矩阵37x37的行列式 [英] determinant of a matrix 37x37

问题描述

我已经做了一个程序来计算矩阵的行列式。我的程序工作，但问题是，它需要很长时间来计算，特别是对于大矩阵。你能告诉我如何执行我的程序，以便在最短的时间内计算行列式？

I have done a program to calculate a determinant of a matrix. My program works, but the problem is that it takes long time to calculate, especially for big matrix. Could you tell me how can a perform my program in order to calculate the determinant in the shortest possible time?

    double Matrice::Determinant(int n)
{
   cout<<"n = "<<n<<endl;
   int i,j,j1,j2;
   double det = 0;
   Matrice tmp(n,n);
   if (n < 1) 
   {
   } 
   else if (n == 1) 
   { 
      det = this->get_el(0,0);
   } else if (n == 2) {
      det = this->get_el(0,0) * this->get_el(1,1) - this->get_el(1,0) * this->get_el(0,1);
   } else {
      det = 0;
      for (j1=0;j1<n;j1++) {
         for (i=1;i<n;i++) {
            j2 = 0;
            for (j=0;j<n;j++) {
               if (j == j1)
                  continue;
               tmp.set_el(i-1,j2,get_el(i,j));
               j2++;
            }
         }
         det += pow(-1.0,1.0+j1+1.0) * get_el(0,j1) * tmp.Determinant(n-1);
      }
   }
   return det;
}

推荐答案

您的算法定义公式的直接实现是在 O（n！）。

Your algorithm, which looks like a straight-forward implementation of the definition formula, is in O(n!).

O（n ^ 3）包括首先使用高斯消除。一旦你这样做，行列式是对角元素的乘积。

A standard algorithm in O(n^3) consists in first transforming it into a triangular matrix using Gauss elimination. Once you have done this, the determinant is the product of the diagonal elements.

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