问题
#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 );
}
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.
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.
回答1:
Each thread should keep its own in and total counts and use a better exit condition. Then add up the in and total values from each thread.
回答2:
#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;
}
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?
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.
来源:https://stackoverflow.com/questions/15376908/how-to-parallelize-monte-carlo-estimation-of-pi-using-pthreads