在上一篇博客裏面,筆者介紹瞭解線性方程組的列主元Guass消元法,這篇將介紹LU分解法及其算法實現.
什麼是LU分解?
對於一個線性方程組Ax=b,其中A是非奇異係數矩陣,b是線性方程組右端項,在列主元Guass消元法裏面我們知道,最後的係數矩陣A將變成一個上三角矩陣,並且是通過一系列的行變換而來的,設最後得到的上三角矩陣爲U,結合高等代數的知識,一個矩陣左乘一個初等矩陣,相當於進行一次行變換,因此設每一次A左乘的初等矩陣爲Li(i=1,2,…,n),則有LnL(n-1)…L1A=U,由於Ln均爲初等矩陣,且均爲下三角單位矩陣(因爲每次A進行消元所做的行變換均是從上面的行消去下面的行),所以設L=LnL(n-1)*…*L1,LA=U,A=L-1U,L-1也爲單位下三角矩陣,然後得到了A=LU,我們可以設Ux=y,Ly=b,由於L爲下三角矩陣,求解難度較小,因此通過求解y向量,再由Ux=y求解x,這便是LU分解全部步驟了,對於LU分解矩陣的詳細計算過程,大家可以參考這個網站link
接下來話不多說,上代碼
初始化矩陣
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 << "請輸入線性方程組對應的增廣矩陣:" << endl;
for (int i = 0; i < r; i++)
{
for (int j = 0; j < d; j++)
{
cin >> p[i][j];
}
}
return p;
}
進行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)//第一行不做處理,單獨算第一列
{
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矩陣爲:" << 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:此時我們得到了LU,由於這兩個矩陣均爲稀疏矩陣,且拼接後大小與A矩陣相同,因此我們一直接將計算的兩個矩陣儲存在A矩陣中(A中最後一列爲右端項,不進行處理),這部分計算大家特別要注意的是數組的行列下邊關係
計算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爲:" << 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 << "所求線性方程組的解向量X爲:" << endl;
for (int i = 0; i < r; i++)
{
printf("%f\t", x[i]);
}
cout << endl;
}
這裏我們將得到的y向量儲存在數組y中
程序完整代碼
#include<iostream>
#include<Windows.h>
using namespace std;
/*
測試數據
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 << "請輸入線性方程組對應的增廣矩陣:" << 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來儲存LU和方程組右邊的常數項 進行LU分解時 右邊常數項不做處理 即最後一列全部不處理
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矩陣爲:" << 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爲:" << 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 << "所求線性方程組的解向量X爲:" << endl;
for (int i = 0; i < r; i++)
{
printf("%f\t", x[i]);
}
cout << endl;
}
void LU_position_main() {
cout << "輸入矩陣的行列:" << 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;
}
歡迎交流探討~~