Selection Sort in Cuda - sorting

So, I'm trying to implement selection sort in Cuda, but so far I haven't been as successful.
__device__ void selection_sort( int *data, int left, int right ){
for( int i = left ; i <= right ; ++i ){
int min_val = data[i];
int min_idx = i;
// Find the smallest value in the range [left, right].
for( int j = i+1 ; j <= right ; ++j ){
int val_j = data[j];
if( val_j < min_val ){
min_idx = j;
min_val = val_j;
}
}
// Swap the values.
if( i != min_idx ){
data[min_idx] = data[i];
data[i] = min_val;
}
}
}
My main attempt here is to find the minimum and parallelize the solution. Now, I realize the code looks very C++ 'ish but I'm nowhere qualified as skilled in Cuda.
Is there a way to parallelize the solution? Are there any more additions to be made?

Selection sort algorithm for N numbers can be roughly described as:
for i from N-1 down to 0
find the maximum element among data[0] ~ data[i]
swap that maximum element with data[i] within the data array
The first part (finding the maximum element) falls into a widely known and well documented class of problems called reduction. However, to perform the second part (swapping), you must track the index of the maximum element while comparing the values, and it is not so natural to do that while performing reduction. This is one of the reasons why selection sort do not port well to parallel architectures.
Also, you can see that the problem size diminishes by one for each loop, and this is another aspect of the selection sort algorithm that does not map well to parallel architectures. In case of CUDA, 32 threads form a warp, which execute at the same time. Although you can tell arbitrary number of threads to run within a warp, it is generally not recommended to do so because it is a loss of computing power.
I've tried to build a CUDA version of selection sort myself, but I stopped doing it because it seems there are better algorithms well suited for CUDA. But I'll just show you what I've done so far to illustrate why selection sort is not good for CUDA.
Firstly, start from a small and simple problem: sorting 32 elements. Since 32 threads form a warp, you can use shuffle instructions to find maximum value. (Full code)
// Finds the maximum element within a warp and gives the maximum element to
// thread with lane id 0. Note that other elements do not get lost but their
// positions are shuffled.
__inline__ __device__ int warpMax(int data, unsigned int threadId)
{
for (int mask = 16; mask > 0; mask /= 2) {
int dual_data = __shfl_xor(data, mask, 32);
if (threadId & mask)
data = min(data, dual_data);
else
data = max(data, dual_data);
}
return data;
}
__global__ void selection32(int* d_data, int* d_data_sorted)
{
unsigned int threadId = blockIdx.x * blockDim.x + threadIdx.x;
unsigned int laneId = threadIdx.x % 32;
int n = N;
while(n-- > 0) {
// get the maximum element among d_data and put it in d_data_sorted[n]
int data = d_data[threadId];
data = warpMax(data, threadId);
d_data[threadId] = data;
// now maximum element is in d_data[0]
if (laneId == 0) {
d_data_sorted[n] = d_data[0];
d_data[0] = INT_MIN; // this element is ignored from now on
}
}
}
int main()
{
// ... build data and trasfer to d_data ...
selection32<<<1, 32>>>(d_data, d_data_sorted);
// ... get the sorted array stored at d_data_sorted ...
}
(Some may argue that this is not exactly a selection sort since 1) the array elements of the unsorted area keep shuffling, and 2) it is not an in-place sort. Please note that I'm just trying to show that selection sort does not fit in for CUDA. Also, note that warpMax has highly divergent branches, making it less optimal for CUDA.)
The case with only 1 warp of elements may look parallel-ish, but the thing gets worse when the problem size increases to multiple warps. Let's see the case for 1024 elements. (I've chosen the number 1024 becuase it is the maximum number limit of threads in a block.) Now there are 32 warps, and after calling warpMax for each warp, we must compare the maximum elements of each warp to get the maximum element among the 1024 elements. This problem of comparing 32 warp-maximum-values cannot be done with warpMax because we need to track in which warp the maximum value came from to swap the maximum value with the last element in the data array. One way I can think of for doing this is using one single thread to compare warp-maximum-values. This is not a good implemenation for CUDA becuase other 1023 threads in the block become idle.
Furthermore, if the problem size grows larger than a block can cover, we need to compare the maximum values of each block, implying that we will have to launch separate kernels since we need to synchronize between blocks. And it is redundant to say that we need to keep track of in which block the maximum value came from. All of these just tells that implementing selection sort for CUDA is not a good idea.

Related

Are the traditional Fletcher/Adler checksum algorithms optimal?

Are the traditional Fletcher/Adler checksum algorithms optimal?
I am not referring to the common optimizations applied to these
algorithms. For example, the controlling of loop iterations to
avoid sum overflows then truncating outside of the main loop.
I am referring to the design itself. I understand the second sum
(sum2) was introduced to make the algorithm position-dependent but
it is truly sub-optimal? Sum2 is just a modification of sum1 (the
sum of the previous sum1 value added to the current sum1 value).
If we take the 16-bit version as our example, sum1 is a 8-bit
product of the input data, while sum2 is a product of sum1, so
therefore the final 16-bit checksum is in fact an 8-bit product
of the data appended to 16-bits to catch out-of-sequence input.
This means our final checksum, an 8-bit sum of our input data,
is a value of only 256 possible values. As we are passing on
a 16-bit value with our data we could have a checksum value that
is one of 65536 possible values.
It occurred to me that if we had a means of ensuring a position-
dependent check without sacrificing any bits of the checksum we
would have an exponentially better validation.
As I understand it, sum2 was introduced solely to identify out-
of-sequence transmission, and so can be discarded if we find an
alternative to producing a position-dependent checksum. And as
it turns out, we do have an alternative and it comes cheap.
It is cheap because it does not add extra coding to the process
compared to current 'sum2' designs - it is the index position
in the sequence that when hashed with each corresponding byte
ensures a position-dependent check.
One final note - the design below is free of overflow checks,
lossless reduction, and possible loop optimizations, just for
clarity, as this is
about a new out-of-sequence error check technique, not about
implementation details. Fletch64 is not demonstrated below as
it may require a more complicated implementation but the
byte/index hash applies the same.
Revision - because a checksum algorithm can process a large data
count the index position check value could itself cause premature
overflow and require a higher number of reduction operations with
lower inner loop count. The fix was to truncate the index position
check to 8 bits. Now the checksum can process a much greater data
count in the inner loop before overflow.
The only down-side to this change is if a contiguous string of the
data of exactly 256 byte is displaced any multiple of 256 positions
from its original position the error would go undetected.
uint8_t Fletch8( uint8_t *data, int count )
{
uint32_t sum = 0;
int index;
for ( index = 0; index < count; ++index )
sum = sum + ( data[index] xor index & ffh );
return sum & ffh;
}
uint16_t Fletch16( uint8_t *data, int count )
{
uint32_t sum = 0;
int index;
for ( index = 0; index < count; ++index )
sum = sum + ( data[index] xor index & ffh );
return sum & ffffh;
}
uint32_t Fletch32( uint8_t *data, int count )
{
uint64_t sum = 0;
int index;
for ( index = 0; index < count; ++index )
sum = sum + ( data[index] xor index & ffh );
return sum & ffffffffh;
}

Using SIMD to find the biggest difference of two elements

I wrote an algorithm to get the biggest difference between two elements in an std::vector where the bigger of the two values must be at a higher index than the lower value.
unsigned short int min = input.front();
unsigned short res = 0;
for (size_t i = 1; i < input.size(); ++i)
{
if (input[i] <= min)
{
min = input[i];
continue;
}
int dif = input[i] - min;
res = dif > res ? dif : res;
}
return res != 0 ? res : -1;
Is it possible to optimize this algorithm using SIMD? I'm new to SIMD and so far I've been unsuccessful with this one
You didn't specify any particular architecture so I'll keep this mostly architecture neutral with an algorithm described in English. But it requires a SIMD ISA that can efficiently branch on SIMD compare results to check a usually-true condition, like x86 but not really ARM NEON.
This won't work well for NEON because it doesn't have a movemask equivalent, and SIMD -> integer causes stalls on many ARM microarchitectures.
The normal case while looping over the array is that an element, or a whole SIMD vector of elements, is not a new min, and not diff candidate. We can quickly fly through those elements, only slowing down to get the details right when there's a new min. This is like a SIMD strlen or SIMD memcmp, except instead of stopping at the first search hit, we just go scalar for one block and then resume.
For each vector v[0..7] of the input array (assuming 8 int16_t elements per vector (16 bytes), but that's arbitrary):
SIMD compare vmin > v[0..7], and check for any elements being true. (e.g. x86 _mm_cmpgt_epi16 / if(_mm_movemask_epi8(cmp) != 0)) If there's a new min somewhere, we have a special case: the old min applies to some elements, but the new min applies to others. And it's possible there are multiple new-min updates within the vector, and new-diff candidates at any of those points.
So handle this vector with scalar code (updating a scalar diff which doesn't need to be in sync with the vector diffmax because we don't need position).
Broadcast the final min to vmin when you're done. Or do a SIMD horizontal min so out-of-order execution of later SIMD iterations can get started without waiting for a vmin from scalar. Should work well if the scalar code is branchless, so there are no mispredicts in the scalar code that cause later vector work to be thrown out.
As an alternative, a SIMD prefix-sum type of thing (actually prefix-min) could produce a vmin where every element is the min up to that point. (parallel prefix (cumulative) sum with SSE). You could always do this to avoid any branching, but if new-min candidates are rare then it's expensive. Still, it could be viable on ARM NEON where branching is hard.
If there's no new min, SIMD packed max diffmax[0..7] = max(diffmax[0..7], v[0..7]-vmin). (Use saturating subtraction so you don't get wrap-around to a large unsigned difference, if you're using unsigned max to handle the full range.)
At the end of the loop, do a SIMD horizontal max of the diffmax vector. Notice that since we don't need the position of the max-difference, we don't need to update all elements inside the loop when one finds a new candidate. We don't even need to keep the scalar special-case diffmax and SIMD vdiffmax in sync with each other, just check at the end to take the max of the scalar and SIMD max diffs.
SIMD min/max is basically the same as a horizontal sum, except you use packed-max instead of packed-add. For x86, see Fastest way to do horizontal float vector sum on x86.
Or on x86 with SSE4.1 for 16-bit integer elements, phminposuw / _mm_minpos_epu16 can be used for min or max, signed or unsigned, with appropriate tweaks to the input. max = -min(-diffmax). You can treat diffmax as unsigned because it's known to be non-negative, but Horizontal minimum and maximum using SSE shows how to flip the sign bit to range-shift signed to unsigned and back.
We probably get a branch mispredict every time we find a new min candidate, or else we're finding new min candidates too often for this to be efficient.
If new min candidates are expected frequently, using shorter vectors could be good. Or on discovering there's a new-min in a current vector, then use narrower vectors to only go scalar over fewer elements. On x86, you might use bsf (bit-scan forward) to find which element had the first new-min. That gives your scalar code a data dependency on the vector compare-mask, but if the branch to it was mispredicted then the compare-mask will be ready. Otherwise if branch-prediction can somehow find a pattern in which vectors need the scalar fallback, prediction+speculative execution will break that data dependency.
Unfinished / broken (by me) example adapted from #harold's deleted answer of a fully branchless version that constructs a vector of min-up-to-that-element on the fly, for x86 SSE2.
(#harold wrote it with suffix-max instead of min, which is I think why he deleted it. I partially converted it from max to min.)
A branchless intrinsics version for x86 could look something like this. But branchy is probably better unless you expect some kind of slope or trend that makes new min values frequent.
// BROKEN, see FIXME comments.
// converted from #harold's suffix-max version
int broken_unfinished_maxDiffSSE(const std::vector<uint16_t> &input) {
const uint16_t *ptr = input.data();
// construct suffix-min
// find max-diff at the same time
__m128i min = _mm_set_epi32(-1);
__m128i maxdiff = _mm_setzero_si128();
size_t i = input.size();
for (; i >= 8; i -= 8) {
__m128i data = _mm_loadu_si128((const __m128i*)(ptr + i - 8));
// FIXME: need to shift in 0xFFFF, not 0, for min.
// or keep the old data, maybe with _mm_alignr_epi8
__m128i d = data;
// link with suffix
d = _mm_min_epu16(d, _mm_slli_si128(max, 14));
// do suffix-min within block.
d = _mm_min_epu16(d, _mm_srli_si128(d, 2));
d = _mm_min_epu16(d, _mm_shuffle_epi32(d, 0xFA));
d = _mm_min_epu16(d, _mm_shuffle_epi32(d, 0xEE));
max = d;
// update max-diff
__m128i diff = _mm_subs_epu16(data, min); // with saturation to 0
maxdiff = _mm_max_epu16(maxdiff, diff);
}
// horizontal max
maxdiff = _mm_max_epu16(maxdiff, _mm_srli_si128(maxdiff, 2));
maxdiff = _mm_max_epu16(maxdiff, _mm_shuffle_epi32(maxdiff, 0xFA));
maxdiff = _mm_max_epu16(maxdiff, _mm_shuffle_epi32(maxdiff, 0xEE));
int res = _mm_cvtsi128_si32(maxdiff) & 0xFFFF;
unsigned scalarmin = _mm_extract_epi16(min, 7); // last element of last vector
for (; i != 0; i--) {
scalarmin = std::min(scalarmin, ptr[i - 1]);
res = std::max(res, ptr[i - 1] - scalarmin);
}
return res != 0 ? res : -1;
}
We could replace the scalar cleanup with a final unaligned vector, if we handle the overlap between the last full vector min.

warp shuffling to reduction of arrays with any length

I am working on a Cuda kernel which performs vector dot product (A x B). I assumed that the length of each vector is multiple of 32 (32,64, ...) and defined the block size to be equal to the length of the array. Each thread in the block multiplies one element of A to the corresponding element of B (thread i ==>psum = A[i]xB[i]). After multiplication, I used the following functions which used warp shuffling technique to perform reduction and calculate the sum all multiplications.
__inline__ __device__
float warpReduceSum(float val) {
int warpSize =32;
for (int offset = warpSize/2; offset > 0; offset /= 2)
val += __shfl_down(val, offset);
return val;
}
__inline__ __device__
float blockReduceSum(float val) {
static __shared__ int shared[32]; // Shared mem for 32 partial sums
int lane = threadIdx.x % warpSize;
int wid = threadIdx.x / warpSize;
val = warpReduceSum(val); // Each warp performs partial reduction
if (lane==0)
shared[wid]=val; // Write reduced value to shared memory
__syncthreads(); // Wait for all partial reductions
//read from shared memory only if that warp existed
val = (threadIdx.x < blockDim.x / warpSize) ? shared[lane] : 0;
if (wid==0)
val = warpReduceSum(val); // Final reduce within first warp
return val;
}
I simply call blockReduceSum(psum) which psum is the multiplication of two elements by a thread.
This approach doesn't work when the length of the array is not multiple of 32, so my question is, can we change this code so that it also works for any length? or is it impossible because if the length of the array is not multiple of 32, some warps have elements belonging more than one array?
First of all, depending on the GPU you are using, performing dot product with just 1 block will probably not be very efficient (as long as you are not batching several dot products in 1 kernel, each done by a single block).
To answer your question: you can reuse the code you have written by just calling your kernel with the number of threads being the closest multiple of 32 higher than N (length of the array) and introducing if statement before calling to blockReduceSum that would like this:
__global__ void kernel(float * A, float * B, int N) {
float psum = 0;
if(threadIdx.x < N) //threadIDx.x because your are using single block, you will need to change it to more general id once you move to multiple blocks
psum = A[threadIdx.x] * B[threadIdx.x];
blockReduceSum(psum);
//The rest of computation
}
That way, threads that do not have array element associated with them, but that need to be there due to use of __shfl, will contribute 0 to the sum.

Algorithm to match sets with overlapping members

Looking for an efficient algorithm to match sets among a group of sets, ordered by the most overlapping members. 2 identical sets for example are the best match, while no overlapping members are the worst.
So, the algorithm takes input a list of sets and returns matching set pairs ordered by the sets with the most overlapping members.
Would be interested in ideas to do this efficiently. Brute force approach is to try all combinations and sort which obviously is not very performant when the number of sets is very large.
Edit: Use case - Assume a large number of sets already exist. When a new set arrives, the algorithm is run and the output includes matching sets (with at least one element overlap) sorted by the most matching to least (doesn't matter how many items are in the new/incoming set). Hope that clarifies my question.
If you can afford an approximation algorithm with a chance of error, then you should probably consider MinHash.
This algorithm allows estimating the similarity between 2 sets in constant time. For any constructed set, a fixed size signature is computed, and then only the signatures are compared when estimating the similarities. The similarity measure being used is Jaccard distance, which ranges from 0 (disjoint sets) to 1 (identical sets). It is defined as the intersection to union ratio of two given sets.
With this approach, any new set has to be compared against all existing ones (in linear time), and then the results can be merged into the top list (you can use a bounded search tree/heap for this purpose).
Since the number of possible different values is not very large, you get a fairly efficient hashing if you simply set the nth bit in a "large integer" when the nth number is present in your set. You can then look for overlap between sets with a simple bitwise AND followed by a "count set bits" operation. On 64 bit architecture, that means that you can look for the similarity between two numbers (out of 1000 possible values) in about 16 cycles, regardless of the number of values in each cluster. As the cluster gets more sparse, this becomes a less efficient algorithm.
Still - I implemented some of the basic functions you might need in some code that I attach here - not documented but reasonably understandable, I think. In this example I made the numbers small so I can check the result by hand - you might want to change some of the #defines to get larger ranges of values, and obviously you will want some dynamic lists etc to keep up with the growing catalog.
#include <stdio.h>
// biggest number you will come across: want this to be much bigger
#define MAXINT 25
// use the biggest type you have - not int
#define BITSPER (8*sizeof(int))
#define NWORDS (MAXINT/BITSPER + 1)
// max number in a cluster
#define CSIZE 5
typedef struct{
unsigned int num[NWORDS]; // want to use longest type but not for demo
int newmatch;
int rank;
} hmap;
// convert number to binary sequence:
void hashIt(int* t, int n, hmap* h) {
int ii;
for(ii=0;ii<n;ii++) {
int a, b;
a = t[ii]%BITSPER;
b = t[ii]/BITSPER;
h->num[b]|=1<<a;
}
}
// print binary number:
void printBinary(int n) {
unsigned int jj;
jj = 1<<31;
while(jj!=0) {
printf("%c",((n&jj)!=0)?'1':'0');
jj>>=1;
}
printf(" ");
}
// print the array of binary numbers:
void printHash(hmap* h) {
unsigned int ii, jj;
for(ii=0; ii<NWORDS; ii++) {
jj = 1<<31;
printf("0x%08x: ", h->num[ii]);
printBinary(h->num[ii]);
}
//printf("\n");
}
// find the maximum overlap for set m of n
int maxOverlap(hmap* h, int m, int n) {
int ii, jj;
int overlap, maxOverlap = -1;
for(ii = 0; ii<n; ii++) {
if(ii == m) continue; // don't compare with yourself
else {
overlap = 0;
for(jj = 0; jj< NWORDS; jj++) {
// just to see what's going on: take these print statements out
printBinary(h->num[ii]);
printBinary(h->num[m]);
int bc = countBits(h->num[ii] & h->num[m]);
printBinary(h->num[ii] & h->num[m]);
printf("%d bits overlap\n", bc);
overlap += bc;
}
if(overlap > maxOverlap) maxOverlap = overlap;
}
}
return maxOverlap;
}
int countBits (unsigned int b) {
int count;
for (count = 0; b != 0; count++) {
b &= b - 1; // this clears the LSB-most set bit
}
return count;
}
int main(void) {
int cluster[20][CSIZE];
int temp[CSIZE];
int ii,jj;
static hmap H[20]; // make them all 0 initially
for(jj=0; jj<20; jj++){
for(ii=0; ii<CSIZE; ii++) {
temp[ii] = rand()%MAXINT;
}
hashIt(temp, CSIZE, &H[jj]);
}
for(ii=0;ii<20;ii++) {
printHash(&H[ii]);
printf("max overlap: %d\n", maxOverlap(H, ii, 20));
}
}
See if this helps at all...

calculating the number of bits using K&R method with infinite memory

I got answer for the question, counting number of sets bits from here.
How to count the number of set bits in a 32-bit integer?
long count_bits(long n) {
unsigned int c; // c accumulates the total bits set in v
for (c = 0; n; c++)
n &= n - 1; // clear the least significant bit set
return c;
}
It is simple to understand also. And found the best answer as Brian Kernighans method, posted by hoyhoy... and he adds the following at the end.
Note that this is an question used during interviews. The interviewer will add the caveat that you have "infinite memory". In that case, you basically create an array of size 232 and fill in the bit counts for the numbers at each location. Then, this function becomes O(1).
Can somebody explain how to do this ? If i have infinite memory ...
The fastest way I have ever seen to populate such an array is ...
array[0] = 0;
for (i = 1; i < NELEMENTS; i++) {
array[i] = array[i >> 1] + (i & 1);
}
Then to count the number of set bits in a given number (provided the given number is less than NELEMENTS) ...
numSetBits = array[givenNumber];
If your memory is not finite, I often see NELEMENTS set to 256 (for one byte's worth) and add the number of set bits in each byte in your integer.
int counts[MAX_LONG];
void init() {
for (int i= 0; i < MAX_LONG; i++)
{
counts[i] = count_bits[i]; // as given
}
}
int count_bits_o1(long number)
{
return counts[number];
}
You can probably pre-populate the array more wiseley, i.e. fill with zeros, then every second index add one, then every fourth index add 1, then every eighth index add 1 etc, which might be a bit faster, although I doubt it...
Also, you might account for unsigned values.

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