Compsci.375 Assignment 1

Group Number: 21
Group Member: Vincent Wing Hei, HO
UPIs: vho003
IDs: 2492512

Answer to questions:

1.1 (m+1) x (m+1) x (m+1) of different colour may be represented.
 
1.2 (m+1) of grey value may be represented.
 
1.3 All the grey values exist from the white point (Gmax, Gmax, Gmax) diagonally across to the black point (0, 0, 0).
 
   
2.1 Hue = arc cos (-0.5) = 120°
Saturation = 1
Intensity = u/3

 \HSI = (120, 0, u/3)
 
2.2 Hue = arc cos (0) = 90°
Saturation = 0
Intensity = u

 \HSI = (90, 0, u)
 
   
3.1  hist(F5,5(f, p, 0)) = 3
 hist(F5,5(f, p, 1)) = 2
 hist(F5,5(f, p, 2)) = 10
 hist(F5,5(f, p, 3)) = 10
 
3.2  HIST(F5,5(f, p, 0)) = 1/25 x 3 = 3/25
 HIST(F5,5(f, p, 1)) = 1/25 x 2 = 2/25
 HIST(F5,5(f, p, 2)) = 1/25 x 10 = 10/25 = 2/5
 HIST(F5,5(f, p, 3)) = 1/25 x 10 = 10/25 = 2/5
 
3.3  cumhist(F5,5(f, p, 0)) =  hist(F5,5(f, p, 0)) = 3
 cumhist(F5,5(f, p, 0)) =  hist(F5,5(f, p, 0)) + hist(F5,5(f, p, 1)) = 5
 cumhist(F5,5(f, p, 0)) =  hist(F5,5(f, p, 0)) + hist(F5,5(f, p, 1))
          + hist(F5,5(f, p, 2)) = 15
 cumhist(F5,5(f, p, 0)) =  hist(F5,5(f, p, 0)) + hist(F5,5(f, p, 1))
          + hist(F5,5(f, p, 2)) + hist(F5,5(f, p, 3)) = 25

 
3.4  CUMHIST(F5,5(f, p, 0)) =  HIST(F5,5(f, p, 0)) = 3/25
 CUMHIST(F5,5(f, p, 0)) =  HIST(F5,5(f, p, 0)) + HIST(F5,5(f, p, 1))
          = 3/25 + 2/25 = 5/25 = 1/5
 CUMHIST(F5,5(f, p, 0)) =  HIST(F5,5(f, p, 0)) + HIST(F5,5(f, p, 1))
          + HIST(F5,5(f, p, 2)) = 1/5 + 2/5 = 15/25 = 3/5
 CUMHIST(F5,5(f, p, 0)) =  HIST(F5,5(f, p, 0)) + HIST(F5,5(f, p, 1))
          + HIST(F5,5(f, p, 2)) + HIST(F5,5(f, p, 3)) = 3/5 + 2/5 = 1

  
3.5 MEAN(F5,5(f, p)) = 1 / (5 X 5) X [(2 + 0 + 0 + 0 + 2) + (3 + 2 + 1 + 2 + 3)
          + (3 + 2 + 1 + 2 + 3) + (3 + 3 + 2 + 3 + 3) + (3 + 2 + 2 + 2 + 3)]
          = 2.08

 
3.6 VARIANCE(F5,5(f, p)) = (0 - 2.08)2 X 3/25 + (1 - 2.08)2 X 2/25
          + (2 - 2.08)2 X 10/25 + (3 - 2.08)2 X 10/25
          = 0.519168 + 0.093312 + 0.00256 + 0.33856
          = 0.9536

 
   
4  If we assume two identical picture in different colour combined together and form a histogram, we can see ourselves having a two-channel image consist of two non-identical channels.

i.e. for such to happened, we need a original image and a processed image combined together.

 
   
5 Click here for the applet.

 
   

Text of Presentation:

Introduction:

Assignment Description:

Expreimental Results and Discussions:

Conslusions:

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