Domination in generalized Petersen graphs Sheng-Chyang Liaw 廖 勝 強 Department of Mathematics National Central University Jhongli 32001, Taiwan scliaw@math.ncu.edu.tw *Joint work with Yuan-Shin Li(李元馨)

Dominating set: A vertex subset S of a graph G is a dominating set 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 . C5

Independent dominating set: S is an independent dominating set if S is not only a dominating set but also an independent set. The independent domination number of G is the cardinality of a minimum independent dominating set of G, denoted by . C5

Total dominating set: S is a total dominating set if each vertex v of G is adjacent to some vertex of S. The total domination number of G is the cardinality of a minimum total dominating set of G, denoted by . C5

Proposition: γ(G) ≤ γi (G). γ(G) ≤ γt (G). If G is k-regular then .

Generalized Petersen graph: A generalized Petersen graph P(n, k) vertex set: ={v1, v2, ..., vn} {u1, u2, ..., un} edge set: with vn+1 = v1, un+1= u1, 1 ≤ i ≤ n, and 1 ≤ k ≤ .

V={v1, v2, ..., vn} U={u1, u2, ..., un} E= {vivi+1, viui, uiui+k} 1 ≤ i ≤ n, and 1 ≤ k ≤ u4 v10 v4 u10 u3 u11 v11 v3 u2 u12 u1 u13 v12 v2 v13 v1 Generlized Petersen graph P(13,3)

Petersen graph = P(5,2):

Previous results: γ(P(2k+1,k)) ≤ : A. Behzad, M. Behzad, and C. E. Praeger, On the domination number of the generalized Petersen graphs, Discrete Mathematics 308(2008), 603~610. γ(P(2k+1,k))  : H. Yan, L. Kang, and G. Xu, The exact domination number of the generalized Petersen graphs, Discrete Mathematics 309(2009), 2596~2607. 8 8

Previous results: γ(P(n,k)), k=1,2,3 : B. J. Ebrahimi, N. Jahanbakht, and E. S. Mahmoodian, Vertex domination of generalized Petersen graphs, Discrete Mathematics 309(2009), 4355~4361. γt(P(n,2)) : J. Cao, W. Lin, and M. Shi, Total domination number of generalized Petersen graphs, Intelligent Information Management 2009, 15~18. 9 9

P(2k + 1, k) ,n ≡ 0,2(mod 3) P(2k, k) P(n, 1) , n≡2 (mod 4) domination number independent domination number total domination number P(2k + 1, k) ,n ≡ 0,2(mod 3) , n ≡1(mod3) P(2k, k) P(n, 1) , n≡2 (mod 4) ,n≡0,1,3 (mod 4) ,n ≡ 1,2 (mod 4) , n ≡ 0,3 (mod 4) , n≡4 (mod 6) P(n, 2) , n ≡ 1(mod3) P(n, 3) n ≡ 2,3 (mod 4) , and n ≠ 11 n ≡ 0,1 (mod 4) , or n=11 ,n ≡ 2,3 (mod 4) , n ≡ 0,1 (mod 4) , n≡6,8 (mod 10) , otherwise 10

γi(P(2k + 1, k)) = : Since γ(G) ≤ γi(G), we have γi(P(2k + 1, k)) ≥ . u1 u8 u7 u6 u5 u4 u3 u2 u9 u15 u12 u13 u14 u10 u11 v1 v2 v3 v4 v5 v6 v7 v8 v10 v9 v11 v12 v13 v14 v15 n = 5a S0 = {V5t+1|0 ≤ t < a} S1 = {U5t+3|0 ≤ t < a} ∪ {U5t+4|0 ≤ t < a}; |S0| + |S1| = a + 2a = 3a =

n = 5a + 3 S0 = {V5t+1|0 ≤ t < } ∪ {V5t| ≤ t ≤ a}; S1 = {U5t+3|0 ≤ t < a} ∪ {U5t+4|0 ≤ t < } ∪ {U5t+2| ≤ t ≤ a}; |S0| + |S1| = a + 1 + a + a + 1 = 3a + 2 = . u1 u3 u2 u4 u5 u6 u7 u8 u9 u10 u11 u12 u13 v1 v2 v3 v4 v5 v6 v7 v8 v9 v10 v11 v12 v13 n = 5a + 3 12 12

Total domination: If G is k-regular then . a connected set D of a 3-regular graph G can dominate at most 2|D|+2 vertices in G with |D| ≥ 2 . Proof. By induction on |D|.

Total domination in a 3-regular and 2n vertices graph: If n = 3a + 1 and γt= |S| = , then we have the following two properties. For any connected subset D ⊂ S, |D| ≤ 3. S is formed by the union of exactly one 3-element connected set and a − 1 2-element connected sets for n = 3a + 1. 14 14

γ(P(2k,k))  : For i = 1, ..., n, we define Si to be an induced subgraph of G. Si P(10,5)

γ(P(2k,k))  : Let D be a minimum dominating set of G = P(2k, k). vi+1 vi ui+k+1 vi-1 ui-1 ui ui+1 ui+k ui+k-1 vi+k-1 vi+k vi+k+1 Let D be a minimum dominating set of G = P(2k, k). Therefore we know . Si We have And each vertex of G belongs to exactly 6 Si So, 16 Hence 6|D| ≥ 4n and 16

γi(P(2k,k))  : (1) n = 3a S0 = {v3t+1|0 ≤ t < a} u12 u11 u10 u9 u8 u7 u6 u5 u4 u3 u2 u1 (1) n = 3a S0 = {v3t+1|0 ≤ t < a} S1 = {u3t+2|0 ≤ t < } ∪ {u3t|1 ≤ t ≤ } |S0| + |S1| = a + a = 2a = 17 17

γi(P(2k,k))  : (2) n = 3a + 1 S0 = {v3t+1|0 ≤ t ≤ } u7 u8 u9 u10 u6 u5 u4 u3 u2 u1 (2) n = 3a + 1 S0 = {v3t+1|0 ≤ t ≤ } ∪ {v3t| < t ≤ a} S1 = {u3t+2|0 ≤ t < a} |S0| + |S1| = 2a + 1 = 18 18

γt(P(2k,k)) = n: An obviously upper bound of γt(P(2k, k)) is n. Let D be the minimum total dominating set of G = P(2k, k). Define Si={vi,vi+k,ui,ui+k} for i = 1, ..., n vi ui ui+k vi+k Si

γt(P(2k,k)) = n: Let Di = D ∩ Si vi ui ui+k vi+k Let Di = D ∩ Si Therefore |Di| ≥ 2. each vertex of G belongs to exactly 2 Si Si So, Hence 2|D| ≥ 2n and |D| ≥ n 20 20

P(2k + 1, k) ,n ≡ 0,2(mod 3) P(2k, k) P(n, 1) , n≡2 (mod 4) domination number independent domination number total domination number P(2k + 1, k) ,n ≡ 0,2(mod 3) , n ≡1(mod3) P(2k, k) P(n, 1) , n≡2 (mod 4) ,n≡0,1,3 (mod 4) ,n ≡ 1,2 (mod 4) , n ≡ 0,3 (mod 4) , n≡4 (mod 6) P(n, 2) , n ≡ 1(mod3) P(n, 3) n ≡ 2,3 (mod 4) , and n ≠ 11 n ≡ 0,1 (mod 4) , or n=11 ,n ≡ 2,3 (mod 4) , n ≡ 0,1 (mod 4) , n≡6,8 (mod 10) , otherwise 21

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