svd奇异值分解 c语言实现
时间: 2023-08-28 18:07:57 浏览: 408
SVD(奇异值分解)是一种矩阵分解方法,可以将矩阵分解为三个矩阵的乘积。在C语言中,可以使用以下代码实现SVD奇异值分解:
```c
#include <stdio.h>
#include <stdlib.h>
#include <math.h>
#define TOLERANCE 1.0e-10 // 精度
void svd(int m, int n, double **a, double **u, double *w, double **v) {
int flag, i, its, j, jj, k, l, nm;
double c, f, h, s, x, y, z;
double anorm = 0.0, g = 0.0, scale = 0.0;
double *rv1;
if (m < n) {
fprintf(stderr, "SVD failed: m < n\n");
return;
}
rv1 = (double *) malloc(n * sizeof(double));
for (i = 0; i < n; i++) {
l = i + 1;
rv1[i] = scale * g;
g = s = scale = 0.0;
if (i < m) {
for (k = i; k < m; k++) {
scale += fabs(a[k][i]);
}
if (scale) {
for (k = i; k < m; k++) {
a[k][i] /= scale;
s += a[k][i] * a[k][i];
}
f = a[i][i];
g = -SIGN(sqrt(s), f);
h = f * g - s;
a[i][i] = f - g;
for (j = l; j < n; j++) {
for (s = 0.0, k = i; k < m; k++) {
s += a[k][i] * a[k][j];
}
f = s / h;
for (k = i; k < m; k++) {
a[k][j] += f * a[k][i];
}
}
for (k = i; k < m; k++) {
a[k][i] *= scale;
}
}
}
w[i] = scale * g;
g = s = scale = 0.0;
if (i < m && i != n - 1) {
for (k = l; k < n; k++) {
scale += fabs(a[i][k]);
}
if (scale) {
for (k = l; k < n; k++) {
a[i][k] /= scale;
s += a[i][k] * a[i][k];
}
f = a[i][l];
g = -SIGN(sqrt(s), f);
h = f * g - s;
a[i][l] = f - g;
for (k = l; k < n; k++) {
rv1[k] = a[i][k] / h;
}
for (j = l; j < m; j++) {
for (s = 0.0, k = l; k < n; k++) {
s += a[j][k] * a[i][k];
}
for (k = l; k < n; k++) {
a[j][k] += s * rv1[k];
}
}
for (k = l; k < n; k++) {
a[i][k] *= scale;
}
}
}
anorm = MAX(anorm, (fabs(w[i]) + fabs(rv1[i])));
}
for (i = n - 1; i >= 0; i--) {
if (i < n - 1) {
if (g) {
for (j = l; j < n; j++) {
v[j][i] = (a[i][j] / a[i][l]) / g;
}
for (j = l; j < n; j++) {
for (s = 0.0, k = l; k < n; k++) {
s += a[i][k] * v[k][j];
}
for (k = l; k < n; k++) {
v[k][j] += s * v[k][i];
}
}
}
for (j = l; j < n; j++) {
v[i][j] = v[j][i] = 0.0;
}
}
v[i][i] = 1.0;
g = rv1[i];
l = i;
}
for (i = MIN(m, n) - 1; i >= 0; i--) {
l = i + 1;
g = w[i];
for (j = l; j < n; j++) {
a[i][j] = 0.0;
}
if (g) {
g = 1.0 / g;
for (j = l; j < n; j++) {
for (s = 0.0, k = l; k < m; k++) {
s += a[k][i] * a[k][j];
}
f = (s / a[i][i]) * g;
for (k = i; k < m; k++) {
a[k][j] += f * a[k][i];
}
}
for (j = i; j < m; j++) {
a[j][i] *= g;
}
} else {
for (j = i; j < m; j++) {
a[j][i] = 0.0;
}
}
++a[i][i];
}
for (k = n - 1; k >= 0; k--) {
for (its = 1; its <= 30; its++) {
flag = 1;
for (l = k; l >= 0; l--) {
nm = l - 1;
if (fabs(rv1[l]) + anorm == anorm) {
flag = 0;
break;
}
if (fabs(w[nm]) + anorm == anorm) {
break;
}
}
if (flag) {
c = 0.0;
s = 1.0;
for (i = l; i <= k; i++) {
f = s * rv1[i];
if (fabs(f) + anorm != anorm) {
g = w[i];
h = sqrt(f * f + g * g);
w[i] = h;
h = 1.0 / h;
c = g * h;
s = (-f * h);
for (j = 0; j < m; j++) {
y = a[j][nm];
z = a[j][i];
a[j][nm] = y * c + z * s;
a[j][i] = z * c - y * s;
}
}
}
}
z = w[k];
if (l == k) {
if (z < 0.0) {
w[k] = -z;
for (j = 0; j < n; j++) {
v[j][k] = (-v[j][k]);
}
}
break;
}
if (its == 30) {
fprintf(stderr, "SVD failed: no convergence after %d iterations\n", its);
return;
}
x = w[l];
nm = k - 1;
y = w[nm];
g = rv1[nm];
h = rv1[k];
f = ((y - z) * (y + z) + (g - h) * (g + h)) / (2.0 * h * y);
g = sqrt(f * f + 1.0);
f = ((x - z) * (x + z) + h * ((y / (f + SIGN(g, f))) - h)) / x;
c = s = 1.0;
for (j = l; j <= nm; j++) {
i = j + 1;
g = rv1[i];
y = w[i];
h = s * g;
g = c * g;
z = sqrt(f * f + h * h);
rv1[j] = z;
c = f / z;
s = h / z;
f = (x * c) + (g * s);
g = (g * c) - (x * s);
h = y * s;
y *= c;
for (jj = 0; jj < n; jj++) {
x = v[jj][j];
z = v[jj][i];
v[jj][j] = x * c + z * s;
v[jj][i] = z * c - x * s;
}
z = sqrt(f * f + h * h);
w[j] = z;
if (z) {
z = 1.0 / z;
c = f * z;
s = h * z;
}
f = (c * g) + (s * y);
x = (c * y) - (s * g);
for (jj = 0; jj < m; jj++) {
y = a[jj][j];
z = a[jj][i];
a[jj][j] = y * c + z * s;
a[jj][i] = z * c - y * s;
}
}
rv1[l] = 0.0;
rv1[k] = f;
w[k] = x;
}
}
free(rv1);
}
```
其中,m和n分别是矩阵A的行数和列数,a是一个m行n列的矩阵,u是一个m行m列的矩阵,v是一个n行n列的矩阵,w是一个长度为n的一维数组,用于存储奇异值。
需要注意的是,这段代码是从Numerical Recipes in C中摘取的,但是有一些宏定义(如SIGN、MAX、MIN)需要自己定义或者修改。此外,代码中还需要使用一些基本的数学函数,如fabs(求绝对值)、sqrt(求平方根)等,需要添加头文件<math.h>。
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