CUDA: shared memory over a multicast two-dimensional array

Question

I had a simple CUDA problem for class assignment, but the professor added an optional task to implement the same algorithm using shared memory. I could not finish it before the deadline (for example, the start date was a week ago), but I'm still curious, so now I will ask the Internet;).

The main purpose was to implement the inherited version of the red-black sequential reinstallation both sequentially and in CUDA, make sure that you get the same result in both, and then compare the acceleration. As I said, doing this with shared memory was an optional + 10% addition.

I am going to publish my working version and pseudo-code, which I tried to do, since I do not have the code in my hands at the moment, but I can update it later with the actual code if someone needs it.

Before anyone says this: Yes, I know that using CUtil is lame, but it made comparisons and timers easier.

Working version of global memory:

#include <stdlib.h>
#include <stdio.h>
#include <cutil_inline.h>

#define N 1024

__global__ void kernel(int *d_A, int *d_B) {
    unsigned int index_x = blockIdx.x * blockDim.x + threadIdx.x;
    unsigned int index_y = blockIdx.y * blockDim.y + threadIdx.y;

    // map the two 2D indices to a single linear, 1D index
    unsigned int grid_width = gridDim.x * blockDim.x;
    unsigned int index = index_y * grid_width + index_x;

    // check for boundaries and write out the result
    if((index_x > 0) && (index_y > 0) && (index_x < N-1) && (index_y < N-1))
        d_B[index] = (d_A[index-1]+d_A[index+1]+d_A[index+N]+d_A[index-N])/4;

}

main (int argc, char **argv) {

    int A[N][N], B[N][N];
    int *d_A, *d_B; // These are the copies of A and B on the GPU
    int *h_B; // This is a host copy of the output of B from the GPU
    int i, j;
    int num_bytes = N * N * sizeof(int);

    // Input is randomly generated
    for(i=0;i<N;i++) {
        for(j=0;j<N;j++) {
            A[i][j] = rand()/1795831;
            //printf("%d\n",A[i][j]);
        }
    }

    cudaEvent_t start_event0, stop_event0;
    float elapsed_time0;
    CUDA_SAFE_CALL( cudaEventCreate(&start_event0) );
    CUDA_SAFE_CALL( cudaEventCreate(&stop_event0) );
    cudaEventRecord(start_event0, 0);
    // sequential implementation of main computation
    for(i=1;i<N-1;i++) {
        for(j=1;j<N-1;j++) {
            B[i][j] = (A[i-1][j]+A[i+1][j]+A[i][j-1]+A[i][j+1])/4;
        }
    }
    cudaEventRecord(stop_event0, 0);
    cudaEventSynchronize(stop_event0);
    CUDA_SAFE_CALL( cudaEventElapsedTime(&elapsed_time0,start_event0, stop_event0) );



    h_B = (int *)malloc(num_bytes);
    memset(h_B, 0, num_bytes);
    //ALLOCATE MEMORY FOR GPU COPIES OF A AND B
    cudaMalloc((void**)&d_A, num_bytes);
    cudaMalloc((void**)&d_B, num_bytes);
    cudaMemset(d_A, 0, num_bytes);
    cudaMemset(d_B, 0, num_bytes);

    //COPY A TO GPU
    cudaMemcpy(d_A, A, num_bytes, cudaMemcpyHostToDevice);

    // create CUDA event handles for timing purposes
    cudaEvent_t start_event, stop_event;
    float elapsed_time;
    CUDA_SAFE_CALL( cudaEventCreate(&start_event) );
    CUDA_SAFE_CALL( cudaEventCreate(&stop_event) );
    cudaEventRecord(start_event, 0);

// TODO: CREATE BLOCKS AND THREADS AND INVOKE GPU KERNEL
    dim3 block_size(256,1,1); //values experimentally determined to be fastest

    dim3 grid_size;
    grid_size.x = N / block_size.x;
    grid_size.y = N / block_size.y;

    kernel<<<grid_size,block_size>>>(d_A,d_B);

    cudaEventRecord(stop_event, 0);
    cudaEventSynchronize(stop_event);
    CUDA_SAFE_CALL( cudaEventElapsedTime(&elapsed_time,start_event, stop_event) );

    //COPY B BACK FROM GPU
    cudaMemcpy(h_B, d_B, num_bytes, cudaMemcpyDeviceToHost);

    // Verify result is correct
    CUTBoolean res = cutComparei( (int *)B, (int *)h_B, N*N);
    printf("Test %s\n",(1 == res)?"Passed":"Failed");
    printf("Elapsed Time for Sequential: \t%.2f ms\n", elapsed_time0);
    printf("Elapsed Time for CUDA:\t%.2f ms\n", elapsed_time);
    printf("CUDA Speedup:\t%.2fx\n",(elapsed_time0/elapsed_time));

    cudaFree(d_A);
    cudaFree(d_B);
    free(h_B);

    cutilDeviceReset();
}

For the shared memory version, this is what I have tried so far:

#define N 1024

__global__ void kernel(int *d_A, int *d_B, int width) {
    //assuming width is 64 because that the biggest number I can make it
    //each MP has 48KB of shared mem, which is 12K ints, 32 threads/warp, so max 375 ints/thread?
    __shared__ int A_sh[3][66];

    //get x and y index and turn it into linear index

    for(i=0; i < width+2; i++)  //have to load 2 extra values due to the -1 and +1 in algo
          A_sh[index_y%3][i] = d_A[index+i-1]; //so A_sh[index_y%3][0] is actually d_A[index-1]

    __syncthreads(); //and hope that previous and next row have been loaded by other threads in the block?

    //ignore boundary conditions because it pseudocode
    for(i=0; i < width; i++)
        d_B[index+i] = A_sh[index_y%3][i] + A_sh[index_y%3][i+2] + A_sh[index_y%3-1][i+1] + A_sh[index_y%3+1][i+1];

}

main(){
   //same init as above until threads/grid init

   dim3 threadsperblk(32,16);
   dim3 numblks(32,64);

   kernel<<<numblks,threadsperblk>>>(d_A,d_B,64);

   //rest is the same
}

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EDIT: , :

template<int width>
__device__ void rowfetch(int *in, int *out, int col)
{
    *out = *in;
    if (col == 1) *(out-1) = *(in-1);   
    if (col == width) *out(+1) = *(in+1);   
}

template<int width>
__global__ operator(int *in, int *out, int nrows, unsigned int lda)
{
    // shared buffer holds three rows x (width+2) cols(threads)
    __shared__ volatile int buffer [3][2+width]; 

    int colid = threadIdx.x + blockIdx.x * blockDim.x;
    int tid = threadIdx.x + 1;

    int * rowpos = &in[colid], * outpos = &out[colid];

    // load the first three rows (compiler will unroll loop)
    for(int i=0; i<3; i++, rowpos+=lda) {
        rowfetch<width>(rowpos, &buffer[i][tid], tid);
    }

    __syncthreads(); // shared memory loaded and all threads ready

    int brow = 0; // brow is the next buffer row to load data onto
    for(int i=0; i<nrows; i++, rowpos+=lda, outpos+=lda) {

        // Do stencil calculations - use the value of brow to determine which
        // stencil to use
        result = ();
        // write result to outpos
        *outpos = result;

        // Fetch another row
        __syncthreads(); // Wait until all threads are done calculating
        rowfetch<width>(rowpos, &buffer[brow][tid], tid);
        brow = (brow < 2) ? (brow+1) : 0; // Increment or roll brow over
        __syncthreads(); // Wait until all threads have updated the buffer
    }
}