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我正在尝试使用 MPI制作并行版本的“ Harmonic Progression Sum ”问题。但我是 MPI 的新手，我不知道如何使用 MPI 运行此方法，因为它不起作用。

并行程序：

//#include "stdafx.h"
#include <stdio.h>
#include <time.h>
#include <stdlib.h>
#include <iostream>
#include <sstream>
#include <mpi.h>

#define d 10    //Numbers of Digits (Example: 5 => 0,xxxxx)
#define n 1000  //Value of N (Example: 5 => 1/1 + 1/2 + 1/3 + 1/4 + 1/5)

using namespace std;

int numProcess, rank, msg, source, dest, tag, qtd_elemento;

int escravo(long unsigned int *digits, int ValueEnd)
{
    MPI_Status status;

    MPI_Recv(digits, (d + 11), MPI_INT, MPI_ANY_SOURCE, MPI_ANY_TAG, MPI_COMM_WORLD, &status);

    for (int i = 1; i <= ValueEnd; ++i) {
        long unsigned int remainder = 1;
        for (long unsigned int digit = 0; digit < d + 11 && remainder; ++digit) {
            long unsigned int div = remainder / i;
            long unsigned int mod = remainder % i;
            digits[digit] += div;
            remainder = mod * 10;
        }
    }

    MPI_Send(&digits, 1, MPI_INT, 0, 1, MPI_COMM_WORLD);
}

void HPSSeguencial(char* output) {
    long unsigned int digits[d + 11];
    int DivN = n / 4; //Limiting slave.

    for (int digit = 0; digit < d + 11; ++digit)
        digits[digit] = 0;

    if (rank != 0){
        escravo(digits, (DivN * 1 ) );
        escravo(digits, (DivN * 2 ) );
        escravo(digits, (DivN * 3 ) );
        escravo(digits, (DivN * 4 ) );
    }

    for (int i = d + 11 - 1; i > 0; --i) {
        digits[i - 1] += digits[i] / 10;
        digits[i] %= 10;
    }
    if (digits[d + 1] >= 5) {
        ++digits[d];
    }


    for (int i = d; i > 0; --i) {
        digits[i - 1] += digits[i] / 10;
        digits[i] %= 10;
    }
    stringstream stringstreamA;
    stringstreamA << digits[0] << ",";


    for (int i = 1; i <= d; ++i) {
        stringstreamA << digits[i];
    }
    string stringA = stringstreamA.str();
    stringA.copy(output, stringA.size());
}

int main() {
    MPI_Init(&argc,&argv);
    MPI_Comm_rank(MPI_COMM_WORLD, &rank);
    MPI_Comm_size(MPI_COMM_WORLD, &numProcess);

    char output[d + 10];
    HPSSeguencial(output);
    cout << output << endl;

    MPI_Finalize();

    system("PAUSE");
    return 0;
}

原始代码

#include "stdafx.h"
#include <iostream>
#include <sstream>
#include <time.h>

#define d 10    //Numbers of Digits (Example: 5 => 0,xxxxx)
#define n 1000  //Value of N (Example: 5 => 1/1 + 1/2 + 1/3 + 1/4 + 1/5)

using namespace std;

void HPS(char* output) {
    long unsigned int digits[d + 11];

    for (int digit = 0; digit < d + 11; ++digit)
        digits[digit] = 0;

    for (int i = 1; i <= n; ++i) {
        long unsigned int remainder = 1;
        for (long unsigned int digit = 0; digit < d + 11 && remainder; ++digit) {
            long unsigned int div = remainder / i;
            long unsigned int mod = remainder % i;
            digits[digit] += div;
            remainder = mod * 10;
        }
    }


    for (int i = d + 11 - 1; i > 0; --i) {
        digits[i - 1] += digits[i] / 10;
        digits[i] %= 10;
    }
    if (digits[d + 1] >= 5) {
        ++digits[d];
    }


    for (int i = d; i > 0; --i) {
        digits[i - 1] += digits[i] / 10;
        digits[i] %= 10;
    }
    stringstream stringstreamA;
    stringstreamA << digits[0] << ",";


    for (int i = 1; i <= d; ++i) {
        stringstreamA << digits[i];
    }
    string stringA = stringstreamA.str();
    stringA.copy(output, stringA.size());
}


int main() {

    char output[d + 10];
    HPS(output);
    cout << output<< endl;

    system("PAUSE");
    return 0;
}

例子：

输入：

#define d 10
#define n 1000

输出：

7,4854708606╠╠╠╠╠╠╠╠╠╠╠╠

输入：

#define d 12
#define n 7

输出：

2,592857142857╠╠╠╠╠╠╠╠╠╠╠╠╠╠ÀÂ♂ü─¨@

问候

原始代码

http://regulus.pcs.usp.br/marathon/current/warmup.pdf

4

1 回答 1

4

我假设你想并行化这部分：

for (int i = 1; i <= ValueEnd; ++i) 
{
        long unsigned int remainder = 1;
        for (long unsigned int digit = 0; digit < d + 11 && remainder; ++digit)
        {
            long unsigned int div = remainder / i;
            long unsigned int mod = remainder % i;
            digits[digit] += div;
            remainder = mod * 10;
        }
}

您可以按每个 MPI 进程划分每个迭代：

int idP = getProcessId(), numP = numberProcess();
for (int i = idP; i <= ValueEnd; i+=numP)
{
  ...
}

) 为您提供getProcessId(进程 ID 并numberProcess()为您提供进程号：

int getProcessId(){
    int rank;
    MPI_Comm_rank(MPI_COMM_WORLD, &rank);
    return rank;
}
// Get number of process
int numberProcess(){
    int numProc;
    MPI_Comm_size(MPI_COMM_WORLD, &numProc);
    return numProc;
}

每个进程都会有一个数组数字的副本；在并行处理之后，主进程使用MPI_reduce从所有从属进程收集结果。或者，如果您想组合来自所有进程的值并将结果分发回所有进程，您可以使用MPI_Allreduce。

 long unsigned int digits[d + 11];
    int DivN = n / 4; //Limiting slave.

    for (int digit = 0; digit < d + 11; ++digit)
        digits[digit] = 0;

    if (rank != 0){
        escravo(digits, (DivN * 1 ) );
        escravo(digits, (DivN * 2 ) );
        escravo(digits, (DivN * 3 ) );
        escravo(digits, (DivN * 4 ) );
    }

根据上面的代码，进程0不会执行该方法escravo。此外，您没有在流程之间正确分配工作。进程 1 将在方法内escravo从 1 到 n/4 执行 out for 循环，但随后进程 2 将从 1 到 2n/4 执行......因此，当你真正想要的是不同的进程执行相同的迭代时在进程之间划分这些迭代。

于 2012-11-22T00:37:29.073 回答