import math

#模 范数
def Norm(x,v):
    if len(x[0]) == 1:
        if v == 1:
            return sum([abs(i[0]) for i in x])
        elif v == 2:
            return (sum([i[0]**2 for i in x]))**0.5
        elif v == float("inf"):
            return max([abs(i[0]) for i in x])
    else:
        if v == 1:
            return max([sum([abs(x[i][j]) for i in range(len(x))]) for j in range(len(x[0]))])
        elif v == float("inf"):
            return max([sum([abs(i) for i in x[j]]) for j in range(len(x))])
    return None

#Jacobi 雅可比迭代
def Jacobi(A,b,x,err,N):
    count = 0
    n = len(A)
    while True:
        count += 1
        x_tt = x.copy()  # 保存上一次迭代的解
        for i in range(n):
            sum1 = sum(A[i][j] * x_tt[j] for j in range(i))
            sum2 = sum(A[i][j] * x_tt[j] for j in range(i+1, n))
            x[i] = (b[i] - sum1 - sum2) / A[i][i]
        r = [[b[i] - sum(A[i][j] * x[j] for j in range(len(A[0])))] for i in range(n)]
        err_now = Norm(r, float("inf"))
        x_t = [round(i,5) for i in x]
        print(f"第{count}次迭代, 误差 = {err_now:.5}, x = {x_t}")
        if err_now < err:
            return x, count, 1
        
        if count > N:
            return None,count, 0
        

if __name__ == "__main__":
    ##########################################################################
    #把矩阵改成题干的矩阵，b改成题干结果，err精度要求修改##########################
    A = [
        [10,-1,2,0],
        [-1,11,-1,3],
        [2,-1,10,-1],
        [0,3,-1,8]
    ]
    b = [6,25,-11,15]

    x = [0,0, 0, 0]   # 初始解
    err = 1e-5              # 精度要求
    x1,k,sta = Jacobi(A, b, x, err, 100)
    print(f"解为: {x1}, 迭代次数: {k}, 状态: {'收敛' if sta == 1 else '未收敛'}")