Question

我想知道如何在c ++中实现矩阵乘法的测试函数。有共同的方法，怎么做？我知道它应该与Inverse有关，但我不确定如何正确实现它。

让我们说这将是我想测试的乘数：

  #include <stdio.h>
  #include <stdlib.h>
  //#include <stdint.h>

  // ---------------------------------------------------------------------------
  // allocate space for empty matrix A[row][col]
  // access to matrix elements possible with:
  // - A[row][col]
  // - A[0][row*col]


  float **alloc_mat(int row, int col)
  {
      float **A1, *A2;

      A1 = (float **)calloc(row, sizeof(float *));      // pointer on rows
      A2 = (float *)calloc(row*col, sizeof(float));    // all matrix elements


      for (int i=0; i<row; i++)
          A1[i] = A2 + i*col;

      return A1;
  }

  // ---------------------------------------------------------------------------
  // random initialisation of matrix with values [0..9]

  void init_mat(float **A, int row, int col)
  {   

      for (int i = 0; i < row*col; i++)
          A[0][i] = (float)(rand() % 10);
  }

  // ---------------------------------------------------------------------------
  // DEBUG FUNCTION: printout of all matrix elements

  void print_mat(float **A, int row, int col, char *tag)
  {
      int i, j;

      printf("Matrix %s:\n", tag);
      for (i = 0; i < row; i++)
      {

          for (j=0; j<col; j++) 
              printf("%6.1f   ", A[i][j]);
          printf("\n"); 
      }
  }

  // ---------------------------------------------------------------------------

  int main(int argc, char *argv[])
  {
      float **A, **B, **C;  // matrices
      int d1, d2, d3;         // dimensions of matrices
      int i, j, k;          // loop variables


      double start, end;
      start = omp_get_wtime();

      /* print user instruction */
      if (argc != 4)
      {
          printf ("Matrix multiplication: C = A x B\n");
          printf ("Usage: %s <NumRowA>; <NumColA> <NumColB>\n",argv[0]); 
           return 0;
       }

       /* read user input */
       d1 = atoi(argv[1]);      // rows of A and C
       d2 = atoi(argv[2]);     // cols of A and rows of B
       d3 = atoi(argv[3]);     // cols of B and C

       printf("Matrix sizes C[%d][%d] = A[%d][%d] x B[%d][%d]\n", 
       d1, d3, d1, d2, d2, d3);

       /* prepare matrices */
       A = alloc_mat(d1, d2);
       init_mat(A, d1, d2); 
       B = alloc_mat(d2, d3);
       init_mat(B, d2, d3);
       C = alloc_mat(d1, d3);   // no initialisation of C, 
   //because it gets filled by matmult

       /* serial version of matmult */
       printf("Perform matrix multiplication...\n");



       int sum;

           for (i = 0; i < d1; i++)
               for (j = 0; j < d3; j++)
                   for (k = 0; k < d2; k++){
                   C[i][j] = C[i][j] + A[i][k] * B[k][j];






       /* test output */
       print_mat(A, d1, d2, "A"); 
       print_mat(B, d2, d3, "B"); 
       print_mat(C, d1, d3, "C"); 

       printf("This task took %f seconds\n", end-start);
       printf ("\nDone.\n");

       return 0;
   }

Answer 1

如何使用openCV库，您可以轻松地使用它来定义不同类型和大小的矩阵。

Answer 2

创建随机向量v并验证C * v == A * (B * v)（但允许一些舍入错误）。

矩阵乘法的数值试验

2 个答案: