3-D Computer Vision CSc 83020

3-D Computer Vision CSc 83020 – Ioannis Stamos3-D Computer Vision CSc 83020 – Ioannis Stamos

3-D Computer Vision3-D Computer VisionCSc 83020CSc 83020

Binary ImagesBinary ImagesHorn (Robot Vision): pages 46-64.Horn (Robot Vision): pages 46-64.


Binary ImagesBinary Images

Binary Image b(x,y): Obtained from Gray-Level (or other image) g(x,y)By Thresholding.

Characteristic Function: 1 g(x,y) < T

b(x,y)= 0 g(x,y) >= T

1) Geometrical Properties2) Discrete Binary Images3) Multiple objects4) Sequential and Iterative Processing.


Selecting the Threshold (T)Selecting the Threshold (T)

g(x,y)

HISTOGRAM

h(i)

T i (intensity)

BIMODAL

b(x,y)


Histogram ThresholdingHistogram Thresholding

T


Geometric PropertiesGeometric PropertiesAssume:

1) b(x,y) is continuous2) Only one object

y

x

b(x,y)




y

x

b(x,y)

Area:

I

dxdyyxbA ),( (Zeroth Moment)




y

x

b(x,y)

Area:

I

dxdyyxbA ),( (Zeroth Moment)

Position: “Center” (x,y) of Area

I

I

dxdyyxbyAy

dxdyyxbxAx

),(*)/1(

),(*)/1(

_

_(First Moment)


OrientationOrientationDifficult to define!We use axis of least second moment

(x,y)

r

ρθ

axisy

x


OrientationOrientationDifficult to define!We use axis of least second momentFor mass distribution: Axis of minimum Inertia.

(x,y)

r

ρθ

axisy

x

Minimize: I

dxdyyxbrE ),(2

OrientationOrientationDifficult to define!We use axis of least second momentFor mass distribution: Axis of minimum Inertia.

(x,y)

r

ρθ

axisy

x

Minimize: I

dxdyyxbrE ),(2

Equation of Axis: y=mx+b ??? 0<= m <= infinity

We use: (ρ, θ) finite.Find (ρ, θ) that minimize E for a given b(x,y)

0)cos()sin( yx


We can show that: 22 ))cos()sin(( yxr

So: dxdyyxbyxEI

),())cos()sin(( 2



So: dxdyyxbyxEI

),())cos()sin(( 2

Using dE/dρ=0 we get: 0))cos()sin((__

yxA



So: dxdyyxbyxEI

),())cos()sin(( 2


yxA

Note: Axis passes through center (x,y)!!

So, change coordinates: x’=x-x, y’=y-y



So: dxdyyxbyxEI

),())cos()sin(( 2


yxA



We get:

I

I

I

dydxyxbyc

dydxyxbyxb

dydxyxbxa

cbaE

''2'

''''

''2'

222

),()(

),()(2

),()(

)(cos)cos()(sin)(sin

Second Momentsw.r.t.(x, y)



So: dxdyyxbyxEI

),())cos()sin(( 2


yxA



We get:

I

I

I

dydxyxbyc

dydxyxbyxb

dydxyxbxa

cbaE

''2'

''''

''2'

222

),()(

),()(2

),()(

)(cos)cos()(sin)(sin

Second Momentsw.r.t.(x, y)

Mistake inyourhandout.


)(cos)cos()(sin)(sin 22 cbaE

Using dE/dθ=0, we get: tan(2θ)=b/(a-c)

2θb

a-c

22 )( cab


)(cos)cos()sin()(sin 22 cbaE


So: 2θb

a-c

22 )( cab

22

22

)(/)()2cos(

)(/)2sin(

cabca

cabb


)(cos)cos()(sin)(sin 222 cbaE


So:

Solutions with positive sign may be used to find θ that minimizesE (why??).

Emin/Emax -> (ROUDNESS of the OBJECT).

2θb

a-c

22 )( cab

22

22

)(/)()2cos(

)(/)2sin(

cabca

cabb


Discrete Binary ImagesDiscrete Binary Imagesbij: value at pixel in row i & column j.

1 11 1 1 1 1

1 1 1 1 11 1 1 1

1

O j (y) m

i(x)

n


Discrete Binary ImagesDiscrete Binary Imagesbij: value at pixel in row i & column j. Assume pixel area is 1.Area:

Position (Center of Area):1 1

1 1 1 1 11 1 1 1 1

1 1 1 11

O j (y) m

i(x)

n

n

i

m

jijbA

1 1

n

i

m

jij

n

i

m

jij

jbAy

ibAx

1 1

_

1 1

_

)/1(

)/1(


Discrete Binary ImagesDiscrete Binary Imagesbij: value at pixel in row i & column j. Assume pixel area is 1.Area:

Position (Center of Area):1 1

1 1 1 1 11 1 1 1 1

1 1 1 11

O j (y) m

i(x)

n

n

i

m

jijbA

1 1

n

i

m

jij

n

i

m

jij

jbAy

ibAx

1 1

_

1 1

_

)/1(

)/1(

n

i

m

j

n

i

m

jijij

n

i

m

jij bjcbijbbia

1 1 1 1

2''

1 1

2' ,2,

Second Moments.


Discrete Binary Images (cont)Discrete Binary Images (cont)

Note: a’,b’,c’ are second moments w.r.t ORIGINa,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.


Multiple ObjectsMultiple ObjectsNeed 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 andB along which b(x,y) is constant.


Connected ComponentsConnected Components

Label all pixels that are connected

4-way connected 8-way connected


Region Growing AlgorithmRegion Growing Algorithm(Connected Component Labeling)(Connected Component Labeling)

Start with “seed” point where bij=1.Start with “seed” point where bij=1. Assign LABEL to seed point.Assign LABEL to seed point. Assign SAME LABEL to its NEIGHBORS (b=1).Assign SAME LABEL to its NEIGHBORS (b=1). Assign SAME LABEL to NEIGHBORS of NEIGHBORS.Assign SAME LABEL to NEIGHBORS of NEIGHBORS.

Terminates when a component is completely labeled.Then pick another UNLABELED seed point.


What do we mean by NEIGHBORS?What do we mean by NEIGHBORS?Connectedness

4-Connectedness (4-c)

8-Connectedness (8-c)

Neither is perfect!


What do we mean by NEIGHBORS?What do we mean by NEIGHBORS?

Jordan’s Curve Theorem: Closed curve -> 2 connected regions

0 1 0

1

1

10

00

B1 O1 B1

O2

O4

O3B2

B1B1(4-c)Hole without closed curve!

What do we mean by NEIGHBORS?What do we mean by NEIGHBORS?

Jordan’s Curve Theorem: Closed curve -> 2 connected regions

0 1 0

1

1

10

00

B1 O1 B1

O2

O4

O3B2

B1B1

B O B

O

O

OB

BB(4-c)Hole without closed curve!

(8-c)Connected background

with closed ring!


Solution: Introduce AssymetrySolution: Introduce AssymetryUse: or

0 1 0

1

1

10

00

B O1 B

O2

O2

O1B

BB

(a) (b)

Using(a)

Two separate linesegments.

HexagonalTesselation

Above assymetry makesSQUARE grid likeHEXAGONAL grid.


Sequential Labeling AlgorithmSequential Labeling Algorithm

A

BD

C

Raster ScanNote: B,C,D are already labeled



A

BD

C


0

XX

X

a. Label(A) = background

1

XD

X

b. Label(A) = label(D)



A

BD

C


1

00

C

c. Label(A) = label(C)

1

B0

0

d. Label(A) = label(B)

1

B0

C

d. If Label(B) = label(C ), then Label(A)=Label(B)= Label(C)



A

BD

C


0

XX

X

a. Label(A) = background

1

XD

X

b. Label(A) = label(D)

1

00

C

c. Label(A) = label(C)

1

B0

0

d. Label(A) = label(B)

1

B0

C

d. If Label(B) = label(C ), then Label(A)=Label(B)= Label(C)


Sequential Labeling (Cont.)Sequential Labeling (Cont.)What if B & C are labeled but label(B) label(C) ?


Sequential Labeling (Cont.)Sequential Labeling (Cont.)What if B & C are labeled but label(B) label(C) ?

2 2 22 2 2 2 2 2

22 2 2 2 2

2 2 2 22 2 2 22 2 2 2 22

1 1 11 1 1

1 1 1 1 1 11 1 11 1 1

1 1 1 1 1 1

11

11

11

1 11 11 11

?


Sequential Labeling (Cont.)Sequential Labeling (Cont.)

Solution: Let: Label(A)=Label(B) =2 & create an EQUIVALENCE TABLE.

Resolve Equivalences in SECOND PASS.

7 3,6,42 1

…

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3-D Computer Vision CSc 83020