An implicit degree condition for k-connected 2-heavy graphs to be hamiltonian☆
Introduction
In this paper, we consider only undirected and finite graphs without loops or multiple edges. Notation and terminology not defined here can be found in [3].
Let G be a graph and H be a subgraph of G. For a vertex , we denote the neighborhood of u in H by , the degree of u in H by , the distance between u and v in H by . If , we can use , and in place of , and , respectively.
Let be k vertex disjoint graphs. The union of , denoted by , is the graph with vertex set and edge set . We use kQ instead of if each is isomorphic to Q. The join of , denoted by , is the graph obtained from by joining each vertex of to each vertex of for .
An induced subgraph of G with vertex set and edge set is called a claw of G, with center u and end-vertices . G is called claw-free if G does not contain an induced subgraph isomorphic to a claw. A vertex v of G is called heavy if . Following [4], a claw of G is called 2-heavy if at least two of its end vertices are heavy. A graph is called 2-heavy if all its claws are 2-heavy.
A cycle in a graph G is called a Hamilton cycle if it contains all vertices of G. G is called hamiltonian if it contains a Hamilton cycle. There are many sufficient conditions for a graph to be hamiltonian, and many of them involve some sort of degree conditions. Dirac's theorem is a classical result among them.
Theorem 1 ([10]) Let G be a graph on vertices. If for every vertex u in G, then G is hamiltonian.
In 1984, Fan [11] proved that only the pairs of vertices that are at distance 2 are essential in Theorem 1 and the condition can be replaced by a weaker one.
Theorem 2 ([11]) Let G be a 2-connected graph on vertices. If for every pair of vertices u and v at distance 2 in G, then G is hamiltonian.
In [8], the authors proved that a k-connected graph is hamiltonian if it contains no independent set S with . In 1980, by considering the degree sums of vertices in independent sets with vertices, Bondy [2] obtained a sufficient condition for a k-connected graph to be hamiltonian.
Theorem 3 ([2]) Let G be a k-connected graph on vertices. If for each independent set S of order , then G is hamiltonian.
Let G be a graph and S be a subset of . S is said to be an essential independent set if S is an independent set and contains two distinct vertices u and v at distance 2 in G. Chen et al. [9] gave a Fan-type sufficient condition for a k-connected graph to be hamiltonian by using degrees of vertices in essential independent set of order k.
Theorem 4 ([9]) Let G be a k-connected graph on vertices (). If for every essential independent set S with k vertices, then G is hamiltonian.
Moreover, the authors in [9] gave an example to show that the condition “ for every essential independent set S with k vertices” in Theorem 4 can not be exchanged by “ for every essential independent set S with vertices”.
In order to generalize some classical results about hamiltonian problems, Zhu, Li and Deng [14] gave the definition of implicit degree of a vertex. We use to denote the vertices which are at distance 2 from v in G.
Definition 1 ([14]) Let v be a vertex of a graph G and . Set and . If and , then let be the degree sequence of vertices of . Define Then the implicit degree of v is defined as . If or , then define .
Clearly, for every vertex v from the definition of implicit degree. The authors [14] gave a sufficient condition for a 2-connected graph to be hamiltonian by considering the relationship between the order of the graph and the implicit degree sum of nonadjacent vertices.
Theorem 5 ([14]) Let G be a 2-connected graph on vertices. If for every pair of nonadjacent vertices u and v in G, then G is hamiltonian.
For more results using implicit degree conditions, we refer to [6], [5], [7] and [12]. Recently, Li, Ning and Cai [13] gave a sufficient condition for a k-connected graph to be hamiltonian by replacing the degree sum in Theorem 3 by implicit degree sum.
Theorem 6 ([13]) Let G be a k-connected graph on vertices. If for each independent set S with vertices, then G is hamiltonian.
The main purpose of this paper is to generalize Theorem 4 in the way as Theorem 3 to Theorem 6 and give an implicit Fan-type sufficient condition for a k-connected graph to be hamiltonian by using implicit degrees of vertices in essential independent sets. We will prove the following result.
Theorem 7 Let G be a k-connected 2-heavy graph on vertices (). If for every essential independent set S with k vertices, then G is hamiltonian.
The proof of Theorem 7 will be given in next section. Now we present the following remarks. The first one shows the sharpness of Theorem 7 and the other one shows a graph which does not satisfy the condition of Theorem 4 but it can be easily verified to be hamiltonian by using Theorem 7.
Remark 1 Let . It is easy to check that G has vertices and G is a k-connected nonhamiltonian graph. And for each essential independent set S with k vertices, and each induced claw with two vertices of degree . This implies that the lower bound of Theorem 7 is best possible.
Remark 2 Let be a nonnegative integer. Set and . We choose a graph G with and (see Fig. 1). It is easy to see that G has vertices and G is a hamiltonian graph not satisfying the conditions of Theorem 4. But since for each , G satisfies the conditions of Theorem 7. This shows that our result in Theorem 7 does strengthen the result in Theorem 4.
Section snippets
Lemma
For a cycle C in G with a given orientation and a vertex x in C, and denote the successor and the predecessor of x in C, respectively. And for any , let and . For two vertices , xCy denotes the subpath of C from x to y. We use for the path from y to x in the reversed direction of C. A similar notation is used for paths.
For a path with and connecting x and y of a graph G, let . A path in G is called
Proof of Theorem 7
Suppose G is a graph satisfying the condition of Theorem 7 and G is not hamiltonian. Set .
Claim 1 There is a cycle C in G containing all vertices of X.
Proof Suppose to the contrary that there is no cycle in G containing all vertices of X. Choose a cycle C in G such that C contains as many vertices of X as possible. Without loss of generality, we give C a clockwise orientation. Then there is a vertex . Since G is 2-connected, by Menger Theorem there are two internally
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2023, Indian Journal of Pure and Applied Mathematics