c - 埃拉托斯特尼筛法 BSP C 实现未找到所有素数

Question

我有一个用于埃拉斯托尼筛法的 C 语言 BSP 实现，请参阅下面的代码。

当使用 ./bspsieve 2 100 执行时，它会给出以下输出:

“过程 0 花费了 0.000045 秒(共 2 秒)。 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97,"

对于 ./bspsieve 1 100 它给出相同的结果，即: “./bspsieve 1 100
1 中的 proc 0 花费了 0.000022 秒。 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97,"

对于 ./bspsieve 8 100(因此使用 8 个处理器)，它会出现段错误。 IE “./bspsieve 8 100 过程 0(共 8 个)花费了 0.000146 秒。 段错误(核心转储)" 这意味着我认为我的界限不太好？

找不到第一个素数!我找不到我的错(对C真的没有经验)。除此之外，你们还可以建议对我的代码进行其他改进吗？该算法不需要很快，但在可理解性和可读性方面的任何改进都是值得欢迎的。

#include <stdio.h>
#include <stdlib.h>
#include <math.h>
#include <mcbsp.h>


/*
Note: To compile, this file has to be in the same folder as mcbsp.h and you need the 2 following commands:
gcc -Iinclude/ -pthread -c -o bspsieve.o bspsieve.c
gcc -o bspsieve bspsieve.o lib/libmcbsp1.1.0.a -lpthread -lrt

*/

int procs;
int upperbound;
int *primes;

//SPMD function
void bspSieve(){
    bsp_begin(procs);

    int p = bsp_nprocs(); // p = number of procs obtained 
    int s = bsp_pid();  // s = proc number

    float blocksize;    // block size to be used, note last proc has a different size!
    if( s != p-1){
        blocksize = ceil(upperbound/p); 
    } else {
        blocksize = upperbound - (p-1)*ceil(upperbound/p);
    }

    // Initialize start time and end time, set start time to now.   
    double start_time,end_time;
    start_time = bsp_time();

    // Create vector that has block of candidates
    int *blockvector;
    blockvector = (int *)malloc(blocksize*sizeof(int));
    int q;
    for(q = 0; q<blocksize; q++){
        //List contains the integers from s*blocksize till blocksize + s*blocksize
        blockvector[q] = q + s*blocksize;
    } 

    //We neglect the first 2 'primes' in processor 0.
    if(s == 0){
        blockvector[0] = 0;
        blockvector[1] = 0;
    }

    // We are using the block distribution. We assume that n is large enough  to
    // assure that n/p is larger than sqrt(n). This means that we will always find the
    // sieving prime in the first block, and so have to broadcast from the first 
    // processor to the others.
    long sieving_prime;
    int i;
    bsp_push_reg( &sieving_prime,sizeof(long) );
    bsp_sync();

    for(i = 2; i * i < upperbound; i++) {
        //Part 1: if first processor, get the newest sieving prime, broadcast. Search for newest prime starting from i.
        if(s == 0){
            int findPrimeNb;
            for(findPrimeNb = i; findPrimeNb < blocksize; findPrimeNb++) {
                if( blockvector[findPrimeNb] != 0) {
                    sieving_prime  = blockvector[findPrimeNb];
                    //broadcast
                    int procNb;
                    for(procNb = 0; procNb < p; ++procNb){
                        bsp_put(procNb, &sieving_prime,&sieving_prime,0,sizeof(long));
                    }
                    break;
                }
            }
        }
        bsp_sync();

        //Part 2: Sieve using the sieving prime
        int sievingNb;
        for(sievingNb = 0; sievingNb < blocksize; sievingNb++){
            //check if element is multiple of sieving prime, if so, pcross out (put to zero)
            if( blockvector[sievingNb] % sieving_prime == 0){
                blockvector[sievingNb] = 0;
            }
        }

    }

    //part 3: get local primes to central area
    int transferNb;
    long transferPrime;
    for(transferNb = 0; transferNb < blocksize; transferNb++){
        transferPrime = blockvector[transferNb];
        primes[transferPrime] = transferPrime;
    }

    // take the end time.
    end_time = bsp_time();

    //Print amount of taken time, only processor 0 has to do this.
    if( s == 0 ){
        printf("It took : %.6lf seconds for proc %d out of %d. \n", end_time-start_time, bsp_pid(), bsp_nprocs());
        fflush(stdout);
    }
    bsp_pop_reg(&sieving_prime);
    bsp_end();
}



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

    if(argc != 3) {
        printf( "Usage: %s <proc count> <upper bound> <n", argv[ 0 ] );
        exit(1);
    }
    //retrieve parameters
    procs = atoi( argv[ 1 ] );
    upperbound = atoi( argv[ 2 ] );

    primes = (int *)malloc(upperbound*sizeof(int));

    // init and call parallel part
    bsp_init(bspSieve, argc, argv); 
    bspSieve();

    //Print all non zeros of candidates, these are the primes.
    // Primes only go to p*p <= n
    int i;
    for(i = 0; i < upperbound; i++) {
        if(primes[i] > 0) {
            printf("%d, ",primes[i]); 
            fflush(stdout);
        }
    }
    return 0;
}

Answer 1

问题可能来自

blockvector[q] = q + s*blocksize;

只要所有进程上的blocksize等于ceil(upperbound/p)，就没有问题。由于 1 和 2 是 100 的约数，因此您的程序运行良好。

正如您在代码中所写，情况并非总是如此......调用 ./bspsieve 8 100 时，最后一个进程的情况并非如此。 blockvector中有些值超过100，写入prime数组时很可能出现段错误。

纠正此行为的方法是:

   blockvector[q] = q + s*ceil(upperbound/p);

(存储ceil(...)以运行得更快。)

在使用 prime 数组之前将其清零可能会更好。 我没有检查它是否有效...尝试一下!

再见，

弗朗西斯