[英]finding global maxima of a function from comparing each processor's local maxima using MPI ring topology
我希望使用MPI環形拓撲,將每個處理器的最大值傳遞給環,比較局部最大值,然后輸出所有處理器的全局最大值。 我正在使用10維蒙特卡洛積分函數。 我的第一個想法是使用每個處理器的局部最大值創建一個數組,然后傳遞該值,進行比較並輸出最高值。 但是我無法優雅地編寫代碼以僅將每個處理器的最大值存儲在對應於處理器等級的數組中,這樣我就可以跟蹤哪個處理器獲得了全局最大值。
我還沒有完成我的代碼,現在我很想看看是否可以創建一個具有處理器最大值的數組。 按照我的編碼方式,這非常耗時,並且如果有很多處理器,那么每次都必須聲明它們,但是我無法生成我要尋找的數組。 我在這里共享代碼:
#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;
}
編譯:
mpiCC -o gd test_code.cpp
mpirun -np 4 ./gd
我將不勝感激的建議:
也可以隨意修改代碼,以便為我提供一個更好的示例。 我將不勝感激任何建議。 謝謝。
對於這種情況,最好將MPI_Reduce()
或MPI_Allreduce()
與MPI_MAX
用作運算符。 前者將計算所有進程公開的值的最大值,並將結果僅提供給“根”進程,而后者將執行相同的操作,但將結果提供給所有進程。
// 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 );
如您所見,您只需在代碼中插入第三個示例即可解決您的問題。
順便說一句,無關的話,但這傷了我的眼睛:
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;
}
像這樣的事情呢:
if ( rank < 4 && rank >= 0 ) {
max_store[rank] = max;
}
聲明:本站的技術帖子網頁,遵循CC BY-SA 4.0協議,如果您需要轉載,請注明本站網址或者原文地址。任何問題請咨詢:yoyou2525@163.com.