3-D Computer Vision CSc 83020

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3-D Computer VisionCSc 83020

• Binary Images
• Horn (Robot Vision) pages 46-64.

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Binary Images

• Binary Image b(x,y) Obtained from Gray-Level (or
  other image) g(x,y)
• By Thresholding.
• Characteristic Function
• 1 g(x,y) lt T
• b(x,y)
• 0 g(x,y) gt T
• Geometrical Properties
• Discrete Binary Images
• Multiple objects
• Sequential and Iterative Processing.

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Selecting the Threshold (T)
g(x,y)
h(i)
BIMODAL
HISTOGRAM
T
i (intensity)
b(x,y)
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Histogram Thresholding
T
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Geometric Properties
b(x,y)
Assume 1) b(x,y) is continuous 2) Only one
object
y
x
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Geometric Properties
b(x,y)
Assume 1) b(x,y) is continuous 2) Only one
object
y
x
Area
(Zeroth Moment)
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Geometric Properties
b(x,y)
Assume 1) b(x,y) is continuous 2) Only one
object
y
x
Area
(Zeroth Moment)
Position Center (x,y) of Area
(First Moment)
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Orientation
Difficult to define! We use axis of least second
moment
y
axis
(x,y)
r
?
x
?
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Orientation
Difficult to define! We use axis of least second
moment For mass distribution Axis of minimum
Inertia.
y
axis
(x,y)
r
?
x
?
Minimize
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Orientation
Difficult to define! We use axis of least second
moment For mass distribution Axis of minimum
Inertia.
y
axis
(x,y)
r
?
x
?
Minimize
Equation of Axis ymxb ??? 0lt m lt
infinity We use (?, ?) finite. Find (?,
?) that minimize E for a given b(x,y)
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We can show that
So
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We can show that
So
Using dE/d?0 we get
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We can show that
So
Using dE/d?0 we get
Note Axis passes through center (x,y)!! So,
change coordinates xx-x, yy-y
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We can show that
So
Using dE/d?0 we get
Note Axis passes through center (x,y)!! So,
change coordinates xx-x, yy-y We get
Second Moments w.r.t. (x, y)
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We can show that
So
Using dE/d?0 we get
Note Axis passes through center (x,y)!! So,
change coordinates xx-x, yy-y We get
Second Moments w.r.t. (x, y)
Mistake in your handout.
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Using dE/d?0, we get tan(2?)b/(a-c)
b
2?
a-c
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Using dE/d?0, we get tan(2?)b/(a-c) So
b
2?
a-c
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Using dE/d?0, we get tan(2?)b/(a-c) So S
olutions with positive sign may be used to find ?
that minimizes E (why??). Emin/Emax -gt (ROUDNESS
of the OBJECT).
b
2?
a-c
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Discrete Binary Images
bij value at pixel in row i column j.
j (y) m
O
i (x) n
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
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Discrete Binary Images
bij value at pixel in row i column j. Assume
pixel area is 1. Area Position (Center of
Area)
j (y) m
O
i (x) n
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
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Discrete Binary Images
bij value at pixel in row i column j. Assume
pixel area is 1. Area Position (Center of
Area)
j (y) m
O
i (x) n
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
Second Moments.
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Discrete Binary Images (cont)
Note a,b,c are second moments w.r.t
ORIGIN a,b,c (w.r.t center) can be found from
(a,b,c,x,y,A). Note UPDATE
(a,b,c,x,y,A) during RASTER SCAN.
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Multiple Objects
Need to SEGMENT image into separate COMPONENTS
(regions). Connected Component Maximal Set of
Connected Points
B
A
Points A and B are connected Path exists between
A and B along which b(x,y) is constant.
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Connected Components
4-way connected
8-way connected
Label all pixels that are connected
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Region Growing Algorithm(Connected Component
Labeling)

• Start with seed point where bij1.
• Assign LABEL to seed point.
• Assign SAME LABEL to its NEIGHBORS (b1).
• Assign SAME LABEL to NEIGHBORS of NEIGHBORS.

Terminates when a component is completely
labeled. Then pick another UNLABELED seed point.
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What do we mean by NEIGHBORS?
Connectedness
8-Connectedness (8-c)
Neither is perfect!
4-Connectedness (4-c)
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What do we mean by NEIGHBORS?
Jordans Curve Theorem Closed curve -gt 2
connected regions
B1
O1
B1
0
1
0
1
1
0
O2
O3
B2
1
0
0
O4
B1
B1
(4-c) Hole without closed curve!
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What do we mean by NEIGHBORS?
Jordans Curve Theorem Closed curve -gt 2
connected regions
B1
O1
B1
B
O
B
0
1
0
1
1
0
O2
O3
B2
O
O
B
(8-c) Connected background with closed ring!
1
0
0
O4
B1
B1
O
B
B
(4-c) Hole without closed curve!
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Solution Introduce Assymetry
Use
or
(a)
(b)
Using (a)
Two separate line segments.
B
O1
B
0
1
0
1
1
0
O2
O1
B
1
0
0
O2
B
B
Hexagonal Tesselation
Above assymetry makes SQUARE grid like HEXAGONAL
grid.
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Sequential Labeling Algorithm
Raster Scan
B
D
C
A
Note B,C,D are already labeled
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Sequential Labeling Algorithm
Raster Scan
B
D
C
A
Note B,C,D are already labeled
Label(A) background
X
X
a.
X
0
Label(A) label(D)
X
D
b.
X
1
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Sequential Labeling Algorithm
Raster Scan
B
D
C
A
Note B,C,D are already labeled
Label(A) label(C)
0
0
c.
C
1
Label(A) label(B)
B
0
d.
0
1
If Label(B) label(C ), then Label(A)Label(B)
Label(C)
B
0
d.
C
1
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Sequential Labeling Algorithm
Raster Scan
B
D
C
A
Note B,C,D are already labeled
Label(A) background
X
X
a.
X
0
Label(A) label(D)
X
D
b.
X
1
Label(A) label(C)
0
0
c.
C
1
Label(A) label(B)
B
0
d.
0
1
If Label(B) label(C ), then Label(A)Label(B)
Label(C)
B
0
d.
C
1
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Sequential Labeling (Cont.)
What if B C are labeled but label(B)
label(C) ?
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Sequential Labeling (Cont.)
What if B C are labeled but label(B)
label(C) ?
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
2
2
2
1
1
1
1
1
1
1
1
2
2
2
2
2
2
1
1
1
1
1
1
1
1
1
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
?
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Sequential Labeling (Cont.)
Solution Let Label(A)Label(B) 2
create an EQUIVALENCE TABLE. Resolve
Equivalences in SECOND PASS.
2
1
7
3,6,4