两个矩阵的 Kronecker 乘积的Python程序

给定一个${m} imes{n}$矩阵 A 和${p} imes{q}$矩阵 B，它们的克罗内克积C = A 张量 B，也称为它们的矩阵直积，是${(mp)} imes{(nq)}$矩阵。

A tensor B =  |a11B   a12B|
              |a21B   a22B|

= |a11b11   a11b12   a12b11  a12b12|
  |a11b21   a11b22   a12b21  a12b22| 
  |a11b31   a11b32   a12b31  a12b32|
  |a21b11   a21b12   a22b11  a22b12|
  |a21b21   a21b22   a22b21  a22b22|
  |a21b31   a21b32   a22b31  a22b32|

例子：

1. The matrix direct(kronecker) product of the 2×2 matrix A 
   and the 2×2 matrix B is given by the 4×4 matrix :

Input : A = 1 2    B = 0 5
            3 4        6 7

Output : C = 0  5  0  10
             6  7  12 14
             0  15 0  20
             18 21 24 28

2. The matrix direct(kronecker) product of the 2×3 matrix A 
   and the 3×2 matrix B is given by the 6×6 matrix :

Input : A = 1 2    B = 0 5 2
            3 4        6 7 3
            1 0

Output : C = 0      5    2    0     10    4    
             6      7    3   12     14    6    
             0     15    6    0     20    8    
            18     21    9   24     28   12    
             0      5    2    0      0    0    
             6      7    3    0      0    0    

下面是查找两个矩阵的克罗内克积并将其存储为矩阵 C 的代码：

Python3

# Python3 code to find the Kronecker Product of two
# matrices and stores it as matrix C
  
# rowa and cola are no of rows and columns
# of matrix A
# rowb and colb are no of rows and columns
# of matrix B
cola = 2
rowa = 3
colb = 3
rowb = 2
  
# Function to computes the Kronecker Product
# of two matrices
 
def Kroneckerproduct( A , B ):
     
    C = [[0 for j in range(cola * colb)] for i in range(rowa * rowb)]
  
    # i loops till rowa
    for i in range(0, rowa):
         
        # k loops till rowb
        for k in range(0, rowb):
  
            # j loops till cola
            for j in range(0, cola):
  
                # l loops till colb
                for l in range(0, colb):
  
                    # Each element of matrix A is
                    # multiplied by whole Matrix B
                    # resp and stored as Matrix C
                    C[i + l + 1][j + k + 1] = A[i][j] * B[k][l]
                    print (C[i + l + 1][j + k + 1],end=' ')
             
             
            print ("
")
         
 
# Driver code.
 
A = [[0 for j in range(2)] for i in range(3)]
B = [[0 for j in range(3)] for i in range(2)]
 
A[0][0] = 1
A[0][1] = 2
A[1][0] = 3
A[1][1] = 4
A[2][0] = 1
A[2][1] = 0
 
B[0][0] = 0
B[0][1] = 5
B[0][2] = 2
B[1][0] = 6
B[1][1] = 7
B[1][2] = 3
 
Kroneckerproduct( A , B )
 
# This code is contributed by Saloni.

输出 ：

0    5    2    0    10    4    
6    7    3    12   14    6    
0    15   6    0    20    8    
18   21   9    24   28    12    
0    5    2    0    0     0    
6    7    3    0    0     0

时间复杂度： O(rowa*rowb*cola*colb)，因为我们使用的是嵌套循环。

辅助空间： O((rowa + colb) * (rowb + cola))，因为我们没有使用任何额外的空间。

有关详细信息，请参阅有关两个矩阵的克罗内克积的完整文章！