如何使用pthread并行化pi的蒙特卡洛估计? [英] How to parallelize monte carlo estimation of pi using pthreads?

问题描述

#include <math.h>
#include <stdio.h>
#include <stdlib.h>
#include <time.h>
#include <unistd.h>

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

    unsigned long long in = 1;
    unsigned long long total = 2;
    double tol , change, new, secs , old = 0.0;
    struct timeval start , end;
    int threads ; /* ignored */

    if ( argc < 2) {
        exit (-1);
    }

    threads = atoi ( argv[1]);
    tol = atof ( argv[2]);
    if (( threads < 1) || ( tol < 0.0)) {
        exit (-1);
    }
    tol = tol *tol;

    srand48(clock());
    gettimeofday (&start , NULL);
    do
    {
        double x, y;
        x = drand48();
        y = drand48();
        total ++;
        if (( x*x + y*y) <= 1.00)
            in ++;
        new = 4.0 * (double)in/( double)total ;
        change = fabs (new - old);
        old = new;
    }while (change > tol );
    gettimeofday (&end, NULL);
    secs = (( double)end.tv_sec - (double)start.tv_sec )
    + (( double)end.tv_usec - (double)start.tv_usec )/1000000.0;
    printf ( "Found estimate of pi of %.12f in %llu iterations , %.6f seconds.n n",
            new, total - 2, secs );
}

上面的代码是一个顺序程序，它接受一个参数来估计pi的接近程度.一旦这些旧值和新值的变化降到容限以下，就会退出.

The code above is a sequential program that takes an argument for the tolerance for how closely to estimate pi. Once the change in these old and new values fall below the tolerance it exits.

我必须在pthreads中并行化此程序.我并不是要有人为我做这件事，而是要想出一些指示和想法，以便我能够做到这一点. pthreads程序将以线程数和容差为参数，并输出估计值.我对并行程序非常陌生，并且不知道从哪里开始，所以我会提一些建议.谢谢.

I have to parallelize this program in pthreads. I am not trying to have someone do it for me but rather to get some pointers and ideas to think about so that I may be able to this. the pthreads program will take number of threads and tolerance as it's argument and output the estimation. I am very new to parallel programs and don't really know where to start so I will take any advice. Thanks.

推荐答案

#include <math.h>
#include <stdio.h>
#include <stdlib.h>
#include <time.h>
#include <unistd.h>
#include <pthread.h>

void* Function(void* i);

#define MAX_THREADS 200
unsigned long long total[MAX_THREADS] = {0};    //total points for threads
unsigned long long in[MAX_THREADS] = {0};       //points inside for threads
double tolerance, change, new, old = 0.0;
long thread_num;
pthread_mutex_t         mutex = PTHREAD_MUTEX_INITIALIZER;



int main(int argc, char **argv)
{
    long i;
    struct timeval start, end;
    double secs;
    unsigned long long master_total;
    pthread_t* threads;

    if (argc != 3){
        printf("\nMust pass 2 arguments:  (Tolerance) (# of Threads)");
        exit(-1);
    }

    thread_num = atoi ( argv[1]);
    tolerance = atof ( argv[2]);

    if (( thread_num < 1) || ( tolerance < 0.0) || (thread_num > 200)) {
        printf("\nIncorrect tolerance or threads.");
        exit (-1);
    }

    threads = malloc(thread_num*sizeof(pthread_t)); //allocating space for threads
    tolerance = tolerance * tolerance;
    change = 0.5;
    srand48(clock());
    gettimeofday (&start, NULL);
    for( i = 0; i < thread_num; i++ ){
        pthread_create(&threads[i], NULL, Function, (void*)i);
    }

    for( i = 0; i < thread_num; i++ ){
        pthread_join(threads[i], NULL);
    }
    gettimeofday (&end, NULL);

    master_total = 0;
    for( i = 0; i < thread_num; i++ ){
        master_total = master_total + total[i];
    }
    secs = (( double)end.tv_sec - (double)start.tv_sec )
    + (( double)end.tv_usec - (double)start.tv_usec )/1000000.0;
    printf ( "Estimation of pi is %.12f in %llu iterations , %.6f seconds.\n", new, master_total, secs );

}
//Each thread will calculate it's own points for in and total
//Per 1000 points it will calculate new and old values and compare to tolerance
//If desired accuracy is met the threads will return. 
void* Function(void* i){
    /*
     rc - return code
     total[i], in[i] - threads own number of calculated points
     my_total, my_in - Each thread calculates global in and total, per 1000 points and calculates tolerance

     */
    long my_spot = (long) i;
    long rc;
    long j;
    unsigned long long my_total;
    unsigned long long my_in;
    do
    {
        double x, y;
        x = drand48();
        y = drand48();
        total[my_spot]++;
        if (( x*x + y*y) <= 1.00){
            in[my_spot]++;
        }
        if(total[my_spot] % 1000 == 0){
            while ( j < thread_num){
                my_total = my_total + total[j];
                my_in = my_in + in[j];
            }
            my_total = my_total;
        //critical section
        //old, new, and change are all global
        rc = pthread_mutex_lock(&mutex);
        new = 4.0 * (double)my_in/( double)my_total;
        change = fabs (new - old);
        old = new;
        rc = pthread_mutex_unlock(&mutex);
        }
    }while (change > tolerance );
    return NULL;
}

这就是我的想法，但是我遇到了错误.它只是停止.我只是让线程跳出一个循环并返回，然后主线程加入它们.关于我在这里做什么的任何建议?

This is what I whipped up but I am getting errors. It just stops. I just have the threads break out of a loop and return then the main thread joins them. Any advice on what I am doing here?

我运行它，似乎所有线程在达到互斥锁时都被锁定了.我让每个线程每1000点检查一次pi的变化.

I run it and it seems that all the threads get locked out when they reach mutex lock. I have each thread check the change in pi every 1000 points.

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