矩阵乘法 |递归的
给定两个矩阵 A 和 B。任务是递归地将矩阵 A 和矩阵 B 相乘。如果矩阵 A 和矩阵 B 不是乘法兼容的，则生成输出“不可能”。
例子 ：
Input: A = 12 56
45 78
B = 2 6
5 8
Output: 304 520
480 894
Input: A = 1 2 3
4 5 6
7 8 9
B = 1 2 3
4 5 6
7 8 9
Output: 30 36 42
66 81 96
102 126 150 建议先参考Iterative Matrix Multiplication。
首先检查矩阵之间的乘法是否可能。为此，检查第一个矩阵的列数是否等于第二个矩阵的行数。如果两者相等，则继续进行，否则生成输出“不可能”。
在递归矩阵乘法中，我们通过递归调用实现了三个迭代循环。 multiplyMatrix()最内部的递归调用是迭代 k（col1 或 row2）。 multiplyMatrix()的第二个递归调用是更改列，最外面的递归调用是更改行。
下面是递归矩阵乘法代码。
C/C++
// Recursive code for Matrix Multiplication
#include
const int MAX = 100;
void multiplyMatrixRec(int row1, int col1, int A[][MAX],
int row2, int col2, int B[][MAX],
int C[][MAX])
{
// Note that below variables are static
// i and j are used to know current cell of
// result matrix C[][]. k is used to know
// current column number of A[][] and row
// number of B[][] to be multiplied
static int i = 0, j = 0, k = 0;
// If all rows traversed.
if (i >= row1)
return;
// If i < row1
if (j < col2)
{
if (k < col1)
{
C[i][j] += A[i][k] * B[k][j];
k++;
multiplyMatrixRec(row1, col1, A, row2, col2,
B, C);
}
k = 0;
j++;
multiplyMatrixRec(row1, col1, A, row2, col2, B, C);
}
j = 0;
i++;
multiplyMatrixRec(row1, col1, A, row2, col2, B, C);
}
// Function to multiply two matrices A[][] and B[][]
void multiplyMatrix(int row1, int col1, int A[][MAX],
int row2, int col2, int B[][MAX])
{
if (row2 != col1)
{
printf("Not Possible\n");
return;
}
int C[MAX][MAX] = {0};
multiplyMatrixRec(row1, col1, A, row2, col2, B, C);
// Print the result
for (int i = 0; i < row1; i++)
{
for (int j = 0; j < col2; j++)
printf("%d ", C[i][j]);
printf("\n");
}
}
// Driven Program
int main()
{
int A[][MAX] = { {1, 2, 3},
{4, 5, 6},
{7, 8, 9}};
int B[][MAX] = { {1, 2, 3},
{4, 5, 6},
{7, 8, 9} };
int row1 = 3, col1 = 3, row2 = 3, col2 = 3;
multiplyMatrix(row1, col1, A, row2, col2, B);
return 0;
}
Java
// Java recursive code for Matrix Multiplication
class GFG
{
public static int MAX = 100;
// Note that below variables are static
// i and j are used to know current cell of
// result matrix C[][]. k is used to know
// current column number of A[][] and row
// number of B[][] to be multiplied
public static int i = 0, j = 0, k = 0;
static void multiplyMatrixRec(int row1, int col1, int A[][],
int row2, int col2, int B[][],
int C[][])
{
// If all rows traversed
if (i >= row1)
return;
// If i < row1
if (j < col2)
{
if (k < col1)
{
C[i][j] += A[i][k] * B[k][j];
k++;
multiplyMatrixRec(row1, col1, A, row2, col2, B, C);
}
k = 0;
j++;
multiplyMatrixRec(row1, col1, A, row2, col2, B, C);
}
j = 0;
i++;
multiplyMatrixRec(row1, col1, A, row2, col2, B, C);
}
// Function to multiply two matrices A[][] and B[][]
static void multiplyMatrix(int row1, int col1, int A[][],
int row2, int col2, int B[][])
{
if (row2 != col1)
{
System.out.println("Not Possible\n");
return;
}
int[][] C = new int[MAX][MAX];
multiplyMatrixRec(row1, col1, A, row2, col2, B, C);
// Print the result
for (int i = 0; i < row1; i++)
{
for (int j = 0; j < col2; j++)
System.out.print(C[i][j]+" ");
System.out.println();
}
}
// driver program
public static void main (String[] args)
{
int row1 = 3, col1 = 3, row2 = 3, col2 = 3;
int A[][] = { {1, 2, 3},
{4, 5, 6},
{7, 8, 9}};
int B[][] = { {1, 2, 3},
{4, 5, 6},
{7, 8, 9} };
multiplyMatrix(row1, col1, A, row2, col2, B);
}
}
// Contributed by Pramod KumarPython3
# Recursive code for Matrix Multiplication
MAX = 100
i = 0
j = 0
k = 0
def multiplyMatrixRec(row1, col1, A,
row2, col2, B, C):
# Note that below variables are static
# i and j are used to know current cell of
# result matrix C[][]. k is used to know
# current column number of A[][] and row
# number of B[][] to be multiplied
global i
global j
global k
# If all rows traversed.
if (i >= row1):
return
# If i < row1
if (j < col2):
if (k < col1):
C[i][j] += A[i][k] * B[k][j]
k += 1
multiplyMatrixRec(row1, col1, A,
row2, col2,B, C)
k = 0
j += 1
multiplyMatrixRec(row1, col1, A,
row2, col2, B, C)
j = 0
i += 1
multiplyMatrixRec(row1, col1, A,
row2, col2, B, C)
# Function to multiply two matrices
# A[][] and B[][]
def multiplyMatrix(row1, col1, A, row2, col2, B):
if (row2 != col1):
print("Not Possible")
return
C = [[0 for i in range(MAX)]
for i in range(MAX)]
multiplyMatrixRec(row1, col1, A,
row2, col2, B, C)
# Print the result
for i in range(row1):
for j in range(col2):
print( C[i][j], end = " ")
print()
# Driver Code
A = [[1, 2, 3],
[4, 5, 6],
[7, 8, 9]]
B = [[1, 2, 3],
[4, 5, 6],
[7, 8, 9]]
row1 = 3
col1 = 3
row2 = 3
col2 = 3
multiplyMatrix(row1, col1, A, row2, col2, B)
# This code is contributed by sahilshelangiaC#
// C# recursive code for
// Matrix Multiplication
using System;
class GFG
{
public static int MAX = 100;
// Note that below variables
// are static i and j are used
// to know current cell of result
// matrix C[][]. k is used to
// know current column number of
// A[][] and row number of B[][]
// to be multiplied
public static int i = 0, j = 0, k = 0;
static void multiplyMatrixRec(int row1, int col1,
int [,]A, int row2,
int col2, int [,]B,
int [,]C)
{
// If all rows traversed
if (i >= row1)
return;
// If i < row1
if (j < col2)
{
if (k < col1)
{
C[i, j] += A[i, k] * B[k, j];
k++;
multiplyMatrixRec(row1, col1, A,
row2, col2, B, C);
}
k = 0;
j++;
multiplyMatrixRec(row1, col1, A,
row2, col2, B, C);
}
j = 0;
i++;
multiplyMatrixRec(row1, col1, A,
row2, col2, B, C);
}
// Function to multiply two
// matrices A[][] and B[][]
static void multiplyMatrix(int row1, int col1,
int [,]A, int row2,
int col2, int [,]B)
{
if (row2 != col1)
{
Console.WriteLine("Not Possible\n");
return;
}
int[,]C = new int[MAX, MAX];
multiplyMatrixRec(row1, col1, A,
row2, col2, B, C);
// Print the result
for (int i = 0; i < row1; i++)
{
for (int j = 0; j < col2; j++)
Console.Write(C[i, j] + " ");
Console.WriteLine();
}
}
// Driver Code
static public void Main ()
{
int row1 = 3, col1 = 3,
row2 = 3, col2 = 3;
int [,]A = {{1, 2, 3},
{4, 5, 6},
{7, 8, 9}};
int [,]B = {{1, 2, 3},
{4, 5, 6},
{7, 8, 9}};
multiplyMatrix(row1, col1, A,
row2, col2, B);
}
}
// This code is contributed by m_kitJavascript
输出 ：
30 36 42
66 81 96
102 126 150