- 设备:Tesla C2050
- 操作系统:Windows 7 企业版
- IDE:VS 2010
- CUDA:5.0(最新)
第一次在这里提问。我在 CUDA 程序中遇到了一些问题。
我有数百万个四面体,其中一个点位于 (0,0,0),因此我可以使用以下公式:
获取四面体的体积。
所以,这是代码:
struct Triangle
{
double x1;
double y1;
double z1;
double x2;
double y2;
double z2;
double x3;
double y3;
double z3;
};
以及 CUDA 代码:
__global__ void getResult(double *d_volume ,Triangle *d_triangles, Origin *d_point)
{
extern __shared__ Triangle s_data[];
int tid = threadIdx.x;
int i = blockDim.x * blockIdx.x + threadIdx.x;
s_data[tid] = d_triangles[i];
__syncthreads();
d_volume[i] =s_data[tid].x1 * s_data[tid].y2 * s_data[tid].z3 + \
s_data[tid].y1 * s_data[tid].z2 * s_data[tid].x3 + \
s_data[tid].x2 * s_data[tid].y3 * s_data[tid].z1 - \
s_data[tid].x3 * s_data[tid].y2 * s_data[tid].z1 - \
s_data[tid].x2 * s_data[tid].y1 * s_data[tid].z3 - \
s_data[tid].y3 * s_data[tid].z2 * s_data[tid].x1;
}
我从其他函数中获得了数百万个四面体作为数组。
// Host
Triangle *h_triangles = triangles;
double *h_volume;
// Device
Triangle *d_triangles;
double *d_volume;
// define grid and block size
int numThreadsPerBlock = numTriangles;
int numBlocks = numTrianges / 512;
// Shard memory size
int sharedMemSize = numThreadsPerBlock * sizeof(Triangle);
// allocate host and device memory
size_t memSize_triangles = numBlocks * numThreadsPerBlock * sizeof(Triangle);
size_t memSize_volume = numBlocks * numThreadsPerBlock * sizeof(double);
cudaMalloc( (void **) &d_triangles, memSize_triangles );
cudaMalloc( (void **) &d_volume, memSize_volume );
// Copy host array to device array
cudaMemcpy( d_triangles, h_triangles, memSize_triangles, cudaMemcpyHostToDevice );
cudaMemcpy( d_point, h_point, memSize_point, cudaMemcpyHostToDevice );
// launch kernel
dim3 dimGrid(numBlocks);
dim3 dimBlock(numThreadsPerBlock);
getResult<<< dimGrid, dimBlock, sharedMemSize >>>( d_volume, d_triangles);
// block until the device has completed
cudaThreadSynchronize();
// device to host copy
cudaMemcpy( h_volume, d_volume, memSize_volume, cudaMemcpyDeviceToHost );
// free device memory
cudaFree(d_triangles);
cudaFree(d_volume);
// free host memory
free(h_triangles);
free(h_volume);
到目前为止,一切正常。但我花了比我想象的更多的时间来获得音量。 我的设备是 Tesla C2050(515Gflops),比我的 CPU(单核,20.25Gflops)快 20 倍。 但仅加速10倍左右(不包括设备和主机之间复制内存的时间)
我想知道如何使其比 CPU 代码快大约 20 倍(for 循环获取音量。)。
谢谢!
PS:也许cudaMallocPitch()会帮助我,但三角形不是矩阵,我不能使用cudaMemcpy2D()代替cudaMemcpy()来复制内存。谁能帮我解答这个问题吗?
最佳答案
与 CPU 相比,GPU 的峰值性能通常更难获得。原因之一是许多内核受带宽限制而不是计算限制。
因为你的内核的计算复杂度是 O(n)。您可能应该使用带宽指标来计算理论峰值性能,如下所示
1024*1024*64 * sizeof(double) * (9 + 1) / (144e9 * 8/9) = 42 ms
#tetrahedron #input #output peak mem bw ECC cost
另一方面,您的内核可以进一步优化。
- 谨慎选择 blockDim/gridDim,错误的数字有时会导致 20% 的性能损失。
- 您可以为每个线程计算多个卷,而不是为每个线程计算一个卷,这将减少线程启动开销。
- 由于您不在线程之间共享数据,因此可以消除
__syncthreads()
。 - 由于非合并内存访问,结构数组 (AoS) 在 GPU 上通常比数组结构 (SoA) 慢。您还可以尝试更改数据结构。
更新
获得了具有大型 L1 缓存设置和最佳 blockDim/gridDim 选择的新内核。速度快了 15%。这是代码和配置文件结果。我的设备是M2090。
#include <stdlib.h>
#include <thrust/transform.h>
#include <thrust/device_vector.h>
#include <thrust/host_vector.h>
#include <iterator>
#include <thrust/inner_product.h>
using namespace thrust::placeholders;
struct Triangle
{
double x1;
double y1;
double z1;
double x2;
double y2;
double z2;
double x3;
double y3;
double z3;
};
__global__ void getResultNoSMem(double *d_volume, Triangle *d_triangles)
{
int i = blockDim.x * blockIdx.x + threadIdx.x;
d_volume[i] = d_triangles[i].x1 * d_triangles[i].y2 * d_triangles[i].z3 +
d_triangles[i].y1 * d_triangles[i].z2 * d_triangles[i].x3 +
d_triangles[i].x2 * d_triangles[i].y3 * d_triangles[i].z1 -
d_triangles[i].x3 * d_triangles[i].y2 * d_triangles[i].z1 -
d_triangles[i].x2 * d_triangles[i].y1 * d_triangles[i].z3 -
d_triangles[i].y3 * d_triangles[i].z2 * d_triangles[i].x1;
}
__global__ void getResult(double *d_volume, Triangle *d_triangles)
{
extern __shared__ Triangle s_data[];
int tid = threadIdx.x;
int i = blockDim.x * blockIdx.x + threadIdx.x;
s_data[tid] = d_triangles[i];
// __syncthreads();
d_volume[i] = s_data[tid].x1 * s_data[tid].y2 * s_data[tid].z3 +
s_data[tid].y1 * s_data[tid].z2 * s_data[tid].x3 +
s_data[tid].x2 * s_data[tid].y3 * s_data[tid].z1 -
s_data[tid].x3 * s_data[tid].y2 * s_data[tid].z1 -
s_data[tid].x2 * s_data[tid].y1 * s_data[tid].z3 -
s_data[tid].y3 * s_data[tid].z2 * s_data[tid].x1;
}
__global__ void getResultOpt(double *d_volume, Triangle *d_triangles, int len)
{
const int gridSize = blockDim.x * gridDim.x;
int i = blockDim.x * blockIdx.x + threadIdx.x;
while (i < len)
{
d_volume[i] = d_triangles[i].x1 * d_triangles[i].y2 * d_triangles[i].z3 +
d_triangles[i].y1 * d_triangles[i].z2 * d_triangles[i].x3 +
d_triangles[i].x2 * d_triangles[i].y3 * d_triangles[i].z1 -
d_triangles[i].x3 * d_triangles[i].y2 * d_triangles[i].z1 -
d_triangles[i].x2 * d_triangles[i].y1 * d_triangles[i].z3 -
d_triangles[i].y3 * d_triangles[i].z2 * d_triangles[i].x1;
i += gridSize;
}
}
int main(void)
{
const int m = 1024 * 1024;
thrust::host_vector<Triangle> data(m);
for (int i = 0; i < m; i++)
{
data[i].x1 = (double) rand() / RAND_MAX;
data[i].y1 = (double) rand() / RAND_MAX;
data[i].z1 = (double) rand() / RAND_MAX;
data[i].x2 = (double) rand() / RAND_MAX;
data[i].y2 = (double) rand() / RAND_MAX;
data[i].z2 = (double) rand() / RAND_MAX;
data[i].x3 = (double) rand() / RAND_MAX;
data[i].y3 = (double) rand() / RAND_MAX;
data[i].z3 = (double) rand() / RAND_MAX;
}
thrust::device_vector<Triangle> triangles = data;
thrust::device_vector<double> volume(m);
thrust::device_vector<double> volumeOpt(m);
Triangle* dTriangles = thrust::raw_pointer_cast(&triangles[0]);
double* dVolume = thrust::raw_pointer_cast(&volume[0]);
double* dVolumeOpt = thrust::raw_pointer_cast(&volumeOpt[0]);
int g;
int b;
int threadUpperLimit = 48 * 1024 / sizeof(Triangle);
//for (b = 32; b <= 1024; b += 32)
{
b = 64;
int gridDim = (m + b - 1) / b;
getResultNoSMem<<<gridDim, b, 0, 0>>>(dVolume, dTriangles);
}
// for (b = 32; b <= threadUpperLimit; b += 32)
{
b = 64;
int gridDim = (m + b - 1) / b;
getResult<<<gridDim, b, b * sizeof(Triangle), 0>>>(dVolume, dTriangles);
}
//for (g = 32; g <= 512; g += 32)
// for (b = 32; b <= 1024; b += 32)
{
b = 64;
g = 64;
getResultOpt<<<g, b, 0, 0>>>(dVolumeOpt, dTriangles, m);
}
//for (g = 32; g <= 512; g += 32)
// for (b = 32; b <= 1024; b += 32)
{
b = 64;
g = 512;
cudaFuncSetCacheConfig(getResultOpt, cudaFuncCachePreferL1);
getResultOpt<<<g, b, 0, 0>>>(dVolumeOpt, dTriangles, m);
}
thrust::device_vector<double> X = volume;
thrust::device_vector<double> Y = volumeOpt;
thrust::transform(X.begin(), X.end(), Y.begin(), X.begin(), _1 - _2);
double result = thrust::inner_product(X.begin(), X.end(), X.begin(), 0.0);
std::cout << "difference: " << result << std::endl;
return 0;
}
关于c++ - CUDA并行计算加速体积计算,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/18480975/