矩阵乘法| 递归的 · GeeksForGeeks 中文系列教程

# 矩阵乘法| 递归的 > 原文： [https://www.geeksforgeeks.org/matrix-multiplication-recursive/](https://www.geeksforgeeks.org/matrix-multiplication-recursive/) 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 不兼容乘法，则生成输出“不可能”。 **示例**： ``` 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 ``` 建议首先参考[迭代矩阵乘法](https://www.geeksforgeeks.org/c-program-multiply-two-matrices/)。首先检查矩阵之间的乘法是否可行。为此，请检查第一矩阵的列数是否等于第二矩阵的行数。如果两者相等，则继续操作，否则生成输出“ Not 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。 *multipleMatrix（）*的最内层递归调用是迭代 k（col1 或 row2）。第二个*乘法矩阵（）*的递归调用将更改列，最外面的递归调用将更改行。以下是递归矩阵乘法代码。 ## C/C++ ``` // Recursive code for Matrix Multiplication #include <stdio.h> 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 // 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 Kumar ``` ## Python3 ```py # 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 sahilshelangia ``` ## C# ```cs // 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_kit ``` **输出**： ``` 30 36 42 66 81 96 102 126 150 ```