Cuda matrix addition - matrix

I have written the following code to sum two 4x4 matrices in cuda.
#include<stdio.h>
#include<stdlib.h>
#include<math.h>
__global__ void Matrix_add(double* a, double* b, double* c,int n)
{
int row = blockIdx.x * blockDim.x + threadIdx.x;
int col = blockIdx.y * blockDim.y + threadIdx.y;
int index = row * n + col;
if(col<n && row <n)
c[index] = a[index] + b[index];
}
int main()
{
int n=4;
double **h_a;
double **h_b;
double **h_c;
double *d_a, *d_b, *d_c;
int size = n*n*sizeof(double);
h_a = (double **) malloc(n*sizeof(double*));
h_b = (double **) malloc(n*sizeof(double*));
h_c = (double **) malloc(n*sizeof(double*));
cudaMalloc((void**)&d_a,size);
cudaMalloc((void**)&d_b,size);
cudaMalloc((void**)&d_c,size);
int t=0;
for (t=0;t<n;t++)
{
h_a[t]= (double *)malloc(n*sizeof(double));
h_b[t]= (double *)malloc(n*sizeof(double));
h_c[t]= (double *)malloc(n*sizeof(double));
}
int i=0,j=0;
for(i=0;i<n;i++)
{
for(j=0;j<n;j++)
{
h_a[i][j]=sin(i)*sin(i);
h_b[i][j]=cos(i)*cos(i);
}
}
cudaMemcpy(d_a,h_a+n,size,cudaMemcpyHostToDevice);
cudaMemcpy(d_b,h_b+n,size,cudaMemcpyHostToDevice);
dim3 dimBlock(4,4);
dim3 dimGrid(1,1);
Matrix_add<<<dimGrid, dimBlock>>>(d_a,d_b,d_c,n);
cudaMemcpy(h_c+n,d_c,size,cudaMemcpyDeviceToHost);
for(i=0;i<n;i++)
{
for( j=0;j<n;j++)
{
printf("%f",h_c[i][j]);
printf("\t");
}
printf("\n");
}
for(i=0;i<n;i++)
{
free(h_a[i]);
free(h_b[i]);
free(h_c[i]);
}
free(h_a);
free(h_b);
free(h_c);
cudaFree(d_a); cudaFree(d_b); cudaFree(d_c);
return 0;
}
Result of this addition should be a 2x2 all-ones matrix but in the result all the elements of matrix are 0. Also I get this message after getting result:
Segmentation fault (core dumped)
Can anyone please help me to find out the problem.
Thank you

Your host arrays (h_a, h_b, h_c) are not contiguous in memory, so your initial cudaMemcpy() calls will read garbage into GPU memory (apparently zeros in your case).
The reason is that your hosts arrays are not actually flat, but instead are represented as arrays of pointers. I guess to fake two-dimensional arrays in C? In any case, you either need to be more careful with your cudaMemcpy()s and copy the host arrays row-by-row, or use a flat representation on the host.

Related

CUDA Initialize Array on Device

I am very new to CUDA and I am trying to initialize an array on the device and return the result back to the host to print out to show if it was correctly initialized. I am doing this because the end goal is a dot product solution in which I multiply two arrays together, storing the results in another array and then summing up the entire thing so that I only need to return the host one value.
In the code I am working on all I am only trying to see if I am initializing the array correctly. I am trying to create an array of size N following the patterns of 1,2,3,4,5,6,7,8,1,2,3....
This is the code that I've written and it compiles without issue but when I run it the terminal is hanging and I have no clue why. Could someone help me out here? I'm so incredibly confused :\
#include <stdio.h>
#include <stdlib.h>
#include <chrono>
#define ARRAY_SIZE 100
#define BLOCK_SIZE 32
__global__ void cu_kernel (int *a_d,int *b_d,int *c_d, int size)
{
int x = blockIdx.x * blockDim.x + threadIdx.x;
__shared__ int temp;
if(temp != 8){
a_d[x] = temp;
temp++;
} else {
a_d[x] = temp;
temp = 1;
}
}
int main (int argc, char *argv[])
{
//declare pointers for arrays
int *a_d, *b_d, *c_d, *sum_h, *sum_d,a_h[ARRAY_SIZE];
//set space for device variables
cudaMalloc((void**) &a_d, sizeof(int) * ARRAY_SIZE);
cudaMalloc((void**) &b_d, sizeof(int) * ARRAY_SIZE);
cudaMalloc((void**) &c_d, sizeof(int) * ARRAY_SIZE);
cudaMalloc((void**) &sum_d, sizeof(int));
// set execution configuration
dim3 dimblock (BLOCK_SIZE);
dim3 dimgrid (ARRAY_SIZE/BLOCK_SIZE);
// actual computation: call the kernel
cu_kernel <<<dimgrid, dimblock>>> (a_d,b_d,c_d,ARRAY_SIZE);
cudaError_t result;
// transfer results back to host
result = cudaMemcpy (a_h, a_d, sizeof(int) * ARRAY_SIZE, cudaMemcpyDeviceToHost);
if (result != cudaSuccess) {
fprintf(stderr, "cudaMemcpy failed.");
exit(1);
}
// print reversed array
printf ("Final state of the array:\n");
for (int i =0; i < ARRAY_SIZE; i++) {
printf ("%d ", a_h[i]);
}
printf ("\n");
}
There are at least 3 issues with your kernel code.
you are using shared memory variable temp without initializing it.
you are not resolving the order in which threads access a shared variable as discussed here.
you are imagining (perhaps) a particular order of thread execution, and CUDA provides no guarantees in that area
The first item seems self-evident, however naive methods to initialize it in a multi-threaded environment like CUDA are not going to work. Firstly we have the multi-threaded access pattern, again, Furthermore, in a multi-block scenario, shared memory in one block is logically distinct from shared memory in another block.
Rather than wrestle with mechanisms unsuited to create the pattern you desire, (informed by notions carried over from a serial processing environment), I would simply do something trivial like this to create the pattern you desire:
__global__ void cu_kernel (int *a_d,int *b_d,int *c_d, int size)
{
int x = blockIdx.x * blockDim.x + threadIdx.x;
if (x < size) a_d[x] = (x&7) + 1;
}
Are there other ways to do it? certainly.
__global__ void cu_kernel (int *a_d,int *b_d,int *c_d, int size)
{
int x = blockIdx.x * blockDim.x + threadIdx.x;
__shared__ int temp;
if (!threadIdx.x) temp = blockIdx.x*blockDim.x;
__syncthreads();
if (x < size) a_d[x] = ((temp+threadIdx.x) & 7) + 1;
}
You can get as fancy as you like.
These changes will still leave a few values at zero at the end of the array, which would require changes to your grid sizing. There are many questions about this already, or study a sample code like vectorAdd.

Binary Matrix Reduction in CUDA

I have to traverse all cells of an imaginary matrix m * n and add + 1 for all cells that meet a certain condition.
My naive solution was as follows:
#include <stdio.h>
__global__ void calculate_pi(int center, int *count) {
int x = threadIdx.x;
int y = blockIdx.x;
if (x*x + y*y <= center*center) {
*count++;
}
}
int main() {
int interactions;
printf("Enter the number of interactions: ");
scanf("%d", &interactions);
int l = sqrt(interactions);
int h_count = 0;
int *d_count;
cudaMalloc(&d_count, sizeof(int));
cudaMemcpy(&d_count, &h_count, sizeof(int), cudaMemcpyHostToDevice);
calculate_pi<<<l,l>>>(l/2, d_count);
cudaMemcpy(&h_count, d_count, sizeof(int), cudaMemcpyDeviceToHost);
cudaFree(d_count);
printf("Sum: %d\n", h_count);
return 0;
}
In my use case, the value of interactions can be very large, making it impossible to allocate l * l of space.
Can someone help me? Any suggestions are welcome.
There are at least 2 problems with your code:
Your kernel code will not work correctly with an ordinary add here:
*count++;
this is because multiple threads are trying to do this at the same time, and CUDA does not automatically sort that out for you. For the purpose of this explanation, we will fix this with an atomicAdd(), although other methods are possible.
The ampersand doesn't belong here:
cudaMemcpy(&d_count, &h_count, sizeof(int), cudaMemcpyHostToDevice);
^
I assume that is just a typo, since you did it correctly on the subsequent cudaMemcpy operation:
cudaMemcpy(&h_count, d_count, sizeof(int), cudaMemcpyDeviceToHost);
This methodology (effectively creating a square array of threads using threadIdx.x for one dimension and blockIdx.x for the other) will only work up to an interactions value that leads to an l value of 1024, or less, because CUDA threadblocks are limited to 1024 threads, and you are using l as the size of the threadblock in your kernel launch. To fix this you would want to learn how to create a CUDA 2D grid of arbitrary dimensions, and adjust your kernel launch and in-kernel indexing calculations appropriately. For now we will just make sure that the calculated l value is in range for your code design.
Here's an example addressing the above issues:
$ cat t1590.cu
#include <stdio.h>
__global__ void calculate_pi(int center, int *count) {
int x = threadIdx.x;
int y = blockIdx.x;
if (x*x + y*y <= center*center) {
atomicAdd(count, 1);
}
}
int main() {
int interactions;
printf("Enter the number of interactions: ");
scanf("%d", &interactions);
int l = sqrt(interactions);
if ((l > 1024) || (l < 1)) {printf("Error: interactions out of range\n"); return 0;}
int h_count = 0;
int *d_count;
cudaMalloc(&d_count, sizeof(int));
cudaMemcpy(d_count, &h_count, sizeof(int), cudaMemcpyHostToDevice);
calculate_pi<<<l,l>>>(l/2, d_count);
cudaMemcpy(&h_count, d_count, sizeof(int), cudaMemcpyDeviceToHost);
cudaFree(d_count);
cudaError_t err = cudaGetLastError();
if (err == cudaSuccess){
printf("Sum: %d\n", h_count);
printf("fraction satisfying test: %f\n", h_count/(float)interactions);
}
else
printf("CUDA error: %s\n", cudaGetErrorString(err));
return 0;
}
$ nvcc -o t1590 t1590.cu
$ ./t1590
Enter the number of interactions: 1048576
Sum: 206381
fraction satisfying test: 0.196820
$
We see that the code indicates a calculated fraction of about 0.2. Does this appear to be correct? I claim that it does appear to be correct based on your test. You are effectively creating a grid that represents dimensions of lxl. Your test is asking, effectively, "which points in that grid are within a circle, with the center at the origin (corner) of the grid, and radius l/2 ?"
Pictorially, that looks something like this:
and it is reasonable to assume the red shaded area is somewhat less than 0.25 of the total area, so 0.2 is a reasonable estimate of that area.
As a bonus, here is a version of the code that reduces the restriction listed in item 3 above:
#include <stdio.h>
__global__ void calculate_pi(int center, int *count) {
int x = threadIdx.x+blockDim.x*blockIdx.x;
int y = threadIdx.y+blockDim.y*blockIdx.y;
if (x*x + y*y <= center*center) {
atomicAdd(count, 1);
}
}
int main() {
int interactions;
printf("Enter the number of interactions: ");
scanf("%d", &interactions);
int l = sqrt(interactions);
int h_count = 0;
int *d_count;
const int bs = 32;
dim3 threads(bs, bs);
dim3 blocks((l+threads.x-1)/threads.x, (l+threads.y-1)/threads.y);
cudaMalloc(&d_count, sizeof(int));
cudaMemcpy(d_count, &h_count, sizeof(int), cudaMemcpyHostToDevice);
calculate_pi<<<blocks,threads>>>(l/2, d_count);
cudaMemcpy(&h_count, d_count, sizeof(int), cudaMemcpyDeviceToHost);
cudaFree(d_count);
cudaError_t err = cudaGetLastError();
if (err == cudaSuccess){
printf("Sum: %d\n", h_count);
printf("fraction satisfying test: %f\n", h_count/(float)interactions);
}
else
printf("CUDA error: %s\n", cudaGetErrorString(err));
return 0;
}
This is launching a 2D grid based on l, and should work up to at least 1 billion interactions .

Conditional reduction in CUDA

I need to sum about 100000 values stored in an array, but with conditions.
Is there a way to do that in CUDA to produce fast results?
Can anyone post a small code to do that?
I think that, to perform conditional reduction, you can directly introduce the condition as a multiplication by 0 (false) or 1 (true) to the addends. In other words, suppose that the condition you would like to meet is that the addends be smaller than 10.f. In this case, borrowing the first code at Optimizing Parallel Reduction in CUDA by M. Harris, then the above would mean
__global__ void reduce0(int *g_idata, int *g_odata) {
extern __shared__ int sdata[];
// each thread loads one element from global to shared mem
unsigned int tid = threadIdx.x;
unsigned int i = blockIdx.x*blockDim.x + threadIdx.x;
sdata[tid] = g_idata[i]*(g_data[i]<10.f);
__syncthreads();
// do reduction in shared mem
for(unsigned int s=1; s < blockDim.x; s *= 2) {
if (tid % (2*s) == 0) {
sdata[tid] += sdata[tid + s];
}
__syncthreads();
}
// write result for this block to global mem
if (tid == 0) g_odata[blockIdx.x] = sdata[0];
}
If you wish to use CUDA Thrust to perform conditional reduction, you can do the same by using thrust::transform_reduce. Alternatively, you can create a new vector d_b copying in that all the elements of d_a satisfying the predicate by thrust::copy_if and then applying thrust::reduce on d_b. I haven't checked which solution performs the best. Perhaps, the second solution will perform better on sparse arrays. Below is an example with an implementation of both the approaches.
#include <thrust/host_vector.h>
#include <thrust/device_vector.h>
#include <thrust/reduce.h>
#include <thrust/count.h>
#include <thrust/copy.h>
// --- Operator for the first approach
struct conditional_operator {
__host__ __device__ float operator()(const float a) const {
return a*(a<10.f);
}
};
// --- Operator for the second approach
struct is_smaller_than_10 {
__host__ __device__ bool operator()(const float a) const {
return (a<10.f);
}
};
void main(void)
{
int N = 20;
// --- Host side allocation and vector initialization
thrust::host_vector<float> h_a(N,1.f);
h_a[0] = 20.f;
h_a[1] = 20.f;
// --- Device side allocation and vector initialization
thrust::device_vector<float> d_a(h_a);
// --- First approach
float sum = thrust::transform_reduce(d_a.begin(), d_a.end(), conditional_operator(), 0.f, thrust::plus<float>());
printf("Result = %f\n",sum);
// --- Second approach
int N_prime = thrust::count_if(d_a.begin(), d_a.end(), is_smaller_than_10());
thrust::device_vector<float> d_b(N_prime);
thrust::copy_if(d_a.begin(), d_a.begin() + N, d_b.begin(), is_smaller_than_10());
sum = thrust::reduce(d_b.begin(), d_b.begin() + N_prime, 0.f);
printf("Result = %f\n",sum);
getchar();
}

Optimizing CUDA interpolation

I have developped the following interpolation with CUDA and I am looking for a way of improving this interpolation. For some reasons, I dont want to use CUDA textures.
The other point that I have noticed that for some unknown reasons, is that the interpolation is not performed on the whole vector in my case if the size of the vector is superior than the number of threads (for example with a vector of size 1000, and a number of threads equal to 512,. A thread does its first job and that’s all. I would like to optimize the singleInterp function.
Here is my code:
__device__ float singleInterp(float* data, float x, int lx_data) {
float res = 0;
int i1=0;
int j=lx_data;
int imid;
while (j>i1+1)
{
imid = (int)(i1+j+1)/2;
if (data[imid]<x)
i1=imid;
else
j=imid;
}
if (i1==j)
res = data[i1+lx_data];
else
res =__fmaf_rn( __fdividef(data[j+lx_data]-data[i1+lx_data],(data[j]-data[i1])),x-data[i1], data[i1+lx_data]);
return res;
}
Kernel:
__global__ void linearInterpolation(float* data, float* x_in, int lx_data) {
int i = threadIdx.x + blockDim.x * blockIdx.x;
int index = i;
if (index < lx_data)
x_in[index] = singleInterp(data, x_in[index], lx_data);
}
It seems that you are interested in 1D linear interpolation. I already had the problem of optimizing such a kind of interpolation and I ended up with the following code
__global__ void linear_interpolation_kernel_function_GPU(double* __restrict__ result_d, const double* __restrict__ data_d, const double* __restrict__ x_out_d, const int M, const int N)
{
int j = threadIdx.x + blockDim.x * blockIdx.x;
if(j<N)
{
double reg_x_out = x_out_d[j/2]+M/2;
int k = floor(reg_x_out);
double a = (reg_x_out)-floor(reg_x_out);
double dk = data_d[2*k+(j&1)];
double dkp1 = data_d[2*k+2+(j&1)];
result_d[j] = a * dkp1 + (-dk * a + dk);
}
}
The data are assumed to be sampled at integer nodes between -M/2 and M/2.
The code is "equivalent" to 1D texture interpolation, as explained at the following web-page. For the 1D linear texture interpolation, see Fig. 13 of the CUDA-Programming-Guide. For comparisons betwee different solutions, please see the following thread.

CUDA Thrust and sort_by_key

I’m looking for a sorting algorithm on CUDA that can sort an array A of elements (double) and returns an array of keys B for that array A.
I know the sort_by_key function in the Thrust library but I want my array of elements A to remain unchanged.
What can I do?
My code is:
void sortCUDA(double V[], int P[], int N) {
real_t *Vcpy = (double*) malloc(N*sizeof(double));
memcpy(Vcpy,V,N*sizeof(double));
thrust::sort_by_key(V, V + N, P);
free(Vcpy);
}
i'm comparing the thrust algorithm against others that i have on sequencial cpu
N mergesort sortCUDA
113 0.000008 0.000010
226 0.000018 0.000016
452 0.000036 0.000020
905 0.000061 0.000034
1810 0.000135 0.000071
3621 0.000297 0.000156
7242 0.000917 0.000338
14484 0.001421 0.000853
28968 0.003069 0.001931
57937 0.006666 0.003939
115874 0.014435 0.008025
231749 0.031059 0.016718
463499 0.067407 0.039848
926999 0.148170 0.118003
1853998 0.329005 0.260837
3707996 0.731768 0.544357
7415992 1.638445 1.073755
14831984 3.668039 2.150179
115035495 39.276560 19.812200
230070990 87.750377 39.762915
460141980 200.940501 74.605219
Thrust performance is not bad, but I think if I use OMP can probably get easily a better CPU time
I think this is because to memcpy
SOLUTION:
void thrustSort(double V[], int P[], int N)
{
thrust::device_vector<int> d_P(N);
thrust::device_vector<double> d_V(V, V + N);
thrust::sequence(d_P.begin(), d_P.end());
thrust::sort_by_key(d_V.begin(), d_V.end(), d_P.begin());
thrust::copy(d_P.begin(),d_P.end(),P);
}
where V is a my double values to sort
You can modify comparison operator to sort keys instead of values. #Robert Crovella correctly pointed that a raw device pointer cannot be assigned from the host. The modified algorithm is below:
struct cmp : public binary_function<int,int,bool>
{
cmp(const double *ptr) : rawA(ptr) { }
__host__ __device__ bool operator()(const int i, const int j) const
{return rawA[i] > rawA[j];}
const double *rawA; // an array in global mem
};
void sortkeys(double *A, int n) {
// move data to the gpu
thrust::device_vector<double> devA(A, A + n);
double *rawA = thrust::raw_pointer_cast(devA.data());
thrust::device_vector<int> B(n);
// initialize keys
thrust::sequence(B.begin(), B.end());
thrust::sort(B.begin(), B.end(), cmp(rawA));
// B now contains the sorted keys
}
And here is alternative with arrayfire. Though I am not sure which one is more efficient since arrayfire solution uses two additional arrays:
void sortkeys(double *A, int n) {
af::array devA(n, A, af::afHost);
af::array vals, indices;
// sort and populate vals/indices arrays
af::sort(vals, indices, devA);
std::cout << devA << "\n" << indices << "\n";
}
How large is this array? The most efficient way, in terms of speed, will likely be to just duplicate the original array before sorting, if the memory is available.
Building on the answer provided by #asm (I wasn't able to get it working), this code seemed to work for me, and does sort only the keys. However, I believe it is limited to the case where the keys are in sequence 0, 1, 2, 3, 4 ... corresponding to the (double) values. Since this is a "index-value" sort, it could be extended to the case of an arbitrary sequence of keys, perhaps by doing an indexed copy. However I'm not sure the process of generating the index sequence and then rearranging the original keys will be any faster than just copying the original value data to a new vector (for the case of arbitrary keys).
#include <iostream>
#include <thrust/device_vector.h>
#include <thrust/host_vector.h>
#include <thrust/sort.h>
using namespace std;
__device__ double *rawA; // an array in global mem
struct cmp : public binary_function<int, int, bool>
{
__host__ __device__ bool operator()(const int i, const int j) const
{return ( rawA[i] < rawA[j]);}
};
void sortkeys(double *A, int n) {
// move data to the gpu
thrust::device_vector<double> devA(A, A + n);
// rawA = thrust::raw_pointer_cast(&(devA[0]));
double *test = raw_pointer_cast(devA.data());
cudaMemcpyToSymbol(rawA, &test, sizeof(double *));
thrust::device_vector<int> B(n);
// initialize keys
thrust::sequence(B.begin(), B.end());
thrust::sort(B.begin(), B.end(), cmp());
// B now contains the sorted keys
thrust::host_vector<int> hostB = B;
for (int i=0; i<hostB.size(); i++)
std::cout << hostB[i] << " ";
std::cout<<std::endl;
for (int i=0; i<hostB.size(); i++)
std::cout << A[hostB[i]] << " ";
std::cout<<std::endl;
}
int main(){
double C[] = {0.7, 0.3, 0.4, 0.2, 0.6, 1.2, -0.5, 0.5, 0.0, 10.0};
sortkeys(C, 9);
std::cout << std::endl;
return 0;
}

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