hough peaks not symmetric for rectangle

hello i use hough transform for rectangle the peaks should be symetric. but they are not symmetric around theta angle based on

[http://citeseerx.ist.psu.edu/viewdoc/download?doi=10.1.1.59.4239&rep=rep1&type=pdf]

%%just test
close all;clear;
m=zeros(200,300);
m=logical(m);
m(50,30:190)=1;
m(100,30:190)=1; 
m(50:100,30)=1;
m(50:100,190)=1;

%m=imrotate(m,45);

stats=regionprops(m,'BoundingBox');
bx = stats.BoundingBox;

m=imcrop(m,bx);
m=padarray(m,[3 3]);

figure(1),imagesc(m)
 [H, T, R] = hough(m);

               figure(2)
               imshow(imadjust(mat2gray(H)),'XData',T,'YData',T,...
                   'InitialMagnification','fit');
               title('Hough Transform of  Image');
               xlabel('\theta'), ylabel('\rho');
               axis on, axis normal, hold on;
               colormap(hot);
               Hv=H(:);
               mxH = 0.2*max(Hv);
               ii=find(Hv>=mxH);
               [aa bb]   =  ind2sub(size(H),ii);
               figure(4),imagesc(T,R,H),hold on,plot(T(bb),R(aa),'gs','MarkerSize',10,'MarkerFaceColor',[0.5,0.5,0.5]);

Answer 1

[Anand]

The symmetry should hold only for rectangles centered at the origin. I just skimmed the paper, but it seems to describe a technique where the search space is centered around the peak you are currently looking at. In MATLAB, the origin is at the top left pixel in the image.

When the rectangle is at the center, rho1=-rho2 and rho3=-rho4. This is not the case when the rectangle isn't centered at the origin.

Answer 2

[Michael scheinfeild]

yes you right i suppose i need to write the hough transform by myself with moving window , i suppose the center should be at center of mass of the shape