Graphs in which G − N[v] is a cycle for each vertex v
Introduction
Let be a finite, simple, undirected graph. As usual, and are called the order and the size of G, respectively. If , then we say that u is a neighbor of v in G and vice versa. For every vertex , the open neighborhood of v is the set and the closed neighborhood of v is the set . The degree of a vertex is . The minimum degree and the maximum degree of a graph G are denoted by and , respectively. For any set , denotes the subgraph of G induced by S and denotes the graph obtained by deleting S from G. As usual, denotes the complete graph, cycle, path of order n, respectively. The complement of a graph G, denoted by , is the graph with vertex set , in which two vertices are adjacent if and only if they are not adjacent in G. The line graph of G, denoted by , is the graph with vertex set , in which two vertices are adjacent if and only if they share a common end vertex in G. For two vertex-disjoint graphs and , the join of and , denoted by , is the graph obtained from by joining each vertex of to all vertices of . For a positive integer k, kG consists of k disjoint copies of G. We refer to [2] for undefined notations and terminology.
A graph H is said to be realizable by a graph G if every neighborhood in G induces a subgraph isomorphic to H. The problem is also referred to as the Trahtenbrot-Zykov problem, see [14], [19]. Agakisieva [1] proved that is realizable by a graph if and only if . But, it was disproved by Chilton, Gould and Polimeni [11]. Laskar and Mulder [16] investigated the graphs in which every neighborhood induces a path. They characterized the graphs G with for . Graphs in which the neighborhood of each vertex is some special graphs are investigated in [6], [7], [8], [9], [12], [13], [15], [16], [17].
In this note, we investigate the graphs with somewhat opposite property to the Trahtenbrot-Zykov problem. We say that G has the property if is a cycle for any vertex . Since for any vertex v, the Petersen graph is such a graph. In addition, since , G has the property if and only if is the complement of a cycle for each vertex v.
Another ingredient why we investigate the graphs with the property comes from so called the isolation number, as the generation of domination, introduced by Caro and Hansberg [10]. It is recently studied by a number of papers [3], [4], [5], [18].
Actually, we can find more graphs with the property by the following operation.
Lemma 1.1 Let G and H be two vertex-disjoint graphs. Then has property if and only if each of G and H has property .
The following lemma is crucial for characterizing all graphs with the property .
Lemma 1.2 Let G be a graph with connected complement. If is a d-regular graph for every , then G is regular, where d is a fixed nonnegative integer. Proof Let u and v be any two nonadjacent vertices of G. By the assumption, both and are d-regular. Since , we have . It means that the degrees of any two adjacent vertices in are equal. Therefore, is regular, because is connected. Thus G is regular. □
By the above lemma, if G is a graph with the property and with the connected complement, it is regular. As we have seen before, since , . The following result is immediate.
Lemma 1.3 Let G be a graph. For a vertex , if and only if .
Lemma 1.4 Assume that a graph G has the property . Then G is disconnected if and only if . Proof It is trivial to see that is a disconnected graph G with the property . Now assume that G is disconnected graph with the property . By the definition of the property , it follows that G has exactly two components, each of which must be . Thus . □
The aim of the note is to characterize all graphs with the property .
Section snippets
The characterization
In view of Lemma 1.2, Lemma 1.4, we may assume that the graphs under consideration are connected.
Proposition 2.1 Assume that a graph G and its complement are connected. Then for any vertex if and only if is a triangle-free cubic graph. Proof First assume that is a triangle-free cubic graph. It is clear that for any vertex , is an independent set of cardinality 3 in . Thus is a clique in G. Moreover, since , . To show its necessity, assume that
Graphs and their complements have the property
Lemma 3.1 Let G be a connected triangle-free graph with the property . If is connected, then is a cycle for any vertex if and only if .
Proof It is straightforward to check that if , then for any vertex . Next, assume that is a cycle for any vertex . By Lemma 1.2, assume that G is k-regular for some integer . Thus , where . Since G is triangle-free, . Fix a vertex x and a neighbor y in G. Since G is triangle-free, .
Further research
Let denote a family of graphs. It is interesting to characterize the graphs G such that for any vertex , is isomorphic to a member in the family . In particular, we are interested to investigate the cases when (1) is the family of all stars; (2) is the family of all paths; (3) is the family of all trees; (4) is the family of all cliques, among others.
Declaration of Competing Interest
The authors declare that they have no known competing financial interests or personal relationships that could have appeared to influence the work reported in this paper.
Acknowledgements
The authors are grateful to the referees for their helpful comments. The work is supported by the Key Laboratory Project of Xinjiang (2018D04017), NSFC (No. 12061073), and XJEDU2019I001.
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