使用Mpi_Scatter和Mpi_Gather进行矩阵乘法

Question

我是mpi编程的新手.我试图写矩阵乘法.通过使用散射和收集例程的关于矩阵乘法的散布集合,通过后MPI矩阵乘法.我尝试修改上面帖子中提供的代码如下...

#define N 4
#include <stdio.h>
#include <math.h>
#include <sys/time.h>
#include <stdlib.h>
#include <stddef.h>
#include "mpi.h"


void print_results(char *prompt, int a[N][N]);

int main(int argc, char *argv[])
{
    int i, j, k, rank, size, tag = 99, blksz, sum = 0;
    int a[N][N]={{1,2,3,4},{5,6,7,8},{9,1,2,3},{4,5,6,7,}};
    int b[N][N]={{1,2,3,4},{5,6,7,8},{9,1,2,3},{4,5,6,7,}};
    int c[N][N];
    int aa[N],cc[N];

    MPI_Init(&argc, &argv);
    MPI_Comm_size(MPI_COMM_WORLD, &size);
    MPI_Comm_rank(MPI_COMM_WORLD, &rank);

    //scatter rows of first matrix to different processes     
    MPI_Scatter(a, N*N/size, MPI_INT, aa, N*N/size, MPI_INT,0,MPI_COMM_WORLD);

    //broadcast second matrix to all processes
    MPI_Bcast(b, N*N, MPI_INT, 0, MPI_COMM_WORLD);

    MPI_Barrier(MPI_COMM_WORLD);

          //perform vector multiplication by all processes
          for (i = 0; i < N; i++)
            {
                    for (j = 0; j < N; j++)
                    {
                            sum = sum + aa[j] * b[i][j];                
                    }
                    cc[i] = sum;
                    sum = 0;
            }

    MPI_Gather(cc, N*N/size, MPI_INT, c, N*N/size, MPI_INT, 0, MPI_COMM_WORLD);

    MPI_Barrier(MPI_COMM_WORLD);        
    MPI_Finalize();
    print_results("C = ", c);
}

void print_results(char *prompt, int a[N][N])
{
    int i, j;

    printf ("\n\n%s\n", prompt);
    for (i = 0; i < N; i++) {
            for (j = 0; j < N; j++) {
                    printf(" %d", a[i][j]);
            }
            printf ("\n");
    }
    printf ("\n\n");
}

我跑到了上面的程序

$mpirun -np 4 ./a.out

对于上述程序,我得到以下不正确的输出..

C = 
 0 0 -562242168 32766
 1 0 4197933 0
 -562242176 32766 0 0
 4197856 0 4196672 0

C = 
 0 0 -1064802792 32765
 1 0 4197933 0
 -1064802800 32765 0 0
 4197856 0 4196672 0

C = 
 30 70 29 60
 70 174 89 148
 29 89 95 74
 60 148 74 126

C = 
 0 0 -1845552920 32765
 1 0 4197933 0
 -1845552928 32765 0 0
 4197856 0 4196672 0

我有以下查询1.为什么结果矩阵C被所有进程打印.它应该只由主要过程打印.2.为什么打印出错误的结果？

在此方面的更正和帮助将不胜感激.

Answer 1

结果矩阵c由所有进程打印,因为每个进程都执行该函数void print_results(char *prompt, int a[N][N]).由于您正在收集排名为0的进程,因此if (rank == 0)在调用该print_results(...)函数之前添加一个语句.此外,由于以下错误的循环逻辑,结果不正确:

                for (j = 0; j < N; j++)
                {
                        sum = sum + aa[j] * b[i][j];                
                }

这应该是:

                for (j = 0; j < N; j++)
                {
                        sum = sum + aa[j] * b[j][i];                
                }

此外,没有必要广播,b因为所有进程已经有了它的副本,你可以避免MPI_Barrier().完整的程序然后变成:

#define N 4
#include <stdio.h>
#include <math.h>
#include <sys/time.h>
#include <stdlib.h>
#include <stddef.h>
#include "mpi.h"


void print_results(char *prompt, int a[N][N]);

int main(int argc, char *argv[])
{
    int i, j, k, rank, size, tag = 99, blksz, sum = 0;
    int a[N][N]={{1,2,3,4},{5,6,7,8},{9,1,2,3},{4,5,6,7,}};
    int b[N][N]={{1,2,3,4},{5,6,7,8},{9,1,2,3},{4,5,6,7,}};
    int c[N][N];
    int aa[N],cc[N];

    MPI_Init(&argc, &argv);
    MPI_Comm_size(MPI_COMM_WORLD, &size);
    MPI_Comm_rank(MPI_COMM_WORLD, &rank);

    //scatter rows of first matrix to different processes     
    MPI_Scatter(a, N*N/size, MPI_INT, aa, N*N/size, MPI_INT,0,MPI_COMM_WORLD);

    //broadcast second matrix to all processes
    MPI_Bcast(b, N*N, MPI_INT, 0, MPI_COMM_WORLD);

    MPI_Barrier(MPI_COMM_WORLD);

          //perform vector multiplication by all processes
          for (i = 0; i < N; i++)
            {
                    for (j = 0; j < N; j++)
                    {
                            sum = sum + aa[j] * b[j][i];  //MISTAKE_WAS_HERE               
                    }
                    cc[i] = sum;
                    sum = 0;
            }

    MPI_Gather(cc, N*N/size, MPI_INT, c, N*N/size, MPI_INT, 0, MPI_COMM_WORLD);

    MPI_Barrier(MPI_COMM_WORLD);        
    MPI_Finalize();
    if (rank == 0)                         //I_ADDED_THIS
        print_results("C = ", c);
}

void print_results(char *prompt, int a[N][N])
{
    int i, j;

    printf ("\n\n%s\n", prompt);
    for (i = 0; i < N; i++) {
            for (j = 0; j < N; j++) {
                    printf(" %d", a[i][j]);
            }
            printf ("\n");
    }
    printf ("\n\n");
}

然后 c =

C = 
 54 37 47 57

 130 93 119 145

 44 41 56 71

 111 79 101 123