The Hub Number, Girth and Mycielski Graphs

Information Processing Letters 114 (2014) 561–563

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Information Processing Letters

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The hub number, girth and Mycielski graphs ∗ Xiaoping Liu a,ZhilanDangb, Baoyindureng Wu b, a Foundation Department, Xinjiang Institute of Engineering, Urumqi, Xinjiang 830000, PR China b College of Mathematics and System Sciences, Xinjiang University, Urumqi, Xinjiang 830046, PR China article info abstract

Article history: In this note, we give a simpler proof of a theorem of Desormeaux et al., which states that Received 25 January 2014 for any connected graph G containing a cycle, γc (G) ≥ g(G) − 2, where γc (G) and g(G) Received in revised form 23 April 2014 are the connected domination number and the girth of G respectively. This enables us to Accepted 23 April 2014 conﬁrm a conjecture of Walsh: for any connected nontree graph G, h(G) ≥ g(G) − 3, where Available online 9 May 2014 h(G) is the hub number of G. We also determine the connected domination number and Communicated by B. Doerr the connected hub number of Mycielski graphs. Keywords: © 2014 Elsevier B.V. All rights reserved. Combinatorial problems Connected domination number Girth Hub number Mycielski graphs

1. Introduction a dominating set D is called a connected dominating set (resp. a total dominating set)ifG[D] is connected (resp. We consider simple undirected graphs only, and refer δ(G[D]) ≥ 1). So, a graph G has a connected dominat- to [2,15] for undeﬁned terminology and notation. Let G be ing set (resp. a total dominating set) if and only if G is a graph, and D ⊆ V (G).Theinduced subgraph G[D] of G connected (resp. G has no isolated vertices). The domina- is the graph with V (G[D]) = D, in which two vertices are tion number γ (G) of G is the minimum cardinality of a adjacent if and only if they are adjacent in G.IfG contains dominating set. Similarly, the connected domination num- acycle,thegirth g(G) of G is minimum length of a cycle ber γc(G) (resp. the total domination number γt (G))ofa in G. As usual, we denote the minimum and maximum graph G is the minimum cardinality of a connected dom- degree of G by δ(G) and (G) respectively. inating set (resp. a total dominating set) of G.So,forany ≥ Walsh [14] introduced the following graph theoretical graph G without isolated vertices γt (G) 2. For a con- ≥ = − concepts. A hub set in a graph G is a set U ⊆ V (G) such nected graph G of order n 2, if (G) n 1, then = = ≥ that any two nonadjacent vertices outside U are connected γc(G) 1 < 2 γt (G), and γc(G) γt (G) otherwise. ≤ by a path whose internal vertices lie in U .Ifahubsetin- It is easy to see that for a connected graph G, h(G) ≤ duces a connected subgraph, then it is called a connected hc(G) γc(G). Grauman et al. [6] and Johnson et al. [9] in- ≤ + hub set.Thehub number h(G) (resp. the connected hub num- dependently proved that γc(G) h(G) 1 for a connected graph G.So, ber hc(G)) is the minimum cardinality of a hub set (resp. a connected hub set) in G. ≤ AsetD ⊆ V (G) is called a dominating set of G if ev- Theorem 1.1. (See [6,9].) For any connected graph G, h(G) ≤ ≤ + ery vertex of V (G) \ D has a neighbor in D.Furthermore, hc(G) γc(G) h(G) 1.

It was shown in [14] that determining a hub number * Corresponding author. or connected hub number of a general graph is NP-hard. E-mail address: [email protected] (B. Wu). So, there are some works devoted to find the hub number http://dx.doi.org/10.1016/j.ipl.2014.04.014 0020-0190/© 2014 Elsevier B.V. All rights reserved. 562 X. Liu et al. / Information Processing Letters 114 (2014) 561–563 or the connected hub number of some special families V μ(G) = X ∪ Y ∪{z} = of graphs. Hamburger et al. [8] showed that h(Lm,n) ={ } = ≤ ≤ x1, x2,...,xn, y1, y2,...,yn, z , hc(Lm,n) γc(Lm,n) for any 3 m n, where Lm,n is the grid lattice, i.e. the cartesian product of the paths Pm E μ(G) ={zy | 1 ≤ i ≤ n} i and Pn.Linetal.[11] determined the hub number of ∪ x x , x y , y x v v ∈ E(G), 1 ≤ i, j ≤ n . Sierpinski-Like´ graphs. i j i j i j i j − In Section 3, we determine the connected hub num- For a graph G,letμ0(G) = G, and μk(G) = μ(μk 1(G)) ber and the connected domination number of Mycielski k for an integer k ≥ 1. Mycielski showed that μ (K2) is graphs. Walsh [14] conjectured that h(G) ≥ g(G) − 3for k triangle-free and χ(μ (K2)) = k + 2, where χ(G) denotes any connected, nontree graph G. In the next section, we the chromatic number of a graph G. will show that the conjecture is true. An obvious inference from the definition of μ(G), the order of μ(G) is 2n + 1anddμ(G)(xi ) = 2dG (vi ), 2. Walsh’s conjecture dμ(G)(yi ) = dG (vi ) + 1, dμ(G)(z) = n + 1. So, δ(μ(G)) = δ(G) + 1. The following result is clear from the definition, We start with the following observation. so we omit its proof. For an edge of G, G/e denotes the graph obtained from G by contracting the edge e. Lemma 3.1. For a graph G, μ(G) is connected if and only if G has no isolated vertices. Lemma 2.1. For any connected graph G, the following holds: Various properties of Mycielski graphs are investigated (1) g(G) − 1 ≤ g(G/e) ≤ g(G), in the literature. Larsen et al. [10] studied the fractional (2) γ (G) − 1 ≤ γ (G/e) ≤ γ (G). c c c chromatic number of Mycielski graphs. Balakrishnan et al. [1] proved that κ(μ(G)) = min{δ(μ(G)), 2κ(G) + 1} for a Proof. Theresultsarestraightforward. 2 connected graph G, where κ(G) denotes the connectivity of G.Indeed,ifG containsanisolatedvertexvi ,thenxi Desormeaux, Haynes, Henning [4] established a lower is an isolated vertex of μ(G), and thus μ(G) is discon- bound of the connected domination number of a connected. If G is disconnected and has no isolated vertices, nected nontree graph in terms of its girth. Next we provide then by Lemma 3.1, μ(G) is connected, but z is a cut a simpler proof of the theorem as follows. vertex of μ(G), and hence κ(μ(G)) = 1. So, the result of Balakrishnan et al. [1] holds for any graph. Guo et al. [7] Theorem 2.2. (See Desormeaux, Haynes, Henning [4].) For any investigated super-connectivity and edge super connectiv- ≥ − connected graph G that contains a cycle, γc(G) g(G) 2. ity of the Mycielski graphs. Among other things, Fisher et al. [5] proved that for a graph G, γ (μ(G)) = γ (G) + 1. Lin Proof. Suppose that the theorem is not true, and let G et al. [12] established a similar relation for the total domi- be a counterexample with minimum order. Then γc(G) ≤ nation number of Mycielski graphs. g(G) − 3, g(G) ≥ 4 and γc(G) ≥ 2. Let S be a connected dominating set of G with |S|=γc(G). Take two adjacent Theorem 3.2. (See Lin et al. [12].) For a graph G without iso- vertices u, v from S.LetG = G/uv and S be the set ob- = + lated vertices, γt (μ(G)) γt (G) 1. tained from S by identifying u and v. It is clear that G is connected and contains cycles, and S is a connected dom- In the next theorem, we show that for a graph G with- inating set of G . By the minimality of G and Lemma 2.1, out isolated vertices, γ (μ(G)) = γ (G) + 1. Thus we find a c t wide family of graphs H with γc(H) = γt (H). |S|−1 = S ≥ γc G ≥ g G − 2 ≥ g(G) − 1 − 2, and thus |S|≥g(G) − 2, a contradiction. 2 Theorem 3.3. If G is a graph without isolated vertices, then γc(μ(G)) = γt (G) + 1. Next corollary confirms the validity of Walsh’s conjecture [14]. Proof. By Lemma 3.1, μ(G) is connected and hence γc(μ(G)) and γt (μ(G)) exists. Since γc(μ(G)) ≥ Corollary 2.3. For any connected, nontree graph G, h(G) ≥ γ (μ(G)) ≥ 2, every connected dominating set of μ(G) is − g(G) 3. a total dominating set of μ(G).So,γc(μ(G)) ≥ γt (μ(G)), and by Theorem 3.2, γc(μ(G)) ≥ γt (G) + 1. ≤ + Proof. By Theorems 1.1 and 2.2, h(G) ≥ γc(G)− 1 ≥ g(G)− To show γc(μ(G)) γt (G) 1, it suffices to find a con- 2 − 1 ≥ g(G) − 3. 2 nected domination set of cardinality γt (G) + 1inμ(G).Let D be a total dominating set of G with |D|=γt (G).Let ={ | ∈ } ={ | ∈ } 3. Mycielski graphs S i vi D and Y S yi vi D .Itisasimpleexer- cise to show that Y S ∪{z} is a connected dominating set Mycielski [13] used a fascinating construction to create of μ(G). 2 triangle-free graphs with large chromatic numbers. For a graph G on vertices V (G) ={v1, v2,...,vn},theMycielski Theorem 3.4. (See Cockayne et al. [3].) If G is a connected graph ≥ ≤ 2n graph μ(G) of G is defined as of order n 3,thenγt (G) 3 . X. Liu et al. / Information Processing Letters 114 (2014) 561–563 563

For the sake of convenience, in what follows, we use xi yi ∈/ E(μ(G)) which contradicts (2) of Claim 1.So,let X S or Y S denote the set of vertices with X S ={xi | i ∈ S} Q ={xq}.By(3)ofClaim 1, W = Nμ(G)(xq).So,thereex- or Y S ={xi | i ∈ S} for S ⊆{1,...,n}. ists S ⊆{1,...,n} such that W = X S ∪ Y S .Furthermore, U ={z}∪{yq}∪XT ∪ Y T , where T ={1,...,n}\({q}∪S). Theorem 3.5. For a graph G of order n without isolated vertices, Since no vertex of XT is adjacent to z or yq, and U hc(μ(G)) < γc(μ(G)) if and only if (G) = n − 1. is a connected hub set of μ(G),eachvertexofXT has a neighbor in Y T .So,U ={z}∪{xq}∪{ys}∪Y T is a ∈ tProof. Le vi ∈ V (G) with dG (vi ) = n − 1. It can be easily connected dominating set of μ(G), where ys Y S .Since | |= ≤| | =∅ checked that U ={yi , z} is a connected hub set of μ(G), U hc(μ(G)) < γc(μ(G)) U , T , which implies = − = − 2 and thus hc(μ(G)) ≤ 2. On the other hand, by Theorem 3.3, that dG (vq) n 1. This proves (G) n 1. γc(μ(G)) = γt (G) + 1 ≥ 3. This proves the sufficiency. To prove the necessity, assume that G is a graph with Corollary 3.6. For a graph G of order n without isolated vertices, ∼ h (μ(G)) < γ (μ(G)).IfG = K , then the result is trivially c c 2 true. So, we may further assume that n ≥ 3. Let U be a γc(μ(G)) − 1 = 2, if (G) = n − 1, hc μ(G) = minimum connected hub set of μ(G), W = N(U ) \ U and γc(μ(G)) = γt (G) + 1, otherwise. Q = V (μ(G)) − (U ∪ W ).Sincehc(μ(G)) < γc(μ(G)),the set U does not dominate μ(G), and hence Q = ∅.Further- Proof. It is immediate from Theorems 1.1, 3.3 and 3.5. 2 more, since μ(G) is connected, W = ∅. Acknowledgements Claim 1. We express our sincere gratitude to the referees for (1) Qisaclique, their many valuable comments. Xiaoping Liu is supported (2) each vertex of Q is adjacent to all vertices of W , by the NSFC (11301452). Baoyindureng Wu is supported by (3) N v = Q ∪ W \{v} for any v ∈ Q. μ(G)( ) ( ) the NSFC (11161046) and by the Scientific and Technologi- cal Innovation Young Personnel Training Project of Xinjiang Proof of Claim 1. Since U is a hub set, (1) and (2) follows. (2013721012). (3) follows from (1), (2) and the definition of U , W , Q . 2 References Recall that V (μ(G)) = X ∪ Y ∪{z}. [1] R. Balakrishnan, S.F. Raj, Connectivity of the Mycielskian of a graph, Claim 2. z ∈ U. Discrete Math. 308 (2008) 2607–2610. [2] J.A. Bondy, U.S.R. Murty, Graph Theory, GTM, vol. 244, Springer, 2008. ∈ [3] E.J. Cockayne, R.M. Dawes, S.T. Hedetniemi, Total domination in Proof of Claim 2. By contradiction, suppose that z / U .Let graphs, Networks 10 (1980) 211–219. us consider two cases. [4] W.J. Desormeaux, T.W. Haynes, M.A. Henning, Bounds on the con- Case 1. Y ∩ U = ∅. nected domination number of a graph, Discrete Appl. 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