Sum two entire vectors with CUDA - vector

Question

I´m trying to learn some cuda and I can't figure out how to solve the following situation:
Consider two groups G1 and G2:
G1 have 2 vectors with 3 elements each a1 = {2,5,8} and b1 =
{8,4,6}
G2 have 2 vectors with 3 elements each a2 = {7,3,1}
and b2 = {4,2,9}
The task is to sum vector a and b from each group and return a sorted c vector, so:
G1 will give c1 = {10,9,14) => (sort algorithm) => c1 = {9,10,14}
G2 will give c2 = {11,5,10) => (sort algorithm) => c1 = {5,10,11}
If I have a gforce with 92 cuda cores I would like to create 92 G groups and make all the sum in parallel so
core 1-> G1 -> c1 = a1 + b1 -> sort c1 -> return c1
core 2-> G2 -> c2 = a2 + b2 -> sort c2 -> return c2
....
core 92-> G92 -> c92 = a92 + b92 -> sort c92 -> return c92
The kernel below sum two vectors in parallel and return another one:
__global__ void add( int*a, int*b, int*c )
{
c[blockIdx.x] = a[blockIdx.x] + b[blockIdx.x];
}
What I can´t understand is how make the kernel handle the entire vector not only one
element of the vector and them return an entire vector.
Something like this:
__global__ void add( int*a, int*b, int*c, int size )
{
for (int i = 0; i < size ; i++)
c[i] = a[i] + b[i];
//sort c
}
Can anyone please explain me if it is possible and how to do it?

Answer 1

This is a small example. It uses cudaMallocPitch and cudaMemcpy2D. I hope it will give you guidelines to solve your particular problem:
#include<stdio.h>
#include<cuda.h>
#include<cuda_runtime.h>
#include<device_launch_parameters.h>
#include<conio.h>
#define N 92
#define M 3
__global__ void test_access(float** d_a,float** d_b,float** d_c,size_t pitch1,size_t pitch2,size_t pitch3)
{
int idx = threadIdx.x;
float* row_a = (float*)((char*)d_a + idx*pitch1);
float* row_b = (float*)((char*)d_b + idx*pitch2);
float* row_c = (float*)((char*)d_c + idx*pitch3);
for (int i=0; i<M; i++) row_c[i] = row_a[i] + row_b[i];
printf("row %i column 0 value %f \n",idx,row_c[0]);
printf("row %i column 1 value %f \n",idx,row_c[1]);
printf("row %i column 2 value %f \n",idx,row_c[2]);
}
/********/
/* MAIN */
/********/
int main()
{
float a[N][M], b[N][M], c[N][M];
float **d_a, **d_b, **d_c;
size_t pitch1,pitch2,pitch3;
cudaMallocPitch(&d_a,&pitch1,M*sizeof(float),N);
cudaMallocPitch(&d_b,&pitch2,M*sizeof(float),N);
cudaMallocPitch(&d_c,&pitch3,M*sizeof(float),N);
for (int i=0; i<N; i++)
for (int j=0; j<M; j++) {
a[i][j] = i*j;
b[i][j] = -i*j+1;
}
cudaMemcpy2D(d_a,pitch1,a,M*sizeof(float),M*sizeof(float),N,cudaMemcpyHostToDevice);
cudaMemcpy2D(d_b,pitch2,b,M*sizeof(float),M*sizeof(float),N,cudaMemcpyHostToDevice);
test_access<<<1,N>>>(d_a,d_b,d_c,pitch1,pitch2,pitch3);
cudaMemcpy2D(c,M*sizeof(float),d_c,pitch3,M*sizeof(float),N,cudaMemcpyDeviceToHost);
for (int i=0; i<N; i++)
for (int j=0; j<M; j++) printf("row %i column %i value %f\n",i,j,c[i][j]);
getch();
return 0;
}

Answer 2

92 3-D vectors can be seen as 1 276-D vector, then you can use the single vector add kernel to add them. Thrust will be a more simple way to do this.
update
If your vector is only 3-D, you could simply sort the elements immediately after they are calculated, using sequential method.
If your vector has higher dimensions, you could consider use cub::BlockRadixSort. The idea is to first add one vector per thread/block, then sort the vector within the block using cub::BlockRadixSort.
http://nvlabs.github.io/cub/classcub_1_1_block_radix_sort.html