Question

我需要计算

(a & b).count()

在大集合（> 10000）位向量（std::bitset<N>）上，其中N在2 ^ 10到2 ^ 16之间。

const size_t N = 2048;
std::vector<std::vector<char>> distances;
std::vector<std::bitset<N>> bits(100000);
load_from_file(bits);
for(int i = 0; i < bits.size(); i++){
    for(int j = 0; j < bits.size(); j++){
        distance[i][j] = (bits[i] & bits[j]).count();
    }
}

目前我依靠分块多线程和SSE / AVX来计算distances。幸运的是，我可以使用AVX中的vpand来计算&，但我的代码仍在使用popcnt (%rax)和一个循环来计算位数。

有没有办法计算GPU上的(a & b).count()功能（nVidia 760m）？理想情况下，我只会传递2个N位的内存块。我正在寻找使用推力，但我找不到popcnt函数。

编辑：

当前的CPU实施。

double validate_pooled(const size_t K) const{                           
    int right = 0;                                                          
    const size_t num_examples = labels.size();                              
    threadpool tp;                                                          
    std::vector<std::future<bool>> futs;                                    
    for(size_t i = 0; i < num_examples; i++){                               
        futs.push_back(tp.enqueue(&kNN<N>::validate_N, this, i, K));       
    }                                                                       
    for(auto& fut : futs)                                                   
        if(fut.get()) right++;                                              

    return right / (double) num_examples;                                   
}      

bool validate_N(const size_t cmp, const size_t n) const{                    
    const size_t num_examples = labels.size();                              
    std::vector<char> dists(num_examples, -1);                              
    for(size_t i = 0; i < num_examples; i++){                               
        if(i == cmp) continue;                                              
        dists[i] = (bits[cmp] & bits[i]).count();                           

    }                                                                       
    typedef std::unordered_map<std::string,size_t> counter;                 
    counter counts;                                                         
    for(size_t i = 0; i < n; i++){                                          
        auto iter = std::max_element(dists.cbegin(), dists.cend());         
        size_t idx = std::distance(dists.cbegin(), iter);                   
        dists[idx] = -1; // Remove the top result.                          
        counts[labels[idx]] += 1;                                           
    }                                                                       
    auto iter = std::max_element(counts.cbegin(), counts.cend(),            
            [](const counter::value_type& a, const counter::value_type& b){ return a.second < b.second; }); 

    return labels[cmp] == iter->first;;                                     
}

编辑：

这就是我想出来的。然而它的残酷。我不确定我做错了什么

template<size_t N>
struct popl 
{
    typedef unsigned long word_type;
    std::bitset<N> _cmp;

    popl(const std::bitset<N>& cmp) : _cmp(cmp) {}

    __device__
    int operator()(const std::bitset<N>& x) const
    {
        int pop_total = 0;
        #pragma unroll
        for(size_t i = 0; i < N/64; i++)
            pop_total += __popcll(x._M_w[i] & _cmp._M_w[i]);

        return pop_total;
    }
}; 

int main(void) {
    const size_t N = 2048;

    thrust::host_vector<std::bitset<N> > h_vec;
    load_bits(h_vec);

    thrust::device_vector<std::bitset<N> > d_vec = h_vec;
    thrust::device_vector<int> r_vec(h_vec.size(), 0);
    for(int i = 0; i < h_vec.size(); i++){
        r_vec[i] = thrust::transform_reduce(d_vec.cbegin(), d_vec.cend(),  popl<N>(d_vec[i]), 0, thrust::maximum<int>());
    }

    return 0;
}

Answer 1

CUDA对于32位和64位类型都有population count intrinsics。（__popc()和__popcll()）

这些可以直接在CUDA内核中使用，也可以通过推力（在仿函数中）传递给thrust::transform_reduce。

如果这是你想在GPU上做的唯一功能，可能很难获得网络＆＃34; win＆＃34;因为＆＃34;成本＆＃34;向/从GPU传输数据。您的整体输入数据集大小约为1GB（位长度为65536的100000个向量），但根据我的计算，输出数据集的大小似乎为10-40GB（每个结果100000 * 100000 * 1-4个字节）

CUDA内核或推力函数和数据布局应该仔细制作，目的是让代码仅受内存带宽限制。数据传输的成本也可能在很大程度上通过复制和计算操作的重叠来缓解，主要是在输出数据集上。

乍一看，这个问题似乎与计算矢量集之间的欧氏距离的问题有些相似，因此从CUDA的角度来看，this question/answer可能是有意义的。

编辑：添加一些我用来调查此问题的代码。我能够在一个天真的单线程CPU实现上获得显着的加速（大约25倍，包括数据复制时间），但我不知道CPU版本将使用多快的速度＆＃34;分块多线程和SSE / AVX＆＃34;，所以看到更多的实现或获得一些性能数字会很有趣。我也不认为我在这里的CUDA代码是高度优化的，它只是一个＆＃34;第一次切割＆＃34;。

在这种情况下，为了概念验证，我专注于一个小问题大小，N = 2048,10000个位集。对于这个小问题的大小，我可以在共享内存中适应足够的位集向量，用于＆＃34;小＆＃34; threadblock大小，利用共享内存。因此，必须针对较大的N修改此特定方法。

$ cat t581.cu
#include <iostream>
#include <vector>
#include <bitset>
#include <stdlib.h>
#include <time.h>
#include <sys/time.h>

#define nTPB 128
#define OUT_CHUNK 250
#define N_bits 2048
#define N_vecs 10000
const size_t N = N_bits;

__global__ void comp_dist(unsigned *in, unsigned *out, unsigned numvecs, unsigned start_idx, unsigned end_idx){
  __shared__ unsigned sdata[(N/32)*nTPB];
  int idx = threadIdx.x+blockDim.x*blockIdx.x;
  if (idx < numvecs)
    for (int i = 0; i < (N/32); i++)
      sdata[(i*nTPB)+threadIdx.x] = in[(i*numvecs)+idx];
  __syncthreads();
  int vidx = start_idx;
  if (idx < numvecs)
    while (vidx < end_idx) {
      unsigned sum = 0;
      for (int i = 0; i < N/32; i++)
        sum += __popc(sdata[(i*nTPB)+ threadIdx.x] & in[(i*numvecs)+vidx]);
      out[((vidx-start_idx)*numvecs)+idx] = sum;
      vidx++;}
}

void cpu_test(std::vector<std::bitset<N> > &in, std::vector<std::vector<unsigned> > &out){

  for (int i=0; i < in.size(); i++)
    for (int j=0; j< in.size(); j++)
      out[i][j] = (in[i] & in[j]).count();
}

int check_data(unsigned *d1, unsigned start_idx, std::vector<std::vector<unsigned> > &d2){
  for (int i = start_idx; i < start_idx+OUT_CHUNK; i++)
    for (int j = 0; j<N_vecs; j++)
      if (d1[((i-start_idx)*N_vecs)+j] != d2[i][j]) {std::cout << "mismatch at " << i << "," << j << " was: " << d1[((i-start_idx)*N_vecs)+j] << " should be: " << d2[i][j] << std::endl;  return 1;}
  return 0;
}

unsigned long long get_time_usec(){
  timeval tv;
  gettimeofday(&tv, 0);
  return (unsigned long long)(((unsigned long long)tv.tv_sec*1000000ULL)+(unsigned long long)tv.tv_usec);
}

int main(){

  unsigned long long t1, t2;
  std::vector<std::vector<unsigned> > distances;
  std::vector<std::bitset<N> > bits;

  for (int i = 0; i < N_vecs; i++){
    std::vector<unsigned> dist_row(N_vecs, 0);
    distances.push_back(dist_row);
    std::bitset<N> data;
    for (int j =0; j < N; j++) data[j] = rand() & 1;
    bits.push_back(data);}
  t1 = get_time_usec();
  cpu_test(bits, distances);
  t1 = get_time_usec() - t1;
  unsigned *h_data = new unsigned[(N/32)*N_vecs];
  memset(h_data, 0, (N/32)*N_vecs*sizeof(unsigned));
  for (int i = 0; i < N_vecs; i++)
    for (int j = 0; j < N; j++)
        if (bits[i][j]) h_data[(i)+((j/32)*N_vecs)] |= 1U<<(31-(j&31));

  unsigned *d_in, *d_out1, *d_out2, *h_out1, *h_out2;
  cudaMalloc(&d_in, (N/32)*N_vecs*sizeof(unsigned));
  cudaMalloc(&d_out1, N_vecs*OUT_CHUNK*sizeof(unsigned));
  cudaMalloc(&d_out2, N_vecs*OUT_CHUNK*sizeof(unsigned));
  cudaStream_t stream1, stream2;
  cudaStreamCreate(&stream1);
  cudaStreamCreate(&stream2);
  h_out1 = new unsigned[N_vecs*OUT_CHUNK];
  h_out2 = new unsigned[N_vecs*OUT_CHUNK];
  t2 = get_time_usec();
  cudaMemcpy(d_in, h_data, (N/32)*N_vecs*sizeof(unsigned), cudaMemcpyHostToDevice);
  for (int i = 0; i < N_vecs; i += 2*OUT_CHUNK){
    comp_dist<<<(N_vecs + nTPB - 1)/nTPB, nTPB, 0, stream1>>>(d_in, d_out1, N_vecs, i, i+OUT_CHUNK);
    cudaStreamSynchronize(stream2);
    if (i > 0) if (check_data(h_out2, i-OUT_CHUNK, distances)) return 1;
    comp_dist<<<(N_vecs + nTPB - 1)/nTPB, nTPB, 0, stream2>>>(d_in, d_out2, N_vecs, i+OUT_CHUNK, i+2*OUT_CHUNK);
    cudaMemcpyAsync(h_out1, d_out1, N_vecs*OUT_CHUNK*sizeof(unsigned), cudaMemcpyDeviceToHost, stream1);
    cudaMemcpyAsync(h_out2, d_out2, N_vecs*OUT_CHUNK*sizeof(unsigned), cudaMemcpyDeviceToHost, stream2);
    cudaStreamSynchronize(stream1);
    if (check_data(h_out1, i, distances)) return 1;
    }
  cudaDeviceSynchronize();
  t2 = get_time_usec() - t2;
  std::cout << "cpu time: " << ((float)t1)/(float)1000 << "ms gpu time: " << ((float) t2)/(float)1000 << "ms" << std::endl;
  return 0;
}
$ nvcc -O3 -arch=sm_20 -o t581 t581.cu
$ ./t581
cpu time: 20324.1ms gpu time: 753.76ms
$

CUDA 6.5，Fedora20，Xeon X5560，Quadro5000（cc2.0）GPU。上述测试用例包括在CPU上产生的距离数据与GPU之间的结果验证。我还将其分解为一个分块算法，结果数据传输（和验证）与计算操作重叠，使其更容易扩展到存在大量输出数据（例如100000个位集）的情况。然而，我还没有通过探查器实现这一点。

编辑2：这是＆＃34; Windows版本＆＃34;代码：

#include <iostream>
#include <vector>
#include <bitset>
#include <stdlib.h>
#include <time.h>


#define nTPB 128
#define OUT_CHUNK 250
#define N_bits 2048
#define N_vecs 10000
const size_t N = N_bits;

#define cudaCheckErrors(msg) \
    do { \
        cudaError_t __err = cudaGetLastError(); \
        if (__err != cudaSuccess) { \
            fprintf(stderr, "Fatal error: %s (%s at %s:%d)\n", \
                msg, cudaGetErrorString(__err), \
                __FILE__, __LINE__); \
            fprintf(stderr, "*** FAILED - ABORTING\n"); \
            exit(1); \
        } \
    } while (0)



__global__ void comp_dist(unsigned *in, unsigned *out, unsigned numvecs, unsigned start_idx, unsigned end_idx){
  __shared__ unsigned sdata[(N/32)*nTPB];
  int idx = threadIdx.x+blockDim.x*blockIdx.x;
  if (idx < numvecs)
    for (int i = 0; i < (N/32); i++)
      sdata[(i*nTPB)+threadIdx.x] = in[(i*numvecs)+idx];
  __syncthreads();
  int vidx = start_idx;
  if (idx < numvecs)
    while (vidx < end_idx) {
      unsigned sum = 0;
      for (int i = 0; i < N/32; i++)
        sum += __popc(sdata[(i*nTPB)+ threadIdx.x] & in[(i*numvecs)+vidx]);
      out[((vidx-start_idx)*numvecs)+idx] = sum;
      vidx++;}
}

void cpu_test(std::vector<std::bitset<N> > &in, std::vector<std::vector<unsigned> > &out){

  for (unsigned i=0; i < in.size(); i++)
    for (unsigned j=0; j< in.size(); j++)
      out[i][j] = (in[i] & in[j]).count();
}

int check_data(unsigned *d1, unsigned start_idx, std::vector<std::vector<unsigned> > &d2){
  for (unsigned i = start_idx; i < start_idx+OUT_CHUNK; i++)
    for (unsigned j = 0; j<N_vecs; j++)
      if (d1[((i-start_idx)*N_vecs)+j] != d2[i][j]) {std::cout << "mismatch at " << i << "," << j << " was: " << d1[((i-start_idx)*N_vecs)+j] << " should be: " << d2[i][j] << std::endl;  return 1;}
  return 0;
}

unsigned long long get_time_usec(){

  return (unsigned long long)((clock()/(float)CLOCKS_PER_SEC)*(1000000ULL));
}

int main(){

  unsigned long long t1, t2;
  std::vector<std::vector<unsigned> > distances;
  std::vector<std::bitset<N> > bits;

  for (int i = 0; i < N_vecs; i++){
    std::vector<unsigned> dist_row(N_vecs, 0);
    distances.push_back(dist_row);
    std::bitset<N> data;
    for (int j =0; j < N; j++) data[j] = rand() & 1;
    bits.push_back(data);}
  t1 = get_time_usec();
  cpu_test(bits, distances);
  t1 = get_time_usec() - t1;
  unsigned *h_data = new unsigned[(N/32)*N_vecs];
  memset(h_data, 0, (N/32)*N_vecs*sizeof(unsigned));
  for (int i = 0; i < N_vecs; i++)
    for (int j = 0; j < N; j++)
        if (bits[i][j]) h_data[(i)+((j/32)*N_vecs)] |= 1U<<(31-(j&31));

  unsigned *d_in, *d_out1, *d_out2, *h_out1, *h_out2;
  cudaMalloc(&d_in, (N/32)*N_vecs*sizeof(unsigned));
  cudaMalloc(&d_out1, N_vecs*OUT_CHUNK*sizeof(unsigned));
  cudaMalloc(&d_out2, N_vecs*OUT_CHUNK*sizeof(unsigned));
  cudaCheckErrors("cudaMalloc fail");
  cudaStream_t stream1, stream2;
  cudaStreamCreate(&stream1);
  cudaStreamCreate(&stream2);
   cudaCheckErrors("cudaStrem fail");
  h_out1 = new unsigned[N_vecs*OUT_CHUNK];
  h_out2 = new unsigned[N_vecs*OUT_CHUNK];
  t2 = get_time_usec();
  cudaMemcpy(d_in, h_data, (N/32)*N_vecs*sizeof(unsigned), cudaMemcpyHostToDevice);
   cudaCheckErrors("cudaMemcpy fail");
  for (int i = 0; i < N_vecs; i += 2*OUT_CHUNK){
    comp_dist<<<(N_vecs + nTPB - 1)/nTPB, nTPB, 0, stream1>>>(d_in, d_out1, N_vecs, i, i+OUT_CHUNK);
    cudaCheckErrors("cuda kernel loop 1 fail");
    cudaStreamSynchronize(stream2);
    if (i > 0) if (check_data(h_out2, i-OUT_CHUNK, distances)) return 1;
    comp_dist<<<(N_vecs + nTPB - 1)/nTPB, nTPB, 0, stream2>>>(d_in, d_out2, N_vecs, i+OUT_CHUNK, i+2*OUT_CHUNK);
    cudaCheckErrors("cuda kernel loop 2 fail");
    cudaMemcpyAsync(h_out1, d_out1, N_vecs*OUT_CHUNK*sizeof(unsigned), cudaMemcpyDeviceToHost, stream1);
    cudaMemcpyAsync(h_out2, d_out2, N_vecs*OUT_CHUNK*sizeof(unsigned), cudaMemcpyDeviceToHost, stream2);
    cudaCheckErrors("cuda kernel loop 3 fail");
    cudaStreamSynchronize(stream1);
    if (check_data(h_out1, i, distances)) return 1;
    }
  cudaDeviceSynchronize();
  cudaCheckErrors("cuda kernel loop 4 fail");
  t2 = get_time_usec() - t2;
  std::cout << "cpu time: " << ((float)t1)/(float)1000 << "ms gpu time: " << ((float) t2)/(float)1000 << "ms" << std::endl;
  return 0;
}

我已将CUDA错误检查添加到此代码中。确保在Visual Studio中构建发布项目，而不是调试。当我在配备Quadro1000M GPU的Windows 7笔记本电脑上运行时，CPU执行大约需要35秒，GPU需要大约1.5秒。

Answer 2

OpenCL 1.2有popcount，它似乎可以做你想要的。它可以在向量上工作，因此最多为ulong16，一次为1024位。请注意，NVIDIA驱动程序仅支持不包含此功能的OpenCL 1.1。

当然，您可以使用函数或表来快速计算它，因此OpenCL 1.1实现也是可行的，并且很可能在设备的内存带宽下运行。

在GPU上使用popcnt

编辑：

编辑：

2 个答案: