Change RGB image to gray without rgb2gray function

Load a RGB Image and convert to grayscale as shown in the following example.

Hint:

0.2989 * R + 0.5870 * G + 0.1140 * B

Reference function rgb2gray( ) in MATLAB

Hint:

Convert uint8 to double datatype before you process the image.

Normalize the final image matrix and convert back to

uint8 before you show the image.

Answer 1

[Ameer Hamza]

Since this is a homework question so I won't give the complete answer, but, you can use the following hints from following code.

Seperate channels in the image

img; % RGB image
img = im2double(img); % convert image to double datatype
r = img(:,:,1);
g = img(:,:,2);
b = img(:,:,3);

r, g, and b are 2D matrices. You can do their weighted sum using formula in your question and you will get the grayscale image. Finally convert the grayscale image back to uint8 format

gray_image; % image you get after adding three channles
gray_image = im2uint8(gray_image)

[David Chiu]

thank you, I try it

[David Chiu]

gray_image; is not a function

[Ameer Hamza]

Yes, as I mentioned, you need to create this variable using the formula in your question.

[David Chiu]

thanks

Answer 2

[Image Analyst]

When they say normalize, I assume they want it rescaled to 0-255 so you can use rescale() or mat2gray(). Sounds like you can use rgb2gray() - it did not explicitly say that you could NOT use it. I imagine you can use other built-in functions like uint8(), rescale(), and mat2gray() also.

How do I get help on homework questions on MATLAB Answers? - MATLAB Answers - MATLAB Central

[David Chiu]

thanks

[Image Analyst]

There is also an imsplit() function that you might be interested in

[R, G, B] = imsplit(double(rgbImage));
grayScaleImage = 0.2989 * R + 0.5870 * G + 0.1140 * B; % The hint given in the problem statement.
% Now normalize with rescale(), cast to uint8() and call imshow().

[Madhusudan]

why are the values of red green blue so?? can the values are choosen random??

[Walter Roberson]

@Madhusudan

The values were determined by psychometric and physiological analysis of how our eyes respond to different wavelengths. For example the high coefficient on green is because our eye is most sensitive to green (the peak output of the Sun is in green; see https://telescoper.wordpress.com/2011/03/11/whys-the-sun-not-green/ )

If you look around in the literature, you will find that the above coefficients are the most common, but you will find balances. That reflects differences in light sources and in "color temperature" .

[DGM]

See also:

https://en.wikipedia.org/wiki/Luma_(video)

Those specific constants are part of the video encoding standards here.

https://www.itu.int/rec/R-REC-BT.470-7-200502-I/en (analog)

https://www.itu.int/rec/R-REC-BT.601-7-201103-I/en (digital)

though other standards exist with slightly differing constants:

https://www.itu.int/rec/R-REC-BT.709-6-201506-I/en

https://www.itu.int/rec/R-REC-BT.2020-2-201510-I/en

... and so on.

Note that all these vectors have a sum of 1 since it's a weighted mean. So even if you wanted to tweak the numbers, there's still that constraint.

Answer 3

[DGM]

Assuming that the input is an RGB image and that the intended grayscale representation is luma and not something else:

% get an RGB image
rgbpict = imread('peppers.png');

% pick luma constants
factors = permute([0.299 0.587 0.114],[1 3 2]); % Rec 470/601 (analog/digital SD video)
%factors = permute([0.2126 0.7152 0.0722],[1 3 2]); % Rec 709 (HDTV video)

% calculate the weighted sum
ypict = sum(bsxfun(@times,im2double(rgbpict),factors),3);
ypict = im2uint8(ypict); % if you want uint8 output

imshow(ypict)