在上一篇博客里面,笔者介绍了解线性方程组的列主元Guass消元法,这篇将介绍LU分解法及其算法实现.
什么是LU分解?
对于一个线性方程组Ax=b,其中A是非奇异系数矩阵,b是线性方程组右端项,在列主元Guass消元法里面我们知道,最后的系数矩阵A将变成一个上三角矩阵,并且是通过一系列的行变换而来的,设最后得到的上三角矩阵为U,结合高等代数的知识,一个矩阵左乘一个初等矩阵,相当于进行一次行变换,因此设每一次A左乘的初等矩阵为Li(i=1,2,…,n),则有LnL(n-1)…L1A=U,由于Ln均为初等矩阵,且均为下三角单位矩阵(因为每次A进行消元所做的行变换均是从上面的行消去下面的行),所以设L=LnL(n-1)*…*L1,LA=U,A=L-1U,L-1也为单位下三角矩阵,然后得到了A=LU,我们可以设Ux=y,Ly=b,由于L为下三角矩阵,求解难度较小,因此通过求解y向量,再由Ux=y求解x,这便是LU分解全部步骤了,对于LU分解矩阵的详细计算过程,大家可以参考这个网站link
接下来话不多说,上代码
初始化矩阵
double** init_Matrix(int r, int c)
{
double** p = new double* [r];
int d = c + 1;
for (int i = 0; i < r; i++)
{
p[i] = new double[d];
memset(p[i], 0, sizeof(double) * d);
}
cout << "请输入线性方程组对应的增广矩阵:" << endl;
for (int i = 0; i < r; i++)
{
for (int j = 0; j < d; j++)
{
cin >> p[i][j];
}
}
return p;
}
进行LU分解
void LU_Position(double**a,int r ,int c)
{
double num1 = 0,num2=0;
for (int i=0;i<r;i++)
{
if (i==0)//第一行不做处理,单独算第一列
{
for (int j = 1; j < c; j++)
{
a[j][i] = a[j][i] / a[0][0];
}
}
else
{
for (int j = i; j < c; j++)
{
num1 = 0;
for (int k = 0; k < i; k++)
{
num1 += a[i][k] * a[k][j];
}
a[i][j] = a[i][j] - num1;
}
for (int j = i+1; j < r; j++)
{
num2 = 0;
for (int k = 0; k < i; k++)
{
num2 += a[j][k] * a[k][i];
}
a[j][i] = (a[j][i] - num2) / a[i][i];
}
}
}
cout << "所得的LU矩阵为:" << endl;
for (int k = 0; k < r; k++)
{
for (int n = 0; n < c; n++)
{
printf("%f\t", a[k][n]);
}
cout << endl;
}
}
注意:此时我们得到了LU,由于这两个矩阵均为稀疏矩阵,且拼接后大小与A矩阵相同,因此我们一直接将计算的两个矩阵储存在A矩阵中(A中最后一列为右端项,不进行处理),这部分计算大家特别要注意的是数组的行列下边关系
计算y向量及x向量
void calculate(double*y, double*x, double**a,int r)
{
y[0] = a[0][r];
double sum=0;
for (int i = 1; i < r; i++)
{
sum = 0;
for (int k = 0; k < i; k++)
{
sum += a[i][k] * y[k];
}
y[i] = a[i][r] - sum;
}
cout << "所求的向量Y为:" << endl;
for (int i = 0; i < r; i++)
{
printf("%f\t", y[i]);
}
cout << endl;
for (int i = r-1; i >=0; i--)
{
sum = 0;
for (int j=i+1;j<r;j++)
{
sum += a[i][j] * x[j];
}
x[i] = (y[i] - sum) / a[i][i];
}
cout << "所求线性方程组的解向量X为:" << endl;
for (int i = 0; i < r; i++)
{
printf("%f\t", x[i]);
}
cout << endl;
}
这里我们将得到的y向量储存在数组y中
程序完整代码
#include<iostream>
#include<Windows.h>
using namespace std;
/*
测试数据
2 2
2 3 5
1 -1 0
3 3
3 2 -3 -2
1 1 1 6
1 2 -1 2
3 3
1 2 -3 1
2 -1 3 5
3 -2 2 1
4 4
4 -3 6 7 11
1 1 3 4 10
-2 9 -7 1 10
3 3 -4 20 25
5 5
28 -3 0 0 0 10
-3 38 -10 0 -5 0
0 -10 25 -15 0 0
0 0 -15 45 0 0
0 -5 0 0 30 0
*/
double** init_Matrix(int r, int c)
{
double** p = new double* [r];
int d = c + 1;
for (int i = 0; i < r; i++)
{
p[i] = new double[d];
memset(p[i], 0, sizeof(double) * d);
}
cout << "请输入线性方程组对应的增广矩阵:" << endl;
for (int i = 0; i < r; i++)
{
for (int j = 0; j < d; j++)
{
cin >> p[i][j];
}
}
return p;
}
//直接用A来储存LU和方程组右边的常数项 进行LU分解时 右边常数项不做处理 即最后一列全部不处理
void LU_Position(double**a,int r ,int c)
{
double num1 = 0,num2=0;
for (int i=0;i<r;i++)
{
if (i==0)
{
for (int j = 1; j < c; j++)
{
a[j][i] = a[j][i] / a[0][0];
}
}
else
{
for (int j = i; j < c; j++)
{
num1 = 0;
for (int k = 0; k < i; k++)
{
num1 += a[i][k] * a[k][j];
}
a[i][j] = a[i][j] - num1;
}
for (int j = i+1; j < r; j++)
{
num2 = 0;
for (int k = 0; k < i; k++)
{
num2 += a[j][k] * a[k][i];
}
a[j][i] = (a[j][i] - num2) / a[i][i];
}
}
}
cout << "所得的LU矩阵为:" << endl;
for (int k = 0; k < r; k++)
{
for (int n = 0; n < c; n++)
{
printf("%f\t", a[k][n]);
}
cout << endl;
}
}
void calculate(double*y, double*x, double**a,int r)
{
y[0] = a[0][r];
double sum=0;
for (int i = 1; i < r; i++)
{
sum = 0;
for (int k = 0; k < i; k++)
{
sum += a[i][k] * y[k];
}
y[i] = a[i][r] - sum;
}
cout << "所求的向量Y为:" << endl;
for (int i = 0; i < r; i++)
{
printf("%f\t", y[i]);
}
cout << endl;
for (int i = r-1; i >=0; i--)
{
sum = 0;
for (int j=i+1;j<r;j++)
{
sum += a[i][j] * x[j];
}
x[i] = (y[i] - sum) / a[i][i];
}
cout << "所求线性方程组的解向量X为:" << endl;
for (int i = 0; i < r; i++)
{
printf("%f\t", x[i]);
}
cout << endl;
}
void LU_position_main() {
cout << "输入矩阵的行列:" << endl;
int i = 0, j = 0;
cin >> i >> j;
double** p = init_Matrix(i, j);
LU_Position(p, i, j);
double* a = new double[i];
memset(a, 0, sizeof(double) * i);
double* b = new double[i];
memset(b, 0, sizeof(double) * i);
calculate(a, b, p, i);
delete[]a;
delete[]b;
for (int i = 0; i < j; i++)
{
delete[]p[i];
}
delete[]p;
}
int main(void) {
LU_position_main();
system("pause");
return 0;
}
欢迎交流探讨~~
