如何使用MPI计算数组的总和？代码实现

先决条件： MPI –简化分布式计算

消息传递接口(MPI)

是例程的库, 可用于在C或Fortran77中创建并行程序。它允许用户通过创建并行进程来构建并行应用程序, 并在这些进程之间交换信息。

MPI使用两个基本的通信例程：

• MPI_Send, 将消息发送到另一个进程。
• MPI_Recv, 以接收来自另一个进程的消息。

MPI_Send和MPI_Recv的语法为：

int MPI_Send(void *data_to_send, int send_count, MPI_Datatype send_type, int destination_ID, int tag, MPI_Comm comm); 

int MPI_Recv(void *received_data, int receive_count, MPI_Datatype receive_type, int sender_ID, int tag, MPI_Comm comm, MPI_Status *status);

为了降低程序的时间复杂度, 子数组的并行执行是通过运行并行进程来计算其部分和, 然后, 主进程(根进程)计算这些部分和的总和, 以返回子数组的总和。数组。

例子：

Input : {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}
Output : Sum of array is 55

Input : {1, 3, 5, 10, 12, 20, 4, 50, 100, 1000}
Output : Sum of array is 1205

注意 -你必须在基于Linux的系统上安装MPI才能执行以下程序。有关详细信息, 请参阅MPI –简化分布式计算

使用以下代码编译并运行程序：

mpicc program_name.c -o object_file
mpirun -np [number of processes] ./object_file

以下是上述主题的实现：

#include <mpi.h>
#include <stdio.h>
#include <stdlib.h>
#include <unistd.h>
  
// size of array
#define n 10
  
int a[] = { 1, 2, 3, 4, 5, 6, 7, 8, 9, 10 };
  
// Temporary array for slave process
int a2[1000];
  
int main( int argc, char * argv[])
{
  
     int pid, np, elements_per_process, n_elements_recieved;
     // np -> no. of processes
     // pid -> process id
  
     MPI_Status status;
  
     // Creation of parallel processes
     MPI_Init(&argc, &argv);
  
     // find out process ID, // and how many processes were started
     MPI_Comm_rank(MPI_COMM_WORLD, &pid);
     MPI_Comm_size(MPI_COMM_WORLD, &np);
  
     // master process
     if (pid == 0) {
         int index, i;
         elements_per_process = n / np;
  
         // check if more than 1 processes are run
         if (np > 1) {
             // distributes the portion of array
             // to child processes to calculate
             // their partial sums
             for (i = 1; i < np - 1; i++) {
                 index = i * elements_per_process;
  
                 MPI_Send(&elements_per_process, 1, MPI_INT, i, 0, MPI_COMM_WORLD);
                 MPI_Send(&a[index], elements_per_process, MPI_INT, i, 0, MPI_COMM_WORLD);
             }
  
             // last process adds remaining elements
             index = i * elements_per_process;
             int elements_left = n - index;
  
             MPI_Send(&elements_left, 1, MPI_INT, i, 0, MPI_COMM_WORLD);
             MPI_Send(&a[index], elements_left, MPI_INT, i, 0, MPI_COMM_WORLD);
         }
  
         // master process add its own sub array
         int sum = 0;
         for (i = 0; i < elements_per_process; i++)
             sum += a[i];
  
         // collects partial sums from other processes
         int tmp;
         for (i = 1; i < np; i++) {
             MPI_Recv(&tmp, 1, MPI_INT, MPI_ANY_SOURCE, 0, MPI_COMM_WORLD, &status);
             int sender = status.MPI_SOURCE;
  
             sum += tmp;
         }
  
         // prints the final sum of array
         printf ( "Sum of array is : %d\n" , sum);
     }
     // slave processes
     else {
         MPI_Recv(&n_elements_recieved, 1, MPI_INT, 0, 0, MPI_COMM_WORLD, &status);
  
         // stores the received array segment
         // in local array a2
         MPI_Recv(&a2, n_elements_recieved, MPI_INT, 0, 0, MPI_COMM_WORLD, &status);
  
         // calculates its partial sum
         int partial_sum = 0;
         for ( int i = 0; i < n_elements_recieved; i++)
             partial_sum += a2[i];
  
         // sends the partial sum to the root process
         MPI_Send(&partial_sum, 1, MPI_INT, 0, 0, MPI_COMM_WORLD);
     }
  
     // cleans up all MPI state before exit of process
     MPI_Finalize();
  
     return 0;
}

输出如下：

Sum of array is 55

以下是计算部分和的过程快照：

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