矩阵乘法代码strassen
时间: 2023-04-05 17:03:59 浏览: 69
我可以回答这个问题。Strassen算法是一种用于矩阵乘法的分治算法,它可以在较短的时间内计算出两个矩阵的乘积。该算法的时间复杂度为O(n^log7),比传统的矩阵乘法算法更快。Strassen算法的实现需要一定的数学基础和编程技巧,可以在网上找到相关的代码实现。
相关问题
生成strassen矩阵乘法matlab代码
以下是Strassen矩阵乘法的Matlab代码实现:
```
function C = strassen(A, B)
% Strassen矩阵乘法算法
% 输入:A、B为n阶矩阵,n为2的幂次方
% 输出:C=A*B
% 确定矩阵A和B的大小
[n, ~] = size(A);
% 当矩阵规模小于等于64时,直接使用传统矩阵乘法
if n <= 64
C = A * B;
return;
end
% 划分A、B、C矩阵
mid = n / 2;
A11 = A(1:mid, 1:mid);
A12 = A(1:mid, mid+1:n);
A21 = A(mid+1:n, 1:mid);
A22 = A(mid+1:n, mid+1:n);
B11 = B(1:mid, 1:mid);
B12 = B(1:mid, mid+1:n);
B21 = B(mid+1:n, 1:mid);
B22 = B(mid+1:n, mid+1:n);
C11 = zeros(mid);
C12 = zeros(mid);
C21 = zeros(mid);
C22 = zeros(mid);
% 递归计算7个矩阵
M1 = strassen(A11+A22, B11+B22);
M2 = strassen(A21+A22, B11);
M3 = strassen(A11, B12-B22);
M4 = strassen(A22, B21-B11);
M5 = strassen(A11+A12, B22);
M6 = strassen(A21-A11, B11+B12);
M7 = strassen(A12-A22, B21+B22);
% 计算C矩阵
C11 = M1 + M4 - M5 + M7;
C12 = M3 + M5;
C21 = M2 + M4;
C22 = M1 + M3 - M2 + M6;
C = [C11, C12; C21, C22];
end
```
其中,我们将矩阵A和B划分成四个大小相等的子矩阵,然后递归计算7个矩阵,最后再根据这7个矩阵计算出矩阵C。当矩阵规模小于等于64时,直接使用传统矩阵乘法。
用C语言写strassen矩阵乘法代码
我可以回答您的问题。以下是用C语言编写的Strassen矩阵乘法代码:
```c
#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 m = n / 2;
int A11[m][m], A12[m][m], A21[m][m], A22[m][m];
int B11[m][m], B12[m][m], B21[m][m], B22[m][m];
int C11[m][m], C12[m][m], C21[m][m], C22[m][m];
int P1[m][m], P2[m][m], P3[m][m], P4[m][m], P5[m][m], P6[m][m], P7[m][m];
int temp1[m][m], temp2[m][m];
// Divide A and B into 4 submatrices
for (int i = 0; i < m; i++) {
for (int j = 0; j < m; j++) {
A11[i][j] = A[i][j];
A12[i][j] = A[i][j + m];
A21[i][j] = A[i + m][j];
A22[i][j] = A[i + m][j + m];
B11[i][j] = B[i][j];
B12[i][j] = B[i][j + m];
B21[i][j] = B[i + m][j];
B22[i][j] = B[i + m][j + m];
}
}
// Compute P1, P2, P3, P4, P5, P6, P7
strassen(m, A11, subtract(m, B12, B22, temp1), P1);
strassen(m, add(m, A11, A12, temp1), B22, P2);
strassen(m, add(m, A21, A22, temp1), B11, P3);
strassen(m, A22, subtract(m, B21, B11, temp1), P4);
strassen(m, add(m, A11, A22, temp1), add(m, B11, B22, temp2), P5);
strassen(m, subtract(m, A12, A22, temp1), add(m, B21, B22, temp2), P6);
strassen(m, subtract(m, A11, A21, temp1), add(m, B11, B12, temp2), P7);
// Compute C11, C12, C21, C22
add(m, subtract(m, add(m, P5, P4, temp1), P2, temp2), P6, C11);
add(m, P1, P2, C12);
add(m, P3, P4, C21);
subtract(m, subtract(m, add(m, P5, P1, temp1), P3, temp2), P7, C22);
// Combine C11, C12, C21, C22 into C
for (int i = 0; i < m; i++) {
for (int j = 0; j < m; j++) {
C[i][j] = C11[i][j];
C[i][j + m] = C12[i][j];
C[i + m][j] = C21[i][j];
C[i + m][j + m] = C22[i][j];
}
}
}
void add(int n, 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] = A[i][j] + B[i][j];
}
}
}
void subtract(int n, 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] = A[i][j] - B[i][j];
}
}
}
int main() {
int n = 4;
int A[n][n] = {{1, 2, 3, 4}, {5, 6, 7, 8}, {9, 10, 11, 12}, {13, 14, 15, 16}};
int B[n][n] = {{17, 18, 19, 20}, {21, 22, 23, 24}, {25, 26, 27, 28}, {29, 30, 31, 32}};
int C[n][n];
strassen(n, A, B, C);
printf("Result:\n");
for (int i = 0; i < n; i++) {
for (int j = 0; j < n; j++) {
printf("%d ", C[i][j]);
}
printf("\n");
}
return 0;
}
```
希望这个代码能够帮到您!