我在 CUDA 中构建了一个基本内核来对两个复向量进行元素向量向量乘法。内核代码插入下面 ( multiplyElementwise
)。它工作正常,但由于我注意到其他看似简单的操作(如缩放向量)在 CUBLAS 或 CULA 等库中进行了优化,我想知道是否可以通过库调用替换我的代码?令我惊讶的是,CUBLAS 和 CULA 都没有这个选项,我试图通过使向量之一成为对角矩阵向量乘积的对角线来伪造它,但结果真的很慢。
作为最后的手段,我尝试通过在共享内存中加载两个向量然后从那里工作来自己优化此代码(参见下面的 multiplyElementwiseFast
),但这比我的原始代码慢。
所以我的问题:
multiplyElementwise
)吗? 任何帮助将不胜感激!
__global__ void multiplyElementwise(cufftComplex* f0, cufftComplex* f1, int size)
{
const int i = blockIdx.x*blockDim.x + threadIdx.x;
if (i < size)
{
float a, b, c, d;
a = f0[i].x;
b = f0[i].y;
c = f1[i].x;
d = f1[i].y;
float k;
k = a * (c + d);
d = d * (a + b);
c = c * (b - a);
f0[i].x = k - d;
f0[i].y = k + c;
}
}
__global__ void multiplyElementwiseFast(cufftComplex* f0, cufftComplex* f1, int size)
{
const int i = blockIdx.x*blockDim.x + threadIdx.x;
if (i < 4*size)
{
const int N = 256;
const int thId = threadIdx.x / 4;
const int rem4 = threadIdx.x % 4;
const int i4 = i / 4;
__shared__ float a[N];
__shared__ float b[N];
__shared__ float c[N];
__shared__ float d[N];
__shared__ float Re[N];
__shared__ float Im[N];
if (rem4 == 0)
{
a[thId] = f0[i4].x;
Re[thId] = 0.f;
}
if (rem4 == 1)
{
b[thId] = f0[i4].y;
Im[thId] = 0.f;
}
if (rem4 == 2)
c[thId] = f1[i4].x;
if (rem4 == 0)
d[thId] = f1[i4].y;
__syncthreads();
if (rem4 == 0)
atomicAdd(&(Re[thId]), a[thId]*c[thId]);
if (rem4 == 1)
atomicAdd(&(Re[thId]), -b[thId]*d[thId]);
if (rem4 == 2)
atomicAdd(&(Im[thId]), b[thId]*c[thId]);
if (rem4 == 3)
atomicAdd(&(Im[thId]), a[thId]*d[thId]);
__syncthreads();
if (rem4 == 0)
f0[i4].x = Re[thId];
if (rem4 == 1)
f0[i4].y = Im[thId];
}
}
最佳答案
如果您要实现的是具有复数的简单元素乘积,那么您似乎确实在 multiplyElementwise
中执行了一些额外的步骤。增加寄存器使用的内核。您尝试计算的是:
f0[i].x = a*c - b*d;
f0[i].y = a*d + b*c;
自
(a + ib)*(c + id) = (a*c - b*d) + i(a*d + b*c)
.通过使用改进的复数乘法,您可以用 1 次乘法换取 3 次加法和一些额外的寄存器。这是否合理可能取决于您使用的硬件。例如,如果您的硬件支持 FMA (融合乘加),这种优化可能效率不高。您应该考虑阅读此文档:“Precision & Performance:
Floating Point and IEEE 754 Compliance for NVIDIA GPUs ”,它也解决了浮点精度问题。不过,您应该考虑使用 Thrust .这个库提供了许多高级工具来操作主机和设备向量。您可以在此处查看一长串示例:https://github.com/thrust/thrust/tree/master/examples .这会让你的生活轻松很多。
更新代码
在你的情况下,你可以使用 this example并将其调整为以下内容:
#include <thrust/host_vector.h>
#include <thrust/device_vector.h>
#include <time.h>
struct ElementWiseProductBasic : public thrust::binary_function<float2,float2,float2>
{
__host__ __device__
float2 operator()(const float2& v1, const float2& v2) const
{
float2 res;
res.x = v1.x * v2.x - v1.y * v2.y;
res.y = v1.x * v2.y + v1.y * v2.x;
return res;
}
};
/**
* See: http://www.embedded.com/design/embedded/4007256/Digital-Signal-Processing-Tricks--Fast-multiplication-of-complex-numbers%5D
*/
struct ElementWiseProductModified : public thrust::binary_function<float2,float2,float2>
{
__host__ __device__
float2 operator()(const float2& v1, const float2& v2) const
{
float2 res;
float a, b, c, d, k;
a = v1.x;
b = v1.y;
c = v2.x;
d = v2.y;
k = a * (c + d);
d = d * (a + b);
c = c * (b - a);
res.x = k -d;
res.y = k + c;
return res;
}
};
int get_random_int(int min, int max)
{
return min + (rand() % (int)(max - min + 1));
}
thrust::host_vector<float2> init_vector(const size_t N)
{
thrust::host_vector<float2> temp(N);
for(size_t i = 0; i < N; i++)
{
temp[i].x = get_random_int(0, 10);
temp[i].y = get_random_int(0, 10);
}
return temp;
}
int main(void)
{
const size_t N = 100000;
const bool compute_basic_product = true;
const bool compute_modified_product = true;
srand(time(NULL));
thrust::host_vector<float2> h_A = init_vector(N);
thrust::host_vector<float2> h_B = init_vector(N);
thrust::device_vector<float2> d_A = h_A;
thrust::device_vector<float2> d_B = h_B;
thrust::host_vector<float2> h_result(N);
thrust::host_vector<float2> h_result_modified(N);
if (compute_basic_product)
{
thrust::device_vector<float2> d_result(N);
thrust::transform(d_A.begin(), d_A.end(),
d_B.begin(), d_result.begin(),
ElementWiseProductBasic());
h_result = d_result;
}
if (compute_modified_product)
{
thrust::device_vector<float2> d_result_modified(N);
thrust::transform(d_A.begin(), d_A.end(),
d_B.begin(), d_result_modified.begin(),
ElementWiseProductModified());
h_result_modified = d_result_modified;
}
std::cout << std::fixed;
for (size_t i = 0; i < 4; i++)
{
float2 a = h_A[i];
float2 b = h_B[i];
std::cout << "(" << a.x << "," << a.y << ")";
std::cout << " * ";
std::cout << "(" << b.x << "," << b.y << ")";
if (compute_basic_product)
{
float2 prod = h_result[i];
std::cout << " = ";
std::cout << "(" << prod.x << "," << prod.y << ")";
}
if (compute_modified_product)
{
float2 prod_modified = h_result_modified[i];
std::cout << " = ";
std::cout << "(" << prod_modified.x << "," << prod_modified.y << ")";
}
std::cout << std::endl;
}
return 0;
}
这将返回:
(6.000000,5.000000) * (0.000000,1.000000) = (-5.000000,6.000000)
(3.000000,2.000000) * (0.000000,4.000000) = (-8.000000,12.000000)
(2.000000,10.000000) * (10.000000,4.000000) = (-20.000000,108.000000)
(4.000000,8.000000) * (10.000000,9.000000) = (-32.000000,116.000000)
然后,您可以比较两种不同乘法策略的时序,并选择最适合您的硬件的策略。
关于cuda - 使用 CUDA 进行逐元素向量乘法,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/16899237/