CODEWITHSUNDEEP
matlabimage-processingcluster-analysis
elMariachi
elMariachi

Reputation: 1

Image clustering - problem with matlab function

Curently I'm writing a project about image recognition and clusterization. In publication which is a basis for my project there is this equation

enter image description here

Variables description is given below

Rj - is a rotation matrix of j-th cluster

t_j - is a translation vector of j-th cluster

p*ij - is a i-th point from j-th cluster

x_i - is a i-th point from the image

I had a little problem with writing this function so I asked the author of publication if he could share with me a source code. Here is what I got

ddx=D.x-Q.translation(1);
 ddy=D.y-Q.translation(2);
 st=sin(Q.theta); ct=cos(Q.theta);  R=[ct -st; st ct]; % rotation matrix
 qq=R*[ppx0; ppy0];
 qqd2=sum(qq.*qq,1);
 Q.scale=sum((ddx.*qq(1,:)+ddy.*qq(2,:)).*Um) / sum(qqd2.*Um);

Here D.x, and D.y are the coordinates of data points

Q.translation (a vector), Q.scale, and Q.theta are the transform parameters

ppx0 and ppy0 are the x- and y- coordinates of *p**ij

Um is the matrix containing [Umij]

However I have a hard time with understanding this solution. First of all I don't understand why he uses array multiplication (operator .*) instead of matrix multiplication (operator *) what is more it seems that he takes only one/first point p*

I hope somebody will help me to try this source code. Thanks in advance

Upvotes: 0

Views: 768

Answers (1)

Phonon
Phonon

Reputation: 12727

it looks like ppx0 and ppy0 are not coordinates, but rather vectors or coordinates. This way,

R*[ppx0; ppy0] =
[ct -st ; st ct] * [x_0 x_1 ... x_N-1 ; y_0 y_1 ... y_N-1]

Therefore qq is a 2xN vector, just like [ppx0; ppy0].

In qqd2=sum(qq.*qq,1), the .* operator is used, because you're actually squaring every value of a matrix to find the square distance of each of its coordinate pairs later on.

In ddx.*qq(1,:)+ddy.*qq(2,:) the .* operator is used as a shortcut to replace a double sum. In other words, istead of taking a sum of each matrix product individually (which in itself is a sum of multiplication), they first performed all necessary multiplications and then took a sum. (I hope this makes sense). You can prove all of these work with some basic matrix algebra.

Upvotes: 1

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