c语言分而治之（施特拉森矩阵乘法）

news2025/1/11 5:00:18

给定两个大小分别为 nxn 的方阵 A 和 B，求它们的乘法矩阵。
朴素方法：以下是两个矩阵相乘的简单方法。

void multiply(int A[][N], int B[][N], int C[][N])
{
for (int i = 0; i < N; i++)
{
for (int j = 0; j < N; j++)
{
C[i][j] = 0;
for (int k = 0; k < N; k++)
{
C[i][j] += A[i][k]*B[k][j];
}
}
}
}

上述方法的时间复杂度为O(N 3 )。

基本的 C 语言实现施特拉森矩阵乘法的示例代码：

#include <stdio.h>

void strassen(int n, int A[][n], int B[][n], int C[][n]) {
if (n == 1) {
C[0][0] = A[0][0] * B[0][0];
return;
}

int newSize = n / 2;
int A11[newSize][newSize], A12[newSize][newSize], A21[newSize][newSize], A22[newSize][newSize];
int B11[newSize][newSize], B12[newSize][newSize], B21[newSize][newSize], B22[newSize][newSize];
int C11[newSize][newSize], C12[newSize][newSize], C21[newSize][newSize], C22[newSize][newSize];
int P1[newSize][newSize], P2[newSize][newSize], P3[newSize][newSize], P4[newSize][newSize], P5[newSize][newSize], P6[newSize][newSize], P7[newSize][newSize];
int temp1[newSize][newSize], temp2[newSize][newSize];

for (int i = 0; i < newSize; i++) {
for (int j = 0; j < newSize; j++) {
A11[i][j] = A[i][j];
A12[i][j] = A[i][j + newSize];
A21[i][j] = A[i + newSize][j];
A22[i][j] = A[i + newSize][j + newSize];

B11[i][j] = B[i][j];
B12[i][j] = B[i][j + newSize];
B21[i][j] = B[i + newSize][j];
B22[i][j] = B[i + newSize][j + newSize];
}
}

// Compute the seven products
// P1 = A11 * (B12 - B22)
// P2 = (A11 + A12) * B22
// P3 = (A21 + A22) * B11
// P4 = A22 * (B21 - B11)
// P5 = (A11 + A22) * (B11 + B22)
// P6 = (A12 - A22) * (B21 + B22)
// P7 = (A11 - A21) * (B11 + B12)

// Calculate the intermediate matrices P1, P2, P3, P4, P5, P6, P7

// Compute submatrices C11, C12, C21, C22 using the products
// C11 = P5 + P4 - P2 + P6
// C12 = P1 + P2
// C21 = P3 + P4
// C22 = P5 + P1 - P3 - P7

// Construct the final matrix C using the submatrices C11, C12, C21, C22
}

int main() {
int n;
printf("Enter the size of the matrices: ");
scanf("%d", &n);

int A[n][n], B[n][n], C[n][n];

printf("Enter matrix A:\n");
for (int i = 0; i < n; i++) {
for (int j = 0; j < n; j++) {
scanf("%d", &A[i][j]);
}
}

printf("Enter matrix B:\n");
for (int i = 0; i < n; i++) {
for (int j = 0; j < n; j++) {
scanf("%d", &B[i][j]);
}
}

strassen(n, A, B, C);

printf("Matrix product C:\n");
for (int i = 0; i < n; i++) {
for (int j = 0; j < n; j++) {
printf("%d ", C[i][j]);
}
printf("\n");
}

return 0;
}
        注意，上述代码主要展示了施特拉森矩阵乘法的基本思路，具体的矩阵拆分、计算以及结果合并部分需要进一步完善。实现施特拉森矩阵乘法需要一定的数学知识和编程技巧，确保正确理解算法原理后再进行实现。

分而治之：
以下是两个方阵相乘的简单分而治之方法。
1、将矩阵 A 和 B 分为 4 个大小为 N/2 x N/2 的子矩阵，如下图所示。
2、递归计算以下值。 ae + bg、af + bh、ce + dg 和 cf + dh。

执行：

#include <stdio.h>

#define MAX_SIZE 10

// Function to multiply two matrices using Divide and Conquer
void strassenMatrixMultiply(int n, int A[MAX_SIZE][MAX_SIZE], int B[MAX_SIZE][MAX_SIZE], int C[MAX_SIZE][MAX_SIZE]) {
if (n == 1) {
C[0][0] = A[0][0] * B[0][0];
return;
}

int newSize = n / 2;
int A11[MAX_SIZE][MAX_SIZE], A12[MAX_SIZE][MAX_SIZE], A21[MAX_SIZE][MAX_SIZE], A22[MAX_SIZE][MAX_SIZE];
int B11[MAX_SIZE][MAX_SIZE], B12[MAX_SIZE][MAX_SIZE], B21[MAX_SIZE][MAX_SIZE], B22[MAX_SIZE][MAX_SIZE];
int C11[MAX_SIZE][MAX_SIZE], C12[MAX_SIZE][MAX_SIZE], C21[MAX_SIZE][MAX_SIZE], C22[MAX_SIZE][MAX_SIZE];

int i, j;

// Divide matrices A and B into submatrices
for (i = 0; i < newSize; i++) {
for (j = 0; j < newSize; j++) {
A11[i][j] = A[i][j];
A12[i][j] = A[i][j + newSize];
A21[i][j] = A[i + newSize][j];
A22[i][j] = A[i + newSize][j + newSize];

B11[i][j] = B[i][j];
B12[i][j] = B[i][j + newSize];
B21[i][j] = B[i + newSize][j];
B22[i][j] = B[i + newSize][j + newSize];
}
}

int M1[MAX_SIZE][MAX_SIZE], M2[MAX_SIZE][MAX_SIZE], M3[MAX_SIZE][MAX_SIZE], M4[MAX_SIZE][MAX_SIZE],
M5[MAX_SIZE][MAX_SIZE], M6[MAX_SIZE][MAX_SIZE], M7[MAX_SIZE][MAX_SIZE];

// Calculate M1, M2, M3, M4, M5, M6, M7 using submatrices
// (Refer to Strassen matrix multiplication algorithm for details)

// Calculate submatrices C11, C12, C21, C22 using M1...M7

// Combine submatrices C11, C12, C21, C22 into final matrix C
}

int main() {
int n;

printf("Enter the size of the matrices: ");
scanf("%d", &n);

int A[MAX_SIZE][MAX_SIZE], B[MAX_SIZE][MAX_SIZE], C[MAX_SIZE][MAX_SIZE];

printf("Enter matrix A:\n");
for (int i = 0; i < n; i++) {
for (int j = 0; j < n; j++) {
scanf("%d", &A[i][j]);
}
}

printf("Enter matrix B:\n");
for (int i = 0; i < n; i++) {
for (int j = 0; j < n; j++) {
scanf("%d", &B[i][j]);
}
}

strassenMatrixMultiply(n, A, B, C);

printf("Matrix product C:\n");
for (int i = 0; i < n; i++) {
for (int j = 0; j < n; j++) {
printf("%d ", C[i][j]);
}
printf("\n");
}

return 0;
}