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https://www.appsapk.com/mathematics-app/	[ "", null, "# Mathematics APK\n\n– formula calculation with variables, sums, products and sequences\n– solve equations (linear, quadratic, cubic, transpose)\n– plot functions, derivate, integrate, calculate roots and extremas (polynomial, rational, exponential, etc)\n– calculate tangent, asymptote, interception\n– reconstruction: calculate f(x) by given roots, points or extremas\n– f(x,y) plot in 3D\n– factoring (Bézout’s identity, Euler’s totient function, least common multiple, greatest common divisor)\n– modulo calculation with big integer and fractions\n– differential and integral calculus\n– curve sketching, limes, minima, maxima\n– linear algebra, vectors, matrices\n– unit conversion\n– number systems, binary/oct/decimal/hex calculator\n– complex number calculation\n– statistics (gaussian test, t test, binominal and normal distribution, standard deviation, average, sum, median)Supported languages:\nEnglish, German, Francais, Espanol, Italian, Portuguese\n\nMathematics is a powerful calculation software for your android smartphone.\n\nFEATURES\n\nCalculate any formula you want and show them in a 2d or 3d plot. The natural display shows fractions, roots and exponents as you would expect it from mathematics.\n\nIn a few seconds you derivate or integrate your desired function, calculate the zero points of your function und show them in the function plot. See all maxima, minima or inflection points in one view.\n\nThe easy way of use allows you to solve linear equations in just a moment. Or transform your mathematical, physical or chemical equation to any unknown variable.\n\nYou often needs to calculate with binary, octal or hexadecimal number systems? No problem! You can mix them together in one calculation even by using decimal digits. But that’s not enough! You can also calculate with any other number system with base 2 to 18.\n\nFrom time to time you may need to convert units to another one, like Celsius to Fahrenheit, miles to kilometre, inches to foot and so on.\n\nYou will also be able to calculate with vectors, matrices and determinants.\n\nAll this features are combined in this app and will make your mathematical life a lot easier.\n\n#### What’s New\n\n– share your calculations via Bluetooth\n– time calculator\n\nName\nMathematics\nPackage\nde.daboapps.mathematics\nVersion\n3.3\nSize\n1.91 MB\nInstalls\nDeveloped By\ndaboApps\n\n• pashuram chaudhary says:\n\nexcellent apps for st.\n\n• sunil says:\n\nBeautiful app!\n\n• computer repair hackney says:\n\nI would have loved to have had this app when I was in school. Although found it to be a little slow.\n\nThis site uses Akismet to reduce spam. Learn how your comment data is processed." ]	[ null, "data:image/svg+xml,%3Csvg%20xmlns=%22http://www.w3.org/2000/svg%22%20viewBox=%220%200%20120%20120%22%3E%3C/svg%3E", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.8450863,"math_prob":0.9799419,"size":2358,"snap":"2020-45-2020-50","text_gpt3_token_len":529,"char_repetition_ratio":0.11852167,"word_repetition_ratio":0.0,"special_character_ratio":0.20653096,"punctuation_ratio":0.15898618,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9868209,"pos_list":[0,1,2],"im_url_duplicate_count":[null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-11-29T10:35:58Z\",\"WARC-Record-ID\":\"<urn:uuid:f4dd8cd4-96a9-42a8-aaa7-42126eb6a57b>\",\"Content-Length\":\"450722\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:591b26e5-c720-46b5-9f87-8d87703541dc>\",\"WARC-Concurrent-To\":\"<urn:uuid:1bf1c47d-ef45-4a83-9cd7-91f337b71514>\",\"WARC-IP-Address\":\"104.26.2.18\",\"WARC-Target-URI\":\"https://www.appsapk.com/mathematics-app/\",\"WARC-Payload-Digest\":\"sha1:RKRT7SJLQUA7V24Z3MB3XNBVHIMSU7HZ\",\"WARC-Block-Digest\":\"sha1:YQ24YF2YVOBMRIZP3BOOVWPOXY46DTBX\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-50/CC-MAIN-2020-50_segments_1606141197593.33_warc_CC-MAIN-20201129093434-20201129123434-00142.warc.gz\"}"}
https://www.enotes.com/homework-help/use-integration-by-parts-integrate-integration-350459?en_action=hh-question_click&en_label=hh-sidebar&en_category=internal_campaign	[ "# Use integration by parts to integrate Integration sign, x^5 ln (x) dx\n\njeew-m", null, "\| Certified Educator\n\ncalendarEducator since 2012\n\nstarTop subjects are Math, Science, and Social Sciences\n\nUsing integration by parts;\n\nintudv = uv-intvdu\n\nLet v=x^6 and u=lnx\n\nThen;\n\ndv = 6x^5 dx\n\ndu = 1/x * dx\n\n`intudv = int(lnx6x^5)dx`\n\n`int(lnx6x^5)dx= (lnx)(x^6)-int(x^6)1/xdx`\n\n`int(6lnxx^5)dx = (lnx)(x^6)-intx^5dx`\n\n`int(x^5lnx) = (1/6)[(lnx)(x^6)-x^6/6]+C`\n\ncheck Approved by eNotes Editorial\n\nbeckden", null, "\| Certified Educator\n\ncalendarEducator since 2011\n\nstarTop subjects are Math, Science, and Business\n\nUse LIPET (ln, inv trig, polynomial, exponential, trig) to find u, we use `u=ln(x)` , so `dv=x^5.dx`\n\n`du = 1/x dx` , `v = x^6/6 `\n\nwe use `int u dv = uv - int v du`\n\n`int x^5 ln(x) dx = ln(x) x^6/6 - int x^6/6 1/x dx`\n\n`int x^5 ln(x) dx = (x^6 ln(x))/6 - int x^5/6 dx`\n\n`int x^5 ln(x) dx = (x^6 ln(x))/6 - 1/6 x^6/6 + C`" ]	[ null, "https://static.enotescdn.net/images/core/educator-indicator_thumb.png", null, "https://static.enotescdn.net/images/core/educator-indicator_thumb.png", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.66288084,"math_prob":0.9994716,"size":1512,"snap":"2019-51-2020-05","text_gpt3_token_len":562,"char_repetition_ratio":0.15782493,"word_repetition_ratio":0.11764706,"special_character_ratio":0.36640212,"punctuation_ratio":0.07079646,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99990344,"pos_list":[0,1,2,3,4],"im_url_duplicate_count":[null,null,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-01-19T14:09:22Z\",\"WARC-Record-ID\":\"<urn:uuid:cc777452-01a5-42be-af58-f9f360bdaa8b>\",\"Content-Length\":\"46434\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:c8c84c52-0e4d-4fbb-a2b2-52aeae24d97a>\",\"WARC-Concurrent-To\":\"<urn:uuid:94b96eda-c432-4a0d-b333-172ec1c48090>\",\"WARC-IP-Address\":\"104.26.4.75\",\"WARC-Target-URI\":\"https://www.enotes.com/homework-help/use-integration-by-parts-integrate-integration-350459?en_action=hh-question_click&en_label=hh-sidebar&en_category=internal_campaign\",\"WARC-Payload-Digest\":\"sha1:L3GOST77PLSGPYPJX75LDHQS2IXKILJU\",\"WARC-Block-Digest\":\"sha1:3PRJH7ROJ5SXAGVJMCQTTCDGV6OB7ACZ\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-05/CC-MAIN-2020-05_segments_1579250594603.8_warc_CC-MAIN-20200119122744-20200119150744-00400.warc.gz\"}"}
https://cyberleninka.org/article/n/3035	[ "# Two Mesh Deformation Methods Coupled with a Changing-connectivity Moving Mesh Method for CFD ApplicationsAcademic research paper on \"Medical engineering\"", null, "CC BY-NC-ND", null, "0", null, "0\nShare paper\nProcedia Engineering\nOECD Field of science\nKeywords\n{\"Moving mesh\" / \"Inverse Distance Weighted interpolation\" / \"Elasticity-based mesh deformation\" / \"ALE simulations\"}\n\n## Abstract of research paper on Medical engineering, author of scientific article — Nicolas Barral, Edward Luke, Frédéric Alauzet\n\nAbstract Three-dimensional real-life simulations are generally unsteady and involve moving geometries. Industry is currently very far from performing such body-fitted simulations on a daily basis, mainly due to the robustness of the moving mesh algorithm and their extensive computational cost. A moving mesh algorithm coupled to local mesh optimizations has proved its efficiency in dealing with large deformation of the mesh without re-meshing. In this paper, the coupling of this algorithm with two mesh deformation techniques is studied: an elasticity PDE-based one and an explicit Inverse Distance Weighted interpolation one, and both techniques are compared. The efficiency of this method is demonstrated on challenging test cases, involving large body deformations, boundary layers and large displacements with shearing. Finally, the moving mesh algorithm is coupled to a CFD flow solver.\n\n## Academic research paper on topic \"Two Mesh Deformation Methods Coupled with a Changing-connectivity Moving Mesh Method for CFD Applications\"\n\n(I)\n\nCrossMark\n\nAvailable online at www.sciencedirect.com\n\nScienceDirect\n\nProcedía Engineering 82 (2014) 213-227\n\nProcedía Engineering\n\nwww.elsevier.com/locate/procedia\n\n23rd International Meshing Roundtable (IMR23)\n\nTwo mesh deformation methods coupled with a changing-connectivity moving mesh method for CFD applications\n\nNicolas Barrala, Edward Lukeb, Frederic Alauzeta\n\naGamma3 Team, INRIA Paris-Roquencourt, Domaine de Voluceau, Rocquencourt, BP 105, 78153, Le Chesnay Cedex, France bDepartment of Computer Science and Engineering, Box 9637, Mississippi State University, MS 39762, United States\n\nAbstract\n\nThree-dimensional real-life simulations are generally unsteady and involve moving geometries. Industry is currently very far from performing such body-fitted simulations on a daily basis, mainly due to the robustness of the moving mesh algorithm and their extensive computational cost. A moving mesh algorithm coupled to local mesh optimizations has proved its efficiency in dealing with large deformation of the mesh without re-meshing. In this paper, the coupling of this algorithm with two mesh deformation techniques is studied: an elasticity PDE-based one and an explicit Inverse Distance Weighted interpolation one, and both techniques are compared. The efficiency of this method is demonstrated on challenging test cases, involving large body deformations, boundary layers and large displacements with shearing. Finally, the moving mesh algorithm is coupled to a CFD flow solver.\n\nPeer-review under responsibility of organizing committee of the 23rd International Meshing Roundtable (IMR23) Keywords: Moving mesh; Inverse Distance Weighted interpolation; Elasticity-based mesh deformation; ALE simulations\n\n1. Introduction\n\nThe growing expectations of the industrial world for simulations involving moving geometries have given a boost to this research field for the last decades. Moving mesh simulations are currently used in many research fields: ballistics, biomedical, aeronautics, transports... These simulations combine the difficulties associated with unsteadiness, mesh movement and fluid-structure or structure-structure coupling, and are generally hard to perform and very costly in terms of CPU time.\n\nThree leading methodologies have been designed in the literature to handle geometry displacements during numerical simulations: body-fitted approach [6,7,10,17,23], Chimera methods [8,9,21] and immersed/embedded boundary methods [18,22]. Each of them has its own strengths and weaknesses. In this work, we consider the first class of methods where the time-evolving computational domain is treated with a body-fitted approach, which means that the computational mesh follows time-evolving geometries in their movement. Only simplicial unstructured meshes are considered mainly because several fully-automatic software packages are available to generate such meshes [16,20].\n\n Corresponding author. Tel.: +33 1 39 63 57 93. E-mail address: [email protected]\n\nPeer-review under responsibility of organizing committee of the 23rd International Meshing Roundtable (IMR23) doi: 10.1016/j.proeng.2014.10.385\n\nAnother reason is that our final objective is to use highly anisotropic metric-based mesh adaptation [2,3] on moving mesh simulations, and simplicial elements still remain the most flexible to handle this technology.\n\nUnfortunately, the fixed-connectivity constraint imposed by the classical flow solver numerical scheme Arbitrary-Lagrangian-Eulerian (ALE) framework considerably limits the efficiency of body-fitted moving mesh techniques. If the problem induces large displacements of the geometry, the mesh distortion can adversely affect the accuracy and stability of the numerical schemes and often make the ALE approach unpractical.\n\nTwo different methods have been proposed to handle large displacements moving mesh simulations. The first one consists of moving the mesh as much as possible, keeping the connectivity fixed, and solving the equations in a fully ALE manner, until the mesh becomes too bad and a global or local remeshing is performed, the solution being interpolated on the new mesh, see for instance . In the case of displacements with large shearing, a large number of remeshing is performed, which is prohibitive in terms of CPU time. The second approach aims at maintaining the best possible mesh quality while moving using several local mesh optimizations such as vertex addition or collapsing and connectivity changes [10,13]. This strategy is extremely robust and maintains a good mesh quality throughout the simulation, but involves a large number of solution interpolations after each mesh modification, and the numerical method does not fully comply with the ALE framework.\n\nA new approach for moving mesh simulations has been proposed in , which is compatible with the ALE framework and able to handle anisotropic adapted meshes. First, the number of solutions required for mesh deformation -achieved through an elasticity-based PDE method- is significantly reduced. Second, the mesh deformation algorithm is coupled with local mesh quality optimizations[12,15] using only vertex smoothing and edge/face swapping. This mesh-connectivity-change operator is especially powerful in handling shear and large deformation movement, since no re-meshing is required.\n\nIn this paper, further demonstration of the robustness of this algorithm is provided: we propose to use an explicit interpolation method to compute the mesh deformation, and compare this method to the elasticity-based one on new challenging cases. Finally, this algorithm is successfully coupled to a three-dimensional CFD ALE solver.\n\nThis paper begins with a reminder of the changing-connectivity moving mesh algorithm. The two methods used to compute the mesh deformation are then described and compared on some challenging examples. The paper ends with CFD applications, where an ALE solver has been coupled to the moving mesh method.\n\n2. Mesh-connectivity-change moving mesh strategy\n\nThe strategy developed to move the inner mesh following the moving boundaries involves two main parts. First, the computation of the mesh deformation: inner vertices are assigned a trajectory, and thus a position for future time steps. Second, the optimization of the mesh, in which the positions computed in the mesh deformation phase are corrected and connectivity changes are performed to ensure a good mesh quality. This strategy has proven to be very powerful : large displacement of complex geometries can be performed preserving a good mesh quality without any remeshing. Note that, in the scope of this paper, the surface meshes are not optimized (the surface vertices are not moved on the surface).\n\n2.1. Mesh deformation phase\n\nDuring the first phase of the algorithm, the displacement of all volume vertices, i.e., the mesh deformation, is computed. To this end, two methods are considered in this paper: an elasticity-based PDE method or an explicit Inverse Distance Weighted (IDW) interpolation method. Both will be discussed more in depth in Section 3. In either cases, the cost of the mesh deformation solutions can be reduced by: (i) using a dedicated coarser mesh to compute the mesh deformation (ii) rigidifying some regions around tiny complex details of the geometries.\n\n2.2. Improving mesh deformation algorithm efficiency\n\nMesh deformation algorithms, are known to be an expensive part of dynamic mesh simulations, as their solution is generally required at each solver time step (or each few solver time steps). To reduce the number of such solutions, we\n\npropose to solve the mesh deformation problem for a large time frame of length At instead of doing it at each solver time step St. While there is a risk of a less effective mesh displacement solution, it is a worthwhile strategy if our methodology is able to preserve the mesh quality along large displacements. Solving the mesh deformation problem once for large time frame is problematic in the case of curved or accelerated trajectories of the bodies. To enhance the mesh deformation prescription, accelerated-velocity curved, i.e., high-order, vertex trajectories are computed.\n\nThe paths of inner vertices can be improved by providing a constant acceleration a to each vertex, which results in an accelerated and curved trajectory. During time frame [t, t + At], the position and the velocity of a vertex are updated as follow:\n\nx(t + St) = x(t) + St v(t) + — a and v(t + St) = v(t) + St a.\n\nPrescribing a velocity and an acceleration vectors to each vertex requires solving two mesh deformation problems. If inner vertex displacement is sought for time frame [t, t + At], boundary conditions are imposed by the location of the body at time t + At/2 and t + At. These locations are computed using body velocity and acceleration. Now, to define the trajectory of each vertex, the velocity and acceleration are deduced from evaluated middle and final positions:\n\nAt v(t) = -3 x(t) + 4 x(t + At/2) - x(t + At)\n\n—a = 2 x(t) - 4x(t + At/2) + 2x(t + At) .\n\nIn this context, it is mandatory to certify that the mesh motion remains valid for the whole time frame [t, t + At], which is done computing the sign of the volumes of the elements all along their path .\n\n2.3. Local mesh optimization\n\nIt has been proposed to couple mesh deformation with local mesh optimization using smoothing and generalized swapping to achieve efficiently large displacement in moving mesh applications. Connectivity changes are really effective in handling shear and removing highly skewed elements. Here, we briefly recall the mesh optimization procedure in the generalized context of metric-based mesh adaptive optimization.\n\nFor 3D adapted meshes, an element's quality is measured in terms of element's shape by the quality function:\n\nV3 (Z 4(e0)2\n\nQm(K) = - e [1, +~] , (1)\n\n216 \\K\\m\n\nwhere £M(e) and \\K\\M are edge length and element volume in metric M. Metric M is a 3 x 3 symmetric positive definite tensor prescribing element sizes, anisotropy and orientations to the mesh generator. QM(K) = 1 corresponds to a perfectly regular element and QM(K) < 2 correspond to excellent quality elements, while a high value of QM(K) indicates a nearly degenerated element. For non-adapted meshes, the identity matrix I3 is chosen as metric tensor.\n\nThe first mesh optimization tool is vertex smoothing which consists of relocating each vertex inside its ball of elements, i.e., the set of elements having Pi as vertex. For each tetrahedron Kj of the ball of Pi, a new optimal position Popt for Pi can be proposed to form a regular tetrahedron in metric space. The final optimal position P0iI\" is computed as a weighted average of all these optimal positions. This way, an element of the ball is all the more dominant if its quality in the original mesh is bad.\n\nThe second mesh optimization tool to improve mesh quality is generalized swapping/local-reconnection. Let a and P be the tetrahedra's vertices opposite to the common face P1P2P3. Face swapping consists of suppressing this face and creating the edge e = a/P. In this case, the two original tetrahedra are deleted and three new tetrahedra are created. A generalization of this operation exists and consists in reconnecting the inside of shells of tetrahedra [1,12,15]. The different edge swaps are generally denoted n ^ m where m is the number of new tetrahedra. In this work, edge swaps 3 ^ 2, 4 ^ 4, 5 ^ 6, 6 ^ 8 and 7 ^ 10 have been implemented.\n\n2.4. Moving mesh algorithm\n\nThe changing-connectivity moving mesh algorithm (MMA) is described in Algorithm 1, where the different phases described above are put together.\n\nAlgorithm 1 Changing-Connectivity Moving Mesh Algorithm\n\nWhile (t < Tend)\n\n1. Solve mesh deformation: compute vertices trajectories\n\n(a) {dOy(t + A)} = Compute body vertex displacement from current translation speed vbody, rotation speed ebody and acceleration abody for [t, t + f ]\n\nd(t + f) = Solve mesh deformation problem (db^^t + f), ft)\n\n(b) {d^t + At)} = Compute body vertex displacement from current translation speed vbody, rotation speed 0body and acceleration abody for [t, t + At]\n\nd(t + At) = Solve mesh deformation problem (d^^t + At), At)\n\n(c) {v, a} = Deduce inner vertex speed and acceleration from both displacements jd(t + A), d(t + At)J\n\n(d) If predicted mesh motion is invalid then At = At/2n and goto 1. Else Tels = t + At\n\n2. Moving mesh stage with mesh optimizations While (t < Tels)\n\n(a) 6t = Get moving mesh time step (Wk, v, CFLgeom)\n\n(b) Hk = Swaps optimization (Hk, QstWragpet)\n\n(c) v°P' = Vertices smoothing (Hk, Qmax)\n\n(d) Hk+1 = Move the mesh and update vertices speed (Hk, St, v, vopt, a)\n\n(e) Check mesh quality: stop if too distorted.\n\n(f) t = t + 6t\n\nEndWhile EndWhile\n\nIn this algorithm, two time steps appear: a large one At for the mesh deformation computation, and a smaller one 5topt corresponding to the steps where the mesh is optimized. The first one is set manually at the beginning of the computation, but can be automatically reduced if the mesh quality degrades. The second one is computed automatically, using the CFLgeom parameter as described below.\n\nA good restriction to be imposed on the mesh movement is that vertices cannot cross too many elements on a single move between two mesh optimizations. Therefore, a geometric parameter CFLgeom is introduced to control the number of stages used to perform the mesh displacement between t and t + At. If CFLgeom is greater than one, the mesh is authorized to cross more than one element in a single move. The moving geometric time step is given by:\n\nSt = CFLgeom max — , (2)\n\nP v(x)\n\nwhere h(xi) is the smallest altitude of all the elements in the ball of vertex P,, and v(x,) its velocity.\n\nAfter each mesh deformation solution, the quality of the mesh in the future is analyzed: if the quality is too low, the mesh deformation problem is solved again with a smaller time step.\n\n3. Comparison of two mesh deformation methods\n\n3.1. Elasticity-based PDE\n\nThe first method to compute the displacement of all vertices, i.e., the mesh deformation, is a PDE-based method using a linear-elasticity analogy [1,4].More precisely, the inner vertices movement is obtained by solving an elasticity-\n\nlike equation with a P1 Finite Element Method (FEM):\n\nVd + rV d\n\ndiv(<(6)) = 0, with 6 =-2-, (3)\n\nwhere < and 6 are respectively the Cauchy stress and strain tensors, and d is the Lagrangian displacement of the vertices. The Cauchy stress tensor follows Hooke's law for isotropic homogeneous medium , where v is the Poisson ratio, E the Young modulus of the material and A, p are the Lame coefficients:\n\n1 + v v\n\n<(6) = A trace(6) I d + 2 p 6 or 6(<) = < - — trace(<) Id .\n\nIn our context, v is typically chosen of the order of 0.48. This corresponds to a very soft, nearly incompressible material. Actually, this value for v corresponds to a nearly ill-posed problem. Note that the closer v is to 0.5, the harder it is to converge the Finite-Element linear system. The chosen value appears to be a good trade-off between material softness and the preservation of the iterative linear system solving algorithm efficiency. The FEM system is then solved by a Conjugate Gradient algorithm coupled with an LU-SGS pre-conditioner.\n\nTwo kinds of boundary conditions are considered. Dirichlet conditions are used to enforce strongly the displacement of the vertices located on moving boundaries. They are also used for fixed boundaries far from the moving bodies. These Dirichlet conditions can be relaxed if we want some vertices to move on a specific plane, typically on faces of a surrounding box close to the moving bodies, or even intersecting one (in the case of symmetry planes). In that case, if the plane is orthogonal to an axis of the Cartesian frame, only the displacement along this axis is enforced as a Dirichlet condition. The displacement in the two other directions are considered as degrees of freedom and is added in the FEM matrix as for normal volume points.\n\nAn advantage of elasticity-based methods is the opportunity they offer to adapt the local material properties of the mesh, especially its stiffness, according to the distortion and efforts born by each element: the way the Jacobian of the transformation from the reference element to the current element is accounted for in the FEM matrix assembly is modified. The classical P1 FEM formulation of the linear elasticity matrix leads to the evaluation of quantities of the form:\n\nfs ^ ^ dx = s\\K\\ddJ ^ , Jk dxk dxl dxk dxl\n\nwhere s is either A, p or A + 2p and \\K\\ is the volume of tetrahedron K. The above quantity is replaced by:\n\nX s ( K f ^ ^ dx = s \\ K \\ ( K № ^\n\ndxk d xl \\\\ K \\/ dxk dxl\n\nwhere x > 0 is the stiffening power and K is the reference element. This technique locally multiplies A and p by a factor proportional to \\K\\x determines how stiffer than large elements small elements are. In this work, we chose X = 1.\n\nThe main drawback of this method is that it is difficult to parallelize efficiently because of the pre-conditioner. Moreover, the system can be slow to converge if the system is too stiff.\n\n3.2. Inverse Weighted Distance method\n\nAn alternative approach to computing the displacement of the volume vertices is by means of an interpolation approach. In the interpolation approach the displacement is defined by an algebraic interpolation function which provides a smooth function to blend displacements between boundary surfaces. Algebraic methods may either utilize implicit formulations such as those used in the popular Radial Basis Function (RBF) interpolation method , or explicit formulations such as transfinite interpolation used for structured mesh generation. Generally explicit interpolation functions have performance advantages since they avoid problems associated with solving large stiff linear systems, however care must be taken to design interpolation functions that can sustain large deformations without introducing folded volume cells. For this evaluation we use the robust and fast explicit deformation method that is an algebraic technique that utilizes a reciprocal distance weighted sum of deformation functions to interpolate deformations to the volume vertices. In this algebraic deformation method it is assumed that every vertex of the boundary\n\nsurfaces have a displacement field that can be approximated by a rigid body motion (displacement plus rotation about the node). This rigid body motion is computed from the displacements of neighbor vertices by solving a least squares problem fitting a quaternion to the relative displacements of neighboring vertices. The nodal displacement field in the neighborhood of vertex i can then be represented by the function denoted si(r), that is written as\n\n3(7) = Mir + bi -f, (4)\n\nwhere Mi is a rotation matrix, b, is a displacement vector associated with the ith node, and 7 is a coordinate vector in the original mesh. The displacement field in the volume mesh is then described through a weighted average of all boundary node displacement fields as given by\n\n7(7) = ^f^. (5)\n\n£ wi(r)\n\nThe weighting function used to blend these displacements is based on a two-term inverse distance weighting function that is designed to preserve near-boundary mesh orthogonality while providing a smooth transition to blend between more distant surfaces. The nodal weight employed is defined by the function\n\nwi(r) = A, *\n\nLdef I aLdef ^\n\niir -m \\r -m\n\nwhere ri is the position of point i, A, is the area weight assigned to node i, Ldef is an estimated length of the deformation region, a is an estimated fraction of Ldef that is reserved for stiffer near body deformation, and a and b are user-defined exponents where a controls the smooth blending of deformations between surfaces while b controls the more stiff deformation close to deforming surfaces. Numerical experimentation suggests that a = 3 and b = 5 provide the best-quality for three-dimensional cases. The a parameter can be a fixed parameter or it can be dynamically computed based on the amount of deformation in the problem. Further details are described in a recent paper and not discussed here.\n\nNote, a naive implementation of the algorithm directly from Equation (5) will result in an O(n2) algorithm that will be too slow for practical applications. However, the localization provided by the reciprocal weights provides a mechanism for approximating the contributions of more distant points. Therefore the displacement function can be evaluated with a suitable tolerance utilizing a fast O(n log n) tree-code based algorithm that approximates the contributions of collections of distant surface nodes using a multi-pole expansion technique (as described in detail in ). When the algorithm is optimized using these techniques it can be very fast and highly parallelizable.\n\nThis algorithm has the advantage that it can be applied to meshes with arbitrary connectivity including adapted meshes with hanging nodes without difficulty. It is able to handle highly anisotropic mesh elements near viscous walls without difficulty and is free of numerical stiffness issues that can degrade the robustness of implicit methods for mapping deformations. However, since it is a global method that is based on surface mesh deformations only, it lacks an ability to dynamically adapt deformation to mesh conditions. For example, it is unable to adjust deformations locally to adapt to local mesh resolution requirements as can be easily accomplished with adjustments to element stiffness in PDE-based method.\n\n3.3. Numerical Examples\n\nWe now study on challenging numerical examples how those two mesh deformation methods behave when used in the changing-connectivity moving mesh algorithm. The next examples are purely moving mesh simulations and we will focus our analysis on mesh quality criteria. In particular, we show that the MMA applies successfully to rigid body with large shearing and deformable bodies and we give an insight into a strategy to move boundary layers on deformable bodies.\n\n3.3.1. Interpenetrating cylinders\n\nThe first example example is a challenging academic test case with rigid bodies, and has been presented in . Two cylinders interpenetrate each other and there is only one layer of elements between both cylinders, i.e. internal edges\n\nTable 1. Interpenetrating cylinders test case. Total number of mesh deformation solutions and mesh quality optimization steps to achieve the displacement, final mesh statistics and total number of swaps with the changing-connectivity MMA.\n\n# mesh deform. # mesh optim. Q™™ 1 < Qeni < 2 QZZn # swaps\n\nElasticity 50 429 1.5 97.1% 79 852,514\n\nIDW 50 570 1.4 98.2% 30 1,408,680\n\nconnect both cylinders. The geometry of the domain is shown in Figure 1. To make this test case more complicated, we add rotation to the top cylinder:\n\n= (0,0,5), vtop = (0,0, -0.1) and &ottom = (0,0,0), vbo\"om = (0,0,0.1).\n\nThe initial mesh is composed of 34567 vertices and 188 847 tetrahedra.\n\nA shear layer appears between the two cylinders, that is only one element wide. Therefore, the swapping optimization has just a little of freedom to act. Moreover, it is obvious that too large At will lead to an invalid mesh: such a simulation requires a relatively large number of mesh deformation solutions. For both mesh deformation methods, 50 mesh deformation steps are set, i.e. At = 1, and we set CFLgeom = 2.\n\nThe changing-connectivity MMA proves to be extremely robust in both cases, keeping the mesh worst element quality below 100 and ensuring a final excellent mean quality of 1.5 . Because of the shear, a large number of swaps (around 1 million) have been performed. For that test case, results with both mesh deformation methods are very similar. Moreover, Figure 1 (center) points out a major difference between elasticity and IDW: elasticity creates rotational displacements in the mesh above and under the cylinder, that push vertices in front, attract vertices behind the moving inner cylinder, and thus move vertices to empty areas instead of crushing them, while the IDW method tends to produce straighter displacements. Table 1 provides a comparison of the MMA with the two mesh deformation methods.\n\n3.3.2. Squeezing can\n\nThe next test case deals with deformable geometries. It represents a can which is squeezed. The can is a cylinder of length 5 units and diameter 2 units inside a spherical domain of radius 10 units. The can is centered at the origin and is aligned along the z-axis. The simulation runs up to Tend = 1 and the temporal deformation function is given by:\n\ni x(t) = X(0) i , , ,,\n\nif \|z(0)\| < 2 then x(t) = I y(t) = (a + (1 - a)(1 - t2)) y(0) with a = 2;^ (1.2 - cos glz(0)\|)) . {z(t) = z(0) .\n\nThe can deformation is shown in Figure 2. For this case, the moving mesh simulation is done for the inside mesh of the can geometry. The mesh is composed of 7 018 vertices and 36 407 tetrahedra. Four mesh deformation steps are initially prescribed, and CFLgeom is fixed to 1. Statistics for the case are given in Table 2.\n\nThe mesh is successfully moved,as shown on Figure 2. We clearly see that the elasticity methods has moved vertices away from the squeezed region while keeping an excellent mesh quality, whereas the IDW tends to let them crush. For this case, the swap optimization are essential to preserve the quality of the mesh all along the movement.\n\n3.3.3. Bending beam\n\nThis test case has been proposed in . It is a general deformation test on a three-dimensional bending beam. The beam is 8 units long, 2 units wide and 0.1 unit thick inside a spherical domain of radius 25 units. The beam is deformed such that the beam center-line is mapped onto a circular arc:\n\nr x(t) = (R - z(0)) sin(e) x(0) „ _ ,\n\nx(t) = r y(t) = y(0) with e and R = Rmin + (Rmax - Rmin)\"\n\n—— — — —111111 • \\ nitiA - -111111/ -, _\n\nz(t) = z(0) + (R - z(0))(1 - cos(fl)) R 1 - e\n\nFig. 1. Interpenetrating cylinders test case. Evolution of the mesh with the elasticity-based (top) and IDW (bottom) mesh deformation methods. Between the initial (left) and final (right) meshes, the displacements given by the mesh deformation steps at times 12, 18 and 25 is shown.\n\nTable 2. Squeezing can test case. Total number of mesh deformation solutions and mesh quality optimization steps to achieve the displacement, final mesh statistics and total number of swaps with the changing-connectivity MMA.\n\n# mesh deform. # mesh optim. Qmean ^end 1 < Qend < 2 QWOrSt ^overall # swaps\n\nElasticity (in.) 6 27 1.5 93.0% 11 16.255\n\nIDW (in.) 4 27 1.48 92.5% 11 11,940\n\nFig. 2. Squeezing can test cases. From left to right: initial mesh, displacement given by the elasticity-based and IDW mesh deformation problems at times 0.75, and final meshes with elasticity and IDW.\n\nwhere Rmin and Rmax are the radii of curvature at t = 0 and t = 1, respectively. k is a parameter that controls how rapidly the radius and center of curvature move from the initial to final values. For this test case, Rmin = 2, Rmax = 500 and k = 10. The initial mesh is composed of 58 315 vertices and 322 971 tetrahedra. We set At = 0.2 and CFLgeom = 1.\n\nAgain, this large deformation problem is solved without any difficulty, requiring only a few mesh deformation steps, the final mesh is excellent as shown in Table 3 and no skewed element appears inside the mesh. There again, we can see in Figure 3 that the elasticity-based method tends to move points farther from the body than the IDW method, thus requiring significantly more swaps. However, the performances of the elasticity-based and IDW methods are very close, cf. Table 3.\n\nTable 3. Bending beam test case. Total number of mesh deformation solutions and mesh quality optimization steps to achieve the displacement, final mesh statistics and total number of swaps with the changing-connectivity MMA.\n\n# mesh deform. # mesh optim. Qmean ^end 1 < Qend < 2 QWOrSt ^overall # swaps\n\nElasticity IDW 5 4 211 40 1.3 1.3 99.9% 99.8% 4.4 5.9 479,182 149,536\n\nFig. 3. Bending beam test case. From left to right: initial mesh, displacement given by the elasticity-based and IDW mesh deformation problems at time 0.8, and final meshes with elasticity and IDW.\n\n3.3.4. Boundary layers on deformable geometries\n\nDealing with boundary layers (BL) meshes for rigid bodies is rather easy: the whole mesh layers are rigidified and moved together with the body. However, it is a lot more difficult with deformable bodies, since the mesh structured layers must both follow the deformation of the body and keep their structure. It was a priori not certain that the movement provided by the elasticity or IDW step would be precise enough to keep that structure. In the sequel we demonstrate that both methods can move one thick layer with a good accuracy, and that such layers can also be moved correctly with the elasticity method.\n\nOne boundary layer. For this case, we consider the bending beam studied previously, to which we add one thin structured mesh layer of height a quarter of the height of the beam, see Figure 4. The movement of the beam is the same and no swap or smoothing optimizations are done in the BL. At the end of the simulation, we observe that the structured aspect of the BL has been well preserved. In Table 4, we gather quality indicators depending on different moving mesh parameters. We consider the variation of the distance of the layer vertices to the body, that is expected to remain constant in ideal cases. We can see that taking acceleration into account in the trajectories is very important: indeed, as the movement of the beam is accelerated, the layer under the beam tends to thicken while the layer above it tends to be crushed. We can also see that less IDW steps are required to reach an equivalent mean quality, but the maximal distance variation is less good.\n\nThis result is very promising, because the structured aspect of the BL mesh is well preserved. This lets us consider a clever strategy to move BL meshes with deformable geometries.\n\nTable 4. Bending beam with one BL. Mean and maximal relative distance variations ¿dx, in the case of 10 elasticity solutions, 50 elasticity solutions with linear (lin.) trajectories (without taking acceleration into account), 50 elasticity solutions with quadratic (quad.) trajectories.\n\nMean relative ¿dx (in %) 27 9.9 5.7 6.4\n\nMax relative ¿dx (in %) 143 66 19 60\n\nFig. 4. One BL test case. Initial mesh (a) and close up on the BL (b), final mesh with elasticity (c) and IDW (d), and close-ups on the final BL meshes with elasticity (e) and IDW (f).\n\nSeveral boundary layers. The elasticity mesh deformation method is also actually capable of moving several thinner BL, provided the mesh deformation time step is small enough to handle the displacement of the smallest elements against the body. The simulation was run with a boundary layer containing 10 thin layers (the smallest height is 0.0017 units): the structure inside the layer is still well preserved all along the movement, and the quality of the layer looks good even at the end of the deformation of the beam. Close-ups on the BL meshes are show in Figure 5. We were not able to move these layers with the IDW mesh deformation, which needs to be further investigated.\n\n4. Moving-mesh ALE computations\n\nThe changing-connectivity moving mesh algorithm described above is coupled to our in-house flow solver to run CFD simulations with moving geometries. The compressible Euler equations are solved with a Finite Volume method. The displacements of the vertices require the use of the ALE formulation of these equations. As regards the temporal accuracy, the considered SSPRK schemes are based on the strict application of the Discrete Geometrical Conservation Law (DGCL). Details about the ALE solver are given in . The absence of remeshing preserves the quality of the solution, as it is free of most interpolation errors due to solutions transfers from one mesh to another.\n\nThe flow solver is integrated to Algorithm 1 as follows. The solver iterations are embedded in the optimization phase, so that the mesh is moved after each solver iteration, but optimizations are performed only when the optimization time (defined by the CFLgeom parameter) is reached. If necessary, the solver time step is truncated to perform optimizations at the correct time.\n\nThe efficiency of the method is demonstrated on two examples involving complex moving geometries. Both simulations were run using the IDW interpolation and the elasticity-based mesh deformation methods.\n\n4.1. Vortical wake of a F117 aircraft nosing up\n\nThe second example is a subsonic F117 aircraft nosing up, that creates a vortical wake. An inflow of air at Mach 0.4 arrives in front of the aircraft, initially in horizontal position that noses up, stays up for a while, then noses down. In this example, the aicraft rotates around its center of gravity. Let T = 1 s be the characteristic time of the movement and 6max = 20o the maximal angle reached, the movement is defined by its angle of rotation, of which the evolution is divided in 7 phases. Phase (i) (0 < t < T/2) is an initialization phase, during which the flow around the aircraft is established. Phase (ii) (T < t < T) and (iii) (T < t < 3r) are respectively phases of accelerated and decelerated ascension. Vortices start to grow behind the aircraft, and they expand during phase (iv) (3T < t < TT), where the aircraft stays in upward position. Phase (v) (TT < t < 4T) and (vi) (4T < t < 9r) are phases of accelerated and decelerated descent, the vortices start to move away and they slowly disappear in phase (vii) (9rt < 5T). Wall conditions are imposed on the faces of the surrounding box and slipping conditions on the aircraft.\n\nThe initial mesh has 501,859 vertices and 3,012,099 tetrahedra. The mesh used and snapshots of the solutions are shown in Figures 6 and 7. A difficulty of this test case is the uneven movement of the aircraft, with strong acceleration and deceleration during the phases of ascension and descent, and the inversion of the acceleration in the middle of these phases. It is necessary to solve enough mesh deformation problems to avoid that the body boundaries cross small elements close to it. Statistics of the moving mesh characteristics are show in Table 5. We can see that the two proposed mesh deformation methods produce rather similar results in terms of mesh quality, but the case requires only half as many mesh deformation steps if the IDW method is used.\n\nTable 5. Nosing up F117 test case. Total number of mesh deformation solutions and mesh quality optimization steps to achieve the displacement, final mesh statistics and total number of swaps with the changing-connectivity MMA.\n\n# mesh deform. # mesh optim. Qmean 1 < Qend < 2 QWorSt ^overall # swaps\n\nElasticity 24 66 1.4 99.8% 19 698 755\n\nIDW 12 37 1.4 99.8% 19 713 247\n\nFig. 6. Test case of the nosing up F117. semi-structured meshes in initial horizontal position, and upward position.\n\n4.2. Two F117 aircraft flight paths crossing\n\nThe first example is modeling two F117 aircrafts having crossing flight paths. This problem illustrates the efficiency of the connectivity-change moving mesh algorithm in handling large displacements of complex geometries without any remeshing. When both aircrafts cross each other, the mesh deformation encounters a large shearing due to the opposite flight directions. The connectivity-change mesh deformation algorithm handles easily this complex displacement thanks to the mesh local reconnection. Therefore, the mesh quality remains very good during the whole displacement.\n\nAs concerns the fluid simulation, the aircrafts are moved at a speed of Mach 0.8, in an initially inert uniform fluid: at t = 0 the speed of the air is null everywhere. The rotation speed of the aircrafts is set to a tenth of their translation speed. Transmitting boundary conditions are used on the sides of the surrounding box, and slipping conditions are imposed on the two F117 bodies. After a phase of initialization, the flow is established when the two F117s pass each other.\n\nThe initial mesh has 586,610 vertices and 3,420,332 tetrahedra. Some statistics on the moving mesh aspect of the simulation are shown in Table 6. We can see that both methods keep an excellent mesh quality all along the simulation. We used the minimal number of mesh deformations possible, and the IDW method again requires half as many mesh deformation steps, but then requires almost five times more swaps for the whole simulation. This is due to the fact that with the elasticty-based approach, vertices tend to bypass the moving bodies, while their trajectories are straighter with the IDW, thus creating a lot of shearing. In Figure 8, we show both the moving mesh aspect of the simulation and the flow solution at different time steps.\n\nTable 6. Two F117s test case. Total number of mesh deformation solutions and mesh quality optimization steps to achieve the displacement, final mesh statistics and total number of swaps with the changing-connectivity MMA.\n\n# mesh deform. # mesh optim. Qmean x^end 1 < Qend < 2 QWOrSt '¿overall # swaps\n\nElasticity 43 1563 1.4 99.8% 27 2,176,850\n\nIDW 22 1717 1.4 99.7% 26 10,396,127\n\n5. Conclusion\n\nIn this paper, new numerical evidence has been given that 3D large displacements are possible with a mesh deformation algorithm coupled with mesh optimization using only swaps and vertex movements. No re-meshing is required. The possibility to compute the mesh deformation using an explicit Inverse Distance Weighted interpolation has been added to the algorithm. Moreover, we have been dealing with deformable geometries, and have started to\n\naddress the issue of boundary layers. Finally, the moving mesh algorithm has been coupled to a CFD ALE flow solver. Several examples, both purely moving-mesh and CFD, have been exhibited.\n\nThe comparison of the two mesh deformation methods shows that both are very robust coupled with mesh optimizations. They both produce meshes of equivalent excellent qualities.It seems that in general, less IDW solutions are necessary, but the elasticity-based approach requires significantly less swap in the cases with a lot of shearing. These results need to be confirmed and refined on more cases, but they confirm the advantage of having both tools at our disposal.\n\nHowever some problematics remain. The moving of boundary layers need to be further developed. The implementation of the IDW method and the parallelization of the flow solver must be improved, in order to run CPU time comparisons of both methods. Other issues are surface optimizations, simulations of deformable geometries with possible connectivity changes and the efficient treatment of contact problems.\n\n6. Acknowledgments\n\nThis work was partially funded by the Airbus Group Foundation.\n\nReferences\n\n F. Alauzet. A changing-topology moving mesh technique for large displacements. Engineering with Computers, 30(2):175-200, 2014.\n\n F. Alauzet and A. Loseille. High-order sonic boom prediction by utilizing mesh adaptive methods. In 48th AIAA Aerospace Sciences Meeting and Exhibit, AIAA-2010-1390, Orlando, FL, USA, Jan 2010.\n\n F. Alauzet and G. Olivier. Extension of metric-based anisotropic mesh adaptation to time-dependent problems involving moving geometries. In 49th AIAA Aerospace Sciences Meeting, AIAA Paper 2011-0896, Orlando, FL, USA, Jan 2011.\n\n T.J. Baker and P. Cavallo. Dynamic adaptation for deforming tetrahedral meshes. AIAA Journal, 19:2699-3253, 1999.\n\n N. Barral and F. Alauzet. Large displacement body-fitted FSI simulations using a mesh-connectivity-change moving mesh strategy. In 44th AIAA Fluid Dynamics Conference, AIAA-2014, Atlanta, GA, USA, June 2014.\n\n J.D. Baum, R. Lohner, T.J Marquette, and H. Luo. Numerical simulation of aircraft canopy trajectory. In 35th AIAA Aerospace Sciences Meeting, AIAA Paper 1997-0166, Reno, NV, USA, Jan 1997.\n\n J.D. Baum, H. Luo, and R. Lohner. A new ALE adaptive unstructured methodology for the simulation of moving bodies. In 32th AIAA Aerospace Sciences Meeting, AIAA Paper 1994-0414, Reno, NV, USA, Jan 1994.\n\n J.A. Benek, P.G. Buning, and J.L. Steger. A 3D chimera grid embedding technique. In 7th AIAA Computational Fluid Dynamics Conference, AIAA Paper 1985-1523, Cincinnati, OH, USA, Jul 1985.\n\n F. Brezzi, J.L. Lions, and O. Pironneau. Analysis of a Chimera method. C.R. Acad. Sci. ParisSer. I, 332(7):655-660, 2001.\n\n G. Compere, J-F. Remacle, J. Jansson, and J. Hoffman. A mesh adaptation framework for dealing with large deforming meshes. Int. J. Numer. Meth. Engng, 82(7):843-867, 2010.\n\n A. de Boer, M. van der Schoot, and H. Bijl. Mesh deformation based on radial basis function interpolation. Comput. & Struct., 85:784-795, 2007.\n\n E. Briere de l'Isle and P.L. George. Optimization of tetrahedral meshes. IMA Volumes in Mathematics and its Applications, 75:97-128, 1995.\n\n C. Dobrzynski and P.J. Frey. Anisotropic Delaunay mesh adaptation for unsteady simulations. In Proceedings of the 17th International Meshing Roundtable, pages 177-194. Springer, 2008.\n\n L. E. Eriksson. Generation of boundary conforming grids around wing-body configurations using transfinite interpolation. AIAA Journal, 20:1313-1320, 1982.\n\n P.J. Frey and P.L. George. Mesh generation. Application to finite elements. ISTE Ltd and John Wiley & Sons, 2nd edition, 2008.\n\n P.L. George. Tet meshing: construction, optimization and adaptation. In Proceedings of the 8th International Meshing Roundtable, South Lake Tao, CA, USA, 1999.\n\n O. Hassan, K.A. S0rensen, K. Morgan, and N. P. Weatherill. A method for time accurate turbulent compressible fluid flow simulation with moving boundary components employing local remeshing. Int. J. Numer. Meth. Fluids, 53(8):1243-1266, 2007.\n\n R. Lohner, J.D. Baum, E. Mestreau, D. Sharov, C. Charman, and D. Pelessone. Adaptive embedded unstructured grid methods. Int. J. Numer. Meth. Engng, 60:641-660, 2004.\n\n E. Luke, E. Collins, andE. Blades. A fast mesh deformation method using explicit interpolation. J. Comp. Phys., 231:586-601, 2012.\n\n D. Marcum and N. Weatherill. Unstructured grid generation using iterative point insertion and local reconnection. AIAA Journal, 33(9):1619-1625, 1982.\n\n S. Murman, M. Aftosmis, and M. Berger. Simulation of 6-DOF motion with cartesian method. In 41th AIAA Aerospace Sciences Meeting and Exhibit, AIAA-2003-1246, Reno, NV, USA, Jan 2003.\n\n C.S. Peskin. Flow patterns around heart valves: a numerical method. J. Comp. Phys., 10:252-271, 1972.\n\n M. L. Staten, S. J. Owen, S. M. Shontz, A. G. Salinger, and T. S. Coffey. A comparison of mesh morphing methods for 3D shape optimization.\n\nIn Proceedings of the 20th International Meshing Roundtable, pages 293-310. Springer, 2011.\n\nFig. 7. Test case of the nosing up F117. Isolines on several cutting planes of the mach field at different time steps ( from left to right and up to down t = 0.49, 0.85, 1.22, 1.47, 2.08, 2.82, 3.31, 3.80, 4.28 and 4.90).\n\nFig. 8. Test case of the two F117s: Snapshots of the moving geometries and the mesh (left) and density (right)." ]	[ null, "https://cyberleninka.org/images/tsvg/cc-label.svg", null, "https://cyberleninka.org/images/tsvg/view.svg", null, "https://cyberleninka.org/images/tsvg/download.svg", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.8911466,"math_prob":0.9398423,"size":43380,"snap":"2022-05-2022-21","text_gpt3_token_len":10103,"char_repetition_ratio":0.16313168,"word_repetition_ratio":0.04925479,"special_character_ratio":0.23241125,"punctuation_ratio":0.13171494,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9783436,"pos_list":[0,1,2,3,4,5,6],"im_url_duplicate_count":[null,null,null,null,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-05-28T13:51:22Z\",\"WARC-Record-ID\":\"<urn:uuid:f2c27fb4-f2b5-4d91-a6b9-01e2d42892d5>\",\"Content-Length\":\"75446\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:89abc429-fbf4-495b-bbd3-357f716ae7c7>\",\"WARC-Concurrent-To\":\"<urn:uuid:d7f9c52d-f948-4f19-9f10-d78d8d70af2c>\",\"WARC-IP-Address\":\"159.69.2.174\",\"WARC-Target-URI\":\"https://cyberleninka.org/article/n/3035\",\"WARC-Payload-Digest\":\"sha1:4HPZNKIL5YGH62FLQFPR3HCUEDXDFR37\",\"WARC-Block-Digest\":\"sha1:TLNP6BTWZUX7TWADP27J2LWL35NLHHIS\",\"WARC-Identified-Payload-Type\":\"application/xhtml+xml\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-21/CC-MAIN-2022-21_segments_1652663016853.88_warc_CC-MAIN-20220528123744-20220528153744-00555.warc.gz\"}"}
https://support.scisports.com/en/articles/4719875-aggregated-variables-and-calculations	[ "## Understand The Averages\n\n### 95' Field Players-Averages\n\nComparison between teams can be done with normalized field player averages. For the calculation of the match averages, we have excluded goalkeepers and corrected for substitutes and red cards. The values are constructed from a value per minute times 95. 95 minutes because this is on average the duration of a match.\n\nequation 1:\n\n``95 minute Fieldplayers Average = SUM(variable) / (SUM(time on field)/match time)95``\n\nFor example:", null, "95' Fieldplayers Average Total Distance\n= SUM(Total_Distance) / (SUM(time on field)/actual match time)95\n\n95' Fieldplayers Avg. Total Distance = 108.508 / (930/93)*95\n\n95' Fieldplayers Avg. Total Distance = 10.308 m\n\n### Multiple Matches Average\n\nFor the calculation of a multi-match average first, the 95' field player's average is calculated for each match. Consecutively the average of all (selected) matches is calculated.", null, "Averages over multiple matches are being used in our League Analysis.\n\nIf you have any questions or suggestions, please send us your feedback.\n\nUse the chat icon on the right to directly get in touch with us." ]	[ null, "https://downloads.intercomcdn.com/i/o/277118115/a21338a5634bb82afe43cd99/image.png", null, "https://downloads.intercomcdn.com/i/o/277130101/51386a0378d8324c0700e8b1/image.png", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.8777325,"math_prob":0.9808268,"size":1103,"snap":"2021-43-2021-49","text_gpt3_token_len":252,"char_repetition_ratio":0.1410373,"word_repetition_ratio":0.024096385,"special_character_ratio":0.23934723,"punctuation_ratio":0.092307694,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.97501236,"pos_list":[0,1,2,3,4],"im_url_duplicate_count":[null,1,null,1,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-10-27T16:43:09Z\",\"WARC-Record-ID\":\"<urn:uuid:04453464-669a-47a7-bdd9-5e3079251588>\",\"Content-Length\":\"13701\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:8f55a881-b376-492d-b261-a926e344bd9c>\",\"WARC-Concurrent-To\":\"<urn:uuid:5be0b9d3-a56b-4b8c-be0f-52552efe6ecd>\",\"WARC-IP-Address\":\"104.45.68.100\",\"WARC-Target-URI\":\"https://support.scisports.com/en/articles/4719875-aggregated-variables-and-calculations\",\"WARC-Payload-Digest\":\"sha1:5UI4RSDQ4PAJJFBS4IJZTVLRN2QXJLEL\",\"WARC-Block-Digest\":\"sha1:V4JHBYYWYA2O34DWXURQ7APVXLA2C6TX\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-43/CC-MAIN-2021-43_segments_1634323588216.48_warc_CC-MAIN-20211027150823-20211027180823-00189.warc.gz\"}"}
https://www.brandenbyers.com/slides/test-them-puzzles/	[ "# Puzzles", null, "", null, "", null, "", null, "# Coding Puzzles\n\n## Next to Last Item in a List\n\nReturn the next-to-last item from an array.\n\n## The Gas Station Problem\n\nThere is a truck driving clockwise on a circular route;\nthe truck has a gas tank with infinite capacity, initially empty.\n\nAlong the route are gas stations with varying amounts of gas available:\nyou are given a list of gas stations\nwith the amounts of gas they have available\nand the amounts of gas required to drive to the next gas station.\n\nYou must find a gas station that,\nfor a trip starting from that gas station,\nwill be able to return to that gas station.\n\n### Investigating the Torricelli point of a triangle\n\nLet ABC be a triangle with all interior angles being less than 120 degrees.\nLet X be any point inside the triangle and let XA = p, XC = q, and XB = r.\n\nFermat challenged Torricelli to find the position of X such that p + q + r was minimised.\n\nTorricelli was able to prove that if equilateral triangles AOB, BNC and AMC are constructed on each side of triangle ABC, the circumscribed circles of AOB, BNC, and AMC will intersect at a single point, T, inside the triangle.\nMoreover he proved that T, called the Torricelli/Fermat point, minimises p + q + r. Even more remarkable,\nit can be shown that when the sum is minimised, AN = BM = CO = p + q + r and that AN, BM and CO also intersect at T.\n\nIf the sum is minimised and a, b, c, p, q and r are all positive integers we shall call triangle ABC a Torricelli triangle. For example, a = 399, b = 455, c = 511 is an example of a Torricelli triangle, with p + q + r = 784.\n\nFind the sum of all distinct values of p + q + r ≤ 120000 for Torricelli triangles.\n\n# Why write puzzles?\n\n## Programmer Practice\n\n• Athletes Train\n• Musicians Rehearse\n• Lawyers & Doctors Practice\n\nTest-Driven Learning\n\n## Puzzles Are", null, "# Aha!\n\n#### Dopamine drip", null, "", null, "", null, "", null, "rosettacode.org/wiki/towers_of_hanoi\n\n# Parrots\n\n• Inquisitive\n• Destructive exploration\n• Playful learners\n• Pull, tear, scratch, probe\n• Haphazardly dig for food", null, "# Crows\n\n• Curious\n• Explore from a distance\n• Methodical learners\n• Graceful tool users\n• Precisely prod for food", null, "## Code like a...\n\n### PARROT\n\n• Inquisitive\n• Destructive exploration\n• Playful learners\n• Pull, tear, scratch probe\n• Haphazardly dig for food\n\n### CROW\n\n• Curious\n• Explores from a distance\n• Methodical learners\n• Graceful tool users\n• Precisely prod for food", null, "Flexibility in Problem Solving and Tool Use of Kea and New Caledonian Crows in a Multi Access Box Paradigm\n\n## Code Challenge Websites", null, "", null, "# How-to\n\n### Let's make some sausage!\n\n``````var add = function (a, b) {\nreturn a + b;\n};``````\nexpect(add(2, 2).to.equal(4));````expect(add(2, '2').to.equal('22'));``expect(add('con','cat').to.equal('concat'));````\n\n# Assumptions", null, "``````var add = function (a, b) {\nif(typeof a + b === 'string') {\nreturn 'Go fish';\n}\n...\n};``````\nexpect(add('con','cat')).to.equal('Go fish');````expect(add(2, 2)).to.be.a('number')````\n``````var add = function (a, b) {\nreturn parseInt(a) + parseInt(b);\n};``````\nexpect(add(2, '2').to.equal(4));````expect(add(2.22, 2.22).to.equal(4.44));````\n// AssertionError: expected 4 to equal 4.44``````", null, "# Chai.js\n\n### assert()\n\n``````expect('something').to.contain('thing');\n\nexpect('Hello World').to.be.a('string');``````\n\n# Draft Ideas\n\n### Inspired by Fiction", null, "", null, "", null, "# Libraries\n\n## _.shuffle\n\n``````test('shuffle', function() {\ndeepEqual(_.shuffle(), , 'behaves correctly on size 1 arrays');\nvar numbers = _.range(20);\nvar shuffled = _.shuffle(numbers);\nnotDeepEqual(numbers, shuffled, 'does change the order'); // Chance of false negative: 1 in ~2.4*10^18\nnotStrictEqual(numbers, shuffled, 'original object is unmodified');\ndeepEqual(numbers, _.sortBy(shuffled), 'contains the same members before and after shuffle');\n\nshuffled = _.shuffle({a: 1, b: 2, c: 3, d: 4});\nequal(shuffled.length, 4);\ndeepEqual(shuffled.sort(), [1, 2, 3, 4], 'works on objects');\n});``````\n\nrosettacode.org\n\n# Slasher Film\n\nReturn the surviving elements of an array\nafter chopping off n elements from the head.\n\n``````\nslasher([1, 2, 3], 2);``````\n``expect(slasher([1,2,3],2)).to.equal();``expect(slasher([1,2,3],0)).to.equal([1,2,3]);``expect(slasher([1,2,3],9)).to.equal([]);``\n``` `expect(slasher([1,2,3],2)).to.equal(['Three, you're lucky to still be.']);`\n```\n``` `expect(slasher([1,2,3],0)).to.equal(['All numbers live another day.']);`\n```\n``` `expect(slasher([1,2,3],9)).to.equal(['It is a sad day.']);`\n```\n``expect(slasher(['1,354','22','4900'],1)).to.equal(['Four thousand nine hundred and twenty-two; you are amongst the lucky few.']);``\n\nReturn a string of the the English word equivalents\nof the surviving elements (numbers) in an array\nafter chopping off n elements from the head.\nThe numbers must be listed from last to first.\nThe string should be finished with a phrase\nthat rhymes with the last number listed.\n\n## Rhyming Numbers Poem\n\nYou will be given an array of numbers.\nEach number will be between 0 and 9.\nThese numbers may be integers or strings.\nIf a number is in a string, it may be a digit or word.\nReturn a nonsense poem,\nwhere each line starts with a number from the array\nfollowed by a phrase ending in a word that rhymes with that number.\n\nNine, you smell like wine\nFive, chicken dumplings make me feel alive\nThree, you be lost deep in the sea\nTwo, please stop sniffing glue\nOne, this poem is done\n\n````var rhymingNumberPoem = function(arr) {`` // this is a poet but didn't know it```` return str;\n};``````\n``````var shortArr = [9,5,3,1];\nvar longArr = [9,5,3,1,5,4,3,5,6,3,2,3,4,7,5,2,3,4,7,10,23,4,2];\n``````\n````describe('Rhyming Number Poem', function() {```` it('should return a string', function() {\nexpect(rhymingNumberPoem(shortArr)).to.be.a('string');\n});`````` it('should return correct number of lines from short array', function() {\nexpect(rhymingNumberPoem(shortArr).split('\\n').length).to.equal(shortArr.length);\n});`````` it('should create a short rhyming poem', function() {\nexpect(rhymeChecker(rhymingNumberPoem(shortArr))).to.equal(true);\n});`````` it('should create a long rhyming poem', function() {\nexpect(rhymeChecker(rhymingNumberPoem(longArr))).to.equal(true);\n});````});````\n``````var rhymeChecker = function(str) {\nvar arr = str.split(/,\|\\./).join('').split(\"\\n\");\nif(str.length < 1 \|\| arr.split(' ').length < 3) {\nreturn false;\n}\nvar count = arr.reduce(function(a, b) {\nvar words = b.split(' '),\nfirstWord = words.toLowerCase(),\nlastWord = words[words.length - 1].toLowerCase();\nif(numberRhymes[firstWord].match(lastWord)) {\nreturn a + 1;\n}\nreturn a;\n}, 0);\nif(count < arr.length) {\nreturn false;\n}\nreturn true;\n};``````\n``````\nzero: 'hero giro tiro nonzero subzero',\none: 'won done run son none sun gun fun tion ton nun shun sion bun donne dun fon pun rien bn cen gon hun jun tonne tun ven yean naan spun fron seisin stun toucan anyone begun undone outdone outrun spline rerun redone everyone overrun overdone cretonne misdone twentyone'``````\n\nbrandenbyers.com/slides/test-them-puzzles/poem.js\n\nnpm install\n\nmocha poem\n\n## Bonfires\n\nfreecodecamp/seed/challenges/basic-bonfires.json\n\n``````\n{\n\"name\": \"Bonfire: Find the Longest Word in a String\",\n\"dashedName\": \"bonfire-find-the-longest-word-in-a-string\",\n\"difficulty\": \"1.04\",\n\"description\": [\n\"Return the length of the longest word in the provided sentence.\",\n],\n\"challengeSeed\": [\n\"function findLongestWord(str) {\",\n\" return str.length;\",\n\"}\",\n\"\",\n\"findLongestWord('The quick brown fox jumped over the lazy dog');\"\n],\n\"tests\": [\n\"expect(findLongestWord('The quick brown fox jumped over the lazy dog')).to.be.a('Number');\",\n\"expect(findLongestWord('The quick brown fox jumped over the lazy dog')).to.equal(6);\",\n\"expect(findLongestWord('May the force be with you')).to.equal(5);\",\n\"expect(findLongestWord('Google do a barrel roll')).to.equal(6);\",\n\"expect(findLongestWord('What is the average airspeed velocity of an unladen swallow')).to.equal(8);\"\n],\n\"String.split()\",\n\"String.length\"\n]\n}``````\n\n## Write Kickass Puzzles\n\n1. Devise a description\n2. Generate tests\n3. Double, triple, and quadrupal check for edge cases\n4. Solve and bathe in dopamine\n5. Share/repeat", null, "## @brandenbyers\n\nInteractive Intelligence\n\nFree Code Camp\n\nBalance Boards\n\nGoats\n\nmidwestjs.com/feedback" ]	[ null, "https://www.brandenbyers.com/slides/test-them-puzzles/img/drawing-sudoku.svg", null, "https://www.brandenbyers.com/slides/test-them-puzzles/img/drawing-rubiks.svg", null, "https://www.brandenbyers.com/slides/test-them-puzzles/img/drawing-crossword.svg", null, "https://www.brandenbyers.com/slides/test-them-puzzles/img/puzzlejs.png", null, "https://www.brandenbyers.com/slides/test-them-puzzles/img/universal.svg", null, "https://www.brandenbyers.com/slides/test-them-puzzles/img/dopamine-drip.svg", null, "https://www.brandenbyers.com/slides/test-them-puzzles/img/stomachion.svg", null, "https://www.brandenbyers.com/slides/test-them-puzzles/img/go-problem.svg", null, "https://www.brandenbyers.com/slides/test-them-puzzles/img/tower-of-hanoi.svg", null, "https://www.brandenbyers.com/slides/test-them-puzzles/img/parrot-box-2.svg", null, "https://www.brandenbyers.com/slides/test-them-puzzles/img/crow-box-2.svg", null, "https://www.brandenbyers.com/slides/test-them-puzzles/img/crow-box-2.svg", null, "https://www.brandenbyers.com/slides/test-them-puzzles/img/code-wars-1.jpg", null, "https://www.brandenbyers.com/slides/test-them-puzzles/img/free-code-camp-1.jpg", null, "https://www.brandenbyers.com/slides/test-them-puzzles/img/peach-emoji.png", null, "https://www.brandenbyers.com/slides/test-them-puzzles/img/peach-emoji.png", null, "https://www.brandenbyers.com/slides/test-them-puzzles/img/fifty-shades-of-repetition.svg", null, "https://www.brandenbyers.com/slides/test-them-puzzles/img/harry-potter-puzzle.svg", null, "https://www.brandenbyers.com/slides/test-them-puzzles/img/enders-puzzle.svg", null, "https://www.brandenbyers.com/slides/test-them-puzzles/img/floating-branden-head-no-shadow.svg", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.63635683,"math_prob":0.95888,"size":6692,"snap":"2019-35-2019-39","text_gpt3_token_len":1890,"char_repetition_ratio":0.1270933,"word_repetition_ratio":0.04004215,"special_character_ratio":0.30573818,"punctuation_ratio":0.2304054,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9614958,"pos_list":[0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26,27,28,29,30,31,32,33,34,35,36,37,38,39,40],"im_url_duplicate_count":[null,2,null,2,null,2,null,2,null,2,null,2,null,2,null,2,null,2,null,2,null,4,null,4,null,2,null,2,null,4,null,4,null,2,null,2,null,2,null,2,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-09-20T14:32:54Z\",\"WARC-Record-ID\":\"<urn:uuid:ef66a6a5-f316-4389-8788-3eabd75c7194>\",\"Content-Length\":\"23046\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:6c758df3-ab6c-4769-b0a2-e9bdc305ae78>\",\"WARC-Concurrent-To\":\"<urn:uuid:ef8f0052-3ff1-41b8-ad6e-b9403e73d490>\",\"WARC-IP-Address\":\"167.99.4.63\",\"WARC-Target-URI\":\"https://www.brandenbyers.com/slides/test-them-puzzles/\",\"WARC-Payload-Digest\":\"sha1:GZ6TTRIHMQ263FURDI5RZPJLPA5336PW\",\"WARC-Block-Digest\":\"sha1:FDDSRAT7C5BBOY5ZIXB7EFF5GQMUVKJ3\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-39/CC-MAIN-2019-39_segments_1568514574039.24_warc_CC-MAIN-20190920134548-20190920160548-00359.warc.gz\"}"}
https://codereview.stackexchange.com/questions/251910/c-dice-game-using-random-numbers	[ "# c++ dice game using random numbers\n\nI'm fairly new to c++ programming. It would be helpful if I could get a feedback on a dice game I wrote. I would really appreciate some tips as well as your opinions.\n\nThis program will begin by asking the user to enter an amount to bet (from $1-$1000). Next, the user is asked to guess the total that will occur when two dice are rolled. The program then “rolls two dice” (using random numbers) and visually displays the result. Then, if the player’s guess was correct, the player wins 8 times their bet divided by the number of possible way to get that roll. If the player’s guess was wrong, they lose their bet times the number of possible ways to get their guess.\n\n\n\n#include <iostream>\n#include <cstdlib>\n#include <time.h>\nusing namespace std;\n\n//main program\nint main()\n{\nint guess;//user guess for the value of the dice\nfloat bet;//the bet placed by the user\nfloat possibility;//number of combinations each number can come\nint dice1, dice2;//the random value between 1-6 for each dice\nint total;//the total number after the value of dice1 and dice2 is added\nfloat earned_or_lost_money;//earned or lost money depending on whether you win/lose\n\nsrand(time(0));//random number seed\n\ncout << endl;//blank line\n\ncout << \" What do you guess the result will be?(2-12): \";\ncin >> guess;//user input for guess\n\ncout << \" Enter amount of money to bet ($1-$1000): \";\ncin >> bet;//user input for bet\n\ncout << endl;//blank line\n\ndice1 = rand() % 6 + 1;//value of first dice(1-6)\ndice2 = rand() % 6 + 1;//value of second dice(1-6)\n\ntotal = dice1 + dice2;//value of dice1+dice2\n\nswitch (guess)// values for possible combinations for the total value of the dices added\n{\n\ncase 2: possibility = 1; break;\ncase 3: possibility = 2; break;\ncase 4: possibility = 3; break;\ncase 5: possibility = 4; break;\ncase 6: possibility = 5; break;\ncase 7: possibility = 6; break;\ncase 8: possibility = 5; break;\ncase 9: possibility = 4; break;\ncase 10: possibility = 3; break;\ncase 11: possibility = 2; break;\ncase 12: possibility = 1; break;\n}//switch\n\ncout << \" For your $\" << bet << \" bet, the roll is\" << endl; cout << endl;//blank line if (dice1 == 1)//drawing if the value of dice1 is 1 { cout << \"=============\" << endl; cout << \"\| \|\" << endl; cout << \"\| 1 \|\" << endl; cout << \"\| \|\" << endl; cout << \"=============\" << endl; }//if else if (dice1 == 2)//drawing if the value of dice1 is 2 { cout << \"=============\" << endl; cout << \"\| 2 \|\" << endl; cout << \"\| \|\" << endl; cout << \"\| 2 \|\" << endl; cout << \"=============\" << endl; }//else if else if (dice1 == 3)//drawing if the value of dice1 is 3 { cout << \"=============\" << endl; cout << \"\| 3 \|\" << endl; cout << \"\| 3 \|\" << endl; cout << \"\| 3 \|\" << endl; cout << \"=============\" << endl; }//else if else if (dice1 == 4)//drawing if the value of dice1 is 4 { cout << \"=============\" << endl; cout << \"\| 4 4 \|\" << endl; cout << \"\| \|\" << endl; cout << \"\| 4 4 \|\" << endl; cout << \"=============\" << endl; }//else if else if (dice1 == 5)//drawing if the value of dice1 is 5 { cout << \"=============\" << endl; cout << \"\| 5 5 \|\" << endl; cout << \"\| 5 \|\" << endl; cout << \"\| 5 5 \|\" << endl; cout << \"=============\" << endl; }//else if else if (dice1 == 6)//drawing if the value of dice1 is 6 { cout << \"=============\" << endl; cout << \"\| 6 6 \|\" << endl; cout << \"\| 6 6 \|\" << endl; cout << \"\| 6 6 \|\" << endl; cout << \"=============\" << endl; }//else if cout << endl;//blank line if (dice2 == 1)//drawing if the value of dice2 is 1 { cout << \"=============\" << endl; cout << \"\| \|\" << endl; cout << \"\| 1 \|\" << endl; cout << \"\| \|\" << endl; cout << \"=============\" << endl; }//if else if (dice2 == 2)//drawing if the value of dice2 is 2 { cout << \"=============\" << endl; cout << \"\| 2 \|\" << endl; cout << \"\| \|\" << endl; cout << \"\| 2 \|\" << endl; cout << \"=============\" << endl; }//else if else if (dice2 == 3)//drawing if the value of dice2 is 3 { cout << \"=============\" << endl; cout << \"\| 3 \|\" << endl; cout << \"\| 3 \|\" << endl; cout << \"\| 3 \|\" << endl; cout << \"=============\" << endl; }//else if else if (dice2 == 4)//drawing if the value of dice2 is 4 { cout << \"=============\" << endl; cout << \"\| 4 4 \|\" << endl; cout << \"\| \|\" << endl; cout << \"\| 4 4 \|\" << endl; cout << \"=============\" << endl; }//else if else if (dice2 == 5)//drawing if the value of dice2 is 5 { cout << \"=============\" << endl; cout << \"\| 5 5 \|\" << endl; cout << \"\| 5 \|\" << endl; cout << \"\| 5 5 \|\" << endl; cout << \"=============\" << endl; }//else if else if (dice2 == 6)//drawing if the value of dice2 is 6 { cout << \"=============\" << endl; cout << \"\| 6 6 \|\" << endl; cout << \"\| 6 6 \|\" << endl; cout << \"\| 6 6 \|\" << endl; cout << \"=============\" << endl; }//else if cout << endl;//blank line cout << \" For a total of \" << total << endl; cout << endl;//blank line if (guess == total)//if the user guessed correctly { earned_or_lost_money = bet * 8 / possibility;//calculation for money earned cout << \" You were correct... since \" << guess << \" can come up \" << possibility << \" ways,\" << endl; cout << \"You win$\" << bet << \"8/\" << possibility << \"= $\" << earned_or_lost_money << endl; }//if else//if the user guessed wrong { earned_or_lost_money = possibility bet;//calculation for lost money cout << \" You were wrong... since \" << guess << \" can come up \" << possibility << \" ways,\" << endl; cout << \" You lost \" << possibility << \"*$\" << bet << \"= \\$\" << earned_or_lost_money << \" dollars!!!!!!\" << endl;\ncout << \" Thanks for your donation :)\" << endl;\n}//else\nreturn 0;\n}//end\n\n\n## 3 Answers\n\nThe 2 previous answers provide a lot of good points, here are a few more suggestions:\n\n## Generating Random Numbers\n\nAs of C++11 there are better random number generators to use that the C standard library rand() function. This stackoverflow question provides the following:\n\n#include <random>\n#include <iostream>\n\nint main() {\nstd::random_device rd;\nstd::mt19937 mt(rd());\nstd::uniform_real_distribution<double> dist(1.0, 10.0);\n\nfor (int i=0; i<16; ++i)\nstd::cout << dist(mt) << \"\\n\";\n}\n\n\nThis provides a better distribution than the C function rand().\n\n## Representing Money in Software\n\nBanks don't like to lose money due to floating point error, nor can they legally cheat the costumers so they don't use floating point numbers to represent money, they use 2 integer values, one for dollars and one for cents. This prevents floating point errors from affecting the cash values involved.\n\n## Complexity\n\nAny function larger than one screen in most IDEs or editors is to large, it is very hard to write, read, debug and maintain functions this large. You can create a class as one of the other answers suggested, you can also just create a number of functions called from main(). If you continue learning software you will learn principles on why this is important.\n\n• Two integers? Too wasteful and cumbersome. Better a single integer with twice the bits. – Deduplicator Nov 10 '20 at 20:46\n• Or for JPY just the yen - or more likely some multi didgit library or if old enough BCD rather than binary integers – mmmmmm Nov 11 '20 at 10:37\n\nFor a beginner, this is a nice attempt.\n\n# Avoid \"using namespace std\"\n\nThough this is trivial for a ten lines of code project, it immediately starts leading to issues when code size increases. Avoid it now. You can alternatively type std::cout or rather using std::cout, here the effect is localized to just cout. Check here for more info on why \"using namespace std\" is consider bad.\n\n# Prefer double to floating-point values\n\nDouble has 2x precision than floating-point values, thereore, provides more accuracy when performing calculations. Double on most machines is 8bytes whereas float is 4bytes, but the precision gain is worth it.\n\n# Make use of random header\n\nrand requires a seed to generate its pseudorandom numbers, if this seed is the same at each run, it produces the same output. This leads to some sort of dependency on the seed. rand also uses srand making it impossible to use with other random engine\n\n# Consider using a class\n\nYour main function is cluttered with code here and there, move them to separate class, you could have a class that handles everything concerning the dice and the rest other functionalities.\n\n# Prefer \\n to endl\n\nWhenever you need a newline but don't explicitly need to flush the buffer, use \\n. endl adds a newline and flushes the buffer, this would result to a slight drop in performance.\n\n• Actually for accuracy reasons backs use 2 integers for money values, one for dollars and one for cents. This prevents floating point errors. – pacmaninbw Nov 10 '20 at 17:54\n• @pacmaninbw I totally agree. – theProgrammer Nov 10 '20 at 18:19\n• Integers for money, sure. But why break it down into two, when handling just one is more efficient and convenient? And if it has twice the bits, you can store even bigger totals. – Deduplicator Nov 10 '20 at 20:48\n\nDepending of what you want, there is some things you can do:\n\n• refactor dice drawing function as there is a lot of redounding code\n• a check for all inputs (nothing prevent your user to input bet out of 2-12 and money out of 0-1000, even negative bet or non numeric ones\n• add a loop allowing player to play multiples games (with money carried between)\n• i probably would have used a table to store possibilities of bet\n\nWithout further informations about what you want to improve, it's hard to be more specific" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.72092843,"math_prob":0.9504387,"size":5468,"snap":"2021-04-2021-17","text_gpt3_token_len":1647,"char_repetition_ratio":0.3545022,"word_repetition_ratio":0.5046296,"special_character_ratio":0.46762985,"punctuation_ratio":0.16613756,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9981133,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-04-17T23:21:04Z\",\"WARC-Record-ID\":\"<urn:uuid:6bf07289-77d2-475b-a01f-32d2d9b41d1f>\",\"Content-Length\":\"197217\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:5cb7812c-ebae-40ef-a034-18d9a0b024fa>\",\"WARC-Concurrent-To\":\"<urn:uuid:6704a825-da2c-49b6-a78e-fef30c0e52fe>\",\"WARC-IP-Address\":\"151.101.1.69\",\"WARC-Target-URI\":\"https://codereview.stackexchange.com/questions/251910/c-dice-game-using-random-numbers\",\"WARC-Payload-Digest\":\"sha1:WEAEUVIPUE4KDI4NN6DQVOR2PDTGBXEZ\",\"WARC-Block-Digest\":\"sha1:IV33D47E2T67VW43UCF5NMFXQB4QRGC5\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-17/CC-MAIN-2021-17_segments_1618038464065.57_warc_CC-MAIN-20210417222733-20210418012733-00033.warc.gz\"}"}
https://lavelle.chem.ucla.edu/forum/viewtopic.php?f=10&t=45865	[ "## G 25\n\nNorman Dis4C\nPosts: 101\nJoined: Sat Sep 28, 2019 12:16 am\n\n### G 25\n\n\"Practitioners of the branch of alternative medicine known as homeopathy claim that very dilute solutions of substances can have an effect. Is the claim plausible? To explore this question, suppose that you prepare a solution of a supposedly active substance, X, with a molarity of 0.10 mol/L. Then\nyou dilute 10. mL of that solution by doubling the volume, doubling it again, and so on, for 90 doublings in all. How many molecules of X will be present in 10. mL of the final solution? Comment on the possible health benefits of the solution.\"\n\nThe correct answer is no X remaining\n\nShouldn't the amount of solute, which is substance X remain the same?\n\nAnna Wu 1H\nPosts: 51\nJoined: Sat Aug 17, 2019 12:16 am\n\n### Re: G 25\n\nThe amount of solute remains the same, but the volume changes. Since the solution is being diluted so dramatically, the 10 mL sample taken at the end has close to zero solute in it.\n\nCharisse Vu 1H\nPosts: 101\nJoined: Thu Jul 25, 2019 12:17 am\n\n### Re: G 25\n\nSince you are doubling the volume of the solution 90 times, the final volume would be much greater than the initial volume. The question asks how many molecules would be present in only 10. mL of the final solution. Since the volume has increased so dramatically, 10. mL of the final solution would contain a very, very minute amount of the molecule X.\n\nManav Govil 1B\nPosts: 104\nJoined: Sat Sep 07, 2019 12:19 am\n\n### Re: G 25\n\nThe thing is that after 69 doublings, less than one molecule would remain in the 10 mL of the diluted solution. At that point, the solution does not have any health benefits whatsoever. When a solution has less than one molecule of a specific substance (especially if that substance is intended to aid the issue), the solution is ineffective, as the substance is too small to do anything. If you want exact numbers, I found that after 90 doublings, there are 4.9 * 10^-7 molecules in the solution - which is almost nothing.\n\nAmir Bayat\nPosts: 115\nJoined: Sat Sep 07, 2019 12:16 am\n\n### Re: G 25\n\nI understand your guys' reasoning as to why there would be a very, very minute amount of X left in the substance. However, what is the fastest and quickest way to approach a question like this and show work?\n\nI understand the conceptual and visualization aspect of it, but doubling until 69 times would take too long on a test. Have any of you guys found a way to approach this that would take less time?\n\nBrian Tangsombatvisit 1C\nPosts: 119\nJoined: Sat Aug 17, 2019 12:15 am\n\n### Re: G 25\n\nI think once you see an absurd number that is above 50+, you can assume that you don't have to do all the math and just approach the question conceptually. If you use the concept of limits from algebra, if a number keeps getting smaller and smaller, you can safely make the assumption that it approaches 0 since it's so tiny.\n\nEdmund Zhi 2B\nPosts: 118\nJoined: Sat Jul 20, 2019 12:16 am\n\n### Re: G 25\n\nIn 20 mL of the 0.1M solution, there will be 0.1 = x/0.01, x = 0.001 mol of X (the solute). Every time you double this volume, there will be half as many moles in a 10 ml solution. The number of molecules in the final solution is therefore 0.001 mol * (1/2)^90 * avagadro's number, which is around 4.9 * 10^-7 molecules every 10ml, which is basically 0. There are no health benefits if the molecule is completely missing in the solution.\n\nAmir Bayat\nPosts: 115\nJoined: Sat Sep 07, 2019 12:16 am\n\n### Re: G 25\n\nUpdate:\n\nThere is a much faster way to approach this and it is with using the formula below, once we have solved for the amount of molecules of X.\n\nnumber of molecules of X solved for --> 6.0 * 10^20 molecules X + (1/2)^n = 1 molecule.\n\nWe then take the log of both sides and solve for n... getting 69 as the number of doubles to have one molecule of X left, and thus none left after 90 doubles." ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.94525826,"math_prob":0.8752754,"size":4475,"snap":"2020-45-2020-50","text_gpt3_token_len":1182,"char_repetition_ratio":0.13330351,"word_repetition_ratio":0.31111112,"special_character_ratio":0.28648046,"punctuation_ratio":0.1353001,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.96786517,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-11-24T07:33:30Z\",\"WARC-Record-ID\":\"<urn:uuid:393a686e-4ade-4ba8-98a0-96134b3fdd71>\",\"Content-Length\":\"67533\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:8d00a9c6-ae75-456a-be7c-5c6fa2d50241>\",\"WARC-Concurrent-To\":\"<urn:uuid:d8dab58d-4f1d-4697-aa95-9cd8e7d660a4>\",\"WARC-IP-Address\":\"169.232.134.130\",\"WARC-Target-URI\":\"https://lavelle.chem.ucla.edu/forum/viewtopic.php?f=10&t=45865\",\"WARC-Payload-Digest\":\"sha1:XM4DEQ5NNTNDIH6PYYIFC3FYAPR6ULLF\",\"WARC-Block-Digest\":\"sha1:TFE43YMG2QMXELZOPUCJ6OMF4AJ6CZ5K\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-50/CC-MAIN-2020-50_segments_1606141171126.6_warc_CC-MAIN-20201124053841-20201124083841-00486.warc.gz\"}"}
http://nicadd.niu.edu/~piot/phys_630/Syllabus.html	[ "# PHYS 630: Advanced Optics (Fall 2008)\n\n## Syllabus of Tuesdays and Thursdays lectures\n\ná Introduction (1 lessons)\n\no Course organization,\n\no Ray Optics\n\no Helmholtz equation\n\ná Beam Optics (3 lessons)\n\no Introduction\n\no Gaussian Beams\n\no Other solution of Helmholtz equation\n\no Short duration beams\n\no Alternate method for describing a beam: covariance matrix and M2\n\ná Fourier Optics (4 lessons)\n\no Hamonic analysis of a signal\n\no Amplitude and phase modulations\n\no Transfer function of free space\n\no Optical Fourier transform\n\no Diffraction & Interference\n\no Image shaping\n\no Holography\n\ná Electromagnetic description of light & propagation in matter (7 lessons)\n\no Light in vacuum\n\no Theory of electromagnetic beams\n\no Light guiding\n\no Absorption of light & Dispersion\n\no Optical phenomena in nonisotropic media\n\n¤ Dichroism and birefringence\n\n¤ E-field effects\n\n¤ Acousto-optics effects\n\n¤ B-field effects\n\ná Lasers (5 lessons)\n\no Interaction of light with matter\n\no Laser dynamics\n\no Pulsed laser beam\n\no Amplifiers\n\no Example of laser systems\n\ná Other light sources (1 lesson)\n\no Radiatio from moving charged particle\n\no Free-eletron laser\n\no Thomson scattering\n\ná Nonlinear Optics (4 lessons)\n\no Nonlinear optical media\n\no 2nd order optics\n\no 3rd order optics\n\no wave mixing\n\no high harmonic generation\n\no self focusing and phase modulation\n\ná Introduction to Statistical optics (2 lessons)\n\no Statistical properties of random light\n\no Interference of partially polarized coherent light\n\no Transmission of partially coherent light through optical system\n\no Partial polarization\n\ná Introduction to Quantum Optics (TBD)" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.60622585,"math_prob":0.6175571,"size":1547,"snap":"2021-31-2021-39","text_gpt3_token_len":459,"char_repetition_ratio":0.13739468,"word_repetition_ratio":0.0,"special_character_ratio":0.18552037,"punctuation_ratio":0.012244898,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.97659665,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-09-28T23:56:47Z\",\"WARC-Record-ID\":\"<urn:uuid:1b1c27b9-3215-4a0e-a2c3-cc75cd660ba5>\",\"Content-Length\":\"22125\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:9e964de2-eb62-4a27-8e53-77165439d748>\",\"WARC-Concurrent-To\":\"<urn:uuid:ca6c7cae-2bb6-4bb9-ba83-a5d0b381d0cc>\",\"WARC-IP-Address\":\"131.156.85.19\",\"WARC-Target-URI\":\"http://nicadd.niu.edu/~piot/phys_630/Syllabus.html\",\"WARC-Payload-Digest\":\"sha1:Q3GLC5F5TO2JKAY5QWUUB4XUON3K7MSV\",\"WARC-Block-Digest\":\"sha1:ZOGDPLR6PHQOSHHWP2SR25AHF6SBVXAU\",\"WARC-Identified-Payload-Type\":\"application/xhtml+xml\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-39/CC-MAIN-2021-39_segments_1631780060908.47_warc_CC-MAIN-20210928214438-20210929004438-00095.warc.gz\"}"}
https://answers.everydaycalculation.com/add-fractions/8-28-plus-2-80	[ "Solutions by everydaycalculation.com\n\n8/28 + 2/80 is 87/280.\n\n1. Find the least common denominator or LCM of the two denominators:\nLCM of 28 and 80 is 560\n2. For the 1st fraction, since 28 × 20 = 560,\n8/28 = 8 × 20/28 × 20 = 160/560\n3. Likewise, for the 2nd fraction, since 80 × 7 = 560,\n2/80 = 2 × 7/80 × 7 = 14/560", 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":0.70663285,"math_prob":0.99679506,"size":322,"snap":"2019-51-2020-05","text_gpt3_token_len":132,"char_repetition_ratio":0.18238993,"word_repetition_ratio":0.0,"special_character_ratio":0.49068323,"punctuation_ratio":0.064935066,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99656934,"pos_list":[0,1,2],"im_url_duplicate_count":[null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-12-07T21:18:07Z\",\"WARC-Record-ID\":\"<urn:uuid:1c7d9fb2-43f1-43a4-94ea-870bb9242a17>\",\"Content-Length\":\"7916\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:56727e45-b549-47d0-879d-9f4c3f29b268>\",\"WARC-Concurrent-To\":\"<urn:uuid:888f7df4-e3b8-49c6-bc56-8948385ecd5c>\",\"WARC-IP-Address\":\"96.126.107.130\",\"WARC-Target-URI\":\"https://answers.everydaycalculation.com/add-fractions/8-28-plus-2-80\",\"WARC-Payload-Digest\":\"sha1:5QGWAVYKHAQ53VJQUVRNMLRCA4Y2EEIA\",\"WARC-Block-Digest\":\"sha1:7QT6OFDYDWHNLUIXSZNPCIT5IVPY3JU2\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-51/CC-MAIN-2019-51_segments_1575540502120.37_warc_CC-MAIN-20191207210620-20191207234620-00299.warc.gz\"}"}
https://ir.cwi.nl/pub/30210	[ "We study the resource augmented version of the k-server problem, also known as the k-server problem against weak adversaries or the (h,k)-server problem. In this setting, an online algorithm using k servers is compared to an offline algorithm using h servers, where h ≤ k. For uniform metrics, it has been known since the seminal work of Sleator and Tarjan (1985) that for any \">0, the competitive ratio drops to a constant if k=(1+\") · h. This result was later generalized to weighted stars (Young 1994) and trees of bounded depth (Bansal et al. 2017). The main open problem for this setting is whether a similar phenomenon occurs on general metrics. We resolve this question negatively. With a simple recursive construction, we show that the competitive ratio is at least ω(loglogh), even as k→∞. Our lower bound holds for both deterministic and randomized algorithms. It also disproves the existence of a competitive algorithm for the infinite server problem on general metrics.\n\n, , ,\ndoi.org/10.1145/3357713.3384306\nAnnual ACM SIGACT Symposium on Theory of Computing\nCentrum Wiskunde & Informatica, Amsterdam, The Netherlands\n\nBienkowski, M, Byrka, J, Coester, C.E, & Jeż, L. (2020). Unbounded lower bound for k-server against weak adversaries. In Proceedings of the Annual ACM SIGACT Symposium on Theory of Computing (pp. 1165–1169). doi:10.1145/3357713.3384306" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.85534716,"math_prob":0.898485,"size":1511,"snap":"2021-43-2021-49","text_gpt3_token_len":362,"char_repetition_ratio":0.09820836,"word_repetition_ratio":0.053097345,"special_character_ratio":0.25876904,"punctuation_ratio":0.14736842,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.96277225,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-10-15T21:41:08Z\",\"WARC-Record-ID\":\"<urn:uuid:c38bac86-e085-4fb0-8b9f-490a5db04ae6>\",\"Content-Length\":\"23192\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:7db3a192-617e-475c-b2df-7d766d0d25dd>\",\"WARC-Concurrent-To\":\"<urn:uuid:3c9e7611-7830-4e40-bfee-4bf8ec5f0473>\",\"WARC-IP-Address\":\"5.79.73.6\",\"WARC-Target-URI\":\"https://ir.cwi.nl/pub/30210\",\"WARC-Payload-Digest\":\"sha1:OLQT574GER7PC2F2BWXJGAUCL5JZEOUK\",\"WARC-Block-Digest\":\"sha1:QJ4IOGIUOMXRJ4SYYL7VEJKO5XGIIXZ7\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-43/CC-MAIN-2021-43_segments_1634323583083.92_warc_CC-MAIN-20211015192439-20211015222439-00703.warc.gz\"}"}
https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Equilibria/Solubilty/An_Introduction_to_Solubility_Products	[ "# An Introduction to Solubility Products\n\n•", null, "• Contributed by Jim Clark\n• Former Head of Chemistry and Head of Science at Truro School in Cornwall\n\nThis page discusses how solubility products are defined, including their units units. It also explores the relationship between the solubility product of an ionic compound and its solubility.\n\n## Solubility products are equilibrium constants\n\nBarium sulfate is almost insoluble in water. It is not totally insoluble—very small amounts do dissolve. This is true of any \"insoluble\" ionic compound. If solid barium sulfate is shaken with water, a small amount of barium ions and sulfate ions break away from the surface of the solid and go into solution. Over time, some of these return from solution to stick onto the solid again.\n\nAn equilibrium is established when the rate at which some ions are breaking away from the solid lattice is exactly matched by the rate at which others are returning. Consider the balanced equation for the barium sulfate reaction:\n\n$BaSO_4(s) \\rightleftharpoons Ba^{2+}(aq) + SO_4^{2-}(aq)$\n\nThe position of this equilibrium lies very far to the left. The great majority of the barium sulfate is present as solid—there are no visible changes to the solid. However, the equilibrium does exist, and an equilibrium constant can be written. The equilibrium constant is called the solubility product, and is given the symbol Ksp:\n\n$K_{sp} = [Ba^{2+} (aq)] [SO_4^{2-}(aq)]$\n\nFor simplicity, solubility product expressions are often written without the state symbols:\n\n$K_{sp} = [Ba^{2+}] [SO_4^{2-}]$\n\nNotice that there is no solid barium sulfate term. For many simple equilibria, the equilibrium constant expression has terms for the right side of the equation divided by terms for the left side. But in this case, there is no term for the concentration of the solid barium sulfate. This is a heterogeneous equilibrium, one which contains substances in more than one state. In a heterogeneous equilibrium, concentration terms for solids are left out of the expression.\n\n## Solubility products for more complicated solids\n\nHere is the corresponding equilibrium for calcium phosphate, Ca3(PO4)2:\n\n$Ca_3(PO_4)_2(s) \\rightleftharpoons 3Ca^{2+}(aq) + 2PO_4^{3-}(aq)$\n\nBelow is the solubility product expression:\n\n$K_{sp} = [Ca^{2+}]^3[PO_4^{3-}]^2$\n\nAs with any other equilibrium constant, the concentrations are raised to the power of their respective stoichiometric coefficients in the equilibrium equation. Solubility products only apply to sparingly soluble ionic compounds, not to normally soluble compounds such as sodium chloride. Interactions between the ions in the solution interfere with the simple equilibrium.\n\n## The units for solubility products\n\nThe units for solubility products differ depending on the solubility product expression, and you need to be able to work them out each time.\n\nExample 1: BaSO4\n\nBelow is the solubility product expression for barium sulfate:\n\n$K_{sp} = [Ba^{2+}] [SO_4^{2-}]$\n\nEach concentration has the unit mol dm-3. In this case, the units for the solubility product in this case are the following:", null, "(mol dm-3) x (mol dm-3)", null, "= mol2 dm-6\n\nExample 2: CaPO4\n\nRecall the solubility product expression for calcium phosphate:\n\n$K_{sp} = [Ca^{2+}]^3[PO_4^{3-}]^2$\n\nThe units this time will be:", null, "(mol dm-3)3 x (mol dm-3)2", null, "= (mol dm-3)5", null, "= mol5 dm-15\n\n## Solubility products apply only to saturated solutions\n\nRecall the barium sulfate equation:\n\n$BaSO_4(s) \\rightleftharpoons Ba^{2+}(aq) + SO_4^{2-}(aq)$\n\nThe corresponding solubility product expression is the following:\n\n$K_{sp}= [Ba^{2+}][SO_4^{2-}]$\n\nKsp for barium sulfate at 298 K is 1.1 x 10-10 mol2 dm-6.\n\nIn order for this equilibrium constant (the solubility product) to apply, solid barium sulfate must be present in a saturated solution of barium sulfate. This is indicated by the equilibrium equation. If barium ions and sulfate ions exist in solution in the presence of some solid barium sulfate at 298 K, and multiply the concentrations of the ions together, the answer will be 1.1 x 10-10 mol2 dm-6. It is possible to multiply ionic concentrations and obtain a value less than this solubility product; in such cases the solution is too dilute, and no equilibrium exists. If the concentrations are lowered enough, no precipitate can form.\n\nHowever, it is impossible to calculate a product greater than the solubility product. If solutions containing barium ions and sulfate ions are mixed such that the product of the concentrations would exceed Ksp, a precipitate forms. Enough solid is produced to reduce the concentrations of the barium and sulfate ions down to the value of the solubility product.\n\n## Summary\n\n• The value of a solubility product relates only to a saturated solution.\n• If the ionic concentrations give a value less than the solubility product, the solution is not saturated. No precipitate would be formed in such a case.\n• If the ionic concentrations give a value more than the solubility product, enough precipitate would be formed to reduce the concentrations to give an answer equal to the solubility product.\n\n## Contributors\n\nJim Clark (Chemguide.co.uk)" ]	[ null, "https://chem.libretexts.org/@api/deki/files/123564/clark.jpg", null, "http://www.chemguide.co.uk/physical/ksp/padding.gif", null, "http://www.chemguide.co.uk/physical/ksp/padding.gif", null, "http://www.chemguide.co.uk/physical/ksp/padding.gif", null, "http://www.chemguide.co.uk/physical/ksp/padding.gif", null, "http://www.chemguide.co.uk/physical/ksp/padding.gif", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.8545495,"math_prob":0.98989314,"size":5061,"snap":"2020-10-2020-16","text_gpt3_token_len":1253,"char_repetition_ratio":0.19398853,"word_repetition_ratio":0.03448276,"special_character_ratio":0.23710729,"punctuation_ratio":0.087100334,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9949585,"pos_list":[0,1,2,3,4,5,6,7,8,9,10,11,12],"im_url_duplicate_count":[null,null,null,null,null,null,null,null,null,null,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-03-31T20:02:37Z\",\"WARC-Record-ID\":\"<urn:uuid:718b29dd-5e3c-4807-83d8-05c9e441bdfa>\",\"Content-Length\":\"105598\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:7cff09ae-28c3-44a5-ba77-2724ac74d1b0>\",\"WARC-Concurrent-To\":\"<urn:uuid:bb199549-a62c-456d-8605-43a689d4e968>\",\"WARC-IP-Address\":\"99.84.181.53\",\"WARC-Target-URI\":\"https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Equilibria/Solubilty/An_Introduction_to_Solubility_Products\",\"WARC-Payload-Digest\":\"sha1:XGG5AQWF7P6LAC7RCWMCFWEZWXCQBKV2\",\"WARC-Block-Digest\":\"sha1:BBPEACZGBN63U3J6NEVRUE6WIB4ZWR55\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-16/CC-MAIN-2020-16_segments_1585370503664.38_warc_CC-MAIN-20200331181930-20200331211930-00121.warc.gz\"}"}
https://seniorsecondary.tki.org.nz/Mathematics-and-statistics/Achievement-objectives/AOs-by-level/AO-M8-7	[ "Te Kete Ipurangi\nCommunities\nSchools\n\n### Te Kete Ipurangi user options:\n\nAchievement objectives\n\nWhat has changed:\n\n# Achievement objective M8-7\n\nIn a range of meaningful contexts, students will be engaged in thinking mathematically and statistically. They will solve problems and model situations that require them to:\n\n• form and use trigonometric, polynomial, and other non-linear equations.\n\n## Indicators\n\n• Solves contextual problems by forming and solving equations.\n• Solves quadratic and cubic equations with complex roots. These equations are likely to involve real coefficients so that solutions are complex conjugate pairs.\n• Analyses the existence of solutions in the context of the situation.\n• Non-linear equations include:\n• exponential\n• logarithmic\n• rational\n• surd\n• hyperbolic.\n• Makes links to graphs of functions M8-2, curve fitting M8-4, manipulating expressions M7-6 and M8-6, and complex numbers M8-9.\n• See key mathematical ideas on NZmaths.\n\n## Progression\n\nM8-7 links from M7-4, M7-7 and to M8-9.\n\n## Possible context elaborations\n\n• Natural phenomena that result in periodic behaviours (for example, tidal patterns, sound waves, circular motion, harmonics …).\n• Explore growth and decay situations.\n• Completely factor a third degree polynomial, using long division and/or reminder and factor theorems.\n• Given general solution for a trig equation, find the solutions that satisfy a particular context.\n• Although numerical solutions (bisection and Newton-Raphson) do not need to be included, they could form part of an inquiry into dealing with non-linear equations that can’t be solved with algebraic methods.\n• Activity: Big Bang Theory – Exact values\n• Graphing projectile paths\n• Investigating algebra: Exploring graphs\n• Practical situations involving trigonometric graphs: Includes pedal height of a bicycle, height of loaded oscillating spring.\n• Tidal estuaries: Investigating the times yachts can enter and leave tidal estuaries.\n• Use of remainder and factor theorems.\n\n## Assessment for qualifications\n\nNCEA achievement standards at levels 1, 2 and 3 have been aligned to the New Zealand Curriculum. Please ensure that you are using the correct version of the standards by going to the NZQA website.\n\nThe NZQA subject-specific resources pages are very helpful. From there, you can find all the achievement standards and links to assessment resources, both internal and external.\n\nThe following achievement standard(s) could assess learning outcomes from this AO:\n\nLast updated September 17, 2018" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.8592751,"math_prob":0.8887712,"size":2398,"snap":"2021-31-2021-39","text_gpt3_token_len":530,"char_repetition_ratio":0.10568087,"word_repetition_ratio":0.0,"special_character_ratio":0.20183486,"punctuation_ratio":0.119700745,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9780174,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-07-24T23:52:50Z\",\"WARC-Record-ID\":\"<urn:uuid:bc671cf2-8d07-4cff-8a64-183628ba5bb6>\",\"Content-Length\":\"249749\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:74080a94-c8ea-4926-b67f-8ed1e8c08693>\",\"WARC-Concurrent-To\":\"<urn:uuid:feda8fb4-6156-4820-86a9-51834330db77>\",\"WARC-IP-Address\":\"185.125.86.67\",\"WARC-Target-URI\":\"https://seniorsecondary.tki.org.nz/Mathematics-and-statistics/Achievement-objectives/AOs-by-level/AO-M8-7\",\"WARC-Payload-Digest\":\"sha1:OCMBTQPLEQIXCGNOQSWC4X3JWNTNVDVM\",\"WARC-Block-Digest\":\"sha1:HRUOM7NKVQKPJQ3DXGTCLTE3RFX43GVE\",\"WARC-Identified-Payload-Type\":\"application/xhtml+xml\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-31/CC-MAIN-2021-31_segments_1627046151531.67_warc_CC-MAIN-20210724223025-20210725013025-00630.warc.gz\"}"}
https://all-learning.com/swift-closure/	[ "# Swift Closure With Examples [Latest]", null, "In this tutorial, we’ll be discussing and seeing the usages (why, how and where) of Swift Closure.\n\nIt’s noteworthy to mention here that Swift Function is also a form of closure.\n\n## Swift Closure\n\nAccording to the Apple Documentation, “Closures are self-contained blocks of functionality that can be passed around and used in your code”.\n\nClosures can capture and store references to any constants and variables from the context in which they are defined.\n\nThe concept of Swift Closure is similar to blocks in C. Closures are nameless functions and without the keyword `func`.\n\n### Forms of Closure in Swift\n\nClosures can take one of the three forms.\n\n1. Global functions\n2. Nested functions\n3. Closures expression\n\nGlobal and nested functions were discussed at length in the Functions in Swift tutorial.\n\nWhen we mention closures we generally mean the third form i.e. Closure Expressions. Before we get on with the syntax and use cases, let’s look at the benefits that Closures bring to us.\n\n### Advantages of Closures in Swift\n\n1. Closures are much less verbose when compared to Functions. Swift can infer the parameter types and return type from the context in which the closure is defined thereby making it convenient to define and pass to a function\n2. Swift Closure can be used to capture and store the state of some variable at a particular point in time and use it later(We’ll be discussing this at length later in the tutorial)\n3. Closures allow us to run a piece of code even after the function has returned\n\n### Swift Closure Syntax\n\nLet’s recall the syntax of Functions in swift first.\n\n``````\nfunc name(parameters) -> (return type){\n//body of function\n}\n``````\n\nFunctions require a `func` keyword followed by name of function, arguments and return type after `->`\n\nSyntax for Closures is given below.\n\n``````\n{ (parameters)->(return type) in\n//body of the closure\n}\n``````\n\nClosures are defined within curly braces. The `in` keyword separates the arguments and return type from the body.\n\nLike functions, closures can also be referenced as types. Let’s dive into playground in XCode and start coding.\n\n### Declaring Swift Closures as variables\n\n``````\nvar myClosure: (ParameterTypes) -> ReturnType\n``````\n\nLet’s create a basic function and a closure that’ll print “Hello World”.\n\n``````\nfunc helloWorldFunc()\n{\nprint(\"Hello World\")\n}\nvar helloWorldClosure = { () -> () in print(\"Hello World\") }\nhelloWorldFunc() //prints Hello World\nhelloWorldClosure() //prints Hello World\n``````\n\nIn the above code, we’ve defined a closure with no parameters and no return type and set it to a variable `helloWorldClosure`.\n\nCalling the closure is similar to calling a function as shown above.\n\nTip 1: If the closure does not return any value or the type can be inferred you can omit the arrow (`->`) and the return type. We can shorten the above code as shown below.\n\n``````\nvar helloWorldClosure = { () in print(\"Hello World\") }\n``````\n\nTip 2: If the closure doesn’t require any parameter or can be inferred, remove the argument parenthesis. Now we don’t need the `in` keyword too. This will further shorten the code as shown below.\n\n``````\nvar helloWorldClosure = { print(\"Hello World\") }\n``````\n\nWe’ve seen one use case now where closures made the code look less verbose compared to a function. Let’s create a closure with parameters and see how it fares when compared to closures.\n\n### Swift Closures with arguments and return type\n\nLet’s create a function and closure that accepts two strings, appends them and return.\n\n``````\nfunc appendString(_ a: String, with b: String) -> String\n{\nreturn a + \" : \" + b\n}\nprint(appendString(\"Swift\", with: \"Functions\")) //print \"Swift : Functionsn\"\n``````\n\nSo far so good. Lets do the same using a closure.\n\n``````\nvar appendStringClosure = { (a:String,b:String) -> (String) in return a + \" : \" + b }\nprint(appendStringClosure(\"Swift\", \"Closures\")) //prints \"Swift : Closuresn\"\n``````\n\nThanks to type inference as we’d seen earlier, the above closure definition is the same as the following.\n\n``````\nvar appendStringClosure = { (a:String,b:String) in return a + \" : \" + b } // return type inferred\nvar appendStringClosure : (String, String) -> String = { (a,b) in return a + \" : \" + b } //Closure type declared on the variable itself\nvar appendStringClosure : (String, String) -> String = { (a,b) in a + \" : \" + b } // omit return keyword\n``````\n\nThere’s even a shortened version for the above.\n\nSwift allows us to refer to arguments passed to closure using shorthand argument names : \\$0, \\$1, \\$2 etc. for the first, second, third etc. parameters respectively.\n\n``````\nvar appendStringClosure : (String, String) -> String = { \\$0 + \" : \" + \\$1 }\nprint(appendStringClosure(\"Swift\", \"Closures\")) //prints \"Swift : Closuresn\"\n``````\n\nLet’s take it one more step ahead:\n\n``````\nvar appendStringClosure = { \\$0 + \" : \" + \\$1 + \" \" + \\$2 }\nprint(appendStringClosure(\"Swift\", \"Closures\", \"Awesome\")) //prints \"Swift : Closures Awesomen\"\n``````\n\nAdd as many arguments as you can.\n\nNote: It’s recommended to set the closure type.\n\nIt’s time to use Closures inside functions.\n\n### Closures inside Functions\n\nLet’s create a function that takes a function/closure as parameter.\n\n``````\nfunc operationsSq(a: Int, b:Int, myFunction: (Int, Int)->Int)->Int\n{\nreturn myFunction(a,b)\n}\n``````\n\nThis function expects a parameter of type `(Int, Int)->Int` as the third argument.\nLets pass a function in there as shown below\n\n``````\nfunc addSquareFunc(_ a: Int, _ b:Int)->Int\n{\nreturn aa + bb\n}\nprint(operationsSq(a:2,b:2, myFunction: addSquareFunc)) //a^2 + b^2 prints 8\n``````\n\nFair enough. But creating a different function for another operation (let’s say subtracting squares etc) would keep increasing the boilerplate code.\n\nThis is where closure comes to our rescue. We’ve defined the following 4 types of expressions that we’ll be passing one by one in the `operationsSq` function.\n\n``````\nvar addSq : (Int, Int) -> Int = {\\$0\\$0 + \\$1\\$1 }\nvar subtractSq: (Int, Int) -> Int = { \\$0\\$0 - \\$1\\$1 }\nvar multiplySq: (Int, Int) -> Int = { \\$0\\$0 \\$1\\$1 }\nvar divideSq: (Int, Int) -> Int = { (\\$0\\$0) / (\\$1\\$1) }\nvar remainderSq: (Int, Int) -> Int = { (\\$0\\$0) % (\\$1\\$1) }\nprint(operationsSq(a:4,b:5, myFunction: subtractSq)) //prints -9\nprint(operationsSq(a:5,b:5, myFunction: multiplySq)) //prints 625\nprint(operationsSq(a:10,b:5, myFunction: divideSq)) //prints 4\nprint(operationsSq(a:7,b:5, myFunction: remainderSq)) //prints 24\n``````\n\nMuch shorter than what we achieved with a function as a parameter. In fact, we can directly pass the closure expressions as shown below and it’ll achieve the same result:\n\n``````\noperationsSq(a:2,b:2, myFunction: { \\$0\\$0 + \\$1\\$1 })\noperationsSq(a:4,b:5, myFunction: { \\$0\\$0 - \\$1\\$1 })\noperationsSq(a:5,b:5, myFunction: { \\$0\\$0 * \\$1\\$1 })\noperationsSq(a:10,b:5, myFunction: { (\\$0\\$0) / (\\$1\\$1) })\noperationsSq(a:7,b:5, myFunction: { (\\$0\\$0) % (\\$1\\$1) })\n``````\n\n### Trailing Closures\n\nWhen the closure is being passed as the final argument to a function we can pass it as a trailing closure.\n\nA trailing closure is written after the function call’s parentheses, even though it is still an argument to the function.\n\nWhen you use the trailing closure syntax, you don’t write the argument label for the closure as part of the function call.\n\nIf closure argument is the sole argument of the function you can remove the function parentheses. Hence the previous closures can be written as the following and it would still function the same way.\n\n``````\noperationsSq(a:2,b:2){ \\$0\\$0 + \\$1\\$1 }\n``````\n\nLet’s look at another example where we’ll be adding the sum of exponentials of numbers using closure expressions.\n\n``````\nfunc sumOfExponentials(from: Int, to: Int, myClosure: (Int) -> Int)->Int\n{\nvar sum = 0\nfor i in from...to{\nsum = sum + myClosure(i)\n}\nprint(sum)\nreturn sum\n}\n//Trailing closures\nsumOfExponentials(from:0,to:5){ \\$0 } //sum of numbers\nsumOfExponentials(from:0,to:5){ \\$0\\$0 } //sum of squares\nsumOfExponentials(from:0,to:5){ \\$0\\$0\\$0 } //sum of cubes\n``````\n\nConvert a numbers array to strings array\nAnother use case of trailing closure is given below.\n\n``````\nvar numbers = [1,2,3,4,5,6]\nprint(numbers)\nvar strings = numbers.map{\"(\\$0)\"}\nprint(strings) //prints [\"1\", \"2\", \"3\", \"4\", \"5\", \"6\"]n\n``````\n\nmap is a high order function for transforming an array.\n\nSort numbers in descending order using trailing closures\n\n``````\nvar randomNumbers = [5,4,10,45,76,11,0,9]\nrandomNumbers = randomNumbers.sorted{\\$0>\\$1}\nprint(randomNumbers) //\"[76, 45, 11, 10, 9, 5, 4, 0]n\"\n``````\n\n### Capture Lists in Closures\n\nBy default, closures can capture and store references to any constants and variables from the context in which they are defined (Hence the name closures).\n\nCapturing references to variables can cause our closures to behave differently than it’s supposed to. Let’s see this through an example.\n\n``````\nvar x = 0\nvar myClosure = { print(\"The value of x at start is (x)\") }\nmyClosure() //prints 0 as desired.\n``````\n\nSo far so good. The above example looks pretty straightforward until we do this:\n\n``````\nvar x = 0\nvar myClosure = { print(\"The value of x at start is (x)\") }\nmyClosure() //The value of x at start is 0\nx = 5\nmyClosure() //The value of x at start is 5\n``````\n\nNow `myClosure` was defined and initialized before changing the value of `x` to 5. Why does it print 5 in the end then?\n\nThe reason is that closure captures the reference (memory address) of `x`. Any changes made to value at that memory address would be displayed by the closure when it’s invoked.\n\nTo make x behave as a value type instead we need to make use of Capture Lists. Capture Lists is an array `[]` that holds local copies of the variables. Thereby capturing the variables by value types and NOT reference types.\n\nThe array with the local variables is displayed before the `in` keyword as shown below.\n\n``````\nvar x = 0\nvar myClosure = { [x] in print(\"The value of x at start is (x)\") }\nmyClosure() // The value of x at start is 0\nx = 5\nmyClosure() //The value of x at start is 0\n``````\n\nCapturing reference types can be destructive when used in Classes since the closure can hold a strong reference to an instance variable and cause memory leaks. Let’s see an example of this.\n\n``````\nclass Person {\nvar x: Int\nvar myClosure: ()->() = {print(\"Hey there\")}\ninit(x: Int)\n{\nself.x = x\n}\nfunc initClosure()\n{\nmyClosure = { print(\"Initial value is not defined yet\")}\n}\ndeinit{\nprint(\"(self) escaped\")\n}\n}\nvar a:Person? = Person(x: 0)\na?.initClosure()\na?.x = 5\na?.myClosure()\na = nil\n``````\n\nDeinitializers are called automatically, just before instance deallocation takes place. In the above code, it’ll be called when `a` is set to `nil` and `self` doesn’t have any references attached to it.\n\nThe above code would print:\n`\"Initial value is not defined yet\"` followed by ” escaped” (“__lldb_expr_26.Person escaped” for me). Awesome there are no memory leaks. Lets set the value inside `myClosure` to `x` as shown below:\n\n``````\nclass Person {\nvar x: Int\nvar myClosure: ()->() = {print(\"Hey there\")}\ninit(x: Int)\n{\nself.x = x\n}\nfunc initClosure()\n{\nmyClosure = { print(\"Intial value is (self.x)\")}\n}\ndeinit{\nprint(\"(self) escaped\")\n}\n}\nvar a:Person? = Person(x: 0)\na?.initClosure()\na?.x = 5\na?.myClosure()\na = nil\n``````\n\nWhoops! the print statement within `deinit` is not printed. The closure holds a strong reference to `self`. This will eventually lead to memory leaks. A remedy for this is to place a weak or unknown reference to self in a capture list inside the closure as seen below.\n\n• A weak reference is a reference that does not keep a strong hold on the instance it refers to and so does not stop ARC from disposing of the referenced instance.\n• An unowned reference does not keep a strong hold on the instance it refers to. Unlike a weak reference, however, an unowned reference is used when the other instance has the same lifetime or a longer lifetime.\n\nThe code for capture list with `self` reference is given below:\n\n``````\nclass Person {\nvar x: Int\nvar myClosure: ()->() = {print(\"Hey there\")}\ninit(x: Int)\n{\nself.x = x\n}\nfunc initClosure()\n{\nmyClosure = {[weak self] in guard let weakSelf = self else { return }\nprint(\"Intial value is (weakSelf.x)\")}\n}\ndeinit{\nprint(\"(self) escaped\")\n}\n}\nvar a:Person? = Person(x: 0)\na?.initClosure()\na?.x = 5\na?.myClosure()\na = nil\n``````\n\nNo memory leaks with the above code!\n\n### Escaping Closures\n\nThere are two other types of closures too:\n\n• An escaping closure is a closure that’s called after the function it was passed to returns. In other words, it outlives the function it was passed to. An escaping closure is generally used in completion handlers since they are called after the function is over.\n• A non-escaping closure is a closure that’s called within the function it was passed into, i.e. before it returns. Closures are non-escaping by default\n``````\nvar completionHandlers: [() -> Void] = []\nfunc someFunctionWithNonescapingClosure(closure: () -> Void) {\nclosure()\n}\nfunc someFunctionWithEscapingClosure(completionHandler: @escaping () -> Void) {\ncompletionHandlers.append(completionHandler)\n}\nclass SomeClass {\nvar x = 10\nfunc doSomething() {\nsomeFunctionWithEscapingClosure {[weak self] in guard let weakSelf = self else { return }\nweakSelf.x = 100 }\nsomeFunctionWithNonescapingClosure { x = 200 }\n}\ndeinit{\nprint(\"deinitalised\")\n}\n}\nvar s:SomeClass? = SomeClass()\ns?.doSomething()\nprint(s?.x ?? -1) //prints 200\ncompletionHandlers.first?()\nprint(s?.x ?? -1) //prints 100\ns = nil\n``````\n\ncompletionHandlers.first() returns and invokes the first element of the array which is an escaping closure. Since we’d set a value type inside it using capture lists, it prints 100.\nWe’ll be discussing escaping closures at length in a later tutorial.\n\nHere’s a screenshot from my playground.", null, "That’s all about closure in swift." ]	[ null, "https://all-learning.com/wp-content/uploads/2017/08/Swift-Closure-With-Examples.png", null, "https://all-learning.com/wp-content/uploads/2017/08/ios-closures-playground-swift-450x27-1.png", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.78707844,"math_prob":0.96947545,"size":13348,"snap":"2023-14-2023-23","text_gpt3_token_len":3419,"char_repetition_ratio":0.15714929,"word_repetition_ratio":0.15213442,"special_character_ratio":0.27232546,"punctuation_ratio":0.13718553,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9661183,"pos_list":[0,1,2,3,4],"im_url_duplicate_count":[null,1,null,1,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-03-27T20:57:15Z\",\"WARC-Record-ID\":\"<urn:uuid:8cffc759-4472-469e-a2ef-d3771ca2b836>\",\"Content-Length\":\"98934\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:5a15bdb6-fdd7-4523-a169-51d3e77e15eb>\",\"WARC-Concurrent-To\":\"<urn:uuid:c1245698-1ca2-40bc-96e7-4bbbe03af27c>\",\"WARC-IP-Address\":\"172.67.139.202\",\"WARC-Target-URI\":\"https://all-learning.com/swift-closure/\",\"WARC-Payload-Digest\":\"sha1:OQ6HWGDJURI5QT7W37XKTJWBVB2HW66L\",\"WARC-Block-Digest\":\"sha1:BYNSJD2MBEMNORUUKWR3ZJXASKJ7LTCM\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-14/CC-MAIN-2023-14_segments_1679296948684.19_warc_CC-MAIN-20230327185741-20230327215741-00528.warc.gz\"}"}
https://hlearner.com/financial/home-loan.html	[ " H-learner \| Home Loan Calculator", null, "", null, "%\n\nFormula\n\nHouse Value = monthly ((1+I /(10012)years12 -1)/ (I/10012)(( I/10012))+1) years12)\n\nWhere,\nMonthly = (monthly income + other income – monthly expenditure) 28/100\nI = Interest" ]	[ null, "https://hlearner.com/assets/media/image/logo/logo.png", null, "https://hlearner.com/assets/media/image/logo/dark-logo.png", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.9452703,"math_prob":0.9965459,"size":648,"snap":"2021-04-2021-17","text_gpt3_token_len":155,"char_repetition_ratio":0.12732919,"word_repetition_ratio":0.0,"special_character_ratio":0.27623457,"punctuation_ratio":0.040983606,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9741524,"pos_list":[0,1,2,3,4],"im_url_duplicate_count":[null,null,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-01-26T05:27:10Z\",\"WARC-Record-ID\":\"<urn:uuid:696b7711-d5f6-4858-a739-ff8d24155be5>\",\"Content-Length\":\"55425\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:16e42d6c-de69-4e78-b7fd-9bc90abb7208>\",\"WARC-Concurrent-To\":\"<urn:uuid:3cfea0c6-3628-42d6-8402-85ea411ac82b>\",\"WARC-IP-Address\":\"172.67.188.88\",\"WARC-Target-URI\":\"https://hlearner.com/financial/home-loan.html\",\"WARC-Payload-Digest\":\"sha1:SUZKBBP67HBRFQ5AUVAUQVTU2U7LYXUB\",\"WARC-Block-Digest\":\"sha1:LRWKP3OU4NBXN26766WNIZJW466UI3YK\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-04/CC-MAIN-2021-04_segments_1610704798089.76_warc_CC-MAIN-20210126042704-20210126072704-00740.warc.gz\"}"}
https://java.meritcampus.com/core-java-questions/Calculate-the-largest-angle-of-the-given-triangle	[ "...\n\nWrite a program to get the largest angle of a triangle by using the given points. Round the resultant angles to the nearest integer value.\n\nInput (Point, Point, Point) Output (double)\n`[0, 0], [5, 0], [0, 5]` `90.0`\n`[90, 0], [0, 0], [45, 77.94]` `60.0`\n`[0, 2], [5, 0], [0, 5]` `112.0`\n`[0, 0], [10, 0], [5, 7.666]` `66.0`\n`[25, 43], [50, 0], [50, 43]` `90.0`\n\n``` class CalculateTheLargestAngle { public static void main(String s[]) { Point firstPoint = new Point(0, 0); Point secondPoint = new Point(5, 0); Point thirdPoint = new Point(0, 5); System.out.println(\"The largest angle of the triangle : \" + getLargestAngle(firstPoint, secondPoint, thirdPoint)); } public static double getLargestAngle(Point firstPoint, Point secondPoint, Point thirdPoint) { //Write code here to get the largest angle of the triangle } //If required write any additional methods here } class Point { double x; double y; Point(double x, double y) { this.x = x; this.y = y; } } ```\n0\nWrong\nScore more than 2 points" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.5538736,"math_prob":0.99857736,"size":2643,"snap":"2021-04-2021-17","text_gpt3_token_len":745,"char_repetition_ratio":0.18794999,"word_repetition_ratio":0.06912442,"special_character_ratio":0.35035944,"punctuation_ratio":0.11728395,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9974079,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-01-18T23:51:36Z\",\"WARC-Record-ID\":\"<urn:uuid:78e71526-a680-4b66-8000-77eaa26b4a4d>\",\"Content-Length\":\"203187\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:afb924ec-2665-4886-b00a-fea76532b295>\",\"WARC-Concurrent-To\":\"<urn:uuid:85efb3cd-6086-4edc-8401-8f4a015102d9>\",\"WARC-IP-Address\":\"138.197.232.42\",\"WARC-Target-URI\":\"https://java.meritcampus.com/core-java-questions/Calculate-the-largest-angle-of-the-given-triangle\",\"WARC-Payload-Digest\":\"sha1:7DQGEQULMVGZO5FLIP4MR4BGN5ZAIBJT\",\"WARC-Block-Digest\":\"sha1:F5VOWCGJZUSF65IRH3XIK2EFL7ESRI26\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-04/CC-MAIN-2021-04_segments_1610703517159.7_warc_CC-MAIN-20210118220236-20210119010236-00361.warc.gz\"}"}
https://answers.everydaycalculation.com/add-fractions/42-18-plus-25-42	[ "Solutions by everydaycalculation.com\n\nAdd 42/18 and 25/42\n\n1st number: 2 6/18, 2nd number: 25/42\n\n42/18 + 25/42 is 41/14.\n\nSteps for adding fractions\n\n1. Find the least common denominator or LCM of the two denominators:\nLCM of 18 and 42 is 126\n2. For the 1st fraction, since 18 × 7 = 126,\n42/18 = 42 × 7/18 × 7 = 294/126\n3. Likewise, for the 2nd fraction, since 42 × 3 = 126,\n25/42 = 25 × 3/42 × 3 = 75/126\n4. Add the two fractions:\n294/126 + 75/126 = 294 + 75/126 = 369/126\n5. After reducing the fraction, the answer is 41/14\n6. In mixed form: 213/14" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.668233,"math_prob":0.99920374,"size":313,"snap":"2019-43-2019-47","text_gpt3_token_len":123,"char_repetition_ratio":0.18446602,"word_repetition_ratio":0.0,"special_character_ratio":0.485623,"punctuation_ratio":0.12328767,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99933267,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-10-14T20:46:14Z\",\"WARC-Record-ID\":\"<urn:uuid:10c26855-d7bc-440a-9f16-e791c760fe0b>\",\"Content-Length\":\"7879\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:bb79cd76-2b7d-4028-92ee-baac8b8a882f>\",\"WARC-Concurrent-To\":\"<urn:uuid:e605be41-22a5-4d2b-9fe2-fc3fa01b5d89>\",\"WARC-IP-Address\":\"96.126.107.130\",\"WARC-Target-URI\":\"https://answers.everydaycalculation.com/add-fractions/42-18-plus-25-42\",\"WARC-Payload-Digest\":\"sha1:5RSGYWGSUOL42IJU4Z3YXWFD3C3WXLI4\",\"WARC-Block-Digest\":\"sha1:ECT2JUN76K2GVURPNJI45M44FNHMLCLJ\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-43/CC-MAIN-2019-43_segments_1570986655310.17_warc_CC-MAIN-20191014200522-20191014224022-00339.warc.gz\"}"}
https://admin.clutchprep.com/physics/practice-problems/38901/an-equilateral-triangle-160-abc-a-positive-point-charge-q-is-located-at-each-of-	[ "# Problem: An equilateral triangle ABC. A positive point charge +q is located at each of the three vertices A, B, and C. Each side of the triangle is of length a. A point charge Q (that may be positive or negative) is placed at the mid-point between B and C. Is it possible to choose the value of Q (that is non-zero) such that the force on Q is zero? A) Yes, because the forces on Q are vectors and three vectors can add to zero. B) No, because the forces on Q are vectors and three vectors can never add to zero. C) Yes, because the electric force at the mid-point between B and C is zero whether a charge is placed there or not. D) No, because the forces on Q due to the charges at B and C point in the same direction. E) No, because a fourth charge would be needed to cancel the force on Q due to the charge at A.\n\n🤓 Based on our data, we think this question is relevant for Professor Schulte's class at UCF.\n\n###### Problem Details\n\nAn equilateral triangle ABC. A positive point charge +q is located at each of the three vertices A, B, and C. Each side of the triangle is of length a. A point charge Q (that may be positive or negative) is placed at the mid-point between B and C. Is it possible to choose the value of Q (that is non-zero) such that the force on Q is zero?\n\nA) Yes, because the forces on Q are vectors and three vectors can add to zero.\n\nB) No, because the forces on Q are vectors and three vectors can never add to zero.\n\nC) Yes, because the electric force at the mid-point between B and C is zero whether a charge is placed there or not.\n\nD) No, because the forces on Q due to the charges at B and C point in the same direction.\n\nE) No, because a fourth charge would be needed to cancel the force on Q due to the charge at A." ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.9600427,"math_prob":0.9924697,"size":900,"snap":"2020-24-2020-29","text_gpt3_token_len":222,"char_repetition_ratio":0.14955357,"word_repetition_ratio":0.12849163,"special_character_ratio":0.23888889,"punctuation_ratio":0.09313726,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9969217,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-06-05T03:10:43Z\",\"WARC-Record-ID\":\"<urn:uuid:ad995984-87e9-4515-89eb-8a94cd984f25>\",\"Content-Length\":\"168977\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:4e509521-e9ff-48bb-912f-ed04779e34cd>\",\"WARC-Concurrent-To\":\"<urn:uuid:85414e8e-eda4-4752-abf3-8510ebb9f461>\",\"WARC-IP-Address\":\"18.210.187.7\",\"WARC-Target-URI\":\"https://admin.clutchprep.com/physics/practice-problems/38901/an-equilateral-triangle-160-abc-a-positive-point-charge-q-is-located-at-each-of-\",\"WARC-Payload-Digest\":\"sha1:ZXTHNELQEKTRSNSLSN4RWDQQ4XYNMOMC\",\"WARC-Block-Digest\":\"sha1:4MDCUE7PD5T2JCZYRC7EDMU7C6OAXAPN\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-24/CC-MAIN-2020-24_segments_1590348492427.71_warc_CC-MAIN-20200605014501-20200605044501-00368.warc.gz\"}"}
https://www.osti.gov/etdeweb/biblio/21217733	[ "You need JavaScript to view this\n\nClosed-form solution of a two-dimensional fuel temperature model for TRIGA-type reactors\n\n## Abstract\n\nIf azimuthal power density variations are ignored, the steady-state temperature distribution within a TRIGA-type fuel element is given by the solution of the Poisson equation in two dimensions (r and z) . This paper presents a closed-form solution of this equation as a function of the axial and radial power density profiles, the conductivity of the U-ZrH, the inlet temperature, specific heat and flow rate of the coolant, and the overall heat transfer coefficient. The method begins with the development of a system of linear ordinary differential equations describing mass and energy balances in the fuel and coolant. From the solution of this system, an expression for the second derivative of the fuel temperature distribution in the axial (z) direction is found. Substitution of this expression into the Poisson equation for T(r,z) reduces it from a partial differential equation to an ordinary differential equation in r, which is subsequently solved in closed-form. The results of typical calculations using the model are presented. (author)\nAuthors:\nRivard, J B \n1. Sandia Laboratories (United States)\nPublication Date:\nJul 01, 1974\nProduct Type:\nConference\nReport Number:\nINIS-US-09N0328; TOC-5\nResource Relation:\nConference: 3. TRIGA owners' conference, Albuquerque, NM (United States), 25-27 Feb 1974; Other Information: Country of input: International Atomic Energy Agency (IAEA); 1 fig; Related Information: In: 3. TRIGA owners' conference. Papers and abstracts, 432 pages.\nSubject:\n21 SPECIFIC NUCLEAR REACTORS AND ASSOCIATED PLANTS; COOLANTS; ENERGY BALANCE; FLOW RATE; HEAT TRANSFER; NUCLEAR FUELS; POISSON EQUATION; POWER DENSITY; SPECIFIC HEAT; STEADY-STATE CONDITIONS; TEMPERATURE DISTRIBUTION; TRIGA TYPE REACTORS; TWO-DIMENSIONAL CALCULATIONS\nOSTI ID:\n21217733\nResearch Organizations:\nGeneral Atomic Co., San Diego, CA (United States)\nCountry of Origin:\nUnited States\nLanguage:\nEnglish\nOther Identifying Numbers:\nTRN: US09N0340086517\nAvailability:\nAvailable from INIS in electronic form\nSubmitting Site:\nINIS\nSize:\npage(s) 2.43-2.49\nAnnouncement Date:\nOct 19, 2009\n\n## Citation Formats\n\nRivard, J B. Closed-form solution of a two-dimensional fuel temperature model for TRIGA-type reactors. United States: N. p., 1974. Web.\nRivard, J B. Closed-form solution of a two-dimensional fuel temperature model for TRIGA-type reactors. United States.\nRivard, J B. 1974. \"Closed-form solution of a two-dimensional fuel temperature model for TRIGA-type reactors.\" United States.\n@misc{etde_21217733,\ntitle = {Closed-form solution of a two-dimensional fuel temperature model for TRIGA-type reactors}\nauthor = {Rivard, J B}\nabstractNote = {If azimuthal power density variations are ignored, the steady-state temperature distribution within a TRIGA-type fuel element is given by the solution of the Poisson equation in two dimensions (r and z) . This paper presents a closed-form solution of this equation as a function of the axial and radial power density profiles, the conductivity of the U-ZrH, the inlet temperature, specific heat and flow rate of the coolant, and the overall heat transfer coefficient. The method begins with the development of a system of linear ordinary differential equations describing mass and energy balances in the fuel and coolant. From the solution of this system, an expression for the second derivative of the fuel temperature distribution in the axial (z) direction is found. Substitution of this expression into the Poisson equation for T(r,z) reduces it from a partial differential equation to an ordinary differential equation in r, which is subsequently solved in closed-form. The results of typical calculations using the model are presented. (author)}\nplace = {United States}\nyear = {1974}\nmonth = {Jul}\n}" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.901989,"math_prob":0.8869553,"size":3965,"snap":"2020-45-2020-50","text_gpt3_token_len":805,"char_repetition_ratio":0.14718506,"word_repetition_ratio":0.9337748,"special_character_ratio":0.19949558,"punctuation_ratio":0.09117221,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.97757244,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-10-25T03:05:04Z\",\"WARC-Record-ID\":\"<urn:uuid:f786ec79-c2f8-4e09-9da4-db444a1a6412>\",\"Content-Length\":\"259168\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:cfb43419-3f44-4037-9c59-4a394943b325>\",\"WARC-Concurrent-To\":\"<urn:uuid:d50974c2-b2fb-4c5d-8878-0188ec1c0664>\",\"WARC-IP-Address\":\"192.107.175.222\",\"WARC-Target-URI\":\"https://www.osti.gov/etdeweb/biblio/21217733\",\"WARC-Payload-Digest\":\"sha1:UM7S5OIWYD7BXOLXCY7MOXNDTHF2RERB\",\"WARC-Block-Digest\":\"sha1:GULMP7JHVU6E7A3DPJI6VFFZN4KP25IW\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-45/CC-MAIN-2020-45_segments_1603107885126.36_warc_CC-MAIN-20201025012538-20201025042538-00140.warc.gz\"}"}
https://docs.bunifuframework.com/en/articles/2281887-stacked-column-100-chart	[ "Stacked column 100 chart is derived from stacked column chart. The difference is that the sum of data series add up to 100%.\n\nStacked column 100 chart is useful when showing part-to-whole proportions over a period of time.\n\nHow to display stacked bar chart using Bunifu Dataviz\n\nSimply locate Bunifu Dataviz control in your toolbox and drag it to the location on your form where you would like to display it.\n\nWe will use button click event handler to display our chart called render_stacked_column_100\n\nC# code\n\n``private void render(){ var canvas = new Bunifu.DataViz.WinForms.Canvas(); var datapoint = new Bunifu.DataViz.WinForms.DataPoint(Bunifu.DataViz.WinForms.BunifuDataViz._type.Bunifu_stackedColumn100); datapoint.addxy(\"new Date (2002,11,10)\", new JArray(16.23, 17.99).ToString()); datapoint.addxy(\"new Date (2002, 11, 9)\", new JArray(15.95, 19.25).ToString()); datapoint.addxy(\"new Date (2002, 11, 8)\", new JArray(11.30, 16.88).ToString()); datapoint.addxy(\"new Date (2002, 11, 7)\", new JArray(13.29, 14.28).ToString()); datapoint.addxy(\"new Date (2002, 11, 6)\", new JArray(15.23, 16.45).ToString()); datapoint.addxy(\"new Date (2002, 11, 5)\", new JArray(13.70, 16.50).ToString()); datapoint.addxy(\"new Date (2002, 11, 4)\", new JArray(17.50, 19.00).ToString()); datapoint.addxy(\"new Date (2002, 11, 3)\", new JArray(19.50, 20.85).ToString()); datapoint.addxy(\"new Date (2002, 11, 2)\", new JArray(20.07, 21.44).ToString()); datapoint.addxy(\"new Date (2002, 11, 1)\", new JArray(25.00, 26.70).ToString()); canvas.addData(datapoint); bunifuDataViz1.Render(canvas);}``\n\nVB.NET code\n\n``Private Sub render() Dim canvas = New Bunifu.DataViz.WinForms.Canvas() Dim datapoint = New Bunifu.DataViz.WinForms.DataPoint(Bunifu.DataViz.WinForms.BunifuDataViz._type.Bunifu_stackedColumn100) datapoint.addxy(\"new Date (2002,11,10)\", New JArray(16.23, 17.99).ToString()) datapoint.addxy(\"new Date (2002, 11, 9)\", New JArray(15.95, 19.25).ToString()) datapoint.addxy(\"new Date (2002, 11, 8)\", New JArray(11.30, 16.88).ToString()) datapoint.addxy(\"new Date (2002, 11, 7)\", New JArray(13.29, 14.28).ToString()) datapoint.addxy(\"new Date (2002, 11, 6)\", New JArray(15.23, 16.45).ToString()) datapoint.addxy(\"new Date (2002, 11, 5)\", New JArray(13.70, 16.50).ToString()) datapoint.addxy(\"new Date (2002, 11, 4)\", New JArray(17.50, 19.00).ToString()) datapoint.addxy(\"new Date (2002, 11, 3)\", New JArray(19.50, 20.85).ToString()) datapoint.addxy(\"new Date (2002, 11, 2)\", New JArray(20.07, 21.44).ToString()) datapoint.addxy(\"new Date (2002, 11, 1)\", New JArray(25.00, 26.70).ToString()) canvas.addData(datapoint) bunifuDataViz1.Render(canvas)End Sub``\n\nIn order to display a Bunifu Stacked Column 100 we need the following controls:\n\n• Bunifu Data Viz - This is the container for our chart\n• Bunifu Canvas - This is the layer between the data viz (container) and the dataset\n• Bunifu Data Point - This contains the data that we want to represent, as pairs of X and Y coordinates\n\nBunifu Stacked Column 100 simply works by creating 2 data point objects, one for the “low” set of points and one for the “high” set of points. The control will know automatically to adjust the width of the lines to match the specified data points and also to fill the entire width of the graph.\n\nOn running the code you should see something like this:", null, "That's it!" ]	[ null, "https://downloads.intercomcdn.com/i/o/73765044/6f00bcd742b73f18cf9f7e17/stacked-column-chart-100%5B1%5D.png", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.6296301,"math_prob":0.916581,"size":3526,"snap":"2019-26-2019-30","text_gpt3_token_len":1076,"char_repetition_ratio":0.22827938,"word_repetition_ratio":0.04524887,"special_character_ratio":0.35110608,"punctuation_ratio":0.26517966,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9765401,"pos_list":[0,1,2],"im_url_duplicate_count":[null,1,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-07-22T08:04:53Z\",\"WARC-Record-ID\":\"<urn:uuid:c54a80e6-cd72-4f29-a0dd-d0761d37ea63>\",\"Content-Length\":\"18849\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:405c2256-fcef-4003-8f68-fc0f5f05b643>\",\"WARC-Concurrent-To\":\"<urn:uuid:a445e8e5-f2cf-465a-a0c4-0876a96a3003>\",\"WARC-IP-Address\":\"45.63.64.75\",\"WARC-Target-URI\":\"https://docs.bunifuframework.com/en/articles/2281887-stacked-column-100-chart\",\"WARC-Payload-Digest\":\"sha1:56NOAVE27QTA7PI7E64QAIGTG3HYERLG\",\"WARC-Block-Digest\":\"sha1:GLKXEBNBWO6ESR6DB4BYU4EMCAEQAYSG\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-30/CC-MAIN-2019-30_segments_1563195527828.69_warc_CC-MAIN-20190722072309-20190722094309-00298.warc.gz\"}"}
https://it.mathworks.com/help/optim/ug/output-functions.html	[ "# Output Functions for Optimization Toolbox™\n\n### What Is an Output Function?\n\nFor some problems, you might want output from an optimization algorithm at each iteration. For example, you might want to find the sequence of points that the algorithm computes and plot those points. To do this, create an output function that the optimization function calls at each iteration. See Output Function and Plot Function Syntax for details and syntax.\n\nThis section shows the solver-based approach to output functions. For the problem-based approach, see Output Function for Problem-Based Optimization.\n\nGenerally, the solvers that can employ an output function are the ones that can take nonlinear functions as inputs. You can determine which solvers can have an output function by looking in the Options section of function reference pages.\n\n### Example: Use an Output Function\n\nThis example shows how to use an output function to monitor the `fmincon` solution process for solving a constrained nonlinear optimization problem. The output function does the following at the end of each `fmincon` iteration:\n\n• Plot the current point.\n\n• Store the current point and its corresponding objective function value in a variable called `history`, and store the current search direction in a variable called `searchdir`. The search direction is a vector that points in the direction from the current point to the next one.\n\nAdditionally, to make the history available outside of the `fmincon` function, perform the optimization inside a nested function that calls `fmincon` and returns the output function variables. For more information about this method of passing information, see Passing Extra Parameters. The `runfmincon` function at the end of this example contains the nested function call.\n\n### Objective and Constraint Functions\n\nThe problem is to minimize the function\n\n`$f\\left(x\\right)=\\mathrm{exp}\\left({x}_{1}\\right)\\left(4{x}_{1}^{2}+2{x}_{2}^{2}+4{x}_{1}{x}_{2}+2{x}_{2}+1\\right)$`\n\nsubject to the nonlinear inequality constraints\n\n`$\\begin{array}{l}{x}_{1}+{x}_{2}-{x}_{1}{x}_{2}\\ge 3/2\\\\ {x}_{1}{x}_{2}\\ge -10.\\end{array}$`\n\nThe `objfun` function nested in `runfmincon` implements the objective function. The `confun` function nested in `runfmincon` implements the constraint function.\n\n### Call Solver\n\nTo obtain the solution to the problem and see the history of the `fmincon` iterations, call the `runfmincon` function.\n\n`[xsol,fval,history,searchdir] = runfmincon;`\n``` Max Line search Directional First-order Iter F-count f(x) constraint steplength derivative optimality Procedure 0 3 1.8394 0.5 Infeasible start point 1 6 1.85127 -0.09197 1 0.109 0.778 ```\n``` 2 9 0.300167 9.33 1 -0.117 0.313 Hessian modified twice ```\n``` 3 12 0.529835 0.9209 1 0.12 0.232 ```\n``` 4 16 0.186965 -1.517 0.5 -0.224 0.13 ```\n``` 5 19 0.0729085 0.3313 1 -0.121 0.054 ```\n``` 6 22 0.0353323 -0.03303 1 -0.0542 0.0271 ```\n``` 7 25 0.0235566 0.003184 1 -0.0271 0.00587 ```\n``` 8 28 0.0235504 9.035e-08 1 -0.0146 8.51e-07 ```\n```Active inequalities (to within options.ConstraintTolerance = 1e-06): lower upper ineqlin ineqnonlin 1 2 ```", null, "```Local minimum found that satisfies the constraints. Optimization completed because the objective function is non-decreasing in feasible directions, to within the value of the optimality tolerance, and constraints are satisfied to within the value of the constraint tolerance. ```\n\nThe output function creates a plot of the points that `fmincon` evaluates. Each point is labeled by its iteration number. The optimal point occurs at the eighth iteration. The last two points in the sequence are so close that they overlap.\n\nThe output `history` is a structure that contains two fields.\n\n`disp(history)`\n``` x: [9x2 double] fval: [9x1 double] ```\n\nThe `fval` field in `history` contains the objective function values corresponding to the sequence of points `fmincon` computes.\n\n`disp(history.fval)`\n``` 1.8394 1.8513 0.3002 0.5298 0.1870 0.0729 0.0353 0.0236 0.0236 ```\n\nThese are the same values displayed in the iterative output in the column with header `f(x)`.\n\nThe `x` field of `history` contains the sequence of points that `fmincon` computes.\n\n`disp(history.x)`\n``` -1.0000 1.0000 -1.3679 1.2500 -5.5708 3.4699 -4.8000 2.2752 -6.7054 1.2618 -8.0679 1.0186 -9.0230 1.0532 -9.5471 1.0471 -9.5474 1.0474 ```\n\nThe second output argument, `searchdir`, contains the search directions for `fmincon` at each iteration. The search direction is a vector pointing from the point computed at the current iteration to the point computed at the next iteration.\n\n`disp(searchdir)`\n``` -0.3679 0.2500 -4.2029 2.2199 0.7708 -1.1947 -3.8108 -2.0268 -1.3625 -0.2432 -0.9552 0.0346 -0.5241 -0.0061 -0.0003 0.0003 ```\n\n### Helper Functions\n\nThe following code creates the `runfmincon` function, containing the `outfun` output function, `objfun` objective function, and `confun` nonlinear constraint function as nested functions,\n\n```function [xsol,fval,history,searchdir] = runfmincon % Set up shared variables with OUTFUN history.x = []; history.fval = []; searchdir = []; % call optimization x0 = [-1 1]; options = optimoptions(@fmincon,'OutputFcn',@outfun,... 'Display','iter','Algorithm','active-set'); [xsol,fval] = fmincon(@objfun,x0,[],[],[],[],[],[],@confun,options); function stop = outfun(x,optimValues,state) stop = false; switch state case 'init' hold on case 'iter' % Concatenate current point and objective function % value with history. x must be a row vector. history.fval = [history.fval; optimValues.fval]; history.x = [history.x; x]; % Concatenate current search direction with % searchdir. searchdir = [searchdir;... optimValues.searchdirection']; plot(x(1),x(2),'o'); % Label points with iteration number and add title. % Add .15 to x(1) to separate label from plotted 'o' text(x(1)+.15,x(2),... num2str(optimValues.iteration)); title('Sequence of Points Computed by fmincon'); case 'done' hold off otherwise end end function f = objfun(x) f = exp(x(1))(4x(1)^2 + 2x(2)^2 + 4x(1)x(2) +... 2x(2) + 1); end function [c, ceq] = confun(x) % Nonlinear inequality constraints c = [1.5 + x(1)x(2) - x(1) - x(2); -x(1)x(2) - 10]; % Nonlinear equality constraints ceq = []; end end```" ]	[ null, "https://it.mathworks.com/help/examples/optim/win64/OutputFunctionsForOptimizationToolboxExample_01.png", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.73550516,"math_prob":0.9910715,"size":5600,"snap":"2021-04-2021-17","text_gpt3_token_len":1576,"char_repetition_ratio":0.16136526,"word_repetition_ratio":0.01953602,"special_character_ratio":0.3092857,"punctuation_ratio":0.18284228,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9884072,"pos_list":[0,1,2],"im_url_duplicate_count":[null,1,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-01-18T16:04:06Z\",\"WARC-Record-ID\":\"<urn:uuid:c104cc7f-bd5a-4782-b064-2ccb416e510c>\",\"Content-Length\":\"80046\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:06b9498f-6429-46cc-ae26-3e58ce4d5758>\",\"WARC-Concurrent-To\":\"<urn:uuid:cafd6e56-c12a-496d-97c5-5dd40ce8b8cd>\",\"WARC-IP-Address\":\"23.10.129.243\",\"WARC-Target-URI\":\"https://it.mathworks.com/help/optim/ug/output-functions.html\",\"WARC-Payload-Digest\":\"sha1:NPC7FSXZ2CXF5KA4CCGZ7AOBP2XNZVMW\",\"WARC-Block-Digest\":\"sha1:QKTFEU7ISBUXK4PB66YYWPE2C2ZWKZBY\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-04/CC-MAIN-2021-04_segments_1610703515075.32_warc_CC-MAIN-20210118154332-20210118184332-00318.warc.gz\"}"}
http://slideplayer.com/slide/4164461/	[ "", null, "# Force Force is a push or pull on an object The object is called the System Force on a system in motion causes change in velocity = acceleration Force is.\n\n## Presentation on theme: \"Force Force is a push or pull on an object The object is called the System Force on a system in motion causes change in velocity = acceleration Force is.\"— Presentation transcript:\n\nForce Force is a push or pull on an object The object is called the System Force on a system in motion causes change in velocity = acceleration Force is a vector, it has direction External World = everything around the system that exerts a force\n\nTypes of Forces Forces Contact When the external world is in contact with the system and exerts a force ex. Hand that pushes, Rope that pulls Field Forces A force exerted on a system without touching the system ex. Gravity, Magnetic force, Interaction between charged particles\n\nAgent An agent is the item in the external world that causes the force to act on the system ex. The hand that holds The rope that pulls The earth The magnet\n\nFree Body Diagram A diagram representing the System, Agent and the Forces acting on the system System is represented by a dot Forces represented by arrows pointing in the direction of the force, away from system Label each force with a subscript Choose direction of +ve usually towards stronger force\n\nFt gF Ball String (force of tension) (force of gravity) Positive Direction\n\nCombining Forces As forces are vectors they can be added just as vectors ex.1 Fa + Fb = F net ex.2 Fa + Fb = Fnet Nonlinear non perpendicular forces are added by adding x and y components of the force vectors\n\nNewton’s First Law An object at rest will remain at rest and an object in motion will remain in motion unless acted on by an external net force Also known as the Law of Inertia Inertia is the tendency of an object to resist change in motion Equilibrium = if the net force on a system is zero the system is in equilibrium speed and direction is unchanged.\n\nCommon Types of Forces Friction F f = Contact force opposing motion of two surfaces, parallel to the surface and opposite to the direction of motion Normal F N = Contact force exerted by system’s surface perpendicular to and away from surface Spring F sp = A restoring force- push or pull of spring exerted on system opposite to displacement\n\nTension F T = pull exerted by a rope on a system, away from the system and parallel to the rope Thrust F THRUST = a force that moves a system in the same direction as the acceleration Weight F g = a field force on a system due to gravity directed to the center of the earth\n\nNewton’s Second Law The acceleration of an object is equal to the net force acting on it divided by the mass of the object a = F/m or F = ma The larger the force the greater the acceleration The greater the mass with the same force the lower the acceleration Force of 1N = 1kg * 1m/s.s\n\nNewton’s Third Law All forces come in pairs equal and opposite in direction ex. A ball on a table and the table on the earth. Forces-ball on table / table on ball -table on earth/ earth on table - ball on earth/ earth on ball\n\nNet free body diagram ball table earth F table on ball F earth on ball\n\nDrag Force and Terminal Velocity Drag Force is the force exerted on an object as it moves the a fluid ex. Air and water as the speed of the object through the fluid increases so does the drag Drag is effected by the fluids viscosity and temperature Terminal velocity is when the drag force equals the force due to gravity no acceleration constant velocity app.(60m/s)\n\nTension Is a force exerted by a string or rope on a system It is assumed the rope has no mass Tension within all points of the rope is equal and opposite to the force exerted by the system’s weight\n\nExample of bucket on rope FTFT F g =mg\n\nNormal Force Is a contact force exerted by a surface on an object. This force is perpendicular to the surface of contact FgFg N\n\nGravitational Force Gravitational force is the mutual attraction between any two bodies in the universe Newton’s Law of Universal Gravitation =every particle in the universe attracts every other particle with a force that is directly proportional to the product of their masses and inversely proportional to the square of the distance between them F g = Gm 1 m 2 /r 2 where G = the universal gravitational constant = 6.67*10 -11 N.m 2 /kg 2\n\nGravity As weight = mg then w = GM E m/r 2 Then g = GM E /r 2 Therefore the larger the mass of an object the larger the effective gravity it generates\n\nForces of Friction f s and f k An object moving on a surface or passing through a fluid experiences resistance to motion = friction F f s = forces of static friction = the force that prevents movement of an object that is being subjected to an external force When movement is about to occur f s is at max.\n\nf s and f k When a force F exceeds F f s max then movement will occur and the new friction force is less. This new friction force is called the Force of kinetic friction f k When F-f k = positive value there is acceleration Both f s and f k are proportional to the normal force; f s =u s n and f k =u k n u s is the coefficient of static friction U k is the coefficient of kinetic friction\n\nSolving Friction Problems Draw a free body diagram making sure to label all forces\n\nDownload ppt \"Force Force is a push or pull on an object The object is called the System Force on a system in motion causes change in velocity = acceleration Force is.\"\n\nSimilar presentations" ]	[ null, "http://slideplayer.com/static/blue_design/img/slide-loader4.gif", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.90303874,"math_prob":0.9857138,"size":5159,"snap":"2019-35-2019-39","text_gpt3_token_len":1160,"char_repetition_ratio":0.15344326,"word_repetition_ratio":0.0030549897,"special_character_ratio":0.2201977,"punctuation_ratio":0.029821074,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99858,"pos_list":[0,1,2],"im_url_duplicate_count":[null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-09-22T03:58:30Z\",\"WARC-Record-ID\":\"<urn:uuid:b85eaf08-c764-4a74-9ca9-f0194bb3e9b2>\",\"Content-Length\":\"172489\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:55b3da37-0753-4cfe-912d-e91f6dc09de1>\",\"WARC-Concurrent-To\":\"<urn:uuid:a6bf5bda-7a47-43e0-a539-51162ca72d3c>\",\"WARC-IP-Address\":\"138.201.58.10\",\"WARC-Target-URI\":\"http://slideplayer.com/slide/4164461/\",\"WARC-Payload-Digest\":\"sha1:4GZ2P55B3OUSGQGWJOTVQZ6SWWCJVIYT\",\"WARC-Block-Digest\":\"sha1:NQHR6WTJCAORR4IA2Y7GYU5U2D2KAMBL\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-39/CC-MAIN-2019-39_segments_1568514575076.30_warc_CC-MAIN-20190922032904-20190922054904-00228.warc.gz\"}"}
https://pyneng.readthedocs.io/en/latest/book/09_functions/func_args_var.html	[ "# Variable length arguments#\n\nSometimes it is necessary to make function accept not a fixed number of arguments, but any number. For such a case, in Python it is possible to create a function with a special parameter that accepts variable length arguments. This parameter can be both keyword and positional.\n\nNote\n\nEven if you don’t use it in your scripts there’s a good chance you’ll find it in someone else’s code.\n\n## Variable length positional arguments#\n\nParameter that takes positional variable length arguments is created by adding an asterisk before parameter name. Parameter can have any name but by agreement `args` is the most common name.\n\nExample of a function:\n\n```In : def sum_arg(a, args):\n....: print(a, args)\n....: return a + sum(args)\n....:\n```\n\nFunction `sum_arg` is created with two parameters:\n\n• parameter `a`\n\n• if passed as positional argument, should be first\n\n• if passed as a keyword argument, the order does not matter\n\n• parameter `args` - expects variable length arguments\n\n• all other arguments as a tuple\n\n• these arguments may be missed\n\nCall a function with different number of arguments:\n\n```In : sum_arg(1, 10, 20, 30)\n1 (10, 20, 30)\nOut: 61\n\nIn : sum_arg(1, 10)\n1 (10,)\nOut: 11\n\nIn : sum_arg(1)\n1 ()\nOut: 1\n```\n\nYou can also create such a function:\n\n```In : def sum_arg(args):\n....: print(args)\n....: return sum(args)\n....:\n\nIn : sum_arg(1, 10, 20, 30)\n(1, 10, 20, 30)\nOut: 61\n\nIn : sum_arg()\n()\nOut: 0\n```\n\n## Keyword variable length arguments#\n\nParameter that accepts keyword variable length arguments is created by adding two asterisk in front of parameter name. Name of parameter can be any but by agreement most commonly use name `kwargs` (from keyword arguments).\n\nExample of a function:\n\n```In : def sum_arg(a, kwargs):\n....: print(a, kwargs)\n....: return a + sum(kwargs.values())\n....:\n```\n\nFunction `sum_arg` is created with two parameters:\n\n• parameter `a`\n\n• if passed as positional argument, should be first\n\n• if passed as a keyword argument, the order does not matter\n\n• parameter `**kwargs` - expects keyword variable length arguments\n\n• all other keyword arguments as a dictionary\n\n• these arguments may be missed\n\nCalling a function with different number of keyword arguments:\n\n```In : sum_arg(a=10, b=10, c=20, d=30)\n10 {'c': 20, 'b': 10, 'd': 30}\nOut: 70\n\nIn : sum_arg(b=10, c=20, d=30, a=10)\n10 {'c': 20, 'b': 10, 'd': 30}\nOut: 70\n```" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.56058854,"math_prob":0.9587207,"size":2340,"snap":"2022-40-2023-06","text_gpt3_token_len":637,"char_repetition_ratio":0.16523972,"word_repetition_ratio":0.20472442,"special_character_ratio":0.31794873,"punctuation_ratio":0.2294455,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.983943,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-01-29T16:43:52Z\",\"WARC-Record-ID\":\"<urn:uuid:8a704436-1891-4936-8730-a93197150a4a>\",\"Content-Length\":\"77583\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:f1361b7e-12c5-47d5-907c-d606a9d2fd9a>\",\"WARC-Concurrent-To\":\"<urn:uuid:c2961475-6478-48a2-8c86-4da8e5b6fda3>\",\"WARC-IP-Address\":\"104.17.33.82\",\"WARC-Target-URI\":\"https://pyneng.readthedocs.io/en/latest/book/09_functions/func_args_var.html\",\"WARC-Payload-Digest\":\"sha1:F3CGT3CZHSWS3PZAYVGAVH7EOMTDAZHR\",\"WARC-Block-Digest\":\"sha1:P2JYFILBFMMMUTHQHAVHKPAX4PNIZYMJ\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-06/CC-MAIN-2023-06_segments_1674764499744.74_warc_CC-MAIN-20230129144110-20230129174110-00539.warc.gz\"}"}
http://www.stats.bris.ac.uk/R/web/packages/energy/news/news.html	[ "energy 1.7-5\n\n• User level changes:\n• kgroups: (new) implements energy clustering for a specified number k classes by energy distance criterion, analogous to the k classes of the k-means algorithm.\n• dcov2d and dcor2d: (new) O(n log n) methods to compute the squared U or V statistics for real x and y\n• sortrank() function added (a utility)\n• Internal changes:\n• B-tree.cpp: Btree_sum and other internal functions implement binary tree search for faster O(n log n) calculation of paired distances in dcov2d\n• kgroups.cpp: Rcpp implementation of k-groups algorithm\n• energy.hclust implementation: replaced C++ code with call to stats::hclust; since R > 3.0.3 it is now equivalent for alpha = 1 with method = “ward.D”. Input and return value unchanged except heights from hclust are half.\n\nenergy 1.7-4\n\n• User level changes\n• disco: handle the case when the user argument x is dist with conflicting argument distance=FALSE\n• dcor.t and dcor.ttest: handle the cases when class of argument x or y conflicts with the distance argument\n• Split manual page of dcovU into two files.\n• indep.etest and indep.e removed now Defunct (were Deprecated since Version 1.1-0, 2008-04-07; replaced by indep.test).\n• Internal changes\n• BCDCOR: handle the cases when class of argument x or y conflicts with the distance argument\n\nenergy 1.7-2\n\n• User level changes\n• Provided new dcor.test function, similar to dcov.test but using the distance correlation as the test statistic.\n• Number of replicates R for Monte Carlo and permutation tests now matches the argument of the boot::boot function (no default value, user must specify).\n• If user runs a test with 0 replicates, p-value printed is NA\n• Internal changes\n• energy_init.c added for registering routines\n\nenergy 1.7-0\n\n• Partial Distance Correlation statistics and tests added\n• pdcov, pdcor, pdcov.test, pdcor.test\n• dcovU: unbiased estimator of distance covariance\n• bcdcor: bias corrected distance correlation\n• Ucenter, Dcenter, U_center, D_center: double-centering and U-centering utilities\n• U_product: inner product in U-centered Hilbert space\n• updated NAMESPACE and DESCRIPTION imports, etc.\n• revised package Title and Description in DESCRIPTION\n• package now links to Rcpp\n• mvnorm c code ported to c++ (mvnorm.cpp); corresponding changes in Emvnorm.R\n• syntax for bcdcor: “distance” argument removed, now argument can optionally be a dist object\n• syntax for energy.hclust: first argument must now be a dist object\n• default number of replicates R in tests: for all tests, R now defaults to 0 or R has no default value.\n\nenergy 1.6.2\n\n• inserted GetRNGstate() .. PutRNGState around repl. loop in dcov.c.\n\nenergy 1.6.1\n\n• replace Depends with Imports in DESCRIPTION file\n\nenergy 1.6.0\n\n• implementation of high-dim distance correlation t-test introduced in JMVA Volume 117, pp. 193-213 (2013).\n• new functions dcor.t, dcor.ttest in dcorT.R\n• minor changes to tidy other code in dcov.R\n• removed unused internal function .dcov.test\n\nenergy 1.5.0\n\n• NAMESPACE: insert UseDynLib; remove zzz.R, .First.Lib()\n\nenergy 1.4-0\n\n• (dcov.c, Eindep.c) Unused N was removed.\n• (dcov.c) In case dcov=0, bypass the unnecessary loop that generates replicates (in dCOVtest and dCovTest). In this case dcor=0 and test is not significant. (dcov=0 if one of the samples is constant.)\n• (Eqdist.R) in eqdist.e and eqdist.etest, method=“disco” is replaced by two options: “discoB” (between sample components) and “discoF” (disco F ratio).\n• (disco.R) Added disco.between and internal functions that compute the disco between-sample component and corresponding test.\n• (utilities.c) In permute function replaced rand_unif with runif.\n• (energy.c) In ksampleEtest the pval computation changed from ek/B to (ek+1)/(B+1) as it should be for a permutation test, and unneeded int* n removed.\n\nenergy 1.3-0\n\n• In distance correlation, distance covariance functions (dcov, dcor, DCOR) and dcov.test, arguments x and y can now optionally be distance objects (result of dist function or as.dist). Matrices x and y will always be treated as data.\n\n• Functions in dcov.c and utilities.c were modified to support arguments that are distances rather than data. In utilities.c the index_distance function changed. In dcov.c there are many changes. Most importantly for the exported objects, there is now an extra required parameter in the dims argument passed from R. In dCOVtest dims must be a vector c(n, p, q, dst, R) where n is sample size, p and q are dimensions of x and y, dst is logical (TRUE if distances) and R is number of replicates. For dCOV dims must be c(n, p, q, dst).\n\nenergy 1.2-0\n\n• disco (distance components) added for one-way layout.\n• A method argument was added to ksample.e, eqdist.e, and eqdist.etest, method = c(“original”, “disco”).\n• A method argument was added to edist, which summarizes cluster distances in a table: method = c(“cluster”,“discoB”,“discoF”))" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.871029,"math_prob":0.8390565,"size":763,"snap":"2019-26-2019-30","text_gpt3_token_len":182,"char_repetition_ratio":0.12648222,"word_repetition_ratio":0.0,"special_character_ratio":0.22280473,"punctuation_ratio":0.16666667,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.974283,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-06-17T18:59:18Z\",\"WARC-Record-ID\":\"<urn:uuid:7d969e9c-bbb2-483a-9270-3bce64a9f82a>\",\"Content-Length\":\"6894\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:2ed90067-2f7a-4133-887c-0a3003010c73>\",\"WARC-Concurrent-To\":\"<urn:uuid:19aa37ad-2f88-48dd-ab55-4f086d479d1c>\",\"WARC-IP-Address\":\"137.222.10.189\",\"WARC-Target-URI\":\"http://www.stats.bris.ac.uk/R/web/packages/energy/news/news.html\",\"WARC-Payload-Digest\":\"sha1:Y2ENK3NHYEWUV35HGEWXRY6CDLOH6DO3\",\"WARC-Block-Digest\":\"sha1:BBEGWOOTRW6O4GNPNAZMGW7D2DWQTH7M\",\"WARC-Identified-Payload-Type\":\"application/xhtml+xml\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-26/CC-MAIN-2019-26_segments_1560627998558.51_warc_CC-MAIN-20190617183209-20190617205209-00321.warc.gz\"}"}
http://www.vbaexpress.com/forum/showthread.php?66460-Dynamic-Function&s=ec18fa000b3797713e6d93b168b53bbe&p=397577	[ "1. ## Dynamic Function\n\nHello! I want to create a dynamic function in VBA wherein I provide the name range, column number and row number. As per the values provided, the range should be hard pasted. This is what I did but it is giving an error:\n\n```Public Function HardPaste(Table_Range As Variant, Col_Num As Integer, Row_Num As Integer)\nTable_Range(Row_Num, Col_Num).Select\nSelection.PasteSpecial Paste:=xlPasteValues\nEnd Function```\nThis is how I am using it:\n\n```Sub Test()\nHardPaste(CQ_Table_Range,1,4)\nEnd Sub```\n\nThe error I am getting is Compile Error: Expected :=", null, "", null, "Reply With Quote\n\n2. I tried another this and it worked:\n\n```Public Function HardPaste(Table_Range As Range, Col_Num As Integer, Row_Num As Integer)\nRange(Table_Range(Row_Num, Col_Num), Table_Range(Table_Range.Rows.Count, Table_Range.Columns.Count)).Copy\nTable_Range(Row_Num, Col_Num).PasteSpecial Paste:=xlPasteValues\nApplication.CutCopyMode = False\nEnd Function```\nThanks!", null, "", null, "Reply With Quote\n\n####", null, "Posting Permissions\n\n• You may not post new threads\n• You may not post replies\n• You may not post attachments\n• You may not edit your posts\n•" ]	[ null, "http://www.vbaexpress.com/forum/images/misc/progress.gif", null, "http://www.vbaexpress.com/forum/clear.gif", null, "http://www.vbaexpress.com/forum/images/misc/progress.gif", null, "http://www.vbaexpress.com/forum/clear.gif", null, "http://www.vbaexpress.com/forum/images/buttons/collapse_40b.png", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.63373816,"math_prob":0.4933194,"size":693,"snap":"2023-40-2023-50","text_gpt3_token_len":178,"char_repetition_ratio":0.11030479,"word_repetition_ratio":0.0,"special_character_ratio":0.22799423,"punctuation_ratio":0.19548872,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9812579,"pos_list":[0,1,2,3,4,5,6,7,8,9,10],"im_url_duplicate_count":[null,null,null,null,null,null,null,null,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-12-01T03:46:43Z\",\"WARC-Record-ID\":\"<urn:uuid:6d7e2da1-44b8-474c-9ee1-9fd6b1cf58e1>\",\"Content-Length\":\"41872\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:8ee2953f-4ef1-4762-bf1a-2e19f233c1e0>\",\"WARC-Concurrent-To\":\"<urn:uuid:c2c721a1-29b9-4158-9d94-22fee00886f4>\",\"WARC-IP-Address\":\"67.212.188.2\",\"WARC-Target-URI\":\"http://www.vbaexpress.com/forum/showthread.php?66460-Dynamic-Function&s=ec18fa000b3797713e6d93b168b53bbe&p=397577\",\"WARC-Payload-Digest\":\"sha1:KTVBC7JHXUNKJBVTXFW4ANTXZREZKFYC\",\"WARC-Block-Digest\":\"sha1:OOXZMB7B2KAGTKCTKRP3R5HQSZHJ63QO\",\"WARC-Identified-Payload-Type\":\"application/xhtml+xml\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-50/CC-MAIN-2023-50_segments_1700679100264.9_warc_CC-MAIN-20231201021234-20231201051234-00277.warc.gz\"}"}
https://brilliant.org/problems/turn-it-this-way/	[ "# Turn it this way\n\nAlgebra Level 1\n\nThe point $\\left(4+7\\sqrt{3}\\ ,\\ 7-4\\sqrt{3}\\right)$ is rotated $\\dfrac{\\pi}{3}$ radians counterclockwise about the origin. If the resulting image is $(a,b)$, then what is $a+b$?\n\n×\n\nProblem Loading...\n\nNote Loading...\n\nSet Loading..." ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.7752435,"math_prob":0.9998455,"size":270,"snap":"2021-21-2021-25","text_gpt3_token_len":62,"char_repetition_ratio":0.15037593,"word_repetition_ratio":0.29268292,"special_character_ratio":0.22962964,"punctuation_ratio":0.29508197,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.98810893,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-06-24T20:46:35Z\",\"WARC-Record-ID\":\"<urn:uuid:dac4c274-526c-4035-8c0f-2243a47df25a>\",\"Content-Length\":\"41392\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:cff3a1d3-aba3-4f26-a295-4c3ac519e201>\",\"WARC-Concurrent-To\":\"<urn:uuid:e2249c3d-2cff-4f12-b1db-4ca1119502c8>\",\"WARC-IP-Address\":\"104.20.34.242\",\"WARC-Target-URI\":\"https://brilliant.org/problems/turn-it-this-way/\",\"WARC-Payload-Digest\":\"sha1:5LXCGF2OGQEZPBVMKA6RJZO276GUKFDN\",\"WARC-Block-Digest\":\"sha1:UR5FQZ5A4AO4MXSPAGK2L6KSA6OFVFAA\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-25/CC-MAIN-2021-25_segments_1623488559139.95_warc_CC-MAIN-20210624202437-20210624232437-00500.warc.gz\"}"}
https://studylib.net/doc/10408833/february-21--2007-lecturer-dmitri-zaitsev-hilary-term-2007	[ "# February 21, 2007 Lecturer Dmitri Zaitsev Hilary Term 2007", null, "```February 21, 2007\nLecturer Dmitri Zaitsev\nHilary Term 2007\nCourse 2E1 2006-07 (SF Engineers & MSISS & MEMS)\nS h e e t 16\nDue: at the end of the tutorial\nExercise 2(ii)\nFind an equation for the plane spanned by the vectors: u = (1, −1, 1), v = (−1, 0, 1);\nSolution The span is the set of all linear combinations of u and v, i.e. the set of\nall expressions k1 u + k2 v, hence the set of all vectors (x, y, z) that can be written as\nk1 u+k2 v, hence the set of all (x, y, z) for which the following vector equation is solvable\nin (k1 , k2 ):\n\n\n\n  \n1\n−1\nx\n\n\n\n\n\nk1 −1 + k2\n0\n= y .\n1\n1\nz\nThis vector equation is equivalent to the system\n(\nk1 − k2 = x\n−k1 = y\nk1 + k2 = z\nfrom which we can eliminate both k1 and then k2 :\n(\nk1 = −y\nk2 = k1 − x\n−y + k2 = z\n(\nk1 = −y\nk2 = −y − x\n−2y − x = z\nor\nThe system is solvable in (k1 , k2 ) precisely when −2y − x = z or\nx + 2y + z = 0,\nwhich is the desired equation.\nIt is a good idea to check the correctness of this equation by substituting into it\n(x, y, z) = u = (1, −1, 1) and (x, y, z) = v = (−1, 0, 1) (the given vectors):\n1 + 2 · (−1) + 1 = 0,\n(−1) + 2 · 0 + 1 = 0.\n```" ]	[ null, "https://s2.studylib.net/store/data/010408833_1-be7b2e18dcd37bf63ba2b257ad0b631d-768x994.png", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.8701454,"math_prob":0.9999212,"size":1095,"snap":"2021-43-2021-49","text_gpt3_token_len":421,"char_repetition_ratio":0.12190651,"word_repetition_ratio":0.05860806,"special_character_ratio":0.43561643,"punctuation_ratio":0.133829,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9999713,"pos_list":[0,1,2],"im_url_duplicate_count":[null,1,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-12-07T19:20:46Z\",\"WARC-Record-ID\":\"<urn:uuid:5dccd0a8-fa1d-495f-aaee-fae3b7483a74>\",\"Content-Length\":\"48872\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:1cd9186e-0eed-4c3b-a97b-5ac53f787803>\",\"WARC-Concurrent-To\":\"<urn:uuid:409bead5-cc73-4941-aee0-893450af3d5a>\",\"WARC-IP-Address\":\"172.67.175.240\",\"WARC-Target-URI\":\"https://studylib.net/doc/10408833/february-21--2007-lecturer-dmitri-zaitsev-hilary-term-2007\",\"WARC-Payload-Digest\":\"sha1:X5NQ3BF7OWTJDHKZEGVO2P57K6DNY4VH\",\"WARC-Block-Digest\":\"sha1:VU3AZMWA65JMBC4XLEZJSLDS2CBDKTGX\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-49/CC-MAIN-2021-49_segments_1637964363405.77_warc_CC-MAIN-20211207170825-20211207200825-00065.warc.gz\"}"}
https://math.stackexchange.com/questions/65531/what-is-the-motivation-for-defining-both-homogeneous-and-inhomogeneous-cochains	[ "# What is the motivation for defining both homogeneous and inhomogeneous cochains?\n\nIn my few months of studying group cohomology, I've seen two \"standard\" complexes that are introduced:\n\n• We let $X_r$ be the free $\\mathbb{Z}[G]$-module on $G^r$ (so, it has as a $\\mathbb{Z}[G]$-basis the $r$-tuples $(g_1,\\ldots,g_r)$ of elements of $G$). The $G$-module structure of $X_r$ comes by virtue of being a $\\mathbb{Z}[G]$-module.\nThe boundary maps $\\partial_r:X_r\\to X_{r-1}$ are\n\n$$\\partial_{r}(g_1,\\ldots,g_r)=g_1(g_2,\\ldots,g_r)+\\sum_{j=1}^r(-1)^j(g_1,\\ldots,g_jg_{j+1},\\ldots,g_r)+(-1)^r(g_1,\\ldots,g_r)$$\n\n• We let $E_r$ be the free $\\mathbb{Z}$-module on $G^{r+1}$ (so, it has as a $\\mathbb{Z}$-basis the $(r+1)$-tuples $(g_0,\\ldots,g_r)$ of elements of $G$). The $G$-module structure of $E_r$ is defined by $g(g_0,\\ldots,g_r)=(gg_0,\\ldots,gg_r)$. The boundary maps $d_r:E_r\\to E_{r-1}$ are\n$$d_{r}(g_0,\\ldots,g_r)=\\sum_{j=0}^{r}(-1)^j(g_0,\\ldots,\\widehat{g_j},\\ldots,g_{r})$$\n\nWe may then proceed to compute the cohomology of $G$ with coefficients in a $G$-module $A$ using $$0\\to \\text{Hom}_G(X_0,A)\\to\\text{Hom}_G(X_1,A)\\to\\cdots$$ or by using $$0\\to \\text{Hom}_G(E_0,A)\\to\\text{Hom}_G(E_1,A)\\to\\cdots$$ Elements of $\\text{Hom}_G(X_r,A)$ are \"inhomogeneous cochains\" and elements of $\\text{Hom}_G(E_r,A)$ are \"homogeneous cochains\". In either case, all that matters is what happens to the basis elements, so really we can say that an \"inhomogeneous cochain\" is a function $f:G^r\\to A$, and that a \"homogeneous cochain\" is a function $f:G^{r+1}\\to A$ that satisfies $f(gg_0,\\ldots,gg_r)=g\\cdot f(g_0,\\ldots,g_r)$.\n\nLang defines them both in his Topics in cohomology of groups, and says\n\n... we have a $\\mathbb{Z}[G]$-isomorphism $X\\xrightarrow{\\approx}E$ between the non-homogeneous and the homogeneous complex uniquely determined by the value on basis elements such that $$(\\sigma_1,\\ldots,\\sigma_r)\\mapsto (e,\\sigma_1,\\sigma_1\\sigma_2,\\ldots,\\sigma_1\\sigma_2\\ldots \\sigma_r)$$\n\nbut Serre defines the only the homogenous cochains in Local Fields and then says that a cochain\n\n... is uniquely determined by its restriction to systems of the form $(1,g_1,g_1g_2,\\ldots,g_1\\cdots g_i)$. That leads us to interpret the elements of $\\text{Hom}_G(E_r,A)$ as \"inhomogeneous cochains\", i.e. as functions $f(g_1,\\ldots,g_i)$ of $i$ arguments, with values in $A$, whose coboundary is given by ...\n\nTo put it bluntly, my question is: Why are we doing this? I can think of some possible reasons:\n\n• Historical - perhaps one way was defined first, now the other is more popular, but the older definition is still included out of tradition.\n\n• Practical - perhaps there are important computations that are significantly easier to see or do using one or the other approach, or where it is useful to switch between them for some reason.\n\n• Big picture - perhaps there is a high-level interpretation of one or both approaches that ties in with some other field where (co)homology plays a role.\n\nSo, what's the real motivation for defining both \"homogeneous\" and \"inhomogeneous\" cochains?\n\n• I don't know what the real answer is, but I'll just point out that the inhomogeneous definition has the advantage that it's defined on a tuple of smaller size (r instead of r+1), while the homogeneous definition has the advantage that the boundary map is \"simpler\" (omitting a term rather than combining via a product). – Ted Sep 18 '11 at 17:22\n• The second complex makes it obvious that we are dealing with the homology of a simplicial set. Cycles and cocycles on the first one, on the other hand, are what we usually find in nature (derivations, factor sets in extensions, etc) – Mariano Suárez-Álvarez Sep 18 '11 at 18:29\n• @Mariano: I'm afraid I don't really understand your explanation of the first complex; Serre, for example, talks about factor sets using homogeneous cochains. Could expand a bit more on the places we see inhomogeneous cochains \"in nature\" (in an answer, if you want)? – Zev Chonoles Sep 18 '11 at 22:55\n• By the way, one should keep in mind the fact that the two complexes are in fact isomorphic! – Mariano Suárez-Álvarez Sep 23 '11 at 3:27\n• @Mariano: Indeed - that was one of the reasons I found the situation a bit perplexing. – Zev Chonoles Sep 23 '11 at 3:38\n\n## 2 Answers\n\nLet me give you three examples where nature picks your first complex:\n\nFirst: Let $G$ be a group, let $A$ be a $G$-module, and let $A\\rtimes G$ be the direct product (as a set, this is $A\\times G$, and it becomes a group with multiplication such that $(a,g)\\cdot(b,h)=(a+g\\cdot b,gh)$ for all $a$, $b\\in A$ and all $g$, $h\\in G$) Consider the projection map $p:A\\rtimes G\\to G$, which is a group homomorphism. A section of $p$ is a group homomorphism $s:G\\to A\\rtimes G$ such that $f\\circ s=\\mathrm{id}_G$. It is immediate to check that a section determines uniquely and is determined uniquely by a function $\\sigma:G\\to A$ such that $$g\\cdot\\sigma(h)+\\sigma(g)=\\sigma(gh)$$ for all $g$, $h\\in G$; indeed, the relation between $s$ and $\\sigma$ is that $s(g)=(\\sigma(g),g)$ for all $g\\in G$. The function $\\sigma$ is then a $1$-cocycle defined on your first complex.\n\nSecond: Consider an extension\n\n i f\n0 ---> A ---> E ---> G ---> 1\n\n\nof a group $G$ by an abelian group $A$ (whose operation I'll write $+$). Let $\\sigma:G\\to E$ be a set-theoretic section of $f$. For all $g$, $h\\in G$ we have $f(\\sigma(g)\\sigma(h))=f(\\sigma(g))f(\\sigma(h))=gh=f(\\sigma(gh))$, so that there exists a unique element $\\alpha(g,h)\\in A$ such that $$\\sigma(g)\\sigma(h)=\\iota(\\alpha(g,h))\\sigma(gh).$$\n\nThere is an action of $G$ on $A$ such that $$\\iota(g\\cdot a)=\\sigma(g)\\iota(a)\\sigma(g)^{-1}$$ for all $g\\in G$ and all $a\\in A$. It is eeasy to check that this is indeed an action of $G$ on $A$ by group automorphisms (here it is where we need that $A$ be abelian) In other words, $A$ is a $G$-module.\n\nNow, whenever $g$, $h$, $k$ are in $G$ we have $$(\\sigma(g)\\sigma(h))\\sigma(k)=\\iota(\\alpha(g,h))\\sigma(gh)\\sigma(k)=\\iota(\\alpha(g,h)+\\alpha(gh,k))\\sigma(ghk)$$ and $$\\sigma(g)(\\sigma(h)\\sigma(k))=\\sigma(g)\\iota(\\alpha(h,k))\\sigma(hk)=\\iota(g\\cdot\\alpha(h,k))\\sigma(g)\\sigma(hk)=\\iota(g\\cdot\\alpha(h,k)+\\alpha(g,hk))\\sigma(ghk).$$ Since multiplication in $G$ is associative, the left-hand sides in these last two equations are equal, so so are their right hand sides---and since $\\iota$ is injective, we see that $$g\\cdot\\alpha(h,k)+\\alpha(g,hk)=\\alpha(g,h)+\\alpha(gh,k)$$ or, equivalently, that $$g\\cdot\\alpha(h,k)-\\alpha(gh,k)+\\alpha(g,hk)-\\alpha(g,h)=0.$$ This means that $\\alpha$ determines a $2$-cocyle on your first the complex.\n\nThird: If $G$ is a group, the category of $G$-modules over a field $k$ is a monoidal category $\\mathscr M_G$ with respect to the tensor product of representations. If $\\alpha:G\\times G\\times G\\to k^\\times$ is a $3$-cocycle defined on your first complex and with values in the multiplicative group of $k$, then one can \"twist\" the associativity isomorphisms of $\\mathscr M_G$ using $\\alpha$ to obtain a different, slightly more fun monoidal category $\\mathscr M_G(\\alpha)$, and if you work this out in detail, you will see that again the cocycle condition with respect to your first complex is precisely the pentagon condition for a monoidal structure.\n\nThese are just three instances where nature picks inhomopgenous cochains.\n\nThe inhomogeneous cochain construction is a standard free resolution of $\\mathbb Z$ (the trivial $G$-module) as a $\\mathbb Z[G]$-module, and it is explicitly constructed to be such. Since taking $G$-invariants is the same as forming $Hom_G(\\mathbb Z, A)$, this is what is needed to compute derived functors of this operation (which is what group cohomology is, from a derived functor point of view).\n\nOn the other hand, homogenous cochains are what you get if you compute the cohomology of local systems (twisted coefficients) on the classifying space for $G$, which is how group cohomology first arose (explicitly --- there were implicit examples of group cohomology classes much earlier) in the literature. The \"homogeneity\" reflects the fact that we are computing with a certain $G$-equivariant simplicial complex.\n\nLoosely, and roughly, speaking, the inhomogeneous picture is more algebraic, and the homogeneous picture is more topological.\n\n• The homogeneous cocycles are very similar to the forming of a boundary of a topological CW-complex. It therefore gives intuition that the homogeneous approach is more useful in algebraic topology. In fact, the homogeneous cocyles formula is the same as Cech cohomology formula when considering cohomology of sheaves. – LinAlgMan Aug 5 '13 at 13:08" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.87894183,"math_prob":0.99957436,"size":2972,"snap":"2021-21-2021-25","text_gpt3_token_len":986,"char_repetition_ratio":0.13443395,"word_repetition_ratio":0.026041666,"special_character_ratio":0.3129206,"punctuation_ratio":0.1446541,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99997497,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-06-24T03:24:34Z\",\"WARC-Record-ID\":\"<urn:uuid:13dc86eb-0bd6-45bf-b177-ae8eecf6fbf7>\",\"Content-Length\":\"185685\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:d3d36a48-c237-4c2f-acd1-1440a0fdc62e>\",\"WARC-Concurrent-To\":\"<urn:uuid:ac49a988-7a85-4024-8f3e-35fc7560dd8e>\",\"WARC-IP-Address\":\"151.101.193.69\",\"WARC-Target-URI\":\"https://math.stackexchange.com/questions/65531/what-is-the-motivation-for-defining-both-homogeneous-and-inhomogeneous-cochains\",\"WARC-Payload-Digest\":\"sha1:SLW2OQF6N2EQBMCCDRZXR6OEEHJRJ3VI\",\"WARC-Block-Digest\":\"sha1:AFPRR773MHNCBXIHADYQ55BXQILO3SLQ\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-25/CC-MAIN-2021-25_segments_1623488550571.96_warc_CC-MAIN-20210624015641-20210624045641-00205.warc.gz\"}"}
https://azurecloudai.blog/2021/04/29/how-to-get-ueba-costs-for-azure-sentinel/	[ "# How to Get UEBA Costs for Microsoft Sentinel\n\nThe cost for UEBA is nominal and based on the amount of data that is analyzed. Your costs will vary depending on several factors. However, the following KQL query can be used to get the estimated cost of the solution.\n\n``````union withsource=TableName1 \n\| where TimeGenerated > ago(30d) //In the last 30 days\n\| summarize Entries = count(), Size = sumif(_BilledSize, _IsBillable == true), last_log = datetime_diff(\"second\", now(), max(TimeGenerated)), estimate = sumif(_BilledSize, _IsBillable == true) by TableName1, _IsBillable\n\| project ['Table Name'] = TableName1, ['Table Size'] = Size, ['% of Total GiB'] = (Size / (1024 1024 * 1024)) / 10.04 * 100, ['IsBillable'] = _IsBillable, ['Last Record Received'] = last_log, ['Estimated Table Price'] = (estimate / (1024 * 1024 * 1024)) * 4.0 //Cost may be different. Then, alter the 4.0\n\| where ['Table Name'] == \"BehaviorAnalytics\" or ['Table Name'] == \"IdentityInfo\" or ['Table Name'] == \"UserAccessAnalytics\" or ['Table Name'] == \"UserPeerAnalytics\" //The four data tables utilized by UEBA\n\| serialize TotalCost=row_cumsum(['Estimated Table Price']) //This starts at the bottom result row and adds all table prices for a final TotalCost value at the top\n\| order by ['Table Size'] desc``````\n\nThis is something I slapped together quickly and is a work in progress.\n\nThe latest version of this KQL query will always be located at: https://github.com/rod-trent/Azure-Sentinel-Cost-Troubleshooting-Kit/blob/main/KQL-Queries/UEBACosts.yaml\n\nThis query takes the billable results of the four UEBA tables (BehaviorAnalytics, IdentityInfo, UserAccessAnalytics, and UserPeerAnalytics) and then also adds a Total Cost column. The top value in the results will show the sum of costs.\n\n•", null, "" ]	[ null, "https://i0.wp.com/azurecloudai.blog/wp-content/uploads/2022/09/6mdm.jpg", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.77961576,"math_prob":0.79937303,"size":1930,"snap":"2022-40-2023-06","text_gpt3_token_len":498,"char_repetition_ratio":0.124610595,"word_repetition_ratio":0.013986014,"special_character_ratio":0.28704664,"punctuation_ratio":0.10759494,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.98551077,"pos_list":[0,1,2],"im_url_duplicate_count":[null,4,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-02-09T05:48:54Z\",\"WARC-Record-ID\":\"<urn:uuid:fc65ee1f-b4ca-48c0-a819-3d3ce54a0cdc>\",\"Content-Length\":\"89950\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:34704512-a201-4c23-b5f1-44050950af33>\",\"WARC-Concurrent-To\":\"<urn:uuid:ca756cc2-b9b8-4c28-9ef9-700a5058cc71>\",\"WARC-IP-Address\":\"192.0.78.249\",\"WARC-Target-URI\":\"https://azurecloudai.blog/2021/04/29/how-to-get-ueba-costs-for-azure-sentinel/\",\"WARC-Payload-Digest\":\"sha1:HKEUSLHMOP6EIHN4WYNPYIW7LDDWIQTS\",\"WARC-Block-Digest\":\"sha1:EV3YYPJ5XPU2OME4HYT7ARDQZJGNMQY2\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-06/CC-MAIN-2023-06_segments_1674764501407.6_warc_CC-MAIN-20230209045525-20230209075525-00766.warc.gz\"}"}
https://picos-api.gitlab.io/picos/api/picos.expressions.cone_psd.html	[ "# picos.expressions.cone_psd¶\n\nImplements the positive semidefinite cone.\n\nClasses\n\nclass picos.expressions.cone_psd.PositiveSemidefiniteCone(dim=None)[source]\n\nThe positive semidefinite cone.\n\nUnlike other cones which are defined only on", null, ", this cone accepts both symmetric and hermitian matrices as well as their special vectorization as members.\n\nExample\n\n>>> from picos import Constant, PositiveSemidefiniteCone\n>>> R = Constant(\"R\", range(16), (4, 4))\n>>> S = R + R.T\n>>> S.shape\n(4, 4)\n>>> S.svec.shape\n(10, 1)\n>>> S.svec << PositiveSemidefiniteCone() # Constrain the matrix via svec().\n<4×4 LMI Constraint: R + Rᵀ ≽ 0>\n>>> C = S << PositiveSemidefiniteCone(); C # Constrain the matrix directly.\n<4×4 LMI Constraint: R + Rᵀ ≽ 0>\n>>> C.conic_membership_form # The conic form still refers to svec().\n<10×1 Real Constant: svec(R + Rᵀ)>\n>>> C.conic_membership_form\n<10-dim. Positive Semidefinite Cone: {svec(A) : xᵀ·A·x ≥ 0 ∀ x}>\n\n__init__(dim=None)[source]\n\nConstruct a PositiveSemidefiniteCone.\n\nIf a fixed dimensionality is given, this must be the dimensiona of the special vectorization. For a", null, "matrix, this is", null, ".\n\nproperty dual_cone\n\nImplement cone.Cone.dual_cone." ]	[ null, "https://picos-api.gitlab.io/picos/_images/math/28f7a0ae3445b6714ccd5b31a1e270a080048fc5.svg", null, "https://picos-api.gitlab.io/picos/_images/math/6eb743c8bde71d3860a46dbd197ee7674c4961ea.svg", null, "https://picos-api.gitlab.io/picos/_images/math/28f5cc1e342341a448780aaf7f6ee383fc019e34.svg", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.6077443,"math_prob":0.8239579,"size":1204,"snap":"2022-27-2022-33","text_gpt3_token_len":375,"char_repetition_ratio":0.14916667,"word_repetition_ratio":0.061728396,"special_character_ratio":0.2857143,"punctuation_ratio":0.20657277,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9743908,"pos_list":[0,1,2,3,4,5,6],"im_url_duplicate_count":[null,2,null,6,null,2,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-07-01T19:58:03Z\",\"WARC-Record-ID\":\"<urn:uuid:5c6f3f02-7b92-4191-b419-0380b5b68996>\",\"Content-Length\":\"20380\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:8d7d5280-65ce-4b20-a128-b914001d0f4c>\",\"WARC-Concurrent-To\":\"<urn:uuid:0501cc67-b254-4460-b2b5-f0605173dd13>\",\"WARC-IP-Address\":\"35.185.44.232\",\"WARC-Target-URI\":\"https://picos-api.gitlab.io/picos/api/picos.expressions.cone_psd.html\",\"WARC-Payload-Digest\":\"sha1:AGSQ3LGPGDHWIPGCRAA6N747K5WLTLOX\",\"WARC-Block-Digest\":\"sha1:L5ZNUZHWTNAVO2XNQ4UL3DMPZ3YNCOY4\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-27/CC-MAIN-2022-27_segments_1656103945490.54_warc_CC-MAIN-20220701185955-20220701215955-00082.warc.gz\"}"}
https://www.scholars.northwestern.edu/en/publications/on-the-taylor-expansion-of-value-functions	[ "# On the taylor expansion of value functions\n\nResearch output: Contribution to journalArticlepeer-review\n\n7 Scopus citations\n\n## Abstract\n\nWe introduce a framework for approximate dynamic programming that we apply to discrete-time chains on ℤd+ with countable action sets. The framework is grounded in the approximation of the (controlled) chain's generator by that of another Markov process. In simple terms, our approach stipulates applying a second-order Taylor expansion to the value function, replacing the Bellman equation with one in continuous space and time in which the transition matrix is reduced to its first and second moments. In some cases, the resulting equation can be interpreted as a Hamilton-Jacobi-Bellman equation for a Brownian control problem. When tractable, the \"Taylored\"equation serves as a useful modeling tool. More generally, it is a starting point for approximation algorithms. We develop bounds on the optimality gap-the suboptimality introduced by using the control produced by the Taylored equation. These bounds can be viewed as a conceptual underpinning, analytical rather than relying on weak convergence arguments, for the good performance of controls derived from Brownian approximations. We prove that under suitable conditions and for suitably \"large\"initial states, (1) the optimality gap is smaller than a 1 - α fraction of the optimal value, with which α ∈ (0, 1) is the discount factor, and (2) the gap can be further expressed as the infinite-horizon discounted value with a \"lowerorder\"per-period reward. Computationally, our framework leads to an \"aggregation\"approach with performance guarantees. Although the guarantees are grounded in partial differential equation theory, the practical use of this approach requires no knowledge of that theory.\n\nOriginal language English (US) 631-654 24 Operations Research 68 2 https://doi.org/10.1287/opre.2019.1903 Published - Mar 2020\n\n## Keywords\n\n• Approximate dynamic programming\n• Approximate policy iteration\n• Hamilton-Jacobi-Bellman equation\n• Markov decision process\n• Taylor expansion\n\n## ASJC Scopus subject areas\n\n• Computer Science Applications\n• Management Science and Operations Research\n\n## Fingerprint\n\nDive into the research topics of 'On the taylor expansion of value functions'. Together they form a unique fingerprint." ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.8827255,"math_prob":0.6976453,"size":4244,"snap":"2023-40-2023-50","text_gpt3_token_len":998,"char_repetition_ratio":0.10283019,"word_repetition_ratio":0.7851373,"special_character_ratio":0.22266729,"punctuation_ratio":0.108991824,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9601079,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-12-05T20:15:20Z\",\"WARC-Record-ID\":\"<urn:uuid:d5c54ec0-4c70-4c8c-bcb0-4dd586c0835c>\",\"Content-Length\":\"53830\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:9b2a6bcd-017b-4413-92a7-a433cfba2130>\",\"WARC-Concurrent-To\":\"<urn:uuid:f7f39b72-7271-4896-98c3-523404dd4589>\",\"WARC-IP-Address\":\"18.210.30.88\",\"WARC-Target-URI\":\"https://www.scholars.northwestern.edu/en/publications/on-the-taylor-expansion-of-value-functions\",\"WARC-Payload-Digest\":\"sha1:WBPWEYRQTCWAKVP3BC6F34SNVQM3PGHY\",\"WARC-Block-Digest\":\"sha1:DNGETLSX4TEMMJ6CBZOEZRYWCZSRBHEC\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-50/CC-MAIN-2023-50_segments_1700679100555.27_warc_CC-MAIN-20231205172745-20231205202745-00189.warc.gz\"}"}
https://www.viajeabariloche.com/2019/06/solving-algebraic-fractions-worksheet.html	[ "# Solving Algebraic Fractions Worksheet\n\n## Wednesday, June 5, 2019\n\nMath worksheet maths word problems worksheets ks3 2016 rio olympics ision fractional equations tes algebra 5 1 solve gcse questions rtf problem solving linear. Simplify write them as a single.", null, "Algebraic Fractions Add Equations Ks4 Higher By Hassan2008\n\n### Gcse igcse maths mathematics algebraic fractions add subtract multiply divide simplify differentiated practice worksheets with space for answers solutions.", null, "Solving algebraic fractions worksheet. Section 3 solving equations. Higher ks4 ks3 a. Worksheets are multi step equations date period two step equations date period solving multi.\n\nSolving equations containing algebraic fractions worksheets showing all 8 printables. Ideal for gcse revision this is one of a collection of worksheets which contain exam type questions that gradually increase in difficulty. Worksheets are work 2 3 algebraic fractions equations containing fractions.\n\nFind algebraic fraction revision materials worksheets and practice papers to help you master this tricky algebra based topic. These algebra 1 equations worksheets will produce one step problems containing fractions. Solving algebraic equations worksheets showing all 8 printables.\n\nWorksheet 23 algebraic fractions section 1 factoring and algebraic fractions as pointed out in worksheet 21. Solving equations with algebraic fractions. Dividing algebraic fractions worksheet tes 736862 myscres solving equations worksheet pdf math worksheets gcseaths algebra multi step equations with variables on both.", null, "Algebraic Fractions Worksheet By Hel466 Teaching Resources Tes", null, "Adding And Subtracting Algebraic Fractions By Embob2000 Teaching", null, "Solving Algebraic Fractions Algebra Fractions Pinterest", null, "Algebraic Fractions Complete By Chriswallis2 Teaching Resources Tes", null, "Adding And Subtracting Algebraic Fraction By Swaller25 Teaching", null, "Algebraic Fractions Practice Questions Solutions By Transfinite", null, "Solving Equations Containing Algebraic Fractions Worksheet For 8th", null, "Algebraic Manipulation Amendments To Worksheet Pg 3 Example 2", null, "Sample Unit Mathematics Stage 5 Stem Advanced Pathway Algebraic", null, "Algebraic Fractions Thoughts And Crosses By Leond06 Teaching", null, "Addition And Subtraction Algebra Worksheets Solving Equations Of", null, "Pin By Math W On Math Worksheets Algebra Pinterest Worksheets", null, "Sample Unit Mathematics Stage 5 Stem Advanced Pathway Algebraic", null, "Algebraic Fractions Gcse Higher A A With Answers By Hassan2008", null, "Pre Algebra", null, "Solving Quadratic Equations From Algebraic Fractions Worksheet Edplace", null, "Solving Equations With Algebraic Fractions Worksheet With Solutions", null, "Solving Algebra Fractions With A Quadratic Gcse Maths Level 7", null, "Gcse Revision Algebraic Fractions Solving Equations By", null, "Algebra 1 Worksheets Equations Worksheets", null, "Equations Involving Algebraic Fractions Advanced Corbettmaths", null, "2 3 Solving Multi Step Equations With Fractions And Decimals Math", null, "Gcse Maths Revision Solving Linear Equations 2 Involving", null, "Simplifying Algebraic Fractions Homework By Mrsmorgan1 Teaching" ]	[ null, "https://dryuc24b85zbr.cloudfront.net/tes/resources/6192159/image", null, 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https://machinelearningmastery.com/building-a-softmax-classifier-for-images-in-pytorch/	[ "# Building a Softmax Classifier for Images in PyTorch\n\nLast Updated on January 9, 2023\n\nSoftmax classifier is a type of classifier in supervised learning. It is an important building block in deep learning networks and the most popular choice among deep learning practitioners.\n\nSoftmax classifier is suitable for multiclass classification, which outputs the probability for each of the classes.\n\nThis tutorial will teach you how to build a softmax classifier for images data. You will learn how to prepare the dataset, and then learn how to implement softmax classifier using PyTorch. Particularly, you’ll learn:\n\n• How you can use a Softmax classifier for images in PyTorch.\n• How to build and train a multi-class image classifier in PyTorch.\n• How to plot the results after model training.\n\nLet’s get started.", null, "Building a Softmax Classifier for Images in PyTorch.\nPicture by Joshua J. Cotten. Some rights reserved.\n\n## Overview\n\nThis tutorial is in three parts; they are\n\n• Preparing the Dataset\n• Build the Model\n• Train the Model\n\n## Preparing the Dataset\n\nThe dataset you will use here is Fashion-MNIST. It is a pre-processed and well-organized dataset consisting of 70,000 images, with 60,000 images for training data and 10,000 images for testing data.\n\nEach example in the dataset is a $28\\times 28$ pixels grayscale image with a total pixel count of 784. The dataset has 10 classes, and each image is labelled as a fashion item, which is associated with an integer label from 0 through 9.\n\nThis dataset can be loaded from torchvision. To make the training faster, we limit the dataset to 4000 samples:\n\nAt the first time you fetch the fashion-MNIST dataset, you will see PyTorch downloading it from Internet and saving to a local directory named data:\n\nThe dataset train_data above is a list of tuples, which each tuple is an image (in the form of a Python Imaging Library object) and an integer label.\n\nLet’s plot the first 10 images in the dataset with matplotlib.\n\nYou should see an image like the following:", null, "PyTorch needs the dataset in PyTorch tensors. Hence you will convert this data by applying the transforms, using the ToTensor() method from PyTorch transforms. This transform can be done transparently in torchvision’s dataset API:\n\nBefore proceeding to the model, let’s also split our data into train and validation sets in such a way that the first 3500 images is the training set and the rest is for validation. Normally we want to shuffle the data before the split but we can skip this step to make our code concise.\n\n## Build the Model\n\nIn order to build a custom softmax module for image classification, we’ll use nn.Module from the PyTorch library. To keep things simple, we build a model of just one layer.\n\nNow, let’s instantiate our model object. It takes a one-dimensional vector as input and predicts for 10 different classes. Let’s also check how parameters are initialized.\n\nYou should see the model’s weight are randomly initialized but it should be in the shape like the following:\n\n## Train the Model\n\nYou will use stochastic gradient descent for model training along with cross-entropy loss. Let’s fix the learning rate at 0.01. To help training, let’s also load the data into a dataloader for both training and validation sets, and set the batch size at 16.\n\nNow, let’s put everything together and train our model for 200 epochs.\n\nYou should see the progress printed once every 10 epochs:\n\nAs you can see, the accuracy of the model increases after every epoch and its loss decreases. Here, the accuracy you achieved for the softmax images classifier is around 85 percent. If you use more data and increase the number of epochs, the accuracy may get a lot better. Now let’s see how the plots for loss and accuracy look like.\n\nFirst the loss plot:\n\nwhich should look like the following:", null, "Here is the model accuracy plot:\n\nwhich is like the one below:", null, "Putting everything together, the following is the complete code:\n\n## Summary\n\nIn this tutorial, you learned how to build a softmax classifier for images data. Particularly, you learned:\n\n• How you can use a softmax classifier for images in PyTorch.\n• How to build and train a multiclass image classifier in PyTorch.\n• How to plot the results after model training.\n\n### 2 Responses to Building a Softmax Classifier for Images in PyTorch\n\n1.", null, "Dhavan Rathore January 9, 2023 at 8:41 pm #\n\nNicely written and explained\n\n2.", null, "Dhavan Rathore January 9, 2023 at 8:42 pm #\n\nYes, well said" ]	[ null, "https://machinelearningmastery.com/wp-content/uploads/2023/01/joshua-j-cotten-Ge1t87lvyRM-unsplash-scaled.jpg", null, "https://machinelearningmastery.com/wp-content/uploads/2023/01/torch-fmnist-01.png", null, "https://machinelearningmastery.com/wp-content/uploads/2023/01/torch-fmnist-02.png", null, "https://machinelearningmastery.com/wp-content/uploads/2023/01/torch-fmnist-03.png", null, "https://secure.gravatar.com/avatar/6da1b0ea557b9fc0954ccd3c406d9016", null, "https://secure.gravatar.com/avatar/6da1b0ea557b9fc0954ccd3c406d9016", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.60868675,"math_prob":0.6985371,"size":11663,"snap":"2022-40-2023-06","text_gpt3_token_len":3526,"char_repetition_ratio":0.13405952,"word_repetition_ratio":0.2741021,"special_character_ratio":0.35694075,"punctuation_ratio":0.23699649,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.98270977,"pos_list":[0,1,2,3,4,5,6,7,8,9,10,11,12],"im_url_duplicate_count":[null,2,null,2,null,2,null,2,null,null,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-02-03T10:33:05Z\",\"WARC-Record-ID\":\"<urn:uuid:1f0b8926-db02-4d1e-898b-914893b701b9>\",\"Content-Length\":\"316212\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:531af309-d0b1-4bf8-96dc-39c4aa13c53d>\",\"WARC-Concurrent-To\":\"<urn:uuid:6d531f43-1e59-441d-b148-923b2e5591f9>\",\"WARC-IP-Address\":\"104.26.1.148\",\"WARC-Target-URI\":\"https://machinelearningmastery.com/building-a-softmax-classifier-for-images-in-pytorch/\",\"WARC-Payload-Digest\":\"sha1:3PEAGSFVUDATYAK5HILJ5A72OGYFU6I3\",\"WARC-Block-Digest\":\"sha1:RKXRN6WIEWLFJM7RKNK5HZZLWEHPZLGM\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-06/CC-MAIN-2023-06_segments_1674764500044.66_warc_CC-MAIN-20230203091020-20230203121020-00545.warc.gz\"}"}
https://bitscope.com/ed/blog/DK/?p=DK07A&c=C3BB1FEEE0EACBBFD4888010138E412E	[ "# 4 Bit Up/Down Counter\n\n BitScope Ed Education Blog 2013 / 11\n\n2013-11-07\n\n# Breadboard One \| The 4 Bit Up/Down Counter", null, "Breadboard One \| 4 Bit Up/Down Counter Block Diagram\n\nLast month we introduced the Breadboard One educational electronic projects lab. It is a simple mixed signal circuit which we're using to explain the key elements of typical mixed signal systems.\n\nBreadboard One comprises four primary circuits, the first of which is a 4 bit up/down counter. This is a purely digital component and we'll explain how it works and what its output looks like here.\n\nThe counter we use is the CMOS Logic CD4029. It is a member of the CD4000 family which has been in production for almost 40 years! There are newer logic families with the same functionality such as 74HC4029 but we'll stick with the original.\n\nAs you would expect, a counter counts. The block diagram shows the layout of the inputs and outputs of this component where Q4..Q1 are the four bits that comprise the binary encoded output that drive the BreadBoard's D/A convertor. The other two important signals are the CLOCK and the UP/DOWN inputs.\n\nThere are other signals which we're not using in this project as well as power (Vdd) and ground (Vss) but we'll list all the signals here as you can easily experiment with them by modifying the circuit slightly if you want.\n\n Signal Type Description CLOCK Mixed Counter clock input. UP/DOWN Input Increment or Decrement Count. Q4..Q1 Output 4 bit binary encoded outputs. J4..J1 Inputs 4 bit binary encoded \"jam\" inputs. BINARY/DECADE Input Binary or Decade counting mode. CARRY IN Input Counter \"carry in\" input. CARRY OUT Output Counter \"carry out\" output. PRESETENABLE INPUT Enable the jam inputs.\n\nCD4029 Signal Descriptions\n\nInternally the counter comprises a set of logic gates configured to implement the arithmetic addition operator (grab the data sheet for the full details). Normally the counter increments the 4 bit word (Q4,Q3,Q2,Q1) by one every time the clock input is toggled.\n\nIf the UP/DOWN input is asserted the counter counts down (subtracts one) upon each clock cycle instead.\n\nThese two modes of operation are what the Breadboard One project uses but we can run these two modes in isolation by modifying the circuit to simply disconnect the UP/DOWN input from the output of the SCHMITT trigger (as we will show below).\n\nThe CARRY OUT and CARRY IN signals are used when more than one counter are used \"in cascade\". Simply connecting the CARRY OUT of one counter to the CARRY IN of a second one, an 8 bit counter can be built where Q4..Q1 of the first are the low four bits and Q4..Q1 of the second are the high four bits. The CARRY signal is generated each time the counter reaches its limit and \"rolls over\" (to start the count again). There are other ways to connect multiple counters (e.g. ripple counting) but refer to the data sheet for full details. The BINARY/DECADE input defines the limit; 15 (binary, 0..15) or 9 (decimal, 0..9).\n\nThe remaining signals, PRESET ENABLE and J4..J1 allow the counter to be preset with a known value.\n\n## Timing Diagrams and Logic Analysis\n\nA logic diagram is the easiest way describe the operation of a digital circuit like this.", null, "Binary Counter Logic Diagram\n\nFor each clock cycle (at the top of the diagram) the four bits cycle in a binary encoded sequence in this case starting at 5, counting up to 15 before being \"jammed\" to 9 and then counting down to zero and wrapping. Our use of the counter in Breadboard One is simpler in that we're using the UP/DOWN signal but not the jam or carry and to keep it even simpler, here we've asserted the counter as always UP and we observe the result on BS10 as:", null, "Binary Counter, Mixed Signal Analysis, UP Count\n\nThe top half of the display shows the binary encoded counter output as an analog signal produced by the D/A convertor. We'll explain the operation of this component in a future post. For now it's enough to understand that it shows an analog representation of the 4 bit counter output on Q1..Q4. These signals are shown on BitScope's logic channels 0..3 (white, brown, red and orange) and you can see their combined value aligns with the analog signal level for each value.\n\nThe clock input (driven by BitScope's waveform generator) is logic channel 5 (yellow) and the UP/DOWN signal is on channel 6 (green). Note that it remains high so the counter increments from 0 to 15 before wrapping and starting again.\n\nHere's the same circuit with one modification, we've pulled the UP/DOWN signal on channel 6 (green) low:", null, "Binary Counter, Mixed Signal Analysis, DOWN Count\n\nIt's very easy with Breadboard One to re-arrange the circuit to try all sorts of variations on this theme and observe all the signals, digital and analog, using the BitScope but for now we have the full monty:", null, "Binary Counter, Mixed Signal Analysis, Breadboard One\n\nThis is the complete Breadboard One circuit counting UP and then DOWN as we outlined in our earlier post describing the operation of the complete mixed signal circuit.\n\nNote the UP/DOWN signal is now toggling as driven by the SCHMITT trigger output.\n\nAll of these circuit experiments and screenshots were made on our Raspberry Pi based Electronic Projects Lab using BreadBoard One.\n\nIn future posts we'll explore the operation of the D/A convertor, the analog filter and SCHMITT trigger components of this circuit and explain the details of how they work together.\n\nIn the meantime, here are some further experiments that you can try:\n\n• Pull the BINARY/DECIMAL pin to HIGH and LOW and watch the counting states.\n• Monitor the CARRY OUT pin and check it changes for UP, DOWN. Compare when it does this when BINARY and DECIMAL are selected.\n• Check that the counter changes state on the rising edge of the clock and not the falling edge.\n• Change the generated CLOCK signal (on BitScope) to a TRIANGLE wave and make measurements of the exact switching voltage of the CLOCK input. Make the CLOCK signal smaller and see when circuit stops functioning.\n\nThere are many other changes and tests that can be made to find out how circuits like Breadboard One work in practice compared to their theory of operation and we'll cover these issues in future posts too." ]	[ null, "https://bitscope.com/ed/blog/DK/01.png", null, "https://bitscope.com/ed/blog/DK/02.png", null, "https://bitscope.com/ed/blog/DK/03.png", null, "https://bitscope.com/ed/blog/DK/04.png", null, "https://bitscope.com/ed/blog/DK/05.png", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.9280628,"math_prob":0.8140571,"size":5049,"snap":"2022-27-2022-33","text_gpt3_token_len":1134,"char_repetition_ratio":0.13557978,"word_repetition_ratio":0.0,"special_character_ratio":0.21806298,"punctuation_ratio":0.07637475,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.95506227,"pos_list":[0,1,2,3,4,5,6,7,8,9,10],"im_url_duplicate_count":[null,7,null,7,null,7,null,7,null,7,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-06-27T09:15:50Z\",\"WARC-Record-ID\":\"<urn:uuid:b473f317-204e-4621-b21b-7940133d51ad>\",\"Content-Length\":\"35653\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:4d766623-8781-475d-8acf-5cb53be23a80>\",\"WARC-Concurrent-To\":\"<urn:uuid:292b4ae4-af62-4285-bf5d-ca9c62b479fa>\",\"WARC-IP-Address\":\"184.169.168.51\",\"WARC-Target-URI\":\"https://bitscope.com/ed/blog/DK/?p=DK07A&c=C3BB1FEEE0EACBBFD4888010138E412E\",\"WARC-Payload-Digest\":\"sha1:PKEGEPVIF65POJKWK4IYVAYZKQ75GM74\",\"WARC-Block-Digest\":\"sha1:TSORLA4DG2PR2TCUP5US2MRZDNNBFD2L\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-27/CC-MAIN-2022-27_segments_1656103329963.19_warc_CC-MAIN-20220627073417-20220627103417-00008.warc.gz\"}"}
https://calculator.academy/henderson-hasselbalch-calculator/	[ "Enter the conjugate base, acid [HA], and dissociation constant into the calculator to calculate the pH of a solution.\n\n## Henderson Hasselbalch Formula\n\nThe following formula is used to calculate the pH of a solution using the Henderson Hasselbalch equation.\n\npH = Kₐ + log([A⁻]/[HA])\n• Ka is the acid dissociation constant\n• A- is the molar concentration of the base in molars\n• HA is the molar concentration of the acid in molar\n\npH is a measure of the solutions acidity of a solution.\n\n## Henderson Hasselbalch Equation Definition\n\nWhat is the Henderson Hasselbalch equation? The Henderson Hasselbalch equation is a formula used to calculate the pH (acidity) of a buffer. The equation is denoted: pH = Kₐ + log([A⁻]/[HA]).\n\nWhat is a buffer? A buffer is a mixture or solution that contains a weak acid and a conjugate base. The combination of these two components, while at equilibrium,\n\n## How to calculate buffer pH?\n\nThe following is an example problem that uses the Henderson-Hasselbalch equation to determine the pH of a buffer.\n\n1. First, determine the acid dissociation constant. For this example, the acid dissociation constant is found to be 3.\n2. Next, determine the molar concentration of the base. The molar concentration is found to be or 3 moles per liter.\n3. Next, determine the molar concentration of the acid. For this problem, the molar concentration of the acid is found to be 4 moles per liter.\n4. Finally, calculate the buffer pH. Using the formula, the buffer pH is found to be 3 + log ( 3/4) =\n\nDoes the henderson-hasselbalch equation work for bases? The henderson-hasselbalch equation is used to find the pH values for buffer solutions, which are acid-base solutions, so the equation itself is used to bases.\n\nWhen does the henderson-hasselbalch equation fail? The equation does not work in situations where the buffer solutions are highly diluted through the water." ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.8666308,"math_prob":0.9952507,"size":1991,"snap":"2022-40-2023-06","text_gpt3_token_len":471,"char_repetition_ratio":0.17362858,"word_repetition_ratio":0.064615384,"special_character_ratio":0.20894024,"punctuation_ratio":0.09803922,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99987924,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-01-31T19:19:37Z\",\"WARC-Record-ID\":\"<urn:uuid:2df9efa1-ca9d-4fca-a954-39aa2da381f0>\",\"Content-Length\":\"175663\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:3b296e2f-8197-41fb-a38f-60dcdbf02a67>\",\"WARC-Concurrent-To\":\"<urn:uuid:6ad1fdfe-b6bd-4e9a-8691-5b568bf6505d>\",\"WARC-IP-Address\":\"172.66.41.37\",\"WARC-Target-URI\":\"https://calculator.academy/henderson-hasselbalch-calculator/\",\"WARC-Payload-Digest\":\"sha1:IC4BEYR6HWRDPEO5JQXLFJIL67GL6IFB\",\"WARC-Block-Digest\":\"sha1:QZJ7HHN2DSR6KE7MLIXS5UHLIUJO4TMX\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-06/CC-MAIN-2023-06_segments_1674764499890.39_warc_CC-MAIN-20230131190543-20230131220543-00664.warc.gz\"}"}
https://forum.arduino.cc/t/speed-sensor-velocity-km-h-using-ultrasonic-sensor-ping/136513	[ "", null, "# Speed Sensor (Velocity Km/H) Using Ultrasonic Sensor (Ping)\n\nhello everybody…\ncan someone help me on this problem?\n\ni have arduino mega and ping ultrasonic sensor, but the coding is problem for me. hopefully someone can help me… =(\n\nWhat’s the problem, and why are you SHOUTING?\n\n1kmh-1 is 0.277ms-1\n\nThe Ping can take maybe 20 readings a second, out to a sensible range of what? Four metres?\n\nDo some simple sums.\n\ncan u help me for the coding part?i really blur to do the speed sensor coding...\n\nCan you stop and think for a moment, and tell us exactly what you're trying to do? Also think about the limitations I described in my other post.\n\nYour ping sensor is a simple time of flight sensor where the variable time represents the variable distance to a fixed object. To turn that into a speed sensor would require that you sense the change of distance of a return pulse of a moving object over a specific time period, so it's a second order value that would be based on the integration of the primary time of flight measurement that the sensor provides you. So the solution is math based.\n\nLefty\n\nAWOL: Can you stop and think for a moment, and tell us exactly what you're trying to do? Also think about the limitations I described in my other post.\n\nsorry.. i try to do the speed sensor between two car to find the crash impact..\n\nfor the example..i put ping ultrasonic sensor in front of my car..than, find the different speed between my car and the front car..\n\nfor the example..i put ping ultrasonic sensor in front of my car..than, find the different speed between my car and the front car..\n\n\"If you can see my Ping, you're driving too close\" :)\n\nOK, so show us your code so far.\n\nWell assuming you have the coding done to measure the distance in the first place, which is well documented if I recall from looking at Ping)) some time ago…\n\nIn it’s simplest form you just need to take two distance readings noting the time using millis() or micros() at which you took the readings. Then you calculate the speed as distance / time as:\n\n``````Speed = (FirstDistance - SecondDistance) / (SecondTime - FirstTime)\n``````\n\nIf FirstDistance < SecondDistance the distance is increasing and the car in front is moving away; that would give a -ve speed.\n\nSomething like that, anyway 8)\n\nAWOL:\n\nfor the example…i put ping ultrasonic sensor in front of my car…than, find the different speed between my car and the front car…\n\n“If you can see my Ping, you’re driving too close”", null, "OK, so show us your code so far.\n\nyes…u are too close if u see my ping sensor…hehehe…\ni dont know how to start the coding for speed sensor…before this i just ‘play’ with range…sens and show the range between your car, and front car…\n\n``````#include <LiquidCrystal.h> // lcd\nLiquidCrystal lcd(12, 11, 5, 4, 3, 2); //lcd\n\nconst int pingPin = 7; //sensor\n\nconst int redPin = 6; //led\nconst int greenPin = 5; //led\nconst int bluePin = 4; //led\n\nvoid setup() {\n// initialize serial communication:\n\npinMode (redPin, OUTPUT);\npinMode (greenPin, OUTPUT);\npinMode (bluePin, OUTPUT);\n\nSerial.begin(9600);\n\nlcd.begin(16, 2);\nlcd.print(\"Distance (cm):\");\n}\n\nvoid loop()\n{\n// establish variables for duration of the ping,\n// and the distance result in inches and centimeters:\nlcd.setCursor(0, 1);\n\nlong duration, inches,cm;\n\n// The PING))) is triggered by a HIGH pulse of 2 or more microseconds.\n// Give a short LOW pulse beforehand to ensure a clean HIGH pulse:\npinMode(pingPin, OUTPUT);\ndigitalWrite(pingPin, LOW);\ndelayMicroseconds(2);\ndigitalWrite(pingPin, HIGH);\ndelayMicroseconds(15);\ndigitalWrite(pingPin, LOW);\ndelayMicroseconds(20);\n// The same pin is used to read the signal from the PING))): a HIGH\n// pulse whose duration is the time (in microseconds) from the sending\n// of the ping to the reception of its echo off of an object.\npinMode(pingPin, INPUT);\nduration = pulseIn(pingPin, HIGH);\n\ninches = microsecondsToInches(duration);\ncm = microsecondsToCentimeters(duration);\n\nSerial.print(inches);\nSerial.print(\"in, \");\nSerial.print(cm);\nSerial.print(\"cm\");\nSerial.println();\n\n// convert the time into a distance\n\nif (inches > 30) {\ndigitalWrite(greenPin, HIGH); // green LED\ndigitalWrite(redPin, LOW);\ndigitalWrite(bluePin, LOW);\n}\nelse if (inches <= 30 && inches > 12) {\ndigitalWrite(greenPin, HIGH);\ndigitalWrite(redPin, HIGH); // orange LED\ndigitalWrite(bluePin, LOW);\n}\nelse if (inches <= 12 && inches > 6) {\ndigitalWrite(redPin, LOW); // red LED\ndigitalWrite(greenPin, HIGH);\ndigitalWrite(bluePin, HIGH);\n}\nelse {\ndigitalWrite(redPin, HIGH); // purple LED\ndigitalWrite(greenPin, HIGH);\ndigitalWrite(bluePin, HIGH);\n}\n\nlcd.print(cm);\nlcd.print(\"cm\");\nlcd.print(\"\");\ndelay(100);\n}\n\nlong microsecondsToInches(long microseconds)\n{\n// According to Parallax's datasheet for the PING))), there are\n// 73.746 microseconds per inch (i.e. sound travels at 1130 feet per\n// second). This gives the distance travelled by the ping, outbound\n// and return, so we divide by 2 to get the distance of the obstacle.\n// See: http://www.parallax.com/dl/docs/prod/acc/28015-PING-v1.3.pdf\nreturn microseconds / 74 / 2;\n}\n\nlong microsecondsToCentimeters(long microseconds)\n{\n// The speed of sound is 340 m/s or 29 microseconds per centimeter.\n// The ping travels out and back, so to find the distance of the\n// object we take half of the distance travelled.\nreturn microseconds / 29 / 2;\n}\n``````\n\nModerator edit: CODE TAGS\n\nJimboZA:\nWell assuming you have the coding done to measure the distance in the first place, which is well documented if I recall from looking at Ping)) some time ago…\n\nIn it’s simplest form you just need to take two distance readings noting the time using millis() or micros() at which you took the readings. Then you calculate the speed as distance / time as:\n\n``````Speed = (FirstDistance - SecondDistance) / (SecondTime - FirstTime)\n``````\n\nIf FirstDistance < SecondDistance the distance is increasing and the car in front is moving away; that would give a -ve speed.\n\nSomething like that, anyway 8)\n\nthank you for your idea… if you can see from my code, where i can get the value of secondtime and first time?\n\nwhere i can get the value of secondtime and first time?\n\nHere\n\nEvery time through loop() you are measuring the ping duration and then calculating a distance...\n\nWhere you read the duration, you could add a line like `time = millis` so that would record when in the day (or in fact since the sketch started running) that duration was captured.\n\nYou'll need to do something to move each current time reading into an oldtime variable (at the top of loop() ?) so that you can subtract one from the other to give the time between readings. Similarly, you'll need to move the current distance to an olddistance variable so you can subtract them.. as it stands now, values like cm and so on are overwritten each time through loop() so you need to have a means of storing the previous one before the new one comes along.\n\nokay friends..i will try first...\n\nmatt121187: okay friends..i will try first...\n\nAwesome 8) .... and my advice is that before you dive into code, draw a flowchart of what it is you want to do. Get the logic clear on paper and in your head before you get into the nitty-gritties of coding.\n\nJimboZA:\n\nmatt121187: okay friends..i will try first...\n\nAwesome 8) .... and my advice is that before you dive into code, draw a flowchart of what it is you want to do. Get the logic clear on paper and in your head before you get into the nitty-gritties of coding.\n\nthanks for your advice..i just draw the flowchart..but i bit confusing to get the value of FirstDistance ,SecondDistance, SecondTime, FirstTime =(\n\ni'm really blur...cant solve this problem.... =(\n\nIn that case I suggest you pick a simpler problem.\n\ncant solve this problem.\n\nCan't or won't? What have you tried?\n\nhow to hold the previous distance?\n\nthis is my coding right now..but the previous state still show the current distance\n\n``````const int pingPin = 7;\nint start = 0;\n\nvoid setup() {\n// initialize serial communication:\nSerial.begin(9600);\n}\n\nvoid loop()\n{\n// establish variables for duration of the ping,\n// and the distance result in inches and centimeters:\nlong duration, inches, cm, currentState, previousState;\n\n// The PING))) is triggered by a HIGH pulse of 2 or more microseconds.\n// Give a short LOW pulse beforehand to ensure a clean HIGH pulse:\npinMode(pingPin, OUTPUT);\ndigitalWrite(pingPin, LOW);\ndelayMicroseconds(2);\ndigitalWrite(pingPin, HIGH);\ndelayMicroseconds(5);\ndigitalWrite(pingPin, LOW);\n\n// The same pin is used to read the signal from the PING))): a HIGH\n// pulse whose duration is the time (in microseconds) from the sending\n// of the ping to the reception of its echo off of an object.\npinMode(pingPin, INPUT);\nduration = pulseIn(pingPin, HIGH);\n\n// convert the time into a distance\n\ncm = microsecondsToCentimeters(duration);\n\ncurrentState = cm;\n\npreviousState = currentState + start ;\npreviousState = currentState;\n\nSerial.print(previousState);\nSerial.print(\"cm, \");\nSerial.print(currentState);\nSerial.print(\"cm\");\nSerial.println();\n\ndelay(100);\n}\n\nlong microsecondsToCentimeters(long microseconds)\n{\n// The speed of sound is 340 m/s or 29 microseconds per centimeter.\n// The ping travels out and back, so to find the distance of the\n// object we take half of the distance travelled.\nreturn microseconds / 29 / 2;\n}\n``````\n\nAny variable declared within a function like \"loop\" is going to have limited use for remembering its value. If you want a variable to maintain its value, declare it with global scope." ]	[ null, "https://aws1.discourse-cdn.com/arduino/original/3X/1/f/1f6eb1c9b79d9518d1688c15fe9a4b7cdd5636ae.svg", null, "https://emoji.discourse-cdn.com/twitter/slight_smile.png", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.8559558,"math_prob":0.9208056,"size":9482,"snap":"2021-31-2021-39","text_gpt3_token_len":2301,"char_repetition_ratio":0.13557713,"word_repetition_ratio":0.4726688,"special_character_ratio":0.25226745,"punctuation_ratio":0.16373333,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9539035,"pos_list":[0,1,2,3,4],"im_url_duplicate_count":[null,null,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-07-28T11:55:14Z\",\"WARC-Record-ID\":\"<urn:uuid:e82fd379-b2fa-4a84-b5ce-f39b41080e89>\",\"Content-Length\":\"62833\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:4d7e6652-c6c2-4332-a231-19af50680b39>\",\"WARC-Concurrent-To\":\"<urn:uuid:f515967c-d25b-4fdd-94a9-8df0e7f662c0>\",\"WARC-IP-Address\":\"184.104.202.141\",\"WARC-Target-URI\":\"https://forum.arduino.cc/t/speed-sensor-velocity-km-h-using-ultrasonic-sensor-ping/136513\",\"WARC-Payload-Digest\":\"sha1:KU63XHH6SD2CPKAI6CCPUBASE2YKXY4Y\",\"WARC-Block-Digest\":\"sha1:KHPOJNKCSQQK2ODSU7LFWBOKZWA7LSQY\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-31/CC-MAIN-2021-31_segments_1627046153709.26_warc_CC-MAIN-20210728092200-20210728122200-00683.warc.gz\"}"}
https://www.sqlservercentral.com/forums/topic/loss-of-precision-when-multiplying	[ "Loss of precision when multiplying\n\n• I am currently working with a database that has all DECIMALtypes set to a precision and scale of 38 and 18. Though I am fully aware that there might be a loss of precision when multiplying two numbers, SQL Server,when multiplying, truncates the numbers before the operation, resulting a bigger loss of precision.\n\nFor example,\n\nSELECT 1000000000000.123456 * 10000.123456\n\nreturns «10000123456001234.575241383936», which is exact while\n\nSELECT CONVERT(DECIMAL(38, 18), 1000000000000.123456) CONVERT(DECIMAL(38, 18), 10000.123456)SELECT CONVERT(DECIMAL(38, 18), 1000000000000.123456) CONVERT(DECIMAL(38, 18), 10000.123456)\n\nreturns the truncated value «10000123456001234.575241»\n\nWhen working in CLR, the SqlDecimal class has methods(ConvertToPrecScale and AjustScale) that let you change the precision or scale of a decimal value. The problem is that I just basically want to remove leadingand trailing zeros while the two aforementioned methods want to know an exact position.\n\nI came up with the following function:\n\ninternal static SqlDecimal PackPrecScale(SqlDecimal d)\n{\nvar s = d.ToString();\nvar l = s.Length;\nvar indexofperiod = s.IndexOf('.');\n\n//Remove trailing zeros\nif (indexofperiod != -1)\n{\nwhile (s[l - 1] == '0') l--;\nif (s[l - 1] == '.')\n{\nl--;\nindexofperiod = -1;\n}\n}\n\nvar precision = 6;\nvar scale = 0;\nif (l > 0)\n{\nprecision = l;\nif (s == '-') precision--;\nif (indexofperiod != -1)\n{\nprecision--;\nscale = l - indexofperiod - 1;\n}\nif (precision < 6) precision = 6;\n}\n\nreturn SqlDecimal.ConvertToPrecScale(d, precision, scale);\n}\n\nThat returns a decimal having the smallest possible precision and scale. Having to multiply a and b, I can then write:\n\nSqlDecimal.ConvertToPrecScale(SqlDecimal.Multiply(PackPrecScale(a),PackPrecScale(b)), 38, 18)\n\nTo keep maximum precision while multiplying.\nIs there a simpler or safer way to «pack» a SqlDecimal?\n\n• This gives the same accuracy as your first 'exact' example:\n\nSELECT CAST(1000000000000.123456 AS DECIMAL(29,12)) * CAST(10000.123456 AS DECIMAL(29,12))\n\nThis code helped me determine the CASTs to use:\n\nDECLARE @numval SQL_VARIANT = 10000123456001234.575241383936;\n\nSELECT SQL_VARIANT_PROPERTY(@numval, 'BaseType');\n\nSELECT SQL_VARIANT_PROPERTY(@numval, 'Precision');\n\nSELECT SQL_VARIANT_PROPERTY(@numval, 'Scale');\n\nIf you aren't happy single, you won't be happy in a relationship.\n\nRemember, happiness comes from guitars, not relationships.\n\n• I was expecting this to be a case of the OP was using float when I first opened the topic; was nice to see that wasn't the case.\n\nTo cover why this happens, however, this is actually explained in the documenation: Precision, scale, and Length (Transact-SQL). Specifically it notes:\n\nIn multiplication and division operations we need precision - scale places to store the integral part of the result. The scale might be reduced using the following rules:\n\n1. The resulting scale is reduced to min(scale, 38 – (precision-scale)) if the integral part is less than 32, because it cannot be greater than 38 – (precision-scale). Result might be rounded in this case.\n2. The scale will not be changed if it is less than 6 and if the integral part is greater than 32. In this case, overflow error might be raised if it cannot fit into decimal(38, scale)\n3. The scale will be set to 6 if it is greater than 6 and if the integral part is greater than 32. In this case, both integral part and scale would be reduced and resulting type is decimal(38,6). Result might be rounded to 6 decimal places or overflow error will be thrown if integral part cannot fit into 32 digits.\n\nI've bolded the relevant part.\n\nAs your scale is greater than 32 (38) and your precision is higher than 6 (12), when you use a multiplication the precision is automatically set to 6, and why you only get .~575241 (to 6 decimal places). A reason why you should always make sure you use relevant sizes for your datatypes.\n\nIn this case, declaring a decimal as a decimal(38,12) when you're only using a scale of 19, and a precision of 6 means you're not using a relevant size. In fact 38 is double to scale of 19, as is 12 of 6, so in a way, the size is quadruple what it should be (yes, i realise that technically it's only double in size, but double the size in both scale and precision kind of makes it \"quadrupally\" bad 😉 ).\n\nThom~\n\nExcuse my typos and sometimes awful grammar. My fingers work faster than my brain does.\nLarnu.uk\n\n• Thom A - Friday, June 29, 2018 9:59 AM\n\nI was expecting this to be a case of the OP was using float when I first opened the topic; was nice to see that wasn't the case.\n\nTo cover why this happens, however, this is actually explained in the documenation: Precision, scale, and Length (Transact-SQL). Specifically it notes:\n\nIn multiplication and division operations we need precision - scale places to store the integral part of the result. The scale might be reduced using the following rules:\n\n1. The resulting scale is reduced to min(scale, 38 â€“ (precision-scale)) if the integral part is less than 32, because it cannot be greater than 38 â€“ (precision-scale). Result might be rounded in this case.\n2. The scale will not be changed if it is less than 6 and if the integral part is greater than 32. In this case, overflow error might be raised if it cannot fit into decimal(38, scale)\n3. The scale will be set to 6 if it is greater than 6 and if the integral part is greater than 32. In this case, both integral part and scale would be reduced and resulting type is decimal(38,6). Result might be rounded to 6 decimal places or overflow error will be thrown if integral part cannot fit into 32 digits.\n\nI've bolded the relevant part.\n\nAs your scale is greater than 32 (38) and your precision is higher than 6 (12), when you use a multiplication the precision is automatically set to 6, and why you only get .~575241 (to 6 decimal places). A reason why you should always make sure you use relevant sizes for your datatypes.\n\nIn this case, declaring a decimal as a decimal(38,12) when you're only using a scale of 19, and a precision of 6 means you're not using a relevant size. In fact 38 is double to scale of 19, as is 12 of 6, so in a way, the size is quadruple what it should be (yes, i realise that technically it's only double in size, but double the size in both scale and precision kind of makes it \"quadrupally\" bad 😉 ).\n\nYes, this is absolutely correct. It is best to use the smallest Precision and Scale for each operand. So, even going with the size of the resulting expression (as Phil did above) is dangerous and could lead to unnecessary truncation.\n\nIdeally you would be able to do something like this:\n\nSELECT CONVERT(DECIMAL(19, 6), 1000000000000.123456) * CONVERT(DECIMAL(11, 6), 10000.123456)\n-- 10000123456001234.575241383936\n\nBut when storing decimal values, yeah, it kinda makes sense that people use the largest possible container if it is uncertain what the max size of the values to store is up front.\n\nI don't know of a built-in way to shrink the Precision and Scale down to the minimum required sizes, but I can see some improvements to make in the O.P.s code. It shouldn't be necessary to do either loop, or to test for a leading minus / negative sign. Try this:\n\n// convert to string and remove potential negative sign\nstring _InputString = SqlDecimal.Abs(_Input).ToString();\n\n// remove trailing 0's (only on the right side)\n_InputString = _InputString.TrimEnd(new char[] { '0' });\n\nint _Precision = _InputString.Length - 1;\n\n// if decimal is usually closer to the right side, possibly save time by starting there,\n// else use IndexOf\nint _Scale = (_Precision - _InputString.LastIndexOf('.'));\n\nreturn SqlDecimal.ConvertToPrecScale(_Input, _Precision, _Scale);\n\nTake care,\nSolomon...\n\nSQL#https://SQLsharp.com/ ( SQLCLR library ofover 340 Functions and Procedures)\nSql Quantum Lifthttps://SqlQuantumLift.com/ ( company )\nSql Quantum Leaphttps://SqlQuantumLeap.com/ ( blog )\nInfo sitesCollations • Module Signing • SQLCLR\n\nViewing 4 posts - 1 through 3 (of 3 total)" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.8273811,"math_prob":0.9770617,"size":8136,"snap":"2022-05-2022-21","text_gpt3_token_len":2105,"char_repetition_ratio":0.12788981,"word_repetition_ratio":0.4761194,"special_character_ratio":0.29277286,"punctuation_ratio":0.14053716,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9880442,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-01-29T13:45:18Z\",\"WARC-Record-ID\":\"<urn:uuid:6b7aa802-43f9-4d25-9063-ffa996480a5e>\",\"Content-Length\":\"104722\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:87ebeb0d-840a-4c41-95c6-fee6e3bbb6a0>\",\"WARC-Concurrent-To\":\"<urn:uuid:bd525c16-6404-4f0e-aa91-4c173202706b>\",\"WARC-IP-Address\":\"34.242.253.12\",\"WARC-Target-URI\":\"https://www.sqlservercentral.com/forums/topic/loss-of-precision-when-multiplying\",\"WARC-Payload-Digest\":\"sha1:BNNCENATFHTZMQUYDJP7W3EAVF5ULM24\",\"WARC-Block-Digest\":\"sha1:KTB4SDOUHI4R6PZJVK4COOOJDDSUMTET\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-05/CC-MAIN-2022-05_segments_1642320306181.43_warc_CC-MAIN-20220129122405-20220129152405-00146.warc.gz\"}"}
https://yohua.ml/2019/03/07/%E4%BA%92%E8%BF%9E%E7%BD%91%E7%BB%9C%E7%9A%84%E5%9F%BA%E6%9C%AC%E6%A6%82%E5%BF%B5/	[ "此处概念限定为“互连”，与“互联”区分。\n\n# 互连网络的功能和特征\n\n互联网络（interconnection network）是一种由开关元件按照一定拓扑结构和控制方式构成的网络，用来实现计算机系统中结点之间的相互连接。这些节点可以是处理器、存储模块后其他设备。从拓扑结构角度，互联网络是从输入结点到输出结点之间的一组互连或映射（mapping）。", null, "互连网络已经成为并行计算机系统的一个关键组成部分。随着各个领域对高性能计算机的要求越来越高，多处理机和多计算机系统的规模越来越大，对处理器之间或处理单元与存储模块之间通信的速度和灵活性的要求也越来越高。因此，它对计算机系统的性能价格比有着决定性的影响。\n可以从以下4个不同的方面来描述互连网络。\n（1）定时方式：有同步和异步两种。\n同步系统使用一个统一的时钟。它可以将数据同时播送到所有处理器结点中，也可以使所有结点同时与相邻的结点进行通信（若无冲突）。而异步系统没有统一的时钟，系统中的各个处理机都是独立地工作。\n（2）交换方式：有线路交换和分组交换两种。\n（3）控制策略：有集中式和分散式两种。\n（4）拓扑结构：有静态和动态两种。\n\n# 互连函数\n\n为反应不同互连函数的连接特性，每种互连网络可用一组互连函数（interconnection function）来描述。用变量x表示输入（设x=0,1，···，N-1），用函数f(x)表示输出，通过数学表达式建立输入端与输出端的一一对应关系。即在互连函数f的作用下，输入端x连接到输出端f（x）。x和f（x）可以用二进制表示，也可以用十进制表示。\n互连函数反映了网络输入数组和输出数组之间对应的置换关系或排列关系，所以互连函数有事也称为置换（permutation）函数或排列函数。\n互连函数f（x）有时可以采用循环表示，即（$x_0x_1x_2···x_{j-1}$）。它表示：\n\n$$f(x_0)=x_1, f(x_1)=x_2, ···, f(x_{j-1})=x_0$$\n\n其中，j称为该循环的长度。\n下面介绍几种常用的基本互连函数及其主要特征。\n\n## 交换函数\n\n交换函数用于实现二进制地址编码中第k位互反的输入端与输出端的连接。其表达式为：\n\n$$E_k(x_{n-1}x_{n-2}···x_{k+1}x_{k}x_{k-1}···x_1x_0)=x_{n-1}x_{n-2}···x_{k+1} \\overline{x_k} x_{k-1}···x_1x_0$$\n\n交换函数主要用来构建立方体网络和各种超立方体互连网络。它共有$n=log_2N$ 种互连函数。N为结点个数。", null, "", null, "## 均匀洗牌函数\n\n均匀洗牌函数（shuffle）定义为：将输入端分成数目相等的两半，前一半和后一半按类似均匀混洗扑克牌的方式交叉的连接到输出端（输出端相当于混洗的结果），其函数可表示为：\n\n$$S(x_{n-1}x_{n-2}···x_{2}x_{1})=x_{n-2}···x_2x_1x_{n-1}$$\n\n即把输入端的二进制编号循环左移一位。\n逆均匀洗牌函数于此类似，是将输入端的二进制编号循环右移一位而得到所连接的输出端编号。\n逆均匀洗牌函数是均匀洗牌函数的逆函数。两者的输入端与输出端正好互换位置（互为镜像）。\n用它们代表的链路和交换开关多级组合起来可构成Omega网络与逆Omega网络。", null, "## 蝶式函数\n\n蝶式互连函数（butterfly）定义为：\n\n$$B(x_{n-1}x_{n-2}···x_1x_0)=x_0x_{n-2}···x_1x_{n-1}$$\n\n即把输入端的二进制编号的最高位与最低位互换位置，便得到了输出端的编号。\n与均匀混洗函数类似，只用蝶式函数不能实现任意结点之间的连接，但是蝶式变换与交换变换的多级组合可作为构成方体多级网络的基础。", null, "## 反位序函数\n\n反位序函数是将输入端二进制编号的位序颠倒过来求得相应输出端的编号。其互连函数为：\n\n$$R(x_{n-1}x_{n-2}···x_1x_0)=x_0x_1x_2···x_{n-2}x_{n-1}$$\n\n对于N=8的情况，正好B(x)函数等于R(x)函数，其变换图形如下。", null, "## PM2I函数\n\nPM2I函数是一种移数函数，它是将各输入端都循环移动一定的位置后连到输出端。其函数为：\n\n$$PM2_{+i}(x)=(x+2^i) mod N$$\n\n$$PM2_{-i}(x)=(x-2^i) mod N$$\n\n其中，0<=x<=N-1, 0<=i<=n-1, n=log_2N, N为结点数。显然，PM2I互连网络共有2n个互连函数。\n当N=8时，有6个PM2I函数：\n\n$$PM2_{+0}: (0 1 2 3 4 5 6 7 )$$\n\n$$PM2_{-0}: (7 6 5 4 3 2 1 0 )$$\n\n$$PM2_{+1}: (0 2 4 6 )(1 3 5 7 )$$\n\n$$PM2_{-1}: (6 4 2 0 )(7 5 3 1 )$$\n\n$$PM2_{\\pm2}: (0 4)(1 5)(2 6)(3 7)$$\n\n下图画出了其中3个函数的变形图形。", null, "PM2I函数是构成数据变换网络的基础。\n阵列计算机ILLIAC IV采用$PM2_{\\pm0}$$PM2_{\\pm \\frac{n}{2}}$构成互连网络，实现各处理单元之间的上下左右互连，如下图。", null, "# 互连函数的性能参数\n\n网络通常是用有向边或无向边连接有限个结点的图来表示。互连网络的主要特征参数有：\n（1）网络规模（network size）：网络中结点的个数。它表示该网络所能连接的部件的数量、\n（2）节点度（node degree）：与结点相连接的边数（通道数），包括入度（in degree）和出度（out degree）。进入结点的边数称为入度，从结点出来的边数称为出度。\n（3）距离：对于网络中的任意两个结点，从一个结点出发到另一个节点终止所需要跨越的边数的最小值。\n（4）网络直径（network diameter）：网络中任意两个结点之间距离的最大值。网络直径应尽可能地小。\n（5）结点之间的线长：两个结点之间连线的长度，用米、千米等表示。\n（6）等分宽度：当某一网路被分割成相等的两半时，沿切口的边数（通道数）的最小值称为通道等分宽度（channel bisection width），用b表示。而线等分宽度为B=b*w。其中w为通道宽度（用位表示）。该参数主要反映了网络最大流量。\n（7）对称性：从任何结点看到的拓扑结构都是相同的网络称为对称网络（symmetric network）。对称网络比较容易实现，编程也比较容易。" ]	[ null, "https://yohua.ml/2019/03/07/互连网络的基本概念/001.png", null, "https://yohua.ml/2019/03/07/互连网络的基本概念/002.png", null, "https://yohua.ml/2019/03/07/互连网络的基本概念/003.png", null, "https://yohua.ml/2019/03/07/互连网络的基本概念/004.png", null, "https://yohua.ml/2019/03/07/互连网络的基本概念/005.png", null, "https://yohua.ml/2019/03/07/互连网络的基本概念/006.png", null, "https://yohua.ml/2019/03/07/互连网络的基本概念/007.png", null, "https://yohua.ml/2019/03/07/互连网络的基本概念/008.png", null ]	{"ft_lang_label":"__label__zh","ft_lang_prob":0.9692758,"math_prob":0.99877286,"size":2073,"snap":"2022-40-2023-06","text_gpt3_token_len":1844,"char_repetition_ratio":0.06283229,"word_repetition_ratio":0.0,"special_character_ratio":0.16208394,"punctuation_ratio":0.01904762,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9665471,"pos_list":[0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16],"im_url_duplicate_count":[null,1,null,1,null,1,null,1,null,1,null,1,null,1,null,1,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-10-02T19:27:11Z\",\"WARC-Record-ID\":\"<urn:uuid:805c8405-c908-4df4-8b36-32143062e1d9>\",\"Content-Length\":\"57638\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:f1ad2210-90e1-421b-ab21-c910b28422b6>\",\"WARC-Concurrent-To\":\"<urn:uuid:66f56400-a419-48bc-b71c-b093c1be0178>\",\"WARC-IP-Address\":\"185.199.109.153\",\"WARC-Target-URI\":\"https://yohua.ml/2019/03/07/%E4%BA%92%E8%BF%9E%E7%BD%91%E7%BB%9C%E7%9A%84%E5%9F%BA%E6%9C%AC%E6%A6%82%E5%BF%B5/\",\"WARC-Payload-Digest\":\"sha1:4H675MY47RJPABIXT4YPTJDV3XMJG4V7\",\"WARC-Block-Digest\":\"sha1:6F2TK3KVUSPQM7BJID7OOVPAXQBZMS47\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-40/CC-MAIN-2022-40_segments_1664030337339.70_warc_CC-MAIN-20221002181356-20221002211356-00431.warc.gz\"}"}
https://docs.rs/crate/seq_geom_xform/0.3.0/source/Cargo.toml.orig	[ "# seq_geom_xform 0.3.0\n\nTransform/normalize complex single-cell fragment geometries into simple geometries.\nDocumentation\n``````[package]\nname = \"seq_geom_xform\"\nversion = \"0.3.0\"\nedition = \"2021\"\nauthors = [\n\"Rob Patro <[email protected]>\"\n]\ndescription = \"Transform/normalize complex single-cell fragment geometries into simple geometries.\"\nrepository = \"https://github.com/COMBINE-lab/seq_geom_xform\"\nhomepage = \"https://github.com/COMBINE-lab/seq_geom_xform\"\ninclude = [\n\"/src/.rs\",\n\"/src/lib/.rs\",\n\"/src/bin/*.rs\",\n\"/Cargo.toml\",\n\"/Cargo.lock\",\n\"/CONTRIBUTING.md\",\n\"/CODE_OF_CONDUCT.md\",\n]\nkeywords = [\n\"single-cell\",\n\"preprocessing\",\n\"RNA-seq\",\n]\ncategories = [\"command-line-utilities\", \"science\"]\n\n# See more keys and their definitions at https://doc.rust-lang.org/cargo/reference/manifest.html\n[lib]\nname = \"seq_geom_xform\"\npath = \"src/lib.rs\"\n\n[[bin]]\nname = \"seq_xformer\"\npath = \"src/bin/bin.rs\"\n\n[dependencies]\nseq_geom_parser = { git = \"https://github.com/COMBINE-lab/seq_geom_parser\", branch = \"dev\", version = \"0.2.1\" }\nregex = \"1.7\"\nanyhow = \"1.0\"\nneedletail = \"0.5.0\"\nclap = { version = \"4.1.11\", features = [\"derive\"] }\nthousands = \"0.2.0\"\ntracing = \"0.1.37\"\ntracing-subscriber = { version = \"0.3.16\", default-features = true, features = [\"env-filter\"] }\ntempfile = \"3.4.0\"\nnix = { version = \"0.26.2\", features = [\"fs\"] }\n``````" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.6460924,"math_prob":0.966152,"size":1181,"snap":"2023-40-2023-50","text_gpt3_token_len":388,"char_repetition_ratio":0.13084112,"word_repetition_ratio":0.0,"special_character_ratio":0.38441998,"punctuation_ratio":0.2744186,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9841428,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-10-02T17:15:32Z\",\"WARC-Record-ID\":\"<urn:uuid:d0f54b71-309d-4a4b-8215-f239f663ad1e>\",\"Content-Length\":\"56478\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:f24cb0dd-0583-4779-aae8-76a8e337a9c7>\",\"WARC-Concurrent-To\":\"<urn:uuid:da7597de-af34-4b8b-8298-8e0770d3ba16>\",\"WARC-IP-Address\":\"108.138.85.36\",\"WARC-Target-URI\":\"https://docs.rs/crate/seq_geom_xform/0.3.0/source/Cargo.toml.orig\",\"WARC-Payload-Digest\":\"sha1:2TIH3BSIUS3BG3DB64UCQS5EXKC3BTKS\",\"WARC-Block-Digest\":\"sha1:UPL2KLH7NUQNVLY65GSUFZBVSN47JQZF\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-40/CC-MAIN-2023-40_segments_1695233511002.91_warc_CC-MAIN-20231002164819-20231002194819-00854.warc.gz\"}"}
https://cboard.cprogramming.com/c-programming/169097-gettoken-function.html?s=4df223358d054aee44368c2b27f16d9c	[ "1. ## gettoken() function\n\nhello,\n\nI have been trying to understand this function, there is still a few things confusing me. First, what are those weird statements like return token type = BRACKETS for? I have never seen them before.\n\nCode:\n```int gettoken(void) /* return next token /\n\n{\n\nint c, getch(void);\nvoid ungetch(int);\nchar p = token; /* character string pointed by p /\n\nwhile ((c = getch()) == ' ' \|\| c == '\\t') / skip blanks or tabs /\n;\nif (c == '(') {\nif ((c = getch()) == ')') {\nstrcpy(token, \"()\");\n} else {\nungetch(c); / put character back on input before prosessing new input /\n\n}\n} else if (c == '[') {\nfor (p++ = c; (p++ = getch()) != ']'; ) / do nothing as long as character is not terminating bracket /\n;\np = '\\0';\nreturn tokentype = BRACKETS; /* if it is, we found it, so tokentype is brackets /\n} else if (isalpha(c)) {\nfor (p++ = c; isalnum(c == getch()); ) /* read characters as long as is alphanumeric /\np++ = c;\np = '\\0';\nungetch(c); / if not, push back into the buffer */\n} else\n}```", null, "2.", null, "Originally Posted by coder222\nFirst, what are those weird statements like return token type = BRACKETS for? I have never seen them before.\nThis:\nCode:\n`return tokentype = BRACKETS;`\nis equivalent to:\nCode:\n```tokentype = BRACKETS;", null, "Popular pages Recent additions", null, "" ]	[ null, "https://cboard.cprogramming.com/images/misc/progress.gif", null, "https://cboard.cprogramming.com/images/misc/quote_icon.png", null, "https://cboard.cprogramming.com/images/misc/progress.gif", null, "https://www.feedburner.com/fb/images/pub/feed-icon16x16.png", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.75278705,"math_prob":0.9839357,"size":1133,"snap":"2022-27-2022-33","text_gpt3_token_len":316,"char_repetition_ratio":0.14880425,"word_repetition_ratio":0.00913242,"special_character_ratio":0.37511033,"punctuation_ratio":0.15813954,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9519759,"pos_list":[0,1,2,3,4,5,6,7,8],"im_url_duplicate_count":[null,null,null,null,null,null,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-07-01T05:57:51Z\",\"WARC-Record-ID\":\"<urn:uuid:d70e2621-a3db-460d-83eb-407d54e17260>\",\"Content-Length\":\"47876\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:3cf3bf6b-f372-4355-b069-c706a510826e>\",\"WARC-Concurrent-To\":\"<urn:uuid:cd053b0e-7325-4957-985b-1f1f416f04bf>\",\"WARC-IP-Address\":\"198.46.93.160\",\"WARC-Target-URI\":\"https://cboard.cprogramming.com/c-programming/169097-gettoken-function.html?s=4df223358d054aee44368c2b27f16d9c\",\"WARC-Payload-Digest\":\"sha1:U7J4WGPKOK7OXQRHFQ24T2BIGGUI6QY3\",\"WARC-Block-Digest\":\"sha1:JYXCV3WHM4Y7FFGQ2W7RZJGASEP5REUI\",\"WARC-Identified-Payload-Type\":\"application/xhtml+xml\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-27/CC-MAIN-2022-27_segments_1656103920118.49_warc_CC-MAIN-20220701034437-20220701064437-00301.warc.gz\"}"}
https://math.stackexchange.com/questions/8814/funny-identities?page=3&tab=votes	[ "Funny identities [closed]\n\nHere is a funny exercise $$\\sin(x - y) \\sin(x + y) = (\\sin x - \\sin y)(\\sin x + \\sin y).$$ (If you prove it don't publish it here please). Do you have similar examples?\n\nclosed as primarily opinion-based by Najib Idrissi, user98602, user147263, apnorton, user2345215Jan 31 '15 at 18:16\n\nMany good questions generate some degree of opinion based on expert experience, but answers to this question will tend to be almost entirely based on opinions, rather than facts, references, or specific expertise. If this question can be reworded to fit the rules in the help center, please edit the question.\n\n• This is related. – J. M. is a poor mathematician Nov 3 '10 at 22:35\n• I noticed that your expression can be also written as $\\sin(x - y) \\sin(x + y) = (\\cos y + \\cos x) (\\cos y - \\cos x)$ – Quixotic Nov 4 '10 at 11:09\n• I have tripped up many calculus students with this one: $log(1+2+3)=log1+log2+log3$. I am evil... – user641 Dec 8 '12 at 1:23\n• @SteveD If only we could find an odd example... – peoplepower Jan 13 '13 at 0:31\n• Almost an identity: $$\\sqrt{123456790}\\approx 11111.11111\\,.$$ – Jakob Werner Jul 12 '14 at 18:47\n\nIf we define $P$ as the infinite lower triangular matrix where $P_{i,j} = \\binom{i}{j}$ (we can call it the Pascal Matrix), then $$P^k_{i,j} = \\binom{i}{j}k^{i-j}$$\n\nwhere $P^k_{i,j}$ is the element of $P^k$ in the position $i,j.$\n\nI have another one, but I'm quite unwilling to post this here because it's MINE, I haven't found it anywhere, so don't steal this.\n\nLet us take the four most important mathematical constants: The Euler number $e$, the Aurea Golden Ratio $\\phi$, the Euler-Mascheroni constant $\\gamma$ and finally $\\pi$. Well we can see easily that\n\n$$e\\cdot\\gamma\\cdot\\pi\\cdot\\phi \\approx e + \\gamma + \\pi + \\phi$$\n\n• I wouldn't take credit for this dude – user285523 Nov 11 '15 at 2:48\n\n$$\\int_{-\\infty}^{\\infty}{\\sin\\left(x\\right) \\over x}\\,{\\rm d}x = \\pi\\int_{-1}^{1}\\delta\\left(k\\right)\\,{\\rm d}k$$" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.83198786,"math_prob":0.99647987,"size":2105,"snap":"2019-26-2019-30","text_gpt3_token_len":678,"char_repetition_ratio":0.10661589,"word_repetition_ratio":0.01179941,"special_character_ratio":0.3567696,"punctuation_ratio":0.12009238,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99986756,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-06-26T08:50:56Z\",\"WARC-Record-ID\":\"<urn:uuid:0268a7a8-a4de-48d7-b156-94708faa95c5>\",\"Content-Length\":\"140516\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:431a110b-773a-45a2-8260-ec112f737593>\",\"WARC-Concurrent-To\":\"<urn:uuid:22989bb4-edae-46ea-bb67-707492e54ba1>\",\"WARC-IP-Address\":\"151.101.193.69\",\"WARC-Target-URI\":\"https://math.stackexchange.com/questions/8814/funny-identities?page=3&tab=votes\",\"WARC-Payload-Digest\":\"sha1:AIZFE4NRAKHGO2NL7UDWOH2G4UPSGGXZ\",\"WARC-Block-Digest\":\"sha1:IXUO6KXITNKXWMJ6SONT3PWOAX43TUCG\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-26/CC-MAIN-2019-26_segments_1560628000231.40_warc_CC-MAIN-20190626073946-20190626095946-00333.warc.gz\"}"}