SOR迭代解线性方程C语言

以下是使用C语言实现SOR迭代解线性方程的示例代码：

#include <stdio.h>
#include <stdlib.h>
#include <math.h>

#define MAX_ITER 1000 // 最大迭代次数
#define TOL 1e-6 // 迭代精度

void sor(double **A, double *b, double *x, int n, double w) {
    int i, j, k;
    double sum;
    for (k = 0; k < MAX_ITER; k++) {
        for (i = 0; i < n; i++) {
            sum = 0.0;
            for (j = 0; j < n; j++) {
                if (j != i) {
                    sum += A[i][j] * x[j];
                }
            }
            x[i] = (1 - w) * x[i] + w * (b[i] - sum) / A[i][i];
        }
        // 判断迭代是否收敛
        double err = 0.0;
        for (i = 0; i < n; i++) {
            sum = 0.0;
            for (j = 0; j < n; j++) {
                sum += A[i][j] * x[j];
            }
            err += fabs(sum - b[i]);
        }
        if (err < TOL) {
            printf("Converged after %d iterations.\n", k + 1);
            return;        }
    }
    printf("Failed to converge after %d iterations.\n", MAX_ITER);
}

int main() {
    int n = 3; // 矩阵维数
    double **A = (double **)malloc(n * sizeof(double *));
    for (int i = 0; i < n; i++) {
        A[i] = (double *)malloc(n * sizeof(double));
    }
    double *b = (double *)malloc(n * sizeof(double));
    double *x = (double *)malloc(n * sizeof(double));
    // 初始化矩阵和向量
    A[0][0] = 4.0; A[0][1] = -1.0; A[0][2] = 0.0;
    A[1][0] = -1.0; A[1][1] = 4.0; A[1][2] = -1.0;
    A[2][0] = 0.0; A[2][1] = -1.0; A[2][2] = 4.0;
    b[0] = 15.0; b[1] = 10.0; b[2] = 10.0;
    for (int i = 0; i < n; i++) {
        x[i] = 0.0;
    }
    double w = 1.25; // 松弛因子
    sor(A, b, x, n, w);
    // 输出结果
    printf("Solution:\n");
    for (int i = 0; i < n; i++) {
        printf("%f\n", x[i]);
    }
    // 释放内存
    for (int i = 0; i < n; i++) {
        free(A[i]);
    }
    free(A);
    free(b);
    free(x);
    return 0;
}