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Every cycle-connected multipartite tournament has a universal arc
Discrete Mathematics 309 (2009) 1013–1017 www.elsevier.com/locate/disc
Note
Every cycle-connected multipartite tournament has a universal arc Lutz Volkmann ∗ , Stefan Winzen Lehrstuhl II f¨ur Mathematik, RWTH Aachen, 52056 Aachen, Germany Received 14 May 2007; received in revised form 17 January 2008; accepted 17 January 2008 Available online 4 March 2008
Abstract A digraph D is cycle-connected if for every pair of vertices u, v ∈ V (D) there exists a directed cycle in D containing both u ´ am [A. Ad´ ´ am, On some cyclic connectivity properties of directed graphs, Acta Cybernet. 14 (1) (1999) 1–12] and v. In 1999, Ad´ posed the following problem. Let D be a cycle-connected digraph. Does there exist a universal arc in D, i.e., an arc e ∈ A(D) such that for every vertex w ∈ V (D) there is a directed cycle in D containing both e and w? A c-partite or multipartite tournament is an orientation of a complete c-partite graph. Recently, Hubenko [A. Hubenko, On a cyclic connectivity property of directed graphs, Discrete Math. 308 (2008) 1018–1024] proved that each cycle-connected bipartite tournament has a universal arc. As an extension of this result, we show in this note that each cycle-connected multipartite tournament has a universal arc. c 2008 Elsevier B.V. All rights reserved.
Keywords: Multipartite tournament; Universal arc; Cycle-connected digraph
1. Terminology and introduction In this paper all digraphs are finite without loops or multiple arcs. The vertex set and arc set of a digraph D are denoted by V (D) and A(D), respectively. If x y is an arc of a digraph D, then we write x → y and say x dominates y. If X and Y are two disjoint subsets of V (D) such that every vertex of X dominates every vertex of Y , then we say that X dominates Y , denoted by X → Y . We call the in-degree of a vertex x the number of vertices that dominate x and the out-degree the number of vertices dominated by x. If δ + = δ + (D) and δ − = δ − (D) are the minimum out-degree and minimum in-degree of D, then δ = δ(D) = min{δ + , δ − } is called the minimum degree of D. By a cycle or path we mean a directed cycle or path. Let C be a cycle and x ∈ V (C). We say that the vertex x + is the successor of x on C if x x + ∈ A(C) and the vertex x − is a predecessor of x on C if x − x ∈ A(C). We call a pair of vertices u, v ∈ V (D) cyclic if there exits a cycle in D containing both u and v. An arc e ∈ A(D) is universal if for every vertex w ∈ V (D) there exists a cycle in D containing both e and w. A digraph is called cycle-connected if any ´ am [1] introduced the following problem. two vertices are cyclic. In 1999, Ad´ ´ am [1] 1999). Let D be a cycle-connected digraph. Does there exist a universal arc in D? Problem (Ad´ ∗ Corresponding author.
E-mail addresses: [email protected] (L. Volkmann), [email protected] (S. Winzen). c 2008 Elsevier B.V. All rights reserved. 0012-365X/$ - see front matter doi:10.1016/j.disc.2008.01.032
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A c-partite or multipartite tournament is an orientation of a complete c-partite graph. Recently, Hubenko [4] proved that each cycle-connected bipartite tournament has a universal arc. As an extension of this result, we show in this note that each cycle-connected multipartite tournament has a universal arc. 2. Preliminary results We start with a well-known result on longest cycles in multipartite tournaments by J. Ayel, which was mentioned in an article of Jackson [5]. A proof of Ayel’s theorem can be found in the perspective paper of Volkmann [6]. Theorem 2.1 (Ayel (cf. [5] 1981)). Let D be a strongly connected multipartite tournament. If C is a longest cycle of D, then D − V (C) contains no cycle. Also the next result on strongly connected multipartite tournaments is well-known. Theorem 2.2 (Guo, Pinkernell, Volkmann [3] 1997). Let D be a non-Hamiltonian strongly connected multipartite tournament with a longest cycle C. Then any vertex x ∈ V (D) − V (C) dominates a vertex of C and is dominated by a vertex of C. Note that our proofs works along the same lines as the proofs in the interesting article by Hubenko [4], but we use longest cycles instead of maximal cycles. By the way, also Hubenko [4] needs longest cycles in the proof of her Lemma 3.10. Observation 2.3. Let D be a digraph containing a cycle C with |V (C)| ≥ 4. Suppose that the four vertices a1 , a2 , a3 , a4 ∈ V (C) are situated on C in this order (not necessarily consecutive). If there is a vertex x ∈ V (D) − V (C) such that x → {a2 , a4 } and {a1 , a3 } → x, then for every arc e of C there is a cycle containing both e and x. Proof. The cycles xa2 Ca1 x and xa4 Ca3 x demonstrate that x is on a cycle with every edge of C.
In the following we assume that D is a strongly connected multipartite tournament with δ(D) ≥ 2, and C is a longest cycle in D. In [4], Hubenko defined so called good vertices and bad vertices. A vertex x ∈ V (D) − V (C) is good if there exist vertices a1 , a2 , a3 , a4 ∈ V (C) with the property of Observation 2.3 and bad otherwise. Clearly, if every vertex in V (D) − V (C) is good, then it follows from Observation 2.3 that every arc of C is universal. The following lemma demonstrates then the direction change between x and C occurs exactly two times. Lemma 2.4. Let x ∈ V (D) − V (C) be a bad vertex. Then there exist y, z ∈ V (C) such that the following hold: (1) (2) (3) (4)
y is in the same partite set as x. y − → x and x → y + . Every vertex on C between y + and z is either dominated by x or is in the same partite set as x. Every vertex on C between z + and y − either dominates x or is in the same partite set as x.
Proof. Let C = u 1 u 2 . . . u k u 1 be a longest cycle of the strongly connected multipartite tournament D. In view of Theorem 2.2, there exists a vertex u i , say u 1 , such that u 1 → x. Applying Theorem 2.2 once more, we notice that x has also an outer neighbor in V (C). Let ` = min{2 ≤ i ≤ k|x → u i }. If u `−1 → x, then D contains the longer cycle u 1 u 2 . . . u `−1 xu ` u `+1 . . . u k u 1 , a contradiction. Hence we conclude that u `−1 and x belong to the same partite set and u `−2 → x. If we set u `−1 = y, then we observe that (1) and (2) are proved. If there is no vertex z on C as described in (3) or (4), then Observation 2.3 is satisfied, making x a good vertex. This contradiction completes the proof of Lemma 2.4. Following the notation in [4], we call a bad vertex x a covering vertex for the arcs y − y, yy + and for the vertices y, y + , if y is as in Lemma 2.4. A covering set is a set of covering vertices. Hubenko proved the following lemma for a cycle-connected bipartite tournament D with a maximal cycle C. Analyzing the proof it is easy to see that the assertion of this lemma is also valid for a strongly connected multipartite tournament D with a longest cycle C. y−,
Lemma 2.5. If an arc e ∈ A(C) is not universal, then there is a bad vertex that covers it.
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Proof. Assume that the arc uv ∈ A(C) is not universal. Consider the path P = vCu. If a vertex z ∈ V (D) − V (C) is dominated by some vertex v1 ∈ V (P) and dominates a vertex v2 ∈ V (P) such that v2 occurs on the path P after v1 , then z is on the cycle zv2 Cv1 z with uv. So there must be a vertex x such that all vertices of P dominated by x precede (on P) all vertices of P that dominates x. Clearly, x covers uv. Remark 2.6. According to the proof of Lemma 2.4, if x is a cover vertex for y − , y, y + , then x and y belong to the same partite set. Moreover, if x covers more than three vertices, then it is straightforward to see that x fulfills the conditions of Observation 2.3 and thus x is a good vertex, a contradiction. Consequently, each bad vertex covers exactly three vertices. On the other hand, according to Lemma 2.4, each arc of C has to be covered by at least one bad vertex, if C does not contain an universal arc. Lemma 2.7. Let C = v1 v2 v3 v4 v5 . . . vk v1 be a longest cycle in D, and let a1 , a2 ∈ V (D) − V (C) such that a1 covers v1 , v2 , v3 , and a2 covers v3 , v4 , v5 . (1) If a1 and a2 are in the same partite set, then a1 → v5 . (2) If a2 → a1 , then a1 → v4 . Proof. If a1 and a2 are in the same partite set, then a1 and v5 are in different partite sets. If we assume that v5 → a1 , then Lemma 2.4 implies that v3 is the only vertex on C dominated by a1 . Since δ(D) ≥ 2, there exists a vertex z ∈ V (D) − V (C) such that a1 → z. Applying Theorem 2.2, there is a vertex u ∈ V (C) such that z → u. If u 6= v4 , v5 , then we observe that u − → a1 or u −− → a1 . Substituting u − u by u − a1 zu or u −− u − u by u −− a1 zu, we increase the length of C, a contradiction. This yields v2 → z. Assume next that u = v4 . If a2 → z, then substituting v3 v4 by v3 a2 zv4 , we increase the length of C, a contradiction. Thus z → a2 . Since δ(D) ≥ 2, there exists a vertex y 6= v5 such that a2 → y. If y ∈ V (C), then y 6= v2 , v3 , v4 , and we deduce that y − → a1 or y −− → a1 . Substituting y − y by y − a1 za2 y or y −− y − y by y −− a1 za2 y, we increase the length of C, a contradiction. This yields y ∈ V (D) − V (C). According to Theorem 2.2, there exists a vertex v ∈ V (C) such that y → v. However, now we observe that v − → a1 , v −− → a1 , v −−− → a1 or v −−−− → a1 . Substituting v − v by v − a1 za2 yv, v −− v − v by v −− a1 za2 yv, v −−− v −− v − v by v −−− a1 za2 yv or v −−−− v −−− v −− v − v by v −−−− a1 za2 yv, we increase the length of C, a contradiction. This leads to v4 → z and u = v5 and so z → v5 . Substituting v4 v5 by v4 zv5 , we increase the length of C. This contradiction completes the proof of (1). If a2 → a1 , then a1 and v4 are in different partite sets. If we assume that v4 → a1 , then Lemma 2.4 again implies that v3 is the only vertex on C dominated by a1 . As above, δ(D) ≥ 2 gives us z ∈ V (D) − V (C) with a1 → z. Theorem 2.2 implies that z → v for some v ∈ V (C). If v 6= v4 , then either v − → a1 or v −− → a1 . We can either replace v − v with v − a1 zv or v −− v − v with v −− a1 zv to create a cycle longer than C, a contradiction. If z → v4 , we replace v3 v4 with v3 a2 a1 za4 to create a cycle longer than C, another contradiction. This completes the proof of (2). Lemma 2.8. Let ck dk ck−1 dk−1 . . . c2 d2 c1 d1 b1 g1 b2 . . . bl gl be an interval of C. Suppose that the vertices ck , dk , ck−1 , dk−1 , . . . , c1 , d1 , b1 are covered by vertices f k , f k−1 , . . . , f 1 such that f k covers ck , dk , ck−1 ; f k−1 covers ck−1 , dk−1 , ck−2 ; . . .; f 2 covers c2 , d2 , c1 ; f 1 covers c1 , d1 , b1 . Denote F = { f k , f k−1 , . . . , f 1 }. Suppose that the vertices d1 , b1 , g1 , b2 , g2 , . . . , bl , gl are covered by vertices a1 , a2 , . . . , al such that a1 covers d1 , b1 , g1 ; a2 covers g1 , b2 , g2 ; . . .; al covers gl−1 , bl , gl . Denote A = {a1 , a2 , . . . , al }. Finally, suppose that F is a monochromatic covering set (that means that F is contained in one partite set) and A is another monochromatic covering set. Then F → A. Proof. First we will prove by induction that f 1 → a j for j ∈ {1, 2, . . . , l}. For j = 1 we have f 1 → a1 , because otherwise, we could replace the arc d1 b1 with the path d1 a1 f 1 b1 that would increase the length of C, a contradiction. Provided, that we already know that f 1 → a j for every j < r , suppose on the contrary that ar → f 1 . From the definition of a covering vertex and Lemma 2.7 follows that ar −1 → gr and therefore we can replace the path gr −1 br gr with gr −1 ar f 1 ar −1 gr and increase the length of C, a contradiction. Now we will prove by induction that f i → a j for every i ∈ {1, 2, . . . , k} and j ∈ {1, 2, . . . , l}. The case i = 1 and j ∈ {1, 2, . . . , l} is shown above. Let q be the smallest index such that aq → fr . Using the same argument, from the definition of a covering vertex and Lemma 2.7 it follows that fr → cr −2 (in the case r = 2: f 2 → b1 ) and therefore
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we can replace the path cr −1 dr −1 cr −2 (in the case r = 2: c1 d1 b1 ) with cr −1 fr −1 aq fr cr −2 (in the case r = 2 with c1 f 1 aq f 2 b1 ) and increase the length of C, a contradiction. Lemma 2.9. Let ck dk ck−1 dk−1 . . . c2 d2 c1 d1 b1 g1 b2 . . . bl gl bl+1 be an interval of the cycle C. Suppose that the vertices ck , dk , ck−1 , dk−1 , . . . , c1 , d1 , b1 are covered by the vertices f k , f k−1 , . . . , f 1 such that f k covers ck , dk , ck−1 ; f k−1 covers ck−1 , dk−1 , ck−2 ; . . .; f 2 covers c2 , d2 , c1 ; f 1 covers c1 , d1 , b1 . Denote F = { f k , f k−1 , . . . , f 1 }. Suppose that the vertices b1 , g1 , b2 , g2 , . . . , bl , gl , bl+1 are covered by vertices a1 , a2 , . . . , al such that a1 covers b1 , g1 , b2 ; a2 covers b2 , g2 , b3 ; . . .; al covers bl , gl , bl+1 . Denote A = {a1 , a2 , . . . , al }. Finally, suppose that F is a monochromatic covering set and A is another monochromatic covering set. Then F → A. Proof. First we will prove by induction that f 1 → a j for j ∈ {1, 2, . . . , l}. For j = 1 suppose that a1 → f 1 . Since f 1 is a covering vertex, Lemma 2.7 implies that f 1 → g1 . Now we can replace the arc b1 g1 with the path b1 a1 f 1 g1 that would increase the length of C, a contradiction. Hence we deduce that f 1 → a1 . Provided, that we already know that f 1 → a j for every j < r , suppose on the contrary that ar → f 1 . From the definition of a covering vertex and Lemma 2.7 follows that ar −1 → br +1 and therefore we can replace the path br gr br +1 with br ar f 1 ar −1 br +1 and increase the length of C, a contradiction. Now we will prove by induction that f i → a j for every i ∈ {1, 2, . . . , k} and j ∈ {1, 2, . . . , l}. The case i = 1 and j ∈ {1, 2, . . . , l} is shown above. Let q be the smallest index such that aq → fr . Using the same argument, from the definition of a covering vertex and Lemma 2.7 it follows that fr → cr −2 (in the case r = 2: f 2 → b1 ) and therefore we can replace the path cr −1 dr −1 cr −2 (in the case r = 2: c1 d1 b1 ) with cr −1 fr −1 aq fr cr −2 (in the case r = 2 with c1 f 1 aq f 2 b1 ) and increase the length of C, a contradiction. Lemma 2.10. All vertices that are elements of a covering set of a cycle C belong to the same partite set. Proof. Assume that there are vertices x, y ∈ V (D) − V (C) from different partite sets that cover some vertices of C. Let us choose a minimal set M of covering vertices such that it covers all vertices of C and x, y ∈ M. Let us partition the cycle C into intervals, each covered with monochromatic vertices from M, such that any two consecutive intervals are covered with different colors. Observe that as we go around C, according to Lemmas 2.8 and 2.9, the covering vertices of each interval dominate the covering vertices of the next interval. Thus there is a cycle in the subdigraph of covering vertices, which is in D − V (C), a contradiction to Theorem 2.1. Lemma 2.11. If there is a monochromatic covering set for C, then every arc of C is universal. Proof. Let M be a set of covering vertices for C such that all vertices of M are from the same partite set, and let x ∈ M be arbitrary such that x covers the path y − yy + . The cycle x y + C y − x shows that x is on a cycle with every arc of C except y − y and yy + . We will construct a cycle C 0 that contains x and the arcs y − y, yy + . Since x is a bad vertex, C a longest cycle and there is a monochromatic covering set for C, Lemma 2.4 implies that there is an interval z − zz + of C such that z + → x, x → z − and z and x are in the same partite set. Let us use the notations z 1 = z −− and z 2 = z ++ . Denote the covering vertex of z + , z 2 and z 2+ by u, the covering vertex of z − , z and z + by v, and the covering vertex of z 1− , z 1 and z − by w. From Lemma 2.7 it follows that v → z 2+ and w → z + . Denote by C 0 the cycle z + x z − vz 2+ C z 1− wz + . Clearly, C 0 contains x, y − y and yy + which completes the proof. 3. Main results Theorem 3.1. Let D be a strongly connected multipartite tournament with δ(D) ≥ 2. Then every longest cycle of D has a universal arc. Proof. Assume on the contrary, that there is no universal arc on the cycle C. Then, by Lemma 2.5, all the arcs of C are covered. By Lemma 2.10, all the covering vertices of C are from the same partite set. It follows from Lemma 2.11 that all the arcs of C are universal, a contradiction. This completes the proof of this theorem. The following example of Bondy [2] shows that Theorem 3.1 is not valid in general for δ(D) = 1. Example 3.2. Let A1 , A2 , . . . , Ac be the partite sets of a c-partite tournament H such that |A1 | = 1, say A1 = {a1 } and |Ai | ≥ 2 for 2 ≤ i ≤ c. If A2 → a1 , a1 → Ai for 3 ≤ i ≤ c and A j → Ai for 2 ≤ i < j ≤ c, then H is strong with δ(H ) = 1, and the longest cycle of H has length c. It is easy to see that H does not contain a universal arc.
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We notice that the multipartite tournament H in Example 3.2 is not cycle-connected, since two vertices ai , bi ∈ Ai for 2 ≤ i ≤ c are not cyclic. However, if a multipartite tournament is cycle-connected, then it contains a universal arc. Theorem 3.3. If D is a cycle-connected multipartite tournament, then D contains a universal arc. Proof. In the case δ(D) = 1 we assume, without loss of generality, that x has out-degree 1 such that x → y. Thus the arc x y is contained in all cycles containing x. Since D is cycle-connected, the arc x y is universal. In the case δ(D) ≥ 2, the desired result follows from Theorem 3.1, since D is strongly connected. Acknowledgements We would like to thank one of the referees for valuable suggestions and comments. References [1] [2] [3] [4] [5] [6]
´ am, On some cyclic connectivity properties of directed graphs, Acta Cybernet. 14 (1) (1999) 1–12. A. Ad´ J.A. Bondy, Disconnected orientation and a conjecture of Las Vergnas, J. London Math. Soc. 14 (1976) 277–282. Y. Guo, A. Pinkernell, L. Volkmann, On cycles through a given vertex in multipartite tournaments, Discrete Math. 164 (1997) 165–170. A. Hubenko, On a cyclic connectivity property of directed graphs, Discrete Math. 308 (2008) 1018–1024. B. Jackson, Long paths and cycles in oriented graphs, J. Graph Theory 5 (1981) 145–157. L. Volkmann, Cycles in multipartite tournaments: Results and problems, Discrete Math. 245 (2002) 19–53.
Note
Every cycle-connected multipartite tournament has a universal arc Lutz Volkmann ∗ , Stefan Winzen Lehrstuhl II f¨ur Mathematik, RWTH Aachen, 52056 Aachen, Germany Received 14 May 2007; received in revised form 17 January 2008; accepted 17 January 2008 Available online 4 March 2008
Abstract A digraph D is cycle-connected if for every pair of vertices u, v ∈ V (D) there exists a directed cycle in D containing both u ´ am [A. Ad´ ´ am, On some cyclic connectivity properties of directed graphs, Acta Cybernet. 14 (1) (1999) 1–12] and v. In 1999, Ad´ posed the following problem. Let D be a cycle-connected digraph. Does there exist a universal arc in D, i.e., an arc e ∈ A(D) such that for every vertex w ∈ V (D) there is a directed cycle in D containing both e and w? A c-partite or multipartite tournament is an orientation of a complete c-partite graph. Recently, Hubenko [A. Hubenko, On a cyclic connectivity property of directed graphs, Discrete Math. 308 (2008) 1018–1024] proved that each cycle-connected bipartite tournament has a universal arc. As an extension of this result, we show in this note that each cycle-connected multipartite tournament has a universal arc. c 2008 Elsevier B.V. All rights reserved.
Keywords: Multipartite tournament; Universal arc; Cycle-connected digraph
1. Terminology and introduction In this paper all digraphs are finite without loops or multiple arcs. The vertex set and arc set of a digraph D are denoted by V (D) and A(D), respectively. If x y is an arc of a digraph D, then we write x → y and say x dominates y. If X and Y are two disjoint subsets of V (D) such that every vertex of X dominates every vertex of Y , then we say that X dominates Y , denoted by X → Y . We call the in-degree of a vertex x the number of vertices that dominate x and the out-degree the number of vertices dominated by x. If δ + = δ + (D) and δ − = δ − (D) are the minimum out-degree and minimum in-degree of D, then δ = δ(D) = min{δ + , δ − } is called the minimum degree of D. By a cycle or path we mean a directed cycle or path. Let C be a cycle and x ∈ V (C). We say that the vertex x + is the successor of x on C if x x + ∈ A(C) and the vertex x − is a predecessor of x on C if x − x ∈ A(C). We call a pair of vertices u, v ∈ V (D) cyclic if there exits a cycle in D containing both u and v. An arc e ∈ A(D) is universal if for every vertex w ∈ V (D) there exists a cycle in D containing both e and w. A digraph is called cycle-connected if any ´ am [1] introduced the following problem. two vertices are cyclic. In 1999, Ad´ ´ am [1] 1999). Let D be a cycle-connected digraph. Does there exist a universal arc in D? Problem (Ad´ ∗ Corresponding author.
E-mail addresses: [email protected] (L. Volkmann), [email protected] (S. Winzen). c 2008 Elsevier B.V. All rights reserved. 0012-365X/$ - see front matter doi:10.1016/j.disc.2008.01.032
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A c-partite or multipartite tournament is an orientation of a complete c-partite graph. Recently, Hubenko [4] proved that each cycle-connected bipartite tournament has a universal arc. As an extension of this result, we show in this note that each cycle-connected multipartite tournament has a universal arc. 2. Preliminary results We start with a well-known result on longest cycles in multipartite tournaments by J. Ayel, which was mentioned in an article of Jackson [5]. A proof of Ayel’s theorem can be found in the perspective paper of Volkmann [6]. Theorem 2.1 (Ayel (cf. [5] 1981)). Let D be a strongly connected multipartite tournament. If C is a longest cycle of D, then D − V (C) contains no cycle. Also the next result on strongly connected multipartite tournaments is well-known. Theorem 2.2 (Guo, Pinkernell, Volkmann [3] 1997). Let D be a non-Hamiltonian strongly connected multipartite tournament with a longest cycle C. Then any vertex x ∈ V (D) − V (C) dominates a vertex of C and is dominated by a vertex of C. Note that our proofs works along the same lines as the proofs in the interesting article by Hubenko [4], but we use longest cycles instead of maximal cycles. By the way, also Hubenko [4] needs longest cycles in the proof of her Lemma 3.10. Observation 2.3. Let D be a digraph containing a cycle C with |V (C)| ≥ 4. Suppose that the four vertices a1 , a2 , a3 , a4 ∈ V (C) are situated on C in this order (not necessarily consecutive). If there is a vertex x ∈ V (D) − V (C) such that x → {a2 , a4 } and {a1 , a3 } → x, then for every arc e of C there is a cycle containing both e and x. Proof. The cycles xa2 Ca1 x and xa4 Ca3 x demonstrate that x is on a cycle with every edge of C.
In the following we assume that D is a strongly connected multipartite tournament with δ(D) ≥ 2, and C is a longest cycle in D. In [4], Hubenko defined so called good vertices and bad vertices. A vertex x ∈ V (D) − V (C) is good if there exist vertices a1 , a2 , a3 , a4 ∈ V (C) with the property of Observation 2.3 and bad otherwise. Clearly, if every vertex in V (D) − V (C) is good, then it follows from Observation 2.3 that every arc of C is universal. The following lemma demonstrates then the direction change between x and C occurs exactly two times. Lemma 2.4. Let x ∈ V (D) − V (C) be a bad vertex. Then there exist y, z ∈ V (C) such that the following hold: (1) (2) (3) (4)
y is in the same partite set as x. y − → x and x → y + . Every vertex on C between y + and z is either dominated by x or is in the same partite set as x. Every vertex on C between z + and y − either dominates x or is in the same partite set as x.
Proof. Let C = u 1 u 2 . . . u k u 1 be a longest cycle of the strongly connected multipartite tournament D. In view of Theorem 2.2, there exists a vertex u i , say u 1 , such that u 1 → x. Applying Theorem 2.2 once more, we notice that x has also an outer neighbor in V (C). Let ` = min{2 ≤ i ≤ k|x → u i }. If u `−1 → x, then D contains the longer cycle u 1 u 2 . . . u `−1 xu ` u `+1 . . . u k u 1 , a contradiction. Hence we conclude that u `−1 and x belong to the same partite set and u `−2 → x. If we set u `−1 = y, then we observe that (1) and (2) are proved. If there is no vertex z on C as described in (3) or (4), then Observation 2.3 is satisfied, making x a good vertex. This contradiction completes the proof of Lemma 2.4. Following the notation in [4], we call a bad vertex x a covering vertex for the arcs y − y, yy + and for the vertices y, y + , if y is as in Lemma 2.4. A covering set is a set of covering vertices. Hubenko proved the following lemma for a cycle-connected bipartite tournament D with a maximal cycle C. Analyzing the proof it is easy to see that the assertion of this lemma is also valid for a strongly connected multipartite tournament D with a longest cycle C. y−,
Lemma 2.5. If an arc e ∈ A(C) is not universal, then there is a bad vertex that covers it.
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Proof. Assume that the arc uv ∈ A(C) is not universal. Consider the path P = vCu. If a vertex z ∈ V (D) − V (C) is dominated by some vertex v1 ∈ V (P) and dominates a vertex v2 ∈ V (P) such that v2 occurs on the path P after v1 , then z is on the cycle zv2 Cv1 z with uv. So there must be a vertex x such that all vertices of P dominated by x precede (on P) all vertices of P that dominates x. Clearly, x covers uv. Remark 2.6. According to the proof of Lemma 2.4, if x is a cover vertex for y − , y, y + , then x and y belong to the same partite set. Moreover, if x covers more than three vertices, then it is straightforward to see that x fulfills the conditions of Observation 2.3 and thus x is a good vertex, a contradiction. Consequently, each bad vertex covers exactly three vertices. On the other hand, according to Lemma 2.4, each arc of C has to be covered by at least one bad vertex, if C does not contain an universal arc. Lemma 2.7. Let C = v1 v2 v3 v4 v5 . . . vk v1 be a longest cycle in D, and let a1 , a2 ∈ V (D) − V (C) such that a1 covers v1 , v2 , v3 , and a2 covers v3 , v4 , v5 . (1) If a1 and a2 are in the same partite set, then a1 → v5 . (2) If a2 → a1 , then a1 → v4 . Proof. If a1 and a2 are in the same partite set, then a1 and v5 are in different partite sets. If we assume that v5 → a1 , then Lemma 2.4 implies that v3 is the only vertex on C dominated by a1 . Since δ(D) ≥ 2, there exists a vertex z ∈ V (D) − V (C) such that a1 → z. Applying Theorem 2.2, there is a vertex u ∈ V (C) such that z → u. If u 6= v4 , v5 , then we observe that u − → a1 or u −− → a1 . Substituting u − u by u − a1 zu or u −− u − u by u −− a1 zu, we increase the length of C, a contradiction. This yields v2 → z. Assume next that u = v4 . If a2 → z, then substituting v3 v4 by v3 a2 zv4 , we increase the length of C, a contradiction. Thus z → a2 . Since δ(D) ≥ 2, there exists a vertex y 6= v5 such that a2 → y. If y ∈ V (C), then y 6= v2 , v3 , v4 , and we deduce that y − → a1 or y −− → a1 . Substituting y − y by y − a1 za2 y or y −− y − y by y −− a1 za2 y, we increase the length of C, a contradiction. This yields y ∈ V (D) − V (C). According to Theorem 2.2, there exists a vertex v ∈ V (C) such that y → v. However, now we observe that v − → a1 , v −− → a1 , v −−− → a1 or v −−−− → a1 . Substituting v − v by v − a1 za2 yv, v −− v − v by v −− a1 za2 yv, v −−− v −− v − v by v −−− a1 za2 yv or v −−−− v −−− v −− v − v by v −−−− a1 za2 yv, we increase the length of C, a contradiction. This leads to v4 → z and u = v5 and so z → v5 . Substituting v4 v5 by v4 zv5 , we increase the length of C. This contradiction completes the proof of (1). If a2 → a1 , then a1 and v4 are in different partite sets. If we assume that v4 → a1 , then Lemma 2.4 again implies that v3 is the only vertex on C dominated by a1 . As above, δ(D) ≥ 2 gives us z ∈ V (D) − V (C) with a1 → z. Theorem 2.2 implies that z → v for some v ∈ V (C). If v 6= v4 , then either v − → a1 or v −− → a1 . We can either replace v − v with v − a1 zv or v −− v − v with v −− a1 zv to create a cycle longer than C, a contradiction. If z → v4 , we replace v3 v4 with v3 a2 a1 za4 to create a cycle longer than C, another contradiction. This completes the proof of (2). Lemma 2.8. Let ck dk ck−1 dk−1 . . . c2 d2 c1 d1 b1 g1 b2 . . . bl gl be an interval of C. Suppose that the vertices ck , dk , ck−1 , dk−1 , . . . , c1 , d1 , b1 are covered by vertices f k , f k−1 , . . . , f 1 such that f k covers ck , dk , ck−1 ; f k−1 covers ck−1 , dk−1 , ck−2 ; . . .; f 2 covers c2 , d2 , c1 ; f 1 covers c1 , d1 , b1 . Denote F = { f k , f k−1 , . . . , f 1 }. Suppose that the vertices d1 , b1 , g1 , b2 , g2 , . . . , bl , gl are covered by vertices a1 , a2 , . . . , al such that a1 covers d1 , b1 , g1 ; a2 covers g1 , b2 , g2 ; . . .; al covers gl−1 , bl , gl . Denote A = {a1 , a2 , . . . , al }. Finally, suppose that F is a monochromatic covering set (that means that F is contained in one partite set) and A is another monochromatic covering set. Then F → A. Proof. First we will prove by induction that f 1 → a j for j ∈ {1, 2, . . . , l}. For j = 1 we have f 1 → a1 , because otherwise, we could replace the arc d1 b1 with the path d1 a1 f 1 b1 that would increase the length of C, a contradiction. Provided, that we already know that f 1 → a j for every j < r , suppose on the contrary that ar → f 1 . From the definition of a covering vertex and Lemma 2.7 follows that ar −1 → gr and therefore we can replace the path gr −1 br gr with gr −1 ar f 1 ar −1 gr and increase the length of C, a contradiction. Now we will prove by induction that f i → a j for every i ∈ {1, 2, . . . , k} and j ∈ {1, 2, . . . , l}. The case i = 1 and j ∈ {1, 2, . . . , l} is shown above. Let q be the smallest index such that aq → fr . Using the same argument, from the definition of a covering vertex and Lemma 2.7 it follows that fr → cr −2 (in the case r = 2: f 2 → b1 ) and therefore
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we can replace the path cr −1 dr −1 cr −2 (in the case r = 2: c1 d1 b1 ) with cr −1 fr −1 aq fr cr −2 (in the case r = 2 with c1 f 1 aq f 2 b1 ) and increase the length of C, a contradiction. Lemma 2.9. Let ck dk ck−1 dk−1 . . . c2 d2 c1 d1 b1 g1 b2 . . . bl gl bl+1 be an interval of the cycle C. Suppose that the vertices ck , dk , ck−1 , dk−1 , . . . , c1 , d1 , b1 are covered by the vertices f k , f k−1 , . . . , f 1 such that f k covers ck , dk , ck−1 ; f k−1 covers ck−1 , dk−1 , ck−2 ; . . .; f 2 covers c2 , d2 , c1 ; f 1 covers c1 , d1 , b1 . Denote F = { f k , f k−1 , . . . , f 1 }. Suppose that the vertices b1 , g1 , b2 , g2 , . . . , bl , gl , bl+1 are covered by vertices a1 , a2 , . . . , al such that a1 covers b1 , g1 , b2 ; a2 covers b2 , g2 , b3 ; . . .; al covers bl , gl , bl+1 . Denote A = {a1 , a2 , . . . , al }. Finally, suppose that F is a monochromatic covering set and A is another monochromatic covering set. Then F → A. Proof. First we will prove by induction that f 1 → a j for j ∈ {1, 2, . . . , l}. For j = 1 suppose that a1 → f 1 . Since f 1 is a covering vertex, Lemma 2.7 implies that f 1 → g1 . Now we can replace the arc b1 g1 with the path b1 a1 f 1 g1 that would increase the length of C, a contradiction. Hence we deduce that f 1 → a1 . Provided, that we already know that f 1 → a j for every j < r , suppose on the contrary that ar → f 1 . From the definition of a covering vertex and Lemma 2.7 follows that ar −1 → br +1 and therefore we can replace the path br gr br +1 with br ar f 1 ar −1 br +1 and increase the length of C, a contradiction. Now we will prove by induction that f i → a j for every i ∈ {1, 2, . . . , k} and j ∈ {1, 2, . . . , l}. The case i = 1 and j ∈ {1, 2, . . . , l} is shown above. Let q be the smallest index such that aq → fr . Using the same argument, from the definition of a covering vertex and Lemma 2.7 it follows that fr → cr −2 (in the case r = 2: f 2 → b1 ) and therefore we can replace the path cr −1 dr −1 cr −2 (in the case r = 2: c1 d1 b1 ) with cr −1 fr −1 aq fr cr −2 (in the case r = 2 with c1 f 1 aq f 2 b1 ) and increase the length of C, a contradiction. Lemma 2.10. All vertices that are elements of a covering set of a cycle C belong to the same partite set. Proof. Assume that there are vertices x, y ∈ V (D) − V (C) from different partite sets that cover some vertices of C. Let us choose a minimal set M of covering vertices such that it covers all vertices of C and x, y ∈ M. Let us partition the cycle C into intervals, each covered with monochromatic vertices from M, such that any two consecutive intervals are covered with different colors. Observe that as we go around C, according to Lemmas 2.8 and 2.9, the covering vertices of each interval dominate the covering vertices of the next interval. Thus there is a cycle in the subdigraph of covering vertices, which is in D − V (C), a contradiction to Theorem 2.1. Lemma 2.11. If there is a monochromatic covering set for C, then every arc of C is universal. Proof. Let M be a set of covering vertices for C such that all vertices of M are from the same partite set, and let x ∈ M be arbitrary such that x covers the path y − yy + . The cycle x y + C y − x shows that x is on a cycle with every arc of C except y − y and yy + . We will construct a cycle C 0 that contains x and the arcs y − y, yy + . Since x is a bad vertex, C a longest cycle and there is a monochromatic covering set for C, Lemma 2.4 implies that there is an interval z − zz + of C such that z + → x, x → z − and z and x are in the same partite set. Let us use the notations z 1 = z −− and z 2 = z ++ . Denote the covering vertex of z + , z 2 and z 2+ by u, the covering vertex of z − , z and z + by v, and the covering vertex of z 1− , z 1 and z − by w. From Lemma 2.7 it follows that v → z 2+ and w → z + . Denote by C 0 the cycle z + x z − vz 2+ C z 1− wz + . Clearly, C 0 contains x, y − y and yy + which completes the proof. 3. Main results Theorem 3.1. Let D be a strongly connected multipartite tournament with δ(D) ≥ 2. Then every longest cycle of D has a universal arc. Proof. Assume on the contrary, that there is no universal arc on the cycle C. Then, by Lemma 2.5, all the arcs of C are covered. By Lemma 2.10, all the covering vertices of C are from the same partite set. It follows from Lemma 2.11 that all the arcs of C are universal, a contradiction. This completes the proof of this theorem. The following example of Bondy [2] shows that Theorem 3.1 is not valid in general for δ(D) = 1. Example 3.2. Let A1 , A2 , . . . , Ac be the partite sets of a c-partite tournament H such that |A1 | = 1, say A1 = {a1 } and |Ai | ≥ 2 for 2 ≤ i ≤ c. If A2 → a1 , a1 → Ai for 3 ≤ i ≤ c and A j → Ai for 2 ≤ i < j ≤ c, then H is strong with δ(H ) = 1, and the longest cycle of H has length c. It is easy to see that H does not contain a universal arc.
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We notice that the multipartite tournament H in Example 3.2 is not cycle-connected, since two vertices ai , bi ∈ Ai for 2 ≤ i ≤ c are not cyclic. However, if a multipartite tournament is cycle-connected, then it contains a universal arc. Theorem 3.3. If D is a cycle-connected multipartite tournament, then D contains a universal arc. Proof. In the case δ(D) = 1 we assume, without loss of generality, that x has out-degree 1 such that x → y. Thus the arc x y is contained in all cycles containing x. Since D is cycle-connected, the arc x y is universal. In the case δ(D) ≥ 2, the desired result follows from Theorem 3.1, since D is strongly connected. Acknowledgements We would like to thank one of the referees for valuable suggestions and comments. References [1] [2] [3] [4] [5] [6]
´ am, On some cyclic connectivity properties of directed graphs, Acta Cybernet. 14 (1) (1999) 1–12. A. Ad´ J.A. Bondy, Disconnected orientation and a conjecture of Las Vergnas, J. London Math. Soc. 14 (1976) 277–282. Y. Guo, A. Pinkernell, L. Volkmann, On cycles through a given vertex in multipartite tournaments, Discrete Math. 164 (1997) 165–170. A. Hubenko, On a cyclic connectivity property of directed graphs, Discrete Math. 308 (2008) 1018–1024. B. Jackson, Long paths and cycles in oriented graphs, J. Graph Theory 5 (1981) 145–157. L. Volkmann, Cycles in multipartite tournaments: Results and problems, Discrete Math. 245 (2002) 19–53.