CS465 Fall 2006
Homework 1 Solution
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1. 1680 x 1050 x 3 x 8 bytes = 40.4 MB

2. 1680 x 1050 x 3 x 1 byte  = 5.05 MB for (a) and (b)
   1680 x 1050 x 3 x 2 bytes = 10.1 MB for (c)

3. 1680 x 1050 x 3 x 8 bits x 60 frames/sec = 2.37 Gb/sec

4a. fbPixel[i] = round(min(255, max(0, 255 * imgPixel[i])))
4b. fbPixel[i] = round(min(255, max(0, 255 * pow(imgPixel[i], 0.45))))
4c. fbPixel[i] = round(min(65535, max(0, 65535 * imgPixel[i])))

5a. 0, 21, 28, 34, 39
5b. 0, 1, 2, 3, 4    
5c. 0, 2, 2, 3, 3    

6a. first 4 steps:  41%; 26%; 23%; 19%
6b. first 4 steps: .05%; .2%; .3%; .5%
6c. first 4 steps:  .2%;  0 ; .3%;  0
All steps are visible for (a).
All steps are invisible for (b) and (c).

7. It would look too dark.

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The morals of the story:

- Pixel format (a) is not a good plan; it will lead to very visible bands in
  the dark tones.

- Gamma can be considered a memory and bandwidth saving trick,
  since we can use 8 bits with gamma quantization where we need
  16 bits if we use linear quantization.

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4: For (a) and (c) We just need to scale the range [0,1] to the
range [0,N].  For (b) we apply gamma correction, since the
framebuffer is supposed to contain gamma-quantized values.  In
all cases we clip values that are too high or too low to the
maximum or minimum allowable pixel value.

5b: We are storing gamma-quantized values, and the monitor
expects gamma-quantized values, so the LUT should do nothing.

5a and 5c: You can get very close to the right answer (and to the
full score) with a simple computation.  Expressed in C/Java
syntax:

  5a: lut[k] = round(255 * pow(k / 255, 0.45))
  5b: lut[k] = round(255 * pow(k / 65535, 0.45))

or, equivalently, by the MATLAB expressions:

  round(255 * ((0:4) / 255).^0.45)
  round(255 * ((0:4) / 65535).^0.45)

but this doesn't quite guarantee you have the best approximation
to the desired luminance level -- instead you have the pixel
value closest to the fractional pixel value that would give the
desired luminance level.  I didn't really expect many students to
catch this subtlety.  To get this last point, you could compute
all the displayable luminances and explicitly find the closest
one to each desired luminance, for instance by using the MATLAB
code:

  disp = ([0:255]/255).^2.2;
  for k = 0:4, [y,i] = min(abs(disp-k/255)); i-1, end
  for k = 0:4, [y,i] = min(abs(disp-k/65535)); i-1, end

6: Once you have the answer to 5, 6 follows simply by mapping the
contents of the LUT through the monitor's transfer function.  The
ouput (with flare) for pixel value k is:

  output_k = ((lut[k] / 255)^2.2 + 0.01) * I_max

and the answers are output_k / output_(k-1) for k = 1,2,3,4,
expressed as percentages.

I computed the answer in MATLAB this way:
  v = [0, 21, 28, 34, 39
       0, 1, 2, 3, 4
       0, 2, 2, 3, 3];
  d = 0.01 + (v/255).^2.2;
  d(:,2:5) ./ d(:,1:4)


Piti's addition:

application:
	0:1
	0:1
	0:1
framebuffer:
	255*(0:1)
	255*(0:1).^0.45
	65535*(0:1)
lookup table entries:
	round(255 * ((0:255) / 255).^0.45)
	0:255
	round(255 * ((0:255) / 65535).^0.45)
observed intensities:
	0.01 + ((0:255)/255).^2.2
	0.01 + ((0:255)/255).^2.2
	0.01 + ((0:255)/255).^2.2