C程序转MPI并行程序

Question

Hello i don't know if this is the right place to ask that. I have this MPI parallel Code that calculates PI and i need help understanding it for an oral exam.

#include <mpi.h>
#include <math.h>

int main(argc,argv)

int argc;

char *argv[];

{

    int done = 0, n, myid, numprocs, i;
    double PI25DT = 3.141592653589793238462643;
    double mypi, pi, h, sum, x;

    MPI_Init(&argc,&argv);
    MPI_Comm_size(MPI_COMM_WORLD,&numprocs);
    MPI_Comm_rank(MPI_COMM_WORLD,&myid);
    while (!done)
    {
    if (myid == 0) {
        printf("Enter the number of intervals: (0 quits) ");
        scanf("%d",&n);
    }
    MPI_Bcast(&n, 1, MPI_INT, 0, MPI_COMM_WORLD);

    if (n == 0) break;

    h   = 1.0 / (double) n;
    sum = 0.0;
    for (i = myid + 1; i <= n; i += numprocs) {
        x = h * ((double)i - 0.5);
        sum += 4.0 / (1.0 + x*x);
    }
    mypi = h * sum;

    MPI_Reduce(&mypi, &pi, 1, MPI_DOUBLE, MPI_SUM, 0,
           MPI_COMM_WORLD);

    if (myid == 0)
        printf("pi is approximately %.16f, Error is %.16f\n",
           pi, fabs(pi - PI25DT));
    }
    MPI_Finalize();
    return 0;
}

Answer 1

Read this: MPI

Starting from top: MPI_Init(&argc,&argv); is used to pass the command line arguments to all processes.

MPI_Comm_size(MPI_COMM_WORLD,&numprocs); MPI_Comm_rank(MPI_COMM_WORLD,&myid); returns the total number of MPI processes in the specified communicator (like MPI_COMM_WORLD) and their rank from pool.

MPI_Bcast Broadcasts (sends) a message from the process with rank "root" to all other processes in the group.(Here sending the number of iteration to every other process)

Between should be proper Monte Carlo pi estimation code.

And then the MPI_Reduce to Collect the areas calculated.

After that root prints the final pi value with error correction PI25DT .

Lastly MPI_Finalize() Terminates the MPI execution environment. This function should be the last MPI routine called in every MPI program - no other MPI routines may be called after it.

Pseudo for Monte Carlo Pi:

 for(i = 0; i < n; ++i) {

 x = (double)rand() / RAND_MAX;

 y = (double)rand() / RAND_MAX;

 z = x * x + y * y;

 if( z <= 1 ) count++;
 }

 pi = (double) count / n * 4;