[英]Trouble implementing Ax = b solver using LU decomposition
我試圖解決形式為Ax = b的線性系統,其中A是實數的nxn矩陣和實數的ba1xn向量,使用A = LU算法。
我已經實現了必要的功能,但我不確定問題所在的功能或功能。
import numpy as np
def forward_sub(L, b):
"""Given a lower triangular matrix L and right-side vector b,
compute the solution vector y solving Ly = b."""
y = []
for i in range(len(b)):
y.append(b[i])
for j in range(i):
y[i]=y[i]-(L[i, j]*y[j])
y[i] = y[i]/L[i, i]
return y
def backward_sub(U, y):
"""Given a lower triangular matrix U and right-side vector y,
compute the solution vector x solving Ux = y."""
x = [0 for i in range(len(U))]
for i in range(len(U)-1, 0, -1):
x[i] = U[i, i]/y[i]
for j in range (i-1, 0, -1):
U[i, i] += U[j, i]*x[i]
return x
def lu_factor(A):
#LU decompostion using Doolittles method
L = np.zeros_like(A)
U = np.zeros_like(A)
N = np.size(A,0)
for k in range(N):
L[k, k] = 1
U[k, k] = (A[k, k] - np.dot(L[k, :k], U[:k, k])) / L[k, k]
for j in range(k+1, N):
U[k, j] = (A[k, j] - np.dot(L[k, :k], U[:k, j])) / L[k, k]
for i in range(k+1, N):
L[i, k] = (A[i, k] - np.dot(L[i, :k], U[:k, k])) / U[k, k]
return (L, U)
def lu_solve(L, U, b):
# Step 1: Solve Uy = b using forward substitution
# Step 2: Solve Lx = y using backward substitution
y = forward_sub(L,b)
x = backward_sub(U,y)
return x
def linear_solve(A, b):
# ...
L, U = lu_factor(A)
x = lu_solve(L,U,b)
return x
b = [6,-4,27]
A = np.matrix([[1,1,1],[0,2,5],[2,5,-1]])
print(linear_solve(A,b))
如上所述選擇A和b得到x = [0,-0.5,-0.42]作為我的解向量,但它應該給出x = [5,3,-2]
A
是整數矩陣。 這也使L
和U
整數矩陣,但正確的結果是:
L:
[[1. 0. 0. ]
[0. 1. 0. ]
[2. 1.5 1. ]]
U:
[[ 1. 1. 1. ]
[ 0. 2. 5. ]
[ 0. 0. -10.5]]
需要一些小數值。 通常,即使輸入由整數組成,LU分解的情況也是如此。 畢竟有一些分裂正在進行中。
更改數據類型修復了該問題。 例如:
A = np.matrix([[1.,1,1],
[0,2,5],
[2,5,-1]])
backward_sub
壞了,我不確定究竟是多少,但無論如何它都是一個奇怪的實現。 這個似乎工作:
def backward_sub(U, y):
"""Given a lower triangular matrix U and right-side vector y,
compute the solution vector x solving Ux = y."""
x = np.zeros_like(y)
for i in range(len(x), 0, -1):
x[i-1] = (y[i-1] - np.dot(U[i-1, i:], x[i:])) / U[i-1, i-1]
return x
結果是[ 5. 3. -2.]
,嘗試一個ideone
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