Journal of Discrete Mathematics

Volume 2016, Article ID 7942192, 5 pages

http://dx.doi.org/10.1155/2016/7942192

## Dicycle Cover of Hamiltonian Oriented Graphs

^{1}Department of Mathematics, College of Science, Qassim University, P.O. Box 6644, Buraydah 51452, Saudi Arabia^{2}Department of Mathematics, West Virginia University, Morgantown, WV 26506, USA^{3}College of Mathematics Sciences, Xinjiang Normal University, Urumqi 830054, China

Received 9 October 2015; Accepted 31 December 2015

Academic Editor: Kinkar Ch Das

Copyright © 2016 Khalid A. Alsatami et al. 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

A dicycle cover of a digraph is a family of dicycles of such that each arc of lies in at least one dicycle in . We investigate the problem of determining the upper bounds for the minimum number of dicycles which cover all arcs in a strong digraph. Best possible upper bounds of dicycle covers are obtained in a number of classes of digraphs including strong tournaments, Hamiltonian oriented graphs, Hamiltonian oriented complete bipartite graphs, and families of possibly non-Hamiltonian digraphs obtained from these digraphs via a sequence of 2-sum operations.

#### 1. The Problem

We consider finite loopless graphs and digraphs, and undefined notations and terms will follow [1] for graphs and [2] for digraphs. In particular, a cycle is a 2-regular connected nontrivial graph. A cycle cover of a graph is a collection of cycles of such that . Bondy [3] conjectured that if is a 2-connected simple graph with vertices, then has a cycle cover with . Bondy [3] showed that this conjecture, if proved, would be best possible. Luo and Chen [4] proved that this conjecture holds for 2-connected simple cubic graphs. It has been shown that, for plane triangulations, serial-parallel graphs, or planar graphs in general, one can have a better bound for the number of cycles used in a cover [5–8]. Barnette [9] proved that if is a 3-connected simple planar graph of order , then the edges of can be covered by at most cycles. Fan [10] settled this conjecture by showing that it holds for all simple 2-connected graphs. The best possible number of cycles needed to cover cubic graphs has been obtained in [11, 12].

A directed path in a digraph from a vertex to a vertex is called a -dipath. To emphasize the distinction between graphs and digraphs, a directed cycle or path in a digraph is often referred to as a dicycle or dipath. It is natural to consider the number of dicycles needed to cover a digraph. Following [2], for a digraph and denote the vertex set and arc set of , respectively. If , then is the subdigraph induced by . Let denote the complete digraph on vertices. Any simple digraph on vertices can be viewed as a subdigraph of . If is an arc subset of , then denotes the digraph .

A digraph is strong if, for any distinct , has a -dipath. As in [2], denotes the arc-strong-connectivity of . Thus a digraph is strong if and only if . We use denoting an arc with tail and head . For , we defineLetWhen , we write and . Let and denote the* out-neighbourhood* and* in-neighbourhood* of in , respectively. We call the vertices in and the* out-neighbours* and the* in-neighbours* of . Thus, for a digraph , if and only if, for any proper nonempty subset , .

A* dicycle cover* of a digraph is a collection of dicycles of such that . If is obtained from a simple undirected graph by assigning an orientation to the edges of , then is an* oriented graph*. The main purpose is to investigate the number of dicycles needed to cover a Hamiltonian oriented graph. We prove the following.

Theorem 1. *Let be an oriented graph on vertices and arcs. If has a Hamiltonian dicycle, then has a dicycle cover with . This bound is best possible.*

In the next section, we will first show that every Hamiltonian oriented graph with vertices and arcs can be covered by at most dicycles. Then we show that, for every Hamiltonian graph with vertices and edges, there exists an orientation of such that any dicycle cover of must have at least dicycles.

#### 2. Proof of the Main Result

In this section, all graphs are assumed to be simple. We start with an observation, stated as lemma below. A digraph is* weakly connected* if the underlying graph of is connected.

Lemma 2. *A weakly connected digraph has a dicycle cover if and only if .*

*Proof. *Suppose that has a dicycle cover . If is not strong, then there exists a proper nonempty subset such that . Since is weakly connected, contains an arc . Since is a dicycle cover of , there exists a dicycle with . Since , we conclude that , contrary to the assumption that . This proves that must be strong.

Conversely, assume that is strong. For any arc , since is strong, there must be a directed -path in . It follows that is a dicycle of containing , and so is a dicycle cover of .

Let be a dicycle and let be an arc not in but with . Then contains a unique dicycle containing . In the following, we call the fundamental dicycle of with respect to .

Lemma 3. *Let be an oriented graph on vertices and arcs. If has a Hamiltonian dicycle, then has a dicycle cover with .*

*Proof. *Let denote the directed Hamiltonian cycle of . For each , let denote the fundamental dicycle of with respect to . Then is a dicycle cover of with .

To prove that Theorem 1 is best possible, we need to construct, for each integer , a Hamiltonian oriented graph on vertices and arcs such that any dicycle cover of must have at least dicycles in .

Let be a Hamiltonian simple graph. We present a construction of such an orientation . Since is Hamiltonian, we may assume that and is a Hamiltonian cycle of .

*Definition 4. *One defines an orientation as follows. (i)Orient the edges in the Hamiltonian cycle as follows:(ii)For each , and for each , assign directions to edges of not in as follows:

We make the following observations stated in the lemma below.

Lemma 5. *Each of the following holds for the digraph : *(i)*The dicycle is a Hamiltonian dicycle of .*(ii)*The digraph is acyclic.*(iii)*; ; .*(iv)*The dicycle is the only dicycle of containing the arc .*(v)*The dicycle is the unique Hamiltonian dicycle of .*(vi)*If is a dicycle of , then contains at most one arc in .*

*Proof. *(i) follows immediately from Definition 4(i).

(ii) By Definition 4, the labels of the vertices satisfy only if . It follows (e.g., Section 2.1 of [2]) that is acyclic, and so (ii) holds.

(iii) This follows immediately from Definition 4.

(iv) Let be a dicycle of with . Since , we choose the largest label , such that . Since , we have . Since is a dicycle, there must be a vertex such that . By the choice of , we must have , and so . By Definition 4(ii), we have , contrary to the fact that is a dicycle of containing . This proves (iv).

(v) Let be a Hamiltonian dicycle of . Since , we have . We claim that . If , then there exists such that . Hence, . By Definition 4(i) and (ii), , contrary to the fact that is a Hamiltonian dicycle of . Thus, . It follows from Lemma 5(iv) that we must have .

(vi) By contradiction, we assume that has a dicycle which contains two arcs: . Since , we assume that and . Without loss of generality and by Lemma 2, we further assume that .

Let be the smallest integer such that . Since is a dicycle of , there must be such that . By Definition 4, either and or and . By the choice of , we can only have and . Choose the largest integer with such that . Since is a dicycle, there must be with such that . By the maximality of and by Definition 4(i), we conclude that . By Definition 4(ii), . By the minimality of , we must have . It follows by that cannot contain , contrary to the assumption. This contradiction justifies (vi).

*To complete the proof of Theorem 1, we present the next lemma.*

*Lemma 6. Let be a Hamiltonian simple graph. There exists an orientation such that every dicycle cover of must have at least dicycles.*

*Proof. *Let be a Hamiltonian graph and let be the orientation of given in Definition 4. For notational convenience, we adopt the notations in Definition 4 and denote . Thus, by Lemma 5(v), is the unique Hamiltonian dicycle of .

Let be a dicycle cover of . By Lemma 5(iv), we must have . For each arc , since is a dicycle cover of , there must be a dicycle such that . By Lemma 5(vi), . It follows that if , then implies in . Thus we have . HenceThis proves the lemma.

*By Lemmas 3 and 6, Theorem 1 follows. We are about to show that Theorem 1 can be applied to obtain dicycle cover bounds for certain families of oriented graphs. Let denote a tournament of order . Then is an oriented graph. Camion [13, 14] proved that every strong tournament is Hamiltonian. Hence the corollary below follows from Theorem 1.*

*Corollary 7. Every strong tournament on vertices has a dicycle cover with . This bound is best possible.*

*A bipartite graph with vertex bipartition is balanced if . If bipartite graph has a Hamiltonian cycle, then is balanced. Let be a complete bipartite graph with vertex bipartition and ; then has Hamiltonian cycle if and only if ; that is, is balanced. Let denote a balanced complete bipartite graph.*

*Corollary 8. Every Hamiltonian orientation of balanced complete bipartite graph has a dicycle cover with . This bound is best possible.*

*Proof. *Since an oriented balanced complete bipartite graph has arcs, so, by Theorem 1, we have .

To prove the bound is best possible, we need to construct, for each integer , a Hamiltonian oriented balanced complete bipartite graph on vertices such that any dicycle cover of must have at least dicycles in . We may assume that and is a Hamiltonian cycle of . We construct an orientation as the orientation of Definition 4; thus, by Lemmas 5 and 6, every dicycle cover of must have at least dicycles. This proves the corollary.

*3. Dicycle Covers of 2 Sums of Digraphs*

*In this section, we will show that Theorem 1 can also be applied to certain non-Hamiltonian digraphs which can be built via 2 sums. We start with 2 sums of digraphs.*

*Definition 9. *Let and be two disjoint digraphs; and The 2-sum of and is obtained from the union of and by identifying the arcs and ; that is, and .

*Definition 10. *Let be disjoint digraphs with vertices, respectively. Let denote a sequence of 2 sums of , that is, .

*Theorem 11. Let be disjoint Hamiltonian oriented graphs on vertices and arcs, respectively, and let . Then has a dicycle cover with . This bound is best possible.*

*Proof. *By Theorem 1, has a dicycle cover with . Let . Then = . By Definition 10, is a dicycle cover of . Thus, has a dicycle cover with .

Let be disjoint Hamiltonian simple graphs for . We may assume that and is a Hamiltonian cycle of , and letFor notational convenience, we adopt the notations in Definition 4 and denote . Thus, by Lemma 5(v), is the unique Hamiltonian dicycle of . Let be an arc of . We construct the 2-sum digraph from the union of by identifying the arcs such that and . We assume that and (the case when is depicted in Figure 1).*Claim 1*. There does not exist a dicycle whose arcs intersect arcs in two or more ’s .

By Definition 9, we have . Without loss of generality, we consider oriented graphs and ; suppose that there exists a dicycle such that Thus, there must exist four different arcswith and , as shown in Figure 2, or four different arcswith and , as shown in Figure 3.

By Definition 9, Lemma 5(iii), and (6), we have , and so or , contrary to the assumption that is a dicycle. This proves Claim 1.

By Claim 1, for every dicycle in , all arcs in (except for the arc () belong to exactly one of oriented graphs By Definition 4 and Lemma 6, every dicycle cover of oriented graph must have at least dicycles. This completes the proof.