Projects

I still have some difficulties understanding sampling lattices in the 4D position-frequency space of images. Here’s a flow of thoughts on this:

A 1D-signal of length N can be interpreted as an element of C^N. (This was always confusing me: an N-dimensional vector is only a 1D-signal!) But in TFA it is actually considered as an infinite periodic vector with period N. Therefore the index set is not just {1,...,N} but rather Z_N:=Z/NZ. This set is a finite group under addition modulo N. The TF-domain of that 1D-signal space is Z_N×Z_N and therefore 2D; it still has the group structure (by components). A sampling subset Λ of this TF-plane takes out certain time-locations and frequencies. If one has a fixed window function g with same signal length, the Gabor family with regard to that sampling subset is the set of those shifts and modulations of the Gabor atom g where the time-locations and frequencies are given by Λ. If Λ has enough elements, namely |Λ|>N, then...??? No, the redundancy doesn’t tell anything about whether the Gabor family is a frame! It is automatically a frame (in the finite setting) as soon as it spans the whole signal space. So what does that redundancy tell us? And why is it automatically large enough as soon as the Gabor family is a frame? Or what?

Another open question is why or when one should have a subgroup as sampling subset. What advances does one get when the sampling is done on a subgroup? If Λ is a subgroup, then there is a dual? Is there no dual when Λ is no subgroup? Or does it just depend on the redundancy then? It might have something to do with the fact that the Gabor frame operator commutes with TF-shifts along that subgroup.

The things get even trickier for 2D-signals. An image of size p×q is also considered as an infinite periodic signal with periods p in one direction and q in the other. The signal space has N=pq dimensions. The index set is Z_p×Z_q. The TF-domain (actually position-frequency domain) becomes (Z_p×Z_q)×(Z_p×Z_q) and is therefore 4D. What’s unclear here is the term of separability with regard to the sampling subset. In the 1D-case (TF-plane is 2D) one names the sampling subset a separable lattice when the shifts and modulations are defined by the multiples of a divisor of N. I.e., one gets a rectangular grid (lattice) in the TF-plane in this case: For every time-shift there is the same set of modulations. A non-separable case could be given by a set of random sampling points. But these don’t form a subgroup in general. A non-separable subgroup could look like a rotated version of a separable lattice. For a rotation by 45° (π/4) one gets the special case of a quincunx lattice (if the correct distances are taken). And now to separabililty of a 4D-lattice: What does a 4D quincunx look like? Or another 4D non-separable lattice? Does this mean that whenever I can split into two 2D-lattices, one talks about separable lattices, independently from the question whether these two are separable again? I think separable means here that one either has a position-lattice ((x₁,x₂)-lattice) and a frequency lattice ((ω₁,ω₂)-lattice) or an (x₁,ω₁)- and an (x₂,ω₂)-lattice, independently from whether these are separable themselves or not. But how does one construct a 4D-set out of two 2D-sets with MATLAB/Octave?

• Find a number s that is not divided by the divisors of N=pq.
• Index vector, method 1:
• idx = mod(1:s:N*s, N);
• nil = find(idx==0); idx(nil)=N;
• Index vector, method 2 (preferred):
• idx = mod((1:s:N*s)-1, N)+1;
• imv(idx) = im(:);

• rim = reshape(imv(idx), p, q);

Let’s try this with our zebra. In my previous tries I simply took p=479 and q=480 to have gcd(479,480)=1, but 479 is a prime number and I therefore have no possibility to divide it by some step size. For GA I have to lay some kind of lattice over the image. So there is actually a second condition forbidding p and q to be prime. I take p=473 instead, it has 11 and 43 as divisors.

As one can see, N has many divisors, but e.g. 19 is not one of them. We can’t take e.g. 9 although it doesn’t occur either, as it is divided by 3, which is a divisor of N. Taking such a wrong step size will lead one back to the first pixel too early and one can never span the whole image.

(zoomed)

Indeed looks like our zebra!

By the way, another try of comparing the FFT of the vector with the FFT2 of the image shows again that one cannot get the FFT2 by doing an FFT of the vector. This is because the line-patterns in the image might not be preserved under that transform, as neighboring pixels might not become neighbors in the resulting vector.

Regarding the previous questions I noticed that GA on LCA groups is too general for my concerns and that the study of GA on finite abelian groups is enough.

In a certain paper HGFei published a result that for p×q-images where p and q are relatively prime there is an isomorphism to a vector of length N=pq. The theory considers the groups and where . One explicit mapping from the matrix indices to the corresponding vector indices is given by

.

Just for a first test I set α=p and β=q and define a primitive mapping function. (I’ll have to find a quicker transform, as in my tests the switching takes too long. But this could be due to the actually large test image. In addition, I currently only manage to do the backward step by storing the indices in an index matrix.)

I have to return N instead of 0 to have a correct index value. Although the mapping is actually an isomorphism I didn’t know how to go back to the tuple (j,k), so I store the indices in an index matrix while building the image vector:

58 seconds!? However, the image looks like this as a vector:

(zoomed)