CODEWITHSUNDEEP
matlabimage-processingsignal-processingshadow
PendingVegan
PendingVegan

Reputation: 137

Implementation of shadow free 1d invariant image

I implemented a method for removing shadows based on invariant color features found in the paper Entropy Minimization for Shadow Removal. My implementation seems to be yielding similar computational results sometimes, but they are always off, and my grayscale image is blocky, maybe as a result of incorrectly taking the geometric mean.

Here is an example plot of the information potential from the horse image in the paper as well as my invariant image. Multiply the x-axis by 3 to get theta(which goes from 0 to 180):

informationpot

And here is the grayscale Image my code outputs for the correct maximum theta (mine is off by 10):

grayscale

You can see the blockiness that their image doesn't have:

Here is their information potential:

informationpot2

When dividing by the geometric mean, I have tried using NaN and tresholding the image so the smallest possible value is .01, but it doesn't seem to change my output.

Here is my code:

I = im2double(imread(strname));
[m,n,d] = size(I);
I = max(I, .01);
chrom = zeros(m, n, 3, 'double');
for i = 1:m
    for j = 1:n
       % if ((I(i,j,1)*I(i,j,2)*I(i,j,3))~= 0)
              chrom(i,j, 1) = I(i,j,1)/((I(i,j,1)*I(i,j,2)*I(i,j, 3))^(1/3));
              chrom(i,j, 2) = I(i,j,2)/((I(i,j,1)*I(i,j,2)*I(i,j, 3))^(1/3));
              chrom(i,j, 3) = I(i,j,3)/((I(i,j,1)*I(i,j,2)*I(i,j, 3))^(1/3));
          % else
          %    chrom(i,j, 1) = 1;
          %    chrom(i,j, 2) = 1;
          %    chrom(i,j, 3) = 1;
        % end
    end
end

p1 = mat2gray(log(chrom(:,:,1)));
p2 = mat2gray(log(chrom(:,:,2)));
p3 = mat2gray(log(chrom(:,:,3)));
X1 = mat2gray(p1*1/(sqrt(2)) - p2*1/(sqrt(2)));
X2 = mat2gray(p1*1/(sqrt(6)) + p2*1/(sqrt(6)) - p3*2/(sqrt(6)));
maxinf = 0;
maxtheta = 0;
data2 = zeros(1, 61);
for theta = 0:3:180
M = X1*cos(theta*pi/180) - X2*sin(theta*pi/180);
s = sqrt(std2(X1)^(2)*cos(theta*pi/180) + std2(X2)^(2)*sin(theta*pi/180));
s = abs(1.06*s*((m*n)^(-1/5)));
[m, n] = size(M);
length = m*n;
sources = zeros(1, length, 'double');
count = 1;
for x=1:m
    for y = 1:n
    sources(1, count) = M(x , y);
    count = count + 1;
end
end
weights = ones(1, length);
sigma = 2*s;
[xc , Ak] = fgt_model(sources , weights , sigma , 10, sqrt(length) , 6  );
sum1 = sum(fgt_predict(sources , xc , Ak , sigma , 10 ));
sum1 = sum1/sqrt(2*pi*2*s*s);
data2(theta/3 + 1) = sum1;
if (sum1 > maxinf)
   maxinf = sum1;
   maxtheta = theta;
end
end
InvariantImage2 = cos(maxtheta*pi/180)*X1 + sin(maxtheta*pi/180)*X2;

Assume the Fast Gauss Transform is correct.

Upvotes: 3

Views: 1677

Answers (1)

Jan Blažíček
Jan Blažíček

Reputation: 53

I don't know whether this makes any difference as it is more than a month now, but the blockiness and different information potential plot is simply caused by compression of the used image. You can't expect to be getting same results using this image as they had, because they have used raw, high resolution uncompressed version of it. I have to say I am fairly impressed with your results, especially with implementing the information potential. That thing went over my head a little.

John.

Upvotes: 2

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