Question

我正在尝试编写一个并行程序，该程序的错误率（即0.01）并返回的PI值比montecarlo模拟的错误更接近PI。我写了一个简单的函数，但是它不会终止，因为错误率始终在11左右。感谢您的评论。

#include "stdio.h"
#include "omp.h"
#include <stdlib.h>
#include <unistd.h>
#include <math.h>

double drand48(void);

double monte_carlo(double epsilon){
    double x,y, pi_estimate = 0.0;
    double drand48(void);
    double error = 10000.0;
    int n = 0; // total number of points
    int i = 0; // total numbers of points inside circle
    int p = omp_get_num_threads();
    while(error>=epsilon){
        #pragma omp parallel private(x, y) reduction(+:i)//OMP parallel directive
        {
            x = drand48();
            y = drand48();
            if((x*x+y*y)<=1.0){i+=1;}
        }
        n+=p;
        printf("%lf\n", error);
        pi_estimate=4.0*(double)i/(double)n;
        error = fabs(M_PI-pi_estimate)/M_PI;
    }
    return pi_estimate;
}

int main(int argc, char* argv[]) {
    double epsilon = 0.01;
    printf("PI estimate: %lf",monte_carlo(epsilon));
    return 0;
}

Answer 1

在并行段之外调用omp_get_num_threads()将始终返回1，因为在调用该函数时只有一个活动线程。下面的代码应该给出正确的结果，但是由于执行非常简单的操作会花费大量的并行化和同步开销，因此它将比串行版本慢得多。

#pragma omp parallel private(x, y) reduction(+:i)//OMP parallel directive
{
    x = drand48();
    y = drand48();
    if((x*x+y*y)<=1.0){i+=1;}
    #pragma omp master
    n+=omp_get_num_threads();
}

以下内容避免了重复生成线程，并且可能会更有效，但可能仍然更慢。

#pragma omp parallel private(x, y)
while(error>=epsilon){
        x = drand48();
        y = drand48();
        if((x*x+y*y)<=1.0){
            #pragma omp atomic
            i++;
        }
    #pragma omp barrier
    #pragma omp single
    {
        n+=omp_get_num_threads();
        pi_estimate=4.0*(double)i/(double)n;
        error = fabs(M_PI-pi_estimate)/M_PI;
        printf("%lf\n", error);
    } // implicit barrier here
}

为了真正更快，应该给出最小的迭代次数，例如：

#define ITER 1000
#pragma omp parallel private(x, y)
while(error>=epsilon){
    #pragma omp for reduction(+:i)
    for (int j=1;j<ITER;j++){
        x = drand48();
        y = drand48();
        if((x*x+y*y)<=1.0) i+=1;
    }
    /* implicit barrier + implicit atomic addition
     * of thread-private accumulator to shared variable i
     */

    #pragma omp single
    {
        n+=ITER;
        pi_estimate=4.0*(double)i/(double)n;
        error = fabs(M_PI-pi_estimate)/M_PI;
        printf("%lf\n", error);
    } // implicit barrier
}

OpenMP Monte_Carlo仿真实现与PI的目标接近度

1 个答案: