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International Journal of Mathematics and Mathematical Sciences
Volume 2012 (2012), Article ID 185346, 25 pages
http://dx.doi.org/10.1155/2012/185346
Research Article

Two Sufficient Conditions for Hamilton and Dominating Cycles

Institute for Informatics and Automation Problems, National Academy of Sciences, Street P. Sevak 1, Yerevan 0014, Armenia

Received 24 March 2012; Accepted 14 August 2012

Academic Editor: Marianna Shubov

Copyright © 2012 Zh. G. Nikoghosyan. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

Abstract

We prove that if is a 2-connect graph of size (the number of edges) and minimum degree with , where when and when , then each longest cycle in is a dominating cycle. The exact analog of this theorem for Hamilton cycles follows easily from two known results according to Dirac and Nash-Williams: each graph with is hamiltonian. Both results are sharp in all respects.

1. Introduction

Only finite undirected graphs without loops or multiple edges are considered. We reserve , , , and to denote the number of vertices (order), the number of edges (size), the minimum degree, and the connectivity of a graph, respectively. A graph is hamiltonian if contains a hamiltonian cycle, that is, a cycle of length . Further, a cycle in is called a dominating cycle if the vertices in are mutually nonadjacent. A good reference for any undefined terms is [1].

The following two well-known theorems provide two classic sufficient conditions for Hamilton and dominating cycles by linking the minimum degree and order .

Theorem A (see [2]). Every graph with is hamiltonian.

Theorem B (see [3]). If is a 2-connect graph with , then each longest cycle in is a dominating cycle.

The exact analog of Theorem A that links the minimum degree and size easily follows from Theorem A and a particular result according to Nash-Williams [4] (see Theorem 1.1 below).

Theorem 1.1. Every graph is hamiltonian if

The hypothesis in Theorem 1.1 is equivalent to and cannot be relaxed to due to the graph consisting of two copies of and having exactly one vertex in common. Hence, Theorem 1.1 is best possible.

The main goal of this paper is to prove the exact analog of Theorem B for dominating cycles based on another similar relation between and .

Theorem 1.2. Let be a 2-connect graph with where when and when . Then each longest cycle in is a dominating cycle.

To show that Theorem 1.2 is sharp, suppose first that , implying that the hypothesis in Theorem 1.2 is equivalent to . The graph shows that the connectivity condition in Theorem 1.2 cannot be relaxed by replacing it with . The graph with vertex set and edge set shows that the size bound cannot be relaxed by replacing it with . Finally, the graph shows that the conclusion “each longest cycle in is a dominating cycle” cannot be strengthened by replacing it with “ is hamiltonian.” Analogously, we can use , , and , respectively, to show that Theorem 1.2 is sharp when . So, Theorem 1.2 is best possible in all respects.

To prove Theorems 1.1 and 1.2, we need two known results, the first of which is belongs Nash-Williams [4].

Theorem C (see [4]). If , then either is hamiltonian or , or , where denote an arbitrary graph on vertices.

The next theorem provides a lower bound for the length of a longest cycle in 2-connected graphs according to Dirac [2].

Theorem D (see [2]). Every 2-connected graph either has a hamiltonian cycle or has a cycle of length at least .

2. Notations and Preliminaries

The set of vertices of a graph is denoted by and the set of edges by . For , a subset of , we denote by the maximum subgraph of with vertex set . We write for the subgraph of induced by . For a subgraph of , we use short for . The neighborhood of a vertex will be denoted by . Set . Furthermore, for a subgraph of and , we define and .

A simple cycle (or just a cycle) of length is a sequence of distinct vertices with for each , where . When , the cycle on two vertices coincides with the edge , and when , the cycle coincides with the vertex . So, all vertices and edges in a graph can be considered as cycles of lengths 1 and 2, respectively.

Paths and cycles in a graph are considered as subgraphs of . If is a path or a cycle, then the length of , denoted by , is . We write with a given orientation by . For , we denote by the subpath of in the chosen direction from to . For , we denote the th successor and the th predecessor of on by and , respectively. We abbreviate and by and , respectively.

Special Definitions
Let be a graph, a longest cycle in , and a longest path in of length . Let be the elements of occurring on in a consecutive order. Set where .(1) We call elementary segments on created by .(2) We call a path an intermediate path between two distinct elementary segments and if (3) The set of all intermediate paths between elementary segments will be denoted by .

Lemma 2.1. Let be a graph, a longest cycle in , and a longest path in of length . If , and , then where and .

Lemma 2.2. Let be a graph, a longest cycle in , and a longest path in of length . If , and , are elementary segments induced by , then(a1) if is an intermediate path between and , then (a2) if and for some , then

Lemma 2.3. Let be a graph, a cut set in , and a connected component of of order . Then where .

Lemma 2.4. Let be a 2-connect graph. If , then either or each longest cycle in is a dominating cycle.

3. Proofs

Proof of Lemma 2.1. Put By the hypothesis, , implying that Let be the elements of occuring on in a consecutive order. Put , where . Clearly, . Since is extreme, . Next, if for some , then . Further, if either , or , , then again .
Case  1. ().
Case  1.1 (). It follows that among there are segments of length at least . Observing also that each of the remaining segments has a length at least 2, we have
Since and , Recalling that , we get Analogously, . So, Case  1.2 (either , or , ). Assume without loss of generality that and , that is, and . Hence, among there are segments of length at least . Taking into account that each of the remaining segments has a length at least 2 and , we get Case  2 (). We first prove that . Since and , there are at least two segments among of length at least . If , then clearly and Otherwise, since , there are at least three elementary segments of length at least , that is, So, in any case, .
To prove that , we distinguish two main cases.
Case  2.1 (). It follows that among there are segments of length at least . Further, since each of the remaining segments has a length at least 2, we get Observing also that we have implying that .
Case  2.2 (either , or , ). Assume without loss of generality that and , that is, and . It follows that among there are segments of length at least . Observing also that , that is, , we get

Proof of Lemma 2.2. Let be the elements of occuring on in a consecutive order. Put , where . To prove , let be an intermediate path between elementary segments and with and . Put Clearly, Since is extreme, we have , implying that . By a symmetric argument, . Hence
The proof of is complete. To prove , let and for some .
Case  1 (). It follows that consists of a unique intermediate edge . By , Case 2 (). It follows that consists of two edges , . Put and , where and .
Case  2.1 ( and ). Assume without loss of generality that and occur in this order on .
Case  2.1.1. and occur in this order on .
Put Clearly, Since is extreme, , implying that . By a symmetric argument, . Hence Case  2.1.2. and occur in this order on .
Putting we can argue as in Case  2.1.1.
Case  2.2 (either , or , ). Assume without loss of generality that , and occur in this order on . Put
Clearly, Since is extreme, and , implying that Hence, Case  3 (). It follows that consists of three edges , , . Let   , where and . If there are two independent edges among , , , then we can argue as in Case  2.1. Otherwise, we can assume without loss of generality that and , , occur in this order on . Put
Clearly, Since is extreme, we have and , implying that Hence,

Proof of Lemma 2.3. Put
Observing that for each , we have   . Therefore,

Proof of Lemma 2.4. Let be a longest cycle in and a longest path in of length . If , then is a dominating cycle and we are done. Let , that is, . By the hypothesis, . Further, by Theorem D, . From these inequalities, we get
Let be the elements of occuring on in a consecutive order. Put where .
Case  1 (). Let be a longest path in with and . Since is extreme, we have and , implying that Since and , we have . By (3.34), , implying that .
Case  2. ().
Case  2.1 (). By (3.32), Case  2.1.1 (). It follows that , where By Lemma 2.1, . Recalling (3.35), we get . If , then by Lemma 2.1, , contradicting (3.35). Let . Clearly, and . Further, if , then , again contradicting (3.35). Let , implying that   . By Lemma 2.2, . Let be the connected components of with and . For each , put Clearly, . By Lemma 2.3, implying that Case  2.1.2 (). Clearly, . If , then we can argue as in Case  2.1.1. Let . Further, if for some distinct , then , contradicting (3.35). Hence Claim 1. and(1)if then ,(2)if then ,(3)if then .
Proof. If then we are done. Otherwise, , for some distinct . By definition, there is an intermediate path between and . If , then by Lemma 2.2, contradicting (3.40). Otherwise, and therefore, . By Lemma 2.2, . Combining this with (3.40), we have Furthermore, if , then by Lemma 2.2, , contradicting (3.42). So, Put . If , then , contradicting (3.35). Further, if , then by Lemma 2.2, . Let .
Case (). It follows that for some distinct and for each . Recalling that and , we have , that is, . By Lemma 2.2, for each distinct . Moreover, we have if either or . So, Case (). It follows that and for some distinct and for each . By (3.42), we can assume without loss of generality that either or , .
Case (). It follows that . By Lemma 2.2, and if , implying that .
Case (, ). It follows that . By Lemma 2.2, we have and for each . Furthermore, if . Thus, .
Case (). It follows that for some and for each . By (3.42), .
Case (). It follows that . By Lemma 2.2, for each , implying that .
Case (). It follows that . By Lemma 2.2, for each and if , that is, .
Case (). It follows that . By Lemma 2.2, for each and if , that is, . Claim 1 is proved.
Let and let , where and for some distinct . Put . Form a graph in the following way. If and in then we take . Next, suppose that and in . Put If , then clearly in , contradicting the hypothesis. Otherwise, for some and we take . Finally, if , then as above, and for some , and we take . Clearly, and . This procedure may be repeated for all edges of . The resulting graph satisfies the following conditions: In fact, where consists of at most appropriate new edges such that is disconnected. Let be the connected components of with    and . For each , put Clearly, . If for some , then contradicting (3.32). Otherwise, . It follows that which is equivalent to Case  2.1.2.1 (). By (3.50) and Lemma 2.3, . Hence Using (3.46) and Claim 1, we have Case  2.1.2.2 (). Assume without loss of generality that , that is, . By (3.50) and Lemma 2.3, and Hence By (3.46) and Claim 1, Case  2.1.2.3 (). Assume without loss of generality that , that is, . By (3.50) and Lemma 2.3, and Hence By (3.46) and Claim 1, Case  2.2 (). According to (3.32), we can distinguish five main cases, namely, , , , , and .
Case  2.2.1 (). It follows that and If , then by (3.59) and Lemma 2.1, , contradicting (3.32). Let . Clearly, and . If , then again contradicting (3.32). Let . It means that , that is, is hamiltonian. By symmetric arguments, for each . Assume that , that is, for some elementary segments and . By the definition, there is an intermediate path between and . If , then by Lemma 2.2 Hence contradicting (3.32). Thus, , that is, . By Lemma 2.2, which yields If , then , contradicting (3.32). Let . Recalling (3.59), we have , that is, and . Hence, . On the other hand, by (3.32) and the fact that , we have . Thus Put . As in Case  2.1.2, form a graph by adding at most new edges in such that and are disconnected. We denote immediately if . Hence Let be the connected components of with and . Since   (i.e., is hamiltonian) and is extreme, we have . Using notation (3.48) for , we have    and . If for some , then By (3.59), , implying that and , contradicting (3.32). Let   . It follows that By Lemma 2.3, , implying that If , then for some distinct . By Lemma 2.2 and hence , contradicting (3.65). So, . By (3.66) and (3.69), Case  2.2.2 (). It follows that . By (3.32), If , then by Lemma 2.1, , contradicting (3.72). Let . Further, if , then again contradicting (3.72). Let . It follows that , that is, is hamiltonian. By symmetric arguments, for each . Clearly, .
Case  2.2.2.1 (). It follows that is disconnected. Let be the connected components of with and . Since is extreme and is hamiltonian, we have . By notation (3.48), and . If for some , then contradicting (3.32). So, . By Lemma 2.3,