mpi 计算 pi 的示例程序 MPI_Bcast MPI_Reduce

1，原理

从 0.0 到 1.0 对 4.0/(1.0+x*x） 进行定积分，便得到了pi 的解析值；

2，代码

//pi_reduce.c
#include 

#include 
#include 

double f(double);
double f(double x)
{
    return (4.0/(1.0+x*x));
}

int main(int argc, char* argv[])
{
    int done =0, n, myid, numprocs, i;
    double PI25DT = 3.141592653589793238462643;// a more accurate PI
    double mypi, pi, h, sum, x;

    double startwtime = 0.0, endwtime;
    int namelen;
    char processor_name[MPI_MAX_PROCESSOR_NAME];//[256] openmpi-4.x

    MPI_Init(&argc, &argv);//& & mutiple processes below:
    MPI_Comm_size(MPI_COMM_WORLD, &numprocs);// total number of processes
    MPI_Comm_rank(MPI_COMM_WORLD, &myid);// which process I am
    MPI_Get_processor_name(processor_name, &namelen);// host name and its lenth

    fprintf(stdout, "Process %d of %d on %s\n", myid, numprocs, processor_name);

    n = 0;
    if(myid == 0)
    {
        fprintf(stdout, "Please give N=\n");
        fflush(stdout);
        scanf("%d",&n);
        fprintf(stdout, "n = %d\n", n);
        //n= 1000000;
        startwtime = MPI_Wtime();
    }
    //MPI_Barrier(MPI_COMM_WORLD);//

    MPI_Bcast(&n, 1, MPI_INT, 0, MPI_COMM_WORLD);// broadcast n to others
    //MPI_Bcast在默认情况下是阻塞调用，它会等待所有进程完成归约操作后才会继续执行。你可以使用非阻塞版本的MPI_Ireduce来进行非阻塞调用

    h = 1.0/(double)n;
    sum = 0.0;

    for(i=myid+1; i<=n; i+=numprocs)// each process would calculate several area of rectangles
    {
        // n proces, N rectangle
        // proc 1-st: 1, n+1, 2n+1, ..., N1; N1 is as big as possible, and N1<=N;
        // proc 2-nd: 2, n+2, 2n+2, ..., N2; N2 is as big as possible, and N2<=N;
        // ...
        // proc n-th: n, n+n, 2n+n, ..., Nn; Nn is as big as possible, and Nn<=N;
        x = h *((double)i - 0.5);// x of this time
        sum += f(x);
        fprintf(stdout, "Process %d of %d on %s, n=%d, mypi=%.6f,  x=%7.3f, h=%7.3f, sum=%7.3f\n", myid, numprocs, processor_name, n, mypi, x, h, sum);
    }
    fflush(stdout);
    mypi = h * sum;

    //fprintf(stdout, "Process %d of %d on %s, n=%d, mypi=%.16f\n", myid, numprocs, processor_name, n, mypi);
    //fflush(stdout);
    //int MPI_Reduce(void* sendbuf, void* recvbuf, int count, PI_Datatype datatype, MPI_Op op, int root, MPI_Comm comm)
    MPI_Reduce(&mypi, &pi, 1, MPI_DOUBLE, MPI_SUM, 0, MPI_COMM_WORLD);
    //MPI_Reduce在默认情况下是阻塞调用，它会等待所有进程完成归约操作后才会继续执行。你可以使用非阻塞版本的MPI_Ireduce来进行非阻塞调用
    //MPI_Barrier(MPI_COMM_WORLD);
    if(myid == 0)
    {
        endwtime = MPI_Wtime();
        printf("wall clock time = %f\n", endwtime -startwtime);
        printf("pi is approximately %.16f, Error is %.16f\n", pi, fabs(pi - PI25DT));
        fflush(stdout);
    }
    MPI_Finalize();

    return 0;
}

3，编译

#FORT := ../local/bin/mpif77
FORT := gfortran
LD_F_FLAGS := -lgfortran -L../local/lib/ -lmpi_mpifh

EXE := pi_red

CC := ../local/bin/mpicxx

all: $(EXE)



pi_red: pi_reduce.c
	../local/bin/mpicc  $< -o $@


jacobi: jacobi.f
	$(FORT) $< -o $@ $(LD_F_FLAGS) -I../local/include

the_jacobi: the_jacobi.cpp
	$(CC) $< -o $@

run_jacobi: jacobi
	../local/bin/mpirun -np 4 ./jacobi

pi_reduce: pi_reduce.c
	../local/bin/mpicc $< -o $@

run_pi_reduce: pi_reduce
	../local/bin/mpirun -np 18 ./pi_reduce

#gcc max_name_openmpi.c -o max.out -I ../local/include/ -L ../local/lib  -lmpi

#gfortran hello_ompi.f -I../local/include/ -L../local/lib/  -lmpi_mpifh -lgfortran

即：

../local/bin/mpicc  pi_reduce.c  -o  pi_red

4，运行

由于cpu是18个物理核心，所以使用了 18：

 ./local/bin/mpirun -np 18 ./pi_red

需要输入n的值，这里输入了36，再大一些的任意整数都可以，理论上讲，越大越精确。

效果：