Error: BFS on CUDA Synchronization

问题

My following code got an error, when it runs, some of the graph weights are being overwritten, but that should not be happening with the Xa array(which keeps which ones have already been visited) and the __syncthreads() function... May someone help?

struct Node 
{
    int begin;     // begining of the substring
    int num;    // size of the sub-string 
};

__global__ void BFS (Node *Va, int *Ea, bool *Fa, bool *Xa, int *Ca, bool *parada)
{
    int tid = threadIdx.x;

    if (Fa[tid] == true && Xa[tid] == false)
    {
        Fa[tid] = false; 
        __syncthreads();

        // Va begin is where it's edges' subarray begins, Va is it's
        // number of elements
        for (int i = Va[tid].begin;  i < (Va[tid].begin + Va[tid].num); i++) 
        {           
            int nid = Ea[i];

            if (Xa[nid] == false)
            {
                Ca[nid] = Ca[tid] + 1;
                Fa[nid] = true;
                *parada = true;
            }   
        }    
        Xa[tid] = true;             
    }
}

// The BFS frontier corresponds to all the nodes being processed 
// at the current level.
int main()
{

    //descrição do grafo
    struct Node node[4]; 
    node[0].begin=0; 
    node[0].num=2; 
    node[1].begin=1; 
    node[1].num=0; 
    node[2].begin=2; 
    node[2].num=2; 
    node[3].begin=1; 
    node[3].num=0; 
    int edges[]={1,2,3,1}; 

    bool frontier[4]={false}; 
    bool visited[4]={false}; 
    int custo[4]={0}; 

    int source=0; 
    frontier[source]=true; 

    Node* Va; 
    cudaMalloc((void**)&Va,sizeof(Node)*4); 
    cudaMemcpy(Va,node,sizeof(Node)*4,cudaMemcpyHostToDevice); 

    int* Ea; 
    cudaMalloc((void**)&Ea,sizeof(Node)*4); 
    cudaMemcpy(Ea,edges,sizeof(Node)*4,cudaMemcpyHostToDevice); 

    bool* Fa; 
    cudaMalloc((void**)&Fa,sizeof(bool)*4); 
    cudaMemcpy(Fa,frontier,sizeof(bool)*4,cudaMemcpyHostToDevice); 

    bool* Xa; 
    cudaMalloc((void**)&Xa,sizeof(bool)*4); 
    cudaMemcpy(Xa,visited,sizeof(bool)*4,cudaMemcpyHostToDevice); 

    int* Ca; 
    cudaMalloc((void**)&Ca,sizeof(int)*4); 
    cudaMemcpy(Ca,custo,sizeof(int)*4,cudaMemcpyHostToDevice); 

    dim3 threads(4,1,1); 

    bool para; 
    bool* parada; 
    cudaMalloc((void**)&parada,sizeof(bool)); 
    printf("\n");
    int n=1;
    do{ 
        para=false; 
        cudaMemcpy(parada,&para,sizeof(bool),cudaMemcpyHostToDevice);       
        BFS <<<1,threads>>>(Va,Ea,Fa,Xa,Ca,parada);     
        CUT_CHECK_ERROR("kernel1 execution failed"); 
        cudaMemcpy(&para,parada,sizeof(bool),cudaMemcpyDeviceToHost); 



        printf("Run number: %d >> ",n); 
        cudaMemcpy(custo,Ca,sizeof(int)*4,cudaMemcpyDeviceToHost);  
        for(int i=0;i<4;i++) 
            printf("%d  ",custo[i]); 
        printf("\n");
        n++;

    }while(para); 


    printf("\nFinal:\n");
    cudaMemcpy(custo,Ca,sizeof(int)*4,cudaMemcpyDeviceToHost); 

    for(int i=0;i<4;i++) 
        printf("%d  ",custo[i]); 
    printf("\n");

}

回答1:

There are a number of pretty major flaws in that device code. Firstly, you have memory races on both Xa and Ca. Secondly, you have a conditionally executed __syncthreads() call, which is illegal and can lead to the kernel hanging if executed by a warp of threads where any branch divergence around the call can occur.

The structure of the algorithm you are using probably isn't going to be correct on CUDA, even if you were to use atomic memory access functions to eliminate the worst pf read-after-write races in the code as posted. Using atomic memory access will effectively serialise the code and cost a great deal of performance.

Breadth first search on CUDA isn't an unsolved problem. There are a number of good papers on implementations, if you care to search for them. I would recommend High Performance and Scalable GPU Graph Traversal, if you have not already seen it. The code for those authors' implementation is also available for download from here.

来源：https://stackoverflow.com/questions/7806882/error-bfs-on-cuda-synchronization