№7 Метод прогонки решения СЛАУ с трёхдиагональной матрицей коэффициентов

import numpy as np


def tridiagonal_algorithm(a, b, c, d):
    n = len(a)
    p = [0] * (n + 1)
    q = [0] * (n + 1)

    # Прямая прогонка
    for i in range(n):
        denominator = b[i] - a[i] * p[i]
        if abs(denominator) < 1e-6:
            raise ValueError("Деление на ноль или почти ноль в прогонке.")
        p[i + 1] = c[i] / denominator
        q[i + 1] = (a[i] * q[i] - d[i]) / denominator

    print("Прямая прогонка:")
    for i in range(1, n):
        print(f"(P{i}, Q{i}) = ({p[i]:.4f}, {q[i]:.4f})")
    print(f"(Q{n}) = ({q[n]:.4f})\n")

    # Обратная прогонка
    res = [0] * n
    res[-1] = q[n]
    for i in range(n - 2, -1, -1):
        res[i] = p[i + 1] * res[i + 1] + q[i + 1]

    return res

n = 5
v = 8.0

A = np.zeros((n, n))
for i in range(n):
    for j in range(max(0, i - 1), min(i + 2, n)):
        A[i][j] = (v + i) / 100
    A[i][i] = v + i

V = np.array([v + i for i in range(n)])

B = np.dot(A, V)

print("Расширенная матрица A|B:")
for i in range(n):
    print("\t".join(f"{A[i][j]:.4f}" for j in range(n)), "|", f"{B[i]:.4f}")
print()

a = [0] * n
b = [0] * n
c = [0] * n
d = B.tolist()
for i in range(n):
    if i > 0:
        a[i] = A[i][i - 1]
    b[i] = -A[i][i]
    if i < n - 1:
        c[i] = A[i][i + 1]

X = tridiagonal_algorithm(a, b, c, d)

print("Решение СЛАУ:")
print("\t".join(f"{x:.4f}" for x in X))