如何做CUBLAS的复数的权力？ [英] how to do power of complex number in CUBLAS?

问题描述

我正在将我的c ++代码移植到CUDA& CUBLAS。我使用stl :: complex的复杂计算（即pow，log，exp等），但我没有看到相同的功能定义在CuComplex库。我不知道如何创建这些函数，但我发现了一些代码在线

I am porting my c++ code to CUDA & CUBLAS. I use stl::complex for complex computation (i.e. pow, log, exp, etc.) but I didn't see the same functions defined in CuComplex library. I don't know how to create those functions but I found some codes online

#include <iostream>
#include <cublas_v2.h>
#include <cuComplex.h>
using namespace std;

typedef cuDoubleComplex Complex;

#define complex(x, y) make_cuDoubleComplex(x, y)

__host__ __device__ double cabs(const Complex& z) {return cuCabs(z);}
__host__ __device__ double carg(const Complex& z) {return atan2(cuCreal(z),   cuCimag(z));}
__host__ __device__ Complex polar(const double &magnitude, const double &angle) {return complex(magnitude*cos(angle), magnitude*sin(angle));}
__host__ __device__ Complex cexp(const Complex& z) {return polar( exp(cuCreal(z)), cuCimag(z));}
__host__ __device__ Complex czlog(const Complex& z) {return complex( ::log(cabs(z)), carg(z) );}
__host__ __device__ Complex cpow(const Complex& z, const int &exponent) {return cexp( cuCmul(czlog(z), complex((double )exponent, 0)) );}

void main(void)
{
  Complex z=complex(0.34, 0.56);
  cout << cuCreal(cpow(z, 2)) << "  " << cuCimag(cpow(z, 2)) << endl;
}

上述结果没有给出正确的答案。这是什么错误的cpow？

the above results didn't give right answer. Is that anything wrong with the cpow? Is that any better to do the power and other function on complex number?

推荐答案

这是不正确的：

__host__ __device__ double carg(const Complex& z) {return atan2(cuCreal(z),   cuCimag(z));}

的第二个参数的比率atan2

因此，您应该使用：

__host__ __device__ double carg(const Complex& z) {return atan2(cuCimag(z),   cuCreal(z));}

$ b b

我不知道你的权力函数（ cpow ）。您是否尝试过 DeMoivre定理？我不知道在计算上是最好的方法，但似乎第一次的业务是得到正确的答案。

I'm not sure about your power function (cpow) either. Have you tried DeMoivre's theorem? I don't know computationally the best method, but it seems like the first order of business is to get the right answer.

其他注意事项：

1. 我不认为这个问题与CUBLAS有什么关系。


2. 当发布这样的问题时，

下面是一个基于DeMoivre定理的工作示例：

Here's a worked example based on DeMoivre's theorem:

$ cat t233.cu
#include <iostream>
#include <math.h>
#include <cuComplex.h>
#include <complex>

typedef double     rtype;
typedef cuDoubleComplex ctype;
#define rpart(x)   (cuCreal(x))
#define ipart(x)   (cuCimag(x))
#define cmplx(x,y) (make_cuDoubleComplex(x,y))

__host__ __device__ rtype carg(const ctype& z) {return (rtype)atan2(ipart(z), rpart(z));} // polar angle
__host__ __device__ rtype cabs(const ctype& z) {return (rtype)cuCabs(z);}
__host__ __device__ ctype cp2c(const rtype d, const rtype a) {return cmplx(d*cos(a), d*sin(a));}
__host__ __device__ ctype cpow(const ctype& z, const int &n) {return cmplx((pow(cabs(z), n)*cos(n*carg(z))), (pow(cabs(z), n)*sin(n*carg(z))));}

int main(){

  double r = 0.34;
  double i = 0.56;
  int n = 2;

  std::complex<double> stl_num(r,i);
  std::complex<double> cn(n,0);
  ctype cu_num = cmplx(r,i);

  std::complex<double> stl_ans = std::pow(stl_num, cn);
  ctype cu_ans = cpow(cu_num, n);

  std::cout << "STL real: " << std::real(stl_ans) << " STL imag: " << std::imag(stl_ans) << std::endl;
  std::cout << "CU  real: " << rpart(cu_ans) << " CU  imag: " << ipart(cu_ans) << std::endl;
  return 0;
}
$ nvcc -arch=sm_20 -O3 -o t233 t233.cu
$ ./t233
STL real: -0.198 STL imag: 0.3808
CU  real: -0.198 CU  imag: 0.3808
$

我不建议这是彻底测试代码，但它似乎是在正确的轨道，并给出正确的答案您的测试用例。

I'm not suggesting this is thoroughly tested code, but it seems to be on the right track and gives the right answer for your test case.

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