I have two sparse matrices "Matrix1" and "Matrix2" of the same size p x n.
By sparse matrix I mean that it contains a lot of exactly zero elements.
I want to show the two matrices under the same colormap and a unique colorbar. Doing this in MATLAB is straightforward:
bottom = min(min(min(Matrix1)),min(min(Matrix2)));
top = max(max(max(Matrix1)),max(max(Matrix2)));
subplot(1,2,1)
imagesc(Matrix1)
colormap(gray)
caxis manual
caxis([bottom top]);
subplot(1,2,2)
imagesc(Matrix2)
colormap(gray)
caxis manual
caxis([bottom top]);
colorbar;
My problem:
In fact, when I show the matrix using imagesc(Matrix), it can ignore the noises (or backgrounds) that always appear with using imagesc(10*log10(Matrix)).
That is why, I want to show the 10*log10 of the matrices. But in this case, the minimum value will be -Inf since the matrices are sparse. In this case caxis will give an error because bottom is equal to -Inf.
What do you suggest me? How can I modify the above code?
Any help will be very appreciated!
A very important point is that the minimum value in your matrix will always be 0. Leveraging this, a very simple way to address your problem is to add 1 inside the log operation so that values that map to 0 in the original matrix also map to 0 in the log operation. This avoids the -Inf error that you're encountering. In fact, this is a very common way of visualizing the Fourier Transform if you will. Adding 1 to the logarithm ensures that the transform has no negative values in the output, yet the derivative or its rate of change remains intact as the effect is simply a translation of the curve by 1 unit to the left.
Therefore, simply do imagesc(10*log10(1 + Matrix));, then the minimum is always bounded at 0 while the maximum is unbounded but subject to the largest value that is seen in Matrix.
Related
I am working in a program that concerns the optimization of some objective function obj over the scalar beta. The true global minimum beta0 is set at beta0=1.
In the mwe below you can see that obj is constructed as the sum of the 100-R (here I use R=3) smallest eigenvalues of the 100x100 symmetric matrix u'*u. While around the true global minimum obj "looks good" when I plot the objective function evaluated at much larger values of beta the objective function becomes very unstable (here or running the mwe you can see that multiple local minima (and maxima) appear, associated with values of obj(beta) smaller than the true global minimum).
My guess is that there is some sort of "numerical instability" going on, but I am unable to find the source.
%Matrix dimensions
N=100;
T=100;
%Reproducibility
rng('default');
%True global minimum
beta0=1;
%Generating data
l=1+randn(N,2);
s=randn(T+1,2);
la=1+randn(N,2);
X(1,:,:)=1+(3*l+la)*(3*s(1:T,:)+s(2:T+1,:))';
s=s(1:T,:);
a=(randn(N,T));
Y=beta0*squeeze(X(1,:,:))+l*s'+a;
%Give "beta" a large value
beta=1e6;
%Compute objective function
u=Y-beta*squeeze(X(1,:,:));
ev=sort(eig(u'*u)); % sort eigenvalues
obj=sum(ev(1:100-3))/(N*T); % "obj" is sum of 97 smallest eigenvalues
This evaluates the objective function at obj(beta=1e6). I have noticed that some of the eigenvalues from eig(u'*u) are negative (see object ev), when by construction the matrix u'*u is positive semidefinite
I am guessing this may have to do with floating point arithmetic issues and may (partly) be the answer to the instability of my function, but I am not sure.
Finally, this is what the objective function obj evaluated at a wide range of values for betalooks like:
% Now plot "obj" for a wide range of values of "beta"
clear obj
betaGrid=-5e5:100:5e5;
for i=1:length(betaGrid)
u=Y-betaGrid(i)*squeeze(X(1,:,:));
ev=sort(eig(u'*u));
obj(i)=sum(ev(1:100-3))/(N*T);
end
plot(betaGrid,obj,"*")
xlabel('\beta')
ylabel('obj')
This gives this figure, which shows how unstable it becomes for extreme values for beta.
The key here is noticing that computing eigenvalues can be a hard problem.
Actually the condition number for this problem is K = norm(A) * norm(inv(A)) (don't compute it this way, use cond(). This means the the an (relative) perturbation in the inpute (i.e. the matrix entries) gets amplified by the condition number when computing the output. I modified your code a little bit to compute and plot the condition number in each step. It turns out that for a large part of the range you are interested in it is greater than 10^17, which is abysmal. (Note that the double floating point numbers are accurate to not quite 16 significant (decimal) digits. This means even the representation error of double floating point numbers will here produce errors that make every digit "insignificant".) This already explains the bad behaviour. You should note that usually we can compute the largest eigenvalues quite accurately, the errors in the smaller (in magnitude) ones usually increase.
If the condition number was better (closer to 1) I would have suggested
computing the singular values, as they happen to be the eigenvalues (due to the symmetry). The svd is numerically more stable, but with this really bad
condition even this will not help. In the following modification of the
final snippet I added a graph that plots the condition number.
The only case where anything is salvageable is for R=0, then we actually
want to compute the sum of all eigenvalues, which happens to be the
trace of our matrix, which can easily be computed by just summing the
diagonal entries.
To summarize: This problem seems to have an inherent bad condition, so it doesn't really matter how you compute it. If you have a completely different formulation for the same problem that might help.
% Now plot "obj" for a wide range of values of "beta"
clear obj
L = 5e5; % decrease to 5e-1 to see that the condition number is still >1e9 around the optimum
betaGrid=linspace(-L,L,1000);
condition = nan(size(betaGrid));
for i=1:length(betaGrid)
disp(i/length(betaGrid))
u=Y-betaGrid(i)*squeeze(X(1,:,:));
A = u'*u;
ev=sort(eig(A));
condition(i) = cond(A);
obj(i)=sum(ev(1:100-3))/(N*t); % for R=0 use trace(A)/(N*T);
end
subplot(1,2,1);
plot(betaGrid,obj,"*")
xlabel('\beta')
ylabel('obj')
subplot(1,2,2);
semilogy(betaGrid, condition);
title('condition number');
I want to normalize images whose pixel can have negative values and found two different ways to do that. Given a two-dimensional matrix X I can do the following:
a) X = 0.5*((X/max(abs(X))+1)
b) X = (X-min(X))/(max(X)-min(X))
Since I'm not an expert, I'm not sure which of the two is the more useful way to normalize images. Does one of the two options have certain advantages?
For GLCM is does not at all matter where the 0 level is, what matters is the differences between intensities. Thus, I would pick the method that linearly stretches between the min and max intensity. This method uses the output range best, and therefore introduced the least quantization error.
When comparing GLCM results across images, it is best if all images are stretched the same way. I would select a global min and max, keep those constant for all images in the set.
Note that for other purposes, the answer will be different.
The second approach will use the full range between 0 and 1, which may be what you want. The first approach will map 0 always to 0.5. When the data is symmetrically spread around 0, also the first approach will use the full range between 0 and 1.
Up to you to decide what you want.
I have a grid map NxN. Each cell may have the value '0' or '1'. I am trying to find the exact number of distinct rectangle sub-blocks of the map that include a specific number of '1' and this number can be between 1 and 6. I have thought of searching for each possible rectangle but this is very slow for a map of size 500x500 and the solution must be ~ 1 sec for a common desktop computer. Can someone tell me a corresponding problem so I can look for a working algorithm or better can someone suggest me a working algorithm for this problem? Thank you all in advance!
I imagine that your search of all the rectangles is slow because you are actually counting on each possible rectangle. The solution to this is not to count all rectangles, but rather create a second array of NxN which contains the count for the rectangle (0,0..x,y), call this OriginCount. Then to calculate the count for any given rectangle, you will not have to go through the rectangle and count. You can simply use
Count(a,b..c,d) = OriginCount(c,d) + OriginCount(a-1,b-1) -
OriginCount(a-1,d) - OriginCount(c,b-1)
That turns the problem of counting the ones in any given rectangle, from an N2 problem to a discrete or constant time problem, and your code gets in the order of thousands times faster (for your 500x500 case)
Mind you, in order to set up the OriginCount array, you can use the same concept, don't just go count the ones for each rectangle, from 0,0 to x,y. Rather, use the formula
OriginCount(x,y) = OriginCount(x-1,y) + OriginCount(x,y-1) - OriginCount(x-1,y-1) +
GridMap(x,y) == 1 ? 1 : 0;
Mind you, you have to account for edge cases - where x=0 or y=0.
I am wondering if I understood the mean normalization of images correctly.
As far as I know, you calculate the mean value over all pixels (lets assume it is in grayscale). Then, for each pixel, you subtract this mean value.
But how should one deal with negative values which could arise? For example, the whole image has a mean value of 100, but one specific pixel has an intensity of 90. After this normalization, the pixel's value would be -10.
This may not be quite what you're looking for but one option that avoids negative numbers in your output would be to normalize to the range of values present rather than to the image mean.
The equation would be: X' = (X - Xmin)/(Xmax - Xmin). This rescales the image to be between 0 and 1 (no negative values involved). If you'd like to save it as an easily view-able greyscale you could multiply values by 255 to rescale it.
It may also be worth noting that unless the entire image has a constant intensity, there is guaranteed to be some negative values after subtracting the mean (not simply a possibility that they could arise).
You don't have to deal with negative inputs, the model can handle them. It is good practice, for a Neural Network for example, to have inputs in the range [-1, 1]
I need to generate a (pseudo) random sequence of N bit integers, where successive integers differ from the previous by only 1 bit, and the sequence never repeats. I know a Gray code will generate non-repeating sequences with only 1 bit difference, and an LFSR will generate non-repeating random-like sequences, but I'm not sure how to combine these ideas to produce what I want.
Practically, N will be very large, say 1000. I want to randomly sample this large space of 2^1000 integers, but I need to generate something like a random walk because the application in mind can only hop from one number to the next by flipping one bit.
Use any random number generator algorithm to generate an integer between 1 and N (or 0 to N-1 depending on the language). Use the result to determine the index of the bit to flip.
In order to satisfy randomness you will need to store previously generated numbers (thanks ShreevatsaR). Additionally, you may run into a scenario where no non-repeating answers are possible so this will require a backtracking algorithm as well.
This makes me think of fractals - following a boundary in a julia set or something along those lines.
If N is 1000, use a 2^500 x 2^500 fractal bitmap (obviously don't generate it in advance - you can derive each pixel on demand, and most won't be needed). Each pixel move is one pixel up, down, left or right following the boundary line between pixels, like a simple bitmap tracing algorithm. So long as you start at the edge of the bitmap, you should return to the edge of the bitmap sooner or later - following a specific "colour" boundary should always give a closed curve with no self-crossings, if you look at the unbounded version of that fractal.
The x and y axes of the bitmap will need "Gray coded" co-ordinates, of course - a bit like oversized Karnaugh maps. Each step in the tracing (one pixel up, down, left or right) equates to a single-bit change in one bitmap co-ordinate, and therefore in one bit of the resulting values in the random walk.
EDIT
I just realised there's a problem. The more wrinkly the boundary, the more likely you are in the tracing to hit a point where you have a choice of directions, such as...
* | .
---+---
. | *
Whichever direction you enter this point, you have a choice of three ways out. Choose the wrong one of the other two and you may return back to this point, therefore this is a possible self-crossing point and possible repeat. You can eliminate the continue-in-the-same-direction choice - whichever way you turn should keep the same boundary colours to the left and right of your boundary path as you trace - but this still leaves a choice of two directions.
I think the problem can be eliminated by making having at least three colours in the fractal, and by always keeping the same colour to one particular side (relative to the trace direction) of the boundary. There may be an "as long as the fractal isn't too wrinkly" proviso, though.
The last resort fix is to keep a record of points where this choice was available. If you return to the same point, backtrack and take the other alternative.
While an algorithm like this:
seed()
i = random(0, n)
repeat:
i ^= >> (i % bitlen)
yield i
…would return a random sequence of integers differing each by 1 bit, it would require a huge array for backtracing to ensure uniqueness of numbers.
Further more your running time would increase exponentially(?) with increasing density of your backtrace, as the chance to hit a new and non-repeating number decreases with every number in the sequence.
To reduce time and space one could try to incorporate one of these:
Bloom Filter
Use a Bloom Filter to drastically reduce the space (and time) needed for uniqueness-backtracing.
As Bloom Filters come with the drawback of producing false positives from time to time a certain rate of falsely detected repeats (sic!) (which thus are skipped) in your sequence would occur.
While the use of a Bloom Filter would reduce the space and time your running time would still increase exponentially(?)…
Hilbert Curve
A Hilbert Curve represents a non-repeating (kind of pseudo-random) walk on a quadratic plane (or in a cube) with each step being of length 1.
Using a Hilbert Curve (on an appropriate distribution of values) one might be able to get rid of the need for a backtrace entirely.
To enable your sequence to get a seed you'd generate n (n being the dimension of your plane/cube/hypercube) random numbers between 0 and s (s being the length of your plane's/cube's/hypercube's sides).
Not only would a Hilbert Curve remove the need for a backtrace, it would also make the sequencer run in O(1) per number (in contrast to the use of a backtrace, which would make your running time increase exponentially(?) over time…)
To seed your sequence you'd wrap-shift your n-dimensional distribution by random displacements in each of its n dimension.
Ps: You might get better answers here: CSTheory # StackExchange (or not, see comments)