import numpy as np
# 矩阵乘法问题
# 问题描述:
# 假设A和B是两个n阶矩阵,求它们的乘积C,n=2*k
# 1、暴力法
# 根据矩阵相乘的性质,直接遍历求解
def matrix_multiply(a, b):
n = a.shape[0]
c = np.zeros((n, n))
for i in range(n):
for j in range(n):
# 累加求和
for k in range(n):
c[i][j] += a[i][k] * b[k][j]
return c
# 2、分治法
# 算法思想:
# 将矩阵的行和列从n/2处分开,化成4个子矩阵
# A = [[A11, A12] B = [[B11, B12],那么C = [[C11, C12]
# [A21, A22]] [B21, B22]] [C21, C22]]
# C11 = A11 * B11 + A12 * B21,C12 = A11 * B12 + A12 * B22
# C21 = A21 * B11 + A22 * B21,C22 = A21 * B12 * A22 * B22
# 再继续划分成4个更小的子矩阵,直到化成1×1的矩阵为止
def divide_matrix(a, number):
n = a.shape[0]
rows, cols = n//2, n//2
b = np.zeros((rows, cols))
# 左上
if number == 1:
for i in range(rows):
for j in range(cols):
b[i][j] = a[i][j]
# 右上
elif number == 2:
for i in range(rows):
for j in range(cols):
b[i][j] = a[i][j+cols]
# 左下
elif number == 3:
for i in range(rows):
for j in range(cols):
b[i][j] = a[i+rows][j]
# 右下
else:
for i in range(rows):
for j in range(cols):
b[i][j] = a[i+rows][j+cols]
return b
def merge_matrix(c11, c12, c21, c22):
n = c11.shape[0]
c = np.zeros((2*n, 2*n), np.int32)
for i in range(2*n):
for j in range(2*n):
if i < n:
if j < n:
c[i][j] = c11[i][j]
else:
c[i][j] = c12[i][j-n]
else:
if j < n:
c[i][j] = c21[i-n][j]
else:
c[i][j] = c22[i-n][j-n]
return c
def matrix_multiply(a, b):
n = a.shape[0]
if n == 1:
c = a * b
return c
if n > 1:
c = np.zeros((n, n))
# 划分矩阵
a11 = divide_matrix(a, 1)
a12 = divide_matrix(a, 2)
a21 = divide_matrix(a, 3)
a22 = divide_matrix(a, 4)
b11 = divide_matrix(b, 1)
b12 = divide_matrix(b, 2)
b21 = divide_matrix(b, 3)
b22 = divide_matrix(b, 4)
c11 = divide_matrix(c, 1)
c12 = divide_matrix(c, 2)
c21 = divide_matrix(c, 3)
c22 = divide_matrix(c, 4)
# 递归求解
c11 = matrix_multiply(a11, b11) + matrix_multiply(a12, b21)
c12 = matrix_multiply(a11, b12) + matrix_multiply(a12, b22)
c21 = matrix_multiply(a21, b11) + matrix_multiply(a22, b21)
c22 = matrix_multiply(a21, b12) + matrix_multiply(a22, b22)
# 合并矩阵
c = merge_matrix(c11, c12, c21, c22)
return c
矩阵乘法问题
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Origin blog.csdn.net/weixin_49346755/article/details/121481609
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