New sufficient condition for Hamiltonian graphs

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1 Applied Mathematics Letters ( ) New sufficient condition for Hamiltonian graphs Kewen Zhao a,b,,hong-jian Lai c,yehong Shao d a Department of Mathematics, Qiongzhou University, Wuzhishan, Hainan, , China b Department of Mathematics, Hainan Normal University, Haikou, Hainan, , China c Department of Mathematics, West Virginia University, Morgantown, WV , United States d Ohio University Southern, Ironton, OH 45638, United States Received 26 December 2003; accepted 17 October 2005 Abstract Let G be a graph, and δ(g) and α(g) be the minimum degree and the independence number of G, respectively. For a vertex v V (G), d(v) and N(v) represent the degree of v and the neighborhood of v in G, respectively. A number of sufficient conditions for a connected simple graph G of order n to be Hamiltonian have been proved. Among them are the well known Dirac condition (1952) (δ(g) n 2 )and Ore condition (1960) (for any pair of independent vertices u and v, d(u) + d(v) n). In 1984 Fan generalized these two conditions and proved that if G is a 2-connected graph of order n and max{d(x), d(y)} n/2for each pair of nonadjacent vertices x, y with distance 2 in G, theng is Hamiltonian. In 1993, Chen proved that if G is a 2-connected graph of order n, andifmax{d(x),d(y)} n/2 for each pair of nonadjacent vertices x, y with 1 N(x) N(y) α(g) 1, then G is Hamiltonian. In 1996, Chen, Egawa, Liu and Saito further showed that if G is a k-connected graph of order n, andif max{d(v) : v S} n/2 foreveryindependent set S of G with S =k which has two distinct vertices x, y S such that the distance between x and y is 2, then G is Hamiltonian. In this paper, we generalize all the above conditions and prove that if G is a k-connected graph of order n, andifmax{d(v) : v S} n/2 foreveryindependent set S of G with S =k which has two distinct vertices x, y S satisfying 1 N(x) N(y) α(g) 1, then G is Hamiltonian. c 2006 Published by Elsevier Ltd Keywords: Hamiltonian graphs; Dirac condition; Ore condition; Fan condition; Chen condition 1. Introduction We consider finite and simple graphs in this paper; undefined notations and terminology can be found in [1]. In particular, we use V (G), E(G), κ(g), δ(g) and α(g) to denote the vertex set, edge set, connectivity, minimum degree and independence number of G,respectively. If G is a graph and u,v V (G),thenapath in G from u to v is called a (u,v)-path of G.Ifv V (G) and H is asubgraph of G,thenN H (v) denotes the set of vertices in H that are adjacent to v in G. Thus, d H (v), thedegree of v relative to H,is N H (v). Wealsowrite d(v) = d G (v) and N(v) = N G (v). If C and H are subgraphs of G, thenn C (H ) = u V (H) N C (u), andg C denotes the subgraph of G induced by V (G) V (C). Forvertices u,v V (G), thedistance between u and v, denoted d(u,v),isthelength of a shortest (u,v)-path in G, or if no such path exists. Corresponding author at: Department of Mathematics, Qiongzhou University, Wuzhishan, Hainan, , China /$ - see front matter c 2006 Published by Elsevier Ltd doi: /j.aml

2 2 K. Zhao et al. / Applied Mathematics Letters ( ) Let C m = x 1 x 2 x m x 1 denote a cycle of order m. Define N + C m (u) ={x i+1 : x i N Cm (u)}, N C m (u) ={x i 1 : x i N Cm (u)}, and define N ± C m (u) = N + C m (u) N C m (u), where subscripts are taken modulo m. Asubset S V (G) is said to be an essential independent set if S is an independent set in G and there exist two distinct vertices x, y S with d(x, y) = 2. The following sufficient conditions to assure the existence of a Hamiltonian cycle in a simple graph G of order n 3arewellknown. Theorem 1.1 (Dirac [4]). If δ(g) n/2,then G is Hamiltonian. Theorem 1.2 (Ore [6]). If d(u)+d(v) n for each pair of nonadjacent vertices u,v V (G),then G is Hamiltonian. Theorem 1.3 (Fan [5]). If G is a 2-connected graph and if max{d(u), d(v)} n/2 for each pair of vertices u,v V (G) with d(u,v) = 2, then G is Hamiltonian. Theorem 1.4 (Chen [2]). If G is a 2-connected graph and if max{d(u), d(v)} n/2 for each pair of vertices u,v V (G) with 1 N(u) N(v) α(g) 1,thenGisHamiltonian. Theorem 1.5 (Chen et al. [3]). If G is a k-connected (k 2) graph and if max{d(v) : v S} n/2 for every independent set S of order k such that S has two distinct vertices x, ywith d(x, y) = 2,thenGisHamiltonian. The purpose of this paper is to unify and extend the theorems above. We shall prove the following result. Theorem 1.6. If G is a k-connected (k 2) graph of order n, and if max{d(v) : v S} n/2 for every independent set S of order k, such that S has two distinct vertices x, ywith 1 N(x) N(y) α(g) 1,thenGisHamiltonian. The proof of Theorem1.6 will be given in Section 2. It is straightforward to verify that if a graph G satisfies the hypothesis of any one of Theorems , then it will also satisfy the hypothesis of Theorem 1.6. InSection 3, we shall show that there exist Hamiltonian graphs satisfying the hypothesis of Theorem 1.6, butwhose Hamiltonicity cannot be assured by any one of Theorems InthissenseTheorem 1.6 extends Theorems Proof of Theorem 1.6 For a cycle C m = x 1 x 2 x m x 1,wewrite [x i, x j ] to denote the section x i x i+1 x j of the cycle C m,where subscripts are taken modulo m. Fornotational convenience, [x i, x j ] will denote the (x i, x j )-path x i x i+1 x j of C m,aswellasthevertexsetofthispath. If C 1 and C 2 are cycles of a graph G such that V (C 1 ) V (C 2 ) and V (C 2 ) > V (C 1 ), thenwesay that C 2 extends C 1.Suppose that P 1 is an (x, y)-path of G and P 2 is a (y, z)- path of G such that V (P 1 ) V (P 2 ) ={y}, thenweusep 1 P 2 to denote the (x, z)-path of G induced by the edges E(P 1 ) E(P 2 ).IfV(P 1 ) V (P 2 ) ={x, y} and x = z, thenp 1 P 2 denotes the cycle of G induced by the edges E(P 1 ) E(P 2 ).Weneed to establish some lemmas. Lemma 2.1. Suppose that C m = x 1 x 2 x m x 1 is a cycle of a graph G, H is a component of G V (C m ), and x i, x j are distinct vertices in N Cm (H ).IfGdoes not have a cycle extending C m,theneach of the following holds. (i) {x i 1, x i+1, x j 1, x j+1 } N Cm (H ) =. (ii) x i+1 x j+1 E(G) and x i 1 x j 1 E(G). (iii) If x t x j+1 E(G) for some vertex x t [x j+2, x i ],thenx t 1 x i+1 E(G) and x t 1 N Cm (H ). (iv) If x t x j+1 E(G) for some vertex x t [x i+1, x j ],thenx t+1 x i+1 E(G). (v) No vertex of G (V (C m ) V (H )) is adjacent to both x i+1 and x j+1. (vi) If x V (H ) such that xx i E(G), then{x} N (H ) must be an independent set. Proof. (i), (ii) and (v) follow immediately from the assumption that G does not have a cycle extending C m.itremains to show that (iii), (iv) and (vi) must also hold. Since x i, x j N Cm (H ), x i, x j V (H ) such that x i x i, x j x j E(G). Let P denote an (x i, x j )-path in H.

3 K. Zhao et al. / Applied Mathematics Letters ( ) 3 Suppose that (iii) fails. Then there exists a vertex x t {x j+2, x j+3,...,x i } satisfying x t x j+1 E(G). If x t 1 x i+1 E(G), thenx i P x j [x j 1, x t 1 ][x t 2, x t ][x t+1, x i ] is alonger cycle than C m,contrary to the assumption that C m is longest. Hence x t x j+1 E(G).Nextweassume that x t 1 is adjacent to some vertex x t 1 V (H ).LetP denote an (x t 1, x j )-path in H.Thenx j P x t 1 [x t 2, x t ][x t+1, x j ] is a cycle extending C m,acontradiction. Hence (iii) must hold. The proof for (iv) is similar, and so it is omitted. To prove (vi), assume to the contrary, that G[{x} N (H )] has an edge e. ByLemma 2.1(i), N (H ) is an independent set. Hence e = xx j+1 for some x j N Cm (H ),andso x j V (H ) such that x j x j E(G).LetP denote an (x j, x)-path in H.Thenx[x j+1, x j ]P is a cycle extending C m.thiscompletes the proof of the lemma. Lemma 2.2. Let G be a 2-connected graph of order n, C m = x 1 x 2 x m x 1 be a cycle of G, and H be a component of G V (C m ).Ifone of the following holds, (i) there exist two distinct vertices x i, x j V (C m ) with x i+1, x j+1 in N (H ) such that d(x i+1 ) n/2 and d(x j+1 ) n/2,or (ii) there exists a vertex x i+1 N (H ) and a vertex y V (H ) such that d(x i+1 ) n/2 and d(y) n/2, then there exists a cycle C extending C m. Proof. Suppose, to the contrary, that G does not have a cycle C extending C m. (1) First we assume that (i) holds. Then x i, x j V (H ) such that x i x i, x j x j E(G). SinceH is connected, H has an (x i, x j )-path P. We define a map f : N G (x j+1 ) {x j } V (G) as follows: v N G (x j+1 ) {x j }, v if v V (C m ) f (v) = v + = x t+1 if v = x t [x i+1, x j 1 ] v = x t 1 if v = x t [x j+2, x i ]. Claim 1. f is an injection, and v N G (x j+1 ) {x j }, f (v) N G (x i+1 ) {x x+1 }. It is straightforward to verify that f is an injection. Firstly, by (1), v V (H ), x i+1 f (v) = x i+1 v E(G). If v = x t [x i+1, x j 1 ] with x t x j+1 E(G), thenbylemma 2.1(iv), x t+1 x i+1 E(G). If x t [x j+2, x i 1 ] with x t x j+1 E(G), thenbylemma 2.1(iii), x t 1 x i+1 E(G). Finally, the definition of f shows that v N G (x j+1 ) {x j }, f (v) x i+1.thisproves Claim 1. By Claim 1, f (N G (x j+1 ) {x j }) N G (x i+1 ) =.ByLemma 2.1(i), f (N G (x j+1 ) {x j }) V (H ) =. Therefore, N G (x i+1 ) ( f (N G (x j+1 ) {x j }) V (H ) {x i+1 }) =.Since f is an injection, f (N G (x j+1 ) {x j }) = N G (x j+1 ) 1 = d(x j+1 ) 1. It follows that d(x i+1 ) = N G (x i+1 ) n f (N G (x j+1 )) V(H) {x i+1 } n d(x j+1 ) V (H ), and so d(x i+1 ) + d(x j+1 ) n 1, contrary to the assumption that both d(x i+1 ) n/2 andd(x j+1 ) n/2. Thus if (i) holds, there exists a cycle C extending C m. The proof for the case when (ii) holds is similar, and so it is omitted. Lemma 2.3. Let G be a k-connected (k 2), C m = x 1 x 2 x m x 1 be a cycle of G with V (C m ) < V (G) such that G does not have a cycle extending C m, and let H be a component of G V (C m ).Letx V (H ) be a vertex satisfying N Cm (x) 1 and d(x) <n/2.then (i) N Cm (H ) k. Moreover, if x i N Cm (x),theneach of the following holds. (ii) 1 N(x) N(x i+1 ) α(g) 1. Proof. Since C m is alongest cycle with V (C m ) < V (G), V (C m ) N Cm (H ),andson Cm (H ) separates V (H ) and V (C m ) N Cm (H ).SinceG is k-connected, N Cm (H ) κ(g) k.thisproves Lemma 2.3(i). By Lemma 2.1(vi), {x} N + C m (H ) must be an independent set. Thus N Cm (H ) = N + C m (H ) α(g) 1, and so 1 N(x) N(x i+1 ) α(g) 1. This proves Lemma 2.3(ii).

4 4 K. Zhao et al. / Applied Mathematics Letters ( ) Lemma 2.4. Let G be a k-connected (k 2), C m = x 1 x 2 x m x 1 be a cycle of G with V (C m ) < V (G) such that G does not have a cycle extending C m, and H be a component of G V (C m ).If Gsatisfies the hypothesis of Theorem 1.6, thend(x) n/2 for every x V (H ) with N Cm (x) 1. Proof. By contradiction, we assume that there exists an x V (H ) satisfying N Cm (x) 1andd(x) <n/2. We assume that x i N Cm (x). ByLemma 2.3(i), V N (H ) with x i+1 V and with V = k 1. By Lemma 2.1(vi), {x} N (H ) is an independent set. It follows by Lemma 2.3(ii) that V {x} is a k-element set satisfying the hypothesis of Theorem 1.6. Sinced(x) <n/2, by the hypothesis of Theorem 1.6, theremust be avertex(sayx h+1 )inv with d(x h+1 ) n/2. By Lemma 2.2(i), every vertex of N (H ) {x h+1 } must have degree less than n/2. Since G is k-connected (k 2), there must be some vertex y V (H ) and x j V (C m ) with x j+1 N (y) {x h+1 } (possibly y = x). By Lemma 2.3 (with y replacing x in Lemma 2.3), we have 1 N(y) N(x j+1 ) α(g) 1. Hence pick a subset X N (H ) {x h+1, x j+1 } with X =k 2andletV = X {y, x j+1 }.ThenV also satisfies the hypothesis of Theorem 1.6. Hence there must be a vertex u V such that d(u) n/2. Since d(x h+1 ) n/2 and d(u) n/2, it follows by Lemma 2.2(i) that G has a cycle C extending C m,contrary to the assumption that C m has no extension in G. Lemma 2.5. Let G be a k-connected (k 2) graph, C m = x 1 x 2 x m x 1 be a cycle of G with V (C m ) < V (G) such that G has no cycleextending C m, and let H be a component of G V (C m ).If d(x) n/2 for every x V (H ) with N Cm (x) 1,thenH = G V (C m ). Proof. By contradiction, we assume that G V (C m ) has at least two distinct components H and H. Since G is connected, there must exist a vertex y V (H ) adjacent to some vertex of C m.bylemma 2.4, d(y) n/2. It follows that d(x)+d(y) n.ontheother hand, if a vertex x V (H ) is adjacent to some x i V (C m ), then x is not adjacent to x i+1 and x i 1.Hence we have d(x) V (C m ) /2 + V (H ) {x}. Similarly, we have d(y) V (C m ) /2 + V (H ) {y}. Itfollows that d(x) + d(y) [ V (C m ) /2 + V (H ) {x} ] + [ V (C m ) /2 + V (H ) {y} ] n 2, a contradiction. Hence we must have H = G V (C m ). Proof of Theorem 1.6. Suppose, to the contrary, that G is not Hamiltonian. Let C m = x 1 x 2 x m x 1 be a longest cycle of G, andleth be a component of subgraph G V (C m ).ByLemma 2.4,wemayassume that d(x) n/2, x V (H ) with N Cm (x) 1. (2) By Lemma 2.5, H = G V (C m ).Without loss of generality, assume that for some x V (H ), thereexists an x i N Cm (x). By Lemma 2.3(i) and since k 2, N Cm (H ) 2. (3) Choose x i, x j N Cm (H ) such that {x i+1, x i+2,...x j 1 } N Cm (H ) =. (4) If V (H ) =1, then C m = C n 1 and N Cm (H ) =d(x) n/2. Then there must exist x i, x i+1 N G (x) V (C n 1 ), and so G is Hamiltonian, a contradiction. Hence we must have V (H ) 2. (5) x V (H ), N G (x) {x} V (H ) N Cm (H ),andso by (2), V (H ) + N Cm (H ) N G (x) {x} d(x) + 1 n/ (6) It follows by (4) (6) that {x i+1, x i+2,...,x j 1 } (n V (H ) N Cm (H ) )/ N Cm (H ) [n ( V (H ) + N Cm (H ) )]/ N Cm (H ) [n (1 + n/2)]/ N Cm (H ) (n/2 1)/ N Cm (H ) ( V (H ) + N Cm (H ) 2)/ N Cm (H ) V (H ) / N Cm (H ) +1 2/ N Cm (H ) < V (H ).

5 K. Zhao et al. / Applied Mathematics Letters ( ) 5 In other words, {x i+1, x i+2,...,x j 1 } < V (H ). Recall that we have chosen x i, x j N Cm (H ) satisfying (3).Letx i, x j V (H ) with x i x i, x j x j E(G),andletP be an (x i, x j )-path of H.ThenL = x j [x j+1, x i ]Px j is a cycle of G.Choose a cycle C of G such that V (L) V (C ) and V (C ) is maximized. (8) By (8) and by applying Lemmas to the cycle C,weobtain the following Claim 2. Claim 2. Let H be a component of G V (C ). (i) Then x V (H ) with N C (x) and with d(x ) n/2. (ii) G V (C ) = H has only one component. Claim 3. V (H ) {x i+1, x i+2,...,x j 1 }=,orv (H ) V (H ) =. By Lemma 2.5, H = G V (C m ),andsov (H ) {x i+1, x i+2,...x j 1 } V (H ). Ifforsomex, x V (H ) such that x {x i+1, x i+2,...x j 1 } and x V (H ), thensinceh is connected, H has an (x, x )-path Q. Since V (H ) {x i+1, x i+2,...x j 1 } V (H ),thereisan edge in Q joining a vertex in {x i+1, x i+2,...,x j 1 } and a vertex in V (H ),contrary to (3).Thisproves Claim 3. If V (H ) {x i+1, x i+2,...,x j 1 }=,thenbyclaim 2(ii), V (C m ) V (C ).By(8), V (H ) V (C ),and so C extends C m,contrary to the assumption that C m is alongest cycle. Thus by Claim 3, V (H ) V (H ) =, and so V (H ) {x i+1, x i+2,...,x j 1 }.By (7), V (H ) > {x i+1, x i+2,...,x j 1 } V (H ). ByLemma 2.5 and Claim 2(ii), we have V (C ) =n V (H ) > n V (H ) = V (C m ), (7) contrary to the assumption that C m is alongest cycle. Therefore, the proof of Theorem 1.6 is complete. 3. Examples The purpose of this section is to show that there exist Hamiltonian graphs satisfying the hypothesis of Theorem 1.6, but whose Hamiltonicity cannot be assured by any one of Theorems Let H and K be two vertex disjoint graphs. As in [1], H K denotes the disjoint union of H and K,andH K denotes the graph obtained from the disjoint union of H and K by adding all the edges in {uv : u V (H ) and v V (K )}. Similarly, for two disjoint vertex subsets Z 1 and Z 2,wedefineZ 1 Z 2 to be the graph whose vertex set is Z 1 Z 2 and whose edge set is {v 1 v 2 : v 1 Z 1 and v 2 Z 2 } (E(Z 1 ) E(Z 2 )). IfZ denotes a vertex subset, let K (Z) denote the complete graph whose vertex set is Z. IfZ E(H ) and Z is an edge set not in H but the two ends of each edge in Z are in H,thenH Z + Z denotes the graph obtained from H by deleting the edges in Z and adding the edges in Z. Let H 1, H 2,...,H 6 be vertex disjoint graphs each of which is isomorphic to a K 3.For each i with 1 i 6, pick avertexx i V (H i ).LetY ={y 1, y 2, y 3, y 4, y 5, y 6 } denote a vertex set disjoint from 6 i=1 V (H i), andletk6 c denote the edgeless graph with V (K6 c ) = Y.Defineedge subsets as follows: W 1 = E({x 6} (Y {y 6 })) W 2 = E({y 6} ( 3 j=1 (V (H j) {x j }))) W 3 = E(K ({x 1, x 2, x 3, x 4, x 5 })) W 4 = E(K ( 6 j=4 V (H j) {x 6 })) W 5 = E((V (H 5) {x 5 }) ( 3 j=1 (V (H j) {x j }))). Define L = (K c 6 ( 6 i=1 H i)) (W 1 W 2 ) + (W 3 + W 4 + W 5 ).

6 6 K. Zhao et al. / Applied Mathematics Letters ( ) Table 1 Degrees of all the vertices in L The vertex v N L (v) d(v) Remark x i (V (H i ) {x i }) Y (X {x i, x 6 }) 12 1 i 3 x i, x i (V (H i ) {v}) (Y {y 6 }) {x 5, x 5 } 9 1 i 3 x 4 ( 6 j=4 V (H j ) Y X) {x 4, x 6 } 16 x 4, x 4 ( 6 j=4 V (H j ) Y ) {v, x 6 } 13 x 5 ( 6 j=4 V (H j ) Y X) {x 5, x 6 } 16 x 5, x 5 ( 6 j=1 V (H j ) Y ) ((X {x 5 }) {v}) 19 x 6 {x 6, x 6, y 6} 3 x 6, x 6 ( 6 j=4 V (H j )) Y {v} 14 y i 6 j=1 V (H j ) {x 6 } 17 1 i 5 y 6 ( 6 i=4 V (H i)) {x 1, x 2, x 3 } 12 For each i with 1 i 6, let V (H i ) ={x i, x i, x i }.Then, for each vertex v V (L), the neighborhood of v can be expressed as in Table 1. We have the following observations. Proposition 3.1. Each of the following holds. (i) V (L) =24 and κ(l) = 3, and α(l) = 6. (ii) Lsatisfies the hypothesis of Theorem 1.6 with k = 3. (iii) L does not satisfy the hypothesis of any one of Theorems Proof. (i) We only need to show that α(l) = 6. Since Y is an independent set in L, wehaveα(l) 6. Let S V (L) be an independent set of L. Note that for each i with 1 i 5, N L (y i ) = 6 i=1 V (H i) {x 6 },and N L (y 6 ) = 6 i=4 V (H i) {x 1, x 2, x 3 }.Therefore, if for some y i S with 1lei 5, then S Y {x 6 },andsoas y 6 x 6 E(L), S 6. Hence we assume that Y S {y 6 }.Ify 6 S, then since each H i is a complete graph, S V (H i ) 1, and so S 6. Hence y 6 S. Hence in this case, ((Y {y 6 }) N L (y 6 )) S =.Butfor i = 1, 2, 3, since H i is complete, S V (H i ) 1, and so S 4. This implies that α(l) = 6, and soproves(i). (ii) As shown in Table 1, ( 3 i=1 V (H i) X) {x 6 } is the set of all vertices of degree less than V (L) /2 = 12 in L.For any two distinct, nonadjacent vertices u,v ( 3 i=1 V (H i) X), wehave,bytable 1, N L (u) N L (v) Y {y 6 } + {x 5, x 5 } 7. Moreover, v ( 3 i=1 V (H i) X) {x 6 },bytable 1, N L (x 6 ) N L (v) =. It follows that if S V (L) is an independent with S = 3suchthatforsomeu,v S, itholds that 1 N L (u) N L (v) α(l) 1 = 5, then one vertex in S must have degree at least 12 = V (L) /2. Hence L satisfies the condition of Theorem 1.6. (iii) By Table 1,itisclear that the hypotheses of Theorems cannot be satisfied. Let S 1 ={x 1, x 6, y 6 }.As N L (x 1 ) N L(y 6 ) ={x 1, x 5, x 5 }, and N L(x 6 ) 3, S 1 is a vertex set of L such that u,v S 1, N L (u) N L (v) α(l) 1. As d L (x 6 ) = 3andd L (x 1 ) = 8, L does not satisfy the condition of Theorem 1.4. Let S 2 ={z 1, z 2, z 3 } ( 3 i=1 V (H i) X) be an independent set of L. Thenasz 1, z 2, z 3 N L (y 1 ),thedistance between z 1, z 2 in L is 2. By Table 1,max{d L (v) : v S 2 } < 11, and so L does not satisfy the condition of Theorem 1.5. The same construction also works for an arbitrary integer k 3. Let H 1, H 2,...,H 2k be vertex disjoint graphs, each of which is isomorphic to a K k.for each i with 1 i 2k 1, pick two distinct vertices x i, x i V (H i ),and

7 K. Zhao et al. / Applied Mathematics Letters ( ) 7 Table 2 Degrees of all the vertices in L k The vertex v N Lk (v) d(v) Remark v V (H i ) {x i, x i } x i, x i v V (H i ) {x i, x i } x i, x i v V (H 2k 1 ) {x 2k 1, x 2k 1 } x 2k 1, x 2k 1 ( 2k 1 j=1 (V (H j ) {x j, x j })) Y (V (H i ) {v}) 2k(k 1) 1 i k (V (H i ) {v}) (Y {y 6 }) {x 2k 1, x 2k 1 } 3k 1 i k ( k j=1 (V (H j ) {x j, x j })) ( 2k j=k+1 V (H j )) Y {v, x 2k } 2(k 2 1) k + 1 i 2k 2 2k j=k+1 V (H j ) Y {v, x 2k } k 2 +2k 2 k + 1 i 2k 2 ( k j=1 (V (H j ) {x j, x j })) ( 2k j=k+1 V (H j )) Y {v, x 2k } 2(k 2 1) ( k j=1 {x j, x j } ( 2k j=k+1 V (H j ) Y ) {v, x 2k }) x 2k {y 6 } V(H 2k ) {x 2k } k v V (H 2k ) {x 2k } y i y 2k k 2 +4k 2 ( 2k j=k+1 V (H j )) Y {v} k 2 +2k 1 2k j=1 V (H j ) {x 2k } 2k i 2k 1 ( 2k j=k+1 (V (H j ))) ( k j=1 (V (H j ) {x j, x j })) 2k2 2k let x 2k V (H 2k ).LetY ={y 1, y 2,...,y 2k } denote a vertex set disjoint from 2k i=1 V (H i), andletk2k c denote the edgeless graph with V (K2k c ) = Y.Defineedge subsets as follows: W 1 = E({x 2k } (Y {y 2k })), W 2 = E({y 2k } ( k j=1 ({x j, x j }))), W 3 = E(K ( 2k 1 j=1 (V (H j) {x j, x j }))), W 4 = E(K ( 2k j=k+1 V (H j) {x 2k })), W 5 = E(({x 2k 1, x 2k 1 }) ( k j=1 {x j, x j })). Define L k = (K2k c ( 2k i=1 H i)) (W 1 W 2 ) + (W 3 + W 4 + W 5 ). Note that L 3 = L above. For each vertex v V (L k ),the neighborhood of v can be expressed as in Table 2. Imitating the proof for Proposition3.1, wehavethefollowing Proposition 3.2. Proposition 3.2. Each of the following holds. (i) V (L k ) =2k(k + 1) and κ(l k ) = k, and α(l k ) = 2k. (ii) L k satisfies the hypothesis of Theorem 1.6 with κ(l k ) = k. (iii) L k does not satisfy the hypothesis of any one of Theorems Acknowledgments The first author s research was partially supported under NFS of Hainan Province (No.10301) and (No.10501). References [1] A.J. Bondy, U.S.R. Murty, Graph Theory with Applications, American Elsevier, New York, [2] G. Chen, Hamiltonian graphs involving neighborhood intersections, Discrete Math. 112 (1993) [3] G. Chen, Y. Egawa, X. Liu, A Saito, Essential independent setand Hamiltonian cycles, J. Graph Theory 21 (1996) [4] G.A. Dirac, Some theorems on abstract graphs, Proc. London Math. Soc. 2 (1952) [5] G. Fan, New sufficient conditions for cycles ingraphs, J. Combin. Theory Ser. B 37 (1984) [6] O. Ore, Note on Hamiltonian circuits, Amer. Math. Monthly 67 (1960) 55.