Finding the contrast ratio of a grayscale image

So as the title says, I want to find the contrast ratio of a grayscale image by Michelson contrast definition. I'm thankful for any help! I'm not sure how to do this. Am I supposed to find the max and min value in each columns or row or the whole matrix?

% Michelson contrast definition
% Imax : the largest value in the image's matrix
% Imin : the smallest value in the image's matrix

contrast = (Imax-Imin)(Imax+Imin)

Answer 1

[Image Analyst]

Do you need loops, like this is a homework question that requires them? If not, just simply do

Imax = max(I(:))
Imin = min(I(:))
contrast = (Imax - Imin) / (Imax + Imin)

Notice that I'm using the correct formula. You are not. They subtract in the numerator, not divide like you have it.

[Image Analyst]

Julie - are you still alive?

[Image Analyst]

And your recent edit to

contrast = (Imax-Imin)(Imax+Imin)

is also not correct, for two reasons. You can't multiply two expressions without a * operator, plus even if you did, that would still be the wrong formula.

[Itzhak Mamistvalov]

Hey, I would like to ask an adition question about contrast ratio.

Im trying to calculate to Contrast Improvment Ratio (CIR) between enhanced and unenhanced images within region of interest R. This is the formula Im using:

where C(x, y) and C¯(x,y)C¯(x,y) are the local contrast values at (x, y) of the unenhanced and enhanced images, respectively.

The Local contrast value C(x, y) computed as

where p and a are the mean values within the center region (3 × 3) pixels and the neighborhood, or surrounding region, (7 × 7) pixels, respectively.

I have a problem to actually write this into a matlab code.

I would appreciate it if you could help, thanks!

[Image Analyst]

@Itzhak Mamistvalov I answered that here:

https://www.mathworks.com/matlabcentral/answers/876238-how-to-calculate-contrast-improvment-ratio-cir#answer_744358

Answer 2

[royed]

if the question is about just comapring the max and min in the images. You can find it quite simply by using (max(max(image)) and the same for the minimum which would return the pixel with the max and minimum intensity value.