Find Max Delta of Sliding 3x3 Window Over Matrix

I have the code below to find the max delta between a point and its 8 neighbors in a sliding 3x3 window. I do this for the whole matrix input, A. This works fine and correct, but I am wondering if this can be accomplished with several uses of conv2 instead of the imerode/imdilation calls. The computation I'm doing is depicted in the attached image.

            % This method works fine...
            A = magic(3); % Test case
            %  This finds the max diff
            % between a pixel and its 8 neighbors.
            drMin = min(A(:)); drMax = max(A(:)); % Phantom (padding) pts outside of domain are assumed the avg.
            padPt = mean([drMin drMax]); % Value to use for padding.
            tmp1 = padarray(A,1,padPt,'both'); tmp2 = padarray(tmp1',1,padPt,'both'); Apad = tmp2'; % Pad matrix.
            maxApad = imdilate(Apad,ones(3)) - Apad; % Local max minus the map. Delta from each pt to max neighbor.
            minApad = imerode(Apad,ones(3)) - Apad; % Local min minus the map. Delta from each pt to min neighbor.
            tmp(:,:,1) = abs(maxApad); tmp(:,:,2) = abs(minApad);
            Ctmp = max(tmp,[],3); % Largest of the two gives the max delta from each point to its neighbors!
            C = Ctmp(2:end-1,2:end-1); % Remove padding...
            

The expected result for the magic(3) input is:

7	7	5
6	4	6
5	7	7

I have tried with the following code, but it's clearly off.

A = magic(3);

arrayTot = zeros(3,3,8);
kernel = zeros([9 1]); kernel(5) = 1;

for k = 1:9

    if k ~=5
        kernel(k) = -1;
        k1 = kernel;
        k1 = reshape(k1,[3 3]);
        tmp = conv2(A,k1,'same');
        arrayTot(:,:,k) = abs(tmp);
        kernel = zeros([9 1]); kernel(5) = 1;
    end
end
tst = max(arrayTot,[],3)

Any advice would be appreciated. Thanks.

Answer 1

[Sandeep Mishra]

Hi Paul,

You can use the MATLAB ‘conv2’ function to compute the Max-Delta of Sliding 3x3 Window by subtracting the center pixel with all 8 neighbours and taking the absolute difference.

Using an appropriate kernel can be helpful while dealing with the ‘conv2’ function, the following kernels can be used for the calculation:

% Kernels of neighbours are as follows: Right, Left, Up, Down, Top-Left, Top-Right, Bottom-Left, Bottom-Right
kernels = {
    [0 0 0; 0 -1 1; 0 0 0],
    [0 0 0; 1 -1 0; 0 0 0],
    [0 1 0; 0 -1 0; 0 0 0],
    [0 0 0; 0 -1 0; 0 1 0],
    [1 0 0; 0 -1 0; 0 0 0],
    [0 0 1; 0 -1 0; 0 0 0],
    [0 0 0; 0 -1 0; 1 0 0],
    [0 0 0; 0 -1 0; 0 0 1] 
    };

To handle edge cases and ensure accurate convolution, you can add the padding to the input matrix as follows:

A = magic(3)

A = 3×3

8 1 6 3 5 7 4 9 2

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padPt = mean([min(A(:)), max(A(:))]);

% Pad the matrix with mean value
Apad = padarray(A, [1, 1], padPt, 'both');

Refer to the following example code snippet to apply Convolution to compute differences and obtain Max-Delta result:

diffs = zeros([size(Apad), 8]);
% Apply each kernel with conv2 and store the absolute difference
for k = 1:8
    diffs(:,:,k) = abs(conv2(Apad, kernels{k}, 'same'));
end

% Using the maximum difference across all 8 neighbors
Ctmp = max(diffs, [], 3);

% Output after removing padding
res = Ctmp(2:end-1, 2:end-1)

res = 3×3

7 7 5 6 4 6 5 7 7

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Refer to the following MathWorks Documentation to learn more about ‘conv2’ function: https://www.mathworks.com/help/releases/R2024a/matlab/ref/conv2.html

I hope this helps you in resolving your query.

[Paul Safier]

Thanks. I see the only difference between my code and yours was the padding. That made the difference!