LU分解如何通过求解Ly=b和Ux=y两个方程组得到解x,可以写一段c++代码示例吗
时间: 2023-06-24 20:05:42 浏览: 38
LU分解是将一个矩阵分解为下三角矩阵L和上三角矩阵U的乘积,因此可以通过LU分解后,分别解Ly=b和Ux=y两个方程组来求解x。
以下是一个简单的C++代码示例:
```c++
#include <iostream>
#include <vector>
// LU分解函数
void LU_Decomposition(std::vector<std::vector<double>>& A, std::vector<std::vector<double>>& L, std::vector<std::vector<double>>& U) {
int n = A.size();
// 初始化L和U
for (int i = 0; i < n; i++) {
L[i][i] = 1;
}
for (int i = 0; i < n; i++) {
for (int j = i; j < n; j++) {
double sum = 0;
for (int k = 0; k < i; k++) {
sum += L[i][k] * U[k][j];
}
U[i][j] = A[i][j] - sum;
}
for (int j = i + 1; j < n; j++) {
double sum = 0;
for (int k = 0; k < i; k++) {
sum += L[j][k] * U[k][i];
}
L[j][i] = (A[j][i] - sum) / U[i][i];
}
}
}
// 解Ly=b方程组
std::vector<double> solve_Ly_b(std::vector<std::vector<double>>& L, std::vector<double>& b) {
int n = L.size();
std::vector<double> y(n, 0);
for (int i = 0; i < n; i++) {
double sum = 0;
for (int j = 0; j < i; j++) {
sum += L[i][j] * y[j];
}
y[i] = b[i] - sum;
}
return y;
}
// 解Ux=y方程组
std::vector<double> solve_Ux_y(std::vector<std::vector<double>>& U, std::vector<double>& y) {
int n = U.size();
std::vector<double> x(n, 0);
for (int i = n - 1; i >= 0; i--) {
double sum = 0;
for (int j = i + 1; j < n; j++) {
sum += U[i][j] * x[j];
}
x[i] = (y[i] - sum) / U[i][i];
}
return x;
}
int main() {
// 初始化矩阵A和向量b
std::vector<std::vector<double>> A = {{2, -1, 0}, {-1, 2, -1}, {0, -1, 2}};
std::vector<double> b = {1, 0, 1};
int n = A.size();
std::vector<std::vector<double>> L(n, std::vector<double>(n, 0));
std::vector<std::vector<double>> U(n, std::vector<double>(n, 0));
// 对A进行LU分解
LU_Decomposition(A, L, U);
// 解Ly=b方程组
std::vector<double> y = solve_Ly_b(L, b);
// 解Ux=y方程组
std::vector<double> x = solve_Ux_y(U, y);
// 输出解x
for (int i = 0; i < n; i++) {
std::cout << "x[" << i << "] = " << x[i] << std::endl;
}
return 0;
}
```
以上代码中,我们首先通过LU_Decomposition()函数对矩阵A进行LU分解,然后分别对Ly=b和Ux=y两个方程组进行求解,最后输出解x。