Tutorial Solutions
Colour Representation

1. Let's assume a frame buffer than can store 8 bits per pixel. Each pixel can thus address one of 256 colours. Let's assume that the DACs (digital-to-analog converters) for each of RGB work with 8-bits, taking an 8-bit number and producing the corresponding analog intensity. The video system is thus {\em capable} of producing any of 2^24 possible colours. The question is, how do we choose which 256 colours to display, from a colour cube of 256x256x256 = 2^24 possible colours?
   • Take 256 evenly-spaced samples in the colour space. This has the advantage that some colour will always be close. If multiple applications must all make use of the same colour map, or no information about the displayable image is known in advance, this may not be a bad choice. Typically, 3 bits will be allocated to each of red and green, with 2 bits being used to represent blue, because the eye is less sensitive to blue. This means that there will be effectively only 8,8, and 4 levels of intensity for R,G, and B, respectively.
   • Choose the 256 colours which occur the most often and place these in the colour map. For any remaining colours, choose the closest entry in the colour map. When would this scheme not do well?
   • Optimize the placement of the colour-map entries such that the maximum distance from all colours to the nearest displayable colour is minimized.
   
   
   
2. If colours R1,G1,B1 and R2,B2,G2 have

      Y1 = Y2, namely:
      0.299 R1 + 0.587 G1 + 0.114 B1  =  0.299 R2 + 0.587 G2 + 0.114 B2
   

   then they will be displayed identically on a black-and-white television.

   
   
   

3. There are (8-1)*16 + 1 = 113 different intensities that can be displayed. The pattern intensity is the sum of all the pixel intensities. The maximum value of this sum is 8*16 (all pixels at intensity 8), and the minimum value is 1*16 (or 7*16 and 0*16).
   
   
4. As an artifical example, consider that we model colours as RGB triples, which correspond roughly to the following wavelengths:

    R = 700 nm
    G = 546 nm
    B = 438 nm
   

   Now suppose that we have a real light which consists of the following light distribution:

    10 W at 700 nm (red)
    20 W at 546 nm (green)
    15 W at 438 nm (blue)
    10 W at 500 nm (perceptually a mix of blue and green -- cyan)
   

   Now suppose this light bounces off a surface which only reflects 500 nm light and absorbs all other wavelengths. A correct model of the light bouncing off the surface should produce a surface which is cyan in colour. If instead we use a three-colour RGB model, a black surface (no light reflected) is predicted.

   Thus, ideally, one should always treat light interactions using a full spectrum model. This is too expensive in reality, though, and light colour and surface properties are typically approximated using only three components, RGB.
   
   

5. Yes. More of the colour gamut could potentially be covered -- it could now be a quadrilateral instead of a triangle. The device would be more complex to build, however.
   
   
6. With each additional spectral colour, the colour gamut could better approximate the full set of visible colours as given in a chromaticity diagram. This is not practical, however.
   
   
7. They could, but it would make for a bad monitor with a small colour gamut.
   
   
8. This is an ongoing area of research. The simplest idea, namely mapping all colour which are outside the gamut to the closest colour within the gamut, produces many undesirable artifacts. A variety of other strategies are possible which produce better continuous mappings from desired colours to the limited set of displayable colours.