計算在兩個不同集合在CUDA點之間的所有點對距離可以通過觀察來解決該
||x-y||^2=||x||^2+||y||^2-2*<x,y>
其中|| ||
是l2
規範和<x,y>
表示x
和y
之間的標量積。
的規範||x||
和||y||
可以通過Reduce matrix rows with CUDA啓發的方法來計算,而標量積<x,y>
然後可以如使用cublas<t>gemm()
的矩陣的矩陣乘法X*Y^T
來計算。
下面是一個完整的實現。請注意,爲了計算規範|| ||
,報告了兩種方法,一種使用cuBLAS
cublas<t>gemv
,另一種使用Thurst的transform
。對於問題大小的興趣的,我已經經歷我的GT540M上以下時序:
Approach nr. 1 0.12ms
Approach nr. 2 0.59ms
include <cublas_v2.h>
#include <thrust/host_vector.h>
#include <thrust/device_vector.h>
#include <thrust/generate.h>
#include <thrust/reduce.h>
#include <thrust/functional.h>
#include <thrust/random.h>
#include <thrust/sequence.h>
#include <stdio.h>
#include <iostream>
#include "Utilities.cuh"
#include "TimingGPU.cuh"
#define BLOCK_SIZE_X 16
#define BLOCK_SIZE_Y 16
/***********************************************************/
/* SQUARED ABSOLUTE VALUE FUNCTOR - NEEDED FOR APPROACH #1 */
/***********************************************************/
struct abs2 {
__host__ __device__ double operator()(const float &x) const { return x * x; }
};
// --- Required for approach #2
__device__ float *vals;
/******************************************/
/* ROW_REDUCTION - NEEDED FOR APPROACH #2 */
/******************************************/
struct row_reduction {
const int Ncols; // --- Number of columns
row_reduction(int _Ncols) : Ncols(_Ncols) {}
__device__ float operator()(float& x, int& y) {
float temp = 0.f;
for (int i = 0; i<Ncols; i++)
temp += vals[i + (y*Ncols)] * vals[i + (y*Ncols)];
return temp;
}
};
/************************************************/
/* KERNEL FUNCTION TO ASSEMBLE THE FINAL RESULT */
/************************************************/
__global__ void assemble_final_result(const float * __restrict__ d_norms_x_2, const float * __restrict__ d_norms_y_2, float * __restrict__ d_dots,
const int NX, const int NY) {
const int i = threadIdx.x + blockIdx.x * gridDim.x;
const int j = threadIdx.y + blockIdx.y * gridDim.y;
if ((i < NY) && (j < NX)) d_dots[i * NX+ j] = d_norms_x_2[j] + d_norms_y_2[i] - 2 * d_dots[i * NX+ j];
}
/********/
/* MAIN */
/********/
int main()
{
//const int Ndims = 128; // --- Number of rows
//const int NX = 1000; // --- Number of columns
//const int NY = 2000; // --- Number of columns
const int Ndims = 3; // --- Number of rows
const int NX = 4; // --- Number of columns
const int NY = 5; // --- Number of columns
// --- Random uniform integer distribution between 10 and 99
thrust::default_random_engine rng;
thrust::uniform_int_distribution<int> dist(10, 99);
// --- Matrices allocation and initialization
thrust::device_vector<float> d_X(Ndims * NX);
thrust::device_vector<float> d_Y(Ndims * NY);
for (size_t i = 0; i < d_X.size(); i++) d_X[i] = (float)dist(rng);
for (size_t i = 0; i < d_Y.size(); i++) d_Y[i] = (float)dist(rng);
TimingGPU timerGPU;
// --- cuBLAS handle creation
cublasHandle_t handle;
cublasSafeCall(cublasCreate(&handle));
/**********************************************/
/* CALCULATING THE NORMS OF THE ELEMENTS OF X */
/**********************************************/
thrust::device_vector<float> d_norms_x_2(NX);
// --- Approach nr. 1
//timerGPU.StartCounter();
thrust::device_vector<float> d_X_2(Ndims * NX);
thrust::transform(d_X.begin(), d_X.end(), d_X_2.begin(), abs2());
thrust::device_vector<float> d_ones(Ndims, 1.f);
float alpha = 1.f;
float beta = 0.f;
cublasSafeCall(cublasSgemv(handle, CUBLAS_OP_T, Ndims, NX, &alpha, thrust::raw_pointer_cast(d_X_2.data()), Ndims,
thrust::raw_pointer_cast(d_ones.data()), 1, &beta, thrust::raw_pointer_cast(d_norms_x_2.data()), 1));
//printf("Timing for approach #1 = %f\n", timerGPU.GetCounter());
// --- Approach nr. 2
//timerGPU.StartCounter();
// float *s_vals = thrust::raw_pointer_cast(&d_X[0]);
// gpuErrchk(cudaMemcpyToSymbol(vals, &s_vals, sizeof(float *)));
// thrust::transform(d_norms_x_2.begin(), d_norms_x_2.end(), thrust::counting_iterator<int>(0), d_norms_x_2.begin(), row_reduction(Ndims));
//printf("Timing for approach #2 = %f\n", timerGPU.GetCounter());
/**********************************************/
/* CALCULATING THE NORMS OF THE ELEMENTS OF Y */
/**********************************************/
thrust::device_vector<float> d_norms_y_2(NX);
thrust::device_vector<float> d_Y_2(Ndims * NX);
thrust::transform(d_Y.begin(), d_Y.end(), d_Y_2.begin(), abs2());
cublasSafeCall(cublasSgemv(handle, CUBLAS_OP_T, Ndims, NY, &alpha, thrust::raw_pointer_cast(d_Y_2.data()), Ndims,
thrust::raw_pointer_cast(d_ones.data()), 1, &beta, thrust::raw_pointer_cast(d_norms_y_2.data()), 1));
/***********************************/
/* CALCULATING THE SCALAR PRODUCTS */
/***********************************/
thrust::device_vector<float> d_dots(NX * NY);
cublasSafeCall(cublasSgemm(handle, CUBLAS_OP_T, CUBLAS_OP_N, NX, NY, Ndims, &alpha,
thrust::raw_pointer_cast(d_X.data()), Ndims, thrust::raw_pointer_cast(d_Y.data()), Ndims, &beta,
thrust::raw_pointer_cast(d_dots.data()), NX));
/*****************************/
/* ASSEMBLE THE FINAL RESULT */
/*****************************/
dim3 dimBlock(BLOCK_SIZE_X, BLOCK_SIZE_Y);
dim3 dimGrid(iDivUp(NX, BLOCK_SIZE_X), iDivUp(NY, BLOCK_SIZE_Y));
assemble_final_result<<<dimGrid, dimBlock>>>(thrust::raw_pointer_cast(d_norms_x_2.data()), thrust::raw_pointer_cast(d_norms_y_2.data()),
thrust::raw_pointer_cast(d_dots.data()), NX, NY);
for(int i = 0; i < NX * NY; i++) std::cout << d_dots[i] << "\n";
return 0;
}
的Utilities.cu
和Utilities.cuh
文件將被保留下來here和這裏不再贅述。 TimingGPU.cu
和TimingGPU.cuh
保持爲here並且也被省略。
應該有可能。但是,您已經制作了一個數據存儲佈局,它是一個結構陣列(AoS)。這對GPU的良好性能並不是特別有利(無論是CUDA還是Thrust)。如果你想高效地完成這項工作,你幾乎可以肯定地將你的數據重新排列成與SoA接近的東西。 –
我意識到,一邊看着你的一些其他職位。我現在正在重構。我會更新線程。 – Luca
我認爲你可以注意到:|| || xy ||^2 = || x ||^2 + || y ||^2-2 *',其中''表示' x'和'y'。如果你假定'x'和'y'向量的行主排序爲'X'和'Y'矩陣,那麼你可以使用類似[使用CUDA減少矩陣行](http://stackoverflow.com/questions/) 17862078/reduce-matrix-rows-with-cuda)來計算所有需要的「|| x ||^2」和「|| y ||^2」。然後可以使用'cublas gemm()'將標量積''計算爲矩陣 - 矩陣乘法'X * Y^T'。 –
JackOLantern