Question

我试图使用MPI来解决蒙特卡罗问题，我们生成x量的兰特。 num在0和1之间，然后向每个处理器发送n长度数字。我正在使用分散功能但我的代码运行不正确，它编译但它不要求输入。我不明白MPI如何在没有循环的情况下自行循环，可以解释一下这个以及我的代码有什么问题吗？

#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#include <time.h>
#include <math.h>
#include "mpi.h"

main(int argc, char* argv[]) {
int         my_rank;       /* rank of process      */
int         p;             /* number of processes  */
int         source;        /* rank of sender       */
int         dest;          /* rank of receiver     */
int         tag = 0;       /* tag for messages     */
char        message[100];  /* storage for message  */
MPI_Status  status;        /* return status for    */
double *total_xr, *p_xr, total_size_xr, p_size_xr;  /* receive              */

/* Start up MPI */
MPI_Init(&argc, &argv);

/* Find out process rank  */
MPI_Comm_rank(MPI_COMM_WORLD, &my_rank);

/* Find out number of processes */
MPI_Comm_size(MPI_COMM_WORLD, &p);

double temp;
int i, partial_sum, x, total_sum, ratio_p, area;
total_size_xr = 0;
partial_sum = 0;
if(my_rank == 0){
    while(total_size_xr <= 0){
        printf("How many random numbers should each process get?: ");
        scanf("%f", &p_size_xr);
    }
    total_size_xr = p*p_size_xr;
    total_xr = malloc(total_size_xr*sizeof(double));

    //xr generator will generate numbers between 1 and 0
    srand(time(NULL));
    for(i=0; i<total_size_xr; i++)
    {
        temp = 2.0 * rand()/(RAND_MAX+1.0) -1.0;
        //this will make sure if any number computer stays in the boundry of 0 and 1, doesn't go over into the negative
        while(temp < 0.0)
        {
            temp = 2.0 * rand()/(RAND_MAX+1.0) -1.0;
        }
        //array set to total random numbers generated to be scatter into processors
        total_xr[i] = temp;
    }

}
else{
//this will be the buffer for the processors to hold their own numbers to add
p_xr = malloc(p_size_xr*sizeof(double));
printf("\n\narray set\n\n");
//scatter xr into processors
MPI_Scatter(total_xr, total_size_xr, MPI_DOUBLE, p_xr, p_size_xr, MPI_DOUBLE, 0, MPI_COMM_WORLD);
//while in processor the partial sum will be caluclated by using xr and the formula sqrt(1-x*x)
for(i=0; i<p_size_xr; i++)
{
    x = p_xr[i];
    temp = sqrt(1 - (x*x));
    partial_sum = partial_sum + temp;
}
//}


//we will send the partial sums to master processor which is processor 0 and add them and place 
//the result in total_sum
MPI_Reduce(&partial_sum, &total_sum, 1, MPI_DOUBLE, MPI_SUM, 0, MPI_COMM_WORLD);

//once we have all of the sums we need to multiply the total sum and multiply it with 1/N
//N being the number of processors, the area should contain the value of pi.
ratio_p = 1/p;
area = total_sum*ratio_p;

printf("\n\nThe area under the curve of f(x) = sqrt(1-x*x), between 0 and 1 is, %f\n\n", area);

/* Shut down MPI */
MPI_Finalize();
} /* main */

Answer 1

一般来说，依靠STDIN / STDOUT来实现MPI程序并不好。 MPI实现可能会将0级放在除您启动作业的节点之外的某个节点上。在这种情况下，您必须担心正确转发。虽然这在大多数情况下都有效，但通常不是一个好主意。

更好的方法是将用户输入放在应用程序可以读取的文件中，或者通过命令行变量。这些将更便携。

我不确定MPI在没有循环的情况下自行循环是什么意思。如果您仍然需要答案，也许您可以澄清该评论。

需要说明MPI_Scatter（）

1 个答案: