求四元一次方程的算法

[aokesha]

本人要通过Excel分析数据，通过4组4个已知数据求4个未知数 ，可是如果出现同列数据相等的情况 就解不出来了 所以求教各位大虾帮忙给算法 要不然这个月工资就没有了 就要喝粥了：（
公式是这样 ：
未知数1 = 常数1 + 未知数2 * 常数2的开方 + 未知数3 * 常数3 + 未知数4 * 常数4;

[suncx001]

如果说有两列相等的话，行列式的值为零，方程无解

[太乙]

[Quote=引用楼主 aokesha 的帖子:]
本人要通过Excel分析数据，通过4组4个已知数据求4个未知数 ，可是如果出现同列数据相等的情况 就解不出来了 所以求教各位大虾帮忙给算法 要不然这个月工资就没有了 就要喝粥了：（
公式是这样 ：
未知数1 = 常数1 + 未知数2 * 常数2的开方 + 未知数3 * 常数3 + 未知数4 * 常数4;
[/Quote]

什么意思？是不是就是：

x = a+y*b+z*c+w*d;
x = e+y*f+z*g+w*h;
x = i+y*j+z*k+w*l;
x = m+y*n+z*o+w*p;

[太乙]

这样，

第一式-第二式
0=a+b*y+c*z+d*w;
同除b：
0=a+y+c*z+d*w;
第三式-第四式
0=e+f*y+g*z+h*w;
同除b：
0=e+y+g*z+h*w;

一直做下去！

[aokesha]

第一式-第二式 不应该是 0=a-e +(b-f)*y + (c-G)*z + (d-h)*w么
0=a+b*y+c*z+d*w;
同除b：
0=a+y+c*z+d*w;
第三式-第四式
0=e+f*y+g*z+h*w;
同除b：
0=e+y+g*z+h*w;

[aokesha]

常熟1不一定相等的

[用户 昵称]

富农不许喝粥

[XiaoG602]

用matlab或者mathmatica吧，很快

[aokesha]

我再看看 找找

[fallening]

不一定非叠代不可的

可以用克来姆法则直接给出
看这里http://hi.baidu.com/limingdai/blog/item/0d7ae0098cfefa206a60fb6c.html

[aokesha]

谢谢楼上的 不过看完了 我头也大了 呵呵

[fallening]

给你以前我写的一段代码看看

//FileName:     det.h
//Author:       Feng Wang
//Date:         Mon Mar 24 21:25:15 CST 2008
//Comment:      return the determinant of a square matrix
//Last Modify:  Tue Mar 25 13:55:44 CST 2008

#ifndef _DET_H
#       define _DET_H

#include <cassert>
#include <vector>
#include <cmath>
//
//USAGE:
//          #include "../algorithm/ran.h"
//          ......
//          vector<long double> s;
//          const unsigned int N = 3;
//
//          for ( unsigned int i = 0; i < N; ++i )
//              for ( unsigned int k = 0; k < N ;++k )
//              s.push_back( _ran() );
//          long double determinant = det( s, N );
//          ......
//

//return the determinant of a square matrix arr whose size is order by order
template< class T >
T det ( const std :: vector<T>& arr, const unsigned int& order = 0 )
{
    unsigned int Order = order;

    //if order is set to default, calculate it
    if ( 0 == Order )
    {
        Order = sqrt ( arr.size() );
    }

    assert ( Order * Order == arr.size() );


    if ( 1 == Order )
        return ( arr[0] ) ;

    T sum = 0;
    std ::vector<T> arr2 ;
    int sign = 1;

    for ( unsigned int i = 0 ; i < Order ; ++i, sign *= -1 )
    {
        /* copy n-1 by n-1 array into another array */
        const unsigned int newsize = ( Order - 1 ) * ( Order - 1 ) ;

        arr2.resize ( newsize );
        for ( unsigned int j = 1 ; j < Order ; ++j )
        {
            for ( unsigned int k = 0, count = 0 ; k < Order ; ++k )
            {
                if ( k == i )
                    continue ;//jump out of k loop

                const unsigned int pos = j * Order + k ;
                const unsigned int newpos = ( j - 1 ) *
                                            ( Order - 1 ) + count ;
                arr2[newpos] = arr[pos] ;
                count++ ;
            }//end of k loop
        }//end of j loop

        /*  find determinant value of n-1 by n-1 array and  add it to sum */
        sum += arr[i] * sign * det ( arr2, Order - 1 ) ;
    }
    return sum;
}
#endif  //_DET_H

这个是计算矩阵的行列式值的，当然，如果超过6阶，后果自负

[fallening]

//FileName:         lq.h
//Author:           Feng Wang
//Date:             Tue Mar 25 13:51:02 CST 2008
//Comment:          Resolve the linear equation
//Last Modify:
//

#ifndef _LQ_H
#       define _LQ_H

#include "det.h"

#include <vector>
#include <cassert>
#include <cstdlib>

#ifdef DEBUG
#include <iostream>
#include <algorithm>
#include <iterator>
using namespace std;
#endif

//
//for solution of the linear equation
//                  A X = B
//
template<class T>
void lq ( const std :: vector<T>& a,    //store A[size][size]
          const std :: vector<T>& b,    //store B[size]
          const unsigned int& order,    //store size
          std :: vector<T>& result      //store X[size]
        )
{
    #ifdef DEBUG
        cout    << "\n>>Currently processing function lq()."
                << "\nThe order is " << order;

        cout << "\nThe incoming vector a is \n";
        copy( a.begin(), a.end(), ostream_iterator<T>( cout, " ") );

        cout << "\nThe incoming vector b is \n";
        copy( b.begin(), b.end(), ostream_iterator<T>( cout, " ") );
    #endif

    unsigned int Order = order;
    //if order is set to default, calculate order
    if ( 0 == Order )
    {
        Order = b.size();
    }

    assert ( Order * Order == a.size() );
    assert ( Order == b.size() );

    result.clear();

    const T _D = det ( a, Order );  //store |D|

    #ifdef DEBUG
        cout << "\nThe base vector det is " << _D;
    #endif

    if ( 0 == _D )
    {
        if ( "Failed to solve the linear equation" )
        {
            #ifdef DEBUG
                cout << "\n the det of vector is 0, failed to solve the equation.";
            #endif

            exit ( EXIT_FAILURE );
        }
    }

    //change colum to get D[i]
    //then solve it
    for ( unsigned int i = 0; i < Order; ++i )
    {
        std :: vector<T> A = a;
        for ( unsigned int j = 0; j < Order; ++j )
        {
            A[i+j*Order] = b[j];
        }
        const T D = det ( A, Order );   //store D[i]

        result.push_back ( D / _D );    //get X[i]

        #ifdef DEBUG
            cout << "\nThe base vector det is " << _D;
            cout << "\nThe current vector det is " << D;
            cout << "\nresult[" << i << "] calculated, whose value is " << D / _D;
        #endif

    }

    #ifdef DEBUG
        cout << "\n<<Stride out of fucntion lq()";
    #endif
}



#endif  //_LQ_H

这个是用克莱姆法则求解的函数，同样，超过6阶后果自负

[aozhi]

高斯消元？

[aokesha]

谢谢 我研究下

[train0714]

matlab真的很简单！

[fallening]

楼上的，一个matlab多少钱能搞定？
这个价格打5折，把钱给我，我帮你搞定这个程序。

[bayuan0414]

建议楼主花点时间看看 线性代数

[phz1985]

http://www.tyut.edu.cn/kecheng/xianxds/TeachMat/1-5.htm
用克莱姆法则计算.