Implementing Max Reduce in Cuda - cuda

I've been learning Cuda and I am still getting to grips with parallelism. The problem I am having at the moment is implementing a max reduce on an array of values. This is my kernel
__global__ void max_reduce(const float* const d_array,
float* d_max,
const size_t elements)
{
extern __shared__ float shared[];
int tid = threadIdx.x;
int gid = (blockDim.x * blockIdx.x) + tid;
if (gid < elements)
shared[tid] = d_array[gid];
__syncthreads();
for (unsigned int s=blockDim.x/2; s>0; s>>=1)
{
if (tid < s && gid < elements)
shared[tid] = max(shared[tid], shared[tid + s]);
__syncthreads();
}
if (gid == 0)
*d_max = shared[tid];
}
I have implemented a min reduce using the same method (replacing the max function with the min) which works fine.
To test the kernel, I found the min and max values using a serial for loop. The min and max values always come out the same in the kernel but only the min reduce matches up.
Is there something obvious I'm missing/doing wrong?

Your main conclusion in your deleted answer was correct: the kernel you have posted doesn't comprehend the fact that at the end of that kernel execution, you have done a good deal of the overall reduction, but the results are not quite complete. The results of each block must be combined (somehow). As pointed out in the comments, there are a few other issues with your code as well. Let's take a look at a modified version of it:
__device__ float atomicMaxf(float* address, float val)
{
int *address_as_int =(int*)address;
int old = *address_as_int, assumed;
while (val > __int_as_float(old)) {
assumed = old;
old = atomicCAS(address_as_int, assumed,
__float_as_int(val));
}
return __int_as_float(old);
}
__global__ void max_reduce(const float* const d_array, float* d_max,
const size_t elements)
{
extern __shared__ float shared[];
int tid = threadIdx.x;
int gid = (blockDim.x * blockIdx.x) + tid;
shared[tid] = -FLOAT_MAX; // 1
if (gid < elements)
shared[tid] = d_array[gid];
__syncthreads();
for (unsigned int s=blockDim.x/2; s>0; s>>=1)
{
if (tid < s && gid < elements)
shared[tid] = max(shared[tid], shared[tid + s]); // 2
__syncthreads();
}
// what to do now?
// option 1: save block result and launch another kernel
if (tid == 0)
d_max[blockIdx.x] = shared[tid]; // 3
// option 2: use atomics
if (tid == 0)
atomicMaxf(d_max, shared[0]);
}
As Pavan indicated, you need to initialize your shared memory array. The last block launched may not be a "full" block, if gridDim.x*blockDim.x is greater than elements.
Note that in this line, even though we are checking that the thread operating (gid) is less than elements, when we add s to gid for indexing into the shared memory we can still index outside of the legitimate values copied into shared memory, in the last block. Therefore we need the shared memory initialization indicated in note 1.
As you already discovered, your last line was not correct. Each block produces it's own result, and we must combine them somehow. One method you might consider if the number of blocks launched is small (more on this later) is to use atomics. Normally we steer people away from using atomics since they are "costly" in terms of execution time. However, the other option we are faced with is saving the block result in global memory, finishing the kernel, and then possibly launching another kernel to combine the individual block results. If I have launched a large number of blocks initially (say more than 1024) then if I follow this methodology I might end up launching two additional kernels. Thus the consideration of atomics. As indicated, there is no native atomicMax function for floats, but as indicated in the documentation, you can use atomicCAS to generate any arbitrary atomic function, and I have provided an example of that in atomicMaxf which provides an atomic max for float.
But is running 1024 or more atomic functions (one per block) the best way? Probably not.
When launching kernels of threadblocks, we really only need to launch enough threadblocks to keep the machine busy. As a rule of thumb we want at least 4-8 warps operating per SM, and somewhat more is probably a good idea. But there's no particular benefit from a machine utilization standpoint to launch thousands of threadblocks initially. If we pick a number like 8 threadblocks per SM, and we have at most, say, 14-16 SMs in our GPU, this gives us a relatively small number of 8*14 = 112 threadblocks. Let's choose 128 (8*16) for a nice round number. There's nothing magical about this, it's just enough to keep the GPU busy. If we make each of these 128 threadblocks do additional work to solve the whole problem, we can then leverage our use of atomics without (perhaps) paying too much of a penalty for doing so, and avoid multiple kernel launches. So how would this look?:
__device__ float atomicMaxf(float* address, float val)
{
int *address_as_int =(int*)address;
int old = *address_as_int, assumed;
while (val > __int_as_float(old)) {
assumed = old;
old = atomicCAS(address_as_int, assumed,
__float_as_int(val));
}
return __int_as_float(old);
}
__global__ void max_reduce(const float* const d_array, float* d_max,
const size_t elements)
{
extern __shared__ float shared[];
int tid = threadIdx.x;
int gid = (blockDim.x * blockIdx.x) + tid;
shared[tid] = -FLOAT_MAX;
while (gid < elements) {
shared[tid] = max(shared[tid], d_array[gid]);
gid += gridDim.x*blockDim.x;
}
__syncthreads();
gid = (blockDim.x * blockIdx.x) + tid; // 1
for (unsigned int s=blockDim.x/2; s>0; s>>=1)
{
if (tid < s && gid < elements)
shared[tid] = max(shared[tid], shared[tid + s]);
__syncthreads();
}
if (tid == 0)
atomicMaxf(d_max, shared[0]);
}
With this modified kernel, when creating the kernel launch, we are not deciding how many threadblocks to launch based on the overall data size (elements). Instead we are launching a fixed number of blocks (say, 128, you can modify this number to find out what runs fastest), and letting each threadblock (and thus the entire grid) loop through memory, computing partial max operations on each element in shared memory. Then, in the line marked with comment 1, we must re-set the gid variable to it's initial value. This is actually unnecessary and the block reduction loop code can be further simplified if we guarantee that the size of the grid (gridDim.x*blockDim.x) is less than elements, which is not difficult to do at kernel launch.
Note that when using this atomic method, it's necessary to initialize the result (*d_max in this case) to an appropriate value, like -FLOAT_MAX.
Again, we normally steer people way from atomic usage, but in this case, it's worth considering if we carefully manage it, and it allows us to save the overhead of an additional kernel launch.
For a ninja-level analysis of how to do fast parallel reductions, take a look at Mark Harris' excellent whitepaper which is available with the relevant CUDA sample.

Here's one that appears naive but isn't. This won't generalize to other functions like sum(), but it works great for min() and max().
__device__ const float float_min = -3.402e+38;
__global__ void maxKernel(float* d_data)
{
// compute max over all threads, store max in d_data[0]
int i = threadIdx.x;
__shared__ float max_value;
if (i == 0) max_value = float_min;
float v = d_data[i];
__syncthreads();
while (max_value < v) max_value = v;
__syncthreads();
if (i == 0) d_data[0] = max_value;
}
Yup, that's right, only syncing once after initialization and once before writing the result. Damn the race conditions! Full speed ahead!
Before you tell me it won't work, please give it a try first. I have tested thoroughly and it works every time on a variety of arbitrary kernel sizes. It turns out that the race condition doesn't matter in this case because the while loop resolves it.
It works significantly faster than a conventional reduction. Another surprise is that the average number of passes for a kernel size of 32 is 4. Yup, that's (log(n)-1), which seems counterintuitive. It's because the race condition gives an opportunity for good luck. This bonus comes in addition to removing the overhead of the conventional reduction.
With larger n, there is no way to avoid at least one iteration per warp, but that iteration only involves one compare operation which is usually immediately false across the warp when max_value is on the high end of the distribution. You could modify it to use multiple SM's, but that would greatly increase the total workload and add a communication cost, so not likely to help.
For terseness I've omitted the size and output arguments. Size is simply the number of threads (which could be 137 or whatever you like). Output is returned in d_data[0].
I've uploaded the working file here: https://github.com/kenseehart/YAMR

Related

why unroll can not accelerate transpose matrix?

I am following a tutorial to learn cuda now and I learn that unroll a kernel function will accelerate the program. And it indeed works when I write a function which used to summarize a array.
But when I write a function used to transpose matrix following tutorial, it dosen't work.
The origin function like below:
__global__ void transform_matrix_read_col(
int* mat_a , int* mat_b , size_t row_num , size_t col_num
){
int ix = threadIdx.x + blockDim.x * blockIdx.x;
int iy = threadIdx.y + blockDim.y * blockIdx.y;
int row_idx = iy*col_num + ix;
int col_idx = ix*row_num + iy;
if(ix < col_num && iy < row_num){
mat_b[row_idx] = mat_a[col_idx];
}
}
and unrool function:
__global__ void transform_matrix_read_col_unrool(
int* mat_a , int* mat_b , size_t row_num , size_t col_num
){
int ix = threadIdx.x +(blockDim.x * blockIdx.x * 4);
int iy = threadIdx.y + blockDim.y * blockIdx.y;
int row_idx = iy*col_num + ix;
int col_idx = ix*row_num + iy;
if(ix < col_num && iy < row_num){
mat_b[row_idx] = mat_a[col_idx];
mat_b[row_idx + blockDim.x*1] = mat_a[col_idx + row_num*blockDim.x*1];
mat_b[row_idx + blockDim.x*2] = mat_a[col_idx + row_num*blockDim.x*2];
mat_b[row_idx + blockDim.x*3] = mat_a[col_idx + row_num*blockDim.x*3];
}
}
and the main function:
size_t width = 128 , height = 128,
array_size = width*height,array_bytes = array_size * sizeof(int);
int* matrix_data = nullptr,*output_data = nullptr;
cudaMallocHost(&matrix_data, array_bytes);
cudaMallocHost(&output_data, array_bytes);
util::init_array_int(matrix_data,array_size);//this func will random generate some integer
int* matrix_data_dev = nullptr,* output_matrix_dev = nullptr;
cudaMalloc(&matrix_data_dev, array_bytes);
cudaMemcpy(matrix_data_dev, matrix_data, array_bytes, cudaMemcpyHostToDevice);
cudaMalloc(&output_matrix_dev, array_bytes);
dim3 block(32,16);
dim3 grid((width-1)/block.x+1,(height-1)/block.y+1);
dim3 gridUnrool4((width-1)/(block.x*4)+1,(height-1)/block.y +1);
transform_matrix_read_col<<<grid,block>>>(matrix_data_dev, output_matrix_dev, height, width);
cudaDeviceSynchronize();
transform_matrix_read_col_unrool<<<gridUnrool4,block>>>(matrix_data_dev, output_matrix_dev, height, width);
cudaDeviceSynchronize();
and the staticstis of nsys(run on linux with a rtx 3090):
CUDA Kernel Statistics:
Time(%) Total Time (ns) Instances Average Minimum Maximum Name
------- --------------- --------- -------- ------- ------- ---------------------------------------------------------------------------
6.3 3,456 1 3,456.0 3,456 3,456 transform_matrix_read_col_unrool(int*, int*, unsigned long, unsigned long)
5.2 2,880 1 2,880.0 2,880 2,880 transform_matrix_read_col(int*, int*, unsigned long, unsigned long)
We can see that unrool version slower a lot.
But on the tutorial , it say that unroll will acclerate transpose actually.
So What cause this problem? And how to accelerate transpose matrix ?
Unrolling only help if the computation is compute bound so that a higher (useful) instruction throughput can decrease the execution time. Memory-bound code tends not to be much faster once unrolled because memory-bound instruction are slowed down by the contention of the memory controller.
A transposition may not seem memory-bound at first glance because of a low apparent memory throughput, but one need to care about cache lines. Indeed, when a single value is requested from memory from the user code, the hardware actually request a pretty big cache line for (subsequent) contiguous accesses to be fast.
Another consideration to take into account is that the code can also be latency bound. Indeed, the inefficient strided accesses can be slow due to the memory latency. The memory controller may not be able to fully saturate the RAM in this case (although this is quite unlikely on GPUs, especially regarding the large cache lines). If so, adding more instruction do not help because they are typically executed in an in-order way as opposed to modern mainstream CPUs. Using larger blocks and more blocks helps to provide more parallelism to the GPUs which can then perform more concurrent memory accesses and possibly better use the memory.
The key with the transposition is to make accesses as contiguous as possible and reuse cache lines. The most critical thing is to operate on small 2D blocks and not on full row/lines (ie. not a 1D kernel) to increase the cache locality. Moreover, one efficient well-known solution is to use the shared memory: each threads of a CUDA block fetch a part of a 2D array block and can then perform the transposition in shared memory possibly more efficiently. It is not so easy due to possible shared memory conflicts that can impact performance. Fortunately, there are few research papers and articles talking about that since the last decades.
The simplest efficient solution to this problem is basically to use cuBLAS which is heavily optimized. This post may also be useful.
Note that a 128x128 transposition is very small for a GPU. GPUs are designed to compute bigger datasets (or far more expensive computations on such small input). If the input array is initially stored on the CPU, then I strongly advise you to do that directly on the CPU as moving data on the GPU will likely be already slower than computing the transposition efficiently on the CPU directly. Indeed, data cannot be moved faster than the main RAM permit and a 128x128 transposition can be implemented in a way it saturate the main RAM (in fact, it can be likely done directly in the CPU caches that are significantly faster than the main RAM).

How to collect individual results of the threads within a block?

In my Kernel, the threads are processing a small part of an array in global memory.
After processing I would also like to set a flag indicating that the result of the calculation is zero for all threads within a block:
__global__ void kernel( int *a, bool *blockIsNull) {
int tid = blockIdx.x * blockDim.x + threadIdx.x;
int result = 0;
// {...} Here calculate result
a[tid] = result;
// some code here, but I don't know, that's my question...
if (condition)
blockIsNull[blockIdx.x] = true; // if all threads have returned result==0
}
Each individual thread owns the information. But I don't find an efficient way to collect it.
For example, I could have a counter in shared memory that is atomically incremented by each thread when result==0. So when the counter reaches blockDim.x it means that all threads have returned zero. Althought not tested, I am afraid that this solution will have a negative impact on performance (atomic functions are slow).
A zero result does not occur very often, so it is very unlikely to have zeros for all threads within a block. I would like to find a solution that has little impact on the performance in the general case.
What would be your recommendation ?
It sounds like you want to perform a block level reduction of the condition value across a block. Just about all CUDA hardware supports a set of very useful warp voting primitives. You could use the __all() warp vote to determine that each warp of threads satisfied the condition, and then use __all() again to check whether all warps satisfy the condition. In code, it might look like this:
__global__ void kernel( int *a, bool *blockIsNull) {
// assume that threads per block is <= 1024
__shared__ volatile int blockcondition[32];
int laneid = threadIdx.x % 32;
int warpid = threadIdx.x / 32;
// Set each condition value to non zero to begin
if (warpid == 0) {
blockcondition[threadIdx.x] = 1;
}
__syncthreads();
//
// your code goes here
//
// warpcondition holds the vote from each warp
int warpcondition = __all(condition);
// First thread in each warp loads the warp vote to shared memory
if (laneid == 0) {
blockcondition[warpid] = warpcondition;
}
__syncthreads();
// First warp reduces all the votes in shared memory
if (warpid == 0) {
int result = __all(blockcondition[threadIdx.x] != 0);
// first thread stores the block result to global memory
if (laneid == 0) {
blockIsNull[blockIdx.x] = (result !=0);
}
}
}
[ Huge disclaimer: written in browser, never compiled or tested, use at own risk ]
This code should (I think) work for any number of threads per block up to 1024. You could, if required, adjust the size of blockcondition to a smaller value if you were confident of an upper block size limit less than 1024. Probably the smartest way would be to use C++ templating and make the warp count limit a template parameter.

Efficient method to check for matrix stability in CUDA

A number of algorithms iterate until a certain convergence criterion is reached (e.g. stability of a particular matrix). In many cases, one CUDA kernel must be launched per iteration. My question is: how then does one efficiently and accurately determine whether a matrix has changed over the course of the last kernel call? Here are three possibilities which seem equally unsatisfying:
Writing a global flag each time the matrix is modified inside the kernel. This works, but is highly inefficient and is not technically thread safe.
Using atomic operations to do the same as above. Again, this seems inefficient since in the worst case scenario one global write per thread occurs.
Using a reduction kernel to compute some parameter of the matrix (e.g. sum, mean, variance). This might be faster in some cases, but still seems like overkill. Also, it is possible to dream up cases where a matrix has changed but the sum/mean/variance haven't (e.g. two elements are swapped).
Is there any of the three options above, or an alternative, that is considered best practice and/or is generally more efficient?
I'll also go back to the answer I would have posted in 2012 but for a browser crash.
The basic idea is that you can use warp voting instructions to perform a simple, cheap reduction and then use zero or one atomic operations per block to update a pinned, mapped flag that the host can read after each kernel launch. Using a mapped flag eliminates the need for an explicit device to host transfer after each kernel launch.
This requires one word of shared memory per warp in the kernel, which is a small overhead, and some templating tricks can allow for loop unrolling if you provide the number of warps per block as a template parameter.
A complete working examplate (with C++ host code, I don't have access to a working PyCUDA installation at the moment) looks like this:
#include <cstdlib>
#include <vector>
#include <algorithm>
#include <assert.h>
__device__ unsigned int process(int & val)
{
return (++val < 10);
}
template<int nwarps>
__global__ void kernel(int *inout, unsigned int *kchanged)
{
__shared__ int wchanged[nwarps];
unsigned int laneid = threadIdx.x % warpSize;
unsigned int warpid = threadIdx.x / warpSize;
// Do calculations then check for change/convergence
// and set tchanged to be !=0 if required
int idx = blockIdx.x * blockDim.x + threadIdx.x;
unsigned int tchanged = process(inout[idx]);
// Simple blockwise reduction using voting primitives
// increments kchanged is any thread in the block
// returned tchanged != 0
tchanged = __any(tchanged != 0);
if (laneid == 0) {
wchanged[warpid] = tchanged;
}
__syncthreads();
if (threadIdx.x == 0) {
int bchanged = 0;
#pragma unroll
for(int i=0; i<nwarps; i++) {
bchanged |= wchanged[i];
}
if (bchanged) {
atomicAdd(kchanged, 1);
}
}
}
int main(void)
{
const int N = 2048;
const int min = 5, max = 15;
std::vector<int> data(N);
for(int i=0; i<N; i++) {
data[i] = min + (std::rand() % (int)(max - min + 1));
}
int* _data;
size_t datasz = sizeof(int) * (size_t)N;
cudaMalloc<int>(&_data, datasz);
cudaMemcpy(_data, &data[0], datasz, cudaMemcpyHostToDevice);
unsigned int *kchanged, *_kchanged;
cudaHostAlloc((void **)&kchanged, sizeof(unsigned int), cudaHostAllocMapped);
cudaHostGetDevicePointer((void **)&_kchanged, kchanged, 0);
const int nwarps = 4;
dim3 blcksz(32*nwarps), grdsz(16);
// Loop while the kernel signals it needs to run again
do {
*kchanged = 0;
kernel<nwarps><<<grdsz, blcksz>>>(_data, _kchanged);
cudaDeviceSynchronize();
} while (*kchanged != 0);
cudaMemcpy(&data[0], _data, datasz, cudaMemcpyDeviceToHost);
cudaDeviceReset();
int minval = *std::min_element(data.begin(), data.end());
assert(minval == 10);
return 0;
}
Here, kchanged is the flag the kernel uses to signal it needs to run again to the host. The kernel runs until each entry in the input has been incremented to above a threshold value. At the end of each threads processing, it participates in a warp vote, after which one thread from each warp loads the vote result to shared memory. One thread reduces the warp result and then atomically updates the kchanged value. The host thread waits until the device is finished, and can then directly read the result from the mapped host variable.
You should be able to adapt this to whatever your application requires
I'll go back to my original suggestion. I've updated the related question with an answer of my own, which I believe is correct.
create a flag in global memory:
__device__ int flag;
at each iteration,
initialize the flag to zero (in host code):
int init_val = 0;
cudaMemcpyToSymbol(flag, &init_val, sizeof(int));
In your kernel device code, modify the flag to 1 if a change is made to the matrix:
__global void iter_kernel(float *matrix){
...
if (new_val[i] != matrix[i]){
matrix[i] = new_val[i];
flag = 1;}
...
}
after calling the kernel, at the end of the iteration (in host code), test for modification:
int modified = 0;
cudaMemcpyFromSymbol(&modified, flag, sizeof(int));
if (modified){
...
}
Even if multiple threads in separate blocks or even separate grids, are writing the flag value, as long as the only thing they do is write the same value (i.e. 1 in this case), there is no hazard. The write will not get "lost" and no spurious values will show up in the flag variable.
Testing float or double quantities for equality in this fashion is questionable, but that doesn't seem to be the point of your question. If you have a preferred method to declare "modification" use that instead (such as testing for equality within a tolerance, perhaps).
Some obvious enhancements to this method would be to create one (local) flag variable per thread, and have each thread update the global flag variable once per kernel, rather than on every modification. This would result in at most one global write per thread per kernel. Another approach would be to keep one flag variable per block in shared memory, and have all threads simply update that variable. At the completion of the block, one write is made to global memory (if necessary) to update the global flag. We don't need to resort to complicated reductions in this case, because there is only one boolean result for the entire kernel, and we can tolerate multiple threads writing to either a shared or global variable, as long as all threads are writing the same value.
I can't see any reason to use atomics, or how it would benefit anything.
A reduction kernel seems like overkill, at least compared to one of the optimized approaches (e.g. a shared flag per block). And it would have the drawbacks you mention, such as the fact that anything less than a CRC or similarly complicated computation might alias two different matrix results as "the same".

CUDA efficient division?

I would like to know if there is, by any chance an efficient way of dividing elements of an array. I am running with matrix values 10000x10000 and it a considerable amount of time in comparison with other kernels. Division are expensive operations, and I can't see how to improve it.
__global__ void division(int N, float* A, int* B){
int row = blockIdx.x * blockDim.x + threadIdx.x;
int col = blockIdx.y * blockDim.y + threadIdx.y;
if((row < N) && (col <= row) ){
if( B[row*N+col] >0 )
A[row*N+col] /= (float)B[row*N+col];
}
}
kernel launched with
int N = 10000;
int threads = 32
int blocks = (N+threads-1)/threads
dim3 t(threads,threads);
dim3 b(blocks, blocks);
division<<< b, t >>>(N, A, B);
cudaThreadSynchronize();
Option B:
__global__ void division(int N, float* A, int* B){
int k = blockIdx.x * blockDim.x + threadIdx.x;
int kmax = N*(N+1)/2
int i,j;
if(k< kmax){
row = (int)(sqrt(0.25+2.0*k)-0.5);
col = k - (row*(row+1))>>1;
if( B[row*N+col] >0 )
A[row*N+col] /= (float)B[row*N+col];
}
}
launched with
int threads =192;
int totalThreadsNeeded = (N*(N+1)/2;
int blocks = ( threads + (totalThreadsNeeded)-1 )/threads;
division<<<blocks, threads >>>(N, A, B);
Why is option B giving a wrong result even if the threadIds are the correct one? what is missing here?
Your basic problem is that you are launching an improbably huge grid (over 100 million threads for your 10000x10000 array example), and then because of the triangular nature of the access pattern in the kernel, fully half of those threads never do anything productive. So a enormous amount of GPU cycles are being wasted for no particularly good reason. Further, the access pattern you are using isn't allowing coalesced memory access, which is going to further reduce the performance of the threads which are actually doing useful work.
If I understand your problem correctly, the kernel is only performing element-wise division on a lower-triangle of a square array. If this is the case, it could be equally done using something like this:
__global__
void division(int N, float* A, int* B)
{
for(int row=blockIdx.x; row<N; row+=gridDim.x) {
for(int col=threadIdx.x; col<=row; col+=blockDim.x) {
int val = max(1,B[row*N+col]);
A[row*N+col] /= (float)val;
}
}
}
[disclaimer: written in browser, never compiled, never tested, use at own risk]
Here, a one dimension grid is used, with each block computing a row at a time. Threads in a block move along the row, so memory access is coalesced. In comments you mention your GPU is a Tesla C2050. That device only requires 112 blocks of 192 threads each to completely "fill" each of the 14 SM with a full complement of 8 blocks each and the maximum number of concurrent threads per SM. So the launch parameters could be something like:
int N = 10000;
int threads = 192;
int blocks = min(8*14, N);
division<<<blocks, threads>>>(N, A, B);
I would expect this to run considerably faster than your current approach. If numerical accuracy isn't that important, you can probably achieve further speed-up by replacing the division with an approximate reciprocal intrinsic and a floating point multiply.
Because threads are executed in groups of 32, called warps, you are paying for the division for all 32 threads in a warp if both if conditions are true for just one of the threads. If the condition is false for many threads, see if you can filter out the values for which the division is not needed in a separate kernel.
The int to float conversion may itself be slow. If so, you might be able to generate floats directly in your earlier step, and pass B in as an array of floats.
You may be able to generate inverted numbers in the earlier step, where you generate the B array. If so, you can use multiplication instead of division in this kernel. (a / b == a * 1 / b).
Depending on your algorithm, maybe you can get away with a lower precision division. There's an intrinsic, __fdividef(x, y), that you can try. There is also a compiler flag, -prec-div=false.
The very first thing to look at should be coalesced memory access. There is no reason for the non-coalesced pattern here, just exchange rows and columns for to avoid wasting a lot of memory bandwidth:
int col = blockIdx.x * blockDim.x + threadIdx.x;
int row = blockIdx.y * blockDim.y + threadIdx.y;
...
A[row*N+col] ...
Even if this is run on compute capability 2.0 or higher, the caches are not large enough to remedy this suboptimal pattern.

Parallel Reduction in CUDA for calculating primes

I have a code to calculate primes which I have parallelized using OpenMP:
#pragma omp parallel for private(i,j) reduction(+:pcount) schedule(dynamic)
for (i = sqrt_limit+1; i < limit; i++)
{
check = 1;
for (j = 2; j <= sqrt_limit; j++)
{
if ( !(j&1) && (i&(j-1)) == 0 )
{
check = 0;
break;
}
if ( j&1 && i%j == 0 )
{
check = 0;
break;
}
}
if (check)
pcount++;
}
I am trying to port it to GPU, and I would want to reduce the count as I did for the OpenMP example above. Following is my code, which apart from giving incorrect results is also slower:
__global__ void sieve ( int *flags, int *o_flags, long int sqrootN, long int N)
{
long int gid = blockIdx.x*blockDim.x+threadIdx.x, tid = threadIdx.x, j;
__shared__ int s_flags[NTHREADS];
if (gid > sqrootN && gid < N)
s_flags[tid] = flags[gid];
else
return;
__syncthreads();
s_flags[tid] = 1;
for (j = 2; j <= sqrootN; j++)
{
if ( gid%j == 0 )
{
s_flags[tid] = 0;
break;
}
}
//reduce
for(unsigned int s=1; s < blockDim.x; s*=2)
{
if( tid % (2*s) == 0 )
{
s_flags[tid] += s_flags[tid + s];
}
__syncthreads();
}
//write results of this block to the global memory
if (tid == 0)
o_flags[blockIdx.x] = s_flags[0];
}
First of all, how do I make this kernel fast, I think the bottleneck is the for loop, and I am not sure how to replace it. And next, my counts are not correct. I did change the '%' operator and noticed some benefit.
In the flags array, I have marked the primes from 2 to sqroot(N), in this kernel I am calculating primes from sqroot(N) to N, but I would need to check whether each number in {sqroot(N),N} is divisible by primes in {2,sqroot(N)}. The o_flags array stores the partial sums for each block.
EDIT: Following the suggestion, I modified my code (I understand about the comment on syncthreads now better); I realized that I do not need the flags array and just the global indexes work in my case. What concerns me at this point is the slowness of the code (more than correctness) that could be attributed to the for loop. Also, after a certain data size (100000), the kernel was producing incorrect results for subsequent data sizes. Even for data sizes less than 100000, the GPU reduction results are incorrect (a member in the NVidia forum pointed out that that may be because my data size is not of a power of 2).
So there are still three (may be related) questions -
How could I make this kernel faster? Is it a good idea to use shared memory in my case where I have to loop over each tid?
Why does it produce correct results only for certain data sizes?
How could I modify the reduction?
__global__ void sieve ( int *o_flags, long int sqrootN, long int N )
{
unsigned int gid = blockIdx.x*blockDim.x+threadIdx.x, tid = threadIdx.x;
volatile __shared__ int s_flags[NTHREADS];
s_flags[tid] = 1;
for (unsigned int j=2; j<=sqrootN; j++)
{
if ( gid % j == 0 )
s_flags[tid] = 0;
}
__syncthreads();
//reduce
reduce(s_flags, tid, o_flags);
}
While I profess to know nothing about sieving for primes, there are a host of correctness problems in your GPU version which will stop it from working correctly irrespective of whether the algorithm you are implementing is correct or not:
__syncthreads() calls must be unconditional. It is incorrect to write code where branch divergence could leave some threads within the same warp unable to execute a __syncthreads() call. The underlying PTX is bar.sync and the PTX guide says this:
Barriers are executed on a per-warp basis as if all the threads in a
warp are active. Thus, if any thread in a warp executes a bar
instruction, it is as if all the threads in the warp have executed the
bar instruction. All threads in the warp are stalled until the barrier
completes, and the arrival count for the barrier is incremented by the
warp size (not the number of active threads in the warp). In
conditionally executed code, a bar instruction should only be used if
it is known that all threads evaluate the condition identically (the
warp does not diverge). Since barriers are executed on a per-warp
basis, the optional thread count must be a multiple of the warp size.
Your code unconditionally sets s_flags to one after conditionally loading some values from global memory. Surely that cannot be the intent of the code?
The code lacks a synchronization barrier between the sieving code and the reduction, this can lead to a shared memory race and incorrect results from the reduction.
If you are planning on running this code on a Fermi class card, the shared memory array should be declared volatile to prevent compiler optimization from potentially breaking the shared memory reduction.
If you fix those things, the code might work. Performance is a completely different issue. Certainly on older hardware, the integer modulo operation was very, very slow and not recommended. I can recall reading some material suggesting that Sieve of Atkin was a useful approach to fast prime generation on GPUs.