Hi,
Thanks for checking it.
I'm not sure to fully understand your comment. The matrix temp_grid is
a matrix of size (threads, intervals). The thread 0 will create an
histogram and will save it on the first column of that matrix, the
thread 1 will create an histogram and will save it on the second row
of that matrix....
Once you have the matrix filled, the reduction part will just sum all
the histograms to produce the final result.
So I think there is no memory conflict between different threads (the
elements filled by one thread are never touched by other threads).
Are you saying that accessing those positions of memory every time, in
the same thread is a low step?
Thanks a lot,
Fran.
El 10/04/2012, a las 22:04, Pazzula, Dominic J escribió:
I'm seeing performance of around a 7.5x speed up on my 8500GT and 1
year old Xeon. 140ms GPU vs 1047ms CPU on 10,000,000 points.
For me, the slower memory throughput kills the idea of using MKL and
numpy for the reduction. MKL can do the reduction step in less
than .01ms and the GPU is doing it in .9ms. But the transfer time
of that 2MB (512x1024 floats) is about 2ms, giving the GPU the edge.
I am guessing here, but my thought on the slower than expected
performance comes from the writing of the bins.
temp_grid[id*interv+bin]+=1.0;
If I am remembering correctly, the GPU cannot write to the same
memory slot simultaneously. If the memory is aligned such that
bin=0 for each core is in the same slot, then if two cores were
attempting to adjust the value in bin=0, it would take 2 cycles
instead of 1. 10 writing means 10 cycles. Etc. Even if the memory
is not aligned, you will be attempting to write to the same block a
large number of times.
Utilizing shared memory to temporarily hold the bin counts and then
putting them into the global output matrix SHOULD increase your
time. If I have a chance, I'll work on it and post an updated kernel.
Dominic
-----Original Message-----
From: [email protected] [mailto:[email protected]] On
Behalf Of Francisco Villaescusa Navarro
Sent: Friday, April 06, 2012 11:26 AM
To: Thomas Wiecki
Cc: [email protected]
Subject: Re: [PyCUDA] Histograms with PyCUDA
Thanks for all the suggestions!
Regarding removing sqrt: it seems that the code only gains about ~1%,
and you lose the capacity to easily define linear intervals...
I have tried with sqrt and sqrtf, but there is not difference in the
total time (or it is very small).
The code to find the histogram of an array with values between 0 and 1
should be something as:
import numpy as np
import time
import pycuda.driver as cuda
import pycuda.autoinit
import pycuda.gpuarray as gpuarray
import pycuda.cumath as cumath
from pycuda.compiler import SourceModule
from pycuda import compiler
grid_gpu_template = """
__global__ void grid(float *values, int size, float *temp_grid)
{
unsigned int id = threadIdx.x;
int i,bin;
const uint interv = %(interv)s;
for(i=id;i<size;i+=blockDim.x){
bin=(int)(values[i]*interv);
if (bin==interv){
bin=interv-1;
}
temp_grid[id*interv+bin]+=1.0;
}
}
"""
reduction_gpu_template = """
__global__ void reduction(float *temp_grid, float *his)
{
unsigned int id = blockIdx.x*blockDim.x+threadIdx.x;
const uint interv = %(interv)s;
const uint threads = %(max_number_of_threads)s;
if(id<interv){
for(int i=0;i<threads;i++){
his[id]+=temp_grid[id+interv*i];
}
}
}
"""
number_of_points=100000000
max_number_of_threads=512
interv=1024
blocks=interv/max_number_of_threads
if interv%max_number_of_threads!=0:
blocks+=1
values=np.random.random(number_of_points).astype(np.float32)
grid_gpu = grid_gpu_template % {
'interv': interv,
}
mod_grid = compiler.SourceModule(grid_gpu)
grid = mod_grid.get_function("grid")
reduction_gpu = reduction_gpu_template % {
'interv': interv,
'max_number_of_threads': max_number_of_threads,
}
mod_redt = compiler.SourceModule(reduction_gpu)
redt = mod_redt.get_function("reduction")
values_gpu=gpuarray.to_gpu(values)
temp_grid_gpu
=gpuarray.zeros((max_number_of_threads,interv),dtype=np.float32)
hist=np.zeros(interv,dtype=np.float32)
hist_gpu=gpuarray.to_gpu(hist)
start=time.clock()*1e3
grid
(values_gpu
,np
.int32
(number_of_points
),temp_grid_gpu,grid=(1,1),block=(max_number_of_threads,1,1))
redt(temp_grid_gpu,hist_gpu,grid=(blocks,
1),block=(max_number_of_threads,1,1))
hist=hist_gpu.get()
print 'Time used to grid with GPU:',time.clock()*1e3-start,' ms'
start=time.clock()*1e3
bins_histo=np.linspace(0.0,1.0,interv+1)
hist_CPU=np.histogram(values,bins=bins_histo)[0]
print 'Time used to grid with CPU:',time.clock()*1e3-start,' ms'
print 'max difference between methods=',np.max(hist_CPU-hist)
################
Results:
Time used to grid with GPU: 680.0 ms
Time used to grid with CPU: 9320.0 ms
max difference between methods= 0.0
So it seems that with this algorithm we can't achieve factors larger
than ~15
Fran.
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