jacobi svd c语言代码
时间: 2023-08-12 19:08:36 浏览: 116
以下是一个使用Jacobi方法实现SVD分解的示例代码:
```c
#include <stdio.h>
#include <math.h>
#define MAX_ITERATIONS 100
#define EPSILON 1e-10
void svd_jacobi(double A[], double U[], double S[], double V[], int m, int n) {
// 初始化U、S、V为单位矩阵
for (int i = 0; i < m; i++) {
for (int j = 0; j < m; j++) {
U[i*m + j] = (i == j) ? 1.0 : 0.0;
}
}
for (int i = 0; i < n; i++) {
for (int j = 0; j < n; j++) {
V[i*n + j] = (i == j) ? 1.0 : 0.0;
}
}
// 迭代更新
int iter = 0;
while (iter < MAX_ITERATIONS) {
double max_offdiag = 0;
int p = 0, q = 0;
// 寻找最大非对角线元素
for (int i = 0; i < m-1; i++) {
for (int j = i+1; j < m; j++) {
double offdiag = fabs(A[i*n + j]);
if (offdiag > max_offdiag) {
max_offdiag = offdiag;
p = i;
q = j;
}
}
}
// 检查收敛条件
if (max_offdiag < EPSILON) {
break;
}
// 计算旋转角度
double theta = 0.5 * atan2(2*A[p*n + q], A[p*n + p] - A[q*n + q]);
double c = cos(theta);
double s = sin(theta);
// 更新A
for (int i = 0; i < m; i++) {
double api = A[p*n + i];
double aqi = A[q*n + i];
A[p*n + i] = c * api + s * aqi;
A[q*n + i] = -s * api + c * aqi;
}
for (int i = 0; i < n; i++) {
double api = A[i*n + p];
double aqi = A[i*n + q];
A[i*n + p] = c * api + s * aqi;
A[i*n + q] = -s * api + c * aqi;
}
// 更新U和V
for (int i = 0; i < m; i++) {
double upi = U[i*m + p];
double uqi = U[i*m + q];
U[i*m + p] = c * upi + s * uqi;
U[i*m + q] = -s * upi + c * uqi;
}
for (int i = 0; i < n; i++) {
double vpi = V[i*n + p];
double vqi = V[i*n + q];
V[i*n + p] = c * vpi + s * vqi;
V[i*n + q] = -s * vpi + c * vqi;
}
iter++;
}
// 提取奇异值并排序
for (int i = 0; i < n; i++) {
S[i] = A[i*n + i];
}
for (int i = 0; i < n-1; i++) {
for (int j = i+1; j < n; j++) {
if (S[i] < S[j]) {
double temp = S[i];
S[i] = S[j];
S[j] = temp;
for (int k = 0; k < m; k++) {
double tempU = U[k*m + i];
U[k*m + i] = U[k*m + j];
U[k*m + j] = tempU;
}
for (int k = 0; k < n; k++) {
double tempV = V[k*n + i];
V[k*n + i] = V[k*n + j];
V[k*n + j] = tempV;
}
}
}
}
}
int main() {
double A[6] = {1, 2, 3, 4, 5, 6}; // 示例输入矩阵
double U[9] = {0}; // 存储左奇异向量
double S[3] = {0}; // 存储奇异值
double V[4] = {0}; // 存储右奇异向量
svd_jacobi(A, U, S, V, 2, 3); // 进行SVD分解
printf("U:\n");
for (int i = 0; i < 2; i++) {
for (int j = 0; j < 2; j++) {
printf("%.2f ", U[i*2 + j]);
}
printf("\n");
}
printf("S:\n");
for (int i = 0; i < 2; i++) {
printf("%.2f ", S[i]);
}
printf("\n");
printf("V:\n");
for (int i = 0; i < 3; i++) {
for (int j = 0; j < 2; j++) {
printf("%.2f ", V[i*2 + j]);
}
printf("\n");
}
return 0;
}
```
这段代码演示了如何使用Jacobi方法进行SVD分解。你可以根据需要修改输入矩阵的大小和元素,然后运行代码以获取相应的左奇异向量、奇异值和右奇异向量。