basic into questions, 1-d diffusion implementation beginner's qustions

I’m just starting out with cuda. Today I worked off of one of the the examples on the Dr. Dobbs site, http://www.ddj.com/hpc-high-performance-computing/207200659.

The problem I’m working can be approached as a 1-d diffusion (or “smoothing”), algorithm. Basically, there’s a large array, a[i], and after each successive time step the new value of a[i] is given by,

da[i] = (a[i+1] -2a[i] + a[i-1])/4

a[i] += da[i]

The computation of da for all elements of the array is parallel, but updating a[i] has to wait until all of the da[i] terms have been computed. I think I have a working implementation, which I’ll post below. I have a few questions about the code I wrote though - I’d appreciate your review of the code if you have the time.

(1) As I wrote the code, I made da a global array, but its only used in the kernel. Is there a good way to define it only in the kernel?

(2) Is my use of __syncthreads() in the kernel appropriate? To be accurate, all da elements need to be computed before a is updated. Is there an easier way?

(3) I also wrote a regular version of the program, also posted below, which runs about SIX times slower on a 2.3GHz Shanghai Opteron than the cuda code runs on a cheap 9500GT card. Is this kind of speedup really possible?

(4) I put no thought at all into the blockSize or the number of blocks used by the kernel - I just stuck with what the example code came with. Any suggestions about how to think about this would be appreciated.

CUDA version of the code

// working off of example at:

//	http://www.ddj.com/hpc-high-performance-computing/207402986

//

// smoothing_device.cu

// idea is to initialize set of vectors and then "smooth" them, 

// by making each vector the average of	its neighbors, eg 

//		du = u(i-1) - 2u(i) + u(i+1)

//

#include <stdio.h>

#include <assert.h>

#include <cuda.h>

#include <time.h>

__global__ void smoothDataOnDevice(float *a, float *da, int N, float one_over_kappa, int repeat)

{

  int idx, j; 

  idx =  blockIdx.x*blockDim.x + threadIdx.x;

  for(j=0;j<repeat;j++) {

	  if(idx==0) {

		  da[idx] = (-2.0*a[idx] + a[idx+1])*one_over_kappa;

	  }

	  else if (idx<(N-1)) {

		  da[idx] = (a[idx-1] - 2.0*a[idx] + a[idx+1])*one_over_kappa;

	  }

	  else if (idx==(N-1)) {

		  da[idx] = (a[idx-1] - 2.0*a[idx])*one_over_kappa;

	  }

	  __syncthreads();

	  if(idx<N){

		  a[idx] = a[idx] + da[idx];

	  }

	  __syncthreads();

  }

}

int main(void)

{

  float *a_h;	// pointers to host memory

  float *a_d, *da_d;	// pointer to device memory

  int i, N = 1000000;

  size_t size = N*sizeof(float);

  //

  // this is a smoothing constant.  For stability, needs to be ~< 0.25

  float one_over_kappa = 1.0/(4.0); 

// initialize the random number generator with the system time 

  time_t t1;

  time(&t1);

  long random_seed;

  random_seed=(long) t1;

  random_seed=100;

  srand(random_seed);

// allocate arrays on host

  a_h = (float *)malloc(size);

// allocate array on device 

  cudaMalloc((void **) &a_d, size);

  cudaMalloc((void **) &da_d, size);

// initialization of host data

  for (i=0; i<N; i++) {

	  a_h[i] = (float)(2.0*drand48()-1.0);

  }

  a_h[0]=0.0; 

  a_h[N-1]=0.0;

// copy data from host to device

  cudaMemcpy(a_d, a_h, sizeof(float)*N, cudaMemcpyHostToDevice);

// do calculation on device:

  // Part 1 of 2. Compute execution configuration

  int blockSize = 512;

  int nBlocks = N/blockSize;

  if(N%blockSize!=0) {

	  nBlocks +=1;

  }

  printf("N = %d\n",N);

  printf("blockSize = %d\n",blockSize);

  printf("nBlocks = %d\n",nBlocks);

int num_steps_to_simulate=5000;

smoothDataOnDevice<<< nBlocks, blockSize >>> (a_d, da_d, N, one_over_kappa,num_steps_to_simulate);

char out_filename[128];

  FILE *OUT;

  sprintf(out_filename,"smoothing.seed.%d.laststep.tsv",random_seed);

  OUT = fopen(out_filename,"w");  

  cudaMemcpy(a_h, a_d, sizeof(float)*N, cudaMemcpyDeviceToHost);

  for(i=0;i<N;i++){

	  fprintf(OUT,"%d\t%f\n",i,a_h[i]);

  }

  fclose(OUT);

// cleanup

  free(a_h); cudaFree(da_d); cudaFree(a_d); 

return 1;

}

[b]traditional code, runs about 6 times slower than Shanghai Opteron

[/b]

// smoothing_host.c

//

// 	idea is to initialize  set of vectors and then "smooth" them, 

//  by making each vector the average of its neighbors, eg 

//		du = u(i-1) - 2u(i) + u(i+1)

//

#include <stdio.h>

#include <stdlib.h>

#include <time.h>

int main(void)

{

  float *a, *da;	// pointer to device memory

  int step, i, N = 1000000;

  size_t size = N*sizeof(float);

// this is a smoothing constant.  For stability, needs to be ~< 0.25

  float one_over_kappa = 1.0/(4.0); 

// allocate arrays 

  a = (float *)malloc(size);

  da = (float *)malloc(size);

// initialize the random number generator with the system time 

  time_t t1;

  time(&t1);

  long random_seed;

  random_seed=(long) t1;

  random_seed=100;

  srand(random_seed);

// initialization of host data

  for (i=0; i<N; i++) {

	  a[i] = (float)(2.0*drand48()-1.0);

  }

  a[0]=0.0; 

  a[N-1]=0.0;

// do calculation on device:

  printf("N = %d\n",N);

int num_steps_to_simulate=5000;

  printf("smoothing iterations: %d\n",num_steps_to_simulate);

  for(step=0;step<num_steps_to_simulate;step++) {

	  for(i=0;i<N;i++){

		  if(i==0) {

			  da[i] = (-2.0*a[i] + a[i+1])*one_over_kappa;

		  }

		  else if (i<(N-1)) {

			  da[i] = (a[i-1] - 2.0*a[i] + a[i+1])*one_over_kappa;

		  }

		  else if (i==(N-1)) {

			  da[i] = (a[i-1] - 2.0*a[i])*one_over_kappa;

		  }						

	  }

	  for(i=0;i<N;i++) {

		  a[i] = a[i] + da[i];

	  }

}

// final output visualization, had to repeat this to include the last smoothing

  char out_filename[128];

  FILE * OUT;

  sprintf(out_filename,"cversion_smoothing.seed.%d.step.%d.tsv",random_seed,step);

  OUT = fopen(out_filename,"w");

  for (i=0; i<N; i++) {

	  fprintf(OUT,"%d\t%f\n",i,a[i]);

  }

  fclose(OUT);

// cleanup

  free(a); free(da);

return 1;

}

Hi!

Ad 1) You may simply use a variable instead of a global array:

__global__ void smoothDataOnDevice(float *a, int N, float one_over_kappa, int repeat)

{

  int idx, j; 

  float da;

  idx =  blockIdx.x*blockDim.x + threadIdx.x;

  for(j=0;j<repeat;j++) {

	  if(idx==0) {

		  da = (-2.0*a[idx] + a[idx+1])*one_over_kappa;

	  }

	  else if (idx<(N-1)) {

		  da = (a[idx-1] - 2.0*a[idx] + a[idx+1])*one_over_kappa;

	  }

	  else if (idx==(N-1)) {

		  da = (a[idx-1] - 2.0*a[idx])*one_over_kappa;

	  }

	  __syncthreads();

	  if(idx<N){

		  a[idx] = a[idx] + da;

	  }

	  __syncthreads();

  }

}

Regards

Navier

Hey navier

I’m going to comment on this from near the exact same experience, seeing I wrote a heat diffusion app in CUDA last week :)
The speed up using the GPU can get better than 6x, but because the kernel of a 1D system is usually quite small (performs little computation) compared to the time it takes to launch a kernel, you don’t see as big a performance improvement. Your code at the moment actually performs a lot of computation in the kernel, but unfortunately has a flaw. If you increase your number of meshpoints you’ll notice things start to go wrong. You see, the kernel will then require so many threadblocks that they cannot all execute on the gpu concurrently. A running kernel thread (as yours is now) will perform an update, wait for all threads (but only those that are in the same BLOCK), and do another update. But this will only be working on a portion of the mesh. Whereas the blocks that handle the other portions of the mesh aren’t running and are waiting to be swapped in for execution on the gpu.
Note that __syncthreads only synchronises threads in a BLOCK not with other blocks. If you’ve been getting correct results thus far, it is due to a low meshsize and luck that all threads are synchronising at the similar times.

Instead your kernel should only perform one update. This kernel can then be repeatedly called. This will ensure all blocks operate on the mesh, and unfortunately this will show the overhead.

One more point, you can achieve speedup by using shared memory and coalescing.

See the forum-thread where i was learning :P : 1d heat equation

rewolf