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

On Equitable Coloring q g for Strong Product of T Two cycles l Advisor : Yung-Ling Lai St d Student : Peng-Wei Wang t P W i W 國立嘉義大學資訊工程系 1

Outline � Terminologies � Motivation � Related works � Related works � Main results 2

⊗ Strong product G G 1 2 vertex set ( ) ( ) × � V G V G 1 2 ( ) ( ) and are adjacent if � x y , x y , 1 1 2 2 ( ) ( ) ( ) ( ) ∈ ∈ and , x x E G , 1. y y E G 1 2 1 1 2 2 ( ) ( ) = ∈ and x x 2. y y , E G 1 2 1 2 2 = ( ( ) ) ( ( ) ) ∈ y y and , 3. x x E G 1 2 1 2 1 3

⊗ Strong product of C C 3 5 4

Proper coloring { } → = L � For a graph , f V : 1,2, , n G ( , V E ) ≠ ∈ is proper coloring if for f v ( ) f u ( ) ( , ) u v E � A graph G is proper k-colorable if G can � A graph G is proper k colorable if G can be proper colored with k colors. � The smallest k is called chromatic number Th ll t k i ll d h ti b χ denoted by . ( ) G 5

Equitable coloring � A proper coloring of G is equitable f { } ≤ ≤ = = if for 1 i n V v f v ( ) i i − ≤ ≠ then for . V V 1 i j i j � The smallest k such that G is equitable Th ll t k h th t G i it bl colorable is called equitable chromatic q χ = number , denoted by . ( ) G 6

Examples Fig 2 i Fig 3 i Fig 1 proper coloring proper coloring equitable colorings not equitable 7

Motivation / Applications � Garbage collection problem � Scheduling � Partitioning � Partitioning � Load balancing problems 8

Previous Result � ( ( ) ) χ = χ ⊗ ⊗ = l ≥ l ≥ l ≥ l ≥ for for and and C C C C 5 5 0 0 2 2 + + 5(2 l 1) 2 l 1 1 2 1 2 9

Lemmas � Lemma 1 : Let , be graphs with at G G 1 2 least one edge each. Then { { } } χ ⊗ ≥ χ χ + ( ) max ( ), ( ) 2. G G G G = 1 2 1 2 � Lemma 2 : The chromatic number of even cycle is 2 and odd cycle is 3. cycle is 2 and odd cycle is 3. 10

Theorem 1 for m, n are both even ( ( ) 4 ) χ ⊗ ⊗ =4 C C C C = m n χ = χ = ( ( C C ) ) ( ( C C ) ) 2 2 Proof: f By Lemma 2, n m ( ( ) ) χ ⊗ ≥ + = C C 2 2 4 By Lemma 1, = m n { } f V G → Define : ( ) 1,2,3,4 + ≤ ≤ − ⎧ 1 ( mod 2), for even , 0 j i i m 1 ⎪ ≤ ≤ − ⎪ ⎪ ( ( ) ) and 0 j j n 1; ; = ⎨ + ⎨ f ( , u v ) ≤ ≤ − i j 3 ( mod 2), for odd , 0 j i i m 1 ⎪ ⎪ ⎪ ≤ ≤ ≤ ≤ − ⎩ ⎩ and 0 d 0 j j n 1; 1 11

Even n and m 1 2 1 2 1 2 3 3 4 4 3 3 4 4 3 3 4 4 1 2 1 2 1 2 3 4 3 4 3 4 1 1 2 2 1 1 2 2 1 1 2 2 3 4 3 4 3 4 ⊗ Coloring matrix of C C 6 6 12

+ m n Theorem 2 for is odd ( ( ) ) χ = ⊗ ⊗ = C C C C 5 5 m n � Proof: Let m be even and n be odd. χ = χ = ( ( C C ) ) 3 3 ( ( C C ) ) 2 2 By lemma 2, , n m ( ( ) ) χ = χ ⊗ ⊗ ≥ + ≥ + = C C C C 3 3 2 2 5 5 By lemma 1, By lemma 1 m n 13

= 5 n k Case1: , for some odd k. 14

= + n 5 k 2 Case2: , for some odd k. 5 3 4 2 1 15

= + Case3: , for some odd k . n 5 k 4 5 4 3 3 2 2 3 1 4 2 1 1 5 5 16

= + 5 6 n k Case4: , for some odd k . 5 2 4 1 1 3 3 3 3 4 5 2 1 5 4 2 2 1 1 3 3 17

= + n 5 k 8 Case5: , for some odd k . 1 5 1 18

= + n 5 k 8 Case5: , for some odd k . 5 4 1 19

= + n 5 k 8 Case5: , for some odd k . 5 3 4 1 20

= + n 5 k 8 Case5: , for some odd k . 3 5 2 4 1 21

10 22 m > Exmaple for case5 when

10 23 m > Exmaple for case5 when

( ( ) ) χ = ⊗ ≤ + C C 5 m n , for is odd m n ( ( ) ) χ = χ ⊗ ⊗ ≥ � We already have We already have C C C C 5 m m n n ( ( ) ) χ χ = ⊗ = C C 5 therefore, , m m n n 24

( ( ) ) χ = ⊗ = C C 9 Theorem 3 3 3 ⊗ = � Theorem 3 is trivial since . C C K 3 3 9 25

n ≥ 7 Theorem 4 for odd n and ( ( ) ) χ = ⊗ ⊗ = C C C C 7 7 3 n ( ) χ ⊗ ≥ χ = ∈ C C ( K ) 6 � Proof: Since , K G 3 n 6 6 ( ) χ = ⊗ ≥ hence . C C 6 3 n � Suppose G is equitable 6-colorable Suppose G is equitable 6 colorable � exactly three colors are used in every column 26

( ( ) ) χ = ⊗ ≥ C C 7 for odd n 3 n 27

28 7 C ⊗ 3 C An equitable coloring of

⊗ Appended columns for C C 3 7 5 4 5 4 3 5 4 2 3 5 4 1 3 5 4 7 1 3 5 2 2 29

30 Examples

( ( ) ) χ = ⊗ ≤ C C 7 n , for is odd 3 n ( ( ) ) χ = χ ⊗ ⊗ ≥ � We already have We already have C C C C 7 3 3 n n ( ( ) ) χ = χ = ⊗ = therefore, , C C 7 3 3 n n 31

Theorem 5 for m, n are both odd ( ( ) 5 ) χ ⊗ ⊗ =5 C C C C = m n χ = χ = Proof: By Lemma 2, ( C ) ( C ) 2 n m ( ) χ ⊗ ≥ + = By Lemma 1, C C 2 2 4 = m n 32

33 5 C ⊗ 7 C An equitable coloring of

34 5 C 4 ⊗ 7 C 2 Appended columns for 3 4 3 4

35 5 C ⊗ 7 C 4 Appended columns for 2 1 3

36 5 C ⊗ 9 C An equitable coloring of

⊗ Appended columns for C C 9 5 2 2 1 1 2 2 1 1 5 5 2 2 4 3 4 3 2 4 37

⊗ Appended columns for C C 9 5 1 1 4 4 5 5 2 2 3 3 1 1 2 2 4 4 38

39 5 C ⊗ 11 C An equitable coloring of

⊗ Appended columns for C C 11 5 2 1 2 1 5 2 5 4 5 4 3 5 4 3 4 3 2 4 40

⊗ Appended columns for C C 11 5 1 4 5 2 4 2 3 5 3 1 2 4 41

42 5 C ⊗ 13 C An equitable coloring of

⊗ Appended columns for C C 13 5 2 1 2 1 5 2 3 3 2 2 3 3 2 2 1 1 3 3 5 5 4 4 5 5 4 4 3 3 5 5 4 3 4 3 2 4 43

⊗ Appended columns for C C 13 5 1 4 5 2 2 2 5 5 1 1 3 3 4 2 3 3 5 5 3 1 2 4 44

45 Examples

46 Thanks for your attention !