Connected Component labeling without using bwlabel or bwconncomp functions ?

I have written a code for labelling and counting number of connected components in binary image. But I am not getting expected results. Can anyone find where I am making mistake ? My code is :

function cca()
A=imread('contours.jpg');
figure,imshow(A);
title('Original Image');
% k indicates number of components in binary image
k = 0;
global B;
B = zeros(size(A,1),size(A,2));
% to make sure boundary conditions, skip first row, column and last row,
% column. These will be taken care by recursive function calls later
for i = 2 : size(A,1) - 1 
  for j = 2 : size(A,2) - 1
      if A(i,j) == 1 && B(i,j) == 0
          k = k + 1;
          rcca(i,j,A,k);
      end
end
for i = 1 : size(A,1)
  if A(i,1) == 1 && B(i,1) == 0
      k = k + 1;
      B(i,1) = k;
  else
      if A(i,size(A,2)) == 1 && B(i,size(A,2)) == 0
          k = k + 1;
          B(i,size(A,2)) = k;
      end
  end
end
for j = 1 : size(A,2)
  if A(1,j) == 1 && B(1,j) == 0
      k = k + 1;
      B(1,j) = k;
  else
      if A(size(A,1),j) == 1 && B(size(A,1),j) == 0
          k = k + 1;
          B(size(A,1),j) = k;
      end
  end
end
fprintf('\ntotal number of components in image = %.0f\n',k);

Function rcca is as follows :

function rcca(x,y,A,k)
global B;
B(x,y) = k;
% dx and dy is used to check for 8 - neighbourhood connectivity
dx = [-1,0,1,1,1,0,-1,-1];
dy = [1,1,1,0,-1,-1,-1,0];
if x > 1 && y > 1 && x < size(A,1) && y < size(A,2)
  for i = 1 : 8
      nx = x + dx(i);
      ny = y + dy(i);
      if A(nx,ny) == 1 && B(nx,ny) == 0
          rcca(nx,ny,A,k);
      end
  end
end

Please help me in finding logical error. Here is my image :

Answer 1

[Thorsten]

A is not a binary image, but a gray scale image with values ranging from 0 to 255. If you use imhist(A) to view the data, you see that the black background data actually spread from 0 to 55. So if you check for A(i,j)==1 in your code, you do not check for a foreground pixel, but for a background pixel of value 1. Assuming that your code is correct, you are labeling the background pixels set to 1 due to jpg artifacts.

So the first step would be to convert the image to binary image:

 A = im2bw(A, 0.5);

Regarding the labeling algorithm, the following page may be useful https://en.wikipedia.org/wiki/Connected-component_labeling.

I implemented a labeling algorithm similar to yours. It works fine, but because of the heavy use of recursion it results in "Maximum recursion limit of 500 reached." for large images.

function B = thlabel
I = imread('contours.jpg');
I = imresize(I, 0.5); % reduce size not to reach Matlab's maximum 
                      % recursion limit 

global A
A = im2bw(I, 0.5);

global B
B = zeros(size(A));
global currentlabel
currentlabel = 1;

sz1 = size(A,1);
global offset
offset = [-sz1-1:-sz1+1 -1 +1 sz1-1:sz1+1];

for ind = 1:numel(A)
  if A(ind) == 1 && B(ind) == 0
    labelconnected(ind)
    currentlabel = currentlabel + 1;
  end
end

function labelconnected(ind)
global currentlabel
global offset
global A
global B
B(ind) = currentlabel;
imshow(B, []), drawnow
neighbors = ind + offset;
neighbors(neighbors <= 0 | neighbors > numel(A)) = [];
for ind = neighbors
  if A(ind) == 1 && B(ind) == 0
    labelconnected(ind)
  end
end

[Sumit Khatri]

Thanks, I got the correct answer by reducing the image size and changing it to black & white. This suggestion helped me a lot to find mistake in my code.

Answer 2

[Image Analyst]

I really don't know why you don't just use the solution I gave you, and you accepted, in your duplicate question. What was wrong with it? It worked. Why do you prefer to go through all that complicated stuff, instead of a single line of code, especially since you say it's not working?

[Sumit Khatri]

Actually I just wanted to implement the algorithm and compare the time for inbuilt function and my function. Now I got to know how we should take care of all possible cases to process images. Sorry for so silly questions but since I am beginner to MatLab, I am getting such questions. But Thanks for your suggestion.