为什么我的并行代码比串行慢？

Question

Issue

Hello everyone, I have got a program (from the net) that I intend to speed up by converting it into its parallel version with the use of pthreads . But surprisingly though, it runs slower than the serial version. Below is the program:

# include <stdio.h>

//fast square root algorithm
double asmSqrt(double x) 
{
  __asm__ ("fsqrt" : "+t" (x));
  return x;
}

//test if a number is prime
bool isPrime(int n)
{   
    if (n <= 1) return false;
    if (n == 2) return true;
    if (n%2 == 0) return false;

    int sqrtn,i;
    sqrtn = asmSqrt(n);

    for (i = 3; i <= sqrtn; i+=2) if (n%i == 0) return false;
    return true;
}

//number generator iterated from 0 to n
int main()
{
    n = 1000000; //maximum number
    int k,j;

    for (j = 0; j<= n; j++)
    {
        if(isPrime(j) == 1) k++;
        if(j == n) printf("Count: %d\n",k);
    }
    return 0;
}

First attempt for parallelization

I let the pthread manage the for loop

# include <stdio.h>
.
.

int main()
{
    .
    .
    //----->pthread code here<----
    for (j = 0; j<= n; j++)
    {
        if(isPrime(j) == 1) k++;
        if(j == n) printf("Count: %d\n",k);
    }
    return 0;
}

Well, it runs slower than the serial one

Second attempt

I divided the for loop into two threads and run them in parallel using pthreads

However, it still runs slower, I am intending that it may run about twice as fast or well faster. But its not!

These is my parallel code by the way:

# include <stdio.h>
# include <pthread.h>
# include <cmath>

# define NTHREADS 2

pthread_mutex_t mutex1 = PTHREAD_MUTEX_INITIALIZER;
int k = 0;

double asmSqrt(double x) 
{
  __asm__ ("fsqrt" : "+t" (x));
  return x;
}

struct arg_struct
{
    int initialPrime;
    int nextPrime;
};

bool isPrime(int n)
{   
    if (n <= 1) return false;

    if (n == 2) return true;

    if (n%2 == 0) return false;

    int sqrtn,i;
    sqrtn = asmSqrt(n);

    for (i = 3; i <= sqrtn; i+=2) if (n%i == 0) return false;

    return true;
}

void *parallel_launcher(void *arguments)
{
    struct arg_struct *args = (struct arg_struct *)arguments;

    int j = args -> initialPrime;
    int n = args -> nextPrime - 1;

    for (j = 0; j<= n; j++)
    {
        if(isPrime(j) == 1)
        {
            printf("This is prime: %d\n",j);
pthread_mutex_lock( &mutex1 );
            k++;
pthread_mutex_unlock( &mutex1 );
        }

        if(j == n) printf("Count: %d\n",k);
    }
pthread_exit(NULL);
}

int main()
{
    int f = 100000000;
    int m;

    pthread_t thread_id[NTHREADS];
    struct arg_struct args;

    int rem = (f+1)%NTHREADS;
    int n = floor((f+1)/NTHREADS);

    for(int h = 0; h < NTHREADS; h++)
    {
        if(rem > 0)
        {
            m = n + 1;
            rem-= 1;
        }
        else if(rem == 0)
        {
            m = n;
        }

        args.initialPrime = args.nextPrime;
        args.nextPrime = args.initialPrime + m;

        pthread_create(&thread_id[h], NULL, &parallel_launcher, (void *)&args);
        pthread_join(thread_id[h], NULL);
    }
   // printf("Count: %d\n",k);
    return 0;
}

Note: OS: Fedora 21 x86_64, Compiler: gcc-4.4, Processor: Intel Core i5 (2 physical core, 4 logical), Mem: 6 Gb, HDD: 340 Gb,

Answer 1

You need to split the range you are examining for primes up into n parts, where n is the number of threads.

The code that each thread runs becomes:

typedef struct start_end {
    int start;
    int end;
} start_end_t;

int find_primes_in_range(void *in) {
    start_end_t *start_end = (start_end_t *) in;

    int num_primes = 0;
    for (int j = start_end->start; j <= start_end->end; j++) {
       if (isPrime(j) == 1)
           num_primes++;
    }
    pthread_exit((void *) num_primes;
}

The main routine first starts all the threads which call find_primes_in_range , then calls pthread_join for each thread. It sums all the values returned by find_primes_in_range . This avoids locking and unlocking a shared count variable.

This will parallelize the work, but the amount of work per thread will not be equal. This can be addressed but is more complicated.

Answer 2

The main design flaw: you must let each thread have its own private counter variable instead of using the shared one. Otherwise they will spend far more time waiting on and handling that mutex, than they will do on the actual calculation. You are essentially forcing the threads to execute in serial.

Instead, sum everything up with a private counter variable and once a thread is done with its work, return the counter variable and sum them up in main().

Also, you should not call printf() from inside the threads. If there is a context switch in the middle of a printf call, you'll end up with crappy output such as This is This is prime: 2 . In which case you must synchronize the printf calls between threads, which will slow the program down again. Also, the printf() calls themselves are likely 90% of the work that the thread is doing. So some sort of re-design of who does the printing might be a good idea, depending on what you want to do with the results.

Answer 3

Summary

Indeed, the use of PThread speed up my code. It was my programming flaw of placing pthread_join right after the first pthread_create and the common counter I have set on arguments. After fixing this up, I tested my parallel code to determine the primality of 100 Million numbers then compared its processing time with a serial code. Below are the results.

http://i.stack.imgur.com/gXFyk.jpg (I could not attach the image as I don't have much reputation yet, instead, I am including a link)

I conducted three trials for each to account for the variations caused by different OS activities. We got speed up for utilizing parallel programming with PThread . What is surprising is a PThread code running in ONE thread was a bit faster than purely serial code. I could not explain this one, nevertheless using PThreads is well, surely worth a try.

Here is the corrected parallel version of the code (gcc-c++):

# include <stdio.h>
# include <pthread.h>
# include <cmath>

# define NTHREADS 4

double asmSqrt(double x) 
{
  __asm__ ("fsqrt" : "+t" (x));
  return x;
}

struct start_end_f
{
    int start;
    int end;
};

//test if a number is prime
bool isPrime(int n)
{
    if (n <= 1) return false;
    if (n == 2) return true;
    if (n%2 == 0) return false;

    int sqrtn = asmSqrt(n);
    for (int i = 3; i <= sqrtn; i+=2) if (n%i == 0) return false;

    return true;
}

//executes the tests for prime in a certain range, other threads will test the next range and so on..
void *find_primes_in_range(void *in) 
{
    int k = 0;

    struct start_end_f *start_end_h = (struct start_end_f *)in;

    for (int j = start_end_h->start; j < (start_end_h->end +1); j++) 
    {
        if(isPrime(j) == 1) k++;
    }

    int *t = new int;
    *t = k;
    pthread_exit(t);
}

int main() 
{
    int f = 100000000; //maximum number to be tested for prime

    pthread_t thread_id[NTHREADS];
    struct start_end_f start_end[NTHREADS];

    int rem = (f+1)%NTHREADS;
    int n = (f+1)/NTHREADS;
    int rem_change = rem;
    int m;

    if(rem>0) m = n+1;
    else if(rem == 0) m = n;

    //distributes task 'evenly' to the number of parallel threads requested
    for(int h = 0; h < NTHREADS; h++)
    {
        if(rem_change > 0)
        {
            start_end[h].start = m*h;
            start_end[h].end = start_end[h].start+m-1;
            rem_change -= 1;
        }
        else if(rem_change<= 0)
        {
            start_end[h].start = m*(h+rem_change)-rem_change*n;
            start_end[h].end = start_end[h].start+n-1;
            rem_change -= 1;
        }
        pthread_create(&thread_id[h], NULL, find_primes_in_range, &start_end[h]);
    }   

    //retreiving returned values
    int *t;
    int c = 0;
    for(int h = 0; h < NTHREADS; h++)
    {
        pthread_join(thread_id[h], (void **)&t);
        int b = *((int *)t);
        c += b;
        b = 0;
    }

    printf("\nNumber of Primes: %d\n",c);
    return 0;
}