International Journal of Mathematics and Mathematical Sciences

Volume 2012 (2012), Article ID 185346, 25 pages

doi:10.1155/2012/185346

Research Article

Two Sufficient Conditions for Hamilton and Dominating Cycles

Zh. G. Nikoghosyan

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,

Hence,

Case  2.2.2.2 (). By definition, there is an intermediate path between and . If , then by Lemma 2.2

contradicting (3.72). Otherwise, . Further, if , then by Lemma 2.2

again contradicting (3.72). Thus .

Case  2.2.2.2.1 (). Put . As in Case  2.1.2, form a graph by adding at most two new edges in such that , is disconnected and . Let be the connected components of with and . Using notation (3.48) for , as in Case  2.2.2.1, we have . By Lemma 2.2, . It means that , that is, . Assume without loss of generality that . By Lemma 2.3,

implying that

Hence

Case  2.2.2.2.2 (). By Lemma 2.2,

Recalling (3.72), we get and . Put . As in Case  2.1.2, form a graph by adding at most four new edges in such that , is disconnected and . Let , and be the connected components of with and . Using notation (3.48) for , we have as in Case  2.2.2.1, . Since and , we can assume without loss of generality that either , or ,  .

Case  2.2.2.2.2.1 (, ). It follows that , and . By Lemma 2.3,

Hence

Case  2.2.2.2.2.2 (, ). Let , where

If and , then as in proof of Lemma 2.2 (Case  2.1),

contradicting (3.72). Let either and or and .

Case  2.2.2.2.2.2.1 ( and ). Assume without loss of generality that occur on in this order. If , then

contradicting (3.72). Let , that is, . Put

Clearly,

By summing and observing that

we get

Hence , again contradicting (3.72).

Case  2.2.2.2.2.2.2 ( and ). Assume without loss of generality that occur on in this order.

Case  2.2.2.2.2.2.2.1 (). If and , then , contradicting the hypothesis. Thus, we can assume without loss of generality that . If , then

contradicting (3.72). Let , that is, . Since , we have . If , then , a contradiction. Otherwise, for some . Put

Clearly,

By summing and observing that , we get

Hence , contradicting (3.72).

Case  2.2.2.2.2.2.2.2 (). It follows that

If either or , then we can argue as in Case  2.2.2.2.2.2.2.1. Otherwise, . If , then

a contradiction. Let . Similarly, , implying that . If , then

a contradiction. So, , again a contradiction, since .

Case  2.2.2.2.2.2.2.3 (). It follows that

Since , we have either or , say . Put

If , then , a contradiction. This means that . If then , a contradiction. Let . Since , we have , that is, . Further, if , then

a contradiction. Let . Since , we have , contradicting the fact that .

Case  2.2.3 (). By (3.32),

It follows that .

Case  2.2.3.1 (). If , then by Lemma 2.1, , contradicting (3.102). Let . If , then

contradicting (3.102). Let . It follows that . Recalling (3.102), we get

Assume that for some and . Besides, we can assume without loss of generality that . Since is extreme, we have

implying that

a contradiction. So, for each . On the other hand, by Lemma 2.2, and hence is disconnected. Let , and be the connected components of with and . Using notation (3.48), we have . By Lemma 2.3,

implying that

Case  2.2.3.2 (either or ). Assume without loss of generality that . Put .

Case  2.2.3.2.1 (). It follows that for some and we can argue as in Case  2.2.3.1.

Case  2.2.3.2.2 (). It follows that

Since , there is an edge such that and . Since , we have . By (3.109), . Then replacing with , we can argue as in Case  2.2.3.1.

Case  2.2.4 (). By (3.32), . Let be a longest path in with . If , then by (3.34), , a contradiction. Let

Case  2.2.4.1 (). It follows that   . If for some , then clearly , contradicting (3.110). Let . Further, if , then again , contradicting (3.110). Let for some . Since , there is a path connecting and such that and . Clearly, . If , then

contradicting (3.110). Let . Further, if , then recalling that we have , a contradiction. Otherwise, and , contradicting (3.110).

Case  2.2.4.2 (). Put . Since , there are two disjoint paths connecting and . Further, since is extreme, we have . Let and , where and . Since is a hamiltonian cycle in , we can assume that is chosen such that . If , then , contradicting (3.110). Let . If for some , then

contradicting (3.110). Otherwise, , implying that for some . Then

again contradicting (3.110).

Case  2.2.5 (). By (3.32), . On the other hand, by Theorem D, , implying that and . Let be a longest path in with . If , then by (3.35), , a contradiction. Let

Case  2.2.5.1 (). Put . Since , there are two disjoint edges and connecting and such that and . Since is a hamiltonian cycle in , we can assume without loss of generality that is chosen such that . If , then , contradicting (3.114). Let . Further, if for some , then

contradicting (3.114). Now let for each . It follows that , that is, for some . But then , contradicting (3.114).

Case  2.2.5.2 (). If then by Theorem C, is hamiltonian and we can argue as in Case  2.2.5.1. Otherwise, , implying that

Assume without loss of generality that .

Case  2.2.5.2.1 (). It follows that . By (3.116), . Put . Since , there is a path connecting and such that and . If , that is, , then is a hamiltonian cycle in and we can argue as in Case  2.2.5.1. Let . Since , we have . Further, since , we have for some . Hence,

contradicting (3.114).

Case  2.2.5.2.2 (). Let . By (3.116), . If either or , then for some and therefore

contradicting (3.114). Otherwise, . Since , there is an edge such that and . Clearly, . Further, we can argue as in Case  2.2.5.1.

Case  2.2.5.2.3 (). Since , we have . Hence , contradicting (3.114).

Proof of Theorem 1.1. Let be a graph satisfying the hypothesis of Theorem 1.1, which is equivalent to

Since

we have which is equivalent to

If is even, that is, for some integer , then

implying that . By Theorem A, is hamiltonian. Let is odd, that is, for some integer . Then implying that . Recalling that is hamiltonian when , we can assume that . By Theorem C, either is hamiltonian or containers at least edges, contradicting (3.119). Theorem 1.1 is proved.

Proof of Theorem 1.2. Let be a 2-connected graph. The hypothesis of Theorem 1.2 is equivalent to

Case  1 ( and ). Let be a longest cycle in and a longest path in of length . If , then is a dominating cycle and we are done. Let . Since , there is a path such that and . Further, since is extreme, we have and therefore, , contradicting the hypothesis.

Case  2 ( and ). Since

we have , which is equivalent to

If for some integer , then

implying that . Next, if for some integer , then

implying that . Finally, if for some integer , then

implying that . So, , in any case. By Lemma 2.4, each longest cycle in is a dominating cycle.

Acknowledgment

The author would like to thank the anonymous referee for careful reading of the paper and valuable suggestions.

References

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C. St. J. A. Nash-Williams, “Edge-disjoint Hamiltonian circuits in graphs with vertices of large valency,” in Studies in Pure Mathematics, pp. 157–183, Academic Press, London, UK, 1971.

C. St. J. A. Nash-Williams, “Valency Sequences which force graphs to have Hamiltonian circuits,” Interim Rep., University of Waterloo, Waterloo, Ontario, 1969.