OpenMP C 程序运行速度比顺序代码慢

我是 OpenMP 的新手，正在尝试并行化 Jarvis 的算法。然而事实证明，与顺序代码相比，并行程序花费的时间要长 2-3 倍。

难道问题本身就不能并行化吗？或者我并行化它的方式有问题。

这是我针对该问题的 openMP 程序，其中有 2 个部分是并行的：

#include <stdio.h>
#include <sys/time.h>
#include <omp.h>

typedef struct Point
{
int x, y;
} Point;

// To find orientation of ordered triplet (p, q, r).
// The function returns
// 0 for colinear points
// 1 as Clockwise
// 2 as Counterclockwise
int orientation(Point p, Point i, Point q)
{
int val = (i.y - p.y) * (q.x - i.x) -
          (i.x - p.x) * (q.y - i.y);
if (val == 0) return 0;  // colinear
return (val > 0)? 1: 2; // clock or counterclock wise
}

// Prints convex hull of a set of n points.
void convexHull(Point points[], int n)
{
// There must be at least 3 points
if (n < 3) return;

// Initialize array to store results
Point results[n];
int count = 0;

// Find the leftmost point
int l = 0,i;

#pragma omg parallel shared (n,l) private (i)
{
    #pragma omp for
    for (i = 1; i < n; i++)
    {
        #pragma omp critical
        {
            if (points[i].x < points[l].x)
            l = i;
        }
    }

}

// Start from leftmost point, keep moving counterclockwise
// until reach the start point again.
int p = l, q;
do
{
    // Add current point to result
    results[count]= points[p];
    count++;

    q = (p+1)%n;
    int k;

    #pragma omp parallel shared (p) private (k)
    {
        #pragma omp for 
        for (k = 0; k < n; k++)
        {
           // If i is more counterclockwise than current q, then
           // update i as new q
           #pragma omp critical
           {
            if (orientation(points[p], points[k], points[q]) == 2)
               q = k;
           }
        }       

    }

    // Now q is the most counterclockwise with respect to p
    // Set p as q for next iteration, to add q to result
    p = q;


} while (p != l);  // While algorithm does not return to first point

// Print Result
int j;
for (j = 0; j < count; j++){
  printf("(%d,%d)\n", results[j].x,results[j].y);
}

}

int main()
{
//declaration for start time, end time
//and total executions for the algorithm
struct timeval start, end;
int i, num_run = 100;

gettimeofday(&start,NULL);

Point points[] = {{0, 3}, {2, 2}, {1, 1}, {2, 1},
                    {3, 0}, {0, 0}, {3, 3}};

int n = sizeof(points)/sizeof(points[0]);

convexHull(points, n);

gettimeofday(&end,NULL);

int cpu_time_used = (((end.tv_sec - start.tv_sec) * 1000000) + (end.tv_usec 
- start.tv_usec));
printf("\n\nExecution time: %d ms\n", cpu_time_used);
return 0;
}

尝试通过添加以下代码行来使输入足够丰富：

Point points[3000];
int i;
for(i=0;i<3000;i++) {
    points[i].x = rand()%100;
    points[i].y = rand()%100;
    int j;
    for(j=i+1;j<3000;j++) {
        if(points[i].x==points[j].x) {
            if(points[i].y==points[j].y) {
            i--; 
            break;
            }
        }
    }
}

但有时会崩溃

在下面的代码片段中，并行的全部内容for循环被包装成critical陈述。这意味着这部分代码永远不会同时由多个线程输入。让多个线程一次工作一个并不会比单个线程完成所有迭代更快。但最重要的是，同步开销会损失一些时间（每个线程必须在进入临界区之前获取互斥体并在之后释放它）。

int l = 0,i;
#pragma omp parallel shared (n,l) private (i)
{
    #pragma omp for
    for (i = 1; i < n; i++)
    {
        #pragma omp critical
        {
            if (points[i].x < points[l].x)
            l = i;
        }
    }
}

串行代码需要进行一些重构以实现并行化。对于简单操作来说，归约通常是一种好方法：让每个线程在迭代的一部分上计算部分结果（例如部分最小值、部分求和），然后将所有结果合并到全局结果中。对于支持的操作，#pragma omp for reduction(op:var)可以使用语法。但在这种情况下，减少必须手动完成。

看看下面的代码如何依赖局部变量来跟踪最小值的索引x，然后使用单个关键部分来计算全局最小索引。

int l = 0,i;
#pragma omp parallel shared (n,l) private (i)
{
    int l_local = 0; //This variable is private to the thread

    #pragma omp for nowait
    for (i = 1; i < n; i++)
    {
        // This part of the code can be executed in parallel
        // since all write operations are on thread-local variables
        if (points[i].x < points[l_local].x)
            l_local = i;
    }

    // The critical section is entered only once by each thread
    #pragma omp critical
    {
    if (points[l_local].x < points[l].x)
        l = l_local;
    }

    #pragma omp barrier
    // a barrier is needed in case some more code follow
    // otherwise there is an implicit barrier at the end of the parallel region
}

同样的原理也应该应用于第二个并行循环，它遇到了同样的问题，即实际上完全由critical陈述。