Domination in generalized Domination in generalized Petersen graphs - PowerPoint PPT Presentation

Domination in generalized Domination in generalized Petersen graphs Petersen graphs Advisor: Sheng-Chyang Liaw Student: Yuan-Shin Li

Introduction • A vertex subset S of a graph G is a dominating set g p g if each vertex in V ( G ) − S is adjacent to at least one vertex in S . • The domination number of G is the cardinality of a minimum dominating set of G , denoted by . i i d i ti t f G d t d b γ γ 5 = 2 （ C （ 5 ） ） C 5

• A dominating set S is called an independent dominating A d i i S i ll d i d d d i i set if S is also an independent set. • The independent domination number of G is the cardinality of a minimum independent dominating set cardinality of a minimum independent dominating set of G , denoted by . γ γ （ ） = 2 （ ） i C 5 5 i C 5 C 5

• A dominating set S is called a total dominating set • A dominating set S is called a total dominating set if each vertex v of G is dominated by some vertex ≠ ≠ u u v v of S of S . • The total domination number of G is the cardinality • The total domination number of G is the cardinality of a minimum total dominating set of G , denoted by . γ γ （ ） = 3 （ ） t C 5 5 t C 5 C 5

By definition γ ( G ) ≤ γ i ( G ) and γ ( G ) ≤ γ t ( G ). A vertex v of G is repeatedly dominated by S if y p y | N ( v ) ∩ S | ≥ 2. ⎡ ⎤ ( ) γ ≥ ⎢ V G ( ) G Proposition 1.1 If G is k-regular then . ⎥ t k • A generalized Petersen graph P ( n , k ) vertex set: ={ v 1 , v 2 , ..., v n } { u 1 , u 2 , ..., u n } V U U U = U edge set: with v n +1 = v 1 , u n +1 = u 1 , E {v v , v u , u u } + + 1 i i i i i i k 1 ≤ i ≤ n , and 1 ≤ k ≤ ⎢ n ⎥ ⎢ ⎥ ⎣ 2 ⎦

v 7 v 8 v 6 V={ v 1 , v 2 , ..., v n } u 7 U={ u 1 , u 2 , ..., u n } u 6 u 8 v 9 v 5 u 5 E= { v i v i +1 , v i u i , u i u i + k } { } 5 U u 9 ⎢ ⎥ n 1 ≤ i ≤ n , and 1 ≤ k ≤ ⎢ ⎥ u 4 ⎣ 2 ⎦ v 4 v 10 v 10 4 u 10 u 3 u 11 v 11 v 3 u 2 u 12 u 12 u 1 u 13 v 2 v 12 v 13 v 1 Generlized Petersen graph P(13,3) Generlized Petersen graph P(13,3)

Petersen graph ＝ P(5 2) Petersen graph ＝ P(5,2)

domination independent total domination number domination number number { { P (2 k + 1, k ) ( , ) , n ≠ 1(mod 3) , ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ( ) 2 n 3 3 n 3 3 n 3 n + , n ≡ 1(mod3) 5 5 ⎡ ⎤ 2 1 ⎢ ⎥ 3 P (2 k k ) P (2 k , k ) ⎡ ⎤ 2 ⎡ ⎤ n 2 n ⎢ ⎥ n ⎢ ⎥ 3 3 { ⎡ ⎤ { ⎡ ⎤ { { ⎡ ⎤ { ⎡ ⎤ , n ≡ 0,3 (mod 4) P ( n , 1) n + , n ≡ 2 (mod 4) , n ≡ 4 (mod 6) n + ⎡ ⎤ 1 ⎡ ⎤ ⎡ ⎤ 1 2 n 2 3 2 , n ≡ 1,2 (mod 4) , n ≡ 4 (mod 6) n ⎡ ⎤ 1 + 2 n n , n ≡ 0,1,3 (mod 4) 2 2 3 n ≡ 0 1 3 (mod 4) 2 { 2 { P ( n , 2) ⎡ ⎤ 2 n , n ≡ 1(mod 3) ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ 3 n 3 n 3 n + 5 5 ⎡ ⎤ ⎡ ⎤ , n ≡ 1(mod3) 2 1 1 ≡ 1( d3) 5 ⎢ ⎥ ⎢ ⎥ n 3 { ⎡ ⎤ 1 { { ⎡ ⎤ { n ≡ 0,1 (mod 4) P ( n , 3) , n ≡ 0,1 (mod 4) n + ⎡ ⎤ ⎧ ,n ≡ 0,2,4 (mod 5) 4 ⎡ ⎤ ⎡ ⎤ 1 n n , or n=11 5 ⎪ ⎪ 2 2 ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ≤ γ ≤ 2 2 ( ( ( ( , 3 3 )) )) 4 4 ⎨ ⎨ n n P n ,n ≡ 1,3 (mod 10) n n + 3 5 t , n ≡ 2,3 (mod 4) n ≡ 2,3 (mod 4) ⎪ ⎡ ⎤ 2 + 1 4 ⎩ n 2 , n ≠ 11 5 ,n ≡ 6,8 (mod 10)

Theorem 2.1 The independent domination number of p f ⎡ ⎤ 3 n P (2 k +1, k ) is for n = 2 k +1 . ⎢ ⎥ ⎢ 5 ⎥ ⎡ 3 n ⎤ Proof Since γ ( G ) ≤ γ ( G ) we have γ ( P (2 k + 1 k )) ≥ Proof. Since γ ( G ) ≤ γ i ( G ), we have γ i ( P (2 k + 1, k )) ≥ . ⎢ ⎢ ⎥ ⎥ ⎢ ⎢ 5 ⎥ ⎥ In the following, we choose the vertices to form an independent dominating set. v 8 v 9 v 7 • n = 5 a u 8 v 10 u 9 S 0 = { V 5 t +1 |0 ≤ t < a } v 6 u 7 u 10 10 S 1 = { U 5 t +3 |0 ≤ t < a } ∪ u 6 { U 5 t +4 |0 ≤ t < a }; v 11 u 11 v 5 3 n | S | + | S | = a + 2 a = 3 a = | S 0 | + | S 1 | = a + 2 a = 3 a = 5 u 5 u 5 u 12 v 12 v 4 u 4 u 13 u 13 u 3 v 13 u 14 v 3 u 2 u 15 u 15 u 1 v 14 v 2 v 15 v 1

• n = 5 a + 3 ⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦ S 0 = { V 5 t +1 |0 ≤ t < } ∪ { V 5 t | ≤ t ≤ a }; a a a 2 2 ⎣ ⎦ ⎣ ⎦ S 1 = { U 5 t +3 |0 ≤ t < a } ∪ { U 5 t +4 |0 ≤ t < } ∪ { U 5 t +2 | ≤ t ≤ a }; a a 2 2 | S | + | S | = a + 1 + a + a + 1 = 3 a + 2 = ⎡ ⎤ ⎡ ⎤ | S 0 | + | S 1 | = a + 1 + a + a + 1 = 3 a + 2 = . 3 n 3 n 5 v 7 v 8 v 6 u 7 u 6 u 8 v 5 v 9 u 5 u 9 u 9 v 10 u 4 v 4 u 10 Therefore Therefore ⎡ ⎤ γ i ( P (2 k +1, k )) ≤ 3 n u 3 5 u 11 v 11 11 v 3 u 2 u 12 u 1 u 13 v 12 v 12 v v 2 v 1 v 13

Lemma 2.2: In the total domination number, a connected set D of a 3 regular graph G can dominate at most 2| D |+2 of a 3-regular graph G can dominate at most 2| D |+2 vertices in G with | D | ≥ 2 . Since G is 3-regular graph Proof. By induction on | D |.

Theorem 2.3 For n = 2 k + 1, ⎧⎡ ⎧⎡ 2 n 2 ⎤ ⎤ + , if (mod 3) ≡ n 1 1 if ( d 3) n 1 ⎪⎢ ⎥ ⎪⎢ 3 ⎥ γ t ( P (2 k + 1, k )) = ⎨ ⎡ ⎤ 2 ≡ , if (mod 3) ⎪ ⎢ n 1 n ⎥ ⎪ ⎢ ⎪ ⎢ 3 3 ⎥ ⎥ ⎩ ⎩ ⎡ ⎤ ⎡ ⎤ S ≥ 2 n 1. By Proposition 1.1 y p 3 3 ⎡ ⎤ 1 γ + ≥ + ≡ 2. To proof when . ( ( 2 1 , )) 1 2 n P k k n 3 t ⎡ ⎤ ⎡ ⎤ γ γ + + = = Suppose we have the following two properties. Suppose ( ( ( ( 2 2 1 1 , )) )) we have the following two properties 2 n P P k k k k 3 t Claim 1. For any connected subset D � S, | D | = d ≤ 3 . Claim 2. S is formed by the union of exactly one 3-element connected set and a − 1 2-element connected sets for n = 3 a + 1 . − Moreover, every vertex in is dominated by exactly one G S vertex in S. S ⎡ ⎤ 1 γ + ≥ + ( ( 2 1 , )) 2 n P k k Proof 3 t S ≤ ⎡ ⎤ 2 n ⎢ ⎥ 3. 3

Theorem 2.3 For n = 2 k + 1, ⎧⎡ ⎧⎡ 2 n 2 ⎤ ⎤ + , if (mod 3) ≡ n 1 1 if ( d 3) n 1 ⎪⎢ ⎡ ⎤ ⎥ S = 2 n ⎪⎢ 3 ⎥ 3 γ t ( P (2 k + 1, k )) = ⎨ ⎡ ⎤ 2 ≡ , if (mod 3) ⎪ ⎢ n 1 n ⎥ ⎪ ⎢ ⎪ ⎢ 3 3 ⎥ ⎥ ⎩ ⎩ ⎡ ⎤ 1 γ + ≥ + ( ( 2 1 , )) 2 (2) If n ≡ 1 (mod 3), proof n P k k 3 t ⎡ ⎤ ⎡ ⎤ γ γ + + = ( ( ( ( 2 2 1 1 , )) )) 2 Suppose , we have the following two properties. Suppose n we have the following two properties P P k k k k 3 t ⊂ = ≤ , 3 Claim 1. For any connected subset D S D d Proof. V(G) ⎡ ⎤ ⎡ ⎤ = = + + 2 2 2 1 1 n S S a a 3 3 S D We have + − + + ≥ 3 ( 2 1 ) ( 2 2 ) 2 and then d ≤ 3. a d d n

Claim 2. S is formed by the union of exactly one 3-element connected set and a − 1 2 element connected sets for n = 3 a + 1 set and a − 1 2-element connected sets for n = 3 a + 1 . Moreover every vertex in G is dominated by exactly one vertex in S. Proof. If there exist two 3-element connected subsets can dominate at most 2 ・ (2 ・ 3+2) = 16 vertices ⎡ ⎤ ⎡ ⎤ = = + 2 1 2 n S a 3 3 then S can dominate at most (2 a +1 − 6)·3+16 = 6 a +1 < 2 n →← →← By counting each connected subset can dominated (2 · 3 + 2) + ( a − 1)(2 · 2 + 2) = 2 n

Let D be a connected subset of S with | D | = 3 = − = − − ( U U ( U U ( ( ( )) )) ( ( ( )) )) and d N N N N D D D D N N N N v N N D D 1 2 1 ∈ D ∈ v v N 1 v k+2 v k+3 v k+1 Case1. | D ∩ V | ≥ 2. v k+4 v k u k+2 u k+3 u k+1 u k+4 u k u u 2k+1 u 4 u 4 u 1 v 2k+1 u 2 u 3 v 1 v 4 v 3 v 2

Case2. | D ∩ V | = 1 and | D ∩ U | = 2 | | | | v k+1 v k+2 v k v k+3 v k 1 v k-1 u k+1 u k+2 u k u k+3 u k-1 u 2k u 3 u 3 u 2k+1 u 1 u 2 v 2k v 3 v 2k+1 v 2 v 1

| D ∩ U | = 3 Case3. v k+2 v k+3 v k+1 k+3 k+1 v k+4 v k u k+2 u k+3 u k+1 u k+4 u With the above cases , we know u k ⎡ ⎤ 1 ⎡ ⎤ 1 γ γ + + ≥ ≥ 3 + + ( ( ( ( 2 2 1 1 , , )) )) 2 n P P k k k k 3 t u 2k+1 u 4 u 1 v 2k+1 u 2 u 3 v 1 v 4 v 3 v 2

Theorem 2.3 For n = 2 k + 1, v 8 ⎧⎡ ⎧⎡ 2 n 2 ⎤ + ⎤ , if (mod 3) ≡ • For n = 3 a : F 3 n 1 1 8 if ( d 3) n 1 ⎪⎢ v 7 ⎥ v 9 ⎪⎢ 3 ⎥ γ t ( P (2 k + 1, k )) = ⎨ ⎡ S 0 = { V 3 t +1 |0 ≤ t < a }; ⎤ 2 ≡ , if (mod 3) ⎪ ⎢ n 1 n ⎥ ⎪ ⎢ ⎪ ⎢ u 8 v 10 3 3 ⎥ ⎥ ⎩ ⎩ u 7 u S 1 = { U 3 t +1 |0 ≤ t < a }; { U |0 ≤ t < } v v 6 S u 9 S ≤ ⎡ ⎤ ⎡ ⎤ 1 2 + n ⎡ 2 n ⎤ ⎢ ⎥ 2 u 6 n (3) or S = | S 0 | + | S 1 | = 2 a = 3 ⎢ ⎥ 3 ⎢ 3 ⎥ u 10 we construct a total dominating set S for required cardinality we construct a total dominating set S for required cardinality. v 11 u 5 v 5 u 11 u 4 v 12 v 4 u 12 u 3 3 u 13 v 13 u 2 v 3 u 14 u 1 u u 15 1 v 2 v 14 v v 1 v v 15