Two-dimensional matched filter

Question

I want to implement two dimensional matched filter for blood vessel extraction according to the paper "Detection of Blood Vessels in Retinal Images Using Two-Dimensional Matched Filters" by Chaudhuri et al., IEEE Trans. on Medical Imaging, 1989 ( there's a PDF on the author's web site<\/a> ).

Such a kernel may be mathematically expressed as:

K(x,y) = -exp(-x^2/2*sigma^2)  for |x|<3*sigma,  |y|<L/2

Answer 1

The reason to build the filter from its analytic expression for each rotation, rather than using imrotate , is that the filter extent is not circular, and therefore rotating brings in "new" pixel values and pushes some other pixels out of the kernel. Furthermore, rotating a kernel constructed as here (smooth transition along one direction, step edge along the other dimension) requires different interpolation methods along each dimension, which imrotate cannot do. The resulting rotated kernel will always be wrong.

Both these issues can be easily seen when displaying the kernel you make together with two rotated versions:

This display brings an additional issues to the front: the kernel is not centered on a pixel, causing it to shift the output by half a pixel.

Note also that, when subtracting the mean, it is important that this mean be computed only over the original domain of the filter, and that any zeros used to pad this domain to a rectangular shape remain zero (these should not become negative).

The rotated kernels can be constructed as follows:

m = max(ceil(3*s),(L-1)/2);
[x,y] = meshgrid(-m:m,-m:m); % non-rotated coordinate system, contains (0,0)
t = pi/6;                    % angle in radian
u = cos(t)*x - sin(t)*y;     % rotated coordinate system
v = sin(t)*x + cos(t)*y;     % rotated coordinate system
N = (abs(u) <= 3*s) & (abs(v) <= L/2);   % domain
k = exp(-u.^2/(2*s.^2));     % kernel
k = k - mean(k(N));
k(~N) = 0;                   % set kernel outside of domain to 0

This is the result for the three rotations used in the example above (the grey around the edges of the kernel corresponds to the value 0, the black pixels have a negative value):

Another issue is that you use conv2 with the default 'full' output shape, you should be using 'same' here, so that the output of the filter matches the input.

Note that, instead of computing all filter responses, and computing the max afterwards, it is much easier to compute the max as you compute each filter response. All of the above leads to the following code:

img = im2double(rgb2gray(img));

s = 1.5; %sigma
L = 7;
theta = 0:15:165; %different rotations

out = zeros(size(img));

m = max(ceil(3*s),(L-1)/2);
[x,y] = meshgrid(-m:m,-m:m); % non-rotated coordinate system, contains (0,0)
for t = theta
   t = t / 180 * pi;        % angle in radian
   u = cos(t)*x - sin(t)*y; % rotated coordinate system
   v = sin(t)*x + cos(t)*y; % rotated coordinate system
   N = (abs(u) <= 3*s) & (abs(v) <= L/2); % domain
   k = exp(-u.^2/(2*s.^2)); % kernel
   k = k - mean(k(N));
   k(~N) = 0;               % set kernel outside of domain to 0

   res = conv2(img,k,'same');
   out = max(out,res);
end

out = out/max(out(:)); % force output to be in [0,1] interval that MATLAB likes
imwrite(out,'so_result.png')

I get the following output:

Answer 2

Nice solution, one more question? how do you put away the big circle produced by the conv around the output. my output<\/a>

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