[1009] | 1 | // |
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| 2 | // Lol Engine - Sample math program: Chebyshev polynomials |
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| 3 | // |
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| 4 | // Copyright: (c) 2005-2011 Sam Hocevar <sam@hocevar.net> |
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| 5 | // This program is free software; you can redistribute it and/or |
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| 6 | // modify it under the terms of the Do What The Fuck You Want To |
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| 7 | // Public License, Version 2, as published by Sam Hocevar. See |
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| 8 | // http://sam.zoy.org/projects/COPYING.WTFPL for more details. |
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| 9 | // |
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| 10 | |
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| 11 | #if !defined __REMEZ_MATRIX_H__ |
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| 12 | #define __REMEZ_MATRIX_H__ |
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| 13 | |
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| 14 | template<int N> struct Matrix |
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| 15 | { |
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| 16 | inline Matrix() {} |
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| 17 | |
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| 18 | Matrix(real x) |
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| 19 | { |
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| 20 | for (int j = 0; j < N; j++) |
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| 21 | for (int i = 0; i < N; i++) |
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| 22 | if (i == j) |
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| 23 | m[i][j] = x; |
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| 24 | else |
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| 25 | m[i][j] = 0; |
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| 26 | } |
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| 27 | |
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| 28 | /* Naive matrix inversion */ |
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| 29 | Matrix<N> inv() const |
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| 30 | { |
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| 31 | Matrix a = *this, b((real)1.0); |
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| 32 | |
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| 33 | /* Inversion method: iterate through all columns and make sure |
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| 34 | * all the terms are 1 on the diagonal and 0 everywhere else */ |
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| 35 | for (int i = 0; i < N; i++) |
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| 36 | { |
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| 37 | /* If the expected coefficient is zero, add one of |
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| 38 | * the other lines. The first we meet will do. */ |
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| 39 | if ((double)a.m[i][i] == 0.0) |
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| 40 | { |
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| 41 | for (int j = i + 1; j < N; j++) |
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| 42 | { |
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| 43 | if ((double)a.m[i][j] == 0.0) |
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| 44 | continue; |
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| 45 | /* Add row j to row i */ |
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| 46 | for (int n = 0; n < N; n++) |
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| 47 | { |
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| 48 | a.m[n][i] += a.m[n][j]; |
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| 49 | b.m[n][i] += b.m[n][j]; |
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| 50 | } |
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| 51 | break; |
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| 52 | } |
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| 53 | } |
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| 54 | |
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| 55 | /* Now we know the diagonal term is non-zero. Get its inverse |
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| 56 | * and use that to nullify all other terms in the column */ |
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| 57 | real x = (real)1.0 / a.m[i][i]; |
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| 58 | for (int j = 0; j < N; j++) |
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| 59 | { |
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| 60 | if (j == i) |
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| 61 | continue; |
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| 62 | real mul = x * a.m[i][j]; |
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| 63 | for (int n = 0; n < N; n++) |
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| 64 | { |
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| 65 | a.m[n][j] -= mul * a.m[n][i]; |
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| 66 | b.m[n][j] -= mul * b.m[n][i]; |
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| 67 | } |
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| 68 | } |
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| 69 | |
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| 70 | /* Finally, ensure the diagonal term is 1 */ |
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| 71 | for (int n = 0; n < N; n++) |
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| 72 | { |
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| 73 | a.m[n][i] *= x; |
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| 74 | b.m[n][i] *= x; |
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| 75 | } |
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| 76 | } |
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| 77 | |
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| 78 | return b; |
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| 79 | } |
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| 80 | |
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| 81 | void print() const |
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| 82 | { |
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| 83 | for (int j = 0; j < N; j++) |
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| 84 | { |
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| 85 | for (int i = 0; i < N; i++) |
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| 86 | printf("%9.5f ", (double)m[j][i]); |
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| 87 | printf("\n"); |
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| 88 | } |
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| 89 | } |
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| 90 | |
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| 91 | real m[N][N]; |
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| 92 | }; |
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| 93 | |
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| 94 | #endif /* __REMEZ_MATRIX_H__ */ |
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| 95 | |
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