A dicycle cover of a digraph D is a family F of dicycles of D such that each arc of D lies in at least one dicycle in F. 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 G is a collection C of cycles of G such that E(G)=⋃C∈CE(C). Bondy [3] conjectured that if G is a 2-connected simple graph with n≥3 vertices, then G has a cycle cover C with C≤2n-3/3. 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 G is a 3-connected simple planar graph of order n, then the edges of G can be covered by at most n+1/2 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 D from a vertex u to a vertex v is called a (u,v)-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 D,V(D) and A(D) denote the vertex set and arc set of D, respectively. If A′⊆A(D), then D[A′] is the subdigraph induced by A′. Let Kn∗ denote the complete digraph on n vertices. Any simple digraph D on n vertices can be viewed as a subdigraph of Kn∗. If W is an arc subset of A(Kn∗), then D+W denotes the digraph Kn∗[A(D)∪W].

A digraph D is strong if, for any distinct u,v∈V(D), D has a (u,v)-dipath. As in [2], λ(D) denotes the arc-strong-connectivity of D. Thus a digraph D is strong if and only if λ(D)≥1. We use (u,v) denoting an arc with tail u and head v. For X,Y⊆V(D), we define(1)X,YD=x,y∈AD:x∈X,y∈Y;∂D+X=X,VD-XD.Let(2)dD+X=∂D+X,dD-X=∂D-X.When X={v}, we write dD+(v)=∂D+v and dD-(v)=|∂D-{v}|. Let ND+(v)={u∈V(D)-v:(v,u)∈A(D)} and ND-(v)={u∈V(D)-v:(u,v)∈A(D)} denote the out-neighbourhood and in-neighbourhood of v in D, respectively. We call the vertices in ND+(v) and ND-(v) the out-neighbours and the in-neighbours of v. Thus, for a digraph D, λ(D)≥1 if and only if, for any proper nonempty subset Ø≠X⊂V(D), ∂D+X≥1.

A dicycle cover of a digraph D is a collection C of dicycles of D such that ⋃C∈CA(C)=A(D). If D is obtained from a simple undirected graph G by assigning an orientation to the edges of G, then D 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 D be an oriented graph on n vertices and m arcs. If D has a Hamiltonian dicycle, then D has a dicycle cover C with C≤m-n+1. This bound is best possible.

In the next section, we will first show that every Hamiltonian oriented graph with n vertices and m arcs can be covered by at most m-n+1 dicycles. Then we show that, for every Hamiltonian graph G with n vertices and m edges, there exists an orientation D=D(G) of G such that any dicycle cover of D must have at least m-n+1 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 D is weakly connected if the underlying graph of D is connected.

Lemma 2.

A weakly connected digraph D has a dicycle cover if and only if λ(D)≥1.

Proof.

Suppose that D has a dicycle cover C. If D is not strong, then there exists a proper nonempty subset Ø≠X⊂V(D) such that |∂D+(X)|=0. Since D is weakly connected, D contains an arc (u,v)∈(V(D)-X),X)D. Since C is a dicycle cover of D, there exists a dicycle C∈C with (u,v)∈A(C). Since (u,v)∈(V(D)-X),X)D, we conclude that Ø≠A(C)∩(X,V(D)-X))D⊆∂D+(X), contrary to the assumption that |∂D+(X)|=0. This proves that D must be strong.

Conversely, assume that D is strong. For any arc a=(u,v)∈A(D), since D is strong, there must be a directed (v,u)-path P in D. It follows that Ca=P+a is a dicycle of D containing a, and so {Ca:a∈A(D)} is a dicycle cover of D.

Let C be a dicycle and let a=(u,v) be an arc not in A(C) but with u,v∈V(C). Then C+a contains a unique dicycle Ca containing a. In the following, we call Ca the fundamental dicycle of a with respect to C.

Lemma 3.

Let D be an oriented graph on n vertices and m arcs. If D has a Hamiltonian dicycle, then D has a dicycle cover C with C≤m-n+1.

Proof.

Let C0 denote the directed Hamiltonian cycle of D. For each a∈A(D)-A(C), let Ca denote the fundamental dicycle of a with respect to C. Then C={C0}∪{Ca:a∈A(D)-A(C)} is a dicycle cover of D with C≤m-n+1.

To prove that Theorem 1 is best possible, we need to construct, for each integer n≥4, a Hamiltonian oriented graph on n vertices and m arcs D such that any dicycle cover C of D must have at least m-n+1 dicycles in C.

Let G be a Hamiltonian simple graph. We present a construction of such an orientation D=D(G). Since G is Hamiltonian, we may assume that V(G)={v1,v2,…,vn} and C=v1v2,…,vnv1 is a Hamiltonian cycle of G.

Definition 4.

One defines an orientation D=D(G) as follows.

Orient the edges in the Hamiltonian cycle C=v1v2,…,vnv1 as follows:(3)vi+1,vi∈AD,i=1,2,…,n-1,v1,vn∈AD.

For each i=2,3,…,n-2, and for each j=i+2,i+3,…,n, assign directions to edges of G not in E(C) as follows:(4)vi,vj∈AD,ifvivj∈EG-EC,i+1<j≤n,v1,vj∈AD,ifv1vj∈EG-EC,i+1<j≤n-1.

We make the following observations stated in the lemma below.

Lemma 5.

Each of the following holds for the digraph D:

The dicycle C0=v1vnvn-1,…,v3v2v1 is a Hamiltonian dicycle of D.

The digraph D-A(C0) is acyclic.

ND+(vn)={vn-1}; ND-(v1)={v2}; ND-(v2)={v3}.

The dicycle C0 is the only dicycle of D containing the arc (v1,vn).

The dicycle C0 is the unique Hamiltonian dicycle of D.

If C′′ is a dicycle of D, then C′′ contains at most one arc in A(D)-A(C0).

Proof.

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

(ii) By Definition 4, the labels of the vertices V(D)={v1,v2,…,vn} satisfy (vi,vj)∈A(D)-A(C0) only if i<j. It follows (e.g., Section 2.1 of [2]) that D-A(C0) is acyclic, and so (ii) holds.

(iii) This follows immediately from Definition 4.

(iv) Let C′ be a dicycle of D with (v1,vn)∈A(C′). Since (v1,vn)∈A(C′)∩A(C0), we choose the largest label i≤n, such that (v1,vn),(vn,vn-1),…,(vi+1,vi)∈A(C′)∩A(C0). Since C′≠C0, we have i≥3. Since C′ is a dicycle, there must be a vertex vj∈V(D) such that (vi,vj)∈A(C′). By the choice of i, we must have (vi,vj)∉A(C0), and so (vi,vj)∈A(D)-A(C0). By Definition 4(ii), we have i+2≤j≤n, contrary to the fact that C′ is a dicycle of D containing (v1,vn). This proves (iv).

(v) Let C′ be a Hamiltonian dicycle of D. Since V(C′)=V(D), we have vn∈V(C′). We claim that (v1,vn)∈A(C′). If (v1,vn)∉A(C′), then there exists vi∈V(C)(i∈{v2,v3,…,vn-1}) such that (vi,vn)∈A(C′). Hence, (vi,vn),(vn,vn-1),…,(vi+2,vi+1)∈A(C′). By Definition 4(i) and (ii), N+(vi+1)⊂{vi+2,vi+3,…,vn}, contrary to the fact that C′ is a Hamiltonian dicycle of D. Thus, (v1,vn)∈A(C′). It follows from Lemma 5(iv) that we must have C′=C0.

(vi) By contradiction, we assume that D has a dicycle C′′ which contains two arcs: a1,a2∈A(D)-A(C0). Since V(D)={v1,v2,…,vn}, we assume that a1=(vi,vi′) and a2=(vj,vj′). Without loss of generality and by Lemma 2, we further assume that 1≤i<j<n.

Let i≥t≥1 be the smallest integer such that vt∈V(C′′). Since C′′ is a dicycle of D, there must be vs∈V(C′′) such that (vs,vt)∈A(C′′). By Definition 4, either (vs,vt)∈A(C0) and s=t+1<j or (vs,vt)∈A(D)-A(C0) and 1<s+1<t. By the choice of t, we can only have s=t+1 and (vt+1,vt)∈A(C′′)∩A(C0). Choose the largest integer h with t+1≤h<j such that (vt+1,vt),(vt+2,vt+1),…,(vh,vh-1)∈A(C′′)∩A(C0). Since C′′ is a dicycle, there must be vk with 1≤k≤n such that (vk,vh)∈A(C′′). By the maximality of h and by Definition 4(i), we conclude that (vk,vh)∉A(C0). By Definition 4(ii), 1≤k≤h-2. By the minimality of t, we must have t≤k≤h-2. It follows by j>h that C′′ cannot contain a2=(vj,vj′), contrary to the assumption. This contradiction justifies (vi).

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

Lemma 6.

Let G be a Hamiltonian simple graph. There exists an orientation D=D(G) such that every dicycle cover of D must have at least m-n+1 dicycles.

Proof.

Let G be a Hamiltonian graph and let D=D(G) be the orientation of G given in Definition 4. For notational convenience, we adopt the notations in Definition 4 and denote V(D)={v1,v2,…,vn}. Thus, by Lemma 5(v), C0=v1vnvn-1,…,v2v1 is the unique Hamiltonian dicycle of D.

Let C be a dicycle cover of D. By Lemma 5(iv), we must have C0∈C. For each arc a∈A(D)-A(C0), since C is a dicycle cover of D, there must be a dicycle C(a)∈C such that a∈A(C(a)). By Lemma 5(vi), A(C(a))∩A(D)-A(C0)={a}. It follows that if a,a′∈A(D)-A(C0), then a≠a′ implies C(a)≠C(a′) in C. Thus we have {C(a)∣a∈A(D)-A(C0)}⊆C. Hence(5)C≥Ca:a∈AD-AC0∪C0=m-n+1.This 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 Tn denote a tournament of order n. Then Tn 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 n vertices has a dicycle cover C with C≤n(n-1)/2-n+1. This bound is best possible.

A bipartite graph G(A,B) with vertex bipartition (A,B) is balanced if A=B. If bipartite graph G(A,B) has a Hamiltonian cycle, then G is balanced. Let Km,n be a complete bipartite graph with vertex bipartition (A,B) and |A|=m,B=n; then Km,n has Hamiltonian cycle if and only if m=n≥2; that is, Km,n is balanced. Let Kn,n denote a balanced complete bipartite graph.

Corollary 8.

Every Hamiltonian orientation of balanced complete bipartite graph Kn,n has a dicycle cover C with C≤(n-1)2. This bound is best possible.

Proof.

Since an oriented balanced complete bipartite graph Kn,n has n2 arcs, so, by Theorem 1, we have |C|≤n2-2n+1=(n-1)2.

To prove the bound is best possible, we need to construct, for each integer n≥2, a Hamiltonian oriented balanced complete bipartite graph on 2n vertices such that any dicycle cover C of Kn,n must have at least (n-1)2 dicycles in C. We may assume that V(Kn,n)={u1,u2,…,un,v1,v2,…,vn} and C=u1v1u2v2,…,unvnu1 is a Hamiltonian cycle of Kn,n. We construct an orientation Dn,n=D(Kn,n) as the orientation of Definition 4; thus, by Lemmas 5 and 6, every dicycle cover C of Dn,n must have at least (n-1)2 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 Dn1=(V(Dn1),A(Dn1)) and Dn2=(V(Dn2),A(Dn2)) be two disjoint digraphs; a1=(v12,v11)∈A(Dn1) and a2=(v22,v21)∈A(Dn2). The 2-sum Dn1⊕2Dn2 of Dn1 and Dn2 is obtained from the union of Dn1 and Dn2 by identifying the arcs a1 and a2; that is, v11=v21 and v12=v22.

Definition 10.

Let Dn1,Dn2,…,Dns be s disjoint digraphs with n1,n2,…,ns vertices, respectively. Let Dn1⊕2Dn2⊕2⋯⊕2Dns denote a sequence of 2 sums of Dn1,Dn2,…,Dns, that is, Dn1⊕2Dn2⊕2Dn3⊕2⋯⊕2Dns.

Theorem 11.

Let Dn1,Dn2,…,Dns be s disjoint Hamiltonian oriented graphs on n1,n2,…,ns vertices and m1,m2,…,ms arcs, respectively, and let D=Dn1⊕2Dn2⊕2⋯⊕2Dns. Then D has a dicycle cover C with C≤|A(D)|-|V(D)|+1. This bound is best possible.

Proof.

By Theorem 1, Dni(i=1,2,…,s) has a dicycle cover Ci with |Ci|≤mi-ni+1. Let C=⋃i=1sCi. Then |C|≤(m1-n1+1)+(m2-n2+1)+⋯+(ms-ns+1)=(m1+m2+⋯+ms)-(n1+n2+⋯+ns)+s = (m1+m2+⋯+ms-(s-1))-(n1+n2+⋯+ns-2(s-1))+1=|A(D)|-|V(D)|+1. By Definition 10, C is a dicycle cover of D. Thus, D has a dicycle cover C with |C|≤|A(D)|-|V(D)|+1.

Let Gni be s disjoint Hamiltonian simple graphs for i∈{1,2,…,s}. We may assume that V(Gni)={vi1,vi2,…,vini} and Ci=vi1vi2,…,vinivi1 is a Hamiltonian cycle of Gni, and let(6)Dni=DGnibetheorientationofGnigiveninDefinition4.For notational convenience, we adopt the notations in Definition 4 and denote V(Dni)={vi1,vi2,…,vini}. Thus, by Lemma 5(v), Ci0=vi1vini,…,vi2vi1 is the unique Hamiltonian dicycle of Dni. Let ai=(vi2,vi1) be an arc of Dni. We construct the 2-sum digraph Dn1⊕2Dn2⊕2⋯⊕2Dns from the union of Dn1,Dn2,…,Dns by identifying the arcs a1,a2,…,as such that v11=v21=⋯=vs1 and v12=v22=⋯=vs2. We assume that v1≔v11=v21=⋯=vs1 and v2≔v12=v22=⋯=vs2 (the case when s=2 is depicted in Figure 1).

Claim 1. There does not exist a dicycle whose arcs intersect arcs in two or more Dni’s (i=1,2,…,s).

By Definition 9, we have V(Dni)∩V(Dnj)={v1,v2}(i≠j). Without loss of generality, we consider oriented graphs Dn1 and Dn2; suppose that there exists a dicycle C0 such that (7)AC0-v2,v1∩ADn1≠Ø,AC0-v2,v1∩ADn2≠Ø.Thus, there must exist four different arcs(8)v1i′,v1,v1,v2i′′,v2j′′,v2,v2,v1j′∈AC0with (v1i′,v1),(v2,v1j′)∈A(Dn1) and (v1,v2i′′),(v2j′′,v2)∈A(Dn2), as shown in Figure 2, or four different arcs(9)v1s′,v2,v2,v2s′′,v2k′′,v1,v1,v1k′∈AC0with (v1s′,v2),(v1,v1k′)∈A(Dn1) and (v2,v2s′′),(v2k′′,v1)∈A(Dn2), as shown in Figure 3.

By Definition 9, Lemma 5(iii), and (6), we have ND-(v1)={v2}, and so v1i′=v2 or v2k′′=v2, contrary to the assumption that C0 is a dicycle. This proves Claim 1.

By Claim 1, for every dicycle C in D, all arcs in C (except for the arc (v2,v1)) belong to exactly one of oriented graphs Dni(i=1,2,…,n). By Definition 4 and Lemma 6, every dicycle cover of oriented graph Dni(i=1,2,…,n) must have at least mi-ni+1 dicycles. This completes the proof.

The 2-sum digraph for Dn1 and Dn2.

By Corollary 7 and Theorem 11, we have the following corollary.

Corollary 12.

Let Dn1,Dn2,…,Dns be s disjoint strong tournaments with n1,n2,…,ns vertices, respectively. Then Dn1⊕2Dn2⊕2⋯⊕2Dns has a dicycle cover C with |C|≤(n1(n1-1)/2+n2(n2-1)/2+⋯+ns(ns-1)/2)-(n1+n2+⋯+ns)+s. This bound is best possible.

Let Gn be a Hamiltonian graph with n vertices and m arcs; let Dni(i is an integer) denote a Hamiltonian orientation of Gn. For a positive integer s, let H(Gn,s) denote the family of all 2-sum generated digraphs Dn1⊕2Dn2⊕2⋯⊕2Dns, as well as a member in the family (for notational convenience). By the definition of H(Gn,s), we have H(Gn,1)=Dn1 and H(Gn,s)=HGn,s-1⊕2Dns. The conclusions of the next corollaries follow from Theorem 1. The sharpness of these corollaries can be demonstrated using similar constructions displayed in Lemma 6 and Corollary 8.

Corollary 13.

Let m,n≥3 be integer, let Gn be a Hamiltonian graph with n vertices and m edges, and let Kn be a complete graph on n≥3 vertices:

Any member in H(Gn,s) has a dicycle cover C with |C|≤s(m-n+1). This bound is best possible.

In particular, any H(Kn,s) has a dicycle cover C with |C|≤s(n(n-1)/2-n+1). This bound is best possible.

Corollary 14.

Let m,n≥3 be integer, let Bn be a Hamiltonian bipartite graph with 2n vertices and m edges, and let Kn,n be a complete bipartite graph:

Any H(Bn,s) has a dicycle cover C with |C|≤s(m-2n+1). This bound is best possible.

In particular, any H(Kn,n,s) has a dicycle cover C with |C|≤s(n-1)2. This bound is best possible.

Conflict of Interests

The authors declare that there is no conflict of interests regarding the publication of this paper.

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