数值分析实验之线性方程组的迭代求解(Python实现)

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  实验内容: 求解如下4元线性方程组的近似解。

      

Jacobi迭代过程

 1 import numpy as np
 2 
 3 A = np.array([[10,-1,2,0],[-1,11,-1,3],[2,-1,10,-1],[0,3,-1,8]])
 4 B = np.array([6, 25, -11, 15])
 5 x0 = np.array([0.0, 0, 0, 0])
 6 x = np.array([0.0, 0, 0, 0])
 7 
 8 times = 0
 9 
10 while True:
11     for i in range(4):
12         temp = 0
13         for j in range(4):
14             if i != j:
15                 temp += x0[j] * A[i][j]
16         x[i] = (B[i] - temp) / A[i][i]
17     calTemp = max(abs(x - x0))
18     times += 1
19     if calTemp < 1e-5:
20         break
21     else:
22         x0 = x.copy()
23 
24 print(times)
25 print(x)

    运行结果:

   

 

Gauss-Seidel迭代

 

 

 1 import numpy as np
 2 
 3 A = np.array([[10,-1,2,0],[-1,11,-1,3],[2,-1,10,-1],[0,3,-1,8]])
 4 B = np.array([6, 25, -11, 15])
 5 x0 = np.array([0.0, 0, 0, 0])
 6 x = np.array([1.0, 2, -1, 1])
 7 times = 0
 8 
 9 while True:
10     for i in range(4):
11         temp = 0
12         tempx = x0.copy()
13         for j in range(4):
14             if i != j:
15                 temp += x0[j] * A[i][j]
16         x[i] = (B[i] - temp) / A[i][i]
17         x0[i] = x[i].copy()
18     calTemp = max(abs(x - tempx))
19     times += 1
20     if calTemp < 1e-5:
21         break
22     else:
23         x0 = x.copy()
24 
25 print(times)
26 print(x)

    运行结果:

   

 

SOR迭代法

 1 import numpy as np
 2 
 3 A = np.array([[10,-1,2,0],[-1,11,-1,3],[2,-1,10,-1],[0,3,-1,8]])
 4 B = np.array([6, 25, -11, 15])
 5 x0 = np.array([0.0, 0, 0, 0])
 6 x = np.array([1.0, 2, -1, 1])
 7 w = 1.2
 8 times, MT = 0, 1000
 9 
10 while times < MT:
11     tempx = x0.copy()
12     for i in range(4):
13         temp = 0
14         for j in range(4):
15             if i != j:
16                 temp += x0[j] * A[i][j]
17         x[i] = (B[i] - temp) / A[i][i]
18         x0[i] = x[i]
19     x = w * x + (1-w) * tempx
20     calTemp = max(abs(x - tempx))
21     times += 1
22     if calTemp < 1e-5:
23         break
24 print(times)
25 print(x)

    运行结果:

   

 

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