56 lines
1.8 KiB
Python
56 lines
1.8 KiB
Python
import math
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#模 范数
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def Norm(x,v):
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if len(x[0]) == 1:
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if v == 1:
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return sum([abs(i[0]) for i in x])
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elif v == 2:
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return (sum([i[0]**2 for i in x]))**0.5
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elif v == float("inf"):
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return max([abs(i[0]) for i in x])
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else:
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if v == 1:
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return max([sum([abs(x[i][j]) for i in range(len(x))]) for j in range(len(x[0]))])
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elif v == float("inf"):
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return max([sum([abs(i) for i in x[j]]) for j in range(len(x))])
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return None
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#Jacobi 雅可比迭代
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def Jacobi(A,b,x,err,N):
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count = 0
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n = len(A)
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while True:
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count += 1
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x_tt = x.copy() # 保存上一次迭代的解
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for i in range(n):
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sum1 = sum(A[i][j] * x_tt[j] for j in range(i))
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sum2 = sum(A[i][j] * x_tt[j] for j in range(i+1, n))
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x[i] = (b[i] - sum1 - sum2) / A[i][i]
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r = [[b[i] - sum(A[i][j] * x[j] for j in range(len(A[0])))] for i in range(n)]
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err_now = Norm(r, float("inf"))
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x_t = [round(i,5) for i in x]
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print(f"第{count}次迭代, 误差 = {err_now:.5}, x = {x_t}")
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if err_now < err:
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return x, count, 1
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if count > N:
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return None,count, 0
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if __name__ == "__main__":
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##########################################################################
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#把矩阵改成题干的矩阵，b改成题干结果，err精度要求修改##########################
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A = [
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[10,-1,2,0],
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[-1,11,-1,3],
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[2,-1,10,-1],
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[0,3,-1,8]
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]
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b = [6,25,-11,15]
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x = [0,0, 0, 0] # 初始解
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err = 1e-5 # 精度要求
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x1,k,sta = Jacobi(A, b, x, err, 100)
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print(f"解为: {x1}, 迭代次数: {k}, 状态: {'收敛' if sta == 1 else '未收敛'}")
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