问题
I posted this on matlab central but didn't get any responses so I figured I'd repost here.
I recently wrote a simple routine in Matlab that uses an FFT in a for-loop; the FFT dominates the calculations. I wrote the same routine in mex just for experimentation purposes and it calls the FFTW 3.3 library. It turns out that the matlab routine runs faster than the mex routine for very large arrays (about twice as fast). The mex routine uses wisdom and and performs the same FFT calculations. I also know matlab uses FFTW, but is it possible their version is slightly more optimized? I even used the FFTW_EXHAUSTIVE flag and its still about twice as slow for large arrays than the MATLAB counterpart. Furthermore I ensured the matlab I used was single threaded with the "-singleCompThread" flag and the mex file I used was not in debug mode. Just curious if this was the case - or if there are some optimizations matlab is using under the hood that I dont know about. Thanks.
Here's the mex portion:
void class_cg_toeplitz::analysis() {
// This method computes CG iterations using FFTs
// Check for wisdom
if(fftw_import_wisdom_from_filename("cd.wis") == 0) {
mexPrintf("wisdom not loaded.\n");
} else {
mexPrintf("wisdom loaded.\n");
}
// Set FFTW Plan - use interleaved FFTW
fftw_plan plan_forward_d_buffer;
fftw_plan plan_forward_A_vec;
fftw_plan plan_backward_Ad_buffer;
fftw_complex *A_vec_fft;
fftw_complex *d_buffer_fft;
A_vec_fft = fftw_alloc_complex(n);
d_buffer_fft = fftw_alloc_complex(n);
// CREATE MASTER PLAN - Do this on an empty vector as creating a plane
// with FFTW_MEASURE will erase the contents;
// Use d_buffer
// This is somewhat dangerous because Ad_buffer is a vector; but it does not
// get resized so &Ad_buffer[0] should work
plan_forward_d_buffer = fftw_plan_dft_r2c_1d(d_buffer.size(),&d_buffer[0],d_buffer_fft,FFTW_EXHAUSTIVE);
plan_forward_A_vec = fftw_plan_dft_r2c_1d(A_vec.height,A_vec.value,A_vec_fft,FFTW_WISDOM_ONLY);
// A_vec_fft.*d_buffer_fft will overwrite d_buffer_fft
plan_backward_Ad_buffer = fftw_plan_dft_c2r_1d(Ad_buffer.size(),d_buffer_fft,&Ad_buffer[0],FFTW_EXHAUSTIVE);
// Get A_vec_fft
fftw_execute(plan_forward_A_vec);
// Find initial direction - this is the initial residual
for (int i=0;i<n;i++) {
d_buffer[i] = b.value[i];
r_buffer[i] = b.value[i];
}
// Start CG iterations
norm_ro = norm(r_buffer);
double fft_reduction = (double)Ad_buffer.size(); // Must divide by size of vector because inverse FFT does not do this
while (norm(r_buffer)/norm_ro > relativeresidual_cutoff) {
// Find Ad - use fft
fftw_execute(plan_forward_d_buffer);
// Get A_vec_fft.*fft(d) - A_vec_fft is only real, but d_buffer_fft
// has complex elements; Overwrite d_buffer_fft
for (int i=0;i<n;i++) {
d_buffer_fft[i][0] = d_buffer_fft[i][0]*A_vec_fft[i][0]/fft_reduction;
d_buffer_fft[i][1] = d_buffer_fft[i][1]*A_vec_fft[i][0]/fft_reduction;
}
fftw_execute(plan_backward_Ad_buffer);
// Calculate r'*r
rtr_buffer = 0;
for (int i=0;i<n;i++) {
rtr_buffer = rtr_buffer + r_buffer[i]*r_buffer[i];
}
// Calculate alpha
alpha = 0;
for (int i=0;i<n;i++) {
alpha = alpha + d_buffer[i]*Ad_buffer[i];
}
alpha = rtr_buffer/alpha;
// Calculate new x
for (int i=0;i<n;i++) {
x[i] = x[i] + alpha*d_buffer[i];
}
// Calculate new residual
for (int i=0;i<n;i++) {
r_buffer[i] = r_buffer[i] - alpha*Ad_buffer[i];
}
// Calculate beta
beta = 0;
for (int i=0;i<n;i++) {
beta = beta + r_buffer[i]*r_buffer[i];
}
beta = beta/rtr_buffer;
// Calculate new direction vector
for (int i=0;i<n;i++) {
d_buffer[i] = r_buffer[i] + beta*d_buffer[i];
}
*total_counter = *total_counter+1;
if(*total_counter >= iteration_cutoff) {
// Set total_counter to -1, this indicates failure
*total_counter = -1;
break;
}
}
// Store Wisdom
fftw_export_wisdom_to_filename("cd.wis");
// Free fft alloc'd memory and plans
fftw_destroy_plan(plan_forward_d_buffer);
fftw_destroy_plan(plan_forward_A_vec);
fftw_destroy_plan(plan_backward_Ad_buffer);
fftw_free(A_vec_fft);
fftw_free(d_buffer_fft);
};
Here's the matlab portion:
% Take FFT of A_vec.
A_vec_fft = fft(A_vec); % Take fft once
% Find initial direction - this is the initial residual
x = zeros(n,1); % search direction
r = zeros(n,1); % residual
d = zeros(n+(n-2),1); % search direction; pad to allow FFT
for i = 1:n
d(i) = b(i);
r(i) = b(i);
end
% Enter CG iterations
total_counter = 0;
rtr_buffer = 0;
alpha = 0;
beta = 0;
Ad_buffer = zeros(n+(n-2),1); % This holds the product of A*d - calculate this once per iteration and using FFT; only 1:n is used
norm_ro = norm(r);
while(norm(r)/norm_ro > 10^-6)
% Find Ad - use fft
Ad_buffer = ifft(A_vec_fft.*fft(d));
% Calculate rtr_buffer
rtr_buffer = r'*r;
% Calculate alpha
alpha = rtr_buffer/(d(1:n)'*Ad_buffer(1:n));
% Calculate new x
x = x + alpha*d(1:n);
% Calculate new residual
r = r - alpha*Ad_buffer(1:n);
% Calculate beta
beta = r'*r/(rtr_buffer);
% Calculate new direction vector
d(1:n) = r + beta*d(1:n);
% Update counter
total_counter = total_counter+1;
end
In terms of time, for N = 50000 and b = 1:n it takes about 10.5 seconds with mex and 4.4 seconds with matlab. I'm using R2011b. Thanks
回答1:
A few observations rather than a definite answer since I do not know any of the specifics of the MATLAB FFT implementation:
- Based on the code you have, I can see two explanations for the speed difference:
- the speed difference is explained by differences in levels of optimization of the FFT
- the while loop in MATLAB is executed a significantly smaller number of times
I will assume you already looked into the second issue and that the number of iterations are comparable. (If they aren't, this is most likely to some accuracy issues and worth further investigations.)
Now, regarding FFT speed comparison:
- Yes, the theory is that FFTW is faster than other high-level FFT implementations but it is only relevant as long as you compare apples to apples: here you are comparing implementations at a level further down, at the assembly level, where not only the selection of the algorithm but its actual optimization for a specific processor and by software developers with varying skills comes at play
- I have optimized or reviewed optimized FFTs in assembly on many processors over the year (I was in the benchmarking industry) and great algorithms are only part of the story. There are considerations that are very specific to the architecture you are coding for (accounting for latencies, scheduling of instructions, optimization of register usage, arrangement of data in memory, accounting for branch taken/not taken latencies, etc.) and that make differences as important as the selection of the algorithm.
- With N=500000, we are also talking about large memory buffers: yet another door for more optimizations that can quickly get pretty specific to the platform you run your code on: how well you manage to avoid cache misses won't be dictated by the algorithm so much as by how the data flow and what optimizations a software developer may have used to bring data in and out of memory efficiently.
- Though I do not know the details of the MATLAB FFT implementation, I am pretty sure that an army of DSP engineers has been (and is still) honing on its optimization as it is key to so many designs. This could very well mean that MATLAB had the right combination of developers to produce a much faster FFT.
回答2:
This is classic performance gain thanks to low-level and architecture-specific optimization.
Matlab uses FFT from the Intel MKL (Math Kernel Library) binary (mkl.dll). These are routines optimized (at assembly level) by Intel for Intel processors. Even on AMD's it seems to give nice performance boosts.
FFTW seems like a normal c library that is not as optimized. Hence the performance gain to use the MKL.
回答3:
I have found the following comment on the MathWorks website [1]:
Note on large powers of 2: For FFT dimensions that are powers of 2, between 2^14 and 2^22, MATLAB software uses special preloaded information in its internal database to optimize the FFT computation. No tuning is performed when the dimension of the FTT is a power of 2, unless you clear the database using the command fftw('wisdom', []).
Although it relates to powers of 2, it may hint upon that MATLAB employs its own 'special wisdom' when using FFTW for certain (large) array sizes. Consider: 2^16 = 65536.
[1] R2013b Documentation available from http://www.mathworks.de/de/help/matlab/ref/fftw.html (accessed on 29 Oct 2013)
回答4:
EDIT: @wakjah 's reply to this answer is accurate: FFTW does support split real and imaginary memory storage via its Guru interface. My claim about hacking is thus not accurate but can very well apply if FFTW's Guru interface is not used - which is the case by default, so beware still!
First, sorry for being a year late. I'm not convinced that the speed increase you see comes from MKL or other optimizations. There is something quite fundamentally different between FFTW and Matlab, and that is how complex data is stored in memory.
In Matlab, the real and imaginary parts of a complex vector X are separate arrays Xre[i] and Xim[i] (linear in memory, efficient when operating on either of them separately).
In FFTW, the real and imaginary parts are interlaced as double[2] by default, i.e. X[i][0] is the real part, and X[i][1] is the imaginary part.
Thus, to use the FFTW library in mex files one cannot use the Matlab array directly, but must allocate new memory first, then pack the input from Matlab into FFTW format, and then unpack the output from FFTW into Matlab format. i.e.
X = (fftw_complex*) fftw_malloc(sizeof(fftw_complex) * N);
Y = (fftw_complex*) fftw_malloc(sizeof(fftw_complex) * N);
then
for (size_t i=0; i<N; ++i) {
X[i][0] = Xre[i];
X[i][1] = Xim[i];
}
then
for (size_t i=0; i<N; ++i) {
Yre[i] = Y[i][0];
Yim[i] = Y[i][1];
}
Hence, this requires 2x memory allocations + 4x reads + 4x writes -- all of size N. This does take a toll speed-wise on large problems.
I have a hunch that Mathworks may have hacked the FFTW3 code to allow it to read input vectors directly in the Matlab format, which avoids all of the above.
In this scenario, one can only allocate X and use X for Y to run FFTW in-place (as fftw_plan_*(N, X, X, ...) instead of fftw_plan_*(N, X, Y, ...)), since it'll be copied to the Yre and Yim Matlab vector, unless the application requires/benefits from keeping X and Y separate.
EDIT: Looking at the memory consumption in real-time when running Matlab's fft2() and my code based on the fftw3 library, it shows that Matlab only allocates only one additional complex array (the output), whereas my code needs two such arrays (the *fftw_complex buffer plus the Matlab output). An in-place conversion between the Matlab and fftw formats is not possible because the Matlab's real and imaginary arrays are not consecutive in memory. This suggests that Mathworks hacked the fftw3 library to read/write the data using the Matlab format.
One other optimization for multiple calls, is to allocate persistently (using mexMakeMemoryPersistent()). I'm not sure if the Matlab implementation does this as well.
Cheers.
p.s. As a side note, the Matlab complex data storage format is more efficient for operating on the real or imaginary vectors separately. On FFTW's format you'd have to do ++2 memory reads.
来源：https://stackoverflow.com/questions/15301426/fftw-vs-matlab-fft