Question

使用c ++ openmp 3.1我实现了最大缩减，它存储了一个对象向量的整数变量（得分）的最大值。但我也想存储矢量索引来访问具有最大分数的（s）对象。我目前的unsuccesfull实现看起来像这样：

//s is a vector of sol objects which contain apart from other variables an  integer     score    variable s[].score            
int bestscore = 0;
int bestant = 0;                
#pragma omp parallel shared(bestant)
{//start parallel session
    #pragma omp for    nowait reduction(max : bestscore)
    for (int ant = 0; ant<maxsols; ++ant) // for all ants
    {
        //procedures on s[ant] object which update the  int s[ant].score
        if (s[ant].score > bestscore)
        {
            //find the object with the highest score
            bestscore = s[ant].score;
            bestant = ant;//i also want know which ant has the highest score
        }
    }
}

代码编译并运行。找到最大的bestscore，但是最好得到一个随机索引。与完成最快的线程相关联的蚂蚁以最佳方式存储。 bestscore以0开头，所以在大多数情况下，s [ant] .score会有更高的分数，而bestscore和bestant会更新。我认为我需要一个减少运算符，以便在更新bestscore＆＃34;时更好。＆＃34;

Answer 1

试试这个

int bestscore = 0;
int bestant = 0;
#pragma omp parallel
{
    int bestscore_private = 0;
    int bestant_private = 0;
    #pragma omp for nowait
    for (int ant = 0; ant<maxsols; ++ant) {         
        if (s[ant].score > bestscore_private) {
            bestscore_private = s[ant].score;
            bestant_private = ant;
        }
    }
    #pragma omp critical 
    {
        if(bestscore_private>bestscore) {
            bestscore = bestscore_private;
            bestant = besant_private;
        }
    }
}

Answer 2

两个观察结果：

我认为您只需将bestscore_private与bestscore的值进行比较即可，这样可以减少比较次数。

此外，至少在今天的操作中，您可以在内部中使用关键部分，如果有条件，则可以在bestscore_private和bestscore之间进行比较并行执行，并且不频繁（希望如此）对bestscore的更新将以严格的方式进行。

int bestscore = 0; int bestant = 0; #pragma omp parallel {

  int bestscore_private = 0;
  int bestant_private = 0;
  #pragma omp for nowait
  for (int ant = 0; ant<maxsols; ++ant) {         
      if (s[ant].score > bestscore_private) {
          bestscore_private = s[ant].score;
          bestant_private = ant;   
          if(bestscore_private>bestscore){   
            #pragma omp critical 
            {

            bestscore = bestscore_private;
            bestant = besant_private;
            }
          }
      }
  }

}

Answer 3

bestant 获得随机索引 i 的原因（如您所料）是因为 bestant 是共享的，并且不像 bestscore 那样从缩减子句中受益。 Z boson 提出的方案很好：critical 指令块被线程只执行一次，这样开销应该是有限的。

您当时使用的是 OpenMP 3.1 运行时。我想发帖解释这个限制自 OpenMP 4.0 以来已得到解决。您现在可以编写 user defined operator（请参阅 2.19.5.7 声明缩减指令）。

在您的情况下，解决方案可以是将两个值打包在一个结构中并定义两个这样的结构元素如何在 #pragma parallel for 循环结束时组合。

//s is a vector of sol objects which contain apart from other variables an  integer     score    variable s[].score

typedef struct {
  int score;
  int ant;
} best_t;

best_t best = { 0, 0 };

// we declare our user reduction operator :
// it is called get_max, return a a value of type best_t.
// omp_out and omp_in are the predefined names of two private elements
// to combine in the end to the final shared variable.

#pragma omp declare reduction(get_max : best_t :\
    omp_out = omp_out.score > omp_in.score ? omp_out : omp_in)\
    initializer (omp_priv=(omp_orig))

                
#pragma omp parallel 
{//start parallel session
    #pragma omp for    nowait reduction(get_max : best)
    for (int ant = 0; ant<maxsols; ++ant) // for all ants
    {
        //procedures on s[ant] object which update the  int s[ant].score
        if (s[ant].score > best.score)
        {
            //find the object with the highest score
            best.score = s[ant].score;
            best.ant = ant;
        }
    }
}

通过存储索引来减少omp max

3 个答案: