# Решение СЛАУ с помощью QR-разложения

import numpy as np
import math

def vec_in_numb(a, b, n):
    count = 0
    for i in range(n):
        count += a[i] * b[i]
    return count

def Gramm_Schmidt(matrix):
    A = matrix.copy()
    B = np.array(np.zeros((A.shape[0], A.shape[1])))
    B[0][...] = A[0][...]
    for i in range(1, A.shape[0]):
        a = A[i][...]
        countR = np.array(np.zeros((A.shape[0])))
        for j in range(0, i):
            b = B[j][...]
            countB = vec_in_numb(a, b, A.shape[0])
            countA = vec_in_numb(b, b, A.shape[0])
            countB = countB / countA
            countR -= countB * b
        B[i][...] = a + countR
    return Orthog(B)

def Orthog(B):
    for i in range(B.shape[0]):
        count = 0
        for k in range(B.shape[0]):
            count += math.fabs(B[i][k])**2
        B[i][...] /= round(math.sqrt(count), 3)
    return B

def find_R(Q, A):
    R = np.array(np.zeros((Q.shape[0], Q.shape[1])))
    for i in range(Q.shape[0]):
        for j in range(Q.shape[0]):
            for k in range(Q.shape[0]):
                R[i][j] += round(Q[i][k] * A[k][j], 3)
    return R



A = np.array([[2.2, 4, -3, 1.5, 0.6, 2, 0.7],
              [4, 3.2, 1.5, -0.7, -0.8, 3, 1],
              [-3, 1.5, 1.8, 0.9, 3, 2, 2],
              [1.5, -0.7, 0.9, 2.2, 4, 3, 1],
              [0.6, -0.8, 3, 4, 3.2, 0.6, 0.7],
              [2, 3, 2, 3, 0.6, 2.2, 4],
              [0.7, 1, 2, 1, 0.7, 4, 3.2]])

Q = Gramm_Schmidt(np.transpose(A))
R = find_R(Q, A)

for i in range(Q.shape[0]):
    for j in range(Q.shape[0]):
        Q[i][j] = round(Q[i][j], 4)
        R[i][j] = round(R[i][j], 4)

#print(Q)
print(R)