I'm attempting to run the halide FFT implementation found here for benchmarking against FTTW. I'm able to run the implementation as is, but I've encountered some issues when digging a little deeper. The routine fails with errors for different values of H and W (the height and width of the random input image). For example, I get the following error with H=W=5:
Error at ./fft.cpp:603:
Cannot vectorize dimension n0 of function v_S1_R5$6 because the function is scheduled inline.
Aborted (core dumped)
I've been attempting to test on small image sizes (i.e. 5x5) to compare the results of the algorithms, but I can't get the algorithm to complete for any values less than 16, which even at that point makes checking the values a long task. The FFT also fails for values greater than 32, seemingly not working for all non-powers of 2.
Has anyone run into this issue before? Are there any other implementations of FFT in halide that work for different sized images?
For reference, I'm running the code on RHEL7 using gcc 4.8.3.
I think there are a few issues going on. First, there looks to be a bug for very small FFTs that only use one pass. I think that's what you hit in your first case.
The second issue is that W and H need to be a multiple of the vector size of your target, not necessarily that W and H need to be a power of 2. For example, W = 48, H = 32 seems to work for me. There's a further complication that for real FFTs, one dimension gets internally cut in half (this is how efficient real FFTs are implemented), so if you are on an AVX machine, that dimension must be a multiple of 16 (2x the vector width of 8 floats).
If you want to run on really small FFTs, you could remove the vectorize scheduling directives, then it should work, at least for learning purposes.
However, I would point out that running 5x5 won't be very interesting, because it will be done in just one radix 5 pass, i.e. just a plain old DFT (this also appears to be broken, as you've found). 4x4 (factored into 2 radix 2 passes) will be the smallest interesting FFT. When debugging it, I often used 8x8 FFTs (radix 4, radix 2).
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Hello stackoverflow community,
I have troubles in understanding a least-square-error-problem in the c++ armadillo package.
I have a matrix A with many more rows than columns (5000 to 100 for example) so it is overdetermined.
I want to find x so that A*x=b gives me the least square error.
If i use the solve function of armadillo on my data like "x = Solve(A,b)" the error of "(A*x-b)^2" is sometimes way to high.
If on the other hand I solve for x with the analytical form by "x = (A^T * A)^-1 *A^T * b" the results are always right.
The results for x in both cases can differ by 10 magnitudes.
I had thought that armadillo would use this analytical form in the background if the system is overdetermined.
Now I would like to understand why these two methods give such different results.
I wanted to give a short example program, but i can't reproduce this behavior with a short program.
I thought about giving the Matrix here, but with 5000 times 100 it's also very big. I can deliver the values for which this happens though if needed.
So as a short background.
The matrix I get from my program is a numerically solved reaction of a nonlinear oscillator in which I put information inside by wiggeling a parameter of this system.
Because the influence of this parameter on the system is small, the values of my different rows are very similar but never the same, otherwise armadillo should throw an error.
I'm still thinking that this is the problem, but the solve function never threw any error.
Another thing that confuses me is that in a short example program with a random matrix, the analytical form is waaay slower than the solve function.
But on my program, both are nearly identically fast.
I guess this has something to do with the numerical convergence of the pseudo inverse and the special case of my matrix, but for that i don't know enough about how armadillo works.
I hope someone can help me with that problem and thanks a lot in advance.
Thanks for the replies. I think i figured the problem out and wanted to give some feedback for everybody who runs into the same problem.
The Armadillo solve function gives me the x that minimizes (A*x-b)^2.
I looked at the values of x and they are sometimes in the magnitude of 10^13.
This comes from the fact that the rows of my matrix only slightly change. (So nearly linear dependent but not exactly).
Because of that i was in the numerical precision of my doubles and as a result my error sometimes jumped around.
If i use the rearranged analytical formular (A^T * A)*x = *A^T * b with the solve function this problem doesn't occur anymore because the fitted values of x are in the magnitude of 10^4. The least square error is a little bit higher but that is okay, as i want to avoid overfitting.
I now additionally added Tikhonov regularization by solving (A^T * A + lambda*Identity_Matrix)*x = *A^T * b with the solve function of armadillo.
Now the weight vectors are in the order of around 1 and the error nearly doesn't change compared to the formular without regularization.
I need to paralize the function numpy.linalg.matrix_power and I use the following code to test how much fast can be the parallel version
def aux_matrix_arg3(A):
aaa=np.linalg.matrix_power(np.random.randn(199,199),100)
return 1
N=10000
processes=4
chunksize=N/processes
poolWorkers=mp.Pool(processes=processes)
ti=t.time()
A=poolWorkers.map(aux_matrix_arg3,range(N),chunksize=chunksize)
print 't_iteration 3',t.time()-ti
I have tried with 1 and 4 processes in my laptop. I got very similar performance
4 processes: t_iteration 3 40.7985408306
1 processes: t_iteration 3 40.6538720131
Any clue why I do not get any improvment with paralle processes?
The docs say:
For positive integers n, the power is computed by repeated matrix squarings and matrix multiplications. If n == 0, the identity matrix of the same shape as M is returned. If n < 0, the inverse is computed and then raised to the abs(n).
If your system is set up correctly, BLAS will be used to parallelize matrix-multiplications and LAPACK (maybe SuperLU, the latter probably only in the sparse-case) for solving systems of linear-equations (used to calculate the inverse). So with a very high probability, your naive code is already very very optimized!
Despite that, you should be careful as naive parallelization copies all the data, which can hurt. (Normaly one would use mmap-arrays to share data instead of copying).
Short version of the question: overflow or timeout in current settings when calculating large int64_t and double, anyway to avoid these?
Test case:
If only demand is 80,000,000,000, solved with correct result. But if it's 800,000,000,000, returned incorrect 0.
If input has two or more demands (means more inequalities need to be calculated), smaller value will also cause incorrectness. e.g., three equal demands of 20,000,000,000 will cause the problem.
I'm using COIN-OR CLP linear programming solver to solve some network flow problems. I use int64_t when representing the link bandwidth. But CLP uses double most of time and cannot transfer to other types easily.
When the values of the variables are not that large (typically smaller than 10,000,000,000) and the constraints (inequalities) are relatively few, it will give the solution I want it to. But if either of the above factors increases, the tool will stop and return a 0 value solution. I think the reason is the calculation complexity is over its maximum, so program breaks at some trivial point (it uses LP simplex method).
The inequality is some kind of:
totalFlowSum <= usePercentage * demand
I changed it to
totalFlowSum - usePercentage * demand <= 0
Since totalFLowSum and demand are very large int64_t, usePercentage is double, if the constraints like this are too many (several or even more), or if the demand is larger than 100,000,000,000, the returned solution will be wrong.
Is there any way to correct this, like increase the break threshold or avoid this level of calculation magnitude?
Decrease some accuracy is acceptable. I have a possible solution is that 1,000 times smaller on inputs and 1,000 time larger on outputs. But this is kind of naïve and may cause too much code modification in the program.
Update:
I have changed the formulation to
totalFlowSum / demand - usePercentage <= 0
but the problem still exists.
Update 2:
I divided usePercentage by 1000, making its coefficient from 1 to 0.001, it worked. But if I also divide totalFlowSum/demand by 1000 simultaneously, still no result. I don't know why...
I changed the rhs of equalities from 0 to 0.1, the problem is then solved! Since the inputs are very large, 0.1 offset won't impact the solution at all.
I think the reason is that previous coeffs are badly scaled, so the complier failed to find an exact answer.
First, please excuse my ignorance in this field, I'm a programmer by trade but have been stuck in a situation a little beyond my expertise (in math and signals processing).
I have a Matlab script that I need to port to a C++ program (without compiling the matlab code into a DLL). It uses the hilbert() function with one argument. I'm trying to find a way to implement the same thing in C++ (i.e. have a function that also takes only one argument, and returns the same values).
I have read up on ways of using FFT and IFFT to build it, but can't seem to get anything as simple as the Matlab version. The main thing is that I need it to work on a 128*2000 matrix, and nothing I've found in my search has showed me how to do that.
I would be OK with either a complex value returned, or just the absolute value. The simpler it is to integrate into the code, the better.
Thank you.
The MatLab function hilbert() does actually not compute the Hilbert transform directly but instead it computes the analytical signal, which is the thing one needs in most cases.
It does it by taking the FFT, deleting the negative frequencies (setting the upper half of the array to zero) and applying the inverse FFT. It would be straight forward in C/C++ (three lines of code) if you've got a decent FFT implementation.
This looks pretty good, as long as you can deal with the GPL license. Part of a much larger numerical computing resource.
Simple code below. (Note: this was part of a bigger project). The value for L is based on the your determination of your order, N. With N = 2L-1. Round N to an odd number. xbar below is based on the signal you define as the input to your designed system. This was implemented in MATLAB.
L = 40;
n = -L:L; % index n from [-40,-39,....,-1,0,1,...,39,40];
h = (1 - (-1).^n)./(pi*n); %impulse response of Hilbert Transform
h(41) = 0; %Corresponds to the 0/0 term (for 41st term, 0, in n vector above)
xhat = conv(h,xbar); %resultant from Hilbert Transform H(w);
plot(abs(xhat))
Not a true answer to your question but maybe a way of making you sleep better. I believe that you won't be able to be much faster than Matlab in the particular case of what is basically ffts on a matrix. That is where Matlab excels!
Matlab FFTs are computed using FFTW, the de-facto fastest FFT algorithm written in C which seem to be also parallelized by Matlab. On top of that, quoting from http://www.mathworks.com/help/matlab/ref/fftw.html:
For FFT dimensions that are powers of 2, between 214 and 222, MATLAB
software uses special preloaded information in its internal database
to optimize the FFT computation.
So don't feel bad if your code is slightly slower...
I need to calculate rank of 4096x4096 sparse matrix, and I use C/C++ code.
I found some libraries (like Armadillo) that do it but they're too slow (almost 5 minutes).
I've also tried two Open Source version of Matlab (Freemat and Octave) but both crashed when I tried to make a test with a script.
5 minutes isn't so much but I must get rank from something like a million of matrix so the faster the better.
Someone knows a fast library for rank computation?
The Eigen library supports sparse matrices, try it out.
Computing the algebraic rank is O(n^3), where n is the matrix size, so it's inherently slow. You need eg. to perform pivoting, and this is slow and inaccurate if your matrix is not well conditioned (for n = 4096, a typical matrix is very ill conditioned).
Now, what is the rank ? It is the dimension of the image. It is very difficult to compute when n is large and it'll be spoiled by any small numerical inaccuracy of the input. For n = 4096, unless you happen to have particularly well conditioned matrices, this will prevent you from doing anything useful with a pivoting algorithm.
The best way is in fact to fix a cutoff epsilon, compute the singular values s_1 > ... > s_n and take as the rank the lowest integer r such that sum(s_i^2, i > r) < epsilon^2 * sum(s_i^2).
You thus need a sparse SVD routine, eg. from there.
This may not be faster, but to the very least it will be correct.
You can ask for less singular values that you need to speed up things. This is a tough problem, and with no info on the background and how you got these matrices, there is nothing more we can do.
Try the following code (the documentation is here).
It is an example for calculating the rank of the matrix A with Eigen library:
MatrixXd A(2,2);
A << 1 , 0, 1, 0;
FullPivLU<MatrixXd> luA(A);
int rank = luA.rank();