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