LAM/MPI
英文原文：
MPI - Message Passing Interface
MPI是一种消息传递库规范。接口规范已经在C/c++和Fortran程序中定义好了。提供的示例使用了C语言和LAM/MPI。LAM/MPI是一种高质量消息传递接口(MPI)的实现。
例1：demo.c
#include "mpi.h"
#include <stdio.h>
int main(int argc,char *argv[])
{
int numtasks, rank, rc;
MPI_Init(&argc,&argv);
MPI_Comm_size(MPI_COMM_WORLD,&numtasks);
MPI_Comm_rank(MPI_COMM_WORLD,&rank);
printf ("Number of tasks= %d My rank= %d
", numtasks,rank);
MPI_Finalize();
}
命令
lamboot
mpicc -o demo demo.c
mpirun -np <number of processes> demo
结果
下一个示例使用MPI来设计矩阵乘法。
一个大小为N的矩阵，该矩阵可以被搬运数整除（比如：一个矩阵的大小为4，那么搬运数也为4，每个搬运工将从矩阵A中领取1行）。控制器给每个搬运工发送同等数量的行的矩阵A，全矩阵B和追查行的位置偏移。每个搬运工接收控制器发送的信息，并完成有关行的矩阵乘法，并创建结果矩阵C的相关行，将它发送给偏移行的位置。控制器从每个搬运工那接收所有矩阵C的结果行，并完成结果矩阵.
例2：
/**********************************************************************
* MPI-based matrix multiplication AxB=C
*********************************************************************/
#include <stdio.h>
#include "mpi.h"
#define N 4 /* number of rows and columns in matrix */
MPI_Status status;
double a[N][N],b[N][N],c[N][N];
main(int argc, char **argv)
{
int numtasks,taskid,numworkers,source,dest,rows,offset,i,j,k;
struct timeval start, stop;
MPI_Init(&argc, &argv);
MPI_Comm_rank(MPI_COMM_WORLD, &taskid);
MPI_Comm_size(MPI_COMM_WORLD, &numtasks);
numworkers = numtasks-1;
/*---------------------------- master ----------------------------*/
if (taskid == 0) {
for (i=0; i<N; i++) {
for (j=0; j<N; j++) {
a[i][j]= 1.0;
b[i][j]= 2.0;
}
}
gettimeofday(&start, 0);
/* send matrix data to the worker tasks */
rows = N/numworkers;
offset = 0;
for (dest=1; dest<=numworkers; dest++)
{
MPI_Send(&offset, 1, MPI_INT, dest, 1, MPI_COMM_WORLD);
MPI_Send(&rows, 1, MPI_INT, dest, 1, MPI_COMM_WORLD);
MPI_Send(&a[offset][0], rows*N, MPI_DOUBLE,dest,1, MPI_COMM_WORLD);
MPI_Send(&b, N*N, MPI_DOUBLE, dest, 1, MPI_COMM_WORLD);
offset = offset + rows;
}
/* wait for results from all worker tasks */
for (i=1; i<=numworkers; i++)
{
source = i;
MPI_Recv(&offset, 1, MPI_INT, source, 2, MPI_COMM_WORLD, &status);
MPI_Recv(&rows, 1, MPI_INT, source, 2, MPI_COMM_WORLD, &status);
MPI_Recv(&c[offset][0], rows*N, MPI_DOUBLE, source, 2, MPI_COMM_WORLD, &status);
}
gettimeofday(&stop, 0);
printf("Here is the result matrix:
");
for (i=0; i<N; i++) {
for (j=0; j<N; j++)
printf("%6.2f ", c[i][j]);
printf ("
");
}
fprintf(stdout,"Time = %.6f
",
(stop.tv_sec+stop.tv_usec*1e-6)-(start.tv_sec+start.tv_usec*1e-6));
}
/*---------------------------- worker----------------------------*/
if (taskid > 0) {
source = 0;
MPI_Recv(&offset, 1, MPI_INT, source, 1, MPI_COMM_WORLD, &status);
MPI_Recv(&rows, 1, MPI_INT, source, 1, MPI_COMM_WORLD, &status);
MPI_Recv(&a, rows*N, MPI_DOUBLE, source, 1, MPI_COMM_WORLD, &status);
MPI_Recv(&b, N*N, MPI_DOUBLE, source, 1, MPI_COMM_WORLD, &status);
/* Matrix multiplication */
for (k=0; k<N; k++)
for (i=0; i<rows; i++) {
c[i][k] = 0.0;
for (j=0; j<N; j++)
c[i][k] = c[i][k] + a[i][j] * b[j][k];
}
MPI_Send(&offset, 1, MPI_INT, 0, 2, MPI_COMM_WORLD);
MPI_Send(&rows, 1, MPI_INT, 0, 2, MPI_COMM_WORLD);
MPI_Send(&c, rows*N, MPI_DOUBLE, 0, 2, MPI_COMM_WORLD);
}
MPI_Finalize();
}
结果
接下来，我们优化代码，让它符合以下要求：
- 应使用#define N 定义矩阵的大小。
- 矩阵中的元素类型应该是“float“。
- 矩阵可以是方形矩阵（例如，N*N），但N不能恰好被处理器的数量整除
- 应检查计算打印结果矩阵的正确性。可使用#define PRINT定义打印代码段
- 应找到这种代码加速（例如：对于N = 100，200等）与连续矩阵乘法代码比较。
例3：解决方案
/**********************************************************************
* MPI-based matrix multiplication AxB=C
*********************************************************************/
#include <stdio.h>
#include <sys/time.h>
#include "mpi.h"
#define N 500 /* number of rows and columns in matrix */
MPI_Status status;
float a[N][N],b[N][N],c[N][N];
main(int argc, char **argv)
{
int numtasks,taskid,numworkers,source,dest,rows,offset,remain,i,j,k;
struct timeval start, stop;
MPI_Init(&argc, &argv);
MPI_Comm_rank(MPI_COMM_WORLD, &taskid);
MPI_Comm_size(MPI_COMM_WORLD, &numtasks);
numworkers = numtasks-1;
/*---------------------------- master ----------------------------*/
if (taskid == 0) {
for (i=0; i<N; i++) {
for (j=0; j<N; j++) {
a[i][j]= 1.0;
b[i][j]= 2.0;
}
}
#ifdef PRINT
/* print matrices */
printf("Matrix A:
");
for (i=0; i<N; i++){
for (j=0; j<N; j++)
printf("%.3f ",a[i][j]);
printf("
");
}
printf("Matrix B:
");
for (i=0; i<N; i++){
for (j=0; j<N; j++)
printf("%.3f ",b[i][j]);
printf("
");
}
#endif
gettimeofday(&start, 0);
/* send matrix data to the worker tasks */
if (N <= numworkers)
{
rows = 1;
}
else
{
if (N%numworkers!=0) // Not divisible by numworkers
{
rows = N/numworkers+1;
remain = N%numworkers;
}
else
{
rows = N/numworkers;
}
}
offset = 0;
for (dest=1; dest<=numworkers; dest++, remain--)
{
MPI_Send(&offset, 1, MPI_INT, dest, 1, MPI_COMM_WORLD);
MPI_Send(&rows, 1, MPI_INT, dest, 1, MPI_COMM_WORLD);
MPI_Send(&a[offset][0], rows*N, MPI_FLOAT,dest,1, MPI_COMM_WORLD);
MPI_Send(&b, N*N, MPI_FLOAT, dest, 1, MPI_COMM_WORLD);
offset = offset + rows;
if(remain==1) rows-=1;
}
/* wait for results from all worker tasks */
for (i=1; i<=numworkers; i++)
{
source = i;
MPI_Recv(&offset, 1, MPI_INT, source, 2, MPI_COMM_WORLD, &status);
MPI_Recv(&rows, 1, MPI_INT, source, 2, MPI_COMM_WORLD, &status);
MPI_Recv(&c[offset][0], rows*N, MPI_FLOAT, source, 2, MPI_COMM_WORLD, &status);
}
gettimeofday(&stop, 0);
#ifdef PRINT
printf("Here is the result matrix:
");
for (i=0; i<N; i++) {
for (j=0; j<N; j++)
printf("%6.2f ", c[i][j]);
printf ("
");
}
#endif
fprintf(stdout,"Time = %.6f
",
(stop.tv_sec+stop.tv_usec*1e-6)-(start.tv_sec+start.tv_usec*1e-6));
}
/*---------------------------- worker----------------------------*/
if (taskid > 0) {
source = 0;
MPI_Recv(&offset, 1, MPI_INT, source, 1, MPI_COMM_WORLD, &status);
MPI_Recv(&rows, 1, MPI_INT, source, 1, MPI_COMM_WORLD, &status);
MPI_Recv(&a, rows*N, MPI_FLOAT, source, 1, MPI_COMM_WORLD, &status);
MPI_Recv(&b, N*N, MPI_FLOAT, source, 1, MPI_COMM_WORLD, &status);
/* Matrix multiplication */
for (k=0; k<N; k++)
for (i=0; i<rows; i++) {
c[i][k] = 0.0;
for (j=0; j<N; j++)
c[i][k] = c[i][k] + a[i][j] * b[j][k];
}
MPI_Send(&offset, 1, MPI_INT, 0, 2, MPI_COMM_WORLD);
MPI_Send(&rows, 1, MPI_INT, 0, 2, MPI_COMM_WORLD);
MPI_Send(&c, rows*N, MPI_FLOAT, 0, 2, MPI_COMM_WORLD);
}
MPI_Finalize();
}
例4：顺序矩阵代码
/* Matrix Multiplication */
#include <stdio.h>
#include <stdlib.h>
#include <math.h>
#include <sys/time.h>
#include <assert.h>
#define RANDLIMIT 5 /* Magnitude limit of generated randno.*/
#define N 500 /* Matrix Size */
#define NUMLIMIT 70.0
float a[N][N];
float b[N][N];
float c[N][N];
int main(int argc, char *argv[])
{
struct timeval start, stop;
int i,j,k;
/* generate mxs */
for (i=0; i<N; i++)
for (j=0; j<N; j++) {
a[i][j] = 1+(int) (NUMLIMIT*rand()/(RAND_MAX+1.0));
/*a[i][j] = 1.0;
b[i][j] = 2.0;*/
b[i][j] = (double) (rand() % RANDLIMIT);
/*c[i][j] = 0.0;*/
}
#ifdef PRINT
/* print matrices */
printf("Matrix A:
");
for (i=0; i<N; i++){
for (j=0; j<N; j++)
printf("%.3f ",a[i][j]);
printf("
");
}
printf("Matrix B:
");
for (i=0; i<N; i++){
for (j=0; j<N; j++)
printf("%.3f ",b[i][j]);
printf("
");
}
printf("Matrix C:
");
for (i=0; i<N; i++){
for (j=0; j<N; j++)
printf("%.3f ",c[i][j]);
printf("
");
}
#endif
gettimeofday(&start, 0);
for (i=0; i<N; i++) {
for (j=0; j<N; j++) {
c[i][j] = 0.0;
for (k=0; k<N; k++)
c[i][j] = c[i][j] + a[i][k]*b[k][j]; /* Working;standard way */
/*c[j][i] = c[j][i] + a[j][k]*b[k][i];*/ /* Working; Makes C column by col */
} /* end j loop */
}
gettimeofday(&stop, 0);
#ifdef PRINT
/* print results*/
printf("Answer c:
");
for (i=0; i<N; i++){
for (j=0; j<N; j++)
printf("%.3f ",c[i][j]);
printf("
");
}
#endif
fprintf(stdout,"Time = %.6f
",
(stop.tv_sec+stop.tv_usec*1e-6)-(start.tv_sec+start.tv_usec*1e-6));
return(0);
}
结果
检查结果，我们可以清楚地看到，当矩阵规模较大时，顺序程序比矩阵乘法的并行程序需要更多的时间。
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