矩阵相乘问题（分治法求解）

古城微笑少年丶 2022-04-15 04:56 221阅读 0赞

采用蛮力+分治进行求解：

矩阵相乘公式：

![20181121202552877.png][]

![watermark_type_ZmFuZ3poZW5naGVpdGk_shadow_10_text_aHR0cHM6Ly9ibG9nLmNzZG4ubmV0L2FpX3l1ZQ_size_16_color_FFFFFF_t_70][]

![watermark_type_ZmFuZ3poZW5naGVpdGk_shadow_10_text_aHR0cHM6Ly9ibG9nLmNzZG4ubmV0L2FpX3l1ZQ_size_16_color_FFFFFF_t_70 1][]

![20181121202726155.png][]

代码：

public class Matrix {
        
        //初始化一个随机nxn阶矩阵
        public static int[][] initializationMatrix(int n){
            int[][] result = new int[n][n];
            for(int i = 0;i < n;i++){
                for(int j = 0;j < n;j++){
                    result[i][j] = (int)(Math.random()*10); //采用随机函数随机生成1~10之间的数
                }
            }            
            return result;            
        }
        
        //蛮力法求解两个nxn和nxn阶矩阵相乘
        public static int[][] BruteForce(int[][] p,int[][] q,int n){
            int[][] result = new int[n][n];
            for(int i=0;i<n;i++){
                for(int j=0;j<n;j++){
                    result[i][j] = 0;
                    for(int k=0;k<n;k++){
                        result[i][j] += p[i][k]*q[k][j];
                    }
                }
            }                
            return result;
        }
        
        //分治法求解两个nxn和nxn阶矩阵相乘
        public static int[][] DivideAndConquer(int[][] p,int[][] q,int n){
            int[][] result = new int[n][n];
            //当n为2时，返回矩阵相乘结果
            if(n == 2){
                result = BruteForce(p,q,n);            
                return result;
            }
            
            //当n大于3时，采用采用分治法，递归求最终结果
            if(n > 2){
                int m = n/2;
                
                int[][] p1 = QuarterMatrix(p,n,1);
                int[][] p2 = QuarterMatrix(p,n,2);
                int[][] p3 = QuarterMatrix(p,n,3);
                int[][] p4 = QuarterMatrix(p,n,4);
    //            System.out.println();
    //            System.out.print("矩阵p1值为：");
    //            PrintfMatrix(p1,m);
    //            System.out.println();
    //            System.out.print("矩阵p2值为：");
    //            PrintfMatrix(p2,m);
    //            System.out.println();
    //            System.out.print("矩阵p3值为：");
    //            PrintfMatrix(p3,m);
    //            System.out.println();
    //            System.out.print("矩阵p4值为：");
    //            PrintfMatrix(p4,m);
                
                int[][] q1 = QuarterMatrix(q,n,1);
                int[][] q2 = QuarterMatrix(q,n,2);
                int[][] q3 = QuarterMatrix(q,n,3);
                int[][] q4 = QuarterMatrix(q,n,4);
                
                int[][] result1 = QuarterMatrix(result,n,1);
                int[][] result2 = QuarterMatrix(result,n,2);
                int[][] result3 = QuarterMatrix(result,n,3);
                int[][] result4 = QuarterMatrix(result,n,4);
                
                
                result1 = AddMatrix(DivideAndConquer(p1,q1,m),DivideAndConquer(p2,q3,m),m);
                result2 = AddMatrix(DivideAndConquer(p1,q2,m),DivideAndConquer(p2,q4,m),m);
                result3 = AddMatrix(DivideAndConquer(p3,q1,m),DivideAndConquer(p4,q3,m),m);
                result4 = AddMatrix(DivideAndConquer(p3,q2,m),DivideAndConquer(p4,q4,m),m);
                
                
                result = TogetherMatrix(result1,result2,result3,result4,m);
            }
            return result;
        }
        
        //获取矩阵的四分之一，并决定返回哪一个四分之一
        public static int[][] QuarterMatrix(int[][] p,int n,int number){
            int rows = n/2;   //行数减半
            int cols = n/2;   //列数减半
            int[][] result = new int[rows][cols];
            switch(number){
               case 1 :
               {
                  // result = new int[rows][cols];
                   for(int i=0;i<rows;i++){
                       for(int j=0;j<cols;j++){
                           result[i][j] = p[i][j];
                       }
                   }
                   break;
               }
                
               case 2 :
               {
                  // result = new int[rows][n-cols];
                   for(int i=0;i<rows;i++){
                       for(int j=0;j<n-cols;j++){
                           result[i][j] = p[i][j+cols];
                       }
                   }
                   break;
               }
               
               case 3 :
               {
                  // result = new int[n-rows][cols];
                   for(int i=0;i<n-rows;i++){
                       for(int j=0;j<cols;j++){
                           result[i][j] = p[i+rows][j];
                       }
                   }
                   break;
               }
               
               case 4 :
               {
                  // result = new int[n-rows][n-cols];
                   for(int i=0;i<n-rows;i++){
                       for(int j=0;j<n-cols;j++){
                           result[i][j] = p[i+rows][j+cols];
                       }
                   }
                   break;
               }
               
               default:
                   break;
            }
            
            return result;
         }
        
        //把均分为四分之一的矩阵，聚合成一个矩阵，其中矩阵a,b,c,d分别对应原完整矩阵的四分中1、2、3、4
        public static int[][] TogetherMatrix(int[][] a,int[][] b,int[][] c,int[][] d,int n){
            int[][] result = new int[2*n][2*n];
            for(int i=0;i<2*n;i++){
                for(int j=0;j<2*n;j++){
                    if(i<n){
                        if(j<n){
                            result[i][j] = a[i][j];
                        }
                        else
                            result[i][j] = b[i][j-n];
                    }
                    else{
                        if(j<n){
                            result[i][j] = c[i-n][j];
                        }
                        else{
                            result[i][j] = d[i-n][j-n];
                        }
                    }
                }
            }
            
            return result;
        }
        
        
        //求两个矩阵相加结果
        public static int[][] AddMatrix(int[][] p,int[][] q,int n){
            int[][] result = new int[n][n];
            for(int i=0;i<n;i++){
                for(int j=0;j<n;j++){
                    result[i][j] = p[i][j]+q[i][j];
                }
            }
            return result;
        }
        
        //控制台输出矩阵
        public static void PrintfMatrix(int[][] matrix,int n){
            for(int i=0;i<n;i++){
                System.out.println();
                for(int j=0;j<n;j++){
                    System.out.print("\t");
                    System.out.print(matrix[i][j]);
                }
            }
            
        }
        
        public static void main(String args[]){
            int[][] p = initializationMatrix(8);
            int[][] q = initializationMatrix(8);
            System.out.print("矩阵p初始化值为：");
            PrintfMatrix(p,8);
            System.out.println();
            System.out.print("矩阵q初始化值为：");
            PrintfMatrix(q,8);
            
            int[][] bf_result = BruteForce(p,q,8);
            System.out.println();
            System.out.print("蛮力法计算矩阵p*q结果为：");
            PrintfMatrix(bf_result,8);
        
            int[][] dac_result = DivideAndConquer(p,q,8);
            System.out.println();
            System.out.print("分治法计算矩阵p*q结果为：");
            PrintfMatrix(dac_result,8);
        }
    
    }

本文转自[https://www.cnblogs.com/liuzhen1995/p/6126460.html][https_www.cnblogs.com_liuzhen1995_p_6126460.html]

[20181121202552877.png]: /images/20220415/9a1a75b859b944dabf75553c2fb4eb37.png
[watermark_type_ZmFuZ3poZW5naGVpdGk_shadow_10_text_aHR0cHM6Ly9ibG9nLmNzZG4ubmV0L2FpX3l1ZQ_size_16_color_FFFFFF_t_70]: /images/20220415/439e11b6cbbc4006baf9e0c818fddd07.png
[watermark_type_ZmFuZ3poZW5naGVpdGk_shadow_10_text_aHR0cHM6Ly9ibG9nLmNzZG4ubmV0L2FpX3l1ZQ_size_16_color_FFFFFF_t_70 1]: /images/20220415/e9bbb42f7a784854b7e670fbd1ddad3a.png
[20181121202726155.png]: /images/20220415/609a670c9be743cca3661cf9f527aabb.png
[https_www.cnblogs.com_liuzhen1995_p_6126460.html]: https://www.cnblogs.com/liuzhen1995/p/6126460.html