# comparing pixels in 3x3 block

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Elysi Cochin on 2 Jan 2020
Edited: Hyeokjin Jho on 3 Jan 2020
Having a matrix as shown below i wanted to take 3x3 pixel and compare the 3 pixels (highlighted in yellow colour) with the pixel (highlighted in green colour),
and if 2 or more pixel (highlighted in yellow) has value greater than the pixel (highlighted in green) and i wanted to assign 1 to it else 0
So in this case i will get 0-1-1-0 and then convert the binary 0110 to its corresponding decimal value = 6
Then the next 3x3 pixel
##### 1 CommentShowHide None
Walter Roberson on 2 Jan 2020
Why 0110? Left edge of the left block, then top edge of the right block, then middle row of the right block, then middle column of the left block??

Hyeokjin Jho on 3 Jan 2020
Edited: Hyeokjin Jho on 3 Jan 2020
Assuming your matrix is A
% collect green pixels
query_right = A(2:end-1,3:end);
query_left = A(2:end-1,1:end-2);
query_down = A(3:end,2:end-1);
query_up = A(1:end-2,2:end-1);
% collect yellow pixels
test_column = zeros(size(A,1)-2,size(A,2)-2,3);
for j = 2:size(A,2)-1
test_column(:,j-1,1:3) = toeplitz(A(3:end,j),A(3:-1:1,j));
end
test_row = zeros(size(A,1)-2,size(A,2)-2,3);
for i = 2:size(A,1)-1
test_row(i-1,:,1:3) = toeplitz(A(i,3:end),A(i,3:-1:1));
end
% evaluate
eval_right = sum(test_column>query_right,3)>=2;
eval_left = sum(test_column>query_left,3) >=2;
eval_down = sum(test_row>query_down,3) >=2;
eval_up = sum(test_row>query_up,3) >=2;
result = eval_right*2^3 + eval_down*2^2 + eval_left*2^1 + eval_up*2^0;
For matrix given in your example, the result would be
6 7 10
2 2 10
5 5 13

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