How to find "rectangular" corners?

Hey,

I have this image:

I want to find the 4 corners of the "rectangular", but I don't want to use the "corner" function. What can I do?

Thanks.

[Cedric]

If your rectangles are not too degenerate, you could get corners with four 2D convolutions using appropriate kernels.

Answer 1

[Matt J]

If the quadrilateral is roughly aligned with the edges of the image, you could also find the corners as follows,

    [I,J]=find(Image>max(Image(:))/2);
    IJ=[I,J];  

    [~,idx]=min(IJ*[1 1; -1 -1; 1 -1; -1 1].');

    corners=IJ(idx,:)

[Mor]

Hi,

It works. Can you explain what this line: [~,idx]=min(IJ*[1 1; -1 -1; 1 -1; -1 1].');?

[Matt J]

It finds where the sums and differences of the coordinates in the shape are maximized and minimized.

[Bálint Udvardy]

Hi, I have just found this thread. I have a similar problem now: I am working on a nonogram-solver and I want to remove the perspective from the image to get the picture of the puzzle, but after extracting the largest connected component from the image (the puzzle grid) and using convex hull of that component I get a pentagram. Unfortunately, all the corner detectors fail to detect the 5 corners. Using your method detects 4 corners, which itself is an improvement, but I need that 5th corner to calculate the theoretical "true" upper left corner, which i need to remove the perspective.

[Matt J]

Why do you need the "upper left" corner? Why can't you use the 4 point correspondences that you already have to compute the perspective transform?

[Bálint Udvardy]

How? I only know that the right and the left vertical lines should be paralell to each other and have the same length in real, but on the photo it can be shorter or longer.

[Matt J]

You know all the little boxes in the puzzle grid are the same size and therefore you know the world coordinates of all corners of those boxes with respect to that grid. That gives you lots of point-to-point correspondence information.

[Syed Haider]

Dear Matt,

Thanks for the code. It did work really well on most of the images. But if I apply the same code to the image attached,

I get the wrong corners. I have attached the output image below. Any suggestions would be highly appreciated.

Thanks,

Irtaza

[Matt J]

A quick fix would be to imrotate() the image, so the rectangle is better aligned with the edges of the image grid. Then find the corners of the rotated rectangle and then rotate back.

[Image Analyst]

Compute the centroid, then the distances to the boundary points. Then use findpeaks() to find the vertices. I have a demo if you want - start your own brand new thread.

[Balkrishna Patankar]

What is the idea behind the code ? Is it a standard kernel ?

  [~,idx]=min(IJ*[1 1; -1 -1; 1 -1; -1 1].') 

[Matt J]

My rationale for the code is described here,

https://www.mathworks.com/matlabcentral/answers/413079-i-am-trying-to-find-the-corners-of-the-rectangular-shapes-this-code-is-working-very-well-but-i-d#answer_331199

[Balkrishna Patankar]

Thank you (y)

[Matt J]

Here is a generalization of the approach to arbitrary convex quadrilaterals. No particular orientation is assumed.

N=360;
theta=linspace(0,360,N);
[I,J]=find(Image);
IJ=[I,J];
c=nan(size(theta));
for i=1:N
   
    [~,c(i)]=max(IJ*[cosd(theta(i));sind(theta(i))]);
    
    
end
H=histcounts(c,1:numel(I)+1);
[~,k] = maxk(H,4);
corners=IJ(k,:)

Answer 2

[Matt J]

Use pgonCorners from the File Exchange (Download). It applies to any convex polyhedron.

numVertices=4;

corners=pgonCorners(Image,numVertices)

Answer 3

[Image Analyst]

Why not try the corner() function in the Image Processing Toolbox. What do you have against using that?

Or else call bwboundaries() and go along the coordinates looking for kinks in the curve as shown by the FAQ: http://matlab.wikia.com/wiki/FAQ#How_do_I_find_.22kinks.22_in_a_curve.3F

[Mor]

Hi,

Thanks for your answer. The reason I don't want to use the corner() function is because I have more pictures and in some of them the corner function doesn't work good (detects 2 corners is the same place) I think it's because the lines are not straight. maybe there is some configuration to make it work for every picture.

Answer 4

[Matt J]

Maybe use edge() followed by houghlines() with an appropriate FillGap selection? The endpoints of the line segments returned by houghlines would be the corners of the quadrilateral.

[Mor]

Hi Matt,

Here is the result:

The top left and the bottom right are not exactly the corner, It's because it's not a straight lines.

[Image Analyst]

If you need the exact corner, use (untested)

boundaries = bwboundaries(binaryImage);
x = boundaries{:,1};
y = boundaries{:,2};

and compare every x and y to see which is closest to the hough points

distances = sqrt((xh - x).^ 2 + (yh - y) .^ 2)
[minDistance, indexOfMin] = min(distances);
xc = x(indexOfMin);
yc = y(indexOfMin);

Do the above for each hough estimated point to find the point in the blob which is closest to the hough point (xh, yh).

[Matt J]

You can be generous with the RhoResolution, given the large size of the quadrilateral. I get a pretty good fit to the edges with the following,

[H,T,R] = hough(E,'RhoResolution',4,'Theta',-90:.5:89);

Similar to what ImageAnalyst was saying, this initial line fit should allow you to segment the boundary points into 4 separate edges. You do this by finding the closest point to each initial line. You can then do a more refined line fit to each edge using each group of points (e.g., using polyfit).

Answer 5

[Image Analyst]

If the quad is roughly aligned with the edges of the image you could also get the distance from the 4 corners. For each corner, take the one point on the white boundary that has the minimum distance. No need to mess with hough in that case.