Dijkstra的算法openmp奇怪行为

Dijkstra's algorithm openmp strange behavior

本文关键字:openmp 算法 Dijkstra      更新时间:2023-10-16

我正在尝试运行Dijkstra算法的openmp实现,我在这里下载了heather.cs.ucdavis.edu/~matlof/openmp/Dijkstra.c

例如,如果我再加一个从5到6的垂直线,使从第0个开始的路径穿过两个顶点,我的程序无法给出正确的结果,说第0个和第6个之间的距离是无限的:^(原因是什么?

#define LARGEINT 2<<30-1  // "infinity"
#define NV 6
// global variables, all shared by all threads by default
int ohd[NV][NV],  // 1-hop distances between vertices
mind[NV],  // min distances found so far
notdone[NV], // vertices not checked yet
nth,  // number of threads
chunk,  // number of vertices handled by each thread
md,  // current min over all threads
mv;  // vertex which achieves that min
void init(int ac, char **av)
{  int i,j;
for (i = 0; i < NV; i++)  
for (j = 0; j < NV; j++)  {
if (j == i) ohd[i][i] = 0;
else ohd[i][j] = LARGEINT;
}
ohd[0][1] = ohd[1][0] = 40;
ohd[0][2] = ohd[2][0] = 15;
ohd[1][2] = ohd[2][1] = 20;
ohd[1][3] = ohd[3][1] = 10;
ohd[1][4] = ohd[4][1] = 25;
ohd[2][3] = ohd[3][2] = 100;
ohd[1][5] = ohd[5][1] = 6;
ohd[4][5] = ohd[5][4] = 8;
for (i = 1; i < NV; i++)  {
notdone[i] = 1;
mind[i] = ohd[0][i];
}
}
// finds closest to 0 among notdone, among s through e
void findmymin(int s, int e, int *d, int *v)
{  int i;
*d = LARGEINT; 
for (i = s; i <= e; i++)
if (notdone[i] && mind[i] < *d)  {
*d = ohd[0][i];
*v = i;
}
}
// for each i in [s,e], ask whether a shorter path to i exists, through
// mv
void updateohd(int s, int e)
{  int i;
for (i = s; i <= e; i++)
if (mind[mv] + ohd[mv][i] < mind[i])  
mind[i] = mind[mv] + ohd[mv][i];
}
void dowork()
{  
#pragma omp parallel  // Note 1  
{  int startv,endv,  // start, end vertices for this thread
step,  // whole procedure goes NV steps
mymd,  // min value found by this thread
mymv,  // vertex which attains that value
me = omp_get_thread_num();  // my thread number
#pragma omp single   // Note 2
{  nth = omp_get_num_threads();  chunk = NV/nth;  
printf("there are %d threadsn",nth);  }
// Note 3
startv = me * chunk; 
endv = startv + chunk - 1;
for (step = 0; step < NV; step++)  {
// find closest vertex to 0 among notdone; each thread finds
// closest in its group, then we find overall closest
#pragma omp single 
{  md = LARGEINT; mv = 0;  }
findmymin(startv,endv,&mymd,&mymv);
// update overall min if mine is smaller
#pragma omp critical  // Note 4
{  if (mymd < md)  
{  md = mymd; mv = mymv;  }
}
// mark new vertex as done 
#pragma omp single 
{  notdone[mv] = 0;  }
// now update my section of ohd
updateohd(startv,endv);
#pragma omp barrier 
}
}
}
int main(int argc, char **argv)
{  int i;
init(argc,argv);
dowork();  
// back to single thread now
printf("minimum distances:n");
for (i = 1; i < NV; i++)
printf("%dn",mind[i]);
}

这里有两个问题:

如果线程的数量不能均匀地除以值的数量,那么这种工作划分

startv = me * chunk;
endv = startv + chunk - 1;

将保留最后的(NV - nth*(NV/nth))元素,这将意味着距离保留在LARGEINT。这可以通过多种方式进行修复;目前最简单的方法是将所有剩余的工作交给最后一个线程

if (me == (nth-1)) endv = NV-1;

(这会导致更多的负载不平衡,但这是让代码正常工作的合理开始。)

另一个问题是在设置notdone[]之前遗漏了一个屏障

#pragma omp barrier
#pragma omp single 
{  notdone[mv] = 0;  }

这确保了只有在每个人都完成了他们的findmymin()并更新了mdmv之后才更新notdone并且启动updateohd()

请注意,在您开始使用的原始代码中很容易引入错误;所使用的全局变量使得推理变得非常困难。John Burkardt在他的网站上有一个更好的版本,可以在这里教授同样的算法,评论几乎非常好,也更容易追踪。