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我目前正在尝试编写一个程序来查找 NxN 矩阵的行列式,但我遇到了 N 大于 2 的递归问题。基本上据我所知,它没有这样做,它只是运行一次函数我的调试选项显示该函数通过列运行,但顺序永远不会下降,然后无论如何它都会给我的行列式为零。我试过到处寻找关于我做错了什么的任何想法,但我似乎找不到任何答案,我什至找到了与我做基本相同事情的例子,并且无论如何使用它们都会给我零,所以我很疑惑在这里)

代码:

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


double det(double **mat, int order);


int main (int argc, char* argv[])
{  


   FILE* input; 
   int row,column,N;
   double **matrix;
   N=3;
   matrix=(double**)malloc(N*sizeof(double));

   input=fopen("matrix.dat", "r");
   if(input !=(FILE*) NULL)
   {

      for(row=0; row<N; row++) 
      { 
     matrix[row]=(double*)malloc(N*sizeof(double));
  }
  for(row=0; row<N; row++) 
  {
        printf("| ");
        for(column=0; column<N; column++)   
        {  
          fscanf(input,"%lf ", &matrix[row][column]); 
      printf("%g ", matrix[row][column]);
    }
     if(row != (N/2))
         { 
       printf("|\n");
         }
         else
         {

           printf("|= %lf \n", det(matrix, N) );
     }
 }
 return(EXIT_SUCCESS);
 }

 else
 { 
  printf("*********************ERROR*********************\n");
  printf("** Cannot open input file 'matrix.dat' make  **\n");
  printf("** sure file is present in working directory **\n");
  printf("***********************************************\n");

  return(EXIT_FAILURE);
 }

}

double det(double **mat, int order)
{
 int debug;
 double cofact[order], determinant, **temp;
 determinant = 0;
 debug=0;

 if(order==1)
 {
  determinant=mat[0][0];

  if(debug==1)
  {
    printf("order 1 if\n");
  }
 }
  else if(order==2)
 {
 determinant= ((mat[0][0]*mat[1][1])-(mat[0][1]*mat[1][0]));

 if(debug==1)
 {
   printf("order 2 if\n");
     }
  }
 else
 {

  int column, rowtemp, coltemp, colread;
  for (column=0; column<order; column++) 
  {
  /* Now create an array of size N-1 to store temporary data used for calculating minors */
     temp= malloc((order-1)*sizeof(*temp)); 
    for(rowtemp=0; rowtemp<(order-1); rowtemp++)
  {
    /* Now asign each element in the array temp as an array of size N-1 itself */
    temp[rowtemp]=malloc((order-1)*sizeof(double));
  }
  for(rowtemp=1; rowtemp<order; rowtemp++)
  {
    /* We now have our empty array, and will now fill it by assinging row and collumn values    from the original mat with the aprroriate elements excluded */
       coltemp=0;
      for(colread=0; colread<order; colread++)
      {
        /* When the collumn of temp is equal to the collumn of the matrix, this indicates this row should be exlcuded and is skiped over */
        if(colread==column)
        {
          continue;
        }
        temp[rowtemp-1][coltemp] = mat[rowtemp][colread];
        coltemp++;
      }

   }
  if(debug==1)
  {
    printf("column =%d, order=%d\n", column, order);
  }
  determinant+=(mat[0][column]*(1 - 2*(column & 1))*det(temp, order-1));
}   

 }  
 return(determinant);   

 }
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1 回答 1

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temp= (double **)malloc((order-1)*sizeof(double));

只要sizeof(double*) <= sizeof(double),这不会导致崩溃,通常的 32 或 64 位系统就是这种情况,但它在概念上是错误的。您正在为 的数组分配空间double*,因此该因子应该是sizeof(double*)or,更好,因为它在类型更改时是不变的,sizeof *temp,

temp = malloc((order-1) * sizeof *temp);

(而且您不需要malloc在 C 中转换结果,最好不要,因为转换可能会隐藏错误,例如忘记#include <stdlib.h>.)

分配也一样

matrix=(double**)malloc(N*sizeof(double));

main.

在行列式的计算中,

         for(coltemp=0; coltemp<order; coltemp++)
         {
           for(colread=0; colread<order; colread++)
               {
               /* When the collumn of temp is equal to the collumn of   the matrix, this indicates this row should be exlcuded and is skiped over */
               if(colread==column)
               {
                continue;
               }
          temp[rowtemp-1][coltemp] = mat[rowtemp][colread];
          coltemp++;
            }
        }

您在列中循环两次,一次 for coltemp == 0,然后在内部循环中,coltemp递增order-1次数,因此内部循环coltemp == order-1在开始时第二次运行。然后coltemp在循环中再次递增多次,并且您正在写入超出分配内存的范围。

应该删除外循环,coltemp = 0;而内循环是您所需要的。

pow(-1,column))

不是确定标志的好方法。

(1 - 2*(column & 1))

比调用pow.

最后,在main

for(row=0; row<N; row++) 
 {
    printf("| ");
    for(column=0; column<N; column++)   
    { 
          fscanf(input,"%lf ", &matrix[row][column]); 
          printf("%g ", matrix[row][column]);
    }
    if(row != (N/2))
    { 
    printf("|\n");
    }
    else
    {

            printf("|= %lf \n", det(matrix, N) );
    }
 }

您正在打印行列式N/2,这看起来不错,但当时您尚未扫描整个矩阵,因此行N/2 + 1包含N-1未初始化的数据,这不太可能全为零。如果将 更改if (row != N/2)if (row != N-1),它将起作用,但是,处理它的正确方法是将矩阵的扫描与计算和打印分开。这些都是应该在各自独立的函数中处理的独立操作。

于 2012-11-17T12:50:48.930 回答