Question

我希望使用MPI环拓扑，将每个处理器的最大值传递到环周围，比较局部最大值，然后输出所有处理器的全局最大值。我正在使用10维蒙特卡罗积分函数。我的第一个想法是使用每个处理器的局部最大值创建一个数组，然后传递该值，比较并输出最高值。但我无法优雅地编写一个数组，只需要每个处理器的最大值并存储它对应于处理器的等级，这样我也可以跟踪哪个处理器获得了全局最大值。

我还没有完成我的代码，现在我有兴趣看看是否可以创建一个具有处理器的局部最大值的数组。我编码的方式，这是非常耗时的，如果有很多处理器，那么我每次都必须声明它们，但我无法生成我正在寻找的数组。我在这里分享代码：

#include <iostream>
#include <fstream>
#include <iomanip>
#include <cmath>
#include <cstdlib>
#include <ctime>
#include <mpi.h>
using namespace std;


//define multivariate function F(x1, x2, ...xk)            

double f(double x[], int n)
{
    double y;
    int j;
    y = 0.0;

    for (j = 0; j < n-1; j = j+1)
      {
         y = y + exp(-pow((1-x[j]),2)-100*(pow((x[j+1] - pow(x[j],2)),2)));

      }     

    y = y;
    return y;
}

//define function for Monte Carlo Multidimensional integration

double int_mcnd(double(*fn)(double[],int),double a[], double b[], int n, int m)

{
    double r, x[n], v;
    int i, j;
    r = 0.0;
    v = 1.0;
    // initial seed value (use system time) 
    //srand(time(NULL)); 


    // step 1: calculate the common factor V
    for (j = 0; j < n; j = j+1)
      {
         v = v*(b[j]-a[j]);
      } 

    // step 2: integration
    for (i = 1; i <= m; i=i+1)
    {
        // calculate random x[] points
        for (j = 0; j < n; j = j+1)
        {
            x[j] = a[j] +  (rand()) /( (RAND_MAX/(b[j]-a[j])));
        }         
        r = r + fn(x,n);
    }
    r = r*v/m;

    return r;
}




double f(double[], int);
double int_mcnd(double(*)(double[],int), double[], double[], int, int); 



int main(int argc, char **argv)
{    

    int rank, size;

    MPI_Init (&argc, &argv);      // initializes MPI
    MPI_Comm_rank (MPI_COMM_WORLD, &rank); // get current MPI-process ID. O, 1, ...
    MPI_Comm_size (MPI_COMM_WORLD, &size); // get the total number of processes


    /* define how many integrals */
    const int n = 10;       

    double b[n] = {5.0, 5.0, 5.0, 5.0, 5.0, 5.0, 5.0, 5.0, 5.0,5.0};                    
    double a[n] = {-5.0, -5.0, -5.0, -5.0, -5.0, -5.0, -5.0, -5.0, -5.0,-5.0};  

    double result, mean;
    int m;

    const unsigned int N = 5;
    double max = -1;
    double max_store[4];

    cout.precision(6);
    cout.setf(ios::fixed | ios::showpoint); 


    srand(time(NULL) * rank);  // each MPI process gets a unique seed

    m = 4;                // initial number of intervals

    // convert command-line input to N = number of points
    //N = atoi( argv[1] );


    for (unsigned int  i=0; i <=N; i++)
    {
        result = int_mcnd(f, a, b, n, m);
        mean = result/(pow(10,10));

        if( mean > max) 
        {
         max = mean;
        }

        //cout << setw(10)  << m << setw(10) << max << setw(10) << mean << setw(10) << rank << setw(10) << size <<endl;
        m = m*4; 
    }

    //cout << setw(30)  << m << setw(30) << result << setw(30) << mean <<endl; 
    printf("Process %d of %d mean = %1.5e\n and local max = %1.5e\n", rank, size, mean, max );
    if (rank==0)
        {
         max_store[0] = max;
        }
        else if (rank==1)
        {
         max_store[1] = max;
        }
        else if (rank ==2)
        {
         max_store[2] = max;
        }
        else if (rank ==3)
        {
         max_store[3] = max;
        }
    for( int k = 0; k < 4; k++ )
    {
     printf( "%1.5e\n", max_store[k]);
    }


    //double max_store[4] = {4.43095e-02, 5.76586e-02, 3.15962e-02, 4.23079e-02}; 

    double send_junk = max_store[0];
    double rec_junk;
    MPI_Status status;


  // This next if-statment implemeents the ring topology
  // the last process ID is size-1, so the ring topology is: 0->1, 1->2, ... size-1->0
  // rank 0 starts the chain of events by passing to rank 1
  if(rank==0) {
    // only the process with rank ID = 0 will be in this block of code.
    MPI_Send(&send_junk, 1, MPI_DOUBLE, 1, 0, MPI_COMM_WORLD); //  send data to process 1
    MPI_Recv(&rec_junk, 1, MPI_DOUBLE, size-1, 0, MPI_COMM_WORLD, &status); // receive data from process size-1
  }
  else if( rank == size-1) { 
    MPI_Recv(&rec_junk, 1, MPI_DOUBLE, rank-1, 0, MPI_COMM_WORLD, &status); // recieve data from process rank-1 (it "left" neighbor")
    MPI_Send(&send_junk, 1, MPI_DOUBLE, 0, 0, MPI_COMM_WORLD); // send data to its "right neighbor", rank 0
  }
  else {
    MPI_Recv(&rec_junk, 1, MPI_DOUBLE, rank-1, 0, MPI_COMM_WORLD, &status); // recieve data from process rank-1 (it "left" neighbor")
    MPI_Send(&send_junk, 1, MPI_DOUBLE, rank+1, 0, MPI_COMM_WORLD); // send data to its "right neighbor" (rank+1)
  }
  printf("Process %d send %1.5e\n and recieved %1.5e\n", rank, send_junk, rec_junk ); 


  MPI_Finalize(); // programs should always perform a "graceful" shutdown
    return 0;
}

编译：

mpiCC -o gd test_code.cpp
 mpirun -np 4 ./gd

我很感激建议：

如果有更优雅的方式制作局部最大值数组？
如何比较局部最大值并在环中传递值时确定全局最大值？

也可以随意修改代码，为我提供更好的示例。我将不胜感激任何建议。感谢。

Answer 1

对于这类事情，最好使用MPI_Reduce()或MPI_Allreduce() MPI_MAX作为运算符。前者将计算所有进程暴露的值的最大值，并将结果提供给＆＃34; root＆＃34;仅处理，而后者将执行相同操作，但将结果提供给所有流程。

// Only process of rank 0 get the global max
MPI_Reduce( &local_max, &global_max, 1, MPI_DOUBLE, MPI_MAX, 0, MPI_COMM_WORLD );
// All processes get the global max
MPI_Allreduce( &local_max, &global_max, 1, MPI_DOUBLE, MPI_MAX, MPI_COMM_WORLD );
// All processes get the global max, stored in place of the local max
// after the call ends - this might be the most interesting one for you
MPI_Allreduce( MPI_IN_PLACE, &max, 1, MPI_DOUBLE, MPI_MAX, MPI_COMM_WORLD );

如您所见，您可以在代码中插入第3个示例来解决问题。

BTW，无关的评论，但这会伤到我的眼睛：

if (rank==0)
    {
     max_store[0] = max;
    }
    else if (rank==1)
    {
     max_store[1] = max;
    }
    else if (rank ==2)
    {
     max_store[2] = max;
    }
    else if (rank ==3)
    {
     max_store[3] = max;
    }

这样的事情：

if ( rank < 4 && rank >= 0 ) {
    max_store[rank] = max;
}

使用MPI环拓扑

1 个答案: