cuda计算两点的距离
cuda calc distance of two points
这里我想计算每两个点的距离，判断它们是否是邻居。这是我在 cuda 中的简单代码。
__global__ void calcNeighbors(const DataPoint* points,
const float doubleRadius, bool* neighbors) {
int tid = threadIdx.x + blockIdx.x * blockDim.x;
float dis = 0.0f;
while (tid < N) {
DataPoint p1 = points[tid];
for (int i=0; i<N; i++) {
DataPoint p2 = points[i];
dis = 0;
dis += (p1.pfDimens[0]-p2.pfDimens[0]) * (p1.pfDimens[0]-p2.pfDimens[0]) +
(p1.pfDimens[1]-p2.pfDimens[1]) * (p1.pfDimens[1]-p2.pfDimens[1]) +
(p1.pfDimens[2]-p2.pfDimens[2]) * (p1.pfDimens[2]-p2.pfDimens[2]);
if (dis <= doubleRadius) {
neighbors[tid*N+i] = true;
} else {
neighbors[tid*N+i] = false;
}
}
tid += blockDim.x * gridDim.x;
}
}
DataPoint 是一个结构是
typedef struct DataPoint {
float pfDimens[3];
} DataPoint;
所以我想减少时间，我该怎么做？我试过使用内存合并和共享内存，但我没有得到很好的加速？
===============使用共享内存==============
__global__ void calcNeighbors2(const DataPoint* points,
const float doubleRadius, bool* neighbors) {
__shared__ DataPoint sharedpoints[threadsPerBlock];
int start = blockIdx.x * blockDim.x;
int len = start+threadIdx.x;
if (len < N) {
sharedpoints[threadIdx.x] = points[len];
}
len = imin(N, blockDim.x + start);
__syncthreads();
int tid = threadIdx.x;
float dis;
while (tid < N) {
DataPoint p1 = points[tid];
for (int i=start; i<len; i++) {
dis = 0;
dis += (p1.pfDimens[0]-sharedpoints[i-start].pfDimens[0]) * (p1.pfDimens[0]-sharedpoints[i-start].pfDimens[0]) +
(p1.pfDimens[1]-sharedpoints[i-start].pfDimens[1]) * (p1.pfDimens[1]-sharedpoints[i-start].pfDimens[1]) +
(p1.pfDimens[2]-sharedpoints[i-start].pfDimens[2]) * (p1.pfDimens[2]-sharedpoints[i-start].pfDimens[2]);
if (dis <= doubleRadius) {
neighbors[i*N+tid] = true;
} else {
neighbors[i*N+tid] = false;
}
}
tid += blockDim.x;
}
}
我在这里将 neighbors[tid*N+i] 更改为 neighbors[i*N+tid]，这让我在 Tesla K10.G2.8GB 上的速度提高了 8 倍。但是当我使用共享内存存储一些点时，它没有用吗？
至少有4个想法，其中一些已经在评论中陈述过：
从 AoS 格式转换您的点距离存储：
struct DataPoint {
float pfDimens[3];
};
转 SoA 格式：
struct DataPoint {
float pfDimens_x[NPTS];
float pfDimens_y[NPTS];
float pfDimens_z[NPTS];
};
这将在加载数据时启用完全合并。事实上，为了帮助解决下面的第 4 点，我只想切换到使用 3 个裸数组，而不是结构。
将计算减少到（略小于）一半：
for (int i=N-1; i>tid; i--) {
然后，无论是在线程代码本身，还是在主机中，您都可以通过复制数据来填充输出矩阵的另一个 "half"。
转置输出矩阵中的存储，这样您就可以编写这样的存储操作：
neighbors[i*N+tid] = true;
这将很好地合并，而不是这个：
neighbors[tid*N+i] = true;
哪个不会。
由于你输入的点数据是只读的，适当标记内核参数：
const float * __restrict__ points_x, const float * __restrict__ points_y, const float * __restrict__ points_z
在某些情况下，在某些 GPU 上，由于使用 the read-only cache，这通常会导致加速。如果你真的想积极使用缓存，并且你的数据数组足够小（4K 或更少 float 点），你可以将点数据的副本放在全局内存中以及 [=21 中的副本=]内存，并通过常量内存加载"uniform"你在这里做的加载：
DataPoint p2 = c_points[i];
因此您可以通过只读缓存执行合并加载，通过常量缓存执行统一加载，以及转到普通全局内存的合并存储。
在 K40c 上，在 linux/CUDA 7 上，对于 N = 4096，这些更改的净效果似乎是内核级别的 3.5 倍加速：
$ cat t749.cu
#include <stdio.h>
#define N 4096
// if N is 16K/3 or less, we can use constant
#define USE_CONSTANT
#define THRESH 0.2f
#define nTPB 256
#define nBLK (N/nTPB+1)
#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)
#include <time.h>
#include <sys/time.h>
#define USECPSEC 1000000ULL
unsigned long long dtime_usec(unsigned long long start){
timeval tv;
gettimeofday(&tv, 0);
return ((tv.tv_sec*USECPSEC)+tv.tv_usec)-start;
}
struct DataPoint {
float pfDimens[3];
};
__global__ void calcNeighbors(const DataPoint* points,
const float doubleRadius, bool* neighbors) {
int tid = threadIdx.x + blockIdx.x * blockDim.x;
float dis = 0.0f;
while (tid < N) {
DataPoint p1 = points[tid];
for (int i=0; i<N; i++) {
DataPoint p2 = points[i];
dis = 0;
dis += (p1.pfDimens[0]-p2.pfDimens[0]) * (p1.pfDimens[0]-p2.pfDimens[0]) +
(p1.pfDimens[1]-p2.pfDimens[1]) * (p1.pfDimens[1]-p2.pfDimens[1]) +
(p1.pfDimens[2]-p2.pfDimens[2]) * (p1.pfDimens[2]-p2.pfDimens[2]);
if (dis <= doubleRadius) {
neighbors[tid*N+i] = true;
} else {
neighbors[tid*N+i] = false;
}
}
tid += blockDim.x * gridDim.x;
}
}
#ifdef USE_CONSTANT
__constant__ float cpx[N];
__constant__ float cpy[N];
__constant__ float cpz[N];
#endif
__global__ void calcNeighbors2(const float * __restrict__ pts_x, const float * __restrict__ pts_y, const float * __restrict__ pts_z, const float doubleRadius, bool * __restrict__ neighbors) {
int tid = threadIdx.x+blockDim.x*blockIdx.x;
while (tid < N) {
float p1x = pts_x[tid];
float p1y = pts_y[tid];
float p1z = pts_z[tid];
for (int i = N-1; i > tid; i--){
float p2x, p2y, p2z;
#ifdef USE_CONSTANT
p2x = cpx[i];
p2y = cpy[i];
p2z = cpz[i];
#else
p2x = pts_x[i];
p2y = pts_y[i];
p2z = pts_z[i];
#endif
float dis = ((p1x-p2x)*(p1x-p2x)) + ((p1y-p2y)*(p1y-p2y)) + ((p1z-p2z)*(p1z-p2z));
neighbors[i*N+tid] = (dis <= doubleRadius);
}
tid += blockDim.x * gridDim.x;
}
}
int main(){
float *dx, *dy, *dz, *hx, *hy, *hz;
DataPoint *dp, *hp;
bool *dn, *hn1, *hn2;
hx =(float *)malloc(N*sizeof(float));
hy =(float *)malloc(N*sizeof(float));
hz =(float *)malloc(N*sizeof(float));
hp =(DataPoint *)malloc(N*sizeof(DataPoint));
hn1=(bool *)malloc(N*N*sizeof(bool));
hn2=(bool *)malloc(N*N*sizeof(bool));
cudaMalloc(&dx, N*sizeof(float));
cudaMalloc(&dy, N*sizeof(float));
cudaMalloc(&dz, N*sizeof(float));
cudaMalloc(&dp, N*sizeof(DataPoint));
cudaMalloc(&dn, N*N*sizeof(bool));
for (int i =0; i < N; i++){
hx[i] = rand()/(float)RAND_MAX;
hy[i] = rand()/(float)RAND_MAX;
hz[i] = rand()/(float)RAND_MAX;
hp[i].pfDimens[0] = hx[i];
hp[i].pfDimens[1] = hy[i];
hp[i].pfDimens[2] = hz[i];}
cudaMemcpy(dx, hx, N*sizeof(float), cudaMemcpyHostToDevice);
cudaMemcpy(dy, hy, N*sizeof(float), cudaMemcpyHostToDevice);
cudaMemcpy(dz, hz, N*sizeof(float), cudaMemcpyHostToDevice);
cudaMemcpy(dp, hp, N*sizeof(DataPoint), cudaMemcpyHostToDevice);
// warm-up
calcNeighbors<<<nBLK, nTPB>>>(dp, THRESH, dn);
cudaDeviceSynchronize();
cudaMemset(dn, 0, N*N*sizeof(bool));
unsigned long long t1 = dtime_usec(0);
calcNeighbors<<<nBLK, nTPB>>>(dp, THRESH, dn);
cudaDeviceSynchronize();
cudaCheckErrors("kernel 1 error");
t1 = dtime_usec(t1);
cudaMemcpy(hn1, dn, N*N*sizeof(bool), cudaMemcpyDeviceToHost);
// warm-up
calcNeighbors2<<<nBLK, nTPB>>>(dx, dy, dz, THRESH, dn);
cudaDeviceSynchronize();
cudaMemset(dn, 0, N*N*sizeof(bool));
unsigned long long t2 = dtime_usec(0);
calcNeighbors2<<<nBLK, nTPB>>>(dx, dy, dz, THRESH, dn);
cudaDeviceSynchronize();
cudaCheckErrors("kernel 2 error");
t2 = dtime_usec(t2);
cudaMemcpy(hn2, dn, N*N*sizeof(bool), cudaMemcpyDeviceToHost);
cudaCheckErrors("some error");
printf("t1: %fs, t2: %fs\n", t1/(float)USECPSEC, t2/(float)USECPSEC);
// results validation
for (int i = 0; i < N; i++)
for (int j = i+1; j < N; j++)
if (hn1[i*N+j] != hn2[j*N+i]) {printf("mismatch at %d, %d, was: %d, should be: %d\n", i, j, hn2[j*N+i], hn1[i*N+j]); return 1;}
return 0;
}
$ nvcc -arch=sm_35 -o t749 t749.cu
$ ./t749
t1: 0.004903s, t2: 0.001395s
$
在 K40c 的情况下，由于延迟，上面启动的块数量有限 (16) 是性能的重大障碍。如果我们注释掉 USE_CONSTANT 定义，并将 N 更改为 16384，我们会观察到改进内核的加速比更高：
$ ./t749
t1: 0.267107s, t2: 0.008209s
$
总计 ~48 个区块足以 "fill" 拥有 15 个 SM 的 K40c。
编辑： 现在您已经发布了一个共享内存内核，我将它作为 calcNeighbors3 添加到我的测试用例中并比较了它的时序性能（如 t3).它几乎和我的内核一样快，而且它似乎提供了正确的结果（与您的原始内核匹配）所以我不确定您的顾虑是什么。
这是更新后的代码和测试用例：
$ cat t749.cu
#include <stdio.h>
#include <math.h>
#define imin(X,Y) ((X)<(Y))?(X):(Y)
#define N 32768
// if N is 16K/3 or less, we can use constant
// #define USE_CONSTANT
#define THRESH 0.2f
#define nTPB 256
#define nBLK (N/nTPB+1)
#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)
#include <time.h>
#include <sys/time.h>
#define USECPSEC 1000000ULL
unsigned long long dtime_usec(unsigned long long start){
timeval tv;
gettimeofday(&tv, 0);
return ((tv.tv_sec*USECPSEC)+tv.tv_usec)-start;
}
struct DataPoint {
float pfDimens[3];
};
__global__ void calcNeighbors(const DataPoint* points,
const float doubleRadius, bool* neighbors) {
int tid = threadIdx.x + blockIdx.x * blockDim.x;
float dis = 0.0f;
while (tid < N) {
DataPoint p1 = points[tid];
for (int i=0; i<N; i++) {
DataPoint p2 = points[i];
dis = 0;
dis += (p1.pfDimens[0]-p2.pfDimens[0]) * (p1.pfDimens[0]-p2.pfDimens[0]) +
(p1.pfDimens[1]-p2.pfDimens[1]) * (p1.pfDimens[1]-p2.pfDimens[1]) +
(p1.pfDimens[2]-p2.pfDimens[2]) * (p1.pfDimens[2]-p2.pfDimens[2]);
if (dis <= doubleRadius) {
neighbors[tid*N+i] = true;
} else {
neighbors[tid*N+i] = false;
}
}
tid += blockDim.x * gridDim.x;
}
}
#ifdef USE_CONSTANT
__constant__ float cpx[N];
__constant__ float cpy[N];
__constant__ float cpz[N];
#endif
__global__ void calcNeighbors2(const float * __restrict__ pts_x, const float * __restrict__ pts_y, const float * __restrict__ pts_z, const float doubleRadius, bool * __restrict__ neighbors) {
int tid = threadIdx.x+blockDim.x*blockIdx.x;
while (tid < N) {
float p1x = pts_x[tid];
float p1y = pts_y[tid];
float p1z = pts_z[tid];
for (int i = N-1; i > tid; i--){
float p2x, p2y, p2z;
#ifdef USE_CONSTANT
p2x = cpx[i];
p2y = cpy[i];
p2z = cpz[i];
#else
p2x = pts_x[i];
p2y = pts_y[i];
p2z = pts_z[i];
#endif
float dis = ((p1x-p2x)*(p1x-p2x)) + ((p1y-p2y)*(p1y-p2y)) + ((p1z-p2z)*(p1z-p2z));
neighbors[i*N+tid] = (dis <= doubleRadius);
}
tid += blockDim.x * gridDim.x;
}
}
__global__ void calcNeighbors3(const DataPoint* points,
const float doubleRadius, bool* neighbors) {
__shared__ DataPoint sharedpoints[nTPB];
int start = blockIdx.x * blockDim.x;
int len = start+threadIdx.x;
if (len < N) {
sharedpoints[threadIdx.x] = points[len];
}
len = imin(N, blockDim.x + start);
__syncthreads();
int tid = threadIdx.x;
float dis;
while (tid < N) {
DataPoint p1 = points[tid];
for (int i=start; i<len; i++) {
dis = 0;
dis += (p1.pfDimens[0]-sharedpoints[i-start].pfDimens[0]) * (p1.pfDimens[0]-sharedpoints[i-start].pfDimens[0]) +
(p1.pfDimens[1]-sharedpoints[i-start].pfDimens[1]) * (p1.pfDimens[1]-sharedpoints[i-start].pfDimens[1]) +
(p1.pfDimens[2]-sharedpoints[i-start].pfDimens[2]) * (p1.pfDimens[2]-sharedpoints[i-start].pfDimens[2]);
if (dis <= doubleRadius) {
neighbors[i*N+tid] = true;
} else {
neighbors[i*N+tid] = false;
}
}
tid += blockDim.x;
}
}
int main(){
float *dx, *dy, *dz, *hx, *hy, *hz;
DataPoint *dp, *hp;
bool *dn, *hn1, *hn2, *hn3;
hx =(float *)malloc(N*sizeof(float));
hy =(float *)malloc(N*sizeof(float));
hz =(float *)malloc(N*sizeof(float));
hp =(DataPoint *)malloc(N*sizeof(DataPoint));
hn1=(bool *)malloc(N*N*sizeof(bool));
hn2=(bool *)malloc(N*N*sizeof(bool));
hn3=(bool *)malloc(N*N*sizeof(bool));
cudaMalloc(&dx, N*sizeof(float));
cudaMalloc(&dy, N*sizeof(float));
cudaMalloc(&dz, N*sizeof(float));
cudaMalloc(&dp, N*sizeof(DataPoint));
cudaMalloc(&dn, N*N*sizeof(bool));
for (int i =0; i < N; i++){
hx[i] = rand()/(float)RAND_MAX;
hy[i] = rand()/(float)RAND_MAX;
hz[i] = rand()/(float)RAND_MAX;
hp[i].pfDimens[0] = hx[i];
hp[i].pfDimens[1] = hy[i];
hp[i].pfDimens[2] = hz[i];}
cudaMemcpy(dx, hx, N*sizeof(float), cudaMemcpyHostToDevice);
cudaMemcpy(dy, hy, N*sizeof(float), cudaMemcpyHostToDevice);
cudaMemcpy(dz, hz, N*sizeof(float), cudaMemcpyHostToDevice);
cudaMemcpy(dp, hp, N*sizeof(DataPoint), cudaMemcpyHostToDevice);
#ifdef USE_CONSTANT
cudaMemcpyToSymbol(cpx, hx, N*sizeof(float));
cudaMemcpyToSymbol(cpy, hy, N*sizeof(float));
cudaMemcpyToSymbol(cpz, hz, N*sizeof(float));
#endif
// warm-up
calcNeighbors<<<nBLK, nTPB>>>(dp, THRESH, dn);
cudaDeviceSynchronize();
cudaMemset(dn, 0, N*N*sizeof(bool));
unsigned long long t1 = dtime_usec(0);
calcNeighbors<<<nBLK, nTPB>>>(dp, THRESH, dn);
cudaDeviceSynchronize();
cudaCheckErrors("kernel 1 error");
t1 = dtime_usec(t1);
cudaMemcpy(hn1, dn, N*N*sizeof(bool), cudaMemcpyDeviceToHost);
// warm-up
calcNeighbors2<<<nBLK, nTPB>>>(dx, dy, dz, THRESH, dn);
cudaDeviceSynchronize();
cudaMemset(dn, 0, N*N*sizeof(bool));
unsigned long long t2 = dtime_usec(0);
calcNeighbors2<<<nBLK, nTPB>>>(dx, dy, dz, THRESH, dn);
cudaDeviceSynchronize();
cudaCheckErrors("kernel 2 error");
t2 = dtime_usec(t2);
cudaMemcpy(hn2, dn, N*N*sizeof(bool), cudaMemcpyDeviceToHost);
// warm-up
calcNeighbors3<<<nBLK, nTPB>>>(dp, THRESH, dn);
cudaDeviceSynchronize();
cudaMemset(dn, 0, N*N*sizeof(bool));
unsigned long long t3 = dtime_usec(0);
calcNeighbors3<<<nBLK, nTPB>>>(dp, THRESH, dn);
cudaDeviceSynchronize();
cudaCheckErrors("kernel 3 error");
t3 = dtime_usec(t3);
cudaMemcpy(hn3, dn, N*N*sizeof(bool), cudaMemcpyDeviceToHost);
cudaCheckErrors("some error");
printf("t1: %fs, t2: %fs, t3: %fs\n", t1/(float)USECPSEC, t2/(float)USECPSEC, t3/(float)USECPSEC);
// results validation
for (int i = 0; i < N; i++)
for (int j = i+1; j < N; j++)
if (hn1[i*N+j] != hn2[j*N+i]) {printf("1:2 mismatch at %d, %d, was: %d, should be: %d\n", i, j, hn2[j*N+i], hn1[i*N+j]); return 1;}
for (int i = 0; i < N*N; i++)
if (hn1[i] != hn3[i]) {printf("1:3 mismatch at %d, was: %d, should be: %d\n", i, hn1[i], hn3[i]); return 1;}
return 0;
}
$ nvcc -arch=sm_35 -o t749 t749.cu
$ ./t749
t1: 1.260010s, t2: 0.022661s, t3: 0.029632s
$
对于此测试，我已将数据集大小更改为 32768，因为这更接近您关心的范围。您的共享内存内核比原始内核显示大约 42 倍的加速，而我的内核在我的 K40c 上显示大约 55 倍的加速。
这里我想计算每两个点的距离，判断它们是否是邻居。这是我在 cuda 中的简单代码。
__global__ void calcNeighbors(const DataPoint* points,
const float doubleRadius, bool* neighbors) {
int tid = threadIdx.x + blockIdx.x * blockDim.x;
float dis = 0.0f;
while (tid < N) {
DataPoint p1 = points[tid];
for (int i=0; i<N; i++) {
DataPoint p2 = points[i];
dis = 0;
dis += (p1.pfDimens[0]-p2.pfDimens[0]) * (p1.pfDimens[0]-p2.pfDimens[0]) +
(p1.pfDimens[1]-p2.pfDimens[1]) * (p1.pfDimens[1]-p2.pfDimens[1]) +
(p1.pfDimens[2]-p2.pfDimens[2]) * (p1.pfDimens[2]-p2.pfDimens[2]);
if (dis <= doubleRadius) {
neighbors[tid*N+i] = true;
} else {
neighbors[tid*N+i] = false;
}
}
tid += blockDim.x * gridDim.x;
}
}
DataPoint 是一个结构是
typedef struct DataPoint {
float pfDimens[3];
} DataPoint;
所以我想减少时间，我该怎么做？我试过使用内存合并和共享内存，但我没有得到很好的加速？
===============使用共享内存==============
__global__ void calcNeighbors2(const DataPoint* points,
const float doubleRadius, bool* neighbors) {
__shared__ DataPoint sharedpoints[threadsPerBlock];
int start = blockIdx.x * blockDim.x;
int len = start+threadIdx.x;
if (len < N) {
sharedpoints[threadIdx.x] = points[len];
}
len = imin(N, blockDim.x + start);
__syncthreads();
int tid = threadIdx.x;
float dis;
while (tid < N) {
DataPoint p1 = points[tid];
for (int i=start; i<len; i++) {
dis = 0;
dis += (p1.pfDimens[0]-sharedpoints[i-start].pfDimens[0]) * (p1.pfDimens[0]-sharedpoints[i-start].pfDimens[0]) +
(p1.pfDimens[1]-sharedpoints[i-start].pfDimens[1]) * (p1.pfDimens[1]-sharedpoints[i-start].pfDimens[1]) +
(p1.pfDimens[2]-sharedpoints[i-start].pfDimens[2]) * (p1.pfDimens[2]-sharedpoints[i-start].pfDimens[2]);
if (dis <= doubleRadius) {
neighbors[i*N+tid] = true;
} else {
neighbors[i*N+tid] = false;
}
}
tid += blockDim.x;
}
}
我在这里将 neighbors[tid*N+i] 更改为 neighbors[i*N+tid]，这让我在 Tesla K10.G2.8GB 上的速度提高了 8 倍。但是当我使用共享内存存储一些点时，它没有用吗？
至少有4个想法，其中一些已经在评论中陈述过：
从 AoS 格式转换您的点距离存储：
struct DataPoint { float pfDimens[3]; };转 SoA 格式：
struct DataPoint { float pfDimens_x[NPTS]; float pfDimens_y[NPTS]; float pfDimens_z[NPTS]; };这将在加载数据时启用完全合并。事实上，为了帮助解决下面的第 4 点，我只想切换到使用 3 个裸数组，而不是结构。
将计算减少到（略小于）一半：
for (int i=N-1; i>tid; i--) {然后，无论是在线程代码本身，还是在主机中，您都可以通过复制数据来填充输出矩阵的另一个 "half"。
转置输出矩阵中的存储，这样您就可以编写这样的存储操作：
neighbors[i*N+tid] = true;这将很好地合并，而不是这个：
neighbors[tid*N+i] = true;哪个不会。
由于你输入的点数据是只读的，适当标记内核参数：
const float * __restrict__ points_x, const float * __restrict__ points_y, const float * __restrict__ points_z在某些情况下，在某些 GPU 上，由于使用 the read-only cache，这通常会导致加速。如果你真的想积极使用缓存，并且你的数据数组足够小（4K 或更少
float点），你可以将点数据的副本放在全局内存中以及 [=21 中的副本=]内存，并通过常量内存加载"uniform"你在这里做的加载：DataPoint p2 = c_points[i];因此您可以通过只读缓存执行合并加载，通过常量缓存执行统一加载，以及转到普通全局内存的合并存储。
在 K40c 上，在 linux/CUDA 7 上，对于 N = 4096，这些更改的净效果似乎是内核级别的 3.5 倍加速：
$ cat t749.cu
#include <stdio.h>
#define N 4096
// if N is 16K/3 or less, we can use constant
#define USE_CONSTANT
#define THRESH 0.2f
#define nTPB 256
#define nBLK (N/nTPB+1)
#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)
#include <time.h>
#include <sys/time.h>
#define USECPSEC 1000000ULL
unsigned long long dtime_usec(unsigned long long start){
timeval tv;
gettimeofday(&tv, 0);
return ((tv.tv_sec*USECPSEC)+tv.tv_usec)-start;
}
struct DataPoint {
float pfDimens[3];
};
__global__ void calcNeighbors(const DataPoint* points,
const float doubleRadius, bool* neighbors) {
int tid = threadIdx.x + blockIdx.x * blockDim.x;
float dis = 0.0f;
while (tid < N) {
DataPoint p1 = points[tid];
for (int i=0; i<N; i++) {
DataPoint p2 = points[i];
dis = 0;
dis += (p1.pfDimens[0]-p2.pfDimens[0]) * (p1.pfDimens[0]-p2.pfDimens[0]) +
(p1.pfDimens[1]-p2.pfDimens[1]) * (p1.pfDimens[1]-p2.pfDimens[1]) +
(p1.pfDimens[2]-p2.pfDimens[2]) * (p1.pfDimens[2]-p2.pfDimens[2]);
if (dis <= doubleRadius) {
neighbors[tid*N+i] = true;
} else {
neighbors[tid*N+i] = false;
}
}
tid += blockDim.x * gridDim.x;
}
}
#ifdef USE_CONSTANT
__constant__ float cpx[N];
__constant__ float cpy[N];
__constant__ float cpz[N];
#endif
__global__ void calcNeighbors2(const float * __restrict__ pts_x, const float * __restrict__ pts_y, const float * __restrict__ pts_z, const float doubleRadius, bool * __restrict__ neighbors) {
int tid = threadIdx.x+blockDim.x*blockIdx.x;
while (tid < N) {
float p1x = pts_x[tid];
float p1y = pts_y[tid];
float p1z = pts_z[tid];
for (int i = N-1; i > tid; i--){
float p2x, p2y, p2z;
#ifdef USE_CONSTANT
p2x = cpx[i];
p2y = cpy[i];
p2z = cpz[i];
#else
p2x = pts_x[i];
p2y = pts_y[i];
p2z = pts_z[i];
#endif
float dis = ((p1x-p2x)*(p1x-p2x)) + ((p1y-p2y)*(p1y-p2y)) + ((p1z-p2z)*(p1z-p2z));
neighbors[i*N+tid] = (dis <= doubleRadius);
}
tid += blockDim.x * gridDim.x;
}
}
int main(){
float *dx, *dy, *dz, *hx, *hy, *hz;
DataPoint *dp, *hp;
bool *dn, *hn1, *hn2;
hx =(float *)malloc(N*sizeof(float));
hy =(float *)malloc(N*sizeof(float));
hz =(float *)malloc(N*sizeof(float));
hp =(DataPoint *)malloc(N*sizeof(DataPoint));
hn1=(bool *)malloc(N*N*sizeof(bool));
hn2=(bool *)malloc(N*N*sizeof(bool));
cudaMalloc(&dx, N*sizeof(float));
cudaMalloc(&dy, N*sizeof(float));
cudaMalloc(&dz, N*sizeof(float));
cudaMalloc(&dp, N*sizeof(DataPoint));
cudaMalloc(&dn, N*N*sizeof(bool));
for (int i =0; i < N; i++){
hx[i] = rand()/(float)RAND_MAX;
hy[i] = rand()/(float)RAND_MAX;
hz[i] = rand()/(float)RAND_MAX;
hp[i].pfDimens[0] = hx[i];
hp[i].pfDimens[1] = hy[i];
hp[i].pfDimens[2] = hz[i];}
cudaMemcpy(dx, hx, N*sizeof(float), cudaMemcpyHostToDevice);
cudaMemcpy(dy, hy, N*sizeof(float), cudaMemcpyHostToDevice);
cudaMemcpy(dz, hz, N*sizeof(float), cudaMemcpyHostToDevice);
cudaMemcpy(dp, hp, N*sizeof(DataPoint), cudaMemcpyHostToDevice);
// warm-up
calcNeighbors<<<nBLK, nTPB>>>(dp, THRESH, dn);
cudaDeviceSynchronize();
cudaMemset(dn, 0, N*N*sizeof(bool));
unsigned long long t1 = dtime_usec(0);
calcNeighbors<<<nBLK, nTPB>>>(dp, THRESH, dn);
cudaDeviceSynchronize();
cudaCheckErrors("kernel 1 error");
t1 = dtime_usec(t1);
cudaMemcpy(hn1, dn, N*N*sizeof(bool), cudaMemcpyDeviceToHost);
// warm-up
calcNeighbors2<<<nBLK, nTPB>>>(dx, dy, dz, THRESH, dn);
cudaDeviceSynchronize();
cudaMemset(dn, 0, N*N*sizeof(bool));
unsigned long long t2 = dtime_usec(0);
calcNeighbors2<<<nBLK, nTPB>>>(dx, dy, dz, THRESH, dn);
cudaDeviceSynchronize();
cudaCheckErrors("kernel 2 error");
t2 = dtime_usec(t2);
cudaMemcpy(hn2, dn, N*N*sizeof(bool), cudaMemcpyDeviceToHost);
cudaCheckErrors("some error");
printf("t1: %fs, t2: %fs\n", t1/(float)USECPSEC, t2/(float)USECPSEC);
// results validation
for (int i = 0; i < N; i++)
for (int j = i+1; j < N; j++)
if (hn1[i*N+j] != hn2[j*N+i]) {printf("mismatch at %d, %d, was: %d, should be: %d\n", i, j, hn2[j*N+i], hn1[i*N+j]); return 1;}
return 0;
}
$ nvcc -arch=sm_35 -o t749 t749.cu
$ ./t749
t1: 0.004903s, t2: 0.001395s
$
在 K40c 的情况下，由于延迟，上面启动的块数量有限 (16) 是性能的重大障碍。如果我们注释掉 USE_CONSTANT 定义，并将 N 更改为 16384，我们会观察到改进内核的加速比更高：
$ ./t749
t1: 0.267107s, t2: 0.008209s
$
总计 ~48 个区块足以 "fill" 拥有 15 个 SM 的 K40c。
编辑： 现在您已经发布了一个共享内存内核，我将它作为 calcNeighbors3 添加到我的测试用例中并比较了它的时序性能（如 t3).它几乎和我的内核一样快，而且它似乎提供了正确的结果（与您的原始内核匹配）所以我不确定您的顾虑是什么。
这是更新后的代码和测试用例：
$ cat t749.cu
#include <stdio.h>
#include <math.h>
#define imin(X,Y) ((X)<(Y))?(X):(Y)
#define N 32768
// if N is 16K/3 or less, we can use constant
// #define USE_CONSTANT
#define THRESH 0.2f
#define nTPB 256
#define nBLK (N/nTPB+1)
#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)
#include <time.h>
#include <sys/time.h>
#define USECPSEC 1000000ULL
unsigned long long dtime_usec(unsigned long long start){
timeval tv;
gettimeofday(&tv, 0);
return ((tv.tv_sec*USECPSEC)+tv.tv_usec)-start;
}
struct DataPoint {
float pfDimens[3];
};
__global__ void calcNeighbors(const DataPoint* points,
const float doubleRadius, bool* neighbors) {
int tid = threadIdx.x + blockIdx.x * blockDim.x;
float dis = 0.0f;
while (tid < N) {
DataPoint p1 = points[tid];
for (int i=0; i<N; i++) {
DataPoint p2 = points[i];
dis = 0;
dis += (p1.pfDimens[0]-p2.pfDimens[0]) * (p1.pfDimens[0]-p2.pfDimens[0]) +
(p1.pfDimens[1]-p2.pfDimens[1]) * (p1.pfDimens[1]-p2.pfDimens[1]) +
(p1.pfDimens[2]-p2.pfDimens[2]) * (p1.pfDimens[2]-p2.pfDimens[2]);
if (dis <= doubleRadius) {
neighbors[tid*N+i] = true;
} else {
neighbors[tid*N+i] = false;
}
}
tid += blockDim.x * gridDim.x;
}
}
#ifdef USE_CONSTANT
__constant__ float cpx[N];
__constant__ float cpy[N];
__constant__ float cpz[N];
#endif
__global__ void calcNeighbors2(const float * __restrict__ pts_x, const float * __restrict__ pts_y, const float * __restrict__ pts_z, const float doubleRadius, bool * __restrict__ neighbors) {
int tid = threadIdx.x+blockDim.x*blockIdx.x;
while (tid < N) {
float p1x = pts_x[tid];
float p1y = pts_y[tid];
float p1z = pts_z[tid];
for (int i = N-1; i > tid; i--){
float p2x, p2y, p2z;
#ifdef USE_CONSTANT
p2x = cpx[i];
p2y = cpy[i];
p2z = cpz[i];
#else
p2x = pts_x[i];
p2y = pts_y[i];
p2z = pts_z[i];
#endif
float dis = ((p1x-p2x)*(p1x-p2x)) + ((p1y-p2y)*(p1y-p2y)) + ((p1z-p2z)*(p1z-p2z));
neighbors[i*N+tid] = (dis <= doubleRadius);
}
tid += blockDim.x * gridDim.x;
}
}
__global__ void calcNeighbors3(const DataPoint* points,
const float doubleRadius, bool* neighbors) {
__shared__ DataPoint sharedpoints[nTPB];
int start = blockIdx.x * blockDim.x;
int len = start+threadIdx.x;
if (len < N) {
sharedpoints[threadIdx.x] = points[len];
}
len = imin(N, blockDim.x + start);
__syncthreads();
int tid = threadIdx.x;
float dis;
while (tid < N) {
DataPoint p1 = points[tid];
for (int i=start; i<len; i++) {
dis = 0;
dis += (p1.pfDimens[0]-sharedpoints[i-start].pfDimens[0]) * (p1.pfDimens[0]-sharedpoints[i-start].pfDimens[0]) +
(p1.pfDimens[1]-sharedpoints[i-start].pfDimens[1]) * (p1.pfDimens[1]-sharedpoints[i-start].pfDimens[1]) +
(p1.pfDimens[2]-sharedpoints[i-start].pfDimens[2]) * (p1.pfDimens[2]-sharedpoints[i-start].pfDimens[2]);
if (dis <= doubleRadius) {
neighbors[i*N+tid] = true;
} else {
neighbors[i*N+tid] = false;
}
}
tid += blockDim.x;
}
}
int main(){
float *dx, *dy, *dz, *hx, *hy, *hz;
DataPoint *dp, *hp;
bool *dn, *hn1, *hn2, *hn3;
hx =(float *)malloc(N*sizeof(float));
hy =(float *)malloc(N*sizeof(float));
hz =(float *)malloc(N*sizeof(float));
hp =(DataPoint *)malloc(N*sizeof(DataPoint));
hn1=(bool *)malloc(N*N*sizeof(bool));
hn2=(bool *)malloc(N*N*sizeof(bool));
hn3=(bool *)malloc(N*N*sizeof(bool));
cudaMalloc(&dx, N*sizeof(float));
cudaMalloc(&dy, N*sizeof(float));
cudaMalloc(&dz, N*sizeof(float));
cudaMalloc(&dp, N*sizeof(DataPoint));
cudaMalloc(&dn, N*N*sizeof(bool));
for (int i =0; i < N; i++){
hx[i] = rand()/(float)RAND_MAX;
hy[i] = rand()/(float)RAND_MAX;
hz[i] = rand()/(float)RAND_MAX;
hp[i].pfDimens[0] = hx[i];
hp[i].pfDimens[1] = hy[i];
hp[i].pfDimens[2] = hz[i];}
cudaMemcpy(dx, hx, N*sizeof(float), cudaMemcpyHostToDevice);
cudaMemcpy(dy, hy, N*sizeof(float), cudaMemcpyHostToDevice);
cudaMemcpy(dz, hz, N*sizeof(float), cudaMemcpyHostToDevice);
cudaMemcpy(dp, hp, N*sizeof(DataPoint), cudaMemcpyHostToDevice);
#ifdef USE_CONSTANT
cudaMemcpyToSymbol(cpx, hx, N*sizeof(float));
cudaMemcpyToSymbol(cpy, hy, N*sizeof(float));
cudaMemcpyToSymbol(cpz, hz, N*sizeof(float));
#endif
// warm-up
calcNeighbors<<<nBLK, nTPB>>>(dp, THRESH, dn);
cudaDeviceSynchronize();
cudaMemset(dn, 0, N*N*sizeof(bool));
unsigned long long t1 = dtime_usec(0);
calcNeighbors<<<nBLK, nTPB>>>(dp, THRESH, dn);
cudaDeviceSynchronize();
cudaCheckErrors("kernel 1 error");
t1 = dtime_usec(t1);
cudaMemcpy(hn1, dn, N*N*sizeof(bool), cudaMemcpyDeviceToHost);
// warm-up
calcNeighbors2<<<nBLK, nTPB>>>(dx, dy, dz, THRESH, dn);
cudaDeviceSynchronize();
cudaMemset(dn, 0, N*N*sizeof(bool));
unsigned long long t2 = dtime_usec(0);
calcNeighbors2<<<nBLK, nTPB>>>(dx, dy, dz, THRESH, dn);
cudaDeviceSynchronize();
cudaCheckErrors("kernel 2 error");
t2 = dtime_usec(t2);
cudaMemcpy(hn2, dn, N*N*sizeof(bool), cudaMemcpyDeviceToHost);
// warm-up
calcNeighbors3<<<nBLK, nTPB>>>(dp, THRESH, dn);
cudaDeviceSynchronize();
cudaMemset(dn, 0, N*N*sizeof(bool));
unsigned long long t3 = dtime_usec(0);
calcNeighbors3<<<nBLK, nTPB>>>(dp, THRESH, dn);
cudaDeviceSynchronize();
cudaCheckErrors("kernel 3 error");
t3 = dtime_usec(t3);
cudaMemcpy(hn3, dn, N*N*sizeof(bool), cudaMemcpyDeviceToHost);
cudaCheckErrors("some error");
printf("t1: %fs, t2: %fs, t3: %fs\n", t1/(float)USECPSEC, t2/(float)USECPSEC, t3/(float)USECPSEC);
// results validation
for (int i = 0; i < N; i++)
for (int j = i+1; j < N; j++)
if (hn1[i*N+j] != hn2[j*N+i]) {printf("1:2 mismatch at %d, %d, was: %d, should be: %d\n", i, j, hn2[j*N+i], hn1[i*N+j]); return 1;}
for (int i = 0; i < N*N; i++)
if (hn1[i] != hn3[i]) {printf("1:3 mismatch at %d, was: %d, should be: %d\n", i, hn1[i], hn3[i]); return 1;}
return 0;
}
$ nvcc -arch=sm_35 -o t749 t749.cu
$ ./t749
t1: 1.260010s, t2: 0.022661s, t3: 0.029632s
$
对于此测试，我已将数据集大小更改为 32768，因为这更接近您关心的范围。您的共享内存内核比原始内核显示大约 42 倍的加速，而我的内核在我的 K40c 上显示大约 55 倍的加速。