3-Factor-criticality in double domination edge critical graphs
aa r X i v : . [ m a t h . C O ] A ug ∗ Haichao Wang , Erfang Shan † , Yancai Zhao School of Mathematics and Physics, Shanghai University of Electric Power,Shanghai 200090, China School of Management, Shanghai University, Shanghai 200444, China Department of Basic Science, Wuxi City College of Vocational Technology,Jiangsu 214153, China
Abstract
A vertex subset S of a graph G is a double dominating set of G if | N [ v ] ∩ S | ≥ v of G , where N [ v ] is the set of the vertex v and vertices adjacent to v .The double domination number of G , denoted by γ × ( G ), is the cardinality of a smallestdouble dominating set of G . A graph G is said to be double domination edge critical if γ × ( G + e ) < γ × ( G ) for any edge e / ∈ E . A double domination edge critical graph G with γ × ( G ) = k is called k - γ × ( G )-critical. A graph G is r -factor-critical if G − S has a perfectmatching for each set S of r vertices in G . In this paper we show that G is 3-factor-criticalif G is a 3-connected claw-free 4- γ × ( G )-critical graph of odd order with minimum degreeat least 4 except a family of graphs. Keywords:
Matching; 3-Factor-criticality; Double domination edge critical graphs, Claw-free
MSC: ∗ This research was partially supported by the National Nature Science Foundation of China (No. 11171207),the Nature Science Foundation of Shanghai (No. 14ZR1417900) and the Key Programs of Wuxi City College ofVocational Technology(WXCY-2012-GZ-007). † Corresponding author. Email address: [email protected] (E. Shan) Introduction
Recently, the matching and factor properties in critical graphs with respect to domination havereceived more attention (see, [1-5, 7, 10, 14-15, 21-24]). A graph G is r -factor-critical if G − S has a perfect matching for each set S of r vertices in G . If r = 1, the graph is said to be factor-critical and if r = 2, the graph G is called bicritical . A double dominating set (DDS) of G is defined in [12] as a subset S of V ( G ) such that | N [ v ] ∩ S | ≥ v of G ,where N [ v ] is the set of the vertex v and vertices adjacent to v in G . The double dominationnumber γ × ( G ) of G is the cardinality of a smallest double dominating set of G . A graph G is called double domination edge critical , or just γ × ( G ) -critical , if γ × ( G + e ) < γ × ( G ) foreach edge e E ( G ). If γ × ( G ) = k , a γ × ( G )-critical graph is said to be k - γ × ( G )- critical . In[22, 23] the matching properties of double domination edge critical graphs were investigated, weproved that G has a perfect matching if G is a connected K , -free 4- γ × ( G )-critical graph ofeven order ≥ G is bicritical if G is either a 2-connectedclaw-free 4- γ × ( G )-critical of even order with minimum degree at least 3 or a 3-connected K , -free 4- γ × ( G )-critical graph of even order with minimum degree at least 4. In this paper weshow that G is 3-factor-critical if G is a 3-connected claw-free 4- γ × ( G )-critical graph of oddorder with minimum degree at least 4 except a family of graphs.For notation and graph theory terminology we in general follow [8]. Specifically, let G bea finite simple graph with vertex set V ( G ) and edge set E ( G ). For a vertex v ∈ V ( G ), the open neighborhood of v is N ( v ) = { u ∈ V ( G ) : uv ∈ E ( G ) } and the closed neighborhood of v is N [ v ] = { v } ∪ N ( v ). The degree of v in G , denoted by d ( v ), is the cardinality of N ( v ). Let δ ( G )represent the minimum degree of G . As usual, K m,n denotes a complete bipartite graph withclasses of cardinality m and n ; K n is the complete graph on n vertices, and C n is the cycle on n vertices. For S ⊆ V ( G ), the subgraph of G induced by S is denoted by G [ S ]. A graph G issaid to be K ,r -free if it contains no K ,r as an induced subgraph. In particular, K , -free isalso called claw-free . The complement of G , denoted by G , is the graph with vertex set V ( G )such that two vertices are adjacent in G if and only if the vertices are not adjacent in G . The diameter of G is the greatest distance between two vertices of G , denoted by diam( G ). A cutset of a connected G is a subset S of V ( G ) such that G − S is disconnected. For S ⊆ V ( G ), we shalldenote by ω ( G − S ), the number of components of G − S and by o ( G − S ), the number of oddcomponents of G − S . A subset S of V ( G ) is called an independent set of G if no two verticesof S are adjacent in G . The independence number of G , denoted by α ( G ), is the cardinality ofa largest independent set of G . A set of pairwise independent edges in G is called a matching of G . A matching is perfect if it is incident with every vertex of G .2or a fixed positive integer k , a k - tuple dominating set of G is a subset S of V ( G ) such that | N [ v ] ∩ S | ≥ k for every vertex v ∈ V ( G ). The k - tuple domination number γ × k ( G ) of G isthe minimum cardinality of a k -tuple dominating set of G . In particular, when k = 1 ,
2, 1-tuple domination and 2-tuple domination are the ordinary domination and double domination,respectively. The concept of k -tuple domination in graphs was introduced and studied in [12].For more results on the k -tuple domination, we refer to [6, 9, 11-13, 16-18, 19-21, 25]. In this section we state some results that are useful in the proof of our main results. For anedge uv ∈ E ( G ), we shall denote by D uv a minimum double dominating set (MDDS) of G + uv throughout this paper.By the definition of γ × -criticality, the following observation follows immediately. Observation 1. If G is a γ × -critical graph and uv ∈ E ( G ) , then D uv contains at least one of u and v . Furthermore, if γ × ( G + uv ) = γ × ( G ) − , then D uv contains both u and v . Lemma 1. ([21]) If G is a connected - γ × ( G ) -critical graph, then diam ( G ) ∈ { , } . Lemma 2. ([21])
A graph G with diam ( G ) = 3 is - γ × ( G ) -critical if and only if G is thesequential join K + K s + K t + K for positive integers s and t . The sequential join , as defined by Akiyama and Harary, for three or more disjoint graphs G , G , . . . , G n , denoted by G + G + · · · + G n , is the graph ( G + G ) ∪ ( G + G ) ∪ · · · ∪ ( G n − + G n ), where G i + G i +1 is obtained from G i ∪ G i +1 by joining each vertex of G i to eachvertex of G i +1 for 1 ≤ i ≤ n − Lemma 3. ([23]) If G is a connected K ,r -free ( k ≥
3) 4 - γ × ( G ) -critical graph, then α ( G ) ≤ r . Lemma 4. ([23])
Let G be a connected - γ × ( G ) -critical graph and S a cutset of G . If ω ( G − S ) ≥ and x and y belong to different components of G − S , then | D xy ∩ { x, y }| = 1 and | D xy | = 3 . By Observation 1, we immediately have the following lemma.
Lemma 5.
Let G be a connected - γ × ( G ) -critical graph and S a cutset of G . If ω ( G − S ) = 2 and each component has at least two vertices, and x and y lie in different components of G − S ,then | D xy | = 3 and D xy ∩ S = ∅ . Theorem 1. ([22]) If G is a connected K , -free - γ × ( G ) -critical graph of odd order and δ ( G ) ≥ , then G is factor-critical. Theorem 2. ([10])
A graph G is k -factor-critical if and only if o ( G − S ) ≤ | S | − k for every S ⊆ V ( G ) and | S | ≥ k . In this section we shall show that G is 3-factor-critical if G is a 3-connected claw-free 4- γ × ( G )-critical graph of odd order with minimum degree at least 4 except a family H of graphs.For convenience, let us introduce more notation and terminology. If S ⊆ V ( G ) is a minimumdouble dominating set (DDS) of G , we call S a γ × ( G )-set. For a vertex v ∈ V ( G ), v is saidto be doubly dominated by S if | N [ v ] ∩ S | ≥
2. For
A, B ⊆ V ( G ), we say that A is doublydominated by B , written B ≻ × A , if each vertex of A is doubly dominated by B . Furthermore,we use B ⊁ × A to present that A is not doubly dominated by B . Lemma 6.
Let G be a - γ × ( G ) -critical graph of odd order with δ ( G ) ≥ . If diam ( G ) = 3 ,then G is -factor-critical. Proof.
If diam( G ) = 3, then, by Lemma 2, G is isomorphic to a sequential join K + K s + K t + K for positive integers s and t . Since δ ( G ) ≥ G has odd order, it follows that s ≥ t ≥
4. Further, s and t must have different parities. It is easy to check that G − D has a perfectmatching for each set D of 3 vertices in G . So the assertion holds. ✷ Lemma 7.
Let G be a -connected claw-free - γ × ( G ) -critical graph of odd order with δ ( G ) ≥ .If G is not -factor-critical, then there exists a subset S ⊆ V ( G ) such that | S | = 3 and G − S contains exactly two odd components. Proof.
Since G is not 3-factor-critical, there exists a subset S ⊆ V ( G ) with | S | ≥ o ( G − S ) > | S | − G is factor-critical, and so o ( G − S ) ≤ | S | −
1. Note that | V ( G ) | is odd, so o ( G − S ) = | S | − | S | ≤ G is 3-connected, 3 ≤ | S | ≤ | S | = 4, by Lemma 3, then G − S has no even components and ω ( G − S ) = 3. Let S = { u , u , u , u } . Choose a ∈ V ( C ) and b ∈ V ( C ). Now consider the graph G + ab . By4emma 4, | D ab ∩ { a, b }| = 1 and | D ab | = 3. Without loss of generality, assume that a ∈ D ab .In order to doubly dominate V ( C ) ∪ V ( C ), 1 ≤ | D ab ∩ S | ≤
2. If | D ab ∩ S | = 2, without lossof generality, say u , u ∈ D ab , then each vertex of V ( C ) ∪ V ( C ) − { b } is adjacent to both u and u . Moreover, b is adjacent to only one of u and u for otherwise { a, u , u } would be aDDS of G , a contradiction. Then | V ( C ) | ≥ d ( b ) ≥ δ ( G ) ≥
4. Since D ab = { a, u , u } is a γ × ( G + ab )-set, a is adjacent to at least one of u and u . This implies that G contains a clawcentered at u or u , a contradiction. Hence | D ab ∩ S | = 1. This implies that | D ab ∩ V ( C ) | = 1and V ( C ) = { b } . Without loss of generality, let u ∈ D ab and c ∈ D ab where c ∈ V ( C ).Then G [ { u , a, b, c } ] is a claw in G , a contradiction. Therefore, | S | ≤
3. Then we have | S | = 3.Moreover, since S is a minimum cutset of G , it follows that each vertex of S is adjacent toa vertex of each component of G − S . Recall that G is claw-free. Thus G − S has no evencomponents, and so G − S contains exactly two odd components. ✷ Let G be defined as that in Lemma 7, S = { s , s , s } and { C , C } be two odd componentsof G − S . Since δ ( G ) ≥ | V ( C ) | ≥ | V ( C ) | ≥
3. Now set A i = V ( C ) ∩ N ( s i ) and B i = V ( C ) ∩ N ( s i ) for 1 ≤ i ≤
3. We have the following lemma.
Lemma 8.
Let G be a -connected claw-free - γ × ( G ) -critical graph of odd order with δ ( G ) ≥ .If G is not -factor-critical and diam ( G ) = 2 , then the following statements are true: (1) For ≤ i ≤ , A i = ∅ and B i = ∅ . Furthermore, both G [ A i ] and G [ B i ] are complete; (2) V ( C ) = ∪ i =1 A i and V ( C ) = ∪ i =1 B i ; (3) There exists at least a pair of A i and A j such that A i ∩ A j = ∅ , where ≤ i = j ≤ ; (4) There exists at least a pair of B i and B j such that B i ∩ B j = ∅ , where ≤ i = j ≤ . Proof. (1) The statement (1) directly follows, because G is claw-free and S is a minimumcutset of G .(2) Suppose not, without loss of generality, let V ( C ) = ∪ i =1 A i . Thus there exists a vertex u ∈ V ( C ) − ∪ i =1 A i . Take any vertex v ∈ V ( C ). Clearly, the distance between u and v ismore than 2, which contradicts our assumption that diam( G ) = 2. So the statement (2) holds.(3) Suppose to the contrary that A i ∩ A j = ∅ for 1 ≤ i = j ≤
3. Choose a ∈ A and b ∈ B . By Observation 1, 1 ≤ | D a b ∩ { a , b }| ≤
2. If | D a b ∩ { a , b }| = 2, by Lemma5, then D a b ∩ S = { s } as A i ∩ A j = ∅ for 1 ≤ i = j ≤
3. However, A and A are notdoubly dominated by D a b , a contradiction. Hence | D a b ∩ { a , b }| = 1. If b ∈ D a b , then s ∈ D a b , so that a can be doubly dominated. But this implies that A and A can not bedoubly dominated by D a b , a contradiction. Thus a ∈ D a b . Clearly, 1 ≤ | D a b ∩ S | ≤
2. If | D a b ∩ S | = 1, then s ∈ D a b because A i ∩ A j = ∅ for 1 ≤ i = j ≤
3. To doubly dominate5 ( C ) − { b } , we have | D a b ∩ ( V ( C ) − { b } ) | = 1. Let x ∈ D a b ∩ ( V ( C ) − { b } ). Then s x ∈ E ( G ). By Claim 1, D a b is also a DDS of G , a contradiction. Hence | D a b ∩ S | = 2.Since a s , a s / ∈ E ( G ), s ∈ D a b . Without loss of generality, assume that s ∈ D a b . For avertex a ∈ A , then a is adjacent to at least one of s and s to doubly dominate a . Thisimplies that A ∩ A = ∅ or A ∩ A = ∅ , a contradiction. So the statement (3) follows.(4) We can show that the statement is also true by a similar argument that used in the proofof the statement (3). ✷ The family H of graphs is defined as follows: For odd integer r ≥
3, let H = K r , H = K and H = K ∪ K . Let H r, , be the graph obtained from ( H + H ) ∪ H by adding 6 edgesbetween V ( H ) and V ( H ) such that each vertex of H has exactly 2 neighbors in H whileeach vertex of H has precisely 2 neighbors in H . By our construction, it is easy to verifiedthat H r, , is a 3-connected claw-free 4- γ × -critical graphs of odd order with minimum degree4. Obviously, H r, , − V ( H ) has no perfect matching, hence H r, , is not 3-factor-critical. Let H = { H r, , : r ≥ } . Lemma 9.
Let G be a -connected claw-free - γ × ( G ) -critical graph of odd order with δ ( G ) ≥ and diam ( G ) = 2 . If G is not -factor-critical and G / ∈ H , then both ∩ i =1 A i = ∅ and ∩ i =1 B i = ∅ . Proof.
Let S , C and C be defined as before. Suppose not, without loss of generality, let ∩ i =1 A i = ∅ . Take a ∈ ∩ i =1 A i . By Lemma 8 (1), N [ a ] = S ∪ V ( C ). We claim that ∩ i =1 B i = ∅ .Otherwise, there exists a vertex b ∈ ∩ i =1 B i . Consider G + ab . By Observation 1, without lossof generality, assume that a ∈ D ab . If b ∈ D ab , then | D ab ∩ S | = 1 by Lemma 5, so D ab is alsoa DDS of G because a and b are adjacent to every vertex of S , a contradiction. Thus b / ∈ D ab .Since D ab ≻ × S ∪ ( V ( C ) − { b } ), | D ab ∩ ( S ∪ ( V ( C ) − { b } )) | = 2. Since b ∈ ∩ i =1 B i , D ab isstill a DDS of G by Lemma 8 (1), a contradiction. Therefore, ∩ i =1 B i = ∅ . By Lemma 8 (4),we may assume that B ∩ B = ∅ . Let x ∈ B ∩ B and consider the graph G + ax . Case 1. | D ax ∩ { a, x }| = 2.By Lemma 5, we have s ∈ D ax , and so D ax = { a, x, s } . Then each vertex of V ( C ) ∪ ( V ( C ) − { x } ) is adjacent to s . Choose b ∈ V ( C ) − { x } . Now consider G + ab . Case 1.1. | D ab ∩ { a, b }| = 2.Without loss of generality, suppose that D ab = { a, s , b } . Then each vertex of V ( C ) isadjacent to s while each vertex of V ( C ) − { b } is adjacent to both s and b . Choose avertex u ∈ V ( C ) − { x, b } . This derives that { s , s , u } is a DDS of G , a contradiction. So6 D ab ∩ { a, y }| = 2 is impossible. Case 1.2. | D ab ∩ { a, b }| = 1.Since N [ a ] = V ( C ) ∪ S , we have a ∈ D ab and b / ∈ D ab . Assume that | D ab ∩ S | = 2. Then D ab = { a, s , s } or { a, s , s } because ∩ i =1 B i = ∅ . However, D ab ⊁ × { x } as xs / ∈ E ( G ), acontradiction. Hence | D ab ∩ S | = 1. Notice that xs / ∈ E ( G ). We deduce that s ∈ D ab or s ∈ D ab . Without loss of generality, suppose that s ∈ D ab . To doubly dominate V ( C ) − { b } ,we see that | D ab ∩ ( V ( C ) − { b } ) | = 1. Let b ∈ D ab ∩ ( V ( C ) − { b } ). Thus each vertexof V ( C ) is adjacent to s while each vertex of V ( C ) − { b } is adjacent to both s and b . If b s ∈ E ( G ), then { s , s , b } is a DDS of G , a contradiction. So b s / ∈ E ( G ). Thus, in orderto doubly dominate b in G + ab , b b ∈ E ( G ). This implies that { s , s , b } is still a DDS of G , a contradiction again. Case 2. | D ax ∩ { a, x }| = 1.If x ∈ D ax , then D ax is also a DDS of G as N [ a ] = S ∪ V ( C ), a contradiction. Hence a ∈ D ax . By Lemma 5, 1 ≤ | D ax ∩ S | ≤
2. Suppose that | D ax ∩ S | = 2. Then D ax = { a, s , s } or { a, s , s } . If D ax = { a, s , s } , then each vertex of V ( C ) − { x } is adjacent to both s and s , i.e., ( V ( C ) − { x } ) ⊆ B ∩ B . Since ∩ i =1 B i = ∅ , s is not adjacent to any vertex of V ( C ) − { x } . Choose x ∈ V ( C ) − { x } . Now consider G + ax . If | D ax ∩ { a, x }| = 2, then D ax = { a, s , x } by Lemma 5, however, D ax ⊁ × V ( C ) − { x, x } . So | D ax ∩ { a, x }| = 1.Since N [ a ] = S ∪ V ( C ), x / ∈ D ax . Then a ∈ D ax . To doubly dominate V ( C ) − { x, x } , itfollows that s / ∈ D ax and | D ax ∩ { s , s } ∪ ( V ( C ) − x ) | = 2. Then, since V ( C ) = B and G [ B ] is complete, D ax is a DDS of G , this is a contradiction. If D ax = { a, s , s } , then wecan reach a contradiction by similar arguments. This implies that | D ax ∩ S | = 2 is impossible.Hence | D ax ∩ S | = 1. Suppose D ax contains s or s . Then | D ax ∩ ( V ( C ) − { x } ) | = 1,so that V ( C ) − { x } can be doubly dominated. Let x ∈ D ax ∩ ( V ( C ) − { x } ). Clearly, xx / ∈ E ( G ) for otherwise D ax would be a DDS of G . But then a claw would occur at s or s in G , a contradiction. Hence, s ∈ D ax . To doubly dominate V ( C ) − { x } , we have | D ax ∩ ( V ( C ) − { x } ) | = 1. Without loss of generality, let x ∈ D ax ∩ ( V ( C ) − { x } ). Then D ax = { a, s , x } . Thus xx ∈ E ( G ) and each vertex of V ( C ) is adjacent to s while eachvertex of V ( C ) − { x, x } is adjacent to both x and s .Now we consider G + ax . By Lemma 5, | D ax | = 3 and D ax ∩ S = ∅ . Case 2.1. | D ax ∩ { a, x }| = 2.In this subcase, we have | D ax ∩ { s , s }| = 1. Without loss of generality, assume that s ∈ D ax . Then D ax = { a, s , x } and each vertex of V ( C ) is adjacent to s while each7ertex of V ( C ) − { x } is adjacent to both s and x . Furthermore, s x / ∈ E ( G ). Note that s s ∈ E ( G ), because D ax = { a, s , x } and D ax doubly dominates s in G + ax . Choose y ∈ V ( C ) − { x, x } and consider G + ay . Suppose | D ay ∩ { a, y }| = 1. Then a ∈ D ay because N [ a ] = S ∪ V ( C ). If | D ay ∩ S | = 2, then D ay = { a, s , s } or { a, s , s } . If D ay = { a, s , s } ,then D ay ⊁ × { x } as s x / ∈ E ( G ). So D ay = { a, s , s } . Since ∩ i =1 B i = ∅ , xs / ∈ E ( G ), and so D ay ⊁ × { x } . Hence | D ay ∩ S | = 1. By similar arguments above, we obtain D ay ∩ { s , s } = ∅ .Hence s ∈ D ay . Since ∩ i =1 B i = ∅ , if | V ( C ) | ≥
5, then D ay ⊁ × V ( C ) − { x, y, x } , acontradiction. So | V ( C ) | = 3, i.e., V ( C ) = { x, y, x } . To doubly dominate y , we have | D ay ∩ { x, x }| = 1. If x ∈ D ay , then x s , s s ∈ E ( G ), which implies that S is a DDS of G , a contradiction. Hence x ∈ D ay . To doubly dominate s , we see that s s ∈ E ( G ) as s x / ∈ E ( G ). This means that S is a DDS of G , a contradiction. Therefore, | D ay ∩ { a, y }| = 2.Then D ay ∩ S = { s } by Lemma 5. Thus each vertex of V ( C ) is adjacent to s while eachvertex of V ( C ) − { y } is adjacent to both y and s . If | V ( C ) | ≥
5, then ∩ i =1 B i = ∅ , acontradiction. So | V ( C ) | = 3, i.e., V ( C ) = { x, y, x } . Clearly, ys / ∈ E ( G ). Further, wecan obtain s s , s s / ∈ E ( G ) for otherwise S would be a DDS of G . By Lemma 8 (1), G isisomorphic to H r, , , a contradiction. Hence | D ax ∩ { a, x }| = 2 is impossible. Case 2.2. | D ax ∩ { a, x }| = 1.In this subcase, clearly a ∈ D ax and x / ∈ D ax . Since xs , ax / ∈ E ( G ), s / ∈ D ax . If | D ax ∩ S | = 2, then D ax = { a, s , s } . Thus each vertex of V ( C ) − { x, x } is adjacent toboth s and s , and so ∩ i =1 B i = V ( C ) − { x, x } 6 = ∅ , a contradiction. Hence | D ax ∩ S | = 1.Without loss of generality, suppose that s ∈ D ax . Then each vertex of V ( C ) ∪ ( V ( C ) − { x } )is adjacent to s and ( V ( C ) − { x, x } ) ⊆ B ∩ B . Since x is adjacent to each vertex of V ( C ) − { x } , s x / ∈ E ( G ) for otherwise D ax is a DDS of G . Thus s s ∈ E ( G ) so that s can be doubly dominated in G + ax . Choose y ∈ V ( C ) − { x, x } and consider G + ay .By an argument similar to that as in Case 2.1, one can arrive at a contradiction. Therefore, | D ax ∩ { a, x }| = 1 is also impossible. ✷ Theorem 3.
Let G be a -connected claw-free - γ × ( G ) -critical graph of odd order with δ ( G ) ≥ . If G / ∈ H , then G is 3-factor-critical. Proof.
By Lemma 1, diam( G ) = 2 or 3. If diam( G ) = 3, then, by Lemma 6, the assertionfollows. We may now assume that diam( G ) = 2.Suppose to the contrary that G is not 3-factor-critical. Then there exists a subset S ⊆ V ( G )such that | S | = 3 and G − S contains exactly two odd components by Lemma 7. Let S , C and C be defined as before. By Lemma 8 (3), without loss of generality, suppose that A ∩ A = ∅ .Further, it follows from Lemma 9 that both ∩ i =1 A i = ∅ and ∩ i =1 B i = ∅ . Thus there exist i and8 such that B i ∩ B j = ∅ by Lemma 8 (4), where 1 ≤ i = j ≤
3. We next consider the followingtwo subcases.
Case 1. B ∩ B = ∅ .Take a ∈ A ∩ A , b ∈ B ∩ B , and consider the graph G + ab . If | D ab ∩ { a, b }| = 2, then | D ab ∩ S | = 1 by Lemma 5. Since ∩ i =1 A i = ∅ and ∩ i =1 B i = ∅ , D ab ∩ S = { s } or { s } . Thisimplies that D ab is a DDS of G , a contradiction. Hence | D ab ∩ { a, b }| = 1. By the symmetry ofstructure of G , without loss of generality, we may assume that a ∈ D ab and b / ∈ D ab . Case 1.1. | D ab ∩ S | = 2.Clearly, we have D ab ∩ S = { s , s } or { s , s } . Because the both cases of D ab ∩ S canbe discussed similarly, thus we may assume that D ab ∩ S = { s , s } . Then each vertex of V ( C ) − { b } is adjacent to s and s . Hence ( V ( C ) − { b } ) ⊆ B ∩ B , and so V ( C ) = B .Take b ∈ V ( C ) − { b } and consider the graph G + ab . Clearly, s b / ∈ E ( G ) as ∩ i =1 B i = ∅ . If | D ab ∩ { a, b }| = 2, then | D ab ∩ S | = 1 by Lemma 5. Furthermore, note that D ab ∩ S = { s } or { s } , because s / ∈ D ab . If s ∈ D ab , then ( V ( C ) − { b, b } ) ⊆ ∩ i =1 B i = ∅ , a contradiction.So s ∈ D ab . But, D ab ⊁ { b } as ab / ∈ E ( G ) and bs / ∈ E ( G ), a contradiction. Therefore, | D ab ∩ { a, b }| = 1.Suppose a ∈ D ab and b / ∈ D ab . If | D ab ∩ S | = 2, then s ∈ D ab . However, D ab ⊁ V ( C ) − { b, b } as ∩ i =1 B i = ∅ . So | D ab ∩ S | = 1. By Lemma 5, in order to doubly dominate V ( C ) −{ b } , we have | D ab ∩ ( V ( C ) −{ b } ) | = 1. Note that s / ∈ D ab , since as / ∈ E ( G ). Recallthat ∩ i =1 B i = ∅ . If s ∈ D ab , then D ab ⊁ V ( C ) − { b, b } . Hence s ∈ D ab . It immediatelyfollows from Lemma 8 (1) that G [ B ] = G [ V ( C )] is complete. This implies that D ab is alsoa DDS of G , a contradiction. Therefore, a / ∈ D ab and b ∈ D ab . Suppose | D ab ∩ S | = 2.Then D ab ∩ S = { s , s } or { s , s } . Note that s b / ∈ E ( G ). If D ab ∩ S = { s , s } , then s s , s s ∈ E ( G ) to doubly dominate s . This means that S is a DDS of G , a contradiction.So D ab ∩ S = { s , s } . To doubly dominate s and s , s s ∈ E ( G ) and s is adjacent to atleast one of s and s , respectively. This implies that S is a DDS of G , a contradiction. Hence | D ab ∩ S | = 1. To doubly dominate V ( C ) − { a } , we see that | D ab ∩ ( V ( C ) − { a } ) | = 1 byLemma 5. Since ∩ i =1 B i = ∅ , bs / ∈ E ( G ). Then s / ∈ D ab . Recall that s b / ∈ E ( G ). Then s / ∈ D ab . So s ∈ D ab . Thus V ( C ) = A . By Lemma 8 (1), G [ V ( C )] is complete, whichimplies that D ab is a DDS of G , a contradiction. Hence Case 1.1 can not occur. Case 1.2. | D ab ∩ S | = 1.Since ∩ i =1 A i = ∅ , as / ∈ E ( G ). Then D ab ∩ S = { s } or { s } . Suppose D ab ∩ S = { s } . Todoubly dominate V ( C ) − { b } , we have | D ab ∩ ( V ( C ) − { b } ) | = 1 by Lemma 5. Then each9ertex of V ( C ) is adjacent to s . Thus V ( C ) = B . By Lemma 8 (1), G [ V ( C )] is complete.This means that D ab is a DDS of G , a contradiction. So D ab ∩ S = { s } . By a similar argument,one reaches the same contradiction. Therefore, Case 1.2 is impossible. Case 2. B ∩ B = ∅ or B ∩ B = ∅ .Suppose first that B ∩ B = ∅ . Choose u ∈ A ∩ A and v ∈ B ∩ B . Now consider G + uv .We distinguish the following two subcases. Case 2.1. | D uv ∩ { u, v }| = 2.Then | D uv ∩ S | = 1 by Lemma 5. Clearly, s / ∈ D uv . Thus D uv ∩ S = { s } or s . First suppose D uv ∩ S = { s } . Then each vertex of V ( C ) is adjacent to s while each vertex of V ( C ) − { v } isadjacent to s and v . By Lemma 8 (1), G [ V ( C )] = G [ A ] is complete. Choose v ∈ V ( C ) −{ v } and consider G + uv . If | D uv ∩ { u, v }| = 2, then | D uv ∩ S | = 1. Obviously, s / ∈ D uv . If s ∈ D uv , then ( V ( C ) − { v, v } ) ⊆ B ∩ B . Thus B ∩ B = ∅ . By a similar argument thatused in the proof of Case 1, one reaches the same contradictions. So s / ∈ D uv and s ∈ D uv .Then each vertex in ( V ( C ) − { u } ) ∪ V ( C ) is adjacent to s . Since us / ∈ E ( G ), s s ∈ E ( G )to doubly dominate s in G + uv . Note that ∩ i =1 B i = ∅ . So s v / ∈ E ( G ) and s s ∈ E ( G )to doubly dominae s in G + uv . This means that S is a DDS of G , a contradiction. Hence | D uv ∩ { u, v }| = 1.Suppose u ∈ D uv and v / ∈ D uv . If | D uv ∩ S | = 2, then s / ∈ D uv as s v / ∈ E ( G ) and uv / ∈ E ( G ). Thus D uv = { u, s , s } and each vertex of V ( C ) − { v, v } is adjacent to both s and s . Then ( V ( C ) − { v, v } ) ⊆ ∩ i =1 B i = ∅ , a contradiction. So | D uv ∩ S | = 1. Todoubly dominate V ( C ) − { v } , we have | D uv ∩ ( V ( C ) − { v } ) | = 1. Recall that s v / ∈ E ( G )and uv / ∈ E ( G ). Then s / ∈ D uv . Further s / ∈ D uv because us / ∈ E ( G ). Thus s ∈ D uv .Then each vertex of V ( C ) − { v, v } is adjacent to s , and so B ∩ B = ∅ . By applying anargument similar to that presented in the proof of Case 1, we can always reach a contradiction.Therefore, u / ∈ D uv and v ∈ D uv . If | D uv ∩ S | = 2, then D uv = { v , s , s } or { v , s , s } .Suppose D uv = { v , s , s } . Then each vertex of V ( C ) − { u } is adjacent to s and s , whichimplies that ∩ i =1 A i = ∅ , a contradiction. Hence D uv = { v , s , s } . If s is adjacent to a vertexin V ( C ) − { v } , then B ∩ B = ∅ . Using a similar argument as in the proof of Case 1, onereaches a contradiction. So s is not adjacent to any vertex in V ( C ) − { v } . Similarly, s is notadjacent to any vertex in V ( C ) − { v } . Thus G − { s , v } is not connected, which contradicts theassumption that G is 3-connected. Hence | D uv ∩ S | = 1. To doubly dominate V ( C ) − { u } , wehave | D uv ∩ ( V ( C ) − { u } ) | = 1. Since G [ V ( C )] is complete, D uv ∩ { s , s } = ∅ for otherwise D uv is a DDS of G . So s ∈ D uv . Then A ∩ A = ∅ and B ∩ B = ∅ . By a similar argumentthat used in the proof of Case 1, we can obtain a contradiction. Hence D uv ∩ S = { s } is10mpossible. Similarly, D uv ∩ S = { s } is also impossible. Therefore, Case 2.1 can not occur. Case 2.2. | D uv ∩ { u, v }| = 1. Case 2.2.1. u ∈ D uv and v / ∈ D uv .If | D uv ∩ S | = 2, then D uv = { u, s , s } or { u, s , s } . Suppose D uv = { u, s , s } . Theneach vertex of V ( C ) − { v } is adjacent to both s and s . Thus B ∩ B = ∅ . By applyingan argument similar to that presented in the proof of Case 1, we can obtain a contradiction.So D uv = { u, s , s } . Then each vertex of V ( C ) − { v } is adjacent to both s and s . Thus( V ( C ) −{ v } ) ⊆ B ∩ B and G [ V ( C )] = G [ B ] is complete by Lemma 8 (1). Since us / ∈ E ( G ), s s ∈ E ( G ). Further, s is adjacent to at least one of s and s because D uv = { u, s , s } ≻ × { s } .Take v ∈ V ( C ) − { v } and consider G + uv . If | D uv ∩ { u, v }| = 2, then | D uv ∩ S | = 1.Note that uv, s v / ∈ E ( G ). So s / ∈ D uv . If s ∈ D uv , then each vertex of V ( C ) − { v, v } isadjacent to s . Thus ∩ i =1 B i = ∅ , a contradiction. Hence s / ∈ D uv and s ∈ D uv . Then eachvertex of V ( C ) − { u } is adjacent to both u and s . If there exists a vertex u ∗ ∈ V ( C ) − { u } such that u ∗ is adjacent to s or s , then A ∩ A = ∅ or A ∩ A = ∅ . Clearly, B ∩ B = ∅ and B ∩ B = ∅ . By a similar argument that used in the proof of Case 1, we can obtain the samecontradictions. Hence s and s are not adjacent to any vertex in V ( C ) −{ u } . Thus G −{ u, s } is not connected, contradicting the fact that G is 3-connected. Hence | D uv ∩ { u, v }| = 1.Suppose u ∈ D uv and v / ∈ D uv . If | D uv ∩ S | = 2, then D uv = { u, s , s } or { u, s , s } .Since uv, s v / ∈ E ( G ), D uv = { u, s , s } is impossible. So D uv = { u, s , s } . Then each vertexof V ( C ) − { v, v } is adjacent to s . Recall that ( V ( C ) − { v } ) ⊆ B ∩ B . Thus ∩ i =1 B i = ∅ , acontradiction. Hence | D uv ∩ S | = 1. Since D uv is a γ × ( G + uv )-set, | D uv ∩ ( V ( C ) − { v } ) | =1. If s ∈ D uv , then vs ∈ E ( G ). This derives that ∩ i =1 B i = ∅ , a contradiction. Thus s / ∈ D uv . Further, s / ∈ D uv as us / ∈ E ( G ). So s ∈ D uv . Then each vertex of V ( C ) −{ v, v } is adjacent to s , which implies that ∩ i =1 B i = ∅ , a contradiction. Therefore, u / ∈ D uv and v ∈ D uv . If | D uv ∩ S | = 2, then D uv = { v , s , s } or { v , s , s } . Thus each vertex of V ( C ) − { u } is adjacent to both s and s or both s and s , respectively. This means that D uv is a DDS of G , a contradiction. Hence | D uv ∩ S | = 1. To doubly dominate V ( C ) − { u } in G + uv , we see that | D uv ∩ ( V ( C ) − { u } ) | = 1, say u ∗ ∈ D uv ∩ ( V ( C ) − { u } ). It is easyto see that s / ∈ D uv as s v / ∈ E ( G ) and u ∗ v / ∈ E ( G ). Since ∩ i =1 B i = ∅ , s v / ∈ E ( G ), andso s / ∈ D uv . Therefore, s ∈ D uv . Then s u ∗ ∈ E ( G ) and s u ∗ ∈ E ( G ). Thus A ∩ A = ∅ .Recall that v ∈ B ∩ B . By a similar argument that used in the proof of Case 1, we can getthe same contradictions. Hence | D uv ∩ S | = 2 is impossible.11hus | D uv ∩ S | = 1. To doubly dominate V ( C ) − { v } in G + uv , | D uv ∩ ( V ( C ) − { v } ) | = 1,say v ∗ ∈ D uv ∩ ( V ( C ) − { v } ). If s ∈ D uv , then V ( C ) = B . By Lemma 8 (1), G [ V ( C )]is complete. This implies that D uv is a DDS of G , a contradiction. So s / ∈ D uv . Since us / ∈ E ( G ), s / ∈ D uv . Hence s ∈ D uv . Then each vertex of V ( C ) ∪ ( V ( C ) − { v } ) is adjacentto s . Since D uv ≻ × { s } , s v ∗ ∈ E ( G ). Thus B ∩ B = ∅ . Further, s is adjacent to a vertexof V ( C ) − { u } because A = ∅ . So A ∩ A = ∅ . Using an argument similar to that presentedin the proof of Case 1, one reaches the same contradictions. Therefore, Case 2.2.1 is impossible. Case 2.2.2. u / ∈ D uv and v ∈ D uv .By applying a similar argument that used in the proof of Case 2.2.1, we can obtain the samecontradictions. So Case 2.2 can not occur. Hence B ∩ B = ∅ . Similarly, we can show that B ∩ B = ∅ . Thus Case 2 can not occur. This completes the proof of Theorem 3. ✷ t tt t t ttt t ❅❅❅❅❅❅ (cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0) ❅❅❅❅❅❅ (cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)❇❇❇❇❇❇❇❇❇ ✂✂✂✂✂✂✂✂✂❜❜❜❜❜❜❜❜❜❜❜❜❜❜ ✧✧✧✧✧✧✧✧✧✧✧✧✧✧✧✧✧✧✧✧✧✧✧✧✧✧✧✧ ❜❜❜❜❜❜❜❜❜❜❜❜❜❜✔✔✔✔✔✔✔✔✔✔✔✔✔✔ ❚❚❚❚❚❚❚❚❚❚❚❚❚❚✂✂✂✂✂✂✂✂✂ ❇❇❇❇❇❇❇❇❇❅❅❅❅❅❅❅❅✏✏✏✏✏✏✏✏✏✏✏✏✏✏✏✏✏✏✏✏✏✏✏✏❳❳❳❳❳❳❳❳❳❳❳❳❳❳❳❳❳❳❳❳❳❳❍❍❍❍❍❍❍❍❍❍❍ z z x x x y y y y Figure 1: The graph H , Remark.
The hypotheses on the connectivity and the minimum degree bounds in Theorem3 are sharp. Indeed, the following results was proved by Favaron in [10]: for all k ≥
0, every k -factor-critical graph of order > k is k -(vertex)-connected and for all k ≥
1, every k -factor-critical graph of order > k is ( k + 1)-(edge)-connected (and hence has minimum degree at least k + 1). Next, to illustrate that the assumption on claw-free is necessary, we construct a graph H ,t as follows. For odd integer t ≥
3, let F = K t , F = C and F = K . Moreover, let V ( F ) = { x , x , . . . x t } , V ( F ) = { y , y , y , y } and V ( F ) = { z , z } . Let H ,t be the graphobtained from the union F ∪ F ∪ F by adding edges x t y and y x i for 1 ≤ i ≤ t −
1, joining z to each vertex of V ( F ) ∪ V ( F ) and joining z to each vertex of ( V ( F ) ∪ V ( F )) − { y , x t } .12t is easy to see that H ,t is a 4-connected 4- γ × -critical graph of order t + 6 with minimumdegree 4, but it contains a claw. Obviously, H ,t − { u, v, y } has no perfect matching. Hence H ,t is not 3-factor-critical. The graph H , is shown in Figure 1. References [1] N. Ananchuen, W. Ananchuen and M.D. Plummer, Matching properties in connecteddomination critical graphs,
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