本文介绍了矩阵37x37的行列式的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着小编来一起学习吧!

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。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.

这篇关于矩阵37x37的行列式的文章就介绍到这了,希望我们推荐的答案对大家有所帮助,也希望大家多多支持!

08-20 00:37