Notes 20110121 CIS 6320 Image Processing

From SnOwy - Ed's Wiki Notebook

Class 3 -- just finished discussing analogue images, functions {domain → domain-of-definition, codomain → range; injection, surjection, bijection}.

Contents 1 Homework 2 Continuing on Functions 3 Analogue Image (Examples) 4 We agree on ... 5 Digital Image ... 6 Digitization 7 Next Time

Homework

• send an e-mail for three transforms (done)
• send slides by Feb. 4th
• this first presentation heavily emphasizes applications to images
• recommend we look at http://www.socs.uoguelph.ca/~matsakis/courses/ -- specifically ...
• http://www.socs.uoguelph.ca/~matsakis/CIS4720/assign.htm -- the first assignment from undergrad
• -- don't do the assignment, just understand it -- we have to focus on our projects

Continuing on Functions

• f (function)
• f = (A, B, G)
• f | A → B (f is a function from A to B)
• f | A → B, x → 1/x
• the above is a partial function -- a total function has a domain equal to the domain-of-definition -- this is missing zero
• the above is missing zero from the range -- the co-domain is real

Analogue Image (Examples)

• partial function ...
• f_a | R² → R₊; R₊ = [0, +∞) (achromatic image)
• total function ...
• f_a | R³ → R₊
• function ...
• f_a | R² → R³ (colour image)
• function ...
• f_a | R² → C
• we agree that f_a is defined for (x, y) -- a rectangular area domain-of-definition
• let's say that the height is m, that the width is n

          x|-
           |
           |
           |
-          |0      N  +
-----------+-------+---
           |       |  y
          M+-------+
           |
           |
           |+

• above -- Domain of definition -- [0, M] × [0, N]
• we can avoid having to deal with the rectangle -- we can create a total function by using ...
  • zero-padding (black everywhere else in the domain)
  • circular indexing (torus, modulus math)
  • reflected indexing (like the above, but edges act as mirrors)

We agree on ...

• N = 0 .. +∞
• Z = -&infin .. +∞
• Q = {p/q} p ∈ Z, q ∈ N^*; N^* = 1 .. +∞
• R = {p/q} p ∈ Z, q ∈ Z -- can q be zero? I think this is an assertion?
• C = R + {i}
• R - Q is irrational

Digital Image ...

• three cases ...
  1. f|Z² → Z₊
  2. f|Z³ → Z
  3. f|Z² → Z³
• a pixel is -- a 2-tuple = ((x, y), f(x, y)) ...
  • (x, y) is a 2-tuple ...
    • x is an integer
    • y is an integer
  • f(x, y) is an integer (read this as the return value) -- greylevel -- case 1 (output is different)
  • (x, y, z) is a 3-tuple -- location in a cube -- case 2 (input is different)
• f(x, y) is a three-tuple of integers -- RGB values -- case 3 (output is different)

   0 .. y .. N-1
  0 +---------|
  . |..........
  . |..........
  x |0≤f(x,y)≤L-1
  . |..........
  . |..........
M-1 -..........

• M -- the number of rows
• N -- the number of columns
• L -- the number of grey levels
• number of bits required M*N*l -- small l because we might not use all grey levels
• M = 2^M
• N = 2^N
• L = 2^l

Digitization

• f_a | R² → R₊
• f | Z² → Z₊
• digitization -- fa to f
• sampling R2 to Z2
• quantization Z2 to R+ or R2 to Z+
• quantization -- can be called an image
• sampling: consider this grid (resolution of area)
• quantization: pick a colour (resolution of grey level)
• sampling with lower M, N leads to the checkerboard effect
• sampling with lower L leads to false contouring

Next Time

• finishing this topic
• image processing approaches
• about presentations
• presentations