DIGITAL IMAGE PROCESSING

CONNECTED COMPONENT LABELING

ALGORITHM

Processing of Images which are Digital in nature by means of Digital Computer

ECEDEPARTMENT

Connected Component Labeling

Ability to assign different labels to variousdisjoint component of an image is calledconnected component labeling.

This labeling is a fundamental step inautomated image analysis:

a) Shape

b) Area

c) Boundary

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ECEDEPARTMENT

Basic Scanning Method

Scan the image from left to right and top tobottom.

Assume 4-adjacency.

Let p be a pixel at any step in the scanningprocess.

Before p the pixel r and t are scanned.

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ECEDEPARTMENT

r

t p

Labeling Algorithm

This algorithm makes two passes over the image:

1. The first pass to assign temporary labels andrecord equivalence classes.

2. The second pass to replace each temporarylabel by the smallest label of its equivalenceclass.

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ECEDEPARTMENT

Steps in First PassConditions to check:

1. Does the pixel to the left (West) have the same value asthe current pixel?

Yes – We are in the same region. Assign the samelabel to the current pixel

No – Check next condition

2. Do both pixels to the North and West of the current pixelhave the same value as the current pixel but not the samelabel?

Yes –Assign the current pixel the minimum of theNorth and West labels, and record theirequivalence relationship

No – Check next condition

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ECEDEPARTMENT

Steps in First Pass..

3. Does the pixel to the left (West) have a differentvalue and the one to the North the same valueas the current pixel?

Yes – Assign the label of the North pixel tothe current pixel

No – Check next condition

4. Do the pixel's North and West neighbors havedifferent pixel values than current pixel?

Yes – Create a new label id and assign it tothe current pixel

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ECEDEPARTMENT

Steps in Second Pass

In the First pass we record some equivalencerelationships.

In Second Pass:

1. Process Equivalence pairs to formequivalent classes.

2. Re-label the element with the labelassigned to its equivalent classes.

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ECEDEPARTMENT

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ECEDEPARTMENT

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ECEDEPARTMENT

1

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ECEDEPARTMENT

1 2

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1 2

1

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1 2

1 2

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1 2

1 2 3

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ECEDEPARTMENT

1 2

1 2 3 3

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1 2

1 2 3 3

1

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1 2

1 2 3 3

1 1

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1 2

1 2 3 3

1 1 1

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1 2

1 2 3 3

1 1 1 1 (1,2) equivalent relation

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ECEDEPARTMENT

1 2

1 2 3 3

1 1 1 1 4

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1 2

1 2 3 3

1 1 1 1 4 3 (3,4) equivalent relation

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ECEDEPARTMENT

1 2

1 2 3 3

1 1 1 1 4 3 3

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ECEDEPARTMENT

1 2

1 2 3 3

1 1 1 1 4 3 3 3

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ECEDEPARTMENT

1 2

1 2 3 3

1 1 1 1 4 3 3 3

1

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1 2

1 2 3 3

1 1 1 1 4 3 3 3

1 1

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1 2

1 2 3 3

1 1 1 1 4 3 3 3

1 1 3

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1 2

1 2 3 3

1 1 1 1 4 3 3 3

1 1 3 3

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ECEDEPARTMENT

1 2

1 2 3 3

1 1 1 1 4 3 3 3

1 1 3 3

5

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ECEDEPARTMENT

1 2

1 2 3 3

1 1 1 1 4 3 3 3

1 1 3 3

5 1 (1,5) equivalent relation

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ECEDEPARTMENT

1 2

1 2 3 3

1 1 1 1 4 3 3 3

1 1 3 3

5 1 3

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ECEDEPARTMENT

1 2

1 2 3 3

1 1 1 1 4 3 3 3

1 1 3 3

5 1 3 3

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ECEDEPARTMENT

1 2

1 2 3 3

1 1 1 1 4 3 3 3

1 1 3 3

5 1 3 3 3

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ECEDEPARTMENT

1 2

1 2 3 3

1 1 1 1 4 3 3 3

1 1 3 3

5 1 3 3 3

5

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ECEDEPARTMENT

1 1

1 2 3 3

1 1 1 1 4 3 3 3

1 1 3 3

5 1 3 3 3

5

1<= 1,2,5

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ECEDEPARTMENT

1 1

1 1 3 3

1 1 1 1 4 3 3 3

1 1 3 3

5 1 3 3 3

5

1<= 1,2,5

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ECEDEPARTMENT

1 1

1 1 3 3

1 1 1 1 3 3 3 3

1 1 3 3

5 1 3 3 3

5

3<= 3,4

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ECEDEPARTMENT

1 1

1 1 3 3

1 1 1 1 3 3 3 3

1 1 3 3

1 1 3 3 3

5

1<= 1,2,5

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ECEDEPARTMENT

1 1

1 1 3 3

1 1 1 1 3 3 3 3

1 1 3 3

1 1 3 3 3

1 1<= 1,2,5

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ECEDEPARTMENT

1 1

1 1 3 3

1 1 1 1 3 3 3 3

1 1 3 3

1 1 3 3 3

1

Here we observed that the image contain two distinct class of regions