我希望使用MPI环形拓 flutter ，将每个处理器的最大值传递给环，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维蒙特卡洛积分函数。我的第一个想法是用每个处理器的局部最大值组成一个数组，然后传递该值，比较并输出最大值。但是我无法优雅地编写代码以仅将每个处理器的最大值存储在对应于处理器等级的数组中，这样我就可以跟踪哪个处理器获得了全局最大值。
我还没有完成代码，现在我很想看看是否可以创建一个具有处理器局部极大值的数组。按照我的编码方式，这非常耗时，并且如果有很多处理器，那么每次都必须声明它们，但是我无法生成我要寻找的数组。
我在这里共享代码:
#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
我将不胜感激的建议:
也可以随意修改代码，以便为我提供一个更好的示例。我将不胜感激任何建议。谢谢。
最佳答案
对于这种情况，最好将MPI_Reduce()或MPI_Allreduce()与MPI_MAX用作运算符。前者将计算所有进程公开的值的最大值，并将结果仅提供给“根”进程，而后者将执行相同的操作，但将结果提供给所有进程。
// 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，您只需在代码中插入第三个示例即可解决您的问题。
顺便说一句，无关的话，但这伤了我的眼睛:
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;
}
那这样的事情呢:
if ( rank < 4 && rank >= 0 ) {
max_store[rank] = max;
}