Chessboard distance in the image matrix - matlab

Question

Given an image matrix, how can I get the locations of the pixels whose chessboard distances from pixel A is less than D. I need to performed this for all pixels.
Using the MATLAB function bwdist I couldn't provide desired result. What's the solution?
[D,idx] = bwdist(Img,'chessboard');

Answer 1

Given an image, pixel and maximum distance:
% Test image
Image = zeros(20,30);
% Maximum chessboard distance from image
maxDist = 7;
% The pixel from which to measure distance
pix = [4,19];
To find the pixels who's chessboard distance from the pix are
less than maxDist and in the image bounds:
Option 1: Using bwdist
% Create a binary image with all pixels zero except 'pix'
bw = zeros(size(Image));
bw(pix(1), pix(2)) = 1;
% Get the chessboard distance transform
[D,idx] = bwdist(bw,'chessboard');
% Get the linear index of 'pix'
pixInd = sub2ind(size(bw), pix(1), pix(2));
% Find linear indices of pixels who's chessboard distance from pixel are
% less than 'maxDist'
pointsInd = find(idx == pixInd & D < maxDist);
% Remove 'pix'
pointsInd(pointsInd == pixInd) = [];
% Get the pairs of (x,y) of the pixels
[pointsX, pointsY] = ind2sub(size(bw), pointsInd);
Option 2: Using meshgrid
% Get the range of x and y indices who's chessboard distance from pixel are
% less than 'maxDist' and in the image bounds
xRange = max((pix(1)-(maxDist-1)),1):min((pix(1)+(maxDist-1)),size(Image,1));
yRange = max((pix(2)-(maxDist-1)),1):min((pix(2)+(maxDist-1)),size(Image,2));
% Create a mesgrid to get the pairs of (x,y) of the pixels
[pointsX, pointsY] = meshgrid(xRange, yRange);
pointsX = pointsX(:);
pointsY = pointsY(:);
% Remove 'pix'
pixIndToRemove = (pointsX == pix(1) & pointsY == pix(2));
pointsX(pixIndToRemove) = [];
pointsY(pixIndToRemove) = [];
Displaying result:
% Get linear indices of pixels
pointsInd = sub2ind(size(Image), pointsX, pointsY);
% To display the result, create a binary image with all found pixels
% colored white
bwPoints = zeros(size(Image));
bwPoints(pointsInd) = 1;
% Show points
imshow(bwPoints, 'InitialMagnification', 2000)
% Show pixel grid lines
hold on
[rows, cols] = size(bwPoints);
for row = 0.5 : 1 : (rows + 0.5)
line([0.5, cols+0.5], [row, row], 'Color', 'r', 'LineWidth', 0.5);
end
for col = 0.5 : 1 : (cols + 0.5)
line([col, col], [0.5, rows+0.5], 'Color', 'r', 'LineWidth', 0.5);
end
Efficiency and running in a loop over all image pixels:
Option 2 is way much faster than Option 1. I wrote first Option 1 because bwdist was mentioned in the question. Running Option 2 in a loop can be improved by calculating the pixels first and than shifting them to the location of each pixel:
% Get the range of x and y indices who's chessboard distance from pixel
% (0,0) are less than 'maxDist'
xRange = (-(maxDist-1)):(maxDist-1);
yRange = (-(maxDist-1)):(maxDist-1);
% Create a mesgrid to get the pairs of (x,y) of the pixels
[pointsX, pointsY] = meshgrid(xRange, yRange);
pointsX = pointsX(:);
pointsY = pointsY(:);
% Remove pixel (0,0)
pixIndToRemove = (pointsX == 0 & pointsY == 0);
pointsX(pixIndToRemove) = [];
pointsY(pixIndToRemove) = [];
for x=1:size(Image, 1)
for y=1:size(Image, 2)
% Get a shifted copy of 'pointsX' and 'pointsY' that is centered
% around (x, y)
pointsX1 = pointsX + x;
pointsY1 = pointsY + y;
% Remove the the pixels that are out of the image bounds
inBounds =...
pointsX1 >= 1 & pointsX1 <= size(Image, 1) &...
pointsY1 >= 1 & pointsY1 <= size(Image, 2);
pointsX1 = pointsX1(inBounds);
pointsY1 = pointsY1(inBounds);
% Do stuff with 'pointsX1' and 'pointsY1'
% ...
end
end

Answer 2

"The aim is to access the location of the pixels whose chessboard distances from pixel A is less than D. The process should be
performed for all pixels..."
Since D is creating a square selection area, just use simple maths..
For example: if D is 3 then from the [x,y] position of pixel A...
//# we minus D by 1 since you want less than D (not equal / higher)
Start-X = pixelA.x - (D-1); //from the left
End-X = pixelA.y + (D-1); //to the right
Start-Y = pixelA.y - (D-1); //from the top
End-Y = pixelA.y + (D-1); //to the bottom
That will give you a square perimeter that represents your required selection area.
Look at this example image below:
Each square is a pixel. If the "crown" icon represents pixel A and D is 3 (where your "less than D" means D has a maximum length of 2 pixels), can you see how the pseudo-code above applies?