How we can check for more than one edge passing through a pixel

Some pixel have more than one edge how can determine the number of edges passing through one pixel.

Answer 1

[Image Analyst]

If the edges are binary, like the pixel is 1 if an edge is at that pixel and 0 if there is no edge there, then you can just identify pixels that have more than 2 neighbors.

kernel = [1,1,1;1,0,1;1,1,1]; % Map of where the 8 neighbors lie.
numNeighbors = conv2(binaryEdgeImage, kernel, 'same'); % Count neighbors
moreThan1 = numNeighbors >= 3;  % Binary image: 1 if 3 or more neighbors at each pixel.
imshow(moreThan1); % Display it.

Answer 2

[Ashish Uthama]

When looking for 3x3 (or 2x2) patterns, BWLOOKUP is usually faster and more flexible:

   function bw = multiedge(bwedge)
   lut = makelut(@getlookup, 3);
   bw  = bwlookup(bwedge,lut);
      function isMultiEdge = getlookup(x)
          % See help for makelut. You can customize what a 'multi edge' 
          % pixel ought to look like.
          x(2,2)      = 0;
          isMultiEdge = sum(x(:))>=3;
      end
  end

Note - You need to define what you mean by multiedge. For example, IA's and the above code snippet will give:

   edge = 
       1     1     0
       0     1     0
       0     0     1
   >> multiedge(a)
   ans =
       0     0     0
       1     1     1
       0     0     0

[Image Analyst]

In some cases it's up for discussion. In that case you gave (which is a good illustration), if the edges were generated as 8 connected lines , then you'd have one line going to the upper left, one line going straight up, and one line going to the lower right. So there are 3 8-connected lines, and it is the case of two edge lines intersecting .

If it's 4-connected , then the 3 1's in the upper left are just one single line, and the pixel in the lower left is not connected . So that is a case of two "endpoints" of 2 lines being in the window. So there are 2 edge lines present in the window, but they do not intersect each other .

[Ashish Uthama]

In either case, a non edge pixel cannot be a multiedge pixel, correct? ((2,1) in ans above).

So Adel should either (preferably) fold his/her definition of multiedge into the computation, or clean up the results by masking in with the original edge image.

[Image Analyst]

True. In my code I should have made sure that the center pixel was 1 instead of "don't care". I guess that's the kind of mistakes that show up sometimes when I just post code off the top of my head and don't test them.