的平方不能整除a,则a不是完全平方数。\n\n证明 由题设可知,a有质因数p,但无因数,可知a分解成标准式时,p的次方为1,而完全平方数分解成标准式时,各质因数的次方均为偶数,可见a不是完全平方数。\n\n性质12:在两个相邻的整数的平方数之间的所有整数都不是完全平方数,即若 n^2 < k^2 < (n+1)^2 则k一定不是整数。\n\n性质13:一个正整数n是完全平方数的充分必要条件是n有奇数个因数(包括1和n本身)。\n\n## 选择城市\n\n#### 轻轻助手",
null,
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https://courses.media.mit.edu/2004fall/mas622j/matlab/ | [
"# MAS 622j matlab help\n\nOriginally written by Tom Minka, and modified by Yuan Qi and Ashish Kapoor Sept. 2002\n\nMatlab on Athena . Matlab variables . Matlab language . Plotting Examples . Useful Subroutines\n\n### Getting started\n\nOpen an editor window next to your matlab window. You'll often find yourself mousing text in the editor window and pasting it into the matlab window.\n\nIf you're a complete matlab novice, type \"intro\" or, if you're real fond of splashy colors, type \"demo\". You can also read help from matlab help menu.\n\nYou can also type \"helpwin\" or \"helpdesk\" to get a help window. The matlab help command is \"help\". Try \"help general\" or just plain \"help\". The apropos command is \"lookfor\". Try, e.g. \"lookfor tangent\" or \"lookfor random\".\n\n### Syntax\n\nThe matlab continuation code is \"...\" i.e. if you have a long formula that is several lines long, end each line with ... and continue the formula on the next line.\n\nThe matlab comment character is \"%\". The semicolon \";\" is useful too; it makes the command on that line operate silently. For example, \"A = B;\" copies matrix B into matrix A. \"A = B\" does the same thing, but prints out the whole contents of matrix B while copying.\n\n### Data structures\n\nThe most useful data strucutre in Matlab is matrix. Scalars is considered 1 by 1 matrices, and strings (delimited by 'single quotes') are considered vectors (which in turn are just skinny matrices). Check out\n` sprintf sscanf num2str int2str`\nMatlab5 and 6 added the very useful \"cell array\" and \"struct\" data types. A cell array is just like a matrix except each entry can be any data type, not just a number. For example:\n```\n>> c = {'joe' 5 [1 2 3]}\n\nc =\n\n'joe' [1x3 double]\n\n>> c{3}\n\nans =\n\n1 2 3```\nA struct is similar except its contents are addressable by name only.\n```>> s = struct('name', 'joe', 'age', 30)\n\ns =\n\nname: 'joe'\nage: 30\n\n>> s.name\n\nans =\n\njoe\n\n>> s.age\n\nans =\n\n30```\n\n### Plotting and I/O\n\nMatlab is very good at plotting your data. See our plotting examples and/or get matlab's help on:\n``` plot grid hold drawnow axis axes orient\nsubplot mesh meshgrid plot3 rotate3d```\nThen you'll want to print your plot to a file. To print the current plot to a postscript file, type\n``` orient tall %% this line is optional\nprint -deps myfilename.ps```\nYou can use C style file I/O. Get help on commands\n` fopen fclose fscanf fprintf printf`\nThe matlab parser is not too clever, so don't put \"-\" in a filename; poor matlab will think you want to subtract!\n\nMatlab has a native data file format, plus it can save and load data from ASCII files. See the stock answers and check out\n\n``` save filename.dat -ascii"
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https://www.xloypaypa.pub/codeforces-round-756-robot-and-candies-solution-java-c/ | [
"## Solution:\n\nFirst of all, we can immediately find that for the grid (x, y), no matter how the robot moves, the parity of the grid x+y is always the same. So it is natural to solve for odd and even numbers separately.\n\nThen we consider how the robot can move in order to minimize the number of times. It is conceivable that the robot must take it in order from left to right.\nTake the following picture as an example, we can see that the robot takes the (3,3) candy first. But as a result, the candy on the left (3,1) must be taken again. This will take 3 times.\n\nTherefore, the strategy in the above figure should become (only 2 times):\n\nSo we came to the conclusion that it must be taken from left to right. Now let us take the following figure as an example, consider the following question: Compared with (1, 6), are the green 1 and red 1 both on the left side of (1, 6)?\n\nStarting from the yellow 1, we can see that the green 1 is actually more \"left\", while the red 1 is actually to the right of the yellow 1. As shown below:\n\nTherefore, according to the conclusion taken from left to right, we define right_top=y+x. So we only need to sort according to right_top to get the points sorted from left to right. That is to say, the smaller the right_top, the more it should be taken away first.\n\nThe final question is how many times it needs to take all candies according to this strategy. We found that in the above picture, the green 1 needs to be taken at one time, while the red 1 and yellow 1 can be taken at one time. Then we consider under when we can take candies together.\nWe define left_top=y-x. So we found that if and only if as x increases, right_top also increases, and left_top decreases, they can be taken together.\n\nAt the same time, we noticed that the larger the right_top-left_top, the larger the x. Therefore, as right_top increases, if left_top decreases, then this 1 must look like the red 1 in the above figure. And if left_top not only does not decrease or even increases, it must fall to the right of green 1, and cannot be taken together.\n\nSo, as right_top continues to move to the right (because we sort from small to large), we only need to check whether there is another 1 on the right side of its left_top each time. If there is, then the two 1s are taken together.\nAnd the x of the 1 on the right must be smaller than the x of the current 1. Therefore, we adjust the left_top of the original 1 to the new left_top so that the subsequent 1s can continue processing.\n\n## Code:\n\nJava\n\nC++",
null,
"蜀ICP备19018968号"
] | [
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https://docs.pyro.ai/en/dev/_modules/pyro/distributions/transforms/batchnorm.html | [
"# Source code for pyro.distributions.transforms.batchnorm\n\n# Copyright (c) 2017-2019 Uber Technologies, Inc.\n\nimport torch\nimport torch.nn as nn\nimport torch.nn.functional as F\n\nfrom .. import constraints\nfrom ..torch_transform import TransformModule\nfrom ..util import copy_docs_from\n\n[docs]@copy_docs_from(TransformModule)\nclass BatchNorm(TransformModule):\nr\"\"\"\nA type of batch normalization that can be used to stabilize training in\nnormalizing flows. The inverse operation is defined as\n\n:math:x = (y - \\hat{\\mu}) \\oslash \\sqrt{\\hat{\\sigma^2}} \\otimes \\gamma + \\beta\n\nthat is, the standard batch norm equation, where :math:x is the input,\n:math:y is the output, :math:\\gamma,\\beta are learnable parameters, and\n:math:\\hat{\\mu}/:math:\\hat{\\sigma^2} are smoothed running averages of\nthe sample mean and variance, respectively. The constraint :math:\\gamma>0 is\nenforced to ease calculation of the log-det-Jacobian term.\n\nThis is an element-wise transform, and when applied to a vector, learns two\nparameters (:math:\\gamma,\\beta) for each dimension of the input.\n\nWhen the module is set to training mode, the moving averages of the sample mean\nand variance are updated every time the inverse operator is called, e.g., when a\nnormalizing flow scores a minibatch with the log_prob method.\n\nAlso, when the module is set to training mode, the sample mean and variance on\nthe current minibatch are used in place of the smoothed averages,\n:math:\\hat{\\mu} and :math:\\hat{\\sigma^2}, for the inverse operator. For\nthis reason it is not the case that :math:x=g(g^{-1}(x)) during training,\ni.e., that the inverse operation is the inverse of the forward one.\n\nExample usage:\n\n>>> from pyro.nn import AutoRegressiveNN\n>>> from pyro.distributions.transforms import AffineAutoregressive\n>>> base_dist = dist.Normal(torch.zeros(10), torch.ones(10))\n>>> iafs = [AffineAutoregressive(AutoRegressiveNN(10, )) for _ in range(2)]\n>>> bn = BatchNorm(10)\n>>> flow_dist = dist.TransformedDistribution(base_dist, [iafs, bn, iafs])\n>>> flow_dist.sample() # doctest: +SKIP\n\n:param input_dim: the dimension of the input\n:type input_dim: int\n:param momentum: momentum parameter for updating moving averages\n:type momentum: float\n:param epsilon: small number to add to variances to ensure numerical stability\n:type epsilon: float\n\nReferences:\n\n Sergey Ioffe and Christian Szegedy. Batch Normalization: Accelerating Deep\nNetwork Training by Reducing Internal Covariate Shift. In International\nConference on Machine Learning, 2015. https://arxiv.org/abs/1502.03167\n\n Laurent Dinh, Jascha Sohl-Dickstein, and Samy Bengio. Density Estimation\nusing Real NVP. In International Conference on Learning Representations, 2017.\nhttps://arxiv.org/abs/1605.08803\n\n George Papamakarios, Theo Pavlakou, and Iain Murray. Masked Autoregressive\nFlow for Density Estimation. In Neural Information Processing Systems, 2017.\nhttps://arxiv.org/abs/1705.07057\n\n\"\"\"\n\ndomain = constraints.real\ncodomain = constraints.real\nbijective = True\n\ndef __init__(self, input_dim, momentum=0.1, epsilon=1e-5):\nsuper().__init__()\n\nself.input_dim = input_dim\nself.gamma = nn.Parameter(torch.ones(input_dim))\nself.beta = nn.Parameter(torch.zeros(input_dim))\nself.momentum = momentum\nself.epsilon = epsilon\n\nself.register_buffer(\"moving_mean\", torch.zeros(input_dim))\nself.register_buffer(\"moving_variance\", torch.ones(input_dim))\n\n@property\ndef constrained_gamma(self):\nreturn F.relu(self.gamma) + 1e-6\n\ndef _call(self, x):\n\"\"\"\n:param x: the input into the bijection\n:type x: torch.Tensor\n\nInvokes the bijection x=>y; in the prototypical context of a\n:class:~pyro.distributions.TransformedDistribution x is a sample from\nthe base distribution (or the output of a previous transform)\n\"\"\"\n# Enforcing the constraint that gamma is positive\nreturn (x - self.beta) / self.constrained_gamma * torch.sqrt(\nself.moving_variance + self.epsilon\n) + self.moving_mean\n\ndef _inverse(self, y):\n\"\"\"\n:param y: the output of the bijection\n:type y: torch.Tensor\n\nInverts y => x.\n\"\"\"\n# During training, keep smoothed average of sample mean and variance\nif self.training:\nmean, var = y.mean(0), y.var(0)\n\n# NOTE: The momentum variable agrees with the definition in e.g. torch.nn.BatchNorm1d\n\n# During test time, use smoothed averages rather than the sample ones\nelse:\nmean, var = self.moving_mean, self.moving_variance\n\nreturn (y - mean) * self.constrained_gamma / torch.sqrt(\nvar + self.epsilon\n) + self.beta\n\n[docs] def log_abs_det_jacobian(self, x, y):\n\"\"\"\nCalculates the elementwise determinant of the log Jacobian, dx/dy\n\"\"\"\nif self.training:\nvar = torch.var(y, dim=0, keepdim=True)\nelse:\n# NOTE: You wouldn't typically run this function in eval mode, but included for gradient tests\nvar = self.moving_variance\nreturn -self.constrained_gamma.log() + 0.5 * torch.log(var + self.epsilon)\n\n[docs]def batchnorm(input_dim, **kwargs):\n\"\"\"\nA helper function to create a :class:~pyro.distributions.transforms.BatchNorm\nobject for consistency with other helpers.\n\n:param input_dim: Dimension of input variable\n:type input_dim: int\n:param momentum: momentum parameter for updating moving averages\n:type momentum: float\n:param epsilon: small number to add to variances to ensure numerical stability\n:type epsilon: float\n\n\"\"\"\nbn = BatchNorm(input_dim, **kwargs)\nreturn bn"
] | [
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.60462946,"math_prob":0.951634,"size":5434,"snap":"2023-14-2023-23","text_gpt3_token_len":1378,"char_repetition_ratio":0.12302026,"word_repetition_ratio":0.09425626,"special_character_ratio":0.25929335,"punctuation_ratio":0.23580366,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9983713,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-06-05T07:16:06Z\",\"WARC-Record-ID\":\"<urn:uuid:0912a42d-c9c3-495c-a172-bb4a618923fa>\",\"Content-Length\":\"29243\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:7a67597a-76ac-4d37-92fa-e3a087a45330>\",\"WARC-Concurrent-To\":\"<urn:uuid:bcffcffe-204f-47f4-98a0-c554fd4e4917>\",\"WARC-IP-Address\":\"104.17.33.82\",\"WARC-Target-URI\":\"https://docs.pyro.ai/en/dev/_modules/pyro/distributions/transforms/batchnorm.html\",\"WARC-Payload-Digest\":\"sha1:537373DHPIWYDR7RWIWVB2UZ4Q3Z6EKF\",\"WARC-Block-Digest\":\"sha1:GSEAK7PV4JLZD6SB4AZJTYWZOGIG7BCC\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-23/CC-MAIN-2023-23_segments_1685224651325.38_warc_CC-MAIN-20230605053432-20230605083432-00779.warc.gz\"}"} |
https://www.colorhexa.com/31de53 | [
"#31de53 Color Information\n\nIn a RGB color space, hex #31de53 is composed of 19.2% red, 87.1% green and 32.5% blue. Whereas in a CMYK color space, it is composed of 77.9% cyan, 0% magenta, 62.6% yellow and 12.9% black. It has a hue angle of 131.8 degrees, a saturation of 72.4% and a lightness of 53.1%. #31de53 color hex could be obtained by blending #62ffa6 with #00bd00. Closest websafe color is: #33cc66.\n\n• R 19\n• G 87\n• B 33\nRGB color chart\n• C 78\n• M 0\n• Y 63\n• K 13\nCMYK color chart\n\n#31de53 color description : Bright lime green.\n\n#31de53 Color Conversion\n\nThe hexadecimal color #31de53 has RGB values of R:49, G:222, B:83 and CMYK values of C:0.78, M:0, Y:0.63, K:0.13. Its decimal value is 3268179.\n\nHex triplet RGB Decimal 31de53 `#31de53` 49, 222, 83 `rgb(49,222,83)` 19.2, 87.1, 32.5 `rgb(19.2%,87.1%,32.5%)` 78, 0, 63, 13 131.8°, 72.4, 53.1 `hsl(131.8,72.4%,53.1%)` 131.8°, 77.9, 87.1 33cc66 `#33cc66`\nCIE-LAB 78.179, -69.541, 54.71 28.947, 53.517, 16.987 0.291, 0.538, 53.517 78.179, 88.483, 141.807 78.179, -67.746, 78.608 73.155, -57.39, 37.441 00110001, 11011110, 01010011\n\nColor Schemes with #31de53\n\n• #31de53\n``#31de53` `rgb(49,222,83)``\n• #de31bc\n``#de31bc` `rgb(222,49,188)``\nComplementary Color\n• #66de31\n``#66de31` `rgb(102,222,49)``\n• #31de53\n``#31de53` `rgb(49,222,83)``\n• #31deaa\n``#31deaa` `rgb(49,222,170)``\nAnalogous Color\n• #de3166\n``#de3166` `rgb(222,49,102)``\n• #31de53\n``#31de53` `rgb(49,222,83)``\n• #aa31de\n``#aa31de` `rgb(170,49,222)``\nSplit Complementary Color\n• #de5331\n``#de5331` `rgb(222,83,49)``\n• #31de53\n``#31de53` `rgb(49,222,83)``\n• #5331de\n``#5331de` `rgb(83,49,222)``\n• #bcde31\n``#bcde31` `rgb(188,222,49)``\n• #31de53\n``#31de53` `rgb(49,222,83)``\n• #5331de\n``#5331de` `rgb(83,49,222)``\n• #de31bc\n``#de31bc` `rgb(222,49,188)``\n• #1ba837\n``#1ba837` `rgb(27,168,55)``\n• #1ebe3e\n``#1ebe3e` `rgb(30,190,62)``\n• #22d445\n``#22d445` `rgb(34,212,69)``\n• #31de53\n``#31de53` `rgb(49,222,83)``\n• #47e265\n``#47e265` `rgb(71,226,101)``\n• #5de578\n``#5de578` `rgb(93,229,120)``\n• #73e98a\n``#73e98a` `rgb(115,233,138)``\nMonochromatic Color\n\nAlternatives to #31de53\n\nBelow, you can see some colors close to #31de53. Having a set of related colors can be useful if you need an inspirational alternative to your original color choice.\n\n``#3ade31` `rgb(58,222,49)``\n• #31de36\n``#31de36` `rgb(49,222,54)``\n• #31de45\n``#31de45` `rgb(49,222,69)``\n• #31de53\n``#31de53` `rgb(49,222,83)``\n• #31de61\n``#31de61` `rgb(49,222,97)``\n• #31de70\n``#31de70` `rgb(49,222,112)``\n• #31de7e\n``#31de7e` `rgb(49,222,126)``\nSimilar Colors\n\n#31de53 Preview\n\nThis text has a font color of #31de53.\n\n``<span style=\"color:#31de53;\">Text here</span>``\n#31de53 background color\n\nThis paragraph has a background color of #31de53.\n\n``<p style=\"background-color:#31de53;\">Content here</p>``\n#31de53 border color\n\nThis element has a border color of #31de53.\n\n``<div style=\"border:1px solid #31de53;\">Content here</div>``\nCSS codes\n``.text {color:#31de53;}``\n``.background {background-color:#31de53;}``\n``.border {border:1px solid #31de53;}``\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, #020e04 is the darkest color, while #fcfffc is the lightest one.\n\n• #020e04\n``#020e04` `rgb(2,14,4)``\n• #051f0a\n``#051f0a` `rgb(5,31,10)``\n• #08300f\n``#08300f` `rgb(8,48,15)``\n• #0a4115\n``#0a4115` `rgb(10,65,21)``\n• #0d511a\n``#0d511a` `rgb(13,81,26)``\n• #106220\n``#106220` `rgb(16,98,32)``\n• #127325\n``#127325` `rgb(18,115,37)``\n• #15842b\n``#15842b` `rgb(21,132,43)``\n• #189530\n``#189530` `rgb(24,149,48)``\n• #1ba636\n``#1ba636` `rgb(27,166,54)``\n• #1db73b\n``#1db73b` `rgb(29,183,59)``\n• #20c841\n``#20c841` `rgb(32,200,65)``\n• #23d946\n``#23d946` `rgb(35,217,70)``\n• #31de53\n``#31de53` `rgb(49,222,83)``\n• #42e161\n``#42e161` `rgb(66,225,97)``\n• #53e36f\n``#53e36f` `rgb(83,227,111)``\n• #64e67d\n``#64e67d` `rgb(100,230,125)``\n• #75e98b\n``#75e98b` `rgb(117,233,139)``\n• #86ec9a\n``#86ec9a` `rgb(134,236,154)``\n• #96eea8\n``#96eea8` `rgb(150,238,168)``\n• #a7f1b6\n``#a7f1b6` `rgb(167,241,182)``\n• #b8f4c4\n``#b8f4c4` `rgb(184,244,196)``\n• #c9f6d2\n``#c9f6d2` `rgb(201,246,210)``\n• #daf9e0\n``#daf9e0` `rgb(218,249,224)``\n• #ebfcee\n``#ebfcee` `rgb(235,252,238)``\n• #fcfffc\n``#fcfffc` `rgb(252,255,252)``\nTint Color Variation\n\nTones of #31de53\n\nA tone is produced by adding gray to any pure hue. In this case, #848b85 is the less saturated color, while #15fa42 is the most saturated one.\n\n• #848b85\n``#848b85` `rgb(132,139,133)``\n• #7b9480\n``#7b9480` `rgb(123,148,128)``\n• #719e7a\n``#719e7a` `rgb(113,158,122)``\n• #68a774\n``#68a774` `rgb(104,167,116)``\n• #5fb06f\n``#5fb06f` `rgb(95,176,111)``\n• #56b969\n``#56b969` `rgb(86,185,105)``\n• #4dc264\n``#4dc264` `rgb(77,194,100)``\n• #43cc5e\n``#43cc5e` `rgb(67,204,94)``\n``#3ad559` `rgb(58,213,89)``\n• #31de53\n``#31de53` `rgb(49,222,83)``\n• #28e74d\n``#28e74d` `rgb(40,231,77)``\n• #1ff048\n``#1ff048` `rgb(31,240,72)``\n• #15fa42\n``#15fa42` `rgb(21,250,66)``\nTone Color Variation\n\nColor Blindness Simulator\n\nBelow, you can see how #31de53 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":0.52738714,"math_prob":0.826696,"size":3707,"snap":"2019-26-2019-30","text_gpt3_token_len":1633,"char_repetition_ratio":0.120712936,"word_repetition_ratio":0.011090573,"special_character_ratio":0.5597518,"punctuation_ratio":0.23783186,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9931343,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-06-17T17:07:08Z\",\"WARC-Record-ID\":\"<urn:uuid:322fcdf2-d34f-444b-9d92-aefa725c133b>\",\"Content-Length\":\"36432\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:4111e068-d27d-41f3-be96-842736ec1968>\",\"WARC-Concurrent-To\":\"<urn:uuid:4fea468c-00d4-4762-9e30-4918d3179183>\",\"WARC-IP-Address\":\"178.32.117.56\",\"WARC-Target-URI\":\"https://www.colorhexa.com/31de53\",\"WARC-Payload-Digest\":\"sha1:ZJVHFQWDS6TMP4AQHUY5FWDYHCENGWSH\",\"WARC-Block-Digest\":\"sha1:OSEGPZN7ACACDU2TS5TSROVAKBEXGR4I\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-26/CC-MAIN-2019-26_segments_1560627998513.14_warc_CC-MAIN-20190617163111-20190617185111-00519.warc.gz\"}"} |
https://math.stackexchange.com/questions/288298/symmetric-random-walk-with-bounds | [
"# Symmetric random walk with bounds\n\ncan anyone help me with this: We are considering a symmetric random walk that ends if level 3 is reached or level -1 is reached. Start=0\n\nWhat is the expected number of walks? So I am looking for: $E[{\\tau}]$ with $\\tau$=the stopping time.\n\n• What are your thoughts on this question? – Did Jan 27 '13 at 20:18\n• I find it difficult to have 2 bounds... – user59871 Jan 27 '13 at 20:21\n• (Actually the question is easier with what you call two bounds than with only one.) What did you try? Which similar problems can you solve? – Did Jan 27 '13 at 20:35\n• I can solve the one with 1 bound, which would yield: $E[\\tau]=infinity$. I tried a similar technique but do not know how to incorporate the 2 bounds... (optional sampling?) – user59871 Jan 27 '13 at 20:39\n• Show how you solve the one with 1 bound. – Did Jan 27 '13 at 20:47\n\nBasic argument: One asks for $t_0=\\mathbb E_0(\\tau)$ where $t_x=\\mathbb E_x(\\tau)$ for every $-1\\leqslant x\\leqslant3$. By the (simple) Markov property after one step, $t_x=1+\\frac12(t_{x-1}+t_{x+1})$ for every $0\\leqslant x\\leqslant2$. By definition, $t_{-1}=t_3=0$. This is an affine system of $5$ equations with $5$ unknowns. Solve it. This yields $t_x=(3-x)(x+1)$ for every $-1\\leqslant x\\leqslant3$. In particular $t_0=3$.\n\nLess basic argument: For every $n\\geqslant0$, let $x_n$ denote the position after $n$ steps, $\\mathfrak X_n$ the sigma-algebra generated by $(x_k)_{0\\leqslant k\\leqslant n}$, and $z_n=(x_n+1)(3-x_n)+n$. Then $z_0=3$, $z_\\tau=\\tau$, and $(z_n)_{n\\geqslant0}$ is a martingale with respect to the filtration $(\\mathfrak X_n)_{n\\geqslant0}$, hence $\\mathbb E(z_0)=\\mathbb E(z_\\tau)$, that is, $\\mathbb E(\\tau)=3$.\n\nIn full generality, the same argument shows that, for every $(a,b)$, the first hitting time of $\\{a,b\\}$ starting from $a\\leqslant x\\leqslant b$ has mean $(b-x)(x-a)$.\n\n• I have a question regarding your basic argument. Why by definition does $u_{-1}=u_{3}=0$? If we are trying to calculated the expected first hitting time for either boundary why do we computer $t_{x}$ at $x=0$? Is this because we assume the process starts at $x=0$? – user75514 Apr 7 '17 at 18:13\n• @user75514 Yes, $\\mathbb E_x$ refers to the random walk starting at $x$. – Did Apr 7 '17 at 18:20\n• So are $u_{-1}$ and $u_{3}$ the expected first arrival at -1 and 3? Why are these 0? – user75514 Apr 7 '17 at 18:23\n• @user75514 Because $\\tau=\\inf\\{n\\geqslant 0\\mid x_n\\in\\{-1,3\\}\\}$ is such that $P_{-1}(\\tau=0)=P_3(\\tau=0)=1$ hence $t_{-1}=E_{-1}(\\tau)=0$ and $t_3=E_3(\\tau)=0$. (Nota: $u_x$ is a typo for $t_x$.) – Did Apr 7 '17 at 18:27\n• @Did In order to use the optimal stopping theorem (I guess this is what you've used here), we need to show that $\\tau$ is a.s. bounded. However, here, we only have that $\\mathrm{P}[\\tau < \\infty] = 1$. I think you need to modify this argument to guarantee that the stopping time is a.s. bounded. – Pantelis Sopasakis Jan 24 '19 at 12:41\n\n$$\\newcommand{\\Z}{\\mathbb{Z}}\\newcommand{\\N}{\\mathbb{N}}\\newcommand{\\E}{\\mathbb{E}}\\newcommand{\\F}{\\mathcal{F}}\\newcommand{\\P}{\\mathrm{P}}$$As @Did pointed out in a comment above, the optional stopping theorem (OST) holds under relaxed uniform integrability assumptions. I was unaware of this more general statement, so I used the more classical statement of OST which requires $$\\tau$$ to be a.s. bounded to compute the expected stopping time. I found these lecture notes and this book where the OST is stated under UI conditions (and the proof is not too difficult).\n\nIn any case, I had already derived the result using the version of the OST I knew, so here it is:\n\nThe a.s. boundedness assumption often fails to hold in practice. However, for every fixed $$N\\in\\N$$, $$\\tau \\wedge N$$ is a bounded stopping time. The idea is to apply OST for $$\\tau \\wedge N$$ and take $$N\\to\\infty$$.\n\nProblem. Let $$(X_n)_n$$ be the standard symmetric random walk with $$X_0=x$$ and let $$a < x < b$$ with $$a,b\\in\\Z$$. Define $$\\tau = \\inf\\{n\\in\\N {}\\mid{} X_n \\in \\{a,b\\}\\}$$. Find $$\\E[\\tau]$$.\n\nStep 1. We will show that $$Y_n = X_n^2 - n$$ is a martingale with respect to the natural filtration of $$X_n$$, $$(\\F_n)_n$$. Indeed,\n\n1. $$Y_n$$ is $$\\F_n$$-measurable\n2. Since $$\\E[|X_n|]\\leq |x| + n$$, we have $$\\E[Y_n] = \\E[X_n^2 - n] \\leq (|x|+n)^2 < \\infty$$\n3. We have $$Y_{n+1} = Y_n + W_n$$, and \\begin{align} \\E[Y_{n+1}\\mid \\F_n] {}={}& \\E[X_{n+1}^2-n-1 {}\\mid{} \\F_n]\\\\ {}={}& \\E[(X_{n}+W_{n+1})^2-n-1 {}\\mid{} \\F_n]\\\\ {}={}& \\E[X_{n}^2+W_{n+1}^2+2X_nW_{n+1}-n-1 {}\\mid{} \\F_n]\\\\ {}={}& X_{n}^2+\\E[W_{n+1}^2\\mid \\F_n] +2X_n\\E[W_{n+1}\\mid \\F_n]-n-1\\\\ {}={}& X_{n}^2-n=Y_n\\\\ \\end{align}\n\nStep 2. Let $$N\\in\\N$$. Since $$Y_n$$ is an $$\\F_n$$-martingale, $$\\tau\\wedge N$$ is a bounded $$\\F_n$$-measurable stopping time (indeed, $$\\tau\\wedge N \\leq N$$), we may employ OST: \\begin{align} {}&{}\\E[Y_{0}] = \\E[Y_{\\tau \\wedge N}]\\\\ \\Leftrightarrow{}&{} x^2 {}={} \\E[X_{\\tau \\wedge N}^2 - \\tau\\wedge N]\\\\ \\Leftrightarrow{}&{} x^2 {}={} \\E[X_{\\tau \\wedge N}^2] - \\E[\\tau\\wedge N] \\tag{1} \\end{align}\n\nStep 3. Note that $$\\tau\\wedge N \\leq \\tau \\wedge (N+1)$$, therefore, by virtue of Lebesgue's monotone convergence theorem, \\begin{align} \\lim_{N\\to\\infty}\\E[\\tau \\wedge N] = \\E[\\lim_{N\\to\\infty} \\tau \\wedge N] = \\E[\\tau]\\tag{2} \\end{align}\n\nStep 4. We need to compute $$\\lim_n \\E[X_{\\tau \\wedge N}^2]$$. We will use a known fact for random walks: for $$x,m\\in \\N$$ let $$\\tau(m) = \\inf\\{n \\in \\N {}\\mid{} X_n = m, \\text{with }X_0 = x\\}$$. Then, for $$a,\n\n\\begin{align} \\P[\\tau(b,x) < \\tau(a, x)] = \\frac{x-a}{b-a}.\\tag{3} \\end{align}\n\nThis is easy to prove. We also have that\n\n\\begin{align} \\E[X_{\\tau\\wedge N}^2] = \\E[X_{\\tau}^2 1_{\\tau < N}] + \\E[X_{N}^2 1_{\\tau \\geq N}]\\tag{4} \\end{align}\n\nFor the second term in (4) we have that $$0 \\leq \\E[X_N^2 1_{\\tau \\geq N}] \\leq \\max\\{a^2, b^2\\} \\E[1_{\\tau \\geq N}] \\to 0$$ as $$N\\to\\infty$$.\n\nFor the first term, we have\n\n\\begin{align} \\lim_n \\E[X_{\\tau}^2 1_{\\tau < N}] {}={}& a^2 \\P[\\tau(a, x) < \\tau(b,x )] + b^2 \\P[\\tau(b, x) < \\tau(a,x )]\\\\ {}={}& a^2 \\left(1 - \\frac{x-a}{b-a}\\right) + b^2 \\frac{x-a}{b-a}\\\\ {}={}& x(a+b) - ab\\tag{5} \\end{align}\n\nTherefore, from all the above we have\n\n\\begin{align} \\E[\\tau] = (b-x)(x-a)\\tag{*} \\end{align}\n\n• For the record, using $\\tau\\wedge n$ and letting $n$ go to infinity is the classical way to prove the most classical version of the OST (which, sorry to say, is definitely not the version for almost surely bounded stopping times, since this restricted version is almost never usable). – Did Jan 25 '19 at 12:24\n\nI asked a similar question to poster and was directed here, but found the level of complexity of the arguments very difficult to sift through for my level of math experience.\n\nIt turns out the answer is incredibly simple - I was shocked.\n\nFor anyone who just needs the solution for application purposes, the answer is A*B, where A is the upper bound and B is the lower bound.\n\nQuote from wikipedia: If a and b are positive integers, then the expected number of steps until a one-dimensional simple random walk starting at 0 first hits b or −a is ab. The probability that this walk will hit b before −a is a/(a+b), which can be derived from the fact that simple random walk is a martingale.\n\n• As explained explicitely for roughly three years in the other answer. Why the shock then? – Did Apr 7 '17 at 18:21"
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https://www.manualslib.com/manual/993135/Hp-Nw280-200x.html?page=249 | [
"# Linear Solver App; Getting Started With The Linear Solver App - HP NW280-200X User Manual\n\nPrime graphing calculator.\n\n## Linear Solver app\n\n### Getting started with the Linear Solver app\n\nOpen the Linear\nSolver app\nLinear Solver app\nThe Linear Solver app enables you to solve a set of linear\nequations. The set can contain two or three linear\nequations.\nIn a two-equation set, each equation must be in the form\n. In a three-equation set, each equation must\nax\n+\nby\n=\nk\nbe in the form\nax\n+\nYou provide values for a, b, and k (and c in three-equation\nsets) for each equation, and the app will attempt to solve\nfor x and y (and z in three-equation sets).\nThe HP Prime will alert you if no solution can be found, or\nif there is an infinite number of solutions.\nThe following example defines the following set of\nequations and then solves for the unknown variables:\n6x\n+\n9y\n+\n6z\n=\n7x\n+\n10y\n+\n8z\n=\n6x\n+\n4y\n=\n6\n1. Open the Linear\nSolver app.\nI\nSelect\nLinear\nSolver\nThe app opens in\nNumeric view.\n.\nby\n+\ncz\n=\nk\n5\n10\n13\n243",
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http://highalphabet.com/staticsprb20/ | [
"statics math problems engineering equations statics basic mechanics school homework engineering math statics formulas statics problems statics problem solutions to statics problems full solution statics engineering statics problem solution statics math problems engineering equations statics basic mechanics school homework engineering math statics formulas statics problems statics problem solutions to statics problems full solution statics engineering statics problem solution\nstatics math problems engineering equations statics basic mechanics school homework engineering math statics formulas statics problems statics problem solutions to statics problems full solution statics engineering statics problem solution statics math problems engineering equations statics basic mechanics school homework engineering math statics formulas statics problems statics problem solutions to statics problems full solution statics engineering statics problem solution\nHighalphabet Home Page statics problem solution statics math Highalphabet Statics Lessons\nDetermine the resultant moment produced by forces F(B) and F(C) about point O. Express the result as a Cartesian vector.",
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"Determine the resultant moment produced by forces F(B) and F(C) about point O. Express the result as a Cartesian vector."
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https://www.queryhome.com/puzzle/23498/what-numbers-calculator-you-are-only-allowed-use-the-and-keys | [
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"# What numbers can you get on a calculator if you are only allowed to use the 3, 5, + and = keys?\n\n118 views\nWhat numbers can you get on a calculator if you are only allowed to use the 3, 5, + and = keys?",
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"posted Sep 23, 2017\n\nHow many times can you use?\n8",
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"answer Oct 2, 2017\n\n`````` 3 + 5 =8\n``````",
null,
"answer Oct 3, 2017\n\nSimilar Puzzles"
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https://www.colorhexa.com/0a0a08 | [
"# #0a0a08 Color Information\n\nIn a RGB color space, hex #0a0a08 is composed of 3.9% red, 3.9% green and 3.1% blue. Whereas in a CMYK color space, it is composed of 0% cyan, 0% magenta, 20% yellow and 96.1% black. It has a hue angle of 60 degrees, a saturation of 11.1% and a lightness of 3.5%. #0a0a08 color hex could be obtained by blending #141410 with #000000. Closest websafe color is: #000000.\n\n• R 4\n• G 4\n• B 3\nRGB color chart\n• C 0\n• M 0\n• Y 20\n• K 96\nCMYK color chart\n\n#0a0a08 color description : Very dark (mostly black) yellow [Olive tone].\n\n# #0a0a08 Color Conversion\n\nThe hexadecimal color #0a0a08 has RGB values of R:10, G:10, B:8 and CMYK values of C:0, M:0, Y:0.2, K:0.96. Its decimal value is 657928.\n\nHex triplet RGB Decimal 0a0a08 `#0a0a08` 10, 10, 8 `rgb(10,10,8)` 3.9, 3.9, 3.1 `rgb(3.9%,3.9%,3.1%)` 0, 0, 20, 96 60°, 11.1, 3.5 `hsl(60,11.1%,3.5%)` 60°, 20, 3.9 000000 `#000000`\nCIE-LAB 2.702, -0.278, 0.756 0.278, 0.299, 0.273 0.327, 0.352, 0.299 2.702, 0.806, 110.192 2.702, 0.035, 0.487 5.469, -0.514, 0.871 00001010, 00001010, 00001000\n\n# Color Schemes with #0a0a08\n\n• #0a0a08\n``#0a0a08` `rgb(10,10,8)``\n• #08080a\n``#08080a` `rgb(8,8,10)``\nComplementary Color\n• #0a0908\n``#0a0908` `rgb(10,9,8)``\n• #0a0a08\n``#0a0a08` `rgb(10,10,8)``\n• #090a08\n``#090a08` `rgb(9,10,8)``\nAnalogous Color\n• #09080a\n``#09080a` `rgb(9,8,10)``\n• #0a0a08\n``#0a0a08` `rgb(10,10,8)``\n• #08090a\n``#08090a` `rgb(8,9,10)``\nSplit Complementary Color\n• #0a080a\n``#0a080a` `rgb(10,8,10)``\n• #0a0a08\n``#0a0a08` `rgb(10,10,8)``\n• #080a0a\n``#080a0a` `rgb(8,10,10)``\n• #0a0808\n``#0a0808` `rgb(10,8,8)``\n• #0a0a08\n``#0a0a08` `rgb(10,10,8)``\n• #080a0a\n``#080a0a` `rgb(8,10,10)``\n• #08080a\n``#08080a` `rgb(8,8,10)``\n• #000000\n``#000000` `rgb(0,0,0)``\n• #000000\n``#000000` `rgb(0,0,0)``\n• #000000\n``#000000` `rgb(0,0,0)``\n• #0a0a08\n``#0a0a08` `rgb(10,10,8)``\n• #181813\n``#181813` `rgb(24,24,19)``\n• #26261f\n``#26261f` `rgb(38,38,31)``\n• #35352a\n``#35352a` `rgb(53,53,42)``\nMonochromatic Color\n\n# Alternatives to #0a0a08\n\nBelow, you can see some colors close to #0a0a08. Having a set of related colors can be useful if you need an inspirational alternative to your original color choice.\n\n• #0a0a08\n``#0a0a08` `rgb(10,10,8)``\n• #0a0a08\n``#0a0a08` `rgb(10,10,8)``\n• #0a0a08\n``#0a0a08` `rgb(10,10,8)``\n• #0a0a08\n``#0a0a08` `rgb(10,10,8)``\n• #0a0a08\n``#0a0a08` `rgb(10,10,8)``\n• #0a0a08\n``#0a0a08` `rgb(10,10,8)``\n• #0a0a08\n``#0a0a08` `rgb(10,10,8)``\nSimilar Colors\n\n# #0a0a08 Preview\n\nThis text has a font color of #0a0a08.\n\n``<span style=\"color:#0a0a08;\">Text here</span>``\n#0a0a08 background color\n\nThis paragraph has a background color of #0a0a08.\n\n``<p style=\"background-color:#0a0a08;\">Content here</p>``\n#0a0a08 border color\n\nThis element has a border color of #0a0a08.\n\n``<div style=\"border:1px solid #0a0a08;\">Content here</div>``\nCSS codes\n``.text {color:#0a0a08;}``\n``.background {background-color:#0a0a08;}``\n``.border {border:1px solid #0a0a08;}``\n\n# Shades and Tints of #0a0a08\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, #0a0a08 is the darkest color, while #fefefe is the lightest one.\n\n• #0a0a08\n``#0a0a08` `rgb(10,10,8)``\n• #151511\n``#151511` `rgb(21,21,17)``\n• #202019\n``#202019` `rgb(32,32,25)``\n• #2b2b22\n``#2b2b22` `rgb(43,43,34)``\n• #36362b\n``#36362b` `rgb(54,54,43)``\n• #404034\n``#404034` `rgb(64,64,52)``\n• #4b4b3c\n``#4b4b3c` `rgb(75,75,60)``\n• #565645\n``#565645` `rgb(86,86,69)``\n• #61614e\n``#61614e` `rgb(97,97,78)``\n• #6c6c56\n``#6c6c56` `rgb(108,108,86)``\n• #77775f\n``#77775f` `rgb(119,119,95)``\n• #828268\n``#828268` `rgb(130,130,104)``\n• #8d8d71\n``#8d8d71` `rgb(141,141,113)``\n• #96967b\n``#96967b` `rgb(150,150,123)``\n• #9e9e86\n``#9e9e86` `rgb(158,158,134)``\n• #a7a791\n``#a7a791` `rgb(167,167,145)``\n• #b0b09c\n``#b0b09c` `rgb(176,176,156)``\n• #b9b9a7\n``#b9b9a7` `rgb(185,185,167)``\n• #c1c1b2\n``#c1c1b2` `rgb(193,193,178)``\n• #cacabd\n``#cacabd` `rgb(202,202,189)``\n• #d3d3c8\n``#d3d3c8` `rgb(211,211,200)``\n• #dbdbd3\n``#dbdbd3` `rgb(219,219,211)``\n• #e4e4dd\n``#e4e4dd` `rgb(228,228,221)``\n• #edede8\n``#edede8` `rgb(237,237,232)``\n• #f6f6f3\n``#f6f6f3` `rgb(246,246,243)``\n• #fefefe\n``#fefefe` `rgb(254,254,254)``\nTint Color Variation\n\n# Tones of #0a0a08\n\nA tone is produced by adding gray to any pure hue. In this case, #090909 is the less saturated color, while #121200 is the most saturated one.\n\n• #090909\n``#090909` `rgb(9,9,9)``\n• #0a0a08\n``#0a0a08` `rgb(10,10,8)``\n• #0b0b07\n``#0b0b07` `rgb(11,11,7)``\n• #0b0b07\n``#0b0b07` `rgb(11,11,7)``\n• #0c0c06\n``#0c0c06` `rgb(12,12,6)``\n• #0d0d05\n``#0d0d05` `rgb(13,13,5)``\n• #0d0d05\n``#0d0d05` `rgb(13,13,5)``\n• #0e0e04\n``#0e0e04` `rgb(14,14,4)``\n• #0f0f03\n``#0f0f03` `rgb(15,15,3)``\n• #101002\n``#101002` `rgb(16,16,2)``\n• #101002\n``#101002` `rgb(16,16,2)``\n• #111101\n``#111101` `rgb(17,17,1)``\n• #121200\n``#121200` `rgb(18,18,0)``\nTone Color Variation\n\n# Color Blindness Simulator\n\nBelow, you can see how #0a0a08 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":0.54146653,"math_prob":0.77727187,"size":3644,"snap":"2021-21-2021-25","text_gpt3_token_len":1696,"char_repetition_ratio":0.12582418,"word_repetition_ratio":0.018348623,"special_character_ratio":0.55214053,"punctuation_ratio":0.23206279,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99024075,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-05-07T17:27:14Z\",\"WARC-Record-ID\":\"<urn:uuid:ddc3dd56-f5b0-4c6e-944f-a818b3c33b6b>\",\"Content-Length\":\"36149\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:762c3347-c591-4207-8dc5-6afac406aa43>\",\"WARC-Concurrent-To\":\"<urn:uuid:9ef894a3-ed87-48aa-bc5c-ab810a9c19ac>\",\"WARC-IP-Address\":\"178.32.117.56\",\"WARC-Target-URI\":\"https://www.colorhexa.com/0a0a08\",\"WARC-Payload-Digest\":\"sha1:X5K6KFWZG2AZ2SZ2NZPRNLVPEIDZC2FC\",\"WARC-Block-Digest\":\"sha1:MHMAUIV6SIO3MFJYOMKG67Z2UAZLVY4Z\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-21/CC-MAIN-2021-21_segments_1620243988796.88_warc_CC-MAIN-20210507150814-20210507180814-00068.warc.gz\"}"} |
https://www.chrisvantienhoven.nl/qa-items/qa-points/qa-p5 | [
"QA-P5: Isotomic Center\n\nThe Isotomic Center is the Perspector of the Reference Quadrangle with the Isotomic Conjugate Quadrangle.\nStated in another way:\nThe Isotomic Center is the common intersection point of lines Pi.Qi, where\n• Pi = ith quadrangle vertice,\n• Qi = Isotomic Conjugate of Pi wrt Pj.Pk.Pl\nfor all permutations of (i,j,k,l) ∈ (1,2,3,4).",
null,
"Construction:\nQA-P5 is the Reflection of the Anticomplement of QA-P1 (wrt QA-Diagonal Triangle) in QA-P1.\n\nCoordinates:\n1st CT-Coordinate:\nq r (q+r) (2p+q+r) (note that this formula is independent of a,b,c)\n1st DT-Coordinate:\n-4 (p4 + q2 r2) + (p2 + q2 + r2)2\n\nProperties:\n• QA-P5 lies on these QA-lines:\nQA-P1.QA-P10 = QA-L3 (4 : -3)\nQA-P29.QA-P34 (3 : -2)\n• QA-P5 lies on these QG-lines:\nQG-P2.QG-P4 (3 : -2)\n• QA-P5 is the Reflection of:\n\nQA-P19 in QA-P21.\n• QA-P5 is the AntiComplement of QA-P20 wrt the QA-Diagonal Triangle.\n• QA-P5.QA-P1 : QA-P1.QA-P10 : QA-P10.QA-P20 = 3 : 1 : 2.\n• QA-P5 forms with the vertices of the QA-Diagonal Triangle a quadrangle that shares the same centroid with the Reference Quadrangle.\n• QA-P5 is the QA-Involutary Conjugate (see QA-Tf2) of QA-P17.\n• QA-P5 lies on the Conic through P1, P2, P3, P4 and QA-P10.\n• QA-P5 lies on the Cubics QA-Cu2 and QA-Cu4.\n• Given Bimedians M12*M34, M13*M24, M14*M23 of the Reference Quadrangle. Let T1.T2.T3 be the AntiComplementary Triangle of the Diagonal Triangle S1.S2.S3. The corresponding lines parallel to the Bimedians through T1, T2 and T3 concur in QA-P5 (Seiichi Kirikami, September 30, 2012).\n• QA-P5 is the center of the circle of points where: sum of quadratic distances to Quadrangle - vertices = sum of quadratic distances to DT - vertices.\n• QA-P5 is the Orthology Center of the Triple Triangles of QG-P14/QG-P15 wrt the Triple triangles of QG-P5/QG-P10/QL-P2. See QA-Tr-1.\n• QA-P5 is the Perspector of the Triple Triangles (see QA-Tr-1) of these pairs (QG-P2, QG-P4), (QG-P14, QL-P15).\n\n#### Plaats reactie",
null,
"Vernieuwen"
] | [
null,
"https://www.chrisvantienhoven.nl/images/EQF/QA-P5-Isotomic-Center-00.png",
null,
"https://www.chrisvantienhoven.nl/component/jcomments/captcha/73225",
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.71478206,"math_prob":0.92956126,"size":1604,"snap":"2019-51-2020-05","text_gpt3_token_len":528,"char_repetition_ratio":0.151875,"word_repetition_ratio":0.030303031,"special_character_ratio":0.31546134,"punctuation_ratio":0.1643454,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99578774,"pos_list":[0,1,2,3,4],"im_url_duplicate_count":[null,6,null,1,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-01-19T18:31:24Z\",\"WARC-Record-ID\":\"<urn:uuid:7908a0b1-4a99-467a-9320-87ae35a9eb0f>\",\"Content-Length\":\"143949\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:9cb436fe-f723-4e05-8629-baed2275a723>\",\"WARC-Concurrent-To\":\"<urn:uuid:89ff4b3d-f8cf-4e38-8a04-f13f8f8f51c8>\",\"WARC-IP-Address\":\"46.249.37.218\",\"WARC-Target-URI\":\"https://www.chrisvantienhoven.nl/qa-items/qa-points/qa-p5\",\"WARC-Payload-Digest\":\"sha1:MS7IIZAIE2MKOF2K6IQG5YG2BYQJOA2V\",\"WARC-Block-Digest\":\"sha1:EZ3XF2CONXGWMLMU2DZUENPKS7XSZRUI\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-05/CC-MAIN-2020-05_segments_1579250594705.17_warc_CC-MAIN-20200119180644-20200119204644-00065.warc.gz\"}"} |
https://www.numberempire.com/85578841 | [
"Home | Menu | Get Involved | Contact webmaster",
null,
"",
null,
"",
null,
"",
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"",
null,
"# Number 85578841\n\neighty five million five hundred seventy eight thousand eight hundred forty one\n\n### Properties of the number 85578841\n\n Factorization 85578841 Divisors 1, 85578841 Count of divisors 2 Sum of divisors 85578842 Previous integer 85578840 Next integer 85578842 Is prime? YES (4975424th prime) Previous prime 85578827 Next prime 85578877 Is a Fibonacci number? NO Is a Bell number? NO Is a Catalan number? NO Is a factorial? NO Is a regular number? NO Is a perfect number? NO Polygonal number (s < 11)? NO Binary 101000110011101010001011001 Octal 506352131 Duodecimal 247b0961 Hexadecimal 519d459 Square 7323738026903281 Square root 9250.8832551276 Natural logarithm 18.264948625977 Decimal logarithm 7.9323664005084 Sine -0.42137028158268 Cosine -0.90688868434827 Tangent 0.46463285831545\nNumber 85578841 is pronounced eighty five million five hundred seventy eight thousand eight hundred forty one. Number 85578841 is a prime number. The prime number before 85578841 is 85578827. The prime number after 85578841 is 85578877. Number 85578841 has 2 divisors: 1, 85578841. Sum of the divisors is 85578842. Number 85578841 is not a Fibonacci number. It is not a Bell number. Number 85578841 is not a Catalan number. Number 85578841 is not a regular number (Hamming number). It is a not factorial of any number. Number 85578841 is a deficient number and therefore is not a perfect number. Binary numeral for number 85578841 is 101000110011101010001011001. Octal numeral is 506352131. Duodecimal value is 247b0961. Hexadecimal representation is 519d459. Square of the number 85578841 is 7323738026903281. Square root of the number 85578841 is 9250.8832551276. Natural logarithm of 85578841 is 18.264948625977 Decimal logarithm of the number 85578841 is 7.9323664005084 Sine of 85578841 is -0.42137028158268. Cosine of the number 85578841 is -0.90688868434827. Tangent of the number 85578841 is 0.46463285831545\n\n### Number properties\n\nExamples: 3628800, 9876543211, 12586269025"
] | [
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"https://www.numberempire.com/images/graystar.png",
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"https://www.numberempire.com/images/graystar.png",
null,
"https://www.numberempire.com/images/graystar.png",
null,
"https://www.numberempire.com/images/graystar.png",
null,
"https://www.numberempire.com/images/graystar.png",
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https://www.kingexcel.info/2021/04/excel-formulas-and-functions-complete.html | [
"Excel Formulas and Functions: The Complete Tutorial for Beginners to Learn and Master Excel Formulas and Functions with Tips & Tricks For Excel 2021 Users - KING OF EXCEL\n\n# Excel Formulas and Functions: The Complete Tutorial for Beginners to Learn and Master Excel Formulas and Functions with Tips & Tricks For Excel 2021 Users\n\n🎁Like Fanpage and Read online bellow⏬\n\n• Length: 81 pages\n• Edition: 1\n• Publication Date: 2021-03-11\n• This book on Excel Formulas and Functions for Beginners 2021 is a perfect guide for complete beginners or casual spreadsheet users. With patience and by following the simple step-by-step process, you will learn how to work with Excel at the office or at home for personal reasons.\nThe purpose of this beginner’s guide is to explain to you how and when to use all the powerful tools and features Excel offers.\nYou shouldn’t be stressing yourself to memorize functions and formulas because Excel can do that heavy lifting for you.\n\nHere are the things you will get the chance of learning in this book:\n\n• Benefits of using Excel\n• Functions and Formulas in Excel\n• How to Enter a Formula?\n• How to Edit a Formula?\n• Precedence of Operators\n• How to Copy or Paste a Formula?\n• How to Insert a Function?\n• The Formula Tab\n• Relative References\n• How to Create and Copy the Formula Using a Relative Reference\n• Absolute References\n• How to Create and Copy Formula Using Absolute Referencing?\n• Mixed Cell Referencing\n• Future Value\n• FVSCHEDULE\n• Present Value\n• Net PresentValue\n• Using the AND function\n• BETWEEN Condition\n• IF Function\n• IFS Function\n• IFERROR Function\n• OR Function\n• Date Function\n• More Functions\n• Statistica lFunctions\n• Computing GDPs\n• Currency\n• Decimal Places\n• Editing Text with Formulas\n• Number of Characters\n• Counting Numbers\n• IF Statements\n• Looking for Information\n• Time and Date\n• AutoFill\n• Text to Columns\n• Transpose\n• Automatically Sum Everything\n• Work on Different Sheets\n• Display Formulas\n• Excel Tips and Tricks\n• And many more….\n\nThis is just a few of what is contained in this book and you can Download FREE with Kindle Unlimited\n\n#evba #kingexcel #etipfree #eama"
] | [
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.8177659,"math_prob":0.48620898,"size":1951,"snap":"2023-40-2023-50","text_gpt3_token_len":449,"char_repetition_ratio":0.1540832,"word_repetition_ratio":0.023255814,"special_character_ratio":0.22552536,"punctuation_ratio":0.055727553,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.95234096,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-12-09T01:49:37Z\",\"WARC-Record-ID\":\"<urn:uuid:632c03d2-be50-41f1-8a50-19f31bd3e0ec>\",\"Content-Length\":\"452295\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:cb9ede37-c7cf-4b9e-9519-813a402c8032>\",\"WARC-Concurrent-To\":\"<urn:uuid:7b86122c-051f-4260-811a-4b11afec7b7d>\",\"WARC-IP-Address\":\"142.251.167.121\",\"WARC-Target-URI\":\"https://www.kingexcel.info/2021/04/excel-formulas-and-functions-complete.html\",\"WARC-Payload-Digest\":\"sha1:PMHC67VJRTHUOS5EGNLMEBB423VTCQBQ\",\"WARC-Block-Digest\":\"sha1:SY6IU7IMHZV2X2A6ROF2YD677K5UELLS\",\"WARC-Identified-Payload-Type\":\"application/xhtml+xml\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-50/CC-MAIN-2023-50_segments_1700679100781.60_warc_CC-MAIN-20231209004202-20231209034202-00399.warc.gz\"}"} |
https://arxiv.org/abs/math/0608032 | [
"math\n\nTitle:Level m stratifications of versal deformations of p-divisible groups\n\nAbstract: Let $k$ be an algebraically closed field of characteristic $p>0$. Let $c,d,m$ be positive integers. Let $D$ be a $p$-divisible group of codimension $c$ and dimension $d$ over $k$. Let $\\scrD$ be a versal deformation of $D$ over a smooth $k$-scheme $\\scrA$ which is equidimensional of dimension $cd$. We show that there exists a reduced, locally closed subscheme $\\grs_D(m)$ of $\\scrA$ that has the following property: a point $y\\in\\scrA(k)$ belongs to $\\grs_D(m)(k)$ if and only if $y^*(\\scrD)[p^m]$ is isomorphic to $D[p^m]$. We prove that $\\grs_D(m)$ is {\\it regular and equidimensional} of {\\it dimension} $cd-\\dim(\\pmb{\\text{Aut}}(D[p^m]))$. We give a proof of {\\it Traverso's formula} which for $m>>0$ computes the codimension of $\\grs_D(m)$ in $\\scrA$ (i.e., $\\dim(\\pmb{\\text{Aut}}(D[p^m]))$) in terms of the Newton polygon of $D$. We also provide a criterion of when $\\grs_D(m)$ satisfies the {\\it purity property} (i.e., it is an affine $\\scrA$-scheme). Similar results are proved for {\\it quasi Shimura $p$-varieties of Hodge type} that generalize the special fibres of good integral models of Shimura varieties of Hodge type in unramified mixed characteristic $(0,p)$."
] | [
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.77991605,"math_prob":0.99998343,"size":1560,"snap":"2019-43-2019-47","text_gpt3_token_len":487,"char_repetition_ratio":0.10796915,"word_repetition_ratio":0.0,"special_character_ratio":0.31346154,"punctuation_ratio":0.106796116,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9999993,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-10-23T14:06:37Z\",\"WARC-Record-ID\":\"<urn:uuid:dcdb6c33-8f6b-45d7-8aff-17f3c2292635>\",\"Content-Length\":\"19967\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:b52c9df4-e244-41ba-89ee-f1657afe0899>\",\"WARC-Concurrent-To\":\"<urn:uuid:aa8656fd-b391-451c-87c4-c9f360392dff>\",\"WARC-IP-Address\":\"128.84.21.199\",\"WARC-Target-URI\":\"https://arxiv.org/abs/math/0608032\",\"WARC-Payload-Digest\":\"sha1:5B2H5AONWAVZDYLDB5V34J3KMYDMPLA4\",\"WARC-Block-Digest\":\"sha1:W6BNE5Z3KB6ML5VL4GRUPVJ6L2ASPNES\",\"WARC-Identified-Payload-Type\":\"application/xhtml+xml\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-43/CC-MAIN-2019-43_segments_1570987833766.94_warc_CC-MAIN-20191023122219-20191023145719-00540.warc.gz\"}"} |
https://scholars.uab.edu/display/pub208123 | [
"# The \"spectral\" decomposition for one-dimensional maps\n\n• We construct the \"spectral\" decomposition of the sets $\\bar{Per\\,f}$, $\\omega(f)=\\cup\\omega(x)$ and $\\Omega(f)$ for a continuous map $f$ of the interval to itself. Several corollaries are obtained; the main ones describe the generic properties of $f$-invariant measures, the structure of the set $\\Omega(f)\\setminus \\bar{Per\\,f}$ and the generic limit behavior of an orbit for maps without wandering intervals. The \"spectral\" decomposition for piecewise-monotone maps is deduced from the Decomposition Theorem. Finally we explain how to extend the results of the present paper for a continuous map of a one-dimensional branched manifold into itself."
] | [
null
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https://www.ams.org/journals/proc/1992-116-01/S0002-9939-1992-1093596-6/home.html | [
"# Proceedings of the American Mathematical Society\n\nPublished by the American Mathematical Society, the Proceedings of the American Mathematical Society (PROC) is devoted to research articles of the highest quality in all areas of pure and applied mathematics.\n\nISSN 1088-6826 (online) ISSN 0002-9939 (print)\n\nThe 2020 MCQ for Proceedings of the American Mathematical Society is 0.85.\n\nWhat is MCQ? The Mathematical Citation Quotient (MCQ) measures journal impact by looking at citations over a five-year period. Subscribers to MathSciNet may click through for more detailed information.\n\n## On the cup product for groupsHTML articles powered by AMS MathViewer\n\nby M. N. Dyer, J. V. Leahy and L. J. Mooney\nProc. Amer. Math. Soc. 116 (1992), 219-227 Request permission\n\n## Abstract:\n\nFor $G$ any group and trivial coefficients in $R$, a commutative ring, the authors analyze the kernel of the cup product $\\cup :{H^1}\\left ( {G,R} \\right ) \\otimes {H^1}\\left ( {G,R} \\right ) \\to {H^2}\\left ( {G,R} \\right )$ by splicing it together with part of the six-term Hom-Ext exact sequence obtained from $0 \\to {I^2}/{I^3} \\to I/{I^3} \\to I/{I^2} \\to 0$.\nSimilar Articles"
] | [
null
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http://academiabookshop.com/product/theory-of-vibrations-with-applications-5th-edition/ | [
"# Theory Of Vibrations With Applications 5th Edition\n\n679.00\n\n• Publisher: Pearson\n• ISBN-13: 9788131704820\n• Pages: 534\n• Binding: Paperback\n• Year of Pub / Reprint Year: 2008\n\n## Description\n\nA thorough treatment of vibration theory and its engineering applications, from simple degree to multi degree-of-freedom system.\n\n1. Oscillatory Motion.\n2. Free Vibration.\n3. Harmonically Excited Vibration.\n4. Transient Vibration.\n5. Systems with Two or More Degrees of Freedom.\n6. Properties of Vibrating Systems.\n7. Lagrange’s Equation.\n8. Computational Methods.\n9. Vibration of Continuous Systems.\n10. Introduction to the Finite Element Method.\n11. Mode-Summation Procedures for Continuous Systems.\n12. Classical Methods.\n13. Random Vibrations.\n14. Nonlinear Vibrations\n\nSalient Features\n\n• Focuses on the physical aspects of the mathematical concepts necessary to describe the vibration phenomena.\n• Provides many example applications, including typical problems faced by practicing engineers.\n• MATLAB… has been introduced where appropriate to take advantage of this industry-standard software for necessary calculations.\n• Mass and stiffness matrices are now defined alongside the discussion of normal mode vibrations, free vibrations, forced vibrations, absorbers, and dampers (Ch. 5).\n• FORTRAN computations have been replaced by MATLAB. FORTRAN plotting has been replaced by MATLAB plotting.\n• Introduction to Finite Element Methods (Ch. 10) now does computations completely in MATLAB. The section on generalized force proportional to displacement has been substantially expanded by detailed computation of rotating helicopter blades.\n• Problem sets have been added throughout."
] | [
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https://www.prodyogi.com/2019/02/surveying-definition-principle-classes.html | [
"# Surveying - Definition ,Principle and Classification\n\nSurveying can be defined as the art of determining the relative positions of a point on or below or above the earth surface by undergoing direct or indirect measurements of distance, elevation, and direction.",
null,
"Surveying - Definition, Principle, and Classes\n\n### Significance of Surveying\n\nBoth Angular and Linear Measurements are determined in order to establish the points. The main objective of surveying is to prepare a map or plan that represents an area held on a horizontal plane.\n\n• The horizontal projection of an area is represented in the plan and maps\n• Vertical distances between the points are shown through contour lines, hachures, and other methods in vertical sections.\n\n### Principles of Surveying\n\nThe fundamental principles of surveying are:\n\n1. Location of a point by measurement from two points of reference\n2. Working from whole to part\n\n### Classes of Surveying\n\nPrimarily surveying can be classified into two classes:\n1. Plane Surveying\n2. Geodetic Surveying\n\n#### Plane Surveying\n\nIn this type of surveying, the spheroidal shape of the earth is neglected. It considers the mean surface of the earth as a plane. Triangles that are formed by the survey lines are considered plane triangles. Level lines are straight and plumb lines are parallel.\n\n#### Geodetic Surveying\n\nIn this type of surveying, the shape of the earth is taken into consideration. Lines laying on the earth's surface are curved lines and the respective triangles formed are spherical triangles. Hence, this surveying takes into consideration, spherical trigonometry."
] | [
null,
"https://4.bp.blogspot.com/-GkAsuaCXxB8/XGlX89LAHWI/AAAAAAAAAxY/cHDcdfpvGdwdeqb_BNyPHp9zj7nH9oFygCLcBGAs/s320/116.jpg",
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.9167157,"math_prob":0.90669364,"size":1611,"snap":"2022-40-2023-06","text_gpt3_token_len":324,"char_repetition_ratio":0.15121344,"word_repetition_ratio":0.032388665,"special_character_ratio":0.17690875,"punctuation_ratio":0.096296296,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9609194,"pos_list":[0,1,2],"im_url_duplicate_count":[null,4,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-02-05T14:52:08Z\",\"WARC-Record-ID\":\"<urn:uuid:77d2a1bc-12dc-47e5-a146-46b82551f994>\",\"Content-Length\":\"501902\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:d2277b8c-1941-4d88-ab6c-210805936bce>\",\"WARC-Concurrent-To\":\"<urn:uuid:af0ee5d1-1d1d-4ccc-bbd7-00266f8b63d1>\",\"WARC-IP-Address\":\"104.21.82.192\",\"WARC-Target-URI\":\"https://www.prodyogi.com/2019/02/surveying-definition-principle-classes.html\",\"WARC-Payload-Digest\":\"sha1:EUXCVHVMHKUMKNBPACHCDKRIPAPUO7VR\",\"WARC-Block-Digest\":\"sha1:IAXD2NOBH6ANDTU7BWZ7UAEU67QZKEHO\",\"WARC-Identified-Payload-Type\":\"application/xhtml+xml\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-06/CC-MAIN-2023-06_segments_1674764500255.78_warc_CC-MAIN-20230205130241-20230205160241-00175.warc.gz\"}"} |
https://homework.cpm.org/category/CC/textbook/CCA2/chapter/Ch4/lesson/4.1.1/problem/4-13 | [
"",
null,
"",
null,
"### Home > CCA2 > Chapter Ch4 > Lesson 4.1.1 > Problem4-13\n\n4-13.\n\nSimplify each of the expressions below. Express your answers as simply as possible.\n\n1. $\\frac { 5 x ^ { 2 } - 11 x + 2 } { x ^ { 2 } + 8 x + 16 } \\cdot \\frac { x ^ { 2 } + 10 x + 24 } { 10 x ^ { 2 } + 13 x - 3 }$\n\nFactor the polynomials, simplify, and then multiply.\n\n$\\frac{(x-2)(x+6)}{(x+4)(2x+3)}$\n\n1. $\\frac { 6 x + 3 } { 2 x - 3 } \\div \\frac { 3 x ^ { 2 } - 12 x - 15 } { 2 x ^ { 2 } - x - 3 }$\n\nFactor the polynomials, simplify, and then divide.\n\nTwo of the polynomials have common factors.\n\n1. $\\frac { 5 m + 18 } { m + 3 } + \\frac { 4 m + 9 } { m + 3 }$\n\nBoth fractions already have a common denominator, so add the fractions and then simplify by factoring.\n\n$9$\n\n1. $\\frac { 3 a ^ { 2 } + a - 1 } { a ^ { 2 } - 2 a + 1 } - \\frac { 2 a ^ { 2 } - a + 2 } { a ^ { 2 } - 2 a + 1 }$\n\nBoth fractions have a common denominator, so subtract the fractions.\n\n$\\frac{(3a^2+a-1)-(2a^2-a+2)}{a^2-2a+1}$\n\n$\\frac{ 3a^2 + a - 1 - 2a^2 +a -2 }{a^2 - 2a + 1}$\n\n$\\frac{a^2 + 2a -3}{a^2 - 2a + 1}$\n\n$\\frac{(a+3)(a-1)}{(a-1)(a-1)}$"
] | [
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":0.8890704,"math_prob":1.0000025,"size":465,"snap":"2020-34-2020-40","text_gpt3_token_len":106,"char_repetition_ratio":0.14750542,"word_repetition_ratio":0.08695652,"special_character_ratio":0.22580644,"punctuation_ratio":0.17977528,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99999654,"pos_list":[0,1,2,3,4],"im_url_duplicate_count":[null,null,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-08-13T03:30:12Z\",\"WARC-Record-ID\":\"<urn:uuid:1b9f5a53-f582-4d75-b7ac-b5a23afe0c8d>\",\"Content-Length\":\"43997\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:0436b5f7-e506-451a-9328-a023451a1882>\",\"WARC-Concurrent-To\":\"<urn:uuid:effbb8bc-73f3-4255-ad78-c8756f6baf90>\",\"WARC-IP-Address\":\"104.26.7.16\",\"WARC-Target-URI\":\"https://homework.cpm.org/category/CC/textbook/CCA2/chapter/Ch4/lesson/4.1.1/problem/4-13\",\"WARC-Payload-Digest\":\"sha1:HESNCV32DEXUFKL5TLJJRZE4V4W2BMGI\",\"WARC-Block-Digest\":\"sha1:J2EH6S2DUEXURUMURRRRZIRB6DONMXA3\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-34/CC-MAIN-2020-34_segments_1596439738950.61_warc_CC-MAIN-20200813014639-20200813044639-00003.warc.gz\"}"} |
https://answers.everydaycalculation.com/subtract-fractions/3-20-minus-25-6 | [
"Solutions by everydaycalculation.com\n\n## Subtract 25/6 from 3/20\n\n1st number: 3/20, 2nd number: 4 1/6\n\n3/20 - 25/6 is -241/60.\n\n#### Steps for subtracting fractions\n\n1. Find the least common denominator or LCM of the two denominators:\nLCM of 20 and 6 is 60\n2. For the 1st fraction, since 20 × 3 = 60,\n3/20 = 3 × 3/20 × 3 = 9/60\n3. Likewise, for the 2nd fraction, since 6 × 10 = 60,\n25/6 = 25 × 10/6 × 10 = 250/60\n4. Subtract the two fractions:\n9/60 - 250/60 = 9 - 250/60 = -241/60\n\nMathStep (Works offline)",
null,
"Download our mobile app and learn to work with fractions in your own time:"
] | [
null,
"https://answers.everydaycalculation.com/mathstep-app-icon.png",
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.63066727,"math_prob":0.99638706,"size":379,"snap":"2020-34-2020-40","text_gpt3_token_len":169,"char_repetition_ratio":0.224,"word_repetition_ratio":0.0,"special_character_ratio":0.52506596,"punctuation_ratio":0.082474224,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9978907,"pos_list":[0,1,2],"im_url_duplicate_count":[null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-08-09T14:31:58Z\",\"WARC-Record-ID\":\"<urn:uuid:c20a0e9b-180d-4f95-9daa-5e95eb0dbb0c>\",\"Content-Length\":\"8074\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:8a70d4a5-943e-4842-92b0-04ddbbcab5f3>\",\"WARC-Concurrent-To\":\"<urn:uuid:c301dd2d-fc7d-4b9d-b3cc-aa1453c506c9>\",\"WARC-IP-Address\":\"96.126.107.130\",\"WARC-Target-URI\":\"https://answers.everydaycalculation.com/subtract-fractions/3-20-minus-25-6\",\"WARC-Payload-Digest\":\"sha1:TEGJBTYKPCA46QNU2LKDZT64XJFGKVIH\",\"WARC-Block-Digest\":\"sha1:NQIDGQAECVKIQKQ7JLJUYJ5HKQ3ZAH4Y\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-34/CC-MAIN-2020-34_segments_1596439738555.33_warc_CC-MAIN-20200809132747-20200809162747-00228.warc.gz\"}"} |
https://techblog.shutl.com/2016/07/property-testing/ | [
"",
null,
"# Incorporate property testing to improve confidence in your code\n\n-\n\nNote: All the examples and libraries use clojure, but the content is portable on any language of your choice.\n\n# Use Case\n\nImagine you need a function which computes the average of values. Using TDD you might write your code likewise:\n\n```(deftest average-unit-test\n(is (= 1 (average )))\n(is (= 10 (average [5 15 10]))))\n```\n```(defn average [values]\n(/ (reduce + values) (count values)))\n```\n\nAnd sure enough, when you run your tests you can see\n\n```Ran 1 tests containing 2 assertions.\n0 failures, 0 errors.\n```\n\nThere are few comments on that simple piece of code. Why did we choose these 2 examples for testing? Where do the “magic numbers” come from. Is that enough testing? – spoiler alert, the answer is NO –\n\nThe most obvious bug in this code is “Division By Zero” — when the values is an empty collection. So after spotting it we can update the code to something like:\n\n```(deftest average-unit-test\n(is (= 1 (average )))\n(is (= 10 (average [5 15 10])))\n(is (nil? (average []))))\n```\n```(defn average [values]\n(if-not (empty? values)\n(/ (reduce + values) (count values))\nnil))\n```\n\nNow we have fixed this obvious issue with this really simple function, we have to ask ourselves “are there any more edge cases in this function?”, “how many more tests should I add to make sure I’ve got everything covered?”. If you strictly follow TDD, we also have to find a failing test first.\n\n# Property testing to the rescue\n\nA property testing library is trying the run the algorithm under test with random values, and check if its properties are satisfied. It’s a bit of a brute force attack.\n\n## Property\n\nIt’s an invariant of your algorithm. For example for a switching the case of every character of a string, it could be, applying twice the function on a string result on the original string. The property takes in parameter one or multiple generators and check the invariant on each generated value.\n\n## Generator\n\nIn order to check your algorithm with a lot of different combinations, you need to be able to generate random data. Every property testing library comes with a way to generator complex data structure, constraining the range of the values.\n\n## Reducing\n\nThe reduce phase is a really important part of the method. When the library detects a combination of values which does not satisfy the invariant, as the data is randomly generated it can contains a lot of “noise”.\n\nSo the reducing phase is there is try to find the minimum combination of values which triggers the bug. For example, reducing a list means trying to remove some values and check if the invariant is still failing.\n\n# Introducing test.check\n\ntest.check is an open source library inspired by QuickCheck. It allows to describe…….\n\nso let’s add the dependency on the library:\n\n```[org.clojure/test.check \"0.9.0\"]\n```\n\nand require it from the test\n\n```(:require [clojure.test.check.clojure-test :refer [defspec]]\n[clojure.test.check :as tc]\n[clojure.test.check.generators :as gen]\n[clojure.test.check.properties :as prop])\n```\n\nTo come back to your average function, we have to answer two questions:\n\n• What are the invariants\n• What’s the input\n\n## Generator\n\nThe input is quite trivial, it’s any vector of numbers, which can be translated in test.check by\n\nBasically we are trying to generate a vector with a random number of large random integer.\n\nWe can test it using the Clojure REPL via the test.check sample function",
null,
"## Invariant\n\nIn term of invariants, we can for example state that the result of the function should not depend on the order of the values in the vector. In terms of code\n\n```(defn- order-independant? [values]\n(let [regular-average (average values)\nshuffled-average (average (shuffle values))]\n(equals-within regular-average shuffled-average 0.1)))\n```\n\nThis function checks the average value of a vector equals the average of the same vector but shuffled.\n\n## All together\n\n```(def vector-generator (gen/vector gen/large-integer))\n\n(defn- order-independant? [values]\n(let [regular-average (average values)\nshuffled-average (average (shuffle values))]\n(equals-within regular-average shuffled-average 0.1)))\n\n(defspec the-average-of-a-list-does-not-depend-on-the-order-of-the-elements number-of-iterations\n(prop/for-all [values vector-generator]\n(order-independant? values)))\n```\n\nThe defspec macro allow to generate a test.unit test, which is going to be executing with the unit tests.\n\nWhen we run the tests, we come across the following issue:\n\n```[clojure.main main \"main.java\" 37]]}, :seed 1467906144689, :failing-size 76, :num-tests 77, :fail [[-8791245474 3169063838 -6055278567293294141 -8543319887 1612827468009 545 27784351334004543 -2885 -107647 -456743012890065 719911797491 7 -15141358 -408950911 -21398427889 -538944277814 24 770462 -7524750475275866215 15236063177479454 3107788752529]], :shrunk {:total-nodes-visited 1069, :depth 46, :result #error {\n:cause \"integer overflow\"\n:via\n[{:type java.lang.ArithmeticException\n:message \"integer overflow\"\n:at [clojure.lang.Numbers throwIntOverflow \"Numbers.java\" 1501]}]\n```\n\nSo basically what’s happening is test.check generated a vector which caused an exception. When using large number, computing the sum of all the values triggers an overflow.\n\nIt might seem like an obvious problem, but that’s an edge case rarely covered straight away by an unit test. Solving this particularly problem is out of the scope of this post, but now you can decide if you won’t support high numbers, or if you need a better algorithm to limit this issue.\n\n# Property testing in your toolbox\n\nSo, hopefully, by now you are convinced that property testing is awesome and you should use it. You now have to think how many “traditional” tests you want to write, and when can you write properties.\n\nOne noticeable thing in the previous example, is that we never test the proper calculation in the spec. We only check the average does not depend on the order of the elements. That’s, I think, were property testing falls short, if you wanted to write a spec for the actual results with random values, you can’t use “magic numbers”, you almost need to write a second implementation of the calculation.\n\nYour mileage might vary, but I think the two techniques are complimentary.\n\n# Further information\n\nThere is a lot of material you can find on this topic, in no particular order, a selection of videos, articles I found useful for me:\n\nHappy testing\n\n•",
null,
"Graham Lyons says:\n•",
null,
"Graham Lyons says:"
] | [
null,
"https://techblog.shutl.com/wp-content/themes/generic-shutl/images/shutl-logo.svg",
null,
"https://techblog.shutl.com/wp-content/uploads/2016/07/Screen-Shot-2016-07-12-at-12.25.53-300x29.png",
null,
"https://secure.gravatar.com/avatar/0639c670fd8e17cd7823be225de7750b",
null,
"https://secure.gravatar.com/avatar/0639c670fd8e17cd7823be225de7750b",
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.8252783,"math_prob":0.8399138,"size":6814,"snap":"2020-24-2020-29","text_gpt3_token_len":1596,"char_repetition_ratio":0.12393539,"word_repetition_ratio":0.045889102,"special_character_ratio":0.25858527,"punctuation_ratio":0.12099922,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.98069924,"pos_list":[0,1,2,3,4,5,6,7,8],"im_url_duplicate_count":[null,null,null,2,null,null,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-06-02T10:01:26Z\",\"WARC-Record-ID\":\"<urn:uuid:de821dc3-babd-4ea7-8d7f-971a118c417a>\",\"Content-Length\":\"45685\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:5741faa1-dcdb-43d2-88d4-cf4f98e3a044>\",\"WARC-Concurrent-To\":\"<urn:uuid:01f3dbdd-8f9d-4786-ac87-cc0c48a2df43>\",\"WARC-IP-Address\":\"35.197.203.135\",\"WARC-Target-URI\":\"https://techblog.shutl.com/2016/07/property-testing/\",\"WARC-Payload-Digest\":\"sha1:BKOJ2JQWGTQP53467Y7TWBCW67MEII3G\",\"WARC-Block-Digest\":\"sha1:CTMDRHLFS666AEWSBDAZAYYNJCQR43LG\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-24/CC-MAIN-2020-24_segments_1590347424174.72_warc_CC-MAIN-20200602100039-20200602130039-00385.warc.gz\"}"} |
https://edurev.in/course/quiz/attempt/18497_Test-Pavement-Design-2/3e5ea331-29ea-4821-b260-664100a30d30 | [
"# Test: Pavement Design - 2\n\n## 20 Questions MCQ Test Topicwise Question Bank for Civil Engineering | Test: Pavement Design - 2\n\nDescription\nAttempt Test: Pavement Design - 2 | 20 questions in 60 minutes | Mock test for Civil Engineering (CE) preparation | Free important questions MCQ to study Topicwise Question Bank for Civil Engineering for Civil Engineering (CE) Exam | Download free PDF with solutions\nQUESTION: 1\n\n### The width of expansion joint gap is 2.5 cm in a cement concrete pavement. The spacing between expansion joint for a maximum rise in temperature of 25°C is (assuming coefficient of thermal expansion of concrete as 10 x 10-6 per °C)\n\nSolution:",
null,
"Coefficient of thermal expansion\nC = 10 X 10-6 per°C\n∴ Spacing of expansion joint,",
null,
"= 50m\n\nQUESTION: 2\n\nSolution:\nQUESTION: 3\n\n### Base course is used in rigid pavements for\n\nSolution:\nQUESTION: 4\n\nThe standard plate size in a plate bearing test for finding modulus of sub-grade reaction (k) value is\n\nSolution:\nQUESTION: 5\n\nConsider the following statements:\n1. In a concrete pavement temperature stress is tensile at bottom during day time.\n2. In a concrete pavement load stress is compressive at bottom.\nWhich of these statements is/are correct?\n\nSolution:\nQUESTION: 6\n\nIt is customary to design the thickness of cement concrete pavement for\n\nSolution:\n\nThe wheel load stress for rigid pavements are computed at interior, edge and corner using Westergaard’s equation. Temperature, stress are developed in cement concrete pavements leading to warping and frictional stress. Critical combination of stress (load and temperature) is evaluated to determine the thickness of cement concrete pavement.\n\nQUESTION: 7\n\nIn reinforced cement concrete pavements, steel is placed\n\nSolution:\n\nThe reinforcement used in cement concrete pavement is in the form of welded wire fabric or bar mats. The function of reinforcement is to hold the cracked slab portions together and thus not to allow them to open up any more. In order to obtain the maximum advantage, it is suggested that greater quantity of the reinforcement should be placed in the longitudinal direction. Further the reinforcement should either be placed in the middepth or towards the top of the pavement for better functioning.\n\nQUESTION: 8\n\nIn cement concrete pavements, tie bars are provided:\n\nSolution:\n\nTie bar are not the load transferring devices. These are provided to ensure the adjacent slabs remain firmly tied together. These are provided in the transverse direction across longitudinal joints, (warping joint) at mid depth.\n\nQUESTION: 9\n\nThe sub-grade soil properties of a sam ple are as follows:\nSoil portion passing 0.075 mm sieve = 50%\nLiquid limit = 40%\nPlasticity index = 20%\nThe group index of the soil is\n\nSolution:\n\nGroup Index\nGl = 0.2 a + 0.005 ac + 0,01 bd\nSoil portion passing 0.075mm sieve\nP = 50%\nLL - 40% PI = 20%\na = P - 3 5 - 15 < 40\nb = P - 1 5 = 35 < 4 0\nc = LL - 40 = 0\nd = PI - 10 = 10\nG.I. = 0.2 x 15 + 0 + 0.01 x 35 x 10\n= 3 + 3.5 = 6,5\n\nQUESTION: 10\n\nConsider the following factors :\n2. Thickness of cement concrete slab\n3. Temperature distribution in the slab\nWhich of these should be taken into reckoning to determine the wheel load stress at critical location in a cement concrete pavement?\n\nSolution:\n\nTo determine wheel load'stress at critical location, - Magnitude of load (P) - Thickness of slab (h) - Radius of wheel Load distribution (a), radius of relative stiffness (I) and radius of resisting section (b) are needed. 'a', 'b' and 'I’ depend upon modulus of sub - grade reaction.\n\nQUESTION: 11\n\nIf the pressure carried by a CBR specim en at 2.5 mm penetration is 3.5 N/mm2, the CBR of the soil is\n\nSolution:\n\nThe pressure 3.5 N/mm2 = 35 kg/cm2",
null,
"QUESTION: 12\n\nConsider the following statements: The failure of sub-grade of a flexible pavement is mainly attributed to\n2. loss of binding action\n3. loss of base course materials\n4. excessive stress concentration\nWhich of the statements are correct?\n\nSolution:\n\nTwo basic reasons for failure of sub-grade are:\n(ii) Excessive stress application.\n\nQUESTION: 13\n\nConsider the following statements: In the design of modern concrete pavements, tie bars are used:\n2. In expansion joints\n3. In contraction joints\n4. In warping joints\nWhich of the statements given above is/are correct?\n\nSolution:\n\nTie bars are used across longitudinal joints (warping joints). They are not used a load transferring devices. Tie bars are thus designed to with stand tensile stresses, and the maximum tensile force in tie bars being equal to the force required to overcome frictional force between the bottom of the adjoining pavement slab and the soil sub-grade. Dowei bars are provided at expansion and contraction joints and they act as load transferring devices.\n\nQUESTION: 14\n\nIf the CBR value obtained at 5 mm penetration is higher than that at 2.5 mm, then the test is repeated for checking: and if the check test reveals a. similar trend, then the CBR value is to be reported as the\n\nSolution:\n\nThe average CBR value of three test specimens is reported to the first decimal place, as the GBR value of the test specimen. If the maximum variation' iii the CBR values of these specimens exceeds specified limits (3% for CB up to 10% to 5% for CBR 10 to 30% and 10% for CBR 30-60%), then the design CBR should be the average of at least six specimens.\n\nQUESTION: 15\n\nIf the load, warping and frictional stresses in a cement concrete slab are 210 N/mm2, 290 N/mm2 and 10 N/mm2 respectively, the critical combination of stresses during summer midday is\n\nSolution:\n\nCritical combination of stresses:\n(i) During Summer mid-day at bottom of slab (Load stress + warpings tress - friction stress), at edge region = 210 + 290 - 10 = 490 N/mm2\n(ii) During winter mid -day at bottom of slab\n(Load stress + warping stress + friction stress); at edge region = 210 + 290 + 10 = 510N/mm2\n(iii) At top Of slab during midnights\n(Load stress + warping stress), at corner region.\n\nQUESTION: 16\n\nIn the design of highway expansion and contraction joints should respectively be provided at\n\nSolution:\nQUESTION: 17\n\nIn the CBR design method recommended by the IRC for the design of flexible pavement, the total thickness depends upon\n\nSolution:\nQUESTION: 18\n\nIf the radius of wheel load distribution is 30 cm and slab thickness is 15 cm, then the equivalent radius of resisting section is\n\nSolution:\n\nBecause the radius of wheel load distribution is more than 1.724 times slab thickness, hence radius of resisting section is equal to the radius of wheel load distribution.\n\nQUESTION: 19\n\nThe critical combination of stresses for corner region in cement concrete roads is\n\nSolution:\nQUESTION: 20\n\nGroup index method of design of flexible pavement is\n\nSolution:\n\nThe Gl method of pavement design is essentially an empirical method based on physical properties of the sub-grade soil. This method does- not consider the strength characteristic of the subgrade soil.",
null,
"Use Code STAYHOME200 and get INR 200 additional OFF Use Coupon Code"
] | [
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null,
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null,
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https://books.google.gr/books?id=4rQ2AAAAMAAJ&dq=editions:UOMDLPabq7928_0001_001&hl=el&output=html_text&lr= | [
"# A Collection of Problems and Examples Adapted to the \"Elementary Course of Mathematics.\": With an Appendix Containing the Questions Proposed During the First Three Days of the Senate-House Examinations in the Years 1848, 1849, 1850, and 1851\n\nJ. Deighton, 1851 - 173 уелЯдет\n\n### ДзмпцйлЮ брпурЬумбфб\n\nУелЯдб 111 - If two triangles have two sides of the one equal to two sides of the...\nУелЯдб 128 - To divide a given straight line into two parts, so that the rectangle contained by the whole and one of the parts, shall be equal to the square of the other part.\nУелЯдб 111 - If a straight line be divided into two equal parts, and also into two unequal parts, the rectangle contained by the unequal parts, together with the square on the line between the points of section, is equal to the square on half the line.\nУелЯдб 112 - EQUAL straight lines in a circle are equally distant from the centre ; and those which are equally distant from the centre, are equal to one another.\nУелЯдб 144 - ... a circle. The angle in a semicircle is a right angle: the angle in a segment greater than a semicircle is less than a right angle ; and the angle in a segment less than a semicircle is greater than a right angle.\nУелЯдб 160 - Triangles upon equal bases, and between the same parallels, are equal to one another.\nУелЯдб 112 - If two triangles have one angle of the one equal to one angle of the other and the sides about these equal angles proportional, the triangles are similar.\nУелЯдб 160 - In any right-angled triangle, the square which is described upon the side subtending the right angle, is equal to the squares described upon the sides which contain the right angle."
] | [
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.68408364,"math_prob":0.9968847,"size":3144,"snap":"2023-40-2023-50","text_gpt3_token_len":887,"char_repetition_ratio":0.12515923,"word_repetition_ratio":0.087649405,"special_character_ratio":0.20642494,"punctuation_ratio":0.08785047,"nsfw_num_words":1,"has_unicode_error":false,"math_prob_llama3":0.99721324,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-09-24T07:16:46Z\",\"WARC-Record-ID\":\"<urn:uuid:996a499c-642a-4c36-b1b6-cc66772cd99b>\",\"Content-Length\":\"63421\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:786c901d-86cb-4d89-8a6b-6f72acfb19c8>\",\"WARC-Concurrent-To\":\"<urn:uuid:74964ee1-747d-4b39-8580-ae73a866873c>\",\"WARC-IP-Address\":\"172.253.115.113\",\"WARC-Target-URI\":\"https://books.google.gr/books?id=4rQ2AAAAMAAJ&dq=editions:UOMDLPabq7928_0001_001&hl=el&output=html_text&lr=\",\"WARC-Payload-Digest\":\"sha1:R5FBZTVCGAUDXPNZUG355LCS3OOD7K4C\",\"WARC-Block-Digest\":\"sha1:R6DVHUAUKHCM7AQ533N46ZVC3ZFVHLXB\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-40/CC-MAIN-2023-40_segments_1695233506623.27_warc_CC-MAIN-20230924055210-20230924085210-00382.warc.gz\"}"} |
https://how-many-ounces-in-a-pound.com/how-many-ounces-in-2-31-lb-2-31 | [
"# How many ounces in 2.31 lb",
null,
"Info\n\n# How many ounces in 2.31 lb\n\n## There are sixteen ounces in one pound!\n\nConvert 2.31 pounds to ounces. Pounds to ounces conversion.\n\nUse the pounds to ounces calculator below to find how many ounces in any quantity of pounds. This calculator accepts fractional values like: 1/2, 2 2/3 etc.\n\n### Pounds to Ounces Converter\n\n Pounds Ounces You can edit any of the fields below: = Detailed result here\n\nTo use this calculator, simply type the value in any box at left or at right. It accepts fractional values.\n\nUsing this converter you can get answers to questions like:\n\n• How many ounces are in 2.31 pounds.?\n• 2.31 pounds is equal to how many ounces?\n• How to convert pounds to ounces? among others.\n\n## How to convert pounds to (avoirdupois) ounces\n\nTo calculate an ounce value to the corresponding value in pounds, just multiply the quantity in pound by 16 (once the are sixteen ounces in a pound).\n\nSee also How many ounces in 9.24 pounds\n\n### Pound to ounces formula\n\nOunces = Pounds x 16\n\n#### Example:\n\nHow many ounces in a quarter pound?\n\nWe know that Ounces = Pounds x 16, so,\n\n1/4 x 16 = 16/4 = 4 ounces.\n\n## How to convert pounds to ounces\n\nTo calculate a pound value to the corresponding value in ounces, just divide the quantity in ounces by 16 once there are sixteen ounces in each pound.\n\n### Ounces to Pound formula\n\nPounds = Ounces / 16\n\n#### Example:\n\nConvert 12 ounces to pounds?\n\nWe know that Pounds = Ounces / 16, so,\n\n12 ounces = 12 / 16 = 6 / 8 = 3 / 4 or 0.75 pound.\n\n## Ounces to pound conversion table\n\nDecimal Fraction Ounces\n.0625 1/16 1\n.125 1/8 2\n.1875 3/16 3\n.25 1/4 4\n.3125 5/16 5\n.375 3/8 6\n.4375 7/16 7\n.5 1/2 8\n.5625 9/16 9\n.625 5/8 10\n.6875 11/16 11\n.75 3/4 12\n.8125 13/16 13\n.875 7/8 14\n.9375 15/16 15\n1 1/1 16\n1.25 1 1/4 20\n1.5 1 1/2 24\n1.75 1 3/4 28\n2 2/2 32\n\nPounds Ounces\n1.91 30.56\n2.01 32.16\n2.11 33.76\n2.21 35.36\n2.31 36.96\n2.41 38.56\n2.51 40.16\n2.61 41.76\n2.71 43.36\n2.81 44.96\n\nRate article"
] | [
null,
"https://how-many-ounces-in-a-pound.com/wp-content/uploads/2021/05/2-31_Lbs_to_Oz-5493562.png",
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.7513333,"math_prob":0.9938319,"size":2370,"snap":"2022-40-2023-06","text_gpt3_token_len":876,"char_repetition_ratio":0.1986475,"word_repetition_ratio":0.07424594,"special_character_ratio":0.4029536,"punctuation_ratio":0.14727272,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9765881,"pos_list":[0,1,2],"im_url_duplicate_count":[null,2,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-09-28T19:36:50Z\",\"WARC-Record-ID\":\"<urn:uuid:462aa485-9b1a-464a-a9e0-5420860f252c>\",\"Content-Length\":\"56423\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:6ca6362b-2ba3-49f3-8bc1-86222a09c0e7>\",\"WARC-Concurrent-To\":\"<urn:uuid:39d678c0-bae4-4d54-bcfa-de5851fd5c8d>\",\"WARC-IP-Address\":\"209.59.169.213\",\"WARC-Target-URI\":\"https://how-many-ounces-in-a-pound.com/how-many-ounces-in-2-31-lb-2-31\",\"WARC-Payload-Digest\":\"sha1:IZIRBCO5H5K3QBVF5KEI6GXMI33PSQOS\",\"WARC-Block-Digest\":\"sha1:HQYQ7ADAHVA3NHVWWHWWSFI6KBPYCRFH\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-40/CC-MAIN-2022-40_segments_1664030335276.85_warc_CC-MAIN-20220928180732-20220928210732-00656.warc.gz\"}"} |
https://socratic.org/questions/how-do-you-simplify-0-10-x-10-5-4-9-x-10-2 | [
"# How do you simplify (0.10 x 10^5)(4.9 x 10^-2)?\n\n$\\left(0.10 \\times {10}^{5}\\right) \\left(4.9 \\times {10}^{- 2}\\right)$\n$= \\left(0.10 \\times 4.9\\right) \\left({10}^{5} \\times {10}^{- 2}\\right)$\n$= 0.49 \\times {10}^{3}$\n$= 4.9 \\times {10}^{2}$"
] | [
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.6384677,"math_prob":1.0000097,"size":302,"snap":"2022-40-2023-06","text_gpt3_token_len":78,"char_repetition_ratio":0.110738255,"word_repetition_ratio":0.0,"special_character_ratio":0.28476822,"punctuation_ratio":0.083333336,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9991955,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-10-04T07:05:34Z\",\"WARC-Record-ID\":\"<urn:uuid:88bcdcf9-90c8-4a69-914e-c1fb17c5871c>\",\"Content-Length\":\"32270\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:57d9c3bd-fb3a-4c08-9856-44b2f6e2c43e>\",\"WARC-Concurrent-To\":\"<urn:uuid:9d6e95f2-bf2a-425b-9233-f42fa62c5770>\",\"WARC-IP-Address\":\"216.239.32.21\",\"WARC-Target-URI\":\"https://socratic.org/questions/how-do-you-simplify-0-10-x-10-5-4-9-x-10-2\",\"WARC-Payload-Digest\":\"sha1:JA4XY7T27BOG5YNY53DTQCWKOQJSOOAF\",\"WARC-Block-Digest\":\"sha1:THFCCLED5P6ZCU7MGDPOHCXE7HKFTNJW\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-40/CC-MAIN-2022-40_segments_1664030337480.10_warc_CC-MAIN-20221004054641-20221004084641-00042.warc.gz\"}"} |
https://cxc.cfa.harvard.edu/ciao/ahelp/grpgetgrpnum.html | [
"AHELP for CIAO 4.13\n\n# grpGetGrpNum\n\nContext: group\n\n## Synopsis\n\nCalculate the group number for each element in the array.\n\n## Syntax\n\n`Integer_Type grpnum = grpGetGrpNum( PyArray_Type grouping )`\n\n## Description\n\nThis function calculates which group each element in the input array belongs to, where the groups are numbered from 1. The return value is an array whose length equals that of the input data (the grouping argument) and each element within a group contains the same value. As an example,\n\n`grpGetGroupNum( numpy.array([1,-1,1,-1,1,1]) )`\n\nwill return the array\n\n`array([1,1,2,2,3,4])`\n\nsince the groups consist of the first two elements, then the next two elements, with the last two elements being un-grouped.\n\nThe group module is not available by default; to use it in a Python program, it must be loaded using the Python import function:\n\n` from group import *, or import group`\n\nIn addition, in order to create arrays, the Python module numpy must be loaded using the Python import function:\n\n` import numpy`\n\n## Examples\n\n### Example 1\n\n`>>> igrp = grpGetGroupNum( grp )`\n\nThe igrp array will contain the group number for each element of the input array, for the grouping scheme stored in the grp array, and has the same size as the grp array.\n\n### Example 2\n\n```>>> x = numpy.arange(0.5, 6.0, 0.05)\n>>> y = 3 + 30 * numpy.exp( - (x-2.0)**2 / 0.1 )\n>>> ( grp, qual ) = grpNumCounts( y, 15 )\n>>> yavg = grpGetGroupSum( y, grp ) / grpGetChansPerGroup( grp )\n>>> igrp = grpGetGroupNum( grp )\n>>> i = numpy.where( grp == 1, qual, grp )\n>>> j = numpy.where(i == 0)```\n\nHere we take the function\n\n`y = 3 + 30 * numpy.exp( - (x-2.0)**2 / 0.1 )`\n\nand group it by 15 counts per group.\n\n### Example 3\n\nTo see more examples of optional parameters and keywords in the parameter list, please see ahelp grpadaptive\n\n## Bugs\n\nSee the bugs page for the group library on the CIAO website for an up-to-date listing of known bugs.\n\nRefer to the CIAO bug pages for an up-to-date listing of known issues."
] | [
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.6018574,"math_prob":0.98305845,"size":1886,"snap":"2021-43-2021-49","text_gpt3_token_len":516,"char_repetition_ratio":0.14612114,"word_repetition_ratio":0.10876133,"special_character_ratio":0.28313893,"punctuation_ratio":0.13648294,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9932117,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-10-23T10:18:03Z\",\"WARC-Record-ID\":\"<urn:uuid:801641cd-ae51-482e-bf0a-40f2e0a18f7a>\",\"Content-Length\":\"63363\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:2b25d701-25fc-44da-b455-165049177c77>\",\"WARC-Concurrent-To\":\"<urn:uuid:bedd1015-752e-4d73-833d-dc64206454d5>\",\"WARC-IP-Address\":\"131.142.198.51\",\"WARC-Target-URI\":\"https://cxc.cfa.harvard.edu/ciao/ahelp/grpgetgrpnum.html\",\"WARC-Payload-Digest\":\"sha1:RS7E6BQEN2GIOHTXY6RTA3BMTTG6IKXV\",\"WARC-Block-Digest\":\"sha1:ICLK6T7W5LHYNNBWPELCJVXPI3FAY7XN\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-43/CC-MAIN-2021-43_segments_1634323585671.36_warc_CC-MAIN-20211023095849-20211023125849-00080.warc.gz\"}"} |
http://people.sc.fsu.edu/~jburkardt/m_src/two_body_simulation/two_body_simulation.html | [
"TWO_BODY_SIMULATION Planar Two Body Problem Simulation\n\nTWO_BODY_SIMULATION, a MATLAB library which simulates the solution of the planar two body problem.\n\nTwo bodies, regarded as point masses, are constrained to lie in a plane. The masses of each body are given, as are the positions and velocities at a starting time T = 0. The bodies move in accordance with the gravitational force between them. One body is assume to be much more massive than the other. Therefore, the common motion of the two bodies about their center of mass can be approximated by assuming that the large body remains fixed.\n\nUnder these assumptions, Newton's equations for (x(t),y(t)), the positition of the lighter body with respect to the heavy body at (0,0), can be written as:\n\n``` r(t) = sqrt ( x(t)^2 + y(t)^2 )\nx''(t) = - x(t) / r^3\ny''(t) = - y(t) / r^3\n```\n\nThese two second order equations can be rewritten as a system of four first order equations using the variable u = [ x(t), x'(t), y(t), y'(t) ], resulting in the equations:\n\n``` r = sqrt ( u(1)^2 + u(3)^2 )\nu'(1) = u(2)\nu'(2) = - u(1) / r^3\nu'(3) = u(4)\nu'(4) = - u(3) / r^3\n```\n\nBy specifying some initial condition for u, the system can then be integrated in time using a standard ODE solver.\n\nLanguages:\n\nTWO_BODY_SIMULATION is available in a FORTRAN90 version and a MATLAB version.\n\nRelated Data and Programs:\n\nBROWNIAN_MOTION_SIMULATION, a MATLAB program which simulates Brownian motion in an M-dimensional region.\n\nCOIN_SIMULATION, a MATLAB library which looks at ways of simulating or visualizing the results of many tosses of a fair or biased coin.\n\nDICE_SIMULATION, a MATLAB program which simulates N tosses of M dice, making a histogram of the results.\n\nDUEL_SIMULATION, a MATLAB program which simulates N repetitions of a duel between two players, each of whom has a known firing accuracy.\n\nGAMBLERS_RUIN_SIMULATION, a MATLAB program which simulates the game of gambler's ruin.\n\nHIGH_CARD_SIMULATION, a MATLAB program which simulates a situation in which you see the cards in a deck one by one, and must select the one you think is the highest and stop.\n\nISING_2D_SIMULATION, a MATLAB program which carries out a Monte Carlo simulation of an Ising model, a 2D array of positive and negative charges, each of which is likely to \"flip\" to be in agreement with neighbors.\n\nLIGHTS_OUT, a MATLAB program which sets up the \"Lights Out\" game and allows a user to try to solve it.\n\nLORENZ_SIMULATION, a MATLAB program which solves the Lorenz equations and displays the solution, for various starting conditions.\n\nPOISSON_SIMULATION, a MATLAB library which simulates a Poisson process in which events randomly occur with an average waiting time of Lambda.\n\nRANDOM_WALK_1D_SIMULATION, a MATLAB program which simulates a random walk in a 1-dimensional region.\n\nRANDOM_WALK_2D_AVOID_SIMULATION, a MATLAB program which simulates a self-avoiding random walk in a 2-dimensional region.\n\nRANDOM_WALK_2D_SIMULATION, a MATLAB program which simulates a random walk in a 2-dimensional region.\n\nRANDOM_WALK_3D_SIMULATION, a MATLAB program which simulates a random walk in a 3-dimensional region.\n\nREACTOR_SIMULATION, a MATLAB program which a simple Monte Carlo simulation of the shielding effect of a slab of a certain thickness in front of a neutron source. This program was provided as an example with the book \"Numerical Methods and Software.\"\n\nRKF45, a MATLAB library which implements the Runge-Kutta-Fehlberg ODE solver.\n\nSIR_SIMULATION, a MATLAB program which simulates the spread of a disease through a hospital room of M by N beds, using the SIR (Susceptible/Infected/Recovered) model.\n\nTHREE_BODY_SIMULATION, a MATLAB program which simulates the behavior of three planets, constrained to lie in a plane, and moving under the influence of gravity, by Walter Gander and Jiri Hrebicek.\n\nTRAFFIC_SIMULATION, a MATLAB program which simulates the cars waiting to get through a traffic light.\n\nTRUEL_SIMULATION, a MATLAB program which simulates N repetitions of a duel between three players, each of whom has a known firing accuracy.\n\nReference:\n\n1. Charles VanLoan,\nIntroduction to Scientific Computing,\nPrentice Hall, 1997,\nISBN: 0-13-125444-8,\nLC: QA76.9.M35V375.\n\nSource Code:\n\n• kepler.m, evaluates the right hand side of the ODE system.\n• timestamp.m, prints the current YMDHMS date as a time stamp."
] | [
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.9083119,"math_prob":0.98389715,"size":4076,"snap":"2019-26-2019-30","text_gpt3_token_len":987,"char_repetition_ratio":0.21782908,"word_repetition_ratio":0.09969325,"special_character_ratio":0.22620216,"punctuation_ratio":0.10386152,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99951804,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-06-19T15:47:07Z\",\"WARC-Record-ID\":\"<urn:uuid:625a87e2-09dc-42e8-a8c8-dde0b00f52dd>\",\"Content-Length\":\"8717\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:9bf6116e-53d8-462e-809d-a0452bd03614>\",\"WARC-Concurrent-To\":\"<urn:uuid:5998cc22-22e9-4d4b-b383-c8e731b35c22>\",\"WARC-IP-Address\":\"144.174.16.100\",\"WARC-Target-URI\":\"http://people.sc.fsu.edu/~jburkardt/m_src/two_body_simulation/two_body_simulation.html\",\"WARC-Payload-Digest\":\"sha1:YHIUS4BIYC5BMX53IXCK3MZOIEM5OMLO\",\"WARC-Block-Digest\":\"sha1:MIBUEJHBWYPCMB4K34TGO2FTVRBCWU44\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-26/CC-MAIN-2019-26_segments_1560627999000.76_warc_CC-MAIN-20190619143832-20190619165832-00314.warc.gz\"}"} |
https://gtsam.org/doxygen/a01292_source.html | [
"gtsam 4.1.0 gtsam\nQPSolver.h\nGo to the documentation of this file.\n1 /* ----------------------------------------------------------------------------\n2\n3 * GTSAM Copyright 2010, Georgia Tech Research Corporation,\n4 * Atlanta, Georgia 30332-0415\n6 * Authors: Frank Dellaert, et al. (see THANKS for the full author list)\n7\n9\n10 * -------------------------------------------------------------------------- */\n11\n23 #include <limits>\n24 #include <algorithm>\n25\n26 namespace gtsam {\n27\n29 struct QPPolicy {\n32 static constexpr double maxAlpha = 1.0;\n33\n35 static const GaussianFactorGraph buildCostFunction(const QP& qp,\n36 const VectorValues& xk = VectorValues()) {\n37 GaussianFactorGraph no_constant_factor;\n38 for (auto factor : qp.cost) {\n39 HessianFactor hf = static_cast<HessianFactor>(*factor);\n40 no_constant_factor.push_back(hf);\n41 }\n42 return no_constant_factor;\n43 }\n44 };\n45\n46 using QPSolver = ActiveSetSolver<QP, QPPolicy, QPInitSolver>;\n47\n48 }\ngtsam::GaussianFactorGraph\nA Linear Factor Graph is a factor graph where all factors are Gaussian, i.e.\nDefinition: GaussianFactorGraph.h:68\ngtsam::QP::cost\nGaussianFactorGraph cost\nDefinition: QP.h:32\ngtsam::QP\nStruct contains factor graphs of a Quadratic Programming problem.\nDefinition: QP.h:31\nActiveSetSolver.h\nActive set method for solving LP, QP problems.\nQPInitSolver.h\nThis finds a feasible solution for a QP problem.\ngtsam::FactorGraph::push_back\nIsDerived< DERIVEDFACTOR > push_back(boost::shared_ptr< DERIVEDFACTOR > factor)\nAdd a factor directly using a shared_ptr.\nDefinition: FactorGraph.h:162\ngtsam::QPPolicy\nPolicy for ActivetSetSolver to solve Linear Programming.\nDefinition: QPSolver.h:29\ngtsam::QPPolicy::maxAlpha\nstatic constexpr double maxAlpha\nMaximum alpha for line search x'=xk + alpha*p, where p is the cost gradient For QP,...\nDefinition: QPSolver.h:32\ngtsam::VectorValues\nThis class represents a collection of vector-valued variables associated each with a unique integer i...\nDefinition: VectorValues.h:74\ngtsam\nGlobal functions in a separate testing namespace.\nDefinition: chartTesting.h:28\nQP.h\nFactor graphs of a Quadratic Programming problem.\ngtsam::HessianFactor\nA Gaussian factor using the canonical parameters (information form)\nDefinition: HessianFactor.h:101\ngtsam::QPPolicy::buildCostFunction\nstatic const GaussianFactorGraph buildCostFunction(const QP &qp, const VectorValues &xk=VectorValues())\nSimply the cost of the QP problem.\nDefinition: QPSolver.h:35"
] | [
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https://blog.learnbay.co/basics-of-functions-in-python-a-glance | [
"",
null,
"# Basics of Functions In Python - A Glance\n\nBy Vineeth Kumar Category Python Reading time 9-10 mins Published on Sep 9, 2022\n\nTransforming the data in an effective way and then utilizing them in an optimized manner while programming is known as Functions in python. Also, executing the block only when it is called is known as a function.\n\nWith the help of 'functions' programs, you can break into smaller chunks of code. When you need to write the same code several times you can use that code in a function and call that function whenever you want to execute that code. In this article let me take you through different functions used in python along with some of the examples.\n\n## Defining a function\n\nA group of reusable code that is called anywhere in a program eliminates the need for writing the same code repeatedly. A function helps programmers for writing modular codes. When you write a function inside a class, it becomes a method. After the function is defined we can use it by calling the function 'fun()'.\n\nA function is used to provide the necessary functionality. Simple rules for defining a function are as follows:\n\n• The 'def' keyword is used to begin a function, which is then followed by the name and parentheses'()'.\n• You can input any parameters or arguments within the parentheses.\n• For documentation, the first function statement can be made optional or docstring.\n• Every function having a code block starts with a colon (:) which is indented.\n• The return(expression) statement retains a function by providing the caller with a returned expression.\n\nExample: Defining a function:\n\n| # A simple python function\n\ndef fun():\n\nprint(\"Hello Data Science\")\n\nExample: Calling a function:\n\nOnce functions are created you can call them by using the name of that function which is followed by a parenthesis that contains parameters for that particular function.\n\n| # A simple python function\n\ndef fun():\n\nprint(\"Hello Data Science\")\n\n# Driver code to call a function\n\nfun()\n\nExample:\n\nDepicting the first call to the function and second call to the same function.\n\ndef my_function(Fun):\nprint(Fun + \"Classes\")\n\nmy_function(\"Hello\")\nmy_function(\"Data\")\nmy_function(\"Science\")\n\nOutput\nHello Classes\nData Classes\nScience Classes\n\n## What are the various types of functions in python?\n\nIn python, you have two important types of functions\n\n• Built-in functions\n• User-defined functions\n\n### Built-in functions\n\nDefault functions which are already programmed in python software called 'Built -in function'. You have a wide range of built-in functions available in python. Some of the built-in functions are\n\n• int\n• list\n• tuple\n• float\n• string\n• set\n• bin\n• oct\n• hex\n• dictionary\n\n### User-defined function\n\nThe functions specifically you write to define certain tasks fall under user-defined functions. It can also be a library for someone else. If you are working on a larger project you can break the workload by maintaining different functions.\n\nExample:\n\n# program for user-defined functions\n\ndef multiply_num(x,y):\nmul = x*y\nreturn mul\n\nnum1 = 5\nnum2 = 6\nprint(\"The Product of: \", multiply_num(num1,num2)\n\nThe Product of: 30",
null,
"### Recursive Functions in Python\n\n#### What do you mean by recursion in python?\n\nThe process of defining anything by itself is known as recursion. A function has the ability to call other functions. Retrieving the own function is possible by using a function.\n\nExample:\n\n| # To find the factorial number\ndef calc_fact(x):\nif x == 1:\nreturn 1\nelse:\nreturn(x * calc_fact(x-1))\nnum = 4\nprint (\"Factorial of\", num, \"is\", calc_fact(num))\n\nIn the above example, calc_fact() is a recursive function in python as it calls the function by itself. The function call multiples with the factorial number 1 until it is equal to 1.\n\nNote : If a recursive function does not provide a base condition that ends the recursion, the function will call itself, indefinitely.\n\n### Lamda Function\n\n#### What are Lambda Functions in python?\n\nLambda Functions in python are also known as anonymous functions which means the function is without a name. As you use the 'def' keyword for normal function in python similarly, the Anonymous function uses the lambda keyword to define the function in python.\n\nThe syntax for lambda = lambda arguments: expression\n\n• Any number of arguments can be there in this function however only one expression gets evaluated and returned.\n• If the function objects are required you can use Lambda functions.\n• Grammatical arrangements are minimized to a single expression in lambda functions.\n• lambda has many uses apart from other types of expressions in a particular area of programming.\n\nExample:\n\nstring = 'HellodataScience'\n#Lambda returns a function object\nprint(lambda string : string)\n\nOutput\n<function at 0x0000016282865820 >\n\n## Take away\n\nAs a programmer a precise and optimal way of writing code would help better rather than having complications. Functions would be a better option as you can break the program into smaller programs.\n\nIf you're interested in learning how to use python effectively, check out the data science course with special python help from an industrial specialist and best-in-class resources. If you're just getting started with Python follow us on Facebook, Youtube, Linkedin, Twitter."
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https://dev.to/thepracticaldev/daily-challenge-17-double-trouble-1ee6 | [
"# Daily Challenge #17 - Double Trouble\n\nGood morning! We're back with another challenge which comes from AlexisHappy on Codewars. Today, you are asked to write a program that will return the name of the person who will drink the n-th Cola at the vending machine\n\nSheldon, Leonard, Penny, Rajesh and Howard are in the queue for a \"Double Cola\" drink vending machine; there are no other people in the queue. The first one in the queue (Sheldon) buys a can, drinks it and doubles! The resulting two Sheldons go to the end of the queue. Then the next in the queue (Leonard) buys a can, drinks it and gets to the end of the queue as two Leonards, and so on.\n\nFor example, Penny drinks the third can of cola and the queue will look like this:\n\nRajesh, Howard, Sheldon, Sheldon, Leonard, Leonard, Penny, Penny\n\nWrite a program that will return the name of the person who will drink the n-th cola. The input data consist of an array which contains at least 1 name, and single integer n which may go as high as the biggest number your language of choice supports (if there's such limit, of course). The output data returns the single line — the name of the person who drinks the n-th can of cola. The cans are numbered starting from 1.\n\nGood luck, and happy coding!\n\nThank you to CodeWars, who has licensed redistribution of this challenge under the 2-Clause BSD License!\n\nWant to propose a challenge for a future post? Email [email protected] with your suggestions!",
null,
"Alvaro Montoro • Edited\n\nJavaScript\n\n``````const cola = (number, people) => {\nconst length = people.length;\nconst iteration = Math.floor(Math.log2(((number - 1) / length) + 1));\nconst post = 2 ** iteration;\n\nreturn people[Math.floor((number - ((post - 1) * length + 1)) / post)]\n}\n``````\n\nThe way it works, the series follows an exponential pattern in which the first iteration has n elements (5 in this case), and from then on each iteration has double the elements than the previous iteration:\n\nIteration(0) = S, L, P, R, H (5 elements)\nIteration(1) = S, S, L, L, P, P, R, R, H, H (10 elements)\nIteration(2) = S, S, S, S, L, L, L, L, P, P, P, P, R, R, R, R, H, H, H, H (20 elements)\nIteration(3) = S, S, S, S, S, S, S, S, S, L, L, L, L, L, L, L, L, ... (40 elements);\nIteration(4) = S, S, S, S, S, S, S, S, S, S, S, S, S, S, S, S, S, S, ... (80 elements)\n\nWith that in mind, it is possible to calculate in which iteration we are by doing log2(number / 5), and from there we can get which element within the actual iteration we need to check. After doing some numbers, trying to figure out the pattern, and all that jazz, I think this may work... although I haven't fully tested it.\n\nLive demo on Codepen.",
null,
"Olivier “Ölbaum” Scherler • Edited\n\nI couldn’t figure out the formula like @alvaromontoro did, but here’s my blunt Erlang version. It was really easy to write and it still runs under a second for N = 10 million:\n\n``````-module( double ).\n\n-include_lib(\"eunit/include/eunit.hrl\").\n\n% when it’s the first person, return the head\ndouble( [ Person | Rest ], 1, More ) ->\nPerson;\n% when it’s not the first person, add the head to the More list twice and decrement N\n% the More list is the rest of the queue, where we can append at the head\n% imagine that the full queue at all times is Queue ++ lists:reverse( More )\ndouble( [ Person | Rest ], N, More ) ->\ndouble( Rest, N - 1, [ Person, Person | More ] );\n% when the queue is empty, take the More list, reverse it and start over\ndouble( [], N, More ) ->\ndouble( lists:reverse( More ), N, [] ).\n\ndouble( Queue, N ) when N > 0 ->\ndouble( Queue, N, [] ).\n\ndouble_test_() ->\nThree = [ sheldon, leonard, penny ],\nEveryone = [ sheldon, leonard, penny, rajesh, howard ],\n[\n?_assertEqual( sheldon, double( [ sheldon ], 1 ) ),\n?_assertEqual( sheldon, double( [ sheldon ], 2 ) ),\n?_assertEqual( sheldon, double( [ sheldon ], 3 ) ),\n?_assertEqual( sheldon, double( [ sheldon ], 4 ) ),\n\n?_assertEqual( sheldon, double( [ sheldon, leonard ], 1 ) ),\n?_assertEqual( leonard, double( [ sheldon, leonard ], 2 ) ),\n?_assertEqual( sheldon, double( [ sheldon, leonard ], 3 ) ),\n?_assertEqual( sheldon, double( [ sheldon, leonard ], 4 ) ),\n?_assertEqual( leonard, double( [ sheldon, leonard ], 5 ) ),\n?_assertEqual( leonard, double( [ sheldon, leonard ], 6 ) ),\n\n?_assertEqual( sheldon, double( Everyone, 1 ) ),\n?_assertEqual( penny, double( Everyone, 3 ) ),\n?_assertEqual( sheldon, double( Everyone, 6 ) ),\n?_assertEqual( leonard, double( Everyone, 8 ) ),\n?_assertEqual( leonard, double( Everyone, 9 ) ),\n?_assertEqual( sheldon, double( Everyone, 19 ) ),\n\n?_assertEqual( sheldon, double( Three, 1 ) ),\n?_assertEqual( leonard, double( Three, 2 ) ),\n?_assertEqual( penny, double( Three, 3 ) ),\n?_assertEqual( sheldon, double( Three, 4 ) ),\n?_assertEqual( sheldon, double( Three, 5 ) ),\n?_assertEqual( leonard, double( Three, 6 ) ),\n?_assertEqual( leonard, double( Three, 7 ) ),\n?_assertEqual( penny, double( Three, 8 ) ),\n?_assertEqual( penny, double( Three, 9 ) ),\n?_assertEqual( sheldon, double( Three, 10 ) ),\n?_assertEqual( sheldon, double( Three, 11 ) ),\n?_assertEqual( sheldon, double( Three, 12 ) ),\n?_assertEqual( sheldon, double( Three, 13 ) ),\n?_assertEqual( leonard, double( Three, 14 ) ),\n?_assertEqual( leonard, double( Three, 15 ) ),\n?_assertEqual( leonard, double( Three, 16 ) ),\n?_assertEqual( leonard, double( Three, 17 ) ),\n?_assertEqual( penny, double( Three, 18 ) ),\n?_assertEqual( penny, double( Three, 19 ) ),\n?_assertEqual( penny, double( Three, 20 ) ),\n?_assertEqual( penny, double( Three, 21 ) ),\n?_assertEqual( sheldon, double( Three, 22 ) )\n].\n``````",
null,
"E. Choroba • Edited\n\nIterating over the whole queue is slow when the number is really large. I wasn't able to find an exact formula, but I was able to find a good upper bound from which I can find the exact solution in linear time.\n\nThe solution is in Perl:\n\n``````#!/usr/bin/perl\nuse warnings;\nuse strict;\n\nsub slow {\nmy (\\$queue, \\$n) = @_;\npush @\\$queue, (shift @\\$queue) x 2 while --\\$n;\nreturn \\$queue->\n}\n\nsub fast {\nmy (\\$queue, \\$n) = @_;\nreturn \\$queue-> if 1 == \\$n;\nmy \\$upper_bound = 5 + int(log(\\$n / @\\$queue) / log(2));\n\nmy \\$x;\ndo {\n--\\$upper_bound;\n\\$x = (2 ** \\$upper_bound - 1) * @\\$queue + 1;\n} until \\$x <= \\$n;\nreturn \\$queue->[(\\$n - \\$x) / (2 ** \\$upper_bound)]\n}\n\nuse Test::More;\n\nis slow([qw[ Sheldon Leonard Penny Rajesh Howard ]], 3), 'Penny';\nis slow([qw[ Sheldon Leonard Penny Rajesh Howard ]], 7), 'Sheldon';\nis slow([qw[ Sheldon Leonard Penny Rajesh Howard ]], 8), 'Leonard';\n\nmy @queue;\nfor my \\$char ('a' .. 'z') {\npush @queue, \\$char;\nis fast(\\@queue, \\$_), slow([@queue], \\$_),\n\"t_\\$_\\_\" . scalar @queue for 2 .. 2000;\n}\ndone_testing();\n``````\n\nBenchmarking is easy using the Benchmark module:\n\n``````use Benchmark qw{ cmpthese };\ncmpthese(-2, {\nslow_1000 => sub { slow(['a' .. 'h'], 1_000) },\nfast_1000 => sub { fast(['a' .. 'h'], 1_000) },\nslow_10_000 => sub { slow(['a' .. 'h'], 10_000) },\nfast_10_000 => sub { fast(['a' .. 'h'], 10_000) },\n});\n``````\n\nThe output shows a table ordered from the slowest to the fastest.\n\n`````` Rate slow_10_000 slow_1000 fast_1000 fast_10_000\nslow_10_000 271/s -- -90% -100% -100%\nslow_1000 2845/s 952% -- -99% -99%\nfast_1000 415531/s 153512% 14505% -- -0%\nfast_10_000 415531/s 153512% 14505% 0% --\n``````\n\nUnfortunately, it breaks for longer sequences, as the rounding error becomes too large. Switching to arbitrary precision is too slow, increasing the upper bound helps - the failures are easy to detect as the function returns `undef` instead of an element of the queue.\n\n``````sub fast {\nmy (\\$queue, \\$n, \\$correction) = @_;\nreturn \\$queue-> if 1 == \\$n;\n\\$correction ||= 4;\nmy \\$upper_bound = \\$correction\n+ int((log \\$n - log @\\$queue) / log 2);\n\nmy \\$x;\ndo {\n\\$x = (2 ** --\\$upper_bound - 1) * @\\$queue + 1;\n} until \\$x <= \\$n;\nmy \\$r = \\$queue->[(\\$n - \\$x) / (2 ** \\$upper_bound)]\n// fast(\\$queue, \\$n, \\$correction + 2);\nreturn \\$r\n}\n``````",
null,
"Yozen Hernandez • Edited\n\nSo I think I was able to get this with an exact calculation. I thought it was going to turn out to be pretty convoluted, but I guess it's not. I think got it right, at least:\n\n``````#!/usr/bin/perl\n\nuse v5.24;\nuse strict;\nuse warnings;\nuse feature qw(signatures);\nno warnings \"experimental::signatures\";\nuse Carp;\nuse List::Util qw(sum0);\nuse POSIX qw(ceil);\n\nsub cola ( \\$n, @queue ) {\nmy \\$q_len = @queue;\nmy \\$i = 0;\nmy \\$last_idx = @queue;\n\n\\$last_idx += ( ( 2**++\\$i ) * @queue ) while ( \\$n > \\$last_idx );\n\nmy \\$prev_last_idx = \\$last_idx - ( ( 2**(\\$i) ) * @queue );\nmy \\$nth_can = ceil( ( \\$n - \\$prev_last_idx ) / 2**\\$i ) - 1;\nreturn \\$queue[\\$nth_can];\n}\n\nmy @queue = qw(Sheldon Leonard Penny Rajesh Howard);\n\nuse Test::More tests => 4;\nis( cola( 4, @queue ), \"Rajesh\" );\nis( cola( 6, @queue ), \"Sheldon\" );\nis( cola( 12, @queue ), \"Rajesh\" );\nis( cola( 27, @queue ), \"Penny\" );\n``````\n\nBenchmarking with `timethis` in `Benchmark` got me this result:\n\n``````use Benchmark qw(timethis);\n\ntimethis( -2, sub{ cola( 12, @queue ) } );\n\n#timethis for 2: 2 wallclock secs ( 2.00 usr + 0.00 sys = 2.00 CPU) @ 859529.50/s (n=1719059)\n``````\n\nBasically, I just keep track of how many nodes there are in a tree structure which represents the doubling of the people in the queue. In each level of the tree, you double the number of nodes in the level before it, but we want the running total so far (which I call `\\$last_index` above because that is the index of the last node in the level where the `\\$n`th cola would be).\n\nOnce you figure out which \"level\" of the tree the `\\$n`th cola would be, you get the last index of the level before so that you can zero out the current level and any indices there. That makes it easy to figure out \"where\" on the current level of the tree the `\\$n`th cola is, and you just divide by the depth of the tree so far to turn that into an index that works on the initial array. Subtract one to get the 0-index value into the original array.\n\nEdit Looks like Corey figured out the same thing, but with more math.",
null,
"K.V.Harish • Edited\n\nNot a better solution compared to @alvaromontoro but here is mine in js\n\n``````\nconst doubleTrouble = (queue, nthCola) => {\nlet person;\nfor(let index = 0; index < nthCola; index++) {\nif(index+1 === nthCola) {\nperson = queue;\nbreak;\n}\nqueue = queue.flat(queue.push(Array(2).fill(queue.shift())));\n}\nreturn person;\n};\n\n``````",
null,
"Bárbara Perdigão • Edited\n\nMy entry, written in Java:\n\n``````\nprotected static void doubleTrouble() {\n\nList<String> people = new ArrayList<>();\n\nfor (int i = 0; i < Integer.MAX_VALUE; i++) {\nString first = people.iterator().next();\nString person = people.get(people.indexOf(first));\npeople.remove(first);\n\n}\n\nSystem.out.println(people.get(people.size() - 1));\n}\n\n``````",
null,
"Philip Hallstrom\n\nRuby.\n\n``````def drinker(queue = [], nth = 1)\ndrinker = nil\ncan = 0\nwhile can < nth\ndrinker = queue.shift\ndrinker = {name: drinker, arity: 1} unless drinker.is_a?(Hash)\ncan += drinker[:arity]\ndrinker[:arity] *= 2\nqueue << drinker\nend\n\ndrinker[:name]\nend\n``````",
null,
"Corey Alexander • Edited\n\nThis was one of my favorites in a while!\n\nI had a feeling I could get this answer in constant time, by just 'mathing' the value out! So I got out my good old pen and paper, and started working through it!\n\nThe repeating pattern of the list made me start thinking of the long list in cycles. Where one cycle was each person in line going as many times as they are supposed to in that cycle. I thought I would be able to determine how long the list would need to be to 'index' into it with the `soda number`, based on how many cycles it took! And we can! Since the cycles lengths follow the patter of: P, 2P, 3P, etc. I knew this could be represented as a summation formula, and I was pretty sure it had an algebraic solution! After a bit of googling I was able to find it! Applied to this problem the formula is as follows:\n\n``````P = Number of People in the Line\nC = Number of Cycles\n\nLength = P / 2 * C * (C+1)\n``````\n\nFrom here, I wanted to kinda reverse that formula. I wanted to know how many cycles it would take to achieve a specific length. After re-arranging that formula I was able to use the quadratic equation to rearrange, and solve for C, given Length and P. The quadratic formula gives two answers, and we are able to throw out one since we can guarantee it is negative and not relevant to our problem.\nSo now we have this!\n\n``````C = (-1 + sqrt(1 + * (8 / P) * Length)) / 2\n``````\n\nSo now if we take the soda bottle number input as the Length, and the number of people in line as P, we can get a discrete value of the number of cycles it will take!\n\nTo figure out which person got the bottle I really only needed to index into the last cycle. And know I was able to figure out which cycle that was!\nFrom here it was just some index math, to figure out which person/index we were looking for!\n\nI wanted to make sure this solution worked, so I also wrote an iterative version to compare! I wrote a quick main function that compares the two results and they were equivalent for everything I tested! I also timed the two versions to see the speed differences, and as expected the constant time solution is orders of magnitudes faster than the iterative solution!\n\n``````Math Took: 2.316656937049712 Iter Took: 77.97191874496184\n``````\n\nHere is my full rust solution and tests!\n\n``````fn find_index_iter(num_people: u64, soda_num: u64) -> u64 {\nlet mut current_group_size = num_people;\nlet mut total_size = num_people;\nlet mut total_groups = 1;\n\nwhile soda_num > total_size {\ncurrent_group_size += num_people;\ntotal_size += current_group_size;\ntotal_groups += 1;\n}\n\nlet local_group_index = current_group_size - (total_size - soda_num) - 1;\n\nlocal_group_index / (total_groups + 0)\n}\n\nfn find_index_math(num_people: usize, soda_num: u64) -> usize {\nlet soda_index = soda_num - 1;\nlet number_of_groups =\n((-1.0 + (1.0 + (8.0 / num_people as f64 * soda_num as f64)).sqrt()) / 2.0).ceil();\n\nlet num_before_last_group =\nnum_people as f64 / 2.0 * number_of_groups * (number_of_groups - 1.0);\n\nlet local_group_index = soda_index as f64 - num_before_last_group;\n\nlet ans = ((local_group_index / number_of_groups).floor()) as usize;\n\nans\n}\n\npub fn whos_soda<'a>(people: &'a [&str], soda_num: u64) -> &'a str {\npeople[find_index_math(people.len(), soda_num)]\n// people[find_index_iter(people.len() as u64, soda_num) as usize]\n}\n\n#[cfg(test)]\nmod tests {\nuse crate::*;\n\n#[test]\nfn it_for_first_few_groups_with_3_names() {\nassert_eq!(whos_soda(&[\"a\", \"b\", \"c\"], 1), \"a\");\nassert_eq!(whos_soda(&[\"a\", \"b\", \"c\"], 2), \"b\");\nassert_eq!(whos_soda(&[\"a\", \"b\", \"c\"], 3), \"c\");\nassert_eq!(whos_soda(&[\"a\", \"b\", \"c\"], 4), \"a\");\nassert_eq!(whos_soda(&[\"a\", \"b\", \"c\"], 5), \"a\");\nassert_eq!(whos_soda(&[\"a\", \"b\", \"c\"], 6), \"b\");\nassert_eq!(whos_soda(&[\"a\", \"b\", \"c\"], 7), \"b\");\nassert_eq!(whos_soda(&[\"a\", \"b\", \"c\"], 8), \"c\");\nassert_eq!(whos_soda(&[\"a\", \"b\", \"c\"], 9), \"c\");\n}\n\n#[test]\nfn it_for_first_few_groups_with_5_names() {\nassert_eq!(whos_soda(&[\"a\", \"b\", \"c\", \"d\", \"e\"], 1), \"a\");\nassert_eq!(whos_soda(&[\"a\", \"b\", \"c\", \"d\", \"e\"], 2), \"b\");\nassert_eq!(whos_soda(&[\"a\", \"b\", \"c\", \"d\", \"e\"], 3), \"c\");\nassert_eq!(whos_soda(&[\"a\", \"b\", \"c\", \"d\", \"e\"], 4), \"d\");\nassert_eq!(whos_soda(&[\"a\", \"b\", \"c\", \"d\", \"e\"], 5), \"e\");\nassert_eq!(whos_soda(&[\"a\", \"b\", \"c\", \"d\", \"e\"], 6), \"a\");\nassert_eq!(whos_soda(&[\"a\", \"b\", \"c\", \"d\", \"e\"], 7), \"a\");\nassert_eq!(whos_soda(&[\"a\", \"b\", \"c\", \"d\", \"e\"], 8), \"b\");\nassert_eq!(whos_soda(&[\"a\", \"b\", \"c\", \"d\", \"e\"], 9), \"b\");\nassert_eq!(whos_soda(&[\"a\", \"b\", \"c\", \"d\", \"e\"], 10), \"c\");\nassert_eq!(whos_soda(&[\"a\", \"b\", \"c\", \"d\", \"e\"], 11), \"c\");\nassert_eq!(whos_soda(&[\"a\", \"b\", \"c\", \"d\", \"e\"], 12), \"d\");\nassert_eq!(whos_soda(&[\"a\", \"b\", \"c\", \"d\", \"e\"], 13), \"d\");\nassert_eq!(whos_soda(&[\"a\", \"b\", \"c\", \"d\", \"e\"], 14), \"e\");\nassert_eq!(whos_soda(&[\"a\", \"b\", \"c\", \"d\", \"e\"], 15), \"e\");\n}\n}\n\npub fn main() {\nlet mut current_timing_iter = 0.0;\nlet mut current_timing_math = 0.0;\nfor people in 1..10 {\nfor i in 1..2000000 {\nuse std::time::Instant;\nlet now = Instant::now();\n\nlet math_ans = find_index_math(people as usize, i);\n\nlet elapsed = now.elapsed();\nlet math_sec =\n(elapsed.as_secs() as f64) + (elapsed.subsec_nanos() as f64 / 1000_000_000.0);\n\nlet now = Instant::now();\n\nlet iter_ans = find_index_iter(people, i) as usize;\n\nlet elapsed = now.elapsed();\nlet iter_sec =\n(elapsed.as_secs() as f64) + (elapsed.subsec_nanos() as f64 / 1000_000_000.0);\n\ncurrent_timing_math += math_sec;\ncurrent_timing_iter += iter_sec;\n\nassert_eq!(iter_ans, math_ans, \"{}x{}\", people, i);\n}\n}\n\nprintln!(\n\"Math Took: {} Iter Took: {}\",\ncurrent_timing_math, current_timing_iter\n);\n}\n``````",
null,
"Oleksii Filonenko\n\nElixir:\n\n``````defmodule DoubleCola do\n\nwhen n > 1,\nend\n``````",
null,
"Payton Bice • Edited\n\nPHP:\n\n``````function cola (int \\$x, array \\$y): string {\n\\$z = count(\\$y);\n\nwhile (\\$x > \\$z) {\n\\$x = round(\\$x / \\$z);\n}\n\nreturn \\$y[\\$x - 1];\n}\n``````",
null,
"E. Choroba\n\nIt returns `b` for `cola(10, array(\"a\", \"b\", \"c\", \"d\"))` which I think should return `c`.",
null,
"Payton Bice\n\nThank you, that's a good catch - I fixed this by replacing `floor()` with `round()`.",
null,
"E. Choroba\n\nI'm still not sure it's correct. `cola(9, array(1, 2))` returns 2 instead of 1.",
null,
"Payton Bice\n\nHm, you're right. It looks like I didn't test it thoroughly enough yesterday, and the tests I did came back with the right answers for the wrong reason. I'm trying to get better at solving math problems so I'll look more into this when I'm off work.",
null,
"Some Smalltalk\nDefining a subclass of Array that would act like an \"infinite collection\" (just because it is fun).\n\n``````Class {\n#name : #DoubleCola,\n#superclass : #Array,\n#type : #variable,\n#category : #''\n}\n\n{ #category : #accessing }\nDoubleCola >> at: anIndex [\n^ anIndex <= self size\nifTrue: [ super at: anIndex ]\nifFalse: [ self at: anIndex inGroup: 1 ]\n]\n\n{ #category : #accessing }\nDoubleCola >> at: anIndex inGroup: anIteration [\n^ anIndex <= (self groupSize: anIteration)\nifTrue: [ (self group: anIteration) at: anIndex ]\nifFalse: [ self\nat: anIndex - (self groupSize: anIteration)\ninGroup: anIteration + 1 ]\n]\n\n{ #category : #accessing }\nDoubleCola >> group: anInteger [\n^ self\nflatCollect: [ :aPlayer | (1 to: 2 ** anInteger) collect: [ :i | aPlayer ] ]\nas: OrderedCollection\n]\n\n{ #category : #accessing }\nDoubleCola >> groupSize: anIteration [\n^ self size * (2 ** (anIteration - 1))\n]\n\n``````\n\nTo execute:\n\n``````(DoubleCola withAll: #( Sheldon Leonard Penny Rajesh Howard )) at: 3. \"#Penny\"\n(DoubleCola withAll: #( Sheldon Leonard Penny Rajesh Howard )) at: 7. \"#Sheldon\"\n(DoubleCola withAll: #( Sheldon Leonard Penny Rajesh Howard )) at: 8. \"#Leonard\"\n``````",
null,
"Tobias SN • Edited\n``````queue = [\"Sheldon\", \"Leonard\", \"Penny\", \"Rajesh\", \"Howard\"];\n\nfor (i=0; i<n; i++) {\ncurr = queue.shift();\nqueue.push(curr, curr);\n}\n\nresult = queue.pop();\n``````",
null,
"Alvaro Montoro\n\nI didn't get the notification for this challenge even when I'm following dev.to staff",
null,
"Corey Alexander\n\nOh huh, I didn't notice but I didn't get my notification either 🤔\n\nAnd while we're on the subject, it would be really cool to be able to follow specific series! So I could just follow this series of posts!"
] | [
null,
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null,
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null,
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null,
"https://res.cloudinary.com/practicaldev/image/fetch/s--XNDmVDYL--/c_fill,f_auto,fl_progressive,h_50,q_auto,w_50/https://dev-to-uploads.s3.amazonaws.com/uploads/user/profile_image/81930/b68a0c47-c454-43e8-b54d-f86b85aef986.jpeg",
null,
"https://res.cloudinary.com/practicaldev/image/fetch/s--Nw5jgx9Y--/c_fill,f_auto,fl_progressive,h_50,q_auto,w_50/https://dev-to-uploads.s3.amazonaws.com/uploads/user/profile_image/57650/ba1f03a9-a1ea-444b-8296-1422f5f0c624.jpeg",
null,
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null,
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null,
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null
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https://luke2375.wordpress.com/tag/english/ | [
"# How to get the absolute position of a control in WinRT\n\nSooner or later you will need to find the absolute position of a control in WinRT.\nHere is how you can achieve it:\n\n// Obtain transform information based off root element\nGeneralTransform gt = this.TransformToVisual(Window.Current.Content);\n\n// Find the four corners of the element\nPoint topLeft = gt.TransformPoint(new Point(0, 0));\nPoint topRight = gt.TransformPoint(new Point(this.RenderSize.Width, 0));\nPoint bottomLeft = gt.TransformPoint(new Point(0, this.RenderSize.Height));\nPoint bottomRight = gt.TransformPoint(new Point(this.RenderSize.Width, this.RenderSize.Height));\n\nAnnunci"
] | [
null
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http://www10.giscafe.com/book/ASIC/CH10/CH10.10.php | [
"10.10 Sequential Statements\n\n## 10.10 Sequential Statements\n\nA sequential statement [VHDL LRM8] is defined as follows:\n\n```sequential_statement ::=\nwait_statement | assertion_statement\n| signal_assignment_statement\n| variable_assignment_statement | procedure_call_statement\n| if_statement | case_statement | loop_statement\n| next_statement | exit_statement\n| return_statement | null_statement | report_statement```\n\nSequential statements may only appear in processes and subprograms. In the following sections I shall describe each of these different types of sequential statements in turn.\n\n### 10.10.1 Wait Statement\n\nThe wait statement is central to VHDL, here are the BNF definitions [VHDL 93LRM8.1]:\n\n```wait_statement ::= [label:] wait [sensitivity_clause]\n[condition_clause] [timeout_clause] ;\nsensitivity_clause ::= on sensitivity_list\nsensitivity_list ::= signal_name { , signal_name }\ncondition_clause ::= until condition\ncondition ::= boolean_expression\ntimeout_clause ::= for time_expression```\n\nA wait statement suspends (stops) a process or procedure (you cannot use a wait statement in a function). The wait statement may be made sensitive to events (changes) on static signals (the value of the signal must be known at analysis time) that appear in the sensitivity list after the keyword on . These signals form the sensitivity set of a wait statement. The process will resume (restart) when an event occurs on any signal (and only signals) in the sensitivity set.\n\nA wait statement may also contain a condition to be met before the process resumes. If there is no sensitivity clause (there is no keyword on ) the sensitivity set is made from signals (and only signals) from the condition clause that appears after the keyword until (the rules are quite complicated [VHDL 93LRM8.1]).\n\nFinally a wait statement may also contain a timeout (following the keyword for ) after which the process will resume. Here is the expanded BNF definition, which makes the structure of the wait statement easier to see (but we lose the definitions of the clauses and the sensitivity list):\n\n```wait_statement ::= [label:]\n\nwait\n[on signal_name {, signal_name}]\n[until boolean_expression]\n[for time_expression] ;```\n\nFor example, the statement, wait on light , makes you wait until a traffic light changes (any change). The statement, wait until light = green , makes you wait (even at a green light) until the traffic signal changes to green. The statement,\n\n```if light = (red or yellow) then wait\nuntil light = green;\nend\nif;```\n\naccurately describes the basic rules at a traffic intersection.\n\nThe most common use of the wait statement is to describe synchronous logic, as in the following model of a D flip-flop:\n\n```entity DFF is port (CLK, D : BIT; Q : out BIT); end;\narchitecture Behave of DFF is\nprocess begin wait\nuntil C\nlk = '1'; Q <= D ; end process;\nend;```\n\nNotice that the statement in line 3 above, wait until C lk = '1', is equivalent to wait on Clk until C lk = '1', and detects a clock edge and not the clock level. Here are some more complex examples of the use of the wait statement:\n\n```entity Wait_1 is port (Clk, s1, s2 :in BIT); end;\narchitecture Behave of Wait_1 is\nsignal x : BIT_VECTOR (0 to 15);\nbegin process variable v : BIT; begin\nwait; -- Wait forever, stops simulation.\nwait on s1 until s2 = '1'; -- Legal, but s1, s2 are signals so\n-- s1 is in sensitivity list, and s2 is not in the sensitivity set.\n-- Sensitivity set is s1 and process will not resume at event on s2.\nwait on s1, s2; -- resumes at event on signal s1 or s2.\nwait on s1 for 10 ns; -- resumes at event on s1 or after 10 ns.\nwait on x; -- resumes when any element of array x has an event.\n-- wait on x(1 to v); -- Illegal, nonstatic name, since v is a variable.\nend process;\nend;\nentity Wait_2 is port (Clk, s1, s2:in BIT); end;\narchitecture Behave of Wait_2 is\nbegin process variable v : BIT; begin\nwait on Clk; -- resumes when Clk has an event: rising or falling.\nwait until Clk = '1'; -- resumes on rising edge.\nwait on Clk until Clk = '1'; -- equivalent to the last statement.\nwait on Clk until v = '1';\n-- The above is legal, but v is a variable so\n-- Clk is in sensitivity list, v is not in the sensitivity set.\n-- Sensitivity set is Clk and process will not resume at event on v.\nwait on Clk until s1 = '1';\n-- The above is legal, but s1 is a signal so\n-- Clk is in sensitivity list, s1 is not in the sensitivity set.\n-- Sensitivity set is Clk, process will not resume at event on s1.\nend process;\nend;```\n\nYou may only use interface signals that may be read (port modes in , inout , and buffer --see Section 10.7) in the sensitivity list of a wait statement.\n\n### 10.10.2 Assertion and Report Statements\n\nYou can use an assertion statement to conditionally issue warnings. The report statement (VHDL-93 only) prints an expression and is useful for debugging.\n\n```assertion_statement ::= [label:] assert\nboolean_expression [report expression] [severity expression] ;\nreport_statement\n::= [label:] report expression [severity expression] ;```\n\nHere is an example of an assertion statement:\n\n```entity Assert_1 is port (I:INTEGER:=0); end;\narchitecture Behave of Assert_1 is\nbegin process begin\nassert (I > 0) report \"I is negative or zero\"; wait;\nend process;\nend;```\n\nThe expression after the keyword report must be of type STRING (the default is \"Assertion violation\" for the assertion statement), and the expression after the keyword severity must be of type SEVERITY_LEVEL (default ERROR for the assertion statement, and NOTE for the report statement) defined in the STANDARD package. The assertion statement prints if the assertion condition (after the keyword assert ) is FALSE . Simulation normally halts for severity of ERROR or FAILURE (you can normally control this threshold in the simulator).\n\n### 10.10.3 Assignment Statements\n\nThere are two sorts of VHDL assignment statements: one for signals and one for variables [VHDL 93LRM8.4-8.5]. The difference is in the timing of the update of the LHS. A variable assignment statement is the closest equivalent to the assignment statement in a computer programming language. Variable assignment statements are always sequential statements and the LHS of a variable assignment statement is always updated immediately. Here is the definition and an example:\n\n```variable_assignment_statement ::=\n[label:] name|aggregate := expression ;\nentity Var_Assignment is end;\narchitecture Behave of Var_Assignment is\nsignal s1 : INTEGER := 0;\nbegin process variable v1,v2 : INTEGER := 0; begin\nassert (v1/=0) report \"v1 is 0\" severity note ; -- this prints\nv1 := v1 + 1; -- after this statement v1 is 1\nassert (v1=0) report \"v1 isn't 0\" severity note ; -- this prints\nv2 := v2 + s1; -- signal and variable types must match\nwait;\nend process;\nend;```\n\nThis is the output from Cadence Leapfrog for the preceding example:\n\n```ASSERT/NOTE (time 0 FS) from : (design unit WORK.VAR_ASSIGNMENT:BEHAVE) v1 is 0\nASSERT/NOTE (time 0 FS) from : (design unit WORK.VAR_ASSIGNMENT:BEHAVE) v1 isn't 0```\n\nA signal assignment statement schedules a future assignment to a signal:\n\n```signal_assignment_statement::=\n[label:] target <=\n[transport | [ reject time_expression ] inertial ] waveform ;```\n\nThe following example shows that, even with no delay, a signal is updated at the end of a simulation cycle after all the other assignments have been scheduled, just before simulation time is advanced:\n\n```entity Sig_Assignment_1 is end;\narchitecture Behave of Sig_Assignment_1 is\nsignal s1,s2,s3 : INTEGER := 0;\nbegin process variable v1 : INTEGER := 1; begin\nassert (s1 /= 0) report \"s1 is 0\" severity note ; -- this prints.\ns1 <= s1 + 1; -- after this statement s1 is still 0.\nassert (s1 /= 0) report \"s1 still 0\" severity note ; -- this prints.\nwait;\nend process;\nend;\nASSERT/NOTE (time 0 FS) from : (design unit WORK.SIG_ASSIGNMENT_1:BEHAVE) s1 is 0\nASSERT/NOTE (time 0 FS) from : (design unit WORK.SIG_ASSIGNMENT_1:BEHAVE) s1 still 0```\n\nHere is an another example to illustrate how time is handled:\n\n```entity Sig_Assignment_2 is end;\narchitecture Behave of Sig_Assignment_2 is\nsignal s1, s2, s3 : INTEGER := 0;\nbegin process variable v1 : INTEGER := 1; begin\n-- s1, s2, s3 are initially 0; now consider the following:\ns1 <= 1 ; -- schedules updates to s1 at end of 0 ns cycle.\ns2 <= s1; -- s2 is 0, not 1.\nwait for 1 ns;\ns3 <= s1; -- now s3 will be 1 at 1 ns.\nwait;\nend process;\nend;```\n\nThe Compass simulator produces the following trace file for this example:\n\n``` Time(fs) + Cycle s1 s2 s3\n---------------------- ------------ ------------ ------------\n0+ 0: 0 0 0\n0+ 1: * 1 * 0 0\n...\n1000000+ 1: 1 0 * 1```\n\nTime is indicated in femtoseconds for each simulation cycle plus the number of delta cycles (we call this delta time, measured in units of delta) needed to calculate all transactions on signals. A transaction consists of a new value for a signal (which may be the same as the old value) and the time delay for the value to take effect. An asterisk '*' before a value in the preceding trace indicates that a transaction has occurred and the corresponding signal updated at that time. A transaction that does result in a change in value is an event. In the preceding simulation trace for Sig_Assignment_2:Behave\n\nAt 0 ns + 0 delta: all signals are 0 .\n\nAt 0 ns + 1 delta: s1 is updated to 1 , s2 is updated to 0 (not to 1 ).\n\nAt 1 ns + 1 delta: s3 is updated to a 1 .\n\nThe following example shows the behavior of the different delay models: transport and inertial (the default):\n\n```entity Transport_1 is end;\narchitecture Behave of Transport_1 is\nsignal s1, SLOW, FAST, WIRE : BIT := '0';\nbegin process begin\ns1 <= '1' after 1 ns, '0' after 2 ns, '1' after 3 ns ;\n-- schedules s1 to be '1' at t+1 ns, '0' at t+2 ns,'1' at t+3 ns\nwait; end process;\n-- inertial delay: SLOW rejects pulsewidths less than 5ns:\nprocess (s1) begin SLOW <= s1 after 5 ns ; end process;\n-- inertial delay: FAST rejects pulsewidths less than 0.5ns:\nprocess (s1) begin FAST <= s1 after 0.5 ns ; end process;\n-- transport delay: WIRE passes all pulsewidths...\nprocess (s1) begin WIRE <= transport s1 after 5 ns ; end process;\nend;```\n\nHere is the trace file from the Compass simulator:\n\n```\nTime(fs) + Cycle s1 slow fast wire\n---------------------- ---- ---- ---- ----\n0+ 0: '0' '0' '0' '0'\n500000+ 0: '0' '0' *'0' '0'\n1000000+ 0: *'1' '0' '0' '0'\n1500000+ 0: '1' '0' *'1' '0'\n2000000+ 0: *'0' '0' '1' '0'\n2500000+ 0: '0' '0' *'0' '0'\n3000000+ 0: *'1' '0' '0' '0'\n3500000+ 0: '1' '0' *'1' '0'\n5000000+ 0: '1' '0' '1' *'0'\n6000000+ 0: '1' '0' '1' *'1'\n7000000+ 0: '1' '0' '1' *'0'\n8000000+ 0: '1' *'1' '1' *'1'```\n\nInertial delay mimics the behavior of real logic gates, whereas transport delay more closely models the behavior of wires. In VHDL-93 you can also add a separate pulse rejection limit for the inertial delay model as in the following example:\n\n`process (s1) begin RJCT <= reject 2 ns s1 after 5 ns ; end process;`\n\n### 10.10.4 Procedure Call\n\nA procedure call in VHDL corresponds to calling a subroutine in a conventional programming language [VHDL LRM8.6]. The parameters in a procedure call statement are the actual procedure parameters (or actuals); the parameters in the procedure definition are the formal procedure parameters (or formals). The two are linked using an association list, which may use either positional or named association (association works just as it does for ports--see Section 10.7.1):\n\n```procedure_call_statement ::=\n[label:] procedure_name [(parameter_association_list)];```\n\nHere is an example:\n\n```package And_Pkg is\nprocedure V_And(a, b : BIT; signal c : out BIT);\nfunction V_And(a, b : BIT) return BIT;\nend;\npackage body And_Pkg is\nprocedure V_And(a, b : BIT; signal c: out BIT) is\nbegin c <= a and b; end;\nfunction V_And(a, b: BIT) return BIT is\nbegin return a and b; end;\nend And_Pkg;\nuse work.And_Pkg.all; entity Proc_Call_1 is end;\narchitecture Behave of Proc_Call_1 is signal A, B, Y: BIT := '0';\nbegin process begin V_And (A, B, Y); wait; end process;\nend;```\n\nTable 10.13 on page 416 explains the rules for formal procedure parameters. There is one other way to call procedures, which we shall cover in Section 10.13.3.\n\n### 10.10.5 If Statement\n\nAn if statement evaluates one or more Boolean expressions and conditionally executes a corresponding sequence of statements [VHDL LRM8.7].\n\n```if_statement ::=\n[if_label:] if boolean_expression then {sequential_statement}\n{elsif boolean_expression then {sequential_statement}}\n[else {sequential_statement}]\nend if [if_label];```\n\nThe simplest form of an if statement is thus:\n\n`if boolean_expression then {sequential_statement} end if;`\n\nHere are some examples of the if statement:\n\n```entity If_Then_Else_1 is end;\narchitecture Behave of If_Then_Else_1 is signal a, b, c: BIT :='1';\nbegin process begin\nif c = '1' then c <= a ; else c <= b; end if; wait;\nend process;\nend;\nentity If_Then_1 is end;\narchitecture Behave of If_Then_1 is signal A, B, Y : BIT :='1';\nbegin process begin\nif A = B then Y <= A; end if; wait;\nend process;\nend;```\n\n### 10.10.6 Case Statement\n\nA case statement [VHDL LRM8.8] is a multiway decision statement that selects a sequence of statements by matching an expression with a list of (locally static [VHDL LRM7.4.1]) choices.\n\n```case_statement ::=\n[case_label:] case expression is\nwhen choice {| choice} => {sequential_statement}\n{when choice {| choice} => {sequential_statement}}\nend case [case_label];```\n\nCase statements are useful to model state machines. Here is an example of a Mealy state machine with an asynchronous reset:\n\n```library IEEE; use IEEE.STD_LOGIC_1164.all;\nentity sm_mealy is\nport (reset, clock, i1, i2 : STD_LOGIC; o1, o2 : out STD_LOGIC);\nend sm_mealy;\narchitecture Behave of sm_mealy is\ntype STATES is (s0, s1, s2, s3); signal current, new : STATES;\nbegin\nsynchronous : process (clock, reset) begin\nif To_X01(reset) = '0' then current <= s0;\nelsif rising_edge(clock) then current <= new; end if;\nend process;\ncombinational : process (current, i1, i2) begin\ncase current is\nwhen s0 =>\nif To_X01(i1) = '1' then o2 <='0'; o1 <='0'; new <= s2;\nelse o2 <= '1'; o1 <= '1'; new <= s1; end if;\nwhen s1 =>\nif To_X01(i2) = '1' then o2 <='1'; o1 <='0'; new <= s1;\nelse o2 <='0'; o1 <='1'; new <= s3; end if;\nwhen s2 =>\nif To_X01(i2) = '1' then o2 <='0'; o1 <='1'; new <= s2;\nelse o2 <= '1'; o1 <= '0'; new <= s0; end if;\nwhen s3 => o2 <= '0'; o1 <= '0'; new <= s0;\nwhen others => o2 <= '0'; o1 <= '0'; new <= s0;\nend case;\nend process;\nend Behave;```\n\nEach possible value of the case expression must be present once, and once only, in the list of choices (or arms) of the case statement (the list must be exhaustive). You can use '|' (that means 'or') or 'to' to denote a range in the expression for choice . You may also use the keyword others as the last, default choice (even if the list is already exhaustive, as in the preceding example).\n\n### 10.10.7 Other Sequential Control Statements\n\nA loop statement repeats execution of a series of sequential statements [VHDL LRM8.9]:\n\n```loop_statement ::=\n[loop_label:]\n[while boolean_expression|for identifier in discrete_range]\nloop\n{sequential_statement}\nend loop [loop_label];```\n\nIf the loop variable (after the keyword for ) is used, it is only visible inside the loop. A while loop evaluates the Boolean expression before each execution of the sequence of statements; if the expression is TRUE , the statements are executed. In a for loop the sequence of statements is executed once for each value of the discrete range.\n\n```package And_Pkg is function V_And(a, b : BIT) return BIT; end;\npackage body And_Pkg is function V_And(a, b : BIT) return BIT is\nbegin return a and b; end; end And_Pkg;\nentity Loop_1 is port (x, y : in BIT := '1'; s : out BIT := '0'); end;\nuse work.And_Pkg.all;\narchitecture Behave of Loop_1 is\nbegin loop\ns <= V_And(x, y); wait on x, y;\nend loop;\nend;```\n\nThe next statement [VHDL LRM8.10] forces completion of the current iteration of a loop (the containing loop unless another loop label is specified). Completion is forced if the condition following the keyword then is TRUE (or if there is no condition).\n\n```next_statement ::=\n[label:] next [loop_label] [when boolean_expression];```\n\nAn exit statement [VHDL LRM8.11] forces an exit from a loop.\n\n```exit_statement ::=\n[label:] exit [loop_label] [when condition] ;```\n\nAs an example:\n\n```loop wait on Clk; exit when Clk = '0'; end loop;\n-- equivalent to: wait until Clk = '0';```\n\nThe return statement [VHDL LRM8.12] completes execution of a procedure or function.\n\n`return_statement ::= [label:] return [expression];`\n\nA null statement [VHDL LRM8.13] does nothing (but is useful in a case statement where all choices must be covered, but for some of the choices you do not want to do anything).\n\n`null_statement ::= [label:] null;`\n\nChapter start Previous page Next page",
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https://jisajournal.springeropen.com/articles/10.1186/s13174-017-0065-0/tables/2 | [
"Accuracy is the percentage of instances (evidence) correctly classified. $$\\frac{TP+ TN}{TP+ TN+ FP+ FN}$$\nRecall is the percentage of instances that were correctly classified as positive. $$\\frac{TP}{TP+ FN}$$\nPrecision is the percentage of instances classified as positive (evidence) that are actually positive. $$\\frac{TP}{TP+ FP}$$\nF Measure is the harmonic mean of precision and recall, meaning it combines the precision and recall. $$\\frac{2\\ x\\ precision\\ \\mathrm{x}\\ recall}{precision+ recall}$$\nError Rate is the proportion of instances that are incorrectly classified. $$\\frac{FP+ FN}{TP+ FN+ FP+ TN}$$"
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http://www.worldlibrary.in/articles/Equatorial_bulge | [
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"This article will be permanently flagged as inappropriate and made unaccessible to everyone. Are you certain this article is inappropriate? Excessive Violence Sexual Content Political / Social Email this Article Email Address:\n\n# Equatorial bulge\n\nArticle Id: WHEBN0000143023\nReproduction Date:\n\n Title: Equatorial bulge",
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"Author: World Heritage Encyclopedia Language: English Subject: Collection: Publisher: World Heritage Encyclopedia Publication Date:\n\n### Equatorial bulge\n\nAn equatorial bulge is a difference between the equatorial and polar diameters of a planet, due to the centrifugal force of its rotation. A rotating body tends to form an oblate spheroid rather than a sphere. The Earth has an equatorial bulge of 42.72 km (26.54 mi): that is, its diameter measured across the equatorial plane (12,756.28 km (7,926.38 mi)) is 42.72 km more than that measured between the poles (12,713.56 km (7,899.84 mi)). An observer standing at sea level on either pole, therefore, is 21.36 km closer to Earth's centrepoint than if standing at sea level on the equator. The value of Earth's radius may be approximated by the average of these radii.\n\nAn often-cited result of Earth's equatorial bulge is that the highest point on Earth, measured from the center outwards, is the peak of Mount Chimborazo in Ecuador, rather than Mount Everest. But since the ocean, like the Earth and the atmosphere, bulges, Chimborazo is not as high above sea level as Everest is.\n\nBasing this on centripetal force, the relationship F_c = Mv^2 / R applies. Viewing the globe as a series of rotating discs, the mass M and radius R at the poles get very small at the same time and thus produce a smaller force for the same velocity. Moving towards the equator, while R gets much bigger, M increases quicker than R thus producing a greater force at the equator. This may be because Earth’s core is included in the cross sectional disc at the equator; the density of the Earth's core is significantly higher than that of the Earth's outer layers, so it contributes more to the mass of the disc. There is a bulge in the water envelope of the oceans surrounding earth. So the fact that water is a fluid and the Earth experiences its greatest centrifugal force at the equator and the fact that the greatest bulge of that water envelope occurs at the equator demonstrates that centrifugal force of earth’s rotation is helping to produce that bulge independent of tides. Sea level at the equator is 21 km higher than sea level at the poles in terms of their distances from the center of the planet.\n\n## The equilibrium as a balance of energies",
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"Fixed to the vertical rod is a spring metal band. When stationary the spring metal band is circular in shape. The top of the metal band can slide along the vertical rod. When spun, the spring-metal band bulges at its equator and flattens at its poles in analogy with the Earth.\n\nGravity tends to contract a celestial body into a perfect sphere, the shape for which all the mass is as close to the center of gravity as possible. Rotation causes a distortion from this spherical shape; a common measure of the distortion is the flattening (sometimes called ellipticity or oblateness), which can depend on a variety of factors including the size, angular velocity, density, and elasticity.\n\nTo get a feel for the type of equilibrium that is involved, imagine someone seated in a spinning swivel chair, with weights in their hands. If the person in the chair pulls the weights towards them, they are doing work and their rotational kinetic energy increases. The increase of rotation rate is so strong that at the faster rotation rate the required centripetal force is larger than with the starting rotation rate.\n\nSomething analogous to this occurs in planet formation. Matter first coalesces into a slowly rotating disk-shaped distribution, and collisions and friction convert kinetic energy to heat, which allows the disk to self-gravitate into a very oblate spheroid.\n\nAs long as the proto-planet is still too oblate to be in equilibrium, the release of gravitational potential energy on contraction keeps driving the increase in rotational kinetic energy. As the contraction proceeds the rotation rate keeps going up, hence the required force for further contraction keeps going up. There is a point where the increase of rotational kinetic energy on further contraction would be larger than the release of gravitational potential energy. The contraction process can only proceed up to that point, so it halts there.\n\nAs long as there is no equilibrium there can be violent convection, and as long as there is violent convection friction can convert kinetic energy to heat, draining rotational kinetic energy from the system. When the equilibrium state has been reached then large scale conversion of kinetic energy to heat ceases. In that sense the equilibrium state is the lowest state of energy that can be reached.\n\nThe Earth's rotation rate is still slowing down, but gradually, about two thousandth of a second per rotation every 100 years. Estimates of how fast the Earth was rotating in the past vary, because it is not known exactly how the moon was formed. Estimates of the Earth's rotation 500 million years ago are around 20 modern hours per \"day\".\n\nThe Earth's rate of rotation is slowing down mainly because of tidal interactions with the Moon and the Sun. Since the solid parts of the Earth are ductile, the Earth's equatorial bulge has been decreasing in step with the decrease in the rate of rotation.\n\n## Differences in gravitational acceleration",
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"The forces at play in the case of a planet with an equatorial bulge due to rotation.\nRed arrow: gravity\nGreen arrow, the normal force\nBlue arrow: the resultant force\n\nThe resultant force provides required centripetal force. Without this centripetal force frictionless objects would slide towards the equator.\n\nIn calculations, when a coordinate system is used that is co-rotating with the Earth, the vector of the fictitious centrifugal force points outward, and is just as large as the vector representing the centripetal force.\n\nBecause of a planet's rotation around its own axis, the gravitational acceleration is less at the equator than at the poles. In the 17th century, following the invention of the pendulum clock, French scientists found that clocks sent to French Guiana, on the northern coast of South America, ran slower than their exact counterparts in Paris. Measurements of the acceleration due to gravity at the equator must also take into account the planet's rotation. Any object that is stationary with respect to the surface of the Earth is actually following a circular trajectory, circumnavigating the Earth's axis. Pulling an object into such a circular trajectory requires a force. The acceleration that is required to circumnavigate the Earth's axis along the equator at one revolution per sidereal day is 0.0339 m/s². Providing this acceleration decreases the effective gravitational acceleration. At the equator, the effective gravitational acceleration is 9.7805 m/s2. This means that the true gravitational acceleration at the equator must be 9.8144 m/s2 (9.7805 + 0.0339 = 9.8144).\n\nAt the poles, the gravitational acceleration is 9.8322 m/s2. The difference of 0.0178 m/s2 between the gravitational acceleration at the poles and the true gravitational acceleration at the equator is because objects located on the equator are about 21 kilometers further away from the center of mass of the Earth than at the poles, which corresponds to a smaller gravitational acceleration.\n\nIn summary, there are two contributions to the fact that the effective gravitational acceleration is less strong at the equator than at the poles. About 70 percent of the difference is contributed by the fact that objects circumnavigate the Earth's axis, and about 30 percent is due to the non-spherical shape of the Earth.\n\nThe diagram illustrates that on all latitudes the effective gravitational acceleration is decreased by the requirement of providing a centripetal force; the decreasing effect is strongest on the equator.\n\n## Satellite orbits\n\nThe fact that the Earth's gravitational field slightly deviates from being spherically symmetrical also affects the orbits of satellites. The principal effect is to cause nodal precession, so that the plane of the orbit does not remain fixed in inertial space. Smaller effects include deviation of orbits away from pure ellipses. This is especially important in the case of the trajectories of GPS satellites.\n\n## Other celestial bodies\n\nGenerally any celestial body that is rotating (and that is sufficiently massive to draw itself into spherical or near spherical shape) will have an equatorial bulge matching its rotation rate. Saturn is the planet with the largest equatorial bulge in the Solar System (11808 km, 7337 miles). However, Haumea is the dwarf planet with the largest equatorial bulge, indeed greater than that of Saturn. This gives it the shape of a tri-axial ellipsoid.\n\nThe following is a table of the equatorial bulge of some major celestial bodies of the Solar System:\n\nBody Equatorial diameter Polar diameter Equatorial bulge Flattening ratio\nEarth 12,756.28 km 12,713.56 km 42.72 km 1:298.2575\nMars 6,805 km 6,754.8 km 50.2 km 1:135.56\nCeres 975 km 909 km 66 km 1:14.77\nJupiter 143,884 km 133,709 km 10,175 km 1:14.14\nSaturn 120,536 km 108,728 km 11,808 km 1:10.21\nUranus 51,118 km 49,946 km 1,172 km 1:43.62\nNeptune 49,528 km 48,682 km 846 km 1:58.54\n\n## Mathematical expression\n\nThe flattening coefficient f for the equilibrium configuration of a self-gravitating spheroid, composed of uniform density incompressible fluid, rotating steadily about some fixed axis, for a small amount of flattening, is approximated by:\n\nf = \\frac{a_e-a_p}{a} = {5 \\over 4} {\\omega^2 a^3 \\over G M} = {15 \\pi \\over 4} {1 \\over G T^2 \\rho}\n\nwhere a_e = a(1+{ f \\over 3} ) and a_p = a(1-{ 2f\\over 3}) are respectively the equatorial and polar radius, a is the mean radius, \\omega = {2 \\pi \\over T} is the angular velocity, T is the rotation period, G is the universal gravitational constant, M \\simeq {4 \\over 3} \\pi \\rho a^3 is the total body mass, and \\rho is the body density."
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https://myokit.readthedocs.io/en/stable/api_core/Variable.html | [
"# `Variable`¶\n\nclass `myokit.``Variable`(parent, name)\n\nRepresents a model variable.\n\nVariables should not be created directly, but only via `Component.add_variable()` or `Variable.add_variable()`.\n\nEach variable has a single defining equation. For state variables, this equation has a derivative on the left-hand side (lhs), for all other variables the lhs of the defining equation is simply the variable’s name.\n\nVariables can be made into state variables using `Variable.promote()`.\n\nVariables can be made to represent an externally defined variable using `Variable.set_binding()`. In this case, the right hand side set for a variable will be used as an alternative in situations where the external input is not available.\n\nMeta-data properties can be accessed via the property `meta`, for example `model.meta['key']= 'value'`.\n\n`add_variable`(name)\n\nAdds a child variable with the given name to this `VarOwner`.\n\n`add_variable_allow_renaming`(name)\n\nAttempts to add a child variable with the given name to this `VarOwner`, but uses a different name if this causes any conflicts.\n\nThe new variable’s name will be modified by appending _1, _2, etc. until the conflict is resolved.\n\nThis method can be used when symbolically manipulating a model in situations where the exact names are unimportant.\n\nReturns the newly created variable.\n\n`binding`()\n\nReturns this variable’s binding label or `None` if no binding is set.\n\n`can_add_variable`(name, variable_whitelist=None)\n\nReturns `True` if a variable can be added to this `VarOwner` under the given `name`.\n\nThis method is automatically called by `add_variable()` and move_variable(), there is no need to call it before using these methods.\n\nTo ignore clashes with known variables, a list `variable_whitelist` can be passed in.\n\n`code`()\n\nReturns this object in `mmt` syntax.\n\n`convert_unit`(new_unit, helpers=None)\n\nConverts the units this variable is expressed in to `new_unit`, and updates the RHS with an appropriate scaling factor.\n\nUnit conversion proceeds in the following steps:\n\n1. A scaling factor is determined (see below).\n2. The variable’s RHS is multiplied by this factor\n3. For state variables, the current state value is also updated.\n4. All references to the variable in the model equations are divided by the scaling factor.\n5. If the time variable is updated, all derivative equations are updated.\n\nThe scaling factor is determined using `myokit.Unit.conversion_factor(old_unit, new_unit, helpers)()`, where `old_unit` is the current variable unit (as determined by `variable.unit(myokit.UNIT_STRICT)`).\n\nFor state variables, as with ordinary variables, the `new_unit` should be the unit of the variable itself (and not of its time derivative).\n\nArguments:\n\n`new_unit`\nThe new unit this variable should be in. Given either as a `myokit.Unit` or as a string that can be converted to a unit using `myokit.parse_unit(new_unit)`.\n`helpers=None`\nAn optional list of conversion factors, which the method will attempt to use if the new and old units are incompatible. Each factor should be specified as a `myokit.Quantity` or something that can be converted to a Quantity e.g. a string `1 [uF/cm^2]` or a `myokit.Number()`.\n\nNote that this method will assume the expression is currently in the unit returned by `Variable.unit()`. It will not check whether the current RHS expression evaluates to the correct units.\n\n`count_equations`(const=None, inter=None, state=None, bound=None, deep=False)\n\nReturns the number of equations matching the given criteria. See `equations()` for an explanation of the arguments.\n\n`count_variables`(const=None, inter=None, state=None, bound=None, deep=False)\n\nReturns the number of variables matching the given criteria. See `variables()` for an explanation of the arguments.\n\n`demote`()\n\nTurns a state variable into an intermediary variable.\n\nThis will reset the validation status of the model this variable belongs to.\n\n`eq`()\n\nReturns this variable’s defining equation.\n\nFor state variables this will be an equation for the variable’s derivative. For ordinary variables the equation for the variable’s value is returned.\n\n`equations`(const=None, inter=None, state=None, bound=None, deep=False)\n\nCreates and returns a filtered iterator over the equations of this object’s variables.\n\nThe returned values can be filtered using the following arguments:\n\n`const=True|False|None`\n\nSet to `True` to return only constants’ equations. `False` to exclude all constants and any other value to ignore this check.\n\nFor a definition of “constant variable” see `variables()`.\n\n`inter=True|False|None`\n\nSet to `True` to return only intermediary variables’ equations, `False` to exclude all intermediary variables and any other value to ignore this check.\n\nFor a definition of “intermediary variable” see `variables()`.\n\n`state=True|False|None`\nSet to `True` to return only state variables’ equations, `False` to exclude all state variables and any other value to ignore this check.\n`bound=True|False|None`\nSet to `True` to return only bound variables’ equations, `False` to exclude all bound variables and any other value to ignore this check.\n`deep=True|False` (by default it’s `False`)\nSet to `True` to include the equations of nested variables meeting all other criteria.\n`eval`()\n\nEvaluates this variable’s defining equation and returns the result.\n\n`get`(name, class_filter=None)\n\nSearches for a variable with the given `qname` and returns it.\n\nTo return only objects of a certain class, pass it in as `class_filter`.\n\nThe qnames are specified relative to the `VarOwner`. For example, in a model with a component `ina` and a variable `ina.h` we expect the following results\n\n```>>> import myokit\n>>> c = m.get('ina') # Retrieves the component <ina>\n>>> h = c.get('h') # Retrieves the variable <ina.h>\n>>> x = c.get('ina.h') # Searches for <ina.ina.h>: KeyError!\nTraceback (most recent call last):\nFile \"<stdin>\", line 1, in <module>\nKeyError: 'ina'\n```\n`has_ancestor`(obj)\n\nReturns `True` if the given object `obj` is an ancestor of this `ModelPart`.\n\n`has_equations`(const=None, inter=None, state=None, bound=None, deep=False)\n\nReturns True if there are any equations that can be returned by calling :meth:`equations` with the same arguments.\n\n`has_variable`(name)\n\nReturns True if the given name corresponds to a variable in this object. Accepts both single names `x` and qualified names `x.y` as input.\n\nThis function performs the same search as `variable`, so in most cases it will be more efficient to call `variable()` in a try-catch block rather than checking for existence explicitly.\n\n`has_variables`(const=None, inter=None, state=None, bound=None, deep=False)\n\nReturns True if there are any variables that can be returned by calling :meth:`variables` with the same arguments.\n\n`indice`()\n\nFor state variables, this will return their index in the state vector. For all other variables, this will raise an exception.\n\n`is_ancestor`(obj)\n\nReturns `True` if this object is an ancestor of the given `ModelPart`` `object`.\n\n`is_bound`()\n\nReturns `True` if a binding label has been added to this variable.\n\n`is_constant`()\n\nReturns `True` if this variable is constant.\n\nMyokit doesn’t discern between mathematical and physical constants, parameters etc. Anything that doesn’t change during a simulation is termed a constant. Note that this specifically excludes variables bound to external inputs.\n\n`is_intermediary`()\n\nReturns `True` if this variable is an intermediary variable, i.e. not a constant or a state variable (and not bound to an external variable).\n\n`is_labelled`()\n\nReturns `True` if a label has been added to this variable.\n\n`is_literal`()\n\nReturns `True` if this variable’s expression contains only literal values.\n\n`is_nested`()\n\nReturns `True` if this variable’s parent is another variable.\n\n`is_referenced`()\n\nReturns `True` if other variables reference this variable’s `lhs`. For states, this means it only returns `True` if other variables depend on this variable’s derivative.\n\n`is_state`()\n\nReturns `True` if this variable is a state variable.\n\n`label`()\n\nReturns this variable’s label or `None` if no label is set.\n\n`lhs`()\n\nReturns the left-hand side of the equation defining this variable.\n\nFor state variables this will be a `myokit.Derivative`, for all others this should be a `myokit.Name`.\n\n`model`()\n\nReturns the `myokit.Model` this object belongs to (if set).\n\n`move_variable`(variable, new_parent, new_name=None)\n\nMoves the given variable to another `VarOwner` `new_parent`. In addition, this method can be used to rename variables (either with or without moving them).\n\n`name`()\n\nReturns this object’s name.\n\n`parent`(kind=None)\n\nReturns this object’s parent.\n\nIf the optional variable `kind` is set, the method will scan until an instance of the requested type is found.\n\n`promote`(state_value=0)\n\nTurns this variable into a state variable with a current state value given by `state_value`.\n\nThis will reset the validation status of the model this variable belongs to.\n\n`pyfunc`(use_numpy=True, arguments=False)\n\nReturns a python function that evaluates this variable as a function of the state variables it depends on.\n\nThe argument names used by the returned function will be the variables’ unames, ordered alphabetically.\n\nIf the function runs in “numpy-mode”, which is enabled by default, the vector ready versions of `log`, `exp` etc are used and piecewise statements are evaluated using an array ready piecewise function. To disable this functionality, set `use_numpy=False`.\n\nTo obtain more information about the arguments in the created function handle, set the optional argument `arguments` to `True`. With this setting, the function will return a tuple `(handle, vars)` where `handle` is the function handle and `vars` is a list of `myokit.LhsExpression` objects in the same order as the function arguments.\n\n`qname`(hide=None)\n\nReturns this object’s fully qualified name. That is, an object named `y` with a parent named `x` will return the string `x.y`. If `y` has a child `z` its qname will be `x.y.z`.\n\nIf the optional argument `hide` is set to this object’s parent the parent qualifier will be omitted.\n\n`refs_by`(state_refs=False)\n\nReturns an iterator over the set of `Variables` that refer to this variable in their defining equation.\n\nNote that only references to this variable’s defining `LhsExpression` (i.e. the one returned by `lhs()`) are returned. For a state variable `x`, this means the returned result contains all variables referring to `dot(x)`. To get an iterator over the variables referring to `x` instead, add the optional attribute `state_refs=True`. For non-state variables this setting will trigger an `Exception`.\n\n`refs_to`(state_refs=False)\n\nReturns an iterator over the set of `Variables` that this variable’s defining equation refers to.\n\nBy default, this will _not_ include references to a state variable’s value. To obtain a list of state variables whose value is referenced, use `state_refs=True`.\n\n`remove_variable`(variable, recursive=False)\n\nRemoves the given variable from this `VarOwner` and from the model.\n\nIf `recursive` is `True`, any child variables will be deleted as well.\n\n`rename`(new_name)\n\nRenames this variable.\n\n`rhs`()\n\nReturns the right-hand side of the equation defining this variable.\n\nFor state variables this will be the expression for their derivative, for all others an expression for their value is returned.\n\n`set_binding`(binding)\n\nAdds a unique binding label to this variable, indicating it can be used as an entry point for external inputs.\n\nTo remove a binding, call `set_binding(None)`.\n\nAdding or removing binding labels resets the model’s validation status.\n\n`set_label`(label=None)\n\nAdds a unique `label` for this variable, indicated that its value can be read by external users.\n\nTo remove a label call `set_label(None)`.\n\n`set_rhs`(rhs)\n\nChanges the expression for this variable’s right hand side (rhs).\n\nFor state variables, this updates their derivative. For all others this will update the expression for their value.\n\nExpressions can be specified using `myokit.Expression` objects, but floats or strings that can be parsed into expressions are also allowed:\n\n```# Ways of setting x = 2\nx.set_rhs(myokit.Number(2))\nx.set_rhs(2)\nx.set_rhs(2.0)\nx.set_rhs('2')\n# Ways of setting x = 1 + y\nx.set_rhs(myokit.Plus(myokit.Number(1), myokit.Name(y)))\nx.set_rhs('1 + y')\n```\n\nCalling set_rhs will reset the validation status of the model this variable belongs to.\n\n`set_state_value`(value)\n\nIf this variable is a state variable, its current value will be updated. For all other variables this raises an exception.\n\n`set_unit`(unit=None)\n\nChanges this variable’s unit. The unit can be set to any valid Unit object, or `None` to remove the unit.\n\n`state_value`()\n\nFor state variables, this will return their current value. For all other variables, this will raise an exception.\n\n`uname`()\n\nReturns a globally unique name for this object.\n\n`unit`(mode=1)\n\nReturns the unit specified for this variable.\n\nIf no unit was set and `mode` is `myokit.UNIT_TOLERANT`, then None is returned. In `myokit.UNIT_STRICT` mode the value `myokit.units.dimensionless` is returned in this case.\n\nVariables’ units are set using `Variable.set_unit()` or using the `in` keyword in `.mmt` files. The unit set for a variable is the unit this variable’s expression _should_ have. This allows expressions to be validated by computing the resulting unit of an expression and comparing it with the value set for the variable.\n\n`validate`()\n\nAttempts to check this variable’s validity, raises errors if it isn’t.\n\n`value`()\n\nWill return the value of this variable’s defining right-hand side expression.\n\n`var`(name)\n\nSearches for the given variable and returns it if found. Accepts both single names `x` and qualified names `x.y` as input.\n\n`variables`(const=None, inter=None, state=None, bound=None, deep=False, sort=False)\n\nCreates and returns a filtered iterator over the contained variables.\n\nThe returned values can be filtered using the following arguments:\n\n`const=True|False|None`\n\nConstants are defined as variables that do not depend on state variables or derivatives. In other words, any variable whose value can be determined before starting an ODE solving routine.\n\nSet to `True` to return only constants, `False` to exclude all constants and any other value to ignore this check.\n\n`inter=True|False|None`\n\nIntermediary variables are those variables that are not constant but are not part of the state. In other words, intermediary variables are the variables that need to be calculated at every step of an ODE solving routine before calculating the ODE derivatives and updating the state.\n\nSet to `True` to return only intermediary variables, `False` to exclude all intermediary variables and any other value to ignore this check.\n\n`state=True|False|None`\nSet to True to return only state variables, False to exclude all state variables and any other value to ignore this check.\n`bound=True|False|None`\nSet to True to return only variables bound to an external value, False to exclude all bound variables and any other value to forgo this check.\n`deep=True|False` (by default it’s `False`)\nSet to True to return nested variables meeting all other criteria.\n`sort=True|False` (by default it’s `False`)\nSet to True to return the variables in a consistent order. (Note that this does _not_ mean alphabetical sorting of all variables, just that the order is consistent between calls!)"
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.6853299,"math_prob":0.93715376,"size":15034,"snap":"2020-34-2020-40","text_gpt3_token_len":3434,"char_repetition_ratio":0.19540918,"word_repetition_ratio":0.14969538,"special_character_ratio":0.21251829,"punctuation_ratio":0.117560074,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99036735,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-10-01T05:43:39Z\",\"WARC-Record-ID\":\"<urn:uuid:640af09a-6833-49b3-a6c8-ee2869a93ada>\",\"Content-Length\":\"55005\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:6066ca76-d902-41d1-90dd-f997dbf6f47e>\",\"WARC-Concurrent-To\":\"<urn:uuid:4e475b63-61cd-4284-8b26-c77698fe6dad>\",\"WARC-IP-Address\":\"104.17.32.82\",\"WARC-Target-URI\":\"https://myokit.readthedocs.io/en/stable/api_core/Variable.html\",\"WARC-Payload-Digest\":\"sha1:5XDAWP4IGW2XDBO7TGX5BADIL5INGBL7\",\"WARC-Block-Digest\":\"sha1:3ZTFY73I5HMARBQ424UXTJD266WPW57Y\",\"WARC-Identified-Payload-Type\":\"application/xhtml+xml\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-40/CC-MAIN-2020-40_segments_1600402130615.94_warc_CC-MAIN-20201001030529-20201001060529-00454.warc.gz\"}"} |
https://www.physicsforums.com/threads/why-does-bringing-n-1-orbital-atoms-together-yield-n-levels.714313/ | [
"# Why does bringing N 1-orbital atoms together yield N levels?\n\nA common example of this is that when bringing N hydrogen atoms together into a ring. Far apart, assume each electron exists in the 1s state. As we bring them together, instead of each electron staying at the original 1s level, or all of them changing by the same amount, the 1s level fans out into N.\n\nFor the case of 2 atoms, I can understand this as bonding or anti-bonding of the atoms. i.e., do the wavefunctions add between the protons, meaning each electron can share in the potential of both protons (bonding) or do the wavefunctions destructively interfere between the protons (anti-bonding).\n\nWith 3 atoms, I can't find 3 levels. Assuming Gaussian shaped wavefunctions, note that the sign of each wavefunction between any two atoms defines the wavefunction on the rest of the ring. Since the signs of the wavefunction are independent, there should be 2^3=8 possibilities since each wavefunction can be + or -. Yet, there are really only 2 energetically distinct arrangements that I see: all have the same sign (two cases) or 2 of 3 have the same sign (2*(3 choose 2)), to account for both sign cases). So I get 3 atoms yield 2 levels.\n\nCan somebody shed light on what I've done incorrectly? Or is 3 too small to work correctly? Is there an argument about the shape of the orbitals I've neglected?\n\nThank you.\n\nSimon Bridge\nHomework Helper\nHint: Pauli exclusion principle.\n\nDrDu\nYet, there are really only 2 energetically distinct arrangements that I see: all have the same sign (two cases) or 2 of 3 have the same sign (2*(3 choose 2)), to account for both sign cases). So I get 3 atoms yield 2 levels.\n\nYou are right in that you only get two distinct energy levels. However there are two independent wavefunctions giving the same energy in case of the upper level.\nIn the case of a ring with N atoms, you get (N+1)/2 distinct energy levels.\n\nHint: Pauli exclusion principle.\n\nYeah, I suppose I'm unclear about what defines a state in this context. If I assume all of the electrons remain in the 1s state, while sort of forming a metal, then I have more electrons than available states. So, the creation of more states solves this problem, but isn't obviously the answer.\n\nYou are right in that you only get two distinct energy levels. However there are two independent wavefunctions giving the same energy in case of the upper level.\nIn the case of a ring with N atoms, you get (N+1)/2 distinct energy levels.\n\nBy the two independent wavefunctions in the upper (anti-bonding) level, do you mean the sign of them? Then this should yield 4 total levels since you could argue the same thing for the constructively interfering wavefunctions.\n\nAlso, the (N+1)/2 argument seems contrary to what I remember reading in Kittel: N atoms yield N levels. I assumed the levels were distinct.\n\nSimon Bridge\nHomework Helper\nLets restrict talk to energy eigenstates of the system then.\n\nWhen you bring a lot of atoms together, the electron-electron interactions mean that there is no longer a well-defined 1s state. The new system has different energy levels. The old 1s state is a superposition of the new system ... and the electrons decay into the new energy levels creating the new ground state.\n\nYou can model this by putting two finite square wells at different distances from each other and look what happens to the first few bound states when they are different distances apart. Also start with just one square well, with one particle, and \"switch on\" a second one.\n\nAnother way of thinking about it: no two electrons in the universe are ever in exactly the same discrete state - its just that the energy levels for far-separated wells are so very similar our instruments cannot tell the difference. When the wells get closer, their interactions force the states to split and you can see the difference.\n\nWhatever - once the combined system has formed, the inital 1s states no longer exist and their energy levels may even be forbidden.\n\nWhen lots of atoms are close together, like in a solid, the energy levels merge to form bands.\n\nNote: I was wondering if DrDu was thinking of molecular energy or, maybe, configuration states."
] | [
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https://math.answers.com/questions/What_is_-3_plus_-4 | [
"",
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"",
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"",
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"",
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"0\n\n# What is -3 plus -4?\n\n-7",
null,
"Study guides\n\n20 cards\n\n## A number a power of a variable or a product of the two is a monomial while a polynomial is the of monomials\n\n➡️\nSee all cards\n3.81\n2270 Reviews\n\nits called a calculator and its 16 twenty-seven *",
null,
"4 + 3 = 7 + 4 = 11 + 3 = 14",
null,
"3 + 3 + 4 = 10",
null,
"(-3) + (-4) = -7",
null,
"4 + 3 = 7",
null,
"23",
null,
"one",
null,
"3 + 4 = 7",
null,
"It is -1.",
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"",
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"Earn +20 pts",
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"",
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""
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https://listengine.tuxfamily.org/lists.tuxfamily.org/eigen/2010/07/msg00412.html | [
"Re: [eigen] Issues regarding Quaternion-alignment and const Maps\n\n```Here comes the patch,\n\ngenericity (is that a word?) is based on Gael's suggestions (I just\nreplaced ei_por by ei_pxor, and used the same mask for both addsub\nreplacements).\n\nI was surprised at first that I only get a speedup of about ~1.6x\nagainst non-vectorized version, but then found out that originally\n-msse2 was actually slower than the version without vectorization.\nAnyways, now -msse2 and -msse3 both run faster than just -O2 (on my\nCore2 Duo).\nOn my (rather archaic) Athlon64 the current sse2 version (it does not\nsupport sse3) is *slower* than just using -O2 :(\nWhich, by the way, reminds me of the original topic of this thread ;)\n\nChristoph\n\nGael Guennebaud schrieb:\n> here a variant using at most as possible generic code:\n>\n> const __m128d mask1 = _mm_castsi128_pd(_mm_set_epi32(0x0,0x0,0x80000000,0x0));\n> const __m128d mask2 = _mm_castsi128_pd(_mm_set_epi32(0x80000000,0x0,0x0,0x0));\n>\n> Quaternion<double> res;\n>\n> typedef ei_packet_traits<double>::type Packet;\n>\n> const double* a = _a.coeffs().data();\n> Packet b_xy = _b.coeffs().template packet<Aligned>(0);\n> Packet b_zw = _b.coeffs().template packet<Aligned>(2);\n> Packet a_xx = ei_pset1(a);\n> Packet a_yy = ei_pset1(a);\n> Packet a_zz = ei_pset1(a);\n> Packet a_ww = ei_pset1(a);\n> Packet t1, t2;\n>\n> t1 = ei_padd(ei_pmul(a_ww, b_xy), ei_pmul(a_yy, b_zw));\n> t2 = ei_psub(ei_pmul(a_zz, b_xy), ei_pmul(a_xx, b_zw));\n>\n> #ifdef __SSE3__\n> #else\n> #endif\n>\n> t1 = ei_psub(ei_pmul(a_ww, b_zw), ei_pmul(a_yy, b_xy));\n> t2 = ei_padd(ei_pmul(a_zz, b_zw), ei_pmul(a_xx, b_xy));\n> #ifdef __SSE3__\n> #else\n> #endif\n>\n> return res;\n>\n> Actually, my recent work on the vectorization of complexes, and this\n> code, let me thought that it would be a good idea to add ei_paddsub\n> and ei_psubadd functions such that we could write generic vectorized\n> code for complex and quaternions (generic in the sense it would work\n> for all vector engine).\n>\n> Here is how I see it. For instance let's take the example of the\n> quaternion multiplication. We could have a generic\n>\n> template<typename Quat> Quat ei_quatmul(Quat& a, Quat& b);\n>\n> function calling a ei_quatmul_selector which would be specialized for\n> the 3 following configurations:\n>\n> 1 - ei_packet_traits<Quat::Scalar>::size == 2 => the above code\n> 2 - ei_packet_traits<Quat::Scalar>::size == 4 => the code we already\n> have but written in a generic way\n> 3 - otherwise => scalar path\n>\n> And we should make sure that one can specialize this function for a\n> given scalar type/vector engine in the case some specific\n> optimizations can be done.\n>\n> gael.\n>\n> On Sat, Jul 10, 2010 at 1:11 AM, Christoph Hertzberg\n> <chtz@xxxxxxxxxxxxxxxxxxxxxxxx> wrote:\n>> Benoit Jacob wrote:\n>>> I have made a patch letting ei_pset1 use _mm_loaddup_pd when we have SSE3:\n>>>\n>>> template<> EIGEN_STRONG_INLINE Packet2d ei_pset1<double>(const double&\n>>> from) {\n>>> #ifdef EIGEN_VECTORIZE_SSE3\n>>> #else\n>>> Packet2d res = _mm_set_sd(from);\n>>> return ei_vec2d_swizzle1(res, 0, 0);\n>>> #endif\n>>> }\n>>>\n>>> But guess what? It's actually not faster (perhaps even a bit slower)\n>>> than our ei_vec2d_swizzle1!\n>>>\n>>> So let's just forget about it.\n>>>\n>>> using ? If yes, by using ei_pset1 instead of _mm_loaddup_pd, you can\n>>> make your code work on SSE2 !\n>> I guess the most important SSE3 instruction is _mm_addsub_pd which adds the\n>> first and subtracts the second element. If there is a code which negates\n>> just one element, this could be replaced.\n>>\n>> Googleing a bit implies that the SSE-way to do it is to XOR with\n>> {-0.0, 0.0} (or the other way around). I will try that ...\n>>\n>> Christoph\n\n--\n----------------------------------------------\nDipl.-Inf. Christoph Hertzberg\nCartesium 0.051\nUniversität Bremen\nEnrique-Schmidt-Straße 5\n28359 Bremen\n\nTel: (+49) 421-218-64252\n----------------------------------------------\n```\n```# HG changeset patch\n# User Christoph Hertzberg <chtz@xxxxxxxxxxxxxxxxxxxxxxxx>\n# Date 1278970247 -7200\n# Node ID eeff20a8804b7cdbf34f8be7c80b69993d1f0dc8\n# Parent a9c28fb21cba3cdf81b1964d2510afc1682569eb\nImplemented SSE optimized double-precision Quaternion multiplication\n\ndiff -r a9c28fb21cba -r eeff20a8804b Eigen/src/Geometry/arch/Geometry_SSE.h\n--- a/Eigen/src/Geometry/arch/Geometry_SSE.h\tSun Jul 11 11:01:17 2010 +0200\n+++ b/Eigen/src/Geometry/arch/Geometry_SSE.h\tMon Jul 12 23:30:47 2010 +0200\n@@ -64,4 +64,58 @@\n}\n};\n\n+\n+\n+\n+template<class Derived, class OtherDerived>\n+struct ei_quat_product<Architecture::SSE, Derived, OtherDerived, double, Aligned>\n+{\n+ inline static Quaternion<double> run(const QuaternionBase<Derived>& _a, const QuaternionBase<OtherDerived>& _b)\n+ {\n+ const Packet2d mask = _mm_castsi128_pd(_mm_set_epi32(0x0,0x0,0x80000000,0x0));\n+\n+ Quaternion<double> res;\n+\n+ const double* a = _a.coeffs().data();\n+ Packet2d b_xy = _b.coeffs().template packet<Aligned>(0);\n+ Packet2d b_zw = _b.coeffs().template packet<Aligned>(2);\n+ Packet2d a_xx = ei_pset1(a);\n+ Packet2d a_yy = ei_pset1(a);\n+ Packet2d a_zz = ei_pset1(a);\n+ Packet2d a_ww = ei_pset1(a);\n+\n+ // two temporaries:\n+ Packet2d t1, t2;\n+\n+ /*\n+ * t1 = ww*xy + yy*zw\n+ * t2 = zz*xy - xx*zw\n+ * res.xy = t1 +/- swap(t2)\n+ */\n+ t1 = ei_padd(ei_pmul(a_ww, b_xy), ei_pmul(a_yy, b_zw));\n+ t2 = ei_psub(ei_pmul(a_zz, b_xy), ei_pmul(a_xx, b_zw));\n+#ifdef __SSE3__\n+#else\n+#endif\n+\n+ /*\n+ * t1 = ww*zw - yy*xy\n+ * t2 = zz*zw + xx*xy\n+ * res.zw = t1 -/+ swap(t2) = swap( swap(t1) +/- t2)\n+ */\n+ t1 = ei_psub(ei_pmul(a_ww, b_zw), ei_pmul(a_yy, b_xy));\n+ t2 = ei_padd(ei_pmul(a_zz, b_zw), ei_pmul(a_xx, b_xy));\n+#ifdef __SSE3__\n+#else"
] | [
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.58994126,"math_prob":0.9406277,"size":6397,"snap":"2020-24-2020-29","text_gpt3_token_len":2143,"char_repetition_ratio":0.13374002,"word_repetition_ratio":0.028477546,"special_character_ratio":0.37689543,"punctuation_ratio":0.1910828,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9739414,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-05-26T07:43:57Z\",\"WARC-Record-ID\":\"<urn:uuid:002511fd-c3f3-402b-b78e-8b7bb6014b7c>\",\"Content-Length\":\"13703\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:9dbf6194-6e24-4956-ac27-0d6b72a31ae9>\",\"WARC-Concurrent-To\":\"<urn:uuid:aefd0d11-d43d-4137-9609-84cc31ff7a26>\",\"WARC-IP-Address\":\"212.85.158.5\",\"WARC-Target-URI\":\"https://listengine.tuxfamily.org/lists.tuxfamily.org/eigen/2010/07/msg00412.html\",\"WARC-Payload-Digest\":\"sha1:D2Y45ZNER2GU2ZUJ2BLV5BMPZVOPEPRL\",\"WARC-Block-Digest\":\"sha1:AIPPDY2RO3CYEQS25QWRN3V2WCTANNMJ\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-24/CC-MAIN-2020-24_segments_1590347390448.11_warc_CC-MAIN-20200526050333-20200526080333-00441.warc.gz\"}"} |
https://cs.stackexchange.com/questions/144270/time-varying-edge-cost-minimum-spaning-tree | [
"# Time-varying edge cost Minimum Spaning Tree\n\nI am having a hard time wrapping my head around the time-varying edge cost of this question :\n\nSuppose we have a connected graph $$G = (V, E)$$. Each edge e now has a time-varying edge cost given by a function $$f_e$$. Thus, at time t, it has cost $$f_e(t)$$. We’ll assume that all these functions are positive over their entire range. Observe that the set of edges constituting the minimum spanning tree of $$G$$ may change over time. Also, of course, the cost of the minimum spanning tree of $$G$$ becomes a function of the time $$t$$; we’ll denote this function $$c_G(t)$$. A natural problem then becomes: find a value of $$t$$ at which $$c_G(t)$$ is minimized. Suppose each function $$f_e$$ is a polynomial of degree 2: $$f_e(t) = a_et^2 + b_et + c_e$$ , where $$a_e$$ is positive for each edge $$e$$. Give an algorithm that takes the graph $$G$$ and the values $$\\{(a_e , b_e , c_e) : e ∈ E\\}$$ and returns a value of the time $$t$$ at which the minimum spanning tree has minimum cost. Your algorithm should run in time polynomial in the number of nodes and edges of the graph $$G$$. You may assume that arithmetic operations on the numbers $$\\{(a_e , b_e , c_e)\\}$$ can be done in constant time per operation.\n\nMy first idea was, because $$a_e$$ is positive, to find out the maximum value of $$t$$ that minimizes the cost of an edge $$e$$ (in other words looping through every edge once and find their minima). Intuitively, the maximum of these minima could be an upper-bound for the value that minimizes $$c_G(t)$$. Then I could run a MST algorithm for every time step from 0 to t and return the t that answers the question. However, this looks like a very brute-forcey method and thus not time-efficient at all.\n\nAny help would be greatly appreciated!\n\nNote that this question was part of the Greedy Algorithm chapter.\n\n• We require you to attribute the original source of all copied material: cs.stackexchange.com/help/referencing. Saying it came from \"the Greedy Algorithm chapter\" is not adequate. Please see the link for our expectations about providing a proper reference: e.g., we'd expect you to provide a full reference, including the textbook name and authors.\n– D.W.\nSep 27, 2021 at 4:11\n\nConsider the edges sorted by $$f_e(t)$$ for some $$t$$.\nTwo quadratic polynomials can intersect at most two times. When the cost function of two edges $$e_1$$ and $$e_2$$ cross at time $$t$$, it means that\n• $$f_{e_1}(t-\\epsilon) < f_{e_2}(t-\\epsilon)$$ and\n• $$f_{e_1}(t+\\epsilon) > f_{e_2}(t+\\epsilon)$$.\nSince there are $$m$$ edges and each pair of edges crosses at most twice, we have at most $$2 \\cdot {m \\choose 2} \\leq m^2$$ many crossings, and therefore at most $$m^2$$ many times two edges can swap order in the sorted list.\nNow you try all possible time intervals that don't contain crossings, $$(t_a, t_b)$$ and pick the $$n-1$$ cheapest edges. What remains is to minimize a sum over $$n-1$$ quadratic functions which is indeed a quadratic function."
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.90650403,"math_prob":0.9998616,"size":1778,"snap":"2022-27-2022-33","text_gpt3_token_len":454,"char_repetition_ratio":0.12401353,"word_repetition_ratio":0.012383901,"special_character_ratio":0.26209223,"punctuation_ratio":0.0989011,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":1.0000032,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-08-09T11:41:01Z\",\"WARC-Record-ID\":\"<urn:uuid:ff675045-b3b7-4caa-a3dc-9297be94be15>\",\"Content-Length\":\"224509\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:0e1333d7-97d0-4aff-af56-1536b6681801>\",\"WARC-Concurrent-To\":\"<urn:uuid:b60929ec-558b-4290-a9ee-58f67ddd88b7>\",\"WARC-IP-Address\":\"151.101.65.69\",\"WARC-Target-URI\":\"https://cs.stackexchange.com/questions/144270/time-varying-edge-cost-minimum-spaning-tree\",\"WARC-Payload-Digest\":\"sha1:7UF5YQR7PRS2YI3JSOEFJHLTNQ5LAH7A\",\"WARC-Block-Digest\":\"sha1:4UBPMFN7QUXRHUEINOG7GMJ5TNGLIGUY\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-33/CC-MAIN-2022-33_segments_1659882570921.9_warc_CC-MAIN-20220809094531-20220809124531-00331.warc.gz\"}"} |
https://math.stackexchange.com/users/614261/maggie?tab=topactivity | [
"### Questions (29)\n\n 6 $T \\in \\mathcal L (V)$ has no real eigenvalues. Prove that every subspace of $V$ invariant under $T$ has even dimension. 3 Need to compute $\\sum_{n=0}^\\infty \\frac{(-1)^n}{(2n+1)(2n+2)3^{n+1}}$. Is my solution correct? 3 Solve the differential equation $\\sin x\\frac {dy}{dx}+(\\cos x)y=\\sin(x^2)$ [closed] 2 Probability of rolling three different dice [duplicate] 2 Evaluate $\\lim_{x\\to \\infty} (\\frac {x+1}{x-1})^x$\n\n### Reputation (353)\n\n -2 Show that $f(x)=\\sum_{n=1}^{\\infty}b_n\\sin nx$ and $f'(x)=\\sum_{n=1}^{\\infty}nb_n\\cos nx$ coverge uniformly for all x. +10 Prove that$|\\int_{D}f(x)dx|\\leq\\int_{D}|f(x)|dx$. +10 $T \\in \\mathcal L (V)$ has no real eigenvalues. Prove that every subspace of $V$ invariant under $T$ has even dimension. +10 Find the curvature of the ellipse by the explicit equation $y=\\pm\\frac{b}{a}\\sqrt{a^2-t^2}$\n\nThis user has not answered any questions\n\n### Tags (30)\n\n 0 integration × 6 0 exponential-function × 2 0 linear-algebra × 6 0 general-topology × 2 0 limits × 3 0 sequences-and-series × 2 0 geometry × 3 0 taylor-expansion 0 eigenvalues-eigenvectors × 2 0 vector-space-isomorphism\n\n### Bookmarks (0)\n\nThis user has no bookmarked questions."
] | [
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http://lists.dep.state.fl.us/pipermail/shrug-l/2011-May/003472.html | [
"# shrug-l: Lat/Long(decimal degrees) to Meters without Geometry service\n\nJanelle Brown Janelle.Brown at citizensfla.com\nThu May 26 14:02:17 EDT 2011\n\n```Hello,\n\nDoes anyone have a javascript code that can convert decimal degrees\n(wkid: 4326) points to meters (wkid: 3857)? I am not able to use the\ngeometry service for a number of reasons, so I believe the only way to\ndo this would be to convert the points mathematically. This is what I\nhave so far and it's kinda working (I'm off about 20-25 miles), but I'm\npretty sure there is some tweaking needed in the equations. If anyone\nhas any ideas or has already written some javascript code for this, I'd\nbe very grateful!\n\nvar xMin,yMin,xMax,yMax;\n\nvar lon ,lat;\n\nvar DEG_TO_RAD = 0.017453292519943296;\n\nvar PI = 3.14159267;\n\nvar NUM = 6356752.3142;\n\nlon = -84.286616;\n\nlat = 30.437574;\n\n//Limit to World\n\nif(lat<=-90)lat=-90;\n\nif(lat>=90)lat=90;\n\nif(lon<=-180)lon=-180;\n\nif(lon>=180)lon=180;\n\nvar Y = Math.round(NUM * Math.log( Math.tan( ((lat * DEG_TO_RAD) +\n(PI/2)) / 2 ) ));\n\nvar X = Math.round(NUM * (lon * DEG_TO_RAD));\n\nThank you in advance!\n\nJanelle Brown\n\nGIS Analyst, Corporate Analytics\n\nCitizens Property Insurance Corporation\n\n-------------------------------------------------------\n\nofc. 850.521.8373 | cell 850.879.2374\n\n-------------- next part --------------\nAn HTML attachment was scrubbed...\nURL: http://lists.dep.state.fl.us/pipermail/shrug-l/attachments/20110526/706f1aaa/attachment.htm\n```"
] | [
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.7392985,"math_prob":0.81001496,"size":1502,"snap":"2019-13-2019-22","text_gpt3_token_len":442,"char_repetition_ratio":0.1188251,"word_repetition_ratio":0.0,"special_character_ratio":0.38215712,"punctuation_ratio":0.19078948,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9639298,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-03-23T23:16:27Z\",\"WARC-Record-ID\":\"<urn:uuid:92f54cff-55b1-4489-86c1-e69b608e125b>\",\"Content-Length\":\"4374\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:5defab2f-ae96-4f21-8ecf-4c12dd2dd683>\",\"WARC-Concurrent-To\":\"<urn:uuid:df727dfc-23f9-420d-a2ff-e8b5cc5dde2f>\",\"WARC-IP-Address\":\"40.79.63.153\",\"WARC-Target-URI\":\"http://lists.dep.state.fl.us/pipermail/shrug-l/2011-May/003472.html\",\"WARC-Payload-Digest\":\"sha1:FIZLY3SQ5OOHO2VVYL33FQW7Z4GADQGI\",\"WARC-Block-Digest\":\"sha1:W7IKR3OAISBJBWZCQVPBUUANR3GQOF2T\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-13/CC-MAIN-2019-13_segments_1552912203093.63_warc_CC-MAIN-20190323221914-20190324003914-00232.warc.gz\"}"} |
https://physics.stackexchange.com/questions/103779/lie-algebra-generators | [
"# Lie Algebra Generators\n\nWe know that rotations are performed via real and orthogonal matrices, $O^{T}=O^{-1}$. We can write $O$ as, (The proper rotations have unit determinant)\n\n$$O = \\exp(A),$$\n\nwhere $A^{T}=-A$. In three space dimensions a real anti-symmetric matrix will require three independent entries.\n\n$$A_{ab} = \\epsilon_{abc}\\theta_c.$$\n\nThis looks like $\\vec{M}\\cdot\\vec{\\theta}$, where the matrix elements of the generators $M_c$ are given as\n\n$$(M_c)_{ab}=\\epsilon_{abc}.$$\n\nLet us remember:\n\n$$[t_a,t_b]= i\\,\\,f_{abc}t_c.$$\n\nWhere $f$ is structure constant. We know that $f$'s should satisfy Jacobi identity\n\n$$f_{bcm}f_{mah} + f_{abh}f_{hcm} + f_{cah}f_{hbm} = 0.$$\n\nWe also know that:\n\n$$t_a\\,t_b - t_b\\,t_c - i\\,f_{abc}t_c = 0.$$\n\nThe question is: How can I find $t_a$ individually in terms of $f$'s? Do you have a suggestion?\n\n• What do $t_a$ mean? If they are the generators, you just have to evaluate the derivate of the group elements. – jinawee Mar 16 '14 at 20:01\n• Yes they are generator. But will I evaluate? – Engin Eren Mar 16 '14 at 20:24\n• Are you given the group elements or just the structure constants. In the latter case, I should edit my answer. – jinawee Mar 16 '14 at 21:27\n• I need to find generators in terms of structure constant $f_{abc}$. – Engin Eren Mar 16 '14 at 21:33\n\n## 1 Answer\n\nIn general you can't find a unique Lie algebra basis $X_j$ from the structure constants because there are in general many Lie groups with the same Lie algebra: $SO(3)$, $SU(2)$ have the same Lie algebra and hence the same structure constants. They do however exponentiate differently to define different global topologies, as I discuss in detail here.\n\nFurthermore, even within the same Lie group, if the $X_j$ all undergo the same similarity transformation $X_j\\mapsto \\sigma\\,X_j\\,\\sigma^{-1}$ then the structure constants are unchanged. This is a literal matrix similarity transformation in a matrix Lie algebra and, by Ado's theorem, every Lie algebra can be faithfully represented as a matrix Lie algebra, although this representation is not unique, as the $SO(3)$, $SU(2)$ example above shows.\n\nAlthough your problem looks as though it should be easy (one feels that one should be able to read the Lie algebra members straight off the structure constants), the general solution to your problem is surprisingly tricky and is in fact the subject matter of Ado's theorem. The best discussion of this I have seen is on Terence Tao's blog here, but it is not constructive. I can't recall whether there is a constructive proof in Bourbaki but in any case the only constructive proof I have ever seen left me with severe bruising around the head and utterly defeated me, so I'm not qualified to talk about it.\n\nHowever, you can get cunning in your $O(N)$ case, by taking heed of the properties of the Adjoint Representation, on which my take is discussed here. $O(N)$, as far as the Lie algebra is concerned, the same as $SO(N)$ because, when we relax the constraint $\\det U = 1$ in $SO(N)$ there are only two, discrete possibilities for the determinant: since $U^T U = 1$, $U$ is real, we must have $\\det U = \\pm 1$. So $O(N)$ is simply two disjoint copies of $SO(N)$. So the identity connected component of $O(N)$ is $SO(N)$. This is unlike the unitary group, where we have $\\det U = e^{i\\,\\phi},\\,\\phi\\in\\mathbb{R}$ and $U(N) = SU(N)\\otimes U(1)$.\n\nNow, $SO(N)$ is simple (in the sense of having no normal Lie subgroups) aside from in the \"fiddly case\" of $SO(4)$, which itself has a universal cover $\\tilde{SO}(3) \\otimes \\tilde{SO}(3)$ where $\\tilde{SO}(3)\\cong SU(2)$ in which case $SO(4)$ is semisimple. In particular $SO(N)$ has no continuous centre. Since $\\ker(\\mathrm{Ad})$ is the centre of a group, the big A adjoint representation annihilates the centre. Therefore in such a lucky case where there is no continuous centre, the image of $SO(N)$ under the big A adjoint representation is either isomorphic to $SO(N)$ itself, or is the quotient group $SO(N)/\\mathbb{Z}_m$, where we have modded a discrete centre out of the latter.\n\nAll of this is a long winded way of saying that in your lucky case, the Lie algebra of the group's image under the adjoint representation is the same as that of the group.\n\nSo now let $[X_j,\\,X_k] = i\\,f_{j,\\,k,\\,\\ell}\\,X_\\ell$ in your notation. This tells you that when you choose a basis for the Lie algebra aligned to the $X_j$, we have:\n\n$$\\mathrm{ad}(X_j) X_k = i\\,\\sum\\limits_{\\ell=1}^N f_{j,\\,k,\\,\\ell}\\,X_\\ell$$\n\nor\n\n$$\\mathrm{ad}(X_j)_{\\ell,\\,k} = i\\, f_{j,\\,k,\\,\\ell}$$\n\nThen, the \"braiding\" relationship $\\mathrm{Ad}(e^{X\\,t}) = \\exp(\\mathrm{ad}(X)\\,t)$ tells you that these matrices are a basis for the Lie algebra $\\mathrm{Lie}(\\mathrm{Ad}(O(N)))$ of $\\mathrm{Ad}(O(N))$. As we have seen,\n\n$$\\mathrm{Lie}(\\mathrm{Ad}(O(N))) = \\mathrm{Lie}(O(N))$$\n\nso you can take the $\\mathrm{ad}(X_j)$ as your \"generators\", i.e. in your notation:\n\n$$(t_a)_{b,\\,c} = \\mathrm{ad}(X_a)_{b,\\,c} = i\\, f_{a,\\,c,\\,b}$$\n\nwill give you an $N\\times N$ faithful matrix representation of your generators. Note that any similarity transformation can be imparted on these generators and you still have a perfectly good set.\n\nSo this answers your question, but a lingering thought is: where does the Jacobi identity come into it and why apparently haven't we used it? The answer is that is has: the Jacobi identity is nothing more than a disguised version of the following general fact: the Lie group homomorphism $\\mathrm{Ad}:\\mathfrak{G} \\to \\mathrm{Aut}(\\mathfrak{G})$ induces a homomorphism $\\mathrm{ad}$ of Lie algebras that respects Lie brackets:\n\n$$\\mathrm{ad}([X,\\,Y]) = \\mathrm{ad}(X)\\,\\mathrm{ad}(Y)-\\mathrm{ad}(Y)\\,\\mathrm{ad}(X) = [\\mathrm{ad}(X),\\,\\mathrm{ad}(Y)]$$\n\nand this last relationship is in fact equivalent to the Jacobi identity."
] | [
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https://discussions.unity.com/t/c-snap-two-squares-together-when-they-touch-stacker-like-game/160302 | [
"# C# Snap two squares together when they touch(Stacker like game)\n\nHello, so i do not know how to make it because i am a beginner.\nA script for 2D Squares to snap together when they touch.\nThe game you try to make is a stacker like game (its just a mini project).\nI would like to make it so when the block hits the ground/another block it snaps to it. Not like exactly one on another but how every it falls. But if the blocks is too much on the edge it shouldn’t snap to it but just fall.\n\nI hope i was clear, if not i can try to explain(English isn’t my first language, sorry)\n\nI think I understand what it is that you are asking for, though I could be wrong.\n\nYou’ll need a snap range\n\n``````private float _snapRange = 0.05f;\n``````\n\nYou’ll then need to know the size of both objects.\n\n`````` float ObjectASize = 10f;\nfloat ObjectBSize = 10f;\n``````\n\nThen you’ll need to get the distance from ObjectA and ObjectB\n\n`````` float distance = Vector3.Distance(ObjectA, ObjectB);\n``````\n\nNow subtract have the size of each object from the distance\n\n``````float netDistance = distance - ((ObjectASize / 2) + (ObjectBSize / 2));\n``````\n\nNow, check if netDistance is within your snap range and if so, snap it\n\n`````` if(Mathf.Abs(netDistance) < _snapRange)\n{\n``````Vector3 targetPos = ObectB.transform.position;"
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http://lambda.jimpryor.net/git/gitweb.cgi?p=lambda.git;a=blobdiff;f=assignment6.mdwn;h=82aee95f9bdcb506e52c1307a54a15ce9340e899;hp=d044c12dcb98bce3fa2fb8101a7926ddd09c4e0b;hb=ce6251343f916dcbfeb7160aff5b1312fa31ce7f;hpb=7834e35d4a2de390ce6a1e9113186dcbb9c07a6f | [
"index d044c12..82aee95 100644 (file)\n@@ -1,5 +1,39 @@\n-1. Test all three monad laws for the intensionality monad. To do\n-this, download the code and load it into your Ocaml evaluator (`# #use\n-\"intensionality-monad.ml\";;`). For instance, what does evaluating\n-`bind (unit 'c') (swap left) 2 == swap left 'c' 2;;` show? Please\n-explain briefly but clearly what you are doing in your discussion.\n+1. **Build a State monad.** Based on the division by zero monad,\n+build a system that will evaluate arithmetic expressions. Instead of\n+returning a simple integer as a result, it will deliver the correct\n+answer along with a count of the number of operations performed during\n+the calculation. That is, the desired behavior should be like this:\n+\n+ # lift2 ( + ) (lift2 ( / ) (unit 20) (unit 2))\n+ (lift2 ( * ) (unit 2) (unit 3)) 0;;\n+ - : int * int = (16, 3)\n+\n+ Here, `lift2` is the function that uses `bind` to prepare an ordinary\n+arithmetic operator (such as addition `( + )`, division `( / )`, or\n+multiplication `( * )`) to recieve objects from the counting monad as\n+arguments. The response of the interpreter says two things: that\n+(20/2) + (2\\*3) = 16, and that the computation took three arithmetic\n+steps. By the way, that zero at the end provides the monadic object\n+with a starting point (0 relevant computations have occurred previous\n+to the current computation).\n+\n+ Assume for the purposes of this excercise that no one ever tries to\n+divide by zero (so there should be no int option types anywhere in\n+your solution).\n+\n+ You'll need to define a computation monad type, unit, bind, and lift2.\n+We encourage you to consider this hint: [[hints/Assignment 6 Hint 1]].\n+\n+ <div style=\"color: red\">See our [commentary](/hints/assignment_6_commentary) on your solutions.</div>\n+\n+\n+2. Prove that your monad satisfies the monad laws. First, give\n+examples illustrating specific cases in which the monad laws are\n+obeyed, then explain (briefly, not exhaustively) why the laws hold in\n+general for your unit and bind operators.\n+\n+3. How would you extend your strategy if you wanted to count\n+arithmetic operations, but you also wanted to be safe from division by\n+zero? This is a deep question: how should you combine two monads into\n+a single system? If you don't arrive at working code, you can still\n+discuss the issues and design choices."
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https://www.bookpeople.com/book/9781138413368 | [
"# Geometry, Topology and Physics (Hardcover)\n\n\\$240.00\nBackordered. Place an order and we will let you know when to expect your book.\n(This book cannot be returned.)\n\n### Description\n\nDifferential geometry and topology have become essential tools for many theoretical physicists. In particular, they are indispensable in theoretical studies of condensed matter physics, gravity, and particle physics. Geometry, Topology and Physics, Second Edition introduces the ideas and techniques of differential geometry and topology at a level suitable for postgraduate students and researchers in these fields.\n\nThe second edition of this popular and established text incorporates a number of changes designed to meet the needs of the reader and reflect the development of the subject. The book features a considerably expanded first chapter, reviewing aspects of path integral quantization and gauge theories. Chapter 2 introduces the mathematical concepts of maps, vector spaces, and topology. The following chapters focus on more elaborate concepts in geometry and topology and discuss the application of these concepts to liquid crystals, superfluid helium, general relativity, and bosonic string theory. Later chapters unify geometry and topology, exploring fiber bundles, characteristic classes, and index theorems. New to this second edition is the proof of the index theorem in terms of supersymmetric quantum mechanics. The final two chapters are devoted to the most fascinating applications of geometry and topology in contemporary physics, namely the study of anomalies in gauge field theories and the analysis of Polakov's bosonic string theory from the geometrical point of view.\n\nGeometry, Topology and Physics, Second Edition is an ideal introduction to differential geometry and topology for postgraduate students and researchers in theoretical and mathematical physics.\n\nProduct Details\nISBN: 9781138413368\nISBN-10: 1138413364\nPublisher: CRC Press\nPublication Date: July 27th, 2017\nLanguage: English"
] | [
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https://physics.stackexchange.com/questions/629067/gravitational-slingshots-and-energy-conservation | [
"# Gravitational slingshots and energy conservation\n\nI was watching YouTube video about gravitational slingshots and conservation of energy was explained as \"planet loses its orbital energy around sun while spacecraft gains energy\"\n\n1. In solar system it explained as decreased orbital energy around sun but its supposed that spacecraft approach from front side (their initial velocities are opposite)\n\n2. What if it approaches from back side, then due to conservation of momentum shouldn't final velocities of both increase? (same direction for initial velocities)\n\nActually when you think with reference to sun, both cases are ok, they don't violate anything, so it doesn't depend from what direction you approach planet with respect to its velocity. But what if there is no any solar/other systems and orbits and planet is stationary?\n\nWhat i was thinking for is, lets consider big mass/planet in space but stationary this time, like really absolute stationary. When spacecraft gets near, due to centrifugal force it changes direction of its initial velocity lets say for matter of simplicity by 180 degrees (u shape turn). But due to conservation of momentum, planet accelerates in opposite direction little amount and thats where i got confused. Looks like total energy is increased.\n\nI was thinking maybe it can be explained with the fact that nothing is stationary in space and there is no absolute speed as everything depends on which reference frame you are using, so there definitely should be some orbital path for that 'stationary planet' but then it violates my first assumption 'absolute stationary'.\n\nWhen you starting do this thought experiment in completely stationary and empty space, things gets weird.\n\nConsider a simpler example: an elastic collision of two bodies when there is a significant difference in masses.\n\nBy going into CoM reference frame, one can show that the velocity of a smaller body after the collision will increase by two velocities of a bigger body $$v_2=v_1+2u$$. What will happen if the larger body were flying away? It's as simple as letting $$u\\to -u$$. The speed of a smaller body will decrease by two velocities of a bigger body.\n\nIn other words, instead of an accelerating slingshot effect, the spacecraft will experience decelerating slingshot effect. (And since the spacecraft lost it's kinetic energy, this energy was spent to increase the angular momentum of the planet, which counterintuitively resulted in an average slowing of the planet. However, it's another story)\n\nIt is convenient to view gravitational interactions as scattering. To illustrate how this works we take a look at the rest frame of the big planet and look only at the planet/spacecraft interaction for now. This gives the following picture",
null,
"In this idealised scenario with only the planet and the spacecraft present we see that the spacecraft follows a hyperbola. This means at large distances the path asymptotically becomes a straight line with constant velocity (momentum). The difference between the initial momentum (right pointing blue arrow) and the final momentum (left pointing blue arrow) is the momentum kick $$\\Delta p$$ that the spacecraft got from this flyby. Similarly the planet got a kick equal to $$-\\Delta p$$ pointing in the other direction. But, as you mentioned, this will have neglible effect on the motion of the planet. If you think in terms of these asymptotic momenta and ignore the time interval where the planet and spaceship are interacting this problem is very similar to a hard sphere collision. The time interval where they are interacting is quite long so this comparison should be taken with a grain of salt.\n\nNow by viewing this problem this way a lot of things should be more clear. In case there are no other planets this problem is the same as a hard sphere collision. So if you want to know why energy is conserved you could easily find a proof of that (for example in Vasily Mitch's answer). If you add back the sun you can reason in terms of this $$\\Delta p$$ to find out if the energy will increase. Take the velocity vector of the spacecraft with respect to the sun. Is $$\\Delta p$$ in the same direction as the velocity vector? The energy will increase and you will get a bigger orbit. Is $$\\Delta p$$ in the opposite direction? The energy will decrease and you will get a smaller orbit.\n\n• I see but it doesnt answer my question.Its helpful to approach it as collision problem but what i was trying to say,momentum-energy should be conserved.Gravitational forces act perpendicular to spacecrafts velocity.So it just changes it direction around 'peak' of that hyperbola.But same long term 'impulse' is acting to planet,which is eventually gains velocity little bit.So in order satisfy conservation of energy we see that final velocity vector of spacraft should lose some of its magnitude besides its directional change.But from where comes that loss if all forces acting perpendicular ? – Tymofei Ermirowich Gyuler May 11 at 10:37\n• @TymofeiErmirowichGyuler \"Gravitational forces act perpendicular to spacecrafts velocity\" This is not true. This is only true at the point of closest approach. But for the two body system the asymptotic momenta still have the same magnitude. To see how the spacecraft gains energy you have to look at the complete system including the sun. If you are in orbit around sun and you can give the spacecraft a momentum kick in any direction you want, how would you orient that kick to increase the orbital energy around the sun? – AccidentalTaylorExpansion May 11 at 11:22\n• I thought like this; on upper and lower areas of hyperbola you drawn,forces will cancel each other out(vertically) and same happens on peak.And you left with net horizontal force (deltaP) pointing to left. \"But for the two body system the asymptotic momenta still have the same magnitude\".Thats the point where i got confused.Beacause as planet is not moving,it cant lose its velocity/energy due to conservation but instead will increase in opposite direction. I would give momentum kick in direction of its initial velocity of coarse. – Tymofei Ermirowich Gyuler May 11 at 11:35\n• But lets say you have stationary \"explosive\" near back of spacecraft and when it explodes,both of them will gain additional velocities (but opposite direction).Here i know that rise in kinetic energy is due to stored explosive energy,but i cant figure out reason for energy rise in first question where you have stationary planet. Maybe i should think like lost potential enery between spacecraft is analogical to electrical potential between to opposite charges that are infinitely far away from each other ? – Tymofei Ermirowich Gyuler May 11 at 11:38\n• @TymofeiErmirowichGyuler You have to remember that both objects are orbiting the sun. In the rest frame of the planet the spaceship doesn't gain any energy. But in rest frame of the sun you do see a gain in energy. In the rest frame of the sun both the planet and spaceship are orbiting at very high velocities. If the momentum kick is in the same direction as this velocity the spacecraft will gain kinetic energy in the sun-frame (and the planet will lose some) – AccidentalTaylorExpansion May 11 at 12:11"
] | [
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"https://i.stack.imgur.com/K4OTc.png",
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