Problem in Gray Level Manipulation

Hi all,

I tried to enhance a gray scale image by using the formula

   E(i,j) = round( I(i,j)^3/ f * M^2)

where

      M = max gray level of the image
      f = 0.8

The output should be a gray scale image, but I get a binary Image.

Thanks for your help in advance.

Code :

[x,y] = size(I);
I1 = repmat(uint8(0),x,y) 
M = max(I(:));
for i = 1:x
    for j = 1:y
        if I(i,j) ~= 0
          v = double(I(i,j));
          I1(i,j) = round( v^3 / f * M^2 );
          % disp(I1(i,j)); ' all manipulated pixels has value 255
        end 
    end
end

Answer 1

[Image Analyst]

What are some values of z? Also, have you looked at I1 in the variable editor to verify that they're all 255 or 1? Have you tried to display I1 with [], like this:

imshow(I1, []);

which you need to do for floating point images? M^2 is in the numerator - is that what you want? The units seem all weird - you'll have an output that is proportional to gray level to the fifth power! Are you missing some parentheses? Like you want M^2 in the denominator? Where did you ever get this formula anyway? Do you have a citation for it?

[Vennila Gangatharan]

Sorry mistakenly I have used z in place of using I and corrected the code.I got this formula from a paper(under the heading "GRAY LEVEL MANIPULATION") in the following link

http://www.ijcaonline.org/journal/number6/pxc387258.pdf

[Image Analyst]

The M squared is a constant so it really doesn't matter - it just affects the range f would be in. But I think it should be in the denominator like

E(i,j) = round( I(i,j)^3/ (f * M^2));

but f is really f(E(i,j)) because they say f depends on the intensity of the original image. For that reason, you should use intlut() to do this operation and not a double loop over i and j.

[Vennila Gangatharan]

I changed the code

I1(i,j) = round( v^3 / f * M^2 );

to

I1(i,j) = round( v^3 / (f * M^2) );

as you said. It worked. Thank you for your help.

[Image Analyst]

Did you make up a lookup table like I said? It will be a lot faster. And remember f has to vary depending on the value of the pixel.

[Vennila Gangatharan]

I have coded for choosing f. I don't know how to use intlut() in my code.

[Image Analyst]

It's a vector of 256 (if uint8) or 65536 (if uint16) that lists the output gray level for the index. For example if the lut(200) is 150, then intlut(grayImage, lut) will send every gray level of grayImage that has gray level 200 to the new gray level of 150. So you make up a lookup table with every input graylevel and its new output gray level. Then you just have to call intlut() and don't have to iterate over every single pixel with a pair of for loops. Search Answers - i just recently uploaded some code using intlut for a 16 bit case.

[Vennila Gangatharan]

ok. I will try.

[Vennila Gangatharan]

I tried using LUT and got answer.

CODE:

lut = uint8(round([1:256].^3/(f*M.^2)));

I1 = intlut(I,lut);

[Image Analyst]

Is f and array? It should be. But you're not using ./, you're using /, so that leads me to think that f is not an array like it needs to be.

[Vennila Gangatharan]

f is not an array.

[Image Analyst]

Then you're not following the algorithm given in the paper.

Answer 2

[Walter Roberson]

Change

M = max(I(:));

to

M = double( max(I(:)) );

[Vennila Gangatharan]

I tried it but no change.