Cuda grid size limitations - cuda

Are there limitations as to what I can set the grid size of a CUDA kernel to be? I ran into a problem where kernels were not launching with a grid size of 33 x 33 but were able to launch when the grid size was 32 x 32. Is there any reason for this to occur? Or is it likely that changing the number of blocks from 32 x 32 to 33 x 33 broke some other constraint?
dim3 blockSize(8, 8);
dim3 gridSize(32, 32);
cudaDeviceSynchronize();
set_start<<<gridSize, blockSize>>>(some_params);
The above works.
dim3 blockSize(8, 8);
dim3 gridSize(33, 33);
cudaDeviceSynchronize();
set_start<<<gridSize, blockSize>>>(some_params);
The above does not work.
kernel & main:
__global__
void set_start(double * const H , double * const HU , double * const HV ,
double * const E , const int Na)
{
int j = threadIdx.x + blockIdx.x*blockDim.x + 1;
int i = threadIdx.y + blockIdx.y*blockDim.y + 1;
if(i >= Na-1 || j >= Na-1)
return;
H[i*Na+j] = 1.0 + exp(-100.0*((E[j-1]-0.75)*(E[j-1]-0.75)+(E[i-1]-0.75)*(E[i-1]-0.75))) + 0.5*exp(-100.0*((E[j-1]-0.75)*(E[j-1]-0.75)+(E[i-1]-0.25)*(E[i-1]-0.25)));
HU[i*Na+j] = 0;
HV[i*Na+j] = 0;
}
int main(int argc, char** argv){
double* E_d;
cudaMalloc(&E_d, sizeof(double) * (Nh+1));
set_E<<<64, (Nh/64) + 1>>>(E_d, dx, Nh);
int Na = 259;
double *H_d, *HU_d, *HV_d, *Ht_d, *HUt_d, *HVt_d;
cudaMalloc(&H_d , sizeof(double) * Na * Na);
cudaMalloc(&HU_d, sizeof(double) * Na * Na);
cudaMalloc(&HV_d, sizeof(double) * Na * Na);
dim3 blockSize(8, 8);
//dim3 gridSize(((Na-1)/blockSize.x) + 1, ((Na-1)/blockSize.y) + 1);
//dim3 gridSize(33, 33);
dim3 gridSize(32, 32);
cudaDeviceSynchronize();
set_start<<<blockSize, gridSize>>>(H_d, HU_d, HV_d, E_d, Na);
}
This was on CUDA 7.0.

You have block size and grid size mixed up when calling the kernel.
set_start<<<blockSize, gridSize>>>(H_d, HU_d, HV_d, E_d, Na);
should read:
set_start<<<gridSize, blockSize>>>(H_d, HU_d, HV_d, E_d, Na);
Because of this bug you are actually trying to launch a grid of size blockSize, and blocks of size gridSize. It would appear that the maximum size of a block on your GPU is 1024 threads, so launching blocks of 33x33 fails.

Related

How to use register memory for each thread in CUDA?

I am trying to flip upside down the array which size is big.(ex. 4096x8192)
At first, I tried with two array for input and output and It works!.
(I will say input is original and output is flipped array)
But I thought it will be easier and much efficient if each thread can hold input elements.
Then I can only use one array!
Could you guys share your knowledge or introduce any documents that help this problem?
Thanks and here is my code.
#include "cuda_runtime.h"
#include "device_launch_parameters.h"
#include <stdio.h>
#define ThreadPB 32 // optimal size
dim3 threadsPerBlock(ThreadPB, ThreadPB);
__global__ void initKernel(int *input, int nx, int ny)
{
int idx_x = blockDim.x * blockIdx.x + threadIdx.x;
int idx_y = blockDim.y * blockIdx.y + threadIdx.y;
int idx = idx_y * nx + idx_x;
if (idx_x < nx && idx_y < ny) {
input[idx] = idx_y;
}
}
__global__ void flipKernel(int *output, int *input, int nx, int ny)
{
int idx_x = blockDim.x * blockIdx.x + threadIdx.x;
int idx_y = blockDim.y * blockIdx.y + threadIdx.y;
int idx = idx_y * nx + idx_x;
// is it possible to use only one array?
if (idx_x < nx && idx_y < ny) {
output[(ny - idx_y - 1) * nx + idx_x] = input[idx_y * nx + idx_x];
}
}
int main()
{
// time check
cudaEvent_t start, stop, start_temp, stop_temp;
cudaEvent_t start_temp2, stop_temp2;
float elapsedTime, elapsedTime_temp, elapsedTime_temp2;
cudaEventCreate(&start); cudaEventCreate(&stop);
cudaEventCreate(&start_temp); cudaEventCreate(&stop_temp);
cudaEventCreate(&start_temp2); cudaEventCreate(&stop_temp2);
const int num_x = 4096;
const int num_y = 8192;
const int arraySize = num_x * num_y;
int *orig, *flip;
orig = (int *)malloc(sizeof(int) * arraySize);
flip = (int *)malloc(sizeof(int) * arraySize);
int *dev_orig = 0;
int *dev_flip = 0;
cudaMalloc((void**)&dev_orig, arraySize * sizeof(int));
cudaMalloc((void**)&dev_flip, arraySize * sizeof(int));
cudaMemcpy(dev_orig, orig, arraySize * sizeof(int), cudaMemcpyHostToDevice);
cudaMemcpy(dev_flip, flip, arraySize * sizeof(int), cudaMemcpyHostToDevice);
dim3 blocksFlip((num_x + threadsPerBlock.x - 1) / threadsPerBlock.x, (num_y + threadsPerBlock.y - 1) / threadsPerBlock.y);
initKernel << <blocksFlip, threadsPerBlock >> > (dev_orig, num_x, num_y);
cudaEventRecord(start, 0);
flipKernel << <blocksFlip, threadsPerBlock >> > (dev_flip, dev_orig, num_x, num_y);
// time check end
cudaEventRecord(stop, 0); cudaEventSynchronize(stop); cudaEventElapsedTime(&elapsedTime, start, stop); printf("flip 1024x2048 처리 시간 = %f ms.\n", elapsedTime);
cudaMemcpy(orig, dev_orig, arraySize * sizeof(int), cudaMemcpyDeviceToHost);
cudaMemcpy(flip, dev_flip, arraySize * sizeof(int), cudaMemcpyDeviceToHost);
// check flip works
printf("FLIP this array { 0, 1, 2, 3, 4 , 5, 6, 7, 8, 9...} \n= { %d, %d, %d, %d, %d, %d, %d, %d, %d, %d...}\n",
flip[num_x * 0], flip[num_x * 1], flip[num_x * 2], flip[num_x * 3], flip[num_x * 4],
flip[num_x * 5], flip[num_x * 6], flip[num_x * 7], flip[num_x * 8], flip[num_x * 9]);
return 0;
}
For an even number of rows in the array, you should be able to do something like this:
__global__ void flipKernel(int *input, int nx, int ny)
{
int idx_x = blockDim.x * blockIdx.x + threadIdx.x;
int idx_y = blockDim.y * blockIdx.y + threadIdx.y;
int idx = idx_y * nx + idx_x;
if (idx_x < nx && idx_y < ny/2) {
int output_temp = input[(ny - idx_y - 1) * nx + idx_x];
input[(ny - idx_y - 1) * nx + idx_x] = input[idx_y * nx + idx_x];
input[idx_y * nx + idx_x] = output_temp;
}
}
You would only need to launch this kernel with half as many threads in y (half as many rows in y). Each thread is updating two values in the matrix.
Rather than thinking about things like "register" or imagining that CUDA is some kind of weird language, if you have C or C++ programming ability, I would encourage you to think about how you might solve the problem if it were framed as an ordinary C or C++ programming challenge. Your intuition from that will often work very well in CUDA.
The core of the routine above is just a swap. The thing you are referring to as a "register" is just an ordinary local variable in C or C++. There is a register keyword in C++, but it serves essentially no purpose in CUDA, and is not needed here anyway.
You can handle an odd number of rows by simply leaving the middle row as-is, and swapping the remaining rows. This would require just a slight change to the indexing calculations.

Problem with response times in CUDA why cpu is faster than gpu?

Hi i have a problem with the time response I am getting a longer response time on GPU than CPU
the algorithm used is a matrix multiplication algorithm
using the next functions:
// Start timers
cudaEvent_t timer1, timer2;
cudaEventCreate(&timer1);
cudaEventCreate(&timer2);
cudaEventRecord(timer1, 0);
cudaEventSynchronize(timer1);
// Stop timers
cudaEventRecord(timer2, 0);
cudaEventSynchronize(timer1);
cudaEventSynchronize(timer2);
float elapsed;
cudaEventElapsedTime(&elapsed, timer1, timer2);
cudaDeviceReset();
return elapsed;
here is my code on GPU:
float Mult_gpu(float* hostPtr, float* hostPtr2, float* hostPtr3, int size, int Ncols, int Nrows, int n) {
size_t pitch;
check("Creating timers");
cudaEvent_t timer1, timer2;
cudaEventCreate(&timer1);
cudaEventCreate(&timer2);
cudaEventRecord(timer1, 0);
cudaEventSynchronize(timer1);
/******************************************/
/***Configuracion de las matrices en gpu***/
/******************************************/
float* devPtr;
cudaMallocPitch(&devPtr, &pitch, n * sizeof(float), Nrows);
cudaMemcpy2D(devPtr, pitch, hostPtr, n * sizeof(float), n * sizeof(float), Nrows, cudaMemcpyHostToDevice);
float* devPtr2;
cudaMallocPitch(&devPtr2, &pitch, Ncols * sizeof(float), n);
cudaMemcpy2D(devPtr2, pitch, hostPtr2, Ncols * sizeof(float), Ncols * sizeof(float), n, cudaMemcpyHostToDevice);
float* devPtr3;
cudaMallocPitch(&devPtr3, &pitch, Ncols * sizeof(float), Nrows);
//dim3 gridSize(iDivUp(Ncols3, BLOCKSIZE_x), iDivUp(Nrows3, BLOCKSIZE_y));
//dim3 blockSize(BLOCKSIZE_y, BLOCKSIZE_x);
dim3 block(32, 32); //hilos por bloque
dim3 grid((size / block.x) + 1, (size / block.y) + 1); //numero de bloques
/**************************/
/**Lanzamiento del kernel**/
/**************************/
Mult << <grid, block >> > (devPtr, devPtr2, devPtr3, pitch, Ncols, Nrows, n);
cudaDeviceSynchronize();
/*********************************/
/***Copiado de devPtr a hosPtr2***/
/*********************************/
cudaMemcpy2D(hostPtr3, Ncols * sizeof(float), devPtr3, pitch, Ncols * sizeof(float), Nrows, cudaMemcpyDeviceToHost);
//cudaMemcpy(hostPtr3, devPtr3, size * sizeof(float), cudaMemcpyDeviceToHost);
cudaFree(devPtr);
cudaFree(devPtr2);
cudaFree(devPtr3);
// Stop timers
cudaEventRecord(timer2, 0);
cudaEventSynchronize(timer1);
cudaEventSynchronize(timer2);
float elapsed;
cudaEventElapsedTime(&elapsed, timer1, timer2);
cudaDeviceReset();
return elapsed;
}
and here is my code on CPU
float Mult_cpu(float* hostPtrA, float* HostPtrB, float* hostPtrC, int Ncols, int Nrows, int n)
{
cudaEvent_t timer1, timer2;
cudaEventCreate(&timer1);
cudaEventCreate(&timer2);
cudaEventRecord(timer1, 0);
cudaEventSynchronize(timer1);
for (int i = 0; i < Nrows; ++i) {
for (int j = 0; j < Ncols; ++j) {
float suma = 0;
for (int k = 0; k < n; ++k) {
suma += hostPtrA[i * n + k] * HostPtrB[k * Ncols + j];
}
hostPtrC[i * Ncols + j] = suma;
}
}
// Stop timers
cudaEventRecord(timer2, 0);
cudaEventSynchronize(timer1);
cudaEventSynchronize(timer2);
float elapsed;
cudaEventElapsedTime(&elapsed, timer1, timer2);
return elapsed;
}
when i use a matrix 500x500 or any matrix the CPU is faster than GPU and i don't understand why i don't know if the problem is my kernel program or the CUDA functions that im using
my kernel code
__global__ void Mult(float* devPtrA, float* devPtrB, float* devPtrC, size_t pitch, int Ncols, int Nrows, int n)
{
float temp;
int r = blockDim.y * blockIdx.y + threadIdx.y; //for (int f = 0; f <= fil - 1; f += 1) equivalencia en for
int c = blockDim.x * blockIdx.x + threadIdx.x; //for (int c = 0; c <= col - 1; c += 1)
if ((r < Ncols) && (c < Nrows)) {
for (int c2 = 0; c2 < n; c2++) {
float* vertical = (float*)((char*)devPtrA + r * pitch);
float element1 = vertical[c2];
float* horizontal = (float*)((char*)devPtrB + c2 * pitch);
float element2 = horizontal[c];
temp += element1 * element2;
}
//printf("\nla fila es: %d la columna es: %d el valor es: %8.4f\n\n", r, c, temp);
float* vertical2 = (float*)((char*)devPtrC + r * pitch);
vertical2[c] = temp;
}
}
You should read on the concept of SIMT architecture, CUDA execution model and branch divergence. Analyze your CUDA kernel performance with a profiler. I suspect that the condition if ((r < Ncols) && (c < Nrows)) in your kernel causes threads in each warp to diverge and hence the reduced performance. Also pitch affects the global memory access pattern in your code which is another factor in the performance of CUDA kernels. Some other excellent optimization tips can be found here.
CudaMalloc is really slow. If you know the size of your matrices beforehand, do the initialization at the beginning of your program.

Solve the Poisson equation using FFT with CUDA

I'm following a tutorial on using the cuFFT library here: http://gpgpu.org/static/sc2007/SC07_CUDA_3_Libraries.pdf
After following line by line of its code, I'm getting really strange results.
I have input data that is an NxN array of floats. The program does a FFT forward transform, solves Poisson's equation, and then does an inverse FFT. The input data (and output data) is referred to as a square image with sidelength N. When I comment out solve_poisson <<<dimGrid, dimBlock>>> (r_complex_d, kx_d, ky_d, N);, it correctly forward transforms the data and then performs an inverse transform, which causes the output data to be the same as the input data. This is supposed to happen.
Here is the output without calling the solve_poisson method.
0 r_initial: 0.00125126 r: 0.00125132
1 r_initial: 0.563585 r: 0.563585
2 r_initial: 0.193304 r: 0.193304
3 r_initial: 0.80874 r: 0.80874
4 r_initial: 0.585009 r: 0.585009
5 r_initial: 0.479873 r: 0.479873
6 r_initial: 0.350291 r: 0.350291
7 r_initial: 0.895962 r: 0.895962
8 r_initial: 0.82284 r: 0.82284
9 r_initial: 0.746605 r: 0.746605
10 r_initial: 0.174108 r: 0.174108
11 r_initial: 0.858943 r: 0.858943
12 r_initial: 0.710501 r: 0.710502
13 r_initial: 0.513535 r: 0.513535
14 r_initial: 0.303995 r: 0.303995
15 r_initial: 0.0149846 r: 0.0149846
Press any key to continue . . .
However, when I uncomment out the solve_poisson method, the output data is inf or nan, which leads me to believe that the scale variable was somehow close to zero in the solve_poisson method.
So I changed float scale = -(kx[idx] * kx[idx] + ky[idy] * ky[idy]); to float scale = -(kx[idx] * kx[idx] + ky[idy] * ky[idy]) + 0.00001f. This change is not in the original tutorial. The results computed here are not supposed to have extreme positive or negative values.
0 r_initial: 0.00125126 r: -11448.1
1 r_initial: 0.563585 r: 11449.3
2 r_initial: 0.193304 r: -11448.3
3 r_initial: 0.80874 r: 11449.2
4 r_initial: 0.585009 r: 11449.4
5 r_initial: 0.479873 r: -11448.4
6 r_initial: 0.350291 r: 11449.5
7 r_initial: 0.895962 r: -11448.6
8 r_initial: 0.82284 r: -11448.5
9 r_initial: 0.746605 r: 11449.4
10 r_initial: 0.174108 r: -11448.3
11 r_initial: 0.858943 r: 11449.3
12 r_initial: 0.710501 r: 11449.2
13 r_initial: 0.513535 r: -11448.4
14 r_initial: 0.303995 r: 11449.3
15 r_initial: 0.0149846 r: -11448.1
Press any key to continue . . .
In the tutorial, a sample calculation on slide 43 on page 22 is computed=0.975879 reference=0.975882, yet my results are completely different and really large.
The following code is what I used.
#include <cuda_runtime.h>
#include <device_launch_parameters.h>
#include <cufft.h>
#include <stdlib.h>
#include <iostream>
#define N 4 //4 X 4 // N is the sidelength of the image -> 16 pixels in entire image
#define block_size_x 2
#define block_size_y 2
__global__ void real2complex(cufftComplex *c, float *a, int n);
__global__ void complex2real_scaled(float *a, cufftComplex *c, float scale, int n);
__global__ void solve_poisson(cufftComplex *c, float *kx, float *ky, int n);
int main()
{
float *kx, *ky, *r;
kx = (float *)malloc(sizeof(float) * N);
ky = (float *)malloc(sizeof(float) * N);
r = (float *)malloc(sizeof(float) * N * N);
float *kx_d, *ky_d, *r_d;
cufftComplex *r_complex_d;
cudaMalloc((void **)&kx_d, sizeof(float) * N);
cudaMalloc((void **)&ky_d, sizeof(float) * N);
cudaMalloc((void **)&r_d, sizeof(float) * N * N);
cudaMalloc((void **)&r_complex_d, sizeof(cufftComplex) * N * N);
for (int y = 0; y < N; y++)
for (int x = 0; x < N; x++)
r[x + y * N] = rand() / (float)RAND_MAX;
//r[x + y * N] = sin(exp(-((x - N / 2.0f) * (x - N / 2.0f) + (N / 2.0f - y) * (N / 2.0f - y)) / (20 * 20))) * 255 / sin(1); //Here is sample data that will high values at the center of the image and low values as you go farther and farther away from the center.
float* r_inital = (float *)malloc(sizeof(float) * N * N);
for (int i = 0; i < N * N; i++)
r_inital[i] = r[i];
for (int i = 0; i < N; i++)
{
kx[i] = i - N / 2.0f; //centers kx values to be at center of image
ky[i] = N / 2.0f - i; //centers ky values to be at center of image
}
cudaMemcpy(kx_d, kx, sizeof(float) * N, cudaMemcpyHostToDevice);
cudaMemcpy(ky_d, ky, sizeof(float) * N, cudaMemcpyHostToDevice);
cudaMemcpy(r_d, r, sizeof(float) * N * N, cudaMemcpyHostToDevice);
cufftHandle plan;
cufftPlan2d(&plan, N, N, CUFFT_C2C);
/* Compute the execution configuration, block_size_x*block_size_y = number of threads */
dim3 dimBlock(block_size_x, block_size_y);
dim3 dimGrid(N / dimBlock.x, N / dimBlock.y);
/* Handle N not multiple of block_size_x or block_size_y */
if (N % block_size_x != 0) dimGrid.x += 1;
if (N % block_size_y != 0) dimGrid.y += 1;
real2complex << < dimGrid, dimBlock >> > (r_complex_d, r_d, N);
cufftExecC2C(plan, r_complex_d, r_complex_d, CUFFT_FORWARD);
solve_poisson << <dimGrid, dimBlock >> > (r_complex_d, kx_d, ky_d, N);
cufftExecC2C(plan, r_complex_d, r_complex_d, CUFFT_INVERSE);
float scale = 1.0f / (N * N);
complex2real_scaled << <dimGrid, dimBlock >> > (r_d, r_complex_d, scale, N);
cudaMemcpy(r, r_d, sizeof(float) * N * N, cudaMemcpyDeviceToHost);
for (int i = 0; i < N * N; i++)
std::cout << i << "\tr_initial: " << r_inital[i] << "\tr: " << r[i] << std::endl;
system("pause");
/* Destroy plan and clean up memory on device*/
free(kx);
free(ky);
free(r);
free(r_inital);
cufftDestroy(plan);
cudaFree(r_complex_d);
cudaFree(kx_d);
}
__global__ void real2complex(cufftComplex *c, float *a, int n)
{
/* compute idx and idy, the location of the element in the original NxN array */
int idx = blockIdx.x * blockDim.x + threadIdx.x;
int idy = blockIdx.y * blockDim.y + threadIdx.y;
if (idx < n && idy < n)
{
int index = idx + idy * n;
c[index].x = a[index];
c[index].y = 0.0f;
}
}
__global__ void complex2real_scaled(float *a, cufftComplex *c, float scale, int n)
{
/* compute idx and idy, the location of the element in the original NxN array */
int idx = blockIdx.x * blockDim.x + threadIdx.x;
int idy = blockIdx.y * blockDim.y + threadIdx.y;
if (idx < n && idy < n)
{
int index = idx + idy * n;
a[index] = scale * c[index].x;
}
}
__global__ void solve_poisson(cufftComplex *c, float *kx, float *ky, int n)
{
/* compute idx and idy, the location of the element in the original NxN array */
int idx = blockIdx.x * blockDim.x + threadIdx.x;
int idy = blockIdx.y * blockDim.y + threadIdx.y;
if (idx < n && idy < n)
{
int index = idx + idy * n;
float scale = -(kx[idx] * kx[idx] + ky[idy] * ky[idy]) + 0.00001f;
if (idx == 0 && idy == 0) scale = 1.0f;
scale = 1.0f / scale;
c[index].x *= scale;
c[index].y *= scale;
}
}
Is there anything I messed up on? I would really appreciate if anyone could help me out.
Although the poster has found the mistake by himself, I want to share my own implementation of the 2D Poisson equation solver.
The implementation slightly differs from the one linked to by the poster.
The theory is reported at Solve Poisson Equation Using FFT.
MATLAB VERSION
I first report the Matlab version for reference:
clear all
close all
clc
M = 64; % --- Number of Fourier harmonics along x (should be a multiple of 2)
N = 32; % --- Number of Fourier harmonics along y (should be a multiple of 2)
Lx = 3; % --- Domain size along x
Ly = 1.5; % --- Domain size along y
sigma = 0.1; % --- Characteristic width of f (make << 1)
% --- Wavenumbers
kx = (2 * pi / Lx) * [0 : (M / 2 - 1) (- M / 2) : (-1)]; % --- Wavenumbers along x
ky = (2 * pi / Ly) * [0 : (N / 2 - 1) (- N / 2) : (-1)]; % --- Wavenumbers along y
[Kx, Ky] = meshgrid(kx, ky);
% --- Right-hand side of differential equation
hx = Lx / M; % --- Grid spacing along x
hy = Ly / N; % --- Grid spacing along y
x = (0 : (M - 1)) * hx;
y = (0 : (N - 1)) * hy;
[X, Y] = meshgrid(x, y);
rSquared = (X - 0.5 * Lx).^2 + (Y - 0.5 * Ly).^2;
sigmaSquared = sigma^2;
f = exp(-rSquared / (2 * sigmaSquared)) .* (rSquared - 2 * sigmaSquared) / (sigmaSquared^2);
fHat = fft2(f);
% --- Denominator of the unknown spectrum
den = -(Kx.^2 + Ky.^2);
den(1, 1) = 1; % --- Avoid division by zero at wavenumber (0, 0)
% --- Unknown determination
uHat = ifft2(fHat ./ den);
% uHat(1, 1) = 0; % --- Force the unknown spectrum at (0, 0) to be zero
u = real(uHat);
u = u - u(1,1); % --- Force arbitrary constant to be zero by forcing u(1, 1) = 0
% --- Plots
uRef = exp(-rSquared / (2 * sigmaSquared));
err = 100 * sqrt(sum(sum(abs(u - uRef).^2)) / sum(sum(abs(uRef).^2)));
errMax = norm(u(:)-uRef(:),inf)
fprintf('Percentage root mean square error = %f\n', err);
fprintf('Maximum error = %f\n', errMax);
surf(X, Y, u)
xlabel('x')
ylabel('y')
zlabel('u')
title('Solution of 2D Poisson equation by spectral method')
CUDA VERSION
Here is the corresponding CUDA version:
#include <stdio.h>
#include <fstream>
#include <iomanip>
// --- Greek pi
#define _USE_MATH_DEFINES
#include <math.h>
#include <cufft.h>
#define BLOCKSIZEX 16
#define BLOCKSIZEY 16
#define prec_save 10
/*******************/
/* iDivUp FUNCTION */
/*******************/
int iDivUp(int a, int b){ return ((a % b) != 0) ? (a / b + 1) : (a / b); }
/********************/
/* CUDA ERROR CHECK */
/********************/
// --- Credit to http://stackoverflow.com/questions/14038589/what-is-the-canonical-way-to-check-for-errors-using-the-cuda-runtime-api
void gpuAssert(cudaError_t code, const char *file, int line, bool abort = true)
{
if (code != cudaSuccess)
{
fprintf(stderr, "GPUassert: %s %s %d\n", cudaGetErrorString(code), file, line);
if (abort) { exit(code); }
}
}
void gpuErrchk(cudaError_t ans) { gpuAssert((ans), __FILE__, __LINE__); }
/**************************************************/
/* COMPUTE RIGHT HAND SIDE OF 2D POISSON EQUATION */
/**************************************************/
__global__ void computeRHS(const float * __restrict__ d_x, const float * __restrict__ d_y,
float2 * __restrict__ d_r, const float Lx, const float Ly, const float sigma,
const int M, const int N) {
const int tidx = threadIdx.x + blockIdx.x * blockDim.x;
const int tidy = threadIdx.y + blockIdx.y * blockDim.y;
if ((tidx >= M) || (tidy >= N)) return;
const float sigmaSquared = sigma * sigma;
const float rSquared = (d_x[tidx] - 0.5f * Lx) * (d_x[tidx] - 0.5f * Lx) +
(d_y[tidy] - 0.5f * Ly) * (d_y[tidy] - 0.5f * Ly);
d_r[tidy * M + tidx].x = expf(-rSquared / (2.f * sigmaSquared)) * (rSquared - 2.f * sigmaSquared) / (sigmaSquared * sigmaSquared);
d_r[tidy * M + tidx].y = 0.f;
}
/****************************************************/
/* SOLVE 2D POISSON EQUATION IN THE SPECTRAL DOMAIN */
/****************************************************/
__global__ void solvePoisson(const float * __restrict__ d_kx, const float * __restrict__ d_ky,
float2 * __restrict__ d_r, const int M, const int N)
{
const int tidx = threadIdx.x + blockIdx.x * blockDim.x;
const int tidy = threadIdx.y + blockIdx.y * blockDim.y;
if ((tidx >= M) || (tidy >= N)) return;
float scale = -(d_kx[tidx] * d_kx[tidx] + d_ky[tidy] * d_ky[tidy]);
if (tidx == 0 && tidy == 0) scale = 1.f;
scale = 1.f / scale;
d_r[M * tidy + tidx].x *= scale;
d_r[M * tidy + tidx].y *= scale;
}
/****************************************************************************/
/* SOLVE 2D POISSON EQUATION IN THE SPECTRAL DOMAIN - SHARED MEMORY VERSION */
/****************************************************************************/
__global__ void solvePoissonShared(const float * __restrict__ d_kx, const float * __restrict__ d_ky,
float2 * __restrict__ d_r, const int M, const int N)
{
const int tidx = threadIdx.x + blockIdx.x * blockDim.x;
const int tidy = threadIdx.y + blockIdx.y * blockDim.y;
if ((tidx >= M) || (tidy >= N)) return;
// --- Use shared memory to minimize multiple access to same spectral coordinate values
__shared__ float kx_s[BLOCKSIZEX], ky_s[BLOCKSIZEY];
kx_s[threadIdx.x] = d_kx[tidx];
ky_s[threadIdx.y] = d_ky[tidy];
__syncthreads();
float scale = -(kx_s[threadIdx.x] * kx_s[threadIdx.x] + ky_s[threadIdx.y] * ky_s[threadIdx.y]);
if (tidx == 0 && tidy == 0) scale = 1.f;
scale = 1.f / scale;
d_r[M * tidy + tidx].x *= scale;
d_r[M * tidy + tidx].y *= scale;
}
/******************************/
/* COMPLEX2REAL SCALED KERNEL */
/******************************/
__global__ void complex2RealScaled(float2 * __restrict__ d_r, float * __restrict__ d_result, const int M, const int N, float scale)
{
const int tidx = threadIdx.x + blockIdx.x * blockDim.x;
const int tidy = threadIdx.y + blockIdx.y * blockDim.y;
if ((tidx >= M) || (tidy >= N)) return;
d_result[tidy * M + tidx] = scale * (d_r[tidy * M + tidx].x - d_r[0].x);
}
/******************************************/
/* COMPLEX2REAL SCALED KERNEL - OPTIMIZED */
/******************************************/
__global__ void complex2RealScaledOptimized(float2 * __restrict__ d_r, float * __restrict__ d_result, const int M, const int N, float scale)
{
const int tidx = threadIdx.x + blockIdx.x * blockDim.x;
const int tidy = threadIdx.y + blockIdx.y * blockDim.y;
if ((tidx >= M) || (tidy >= N)) return;
__shared__ float d_r_0[1];
if (threadIdx.x == 0) d_r_0[0] = d_r[0].x;
volatile float2 c;
c.x = d_r[tidy * M + tidx].x;
c.y = d_r[tidy * M + tidx].y;
d_result[tidy * M + tidx] = scale * (c.x - d_r_0[0]);
}
/**************************************/
/* SAVE FLOAT2 ARRAY FROM GPU TO FILE */
/**************************************/
void saveGPUcomplextxt(const float2 * d_in, const char *filename, const int M) {
float2 *h_in = (float2 *)malloc(M * sizeof(float2));
gpuErrchk(cudaMemcpy(h_in, d_in, M * sizeof(float2), cudaMemcpyDeviceToHost));
std::ofstream outfile;
outfile.open(filename);
for (int i = 0; i < M; i++) {
//printf("%f %f\n", h_in[i].c.x, h_in[i].c.y);
outfile << std::setprecision(prec_save) << h_in[i].x << "\n"; outfile << std::setprecision(prec_save) << h_in[i].y << "\n";
}
outfile.close();
}
/*************************************/
/* SAVE FLOAT ARRAY FROM GPU TO FILE */
/*************************************/
template <class T>
void saveGPUrealtxt(const T * d_in, const char *filename, const int M) {
T *h_in = (T *)malloc(M * sizeof(T));
gpuErrchk(cudaMemcpy(h_in, d_in, M * sizeof(T), cudaMemcpyDeviceToHost));
std::ofstream outfile;
outfile.open(filename);
for (int i = 0; i < M; i++) outfile << std::setprecision(prec_save) << h_in[i] << "\n";
outfile.close();
}
/********/
/* MAIN */
/********/
int main()
{
const int M = 64; // --- Number of Fourier harmonics along x (should be a multiple of 2)
const int N = 32; // --- Number of Fourier harmonics along y(should be a multiple of 2)
const float Lx = 3.f; // --- Domain size along x
const float Ly = 1.5f; // --- Domain size along y
const float sigma = 0.1f; // --- Characteristic width of f(make << 1)
// --- Wavenumbers on the host
float *h_kx = (float *)malloc(M * sizeof(float));
float *h_ky = (float *)malloc(N * sizeof(float));
for (int k = 0; k < M / 2; k++) h_kx[k] = (2.f * M_PI / Lx) * k;
for (int k = -M / 2; k < 0; k++) h_kx[k + M] = (2.f * M_PI / Lx) * k;
for (int k = 0; k < N / 2; k++) h_ky[k] = (2.f * M_PI / Ly) * k;
for (int k = -N / 2; k < 0; k++) h_ky[k + N] = (2.f * M_PI / Ly) * k;
// --- Wavenumbers on the device
float *d_kx; gpuErrchk(cudaMalloc(&d_kx, M * sizeof(float)));
float *d_ky; gpuErrchk(cudaMalloc(&d_ky, N * sizeof(float)));
gpuErrchk(cudaMemcpy(d_kx, h_kx, M * sizeof(float), cudaMemcpyHostToDevice));
gpuErrchk(cudaMemcpy(d_ky, h_ky, N * sizeof(float), cudaMemcpyHostToDevice));
// --- Domain discretization on the host
float *h_x = (float *)malloc(M * sizeof(float));
float *h_y = (float *)malloc(N * sizeof(float));
for (int k = 0; k < M; k++) h_x[k] = Lx / (float)M * k;
for (int k = 0; k < N; k++) h_y[k] = Ly / (float)N * k;
// --- Domain discretization on the device
float *d_x; gpuErrchk(cudaMalloc(&d_x, M * sizeof(float)));
float *d_y; gpuErrchk(cudaMalloc(&d_y, N * sizeof(float)));
gpuErrchk(cudaMemcpy(d_x, h_x, M * sizeof(float), cudaMemcpyHostToDevice));
gpuErrchk(cudaMemcpy(d_y, h_y, N * sizeof(float), cudaMemcpyHostToDevice));
// --- Compute right-hand side of the equation on the device
float2 *d_r; gpuErrchk(cudaMalloc(&d_r, M * N * sizeof(float2)));
dim3 dimBlock(BLOCKSIZEX, BLOCKSIZEY);
dim3 dimGrid(iDivUp(M, BLOCKSIZEX), iDivUp(N, BLOCKSIZEY));
computeRHS << <dimGrid, dimBlock >> >(d_x, d_y, d_r, Lx, Ly, sigma, M, N);
gpuErrchk(cudaPeekAtLastError());
gpuErrchk(cudaDeviceSynchronize());
// --- Create plan for CUDA FFT
cufftHandle plan;
cufftPlan2d(&plan, N, M, CUFFT_C2C);
// --- Compute in place forward FFT of right-hand side
cufftExecC2C(plan, d_r, d_r, CUFFT_FORWARD);
// --- Solve Poisson equation in Fourier space
//solvePoisson << <dimGrid, dimBlock >> > (d_kx, d_ky, d_r, M, N);
solvePoissonShared << <dimGrid, dimBlock >> > (d_kx, d_ky, d_r, M, N);
// --- Compute in place inverse FFT
cufftExecC2C(plan, d_r, d_r, CUFFT_INVERSE);
//saveGPUcomplextxt(d_r, "D:\\Project\\poisson2DFFT\\poisson2DFFT\\d_r.txt", M * N);
// --- With cuFFT, an FFT followed by an IFFT will return the same array times the size of the transform
// --- Accordingly, we need to scale the result.
const float scale = 1.f / ((float)M * (float)N);
float *d_result; gpuErrchk(cudaMalloc(&d_result, M * N * sizeof(float)));
//complex2RealScaled << <dimGrid, dimBlock >> > (d_r, d_result, M, N, scale);
complex2RealScaledOptimized << <dimGrid, dimBlock >> > (d_r, d_result, M, N, scale);
//saveGPUrealtxt(d_result, "D:\\Project\\poisson2DFFT\\poisson2DFFT\\d_result.txt", M * N);
// --- Transfer data from device to host
float *h_result = (float *)malloc(M * N * sizeof(float));
gpuErrchk(cudaMemcpy(h_result, d_result, M * N * sizeof(float), cudaMemcpyDeviceToHost));
return 0;
}
As it shows in the tutorial, the Matlab implementation on slide 33 on page 17 shows that the Poisson calculations are based on the top left corner of the screen as the origin. The x and y data values are then x = (0:(N-1))*h; and y = (0:(N-1))*h;, which is why the meshgrid created from these x and y values both start from 0 and increase, as shown on the graph's x and y axes on slide 31 on page 16. In this case, where the image length was 1 (I refer to the input data of the NxN float array or the meshgrid as an image), the center of the image is actually (0.5, 0.5). I wanted to translate these points, so the center point would instead be (0,0) and followed a typical representation of the Cartesian Plane.
So in my code, instead of the Matlab code of
x = (0:(N-1))*h;
y = (0:(N-1))*h;
which could be implemented as
for (int i = 0; i < N; i++)
{
kx[i] = i;
ky[i] = i;
}
I replaced it with
for (int i = 0; i < N; i++)
{
kx[i] = i - N / 2.0f; //centers kx values to be at center of image
ky[i] = N / 2.0f - i; //centers ky values to be at center of image
}
However, I had forgot to change the Poisson calculation so it recognizes the center of the image as the origin instead of the top right corner as the origin. So as Mr. Robert Crovella said, I would have to
change this line: if (idx == 0 && idy == 0) scale = 1.0f; to this: if
(idx == 2 && idy == 2) scale = 1.0f;
for the case where the image length, or N, is 4.
To generalize this for any image length, this line of code could be then changed to if (idx == n/2 && idy == n/2) scale = 1.0f;

Instructions Per Count (IPC) and Instruction Level Parallelism (ILP) in CUDA

I observe IPC drops as ILP goes up for 32-bit int operations when trying to speed up my cryptographic kernel. The kernel consists of fairly unrolled loops of long sequence of ADD and XOR operations, which should have a throughput of 160 ops per 192 cores per cycle on Kepler (GTX Titan/780).
IPC for my kernel hits the upper bound of 3.28. Using ILP even drops IPC. Apparently ILP fails to help achieve my goal -- fully utilize the pipeline, so I wrote some little experiments. I put the code for ILP 4 at the end.
Profiler Measurements
Results are measured on GTX Titan.
cubin outputs are examined to make sure no instructions are eliminated during optimization.
Executed IPC is almost the same as issued IPC, so I just list one of them.
ADD instructions (XORs have identical behavior)
| ILP 1 | ILP 2 | ILP 4 | ILP 8
--------------------------------------------------
IPC | 4.00 | 3.32 | 2.72 | 3.44
--------------------------------------------------
Issue Slot | 99.17% | 59.34% | 48.61% | 61.71%
Utilization | | | |
I expect ILP 2, 4 and 8 would give better performance, but not.
Recall the integer throughput is 160. The 4 warp scheduler per SM should dual issue up to 5 instructions per cycle, so that IPC should go up towards 5. How can I explain what I observed? Why is the issue slot 99% utilized when IPC = 4?
Float / Int ADD instruction mix
If I modify the code for ILP 4 to do two int ADDs and two float ADDs:
IPC: 5.1
Issue slot utilization: 99.12%
Strangely enough, it seems that the warp scheduler does a better job to issue floating operations.
Discussion
Available literature suggests using ILP help reach the peak performance for floating point operations. Why doesn't ILP apply to integers? How can I do this for integer operations?
My kernel theoretically should do 2.25 integer operations per candidate. This is consistent with what I observed in cuobjdump. There are 2^48 candidates, so the minimun runtime on GTX Titan should be 2.25 * 2^48 / (2688 * 160/192) / 876 MHz = 322.75s. Is this estimation reasonable?
The measured performance for my kernel is 523s. This does imply that integer throughput is only about 160 * 3.28 (measure IPC) / 5 (max IPC).
ILP test code
__device__ int x[10];
__global__ void test(int flag = 0)
{
int a = x[0], b = x[1], c = x[2], d = x[3];
int _a = x[4], _b = x[5], _c = x[6], _d = x[7];
#pragma unroll 128
for (int i = 0; i < 51200; ++i)
{
asm volatile("add.u32 %0, %0, %1;": "+r"(a): "r"(_a));
asm volatile("add.u32 %0, %0, %1;": "+r"(b): "r"(_b));
asm volatile("add.u32 %0, %0, %1;": "+r"(c): "r"(_c));
asm volatile("add.u32 %0, %0, %1;": "+r"(d): "r"(_d));
}
int v = a + b + c + d;
if (flag * v == 1)
x[0] = v;
}
Code fragment for 4 candidates
Each candidate takes 9 / 4 = 2.25 ops. Cuobjdump also verifies this.
d ^= d2(1, 3); // d2 is located in constant memory
s ^= d;
t ^= d2(1, 16);
u ^= d2(1, 17);
v ^= some_const;
flag_s = min(flag_s, s); // int min has throughput of 160
flag_t = flag_t || (s == t); // setp.or should be the same
flag_u = flag_u || (s == u);
flag_v = flag_v || (s == v);
I'm providing an answer to remove this question from the unanswered list.
I do not observe a change in executed Instructions Per Count (IPC) with Instruction Level Parallelism. Overall, it is difficult to argue the reason for the effect observed by the OP without knowing any further information but that provided by the OP himself (f.i., the launch configuration).
In the code below, I'm considering an example using floats, although I have tested the same code with ints without changing the conceptual results. The code implements cyclical Multiply Add (MAD) operations with ILP=1, ILP=2 and ILP=4.
The executed IPC has been the following
ILP IPC FLOPs
1 3.924 67108864
2 4.323 67108864
4 4.016 67108864
for N=8192. The code has been compiled with CUDA 8.0 and run on an NVIDIA GT920M. As it can be seen, IPC keeps almost constant for the differently considered values of ILP. The Floating Point Operations (FLOPs) as estimated by the code assuming 2 FLOPs per MAD coincides with that measured by the Visual Profiler.
THE CODE
#include<stdio.h>
#define N_ITERATIONS 8192
#include "Utilities.cuh"
#include "TimingGPU.cuh"
#define BLOCKSIZE 512
//#define DEBUG
/********************************************************/
/* KERNEL0 - NO INSTRUCTION LEVEL PARALLELISM (ILP = 0) */
/********************************************************/
__global__ void kernel0(float * __restrict__ d_a, const float * __restrict__ d_b, const float * __restrict__ d_c, const int N) {
const int tid = threadIdx.x + blockIdx.x * blockDim.x;
if (tid < N) {
float a = d_a[tid];
float b = d_b[tid];
float c = d_c[tid];
for (unsigned int i = 0; i < N_ITERATIONS; i++) {
a = a * b + c;
}
d_a[tid] = a;
}
}
/*****************************************************/
/* KERNEL1 - INSTRUCTION LEVEL PARALLELISM (ILP = 2) */
/*****************************************************/
__global__ void kernel1(float * __restrict__ d_a, const float * __restrict__ d_b, const float * __restrict__ d_c, const int N) {
const int tid = threadIdx.x + blockIdx.x * blockDim.x;
if (tid < N / 2) {
float a1 = d_a[tid];
float b1 = d_b[tid];
float c1 = d_c[tid];
float a2 = d_a[tid + N / 2];
float b2 = d_b[tid + N / 2];
float c2 = d_c[tid + N / 2];
for (unsigned int i = 0; i < N_ITERATIONS; i++) {
a1 = a1 * b1 + c1;
a2 = a2 * b2 + c2;
}
d_a[tid] = a1;
d_a[tid + N / 2] = a2;
}
}
/*****************************************************/
/* KERNEL2 - INSTRUCTION LEVEL PARALLELISM (ILP = 4) */
/*****************************************************/
__global__ void kernel2(float * __restrict__ d_a, const float * __restrict__ d_b, const float * __restrict__ d_c, const int N) {
const int tid = threadIdx.x + blockIdx.x * blockDim.x;
if (tid < N / 4) {
float a1 = d_a[tid];
float b1 = d_b[tid];
float c1 = d_c[tid];
float a2 = d_a[tid + N / 4];
float b2 = d_b[tid + N / 4];
float c2 = d_c[tid + N / 4];
float a3 = d_a[tid + N / 2];
float b3 = d_b[tid + N / 2];
float c3 = d_c[tid + N / 2];
float a4 = d_a[tid + 3 * N / 4];
float b4 = d_b[tid + 3 * N / 4];
float c4 = d_c[tid + 3 * N / 4];
for (unsigned int i = 0; i < N_ITERATIONS; i++) {
a1 = a1 * b1 + c1;
a2 = a2 * b2 + c2;
a3 = a3 * b3 + c3;
a4 = a4 * b4 + c4;
}
d_a[tid] = a1;
d_a[tid + N / 4] = a2;
d_a[tid + N / 2] = a3;
d_a[tid + 3 * N / 4] = a4;
}
}
/********/
/* MAIN */
/********/
int main() {
//const int N = 8192 * 64;
const int N = 8192;
//const int N = 1024;
TimingGPU timerGPU;
float *h_a = (float*)malloc(N*sizeof(float));
float *h_a_result_host = (float*)malloc(N*sizeof(float));
float *h_a_result_device = (float*)malloc(N*sizeof(float));
float *h_b = (float*)malloc(N*sizeof(float));
float *h_c = (float*)malloc(N*sizeof(float));
for (int i = 0; i<N; i++) {
h_a[i] = 2.;
h_b[i] = 1.;
h_c[i] = 2.;
h_a_result_host[i] = h_a[i];
for (unsigned int k = 0; k < N_ITERATIONS; k++) {
h_a_result_host[i] = h_a_result_host[i] * h_b[i] + h_c[i];
}
}
float *d_a; gpuErrchk(cudaMalloc((void**)&d_a, N*sizeof(float)));
float *d_b; gpuErrchk(cudaMalloc((void**)&d_b, N*sizeof(float)));
float *d_c; gpuErrchk(cudaMalloc((void**)&d_c, N*sizeof(float)));
gpuErrchk(cudaMemcpy(d_a, h_a, N*sizeof(float), cudaMemcpyHostToDevice));
gpuErrchk(cudaMemcpy(d_b, h_b, N*sizeof(float), cudaMemcpyHostToDevice));
gpuErrchk(cudaMemcpy(d_c, h_c, N*sizeof(float), cudaMemcpyHostToDevice));
/***********/
/* KERNEL0 */
/***********/
timerGPU.StartCounter();
kernel0 << <iDivUp(N, BLOCKSIZE), BLOCKSIZE >> >(d_a, d_b, d_c, N);
#ifdef DEBUG
gpuErrchk(cudaPeekAtLastError());
gpuErrchk(cudaDeviceSynchronize());
#endif
// --- Remember: timing is in ms
printf("Number of operations = %f; GFlops = %f\n", (float)N*(float)N_ITERATIONS, (1.e-6)*((float)N*(float)N_ITERATIONS) / timerGPU.GetCounter());
gpuErrchk(cudaMemcpy(h_a_result_device, d_a, N*sizeof(float), cudaMemcpyDeviceToHost));
for (int i = 0; i<N; i++) if (h_a_result_device[i] != h_a_result_host[i]) { printf("Error at i=%i! Host = %f; Device = %f\n", i, h_a_result_host[i], h_a_result_device[i]); return 1; }
/***********/
/* KERNEL1 */
/***********/
gpuErrchk(cudaMemcpy(d_a, h_a, N*sizeof(float), cudaMemcpyHostToDevice));
timerGPU.StartCounter();
kernel1 << <iDivUp(N / 2, BLOCKSIZE), BLOCKSIZE >> >(d_a, d_b, d_c, N);
#ifdef DEBUG
gpuErrchk(cudaPeekAtLastError());
gpuErrchk(cudaDeviceSynchronize());
#endif
// --- Remember: timing is in ms
printf("Number of operations = %f; GFlops = %f\n", (float)N*(float)N_ITERATIONS, (1.e-6)*((float)N*(float)N_ITERATIONS) / timerGPU.GetCounter());
gpuErrchk(cudaMemcpy(h_a_result_device, d_a, N*sizeof(float), cudaMemcpyDeviceToHost));
for (int i = 0; i<N; i++) if (h_a_result_device[i] != h_a_result_host[i]) { printf("Error at i=%i! Host = %f; Device = %f\n", i, h_a_result_host[i], h_a_result_device[i]); return 1; }
/***********/
/* KERNEL2 */
/***********/
gpuErrchk(cudaMemcpy(d_a, h_a, N*sizeof(float), cudaMemcpyHostToDevice));
timerGPU.StartCounter();
kernel2 << <iDivUp(N / 4, BLOCKSIZE), BLOCKSIZE >> >(d_a, d_b, d_c, N);
#ifdef DEBUG
gpuErrchk(cudaPeekAtLastError());
gpuErrchk(cudaDeviceSynchronize());
#endif
// --- Remember: timing is in ms
printf("Number of operations = %f; GFlops = %f\n", (float)N*(float)N_ITERATIONS, (1.e-6)*((float)N*(float)N_ITERATIONS) / timerGPU.GetCounter());
gpuErrchk(cudaMemcpy(h_a_result_device, d_a, N*sizeof(float), cudaMemcpyDeviceToHost));
for (int i = 0; i<N; i++) if (h_a_result_device[i] != h_a_result_host[i]) { printf("Error at i=%i! Host = %f; Device = %f\n", i, h_a_result_host[i], h_a_result_device[i]); return 1; }
cudaDeviceReset();
return 0;
}

Unable to execute device kernel in CUDA

I am trying to call a device kernel within a global kernel. My global kernel is a Matrix Multiplication and my device kernel is finding the maximum value and the index in each column of the product matrix. Following is the code :
__device__ void MaxFunction(float* Pd, float* max)
{
int x = (threadIdx.x + blockIdx.x * blockDim.x);
int y = (threadIdx.y + blockIdx.y * blockDim.y);
int k = 0;
int temp = 0; int temp_idx = 0;
for (k = 0; k < wB; ++k) {
if(Pd[x*wB + y] > temp){
temp = Pd[x*wB + y];
temp_idx = x*wB + y;
}
max[y*2 + 0] = temp;
max[y*2 + 1] = temp_idx;
}
}
__global__ void MatrixMulKernel(float* Md, float* Nd, float* Pd, float* max)
{
// declare cache in the shared memory
__shared__ float Mds[blockD][blockD];
__shared__ float Nds[blockD][blockD];
float Pvalue = 0;
// Loop over the Md and Nd block dimension required to compute the Pd element
for (int m = (wA * blockD * blockIdx.y), n = (blockD * blockIdx.x);
m < ((wA * blockD * blockIdx.y)+wA-1);
m += blockD, n += (blockD*hB)){
// collaboratively loading of Md and Nd blocks into shared memory
Mds[threadIdx.y][threadIdx.x] = Md[m + wA * threadIdx.y + threadIdx.x];
Nds[threadIdx.y][threadIdx.x] = Nd[n + wA * threadIdx.y + threadIdx.x];
__syncthreads();
// keep track of the running sum
for (int k = 0; k < blockD; k++)
Pvalue += Mds[threadIdx.y][k] * Nds[k][threadIdx.x];
__syncthreads();
}
// write back to the global memory
int p = hB * blockD * blockIdx.y + blockD * blockIdx.x;
Pd[p + hB * threadIdx.y + threadIdx.x] = Pvalue;
__syncthreads();
MaxFunction(Pd, max);
}
The Main code :
#include<stdio.h>
#include "cuda.h"
#include<stdlib.h>
#define blockD 32
const int wA = 128;
const int hA = 1024;
const int wB = 128;
const int hB = wA;
main(void){
void MatrixMultiplication(float *, float *, float *, float *);
int size_A = wA * hA * sizeof(float);
int size_B = wB * hB * sizeof(float);
int size_C = wB * hA * sizeof(float);
int size_max = 2 * wB * sizeof(float);
float *M, *N, *P, *C;
// allocate memory on the CPU
M = (float*)malloc(size_A);
N = (float*)malloc(size_B);
P = (float*)malloc(size_max);
C = (float*)malloc(size_C);
// initialize the matrices
for (int y=0; y < hA; y++) {
for (int x=0; x < wA; x++){
M[y*wA + x] = x;
}
}
for (int y=0; y<hB; y++) {
for (int x=0; x<wB; x++){
N[y*wB + x] = x;
}
}
MatrixMultiplication(M, N, P, C);
//Write
FILE *f1;
int i, j;
f1 = fopen("max_val.txt","w");
for(i=0; i < (wB * 2); i+=2){
fprintf(f1,"%d\t%d\n",int(P[i]),int(P[i+1]));
}
fclose(f1);
f1 = fopen("Prod_mat.txt","w");
for(i=0; i < 2; i++){
for(j=0; j < wB; j++){
fprintf(f1,"%d\t",int(C[i*wB + j]));
}
fprintf(f1,"\n");
}
fclose(f1);
free( M );
free( N );
free( P );
free( C );
cudaDeviceReset();
return 0;
}
void MatrixMultiplication(float *M, float *N, float *P, float *C) {
int size_A = wA * hA * sizeof(float);
int size_B = wB * hB * sizeof(float);
int size_C = wB * hA * sizeof(float);
int size_max = 2 * wB * sizeof(float);
float *Md, *Nd, *Pd, *max;
// allocate memory on the GPU
cudaMalloc((void**)&Md, size_A);
cudaMalloc((void**)&Nd, size_B);
cudaMalloc((void**)&Pd, size_C);
cudaMalloc((void**)&max, size_max);
// transfer M and N to device memory
cudaMemcpy(Md, M, size_A, cudaMemcpyHostToDevice);
cudaMemcpy(Nd, N, size_B, cudaMemcpyHostToDevice);
// kernel invocation code
dim3 dimBlock(blockD, blockD);
dim3 dimGrid(wA/blockD, hB/blockD);
//Execute Kernel
MatrixMulKernel<<<dimGrid, dimBlock>>>( Md, Nd, Pd, max);
// transfer P from device
cudaMemcpy(P, max, size_max, cudaMemcpyDeviceToHost);
cudaMemcpy(C, Pd, size_C, cudaMemcpyDeviceToHost);
cudaFree(Md);
cudaFree(Nd);
cudaFree(Pd);
cudaFree(max);
}
The Matrix Multiplication result is fine (Verified using Matlab), but I am not able to get the max values and their corresponding index. I would appreciate if anyone can kindly point out at what I am doing wrong. The max variable has only garbage when I run the above code.
Apparently you are attempting to find the maximum value in each column, as well as the offset to that value.
But all of your threads in y are hammering on the same location for max value (max[x*2 + 0]). This isn't recommended, as there is no way to sort out a race condition. You should use atomic operations, or other methods (e.g. reduction) to handle multiple threads updating a single max value this way.
Since you have a need to update two values atomically (the max value and it's location), it's not a simple matter of replacing your plain access with a standard atomic function. However, since you are dealing with two 32-bit adjacent quantities, you may be interested in my answer here.
By the way I think matlab's native matrix multiply on gpuArray should be faster than any matrix multiply code you write. But it would require the Parallel Compute Toolbox.