Some results related to the toughness of3-domination critical graphs

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Discrete Mathematics 272 (2003) 5 15 www.elsevier.com/locate/disc ome results related to the toughness of3-domination critical graphs Nawarat Ananchuen a;1, Michael D.

1 Discrete Mathematics 272 (2003) ome results related to the toughness of3-domination critical graphs Nawarat Ananchuen a;1, Michael D. Plummer b;2 a Department of Mathematics, ilpakorn University, Nakorn Pathom, Thailand b Department of Mathematics, Vanderbilt University, Nashville, TN 37240, UA Received 31 May 2001; received in revised form 3 October 2001; accepted 30 November 2001 Dedicated to Frank Harary on the occasion ofhis 80th birthday Abstract A graph G is said to be k -critical ifthe size ofany minimum dominating set ofvertices is k, but ifany edge is added to G the resulting graph can be dominated with k 1 vertices. The structure of k -critical graphs remains far from completely understood, even in the special case when the domination number = 3. In a 1983 paper, umner and Blitch proved a theorem which may regarded as a result related to the toughness of3--critical graphs which says that if is any vertex cutset ofsuch a graph, then G has at most + 1 components. In the present paper, we improve and extend this result considerably. c 2003 Elsevier B.V. All rights reserved. Keywords: Domination; Critical edge; Toughness 1. Introduction Let G denote a nite undirected graph with vertex set V (G) and edge set E(G). A set V (G) isa(vertex) dominating set for G ifevery vertex ofg either belongs to or is adjacent to a vertex of. The minimum cardinality ofa vertex dominating set in G is called the vertex domination number (or simply the domination number) of G and is denoted by (G). Graph G is said to be k-domination critical (or simply k -critical) if(g)=k, but (G + e)=k 1 for each edge e E(G). In this paper, 1 upported by The Thailand Research Fund Grant #BRG upported by NF Grant # INT addresses: nawarat@kanate.su.ac.th (N. Ananchuen), michael.d.plummer@vanderbilt.edu (M.D. Plummer) X/03/$ - see front matter c 2003 Elsevier B.V. All rights reserved. doi: / x(03)

2 6 N. Ananchuen, M.D. Plummer / Discrete Mathematics 272 (2003) 5 15 we will be concerned only with the case k = 3. Although a number ofresults exist concerning 3--critical graphs, their structure is far from completely understood. (For summaries ofmost known results, see [5, Chapter 16] as well as [4].) If G is 3--critical and disconnected, then G is the disjoint union ofa 2--critical graph and a complete graph. (ee [6].) Hence we will consider only connected 3-critical graphs. umner and Blitch [6] were the rst to study 3--critical graphs. Our main purpose in this paper is to extend their result which follows. It can be viewed as a toughness result for 3--critical graphs. Theorem 1.1. Let G be a connected 3--critical graph. Then if is a vertex cutset in G; G has at most +1 components. This result was recently extended by Flandrin et al. [4] as follows. We denote the number ofcomponents ofg by!(g ). Theorem 1.2. Let G be a connected 3--critical graph. If is a vertex cutset in G such that!(g )= +1, then each vertex v is a cutvertex of G. If u; v and w are vertices of G and u and v dominate G w, we will follow the previously accepted notation and write [u; v] w. uppose G is 3--critical. If u and v are non-adjacent vertices of G, then (G + uv) = 2 and so there is a vertex x V (G) such that either [u; x] v or [v; x] u. In addition to Theorem 1.1, umner and Blitch [6] also proved the following lemma for the case n 4. The cases n = 2 and 3 were proved in [4, Lemma 1]. This lemma will be used repeatedly throughout our paper. Lemma 1.3. Let G be a connected 3--critical graph and let be an independent set of n 2 vertices in V (G). (i) Then the vertices of can be ordered a 1 ;a 2 ;:::;a n in such a way that there exists a sequence of distinct vertices x 1 ;x 2 ;:::;x n 1 so that [a i ;x i ] a i+1 for i = 1; 2;:::;n 1. (ii) If, in addition, n 4, then the x i s can be chosen so that x 1 x 2 x n 1 is a path and {x 1 ;:::;x n 1 } =. Two additional results from [6] which will be ofhelp to us are the next two lemmas. Lemma 1.4. If G is a connected 3--critical graph, then no two endvertices of G have a common neighbor. Lemma 1.5. The diameter of any connected 3--critical graph is at most three. Blitch [2] proved the next result.

3 N. Ananchuen, M.D. Plummer / Discrete Mathematics 272 (2003) Lemma 1.6. If G is a connected 3--critical graph and v is a cutvertex of G, then v is adjacent to an endvertex of G. The following two results of Wojcicka [7] will also prove useful to us. Theorem 1.7. If G is a connected 3--critical graph with more than six vertices, then G has a Hamiltonian path. To state the next theorem, we make use ofthe concept ofa full 3--critical graph. For any p 6, let a + b + c = p 3 be any partition of p 3. Let H be a complete graph on p 3 vertices and let A B C = V (H) be a partition ofthe vertices ofh with A = a; B = b and C = c. Form a new graph G by adding to H three new vertices u; v and w with N (u) =A; N (v) =B and N(w) =C. (Here as usual, N(v) denotes the neighborhood ofvertex v.) Then G is clearly 3--critical and is said to be full. Theorem 1.8. Let G be a connected 3--critical graph having two endvertices. Then G is full. Finally, in what is to follow, we shall also make frequent use of the following easy result. Lemma 1.9. Let G be a 3--critical graph and let u and v be non-adjacent vertices of G. If x is a vertex of G such that [u; x] v, then xv E(G) and if x is a vertex of G with [v; x] u then xu E(G). Proof. uppose [u; x] v. Ifxv E(G), then u and x dominate G, contradicting the assumption that (G) = 3. imilarly, if[v; x] u, then xu E(G). 2. The main theorem Theorem 2.1. Let G be a connected 3--critical graph and let be a vertex cutset in G. Then (a) if 4; G has at most 1 components, (b) if =3, then G contains at most components, and if G has exactly three components, then each component is complete and at least one is a singleton, (c) if =2, then G has at most three components and if G has exactly three components, then G must have the structure shown below in Fig. 1, (d) if =1, then G has two components, exactly one of which is a singleton. Furthermore, G has exactly one or two cutvertices and if it has two, G is isomorphic to a graph of the type shown in Fig. 1. Proof. Part (a) follows immediately from Lemma 6 of [6] and Lemma 3 of[3].

4 8 N. Ananchuen, M.D. Plummer / Discrete Mathematics 272 (2003) K 1 K 2 + K n-1 n 2 + K 1 Fig. 1. We turn now to part (b). uppose is a vertex cutset in G and = 3. We want to show that G contains at most components. uppose, to the contrary, that G contains at least +1=4 components. Then by Theorem 1.1, G contains exactly four components. Let H i denote these four components, i =1;:::;4. Choose a vertex w i V (H i ); 1 6 i 6 4. Clearly W = {w 1 ;w 2 ;w 3 ;w 4 } is an independent set. By Lemma 1.3, the vertices in W may be ordered as a 1 ;a 2 ;a 3 ;a 4 in such a way that there exists a path x 1 x 2 x 3 in G W such that [a i ;x i ] a i+1,fori =1; 2; 3. By Lemma 1.9, x i a i+1 E(G), for each i =1; 2; 3. Clearly x i for i =1; 2; 3. ince [a i ;x i ] a i+1 and x i a i+1 E(G); x i is adjacent to every vertex of 4 j=1 V (H j) (V (H i ) {a i+1 }), for i =1; 2; 3. o x i x j for all 1 6 i j 6 3. Thus {x 1 ;x 2 ;x 3 } =. Consider now G + a 1 a 3. There must exist a vertex y G {a 1 ;a 3 } such that either [a 3 ;y] a 1 or [a 1 ;y] a 3. Clearly in either case y. By Lemma 1.9 and the fact that x i a i+1 E(G), for i =1; 2; 3, the case [a 3 ;y] a 1 is impossible. Hence [a 1 ;y] a 3. But then y = x 2. Now x 2 is adjacent to every vertex of 4 i=1 V (H i) {a 3 }.By

5 N. Ananchuen, M.D. Plummer / Discrete Mathematics 272 (2003) this fact and the fact that x 1 x 2 x 3 is a path, vertices x 2 and a 3 together dominate G, a contradiction. o G contains at most three components as claimed. Now suppose G contains exactly three components. We now show that in this instance, each ofthe three components must be complete. uppose to the contrary that there exists a component C of G such that V (C) 2 and C is not complete. Then there exist two non-adjacent vertices w 1 and w 2 in C. Let C 1 and C 2 be the other two components of G. Choose w 3 V (C 1 ) and w 4 V (C 2 ). Then W = {w 1 ;w 2 ;w 3 ;w 4 } is an independent set. o by Lemma 1.3 the vertices of W may be ordered a 1 ;a 2 ;a 3 ;a 4 in such a way that there exists a path x 1 x 2 x 3 in G W such that [a i ;x i ] a i+1 for i =1; 2; 3. Clearly, x i x j for 1 6 i j 6 3 and by Lemma 1.9, x i a i+1 E(G). Claim 1. {x 1 ;x 2 ;x 3 }. uppose to the contrary that {x 1 ;x 2 ;x 3 } = and consider G + a 1 a 4. ince (G + a 1 a 4 ) = 2, there exists a vertex z V (G) {a 1 ;a 4 } such that [a 4 ;z] a 1 or [a 1 ;z] a 4. uppose [a 4 ;z] a 1.Ifz, then z = x 3 since x 1 a 2 E(G) and x 2 a 3 E(G). But this contradicts Lemma 1.9 since [a 3 ;x 3 ] a 4 and a 1 a 3 E(G). Thus z. But this implies that a 2 and a 3 are in the same component of G ; call it H. But then z V (H). Hence z is adjacent to every vertex of H {z}. ince [a 2 ;x 2 ] a 3, vertex x 2 is adjacent to every vertex of H 1 H 2 where H 1 and H 2 are the components of G containing vertices a 1 and a 4, respectively. ince x 1 x 2 x 3 is a path, vertex x 2 is adjacent to x 1 and x 3. Hence {x 2 ;z} dominates G, a contradiction. Hence [a 1 ;z] a 4. uppose z. ince x 1 a 2 E(G) and x 2 a 3 E(G); z= x 3.By using this fact and the fact that [a 3 ;x 3 ] a 4, vertex x 3 is adjacent to every vertex of G (N (a 1 ) N (a 3 )), except a 4. ince [a 2 ;x 2 ] a 3, edge x 2 a 4 E(G). Because x 1 x 2 x 3 is a path, edge x 1 x 2 E(G). Hence {x 2 ;x 3 } dominates G, a contradiction. Hence z. By applying the same argument as above, one can show that {x 2 ;z} dominates G, again a contradiction. This completes the proofofclaim 1. Claim 2. {x 1 ;x 2 ;x 3 } 2. uppose to the contrary that {x 1 ;x 2 ;x 3 } 2. Then by Claim 1, {x 1 ;x 2 ;x 3 } =2. Case 2.1: uppose x 1 and x 2. ince [a 3 ;x 3 ] a 4, and {a 1 ;a 2 ;a 3 ;a 4 } is independent, edge x 3 a 1 E(G) and edge x 3 a 2 E(G). Thus a 1 ;a 2 and x 3 belong to the same component of G, say H 1. Moreover, then, vertex x 3 dominates all of H 1. Let H 2 and H 3 be the components of G containing a 3 and a 4, respectively. Clearly H i H j, for 1 6 i j 6 3. Furthermore, V (H 3 )={a 4 }. Choose w {x 1 ;x 2 }.Ifx 1 w E(G), then {x 1 ;x 3 } dominates G since x 3 dominates H 1 and x 1 dominates H 2 H 3 {x 2 ;w}, a contradiction. Hence x 1 w E(G). imilarly, x 3 w E(G). ince [a 1 ;x 1 ] a 2 and [a 3 ;x 3 ] a 4, we have a 1 w E(G) and a 3 w E(G). Fig. 2 depicts this situation. Recall that x i a i+1 E(G), for i=1; 2; 3. Now consider G+a 1 a 4. ince (G+a 1 a 4 )=2, there exists a vertex z G {a 1 ;a 4 } such that [a 4 ;z] a 1 or [a 1 ;z] a 4. uppose

6 10 N. Ananchuen, M.D. Plummer / Discrete Mathematics 272 (2003) 5 15 x 1 x 2 w a 1 x 3 a3 a 4 a 2 H 3 H 1 H2 Fig. 2. rst that [a 4 ;z] a 1. ince a 2 and a 3 are in dierent components of G ; z. But z x 1 since x 1 a 2 E(G), and z x 2 since x 2 a 3 E(G). Thus z = w. But this is also impossible since wx 3 E(G). Hence [a 1 ;z] a 4. Because a 1 a 2 E(G) and a 2 and a 3 are in dierent components of G, it follows that z. By Lemma 1.9, z x 1 and z x 2 since x 1 a 4 E(G) and x 2 a 4 E(G). Thus z = w. But then wa 4 E(G) by Lemma 1.9. Now consider G+a 2 a 4. ince (G+a 2 a 4 )=2, there exists a vertex z V (G) {a 2 ;a 4 } such that [a 2 ;z] a 4 or [a 4 ;z] a 2. uppose [a 2 ;z] a 4. ince a 1 a 2 E(G) and a 1 and a 3 are in dierent components of G, it follows that z. ince x 1 a 4 and x 2 a 4 E(G), by Lemma 1.9 it follows that z {x 1 ;x 2 }. But then z = w. However, this is also impossible since wx 1 E(G) and a 2 x 1 E(G). Hence [a 4 ;z] a 2. Because a 1 and a 3 are in dierent components of G, it follows that z. Clearly, z x 2 and z w since x 2 a 3 E(G) and wx 3 E(G). Thus z = x 1. But this is also impossible since x 1 w E(G) and a 4 w E(G). This contradiction proves Case 2.1. Case 2.2: uppose x 1 and x 3. ince [a 2 ;x 2 ] a 3 and {a 1 ;a 2 ;a 3 ;a 4 } is independent, edge a 1 x 2 E(G) and edge a 4 x 2 E(G). Thus, a 1 ;a 4 and x 2 all belong to the same component of G, say H 1. Let H 2 and H 3 be the components of G containing a 2 and a 3, respectively.

7 N. Ananchuen, M.D. Plummer / Discrete Mathematics 272 (2003) x 1 x 3 w a 1 x 2 a 2 a 3 a 4 H 3 H 1 H 2 Fig. 3. Clearly H i H j for 1 6 i j 6 3. Furthermore, V (H 3 )={a 3 }. Choose w {x 1 ;x 3 }.Ifx 1 w E(G), then {x 1 ;x 3 } dominates G since x 3 dominates (H 1 {a 4 }) H 2 and x 1 dominates (H 2 {a 2 }) H 3 {a 4 ;w}, a contradiction. Hence x 1 w E(G). imilarly, x 3 w E(G). ince [a 1 ;x 1 ] a 2, edge a 1 w E(G) and since [a 3 ;x 3 ] a 4, edge a 3 w E(G). Fig. 3 illustrates this situation. Now consider G + a 1 a 3. since (G + a 1 a 3 ) = 2, there exists a vertex z G {a 1 ;a 3 } such that [a 1 ;z] a 3 or [a 3 ;z] a 1. uppose [a 1 ;z] a 3. ince a 1 a 4 E(G) and a 2 and a 4 are in dierent components of G, it follows that vertex z. Because x 1 a 3 and wa 3 E(G), vertex z {x 1 ;w} by Lemma 1.9. Thus z = x 3. But this is impossible since x 3 a 4 and a 1 a 4 E(G). Hence [a 3 ;z] a 1. Because a 2 and a 4 are in dierent components of G, it follows that vertex z. By Lemma 1.9, vertex z {x 3 ;w}, since x 3 a 1 and wa 1 E(G). Thus z = x 1. But this is also impossible since x 1 a 2 E(G). Thus Case 2.2 is settled. Case 2.3: uppose x 2 and x 3. ince [a 1 ;x 1 ] a 2 and {a 1 ;a 2 ;a 3 ;a 4 } is independent, edge a 3 x 1 E(G) and edge a 4 x 1 E(G). Thus, a 3 ;a 4 and x 1 belong to the same component of G, say H 3. Moreover, vertex x 1 dominates all of H 3. Let H 1 and H 2 be the components of G containing a 1 and a 2, respectively. Clearly H i H j, for 1 6 i j 6 3. Furthermore, V (H 2 )={a 2 }. Choose w {x 2 ;x 3 }.Ifx 3 w E(G), then {x 1 ;x 3 } dominates G since x 3 dominates H 1 H 2 {x 2 ;w} and x 1 dominates H 3, a contradiction.

8 12 N. Ananchuen, M.D. Plummer / Discrete Mathematics 272 (2003) 5 15 x 2 x 3 w a 3 a 4 a 1 a 2 x 1 H 1 H 2 H 3 Fig. 4. Thus x 3 w E(G). imilarly, x 1 w E(G). ince [a 1 ;x 1 ] a 2, edge a 1 w E(G) and since [a 3 ;x 3 ] a 4, edge a 3 w E(G). Fig. 4 shows this situation. Now consider G + a 1 a 3. ince (G + a 1 a 3 ) = 2, there exists a vertex z G {a 1 ;a 3 } such that [a 1 ;z] a 3 or [a 3 ;z] a 1. uppose rst that [a 3 ;z] a 1. ince a 2 and a 4 are in dierent components of G, vertex z. Because [a i ;x i ] a i+1 for i =1; 2; 3, edges a 1 x 2 and a 1 x 3 E(G). ince a 1 w E(G), by Lemma 1.9 vertex z {x 2 ;x 3 ;w} =, a contradiction. Hence [a 1 ;z] a 3. By the same argument as above, vertex z. By Lemma 1.9, z w since wa 3 E(G). Clearly z x 3, since x 3 a 4 E(G). Thus z = x 2. By this fact and the fact that [a 2 ;x 2 ] a 3, vertex x 2 dominates (H 1 H 2 H 3 {x 3 }) {a 3 }. But then, since a 3 w E(G); {w; x 2 } dominates G, a contradiction. This completes the proofin Case 2.3 and hence Claim 2 is proved. By Claims 1 and 2, {x 1 ;x 2 ;x 3 } 6 1. uppose x 1. Then x 2 and x 3 are in some component of G. Because {a 1 ;a 2 ;a 3 ;a 4 } is independent and [a 2 ;x 2 ] a 3, vertex x 2 is adjacent to both a 1 and a 4. imilarly, vertex x 3 is adjacent to both a 1 and a 2. Hence a 1 ;a 2 and a 4 are in the same component. But this contradicts our choice ofthe a i. Hence x 1. imilarly, x 3. By applying a similar argument, if x 2, then a 1 and a 2 are in the same component of G and a 3 and a 4 are in the same component of G which again contradicts the choice of a i. Hence {x 1 ;x 2 ;x 3 }=. This implies that each x i belongs to some component of G. ince [a i ;x i ] a i+1,fori =1; 2; 3, it follows that x 1 is adjacent to a 3 and a 4 ; x 2 is adjacent to a 1 and a 4 and x 3 is adjacent to a 1 and a 2. This implies that a 1 ;a 2 ;a 3 and a 4 are in the same component of G, again contradicting the choice ofthe a i. Hence each component of G is complete. Next, we show that at least one ofthe three complete components must be a singleton. uppose to the contrary that each component of G has at least two vertices. Let

9 N. Ananchuen, M.D. Plummer / Discrete Mathematics 272 (2003) H i be the components of G ; i=1; 2; 3. For each i=1; 2; 3, choose w i V (H i ). Clearly {w 1 ;w 2 ;w 3 } is an independent set. By Lemma 1.3 the vertices in W may be ordered as a 1 ;a 2 ;a 3 in such a way that there exist vertices x 1 and x 2 such that [a i ;x i ] a i+1 for i =1; 2. Without loss ofgenerality, we may renumber the components ofg in such a way that a i V (H i ). ince each component of G has at least two vertices, x 1 and x 2 must belong to. Clearly x 1 x 2. Let {x 1 ;x 2 }={w}. Choose a 1 V (H 1) {a 1 } and a 3 V (H 3) {a 3 }. Consider G + a 1 a 3. ince (G + a 1 a 3 ) = 2, there exists a vertex z V (G) {a 1 ;a 3 } such that [a 1 ;z] a 3 or [a 3 ;z] a 1. In either case, z. uppose rst that [a 1 ;z] a 3. ince x 1a 2 E(G) and x 2 a 3 E(G); z x 1 and z x 2. Thus z =w. This implies that w dominates (H 2 H 3 ) {a 3 }. ince x 1a 2 E(G) and [a 2 ;x 2 ] a 3, vertex x 2 dominates (H 1 H 3 {x 1 }) {a 3 }. Thus {w; x 2 } dominates G, a contradiction. Hence [a 3 ;z] a 1. Because x 1a 2 E(G), z x 1. ince x 2 a 1 E(G), by Lemma 1.9 z x 2. Thus z = w. This implies that w dominates (H 1 H 2 ) {a 1 }. Now consider G + a 1 a 3. ince (G + a 1a 3 ) = 2, there exists a vertex z 1 G {a 1 ;a 3 } such that [a 1 ;z 1 ] a 3 or [a 3 ;z 1] a 1. In either case z 1. uppose [a 1 ;z 1 ] a 3. ince x 1a 2 E(G) and x 2 a 3 E(G); z 1 x 1 and z 1 x 2. Thus z 1 =w. By using this fact and the fact that [a 3 ;w] a 1, we see that w dominates (H 1 H 2 H 3 ) {a 1 ;a 3 }. ince x 1a 2 E(G) and [a 2 ;x 2 ] a 3, vertex x 2 dominates (H 1 H 3 {x 1 }) {a 3 }. Thus {x 2 ;w} dominates G, a contradiction. Hence [a 3 ;z 1] a 1. ince x 1 a 2 E(G) and wa 1 E(G); z 1 x 1 and z 1 w. Thus z 1 = x 2. But this contradicts Lemma 1.9 since a 1 x 2 E(G). This completes the proof ofpart (b). Next we turn to part (c). uppose therefore that is a vertex cutset with =2. Then by the Theorem 1.1,!(G ) 6 3. uppose that G has precisely three components. Let s 1 and s 2 be the vertices of and let H i ; i =1; 2; 3, be the three components of G. By Theorem 1.2, both s 1 and s 2 are cutvertices. o each s i ; i=1; 2, is adjacent to an endvertex of G by Lemma 1.6. og has at least two endvertices, say a 1 and a 2. Furthermore, neither a 1 nor a 2 is in. ince no two endvertices of G have a common neighbor by Lemma 1.4, we may assume, without loss ofgenerality, that a i s i E(G) and a i V (H i )fori =1; 2. If V (G) 7, then G must have exactly two endvertices since G has a Hamiltonian path by Theorem 1.7. Hence by Theorem 1.8, graph G is ofthe type shown in Fig. 1, where n 3. o now let us assume that V (G) 6 6. ince a i is an endvertex of G and a i s i E(G), for i =1; 2;V(H i )={a i }. If V (H 3 ) = 1, then, since G is connected, (G) =2, a contradiction. Hence V (H 3 ) = 2. Let V (H 3 )={a 3 ;a 4 }. ince G is connected, we may assume that a 3 s 1 E(G). But then a 4 s i E(G) f or i = 1 and 2; otherwise {s 1 ;s 2 } dominates G. Thus a 4 is an endvertex of G. uppose a 3 s 2 E(G) and consider G + a 3 s 2. ince (G + a 3 s 2 ) = 2, there is a vertex z of G {a 3 ;s 2 } such that either [a 3 ;z] s 2 or [s 2 ;z] a 3. uppose [a 3 ;z] s 2. Then by Lemma 1.9, z a 2 since a 2 s 2 E(G). Thus [a 3 ;z] s 2 is impossible since N(a 2 )={s 2 }. imilarly, [s 2 ;z] a 3 is also impossible. This contradiction proves that s 2 a 3 E(G). By applying a similar argument, edge s 1 s 2 E(G). Hence G is a graph ofthe type shown in Fig. 1 where n =2.

10 14 N. Ananchuen, M.D. Plummer / Discrete Mathematics 272 (2003) 5 15 K n (a) (b) (c) n 2 Fig. 5. This completes the proofofpart (c). Finally, suppose = 1. Let c be a cutvertex of G. Then G c has exactly two components by the Theorem 1.1. Moreover, one ofthe two components is a singleton by Lemma 1.6. (Clearly it cannot happen that both components of G c are singletons, for then V (G) =3 and so (G) 3 since G is connected.) If V (G) 6 6, it is easy to see that G must be isomorphic to the six-vertex graph shown in Fig. 5(a). o suppose V (G) 7. Then G has a Hamiltonian path by Theorem 1.7. But then G has at most two endedges. Ifit has two endedges, then by [6, Remark, p. 70] it must be isomorphic to a graph ofthe type shown in Fig. 1. This completes the proofofpart (d) and hence the theorem is proved. Pertaining to part (b) ofthe preceding theorem, we point out that it is not possible to say more about the number ofsingleton components, for in Fig. 5 we present examples in which G has three, two and one singleton components. On the other hand, ifone assumes that is a cutset with =3;!(G ) = 3 and, in addition, (G) = 3, it is shown in [3] that one can say a bit more. Namely, in this case each ofthe three components ofg is complete and ifone orders the three components by size as V (C 1 ) 6 V (C 2 ) 6 V (C 3 ), then either (i) C 1 = C 2 = K 1 and G belongs to an innite family of graphs G 1 or (ii) C 1 = K 1 ; C 2 = K 3 and G belongs to a second innite family of graphs G 2. (The graph families G 1 and G 2 are described in [3].) However, if1 6 (G) 6 2, Lemmas 4 6 of[3] do not apply, whereas, on the other hand, parts (c) and (d) oftheorem 2.1 hold. The following result is an immediate corollary of Theorem 2.1. Corollary 2.2. Let G be a connected 3--critical graph and let be a cutset in G. If G has exactly +1 components, then 6 2. The results ofthe present paper, together with those in [3], can be applied to the study oftoughness in a 3--critical graph. Recall that the toughness ofa (connected)

11 N. Ananchuen, M.D. Plummer / Discrete Mathematics 272 (2003) graph G; (G), is dened as follows. (G) = min{ =!(G )} where the minimum is taken over all cutsets of G. It follows from Theorem 2.1(d) that every connected 3--critical graph has toughness at least 1 2 and it was shown in [4] that every 2-connected 3--critical graph has toughness at least 1. If G isa3--critical graph with (G) 3, then (G) 1 and (G) = 1 ifand only ifg belongs to a special innite family described in [3]. If G isa3--critical graph with (G) = 2, then by Theorem 3of[3], (G) = 1. Finally, if G isa3--critical graph with (G) = 1, then G has the structure described in part (d) oftheorem 2.1 and in particular, (G)= 1 2. Remarks. (1) We use the results ofthe present work in [1] in which we obtain new information on matchings in 3--critical graphs. (2) We are very grateful to one of the referees who, while refereeing our paper, pointed out the existence ofpaper [3]. Although we have an independent proofofpart (a) oftheorem 2.1, we have omitted it here in favor of [3] which has publication precedence. References [1] N. Ananchuen, M.D. Plummer, Matching properties in domination critical graphs, Discrete Math., to be published. [2] P. Blitch, Domination in Graphs, Ph.D. Dissertation, University ofouth Carolina, [3] Y. Chen, F. Tian, B. Wei, The 3-domination-critical graphs with toughness one, Utilitas Math. 61 (2002) [4] E. Flandrin, F. Tian, B. Wei, L. Zhang, ome properties of3-domination-critical graphs, Discrete Math. 205 (1999) [5] T.W. Haynes,.T. Hedetniemi, P.J. later, Domination in Graphs; Advanced Topics, Marcel Dekker, New York, [6] D.P. umner, P. Blitch, Domination critical graphs, J. Combin. Theory er. B 34 (1983) [7] E. Wojcicka, Hamiltonian properties ofdomination-critical graphs, J. Graph Theory 14 (1990)

Abstract. 1. Introduction

Abstract. 1. Introduction MATCHINGS IN 3-DOMINATION-CRITICAL GRAPHS: A SURVEY by Nawarat Ananchuen * Department of Mathematics Silpaorn University Naorn Pathom, Thailand email: nawarat@su.ac.th Abstract A subset of vertices D of