【计算方法】实验二：python实现高斯消去、列主元高斯消去，LU分解分别求解线性方程组
文章目录
- 题目
- 方法一：高斯消去法
-
-
-
- 结果截图
-
-
- 方法二：列主元素高斯消元法
-
-
-
- 结果截图
-
-
- 方法三：LU分解
- 结果截图
-
-
- 结果总结
-
题目
实现高斯消去、列主元高斯消去，LU分解分别求解线性方程组
方法一：高斯消去法
消去过程：
l i k = a i k ( k − 1 ) / a k k ( k − 1 ) ( i = k + 1 , k + 2 , . . . , n ) l_{ik} = a_{ik}^{(k-1)}/a_{kk}^{(k-1)} (i=k+1,k+2,...,n) lik=aik(k−1)/akk(k−1)(i=k+1,k+2,...,n)
a i j ( k ) = a i j ( k − 1 ) − l i k a j k ( k − 1 ) ( i , j = k + 1 , k + 2 , . . . , n ) a_{ij}^{(k)} = a_{ij}^{(k-1)}-l_{ik}a_{jk}^{(k-1)}(i,j=k+1,k+2,...,n) aij(k)=aij(k−1)−likajk(k−1)(i,j=k+1,k+2,...,n)
b i ( k ) = b i ( k − 1 ) − l i k b k ( k − 1 ) ( i = k + 1 , k + 2 , . . . , n ) b_{i}^{(k)} = b_{i}^{(k-1)}-l_{ik}b_{k}^{(k-1)}(i=k+1,k+2,...,n) bi(k)=bi(k−1)−likbk(k−1)(i=k+1,k+2,...,n)
回代过程：
x n = b n ( n − 1 ) / a n n ( n − 1 ) x_{n}= b_{n}^{(n-1)}/a_{nn}^{(n-1)} xn=bn(n−1)/ann(n−1)
x k = ( b k ( k − 1 ) − ∑ j = k + 1 n a k j ( k − 1 ) x j ) / a k k ( k − 1 ) x_{k}= (b_{k}^{(k-1)}-\sum_{j=k+1}^{n}a_{kj}^{(k-1)}x_j )/a_{kk}^{(k-1)} xk=(bk(k−1)−j=k+1∑nakj(k−1)xj)/akk(k−1)
在消元法第k次迭代时，首先 ，寻找第k列中从k到n处的绝对值最大的一个，然后互换该行和第k行的所有位置。 然后再消元。这样的好处是能控制舍入误差的扩大和传播。 基本算法如下：
import numpy as np
A = np.array([[10,-7,0,1],[-3, 2.099999,6,2 ],[5,-1,5,-1],[2,1,0,2]])
b = np.array([[8],[5.900001],[5],[1]])
def guass(A,b):
n = b.shape[0]
A=A.astype('float')
b=b.astype('float')
#消元
for k in range(n-1): #k第一层循环，第（0~n-1）行
# print('第',k+1,'次的初值')
# print(A)
# print(b)
for i in range(k+1,n): #i第二层循环，第（k+1,n）行
mik = A[i,k]/A[k,k]
for j in range(k+1,n):
A[i,j] = A[i,j] - mik*A[k,j]
b[i] = b[i] - mik*b[k]
# print('第',k+1,'次消元后的值')
# print(A)
# print(b)
#回代
solution = np.zeros(n)
solution[n-1] = b[n-1]/A[n-1,n-1] #改为n-1
for i in range(n-2,-1,-1):
for j in range(i+1,n):
solution[i] = solution[i] + A[i,j]*solution[j]
solution[i] = (b[i] - solution[i])/A[i,i]
return solution
guass(A,b)
结果截图
方法二：列主元素高斯消元法
def guass_zhuyuan(A,b):
n = b.shape[0]
A=A.astype('float')
b=b.astype('float')
#寻找列主元素
for k in range(n-1): #k第一层循环，第（0~n-2）行
# print('第',k+1,'次的初值')
# print(A)
# print(b)
arr = abs(A[k:n,k]) #每一列最大值的数组
value = max(arr) #最大值
position = np.argmax(arr) #最大值索引
#print(position)
if position != 0:
# A[k,k:n-1],A[position+k,k:n-1] = A[position+k,k:n-1],A[k,k:n-1]
# b[k],b[position+k] = b[position+k],b[k]
A[[k,position+k],k:n] = A[[position+k,k],k:n]
b[[k,position+k],:] = b[[position+k,k],:]
# print('第',k+1,'次交换后的值')
# print(A)
# print(b)
#消元
for i in range(k+1,n): #i第二层循环，第（k+1,n-1）行
mik = A[i,k]/A[k,k]
for j in range(k+1,n): #j从k开始循环，不是k+1
A[i,j] -= mik*A[k,j]
b[i] = b[i] - mik*b[k]
# print('第',k+1,'次消元后的值')
# print(A)
# print(b)
#回代
solution = np.zeros(n)
solution[n-1] = b[n-1]/A[n-1,n-1] #改为n-1
for i in range(n-2,-1,-1):
for j in range(i+1,n):
solution[i] = solution[i] + A[i,j]*solution[j]
solution[i] = (b[i] - solution[i])/A[i,i]
return solution
结果截图
方法三：LU分解
def LUfenjie(A,b):
n = b.shape[0]
A = A.astype('float')
b = b.astype('float')
U = np.zeros(A.shape)
L = np.zeros(A.shape)
U[0,:] = A[0,:] #U的第一行元素和A的第一行元素相等
L[1:n,0] = A[1:n,0]/U[0,0] #L的每一行的第一列元素和U第一个元素相乘=A的对应元素
#分解
#U为上三角元素 k=1~n-1行 j=k~n-1列 r=0~k-2行 i=k+1~n-1行
# for k in range(1,n):
# for j in range(k,n):
# U[k,j] = A[k,j] - L[k,0:k-1] * U[0:k-1,j]
# for i in range(k+1,n):
# L[i,k] = (A[i,k]-L[i,0:k-1] * U[0:k-1,k])/U[k,k]
#分解
#U为上三角元素 k=1~n-1行 j=k~n-1列 r=0~k-2行 i=k+1~n-1行
for k in range(1,n):
for j in range(k,n):
delta = 0
for r in range(0,k):
delta += L[k,r] * U[r,j]
U[k,j] = A[k,j] - delta
for i in range(k+1,n):
theta = 0
for r in range(0,k):
theta += L[i,r] * U[r,k]
L[i,k] = (A[i,k] - theta)/U[k,k]
#求解y
y = np.zeros(b.shape[0])
#忘了初始化……
y[0] = b[0]
for i in range(1,n):
alpha = 0
for j in range(0,i):
alpha += L[i,j]*y[j].T
y[i] = b[i] - alpha
#求解x
x = np.zeros(b.shape[0])
x[n-1] = y[n-1]/U[n-1,n-1]
for i in range(n-2,-1,-1):
gama = 0
for j in range(i+1,n):
gama += U[i,j]*x[j].T
x[i] = (y[i] - gama )/U[i,i]
return (x)
结果截图
结果总结
| 方法 | 结果 |
|---|---|
| 高斯迭代法 | [-6.03391825e-10, -1.00000000e+00, 1.00000000e+00, 1.00000000e+00] |
| 列主元素高斯 | [ 2.66453526e-16, -1.00000000e+00, 1.00000000e+00, 1.00000000e+00] |
| LU分解 | [-6.03391825e-10, -1.00000000e+00, 1.00000000e+00, 1.00000000e+00] |