Welcome to OStack Knowledge Sharing Community for programmer and developer-Open, Learning and Share
Welcome To Ask or Share your Answers For Others

Categories

0 votes
353 views
in Technique[技术] by (71.8m points)

c - MPI partition matrix into blocks

I want to partition matrix into blocks (not stripes) and then distribute this blocks using MPI_Scatter.

I came up with solution which works, but I think it is far from "best practice". I have 8x8 matrix, filled with numbers from 0 to 63. Then I divide it into 4 4x4 blocks, using MPI_Type_vector and distribute it via MPI_Send, but this require some extra computation since i have to compute offsets for each block in big matrix.

If I use scatter, first (top left) block is transfered OK, but other blocks are not (wrong offset for start of block).

So is it possible to transfer blocks of matrix using MPI_Scatter, or what is the best way to do desired decomposition?

This is my code:

#include <stdio.h>
#include <stdlib.h>
#include <mpi.h>

#define SIZE 8


int main(void) {

        MPI_Init(NULL, NULL);
        int p, rank;
        MPI_Comm_size(MPI_COMM_WORLD, &p);
        MPI_Comm_rank(MPI_COMM_WORLD, &rank);
        char i;

        char a[SIZE*SIZE];
        char b[(SIZE/2)*(SIZE/2)];

        MPI_Datatype columntype;
        MPI_Datatype columntype2;

        MPI_Type_vector(4, 4, SIZE, MPI_CHAR, &columntype2);
        MPI_Type_create_resized( columntype2, 0, sizeof(MPI_CHAR), &columntype );
        MPI_Type_commit(&columntype);

        if(rank == 0) {
                for( i = 0; i < SIZE*SIZE; i++) {
                        a[i] = i;
                }

                for(int rec=0; rec < p; rec++) {
                        int offset = (rec%2)*4 + (rec/2)*32;
                      MPI_Send (a+offset, 1, columntype, rec, 0, MPI_COMM_WORLD);
                }
        }
        MPI_Recv (b, 16, MPI_CHAR, 0, 0, MPI_COMM_WORLD, MPI_STATUS_IGNORE);
        //MPI_Scatter(&a, 1, boki, &b, 16, MPI_CHAR , 0, MPI_COMM_WORLD);

        printf("rank= %d  b= 
%d %d %d %d
%d %d %d %d
%d %d %d %d
%d %d %d %d
", rank, b[0], b[1], b[2], b[3], b[4], b[5], b[6], b[7], b[8], b[9], b[10], b[11], b[12], b[13], b[14], b[15]);

        MPI_Finalize();

        return 0;
}
See Question&Answers more detail:os

与恶龙缠斗过久,自身亦成为恶龙;凝视深渊过久,深渊将回以凝视…
Welcome To Ask or Share your Answers For Others

1 Answer

0 votes
by (71.8m points)

What you've got is pretty much "best practice"; it's just a bit confusing until you get used to it.

Two things, though:

First, be careful with this: sizeof(MPI_CHAR) is, I assume, 4 bytes, not 1. MPI_CHAR is an (integer) constant that describes (to the MPI library) a character. You probably want sizeof(char), or SIZE/2*sizeof(char), or anything else convenient. But the basic idea of doing a resize is right.

Second, I think you're stuck using MPI_Scatterv, though, because there's no easy way to make the offset between each block the same size. That is, the first element in the first block is at a[0], the second is at a[SIZE/2] (jump of size/2), the next is at a[SIZE*(SIZE/2)] (jump of (SIZE-1)*(SIZE/2)). So you need to be able to manually generate the offsets.

The following seems to work for me (I generalized it a little bit to make it clearer when "size" means "number of rows" vs "number of columns", etc):

#include <stdio.h>
#include <stdlib.h>
#include <mpi.h>

#define COLS  12
#define ROWS  8

int main(int argc, char **argv) {

    MPI_Init(&argc, &argv);
    int p, rank;
    MPI_Comm_size(MPI_COMM_WORLD, &p);
    MPI_Comm_rank(MPI_COMM_WORLD, &rank);
    char i;

    char a[ROWS*COLS];
    const int NPROWS=2;  /* number of rows in _decomposition_ */
    const int NPCOLS=3;  /* number of cols in _decomposition_ */
    const int BLOCKROWS = ROWS/NPROWS;  /* number of rows in _block_ */
    const int BLOCKCOLS = COLS/NPCOLS; /* number of cols in _block_ */

    if (rank == 0) {
        for (int ii=0; ii<ROWS*COLS; ii++) {
            a[ii] = (char)ii;
        }
    }

    if (p != NPROWS*NPCOLS) {
        fprintf(stderr,"Error: number of PEs %d != %d x %d
", p, NPROWS, NPCOLS);
        MPI_Finalize();
        exit(-1);
    }
    char b[BLOCKROWS*BLOCKCOLS];
    for (int ii=0; ii<BLOCKROWS*BLOCKCOLS; ii++) b[ii] = 0;

    MPI_Datatype blocktype;
    MPI_Datatype blocktype2;

    MPI_Type_vector(BLOCKROWS, BLOCKCOLS, COLS, MPI_CHAR, &blocktype2);
    MPI_Type_create_resized( blocktype2, 0, sizeof(char), &blocktype);
    MPI_Type_commit(&blocktype);

    int disps[NPROWS*NPCOLS];
    int counts[NPROWS*NPCOLS];
    for (int ii=0; ii<NPROWS; ii++) {
        for (int jj=0; jj<NPCOLS; jj++) {
            disps[ii*NPCOLS+jj] = ii*COLS*BLOCKROWS+jj*BLOCKCOLS;
            counts [ii*NPCOLS+jj] = 1;
        }
    }

    MPI_Scatterv(a, counts, disps, blocktype, b, BLOCKROWS*BLOCKCOLS, MPI_CHAR, 0, MPI_COMM_WORLD);
    /* each proc prints it's "b" out, in order */
    for (int proc=0; proc<p; proc++) {
        if (proc == rank) {
            printf("Rank = %d
", rank);
            if (rank == 0) {
                printf("Global matrix: 
");
                for (int ii=0; ii<ROWS; ii++) {
                    for (int jj=0; jj<COLS; jj++) {
                        printf("%3d ",(int)a[ii*COLS+jj]);
                    }
                    printf("
");
                }
            }
            printf("Local Matrix:
");
            for (int ii=0; ii<BLOCKROWS; ii++) {
                for (int jj=0; jj<BLOCKCOLS; jj++) {
                    printf("%3d ",(int)b[ii*BLOCKCOLS+jj]);
                }
                printf("
");
            }
            printf("
");
        }
        MPI_Barrier(MPI_COMM_WORLD);
    }

    MPI_Finalize();

    return 0;
}

Running:

$ mpirun -np 6 ./matrix

Rank = 0
Global matrix: 
  0   1   2   3   4   5   6   7   8   9  10  11 
 12  13  14  15  16  17  18  19  20  21  22  23 
 24  25  26  27  28  29  30  31  32  33  34  35 
 36  37  38  39  40  41  42  43  44  45  46  47 
 48  49  50  51  52  53  54  55  56  57  58  59 
 60  61  62  63  64  65  66  67  68  69  70  71 
 72  73  74  75  76  77  78  79  80  81  82  83 
 84  85  86  87  88  89  90  91  92  93  94  95 
Local Matrix:
  0   1   2   3 
 12  13  14  15 
 24  25  26  27 
 36  37  38  39 

Rank = 1
Local Matrix:
  4   5   6   7 
 16  17  18  19 
 28  29  30  31 
 40  41  42  43 

Rank = 2
Local Matrix:
  8   9  10  11 
 20  21  22  23 
 32  33  34  35 
 44  45  46  47 

Rank = 3
Local Matrix:
 48  49  50  51 
 60  61  62  63 
 72  73  74  75 
 84  85  86  87 

Rank = 4
Local Matrix:
 52  53  54  55 
 64  65  66  67 
 76  77  78  79 
 88  89  90  91 

Rank = 5
Local Matrix:
 56  57  58  59 
 68  69  70  71 
 80  81  82  83 
 92  93  94  95 

与恶龙缠斗过久,自身亦成为恶龙;凝视深渊过久,深渊将回以凝视…
Welcome to OStack Knowledge Sharing Community for programmer and developer-Open, Learning and Share
Click Here to Ask a Question

...