For my research, I have a lot of different images A, which I want to convolve with kernel B as fast as possible. The images are (M x N) and the kernel (M x P), in the normal convolution (which I have implemented right now) I slide them over each other in the 'x' direction resulting in a 1-dimensional image (1 x (N + P - 1)). Which produces the correct result. N and P are quite big (order 5000) and I want to speed up this process as I have to repeat it with a lot of different images.
One way I was thinking about is taking the FFT along the 'x' direction of both the kernel and image (after padding), multiplying them (element wise), IFFT and taking the sum over the columns. This should work, but I was thinking if I could also do a 2D convolution. Could taking the 2D FFT and than only multiplying the middle row? Which could maybe work, but how would you then do the padding, also which rows do you multiply exactly ( the zero'th frequencies?).
I'm currently working on some test cases, but have not found a proper answer yet, will update if I know more.
Would love to know what you guys think.
PS: the test case I'm building in python, but in the end I want to implement it with CUDA in C++ to really make it fast.
Related
I am writing some CUDA code for finding the 3 parameters of a circle (centre X,Y & radius) from many (m) measurements of positions around the perimeter.
As m > 3 I am (successfully) using Singular Value Decomposition (SVD) for this purpose (using the cuSolver library). Effectively I am solving m simulaneous equations with 3 unknowns.
However, not all of my perimeter positions are valid (say q of them), and so I have to go through my initial set of m measurements and remove the q invalid ones. This involves moving the size m data array from the card to the host, processing linearly to remove the q invalid entries and then re loading the smaller (m-q) array back onto the card...
My question is; if I were to set all terms on both sides of the q invalid equations to zero, could I just run the m equations (including the zeros) through my SVD analysis (without the data transfer etc) or would this cause other problems?
My instinct tells me that this is a bit like applying weights to the data but instinct and SVD are not terms that sit well together in my experience...
I am hesitant just to try this as I don't know if it will work in some cases and not in others...
I have tested the idea by inserting rows of zeros into my matrix. The solution that I am getting is not significantly affected by this.
So I am answering my own question with a non-rigorous Yes it is OK do do this.
If anybody has a more rigorous or more considered answer I would very much like to hear it.
I have a compute shader which simulates some fluid as particle. Particles are read from a buffer. Each particle is handled in one thread. During the execution of the thread, one particle moves its uv position and adds to pixel of a UAV named Water . Therefore each thread leaves a trail of its movement on the Water texture.
_watTx[texID] += watAddition * cellArea.x;
The problem is there are lots of particles moving around and most often multiples are present at the same texID. It seems there is a race condition since every time I run the simulation the results are slightly different. Is there a way to enforce mutual exclusion so the writes do not happen at the same time and the results become predictable?
I found a way to resolve this issue. InterlockedAdd adds to the pixel in an atomic fashion. But it only works on int and unit UAVs.
In my case the values are floating point but the range is quite limited (like 0 to 10). So the solution is to use an int UAV. We multiply the calculation result by a huge number (like 10000) and then write to the UAV:
InterlockedAdd(_watTx[texID], (watAddition * cellArea.x * 10000));
The results will have a 0.0001 precision which is perfectly fine in my case. After this in another pixel or compute shader we can multiply values from the int UAV by 0.0001 and write to the desired floating point render target.
This process eliminates the concurrent write problem and the results are identical in each run.
Recently I learn DM_Script for TEM image processing
I needed Gaussian blur process and I found one whose name is 'Gaussian Blur' in http://www.dmscripting.com/recent_updates.html
This code implements Gaussian blur algorithm by multiplying the fast fourier transform(FFT) of source image by the FFT of Gaussian-kernel image and finally doing inverse fourier transform of it.
Here is the part of the code,
// Carry out the convolution in Fourier space
compleximage fftkernelimg:=realFFT(kernelimg) (-> FFT of Gaussian-kernel image)
compleximage FFTSource:=realfft(warpimg) (-> FFT of source image)
compleximage FFTProduct:=FFTSource*fftkernelimg.modulus().sqrt()
realimage invFFT:=realIFFT(FFTProduct)
The point I want to ask is this
compleximage FFTProduct:=FFTSource*fftkernelimg.modulus().sqrt()
Why does the FFT of Gaussian-kernel need '.modulus().sqrt()' for the convolution?
It is related to the fact that the fourier transform of a Gaussian function becomes another Gaussian function?
Or It is related to a sort of limitation of discrete fourier transform?
Please answer me
Thanks
This is related to the general precision limitation of any floating point numeric computing. (see f.e. here, or more in depth here)
A rotational (real-valued) Gaussian of stand.dev. sigma should be transformed into a 100% real-values rotational Gaussioan of 1/sigma. However, doing this numerically will show you deviations: Just try the following:
number sigma = 30
number A0 = 1
realimage first := RealImage( "First", 8, 256, 256 )
first = A0 * exp( - (iradius**2/(2*sigma*sigma) ))
first.showimage()
complexImage second := FFT(first)
second.Showimage()
image nonZeroImaginaryMask = ( 0 != second.Imaginary() )
nonZeroImaginaryMask.Showimage()
nonZeroImaginaryMask.SetLimits(0,1)
When you then multiply these complex images (before back-transferring) you are introducing even more errors. By using modulus, one ensures that the forward transformed kernel is purely real and hence a better "damping" curve.
A better implementation of a FFT filtering code would actually create the FFT(Gaussian) directly with a std.dev of 1/sigma, as this is the analytically correct result. Doing a FFT of the kernel only makes sense if the kernel (or its FFT) is not analytically known.
In general: When implementing any "maths" into a program code, it can pay hugely to think it through with numerical computation limits in the back of your head. Reduce actual computation whenever possible (i.e. compute analytically and use the result instead of relying on brute force numerical computation) and try to "reshape" equations when possible, f.e. avoid large sums over many small numbers, be careful about checks against exact numeric values, try to avoid expressions which are very sensitive on small numerica errors etc.
How should stereo (2 channel) audio data be represented for FFT? Do you
A. Take the average of the two channels and assign it to the real component of a number and leave the imaginary component 0.
B. Assign one channel to the real component and the other channel to the imag component.
Is there a reason to do one or the other? I searched the web but could not find any definite answers on this.
I'm doing some simple spectrum analysis and, not knowing any better, used option A). This gave me an unexpected result, whereas option B) went as expected. Here are some more details:
I have a WAV file of a piano "middle-C". By definition, middle-C is 260Hz, so I would expect the peak frequency to be at 260Hz and smaller peaks at harmonics. I confirmed this by viewing the spectrum via an audio editing software (Sound Forge). But when I took the FFT myself, with option A), the peak was at 520Hz. With option B), the peak was at 260Hz.
Am I missing something? The explanation that I came up with so far is that representing stereo data using a real and imag component implies that the two channels are independent, which, I suppose they're not, and hence the mess-up.
I don't think you're taking the average correctly. :-)
C. Process each channel separately, assigning the amplitude to the real component and leaving the imaginary component as 0.
Option B does not make sense. Option A, which amounts to convert the signal to mono, is OK (if you are interested in a global spectrum).
Your problem (double freq) is surely related to some misunderstanding in the use of your FFT routines.
Once you take the FFT you need to get the Magnitude of the complex frequency spectrum. To get the magnitude you take the absolute of the complex spectrum |X(w)|. If you want to look at the power spectrum you square the magnitude spectrum, |X(w)|^2.
In terms of your frequency shift I think it has to do with you setting the imaginary parts to zero.
If you imagine the complex Frequency spectrum as a series of complex vectors or position vectors in a cartesian space. If you took one discrete frequency bin X(w), there would be one real component representing its direction in the real axis (x -direction), and one imaginary component in the in the imaginary axis (y - direction). There are four important values about this discrete frequency, 1. real value, 2. imaginary value, 3. Magnitude and, 4. phase. If you just take the real value and set imaginary to 0, you are setting Magnitude = real and phase = 0deg or 90deg. You have hence forth modified the resulting spectrum, and applied a bias to every frequency bin. Take a look at the wiki on Magnitude of a vector, also called the Euclidean norm of a vector to brush up on your understanding. Leonbloy was correct, but I hope this was more informative.
Think of the FFT as a way to get information from a single signal. What you are asking is what is the best way to display data from two signals. My answer would be to treat each independently, and display an FFT for each.
If you want a really fast streaming FFT you can read about an algorithm I wrote here: www.depthcharged.us/?p=176
I want to invert a 4x4 matrix. My numbers are stored in fixed-point format (1.15.16 to be exact).
With floating-point arithmetic I usually just build the adjoint matrix and divide by the determinant (e.g. brute force the solution). That worked for me so far, but when dealing with fixed point numbers I get an unacceptable precision loss due to all of the multiplications used.
Note: In fixed point arithmetic I always throw away some of the least significant bits of immediate results.
So - What's the most numerical stable way to invert a matrix? I don't mind much about the performance, but simply going to floating-point would be to slow on my target architecture.
Meta-answer: Is it really a general 4x4 matrix? If your matrix has a special form, then there are direct formulas for inverting that would be fast and keep your operation count down.
For example, if it's a standard homogenous coordinate transform from graphics, like:
[ux vx wx tx]
[uy vy wy ty]
[uz vz wz tz]
[ 0 0 0 1]
(assuming a composition of rotation, scale, translation matrices)
then there's an easily-derivable direct formula, which is
[ux uy uz -dot(u,t)]
[vx vy vz -dot(v,t)]
[wx wy wz -dot(w,t)]
[ 0 0 0 1 ]
(ASCII matrices stolen from the linked page.)
You probably can't beat that for loss of precision in fixed point.
If your matrix comes from some domain where you know it has more structure, then there's likely to be an easy answer.
I think the answer to this depends on the exact form of the matrix. A standard decomposition method (LU, QR, Cholesky etc.) with pivoting (an essential) is fairly good on fixed point, especially for a small 4x4 matrix. See the book 'Numerical Recipes' by Press et al. for a description of these methods.
This paper gives some useful algorithms, but is behind a paywall unfortunately. They recommend a (pivoted) Cholesky decomposition with some additional features too complicated to list here.
I'd like to second the question Jason S raised: are you certain that you need to invert your matrix? This is almost never necessary. Not only that, it is often a bad idea. If you need to solve Ax = b, it is more numerically stable to solve the system directly than to multiply b by A inverse.
Even if you have to solve Ax = b over and over for many values of b, it's still not a good idea to invert A. You can factor A (say LU factorization or Cholesky factorization) and save the factors so you're not redoing that work every time, but you'd still solve the system each time using the factorization.
You might consider doubling to 1.31 before doing your normal algorithm. It'll double the number of multiplications, but you're doing a matrix invert and anything you do is going to be pretty tied to the multiplier in your processor.
For anyone interested in finding the equations for a 4x4 invert, you can use a symbolic math package to resolve them for you. The TI-89 will do it even, although it'll take several minutes.
If you give us an idea of what the matrix invert does for you, and how it fits in with the rest of your processing we might be able to suggest alternatives.
-Adam
Let me ask a different question: do you definitely need to invert the matrix (call it M), or do you need to use the matrix inverse to solve other equations? (e.g. Mx = b for known M, b) Often there are other ways to do this w/o explicitly needing to calculate the inverse. Or if the matrix M is a function of time & it changes slowly then you could calculate the full inverse once, & there are iterative ways to update it.
If the matrix represents an affine transformation (many times this is the case with 4x4 matrices so long as you don't introduce a scaling component) the inverse is simply the transpose of the upper 3x3 rotation part with the last column negated. Obviously if you require a generalized solution then looking into Gaussian elimination is probably the easiest.