In a project using Matlab's C++ MEX API, I have to compute the value exp(j * 2pi * x) for over 100,000 values of x where x is always a positive double. I've written some helper functions that breakdown the computation into sin/cos using euler's formula. I then apply the method of range reduction to reduce my values to their corresponding points in the domain [0,T/4] where T is the period of the exponential I'm computing. I keep track of which quadrant in [0, T] the original value would have fallen into for later. I can then compute the trig function using a taylor series polynomial in horner form and apply the appropriate shift depending on which quadrant the original value was in. For further information on some of the concepts in this technique, check out this answer. Here is the code for this function:
Eigen::VectorXcd calcRot2(const Eigen::Ref<const Eigen::VectorXd>& idxt) {
Eigen::VectorXd vidxt = idxt.array() - idxt.array().floor();
Eigen::VectorXd quadrant = (vidxt.array()*2+0.5).floor();
vidxt.array() -= (quadrant.array()*0.5);
vidxt.array() *= 2*3.14159265358979;
const Eigen::VectorXd sq = vidxt.array()*vidxt.array();
Eigen::VectorXcd M(vidxt.size());
M.real() = fastCos2(sq);
M.imag() = fastSin2(vidxt,sq);
M = (quadrant.array() == 1).select(-M,M);
return M;
}
I profiled the code segment in which this function is called using std::chrono and averaged over 500 calls to the function (where each call to the mex function processes all 100,000+ values by calling calcRot2 in a loop. Each iteration passes about 200 values to calcRot2). I find the following average runtimes:
runtime with calcRot2: 75.4694 ms
runtime with fastSin/Cos commented out: 50.2409 ms
runtime with calcRot2 commented out: 30.2547 ms
Looking at the difference between the two extreme cases, it seems like calcRot has a large contribution to the runtime. However, only a portion of that comes from the sin/cos calculation. I would assume Eigen's implicit vectorization and the compiler would make the runtime of the other operations in the function effectively negligible. (floor shouldn't be a problem!) Where exactly is the performance bottleneck here?
This is the compilation command I'm performing (It uses MinGW64 which I think is the same as gcc):
mex(ipath,'CFLAGS="$CFLAGS -O3 -fno-math-errno -ffast-math -fopenmp -mavx2"','LDFLAGS="$LDFLAGS -fopenmp"','DAS.cpp','DAShelper.cpp')
Reference Code
For reference, here is the code segment in the main mex function where the timer is called, and the helper function that calls calcRot2():
MEX function call:
chk1 = std::chrono::steady_clock::now();
// Calculate beamformed signal at each point
Eigen::MatrixXcd bfVec(p.nPoints,1);
#pragma omp parallel for
for (int i = 0; i < p.nPoints; i++) {
calcPoint(idxt.col(i),SIG,p,bfVec(i));
}
chk2 = std::chrono::steady_clock::now();
auto diff3 = chk2 - chk1;
calcPoint:
void calcPoint(const Eigen::Ref<const Eigen::VectorXd>& idxt,
const Eigen::Ref<const Eigen::MatrixXcd>& SIG,
Parameters& p, std::complex<double>& bfVal) {
Eigen::VectorXcd pRot = calcRot2(idxt*p.fc/p.fs);
int j = 0;
for (auto x : idxt) {
if(x >= 0) {
int vIDX = static_cast<int>(x);
bfVal += (SIG(vIDX,j)*(vIDX + 1 - x) + SIG(vIDX+1,j)*(x - vIDX))*pRot(j);
}
j++;
}
}
Clarification
To clarify, the line
(vidxt.array()*2+0.5).floor()
is meant to yield:
0 if vidxt is between [0,0.25]
1 if vidxt is between [0.25,0.75]
2 if vidxt is between [0.75,1]
The idea here is that when vidxt is in the second interval (quadrants 2 and 3 on the unit circle for functions with period 2pi), then the value needs to map to its negative value. Otherwise, the range reduction maps the values to the correct values.