About one conjecture of twin domination number of oriented graphs

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1 arxiv: v [math.co] 6 Feb 017 About one conjecture of twin domination number of oriented graphs Dorota Urbańska-Osula 1 and Rita Zuazua 1 Faculty of Electronics, Telecommunications and Informatics Gdańsk University of Technology, Gdańsk, Poland dorurban@student.pg.gda.pl Universidad Nacional Autónoma de México, Mexico ritazuazua@ciencias.unam.com Abstract Let D = (V, A) be a digraph. A subset S of V is called a twin dominating set of D if for every vertex v V S, there exists vertices u 1, u S such that (v, u 1 ) and (u, v) are arcs in D. The minimum cardinality of a twin dominating set in D is called the twin domination number of D and is denoted by γ (D). In [], is defined the concept of upper orientable twin domination number of a graph G, DOM (G) = max{γ (D) D is an orientation of G}. In [1], it is conjectured that for the complete graph K n with n 8, DOM (K n ) = n+1. In this work we prove that DOM (K n ) n for all even number n 8. Keywords: Twin domination; upper orientable twin domination, tournaments. AMS Subject Classification Numbers: 05C69; 05C05; 05C99. 1 Introduction Let D = (V, A) be a digraph. For any vertex v V, the sets I(v) = {u (u, v) A} and O(v) = {u (v, u) A} are called the inset and outset of v. The indegree and outdegree of v are defined by id(v) = I(v) and od(v) = O(v). 1

2 Definition 1 Let D = (V, A) be a digraph. A subset S of V is called a twin dominating set of D if for every vertex v V S, there exists vertices u 1, u S (u 1, u may coincide) such that (v, u 1 ) and (u, v) are arcs in D. The minimum cardinality of a twin dominating set in D is called the twin domination number of D and is denoted by γ (D). For different orientations D 1 and D of a graph G, it is possible to have γ (D 1 ) γ (D ). In [], the authors defined the concept of upper orientable twin domination number of a graph G, n+1 DOM (G) = max{γ (D) D is an orientation of G}. In [1], it is conjectured that for n 8 and the complete graph K n, DOM (K n ) =. We will prove in section 3 that for n = 8, DOM (K 8 ) = 4. In section 4, we prove that DOM (K n ) n for all even number n 8. Previous result and lemmas The following observation is a consequence of the results given in [1]. Observation 1 Let T be a tournament (one orientation of the complete graph) of order n 3. From Theorem.6 [1], if T contains at least one vertex u V (T ) such that id(u) = 0 or od(u) = 0, then γ (T ) log (n 1) + 1. In the case of n 8, γ (T ) n. Theorem Let T be an orientation of K 8. If there exists v V (T ) such that id(v) = or od(v) =, then γ (T ) 4. Proof. Suppose there exists a vertex v V (T ) such that id(v) =, I(v) = {i, i } and O(v) = {o 1, o 1, o, o, z}. Without loss of generality we can assume that the arcs (i, i ), (o 1, o 1 ), (o, o ) A(T ). 1. If the arcs (z, i), (z, o 1 ) or (z, o ) are in A(T ), then S = {v, i, o 1, o } is a twin dominating set of T. So we can assume that the arcs (i, z), (o 1, z), (o, z) A(T ). See Figure 1.. If the arc (o 1, o ) A(T ), then the set S = {v, i, z, o } is a twin dominating set of T. If the arc (o, o 1) A(T ), then the set S = {v, i, z, o 1} is a twin dominating set of T.

3 Therefore, if id(v) =, γ (T ) 4. The case od(v) =, is symmetric. i o 1 i o 1 v o z o Figure 1: Illustration of Step 1 from Theorem for the tournament T of order 8, which has a vertex v of id(v) = ; dashed arrows denote new added edges in Step 1. Theorem 3 Let T be an orientation of K 8. If there exists v V (T ) such that id(v) = 1 or od(v) = 1, then γ (T ) 4. Proof. Let v V (T ) such that id(v) = 1 with I(v) = {z} and O(v) = {o 1, o 1, o, o, o 3, o 3}. Without loss of generality we can assume that the arcs (o 1, o 1 ), (o, o ), (o 3, o 3 ) A(T ). 1. If (o 1, z), (o, z) or (o 3, z) are in A(T ), then S = {v, o 1, o, o 3 } is a twin dominating set of T. So {v, o 1, o, o 3 } O(z).. By the Observation 1, id(z) 0, in other case, γ (T ) 4. Assume, without lost of generality, that (o 3, z) A(T ). If one of the arcs (o 3, o 1 ) or (o 3, o ) are in T, then S = {v, z, o 1, o } is a twin dominating set of T. So let us assume that (o 1, o 3 ), (o, o 3 ) A(T ), see Figure. 3. If (o 3, o 1) and (o 3, o ) are arcs in T, then od(o 3 ) = and from Theorem, γ (T ) 4. On the other hand, if (o 1, o 3 ) (resp. (o, o 3 )) is an arc in T, then S = {v, z, o, o 3 } (resp. S = {v, z, o 1, o 3 }) is a γ -set of T. The case od(v) = 1 is symmetric. 3

4 o 1 o o 1 o v o 3 z o 3 Figure : Illustration of Step 1- from Theorem 3 for the tournament T of order 8, which has a vertex v of id(v) = 1; dashed arrows denote new added edges in Step 1 and gray arrows denote new added edges in Step. Observation 4 Denote by T 6 n the tournament of order n + 6 of the Figure 3. In [1] is proved that γ (T 6 0 ) = γ (T 6 1 ) = 4. So for any n 0, γ (T 6 n) = 4 and 4 DOM (K n ). K n Figure 3: Illustration for Observation 4; boxes denote the γ -set of T 6 n of size 4, for n 0. Theorem 5 The upper orientable twin domination number of K 8, DOM (K 8 ) = 4. Proof. Let T be an orientation of K 8. By the previous observation, if there exist v V (T ) such that id(v) {0, 1,, 5, 6} or od(v) {0, 1,, 5, 6}, then γ (T ) 4. So, we can suppose that for every v V (T ), id(v) = 3 or id(v) = 4. 4

5 Let v V (T ) such that O(v) = {o 1, o 1, o, o } and I(v) = {i, i, z}. Without lost of generality, we can suppose that the arcs (i, i ), (o 1, o 1 ), (o, o ) A(T ). If one of the arcs (i, z), (o 1, z) or (o, z) are in A(T ), then S = {v, i, o 1, o } is a twin dominating set of T. So, we can suppose that O(z) = {v, i, o 1, o } and I(z) = {i, o 1, o }. 1. If (o 1, o ) A(T ), then S = {v, z, i, o } is a twin dominating set of T.. If (o, o 1 ) A(T ), then S = {v, z, i, o 1 } is a twin dominating set of T. The case of v V (T ) such that id(4) = 4 is symmetric. Therefore, DOM (K 8 ) 4. By the previous observation, we can conclude that DOM (K 8 ) = 4. 3 Main result In this section we proof that for any even number n 8, the conjecture given in [1] is false. Theorem 6 For every even natural number n 8, the upper orientable twin domination number of K n, DOM (K n ) n. Proof. Let n = k, k 4. By use induction on k. If k = 4, the theorem holds by Theorem 5. Suppose the theorem is true for any tournament T 1 of order k. Let T to be a tournament of order k +. Consider v 1, v V (T ), (v 1, v ) A(T ) and T 1 the subtournament of T induced by V (T 1 ) = V (T ) {v 1, v }. By our hypothesis induction, there exist a γ -set S 1 of T 1, such that S 1 k. If S 1 < k, then S = S 1 {v 1, v } is a twin dominating set of T with S k+1. So, suppose S 1 = k. 1. If for one vertex v S 1, (v, v 1 ) A(T ) or (v, v) A(T ), then the set S = S 1 {v } or S = S 1 {v 1 }, respectively, is a twin dominating set of T with cardinality S = k + 1. Therefore we can assume that v 1 is a source and v a sink with respect to the set S 1. Notice that {v 1, v } is a twin dominating set of S 1. 5

6 . If one of the vertices in T is source or sink the proof is finished according to Observation 1. So there exists o 1, o V (T ) such that the arcs (o 1, v 1 ), (v, o ) A(T ). If o 1 = o, v 1 v o or o 1 v 1 v is an oriented cycle in T, then S = V (T ) \ {S 1 {o i }} with i {1, } is a twin dominating set of T with S = k + 1. Thus, the arcs (o 1, v ), (v 1, o ) A(T ). It is clear, that the set S = V (T ) \ S 1 is a twin dominating set of T with cardinality S = k +. We will prove that S it is not minimum. Let v S \ {v 1, v, o 1, o }. (a) If the arc (o, o 1 ) A(T ), then S {o } is a twin dominating set of T with cardinality k + 1. So, we can suppose we have the arc (o 1, o ) A(T ), see Figure 4. (b) If the arc (v, o 1 ) or (o, v) is in A(T ), then S {o 1 } or S {o } are twin dominating set of T, respectively. So we can assume that in T, we have the arcs (o 1, v), (v, o ), which implies that S {v} is a twin dominating set of T with cardinality k + 1. S 1 v 1 v o 1 o Figure 4: Tournament T ; black dots denote vertices from V (T ), where V (T 1 ) = V (T ) \ {v 1, v } and S 1 denotes the γ -set of T 1. Acknowledgments The authors thanks the financial support received from Grant UNAM-PAPIIT- IN and by the National Science Centre, Poland, Grant number 015/17/B/ST6/

7 References [1] S. Arumugam, Karam Ebadi, L. Sathikala, Twin domination and twin irredundance in digraphs, Appl. Anal. Discrete Math 7 (013), [] G. Chartrand, P. Dankelmann, M. Schultz, H. C. Swart, Twin domination in digraphs, Ars Comb., 67 (003),