What is the best practice to do reduce in OpenCL? - opencl

Question

Imagine a binary operation (lets name it "+") with associative property. When you can compute a1 + a2 + a3 + a4 + ... in parallel, first computing
b1 = a1 + a2
b2 = a3 + a4
then
c1 = b1 + b2
c2 = b3 + b4
then doing the same thing for results of previous step, and so on, until there is one element left.
I'am learning OpenCL and trying to implement this approach to summarize all elements in array. I am a total newbie in this technology, so the program might look something weird.
This is the kernel:
__kernel void reduce (__global float *input, __global float *output)
{
size_t gl = get_global_id (0);
size_t s = get_local_size (0);
int i;
float accum = 0;
for (i=0; i<s; i++) {
accum += input[s*gl+i];
}
output[gl] = accum;
}
This is the main program:
#include <stdio.h>
#include <stdlib.h>
#include <fcntl.h>
#include <unistd.h>
#include <sys/mman.h>
#include <sys/stat.h>
#include <CL/cl.h>
#define N (64*64*64*64)
#include <sys/time.h>
#include <stdlib.h>
double gettime ()
{
struct timeval tv;
gettimeofday (&tv, NULL);
return (double)tv.tv_sec + (0.000001 * (double)tv.tv_usec);
}
int main()
{
int i, fd, res = 0;
void* kernel_source = MAP_FAILED;
cl_context context;
cl_context_properties properties[3];
cl_kernel kernel;
cl_command_queue command_queue;
cl_program program;
cl_int err;
cl_uint num_of_platforms=0;
cl_platform_id platform_id;
cl_device_id device_id;
cl_uint num_of_devices=0;
cl_mem input, output;
size_t global, local;
cl_float *array = malloc (sizeof (cl_float)*N);
cl_float *array2 = malloc (sizeof (cl_float)*N);
for (i=0; i<N; i++) array[i] = i;
fd = open ("kernel.cl", O_RDONLY);
if (fd == -1) {
perror ("Cannot open kernel");
res = 1;
goto cleanup;
}
struct stat s;
res = fstat (fd, &s);
if (res == -1) {
perror ("Cannot stat() kernel");
res = 1;
goto cleanup;
}
kernel_source = mmap (NULL, s.st_size, PROT_READ, MAP_PRIVATE, fd, 0);
if (kernel_source == MAP_FAILED) {
perror ("Cannot map() kernel");
res = 1;
goto cleanup;
}
if (clGetPlatformIDs (1, &platform_id, &num_of_platforms) != CL_SUCCESS) {
printf("Unable to get platform_id\n");
res = 1;
goto cleanup;
}
if (clGetDeviceIDs(platform_id, CL_DEVICE_TYPE_GPU, 1, &device_id,
&num_of_devices) != CL_SUCCESS)
{
printf("Unable to get device_id\n");
res = 1;
goto cleanup;
}
properties[0]= CL_CONTEXT_PLATFORM;
properties[1]= (cl_context_properties) platform_id;
properties[2]= 0;
context = clCreateContext(properties,1,&device_id,NULL,NULL,&err);
command_queue = clCreateCommandQueue(context, device_id, 0, &err);
program = clCreateProgramWithSource(context, 1, (const char**)&kernel_source, NULL, &err);
if (clBuildProgram(program, 0, NULL, NULL, NULL, NULL) != CL_SUCCESS) {
char buffer[4096];
size_t len;
printf("Error building program\n");
clGetProgramBuildInfo (program, device_id, CL_PROGRAM_BUILD_LOG, sizeof (buffer), buffer, &len);
printf ("%s\n", buffer);
res = 1;
goto cleanup;
}
kernel = clCreateKernel(program, "reduce", &err);
if (err != CL_SUCCESS) {
printf("Unable to create kernel\n");
res = 1;
goto cleanup;
}
// create buffers for the input and ouput
input = clCreateBuffer(context, CL_MEM_READ_ONLY,
sizeof(cl_float) * N, NULL, NULL);
output = clCreateBuffer(context, CL_MEM_WRITE_ONLY,
sizeof(cl_float) * N, NULL, NULL);
// load data into the input buffer
clEnqueueWriteBuffer(command_queue, input, CL_TRUE, 0,
sizeof(cl_float) * N, array, 0, NULL, NULL);
size_t size = N;
cl_mem tmp;
double time = gettime();
while (size > 1)
{
// set the argument list for the kernel command
clSetKernelArg(kernel, 0, sizeof(cl_mem), &input);
clSetKernelArg(kernel, 1, sizeof(cl_mem), &output);
global = size;
local = 64;
// enqueue the kernel command for execution
clEnqueueNDRangeKernel(command_queue, kernel, 1, NULL, &global,
&local, 0, NULL, NULL);
clFinish(command_queue);
size = size/64;
tmp = output;
output = input;
input = tmp;
}
cl_float answer[1];
clEnqueueReadBuffer(command_queue, tmp, CL_TRUE, 0,
sizeof(cl_float), array, 0, NULL, NULL);
time = gettime() - time;
printf ("%f %f\n", array[0], time);
cleanup:
free (array);
free (array2);
clReleaseMemObject(input);
clReleaseMemObject(output);
clReleaseProgram(program);
clReleaseKernel(kernel);
clReleaseCommandQueue(command_queue);
clReleaseContext(context);
if (kernel_source != MAP_FAILED) munmap (kernel_source, s.st_size);
if (fd != -1) close (fd);
_Exit (res); // Kludge
return res;
}
So I re-run kernel until there is only one element in the buffer. Is this correct approach to compute sum of elements in OpenCL? The time which I measure with gettime is about 10 times slower when execution time of a simple loop on CPU (compiled clang 4.0.0 and -O2 -ffast-math flags). Hardware I use: Amd Ryzen 5 1600X and Amd Radeon HD 6950.

Answer 1

There's a couple of things you can do to try to improve performance.
Firstly, get rid of the clFinish call inside your loop. This forces individual executions of the kernels to be dependent on the entire state of the Command Queue reaching a synchronization point with the Host before continuing, which is unnecessary. The only synchronization required is that the kernels execute in order, and even if you have an out-of-order queue (which your program isn't requesting anyways), you can guarantee that with simple use of event objects.
size_t size = N;
size_t total_expected_events = 0;
for(size_t event_count = size; event_count > 1; event_count /= 64)
total_expected_events++;
cl_event * events = malloc(total_expected_events * sizeof(cl_event));
cl_mem tmp;
double time = gettime();
size_t event_index = 0;
while (size > 1)
{
// set the argument list for the kernel command
clSetKernelArg(kernel, 0, sizeof(cl_mem), &input);
clSetKernelArg(kernel, 1, sizeof(cl_mem), &output);
global = size;
local = 64;
if(event_index == 0)
// enqueue the kernel command for execution
clEnqueueNDRangeKernel(command_queue, kernel, 1, NULL, &global,
&local, 0, NULL, events);
else
clEnqueueNDRangeKernel(command_queue, kernel, 1, NULL, &global,
&local, 1, events + (event_index - 1), events + event_index);
size = size/64;
tmp = output;
output = input;
input = tmp;
event_index++;
}
clFinish(command_queue);
for(; event_index > 0; event_index--)
clReleaseEvent(events[event_index-1]);
free(events);
cl_float answer[1];
clEnqueueReadBuffer(command_queue, tmp, CL_TRUE, 0,
sizeof(cl_float), array, 0, NULL, NULL);
The other thing to potentially look into is performing the reduction all in one kernel, instead of spreading it out over multiple invocations of the same kernel. This is one potential example, though it may be more complicated than you need it to be.