這里我有一些 python 腳本，它使用 Gauss-Seidel 方法求解線性方程組：import numpy as npITERATION_LIMIT = 1000#systemA = np.array([[15., -4., -3., 8.],          [-4., 10., -4., 2.],          [-3., -4., 10., 2.],          [8., 2., 2., 12.]          ])# vector bb = np.array([2., -12., -4., 6.])print("System of equations:")for i in range(A.shape[0]):    row = ["{0:3g}*x{1}".format(A[i, j], j + 1) for j in range(A.shape[1])]    print("[{0}] = [{1:3g}]".format(" + ".join(row), b[i]))x = np.zeros_like(b)for it_count in range(1, ITERATION_LIMIT):    x_new = np.zeros_like(x)    print("Iteration {0}: {1}".format(it_count, x))    for i in range(A.shape[0]):        s1 = np.dot(A[i, :i], x_new[:i])        s2 = np.dot(A[i, i + 1:], x[i + 1:])        x_new[i] = (b[i] - s1 - s2) / A[i, i]    if np.allclose(x, x_new, rtol=1e-8):        break    x = x_new它輸出的是：Iteration 379: [-21.36409652 -22.09743    -19.9999946   21.75896845]Iteration 380: [-21.36409676 -22.09743023 -19.99999481  21.75896868]Iteration 381: [-21.36409698 -22.09743045 -19.99999501  21.7589689 ]我的任務(wù)是利用此方法制作連續(xù)過度松弛 (SOR) 方法，該方法使用 omega 值來減少迭代次數(shù)。如果omega = 1，則變?yōu)?Gauss-Seidel 方法、if < 1- 簡單迭代法> 1和< 2- SOR。顯然，隨著 omega 值的增加，迭代次數(shù)應(yīng)該減少。這是維基百科提供的算法：Inputs: A, b, omegaOutput: phi (roots for linear equations)Choose an initial guess phi to the solutionrepeat until convergence  for i from 1 until n do    sigma <- 0    for j from 1 until n do      if j ≠ i then        sigma <- sigma + A[i][j]*phi[j]      end if    end (j-loop)    phi[i] = phi[i] + omega*((b[i] - sigma)/A[i][i]) - phi[i]  end (i-loop)  check if convergence is reachedend (repeat)有人在python上有工作算法嗎？如果您可以對代碼進行一些評論或幫助我如何更改此代碼，那就太好了。謝謝！