# I am trying to find the corners of the "rectangular" shapes. This code is working very well. But I dont know exactly how it works. Can you explain me ??

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ahmet ilhan on 1 Aug 2018
Answered: jim as on 14 Jan 2021
[I,J] = find(binaryimg>max(binaryimg(:))/2);
IJ = [I,J];
[~,idx] = min(IJ*[1 1; -1 1; 1 -1; -1 -1].');
Corners = IJ(idx,:);

Matt J on 1 Aug 2018
Edited: Matt J on 14 Apr 2019
Maybe this image will help. The corners of a polyhedron will maximize/minimize the intercept of lines of a certain slope - it's a linear programming principle. In the code shown, the lines used x+y, x-y, -x+y,-x-y are those oriented at 45 degrees to the x,y axes. As long as the rectangle edges are approximately parallel/perpendicular to the x,y axes, its corners will minimize the intercept of one of these lines.
Matt J on 6 Feb 2020
Incidentally, I have made a File Exchange submission which generalizes the method to any convex polygon:

### More Answers (1)

jim as on 14 Jan 2021
An algorithm addressing this problem that you have raised is escribed in https://arxiv.org/abs/2011.14035
DC