Question

这是一项简单的方案:

   void multiply(const int* v_in, const int* w_in, int n_v, int n_w, int* w_out)
   {
      for(int i=0; i<n_w; i++)
      {
         int sum=0;
         for(int j=0; j<n_v; j++)
            sum += (w_in[i]*v_in[j])>>1;
         w_out[i]=sum;
      }
   }

Presume n_v, n_w ~10^6。显然,在《世界人权宣言》中至少有十种类似方式这样做,有不同的方式将(n_v*n_w)行动注入深层,并不会分享记忆。理论上说,什么方式应该最快?

Answer 1

最简单:

   void multiply(const int* v_in, const int* w_in, int n_v, int n_w, int* w_out)
   {
      int *v = shared; // dynamic
      for(int i = block.rank; i < n_w; i += block.size)
      {
         int w = w_in[i]; // coalesced
         int sum=0;
         for(int j=0; j<n_v; j += block.size) { // assumption
            v[block.rank] = v_in[j+block.rank];
            __synch();
            for(int k = 0; k < block.size; ++k) 
                sum += (w*v[k])>>1;  // 
            __synch(); // ouch
         }
         w_out[i] = sum; // ditto
      }
   }

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