Kishoe Euler

7/30/2019 Kishoe Euler

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Euler Number Computation

ByKishore Kulkarni

(Under the Guidance of Dr. Longin Jan Latecki)


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Outline

Connected Components & Holes

Euler Number & its significance

For Binary Images For Graphs Analogy between Euler number for graphs & binary

images

Concavities and Convexities Relation between Euler Number, Concavities & Convexities

Euler Number computation for test images.

Justification of correctness of algorithm used for computing

Euler Number


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Euler Number

The Euler Number (also called as the connectivityfactor) is defined as the difference between

connected components and holes.Formally, Euler Number is given by

ncomp

E = ncomp

- nihole

where

i = 1

ncomp=> number of foreground connected components

nihole => number of holes for ith connected component


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Euler Number

E = 5 11 = -6E = 11 4 = 7


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Significance of Euler Number

Euler Number is a topological parameter, whichdescribes the structure of an object, regardless of its

specific geometric shape. Euler number (for discrete case) is a local measure

and is obtained as the sum of measures in the localneighbourhoods.

Although it is a local measure , it can be used toobtain global properties.

For objects with no holes, it can be used to calculatethe number of connected components.


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Euler Number for Graphs

IfV, E, and F represent the set of vertices, edges and faces

respectively then for any graph we have

|V| + |F| = |E|+ Rwhere R is the Euler Number for graphs

|V| = 7, |F| = 3(or 4), |E| = 9

Euler Number ,

R = 7 + 3(or 4) 9 = 1(or 2)

depending upon whether or

not we treat the area outside the

graph as a face.


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Equivalence of Euler Number for

graphs and Binary images

Any binary image can be represented as agraphs where Every black pixel (1) is represents the vertex of

the graph Two neighbouring black pixels (vertices of the

graphs) are joined to form an edge The Euler Number for the binary image and

its corresponding graph is the same.


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Euler Number Contd

Consider the following image.

Euler Number E = 2 0 = 2.

For the corresponding graph,

we have

|V

| = 28, |E| = 30, |F| = 4R = 28 30 + 4 = 2


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Convexities & Concavities

A region Ris said to be convex iff, for every

x1, x2 inR,

[x1, x2] belongs toR.

A convex hull for a region Ris defined as

the convex closure of region R.

Convexity (or the convex number) N2+

(X, )is defined as the number of first entries in the

object in a given direction .


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Convexity & concavity.

Concavity (or the concave number) N2-(X, ) is

defined as the number of exits in a given direction .

In the following figure,

N2+(X, 90) = 3

N2

-(X, 90) = 2


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Convexities and Concavities

For a binary image with 4-connectivity the convexities and

concavities can be represented by the following 2*2 masks.

Convexities Three 0s and one 1. Concavities Three 1s and one 0.


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Euler Number in terms of

convexity and concavity The relationship between Euler Number,

Convexities and concavities is given by

N2(X) = N2+(X) - N2-(X)

So now we have the following relation

E = c h = v e + f = N2+(X) - N2-(X) The upstream convexity [ ] -------- (1)

The upstream concavity [ ] ----------(2)

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Algorithm for computing Euler Number(for 4-connectivity)

For every pixel (x, y) of the image,

Do

if p(x,y) = 1 and p(x-1, y) = p(x , y-1) = p(x-1 , y-1) = 0

convexities ++

if p(x,y) = 0 & p(x+1, y) = p(x , y+1) = p(x+1 , y+1) =1

concavities ++end

Euler number = convexities - concavities


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Euler Number for test images

Image 1:

Number of Convexities

= 74

Number of Concavities

= 74

Euler Number = 0


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Image 2:


= 86


= 85

Euler Number = 1


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Image 3:


= 135


= ?

Euler Number = ?


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Image 4:


= 74


= 73

Euler Number = 1


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Image 5:


= 27


= 16

Euler Number = 11

= Number of Objects


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Image 6:


= 51


= 61

Euler Number = -10Number of Holes

= 1-(-10) = 11


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Why does this approach always

returns Euler Number ?

To prove:

Euler Number for an image = Number of upstream

convexities - Number of upstream concavities

Proof:

First we prove that Euler Number for the whole

image is the sum of Euler numbers for everycomponent. Then we prove the above equality for

one component to complete the proof.


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Proof - Contd.

Euler number for component is given by

Ei= 1 nihole

where nihole is the number of holes in component i

Consider ncomp ncomp

Ei = (1 - nihole)

i = 1 i = 1ncomp

= ncomp - nihole

i = 1

= Eimage


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Proof - Contd.

The upstream convexity [ ] -------- (1)

The upstream concavity [ ] ----------(2)

The first pixel with value 1 in the image

should be of type (1). Hence this convexity

refers to the connected component. Now any

pixel with the mask [ ] can(4) be a Hole

(5) be a Background

If (1) then we are done..

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Proof - Contd.

If it is a background pixel then we have an extra concavitycount. Hence we need to prove that there exists an equivalentconvexity to balanced the same. Now if it a background pixel,then the two edges in the 2*2 mask (referring to graphicalrepresentation of a binary image) should not form a cyclewith any of the previously visited black pixels (otherwise weget a hole). If this is the case there should be a pixel with the

mask [ ]. Hence we can prove that number of concavitiesexceeding number of holes and convexities exceeding 1 areequal by similar argument.

Thus they cancel out. Hence the result.

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