[PPT] - On Equitable Coloring q g for Strong Product of T Two cycles l PowerPoint Presentation

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On Equitable Coloring q g for Strong Product of T l Two cycles

Advisor : Yung-Ling Lai St d t P W i W Student : Peng-Wei Wang 國立嘉義大學資訊工程系

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Outline

Terminologies Motivation Related works Related works Main results

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Strong product

1 2

G G ⊗

vertex set (

) ( )

1 2

V G V G ×

and are adjacent if

( )

2 2

, x y

( )

1 1

, x y

1.

and

( ) ( )

1 2 1

, x x E G ∈

( ) ( )

1 2 2

, y y E G ∈

2.

and

1 2

x x =

( ) ( )

1 2 2

, y y E G ∈

( ) ( )

3.

and

1 2

y y =

( ) ( )

1 2 1

, x x E G ∈

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Strong product of

3 5

C C ⊗

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Proper coloring

For a graph ,

( , ) G V E =

{ }

: 1,2, , f V n → L

is proper coloring if for

A graph G is proper k-colorable if G can

( ) ( ) f v f u ≠ ( , ) u v E ∈

A graph G is proper k colorable if G can

be proper colored with k colors. Th ll t k i ll d h ti b

The smallest k is called chromatic number

denoted by .

( ) G χ

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Equitable coloring

A proper coloring of G is equitable

f

if for

{ }

( )

i

V v f v i = =

1 i n ≤ ≤

then for . Th ll t k h th t G i it bl

1

i j

V V − ≤ i j ≠

The smallest k such that G is equitable

colorable is called equitable chromatic q number, denoted by .

( ) G χ=

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Examples

i i Fig 1 Fig 2 Fig 3

proper coloring proper coloring not equitable equitable colorings

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Motivation / Applications

Garbage collection problem Scheduling Partitioning Partitioning Load balancing problems

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Previous Result

for

and

l ≥

2 l ≥

( )

5 C C χ ⊗

for and

1

l ≥

2

2 l ≥

( )

1 2

5(2 1) 2 1

5

l l

C C χ=

+ +

⊗ =

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Lemmas

Lemma 1 : Let , be graphs with at

1

G

2

G

least one edge each. Then

{ } { }

1 2 1 2

( ) max ( ), ( ) 2. G G G G χ χ χ

=

⊗ ≥ +

Lemma 2 : The chromatic number of even

cycle is 2 and odd cycle is 3. cycle is 2 and odd cycle is 3.

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Theorem 1 for m, n are both even

( ) 4

C C ⊗

f

( )

=

=4

m n

C C χ ⊗

( ) ( ) 2 C C Proof: By Lemma 2, By Lemma 1, ( ) ( ) 2

n m

C C χ χ = =

( )

=

2 2 4

m n

C C χ ⊗ ≥ + = Define

( ) { }

: ( ) 1,2,3,4 f V G →

( )

1 ( mod 2), for even , 0 1 and 0 1; j i i m j n + ≤ ≤ − ⎧ ⎪ ≤ ≤ − ⎪ ⎨

( )

; ( , ) 3 ( mod 2), for odd , 0 1 d 0 1

i j

j f u v j i i m j ⎪ = ⎨ + ≤ ≤ − ⎪ ⎪ ≤ ≤ ⎩

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and 0 1; j n ⎪ ≤ ≤ − ⎩

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Even n and m

1 2 1 2 1 2 3 4 3 4 3 4 3 4 3 4 3 4 1 2 1 2 1 2 3 4 3 4 3 4 1 2 1 2 1 2 1 2 1 2 1 2 3 4 3 4 3 4 Coloring matrix of

6 6

C C ⊗

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Theorem 2 for is odd

( )

5 C C ⊗

m n +

( )

5

m n

C C χ= ⊗ =

Proof: Let m be even and n be odd.

( ) 2 C ( ) 3 C

By lemma 2, , By lemma 1

( )

3 2 5 C C χ ⊗ ≥ + ( ) 2

m

C χ = ( ) 3

n

C χ =

By lemma 1,

( )

3 2 5

m n

C C χ= ⊗ ≥ + =

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Case1: , for some odd k.

5 n k =

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Case2: , for some odd k.

5 2 n k = +

5 3 4 2 1

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Case3: , for some odd k.

5 4 n k = +

5 3 4 2 3 3 2 1 4 1 2 5 1 5

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Case4: , for some odd k.

5 6 n k = +

2 1 4 3 5 3 1 2 3 5 3 4 4 2 5 3 1 1 2 3 1

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Case5: , for some odd k.

5 8 n k = +

1 5 1

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Case5: , for some odd k.

5 8 n k = +

5 4 1

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Case5: , for some odd k.

5 8 n k = +

5 3 4 1

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Case5: , for some odd k.

5 8 n k = +

3 5 2 4 1

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Exmaple for case5 when

10 m >

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Exmaple for case5 when

10 m >

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( )

, for is odd

( )

5

m n

C C χ= ⊗ ≤

m n +

We already have

( )

5

m n

C C χ ⊗ ≥ We already have therefore,

( )

m n

C C χ= ⊗

( )

5

m n

C C χ= ⊗ = ,

( )

m n

χ

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( )

Theorem 3

( )

3 3

9 C C χ= ⊗ =

Theorem 3 is trivial since

.

3 3 9

C C K ⊗ =

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Theorem 4 for odd n and

7 n ≥ ( )

7 C C ⊗

( )

3

7

n

C C χ= ⊗ =

Proof: Since

,

6

K G ∈

( )

3 6

( ) 6

n

C C K χ χ ⊗ ≥ = hence .

( )

3

6

n

C C χ= ⊗ ≥

Suppose G is equitable 6-colorable

Suppose G is equitable 6 colorable

exactly three colors are used in every column

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( ) ( )

3

7

n

C C χ= ⊗ ≥

for odd n

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An equitable coloring of

3 7

C C ⊗

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Appended columns for

3 7

C C ⊗

5 4 5 4 3 5 4 2 3 5 4 2 1 3 5 4 2 7 1 3 5

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Examples

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( )

, for is odd

( )

3

7

n

C C χ= ⊗ ≤

n

We already have

( )

3

7

n

C C χ ⊗ ≥ We already have therefore,

( )

3 n

C C χ= ⊗

( )

3

7

n

C C χ= ⊗ = ,

( )

3 n

χ=

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Theorem 5 for m, n are both odd

( ) 5

C C ⊗

( )

=

=5

m n

C C χ ⊗

Proof: By Lemma 2, ( ) ( ) 2

n m

C C χ χ = = By Lemma 1,

( )

=

2 2 4

m n

C C χ ⊗ ≥ + =

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An equitable coloring of

7 5

C C ⊗

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Appended columns for

7 5

C C ⊗

4 4 3 4 3 2

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Appended columns for

7 5

C C ⊗

2 3 1 4

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An equitable coloring of

9 5

C C ⊗

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Appended columns for

9 5

C C ⊗

2 2 1 1 2 5 4 2 4 2 3 1 3 1 4 2 2 5

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Appended columns for

9 5

C C ⊗

4 2 5 1 1 4 4 2 2 5 3 1 1 4 2 3

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An equitable coloring of

11 5

C C ⊗

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Appended columns for

11 5

C C ⊗

2 1 2 1 5 2 5 4 5 4 3 5 4 3 4 3 2 4

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Appended columns for

11 5

C C ⊗

2 5 4 1 5 3 2 4 4 2 1 3

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An equitable coloring of

13 5

C C ⊗

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Appended columns for

13 5

C C ⊗

3 2 3 2 3 2 1 5 2 1 2 1 5 3 5 3 5 3 3 1 4 2 4 2 4 5 4 5 4 5 2 3 3 4 3 4

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Appended columns for

13 5

C C ⊗

3 2 1 5 5 4 2 1 5 3 3 1 2 5 4 2 4 5 2 3 1 3

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Examples

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Thanks for your attention !

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