Image compression

Question

I have a colour palette such that there are 4 colours. These colours are white, green, yellow and orange. The RBG values for these colours respectively are (255, 255, 255) (0, 255, 0) (255, 120, 0) (255, 255, 0). An image is an 8 x 8 pixel RGB image. I want to calculate how many bits are needed to store this compressed image.

I am not sure, since we have four colours if 2 or 3 bits would be needed to store the code. I originally thought it was 2 because 2^2=4 but now because orange and yellow aren't definitive RGB colours we need 3 bits?

from then on I know how to do the compression calculation I would just like explanation as to how many bits are needed to store the code

score 0 · Accepted Answer · answered Nov 02 '19 at 22:23

There are 4 possible colors, so you need 2 bits per pixel.

Your original thought it correct, 2^2=4, and 2 bits can represent 4 different values: In binary: 00, 01, 10, 11 (in decimal: 0, 1, 2, 3).

Each value represents an index in a color map.
The color map is used for mapping 0 to (255, 255, 255), 1 tp (0, 255, 0) 2 to (255, 120, 0) and 3 to (255, 255, 0).

Image is 8x8 pixels, so total bits needed are 8*8*2 = 128 bits (16 bytes).

When you need do decompress the image, there are two possibilities:

First option: the decoder knows the 4 colors from advance - no need to store the color map with the compressed image.
Second option: You need to store the color map with the compressed image, so additional bits are needed for storing the color map.

As your question is formulated, I am sure, the first option is true (you don't need to store the additional "bits" for the color map).

To make thinks more interesting, I coded the following MATLAB sample:

RGB = imread('peppers.png');   %Read input RGB image.
RGB = imresize(RGB, [64, 64]); %Reduce size to 64x64 (jsut for the example).

%Convert image to indexed image with 4 color.
[X, cmap] = rgb2ind(RGB, 4);
J = ind2rgb(X, cmap);

%Replace indices of color map:
cmap(1, 1:3) = [255, 255, 255]/255; %Fist color is white (255, 255, 255)
cmap(2, 1:3) = [0, 255, 0]/255; %Second color is green (0, 255, 0)
cmap(3, 1:3) = [255, 255, 0]/255; %Second color is yellow (255, 255, 0)
cmap(4, 1:3) = [255, 120, 0]/255; %Second color is orange (255, 120, 0)

K = ind2rgb(X, cmap);

figure;imshow(RGB);
figure;imshow(J);
figure;imshow(K);

RGB input image (true color):

Indexed image with only 4 color ("compressed" image):

Image after replacing color to white, green, yellow and orange (with arbitrary order):

Illustration for 8x8 image with 4 colors, as 8x8 matrix:

3,3,3,3,3,3,3,3
3,3,3,3,3,0,3,3
3,0,0,0,0,0,0,0
3,0,0,2,2,2,1,1
1,2,1,2,2,1,1,1
1,1,2,1,0,1,1,2
1,0,0,3,3,3,3,2
3,3,3,3,3,3,3,3

Illustration of color map:

0 -->    (255   255   255)
1 -->    (  0   255     0)
2 -->    (255   255     0)
3 -->    (255   120     0)

For storing the color map, you need 4*3*8 = 96 bits (assuming a value like 255 requires 8 bits).

Thank you, this helps my understanding. Would this therefore mean the 3 colours would also require 2 bits to store the code? — , Nov 03 '19 at 14:50
Yes, for storing 3 different colors, you need at least 2 bits (only 3 out of 4 combinations are utilized). Sure there are more sophisticated compression methods than indexed images that are able to take advantage of the fact that there are only 3 colors. Subject is related to the [Entropy](https://en.wikipedia.org/wiki/Entropy_(information_theory)) of the data. I am not an expert.... For indexed images, you do need **2 bits** for 3 colors. — Rotem, Nov 03 '19 at 15:02

Image compression

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