Okay, so I'm working on a problem related to quantum chaos and one of the things I need to do is to map the unit cube in n-dimensions to a parallelepiped in n-dimensions and find all integer points in the interior of this parallelepiped. I have been trying to do this using the following scheme:
Given the linear map B and the dimension of the cube n, we find the coordinates of the corners of the unit hypercube by converting numbers j from 0 to (2^n -1) into their binary representation and turning them into vectors that describe the vertices of the cube.
The next step was to apply the map B to each of these vectors, which gives me a set of 2^n vectors describing the coordinates of the vertices of the parallelepiped in n dimensions
Now, we take the maximum and minimum value attained by any of these vertices in each coordinate direction, i.e the first element of my vectors might have a maximum value of 4 across all of the vertices and a minimum value of -3 etc. This gives me an n-dimensional rectangular prism that contains my parallelepiped and some extra unwanted space.
I now find all points with integer coordinates in this bounding rectangular prism described as vectors in n dimensions
Finally, I apply the inverse of the map B to each of the points and throw away any points that have any coefficients greater than 1 as they must originally have lain outside my unit hypercube.
My issue arises in step 4, I'm struggling to come up with a way of generating all vectors with integer coordinates in my rectangular hyper-prism such that I can change the number of dimensions n on the fly. Ideally, i'd like to be able to increase n at will until it becomes too computationally heavy to do so, but every method of finding all integer points in the prism i've tried so far has relied on n for loops to permute each element and thus I need to rewrite the code every time.
So I guess my question is this, is there any way to code this up so that I can change n on the fly? Also, any thoughts on the idea of the algorithm itself would be appreciated :) It wouldn't surprise me if i've overcomplicated things massively...
EDIT:
Of course as soon as I post the question I see a lovely little link in the side-bar where a clever method has been given already for how to do this: Generate a matrix containing all combinations of elements taken from n vectors
I'll leave this up for the moment just in case anyone has any comments on the method in general, but otherwise (since I can't upvote yet I'll just say it here) Luis Mendo, you are a hero!
Related
I have an image of a cytoskeleton. There are a lot of small objects inside and I want to calculate the length between all of them in every axis and to get a matrix with all this data. I am trying to do this in matlab.
My final aim is to figure out if there is any axis with a constant distance between the object.
I've tried bwdist and to use connected components without any luck.
Do you have any other ideas?
So, the end goal is that you want to globally stretch this image in a certain direction (linearly) so that the distances between nearest pairs end up the closest together, hopefully the same? Or may you do more complex stretching ? (note that with arbitrarily complex one you can always make it work :) )
If linear global one, distance in x' and y' is going to be a simple multiplication of the old distance in x and y, applied to every pair of points. So, the final euclidean distance will end up being sqrt((SX*x)^2 + (SY*y)^2), with SX being stretch in x and SY stretch in y; X and Y are distances in X and Y between pairs of points.
If you are interested in just "the same" part, solution is not so difficult:
Find all objects of interest and put their X and Y coordinates in a N*2 matrix.
Calculate distances between all pairs of objects in X and Y. You will end up with 2 matrices sized N*N (with 0 on the diagonal, symmetric and real, not sure what is the name for that type of matrix).
Find minimum distance (say this is between A an B).
You probably already have this. Now:
Take C. Make N-1 transformations, which all end up in C->nearestToC = A->B. It is a simple system of equations, you have X1^2*SX^2+Y1^2*SY^2 = X2^2*SX^2+Y2*SY^2.
So, first say A->B = C->A, then A->B = C->B, then A->B = C->D etc etc. Make sure transformation is normalized => SX^2 + SY^2 = 1. If it cannot be found, the only valid transformation is SX = SY = 0 which means you don't have solution here. Obviously, SX and SY need to be real.
Note that this solution is unique except in case where X1 = X2 and Y1 = Y2. In this case, grab some other point than C to find this transformation.
For each transformation check the remaining points and find all nearest neighbours of them. If distance is always the same as these 2 (to a given tolerance), great, you found your transformation. If not, this transformation does not work and you should continue with the next one.
If you want a transformation that minimizes variations between distances (but doesn't require them to be nearly equal), I would do some optimization method and search for a minimum - I don't know how to find an exact solution otherwise. I would pick this also in case you don't have linear or global stretch.
If i understand your question correctly, the first step is to obtain all of the objects center of mass points in the image as (x,y) coordinates. Then, you can easily compute all of the distances between all points. I suggest taking a look on a histogram of those distances which may provide some information as to the nature of distance distribution (for example if it is uniformly random, or are there any patterns that appear).
Obtaining the center of mass points is not an easy task, consider transforming the image into a binary one, or some sort of background subtraction with blob detection or/and edge detector.
For building a histogram you can use histogram.
I have a matrix M of size NxP. Every P columns are orthogonal (M is a basis). I also have a vector V of size N.
My objective is to transform the first vector of M into V and to update the others in order to conservate their orthogonality. I know that the origins of V and M are the same, so it is basically a rotation from a certain angle. I assume we can find a matrix T such that T*M = M'. However, I can't figure out the details of how to do it (with MATLAB).
Also, I know there might be an infinite number of transforms doing that, but I'd like to get the simplest one (in which others vectors of M approximately remain the same, i.e no rotation around the first vector).
A small picture to illustrate. In my actual case, N and P can be large integers (not necessarily 3):
Thanks in advance for your help!
[EDIT] Alternative solution to Gram-Schmidt (accepted answer)
I managed to get a correct solution by retrieving a rotation matrix R by solving an optimization problem minimizing the 2-norm between M and R*M, under the constraints:
V is orthogonal to R*M[1] ... R*M[P-1] (i.e V'*(R*M[i]) = 0)
R*M[0] = V
Due to the solver constraints, I couldn't indicate that R*M[0] ... R*M[P-1] are all pairwise orthogonal (i.e (R*M)' * (R*M) = I).
Luckily, it seems that with this problem and with my solver (CVX using SDPT3), the resulting R*M[0] ... R*M[P-1] are also pairwise orthogonal.
I believe you want to use the Gram-Schmidt process here, which finds an orthogonal basis for a set of vectors. If V is not orthogonal to M[0], you can simply change M[0] to V and run Gram-Schmidt, to arrive at an orthogonal basis. If it is orthogonal to M[0], instead change another, non-orthogonal vector such as M[1] to V and swap the columns to make it first.
Mind you, the vector V needs to be in the column space of M, or you will always have a different basis than you had before.
Matlab doesn't have a built-in Gram-Schmidt command, although you can use the qr command to get an orthogonal basis. However, this won't work if you need V to be one of the vectors.
Option # 1 : if you have some vector and after some changes you want to rotate matrix to restore its orthogonality then, I believe, this method should work for you in Matlab
http://www.mathworks.com/help/symbolic/mupad_ref/numeric-rotationmatrix.html
(edit by another user: above link is broken, possible redirect: Matrix Rotations and Transformations)
If it does not, then ...
Option # 2 : I did not do this in Matlab but a part of another task was to find Eigenvalues and Eigenvectors of the matrix. To achieve this I used SVD. Part of SVD algorithm was Jacobi Rotation. It says to rotate the matrix until it is almost diagonalizable with some precision and invertible.
https://math.stackexchange.com/questions/222171/what-is-the-difference-between-diagonalization-and-orthogonal-diagonalization
Approximate algorithm of Jacobi rotation in your case should be similar to this one. I may be wrong at some point so you will need to double check this in relevant docs :
1) change values in existing vector
2) compute angle between actual and new vector
3) create rotation matrix and ...
put Cosine(angle) to diagonal of rotation matrix
put Sin(angle) to the top left corner of the matric
put minus -Sin(angle) to the right bottom corner of the matrix
4) multiple vector or matrix of vectors by rotation matrix in a loop until your vector matrix is invertible and diagonalizable, ability to invert can be calculated by determinant (check for singularity) and orthogonality (matrix is diagonalized) can be tested with this check - if Max value in LU matrix is less then some constant then stop rotation, at this point new matrix should contain only orthogonal vectors.
Unfortunately, I am not able to find exact pseudo code that I was referring to in the past but these links may help you to understand Jacobi Rotation :
http://www.physik.uni-freiburg.de/~severin/fulltext.pdf
http://web.stanford.edu/class/cme335/lecture7.pdf
https://www.nada.kth.se/utbildning/grukth/exjobb/rapportlistor/2003/rapporter03/maleko_mercy_03003.pdf
So I was trying to spread one matrix elements, which were generated with poissrnd, to another with using some bigger (wider?) probability function (for example 100 different possibilities with different weights) to plot both of them and see if the fluctuations after spread went down. After seeing it doesn't work right (fluctuations got bigger) I tried to identify what I did wrong on a really simple example. After testing it for a really long time I still can't understand what's wrong. The example goes like this:
I generate vector with poissrnd and vector for spreading (filled with zeros at the start)
Each element from the poiss vector tells me how many numbers (0.1 of the element value) to generate from the possible options which are: [1,2,3] with corresponding weights [0.2,0.5,0.2]
I spread what I got to my another vector on 3 elements: the corresponding (k-th one), one bofore the corresponding one and one after the corresponding one (so for example if k=3, the elements should be spread like this: most should go into 3rd element of another vector, and rest should go to 2nd and 1st element)
Plot both 0.1*poiss vector and vector after spreading to compare if fluctuations went down
The way I generate weighted numbers is from this thread: Weighted random numbers in MATLAB
and this is the code I'm using:
clear all
clc
eta=0.1;
N=200;
fot=10000000;
ix=linspace(-100,100,N);
mn =poissrnd(fot/N, 1, N);
dataw=zeros(1,N);
a=1:3;
w=[.25,.5,.25];
for k=1:N
[~,R] = histc(rand(1,eta*mn(1,k)),cumsum([0;w(:)./sum(w)]));
R = a(R);
przydz=histc(R,a);
if (k>1) && (k<N)
dataw(1,k)=dataw(1,k)+przydz(1,2);
dataw(1,k-1)=dataw(1,k-1)+przydz(1,1);
dataw(1,k+1)=dataw(1,k+1)+przydz(1,3);
elseif k==1
dataw(1,k)=dataw(1,k)+przydz(1,2);
dataw(1,N)=dataw(1,N)+przydz(1,1);
dataw(1,k+1)=dataw(1,k+1)+przydz(1,3);
else
dataw(1,k)=dataw(1,k)+przydz(1,2);
dataw(1,k-1)=dataw(1,k-1)+przydz(1,1);
dataw(1,1)=dataw(1,1)+przydz(1,3);
end
end
plot(ix,eta*mn,'g',ix,dataw,'r')
The fluctuations are still bigger, and I can't identify what's wrong... Is the method for generating weighted numbers wrong in this situation? Cause it doesn't seem so. The way I'm accumulating data from the first vector seems fine too. Is there another way I could do it (so I could then optimize it for using 'bigger' probability functions)?
Sorry for my terrible English.
[EDIT]:
Here is simple pic to show what I meant (I hope it's understandable)
How about trying negative binomial distribution? It is often used as a hyper-dispersed analogue of Poisson distribution. Additional links can be found in this paper, as well as some apparatus in supplement.
Say, I have a cube of dimensions 1x1x1 spanning between coordinates (0,0,0) and (1,1,1). I want to generate a random set of points (assume 10 points) within this cube which are somewhat uniformly distributed (i.e. within certain minimum and maximum distance from each other and also not too close to the boundaries). How do I go about this without using loops? If this is not possible using vector/matrix operations then the solution with loops will also do.
Let me provide some more background details about my problem (This will help in terms of what I exactly need and why). I want to integrate a function, F(x,y,z), inside a polyhedron. I want to do it numerically as follows:
$F(x,y,z) = \sum_{i} F(x_i,y_i,z_i) \times V_i(x_i,y_i,z_i)$
Here, $F(x_i,y_i,z_i)$ is the value of function at point $(x_i,y_i,z_i)$ and $V_i$ is the weight. So to calculate the integral accurately, I need to identify set of random points which are not too close to each other or not too far from each other (Sorry but I myself don't know what this range is. I will be able to figure this out using parametric study only after I have a working code). Also, I need to do this for a 3D mesh which has multiple polyhedrons, hence I want to avoid loops to speed things out.
Check out this nice random vectors generator with fixed sum FEX file.
The code "generates m random n-element column vectors of values, [x1;x2;...;xn], each with a fixed sum, s, and subject to a restriction a<=xi<=b. The vectors are randomly and uniformly distributed in the n-1 dimensional space of solutions. This is accomplished by decomposing that space into a number of different types of simplexes (the many-dimensional generalizations of line segments, triangles, and tetrahedra.) The 'rand' function is used to distribute vectors within each simplex uniformly, and further calls on 'rand' serve to select different types of simplexes with probabilities proportional to their respective n-1 dimensional volumes. This algorithm does not perform any rejection of solutions - all are generated so as to already fit within the prescribed hypercube."
Use i=rand(3,10) where each column corresponds to one point, and each row corresponds to the coordinate in one axis (x,y,z)
As i've explained in a previous question: I have a dataset consisting of a large semi-random collection of points in three dimensional euclidian space. In this collection of points, i am trying to find the point that is closest to the area with the highest density of points.
As high performance mark answered;
the most straightforward thing to do would be to divide your subset of
Euclidean space into lots of little unit volumes (voxels) and count
how many points there are in each one. The voxel with the most points
is where the density of points is at its highest. Perhaps initially
dividing your space into 2 x 2 x 2 voxels, then choosing the voxel
with most points and sub-dividing that in turn until your criteria are
satisfied.
Mark suggested i use triplequad for this, but this is not a function i am familiar with, or understand very well. Does anyone have any pointers on how i could go about using this function in Matlab for what i am trying to do?
For example, say i have a random normally distributed matrix A = randn([300,300,300]), how could i use triplequad to find the point i am looking for? Because as i understand currently, i also have to provide triplequad with a function fun when using it. Which function should that be for this problem?
Here's an answer which doesn't use triplequad.
For the purposes of exposition I define an array of data like this:
A = rand([30,3])*10;
which gives me 30 points uniformly distributed in the box (0:10,0:10,0:10). Note that in this explanation a point in 3D space is represented by each row in A. Now define a 3D array for the counts of points in each voxel:
counts = zeros(10,10,10)
Here I've chosen to have a 10x10x10 array of voxels, but this is just for convenience, it would be only a little more difficult to have chosen some other number of voxels in each dimension, and there don't have to be the same number of voxels along each axis. Then the code
for ix = 1:size(A,1)
counts(ceil(A(ix,1)),ceil(A(ix,2)),ceil(A(ix,3))) = counts(ceil(A(ix,1)),ceil(A(ix,2)),ceil(A(ix,3)))+1
end
will count up the number of points in each of the voxels in counts.
EDIT
Unfortunately I have to do some work this afternoon and won't be able to get back to wrestling with the triplequad solution until later. Hope this is OK in the meantime.