矩阵的加法或减法在没有线程的情况下花费O(n ^ 2)时间，但使用线程不会降低程序的时间复杂度，而是将任务划分为多个核心，例如如果我们有4个核心，然后将矩阵划分为4个部分核心占用矩阵的一部分并计算操作，最后在完成每个任务后，所有4个线程将加入主程序并显示最终输出。
注–仅当我们只有一个CPU内核时，该方法才有效。如果我们只有一个内核，则不要创建线程，因为我们将任务划分为仅在一个CPU上执行，因此浪费时间来划分任务，然后计算。
例子：
Input :
Matrix A:
3 7 3 6
9 2 0 3
0 2 1 7
2 2 7 9
Matrix B:
6 5 5 2
1 7 9 6
6 6 8 9
0 3 5 2
Output :
Sum of Matrix A and B:
9 12 8 8
10 9 9 9
6 8 9 16
2 5 12 11
Subtraction of Matrix A and B:
-3 2 -2 4
8 -5 -9 -3
-6 -4 -7 -2
2 -1 2 7
代码–建议在基于Linux的系统中执行程序。
#include
#include
#include
// Value depend on System core
#define CORE 4
// Maximum matrix size
#define MAX 4
// Maximum threads is equal to total core of system
pthread_t thread[CORE * 2];
int mat_A[MAX][MAX], mat_B[MAX][MAX], sum[MAX][MAX], sub[MAX][MAX];
// Addition of a Matrix
void* addition(void* arg)
{
int i, j;
int core = (int)arg;
// Each thread computes 1/4th of matrix addition
for (i = core * MAX / 4; i < (core + 1) * MAX / 4; i++) {
for (j = 0; j < MAX; j++) {
// Compute Sum Row wise
sum[i][j] = mat_A[i][j] + mat_B[i][j];
}
}
}
// Subtraction of a Matrix
void* subtraction(void* arg)
{
int i, j;
int core = (int)arg;
// Each thread computes 1/4th of matrix subtraction
for (i = core * MAX / 4; i < (core + 1) * MAX / 4; i++) {
for (j = 0; j < MAX; j++) {
// Compute Subtract row wise
sub[i][j] = mat_A[i][j] - mat_B[i][j];
}
}
}
// Driver Code
int main()
{
int i, j, step = 0;
// Generating random values in mat_A and mat_B
for (i = 0; i < MAX; i++) {
for (j = 0; j < MAX; j++) {
mat_A[i][j] = rand() % 10;
mat_B[i][j] = rand() % 10;
}
}
// Displaying mat_A
printf("\nMatrix A:\n");
for (i = 0; i < MAX; i++) {
for (j = 0; j < MAX; j++) {
printf("%d ", mat_A[i][j]);
}
printf("\n");
}
// Displaying mat_B
printf("\nMatrix B:\n");
for (i = 0; i < MAX; i++) {
for (j = 0; j < MAX; j++) {
printf("%d ", mat_B[i][j]);
}
printf("\n");
}
// Creating threads equal
// to core size and compute matrix row
for (i = 0; i < CORE; i++) {
pthread_create(&thread[i], NULL, &addition, (void*)step);
pthread_create(&thread[i + CORE], NULL, &subtraction, (void*)step);
step++;
}
// Waiting for join threads after compute
for (i = 0; i < CORE * 2; i++) {
pthread_join(thread[i], NULL);
}
// Display Addition of mat_A and mat_B
printf("\n Sum of Matrix A and B:\n");
for (i = 0; i < MAX; i++) {
for (j = 0; j < MAX; j++) {
printf("%d ", sum[i][j]);
}
printf("\n");
}
// Display Subtraction of mat_A and mat_B
printf("\n Subtraction of Matrix A and B:\n");
for (i = 0; i < MAX; i++) {
for (j = 0; j < MAX; j++) {
printf("%d ", sub[i][j]);
}
printf("\n");
}
return 0;
}
输出：
Matrix A:
3 7 3 6
9 2 0 3
0 2 1 7
2 2 7 9
Matrix B:
6 5 5 2
1 7 9 6
6 6 8 9
0 3 5 2
Sum of Matrix A and B:
9 12 8 8
10 9 9 9
6 8 9 16
2 5 12 11
Subtraction of Matrix A and B:
-3 2 -2 4
8 -5 -9 -3
-6 -4 -7 -2
2 -1 2 7