Discrete Applied Mathematics 160 (2012) 488–493

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Discrete Applied Mathematics

journal homepage: www.elsevier.com/locate/dam

Constructing connected bicritical graphs with edge-connectivity 2✩

Xue-gang Chen a,∗, Shinya Fujita b, Michitaka Furuya c, Moo Young Sohn d

a Department of Mathematics, North China Electric Power University, Beijing 102206, Chinab Department of Mathematics, Gunma National College of Technology, Maebashi 371-8530, Japanc Department of Mathematical Information Science, Tokyo University of Science 1-3 Kagurazaka, Sinjuku-ku, Tokyo 162-8601, Japand Department of Mathematics, Changwon National University, Changwon, 641-773, Republic of Korea

a r t i c l e i n f o

Article history:Received 15 June 2010Received in revised form 10 March 2011Accepted 15 March 2011Available online 22 April 2011

Keywords:Domination numberDomination bicritical graph

a b s t r a c t

A graph G is said to be bicritical if the removal of any pair of vertices decreases thedomination number of G. For a bicritical graph G with the domination number t , we saythatG is t-bicritical. Let λ(G) denote the edge-connectivity ofG. In [2], Brigham et al. (2005)posed the following question: If G is a connected bicritical graph, is it true that λ(G) ≥ 3?

In this paper, we give a negative answer toward this question; namely, we give aconstruction of infinitely many connected t-bicritical graphs with edge-connectivity 2 forevery integer t ≥ 5. Furthermore, we give some sufficient conditions for a connected 5-bicritical graph to have λ(G) ≥ 3.

© 2011 Elsevier B.V. All rights reserved.

1. Introduction

Graph theory terminology not presented here can be found in [2]. Let G = (V , E) be a graph with |V | = n. The degree,neighborhood and closed neighborhood of a vertex v in the graph G are denoted by d(v),N(v) and N[v] = N(v) ∪ {v},respectively. The minimum degree and maximum degree of the graph G are denoted by δ(G) and ∆(G), respectively. Thegraph induced by S ⊆ V is denoted by G[S]. Let G denote the complement of G. For any two graphs G andH , let G+H denotethe graph obtained from G and H by joining edges from every vertex of G to every vertex of H . A set S is called a dominatingset if, for every vertex u ∈ V (G) − S, there exists v ∈ S such that uv ∈ E. The domination number of G, denoted by γ (G), isthe minimum cardinality of a dominating set of G. A minimum dominating set of a graph G is called a γ (G)-set.

For many graph parameters, criticality is a fundamental issue. Brigham et al. [1], Sumner [5], and Sumner andBlitch [6] began the study of graphs where the domination number decreases on the removal of any vertex. Furtherproperties of these graphs were explored in [3,4]. Note that removing a vertex can increase the domination number bymore than one, but can decrease it by at most one. It is useful to write the vertex set of a graph as a disjoint union of threesets according to how their removal affects γ (G). Let V (G) = V 0

∪ V+∪ V− where V 0

= {v ∈ V |γ (G − v) = γ (G)},V+

= {v ∈ V |γ (G − v) > γ (G)} and V−= {v ∈ V |γ (G − v) < γ (G)}.

As defined in [1], a vertex v is said to be critical if v ∈ V−, and a graph G is said to be domination critical if every vertexof G is critical. Brigham et al. [2] gave a generalization of this concept. A graph G is said to be domination (γ , k)-critical, ifγ (G − S) < γ (G) for any set S of k vertices. Obviously, a domination (γ , k)-critical G has γ (G) ≥ 2. In the special case ofk = 2, we say that G is domination bicritical, or just call it bicritical. For a bicritical graph G with γ (G) = t , we say that G ist-bicritical.

Brigham et al. [2] showed the following results concerning the edge-connectivity λ(G) of a bicritical graph G.

✩ Research supported by the Fundamental Research Funds for the Central Universities (10ML39) (to X.C), JSPS Grant 20740068 (to S.F). The fourth authorwas supported by Changwon National University in 2009–2010.∗ Corresponding author. Fax: +86 010 51963871.

E-mail address: gxc_xdm@163.com (X.-g. Chen).

0166-218X/$ – see front matter© 2011 Elsevier B.V. All rights reserved.doi:10.1016/j.dam.2011.03.012

X.-g. Chen et al. / Discrete Applied Mathematics 160 (2012) 488–493 489

Fig. 1. Connected 5-bicritical graph Gwith λ(G) = 2.

Theorem 1. If G is a connected bicritical graph, then δ(G) ≥ 3 and λ(G) ≥ 2.

Theorem 2. If G is a connected graph that is 3-bicritical or 4-bicritical, then λ(G) ≥ 3.

Motivated by the above theorems, they proposed the following open question.

Question 3. If G is a connected bicritical graph, is it true that λ(G) ≥ 3? In particular, if G is a connected 5-bicritical graph, isit true that λ(G) ≥ 3?

In this paper, we give a negative answer toward this question. We give a construction of infinitely many connectedt-bicritical graph with edge-connectivity 2 for every integer t ≥ 5. Moreover, we give some sufficient conditions for aconnected 5-bicritical graph G to have λ(G) ≥ 3 in view of forbidden subgraph conditions andminimum degree conditions.

2. Main results

First,we construct a connected 5-bicritical graphwith edge-connectivity 2. LetX = K5 and setV (X) = {v5, u5, w5, s5, t5}.Let Y ′

1 = Y ′

2 = C10, say Y ′

1 = v1u1w1s1t1v2u2w2s2t2 and Y ′

2 = v3w3t3u3s3v4w4t4u4s4. Let Y1 = Y ′

1 and Y2 = Y ′

2.Let F = (Y1 + Y2) + X . Let Sv = {vi|i = 1, 2, . . . , 5}, Sw = {wi|i = 1, 2, . . . , 5}, Su = {ui|i = 1, 2, . . . , 5},Ss = {si|i = 1, 2, . . . , 5} and St = {ti|i = 1, 2, . . . , 5}. Let H = F \ (E(F [Sv]) ∪ E(F [Su]) ∪ E(F [Sw]) ∪ E(F [Ss]) ∪ E(F [St ])).Let G1 be the graph obtained from H and P2 = ac by joining a to every vertex of Y1 with edges and joining c to every vertexof Y2 with edges. Let G2 be a graph as in Fig. 1.

Let G be the graph obtained from G1 and G2 by joining edges ab and cd. Assume that these vertices are located as in Fig. 1.Here dashed lines in Fig. 1 denote non-edges in Yi for i = 1, 2.

By the construction of G1 and G2, we have the following lemmas.

Lemma 4. γ (G1) = 3, γ (G2) = 2 and γ (G2 − b) = γ (G2 − d) = 2.

Lemma 5. Let G be the graph as in Fig. 1. Then G is a connected 5-bicritical graph with λ(G) = 2.

Proof. By the construction of G, it is obvious that λ(G) = 2. We first prove that γ (G) = 5. Let S be a γ (G)-set. In order todominate V (X), |S ∩ (V (X) ∪ V (Y1) ∪ V (Y2))| ≥ 2. If a, c ∈ S, since S ∩ N[x] = ∅, it follows that |S| ≥ 5. If a ∈ S and c ∈ S,then S ∩ V (G2) is a dominating set of G2 − b. By Lemma 4, |S| ≥ 5. Suppose that a, c ∈ S. If b, d ∈ S, then |S| ≥ 5. If b ∈ Sand d ∈ S, then for any S ′

⊆ N[b], since γ (G2[V (G2) − S ′]) ≥ 2, it follows that |S| ≥ 5. If b, d ∈ S, then S ∩ V (G1) and

S ∩ V (G2) are dominating sets of G1 and G2, respectively. By Lemma 4, |S ∩ V (G1)| ≥ 3 and |S ∩ V (G2)| ≥ 2. This implies|S| ≥ 5, and hence γ (G) ≥ 5. Since {v1, u1, c, x, g} is a dominating set of G, γ (G) ≤ 5. Therefore, γ (G) = 5. In what follows,we will prove that G is a bicritical graph. To see this, for any u, v ∈ V (G), we divide the proof into four cases.

Case 1. {u, v} ∩ (V (Y1) ∪ V (Y2)) = ∅, say v = v1. If u ∈ {x, g}, then {t2, v3, x, g} or {u1, v4, x, g} is a dominating set ofG − {u, v}. If u = x, then {u1, t2, b, d} is a dominating set of G − {u, v}. If u = g , then {u1, t2, f , h} is a dominating set ofG − {u, v}.

490 X.-g. Chen et al. / Discrete Applied Mathematics 160 (2012) 488–493

Case 2. {u, v} ∩ (V (Y1) ∪ V (Y2)) = ∅ and {u, v} ∩ V (X) = ∅, say v = v5. If u ∈ {a, c, x}, then {v1, a, c, x} is a dominatingset of G − {u, v}. If u ∈ {a, c}, say u = a, then {v3, w3, x, g} is a dominating set of G − {u, v}. If u = x, then {v1, u1, b, d} is adominating set of G − {u, v}.

Case 3. {u, v} ∩ (V (X) ∪ V (Y1) ∪ V (Y2)) = ∅ and {u, v} ∩ {a, c} = ∅, say v = a. If u ∈ {x, g}, then {v3, w3, x, g} is adominating set of G − {u, v}. If u = x, then {v3, w3, b, d} is a dominating set of G − {u, v}. If u = g , then {v3, w3, f , h} is adominating set of G − {u, v}.

Case 4. v, u ∈ V (G2). Suppose that {u, v} ∩ {b, d} = ∅, say v = d. Then {x, a, v3, w3} or {b, f , v3, w3} is a dominatingset of G − {u, v}. Without loss of generality, we can assume that {u, v} ∩ {b, d} = ∅. If x ∈ {u, v}, then {b, d, v1, u1} is adominating set of G − {u, v}. Hence, we can assume that u, v ∈ N(x). Suppose that {u, v} ∩ {e, t} = ∅, say v = e. If u = f ,then {f , d, v1, u1} is a dominating set of G − {u, v}. If u = f , then {g, d, v1, u1} is a dominating set of G − {u, v}. Hence, wecan assume that {u, v} ⊆ {g, h, f }. Then {e, d, v1, u1} or {t, b, v3, w3} is a dominating set of G − {u, v}.

By Cases 1–4, for any u, v ∈ V (G), there exists a dominating set of G − {u, v} with cardinality at most 4. Hence, G is abicritical graph. So, G is a connected 5-bicritical graph with λ(G) = 2. �

Next we investigate the structure of connected bicritical graphs with edge-connectivity 2. For a connected bicriticalgraph Gwith λ(G) = 2, there is an edge cut {ab, cd} in G. In this paper, we would like to characterize such graphs, althoughit seems difficult in general. As a first step, we are concernedwith such a graph Gwhose component of G−{ab, cd} has smalldomination number. We give the following lemmas.

Lemma 6. Let G be a graph with γ (G) ≥ 2. If V (G) = A ∪ B, A ∩ B = ∅ and |A| ≥ 4, then there exist A1 ⊆ A and B1 ⊆ B suchthat |A1| = |A| − 2 and γ (G[A1 ∪ B1]) ≥ 2.

Proof. Let v1 ∈ A. Since γ (G) ≥ 2, there exists a vertex u1 such that v1u1 ∈ E(G). Let S0 = ∅ and S1 = {v1, u1}.Let v2 ∈ A − (A ∩ S1). If γ (G[S1 ∪ {v2}]) ≥ 2, then we let S2 = S1 ∪ {v2}. If γ (G[S1 ∪ {v2}]) = 1, then there exists

u2 ∈ V (G) − S1 such that v2u2 ∈ E(G). Then we let S2 = S1 ∪ {v2, u2}. Choose v3 ∈ A − (A ∩ S2). If γ (G[S2 ∪ {v3}]) ≥ 2,then we let S3 = S2 ∪ {v3}. If γ (G[S2 ∪ {v3}]) = 1, then there exists u3 ∈ V (G) − S2 such that v3u3 ∈ E(G). Now letS3 = S2 ∪ {v3, u3}. Repeat this process to obtain a sequence {S1, S2, . . . , Sk} such that γ (G[Si]) ≥ 2 for i = 1, 2, . . . , k and|A−(A∩Sk)| = 2, 3. If |A−(A∩Sk)| = 2, thenwe let A1 = A∩Sk and B1 = B∩Sk, as desired. Suppose that |A−(A∩Sk)| = 3.Let u, v, w ∈ A− (A∩ Sk). If there exists a vertex, say w, such that w does not dominate Sk, then we let A1 = (A∩ Sk) ∪ {w}

and B1 = B ∩ Sk, as desired. If there exists a vertex, say w, such that w does not dominate a vertex of B − B ∩ Sk, say t , thenwe let A1 = (A∩Sk)∪{w} and B1 = (B∩Sk)∪{t}, as desired. Suppose that each vertex of A− (A∩Sk) dominates Sk ∪B. Thenγ (G[{u, v, w}]) ≥ 2, say uv ∈ E(G). If |A ∩ Sk| − |A ∩ Sk−1| = 2, then we let A1 = (A ∩ Sk−1) ∪ {v, w, u} and B1 = B ∩ Sk−1.Then the result holds.

Suppose that |A ∩ Sk| − |A ∩ Sk−1| = 1. Let A1 = (A ∩ Sk−1) ∪ {v, u} and B1 = B ∩ Sk−1. Then the result holds. �

By Lemma 6, we have the following corollary.

Corollary 7. Let G be a graph with γ (G) ≥ 2 and n ≥ 4. Then there exists a subset S ⊆ V (G) such that |S| = n − 2 andγ (G[S]) ≥ 2.

Lemma 8 (Brigham et al. [2]). Suppose that G is a connected bicritical graph with λ(G) = 2 and an edge cut {ab, cd}. Let G1 andG2 be the two components of G − ab − cd, with a, c ∈ V (G1), b, d ∈ V (G2) and a = c. Then all of the following must be true.

(i) γ (G) = γ (G1) + γ (G2).(ii) a, c ∈ V+(G1) and b, d ∈ V+(G2).(iii) b = d.(iv) Without loss of generality, a, c ∈ V−(G1) and b, d ∈ V 0(G2).(v) Neither b nor d is in a γ (G2)-set.(vi) γ (G2 − {b, d}) = γ (G2) − 1 and a γ (G2 − {b, d})-set dominates neither b nor d.(vii) There is a γ (G2 − d)-set containing b, and there is a γ (G2 − b)-set containing d.(viii) There is no γ (G1)-set containing both a and c.(ix) There is no γ (G1 − a)-set containing c, and there is no γ (G1 − c)-set containing a.(x) γ (G1) ≥ 3.

Let G21 = G2,G22 = G2 + gf ,G23 = G2 + ge, G24 = G22 + gt,G25 = G22 + gh and G26 = G23 + gt . LetG27 = G21 + et,G28 = G27 + gf and G29 = G28 + gt . Notice that by Lemma 8(vi), we have γ (G2) ≥ 2. Now we provethe following theorem.

Theorem 9. Let G be a connected t-bicritical graph with λ(G) = 2, and let {ab, cd} be an edge cut of G. Suppose that there existsa connected component G2 of G − ab − cd such that γ (G2) = 2. Then G2 is isomorphic to G2i for some i with 1 ≤ i ≤ 9.

X.-g. Chen et al. / Discrete Applied Mathematics 160 (2012) 488–493 491

Proof. Suppose that G1 and G2 are the two components of G − ab − cd, with a, c ∈ V (G1), b, d ∈ V (G2) and γ (G2) = 2. ByLemma 8(vi) and (x), γ (G1) = t −2. Then γ (G2 −{b, d}) = 1. By Lemma 8(v) and (vi), bd ∈ E(G). Let {x} be a γ (G2 −{b, d})-set. Let A1 = {v|v ∈ N(x) ∩ N(b), v ∈ N(d)}, A2 = {v|v ∈ N(x) ∩ N(b) ∩ N(d)}, A3 = {v|v ∈ N(x) ∩ N(d), v ∈ N(b)}and A4 = {v|v ∈ N(x), v ∈ N(b) ∪ N(d)}. By Lemma 8(v), note that N(x) ∩ {b, d} = ∅. Hence, it follows thatV (G2) = {b, d, x} ∪ A1 ∪ A2 ∪ A3 ∪ A4.

Claim 1. For any u, v ∈ N(x), let D be a γ (G − {u, v})-set. Let D1 = D ∩ V (G1) and D2 = D ∩ V (G2 − {u, v}).

(1) If |D2| = 2, then D2 ∩ {b, d} = ∅ and D2 dominates both b and d.(2) If |D2| = 1, then D2 dominates at least one vertex of b and d.(3) If u, v ∈ N(x) ∩ N(b) and |D2| = 1, then D2 ∩ N(b) = ∅.(4) If u, v ∈ N(x) ∩ N(b) and |D2| = 2, then D2 = {w, d}, where w ∈ (N(b) ∩ N(x)) − {u, v}.

Proof. Since G is bicritical, |D| = γ (G − {u, v}) ≤ γ (G) − 1 = t − 1.

(1) If D2 ∩ {b, d} = ∅, then D1 is a dominating set of G1. Hence, |D1| ≥ γ (G1) = t − 2. So, |D| = |D1| + |D2| ≥ t , which isa contradiction. Hence, D2 ∩ {b, d} = ∅, say d ∈ D2. Suppose that b is not dominated by D2. Then b is dominated by D1.So, a ∈ D1. Then D1 ∪ {c} is a dominating set of G1. Since there is no γ (G1)-set containing both a and c , it follows that|D1| ≥ t − 2. Hence, |D| = |D1| + |D2| ≥ t , which is a contradiction. So, D2 dominates both b and d.

(2) Suppose that |D2| = 1. If D2 dominates neither b nor d, then D1 is a dominating set of G1, and D1 dominates bothb and d. Then a, c ∈ D1. Since there is no γ (G1)-set containing both a and c , it follows that |D1| > t − 2. Hence,|D| = |D1| + |D2| ≥ t , which is a contradiction. So, D2 dominates at least one vertex of b and d.

(3) Let D2 = {w}. By (2), it follows that w ∈ A1 ∪ A2 ∪ A3. If w ∈ A3, then w dominates A3 ∪ A4. So, {w, b} is a γ (G2)-setcontaining b, which contradicts Lemma 8(v). Hence, w ∈ A1 ∪ A2. This shows w ∈ N(b).

(4) Since bu, bv ∈ E(G), b ∈ D2. By (1), D2 = {w, d}, where w ∈ (N(b) ∩ N(x)) − {u, v}. �

By the symmetry of the roles of b and d, Claim 1(3), (4) holds if we interchange bwith d.

Claim 2. |A2| ≤ 1.

Proof. Suppose that |A2| ≥ 2. Let u, v ∈ A2. Let D be a γ (G − {u, v})-set and D2 = D ∩ V (G2 − {u, v}). Then |D| ≤ t − 1.Since u, v ∈ N(x) ∩ N(b) ∩ N(d), then x, b, d ∈ D2. By Claim 1, |D2| = 1,D2 ∩ N(b) = ∅ and D2 ∩ N(d) = ∅, say D2 = {w}.Then it follows that w ∈ A2 and w dominates N[x] − {u, v}. Hence {w, b} is a γ (G2)-set containing b, which contradictsLemma 8(v). �

Claim 3. If |A2| = 1, then A4 = ∅.

Proof. Suppose that A4 = ∅. Let u ∈ A2. Let D be a γ (G − {u, x})-set and D2 = D ∩ V (G2 − {u, x}). In order to dominateA4,D ∩ (N(x) − {u}) = ∅, say w ∈ D ∩ (N(x) − {u}). Since wx, bu, du ∈ E(G), b, d ∈ D. Otherwise, D is a dominating setof G, which is a contradiction. Then |D ∩ V (G1)| ≥ t − 2. Hence, |D2| = 1. This shows D2 = {w}. Since D dominates {b, d},by Lemma 8(viii), it follows that w ∈ (A1 ∪ A3). Then {w, d} or {w, b} is a γ (G2)-set, which contradicts Lemma 8(v). So wehave A4 = ∅. �

Claim 4. If |A2| = 1 and |A1| ≥ 3, then for any v ∈ A1, there exists a vertex w ∈ A1 − {v} such that w dominates A1 − {v} andvw ∈ E(G).

Proof. Letu ∈ A2, and letDbe aγ (G−{u, v})-set. Since xu, xv, bu, bv ∈ E(G), it follows that b, x ∈ D2 andD2∩(N(x)−{u}) =

∅. Suppose that |D2| = 1, say D2 = {w}. By Claim 1, w ∈ A1 − {v} and w dominates A1 − {v}. If vw ∈ E(G), then {w, d} is aγ (G2)-set containing d, which contradicts Lemma 8(v). Hence vw ∈ E(G). Suppose that |D2| = 2. By Claim 1, D2 = {w, d},where w ∈ A1 − {v}. Then w dominates A1 − {v}. If vw ∈ E(G), then {w, d} is a γ (G2)-set containing d, which contradictsLemma 8(v). Hence vw ∈ E(G). �

By Claims 2 and 3, we divide the proof into the following cases.Case 1. |A2| = 1. By Claim 3, A4 = ∅. Let A2 = {u}. If γ (G[A1]) = 1, then there exists a γ (G2)-set containing d, which is

a contradiction. Hence γ (G[A1]) ≥ 2. Similarly, γ (G[A3]) ≥ 2. That is |A1| ≥ 2 and |A3| ≥ 2. Let A1 = {v1, . . . , v|A1|} andA3 = {w1, . . . , w|A3|}. We further divide the proof into the following subcases.

Case1.1. |A1| ≥ 4 is even. By Claim4,we can assume that vivi+1 ∈ E(G) for i = 1, 3, . . . , |A1|−1. LetDbe aγ (G−{v1, v2})-set. Since vivi+1 ∈ E(G) for i = 3, . . . , |A1| − 1, by Claim 1, it follows that u ∈ D and u dominates v3 and v4. Let D′ be aγ (G − {v3, v4})-set. Arguing similarly, we have u ∈ D′. Then D′ is a dominating set of G, which is a contradiction.

Case 1.2. |A1| ≥ 3 is odd. By Claim 4, we can assume that vivi+1 ∈ E(G) for i = 1, 3, . . . , |A1| − 2. Let v|A1| ∈

A1 − {v1, . . . , v|A1|−1}. Let D be a γ (G − {u, v|A1|})-set. By Claim 1, there exists a vertex vi ∈ {v1, . . . , v|A1|−1} such thatvi dominates A1 − {v|A1|}, which is a contradiction.

By Cases 1.1 and 1.2, without loss of generality, we may assume that |A1| = 2 and |A3| = 2.Case 1.3. |A1| = 2 and |A3| = 2. Since γ (G[A1]) ≥ 2 and γ (G[A3]) ≥ 2, v1v2 ∈ E(G) and w1w2 ∈ E(G). Let D be a

γ (G − {b, x})-set. Since (A1 ∪ A2) ∩ D2 = ∅, we can assume that D2 ⊆ {w1, d}. Then w1v1, w1v2 ∈ E(G). Arguing similarly,

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we may assume that v1w1, v1w2 ∈ E(G). If u is adjacent to both v1 and v2, then {u, d} is a dominating set of G2 containingd, which is a contradiction. So, u is adjacent to at most one vertex in {v1, v2}. Similarly, u is adjacent to at most one vertexin {w1, w2}. Then there exists an i such that G2 is isomorphic to G2i.

Case 2. A2 = ∅ and A4 = ∅. By Theorem 1 and Claim 1, |A1| ≥ 3 and |A3| ≥ 3. Suppose that |A1| ≥ 4 or |A3| ≥ 4, say|A1| ≥ 4. Since no vertex of A1 dominates A1 ∪ A4, arguing similarly as in the proof of Lemma 6, there exists B1 ⊆ A1 andB4 ⊆ A4 such that γ (G[B1 ∪ B4]) ≥ 2 and |A1 − B1| = 2, say A1 − B1 = {u, v}. Let D be a γ (G− {u, v})-set. By Claim 1, thereexists a vertex w ∈ A1 such that w dominates (A1 ∪ A4) − {u, v}, which is a contradiction.

Suppose that |A1| = |A3| = 3. If there exists a vertex of A1 such that it does not dominate A4, arguing similarly as above,we get a contradiction. So we can assume that every vertex of A1 dominates A4. Then γ (G[A1]) ≥ 2 and γ (G[A3]) ≥ 2, sayAi = {ui, vi, wi} and uivi ∈ E(G) for i = 1, 3. Let D be a γ (G − {w1, w3})-set. By Claim 1, there exists w ∈ D such thatw ∈ (A1 ∪ A3) − {w1, w3} and w dominates (A1 ∪ A3) − {w1, w3}, which is a contradiction.

Case 3. A2 = A4 = ∅. By Theorem 1, Claim 1 and Lemma 8(v), |A1| ≥ 3, |A3| ≥ 3, γ (G[A1]) ≥ 2 and γ (G[A3]) ≥ 2.If |A1| ≥ 4 or |A3| ≥ 4, say |A1| ≥ 4, by Corollary 7, there exists B1 ⊆ A1 such that γ (G[B1)] ≥ 2 and |A1 − B1| = 2,say A1 − B1 = {u, v}. Let D be a γ (G − {u, v})-set. Arguing similarly as in Case 2, we get a contradiction. Suppose that|A1| = |A3| = 3, say Ai = {ui, vi, wi} and uivi ∈ E(G) for i = 1, 3. Let D be a γ (G − {w1, w3})-set. Arguing similarly as inCase 2, we get a contradiction. �

Utilizing Theorem 9, we can easily get the following sufficient condition for a connected 5-bicritical graph G to satisfyλ(G) ≥ 3 in view of forbidden subgraph conditions and minimum degree conditions.

Corollary 10. Let G be a connected 5-bicritical graph. If G satisfies one of the following conditions, we have λ(G) ≥ 3.(i) G does not contain G2i as an induced subgraph for each i with 1 ≤ i ≤ 9.(ii) δ(G) ≥ 5.

Proof. SupposeG is a connected5-bicritical graphwithλ(G) = 2. Let {ab, cd}be an edge cut ofG. LetG1,G2 be two connectedcomponents of G − ab − cd with γ (G1) ≥ γ (G2). Then in view of Lemma 8, we have γ (G1) = 3, γ (G2) = 2. However, if Gsatisfies one of the conditions (i) and (ii), it is easy to see that we cannot have a structure described as in Theorem 9 for G2.This is a contradiction. �

Also, with a slight change of G in Fig. 1, we can construct a connected 5-bicritical graph with arbitrary many vertices.To see this, for any integer z ≥ 1, let C25z−10 = v3w3t3u3s3v4w4t4u4s4, . . . , v5zw5z t5zu5zs5zv3. Let X ′

= K5 and setV (X ′) = {v5z+1, u5z+1, w5z+1, s5z+1, t5z+1}, and reset Sv = {vi|i = 1, 2, . . . , 5z}, Sw = {wi|i = 1, 2, . . . , 5z}, Su = {ui|i =

1, 2, . . . , 5z}, Ss = {si|i = 1, 2, . . . , 5z}, St = {ti|i = 1, 2, . . . , 5z}.Let G′ be the graph obtained from G by replacing Y2 and X by C25z−10 and X ′, respectively (also use new sets Sv, . . . , St

re-defined above). Then, arguing similarly as in the proof of Lemma 5, we have the following.

Lemma 11. The graph G′ is a connected 5-bicritical graph with λ(G) = 2.

In what follows, we construct a connected l-bicritical graph with edge-connectivity exactly 2 for any l with l > 6. Let Fbe the connected t-critical and t-bicritical graph where t ≥ 3. According to Brighams’ paper [2], such a graph is obtained(see Proposition 2 in [2]). Let y be a vertex in F . Let G be the graph as in Fig. 1 and let x be the vertex as in Fig. 1. LetG′′

= (F ◦ G)(y, x : w). (Here this graph means the graph obtained from F and G by identifying y and x. Let w be theidentified vertex in G′′.)

Lemma 12. For t ≥ 3, the graph G′′ is a connected (t + 4)-bicritical graph with λ(G′′) = 2.

Proof. It is obvious that λ(G′′) = 2. By the construction of G, x is a critical vertex of G. We first prove that γ (G) = t + 4. LetSG be a γ (G)-set containing x, and let SF be a γ (F)-set containing y. Then (SG ∪ SF ∪ {w}) \ {x, y} is a dominating set of G′′.Hence, γ (G′′) ≤ |(SG ∪ SF ∪ {w}) \ {x, y}| = γ (G) + γ (F) − 1 = t + 4.

Let S be a γ (G′′)-set. Ifw ∈ S, then (S∪{x}\{w})∩V (G) is a dominating set ofG and (S∪{y}\{w})∩V (F) is a dominating setof F . So,γ (G′′) = |S| = |(S∪{x}\{w})∩V (G)|+|(S∪{y}\{w})∩V (F)|−1 ≥ γ (G)+γ (F)−1. Suppose thatw ∈ S. Then S∩V (G)is a dominating set of G and S ∩V (F) is a dominating set of F . So, γ (G′′) = |S| = |S ∩V (G)|+ |S ∩V (F)| ≥ γ (G)+γ (F)−1.Hence, γ (G′′) = γ (G) + γ (F) − 1 = t + 4.

In what follows, we will prove that G′′ is a bicritical graph. For any u, v ∈ V (G′′), we divide the proof into the followingcases.

Case 1. v, u ∈ (V (F) \ {y})∪{w}. Let SF be a γ (F −v −u)-set and SG be a γ (G− x)-set. Then SF ∪ SG is a dominating set ofG′′

−u−v. Since F is a bicritical graph, γ (F −v−u) ≤ γ (F)−1. So, γ (G′′−u−v) ≤ |SF ∪SG| ≤ (γ (F)−1)+ (γ (G)−1) =

γ (G′′) − 1.Case 2. v, u ∈ (V (G)\ {x})∪{w}. Let SF be a γ (F − y)-set and SG be a γ (G−u−v)-set. Then SF ∪ SG is a dominating set of

G′′−u−v. Since F is a critical graph andG is a bicritical graph, it follows that γ (F−y) ≤ γ (F)−1 and γ (G−u−v) ≤ γ (G)−1.

So, γ (G′′− u − v) ≤ |SF ∪ SG| ≤ (γ (F) − 1) + (γ (G) − 1) = γ (G′′) − 1.

Case 3. v ∈ V (F)\{y} and u ∈ V (G)\{x}. Let SF be a γ (F−v)-set and SG be a γ (G−x−u)-set. Then SF∪SG is a dominating setofG′′

−u−v. Since F is a critical graph,γ (F−v) ≤ γ (F)−1. So,γ (G′′−u−v) ≤ |SF∪SG| ≤ (γ (F)−1)+(γ (G)−1) = γ (G′′)−1.

Hence, G′′ is a bicritical graph with λ(G′′) = 2. �

X.-g. Chen et al. / Discrete Applied Mathematics 160 (2012) 488–493 493

Finally, we construct such an example for t = 6. Consider the graph G as in Fig. 1. Now we construct a connected 6-bicritical graph from G−V (X) by adding some vertices. Take three vertices p, q, r from Y1 ∪Y2 such that {p, q, r} dominatesV (Y1) ∪ V (Y2) ∪ {a, c}. For such a triple {p, q, r}, we introduce a new vertex v{p,q,r}. For the vertex v{p,q,r}, join v{p,q,r}to (V (Y1) ∪ V (Y2)) \ {p, q, r} completely with edges. Define those new vertices for all the triples satisfying the aboveproperties, and repeat this operation to all the new vertices. Let G∗ be the resulting graph. Let G′

1 = G∗− V (G2) and

X ′= {v{p,q,r}|p, q, r ∈ V (Y1 ∪ Y2), {p, q, r} dominates V (Y1 ∪ Y2) ∪ {a, c}}. By the construction, we can easily show the

following.

Lemma 13. For any two vertices y1, y2 ∈ V (Y1)∪V (Y2), there exist two vertex y3, y4 ∈ V (Y1)∪V (Y2) such that each {y1, y2, yj}with j = 3, 4 dominates V (Y1) ∪ V (Y2) ∪ {a, c}.

Lemma 14. The graph G′

1 satisfies γ (G′

1) = 4.

Proof. Since {v1, u1, w1, c} is a dominating set of G′

1, γ (G′

1) ≤ 4. Suppose γ (G′

1) ≤ 3, and let S be a γ (G′

1)-set. ByLemma 13 and |V (X ′)| ≥ 3, since there exists no set of two vertices in V (Y1) ∪ V (Y2) ∪ V (X ′) which dominatesV (X ′), S ∩ ({a, c} ∪ V (X ′)) = φ. S dominates V (Y1) ∪ V (Y2) ∪ {a, c}, and hence there exists a vertex in X ′ not dominated byS, which is a contradiction. �

Lemma 15. The graph G∗ is a connected 6-bicritical with λ(G∗) = 2.

Proof. By the construction of G∗, it is obvious that λ(G∗) = 2.We first prove that γ (G∗) = 6. Let S be a γ (G∗)-set. In order todominate V (X ′), |S∩ (V (X ′)∪V (Y1)∪V (Y2))| ≥ 3. If a, c ∈ S, then we see from S∩N[x] = ∅ that |S| ≥ 6. If |{a, c}∩S| = 1,then S∩V (G2) is a dominating set of G2 −b or G2 −d. By Lemma 4, it follows that |S| ≥ 6. Thus wemay assume that a, c ∈ S.If b, d ∈ S, then |S| ≥ 6. If b ∈ S and d ∈ S, then for any S ′

⊆ N[b], since γ (G2[V (G2) − S ′]) ≥ 2, it follows that |S| ≥ 6. If

b, d ∈ S, then S ∩ V (G′

1) and S ∩ V (G2) are dominating sets of G′

1 and G2, respectively. By Lemmas 4 and 14, |S ∩ V (G′

1)| ≥ 4and |S∩V (G2)| ≥ 2. Hence |S| ≥ 6. Thus we have γ (G∗) ≥ 6. Since {v1, u1, w1, c, x, g} is a dominating set of G∗, γ (G∗) ≤ 6.Therefore, γ (G∗) = 6. In the following, we will prove that G∗ is a bicritical graph. For any u, v ∈ V (G∗), we divide the proofinto the following cases.

Case 1. {u, v}∩(V (Y1)∪V (Y2)) = ∅, say v = v1. If u ∈ {x, g}, then {u1, t2, v3, x, g}, {u1, t2, v4, x, g}, or {v2, v3, v4, x, g} isa dominating set of G∗

−{u, v}. If u = x, then {u1, t2, v4, b, d} is a dominating set of G∗−{u, v}. If u = g , then {u1, t2, v4, f , h}

is a dominating set of G∗− {u, v}.

Case 2. {u, v}∩(V (Y1)∪V (Y2)) = ∅ and {u, v}∩V (X ′) = ∅, say v ∈ V (X ′). Then v = vp,q,r for some p, q, r ∈ V (Y1)∪V (Y2).If u ∈ {x, g}, then {p, q, r, x, g} is a dominating set ofG∗

−{u, v}. If u = x, then {p, q, r, b, d} is a dominating set ofG∗−{u, v}.

If u = g , then {p, q, r, f , h} is a dominating set of G∗− {u, v}.

Case 3. {u, v} ∩ (V (X ′) ∪ V (Y1) ∪ V (Y2)) = ∅ and {u, v} ∩ {a, c} = ∅, say v = a. If u ∈ {x, g}, then {v3, w3, t3, x, g} is adominating set of G∗

−{u, v}. If u = x, then {v3, w3, t3, b, d} is a dominating set of G∗−{u, v}. If u = g , then {v3, w3, t3, f , h}

is a dominating set of G∗− {u, v}.

Case 4. v, u ∈ V (G2). Suppose that {u, v}∩{b, d} = ∅, say v = d. Then {x, a, u3, w3, t3} or {b, f , u3, w3, t3} is a dominatingset of G∗

− {u, v}. Without loss of generality, we can assume that {u, v} ∩ {b, d} = ∅. If x ∈ {u, v}, then {b, d, v1, u1, w1} is adominating set of G∗

− {u, v}. Hence, we can assume that u, v ∈ N(x). Suppose that {u, v} ∩ {e, t} = ∅, say v = e. If u = f ,then {f , d, v1, u1, w1} is a dominating set of G∗

− {u, v}. If u = f , then {g, d, v1, u1, w1} is a dominating set of G∗− {u, v}.

Hence, we can assume that {u, v} ⊆ {g, h, f }. Then {e, d, v1, u1, w1} or {t, b, v3, w3, t3} is a dominating set of G∗− {u, v}.

By Cases 1–4, for any u, v ∈ V (G∗), there exists a dominating set of G∗− {u, v} with cardinality at most 5. Hence G∗ is a

bicritical graph. So G∗ is a connected 6-bicritical graph with λ(G∗) = 2. �

Notice that we can also construct such 6-bicritical graphs with arbitrary many vertices in a similar manner as observedin connected 5-bicritical graphs in the previous argument. Combining Lemmas 11, 12 and 15, we finally obtain the followingresult.

Theorem 16. Let t be an integer with t ≥ 5. There exist infinitely many connected t-bicritical graphs G with λ(G) = 2.

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