comparing pixels in 3x3 block

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

[Walter Roberson]

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??

Answer 1

[Hyeokjin Jho]

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