使用向前和向后替换从LU分解求解系统的Python代码

import numpy as np

def forward_substitution(L, b):
    n = len(b)
    x = np.zeros_like(b)
    for i in range(n):
        x[i] = b[i]
        for j in range(i):
            x[i] -= L[i, j] * x[j]
        x[i] /= L[i, i]
    return x

def backward_substitution(U, y):
    n = len(y)
    x = np.zeros_like(y)
    for i in range(n-1, -1, -1):
        x[i] = y[i]
        for j in range(i+1, n):
            x[i] -= U[i, j] * x[j]
        x[i] /= U[i, i]
    return x

def lu_solve(A, b):
    n = len(b)
    L = np.zeros((n, n))
    U = np.zeros((n, n))

    for i in range(n):
        L[i, i] = 1.0

        for j in range(i, n):
            U[i, j] = A[i, j] - np.dot(L[i, :i], U[:i, j])

        for j in range(i+1, n):
            L[j, i] = (A[j, i] - np.dot(L[j, :i], U[:i, i])) / U[i, i]

    y = forward_substitution(L, b)
    x = backward_substitution(U, y)
    return x

# 示例用法
A = np.array([[2, -1, 0], [-1, 2, -1], [0, -1, 2]])
b = np.array([1, 0, 1])

x = lu_solve(A, b)
print("Solution:", x)