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I have 2 questions, firstly, why is my output giving me 3 images as a result of the local equalization and then why is my resultant image in grayscale when i entered an rgb image?

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Abdul Raffay
Abdul Raffay on 9 Jul 2020
Edited: Abdul Raffay on 11 Jul 2020
function myThirdAssignment(I,WindowSize,K0,K1,K2)
if size(I,3)==1
x= im2double(imread(I));
[r,c]=size(x);
elseif size(I,3)==3
x= im2double(imread(I));
gray1=rgb2gray(x);
[r,c]=size(gray1);
figure(1); imshow(x); title('ORIGINAL IMAGE');
imwrite(x,'OriginalImage.bmp.bmp'); %writing data of original image in to current directory.
% GIVING B THE SAME ROWS AND COLUMS AS THE ORIGINAL IMAGE
B=zeros(r,c);
% CALCULATING CEIL & FLOOR VALUES TO MAKE PROGRAM MORE GENERAL
p= ceil((Size / 2)); %3/2= 1.5=2
s= floor((Size / 2)); %3/2=1.5=1
for i=p:r-s
for j=p:c-s
sum=0;
temp=0;
% CHECKING IF BOTH CRITERIAS ARE FULFILLED
%--------------------------------------------
if avg <= K0*(meanIntensity) && (K1*(stdG) <= std) && (std <= K2*(stdG))
% only enhance an area of defined window size when its mean/average is
% lesser than or equal to the mean of the image by some constant
% K0 AND its standard deviation is lesser than the value of the
% standard deviation by a constant K2 and greater than the
% global standard deviation by a constant K1.
B(i,j)= 2*x(i,j);
else
B(i,j)= x(i,j);
end
%--------------------------------------------
end
end
%RGB = cat(3, B, B, B);
figure(2);imshow(B); title('IMAGE AFTER LOCAL HISTOGRAM EQUALIZATION');
imwrite(B,'enhancedImage.jpeg.jpeg'); %writing data of enhanced image in to current directory
end
Im attaching the original and resultant images here:
Original image:
Resultant/enhanced image:

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Accepted Answer

Image Analyst
Image Analyst on 9 Jul 2020
For the first question:
When you do this:
x= im2double(imread(I));
[r,c]=size(x);
x is still an RGB image that you later go on to use instead of gray1. And c is not the number of columns in the image, it's the number of columns times the number of color channels. Why? See Steve's article: http://blogs.mathworks.com/steve/2011/03/22/too-much-information-about-the-size-function/ You should never use size like that for an image. You should always do it like this when you're unsure if the image might be RGB or grayscale:
[rows, columns, numberOfColorChannels] = size(rgbImage);
So when you're looping over c, you're actually looping over three times as much as you thought you were. So when you do this:
B(i,j)= 2*x(i,j);
you're adding columns because the badly-named j is 3 times as big as you thought. Here, let's take an example. Let's instantiate B as a 2-by-2 matrix:
>> B = rand(2,2)
B =
0.96489 0.97059
0.15761 0.95717
and now let's set B(2, 6) and see what happens to B:
>> B(2, 6) = 100
B =
0.96489 0.97059 0 0 0 0
0.15761 0.95717 0 0 0 100
See? It made B three times as wide. That's why you're seeing B three times as wide.
---------------------------------------------------------------------------------------------------------------------
Now for the secnod question (why is B grayscale?):
B is grayscale because you instantiated it with 2 dimensions, not three.
% GIVING B THE SAME ROWS AND COLUMS AS THE ORIGINAL IMAGE
B=zeros(r,c);
Gray scale images are 2-D while RGB images are 3-D. You did not make B an RGB image, like you'd get if you did this:
[rows, columns, numberOfColorChannels] = size(x);
B = zeros(rows, columns, numberOfColorChannels, 'uint8');
so you got a grayscale image.

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Image Analyst
Image Analyst on 9 Jul 2020
If you want floating point, don't tell it to make zeros 'uint8'.
If your image is floating point, it will need to be in the range 0-1 to see it, unless you use [] in imshow():
imshow(B, []); % Scales max to min
otherwise anything above 1 will appear white and anything below 0 will appear black.

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