Independence Number, Neighborhood Intersection and Hamiltonian Properties

Let G be a 2-connected simple graph of order n with the independence number\alpha. We show here that \forall u; v \in V (G)\backslash\{u,v\} and any z \in \{u,v\}; w \in V (G)\backslash \{u,v\}; with d(w; z) = 2, if |N(u) \ cap N(w)| \geq \alpha - 1 or |N(v) \cap N(w)| \geq \alpha - 1, then G is Hamiltonian, unless G belongs to a kind of special graphs.


Introduction 43
2 The proof of Theorem 1 44

Introduction
Hamiltonian graph is a very useful graph class in graph theory and many applications [1,2,3].The research on sufficient conditions of Hamiltonian graphs is very active.Here we establish a new sufficient condition for Hamiltonian graphs using only independence number and neighborhood intersection properties.The result is very useful for the research of Hamiltonian graphs.By the new condition, it does not need to check all pairs of nonadjacent vertices in G.The following is our main result.
Theorem 1 Let G be a 2-connected simple graph of order n with the independence number α.For any three vertices u, v, w ∈ V (G) with d(u, w) The outline of the paper is as follows.We propose our main result in the current section.The proof of the main result is given in the next section.For the proof, we shall prove the six useful lemmas.With several claims and these lemmas, we complete our demonstration.
For the simplicity, we shall use following terms and notations throughout this paper.G = (V, E) denotes an undirected connected simple graph of order n(≥ 3) Specially, if C = V (G), we simply write it as N (x) and N (B).If no ambiguity can rise,we sometimes write B instead of V (B).
Let G (r, t) be a kind of special graph, V 1 , V 2 is a pair of sets of vertices with

The proof of Theorem 1
Proof: The theorem is true for α = 1 because G is complete.Now we assume that α 2. With the conditions of the theorem, we shall show that if The same vertices, in reverse order are given by v ← − C u.We here consider u − → C v and v ← − C u both as paths and as vertex sets.uBv stands for the path from u via B to v. We use u + and u − to denote the successor and predecessor respectively of u on Then the following claims are obvious from results in [2,3].
Claim 2: Let x be any vertex in B, then N + ∪ {x} and N − ∪ {x} are independent sets.
By (1) and the conditions of theorem, we have: It follows from claims that there are the following two cases: Case 1: By ( 2) and (3),α − 1 ≤ m and Case 2: Combining Cases 1 and 2, we have that Based on claims above, we shall prove the following 6 lemmas to complete the proof of the theorem. Then, Remark: Similarly, Lemma 1 holds as well for N + when N + is substituted by N − in Lemma 1. Lemma 2: Assume that there exist vertices v − i ∈ N − and v + j ∈ N + with i = j + 1, and

Lemma 2 For any v
Then we can prove the following lemma.
Lemma 3 For any j with 0 ≤ j ≤ m, G[C j ] is complete graph.Proof of Lemma 3: It follows immediately from Claim 2 and 4 that G[C 0 ] is complete.For any j = 0, while |C j | = 1, 2, Lemma 3 holds.We here consider only Similarly, we have that v + j is adjacent to each vertex of C j , by symmetry, v − j+1 is adjacent to each vertex of C j .Up to now, if G[C j ] is not complete yet, we take vertex s and t from C j , such that st / ∈ E(G) and the s − → C t as long as possibly.By the choice of s, t, This is a contradiction as well.Therefore, (N + \ {v + j }) ∪ {s, t, x j } is an independent set of cardinality m + 2 which contradicts to.α = m + 1. 2 Lemma 4 For any u ∈ C i , v ∈ C j with i = j, uv / ∈ E(G).Proof of Lemma 4: If there exists a vertex u ∈ C i and v ∈ C j , ( we may let i < j ), then uv ∈ E(G) It follows from Claim 2 and Lemma 2, without loss of generality, Note that Lemma 3 and Claim 4 imply v + j+1 v j ∈ E(G).Hence, the cycle Proof of Lemma 5: Assume that the lemma is not true.Suppose that B 1 is another component of G\V (C).Then f roallx ∈ B, ∀y ∈ B 1 , there is yx / ∈ E(G).It follows from Claim 2 and and α = m + 1 that there exists vertices v In terms of the definition of C j , there is Lemma 6 For any x ∈ C 0 , if there exists Proof of Lemma 6: Suppose that v j ∈ V 1 and xv j ∈ E(G).From the definition of N C (B) and x j (1 ≤ j ≤ m), there is x = x j .By Claim 4, N C (B) ⊆ N (x).Hence, xv k ∈ E(G), for any k (1 ≤ j ≤ m) 2 By symmetry of C 0 and C j , by replacement of C j (1 ≤ j ≤ m) instead of C 0 in Lemma 6, Lemma 6 is true for C j . Set Finally, G ∼ = G (α − 1, α) follows from ( 5)-(7) and Lemmas 3, 4, and 6.The proof of the theorem is complete. 2