CODEWITHSUNDEEP
c++cudaiteration
user1547027
user1547027

Reputation: 13

Iteration second cycle with sum reduction in CUDA

I have to parallelize this code from c ++ to CUDA C 

  for(ihist = 0; ihist < numhist; ihist++){ 
      for(iwin = 0; iwin<numwin; iwin++){
          denwham[ihist] += (numbinwin[iwin]/g[iwin])*exp(F[iwin]-U[ihist]); 
          }
          Punnorm[ihist] = numwham[ihist]/denwham[ihist];
        }

In CUDA C, using the sum reduction :

extern __shared__ float sdata[];
  int tx = threadIdx.x;
  int i=blockIdx.x;
  int j=blockIdx.y;
  float sum=0.0;
  float temp=0.0;
  temp=U[j];


   if(tx<numwin)
   {
    sum=(numbinwin[tx]/g[tx])*exp(F[tx]- temp); 
    sdata[tx] = sum;
     __syncthreads();  
   }


  for(int offset = blockDim.x / 2;offset > 0;offset >>= 1)
  {
   if(tx < offset)
   {
    // add a partial sum upstream to our own
    sdata[tx] += sdata[tx + offset];
   }
   __syncthreads();
  }

   // finally, thread 0 writes the result
  if(threadIdx.x == 0)
  {
   // note that the result is per-block
   // not per-thread
   denwham[i] = sdata[0];

    for(int k=0;k<numhist;k++)
    Punnorm[k] = numwham[k]/denwham[k];
  }

And initialize it in this way:

 int smem_sz = (256)*sizeof(float);
  dim3 Block(numhist,numhist,1);
  NewProbabilitiesKernel<<<Block,256,smem_sz>>>(...);

My problem is that I cannot iterate over U using exp , I have tried the following:

a) loop for/while inside the kernel that iterates over U 
b) iterate by thread
c) iterate to block

All these attempts lead me to different results between C++ code and code cuda.The code works fine if instead of U [i] I put a constant!

have you any idea to help me ?

thanks.

Upvotes: 0

Views: 152

Answers (1)

kangshiyin
kangshiyin

Reputation: 9789

It looks like you could move the U out of the inner loop by

for(iwin = 0; iwin<numwin; iwin++){
    denwham += numbinwin[iwin] / g[iwin] * exp(F[iwin]); 
}
for(ihist = 0; ihist < numhist; ihist++){ 
    Punnorm[ihist] = numwham[ihist] / denwham * exp(U[ihist]);
}

Update

After that you could use 2 simple kernels instead of 1 complex one to finish the task.

  1. reduction kernel to compute denwham;
  2. 1-D transform kernel to compute Punnorm;

Upvotes: 1

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