Afdeling Toegepaste Wiskunde / Division of Applied Mathematics Image segmentation (10.1 to 10.2.4) SLIDE 1/15 CHAPTER 10: IMAGE SEGMENTATION Image Analysis: Chapter 10: Segmentation Chapter 11: Representation and description Chapter 12: Object recognition Segmentation: Subdivision of an image into its constituent parts The level of segmentation depends on the application 10.1 Fundamentals Segmentation based on: (1) Discontinuities (1) Isolated points (2) Lines (3) Edges (2) Similarity (1) Thresholding (2) Region growing (3) Region splitting/merging (1) Edge-based segmentation (2) Region-based segmentation
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Afdeling Toegepaste Wiskunde / Division of Applied Mathematics
Image segmentation (10.1 to 10.2.4) SLIDE 1/15
CHAPTER 10: IMAGE SEGMENTATION
Image Analysis:
Chapter 10: SegmentationChapter 11: Representation and descriptionChapter 12: Object recognition
Segmentation: Subdivision of an image into its constituent parts
The level of segmentation depends on the application
10.1 Fundamentals
Segmentation based on:
(1) Discontinuities
(1) Isolated points(2) Lines(3) Edges
(2) Similarity
(1) Thresholding(2) Region growing(3) Region splitting/merging
(1) Edge-based segmentation
(2) Region-based segmentation
Afdeling Toegepaste Wiskunde / Division of Applied Mathematics
Image segmentation (10.1 to 10.2.4) SLIDE 2/15
Let R represent the entire image regionThe segmentation process partitions R into n subregions, R1, R2, . . . , Rn,such that...
(a)n⋃
i=1
Ri = R
(b) Ri is a connected set, i = 1, 2, . . . , n
(c) Ri⋂Rj = ∅ for all i and j, i �= j
(d) Q(Ri) = TRUE for i = 1, 2, . . . , n
(e) Q(Ri⋃Rj) = FALSE for any adjacent regions Ri and Rj
Here Q(Rk) is logical predicate defined over all points in Rk
(a) Every pixel must be in a region
(b) All the points in a region must be “connected”
(c) Regions must be disjoint
(d) For example Q(Ri) = TRUE if all the pixels in Ri have the same graylevel
(e) Regions Ri and Rj are different in some sense
Afdeling Toegepaste Wiskunde / Division of Applied Mathematics
Image segmentation (10.1 to 10.2.4) SLIDE 3/15
Example
Afdeling Toegepaste Wiskunde / Division of Applied Mathematics
Image segmentation (10.1 to 10.2.4) SLIDE 4/15
10.2 Point, line and edge detection
10.2.1 Background∂f
∂x= f ′(x) = f(x + 1)− f(x)
∂2f
∂x2=∂f ′(x)
∂x
= f ′(x + 1)− f ′(x)
= f (x + 2)− f (x + 1)− f (x + 1) + f(x)
= f (x + 2)− 2f (x + 1) + f (x)
The above represents a truncated Taylor expansion about the point x + 1
Expansion about point x:
∂2f
∂x2= f ′′(x) = f (x + 1) + f(x− 1)− 2f(x)
Afdeling Toegepaste Wiskunde / Division of Applied Mathematics
Image segmentation (10.1 to 10.2.4) SLIDE 5/15
Afdeling Toegepaste Wiskunde / Division of Applied Mathematics
Image segmentation (10.1 to 10.2.4) SLIDE 6/15
General linear filter: (Response) R = w1z1 + w2z2 + . . . + wmnzmn