Search Results

Search found 3425 results on 137 pages for 'polynomial math'.

Page 26/137 | < Previous Page | 22 23 24 25 26 27 28 29 30 31 32 33  | Next Page >

  • Move penetrating OBB out of another OBB to resolve collision

    - by Milo
    I'm working on collision resolution for my game. I just need a good way to get an object out of another object if it gets stuck. In this case a car. Here is a typical scenario. The red car is in the green object. How do I correctly get it out so the car can slide along the edge of the object as it should. I tried: if(buildings.size() > 0) { Entity e = buildings.get(0); Vector2D vel = new Vector2D(); vel.x = vehicle.getVelocity().x; vel.y = vehicle.getVelocity().y; vel.normalize(); while(vehicle.getRect().overlaps(e.getRect())) { vehicle.setCenter(vehicle.getCenterX() - vel.x * 0.1f, vehicle.getCenterY() - vel.y * 0.1f); } colided = true; } But that does not work too well. Is there some sort of vector I could calculate to use as the vector to move the car away from the object? Thanks Here is my OBB2D class: public class OBB2D { // Corners of the box, where 0 is the lower left. private Vector2D corner[] = new Vector2D[4]; private Vector2D center = new Vector2D(); private Vector2D extents = new Vector2D(); private RectF boundingRect = new RectF(); private float angle; //Two edges of the box extended away from corner[0]. private Vector2D axis[] = new Vector2D[2]; private double origin[] = new double[2]; public OBB2D(Vector2D center, float w, float h, float angle) { set(center,w,h,angle); } public OBB2D(float left, float top, float width, float height) { set(new Vector2D(left + (width / 2), top + (height / 2)),width,height,0.0f); } public void set(Vector2D center,float w, float h,float angle) { Vector2D X = new Vector2D( (float)Math.cos(angle), (float)Math.sin(angle)); Vector2D Y = new Vector2D((float)-Math.sin(angle), (float)Math.cos(angle)); X = X.multiply( w / 2); Y = Y.multiply( h / 2); corner[0] = center.subtract(X).subtract(Y); corner[1] = center.add(X).subtract(Y); corner[2] = center.add(X).add(Y); corner[3] = center.subtract(X).add(Y); computeAxes(); extents.x = w / 2; extents.y = h / 2; computeDimensions(center,angle); } private void computeDimensions(Vector2D center,float angle) { this.center.x = center.x; this.center.y = center.y; this.angle = angle; boundingRect.left = Math.min(Math.min(corner[0].x, corner[3].x), Math.min(corner[1].x, corner[2].x)); boundingRect.top = Math.min(Math.min(corner[0].y, corner[1].y),Math.min(corner[2].y, corner[3].y)); boundingRect.right = Math.max(Math.max(corner[1].x, corner[2].x), Math.max(corner[0].x, corner[3].x)); boundingRect.bottom = Math.max(Math.max(corner[2].y, corner[3].y),Math.max(corner[0].y, corner[1].y)); } public void set(RectF rect) { set(new Vector2D(rect.centerX(),rect.centerY()),rect.width(),rect.height(),0.0f); } // Returns true if other overlaps one dimension of this. private boolean overlaps1Way(OBB2D other) { for (int a = 0; a < axis.length; ++a) { double t = other.corner[0].dot(axis[a]); // Find the extent of box 2 on axis a double tMin = t; double tMax = t; for (int c = 1; c < corner.length; ++c) { t = other.corner[c].dot(axis[a]); if (t < tMin) { tMin = t; } else if (t > tMax) { tMax = t; } } // We have to subtract off the origin // See if [tMin, tMax] intersects [0, 1] if ((tMin > 1 + origin[a]) || (tMax < origin[a])) { // There was no intersection along this dimension; // the boxes cannot possibly overlap. return false; } } // There was no dimension along which there is no intersection. // Therefore the boxes overlap. return true; } //Updates the axes after the corners move. Assumes the //corners actually form a rectangle. private void computeAxes() { axis[0] = corner[1].subtract(corner[0]); axis[1] = corner[3].subtract(corner[0]); // Make the length of each axis 1/edge length so we know any // dot product must be less than 1 to fall within the edge. for (int a = 0; a < axis.length; ++a) { axis[a] = axis[a].divide((axis[a].length() * axis[a].length())); origin[a] = corner[0].dot(axis[a]); } } public void moveTo(Vector2D center) { Vector2D centroid = (corner[0].add(corner[1]).add(corner[2]).add(corner[3])).divide(4.0f); Vector2D translation = center.subtract(centroid); for (int c = 0; c < 4; ++c) { corner[c] = corner[c].add(translation); } computeAxes(); computeDimensions(center,angle); } // Returns true if the intersection of the boxes is non-empty. public boolean overlaps(OBB2D other) { if(right() < other.left()) { return false; } if(bottom() < other.top()) { return false; } if(left() > other.right()) { return false; } if(top() > other.bottom()) { return false; } if(other.getAngle() == 0.0f && getAngle() == 0.0f) { return true; } return overlaps1Way(other) && other.overlaps1Way(this); } public Vector2D getCenter() { return center; } public float getWidth() { return extents.x * 2; } public float getHeight() { return extents.y * 2; } public void setAngle(float angle) { set(center,getWidth(),getHeight(),angle); } public float getAngle() { return angle; } public void setSize(float w,float h) { set(center,w,h,angle); } public float left() { return boundingRect.left; } public float right() { return boundingRect.right; } public float bottom() { return boundingRect.bottom; } public float top() { return boundingRect.top; } public RectF getBoundingRect() { return boundingRect; } public boolean overlaps(float left, float top, float right, float bottom) { if(right() < left) { return false; } if(bottom() < top) { return false; } if(left() > right) { return false; } if(top() > bottom) { return false; } return true; } };

    Read the article

  • A Taxonomy of Numerical Methods v1

    - by JoshReuben
    Numerical Analysis – When, What, (but not how) Once you understand the Math & know C++, Numerical Methods are basically blocks of iterative & conditional math code. I found the real trick was seeing the forest for the trees – knowing which method to use for which situation. Its pretty easy to get lost in the details – so I’ve tried to organize these methods in a way that I can quickly look this up. I’ve included links to detailed explanations and to C++ code examples. I’ve tried to classify Numerical methods in the following broad categories: Solving Systems of Linear Equations Solving Non-Linear Equations Iteratively Interpolation Curve Fitting Optimization Numerical Differentiation & Integration Solving ODEs Boundary Problems Solving EigenValue problems Enjoy – I did ! Solving Systems of Linear Equations Overview Solve sets of algebraic equations with x unknowns The set is commonly in matrix form Gauss-Jordan Elimination http://en.wikipedia.org/wiki/Gauss%E2%80%93Jordan_elimination C++: http://www.codekeep.net/snippets/623f1923-e03c-4636-8c92-c9dc7aa0d3c0.aspx Produces solution of the equations & the coefficient matrix Efficient, stable 2 steps: · Forward Elimination – matrix decomposition: reduce set to triangular form (0s below the diagonal) or row echelon form. If degenerate, then there is no solution · Backward Elimination –write the original matrix as the product of ints inverse matrix & its reduced row-echelon matrix à reduce set to row canonical form & use back-substitution to find the solution to the set Elementary ops for matrix decomposition: · Row multiplication · Row switching · Add multiples of rows to other rows Use pivoting to ensure rows are ordered for achieving triangular form LU Decomposition http://en.wikipedia.org/wiki/LU_decomposition C++: http://ganeshtiwaridotcomdotnp.blogspot.co.il/2009/12/c-c-code-lu-decomposition-for-solving.html Represent the matrix as a product of lower & upper triangular matrices A modified version of GJ Elimination Advantage – can easily apply forward & backward elimination to solve triangular matrices Techniques: · Doolittle Method – sets the L matrix diagonal to unity · Crout Method - sets the U matrix diagonal to unity Note: both the L & U matrices share the same unity diagonal & can be stored compactly in the same matrix Gauss-Seidel Iteration http://en.wikipedia.org/wiki/Gauss%E2%80%93Seidel_method C++: http://www.nr.com/forum/showthread.php?t=722 Transform the linear set of equations into a single equation & then use numerical integration (as integration formulas have Sums, it is implemented iteratively). an optimization of Gauss-Jacobi: 1.5 times faster, requires 0.25 iterations to achieve the same tolerance Solving Non-Linear Equations Iteratively find roots of polynomials – there may be 0, 1 or n solutions for an n order polynomial use iterative techniques Iterative methods · used when there are no known analytical techniques · Requires set functions to be continuous & differentiable · Requires an initial seed value – choice is critical to convergence à conduct multiple runs with different starting points & then select best result · Systematic - iterate until diminishing returns, tolerance or max iteration conditions are met · bracketing techniques will always yield convergent solutions, non-bracketing methods may fail to converge Incremental method if a nonlinear function has opposite signs at 2 ends of a small interval x1 & x2, then there is likely to be a solution in their interval – solutions are detected by evaluating a function over interval steps, for a change in sign, adjusting the step size dynamically. Limitations – can miss closely spaced solutions in large intervals, cannot detect degenerate (coinciding) solutions, limited to functions that cross the x-axis, gives false positives for singularities Fixed point method http://en.wikipedia.org/wiki/Fixed-point_iteration C++: http://books.google.co.il/books?id=weYj75E_t6MC&pg=PA79&lpg=PA79&dq=fixed+point+method++c%2B%2B&source=bl&ots=LQ-5P_taoC&sig=lENUUIYBK53tZtTwNfHLy5PEWDk&hl=en&sa=X&ei=wezDUPW1J5DptQaMsIHQCw&redir_esc=y#v=onepage&q=fixed%20point%20method%20%20c%2B%2B&f=false Algebraically rearrange a solution to isolate a variable then apply incremental method Bisection method http://en.wikipedia.org/wiki/Bisection_method C++: http://numericalcomputing.wordpress.com/category/algorithms/ Bracketed - Select an initial interval, keep bisecting it ad midpoint into sub-intervals and then apply incremental method on smaller & smaller intervals – zoom in Adv: unaffected by function gradient à reliable Disadv: slow convergence False Position Method http://en.wikipedia.org/wiki/False_position_method C++: http://www.dreamincode.net/forums/topic/126100-bisection-and-false-position-methods/ Bracketed - Select an initial interval , & use the relative value of function at interval end points to select next sub-intervals (estimate how far between the end points the solution might be & subdivide based on this) Newton-Raphson method http://en.wikipedia.org/wiki/Newton's_method C++: http://www-users.cselabs.umn.edu/classes/Summer-2012/csci1113/index.php?page=./newt3 Also known as Newton's method Convenient, efficient Not bracketed – only a single initial guess is required to start iteration – requires an analytical expression for the first derivative of the function as input. Evaluates the function & its derivative at each step. Can be extended to the Newton MutiRoot method for solving multiple roots Can be easily applied to an of n-coupled set of non-linear equations – conduct a Taylor Series expansion of a function, dropping terms of order n, rewrite as a Jacobian matrix of PDs & convert to simultaneous linear equations !!! Secant Method http://en.wikipedia.org/wiki/Secant_method C++: http://forum.vcoderz.com/showthread.php?p=205230 Unlike N-R, can estimate first derivative from an initial interval (does not require root to be bracketed) instead of inputting it Since derivative is approximated, may converge slower. Is fast in practice as it does not have to evaluate the derivative at each step. Similar implementation to False Positive method Birge-Vieta Method http://mat.iitm.ac.in/home/sryedida/public_html/caimna/transcendental/polynomial%20methods/bv%20method.html C++: http://books.google.co.il/books?id=cL1boM2uyQwC&pg=SA3-PA51&lpg=SA3-PA51&dq=Birge-Vieta+Method+c%2B%2B&source=bl&ots=QZmnDTK3rC&sig=BPNcHHbpR_DKVoZXrLi4nVXD-gg&hl=en&sa=X&ei=R-_DUK2iNIjzsgbE5ID4Dg&redir_esc=y#v=onepage&q=Birge-Vieta%20Method%20c%2B%2B&f=false combines Horner's method of polynomial evaluation (transforming into lesser degree polynomials that are more computationally efficient to process) with Newton-Raphson to provide a computational speed-up Interpolation Overview Construct new data points for as close as possible fit within range of a discrete set of known points (that were obtained via sampling, experimentation) Use Taylor Series Expansion of a function f(x) around a specific value for x Linear Interpolation http://en.wikipedia.org/wiki/Linear_interpolation C++: http://www.hamaluik.com/?p=289 Straight line between 2 points à concatenate interpolants between each pair of data points Bilinear Interpolation http://en.wikipedia.org/wiki/Bilinear_interpolation C++: http://supercomputingblog.com/graphics/coding-bilinear-interpolation/2/ Extension of the linear function for interpolating functions of 2 variables – perform linear interpolation first in 1 direction, then in another. Used in image processing – e.g. texture mapping filter. Uses 4 vertices to interpolate a value within a unit cell. Lagrange Interpolation http://en.wikipedia.org/wiki/Lagrange_polynomial C++: http://www.codecogs.com/code/maths/approximation/interpolation/lagrange.php For polynomials Requires recomputation for all terms for each distinct x value – can only be applied for small number of nodes Numerically unstable Barycentric Interpolation http://epubs.siam.org/doi/pdf/10.1137/S0036144502417715 C++: http://www.gamedev.net/topic/621445-barycentric-coordinates-c-code-check/ Rearrange the terms in the equation of the Legrange interpolation by defining weight functions that are independent of the interpolated value of x Newton Divided Difference Interpolation http://en.wikipedia.org/wiki/Newton_polynomial C++: http://jee-appy.blogspot.co.il/2011/12/newton-divided-difference-interpolation.html Hermite Divided Differences: Interpolation polynomial approximation for a given set of data points in the NR form - divided differences are used to approximately calculate the various differences. For a given set of 3 data points , fit a quadratic interpolant through the data Bracketed functions allow Newton divided differences to be calculated recursively Difference table Cubic Spline Interpolation http://en.wikipedia.org/wiki/Spline_interpolation C++: https://www.marcusbannerman.co.uk/index.php/home/latestarticles/42-articles/96-cubic-spline-class.html Spline is a piecewise polynomial Provides smoothness – for interpolations with significantly varying data Use weighted coefficients to bend the function to be smooth & its 1st & 2nd derivatives are continuous through the edge points in the interval Curve Fitting A generalization of interpolating whereby given data points may contain noise à the curve does not necessarily pass through all the points Least Squares Fit http://en.wikipedia.org/wiki/Least_squares C++: http://www.ccas.ru/mmes/educat/lab04k/02/least-squares.c Residual – difference between observed value & expected value Model function is often chosen as a linear combination of the specified functions Determines: A) The model instance in which the sum of squared residuals has the least value B) param values for which model best fits data Straight Line Fit Linear correlation between independent variable and dependent variable Linear Regression http://en.wikipedia.org/wiki/Linear_regression C++: http://www.oocities.org/david_swaim/cpp/linregc.htm Special case of statistically exact extrapolation Leverage least squares Given a basis function, the sum of the residuals is determined and the corresponding gradient equation is expressed as a set of normal linear equations in matrix form that can be solved (e.g. using LU Decomposition) Can be weighted - Drop the assumption that all errors have the same significance –-> confidence of accuracy is different for each data point. Fit the function closer to points with higher weights Polynomial Fit - use a polynomial basis function Moving Average http://en.wikipedia.org/wiki/Moving_average C++: http://www.codeproject.com/Articles/17860/A-Simple-Moving-Average-Algorithm Used for smoothing (cancel fluctuations to highlight longer-term trends & cycles), time series data analysis, signal processing filters Replace each data point with average of neighbors. Can be simple (SMA), weighted (WMA), exponential (EMA). Lags behind latest data points – extra weight can be given to more recent data points. Weights can decrease arithmetically or exponentially according to distance from point. Parameters: smoothing factor, period, weight basis Optimization Overview Given function with multiple variables, find Min (or max by minimizing –f(x)) Iterative approach Efficient, but not necessarily reliable Conditions: noisy data, constraints, non-linear models Detection via sign of first derivative - Derivative of saddle points will be 0 Local minima Bisection method Similar method for finding a root for a non-linear equation Start with an interval that contains a minimum Golden Search method http://en.wikipedia.org/wiki/Golden_section_search C++: http://www.codecogs.com/code/maths/optimization/golden.php Bisect intervals according to golden ratio 0.618.. Achieves reduction by evaluating a single function instead of 2 Newton-Raphson Method Brent method http://en.wikipedia.org/wiki/Brent's_method C++: http://people.sc.fsu.edu/~jburkardt/cpp_src/brent/brent.cpp Based on quadratic or parabolic interpolation – if the function is smooth & parabolic near to the minimum, then a parabola fitted through any 3 points should approximate the minima – fails when the 3 points are collinear , in which case the denominator is 0 Simplex Method http://en.wikipedia.org/wiki/Simplex_algorithm C++: http://www.codeguru.com/cpp/article.php/c17505/Simplex-Optimization-Algorithm-and-Implemetation-in-C-Programming.htm Find the global minima of any multi-variable function Direct search – no derivatives required At each step it maintains a non-degenerative simplex – a convex hull of n+1 vertices. Obtains the minimum for a function with n variables by evaluating the function at n-1 points, iteratively replacing the point of worst result with the point of best result, shrinking the multidimensional simplex around the best point. Point replacement involves expanding & contracting the simplex near the worst value point to determine a better replacement point Oscillation can be avoided by choosing the 2nd worst result Restart if it gets stuck Parameters: contraction & expansion factors Simulated Annealing http://en.wikipedia.org/wiki/Simulated_annealing C++: http://code.google.com/p/cppsimulatedannealing/ Analogy to heating & cooling metal to strengthen its structure Stochastic method – apply random permutation search for global minima - Avoid entrapment in local minima via hill climbing Heating schedule - Annealing schedule params: temperature, iterations at each temp, temperature delta Cooling schedule – can be linear, step-wise or exponential Differential Evolution http://en.wikipedia.org/wiki/Differential_evolution C++: http://www.amichel.com/de/doc/html/ More advanced stochastic methods analogous to biological processes: Genetic algorithms, evolution strategies Parallel direct search method against multiple discrete or continuous variables Initial population of variable vectors chosen randomly – if weighted difference vector of 2 vectors yields a lower objective function value then it replaces the comparison vector Many params: #parents, #variables, step size, crossover constant etc Convergence is slow – many more function evaluations than simulated annealing Numerical Differentiation Overview 2 approaches to finite difference methods: · A) approximate function via polynomial interpolation then differentiate · B) Taylor series approximation – additionally provides error estimate Finite Difference methods http://en.wikipedia.org/wiki/Finite_difference_method C++: http://www.wpi.edu/Pubs/ETD/Available/etd-051807-164436/unrestricted/EAMPADU.pdf Find differences between high order derivative values - Approximate differential equations by finite differences at evenly spaced data points Based on forward & backward Taylor series expansion of f(x) about x plus or minus multiples of delta h. Forward / backward difference - the sums of the series contains even derivatives and the difference of the series contains odd derivatives – coupled equations that can be solved. Provide an approximation of the derivative within a O(h^2) accuracy There is also central difference & extended central difference which has a O(h^4) accuracy Richardson Extrapolation http://en.wikipedia.org/wiki/Richardson_extrapolation C++: http://mathscoding.blogspot.co.il/2012/02/introduction-richardson-extrapolation.html A sequence acceleration method applied to finite differences Fast convergence, high accuracy O(h^4) Derivatives via Interpolation Cannot apply Finite Difference method to discrete data points at uneven intervals – so need to approximate the derivative of f(x) using the derivative of the interpolant via 3 point Lagrange Interpolation Note: the higher the order of the derivative, the lower the approximation precision Numerical Integration Estimate finite & infinite integrals of functions More accurate procedure than numerical differentiation Use when it is not possible to obtain an integral of a function analytically or when the function is not given, only the data points are Newton Cotes Methods http://en.wikipedia.org/wiki/Newton%E2%80%93Cotes_formulas C++: http://www.siafoo.net/snippet/324 For equally spaced data points Computationally easy – based on local interpolation of n rectangular strip areas that is piecewise fitted to a polynomial to get the sum total area Evaluate the integrand at n+1 evenly spaced points – approximate definite integral by Sum Weights are derived from Lagrange Basis polynomials Leverage Trapezoidal Rule for default 2nd formulas, Simpson 1/3 Rule for substituting 3 point formulas, Simpson 3/8 Rule for 4 point formulas. For 4 point formulas use Bodes Rule. Higher orders obtain more accurate results Trapezoidal Rule uses simple area, Simpsons Rule replaces the integrand f(x) with a quadratic polynomial p(x) that uses the same values as f(x) for its end points, but adds a midpoint Romberg Integration http://en.wikipedia.org/wiki/Romberg's_method C++: http://code.google.com/p/romberg-integration/downloads/detail?name=romberg.cpp&can=2&q= Combines trapezoidal rule with Richardson Extrapolation Evaluates the integrand at equally spaced points The integrand must have continuous derivatives Each R(n,m) extrapolation uses a higher order integrand polynomial replacement rule (zeroth starts with trapezoidal) à a lower triangular matrix set of equation coefficients where the bottom right term has the most accurate approximation. The process continues until the difference between 2 successive diagonal terms becomes sufficiently small. Gaussian Quadrature http://en.wikipedia.org/wiki/Gaussian_quadrature C++: http://www.alglib.net/integration/gaussianquadratures.php Data points are chosen to yield best possible accuracy – requires fewer evaluations Ability to handle singularities, functions that are difficult to evaluate The integrand can include a weighting function determined by a set of orthogonal polynomials. Points & weights are selected so that the integrand yields the exact integral if f(x) is a polynomial of degree <= 2n+1 Techniques (basically different weighting functions): · Gauss-Legendre Integration w(x)=1 · Gauss-Laguerre Integration w(x)=e^-x · Gauss-Hermite Integration w(x)=e^-x^2 · Gauss-Chebyshev Integration w(x)= 1 / Sqrt(1-x^2) Solving ODEs Use when high order differential equations cannot be solved analytically Evaluated under boundary conditions RK for systems – a high order differential equation can always be transformed into a coupled first order system of equations Euler method http://en.wikipedia.org/wiki/Euler_method C++: http://rosettacode.org/wiki/Euler_method First order Runge–Kutta method. Simple recursive method – given an initial value, calculate derivative deltas. Unstable & not very accurate (O(h) error) – not used in practice A first-order method - the local error (truncation error per step) is proportional to the square of the step size, and the global error (error at a given time) is proportional to the step size In evolving solution between data points xn & xn+1, only evaluates derivatives at beginning of interval xn à asymmetric at boundaries Higher order Runge Kutta http://en.wikipedia.org/wiki/Runge%E2%80%93Kutta_methods C++: http://www.dreamincode.net/code/snippet1441.htm 2nd & 4th order RK - Introduces parameterized midpoints for more symmetric solutions à accuracy at higher computational cost Adaptive RK – RK-Fehlberg – estimate the truncation at each integration step & automatically adjust the step size to keep error within prescribed limits. At each step 2 approximations are compared – if in disagreement to a specific accuracy, the step size is reduced Boundary Value Problems Where solution of differential equations are located at 2 different values of the independent variable x à more difficult, because cannot just start at point of initial value – there may not be enough starting conditions available at the end points to produce a unique solution An n-order equation will require n boundary conditions – need to determine the missing n-1 conditions which cause the given conditions at the other boundary to be satisfied Shooting Method http://en.wikipedia.org/wiki/Shooting_method C++: http://ganeshtiwaridotcomdotnp.blogspot.co.il/2009/12/c-c-code-shooting-method-for-solving.html Iteratively guess the missing values for one end & integrate, then inspect the discrepancy with the boundary values of the other end to adjust the estimate Given the starting boundary values u1 & u2 which contain the root u, solve u given the false position method (solving the differential equation as an initial value problem via 4th order RK), then use u to solve the differential equations. Finite Difference Method For linear & non-linear systems Higher order derivatives require more computational steps – some combinations for boundary conditions may not work though Improve the accuracy by increasing the number of mesh points Solving EigenValue Problems An eigenvalue can substitute a matrix when doing matrix multiplication à convert matrix multiplication into a polynomial EigenValue For a given set of equations in matrix form, determine what are the solution eigenvalue & eigenvectors Similar Matrices - have same eigenvalues. Use orthogonal similarity transforms to reduce a matrix to diagonal form from which eigenvalue(s) & eigenvectors can be computed iteratively Jacobi method http://en.wikipedia.org/wiki/Jacobi_method C++: http://people.sc.fsu.edu/~jburkardt/classes/acs2_2008/openmp/jacobi/jacobi.html Robust but Computationally intense – use for small matrices < 10x10 Power Iteration http://en.wikipedia.org/wiki/Power_iteration For any given real symmetric matrix, generate the largest single eigenvalue & its eigenvectors Simplest method – does not compute matrix decomposition à suitable for large, sparse matrices Inverse Iteration Variation of power iteration method – generates the smallest eigenvalue from the inverse matrix Rayleigh Method http://en.wikipedia.org/wiki/Rayleigh's_method_of_dimensional_analysis Variation of power iteration method Rayleigh Quotient Method Variation of inverse iteration method Matrix Tri-diagonalization Method Use householder algorithm to reduce an NxN symmetric matrix to a tridiagonal real symmetric matrix vua N-2 orthogonal transforms     Whats Next Outside of Numerical Methods there are lots of different types of algorithms that I’ve learned over the decades: Data Mining – (I covered this briefly in a previous post: http://geekswithblogs.net/JoshReuben/archive/2007/12/31/ssas-dm-algorithms.aspx ) Search & Sort Routing Problem Solving Logical Theorem Proving Planning Probabilistic Reasoning Machine Learning Solvers (eg MIP) Bioinformatics (Sequence Alignment, Protein Folding) Quant Finance (I read Wilmott’s books – interesting) Sooner or later, I’ll cover the above topics as well.

    Read the article

  • How to calculate an angle from three points?

    - by HelloMoon
    Lets say you have this: P1 = (x=2, y=50) P2 = (x=9, y=40) P3 = (x=5, y=20) Assume that P1 is the center point of a circle. It is always the same. I want the angle that is made up by P2 and P3, or in other words the angle that is next to P1. The inner angle to be precise. It will be always a sharp angle, so less than -90 degrees. I thought: Man, that's simplest geometry maths. But I looked for a formula for like 6 hours now and people talk about most complicated NASA stuff like arcos and vector scalar product stuff. My head feels like in a fridge. Some math gurus here that think this is a simple problem? I think the programing language doesn't matter here but for those who think it does: java and objective-c. need that for both. haven't tagged it for these, though.

    Read the article

  • Is there a calculator with LaTeX-syntax?

    - by Jørgen Fogh
    When I write math in LaTeX I often need to perform simple arithmetic on numbers in my LaTeX source, like 515.1544 + 454 = ???. I usually copy-paste the LaTeX code into Google to get the result, but I still have to manually change the syntax, e.g. \frac{154,7}{25} - (289 - \frac{1337}{42}) must be changed to 154,7/25 - (289 - 1337/42) It seems trivial to write a program to do this for the most commonly used operations. Is there a calculator which understand this syntax? EDIT: I know that doing this perfectly is impossible (because of the halting problem). Doing it for the simple cases I need is trivial. \frac, \cdot, \sqrt and a few other tags would do the trick. The program could just return an error for cases it does not understand.

    Read the article

  • How do i break a number down into a percentage (0 - 100%)? Details inside...

    - by AJ
    I am using a JS progress bar that is set using a percentage: 0 to 100 (percent). I need the progress bar to reach 100% when 160,000 people have signed a certain form. I have the total number of signers set in a PHP variable but am lost on how to do the math to convert that into a percentage that fits within 1 - 100 (so that the progress bar actually reflects the goal of 160,000). I may be missing something obvious here (i suck at anything number-related) so does anyone here have a clue as to how to do this?

    Read the article

  • Mapping A Sphere To A Cube

    - by petrocket
    There is a special way of mapping a cube to a sphere described here: http://mathproofs.blogspot.com/2005/07/mapping-cube-to-sphere.html It is not your basic "normalize the point and your done" approach and gives a much more evenly spaced mapping. I've tried to do the inverse of the mapping going from sphere coords to cube coords and have been unable to come up the working equations. It's a rather complex system of equations with lots of square roots. Any math geniuses want to take a crack at it? Here's the equations in c++ code: sx = x * sqrtf(1.0f - y * y * 0.5f - z * z * 0.5f + y * y * z * z / 3.0f); sy = y * sqrtf(1.0f - z * z * 0.5f - x * x * 0.5f + z * z * x * x / 3.0f); sz = z * sqrtf(1.0f - x * x * 0.5f - y * y * 0.5f + x * x * y * y / 3.0f); sx,sy,sz are the sphere coords and x,y,z are the cube coords.

    Read the article

  • Inverse Logistic Function / Reverse Sigmoid Function

    - by Chanq
    I am currently coding up a fuzzy logic library in java. I have found the equations for all the standard functions - Grade, inverseGrade, Triangle, Trapezoid, Gaussian. However, I can't find the inverse of the sigmoid/ logistic function. The way I have written the logistic function is java is : //f(x) = 1/(1+e(-x)) public double logistic(double x){ return (1/(1+(Math.exp(-x))); } But I can't work out or find the inverse anywhere. My algebraic/calculus abilities are fairly limited, hence why I haven't been able to work out the inverse of the function. Any hints or pointers would be a big help. Thanks

    Read the article

  • width of a frustum at a given distance from the near plane

    - by structolite
    I'm using CML to manage the 3D math in an OpenGL-based interface project I'm making for work. I need to know the width of the viewing frustum at a given distance from the eye point, which is kept as a part of a 4x4 matrix that represents the camera. My goal is to position gui objects along the apparent edge of the viewport, but at some distance into the screen from the near clipping plane. CML has a function to extract the planes of the frustum, giving them back in Ax + By + Cz + D = 0 form. This frustum is perpendicular to the camera, which isn't necessarily aligned with the z axis of the perspective projection. I'd like to extract x and z coordinates so as to pin graphical elements to the sides of the screen at different distances from the camera. What is the best way to go about doing it? Thanks!

    Read the article

  • Union of complex polygons

    - by grenade
    Given two polygons: POLYGON((1 0, 1 8, 6 4, 1 0)) POLYGON((4 1, 3 5, 4 9, 9 5, 4 1),(4 5, 5 7, 6 7, 4 4, 4 5)) How can I calculate the union (combined polygon)? Dave's example uses SQL server to produce the union, but I need to accomplish the same in code. I'm looking for a mathematical formula or code example in any language that exposes the actual math. I am attempting to produce maps that combine countries dynamically into regions. I asked a related question here: http://stackoverflow.com/questions/2653812/grouping-geographical-shapes

    Read the article

  • Make character escape from shot

    - by M28
    Hello all math masters, I got a problem for you: I have a 2D game (top down), and I would like to make the character escape from a shot, but not just walk away from the shot (I mean, don't be pushed by the shot), I want it to have a good dodging skills. The variables are: shotX - shot x position shotY - shot y position shotSpeedX - shot x speed shotSpeedY - shot x speed charX - character x position charY - character y position keyLeft - Set to true to make the character press the to left key keyRight - Set to true to make the character press the to right key keyUp - Set to true to make the character press the to up key keyDown - Set to true to make the character press the down key I can understand the following languages: C/C++ Java Actionscript 2/3 Javascript

    Read the article

  • Java: confirm method Binary division and find remainder is correct?

    - by cadwag
    I am parsing binary files and have to implement a CRC algorithm to ensure the file is not corrupted. Problem is that I can't seem to get the binary math working when using larger numbers. The example I'm trying to get working: BigInteger G = new BigInteger("11001", 2); BigInteger M = new BigInteger("1110010000", 2); BigInteger R = M.remainder(G); I am expecting: R = "0101" But I am getting: R = "1100" I am assuming the remainder of 0101 is correct since it is given to me in this book I am using as a reference for the CRC algorithm (it's not based in Java), but I can't seem to get it working. I can get small binary calculations to work that I have solved by hand, but not the larger ones. I'll admit that I haven't worked the larger ones by hand yet, that is my next step, but I wanted to see if someone could point out a glaring flaw I have in my code. Can anyone confirm or deny that my methodology is correct? Thanks

    Read the article

  • Inaccurate Logarithm in Python

    - by Avihu Turzion
    I work daily with Python 2.4 at my company. I used the versatile logarithm function 'log' from the standard math library, and when I entered log(2**31, 2) it returned 31.000000000000004, which struck me as a bit odd. I did the same thing with other powers of 2, and it worked perfectly. I ran 'log10(2**31) / log10(2)' and I got a round 31.0 I tried running the same original function in Python 3.0.1, assuming that it was fixed in a more advanced version. Why does this happen? Is it possible that there are some inaccuracies in mathematical functions in Python?

    Read the article

  • Factorising program not working. Help required.

    - by Ender
    I am working on a factorisation problem using Fermat's Factorization and for small numbers it is working well. I've been able to calculate the factors (getting the answers from Wolfram Alpha) for small numbers, like the one on the Wikipedia page (5959). Just when I thought I had the problem licked I soon realised that my program was not working when it came to larger numbers. The program follows through the examples from the Wikipedia page, printing out the values a, b, a2 and b2; the results printed for large numbers are not correct. I've followed the pseudocode provided on the Wikipedia page, but am struggling to understand where to go next. Along with the Wikipedia page I have been following this guide. Once again, as my Math knowledge is pretty poor I cannot follow what I need to do next. The code I am using so far is as follows: import java.math.BigInteger; /** * * @author AlexT */ public class Fermat { private BigInteger a, b; private BigInteger b2; private static final BigInteger TWO = BigInteger.valueOf(2); public void fermat(BigInteger N) { // floor(sqrt(N)) BigInteger tmp = getIntSqrt(N); // a <- ceil(sqrt(N)) a = tmp.add(BigInteger.ONE); // b2 <- a*a-N b2 = (a.multiply(a)).subtract(N); final int bitLength = N.bitLength(); BigInteger root = BigInteger.ONE.shiftLeft(bitLength / 2); root = root.add(b2.divide(root)).divide(TWO); // while b2 not square root while(!(isSqrt(b2, root))) { // a <- a + 1 a = a.add(BigInteger.ONE); // b2 <- (a * a) - N b2 = (a.multiply(a)).subtract(N); root = root.add(b2.divide(root)).divide(TWO); } b = getIntSqrt(b2); BigInteger a2 = a.pow(2); // Wrong BigInteger sum = (a.subtract(b)).multiply((a.add(b))); //if(sum.compareTo(N) == 0) { System.out.println("A: " + a + "\nB: " + b); System.out.println("A^2: " + a2 + "\nB^2: " + b2); //} } /** * Is the number provided a perfect Square Root? * @param n * @param root * @return */ private static boolean isSqrt(BigInteger n, BigInteger root) { final BigInteger lowerBound = root.pow(2); final BigInteger upperBound = root.add(BigInteger.ONE).pow(2); return lowerBound.compareTo(n) <= 0 && n.compareTo(upperBound) < 0; } public BigInteger getIntSqrt(BigInteger x) { // It returns s where s^2 < x < (s+1)^2 BigInteger s; // final result BigInteger currentRes = BigInteger.valueOf(0); // init value is 0 BigInteger currentSum = BigInteger.valueOf(0); // init value is 0 BigInteger sum = BigInteger.valueOf(0); String xS = x.toString(); // change input x to a string xS int lengthOfxS = xS.length(); int currentTwoBits; int i=0; // index if(lengthOfxS % 2 != 0) {// if odd length, add a dummy bit xS = "0".concat(xS); // add 0 to the front of string xS lengthOfxS++; } while(i < lengthOfxS){ // go through xS two by two, left to right currentTwoBits = Integer.valueOf(xS.substring(i,i+2)); i += 2; // sum = currentSum*100 + currentTwoBits sum = currentSum.multiply(BigInteger.valueOf(100)); sum = sum.add(BigInteger.valueOf(currentTwoBits)); // subtraction loop do { currentSum = sum; // remember the value before subtract // in next 3 lines, we work out // currentRes = sum - 2*currentRes - 1 sum = sum.subtract(currentRes); // currentRes++ currentRes = currentRes.add(BigInteger.valueOf(1)); sum = sum.subtract(currentRes); } while(sum.compareTo(BigInteger.valueOf(0)) >= 0); // the loop stops when sum < 0 // go one step back currentRes = currentRes.subtract(BigInteger.valueOf(1)); currentRes = currentRes.multiply(BigInteger.valueOf(10)); } s = currentRes.divide(BigInteger.valueOf(10)); // go one step back return s; } /** * @param args the command line arguments */ public static void main(String[] args) { Fermat fermat = new Fermat(); //Works //fermat.fermat(new BigInteger("5959")); // Doesn't Work fermat.fermat(new BigInteger("90283")); } } If anyone can help me out with this problem I'll be eternally grateful.

    Read the article

  • CODE1 at SPOJ - cannot solve it

    - by VaioIsBorn
    I am trying to solve the problem Secret Code on SPOJ, and it's obviously a math problem. The full problem For those who are lazy to go and read, it's like this: a0, a1, a2, ..., an - sequence of N numbers B - a Complex Number (has both real and imaginary components) X = a0 + a1*B + a2*(B^2) + a3*(B^3) + ... + an*(B^n) So if you are given B and X, you should find a0, a1, ..an. I don't know how or where to start, because not even N is known, just X and B. The problem is not as easy as expressing a number in a base B, because B is a complex number. How can it be solved?

    Read the article

  • 3D effect to distort paper

    - by donpal
    This may be a little hard to describe since I don't have a sample. I'm trying to find a math function or full 3d function in php or a similar language that can help me with the following effect: imagine if you were to take a flat sheet or paper and glue it on a glass of water. It wouldn't be flat any more. It would have a curve, and one of its sides might end up being slightly hidden. Anyone can refer me to a good library or resource on the web where such functions can be found?

    Read the article

  • Are all the system's floating points operations the same?

    - by Jj
    We're making this web app in PHP and when working in the reports we have Excel files to compare our results to make sure our coding is doing the right operations. Now we're running into some differences due floating point arithmetics. We're doing the same divisions and multiplications and running into slightly different numbers, that add up to a notable difference. My question is if Excel is delegating it's floating point arithmetic to the CPU and PHP is also relying in the CPU for it's operations. Or does each application implements its own set of math algorithms?

    Read the article

  • Explicitly multiplying values as longs

    - by Bill Szerdy
    I understand that all math is done as the largest data type required to handle the current values but when you transverse a loop how do you explicitly multiply longs? The following code returns 0, I suspect, because of an overflow. long result = 0L; List<Long> temp = (List<Long>) getListOfIntegers(); for (int i = 0; i < temp.size(); i++) { result *= temp.get(i).longValue(); } System.out.println(result);

    Read the article

  • What's a better choice for SQL-backed number crunching - Ruby 1.9, Python 2, Python 3, or PHP 5.3?

    - by Ivan
    Crterias of 'better': fast im math and simple (little of fields, many records) db transactions, convenient to develop/read/extend, flexible, connectible. The task is to use a common web development scripting language to process and calculate long time series and multidimensional surfaces (mostly selectint/inserting sets of floats and dong maths with rhem). The choice is Ruby 1.9, Python 2, Python 3, PHP 5.3, Perl 5.12, JavaScript (node.js). All the data is to be stored in a relational database (due to its heavily multidimensional nature), all the communication with outer world is to be done by means of web services.

    Read the article

  • Calculate proportional width of object (proportion: 1600x1080)

    - by Hans Stauden
    Hello, this jquery question: when I set a specific height to a "#div", i want to set the width of a inner object automatically too (cause i need a width to read it out) [ "#div" ["object"] ] example: (object).css(width: [ CALCULATION ], height: ($("#div").height())+'px' ) the original proportion of the image is: 1600x1080 here's the link to the attachment, take a look at it (tinypic): link text the heights "500px", "600px" and "700px" you can see in the attachment are just examples, the heigth could also be "711px", "623px", "998px" etc. cause the "#div" scales with the browser (i can read out the height of window, that works) my math skills aren't really good, would be great if someone could help me out :-)

    Read the article

  • How do you create a formula that has diminishing returns?

    - by egervari
    I guess this is a math question and not a programming question, but what is a good way to create a formula that has diminishing returns? Here are some example points on how I want the curve to look like. f(1) = 1 f(1.5)= .98 f(2) = .95 f(2.5) = .9 f(3) = .8 f(4) = .7 f(5) = .6 f(10) = .5 f(20) = .25 Notice that as the input gets higher, the percentage decreases rapidly. Is there any way to model a function that has a very smooth and accurate curve that says this? Another way to say it is by using a real example. You know in Diablo II they have Magic Find? There are diminishing returns for magic find. If you get 100%, the real magic find is still 100%. But the more get, your actual magic find goes down. So much that say if you had 1200, your real magic find is probably 450%. So they have a function like: actualMagicFind(magicFind) = // some way to reduced magic find

    Read the article

  • Obtain Latitude and Longitude from a GeoTIFF File

    - by Mikee
    Using GDAL in Python, how do you get the latitude and longitude of a GeoTIFF file? GeoTIFF's do not appear to store any coordinate information. Instead, they store the XY Origin coordinates. However, the XY coordinates do not provide the latitude and longitude of the top left corner and bottom left corner. It appears I will need to do some math to solve this problem, but I don't have a clue on where to start. What procedure is required to have this performed? I know that the GetGeoTransform() method is important for this, however, I don't know what to do with it from there.

    Read the article

  • Fastest implementation of the frac function in C#

    - by user349937
    I would like to implement a frac function in C# (just like the one in hsl here http://msdn.microsoft.com/en-us/library/bb509603%28VS.85%29.aspx) but since it is for a very processor intensive application i would like the best version possible. I was using something like public float Frac(float value) { return value - (float)Math.Truncate(value); } but I'm having precision problems, for example for 2.6f it's returning in the unit test Expected: 0.600000024f But was: 0.599999905f I know that I can convert to decimal the value and then at the end convert to float to obtain the correct result something like this: public float Frac(float value) { return (float)((decimal)value - Decimal.Truncate((decimal)value)); } But I wonder if there is a better way without resorting to decimals...

    Read the article

  • Using exponential smoothing with NaN values

    - by Eric
    I have a sample of some kind that can create somewhat noisy output. The sample is the result of some image processing from a camera, which indicates the heading of a blob of a certain color. It is an angle from around -45° to +45°, or a NaN, which means that the blob is not actually in view. In order to combat the noisy data, I felt that exponential smoothing would do the trick. However, I'm not sure how to handle the NaN values. On the one hand, involving them in the math would result in a NaN average, which would then prevent any meaningful results. On the other hand, ignoring NaN values completely would mean that a "no-detection" scenario would never be reported. And just to complicate things, the data is also noisy in that it can get false NaN value, which ideally would be smoothed somehow to prevent random noise. Any ideas about how I could implement such an exponential smoother?

    Read the article

  • Android: Constructing a triangle based on Geographical information

    - by Aidan
    Hi Guys, I'm constructing a geolocation based application and I'm trying to figure out a way to make my application realise when a user is facing the direction of the given location (a particular long / lat co-ord). I've got the math figured, I just have the triangle to construct. Here's a further clarification of what I want to do.. I just want to know is there a way to get java to construct 2 other co-ordinates based on my orientation in relation to true north and my current co-ordinate? I'd like to construct a tri-angle, 45 degrees out each way of my current location (one of the points) and 1 kilometer in that direction. The problem is I don't know how to make Android/Java recognise that I want to find that point in the direction I'm currently facing.. Anyone got any ideas?

    Read the article

  • Determining if and where a photon will collide with a polygon in 3D space.

    - by Peter
    The problem is straight forward: 1) We have a photon traveling from Point 1 (x,y,z) to Point 2 (x,y,z), both of which could be located anywhere in 3D space. 2) We have a polygon that is both rotated randomly on the x-axis and/or y-axis and also located anywhere in 3D space. 3) We want to find: a) if the photon will collide with the polygon at all and b) if it does where will that be (x,y,z)? An image of the problem: http://dl.dropbox.com/u/3150177/Programming/3D/Math/Photon%20Path/Photon%20Path.png The aim of this is to calculate how the photon's path should be altered from an interaction(s) with the polygon(s). I am reading up on this subject now but I was wondering if anyone could give me a head start. Thanks in advance.

    Read the article

< Previous Page | 22 23 24 25 26 27 28 29 30 31 32 33  | Next Page >