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Vertex disjoint cycles of different lengths in d-arc-dominated digraphs
Operations Research Letters 42 (2014) 351–354
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Vertex disjoint cycles of different lengths in d-arc-dominated digraphs Ngo Dac Tan ∗ Institute of Mathematics, Vietnam Academy of Science and Technology, 18 Hoang Quoc Viet Road, 10307 Hanoi, Viet Nam
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Article history: Received 3 October 2013 Received in revised form 16 March 2014 Accepted 8 June 2014 Available online 14 June 2014
abstract Gao and Ma (2013) have proved that every 4-arc-dominated digraph contains two vertex disjoint directed cycles of different lengths. In this paper, by arguments similar to those used by Gao and Ma in the abovementioned paper, we show that every d-arc-dominated digraph with d ≥ 4 contains two vertex disjoint directed cycles of different lengths. © 2014 Elsevier B.V. All rights reserved.
Keywords: Vertex disjoint cycles Minimum outdegree d-regular digraph d-arc-dominated digraph
1. Introduction In this paper, the term digraph always means a finite simple digraph, i.e., a digraph that has a finite number of vertices, no loops and no multiple arcs. Unless otherwise indicated, our graphtheoretic terminology will follow [1]. Let D be a digraph. Then the vertex set and the arc set of D are denoted by V (D) and A(D) (or by V and A for short), respectively. The order and the size of D are |V | and |A|, respectively. A vertex v ∈ V is called an outneighbor of a vertex u ∈ V if (u, v) ∈ A. The set of all outneighbors of u, denoted by ND+ (u), is called the outneighborhood of u. The outdegree of u ∈ V , denoted by d+ D (u), is |ND+ (u)|. The number d = min{d+ D (u) | u ∈ V } is called the minimum outdegree of the digraph D. Similarly, a vertex w ∈ V is called an inneighbor of a vertex u ∈ V if (w, u) ∈ A. The set of all inneighbors of u, denoted by ND− (u), is called the inneighborhood of u. The − indegree of u ∈ V , denoted by d− D (u), is |ND (u)|. If W ⊆ V , then the subdigraph of D induced by W is denoted by D[W ]. By a cycle (resp., path) in a digraph D = (V , A) we always mean a directed cycle (resp., directed path). By disjoint cycles in D we always mean vertex disjoint cycles. A digraph D = (V , A) is called an oriented graph if D has no cycles of length 2. For a natural number d, a digraph D = (V , A) is − called d-regular if d+ D (u) = dD (u) = d for every vertex u ∈ V .
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http://dx.doi.org/10.1016/j.orl.2014.06.004 0167-6377/© 2014 Elsevier B.V. All rights reserved.
Let D = (V , A) be a digraph with minimum outdegree d. Then D is called d-arc-dominated if for every arc (x, y) ∈ A there is a vertex u ∈ V of outdegree d such that both (u, x) ∈ A and (u, y) ∈ A hold. At the 20th British Combinatorial Conference held at the University of Durham from 10–15, 2005, Lichiardopol posed the problem of characterizing d-arc-dominated oriented graphs (see Problem 912 (BCC20.4) in [3]). These digraphs are interesting for the Bermond–Thomassen conjecture, which states that a digraph with minimum outdegree d ≥ 2k − 1 contains k disjoint cycles (see [1] or [2]), because no arcs in them can be path-contracted (see [1] for a definition) to maintain the outdegree condition in the conjecture—thus minimal counterexamples are d-arc-dominated. Some partial results for the Lichiardopol’s problem have been obtained recently by Ngo Dac Tan in [8,9]. In [5], Henning and Yeo considered the problem about existence of two disjoint cycles of different lengths in digraphs. In particular, they have proved in [5] that every 4-regular digraph contains two disjoint cycles of different lengths. They also conjectured in [5] that a digraph with minimum outdegree at least 4 contains two disjoint cycles of different lengths. Recently in [4], Gao and Ma have verified this conjecture for 4-arc-dominated digraphs. In this paper, by using arguments similar to those used by Gao and Ma in [4], we will verify this conjecture of Henning and Yeo for any d-arc-dominated digraphs with d ≥ 4. Namely, we will prove the following result. Theorem 1. Every d-arc-dominated digraph with d ≥ 4 contains two disjoint cycles of different lengths.
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N.D. Tan / Operations Research Letters 42 (2014) 351–354
2. Preliminaries If ℓ is a natural number, then integers modulo ℓ are 1, 2, . . . , ℓ. Let D = (V , A) be a digraph. Then for short, we will write uv for an arc (u, v) ∈ A. If C = v1 , v2 , . . . , vℓ , v1 is a cycle in D and vi , vj ∈ V (C ), then vi C vj denotes the sequence vi , vi+1 , vi+2 , . . . , vj , where all indices are taken modulo ℓ. We will consider vi C vj both as a path and as a vertex set. If w ∈ V (C ), then wC− denotes the predecessor of w and wC+ denotes its successor on C . We recall that a cycle factor of a digraph D = (V , A) is a spanning subdigraph F = (V (F ), A(F )) of D with A(F ) ̸= ∅ such that every connected component of F is a cycle. First, we prove the following result. Theorem 2. Every d-regular digraph with d ≥ 4 contains two disjoint cycles of different lengths. Proof. We prove the statement of Theorem 2 by induction on d. If d = 4, then the statement of Theorem 2 has been proved by Henning and Yeo in [5]. Suppose that the statement of Theorem 2 has been proved to be true for all k-regular digraphs with 4 ≤ k < d and let D = (V , A) be a d-regular digraph. Since D is dregular, by [7] it contains a cycle factor F = (V (F ), A(F )). Then D′ = D − A(F ) is a (d − 1)-regular digraph. By the induction hypothesis, D′ contains two disjoint cycles C1 and C2 of different lengths. Since D′ is a subdigraph of D, the cycles C1 and C2 are also two disjoint cycles of different lengths in D. Theorem 2 is proved. For the proof of our main Theorem 1, we need the following result obtained in [6]. Proposition 3 ([6]). Let D = (V , A) be a d-arc-dominated oriented graph and let X be a subset of V such that D[X ] is either acyclic or an induced cycle of D. Then there exists a cycle C disjoint from D[X ] such that every vertex of C has at least one outneighbor in X . Now, we prove the following Proposition 4 which we will use frequently in the proof of our main Theorem 1. Proposition 4. Let C1 , C2 , . . . , Ct with t ≥ 2 be a sequence of mutually disjoint cycles of a digraph D = (V , A) such that for any i ∈ {1, 2, . . . , t − 1} every vertex of Ci+1 has at least one outneighbor in V (Ci ). Then (i) if there exists an arc from a vertex of V (Ci ) to a vertex of V (Cj ), j
where 1 ≤ i < j ≤ t, then D[∪m=i V (Cm )] contains two cycles of different lengths (not necessarily disjoint). j (ii) if a vertex x ̸∈ ∪m=i V (Cm ) has an outneighbor in V (Cj ) and an inneighbor in V (Ci ), where 1 ≤ i < j ≤ t, then j D[{x} ∪ (∪m=i V (Cm ))] contains two cycles of different lengths (not necessarily disjoint). Proof. Let ui uj be an arc in D, where ui ∈ V (Ci ) and uj ∈ V (Cj ). Further, let ux−1 be an outneighbor in V (Cx−1 ) of the vertex (ux )− Cx , where x = j, j − 1, . . . , i + 2. Then − − P = (ui )+ Ci Ci ui , uj Cj (uj )Cj , uj−1 Cj−1 (uj−1 )Cj−1 , − uj−2 Cj−2 (uj−2 )− Cj−2 , . . . , ui+1 Ci+1 (ui+1 )Ci+1 j
is a Hamilton path of D′ = D[∪m=i V (Cm )]. Since (ui+1 )− Ci+1 has at least two outneighbors in D′ , it is clear that D′ has two cycles of different lengths. Assertion (i) is proved. Assertion (ii) can be proved similarly. 3. Proof of Theorem 1 Suppose, on the contrary, that Theorem 1 is false and let D = (V , A) be a d-arc-dominated digraph of minimum size such that
every two disjoint cycles in D have the same length, where d ≥ 4. Below, we prove several claims to clarify the structure of D. Claim 5. Every vertex of D has outdegree d. Proof. Suppose, on the contrary, that a vertex u ∈ V has d+ D ( u) > d. For an outneighbor v of u, the digraph D − uv is a spanning darc-dominated subdigraph of D. Being a subdigraph of D, every two disjoint cycles in D − uv have the same length. This contradicts the minimality of size of D because D − uv is a proper subdigraph of D. Thus, every vertex of D must have outdegree d and Claim 5 is proved. Claim 6. The following hold in D: (i) D is an oriented graph. (ii) The subdigraph induced by the inneighborhood of every vertex of D contains a cycle of length at least 3. (iii) The girth of D is less than d. Proof. First, we prove Assertion (i). Suppose, on the contrary, that D is not an oriented graph. Let C = u, v, u be a cycle of length 2 in D. Then the digraph D′ = D[V \ {u, v}] is a subdigraph of D with minimum outdegree at least 2. Therefore, it is not difficult to see that D′ contains a cycle C ′ of length at least 3. We have got two disjoint cycles C and C ′ of different lengths in D, a contradiction. Thus, D must be an oriented graph. Next, we prove Assertion (ii). Let u be any vertex in D. Then by Claim 5, d+ D (u) = d ≥ 4. Let v be an outneighbor of u. Since D is d-arc-dominated, there exists a vertex w of outdegree d such that both w u ∈ A and wv ∈ A hold. This means that ND− (u) ̸= ∅. Further, for every vertex y ∈ ND− (u), there exists a vertex z of outdegree d in D such that both zy and zu are arcs of D. This means that z is an inneighbor of y in D[ND− (u)]. So, every vertex of ND− (u) is of indegree at least 1 in D[ND− (u)]. Then D[ND− (u)] contains a cycle of length at least 3 and Assertion (ii) is proved. Finally, we prove Assertion (iii). Let g be the girth of D. Then by [9], g ≤ d. If g = d, then D must be d-regular [9]. Therefore, by Theorem 2, D contains two disjoint cycles of different lengths, a contradiction. Thus, g < d. The proof of Claim 6 is complete. By Claim 6(iii), the girth g of D is less than d. It is clear that any cycle of length g is chordless, i.e., it is an induced cycle in D. Let C1 be any cycle of length g in D. Then by Proposition 3, there exists a cycle C2 disjoint from the cycle C1 such that every vertex of C2 has at least one outneighbor in V (C1 ). By our assumption about D, the cycle C2 also has length g. Moreover, we can prove the following. Claim 7. There are no arcs in D from V (C1 ) to V (C2 ). Proof. Suppose, on the contrary, that there is an arc from V (C1 ) to V (C2 ), say w1 w2 with w1 ∈ V (C1 ) and w2 ∈ V (C2 ). If D[V (C1 ) ∪ V (C2 )] = D, then every vertex of C1 (resp., of C2 ) is an outneighbor of every vertex of C2 (resp., of C1 ) because every vertex of D has d outneighbors, and both C1 and C2 are chordless and |V (C1 )| = |V (C2 )| = g < d. This contradicts the fact that D is oriented by Claim 6(i). Thus, D[V (C1 ) ∪ V (C2 )] ̸= D. By Proposition 4(i), D[V (C1 ) ∪ V (C2 )] contains two cycles C ′ and C ′′ of different lengths. Therefore, if D∗ = D[V \ (V (C1 ) ∪ V (C2 ))] contains a cycle C , then either C and C ′ or C and C ′′ are two disjoint cycles of different lengths in D, a contradiction. So, D∗ is acyclic. This implies that there exists a vertex u ∈ V (D∗ ) such that all d outneighbors of u are in V (C1 ) ∪ V (C2 ). Since d > g = |V (C1 )| = |V (C2 )|, both V (C1 ) and V (C2 ) have to contain an outneighbor of u and at least one of them, say V (C1 ), contains two outneighbors of u.
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Consider D[ND− (u)]. By Claim 6(ii), D[ND− (u)] contains a cycle T of length at least 3. Since D∗ is acyclic, T has to contain a vertex of V (C1 ) ∪ V (C2 ). Let V (T ) ∩ V (C1 ) ̸= ∅ and v1 ∈ V (T ) ∩ V (C1 ). Further, let u1 and u2 be two outneighbors of u in V (C1 ). Since D is oriented by Claim 6(i), both v1 ̸= u1 and v1 ̸= u2 hold. Then C ′ = u, u1 C1 v1 , u and C ′′ = u, u2 C1 v1 , u are two cycles of different lengths, which are disjoint from C2 . Therefore, either C ′ and C2 or C ′′ and C2 are two disjoint cycles of different lengths in D, a contradiction. Thus, V (T ) ∩ V (C1 ) = ∅ and therefore V (T ) ∩ V (C2 ) ̸= ∅. Let v2 ∈ V (T ) ∩ V (C2 ) and z be an outneighbor of u in V (C2 ). Then z ̸= v2 and z ̸= (v2 )+ T because D is oriented. Moreover, ∗ since C2 is chordless, (v2 )+ = u, zC2 v2 , u T ̸∈ zC2 v2 . Therefore, C and C ∗∗ = u, zC2 v2 , (v2 )+ , u are two cycles of different lengths, T which are disjoint from C1 . So, either C ∗ and C1 or C ∗∗ and C1 are two disjoint cycles of different lengths in D, a contradiction again. Thus, Claim 7 must be true. Since the cycle C2 has length g, it is an induced cycle in D. So, again by Proposition 3, there exists a cycle C3 in D disjoint from C2 such that every vertex of C3 has at least one outneighbor in C2 . Hence, the length of C3 is g by our assumption about D. The cycle C3 is also disjoint from C1 because there are no arcs from V (C1 ) to V (C2 ) by Claim 7. Thus, C1 , C2 and C3 are mutually disjoint cycles of the same length g. By continuing this process, we can define a sequence of cycles C1 , C2 , . . . , Cℓ in D with ℓ ≥ 3 such that: (1) C1 , C2 , . . . , Cℓ are mutually disjoint and all of them have length g; (2) for each i ∈ {1, 2, . . . , ℓ − 1}, every vertex in V (Ci+1 ) has at least one outneighbor in V (Ci ); and (3) for each i ∈ {1, 2, . . . , ℓ − 1}, there are no arcs from V (Ci ) to V (Ci+1 ). Subject to the above (1)–(3), we choose C1 , C2 , . . . , Cℓ so that ℓ is maximum. Consider the cycle Cℓ . Since Cℓ is an induced cycle of D, again by Proposition 3, there exists a cycle C disjoint from Cℓ such that every vertex of C has at least one outneighbor in Cℓ . By our assumption about D, C has length g. By the maximality of ℓ, V (C )∩ 1 (∪ℓ− m=1 V (Cm )) ̸= ∅. If there exists i ∈ {2, 3, . . . , ℓ − 1} such that V (C ) ∩ V (Ci ) ̸= ∅, then by Proposition 4(i) D[∪ℓm=i V (Cm )] contains two cycles C ′ and C ′′ of different lengths, which are disjoint from C1 . Therefore, either C ′ and C1 or C ′′ and C1 are two disjoint cycles of different lengths in D, a contradiction. Thus, ℓ−1 V (C ) ∩ (∪m =2 V (Cm )) = ∅ and therefore V (C ) ∩ V (C1 ) ̸= ∅.
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D[∪ℓm=1 V (Cm )] contains two cycles C ∗ and C ∗∗ of different lengths, which are disjoint from T . Therefore, either C ∗ and T or C ∗∗ and T are two disjoint cycles of different lengths in D, a contradiction. Thus, we must have V (T ) ∩ V (C1 ) ̸= ∅. First, suppose that v1 is also a vertex in V (T ) ∩ V (C1 ). Then + ′ (v1 )+ T ̸∈ V (C ) because C is chordless. Therefore, C = v1 , (v1 )T , xC v1 is a cycle. Moreover, it has length greater than g and is disjoint from Cℓ , a contradiction. Next, suppose that v1 is not a vertex in V (T ) ∩ V (C1 ). Let v2 ∈ V (T ) ∩ V (C1 ). Then v2 ̸= v1 . Now we go along the cycle C from the vertex x in the direction specified by the direction of its arcs. Since V (C ) ∩ V (C1 ) ̸= ∅, we must encounter some vertices of C1 that lie on C . Let y be the first vertex of C1 which we encounter by this travelling along C from x. Then C ∗ = v2 , xCy, y+ C1 C1 v2 is a cycle. If y = v1 , then C ∗ has length greater than g because v2 ̸= v1 . Since C ∗ is disjoint from Cℓ , this contradicts our assumption about D. Thus, y ̸= v1 . Then (v2 )+ T ̸∈ xCy because C is chordless. + Similarly, (v2 )+ ∈ ̸ y C v because C1 is chordless. Therefore, T C1 1 2 + C ∗∗ = v2 , (v2 )+ T , xCy, yC1 C1 v2 is a cycle with length greater than ∗ the length of C . It is clear that both C ∗ and C ∗∗ are disjoint from Cℓ . Therefore, either C ∗ and Cℓ or C ∗∗ and Cℓ are two disjoint cycles of different lengths in D. This final contradiction shows that Claim 8 must be true. By Claim 8, every vertex of C1 has at least one outneighbor in Cℓ . Therefore, we can apply Proposition 4 to any sequence Ci , Ci+1 , . . . , Cℓ , C1 , C2 , . . . , Ct , where 2 ≤ i ≤ ℓ and 1 ≤ t ≤ i − 1.
Claim 9. F = C1 ∪ C2 ∪ · · · ∪ Cℓ is a cycle factor of D. Proof. Since C1 , C2 , . . . , Cℓ are mutually disjoint cycles, it remains to show that V (F ) = V (D). We prove this by contradiction. So, suppose on the contrary that V (F ) ̸= V (D). Set D′ = D[V (D) \ V (F )]. We consider separately two cases. Case 1: D′ is acyclic. Then there exists a vertex x ∈ V (D′ ) such that all outneighbors of x are in V (F ). In fact, if |V (D′ )| = 1, then this is trivial. If |V (D′ )| > 1 and every vertex of V (D′ ) has an outneighbor in V (D′ ), then it is not difficult to see that D′ contains a cycle. This contradicts our assumption about D′ in this case. First, suppose that x has outneighbors in all three different cycles Ci , Cj and Ck . Without loss of generality, we may assume that 1 ≤ i < j < k ≤ ℓ. Consider ND− (x). If ND− (x) ∩ V (Ca ) ̸= ∅ for j
Claim 8. C = C1 . Proof. Suppose, on the contrary, that C ̸= C1 . Since V (C )∩V (C1 ) ̸= ∅ and C ̸= C1 , we can choose a vertex v1 ∈ V (C ) ∩ V (C1 ) such that x = (v1 )+ C ̸∈ V (C1 ). Since x is a vertex of C , it has an outneighbor uℓ ∈ V (Cℓ ). Consider ND− (x). Since C2 is disjoint from both Cℓ and C and every vertex of C has an outneighbor in Cℓ , by using Proposition 4(i), it is not difficult to see that there are no arcs from a vertex of Cℓ to a vertex of C . This means that ND− (x)∩ V (Cℓ ) = ∅. If ND− (x)∩ V (Ci ) ̸= ∅ for i ∈ {2, . . . , ℓ− 1}, then by Proposition 4(ii), D[{x} ∪ (∪ℓm=i V (Cm ))] contains two cycles C ′ and C ′′ of different lengths, which are disjoint from C1 . Therefore, either C ′ and C1 or C ′′ and C1 are two disjoint cycles of different lengths in D, a contradiction. Thus, we have ND− (x) ∩ V (Ci ) = ∅ for any i ∈ {2, . . . , ℓ − 1}. Therefore, ND− (x) ∩ (∪ℓm=2 V (Cm )) = ∅. By Claim 6(ii), D[ND− (x)] contains a cycle T of length at least 3. Since ND− (x) ∩ V (Cℓ ) = ∅, by our assumption about D, the cycle T must have length g. Suppose that V (T ) ∩ V (C1 ) = ∅. Since v1 is a vertex in V (C ) ∩ V (C1 ), v1 has an outneighbor in V (Cℓ ). Hence, by Proposition 4(i),
a ∈ {1, . . . , j − 1}, then by Proposition 4(ii) D[{x} ∪ (∪m=a V (Cm ))] contains two cycles C ′ and C ′′ of different lengths. Since both C ′ and C ′′ are disjoint from Ck , either C ′ and Ck or C ′′ and Ck are two disjoint cycles of different lengths in D, a contradiction. If ND− (x) ∩ V (Ca ) ̸= ∅ for a ∈ {j + 1, . . . , ℓ}, then by applying Proposition 4(ii) to the sequence Cj+1 , Cj+2 , . . . , Cℓ , C1 , . . . , Ci , . . . , Cj we see that D[{x} ∪ (∪ℓm=a V (Cm )) ∪ (∪im=1 V (Cm ))] contains two cycles C ′ and C ′′ of different lengths, which are disjoint from Cj . Therefore, either C ′ and Cj or C ′′ and Cj are two disjoint cycles of different lengths in D, a contradiction again. Finally, if ND− (x) ∩ V (Cj ) ̸= ∅, then by Proposition 4(ii) D[{x} ∪ (∪km=j V (Cm ))] contains two cycles C ′ and C ′′ of different lengths which are disjoint from Ci . Therefore, either C ′ and Ci or C ′′ and Ci are two disjoint cycles of different lengths in D, a contradiction. The above consideration shows that ND− (x) ∩ V (F ) = ∅, i.e., ND− (x) ⊆ V (D′ ). On the other hand, by Claim 6(ii) D[ND− (x)] contains a cycle of length at least 3. This contradicts the fact that D′ is acyclic in Case 1. Thus, x has outneighbors in at most two different cycles of the sequence C1 , C2 , . . . , Cℓ . But the number of outneighbors of x is d > g = |V (C1 )| = · · · = |V (Cℓ )|. Therefore, we can conclude that x has outneighbors in exactly two different cycles of the sequence C1 , C2 , . . . , Cℓ , say Ci and Cj with 1 ≤ i < j ≤ ℓ. Moreover, in
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at least one of these cycles the number of outneighbors of x is at least 2. We show now that inneighbors of x in V (F ), if any, can be only in V (Ci ) or V (Cj ). In fact, if x has an inneighbor in V (Ca ) with a ∈ {1, . . . , i − 1}, then by Proposition 4(ii) D[{x} ∪ (∪im=a V (Cm ))] has two cycles C ′ and C ′′ of different lengths, which are disjoint from Cj . Therefore, either C ′ and Cj or C ′′ and Cj are two disjoint cycles of different lengths in D, a contradiction. If x has an inneighbor in V (Ca ) with a ∈ {i + 1, . . . , j − 1}, then j again by Proposition 4(ii) D[{x} ∪ (∪m=a V (Cm ))] has two cycles C ′ ′′ and C of different lengths, which are disjoint from Ci . Therefore, either C ′ and Ci or C ′′ and Ci are two disjoint cycles of different lengths in D, a contradiction again. Finally, if x has an inneighbor in V (Ca ) with a ∈ {j + 1, . . . , ℓ}, then by applying Proposition 4(ii) to the sequence Cj+1 , Cj+2 , . . . , Cℓ , C1 , . . . , Ci , . . . , Cj we see that D[{x} ∪ (∪ℓm=a V (Cm )) ∪ (∪im=1 V (Cm ))] contains two cycles C ′ and C ′′ of different lengths, which are disjoint from Cj . Therefore, either C ′ and Cj or C ′′ and Cj are two disjoint cycles of different lengths in D, a contradiction again. Thus, ND+ (x) ⊆ V (Ci ) ∪ V (Cj ) and ND− (x) ⊆ V (D′ ) ∪ V (Ci ) ∪ V (Cj ). Moreover, as we have noted before, at least one of V (Ci ) and V (Cj ) contains two outneighbors of x. We consider separately these possibilities. First, suppose that x has two outneighbors in V (Ci ), say u1 and u2 . If x also has an inneighbor v ∈ V (Ci ), then both v ̸= u1 and v ̸= u2 hold because D is oriented. Further, C ∗ = x, u1 Ci v, x and C ∗∗ = x, u2 Ci v, x are two cycles of different lengths, which are disjoint from Cj . Therefore, either C ∗ and Cj or C ∗∗ and Cj are two disjoint cycles of different lengths in D, a contradiction. This implies that ND− (x) ⊆ V (D′ ) ∪ V (Cj ). Now, consider ND− (x). By Claim 6(ii), D[ND− (x)] contains a cycle T of length at least 3. Since D′ is acyclic, V (T ) ∩ V (Cj ) ̸= ∅. Let y ∈ V (T ) ∩ V (Cj ) and w be an outneighbor of x in V (Cj ). Since D is oriented, both y ̸= w + and y+ T ̸= w hold. Moreover, yT ̸∈ w Cj y because Cj is chordless. ∗ Therefore, T = x, w Cj y, x and T ∗∗ = x, w Cj y, y+ T , x are two cycles of different lengths in D[V (D′ ) ∪ V (Cj )]. Therefore, either T ∗ and Ci or T ∗∗ and Ci are two disjoint cycles of different lengths in D, a contradiction. Next, suppose that x has two outneighbors in V (Cj ). By arguments similar to those used in the previous paragraph we can see that D contains two disjoint cycles of different lengths, a contradiction again. Case 2: D′ contains a cycle C . Since every vertex of C1 has an outneighbor in V (Cℓ ), by Proposition 4(i) D[V (F )] contains two cycles C ′ and C ′′ of different lengths. Therefore, either C and C ′ or C and C ′′ are two disjoint cycles of different lengths in D, a contradiction. We have got a contradiction in both cases. Thus, Claim 9 must be true. Now, we complete the proof of Theorem 1. By Claim 9, F = C1 ∪ C2 ∪ · · · ∪ Cℓ is a cycle factor of D. Therefore, outneighbors of every vertex in D are in one of V (Cj ) with j ∈ {1, 2, . . . , ℓ}. For a vertex ui ∈ V (Ci ), it is clear that (ui )+ Ci is the only outneighbor in V (Ci ) of ui
because Ci is chordless. If ui ∈ V (Ci ) has an outneighbor v ∈ V (Cj ) with j ∈ {1, . . . , i − 2}, then by applying Proposition 4(i) to the sequence Ci , Ci+1 , . . . , Cℓ , C1 , . . . , Ci−1 we see that D[(∪ℓm=i V (Cm ))∪
(∪jm=1 V (Cm ))] contains two cycles C ′ and C ′′ of different lengths, which are disjoint from Ci−1 . Similarly, if ui ∈ V (Ci ) has an outneighbor v ∈ V (Cj ) with j ∈ {i + 1, . . . , ℓ}, then by Proposition 4(i) j D[(∪m=i V (Cm ))] contains two cycles C ′ and C ′′ of different lengths,
which are disjoint from Ci−1 . Therefore, in both situations, either C ′ and Ci−1 or C ′′ and Ci−1 are two disjoint cycles of different lengths in D, a contradiction. Thus, for any i ∈ {1, 2, . . . , ℓ} all outneighbors of every ui ∈ V (Ci ), which are not in V (Ci ), are in V (Ci−1 ), where indices are taken modulo ℓ. This implies, since every vertex of D has outdegree d, the cycles C1 , C2 , . . . , Cℓ are chordless and |V (C1 )| = |V (C2 )| = · · · = |V (Cℓ )| = g < d, that g = d − 1 and every vertex of V (Ci ) is an outneighbor of every vertex of V (Ci+1 ). Let V (Ci ) = {ui1 , ui2 , . . . , uig } and Ci = ui1 , ui2 , . . . , uig , ui1 . Then Q1 = uℓ1 , u1ℓ−1 , . . . , u21 , u11 , uℓ1 , Q2 = uℓ2 Cℓ uℓg , u2ℓ−1 Cℓ−1 ugℓ−1 , . . . , u22 C2 u2g , u12 C1 u1g , uℓ2 are two disjoint cycles of different lengths in D. This final contradiction shows that Theorem 1 must be true. The proof of Theorem 1 is complete. Remark. One of the anonymous referees of the paper has informed in his referee’s report that in a manuscript, submitted now to SIAM J. Discrete Math., Nicolas Lichiardopol has proved the conjecture of Henning and Yeo, mentioned in Introduction of this paper, in its totality. My result in this paper is obtained independently from the Lichiardopol’s one. Acknowledgements The author would like to thank the referees and area editor for their useful remarks and suggestions which helped me to improve the paper. References [1] J. Bang-Jensen, G. Gutin, Digraphs: Theory, Algorithms and Applications, Springer, London, 2001. [2] J.-C. Bermond, C. Thomassen, Cycles in digraphs—a survey, J. Graph Theory 5 (1981) 1–43. [3] P.J. Cameron, D.B. West, Research problems from the 20th British Combinatorial Conference, Discrete Math. 308 (2008) 621–630. [4] Y. Gao, D. Ma, Disjoint cycles with different length in 4-arc-dominated digraphs, Oper. Res. Lett. 41 (2013) 650–653. [5] M.A. Henning, A. Yeo, Vertex disjoint cycles of different length in digraphs, SIAM J. Discrete Math. 26 (2012) 687–694. [6] N. Lichiardopol, A. Pór, J.-S. Sereni, A step toward the Bermond–Thomassen conjecture about disjoint cycles in digraphs, SIAM J. Discrete Math. 23 (2009) 979–992. [7] O. Ore, Theory of Graphs, in: Amer. Math. Soc. Transl., vol. 38, AMS, Providence, RI, 1962. [8] N.D. Tan, 3-arc-dominated digraphs, SIAM J. Discrete Math. 24 (2010) 1153–1161. [9] N.D. Tan, On d-arc-dominated oriented graphs, Graphs Combin. (2013) http://dx.doi.org/10.1007/s00373-013-1313-0.
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Vertex disjoint cycles of different lengths in d-arc-dominated digraphs Ngo Dac Tan ∗ Institute of Mathematics, Vietnam Academy of Science and Technology, 18 Hoang Quoc Viet Road, 10307 Hanoi, Viet Nam
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Article history: Received 3 October 2013 Received in revised form 16 March 2014 Accepted 8 June 2014 Available online 14 June 2014
abstract Gao and Ma (2013) have proved that every 4-arc-dominated digraph contains two vertex disjoint directed cycles of different lengths. In this paper, by arguments similar to those used by Gao and Ma in the abovementioned paper, we show that every d-arc-dominated digraph with d ≥ 4 contains two vertex disjoint directed cycles of different lengths. © 2014 Elsevier B.V. All rights reserved.
Keywords: Vertex disjoint cycles Minimum outdegree d-regular digraph d-arc-dominated digraph
1. Introduction In this paper, the term digraph always means a finite simple digraph, i.e., a digraph that has a finite number of vertices, no loops and no multiple arcs. Unless otherwise indicated, our graphtheoretic terminology will follow [1]. Let D be a digraph. Then the vertex set and the arc set of D are denoted by V (D) and A(D) (or by V and A for short), respectively. The order and the size of D are |V | and |A|, respectively. A vertex v ∈ V is called an outneighbor of a vertex u ∈ V if (u, v) ∈ A. The set of all outneighbors of u, denoted by ND+ (u), is called the outneighborhood of u. The outdegree of u ∈ V , denoted by d+ D (u), is |ND+ (u)|. The number d = min{d+ D (u) | u ∈ V } is called the minimum outdegree of the digraph D. Similarly, a vertex w ∈ V is called an inneighbor of a vertex u ∈ V if (w, u) ∈ A. The set of all inneighbors of u, denoted by ND− (u), is called the inneighborhood of u. The − indegree of u ∈ V , denoted by d− D (u), is |ND (u)|. If W ⊆ V , then the subdigraph of D induced by W is denoted by D[W ]. By a cycle (resp., path) in a digraph D = (V , A) we always mean a directed cycle (resp., directed path). By disjoint cycles in D we always mean vertex disjoint cycles. A digraph D = (V , A) is called an oriented graph if D has no cycles of length 2. For a natural number d, a digraph D = (V , A) is − called d-regular if d+ D (u) = dD (u) = d for every vertex u ∈ V .
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http://dx.doi.org/10.1016/j.orl.2014.06.004 0167-6377/© 2014 Elsevier B.V. All rights reserved.
Let D = (V , A) be a digraph with minimum outdegree d. Then D is called d-arc-dominated if for every arc (x, y) ∈ A there is a vertex u ∈ V of outdegree d such that both (u, x) ∈ A and (u, y) ∈ A hold. At the 20th British Combinatorial Conference held at the University of Durham from 10–15, 2005, Lichiardopol posed the problem of characterizing d-arc-dominated oriented graphs (see Problem 912 (BCC20.4) in [3]). These digraphs are interesting for the Bermond–Thomassen conjecture, which states that a digraph with minimum outdegree d ≥ 2k − 1 contains k disjoint cycles (see [1] or [2]), because no arcs in them can be path-contracted (see [1] for a definition) to maintain the outdegree condition in the conjecture—thus minimal counterexamples are d-arc-dominated. Some partial results for the Lichiardopol’s problem have been obtained recently by Ngo Dac Tan in [8,9]. In [5], Henning and Yeo considered the problem about existence of two disjoint cycles of different lengths in digraphs. In particular, they have proved in [5] that every 4-regular digraph contains two disjoint cycles of different lengths. They also conjectured in [5] that a digraph with minimum outdegree at least 4 contains two disjoint cycles of different lengths. Recently in [4], Gao and Ma have verified this conjecture for 4-arc-dominated digraphs. In this paper, by using arguments similar to those used by Gao and Ma in [4], we will verify this conjecture of Henning and Yeo for any d-arc-dominated digraphs with d ≥ 4. Namely, we will prove the following result. Theorem 1. Every d-arc-dominated digraph with d ≥ 4 contains two disjoint cycles of different lengths.
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2. Preliminaries If ℓ is a natural number, then integers modulo ℓ are 1, 2, . . . , ℓ. Let D = (V , A) be a digraph. Then for short, we will write uv for an arc (u, v) ∈ A. If C = v1 , v2 , . . . , vℓ , v1 is a cycle in D and vi , vj ∈ V (C ), then vi C vj denotes the sequence vi , vi+1 , vi+2 , . . . , vj , where all indices are taken modulo ℓ. We will consider vi C vj both as a path and as a vertex set. If w ∈ V (C ), then wC− denotes the predecessor of w and wC+ denotes its successor on C . We recall that a cycle factor of a digraph D = (V , A) is a spanning subdigraph F = (V (F ), A(F )) of D with A(F ) ̸= ∅ such that every connected component of F is a cycle. First, we prove the following result. Theorem 2. Every d-regular digraph with d ≥ 4 contains two disjoint cycles of different lengths. Proof. We prove the statement of Theorem 2 by induction on d. If d = 4, then the statement of Theorem 2 has been proved by Henning and Yeo in [5]. Suppose that the statement of Theorem 2 has been proved to be true for all k-regular digraphs with 4 ≤ k < d and let D = (V , A) be a d-regular digraph. Since D is dregular, by [7] it contains a cycle factor F = (V (F ), A(F )). Then D′ = D − A(F ) is a (d − 1)-regular digraph. By the induction hypothesis, D′ contains two disjoint cycles C1 and C2 of different lengths. Since D′ is a subdigraph of D, the cycles C1 and C2 are also two disjoint cycles of different lengths in D. Theorem 2 is proved. For the proof of our main Theorem 1, we need the following result obtained in [6]. Proposition 3 ([6]). Let D = (V , A) be a d-arc-dominated oriented graph and let X be a subset of V such that D[X ] is either acyclic or an induced cycle of D. Then there exists a cycle C disjoint from D[X ] such that every vertex of C has at least one outneighbor in X . Now, we prove the following Proposition 4 which we will use frequently in the proof of our main Theorem 1. Proposition 4. Let C1 , C2 , . . . , Ct with t ≥ 2 be a sequence of mutually disjoint cycles of a digraph D = (V , A) such that for any i ∈ {1, 2, . . . , t − 1} every vertex of Ci+1 has at least one outneighbor in V (Ci ). Then (i) if there exists an arc from a vertex of V (Ci ) to a vertex of V (Cj ), j
where 1 ≤ i < j ≤ t, then D[∪m=i V (Cm )] contains two cycles of different lengths (not necessarily disjoint). j (ii) if a vertex x ̸∈ ∪m=i V (Cm ) has an outneighbor in V (Cj ) and an inneighbor in V (Ci ), where 1 ≤ i < j ≤ t, then j D[{x} ∪ (∪m=i V (Cm ))] contains two cycles of different lengths (not necessarily disjoint). Proof. Let ui uj be an arc in D, where ui ∈ V (Ci ) and uj ∈ V (Cj ). Further, let ux−1 be an outneighbor in V (Cx−1 ) of the vertex (ux )− Cx , where x = j, j − 1, . . . , i + 2. Then − − P = (ui )+ Ci Ci ui , uj Cj (uj )Cj , uj−1 Cj−1 (uj−1 )Cj−1 , − uj−2 Cj−2 (uj−2 )− Cj−2 , . . . , ui+1 Ci+1 (ui+1 )Ci+1 j
is a Hamilton path of D′ = D[∪m=i V (Cm )]. Since (ui+1 )− Ci+1 has at least two outneighbors in D′ , it is clear that D′ has two cycles of different lengths. Assertion (i) is proved. Assertion (ii) can be proved similarly. 3. Proof of Theorem 1 Suppose, on the contrary, that Theorem 1 is false and let D = (V , A) be a d-arc-dominated digraph of minimum size such that
every two disjoint cycles in D have the same length, where d ≥ 4. Below, we prove several claims to clarify the structure of D. Claim 5. Every vertex of D has outdegree d. Proof. Suppose, on the contrary, that a vertex u ∈ V has d+ D ( u) > d. For an outneighbor v of u, the digraph D − uv is a spanning darc-dominated subdigraph of D. Being a subdigraph of D, every two disjoint cycles in D − uv have the same length. This contradicts the minimality of size of D because D − uv is a proper subdigraph of D. Thus, every vertex of D must have outdegree d and Claim 5 is proved. Claim 6. The following hold in D: (i) D is an oriented graph. (ii) The subdigraph induced by the inneighborhood of every vertex of D contains a cycle of length at least 3. (iii) The girth of D is less than d. Proof. First, we prove Assertion (i). Suppose, on the contrary, that D is not an oriented graph. Let C = u, v, u be a cycle of length 2 in D. Then the digraph D′ = D[V \ {u, v}] is a subdigraph of D with minimum outdegree at least 2. Therefore, it is not difficult to see that D′ contains a cycle C ′ of length at least 3. We have got two disjoint cycles C and C ′ of different lengths in D, a contradiction. Thus, D must be an oriented graph. Next, we prove Assertion (ii). Let u be any vertex in D. Then by Claim 5, d+ D (u) = d ≥ 4. Let v be an outneighbor of u. Since D is d-arc-dominated, there exists a vertex w of outdegree d such that both w u ∈ A and wv ∈ A hold. This means that ND− (u) ̸= ∅. Further, for every vertex y ∈ ND− (u), there exists a vertex z of outdegree d in D such that both zy and zu are arcs of D. This means that z is an inneighbor of y in D[ND− (u)]. So, every vertex of ND− (u) is of indegree at least 1 in D[ND− (u)]. Then D[ND− (u)] contains a cycle of length at least 3 and Assertion (ii) is proved. Finally, we prove Assertion (iii). Let g be the girth of D. Then by [9], g ≤ d. If g = d, then D must be d-regular [9]. Therefore, by Theorem 2, D contains two disjoint cycles of different lengths, a contradiction. Thus, g < d. The proof of Claim 6 is complete. By Claim 6(iii), the girth g of D is less than d. It is clear that any cycle of length g is chordless, i.e., it is an induced cycle in D. Let C1 be any cycle of length g in D. Then by Proposition 3, there exists a cycle C2 disjoint from the cycle C1 such that every vertex of C2 has at least one outneighbor in V (C1 ). By our assumption about D, the cycle C2 also has length g. Moreover, we can prove the following. Claim 7. There are no arcs in D from V (C1 ) to V (C2 ). Proof. Suppose, on the contrary, that there is an arc from V (C1 ) to V (C2 ), say w1 w2 with w1 ∈ V (C1 ) and w2 ∈ V (C2 ). If D[V (C1 ) ∪ V (C2 )] = D, then every vertex of C1 (resp., of C2 ) is an outneighbor of every vertex of C2 (resp., of C1 ) because every vertex of D has d outneighbors, and both C1 and C2 are chordless and |V (C1 )| = |V (C2 )| = g < d. This contradicts the fact that D is oriented by Claim 6(i). Thus, D[V (C1 ) ∪ V (C2 )] ̸= D. By Proposition 4(i), D[V (C1 ) ∪ V (C2 )] contains two cycles C ′ and C ′′ of different lengths. Therefore, if D∗ = D[V \ (V (C1 ) ∪ V (C2 ))] contains a cycle C , then either C and C ′ or C and C ′′ are two disjoint cycles of different lengths in D, a contradiction. So, D∗ is acyclic. This implies that there exists a vertex u ∈ V (D∗ ) such that all d outneighbors of u are in V (C1 ) ∪ V (C2 ). Since d > g = |V (C1 )| = |V (C2 )|, both V (C1 ) and V (C2 ) have to contain an outneighbor of u and at least one of them, say V (C1 ), contains two outneighbors of u.
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Consider D[ND− (u)]. By Claim 6(ii), D[ND− (u)] contains a cycle T of length at least 3. Since D∗ is acyclic, T has to contain a vertex of V (C1 ) ∪ V (C2 ). Let V (T ) ∩ V (C1 ) ̸= ∅ and v1 ∈ V (T ) ∩ V (C1 ). Further, let u1 and u2 be two outneighbors of u in V (C1 ). Since D is oriented by Claim 6(i), both v1 ̸= u1 and v1 ̸= u2 hold. Then C ′ = u, u1 C1 v1 , u and C ′′ = u, u2 C1 v1 , u are two cycles of different lengths, which are disjoint from C2 . Therefore, either C ′ and C2 or C ′′ and C2 are two disjoint cycles of different lengths in D, a contradiction. Thus, V (T ) ∩ V (C1 ) = ∅ and therefore V (T ) ∩ V (C2 ) ̸= ∅. Let v2 ∈ V (T ) ∩ V (C2 ) and z be an outneighbor of u in V (C2 ). Then z ̸= v2 and z ̸= (v2 )+ T because D is oriented. Moreover, ∗ since C2 is chordless, (v2 )+ = u, zC2 v2 , u T ̸∈ zC2 v2 . Therefore, C and C ∗∗ = u, zC2 v2 , (v2 )+ , u are two cycles of different lengths, T which are disjoint from C1 . So, either C ∗ and C1 or C ∗∗ and C1 are two disjoint cycles of different lengths in D, a contradiction again. Thus, Claim 7 must be true. Since the cycle C2 has length g, it is an induced cycle in D. So, again by Proposition 3, there exists a cycle C3 in D disjoint from C2 such that every vertex of C3 has at least one outneighbor in C2 . Hence, the length of C3 is g by our assumption about D. The cycle C3 is also disjoint from C1 because there are no arcs from V (C1 ) to V (C2 ) by Claim 7. Thus, C1 , C2 and C3 are mutually disjoint cycles of the same length g. By continuing this process, we can define a sequence of cycles C1 , C2 , . . . , Cℓ in D with ℓ ≥ 3 such that: (1) C1 , C2 , . . . , Cℓ are mutually disjoint and all of them have length g; (2) for each i ∈ {1, 2, . . . , ℓ − 1}, every vertex in V (Ci+1 ) has at least one outneighbor in V (Ci ); and (3) for each i ∈ {1, 2, . . . , ℓ − 1}, there are no arcs from V (Ci ) to V (Ci+1 ). Subject to the above (1)–(3), we choose C1 , C2 , . . . , Cℓ so that ℓ is maximum. Consider the cycle Cℓ . Since Cℓ is an induced cycle of D, again by Proposition 3, there exists a cycle C disjoint from Cℓ such that every vertex of C has at least one outneighbor in Cℓ . By our assumption about D, C has length g. By the maximality of ℓ, V (C )∩ 1 (∪ℓ− m=1 V (Cm )) ̸= ∅. If there exists i ∈ {2, 3, . . . , ℓ − 1} such that V (C ) ∩ V (Ci ) ̸= ∅, then by Proposition 4(i) D[∪ℓm=i V (Cm )] contains two cycles C ′ and C ′′ of different lengths, which are disjoint from C1 . Therefore, either C ′ and C1 or C ′′ and C1 are two disjoint cycles of different lengths in D, a contradiction. Thus, ℓ−1 V (C ) ∩ (∪m =2 V (Cm )) = ∅ and therefore V (C ) ∩ V (C1 ) ̸= ∅.
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D[∪ℓm=1 V (Cm )] contains two cycles C ∗ and C ∗∗ of different lengths, which are disjoint from T . Therefore, either C ∗ and T or C ∗∗ and T are two disjoint cycles of different lengths in D, a contradiction. Thus, we must have V (T ) ∩ V (C1 ) ̸= ∅. First, suppose that v1 is also a vertex in V (T ) ∩ V (C1 ). Then + ′ (v1 )+ T ̸∈ V (C ) because C is chordless. Therefore, C = v1 , (v1 )T , xC v1 is a cycle. Moreover, it has length greater than g and is disjoint from Cℓ , a contradiction. Next, suppose that v1 is not a vertex in V (T ) ∩ V (C1 ). Let v2 ∈ V (T ) ∩ V (C1 ). Then v2 ̸= v1 . Now we go along the cycle C from the vertex x in the direction specified by the direction of its arcs. Since V (C ) ∩ V (C1 ) ̸= ∅, we must encounter some vertices of C1 that lie on C . Let y be the first vertex of C1 which we encounter by this travelling along C from x. Then C ∗ = v2 , xCy, y+ C1 C1 v2 is a cycle. If y = v1 , then C ∗ has length greater than g because v2 ̸= v1 . Since C ∗ is disjoint from Cℓ , this contradicts our assumption about D. Thus, y ̸= v1 . Then (v2 )+ T ̸∈ xCy because C is chordless. + Similarly, (v2 )+ ∈ ̸ y C v because C1 is chordless. Therefore, T C1 1 2 + C ∗∗ = v2 , (v2 )+ T , xCy, yC1 C1 v2 is a cycle with length greater than ∗ the length of C . It is clear that both C ∗ and C ∗∗ are disjoint from Cℓ . Therefore, either C ∗ and Cℓ or C ∗∗ and Cℓ are two disjoint cycles of different lengths in D. This final contradiction shows that Claim 8 must be true. By Claim 8, every vertex of C1 has at least one outneighbor in Cℓ . Therefore, we can apply Proposition 4 to any sequence Ci , Ci+1 , . . . , Cℓ , C1 , C2 , . . . , Ct , where 2 ≤ i ≤ ℓ and 1 ≤ t ≤ i − 1.
Claim 9. F = C1 ∪ C2 ∪ · · · ∪ Cℓ is a cycle factor of D. Proof. Since C1 , C2 , . . . , Cℓ are mutually disjoint cycles, it remains to show that V (F ) = V (D). We prove this by contradiction. So, suppose on the contrary that V (F ) ̸= V (D). Set D′ = D[V (D) \ V (F )]. We consider separately two cases. Case 1: D′ is acyclic. Then there exists a vertex x ∈ V (D′ ) such that all outneighbors of x are in V (F ). In fact, if |V (D′ )| = 1, then this is trivial. If |V (D′ )| > 1 and every vertex of V (D′ ) has an outneighbor in V (D′ ), then it is not difficult to see that D′ contains a cycle. This contradicts our assumption about D′ in this case. First, suppose that x has outneighbors in all three different cycles Ci , Cj and Ck . Without loss of generality, we may assume that 1 ≤ i < j < k ≤ ℓ. Consider ND− (x). If ND− (x) ∩ V (Ca ) ̸= ∅ for j
Claim 8. C = C1 . Proof. Suppose, on the contrary, that C ̸= C1 . Since V (C )∩V (C1 ) ̸= ∅ and C ̸= C1 , we can choose a vertex v1 ∈ V (C ) ∩ V (C1 ) such that x = (v1 )+ C ̸∈ V (C1 ). Since x is a vertex of C , it has an outneighbor uℓ ∈ V (Cℓ ). Consider ND− (x). Since C2 is disjoint from both Cℓ and C and every vertex of C has an outneighbor in Cℓ , by using Proposition 4(i), it is not difficult to see that there are no arcs from a vertex of Cℓ to a vertex of C . This means that ND− (x)∩ V (Cℓ ) = ∅. If ND− (x)∩ V (Ci ) ̸= ∅ for i ∈ {2, . . . , ℓ− 1}, then by Proposition 4(ii), D[{x} ∪ (∪ℓm=i V (Cm ))] contains two cycles C ′ and C ′′ of different lengths, which are disjoint from C1 . Therefore, either C ′ and C1 or C ′′ and C1 are two disjoint cycles of different lengths in D, a contradiction. Thus, we have ND− (x) ∩ V (Ci ) = ∅ for any i ∈ {2, . . . , ℓ − 1}. Therefore, ND− (x) ∩ (∪ℓm=2 V (Cm )) = ∅. By Claim 6(ii), D[ND− (x)] contains a cycle T of length at least 3. Since ND− (x) ∩ V (Cℓ ) = ∅, by our assumption about D, the cycle T must have length g. Suppose that V (T ) ∩ V (C1 ) = ∅. Since v1 is a vertex in V (C ) ∩ V (C1 ), v1 has an outneighbor in V (Cℓ ). Hence, by Proposition 4(i),
a ∈ {1, . . . , j − 1}, then by Proposition 4(ii) D[{x} ∪ (∪m=a V (Cm ))] contains two cycles C ′ and C ′′ of different lengths. Since both C ′ and C ′′ are disjoint from Ck , either C ′ and Ck or C ′′ and Ck are two disjoint cycles of different lengths in D, a contradiction. If ND− (x) ∩ V (Ca ) ̸= ∅ for a ∈ {j + 1, . . . , ℓ}, then by applying Proposition 4(ii) to the sequence Cj+1 , Cj+2 , . . . , Cℓ , C1 , . . . , Ci , . . . , Cj we see that D[{x} ∪ (∪ℓm=a V (Cm )) ∪ (∪im=1 V (Cm ))] contains two cycles C ′ and C ′′ of different lengths, which are disjoint from Cj . Therefore, either C ′ and Cj or C ′′ and Cj are two disjoint cycles of different lengths in D, a contradiction again. Finally, if ND− (x) ∩ V (Cj ) ̸= ∅, then by Proposition 4(ii) D[{x} ∪ (∪km=j V (Cm ))] contains two cycles C ′ and C ′′ of different lengths which are disjoint from Ci . Therefore, either C ′ and Ci or C ′′ and Ci are two disjoint cycles of different lengths in D, a contradiction. The above consideration shows that ND− (x) ∩ V (F ) = ∅, i.e., ND− (x) ⊆ V (D′ ). On the other hand, by Claim 6(ii) D[ND− (x)] contains a cycle of length at least 3. This contradicts the fact that D′ is acyclic in Case 1. Thus, x has outneighbors in at most two different cycles of the sequence C1 , C2 , . . . , Cℓ . But the number of outneighbors of x is d > g = |V (C1 )| = · · · = |V (Cℓ )|. Therefore, we can conclude that x has outneighbors in exactly two different cycles of the sequence C1 , C2 , . . . , Cℓ , say Ci and Cj with 1 ≤ i < j ≤ ℓ. Moreover, in
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at least one of these cycles the number of outneighbors of x is at least 2. We show now that inneighbors of x in V (F ), if any, can be only in V (Ci ) or V (Cj ). In fact, if x has an inneighbor in V (Ca ) with a ∈ {1, . . . , i − 1}, then by Proposition 4(ii) D[{x} ∪ (∪im=a V (Cm ))] has two cycles C ′ and C ′′ of different lengths, which are disjoint from Cj . Therefore, either C ′ and Cj or C ′′ and Cj are two disjoint cycles of different lengths in D, a contradiction. If x has an inneighbor in V (Ca ) with a ∈ {i + 1, . . . , j − 1}, then j again by Proposition 4(ii) D[{x} ∪ (∪m=a V (Cm ))] has two cycles C ′ ′′ and C of different lengths, which are disjoint from Ci . Therefore, either C ′ and Ci or C ′′ and Ci are two disjoint cycles of different lengths in D, a contradiction again. Finally, if x has an inneighbor in V (Ca ) with a ∈ {j + 1, . . . , ℓ}, then by applying Proposition 4(ii) to the sequence Cj+1 , Cj+2 , . . . , Cℓ , C1 , . . . , Ci , . . . , Cj we see that D[{x} ∪ (∪ℓm=a V (Cm )) ∪ (∪im=1 V (Cm ))] contains two cycles C ′ and C ′′ of different lengths, which are disjoint from Cj . Therefore, either C ′ and Cj or C ′′ and Cj are two disjoint cycles of different lengths in D, a contradiction again. Thus, ND+ (x) ⊆ V (Ci ) ∪ V (Cj ) and ND− (x) ⊆ V (D′ ) ∪ V (Ci ) ∪ V (Cj ). Moreover, as we have noted before, at least one of V (Ci ) and V (Cj ) contains two outneighbors of x. We consider separately these possibilities. First, suppose that x has two outneighbors in V (Ci ), say u1 and u2 . If x also has an inneighbor v ∈ V (Ci ), then both v ̸= u1 and v ̸= u2 hold because D is oriented. Further, C ∗ = x, u1 Ci v, x and C ∗∗ = x, u2 Ci v, x are two cycles of different lengths, which are disjoint from Cj . Therefore, either C ∗ and Cj or C ∗∗ and Cj are two disjoint cycles of different lengths in D, a contradiction. This implies that ND− (x) ⊆ V (D′ ) ∪ V (Cj ). Now, consider ND− (x). By Claim 6(ii), D[ND− (x)] contains a cycle T of length at least 3. Since D′ is acyclic, V (T ) ∩ V (Cj ) ̸= ∅. Let y ∈ V (T ) ∩ V (Cj ) and w be an outneighbor of x in V (Cj ). Since D is oriented, both y ̸= w + and y+ T ̸= w hold. Moreover, yT ̸∈ w Cj y because Cj is chordless. ∗ Therefore, T = x, w Cj y, x and T ∗∗ = x, w Cj y, y+ T , x are two cycles of different lengths in D[V (D′ ) ∪ V (Cj )]. Therefore, either T ∗ and Ci or T ∗∗ and Ci are two disjoint cycles of different lengths in D, a contradiction. Next, suppose that x has two outneighbors in V (Cj ). By arguments similar to those used in the previous paragraph we can see that D contains two disjoint cycles of different lengths, a contradiction again. Case 2: D′ contains a cycle C . Since every vertex of C1 has an outneighbor in V (Cℓ ), by Proposition 4(i) D[V (F )] contains two cycles C ′ and C ′′ of different lengths. Therefore, either C and C ′ or C and C ′′ are two disjoint cycles of different lengths in D, a contradiction. We have got a contradiction in both cases. Thus, Claim 9 must be true. Now, we complete the proof of Theorem 1. By Claim 9, F = C1 ∪ C2 ∪ · · · ∪ Cℓ is a cycle factor of D. Therefore, outneighbors of every vertex in D are in one of V (Cj ) with j ∈ {1, 2, . . . , ℓ}. For a vertex ui ∈ V (Ci ), it is clear that (ui )+ Ci is the only outneighbor in V (Ci ) of ui
because Ci is chordless. If ui ∈ V (Ci ) has an outneighbor v ∈ V (Cj ) with j ∈ {1, . . . , i − 2}, then by applying Proposition 4(i) to the sequence Ci , Ci+1 , . . . , Cℓ , C1 , . . . , Ci−1 we see that D[(∪ℓm=i V (Cm ))∪
(∪jm=1 V (Cm ))] contains two cycles C ′ and C ′′ of different lengths, which are disjoint from Ci−1 . Similarly, if ui ∈ V (Ci ) has an outneighbor v ∈ V (Cj ) with j ∈ {i + 1, . . . , ℓ}, then by Proposition 4(i) j D[(∪m=i V (Cm ))] contains two cycles C ′ and C ′′ of different lengths,
which are disjoint from Ci−1 . Therefore, in both situations, either C ′ and Ci−1 or C ′′ and Ci−1 are two disjoint cycles of different lengths in D, a contradiction. Thus, for any i ∈ {1, 2, . . . , ℓ} all outneighbors of every ui ∈ V (Ci ), which are not in V (Ci ), are in V (Ci−1 ), where indices are taken modulo ℓ. This implies, since every vertex of D has outdegree d, the cycles C1 , C2 , . . . , Cℓ are chordless and |V (C1 )| = |V (C2 )| = · · · = |V (Cℓ )| = g < d, that g = d − 1 and every vertex of V (Ci ) is an outneighbor of every vertex of V (Ci+1 ). Let V (Ci ) = {ui1 , ui2 , . . . , uig } and Ci = ui1 , ui2 , . . . , uig , ui1 . Then Q1 = uℓ1 , u1ℓ−1 , . . . , u21 , u11 , uℓ1 , Q2 = uℓ2 Cℓ uℓg , u2ℓ−1 Cℓ−1 ugℓ−1 , . . . , u22 C2 u2g , u12 C1 u1g , uℓ2 are two disjoint cycles of different lengths in D. This final contradiction shows that Theorem 1 must be true. The proof of Theorem 1 is complete. Remark. One of the anonymous referees of the paper has informed in his referee’s report that in a manuscript, submitted now to SIAM J. Discrete Math., Nicolas Lichiardopol has proved the conjecture of Henning and Yeo, mentioned in Introduction of this paper, in its totality. My result in this paper is obtained independently from the Lichiardopol’s one. Acknowledgements The author would like to thank the referees and area editor for their useful remarks and suggestions which helped me to improve the paper. References [1] J. Bang-Jensen, G. Gutin, Digraphs: Theory, Algorithms and Applications, Springer, London, 2001. [2] J.-C. Bermond, C. Thomassen, Cycles in digraphs—a survey, J. Graph Theory 5 (1981) 1–43. [3] P.J. Cameron, D.B. West, Research problems from the 20th British Combinatorial Conference, Discrete Math. 308 (2008) 621–630. [4] Y. Gao, D. Ma, Disjoint cycles with different length in 4-arc-dominated digraphs, Oper. Res. Lett. 41 (2013) 650–653. [5] M.A. Henning, A. Yeo, Vertex disjoint cycles of different length in digraphs, SIAM J. Discrete Math. 26 (2012) 687–694. [6] N. Lichiardopol, A. Pór, J.-S. Sereni, A step toward the Bermond–Thomassen conjecture about disjoint cycles in digraphs, SIAM J. Discrete Math. 23 (2009) 979–992. [7] O. Ore, Theory of Graphs, in: Amer. Math. Soc. Transl., vol. 38, AMS, Providence, RI, 1962. [8] N.D. Tan, 3-arc-dominated digraphs, SIAM J. Discrete Math. 24 (2010) 1153–1161. [9] N.D. Tan, On d-arc-dominated oriented graphs, Graphs Combin. (2013) http://dx.doi.org/10.1007/s00373-013-1313-0.