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On cycles in regular 3-partite tournaments
Discrete Mathematics 342 (2019) 1223–1232
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Discrete Mathematics journal homepage: www.elsevier.com/locate/disc
On cycles in regular 3-partite tournaments✩ ∗
Qiaoping Guo , Wei Meng School of Mathematical Sciences, Shanxi University, Taiyuan, 030006, China
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Article history: Received 3 April 2018 Received in revised form 8 July 2018 Accepted 12 December 2018 Available online 30 January 2019 Keywords: Multipartite tournament Regular 3-partite tournament Cycle
a b s t r a c t A c-partite tournament is an orientation of a complete c-partite graph. In 2006, Volkmann conjectured that every arc of a regular 3-partite tournament D is contained in an m-, (m+1)or (m + 2)-cycle for each m ∈ {3, 4, . . . , |V (D)| − 2}, and this conjecture was proved to be correct for 3 ≤ m ≤ 7. In 2012, Xu et al. conjectured that every arc of an r-regular 3-partite tournament D with r ≥ 2 is contained in a (3k − 1)- or 3k-cycle for k = 2, 3, . . . , r. They proved that this conjecture is true for k = 2. In this paper, we confirm this conjecture for k = 3, which also implies that Volkmann’s conjecture is correct for m = 7, 8. © 2019 Elsevier Ltd. All rights reserved.
1. Terminology and introduction We consider finite digraphs without loops or multiple arcs. The vertex set and the arc set of a digraph D are denoted by V (D) and A(D), respectively. If xy is an arc of D, then we say x dominates y and write x → y. More generally, if A and B are two disjoint subdigraphs of D such that every vertex of A dominates every vertex of B, then we say that A dominates B and denote it by A → B. Otherwise, we denote it by A ↛ B. Let D′ be a subdigraph of D. We use |D′ | or |V (D′ )| to stand for the number of the vertices of D′ . The outset ND+′ (x) of a vertex x is the set of vertices of D′ dominated by x and the inset ND−′ (x) is the set of vertices of D′ dominating x. We call the numbers d+ (x) = |ND+′ (x)| and d− (x) = |ND−′ (x)| the out-degree and in-degree of x D′ D′ ′ ′ + − + − in D , respectively. When D = D, we usually use N (x), N (x), d (x) and d (x) instead of ND+′ (x), ND−′ (x), d+ (x) and d− (x), D′ D′ respectively. A digraph D is said to be k-regular or regular, if there is an integer k such that d+ (x) = d− (x) = k holds for every x ∈ V (D). A c-partite or multipartite tournament is an orientation of a complete c-partite graph. An m-cycle is a directed cycle of length m. The structure of cycles in multipartite tournaments has been extensively studied; see for example [4]. However, many statements on cycles in c-partite tournaments are only valid for c ≥ 4. At present, the results on cycles in 3-partite tournaments are still very few. For cycles through an arc in a regular 3-partite tournament, Volkmann[3] posed the following conjecture. Conjecture 1.1 ([3]). If D is a regular 3-partite tournament, then every arc of D is contained in an m-, (m + 1)- or (m + 2)-cycle for each m ∈ {3, 4, . . . , |V (D)| − 2}. In [2,3], Volkmann and Stella confirmed Conjecture 1.1 for m ∈ {3, 4, . . . , 7}. In 2012, Xu et al. posed the following two conjectures. ✩ This work is supported by the National Natural Science Foundation of China (No. 11701349) and by the Natural Science Foundation of Shanxi Province, China (No. 201601D011005) and by Shanxi Scholarship Council of China (2017-018). ∗ Corresponding author. E-mail address: [email protected] (Q. Guo). https://doi.org/10.1016/j.disc.2018.12.007 0012-365X/© 2019 Elsevier Ltd. All rights reserved.
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Conjecture 1.2 ([5]). If D is an r-regular 3-partite tournament with r ≥ 2, then every arc of D is contained in a 3k- or (3k + 1)-cycle for k = 1, 2, . . . , r − 1. Conjecture 1.3 ([5]). If D is an r-regular 3-partite tournament with r ≥ 2, then every arc of D is contained in a (3k − 1)- or 3k-cycle for k = 2, 3, . . . , r. In [3], Volkmann proved that every arc of a regular 3-partite tournament D is contained in a 3- or 4-cycle, which implies that Conjecture 1.2 is true for k = 1. It is also proved that Conjecture 1.2 is true for k = 2 in [1]. Xu et al. [5] proved that every arc of a regular 3-partite tournament is contained in a 5- or 6-cycle, which also implies that Conjecture 1.3 is correct for k = 2. In this paper, we confirm Conjecture 1.3 for k = 3, which also implies that Conjecture 1.1 is true for m = 7,8. 2. Main result The following two lemmas are important to prove our main results. Lemma 2.1. If D is an r-regular 3-partite tournament with partite sets V1 , V2 , V3 and v is a vertex of D, then |V1 | = |V2 | = |V3 | = r and d+ (v ) = d− (v ) = r. − Lemma 2.2 ([5]). If D is an r-regular 3-partite tournament with partite sets V1 , V2 , V3 and u is a vertex of V1 , then d+ V2 (u) = dV3 (u) + − and dV (u) = dV (u). 2
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Theorem 2.1. Let D be an r-regular 3-partite tournament with r ≥ 3 and let V1 , V2 , V3 be the partite sets of D. If ab is an arc from V1 to V2 satisfying V3 → a → V2 , then ab is contained in a 8- or 9-cycle. Proof. By Lemma 2.1, we have |V1 | = |V2 | = |V3 | = r and d+ (v ) = d− (v ) = r for each vertex v of D. By a → b and Lemma 2.2, there exists a vertex x ∈ V3 such that b → x. We distinguish the following two cases. Case 1 V2 − {b} ↛ x. By the hypothesis, there is a vertex y ∈ V2 − {b} such that x → y. By Lemma 2.2, there is a vertex u ∈ V1 such that y → u. Since a → V2 , we have a → y and u ̸ = a. By a → y and Lemma 2.2, there exists a vertex v ∈ V3 such that y → v . Obviously, we have v ̸ = x and v → a. Subcase 1.1 V2 − {b, y} ↛ v . In this case, there exists w ∈ V2 − {b, y} such that v → w . Suppose first that V3 − {x, v} ↛ w . Then there is a vertex v ′ ∈ V3 − {x, v} such that w → v ′ . If u → v , we get a 8-cycle abxyuvwv ′ a through ab. Assume v → u. If u → w , then abxyv uwv ′ a is a 8-cycle through ab. If w → u, then by {y, w} → u and Lemma 2.2, there exists a vertex z ′ ∈ V3 − {x} such that u → z ′ (z ′ may be equal to v ′ , but not equal to v ). Therefore, abxyvw uz ′ a is a 8-cycle through ab. Suppose now that V3 − {x, v} → w . Combining with {a, v} → w , we have N − (w ) = (V3 − {x}) ∪ {a} and N + (w ) = (V1 − {a}) ∪ {x}. Let c ∈ V3 − {x, v}. Then c → w . By Lemma 2.2, there exists a vertex z ∈ V1 (z may be equal to u) such that z → c. Note that z ̸ = a. Then w → z and abxyvw zca is a desired 8-cycle. Subcase 1.2 V2 − {b, y} → v and u → v . In this case, we have N − (v ) = (V2 − {b}) ∪ {u} and N + (v ) = (V1 − {u}) ∪ {b}. If V3 − {x, v} ↛ V1 − {a, u}, then there is an arc z w from V1 − {a, u} to V3 − {x, v}. So v → z and w → a. Thus, abxyuv z w a is a 8-cycle through ab. Assume V3 − {x, v} → V1 − {a, u}. Let z ∈ V1 − {a, u}. Then V3 − {x} → z. If V2 − {b, y} ↛ z, then there is an arc zy′ for y′ ∈ V2 − {b, y}. Obviously, y′ → v . Let w ∈ V3 − {x, v}. If y′ → w , then abxyv zy′ w a is a 8-cycle through ab. Assume w → y′ . Then N + (w ) = (V1 − {u}) ∪ {y′ } and so u → w . Thus, abxyuw y′ v a is a 8-cycle through ab. Assume V2 −{b, y} → z. Note that V3 −{x} → z and d− (z) = r. We have |V2 − {b, y}| = 1 and r = 3. Let V2 −{b, y} = {y′ } and V3 − {x, v} = {v ′ }. Then {y′ , v, v ′ } → z → {b, x, y}. If b → v ′ , then abv ′ zxyuv a is a 8-cycle through ab. Assume v ′ → b. Then N + (v ′ ) = {a, b, z }, N − (v ′ ) = {u, y, y′ }, N − (b) = {a, z , v ′ }, N + (b) = {u, x, v}, N − (v ) = {u, b, y} and N + (v ) = {a, z , y′ }. Thus abxyuv y′ v ′ a is a 8-cycle through ab. Subcase 1.3 V2 − {b, y} → v → u. Suppose first that V2 − {b, y} ↛ u. Then there exists a vertex y′ ∈ V2 − {b, y} such that u → y′ . If V3 − {x, v} ↛ y′ , then there is a vertex v ′ ∈ V3 − {x, v} such that y′ → v ′ and abxyv uy′ v ′ a is a 8-cycle through ab. Assume V3 − {x, v} → y′ . Note that {a, u} → y′ . We have N − (y′ ) = (V3 − {x, v}) ∪ {a, u} and N + (y′ ) = (V1 − {a, u}) ∪ {x, v}. If V3 − {x, v} ↛ V1 − {a, u}, then there is an arc u′ v ′ from V1 − {a, u} to V3 − {x, v}. Obviously, y′ → u′ and v ′ → a. Thus, abxyuy′ u′ v ′ a is a desired 8-cycle. Assume V3 − {x, v} → V1 − {a, u}. Note that V3 − {x, v} → {a, y′ }. We get V3 − {x, v} → (V1 − {u}) ∪ {y′ } and (V2 − {y′ }) ∪ {u} → V3 − {x, v}. Let z ∈ V3 − {x, v}. Then u → z → y′ and abxyuzy′ v a is a 8-cycle through ab. Suppose now that V2 − {b, y} → u. Since {y, v} → u, we have (V2 − {b}) ∪ {v} → u → (V3 − {v}) ∪ {b}. If V2 −{b, y} ↛ V3 −{x, v}, then there is an arc v ′ y′ from V3 −{x, v} and V2 −{b, y}. Obviously, u → v ′ and y′ → v . So abxyuv ′ y′ v a is a 8-cycle through ab. Assume V2 −{b, y} → V3 −{x, v}. Note that V2 −{b, y} → {u, v}. We have V2 −{b, y} → (V3 −{x}) ∪{u}
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and (V1 − {u}) ∪ {x} → V2 − {b, y}. Let y1 ∈ V2 − {b, y} and v1 ∈ V3 − {x, v}. Then x → y1 → {u, v, v1 } and u → v1 . If v1 → y, then abxy1 uv1 yv a is a 8-cycle through ab. If y → v1 , then by {y, y1 } → v1 and Lemma 2.2, there exists a vertex u1 ∈ V1 − {a} such that v1 → u1 . It is easy to see that u1 ̸ = u and u1 → y1 . Now, abxyv1 u1 y1 v a is a desired 8-cycle. Case 2 V2 − {b} → x. By b → x, we have V2 → x → V1 . Subcase 2.1 V1 − {a} ↛ b. By the hypothesis, there exists u ∈ V1 − {a} such that b → u. Obviously, x → u. By {b, x} → u and Lemma 2.2, there exist two vertices w ∈ V3 − {x} and v ∈ V2 − {b} such that u → {w, v}. Suppose first that V3 − {x, w} ↛ V1 − {a, u}. Then there exists an arc u0 w0 from V1 − {a, u} to V3 − {x, w}. If V2 − {b} ↛ w , then there exists a vertex v0 ∈ V2 − {b} (v0 may be equal to v ) such that w → v0 . Thus, v0 → x → u0 and abuwv0 xu0 w0 a is a 8-cycle through ab. Assume V2 − {b} → w . Then N − (w ) = {u} ∪ (V2 − {b}) and N + (w ) = (V1 − {u}) ∪ {b}. So v → w → u0 and abxuvw u0 w0 a is a 8-cycle through ab. Suppose now that V3 − {x, w} → V1 − {a, u}. By V3 → a and x → V1 , we have V3 − {w} → V1 − {u}. Let u′ ∈ V1 − {a, u}. If w → u′ , then V3 → u′ → V2 . Let v ′ ∈ V2 − {b, v}. Then {a, u′ } → v ′ . By Lemma 2.2, there is a vertex w ′ ∈ V3 − {x} (w ′ may be equal to w ) such that v ′ → w ′ . Then abuv xu′ v ′ w ′ a is a 8-cycle through ab. Assume u′ → w . If V3 − {x, w} ↛ v , then there exists an arc vw ′ for some w ′ ∈ V3 − {x, w}. Thus, w ′ → u′ and abxuvw ′ u′ w a is a 8-cycle through ab. Assume V3 − {x, w} → v . Note that {a, u} → v . We have N − (v ) = (V3 − {x, w}) ∪ {a, u} and v → {u′ , w}. By {u, u′ } → w and Lemma 2.2, there is a vertex v ′ ∈ V2 − {b} such that w → v ′ . Since v → w , we have v ′ ̸ = v . Thus, v ′ → x → a and abuv u′ wv ′ xa is a 8-cycle through ab. Subcase 2.2 V1 − {a} → b. By a → b, we obtain V1 → b → V3 . Now we have V3 → a → V2 → x → V1 → b → V3 . Let w ∈ V3 − {x}. Then b → w . Subcase 2.2.1 V2 − {b} ↛ w . In this case, there exists a vertex v ∈ V2 − {b} such that w → v . By Lemma 2.2, there exists a vertex u ∈ V1 − {a} such that v → u. Suppose first that V2 − {b, v} ↛ u. Then there exists a vertex y ∈ V2 − {b, v} such that u → y. If V3 − {x, w} ↛ y, then there is a vertex z ∈ V3 − {x, w} such that y → z and abwv xuyza is a 8-cycle through ab. Assume V3 − {x, w} → y. By {a, u} → y, we have N − (y) = {a, u} ∪ (V3 − {x, w}) and N + (y) = (V1 − {a, u}) ∪ {x, w}. Let z ∈ V3 − {x, w}. Then abz v xuywa (when z → v ) or abxuywv za (when v → z) is a desired 8-cycle. Suppose now that V2 − {b, v} → u. By {x, v} → u, we get N − (u) = (V2 − {b}) ∪ {x} and N + (u) = {b} ∪ (V3 − {x}). If V2 − {b, v} ↛ V3 − {x, w}, then there is an arc w ′ y where w ′ ∈ V3 − {x, w} and y ∈ V2 − {b, v}. Thus, u → w ′ , y → x and ab is contained in a 8-cycle abwv uw ′ yxa. Assume V2 − {b, v} → V3 − {x, w}. Note that {b, u} → V3 − {x, w} and V2 − {b, v} → {u, x}. We have {u} ∪ (V2 − {v}) → V3 − {x, w} → (V1 − {u}) ∪ {v}, V2 − {b, v} → (V3 − {w}) ∪ {u} and {w} ∪ (V1 − {u}) → V2 − {b, v}. Let u′ ∈ V1 − {a, u}, y′ ∈ V2 − {b, v} and w ′ ∈ V3 − {x, w}. Then abwv uw ′ u′ y′ xa is a 9-cycle through ab. Subcase 2.2.2 V2 − {b} → w . By b → w , we have V2 → w → V1 . Let y, u ∈ V1 − {a} be two distinct vertices. Then {x, w} → y. Lemma 2.2 implies that there exists a vertex z ∈ V2 − {b} such that y → z. Obviously, z → w → u. If V3 − {x, w} ↛ u, then there is a vertex w′ ∈ V3 − {x, w} such that u → w′ . Thus, abxyz wuw′ a is a 8-cycle through ab. Assume V3 − {x, w} → u. By {x, w} → u, we have V3 → u → V2 and so u → z. By {a, y, u} → z and Lemma 2.2, there is a vertex w0 ∈ V3 − {x, w} such that z → w0 . Obviously w0 → u. Let z0 ∈ V2 − {b, z }. Then u → z0 → w and abxyz w0 uz0 w a is a desired 9-cycle. ■ Theorem 2.2. Let D be an r-regular 3-partite tournament with r ≥ 3 and let V1 , V2 , V3 be the partite sets of D. If ab is an arc from V1 to V2 satisfying V3 ↛ a ↛ V2 , then ab is contained in a 8- or 9-cycle. Proof. By Lemma 2.1, we have |V1 | = |V2 | = |V3 | = r and d+ (v ) = d− (v ) = r for each vertex v of D. We divide the partite set V2 into two nonempty parts V2+ , V2− such that V2− → a → V2+ . Similarly, V3 can be divided into two nonempty parts V3+ , V3− such that V3− → a → V3+ . Let V ′ = V2+ ∪ V3+ and V ′′ = V2− ∪ V3− . Observe that N + (a) = V ′ , N − (a) = V ′′ and |V ′ | = |V ′′ | = r. By Lemma 2.1, we also have |V2+ | = |V3− | and |V2− | = |V3+ |. Furthermore, we have the following claim. Claim 1. Let y ∈ V1 be arbitrary. Then d+ (y) = d− (y) and d− (y) = d+ (y). V′ V ′′ V′ V ′′ Proof. By Lemma 2.1, we have d+ (y) + d− (y) = |V ′ | = r , d+ (y) + d− (y) = |V ′′ | = r , d+ (y) + d+ (y) = d+ (y) = r and V′ V′ V ′′ V ′′ V′ V ′′ − − − + + − − dV ′ (y) + dV ′′ (y) = d (y) = r. So dV ′ (y) = dV ′′ (y) and dV ′ (y) = dV ′′ (y). □ Below, we prove this theorem in four cases: Case A1. |V2+ | = 1 and V3+ ↛ b. Case A2. |V2+ | = 1 and V3+ → b. Case B1. 2 ≤ |V2+ | ≤ r − 1 and V3+ ↛ b. Case B2. 2 ≤ |V2+ | ≤ r − 1 and V3+ → b.
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Proof of the theorem under Case A1. In this case, we have |V2+ | = |V3− | = 1 and |V2− | = |V3+ | = r − 1. Obviously, V2+ = {b}. Since V3+ ↛ b, there is a vertex x ∈ V3+ such that b → x. Clearly, we also have a → x. By {a, b} → x and Lemma 2.2, there are a vertex y ∈ V1 − {a} and a vertex u ∈ V2 − {b} such that x → {y, u}. Let V3− = {v}. Obviously, (V2 − {b}) ∪ {v} → a → {b} ∪ (V3 − {v}). Subcase 1 y → u. Subcase 1.1 V3 − {x, v} → u. In this case, we have (V3 − {v}) ∪ {y} → u → (V1 − {y}) ∪ {v}. Let y′ and w be two arbitrary vertices in V1 − {a, y} and + V3 respectively. Then a → w → u → {y′ , v}. Suppose first that y′ → v . If y → w , then abxyw uy′ v a is a desired 8-cycle. Assume w → y. If v → y, then {x, w, v} → y. Lemma 2.2 implies that there exists a vertex z ∈ V2 − {b, u} such that y → z. Thus, z → a and ab is contained in a 8-cycle abxuy′ v yza. If y → v , then by {y, y′ } → v and Lemma 2.2, there exists a vertex u′ ∈ V2 − {b} such that v → u′ . Obviously, u′ ̸ = u and abxyuy′ v u′ a is a desired 8-cycle. Suppose now that v → y′ . If V2− − {u} ↛ y′ , then there exists a vertex u′ ∈ V2− − {u} such that y′ → u′ . Thus, abxyuv y′ u′ a is a 8-cycle through ab. Assume V2− − {u} → y′ . Since {u, v} → y′ , we have that N − (y′ ) = V2− ∪ {v} and N + (y′ ) = V3+ ∪ {b} and so y′ → w . If w → y, then by {x, w} → y and Claim 1, there exists a vertex z ∈ V ′′ − {u} (z may be equal to v ) such that y → z. Thus, abxuy′ w yza is a 8-cycle through ab. If y → w , then by {a, y, y′ } → w and Lemma 2.2, there exists a vertex u1 ∈ V2 − {b, u} such that w → u1 . Thus, u1 → a and abxyuy′ w u1 a is a 8-cycle through ab. Subcase 1.2 V3 − {x, v} ↛ u. In this case, there exists a vertex w ∈ V3 − {x, v} such that u → w . Obviously, we have a → w . Suppose first that V1 − {a, y} → w . Since {a, u} → w , we have N − (w ) = (V1 − {y}) ∪ {u} and N + (w ) = {y} ∪ (V2 − {u}). Let u′ ∈ V2− − {u}. Then w → u′ . If V1 − {a, y} ↛ u, then there is an arc uy′ for some y′ ∈ V1 − {a, y}. Obviously, y′ → w and abxyuy′ w u′ a is a 8-cycle through ab. Assume V1 − {a, y} → u. It follows from {x, y} → u that N − (u) = (V1 − {a}) ∪ {x} and so u → v . If V1 − {a, y} ↛ v , then there is an arc v y1 for some y1 ∈ V1 − {a, y}. Thus, y1 → w and abxuv y1 w u′ a is a 8-cycle through ab. If V1 − {a, y} → v , then V1 − {a, y} ↛ V2 − {b, u} (as otherwise, V1 − {a, y} → (V2 − {b, u}) ∪ {w, u, v}, this is impossible.). Therefore, there is an arc u0 y0 from V2 − {b, u} to V1 − {a, y}. Thus, w → u0 , y0 → v and abxuw u0 y0 v a is a 8-cycle through ab. Suppose now that V1 − {a, y} ↛ w . Then there is a vertex y′ ∈ V1 − {a, y} such that w → y′ . If V ′′ − {u} ↛ y′ , then there exists a vertex u0 ∈ V ′′ − {u} such that y′ → u0 and abxyuw y′ u0 a is a 8-cycle through ab. Assume V ′′ − {u} → y′ . Since w → y′ , we get N − (y′ ) = (V ′′ − {u}) ∪ {w} and N + (y′ ) = {u} ∪ (V ′ − {w}). So {w, v} → y′ → {u, x}. If y → w , then by {a, y} → w and Lemma 2.2, there exists a vertex u′ ∈ V2 − {b} such that w → u′ . It is easy to see that ′ u ̸ = u, u′ → y′ and abxyw u′ y′ ua is a 8-cycle through ab. Assume w → y. By {a, y′ } → x and Lemma 2.2, there exists a vertex v ′ ∈ V2 − {u} such that x → v ′ . Then v ′ ̸ = b and v ′ → y′ . If y → v , then abxv ′ y′ uw yv a is a 9-cycle through ab. If v → y, then {x, w, v} → y. Lemma 2.2 implies that there exists a vertex z ∈ V2 − {b, u} (z may be equal to v ′ ) such that y → z. If z → v , then abxuw yz v a is a 8-cycle through ab. If v → z, then v → {a, y, y′ , z } and r ≥ 4. Let w′ ∈ V3+ − {x, w}. Then {a, y′ } → w′ . By Lemma 2.2, there is a vertex u1 ∈ V2 − {b} such that w′ → u1 . Thus, abxyzy′ w′ u1 a (when u1 ̸= z) or abxuw y′ w′ u1 a (when u1 = z) is a 8-cycle through ab. Subcase 2 u → y. Subcase 2.1 V2− → y. In this case, we get N − (y) = V2− ∪ {x} and N + (y) = {b} ∪ (V3 − {x}). Let w ∈ V3+ − {x}. Then {a, y} → w . Suppose first that V1 − {a, y} → w . In this case, we have V1 → w → V2 and so w → u. If v ↛ V2 − {b, u}, then there is a vertex u′ ∈ V2 − {b, u} such that u′ → v . Thus, we have that w → u′ and abxuyw u′ v a is a 8-cycle through ab. Assume v → V2 − {b, u}. Let u0 ∈ V2 − {b, u}. If u → v , then abxyw uv u0 a is a desired 8-cycle. If v → u, then N + (v ) = (V2 − {b, u}) ∪ {a, u} = (V2 − {b}) ∪ {a} and N − (v ) = (V1 − {a}) ∪ {b}. By {x, w, v} → u and Lemma 2.2, there exists a vertex y′ ∈ V1 − {a, y} such that u → y′ . Then y′ → v and abxyw uy′ v a is a desired 8-cycle. Suppose now that V1 − {a, y} ↛ w . Then there is a vertex y′ ∈ V1 − {a, y} such that w → y′ . If V ′′ − {u} ↛ y′ , then there is an arc y′ u′ for some u′ ∈ V ′′ − {u}. Thus, abxuyw y′ u′ a is a 8-cycle through ab. Assume V ′′ − {u} → y′ . Then N − (y′ ) = (V ′′ − {u}) ∪ {w} and N + (y′ ) = {u} ∪ (V ′ − {w}). Let u0 be an arbitrary vertex in V2 − {b, u}. Then u0 → {a, y′ }. By Lemma 2.2, there exists a vertex w ′ ∈ V3 − {x} (w ′ may be equal to w or v ) such that w ′ → u0 . Recalling that N + (y) = {b} ∪ (V3 − {x}), we have that y → w ′ and ab is contained in a 8-cycle abxyw ′ u0 y′ ua. Subcase 2.2 V2− ↛ y. In this case, there is a vertex u′ ∈ V2− − {u} such that y → u′ . Subcase 2.2.1 V3+ − {x} ↛ u′ . In this case, there exists a vertex w ∈ V3+ − {x} such that u′ → w . Clearly, a → w . Suppose first that V1 − {a, y} → w . It follows from {a, u′ } → w that N − (w ) = (V1 − {y}) ∪ {u′ } and so w → {u, y}. If u → v , then abxyu′ w uv a is a 8-cycle through ab. Assume v → u. By {x, w, v} → u and Lemma 2.2, there is a vertex y′ ∈ V1 − {a, y} such that u → y′ . Thus, y′ → w and abxuy′ w yu′ a is a 8-cycle through ab. Suppose now that V1 − {a, y} ↛ w . Then there exists a vertex y0 ∈ V1 − {a, y} such that w → y0 . If V ′′ − {u′ } ↛ y0 , then there is a vertex u0 ∈ V ′′ − {u′ } such that y0 → u0 . Thus, abxyu′ w y0 u0 a is a 8-cycle through ab. Assume V ′′ − {u′ } → y0 . Since
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w → y0 , we have N − (y0 ) = (V ′′ − {u′ }) ∪ {w} and so y0 → u′ . If y → w, then abxuywy0 u′ a is a 8-cycle through ab. Assume w → y. By {x, w} → y and Claim 1, there exists a vertex z ∈ V ′′ − {u′ } such that y → z. Obviously, z ̸= u and z → y0 . Thus,
abxuyzy0 u′ a is a 8-cycle through ab.
Subcase 2.2.2 V3+ − {x} → u′ . Suppose first that V1 −{a, y} → u′ . Since (V3+ −{x}) ∪{y} → u′ and d− (u′ ) = r, we have |V3+ − {x}| = 1. Let V3+ −{x} = {w}. Then V3 = {x, w, v} and r = 3. Let V1 = {a, y, y′ }. Then {y, y′ , w} → u′ → {a, x, v} and {a, b, u′ } → x → {y, y′ , u}. If y′ → u, then abxy′ uyu′ v a is a 8-cycle through ab. Assume u → y′ . Then u → V1 and so v → u. If y → v , then abxyv uy′ u′ a is a 8-cycle through ab. If v → y, then v → {a, u, y} and so y′ → v . Thus, abxuy′ v yu′ a is a 8-cycle through ab. Suppose now that V1 − {a, y} ↛ u′ . Then there exists a vertex y′ ∈ V1 − {a, y} such that u′ → y′ . If V ′′ − {u} ↛ y′ , then there is an arc y′ z for some z ∈ V ′′ − {u}. Obviously, z ̸ = u′ and abxuyu′ y′ za is a 8-cycle through ab. Assume V ′′ − {u} → y′ . If u → y′ , then V ′′ → y′ → V ′ . By u → {a, y, y′ } and Lemma 2.2, there is a vertex w ∈ V3 − {x, v} such that w → u. Then y′ → w and abxyu′ y′ w ua is a 8-cycle through ab. Assume y′ → u. If V3+ − {x} ↛ y, then there exists a vertex w ∈ V3+ − {x} such that y → w . Recalling that V3+ − {x} → u′ , we have that w → u′ and abxywu′ y′ ua is a desired 8-cycle. Assume V3+ − {x} → y. Since {x, u} → y, we get N − (y) = V3+ ∪ {u} and so y → v . Then abxyu′ v y′ ua (when u′ → v ) or abxyv u′ y′ ua (when v → u′ ) is a desired 8-cycle. □ Proof of the theorem under Case A2. In this case, we have |V2+ | = |V3− | = 1, |V2− | = |V3+ | = r − 1 and V3+ → b. Obviously, V2+ = {b}, N − (b) = V3+ ∪ {a} and N + (b) = (V1 − {a}) ∪ {v}, where v ∈ V3− . If V3+ → V1 − {a}, then V3+ ∪ {b} → V1 − {a} → (V2 − {b}) ∪ {v} and {a} ∪ (V2 − {b}) → V3+ → (V1 − {a}) ∪ {b}. Let y, y′ ∈ V1 − {a}, u, u′ ∈ V2 − {b} be four distinct vertices and let x ∈ V3+ . Then abyuxy′ u′ v a (when u′ → v ) or abyuxy′ v u′ a (when v → u′ ) is a 8-cycle through ab. Assume V3+ ↛ V1 − {a}. Then there is an arc yx from V1 − {a} to V3+ . Moreover, we have b → y. Consider the following two cases. Subcase 1 V1 − {a, y} ↛ x. In this case, there exists a vertex y′ ∈ V1 − {a, y} such that x → y′ . By Lemma 2.2, there is a vertex u ∈ V2 such that y′ → u. Obviously, b → y′ and u ∈ V2− . Suppose first that V3+ − {x} ↛ u. Then there exists a vertex w ∈ V3+ − {x} such that u → w . If V2− ↛ w , then there is an arc w u′ for some u′ ∈ V2− − {u} and abyxy′ uw u′ a is a 8-cycle through ab. If V2− → w , then N − (w ) = V2− ∪ {a} and N + (w ) = (V1 − {a}) ∪ {b}. By {a, y} → x and Lemma 2.2, there exists a vertex u0 ∈ V2− (u0 may be equal to u) such that x → u0 . Then u0 → w → y′ . By {b, x} → y′ and Claim 1, there exists a vertex v ′ ∈ V ′′ − {u0 } such that y′ → v ′ . Then abyxu0 w y′ v ′ a is a 8-cycle through ab. Suppose now that V3+ − {x} → u. If v → u, then we get N − (u) = {y′ } ∪ (V3 − {x}) and N + (u) = (V1 − {y′ }) ∪ {x}. By {a, y} → x and Lemma 2.2, there exists a vertex u0 ∈ V2 − {b} such that x → u0 . Since u → x, we get u0 ̸= u. Recalling that N + (b) = (V1 − {a}) ∪ {v}, then b → {y′ , v} and aby′ v uyxu0 a (when y′ → v ) or abv y′ uyxu0 a (when v → y′ ) is a 8-cycle through ab. Assume u → v . If V2− −{u} ↛ v , then there exists an arc v u′ for some u′ ∈ V2− −{u}. Thus, abyxy′ uv u′ a is a 8-cycle through ab. If V2− − {u} → v , then V2 → v since {b, u} → v . So v → V1 . By b → y′ and Lemma 2.2, there exists a vertex w ∈ V3 such that y′ → w. Since {x, v} → y′ , we have w ̸= x, v . Therefore, w ∈ V3+ − {x} and w → u. Thus, abyxy′ wuv a is a 8-cycle through ab. Subcase 2 V1 − {a, y} → x. In this case, we have V1 → x → V2 since {a, y} → x. Suppose first that V1 − {a} → V2− . Then V1 − {a} → V2− ∪ {x} and {b} ∪ (V3 − {x}) → V1 − {a}. In addition, (V1 − {a}) ∪ {x} → V2− → {a} ∪ (V3 − {x}). Let y′ ∈ V1 − {a, y} and let u, u′ ∈ V2 − {b} be two different vertices. Then abyxuv y′ u′ a is a desired 8-cycle. Suppose now that V1 −{a} ↛ V2− . Then there exists an arc uy′ from V2− to V1 −{a} (y′ may be equal to y). Let y1 ∈ V1 −{a, y′ }. Then b → y1 → x → u. If V3+ − {x} → y′ , then N − (y′ ) = {b, u} ∪ (V3+ − {x}) and N + (y′ ) = (V2 − {b, u}) ∪ {x, v}. Let w ∈ V2 − {b, u}. Then y′ → {w, v}, and aby1 xuy′ vw a (when v → w ) or aby1 xuy′ wv a (when w → v ) is a 8-cycle through ab. If V3+ − {x} ↛ y′ , then there is an arc y′ z for some z ∈ V3+ − {x}. In the case when V2− − {u} ↛ z, there exists an arc zu′ for some u′ ∈ V2− − {u} and aby1 xuy′ zu′ a is a 8-cycle through ab. In the other case when V2− − {u} → z, we have N − (z) = {a, y′ } ∪ (V2− − {u}) and N + (z) = (V1 − {a, y′ }) ∪ {b, u}. Let u0 ∈ V2− − {u}. Then x → u0 → z. By {b, z } → y1 and Claim 1, there exists a vertex w ∈ V ′′ − {u0 } such that y1 → w . Thus, aby′ xu0 zy1 w a is a 8-cycle through ab. □ Proof of the theorem under Case B1. In this case, we have 2 ≤ |V2+ | = |V3− | ≤ r − 1, 1 ≤ |V2− | = |V3+ | ≤ r − 2 and V3+ ↛ b. Then there exists a vertex x ∈ V3+ such that b → x. Subcase 1 V2+ ↛ x. By the hypothesis, there exists a vertex u ∈ V2+ − {b} such that x → u. By Lemma 2.2, there is a vertex y ∈ V1 such that u → y. Since a → u, we know y ̸ = a. We first prove the following claim. Claim 2. If |V3+ | ≥ 2 and V3+ − {x} ↛ y, then ab is contained in a 8- or 9-cycle.
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Proof. Since |V3+ | ≥ 2 and V3+ − {x} ↛ y, we have that |V2− | ≥ 2 and there is a vertex x′ ∈ V3+ − {x} such that y → x′ . If V1 − {a, y} ↛ x′ , then there is a vertex y′ ∈ V1 − {a, y} such that x′ → y′ . By Claim 1, there exists a vertex z ∈ V ′′ such that y′ → z and abxuyx′ y′ za is a 8-cycle through ab. Assume V1 − {a, y} → x′ . Since {a, y} → x′ , we get V1 → x′ → V2 . Let w ∈ V2− . Then x′ → w. If V3− ↛ w , then there is an arc w z for some z ∈ V3− and abxuyx′ w za is a 8-cycle through ab. Assume V3− → w . By V3− ∪ {x′ } → w and Lemma 2.2, there is a vertex y′ ∈ V1 − {a, y} such that w → y′ . If V ′′ ↛ y′ , then there exists an arc y′ v for some v ∈ V ′′ . Obviously, v ̸ = w and abxuyx′ w y′ v a is a desired 9-cycle. If V ′′ → y′ , then y′ → V ′ . By u → y and Claim 1, there exists a vertex v0 ∈ V ′′ (v0 may be equal to w ) such that y → v0 and so v0 → y′ → x′ . Let w ′ ∈ V2− − {v0 }. Then x′ → w ′ and abxuyv0 y′ x′ w ′ a is a desired 9-cycle. □ By Claim 2, we may assume V3+ − {x} → y when |V3+ | ≥ 2. If V3− → y, then by the assumption above, we always have V3 − {x} → y. Now, N − (y) = (V3 − {x}) ∪ {u} and + N (y) = (V2 − {u}) ∪ {x}. By a → u and Lemma 2.2, there exists a vertex z ∈ V3 − {x} such that u → z. Let w ∈ V2− . Then z → y → {w, x}. By b → x and Lemma 2.2, there exists a vertex y0 ∈ V1 − {a, y} such that x → y0 . If y0 → u, then abxy0 uzyw a is a desired 8-cycle. If u → y0 , then by Lemma 2.2, there exists a vertex x′ ∈ V3 − {x} such that y0 → x′ . Thus, x′ → y and abxuy0 x′ yw a is a desired 8-cycle. Now, we assume V3− ↛ y. Then there is a vertex v ∈ V3− such that y → v . Subcase 1.1 V1 − {a, y} ↛ v . In this case, there exists a vertex y′ ∈ V1 − {a, y} such that v → y′ . If V ′′ ↛ y′ , then there exists a vertex w ∈ V ′′ − {v} such that y′ → w and abxuyv y′ w a is a 8-cycle through ab. Assume V ′′ → y′ → V ′ . By {a, y′ } → u and Lemma 2.2, there exists a vertex v ′ ∈ V3 − {v} such that u → v ′ . Obviously, v ′ ̸ = x. Suppose first that v ′ ∈ V3+ . Then |V3+ | ≥ 2 and y′ → v ′ → y. If V2− ↛ v ′ , then there is an arc v ′ w for some w ∈ V2− and abxuyv y′ v ′ w a is a desired 9-cycle. Assume V2− → v ′ . By {a, y′ } → v ′ and Lemma 2.2, there exists a vertex u′ ∈ V2+ − {b, u} such that v ′ → u′ . Then y′ → u′ . By v ′ → u′ and Lemma 2.2, there is a vertex y0 ∈ V1 − {a, y′ } (y0 may be equal to y) such that u′ → y0 . In addition, by u′ → y0 and Claim 1, there exists a vertex z ∈ V ′′ (z may be equal to v ) such that y0 → z. Then abxuv ′ u′ y0 za is a desired 8-cycle. Suppose now that v ′ ∈ V3− . Then v ′ → y′ . If V2− ↛ x, then there is an arc xw for some w ∈ V2− . Thus, w → y′ and abxw y′ uyv a is a desired 8-cycle. Assume V2− → x. By {a, y′ } → x and Lemma 2.2, there exists a vertex u′ ∈ V2+ − {b, u} such that x → u′ . Then y′ → u′ . If u′ → v , then abxuv ′ y′ u′ v a is a desired 8-cycle. If v → u′ , then by {x, v} → u′ and Lemma 2.2, there is a vertex y0 ∈ V1 − {y} such that u′ → y0 . Since {a, y′ } → u′ , we get y0 ̸ = a, y′ . By u′ → y0 and Claim 1, there exists a vertex z0 ∈ V ′′ such that y0 → z0 . Then abxu′ y0 z0 y′ uv ′ a (when z0 = v ) or abxuyv u′ y0 z0 a (when z0 ̸ = v ) is a desired 9-cycle. Subcase 1.2 V1 − {a, y} → v . By a → u and Lemma 2.2, there exists a vertex z ∈ V3 − {x} (z may be equal to v ) such that u → z. Suppose first that z → y. Obviously, z ̸ = v . If V2− ↛ v , then there exists an arc vw for some w ∈ V2− . Thus, abxuzyvw a is a 8-cycle through ab. Assume V2− → v . Since d− (v ) = r, we have N − (v ) = (V1 − {a}) ∪ V2− and |V2− | = |V3+ | = 1. Then N + (v ) = {a} ∪ V2+ , V3+ = {x} and z ∈ V3− . Let V2− = {w}. Then w → v . If y → w , then ab is contained in a 8-cycle abxuzywv a. Assume w → y. If z → w , then abxuz w yv a is a 8-cycle through ab. Assume w → z. Then w → {a, y, z , v}, r ≥ 4 and |V2+ | ≥ 3. Let u′ ∈ V2+ − {b, u}. Then v → u′ . If u′ → z, then abxuyv u′ za is a desired 8-cycle. If z → u′ , then by {v, z } → u′ and Lemma 2.2, there exists a vertex y′ ∈ V1 − {a, y} such that u′ → y′ . Thus, y′ → v and abxuzu′ y′ v a is a desired 8-cycle. Suppose now that y → z. If V1 −{a, y} ↛ z, then using z instead of v in the subcase 1.1, we may prove that ab is contained in a 8- or 9-cycle. So we assume V1 − {a, y} → z. Now, N − (z) = (V1 − {a}) ∪ {u} and N + (z) = {a} ∪ (V2 − {u}). Let w ∈ V2− . We have z → w . If V3− ↛ w , then there exists a vertex v0 ∈ V3− − {z } such that w → v0 , and abxuyz wv0 a is a desired 8-cycle. Assume V3− → w . If y → w , then by V3− → w and Lemma 2.2, there exists a vertex y′ ∈ V1 − {a, y} such that w → y′ . Now, y′ → z and abxuyw y′ za is a 8-cycle through ab. Assume w → y. If V ′′ − {z } ↛ y, then there is an arc yv ′ for some v ′ ∈ V ′′ − {z , w} and abxuz w yv ′ a is a desired 8-cycle. Assume V ′′ − {z } → y. Then N − (y) = (V ′′ − {z }) ∪ {u} and N + (y) = {z } ∪ (V ′ − {u}) (now, z = v .). Let v ′ ∈ V3− − {z }. If u → v ′ , then abxuv ′ w yza is a desired 8-cycle. If v ′ → u, then v ′ → {a, y, w, u} and r ≥ 4. Note that + N (y) = {z } ∪ (V ′ − {u}) and V3+ − {x} → y when |V3+ | ≥ 2. We have V3+ = {x} and |V2+ | ≥ 3. Let u′ ∈ V2+ − {b, u}. Then {a, y} → u′ . Lemma 2.2 implies that there exists a vertex z ′ ∈ V3 − {x} such that u′ → z ′ . Then z ′ ∈ V3− and z ′ → w . Now, abxuyu′ z ′ w a is a 8-cycle through ab. Subcase 2 V2+ → x. By b → x and Lemma 2.2, there is a vertex y ∈ V1 − {a} such that x → y. In addition, by a → x and Lemma 2.2, there is a vertex u′ ∈ V2 such that x → u′ . Obviously, u′ ∈ V2− . Subcase 2.1 V2+ − {b} ↛ y. In this case, there exists a vertex u ∈ V2+ − {b} such that y → u. Obviously, u → x. Claim 3. If |V3+ | ≥ 2 and V3+ − {x} ↛ u, then ab is contained in a 8- or 9-cycle.
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Proof. By the hypothesis, there is a vertex x′ ∈ V3+ − {x} such that u → x′ . If V1 − {a, y} ↛ x′ , then there exists a vertex y′ ∈ V1 − {a, y} such that x′ → y′ . By Claim 1, there is a vertex w ∈ V ′′ such that y′ → w and ab is contained in a 8-cycle abxyux′ y′ w a. Assume V1 − {a, y} → x′ . Since {a, u} → x′ , we have N − (x′ ) = (V1 − {y}) ∪ {u} and N + (x′ ) = {y} ∪ (V2 − {u}). So x′ → {y, u′ }. If V3− ↛ u′ , then there is an arc u′ w ′ for some w ′ ∈ V3− and abxyux′ u′ w ′ a is a 8-cycle through ab. If V3− → u′ , then V3− ∪ {x, x′ } → u′ . Lemma 2.2 implies that there is a vertex y′ ∈ V1 − {a, y} such that u′ → y′ . Obviously, y′ → x′ . By {x, x′ } → y and Claim 1, there is a vertex z ∈ V ′′ − {u′ } such that y → z. Thus, abxu′ y′ x′ yza is a desired 8-cycle. □ By Claim 3, we may assume V3+ − {x} → u when |V3+ | ≥ 2 and then NV+ −{x} (u) ⊆ V3− . By {a, y} → u and Lemma 2.2, there 3 exists a vertex v ∈ V3 − {x} such that u → v . Then v ∈ V3− . Suppose first that V1 − {a, y} ↛ v . Then there exists an arc v y′ for some y′ ∈ V1 − {a, y}. If V ′′ ↛ y′ , then there is a vertex z ∈ V ′′ − {v} such that y′ → z, and abxyuv y′ za is a 8-cycle through ab. If V ′′ → y′ , then y′ → V ′ . By {a, y′ } → x and Lemma 2.2, there exists a vertex w ∈ V2 − {u′ } such that x → w . Recalling that V2+ → x, we have w ∈ V2− − {u′ }. So |V2− | = |V3+ | ≥ 2. Let x′ ∈ V3+ − {x}. Then u′ → y′ → x′ → u and abxu′ y′ x′ uv a is a 8-cycle through ab. Suppose now that V1 − {a, y} → v . If u′ → v , then N − (v ) = (V1 − {a, y}) ∪ {u, u′ } and v → y. In the case when V3 − {x, v} ↛ u, there is an arc uv ′ for some v ′ ∈ V3 − {x, v}. By the assumption above, v ′ ∈ V3− − {v} and abxu′ v yuv ′ a is a desired 8-cycle. In the other case when V3 − {x, v} → u, we have N − (u) = {a, y} ∪ (V3 − {x, v}) and N + (u) = (V1 − {a, y}) ∪ {x, v}. By V2+ → x and Lemma 2.2, there exists a vertex y′ ∈ V1 − {a, y} such that x → y′ , and then u → y′ . When y′ → u′ , we see that ab is contained in a 8-cycle abxyuy′ u′ v a. When u′ → y′ , by {x, u} → y′ and Claim 1, there exists a vertex z ∈ V ′′ − {v} such that y′ → z. Then z ̸ = u′ and abxu′ v yuy′ za is a desired 9-cycle. Assume v → u′ . If V3− ↛ u′ , then there is an arc u′ w ′ for some w ′ ∈ V3− − {v}, and abxyuv u′ w ′ a is a 8-cycle through ab. If − V3 → u′ , then V3− ∪ {x} → u′ . Lemma 2.2 implies that there is a vertex y′ ∈ V1 − {a, y} such that u′ → y′ . Obviously, y′ → v . In the case when u → y′ , we know that abxyuy′ v u′ a is a 8-cycle through ab. In the other case when y′ → u, by {a, y, y′ } → u and Lemma 2.2, there is a vertex v ′ ∈ V3 − {x, v} such that u → v ′ . Recalling NV+ −{x} (u) ⊆ V3− , we have v ′ ∈ V3− and v ′ → u′ . 3 Then abxyuv ′ u′ y′ v a is a desired 9-cycle. Subcase 2.2 V2+ − {b} → y. Let u ∈ V2+ − {b}. Then u → {x, y}. Subcase 2.2.1 y → u′ . By {u, x} → y and Claim 1, there exists a vertex w ∈ V ′′ − {u′ } such that y → w . First suppose that V3+ ↛ u′ . Then there is a vertex x′ ∈ V3+ − {x} such that u′ → x′ . If x′ → u, then abxu′ x′ uyw a is a 8-cycle through ab. Assume u → x′ . By {u, u′ } → x′ and Lemma 2.2, there exists a vertex y′ ∈ V1 − {a, y} such that x′ → y′ . If y′ → u, then abxu′ x′ y′ uyw a is a 9-cycle through ab. Assume u → y′ . By {u, x′ } → y′ and Claim 1, there exists a vertex w ′ ∈ V ′′ − {u′ } such that y′ → w ′ . Then abxyu′ x′ y′ w ′ a is a 8-cycle through ab. Now suppose that V3+ → u′ . By y → u′ and Lemma 2.2, there exists a vertex v ∈ V3− (v may be equal to w ) such that u′ → v . Assume V1 − {a, y} → v . By V2+ → x and Lemma 2.2, there exists a vertex y′ ∈ V1 − {a} such that x → y′ . Obviously, ′ y ̸ = y and y′ → v . If v → u, then abxy′ v uyu′ a is a 8-cycle through ab. If u → v , then N − (v ) = (V1 − {a, y}) ∪ {u, u′ } and N + (v ) = {a, y} ∪ (V2 − {u, u′ }). Recalling y → w and w ∈ V ′′ − {u′ }, we have v ̸ = w . Then abxy′ u′ v yw a (when y′ → u′ ) or abxu′ y′ v yw a (when u′ → y′ ) is a desired 8-cycle. Assume V1 − {a, y} ↛ v . Then there exists a vertex y0 ∈ V1 − {a, y} such that v → y0 . If V ′′ − {u′ } ↛ y0 , then there is a vertex z ′ ∈ V ′′ − {u′ } such that y0 → z ′ . Obviously, z ′ ̸ = v and abxyu′ v y0 z ′ a is a 8-cycle through ab. If V ′′ − {u′ } → y0 , then w → y0 . In the case when u′ → y0 , we have V ′′ → y0 → V ′ and abxu′ y0 uyw a is a 8-cycle through ab. In the other case when y0 → u′ , by {y, y0 } → u′ and Lemma 2.2, there exists a vertex v ′ ∈ V3 − {v} such that u′ → v ′ . Since V3+ → u′ , we get v ′ ∈ V3− − {v}. Thus, abxywy0 u′ v ′ a (when w = v ) or abxywy0 u′ v a (when w ̸= v ) is a 8-cycle through ab. Subcase 2.2.2 u′ → y. Suppose first that V3+ ↛ y. Then there exists a vertex x′ ∈ V3+ − {x} such that y → x′ . If x′ → u, then in the case when V3− ↛ u, there exists a vertex w ∈ V3− such that u → w and abxu′ yx′ uw a is a 8-cycle through ab. In the other case when V3− → u, by V3− ∪ {x′ } → u and Lemma 2.2, there exists a vertex y′ ∈ V1 − {a, y} such that u → y′ . Claim 1 yields that there is a vertex w ′ ∈ V ′′ (w ′ may be equal to u′ ) such that y′ → w ′ . Thus, abxyx′ uy′ w ′ a is a 8-cycle through ab. If u → x′ , then by Lemma 2.2, there exists a vertex y′ ∈ V1 such that x′ → y′ . Since {a, y} → x′ , we know y′ ̸ = a, y. In the case when V ′′ − {u′ } ↛ y′ , there is an arc y′ w for some w ∈ V ′′ − {u′ }, and abxu′ yx′ y′ w a is a 8-cycle through ab. In the other case when V ′′ − {u′ } → y′ , we get N − (y′ ) = (V ′′ − {u′ }) ∪ {x′ } and N + (y′ ) = {u′ } ∪ (V ′ − {x′ }). Then y′ → {u, u′ }. By {u, x} → y and Claim 1, there is a vertex z ∈ V ′′ − {u′ } such that y → z. Thus z → {y′ , a}, and abxu′ x′ y′ uyza (when u′ → x′ ) or abxyzy′ ux′ u′ a (when x′ → u′ ) is a 9-cycle through ab. Suppose now that V3+ → y. Since (V2+ − {b}) ∪ {u′ } → y, we get N − (y) = (V ′ − {b}) ∪ {u′ } and N + (y) = {b} ∪ (V ′′ − {u′ }). Let v, v ′ ∈ V3− be two distinct vertices. Then y → {v, v ′ }. If V1 − {a, y} → v , then by V2+ → x and Lemma 2.2, there exists a vertex y′ ∈ V1 − {y} such that x → y′ . Obviously, y′ ̸ = a and y′ → v . When v → u′ , we know that abxy′ v u′ yv ′ a is a 8-cycle through ab. When u′ → v , we get N − (v ) = (V1 −{a}) ∪{u′ } and N + (v ) = {a} ∪ (V2 − {u′ }). Thus, v → u and abxu′ v uyv ′ a is a 8-cycle through ab.
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If V1 − {a, y} ↛ v , then there is an arc v y′ for some y′ ∈ V1 − {a, y}. When V ′′ − {u′ } ↛ y′ , then there is an arc y′ z for some z ∈ V ′′ − {u′ , v}, and abxu′ yv y′ za is a 8-cycle through ab. When V ′′ − {u′ } → y′ and u′ → y′ , we have V ′′ → y′ → V ′ and abxu′ y′ uyv a is a desired 8-cycle. Assume now that V ′′ − {u′ } → y′ → u′ . In the case when y′ → x, we have {a, y′ } → x. Lemma 2.2 implies that there is a vertex w ∈ V2 − {u′ } such that x → w . Recalling that V2+ → x, we have w ∈ V2− − {u′ }. Therefore, y → w → y′ and abxywv y′ u′ a (when w → v ) or abxyvw y′ u′ a (when v → w ) is a desired 8-cycle. In the other case when x → y′ , we know N − (y′ ) = (V ′′ − {u′ }) ∪ {x} and N + (y′ ) = {u′ } ∪ (V ′ − {x}). Then v ′ → y′ → u and abxyv ′ y′ u′ v a (when u′ → v ) or abxy′ uyv u′ a (when v → u′ ) is a desired 8-cycle. □ Proof of the theorem under Case B2. In this case, we have 2 ≤ |V2+ | = |V3− | ≤ r − 1, 1 ≤ |V2− | = |V3+ | ≤ r − 2 and V3+ → b. By Lemma 2.2, there is an arc by for some y ∈ V1 − {a}. Similarly, by a → b and Lemma 2.2, there exists a vertex c ∈ V3− such that b → c. Subcase 1 V3+ ↛ y. By the hypothesis, there is a vertex x ∈ V3+ such that y → x. Subcase 1.1 V1 − {a, y} → x. In this case, we have V1 → x → V2 . Let u ∈ V2+ − {b} and w ∈ V2− . Then x → {u, w}. By a → u and Lemma 2.2, there is a vertex z ∈ V3 − {x} (z may be equal to c) such that u → z. Suppose first that V1 − {a, y} ↛ z. Then there exists a vertex y′ ∈ V1 − {a, y} such that z → y′ . If V ′′ ↛ y′ , then there is a vertex z ′ ∈ V ′′ − {z } such that y′ → z ′ , and abyxuzy′ z ′ a is a 8-cycle through ab. If V ′′ → y′ , then y′ → V ′ , z ∈ V ′′ and abyxw y′ uza is a desired 8-cycle. Suppose now that V1 − {a, y} → z. If w → z, then N − (z) = (V1 − {a, y}) ∪ {w, u} and N + (z) = {a, y} ∪ (V2 − {u, w}). So z → {b, y} and z ̸ = c. By {x, z } → b and Lemma 2.2, there exists a vertex y0 ∈ V1 − {a, y} such that b → y0 . Thus, y0 → z and abcuzyxw a (when c → u) or aby0 zyxuca (when u → c) is a 8-cycle through ab. Assume z → w . If V3− ↛ w , then there exists a vertex z ′ ∈ V3− − {z } such that w → z ′ , and abyxuz w z ′ a is a 8-cycle through ab. If V3− → w , then by V3− ∪ {x} → w and Lemma 2.2, there exists a vertex y′ ∈ V1 − {a, y} such that w → y′ . So y′ → z. When z ∈ V3− , we get that abyxw y′ uza (when y′ → u) or abyxuy′ z w a (when u → y′ ) is a 8-cycle through ab. When z ∈ V3+ , we have |V3+ | = |V2− | ≥ 2, a → z, N − (z) = (V1 − {y}) ∪ {u} and N + (z) = {y} ∪ (V2 − {u}). Let v ∈ V2− − {w}. Then y′ → z → v and abyxw y′ z v a is a 8-cycle through ab. Subcase 1.2 V1 − {a, y} ↛ x. By the hypothesis, there exists a vertex y′ ∈ V1 − {a, y} such that x → y′ . Subcase 1.2.1 V2+ − {b} ↛ y′ . In this case, there exists a vertex u ∈ V2+ − {b} such that y′ → u. If V1 − {y} ↛ u, then there is an arc uy1 for some y1 ∈ V1 − {a, y, y′ }. By u → y1 , u ∈ V ′ and Claim 1, there exists a vertex v ∈ V ′′ such that y1 → v . Then abyxy′ uy1 v a is a desired 8-cycle. Assume V1 − {y} → u. If c → u, then N − (u) = (V1 − {y}) ∪ {c } and N + (u) = {y} ∪ (V3 − {c }). When y′ → c, we have that abyxy′ cuv a is a desired 8-cycle where v ∈ V3− − {c }. When c → y′ , by x → y′ and Claim 1, there is a vertex w ′ ∈ V ′′ − {c } such that y′ → w ′ , and abcuyxy′ w ′ a is a desired 8-cycle. Assume u → c. If V2− ↛ c, then there is an arc c w for some w ∈ V2− and abyxy′ uc w a is a desired 8-cycle. So assume V2− → c. If V3− − {c } ↛ u, then there exists an arc uv for some v ∈ V3− − {c }. When c → y, we know that abcyxy′ uv a is a desired 8-cycle. Assume y → c. By {b, u} ∪ V2− → c and Lemma 2.2, there exists a vertex y1 ∈ V1 − {a, y, y′ } such that c → y1 . If V ′′ ↛ y1 , then there is an arc y1 z1 for some z1 ∈ V ′′ − {c }, and abyxy′ ucy1 z1 a is a desired 9-cycle. If V ′′ → y1 , then y1 → V ′ and abcy1 xy′ uv a is a desired 8-cycle. Assume V3− − {c } → u. Recalling that V1 − {y} → u and d− (u) = r, we get |V3− − {c }| = 1 and u → {y, c } ∪ V3+ . Let V3− − {c } = {v}. If |V3+ | = |V2− | ≥ 2, then let x′ ∈ V3+ − {x}. When V2− ↛ x′ , then there is an arc x′ w for some w ∈ V2− and abyxy′ ux′ w a is a 8-cycle through ab. When V2− → x′ , by V2− ∪ {u} → x′ and Lemma 2.2, there is a vertex y1 ∈ V1 − {a, y, y′ } such that x′ → y1 . By x′ → y1 and Claim 1, there exists a vertex v ′ ∈ V ′′ (v ′ may be equal to c or v ) such that y1 → v ′ . Thus, abyxy′ ux′ y1 v ′ a is a 9-cycle through ab. If |V3+ | = |V2− | = 1, then V3+ = {x} and r = 3. Let V2− = {w}. When y′ → v , abyxy′ v uca is a 8-cycle through ab. When v → y′ , we have v → {a, y′ , u}, {y, b, w} → v and {a, y, u} → x → {b, y′ , w}. So abyxwv uca is a desired 8-cycle. Subcase 1.2.2 V3+ ↛ y′ . By the hypothesis, there exists a vertex x′ ∈ V3+ − {x} such that y′ → x′ . So |V3+ | = |V2− | ≥ 2 and r ≥ 4. If V1 − {y} ↛ x′ , then there exists a vertex y1 ∈ V1 − {a, y, y′ } such that x′ → y1 . By Claim 1, there is a vertex w ∈ V ′′ such that y1 → w . Thus, abyxy′ x′ y1 w a is a 8-cycle through ab. Assume V1 − {y} → x′ . By Lemma 2.2, there are two vertices u, u′ ∈ V2 − {b} such that x′ → {u, u′ }. Furthermore, V2− ↛ x′ (as otherwise, (V1 − {y}) ∪ V2− → x′ and d− (x′ ) ≥ r + 1, a contradiction.). Without loss of generality, we suppose u′ ∈ V2− . If V3− ↛ u, then there is an arc uv ′ for some v ′ ∈ V3− , and abyxy′ x′ uv ′ a is a 8-cycle through ab. Assume V3− → u. If V1 − {a, y, y′ } → u, then (V1 − {a, y, y′ }) ∪ V3− ∪ {x′ } → u. By |V3− | ≥ 2 and d− (u) = r, we have |V3− | = 2 and N − (u) = (V1 − {a, y, y′ }) ∪ V3− ∪ {x′ }. So c → u → x and abcuxy′ x′ u′ a is a 8-cycle through ab. Assume V1 − {a, y, y′ } ↛ u.
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Then there exists a vertex y1 ∈ V1 − {a, y, y′ } such that u → y1 . If V ′′ ↛ y1 , then there is an arc y1 z1 for some z1 ∈ V ′′ − {u} and abyxy′ x′ uy1 z1 a is a 9-cycle through ab. If V ′′ → y1 , then y1 → V ′ and abcy1 xy′ x′ u′ a is a 8-cycle through ab. Subcase 1.2.3 V2+ − {b} → y′ and V3+ → y′ . In this case, we have V ′ − {b} → y′ . Suppose first that V2− → y′ . Then |V2− | = 1 and N − (y′ ) = (V ′ − {b}) ∪ V2− , which lead to |V3− | = r − 1, V3+ = {x} and + ′ N (y ) = {b} ∪ V3− . Also, we have V2− ↛ V3− (as otherwise, V2− → {a, y′ } ∪ V3− , this is impossible.). Then there exists an arc z w from V3− to V2− and furthermore, y′ → z. If V3− ↛ w , then there exists an arc w z ′ for some z ′ ∈ V3− − {z }, and abyxy′ z w z ′ a is a 8-cycle through ab. Assume V3− → w . If x → w , then V3 → w → V1 and abc w yxy′ v a is a desired 8-cycle for any vertex v ∈ V3− − {c }. Assume w → x. By {a, y} → x and Lemma 2.2, there exists a vertex u ∈ V2 − {b} such that x → u. Obviously, u ̸ = w and abyxuy′ c w a is a desired 8-cycle. Suppose now that V2− ↛ y′ . Then there is an arc y′ w for some w ∈ V2− . By V2+ − {b} → y′ and Lemma 2.2, there exists a vertex v ∈ V3 (v may be equal to c) such that y′ → v . Note that V3+ → y′ . So v ∈ V3− . Let u ∈ V2+ − {b}. If x → u, then abyxuy′ wv a (when w → v ) or abyxuy′ vw a (when v → w ) is a 8-cycle through ab. Assume u → x. If y → u, then abyuxy′ wv a (when w → v ) or abyuxy′ vw a (when v → w ) is a 8-cycle through ab. Assume u → y. If c → u, then abcuyxy′ w a is a desired 8-cycle. Assume u → c. Now, we have u → {y, y′ x, c } and r ≥ 4. By the arbitrariness of u, we may assume V2+ − {b} → {y, y′ , x, c }. If V1 − {a, y} ↛ w , then there exists a vertex y1 ∈ V1 − {a, y, y′ } such that w → y1 . When V ′′ ↛ y1 , then there is an arc y1 z1 for some z1 ∈ V ′′ . Thus, z1 ̸ = w and abyxy′ w y1 z1 a is a 8-cycle through ab. When V ′′ → y1 , we have y1 → V ′ . Thus, c → y1 → u and abcy1 uxy′ w a is a 8-cycle through ab. Assume V1 − {a, y} → w . Similarly, we may assume V1 − {a, y} → v . Consider the case v → w . If V3− ↛ w , then there exists an arc wv ′ for some v ′ ∈ V3− . Thus, v ′ ̸ = v and abyxy′ vwv ′ a is a 8cycle through ab. Assume V3− → w . Note that V1 −{a, y} → w and d− (w ) = r. We have |V3− | = 2, N − (w ) = (V1 −{a, y}) ∪ V3− and N + (w ) = {a, y} ∪ V3+ . So c → w → y. By V ′ − {b} → y′ and Claim 1, we get d+ (y′ ) ≥ r − 1 ≥ 3. Then there is a vertex V ′′ ′ ′′ ′ ′ ′ w ∈ V − {c , w} such that y → w and abc wyxy w a is a 8-cycle through ab. Consider the other case w → v . If V2− ↛ v , then there exists an arc v z ′ for some z ′ ∈ V2− . Thus, z ′ ̸ = w and abyxy′ wv z ′ a is a 8-cycle through ab. Assume V2− → v . If V2+ − {b} → v , then (V1 − {a, y}) ∪ (V2 − {b}) → v . Note that d− (v ) = r. We have |V1 − {a, y}| = 1 and r = 3, which contradicts r ≥ 4. So we have V2+ − {b} ↛ v and there exists a vertex u′ ∈ V2+ − {b} such that v → u′ . Recalling V2+ − {b} → {y, y′ , x, c }, we have u′ → {y, y′ , x, c } and v ̸ = c. Thus, abyxy′ v u′ ca is a 8-cycle through ab. Subcase 2 V3+ → y. Let x ∈ V3+ be arbitrary. Subcase 2.1 V2+ → y. In this case, we have V ′ → y → V ′′ . Note that {b, y} → c → a and a → x → {b, y}. Then there exists a vertex u ∈ (V1 −{a, y}) ∪ (V2 −{b}) such that c → u → x (as otherwise, we have d+ (c) < d+ (x), this is impossible.). Let v ∈ V3− −{c }. If u ∈ (V1 − {a, y}) ∪ (V2+ − {b}), then abcuxywv a (when w → v ) or abcuxyvw a (when v → w ) is a 8-cycle through ab for any vertex w ∈ V2− . If u ∈ V2− and |V2− | ≥ 2, then abcuxywv a (when w → v ) or abcuxyvw a (when v → w ) is a desired 8-cycle for any vertex w ∈ V2− − {u}. Assume that u ∈ V2− , |V2− | = 1 and x dominates the outset of c in (V1 − {a, y}) ∪ (V2+ − {b}). Now, we have |V2+ | = r − 1. Note that {b, y} → c. We have (V1 − {a, y}) ∪ (V2+ − {b}) ↛ c. Otherwise, d− (c) ≥ 2r − 2 > r, a contradiction. Therefore, there exists an arc cu1 for some u1 ∈ (V1 − {a, y}) ∪ (V2+ − {b}). By the assumption above, we have x → u1 . If u1 ∈ V2+ − {b}, then u1 → y and abcuxu1 yv a is a desired 8-cycle through ab. If u1 ∈ V1 − {a, y}, then by {x, c } → u1 and Lemma 2.2, there is a vertex w ∈ V2 − {b} such that u1 → w . When w = u, we have that abcu1 uxyv a is a desired 8-cycle. When w ̸ = u, we have w ∈ V2+ − {b} and w → y. Now, abcuxu1 w yv a is a 9-cycle through ab. Subcase 2.2 V2+ ↛ y. By the hypothesis, there exists a vertex u ∈ V2+ − {b} such that y → u. Suppose first that V1 −{a, y} → u. Then V1 → u → V3 . If V1 −{a, y} ↛ c, then there is an arc cy1 for some y1 ∈ V1 −{a, y}. By V3+ ∪ {b} → y and Claim 1, there is a vertex v ∈ V ′′ − {c } such that y → v . Thus, abcy1 uxyv a is a 8-cycle through ab. Assume V1 − {a, y} → c. Then N − (c) = (V1 − {a, y}) ∪ {b, u} and N + (c) = {a, y} ∪ (V2 − {b, u}). Let y′ ∈ V1 − {a, y} and w ∈ V2− . Then y′ → c → w. If x → y′ , then abyuxy′ c wa is a desired 8-cycle. If y′ → x, then y′ → {x, c }. Lemma 2.2 implies that there is a vertex u′ ∈ V2 − {b} such that u′ → y′ . Obviously, u′ ̸ = u and c → u′ . By V3+ ∪ {b} → y and Claim 1, there exists a vertex c ′ ∈ V ′′ − {u′ } such that y → c ′ . Since c → y, we have that c ′ ̸ = c and abcu′ y′ xyc ′ a is a 8-cycle through ab. Suppose now that V1 − {a, y} ↛ u. Then there exists a vertex y′ ∈ V1 − {a, y} such that u → y′ . Subcase 2.2.1 V3+ ↛ y′ . By the hypothesis, there is an arc y′ x′ for some x′ ∈ V3+ (x′ may be equal to x). Recalling that V3+ → y, we get x′ → y. If V1 − {y} ↛ x′ , then there exists a vertex y1 ∈ V1 − {a, y, y′ } such that x′ → y1 . By x′ → y1 and Claim 1, there is a vertex w ∈ V ′′ such that y1 → w . Thus, abyuy′ x′ y1 w a is a 8-cycle through ab. Assume V1 − {y} → x′ . If V2 − {b, u} ↛ x′ , then there exists an arc x′ u′ for some u′ ∈ V2 − {b, u}. When V3− ↛ u′ , then there is an arc u′ w ′ for some w ′ ∈ V3− . Thus, abyuy′ x′ u′ w ′ a is a 8-cycle through ab. When V3− → u′ , by V3− ∪ {x′ } → u′ and Lemma 2.2, there exists a vertex y1 ∈ V1 − {a, y} (y1 may be equal to y′ ) such that u′ → y1 . Then y1 → x′ . If |V3+ | ≥ 2, then by V3+ ∪ {b} → y and Claim 1, there exists a vertex v ′ ∈ V ′′ − {u′ , c } such that y → v ′ . Now, abcu′ y1 x′ yv ′ a is a 8-cycle through ab. If |V3+ | = 1, then
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V3+ = {x′ }, V3 → u′ → V1 and so u′ → y. By V3+ ∪ {b} → y and Claim 1, there exists a vertex v ′ ∈ V ′′ − {c } such that y → v ′ . Obviously, v ′ ̸ = u′ and abcu′ y1 x′ yv ′ a is a 8-cycle through ab. If V2 − {b, u} → x′ , then |V2 − {b, u}| = 1 since V1 − {y} → x′ and d− (x′ ) = r. Let V2 − {b, u} = {w}. Then r = 3 and {a, y′ , w} → x′ → {y, b, u}. So {a, y, x′ } → u → {y′ } ∪ V3− . By {b, u} → c, we get c → y′ or c → w. Let v ′ ∈ V3− − {c }. Then u → v ′ and abcy′ x′ yuv ′ a or abc w x′ yuv ′ a is a 8-cycle through ab. Subcase 2.2.2 V3+ → y′ . In this case, we have x → y′ . Suppose first that V2+ − {b} ↛ y′ . Then there exists a vertex u′ ∈ V2+ − {b, u} such that y′ → u′ . If V1 − {y} ↛ u′ , then there exists a vertex y1 ∈ V1 − {a, y, y′ } such that u′ → y1 . By u′ → y1 and Claim 1, there is a vertex w1 ∈ V ′′ such that y1 → w1 and abyuy′ u′ y1 w1 a is a 8-cycle through ab. Assume V1 − {y} → u′ . By Lemma 2.2, we have |NV+ (u′ )| ≥ r − 1. Note 3 that |V3− | ≥ 2. Then there exists a vertex v ′ ∈ V3− such that u′ → v ′ . If u → x, then ab is contained in a 8-cycle abyuxy′ u′ v ′ a. Assume x → u. Then x → {y, y′ , b, u} and r ≥ 4. Let w ∈ V2− . If w ↛ NV+ (u′ ), then there exists an arc v0 w for some v0 ∈ NV+ (u′ ). Now, abyuy′ u′ v0 w a is a desired 8-cycle. Assume 3 3 w → NV+3 (u′ ). Note that w → a and |NV+3 (u′ )| ≥ r − 1. We have N + (w) = NV+3 (u′ ) ∪ {a}, |NV+3 (u′ )| = r − 1 and V1 − {a} → w. Let y1 ∈ V1 − {a, y, y′ }. Then y1 → {u′ , w}. By Lemma 2.2, there exist two vertices v1 , v2 in V3 such that {v1 , v2 } → y1 . Since |NV+ (u′ )| = r − 1, there is at least one vertex in {v1 , v2 } which belongs to NV+ (u′ ). Without loss of generality, assume 3 3 v1 ∈ NV+3 (u′ ). Then abyuy′ u′ v1 y1 wa is a desired 9-cycle. Suppose now that V2+ − {b} → y′ . Note that V3+ → y′ . We have V ′ − {b} → y′ . If V2− → y′ , then it is easy to see that |V2− | = |V3+ | = 1 and N − (y′ ) = (V ′ − {b}) ∪ V2− . So |V3− | = r − 1, V3+ = {x} and N + (y′ ) = {b} ∪ V3− . Furthermore, we have V2− ↛ V3− (as otherwise, V2− → {a, y′ } ∪ V3− , this is impossible.). Then there exists an arc z w from V3− to V2− (z may be equal to c). If V3− ↛ w , then there exists an arc w z ′ for some z ′ ∈ V3− − {z }, and abyuy′ z w z ′ a is a 8-cycle through ab. Assume V3− → w . Let v ∈ V3− − {c }. If x → w , then V3 → w → V1 and abc w yuy′ v a is a desired 8-cycle. If w → x, then abc w xyuy′ v a is a desired 9-cycle. If V2− ↛ y′ , then there is an arc y′ w for some w ∈ V2− . By u → y′ and Lemma 2.2, there exists a vertex v ∈ V3 such that y′ → v . Note that V3+ → y′ . We have v ∈ V3− . If u → x, then abyuxy′ wv a (when w → v ) or abyuxy′ vw a (when v → w ) is a desired 8-cycle. Assume x → u. Now, x → {y, y′ , b, u} and r ≥ 4. + ′ Let X = (V1 ∪ V2 ) − {a, y, y′ , b, u}. When NX+ (c) ⊆ NX+ (x), we have d+ X (c) ≤ dX (x). Note that x → {y, y , b, u}. Then c → {a, y, y′ , u} (as otherwise, d+ (c) < d+ (x), this is impossible.). Since y′ → v , we get that v ̸ = c and abcyuy′ wv a (when w → v ) or abcyuy′ vwa (when v → w) is a desired 8-cycle. When NX+ (c) ⊊ NX+ (x), then there exists a vertex u0 ∈ X such that c → u0 → x. If u0 ̸ = w , then abcu0 xyuy′ w a is a 9-cycle through ab. Assume u0 = w . If V ′′ − {w, c } ↛ y′ , then there is an arc y′ v ′ for some v ′ ∈ V ′′ −{w, c } and abc w xyuy′ v ′ a is a desired 9-cycle. If V ′′ −{w, c } → y′ , then N − (y′ ) = (V ′ −{b})∪(V ′′ −{w, c }) and |V ′′ − {w, c }| = 1. Now, r = 3 which contradicts with r ≥ 4. □ The proof of Theorem 2.2 is complete. ■ Combining Theorem 2.1 with Theorem 2.2, we have the following theorem. Theorem 2.3. If D is an r-regular 3-partite tournament with r ≥ 3, then each arc of D is contained in a 8- or 9-cycle. Acknowledgments We would like to thank the referees for their comments and suggestions which improved the presentation. References [1] [2] [3] [4] [5]
Q. Guo, L. Cui, W. Meng, On cycles through an arc in regular 3-partite tournaments, Discrete Math. 339 (2016) 17–20. I. Stella, L. Volkmann, Each regular 3-partite tournaments has a 9-cycle, An. Univ. Bucuresti Mat. 56 (2007) 91–120. L. Volkmann, On cycles in regular 3-partite tournaments, Discrete Math. 306 (2006) 1198–1206. L. Volkmann, Multipartite tournaments: a survey, Discrete Math. 307 (2007) 3097–3129. G. Xu, S. Li, Q. Guo, H. Li, Notes on cycles through a vertex or an arc in regular 3-partite tournaments, Appl. Math. Lett. 25 (2012) 662–664.
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On cycles in regular 3-partite tournaments✩ ∗
Qiaoping Guo , Wei Meng School of Mathematical Sciences, Shanxi University, Taiyuan, 030006, China
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Article history: Received 3 April 2018 Received in revised form 8 July 2018 Accepted 12 December 2018 Available online 30 January 2019 Keywords: Multipartite tournament Regular 3-partite tournament Cycle
a b s t r a c t A c-partite tournament is an orientation of a complete c-partite graph. In 2006, Volkmann conjectured that every arc of a regular 3-partite tournament D is contained in an m-, (m+1)or (m + 2)-cycle for each m ∈ {3, 4, . . . , |V (D)| − 2}, and this conjecture was proved to be correct for 3 ≤ m ≤ 7. In 2012, Xu et al. conjectured that every arc of an r-regular 3-partite tournament D with r ≥ 2 is contained in a (3k − 1)- or 3k-cycle for k = 2, 3, . . . , r. They proved that this conjecture is true for k = 2. In this paper, we confirm this conjecture for k = 3, which also implies that Volkmann’s conjecture is correct for m = 7, 8. © 2019 Elsevier Ltd. All rights reserved.
1. Terminology and introduction We consider finite digraphs without loops or multiple arcs. The vertex set and the arc set of a digraph D are denoted by V (D) and A(D), respectively. If xy is an arc of D, then we say x dominates y and write x → y. More generally, if A and B are two disjoint subdigraphs of D such that every vertex of A dominates every vertex of B, then we say that A dominates B and denote it by A → B. Otherwise, we denote it by A ↛ B. Let D′ be a subdigraph of D. We use |D′ | or |V (D′ )| to stand for the number of the vertices of D′ . The outset ND+′ (x) of a vertex x is the set of vertices of D′ dominated by x and the inset ND−′ (x) is the set of vertices of D′ dominating x. We call the numbers d+ (x) = |ND+′ (x)| and d− (x) = |ND−′ (x)| the out-degree and in-degree of x D′ D′ ′ ′ + − + − in D , respectively. When D = D, we usually use N (x), N (x), d (x) and d (x) instead of ND+′ (x), ND−′ (x), d+ (x) and d− (x), D′ D′ respectively. A digraph D is said to be k-regular or regular, if there is an integer k such that d+ (x) = d− (x) = k holds for every x ∈ V (D). A c-partite or multipartite tournament is an orientation of a complete c-partite graph. An m-cycle is a directed cycle of length m. The structure of cycles in multipartite tournaments has been extensively studied; see for example [4]. However, many statements on cycles in c-partite tournaments are only valid for c ≥ 4. At present, the results on cycles in 3-partite tournaments are still very few. For cycles through an arc in a regular 3-partite tournament, Volkmann[3] posed the following conjecture. Conjecture 1.1 ([3]). If D is a regular 3-partite tournament, then every arc of D is contained in an m-, (m + 1)- or (m + 2)-cycle for each m ∈ {3, 4, . . . , |V (D)| − 2}. In [2,3], Volkmann and Stella confirmed Conjecture 1.1 for m ∈ {3, 4, . . . , 7}. In 2012, Xu et al. posed the following two conjectures. ✩ This work is supported by the National Natural Science Foundation of China (No. 11701349) and by the Natural Science Foundation of Shanxi Province, China (No. 201601D011005) and by Shanxi Scholarship Council of China (2017-018). ∗ Corresponding author. E-mail address: [email protected] (Q. Guo). https://doi.org/10.1016/j.disc.2018.12.007 0012-365X/© 2019 Elsevier Ltd. All rights reserved.
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Conjecture 1.2 ([5]). If D is an r-regular 3-partite tournament with r ≥ 2, then every arc of D is contained in a 3k- or (3k + 1)-cycle for k = 1, 2, . . . , r − 1. Conjecture 1.3 ([5]). If D is an r-regular 3-partite tournament with r ≥ 2, then every arc of D is contained in a (3k − 1)- or 3k-cycle for k = 2, 3, . . . , r. In [3], Volkmann proved that every arc of a regular 3-partite tournament D is contained in a 3- or 4-cycle, which implies that Conjecture 1.2 is true for k = 1. It is also proved that Conjecture 1.2 is true for k = 2 in [1]. Xu et al. [5] proved that every arc of a regular 3-partite tournament is contained in a 5- or 6-cycle, which also implies that Conjecture 1.3 is correct for k = 2. In this paper, we confirm Conjecture 1.3 for k = 3, which also implies that Conjecture 1.1 is true for m = 7,8. 2. Main result The following two lemmas are important to prove our main results. Lemma 2.1. If D is an r-regular 3-partite tournament with partite sets V1 , V2 , V3 and v is a vertex of D, then |V1 | = |V2 | = |V3 | = r and d+ (v ) = d− (v ) = r. − Lemma 2.2 ([5]). If D is an r-regular 3-partite tournament with partite sets V1 , V2 , V3 and u is a vertex of V1 , then d+ V2 (u) = dV3 (u) + − and dV (u) = dV (u). 2
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Theorem 2.1. Let D be an r-regular 3-partite tournament with r ≥ 3 and let V1 , V2 , V3 be the partite sets of D. If ab is an arc from V1 to V2 satisfying V3 → a → V2 , then ab is contained in a 8- or 9-cycle. Proof. By Lemma 2.1, we have |V1 | = |V2 | = |V3 | = r and d+ (v ) = d− (v ) = r for each vertex v of D. By a → b and Lemma 2.2, there exists a vertex x ∈ V3 such that b → x. We distinguish the following two cases. Case 1 V2 − {b} ↛ x. By the hypothesis, there is a vertex y ∈ V2 − {b} such that x → y. By Lemma 2.2, there is a vertex u ∈ V1 such that y → u. Since a → V2 , we have a → y and u ̸ = a. By a → y and Lemma 2.2, there exists a vertex v ∈ V3 such that y → v . Obviously, we have v ̸ = x and v → a. Subcase 1.1 V2 − {b, y} ↛ v . In this case, there exists w ∈ V2 − {b, y} such that v → w . Suppose first that V3 − {x, v} ↛ w . Then there is a vertex v ′ ∈ V3 − {x, v} such that w → v ′ . If u → v , we get a 8-cycle abxyuvwv ′ a through ab. Assume v → u. If u → w , then abxyv uwv ′ a is a 8-cycle through ab. If w → u, then by {y, w} → u and Lemma 2.2, there exists a vertex z ′ ∈ V3 − {x} such that u → z ′ (z ′ may be equal to v ′ , but not equal to v ). Therefore, abxyvw uz ′ a is a 8-cycle through ab. Suppose now that V3 − {x, v} → w . Combining with {a, v} → w , we have N − (w ) = (V3 − {x}) ∪ {a} and N + (w ) = (V1 − {a}) ∪ {x}. Let c ∈ V3 − {x, v}. Then c → w . By Lemma 2.2, there exists a vertex z ∈ V1 (z may be equal to u) such that z → c. Note that z ̸ = a. Then w → z and abxyvw zca is a desired 8-cycle. Subcase 1.2 V2 − {b, y} → v and u → v . In this case, we have N − (v ) = (V2 − {b}) ∪ {u} and N + (v ) = (V1 − {u}) ∪ {b}. If V3 − {x, v} ↛ V1 − {a, u}, then there is an arc z w from V1 − {a, u} to V3 − {x, v}. So v → z and w → a. Thus, abxyuv z w a is a 8-cycle through ab. Assume V3 − {x, v} → V1 − {a, u}. Let z ∈ V1 − {a, u}. Then V3 − {x} → z. If V2 − {b, y} ↛ z, then there is an arc zy′ for y′ ∈ V2 − {b, y}. Obviously, y′ → v . Let w ∈ V3 − {x, v}. If y′ → w , then abxyv zy′ w a is a 8-cycle through ab. Assume w → y′ . Then N + (w ) = (V1 − {u}) ∪ {y′ } and so u → w . Thus, abxyuw y′ v a is a 8-cycle through ab. Assume V2 −{b, y} → z. Note that V3 −{x} → z and d− (z) = r. We have |V2 − {b, y}| = 1 and r = 3. Let V2 −{b, y} = {y′ } and V3 − {x, v} = {v ′ }. Then {y′ , v, v ′ } → z → {b, x, y}. If b → v ′ , then abv ′ zxyuv a is a 8-cycle through ab. Assume v ′ → b. Then N + (v ′ ) = {a, b, z }, N − (v ′ ) = {u, y, y′ }, N − (b) = {a, z , v ′ }, N + (b) = {u, x, v}, N − (v ) = {u, b, y} and N + (v ) = {a, z , y′ }. Thus abxyuv y′ v ′ a is a 8-cycle through ab. Subcase 1.3 V2 − {b, y} → v → u. Suppose first that V2 − {b, y} ↛ u. Then there exists a vertex y′ ∈ V2 − {b, y} such that u → y′ . If V3 − {x, v} ↛ y′ , then there is a vertex v ′ ∈ V3 − {x, v} such that y′ → v ′ and abxyv uy′ v ′ a is a 8-cycle through ab. Assume V3 − {x, v} → y′ . Note that {a, u} → y′ . We have N − (y′ ) = (V3 − {x, v}) ∪ {a, u} and N + (y′ ) = (V1 − {a, u}) ∪ {x, v}. If V3 − {x, v} ↛ V1 − {a, u}, then there is an arc u′ v ′ from V1 − {a, u} to V3 − {x, v}. Obviously, y′ → u′ and v ′ → a. Thus, abxyuy′ u′ v ′ a is a desired 8-cycle. Assume V3 − {x, v} → V1 − {a, u}. Note that V3 − {x, v} → {a, y′ }. We get V3 − {x, v} → (V1 − {u}) ∪ {y′ } and (V2 − {y′ }) ∪ {u} → V3 − {x, v}. Let z ∈ V3 − {x, v}. Then u → z → y′ and abxyuzy′ v a is a 8-cycle through ab. Suppose now that V2 − {b, y} → u. Since {y, v} → u, we have (V2 − {b}) ∪ {v} → u → (V3 − {v}) ∪ {b}. If V2 −{b, y} ↛ V3 −{x, v}, then there is an arc v ′ y′ from V3 −{x, v} and V2 −{b, y}. Obviously, u → v ′ and y′ → v . So abxyuv ′ y′ v a is a 8-cycle through ab. Assume V2 −{b, y} → V3 −{x, v}. Note that V2 −{b, y} → {u, v}. We have V2 −{b, y} → (V3 −{x}) ∪{u}
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and (V1 − {u}) ∪ {x} → V2 − {b, y}. Let y1 ∈ V2 − {b, y} and v1 ∈ V3 − {x, v}. Then x → y1 → {u, v, v1 } and u → v1 . If v1 → y, then abxy1 uv1 yv a is a 8-cycle through ab. If y → v1 , then by {y, y1 } → v1 and Lemma 2.2, there exists a vertex u1 ∈ V1 − {a} such that v1 → u1 . It is easy to see that u1 ̸ = u and u1 → y1 . Now, abxyv1 u1 y1 v a is a desired 8-cycle. Case 2 V2 − {b} → x. By b → x, we have V2 → x → V1 . Subcase 2.1 V1 − {a} ↛ b. By the hypothesis, there exists u ∈ V1 − {a} such that b → u. Obviously, x → u. By {b, x} → u and Lemma 2.2, there exist two vertices w ∈ V3 − {x} and v ∈ V2 − {b} such that u → {w, v}. Suppose first that V3 − {x, w} ↛ V1 − {a, u}. Then there exists an arc u0 w0 from V1 − {a, u} to V3 − {x, w}. If V2 − {b} ↛ w , then there exists a vertex v0 ∈ V2 − {b} (v0 may be equal to v ) such that w → v0 . Thus, v0 → x → u0 and abuwv0 xu0 w0 a is a 8-cycle through ab. Assume V2 − {b} → w . Then N − (w ) = {u} ∪ (V2 − {b}) and N + (w ) = (V1 − {u}) ∪ {b}. So v → w → u0 and abxuvw u0 w0 a is a 8-cycle through ab. Suppose now that V3 − {x, w} → V1 − {a, u}. By V3 → a and x → V1 , we have V3 − {w} → V1 − {u}. Let u′ ∈ V1 − {a, u}. If w → u′ , then V3 → u′ → V2 . Let v ′ ∈ V2 − {b, v}. Then {a, u′ } → v ′ . By Lemma 2.2, there is a vertex w ′ ∈ V3 − {x} (w ′ may be equal to w ) such that v ′ → w ′ . Then abuv xu′ v ′ w ′ a is a 8-cycle through ab. Assume u′ → w . If V3 − {x, w} ↛ v , then there exists an arc vw ′ for some w ′ ∈ V3 − {x, w}. Thus, w ′ → u′ and abxuvw ′ u′ w a is a 8-cycle through ab. Assume V3 − {x, w} → v . Note that {a, u} → v . We have N − (v ) = (V3 − {x, w}) ∪ {a, u} and v → {u′ , w}. By {u, u′ } → w and Lemma 2.2, there is a vertex v ′ ∈ V2 − {b} such that w → v ′ . Since v → w , we have v ′ ̸ = v . Thus, v ′ → x → a and abuv u′ wv ′ xa is a 8-cycle through ab. Subcase 2.2 V1 − {a} → b. By a → b, we obtain V1 → b → V3 . Now we have V3 → a → V2 → x → V1 → b → V3 . Let w ∈ V3 − {x}. Then b → w . Subcase 2.2.1 V2 − {b} ↛ w . In this case, there exists a vertex v ∈ V2 − {b} such that w → v . By Lemma 2.2, there exists a vertex u ∈ V1 − {a} such that v → u. Suppose first that V2 − {b, v} ↛ u. Then there exists a vertex y ∈ V2 − {b, v} such that u → y. If V3 − {x, w} ↛ y, then there is a vertex z ∈ V3 − {x, w} such that y → z and abwv xuyza is a 8-cycle through ab. Assume V3 − {x, w} → y. By {a, u} → y, we have N − (y) = {a, u} ∪ (V3 − {x, w}) and N + (y) = (V1 − {a, u}) ∪ {x, w}. Let z ∈ V3 − {x, w}. Then abz v xuywa (when z → v ) or abxuywv za (when v → z) is a desired 8-cycle. Suppose now that V2 − {b, v} → u. By {x, v} → u, we get N − (u) = (V2 − {b}) ∪ {x} and N + (u) = {b} ∪ (V3 − {x}). If V2 − {b, v} ↛ V3 − {x, w}, then there is an arc w ′ y where w ′ ∈ V3 − {x, w} and y ∈ V2 − {b, v}. Thus, u → w ′ , y → x and ab is contained in a 8-cycle abwv uw ′ yxa. Assume V2 − {b, v} → V3 − {x, w}. Note that {b, u} → V3 − {x, w} and V2 − {b, v} → {u, x}. We have {u} ∪ (V2 − {v}) → V3 − {x, w} → (V1 − {u}) ∪ {v}, V2 − {b, v} → (V3 − {w}) ∪ {u} and {w} ∪ (V1 − {u}) → V2 − {b, v}. Let u′ ∈ V1 − {a, u}, y′ ∈ V2 − {b, v} and w ′ ∈ V3 − {x, w}. Then abwv uw ′ u′ y′ xa is a 9-cycle through ab. Subcase 2.2.2 V2 − {b} → w . By b → w , we have V2 → w → V1 . Let y, u ∈ V1 − {a} be two distinct vertices. Then {x, w} → y. Lemma 2.2 implies that there exists a vertex z ∈ V2 − {b} such that y → z. Obviously, z → w → u. If V3 − {x, w} ↛ u, then there is a vertex w′ ∈ V3 − {x, w} such that u → w′ . Thus, abxyz wuw′ a is a 8-cycle through ab. Assume V3 − {x, w} → u. By {x, w} → u, we have V3 → u → V2 and so u → z. By {a, y, u} → z and Lemma 2.2, there is a vertex w0 ∈ V3 − {x, w} such that z → w0 . Obviously w0 → u. Let z0 ∈ V2 − {b, z }. Then u → z0 → w and abxyz w0 uz0 w a is a desired 9-cycle. ■ Theorem 2.2. Let D be an r-regular 3-partite tournament with r ≥ 3 and let V1 , V2 , V3 be the partite sets of D. If ab is an arc from V1 to V2 satisfying V3 ↛ a ↛ V2 , then ab is contained in a 8- or 9-cycle. Proof. By Lemma 2.1, we have |V1 | = |V2 | = |V3 | = r and d+ (v ) = d− (v ) = r for each vertex v of D. We divide the partite set V2 into two nonempty parts V2+ , V2− such that V2− → a → V2+ . Similarly, V3 can be divided into two nonempty parts V3+ , V3− such that V3− → a → V3+ . Let V ′ = V2+ ∪ V3+ and V ′′ = V2− ∪ V3− . Observe that N + (a) = V ′ , N − (a) = V ′′ and |V ′ | = |V ′′ | = r. By Lemma 2.1, we also have |V2+ | = |V3− | and |V2− | = |V3+ |. Furthermore, we have the following claim. Claim 1. Let y ∈ V1 be arbitrary. Then d+ (y) = d− (y) and d− (y) = d+ (y). V′ V ′′ V′ V ′′ Proof. By Lemma 2.1, we have d+ (y) + d− (y) = |V ′ | = r , d+ (y) + d− (y) = |V ′′ | = r , d+ (y) + d+ (y) = d+ (y) = r and V′ V′ V ′′ V ′′ V′ V ′′ − − − + + − − dV ′ (y) + dV ′′ (y) = d (y) = r. So dV ′ (y) = dV ′′ (y) and dV ′ (y) = dV ′′ (y). □ Below, we prove this theorem in four cases: Case A1. |V2+ | = 1 and V3+ ↛ b. Case A2. |V2+ | = 1 and V3+ → b. Case B1. 2 ≤ |V2+ | ≤ r − 1 and V3+ ↛ b. Case B2. 2 ≤ |V2+ | ≤ r − 1 and V3+ → b.
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Proof of the theorem under Case A1. In this case, we have |V2+ | = |V3− | = 1 and |V2− | = |V3+ | = r − 1. Obviously, V2+ = {b}. Since V3+ ↛ b, there is a vertex x ∈ V3+ such that b → x. Clearly, we also have a → x. By {a, b} → x and Lemma 2.2, there are a vertex y ∈ V1 − {a} and a vertex u ∈ V2 − {b} such that x → {y, u}. Let V3− = {v}. Obviously, (V2 − {b}) ∪ {v} → a → {b} ∪ (V3 − {v}). Subcase 1 y → u. Subcase 1.1 V3 − {x, v} → u. In this case, we have (V3 − {v}) ∪ {y} → u → (V1 − {y}) ∪ {v}. Let y′ and w be two arbitrary vertices in V1 − {a, y} and + V3 respectively. Then a → w → u → {y′ , v}. Suppose first that y′ → v . If y → w , then abxyw uy′ v a is a desired 8-cycle. Assume w → y. If v → y, then {x, w, v} → y. Lemma 2.2 implies that there exists a vertex z ∈ V2 − {b, u} such that y → z. Thus, z → a and ab is contained in a 8-cycle abxuy′ v yza. If y → v , then by {y, y′ } → v and Lemma 2.2, there exists a vertex u′ ∈ V2 − {b} such that v → u′ . Obviously, u′ ̸ = u and abxyuy′ v u′ a is a desired 8-cycle. Suppose now that v → y′ . If V2− − {u} ↛ y′ , then there exists a vertex u′ ∈ V2− − {u} such that y′ → u′ . Thus, abxyuv y′ u′ a is a 8-cycle through ab. Assume V2− − {u} → y′ . Since {u, v} → y′ , we have that N − (y′ ) = V2− ∪ {v} and N + (y′ ) = V3+ ∪ {b} and so y′ → w . If w → y, then by {x, w} → y and Claim 1, there exists a vertex z ∈ V ′′ − {u} (z may be equal to v ) such that y → z. Thus, abxuy′ w yza is a 8-cycle through ab. If y → w , then by {a, y, y′ } → w and Lemma 2.2, there exists a vertex u1 ∈ V2 − {b, u} such that w → u1 . Thus, u1 → a and abxyuy′ w u1 a is a 8-cycle through ab. Subcase 1.2 V3 − {x, v} ↛ u. In this case, there exists a vertex w ∈ V3 − {x, v} such that u → w . Obviously, we have a → w . Suppose first that V1 − {a, y} → w . Since {a, u} → w , we have N − (w ) = (V1 − {y}) ∪ {u} and N + (w ) = {y} ∪ (V2 − {u}). Let u′ ∈ V2− − {u}. Then w → u′ . If V1 − {a, y} ↛ u, then there is an arc uy′ for some y′ ∈ V1 − {a, y}. Obviously, y′ → w and abxyuy′ w u′ a is a 8-cycle through ab. Assume V1 − {a, y} → u. It follows from {x, y} → u that N − (u) = (V1 − {a}) ∪ {x} and so u → v . If V1 − {a, y} ↛ v , then there is an arc v y1 for some y1 ∈ V1 − {a, y}. Thus, y1 → w and abxuv y1 w u′ a is a 8-cycle through ab. If V1 − {a, y} → v , then V1 − {a, y} ↛ V2 − {b, u} (as otherwise, V1 − {a, y} → (V2 − {b, u}) ∪ {w, u, v}, this is impossible.). Therefore, there is an arc u0 y0 from V2 − {b, u} to V1 − {a, y}. Thus, w → u0 , y0 → v and abxuw u0 y0 v a is a 8-cycle through ab. Suppose now that V1 − {a, y} ↛ w . Then there is a vertex y′ ∈ V1 − {a, y} such that w → y′ . If V ′′ − {u} ↛ y′ , then there exists a vertex u0 ∈ V ′′ − {u} such that y′ → u0 and abxyuw y′ u0 a is a 8-cycle through ab. Assume V ′′ − {u} → y′ . Since w → y′ , we get N − (y′ ) = (V ′′ − {u}) ∪ {w} and N + (y′ ) = {u} ∪ (V ′ − {w}). So {w, v} → y′ → {u, x}. If y → w , then by {a, y} → w and Lemma 2.2, there exists a vertex u′ ∈ V2 − {b} such that w → u′ . It is easy to see that ′ u ̸ = u, u′ → y′ and abxyw u′ y′ ua is a 8-cycle through ab. Assume w → y. By {a, y′ } → x and Lemma 2.2, there exists a vertex v ′ ∈ V2 − {u} such that x → v ′ . Then v ′ ̸ = b and v ′ → y′ . If y → v , then abxv ′ y′ uw yv a is a 9-cycle through ab. If v → y, then {x, w, v} → y. Lemma 2.2 implies that there exists a vertex z ∈ V2 − {b, u} (z may be equal to v ′ ) such that y → z. If z → v , then abxuw yz v a is a 8-cycle through ab. If v → z, then v → {a, y, y′ , z } and r ≥ 4. Let w′ ∈ V3+ − {x, w}. Then {a, y′ } → w′ . By Lemma 2.2, there is a vertex u1 ∈ V2 − {b} such that w′ → u1 . Thus, abxyzy′ w′ u1 a (when u1 ̸= z) or abxuw y′ w′ u1 a (when u1 = z) is a 8-cycle through ab. Subcase 2 u → y. Subcase 2.1 V2− → y. In this case, we get N − (y) = V2− ∪ {x} and N + (y) = {b} ∪ (V3 − {x}). Let w ∈ V3+ − {x}. Then {a, y} → w . Suppose first that V1 − {a, y} → w . In this case, we have V1 → w → V2 and so w → u. If v ↛ V2 − {b, u}, then there is a vertex u′ ∈ V2 − {b, u} such that u′ → v . Thus, we have that w → u′ and abxuyw u′ v a is a 8-cycle through ab. Assume v → V2 − {b, u}. Let u0 ∈ V2 − {b, u}. If u → v , then abxyw uv u0 a is a desired 8-cycle. If v → u, then N + (v ) = (V2 − {b, u}) ∪ {a, u} = (V2 − {b}) ∪ {a} and N − (v ) = (V1 − {a}) ∪ {b}. By {x, w, v} → u and Lemma 2.2, there exists a vertex y′ ∈ V1 − {a, y} such that u → y′ . Then y′ → v and abxyw uy′ v a is a desired 8-cycle. Suppose now that V1 − {a, y} ↛ w . Then there is a vertex y′ ∈ V1 − {a, y} such that w → y′ . If V ′′ − {u} ↛ y′ , then there is an arc y′ u′ for some u′ ∈ V ′′ − {u}. Thus, abxuyw y′ u′ a is a 8-cycle through ab. Assume V ′′ − {u} → y′ . Then N − (y′ ) = (V ′′ − {u}) ∪ {w} and N + (y′ ) = {u} ∪ (V ′ − {w}). Let u0 be an arbitrary vertex in V2 − {b, u}. Then u0 → {a, y′ }. By Lemma 2.2, there exists a vertex w ′ ∈ V3 − {x} (w ′ may be equal to w or v ) such that w ′ → u0 . Recalling that N + (y) = {b} ∪ (V3 − {x}), we have that y → w ′ and ab is contained in a 8-cycle abxyw ′ u0 y′ ua. Subcase 2.2 V2− ↛ y. In this case, there is a vertex u′ ∈ V2− − {u} such that y → u′ . Subcase 2.2.1 V3+ − {x} ↛ u′ . In this case, there exists a vertex w ∈ V3+ − {x} such that u′ → w . Clearly, a → w . Suppose first that V1 − {a, y} → w . It follows from {a, u′ } → w that N − (w ) = (V1 − {y}) ∪ {u′ } and so w → {u, y}. If u → v , then abxyu′ w uv a is a 8-cycle through ab. Assume v → u. By {x, w, v} → u and Lemma 2.2, there is a vertex y′ ∈ V1 − {a, y} such that u → y′ . Thus, y′ → w and abxuy′ w yu′ a is a 8-cycle through ab. Suppose now that V1 − {a, y} ↛ w . Then there exists a vertex y0 ∈ V1 − {a, y} such that w → y0 . If V ′′ − {u′ } ↛ y0 , then there is a vertex u0 ∈ V ′′ − {u′ } such that y0 → u0 . Thus, abxyu′ w y0 u0 a is a 8-cycle through ab. Assume V ′′ − {u′ } → y0 . Since
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w → y0 , we have N − (y0 ) = (V ′′ − {u′ }) ∪ {w} and so y0 → u′ . If y → w, then abxuywy0 u′ a is a 8-cycle through ab. Assume w → y. By {x, w} → y and Claim 1, there exists a vertex z ∈ V ′′ − {u′ } such that y → z. Obviously, z ̸= u and z → y0 . Thus,
abxuyzy0 u′ a is a 8-cycle through ab.
Subcase 2.2.2 V3+ − {x} → u′ . Suppose first that V1 −{a, y} → u′ . Since (V3+ −{x}) ∪{y} → u′ and d− (u′ ) = r, we have |V3+ − {x}| = 1. Let V3+ −{x} = {w}. Then V3 = {x, w, v} and r = 3. Let V1 = {a, y, y′ }. Then {y, y′ , w} → u′ → {a, x, v} and {a, b, u′ } → x → {y, y′ , u}. If y′ → u, then abxy′ uyu′ v a is a 8-cycle through ab. Assume u → y′ . Then u → V1 and so v → u. If y → v , then abxyv uy′ u′ a is a 8-cycle through ab. If v → y, then v → {a, u, y} and so y′ → v . Thus, abxuy′ v yu′ a is a 8-cycle through ab. Suppose now that V1 − {a, y} ↛ u′ . Then there exists a vertex y′ ∈ V1 − {a, y} such that u′ → y′ . If V ′′ − {u} ↛ y′ , then there is an arc y′ z for some z ∈ V ′′ − {u}. Obviously, z ̸ = u′ and abxuyu′ y′ za is a 8-cycle through ab. Assume V ′′ − {u} → y′ . If u → y′ , then V ′′ → y′ → V ′ . By u → {a, y, y′ } and Lemma 2.2, there is a vertex w ∈ V3 − {x, v} such that w → u. Then y′ → w and abxyu′ y′ w ua is a 8-cycle through ab. Assume y′ → u. If V3+ − {x} ↛ y, then there exists a vertex w ∈ V3+ − {x} such that y → w . Recalling that V3+ − {x} → u′ , we have that w → u′ and abxywu′ y′ ua is a desired 8-cycle. Assume V3+ − {x} → y. Since {x, u} → y, we get N − (y) = V3+ ∪ {u} and so y → v . Then abxyu′ v y′ ua (when u′ → v ) or abxyv u′ y′ ua (when v → u′ ) is a desired 8-cycle. □ Proof of the theorem under Case A2. In this case, we have |V2+ | = |V3− | = 1, |V2− | = |V3+ | = r − 1 and V3+ → b. Obviously, V2+ = {b}, N − (b) = V3+ ∪ {a} and N + (b) = (V1 − {a}) ∪ {v}, where v ∈ V3− . If V3+ → V1 − {a}, then V3+ ∪ {b} → V1 − {a} → (V2 − {b}) ∪ {v} and {a} ∪ (V2 − {b}) → V3+ → (V1 − {a}) ∪ {b}. Let y, y′ ∈ V1 − {a}, u, u′ ∈ V2 − {b} be four distinct vertices and let x ∈ V3+ . Then abyuxy′ u′ v a (when u′ → v ) or abyuxy′ v u′ a (when v → u′ ) is a 8-cycle through ab. Assume V3+ ↛ V1 − {a}. Then there is an arc yx from V1 − {a} to V3+ . Moreover, we have b → y. Consider the following two cases. Subcase 1 V1 − {a, y} ↛ x. In this case, there exists a vertex y′ ∈ V1 − {a, y} such that x → y′ . By Lemma 2.2, there is a vertex u ∈ V2 such that y′ → u. Obviously, b → y′ and u ∈ V2− . Suppose first that V3+ − {x} ↛ u. Then there exists a vertex w ∈ V3+ − {x} such that u → w . If V2− ↛ w , then there is an arc w u′ for some u′ ∈ V2− − {u} and abyxy′ uw u′ a is a 8-cycle through ab. If V2− → w , then N − (w ) = V2− ∪ {a} and N + (w ) = (V1 − {a}) ∪ {b}. By {a, y} → x and Lemma 2.2, there exists a vertex u0 ∈ V2− (u0 may be equal to u) such that x → u0 . Then u0 → w → y′ . By {b, x} → y′ and Claim 1, there exists a vertex v ′ ∈ V ′′ − {u0 } such that y′ → v ′ . Then abyxu0 w y′ v ′ a is a 8-cycle through ab. Suppose now that V3+ − {x} → u. If v → u, then we get N − (u) = {y′ } ∪ (V3 − {x}) and N + (u) = (V1 − {y′ }) ∪ {x}. By {a, y} → x and Lemma 2.2, there exists a vertex u0 ∈ V2 − {b} such that x → u0 . Since u → x, we get u0 ̸= u. Recalling that N + (b) = (V1 − {a}) ∪ {v}, then b → {y′ , v} and aby′ v uyxu0 a (when y′ → v ) or abv y′ uyxu0 a (when v → y′ ) is a 8-cycle through ab. Assume u → v . If V2− −{u} ↛ v , then there exists an arc v u′ for some u′ ∈ V2− −{u}. Thus, abyxy′ uv u′ a is a 8-cycle through ab. If V2− − {u} → v , then V2 → v since {b, u} → v . So v → V1 . By b → y′ and Lemma 2.2, there exists a vertex w ∈ V3 such that y′ → w. Since {x, v} → y′ , we have w ̸= x, v . Therefore, w ∈ V3+ − {x} and w → u. Thus, abyxy′ wuv a is a 8-cycle through ab. Subcase 2 V1 − {a, y} → x. In this case, we have V1 → x → V2 since {a, y} → x. Suppose first that V1 − {a} → V2− . Then V1 − {a} → V2− ∪ {x} and {b} ∪ (V3 − {x}) → V1 − {a}. In addition, (V1 − {a}) ∪ {x} → V2− → {a} ∪ (V3 − {x}). Let y′ ∈ V1 − {a, y} and let u, u′ ∈ V2 − {b} be two different vertices. Then abyxuv y′ u′ a is a desired 8-cycle. Suppose now that V1 −{a} ↛ V2− . Then there exists an arc uy′ from V2− to V1 −{a} (y′ may be equal to y). Let y1 ∈ V1 −{a, y′ }. Then b → y1 → x → u. If V3+ − {x} → y′ , then N − (y′ ) = {b, u} ∪ (V3+ − {x}) and N + (y′ ) = (V2 − {b, u}) ∪ {x, v}. Let w ∈ V2 − {b, u}. Then y′ → {w, v}, and aby1 xuy′ vw a (when v → w ) or aby1 xuy′ wv a (when w → v ) is a 8-cycle through ab. If V3+ − {x} ↛ y′ , then there is an arc y′ z for some z ∈ V3+ − {x}. In the case when V2− − {u} ↛ z, there exists an arc zu′ for some u′ ∈ V2− − {u} and aby1 xuy′ zu′ a is a 8-cycle through ab. In the other case when V2− − {u} → z, we have N − (z) = {a, y′ } ∪ (V2− − {u}) and N + (z) = (V1 − {a, y′ }) ∪ {b, u}. Let u0 ∈ V2− − {u}. Then x → u0 → z. By {b, z } → y1 and Claim 1, there exists a vertex w ∈ V ′′ − {u0 } such that y1 → w . Thus, aby′ xu0 zy1 w a is a 8-cycle through ab. □ Proof of the theorem under Case B1. In this case, we have 2 ≤ |V2+ | = |V3− | ≤ r − 1, 1 ≤ |V2− | = |V3+ | ≤ r − 2 and V3+ ↛ b. Then there exists a vertex x ∈ V3+ such that b → x. Subcase 1 V2+ ↛ x. By the hypothesis, there exists a vertex u ∈ V2+ − {b} such that x → u. By Lemma 2.2, there is a vertex y ∈ V1 such that u → y. Since a → u, we know y ̸ = a. We first prove the following claim. Claim 2. If |V3+ | ≥ 2 and V3+ − {x} ↛ y, then ab is contained in a 8- or 9-cycle.
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Proof. Since |V3+ | ≥ 2 and V3+ − {x} ↛ y, we have that |V2− | ≥ 2 and there is a vertex x′ ∈ V3+ − {x} such that y → x′ . If V1 − {a, y} ↛ x′ , then there is a vertex y′ ∈ V1 − {a, y} such that x′ → y′ . By Claim 1, there exists a vertex z ∈ V ′′ such that y′ → z and abxuyx′ y′ za is a 8-cycle through ab. Assume V1 − {a, y} → x′ . Since {a, y} → x′ , we get V1 → x′ → V2 . Let w ∈ V2− . Then x′ → w. If V3− ↛ w , then there is an arc w z for some z ∈ V3− and abxuyx′ w za is a 8-cycle through ab. Assume V3− → w . By V3− ∪ {x′ } → w and Lemma 2.2, there is a vertex y′ ∈ V1 − {a, y} such that w → y′ . If V ′′ ↛ y′ , then there exists an arc y′ v for some v ∈ V ′′ . Obviously, v ̸ = w and abxuyx′ w y′ v a is a desired 9-cycle. If V ′′ → y′ , then y′ → V ′ . By u → y and Claim 1, there exists a vertex v0 ∈ V ′′ (v0 may be equal to w ) such that y → v0 and so v0 → y′ → x′ . Let w ′ ∈ V2− − {v0 }. Then x′ → w ′ and abxuyv0 y′ x′ w ′ a is a desired 9-cycle. □ By Claim 2, we may assume V3+ − {x} → y when |V3+ | ≥ 2. If V3− → y, then by the assumption above, we always have V3 − {x} → y. Now, N − (y) = (V3 − {x}) ∪ {u} and + N (y) = (V2 − {u}) ∪ {x}. By a → u and Lemma 2.2, there exists a vertex z ∈ V3 − {x} such that u → z. Let w ∈ V2− . Then z → y → {w, x}. By b → x and Lemma 2.2, there exists a vertex y0 ∈ V1 − {a, y} such that x → y0 . If y0 → u, then abxy0 uzyw a is a desired 8-cycle. If u → y0 , then by Lemma 2.2, there exists a vertex x′ ∈ V3 − {x} such that y0 → x′ . Thus, x′ → y and abxuy0 x′ yw a is a desired 8-cycle. Now, we assume V3− ↛ y. Then there is a vertex v ∈ V3− such that y → v . Subcase 1.1 V1 − {a, y} ↛ v . In this case, there exists a vertex y′ ∈ V1 − {a, y} such that v → y′ . If V ′′ ↛ y′ , then there exists a vertex w ∈ V ′′ − {v} such that y′ → w and abxuyv y′ w a is a 8-cycle through ab. Assume V ′′ → y′ → V ′ . By {a, y′ } → u and Lemma 2.2, there exists a vertex v ′ ∈ V3 − {v} such that u → v ′ . Obviously, v ′ ̸ = x. Suppose first that v ′ ∈ V3+ . Then |V3+ | ≥ 2 and y′ → v ′ → y. If V2− ↛ v ′ , then there is an arc v ′ w for some w ∈ V2− and abxuyv y′ v ′ w a is a desired 9-cycle. Assume V2− → v ′ . By {a, y′ } → v ′ and Lemma 2.2, there exists a vertex u′ ∈ V2+ − {b, u} such that v ′ → u′ . Then y′ → u′ . By v ′ → u′ and Lemma 2.2, there is a vertex y0 ∈ V1 − {a, y′ } (y0 may be equal to y) such that u′ → y0 . In addition, by u′ → y0 and Claim 1, there exists a vertex z ∈ V ′′ (z may be equal to v ) such that y0 → z. Then abxuv ′ u′ y0 za is a desired 8-cycle. Suppose now that v ′ ∈ V3− . Then v ′ → y′ . If V2− ↛ x, then there is an arc xw for some w ∈ V2− . Thus, w → y′ and abxw y′ uyv a is a desired 8-cycle. Assume V2− → x. By {a, y′ } → x and Lemma 2.2, there exists a vertex u′ ∈ V2+ − {b, u} such that x → u′ . Then y′ → u′ . If u′ → v , then abxuv ′ y′ u′ v a is a desired 8-cycle. If v → u′ , then by {x, v} → u′ and Lemma 2.2, there is a vertex y0 ∈ V1 − {y} such that u′ → y0 . Since {a, y′ } → u′ , we get y0 ̸ = a, y′ . By u′ → y0 and Claim 1, there exists a vertex z0 ∈ V ′′ such that y0 → z0 . Then abxu′ y0 z0 y′ uv ′ a (when z0 = v ) or abxuyv u′ y0 z0 a (when z0 ̸ = v ) is a desired 9-cycle. Subcase 1.2 V1 − {a, y} → v . By a → u and Lemma 2.2, there exists a vertex z ∈ V3 − {x} (z may be equal to v ) such that u → z. Suppose first that z → y. Obviously, z ̸ = v . If V2− ↛ v , then there exists an arc vw for some w ∈ V2− . Thus, abxuzyvw a is a 8-cycle through ab. Assume V2− → v . Since d− (v ) = r, we have N − (v ) = (V1 − {a}) ∪ V2− and |V2− | = |V3+ | = 1. Then N + (v ) = {a} ∪ V2+ , V3+ = {x} and z ∈ V3− . Let V2− = {w}. Then w → v . If y → w , then ab is contained in a 8-cycle abxuzywv a. Assume w → y. If z → w , then abxuz w yv a is a 8-cycle through ab. Assume w → z. Then w → {a, y, z , v}, r ≥ 4 and |V2+ | ≥ 3. Let u′ ∈ V2+ − {b, u}. Then v → u′ . If u′ → z, then abxuyv u′ za is a desired 8-cycle. If z → u′ , then by {v, z } → u′ and Lemma 2.2, there exists a vertex y′ ∈ V1 − {a, y} such that u′ → y′ . Thus, y′ → v and abxuzu′ y′ v a is a desired 8-cycle. Suppose now that y → z. If V1 −{a, y} ↛ z, then using z instead of v in the subcase 1.1, we may prove that ab is contained in a 8- or 9-cycle. So we assume V1 − {a, y} → z. Now, N − (z) = (V1 − {a}) ∪ {u} and N + (z) = {a} ∪ (V2 − {u}). Let w ∈ V2− . We have z → w . If V3− ↛ w , then there exists a vertex v0 ∈ V3− − {z } such that w → v0 , and abxuyz wv0 a is a desired 8-cycle. Assume V3− → w . If y → w , then by V3− → w and Lemma 2.2, there exists a vertex y′ ∈ V1 − {a, y} such that w → y′ . Now, y′ → z and abxuyw y′ za is a 8-cycle through ab. Assume w → y. If V ′′ − {z } ↛ y, then there is an arc yv ′ for some v ′ ∈ V ′′ − {z , w} and abxuz w yv ′ a is a desired 8-cycle. Assume V ′′ − {z } → y. Then N − (y) = (V ′′ − {z }) ∪ {u} and N + (y) = {z } ∪ (V ′ − {u}) (now, z = v .). Let v ′ ∈ V3− − {z }. If u → v ′ , then abxuv ′ w yza is a desired 8-cycle. If v ′ → u, then v ′ → {a, y, w, u} and r ≥ 4. Note that + N (y) = {z } ∪ (V ′ − {u}) and V3+ − {x} → y when |V3+ | ≥ 2. We have V3+ = {x} and |V2+ | ≥ 3. Let u′ ∈ V2+ − {b, u}. Then {a, y} → u′ . Lemma 2.2 implies that there exists a vertex z ′ ∈ V3 − {x} such that u′ → z ′ . Then z ′ ∈ V3− and z ′ → w . Now, abxuyu′ z ′ w a is a 8-cycle through ab. Subcase 2 V2+ → x. By b → x and Lemma 2.2, there is a vertex y ∈ V1 − {a} such that x → y. In addition, by a → x and Lemma 2.2, there is a vertex u′ ∈ V2 such that x → u′ . Obviously, u′ ∈ V2− . Subcase 2.1 V2+ − {b} ↛ y. In this case, there exists a vertex u ∈ V2+ − {b} such that y → u. Obviously, u → x. Claim 3. If |V3+ | ≥ 2 and V3+ − {x} ↛ u, then ab is contained in a 8- or 9-cycle.
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Proof. By the hypothesis, there is a vertex x′ ∈ V3+ − {x} such that u → x′ . If V1 − {a, y} ↛ x′ , then there exists a vertex y′ ∈ V1 − {a, y} such that x′ → y′ . By Claim 1, there is a vertex w ∈ V ′′ such that y′ → w and ab is contained in a 8-cycle abxyux′ y′ w a. Assume V1 − {a, y} → x′ . Since {a, u} → x′ , we have N − (x′ ) = (V1 − {y}) ∪ {u} and N + (x′ ) = {y} ∪ (V2 − {u}). So x′ → {y, u′ }. If V3− ↛ u′ , then there is an arc u′ w ′ for some w ′ ∈ V3− and abxyux′ u′ w ′ a is a 8-cycle through ab. If V3− → u′ , then V3− ∪ {x, x′ } → u′ . Lemma 2.2 implies that there is a vertex y′ ∈ V1 − {a, y} such that u′ → y′ . Obviously, y′ → x′ . By {x, x′ } → y and Claim 1, there is a vertex z ∈ V ′′ − {u′ } such that y → z. Thus, abxu′ y′ x′ yza is a desired 8-cycle. □ By Claim 3, we may assume V3+ − {x} → u when |V3+ | ≥ 2 and then NV+ −{x} (u) ⊆ V3− . By {a, y} → u and Lemma 2.2, there 3 exists a vertex v ∈ V3 − {x} such that u → v . Then v ∈ V3− . Suppose first that V1 − {a, y} ↛ v . Then there exists an arc v y′ for some y′ ∈ V1 − {a, y}. If V ′′ ↛ y′ , then there is a vertex z ∈ V ′′ − {v} such that y′ → z, and abxyuv y′ za is a 8-cycle through ab. If V ′′ → y′ , then y′ → V ′ . By {a, y′ } → x and Lemma 2.2, there exists a vertex w ∈ V2 − {u′ } such that x → w . Recalling that V2+ → x, we have w ∈ V2− − {u′ }. So |V2− | = |V3+ | ≥ 2. Let x′ ∈ V3+ − {x}. Then u′ → y′ → x′ → u and abxu′ y′ x′ uv a is a 8-cycle through ab. Suppose now that V1 − {a, y} → v . If u′ → v , then N − (v ) = (V1 − {a, y}) ∪ {u, u′ } and v → y. In the case when V3 − {x, v} ↛ u, there is an arc uv ′ for some v ′ ∈ V3 − {x, v}. By the assumption above, v ′ ∈ V3− − {v} and abxu′ v yuv ′ a is a desired 8-cycle. In the other case when V3 − {x, v} → u, we have N − (u) = {a, y} ∪ (V3 − {x, v}) and N + (u) = (V1 − {a, y}) ∪ {x, v}. By V2+ → x and Lemma 2.2, there exists a vertex y′ ∈ V1 − {a, y} such that x → y′ , and then u → y′ . When y′ → u′ , we see that ab is contained in a 8-cycle abxyuy′ u′ v a. When u′ → y′ , by {x, u} → y′ and Claim 1, there exists a vertex z ∈ V ′′ − {v} such that y′ → z. Then z ̸ = u′ and abxu′ v yuy′ za is a desired 9-cycle. Assume v → u′ . If V3− ↛ u′ , then there is an arc u′ w ′ for some w ′ ∈ V3− − {v}, and abxyuv u′ w ′ a is a 8-cycle through ab. If − V3 → u′ , then V3− ∪ {x} → u′ . Lemma 2.2 implies that there is a vertex y′ ∈ V1 − {a, y} such that u′ → y′ . Obviously, y′ → v . In the case when u → y′ , we know that abxyuy′ v u′ a is a 8-cycle through ab. In the other case when y′ → u, by {a, y, y′ } → u and Lemma 2.2, there is a vertex v ′ ∈ V3 − {x, v} such that u → v ′ . Recalling NV+ −{x} (u) ⊆ V3− , we have v ′ ∈ V3− and v ′ → u′ . 3 Then abxyuv ′ u′ y′ v a is a desired 9-cycle. Subcase 2.2 V2+ − {b} → y. Let u ∈ V2+ − {b}. Then u → {x, y}. Subcase 2.2.1 y → u′ . By {u, x} → y and Claim 1, there exists a vertex w ∈ V ′′ − {u′ } such that y → w . First suppose that V3+ ↛ u′ . Then there is a vertex x′ ∈ V3+ − {x} such that u′ → x′ . If x′ → u, then abxu′ x′ uyw a is a 8-cycle through ab. Assume u → x′ . By {u, u′ } → x′ and Lemma 2.2, there exists a vertex y′ ∈ V1 − {a, y} such that x′ → y′ . If y′ → u, then abxu′ x′ y′ uyw a is a 9-cycle through ab. Assume u → y′ . By {u, x′ } → y′ and Claim 1, there exists a vertex w ′ ∈ V ′′ − {u′ } such that y′ → w ′ . Then abxyu′ x′ y′ w ′ a is a 8-cycle through ab. Now suppose that V3+ → u′ . By y → u′ and Lemma 2.2, there exists a vertex v ∈ V3− (v may be equal to w ) such that u′ → v . Assume V1 − {a, y} → v . By V2+ → x and Lemma 2.2, there exists a vertex y′ ∈ V1 − {a} such that x → y′ . Obviously, ′ y ̸ = y and y′ → v . If v → u, then abxy′ v uyu′ a is a 8-cycle through ab. If u → v , then N − (v ) = (V1 − {a, y}) ∪ {u, u′ } and N + (v ) = {a, y} ∪ (V2 − {u, u′ }). Recalling y → w and w ∈ V ′′ − {u′ }, we have v ̸ = w . Then abxy′ u′ v yw a (when y′ → u′ ) or abxu′ y′ v yw a (when u′ → y′ ) is a desired 8-cycle. Assume V1 − {a, y} ↛ v . Then there exists a vertex y0 ∈ V1 − {a, y} such that v → y0 . If V ′′ − {u′ } ↛ y0 , then there is a vertex z ′ ∈ V ′′ − {u′ } such that y0 → z ′ . Obviously, z ′ ̸ = v and abxyu′ v y0 z ′ a is a 8-cycle through ab. If V ′′ − {u′ } → y0 , then w → y0 . In the case when u′ → y0 , we have V ′′ → y0 → V ′ and abxu′ y0 uyw a is a 8-cycle through ab. In the other case when y0 → u′ , by {y, y0 } → u′ and Lemma 2.2, there exists a vertex v ′ ∈ V3 − {v} such that u′ → v ′ . Since V3+ → u′ , we get v ′ ∈ V3− − {v}. Thus, abxywy0 u′ v ′ a (when w = v ) or abxywy0 u′ v a (when w ̸= v ) is a 8-cycle through ab. Subcase 2.2.2 u′ → y. Suppose first that V3+ ↛ y. Then there exists a vertex x′ ∈ V3+ − {x} such that y → x′ . If x′ → u, then in the case when V3− ↛ u, there exists a vertex w ∈ V3− such that u → w and abxu′ yx′ uw a is a 8-cycle through ab. In the other case when V3− → u, by V3− ∪ {x′ } → u and Lemma 2.2, there exists a vertex y′ ∈ V1 − {a, y} such that u → y′ . Claim 1 yields that there is a vertex w ′ ∈ V ′′ (w ′ may be equal to u′ ) such that y′ → w ′ . Thus, abxyx′ uy′ w ′ a is a 8-cycle through ab. If u → x′ , then by Lemma 2.2, there exists a vertex y′ ∈ V1 such that x′ → y′ . Since {a, y} → x′ , we know y′ ̸ = a, y. In the case when V ′′ − {u′ } ↛ y′ , there is an arc y′ w for some w ∈ V ′′ − {u′ }, and abxu′ yx′ y′ w a is a 8-cycle through ab. In the other case when V ′′ − {u′ } → y′ , we get N − (y′ ) = (V ′′ − {u′ }) ∪ {x′ } and N + (y′ ) = {u′ } ∪ (V ′ − {x′ }). Then y′ → {u, u′ }. By {u, x} → y and Claim 1, there is a vertex z ∈ V ′′ − {u′ } such that y → z. Thus z → {y′ , a}, and abxu′ x′ y′ uyza (when u′ → x′ ) or abxyzy′ ux′ u′ a (when x′ → u′ ) is a 9-cycle through ab. Suppose now that V3+ → y. Since (V2+ − {b}) ∪ {u′ } → y, we get N − (y) = (V ′ − {b}) ∪ {u′ } and N + (y) = {b} ∪ (V ′′ − {u′ }). Let v, v ′ ∈ V3− be two distinct vertices. Then y → {v, v ′ }. If V1 − {a, y} → v , then by V2+ → x and Lemma 2.2, there exists a vertex y′ ∈ V1 − {y} such that x → y′ . Obviously, y′ ̸ = a and y′ → v . When v → u′ , we know that abxy′ v u′ yv ′ a is a 8-cycle through ab. When u′ → v , we get N − (v ) = (V1 −{a}) ∪{u′ } and N + (v ) = {a} ∪ (V2 − {u′ }). Thus, v → u and abxu′ v uyv ′ a is a 8-cycle through ab.
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If V1 − {a, y} ↛ v , then there is an arc v y′ for some y′ ∈ V1 − {a, y}. When V ′′ − {u′ } ↛ y′ , then there is an arc y′ z for some z ∈ V ′′ − {u′ , v}, and abxu′ yv y′ za is a 8-cycle through ab. When V ′′ − {u′ } → y′ and u′ → y′ , we have V ′′ → y′ → V ′ and abxu′ y′ uyv a is a desired 8-cycle. Assume now that V ′′ − {u′ } → y′ → u′ . In the case when y′ → x, we have {a, y′ } → x. Lemma 2.2 implies that there is a vertex w ∈ V2 − {u′ } such that x → w . Recalling that V2+ → x, we have w ∈ V2− − {u′ }. Therefore, y → w → y′ and abxywv y′ u′ a (when w → v ) or abxyvw y′ u′ a (when v → w ) is a desired 8-cycle. In the other case when x → y′ , we know N − (y′ ) = (V ′′ − {u′ }) ∪ {x} and N + (y′ ) = {u′ } ∪ (V ′ − {x}). Then v ′ → y′ → u and abxyv ′ y′ u′ v a (when u′ → v ) or abxy′ uyv u′ a (when v → u′ ) is a desired 8-cycle. □ Proof of the theorem under Case B2. In this case, we have 2 ≤ |V2+ | = |V3− | ≤ r − 1, 1 ≤ |V2− | = |V3+ | ≤ r − 2 and V3+ → b. By Lemma 2.2, there is an arc by for some y ∈ V1 − {a}. Similarly, by a → b and Lemma 2.2, there exists a vertex c ∈ V3− such that b → c. Subcase 1 V3+ ↛ y. By the hypothesis, there is a vertex x ∈ V3+ such that y → x. Subcase 1.1 V1 − {a, y} → x. In this case, we have V1 → x → V2 . Let u ∈ V2+ − {b} and w ∈ V2− . Then x → {u, w}. By a → u and Lemma 2.2, there is a vertex z ∈ V3 − {x} (z may be equal to c) such that u → z. Suppose first that V1 − {a, y} ↛ z. Then there exists a vertex y′ ∈ V1 − {a, y} such that z → y′ . If V ′′ ↛ y′ , then there is a vertex z ′ ∈ V ′′ − {z } such that y′ → z ′ , and abyxuzy′ z ′ a is a 8-cycle through ab. If V ′′ → y′ , then y′ → V ′ , z ∈ V ′′ and abyxw y′ uza is a desired 8-cycle. Suppose now that V1 − {a, y} → z. If w → z, then N − (z) = (V1 − {a, y}) ∪ {w, u} and N + (z) = {a, y} ∪ (V2 − {u, w}). So z → {b, y} and z ̸ = c. By {x, z } → b and Lemma 2.2, there exists a vertex y0 ∈ V1 − {a, y} such that b → y0 . Thus, y0 → z and abcuzyxw a (when c → u) or aby0 zyxuca (when u → c) is a 8-cycle through ab. Assume z → w . If V3− ↛ w , then there exists a vertex z ′ ∈ V3− − {z } such that w → z ′ , and abyxuz w z ′ a is a 8-cycle through ab. If V3− → w , then by V3− ∪ {x} → w and Lemma 2.2, there exists a vertex y′ ∈ V1 − {a, y} such that w → y′ . So y′ → z. When z ∈ V3− , we get that abyxw y′ uza (when y′ → u) or abyxuy′ z w a (when u → y′ ) is a 8-cycle through ab. When z ∈ V3+ , we have |V3+ | = |V2− | ≥ 2, a → z, N − (z) = (V1 − {y}) ∪ {u} and N + (z) = {y} ∪ (V2 − {u}). Let v ∈ V2− − {w}. Then y′ → z → v and abyxw y′ z v a is a 8-cycle through ab. Subcase 1.2 V1 − {a, y} ↛ x. By the hypothesis, there exists a vertex y′ ∈ V1 − {a, y} such that x → y′ . Subcase 1.2.1 V2+ − {b} ↛ y′ . In this case, there exists a vertex u ∈ V2+ − {b} such that y′ → u. If V1 − {y} ↛ u, then there is an arc uy1 for some y1 ∈ V1 − {a, y, y′ }. By u → y1 , u ∈ V ′ and Claim 1, there exists a vertex v ∈ V ′′ such that y1 → v . Then abyxy′ uy1 v a is a desired 8-cycle. Assume V1 − {y} → u. If c → u, then N − (u) = (V1 − {y}) ∪ {c } and N + (u) = {y} ∪ (V3 − {c }). When y′ → c, we have that abyxy′ cuv a is a desired 8-cycle where v ∈ V3− − {c }. When c → y′ , by x → y′ and Claim 1, there is a vertex w ′ ∈ V ′′ − {c } such that y′ → w ′ , and abcuyxy′ w ′ a is a desired 8-cycle. Assume u → c. If V2− ↛ c, then there is an arc c w for some w ∈ V2− and abyxy′ uc w a is a desired 8-cycle. So assume V2− → c. If V3− − {c } ↛ u, then there exists an arc uv for some v ∈ V3− − {c }. When c → y, we know that abcyxy′ uv a is a desired 8-cycle. Assume y → c. By {b, u} ∪ V2− → c and Lemma 2.2, there exists a vertex y1 ∈ V1 − {a, y, y′ } such that c → y1 . If V ′′ ↛ y1 , then there is an arc y1 z1 for some z1 ∈ V ′′ − {c }, and abyxy′ ucy1 z1 a is a desired 9-cycle. If V ′′ → y1 , then y1 → V ′ and abcy1 xy′ uv a is a desired 8-cycle. Assume V3− − {c } → u. Recalling that V1 − {y} → u and d− (u) = r, we get |V3− − {c }| = 1 and u → {y, c } ∪ V3+ . Let V3− − {c } = {v}. If |V3+ | = |V2− | ≥ 2, then let x′ ∈ V3+ − {x}. When V2− ↛ x′ , then there is an arc x′ w for some w ∈ V2− and abyxy′ ux′ w a is a 8-cycle through ab. When V2− → x′ , by V2− ∪ {u} → x′ and Lemma 2.2, there is a vertex y1 ∈ V1 − {a, y, y′ } such that x′ → y1 . By x′ → y1 and Claim 1, there exists a vertex v ′ ∈ V ′′ (v ′ may be equal to c or v ) such that y1 → v ′ . Thus, abyxy′ ux′ y1 v ′ a is a 9-cycle through ab. If |V3+ | = |V2− | = 1, then V3+ = {x} and r = 3. Let V2− = {w}. When y′ → v , abyxy′ v uca is a 8-cycle through ab. When v → y′ , we have v → {a, y′ , u}, {y, b, w} → v and {a, y, u} → x → {b, y′ , w}. So abyxwv uca is a desired 8-cycle. Subcase 1.2.2 V3+ ↛ y′ . By the hypothesis, there exists a vertex x′ ∈ V3+ − {x} such that y′ → x′ . So |V3+ | = |V2− | ≥ 2 and r ≥ 4. If V1 − {y} ↛ x′ , then there exists a vertex y1 ∈ V1 − {a, y, y′ } such that x′ → y1 . By Claim 1, there is a vertex w ∈ V ′′ such that y1 → w . Thus, abyxy′ x′ y1 w a is a 8-cycle through ab. Assume V1 − {y} → x′ . By Lemma 2.2, there are two vertices u, u′ ∈ V2 − {b} such that x′ → {u, u′ }. Furthermore, V2− ↛ x′ (as otherwise, (V1 − {y}) ∪ V2− → x′ and d− (x′ ) ≥ r + 1, a contradiction.). Without loss of generality, we suppose u′ ∈ V2− . If V3− ↛ u, then there is an arc uv ′ for some v ′ ∈ V3− , and abyxy′ x′ uv ′ a is a 8-cycle through ab. Assume V3− → u. If V1 − {a, y, y′ } → u, then (V1 − {a, y, y′ }) ∪ V3− ∪ {x′ } → u. By |V3− | ≥ 2 and d− (u) = r, we have |V3− | = 2 and N − (u) = (V1 − {a, y, y′ }) ∪ V3− ∪ {x′ }. So c → u → x and abcuxy′ x′ u′ a is a 8-cycle through ab. Assume V1 − {a, y, y′ } ↛ u.
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Then there exists a vertex y1 ∈ V1 − {a, y, y′ } such that u → y1 . If V ′′ ↛ y1 , then there is an arc y1 z1 for some z1 ∈ V ′′ − {u} and abyxy′ x′ uy1 z1 a is a 9-cycle through ab. If V ′′ → y1 , then y1 → V ′ and abcy1 xy′ x′ u′ a is a 8-cycle through ab. Subcase 1.2.3 V2+ − {b} → y′ and V3+ → y′ . In this case, we have V ′ − {b} → y′ . Suppose first that V2− → y′ . Then |V2− | = 1 and N − (y′ ) = (V ′ − {b}) ∪ V2− , which lead to |V3− | = r − 1, V3+ = {x} and + ′ N (y ) = {b} ∪ V3− . Also, we have V2− ↛ V3− (as otherwise, V2− → {a, y′ } ∪ V3− , this is impossible.). Then there exists an arc z w from V3− to V2− and furthermore, y′ → z. If V3− ↛ w , then there exists an arc w z ′ for some z ′ ∈ V3− − {z }, and abyxy′ z w z ′ a is a 8-cycle through ab. Assume V3− → w . If x → w , then V3 → w → V1 and abc w yxy′ v a is a desired 8-cycle for any vertex v ∈ V3− − {c }. Assume w → x. By {a, y} → x and Lemma 2.2, there exists a vertex u ∈ V2 − {b} such that x → u. Obviously, u ̸ = w and abyxuy′ c w a is a desired 8-cycle. Suppose now that V2− ↛ y′ . Then there is an arc y′ w for some w ∈ V2− . By V2+ − {b} → y′ and Lemma 2.2, there exists a vertex v ∈ V3 (v may be equal to c) such that y′ → v . Note that V3+ → y′ . So v ∈ V3− . Let u ∈ V2+ − {b}. If x → u, then abyxuy′ wv a (when w → v ) or abyxuy′ vw a (when v → w ) is a 8-cycle through ab. Assume u → x. If y → u, then abyuxy′ wv a (when w → v ) or abyuxy′ vw a (when v → w ) is a 8-cycle through ab. Assume u → y. If c → u, then abcuyxy′ w a is a desired 8-cycle. Assume u → c. Now, we have u → {y, y′ x, c } and r ≥ 4. By the arbitrariness of u, we may assume V2+ − {b} → {y, y′ , x, c }. If V1 − {a, y} ↛ w , then there exists a vertex y1 ∈ V1 − {a, y, y′ } such that w → y1 . When V ′′ ↛ y1 , then there is an arc y1 z1 for some z1 ∈ V ′′ . Thus, z1 ̸ = w and abyxy′ w y1 z1 a is a 8-cycle through ab. When V ′′ → y1 , we have y1 → V ′ . Thus, c → y1 → u and abcy1 uxy′ w a is a 8-cycle through ab. Assume V1 − {a, y} → w . Similarly, we may assume V1 − {a, y} → v . Consider the case v → w . If V3− ↛ w , then there exists an arc wv ′ for some v ′ ∈ V3− . Thus, v ′ ̸ = v and abyxy′ vwv ′ a is a 8cycle through ab. Assume V3− → w . Note that V1 −{a, y} → w and d− (w ) = r. We have |V3− | = 2, N − (w ) = (V1 −{a, y}) ∪ V3− and N + (w ) = {a, y} ∪ V3+ . So c → w → y. By V ′ − {b} → y′ and Claim 1, we get d+ (y′ ) ≥ r − 1 ≥ 3. Then there is a vertex V ′′ ′ ′′ ′ ′ ′ w ∈ V − {c , w} such that y → w and abc wyxy w a is a 8-cycle through ab. Consider the other case w → v . If V2− ↛ v , then there exists an arc v z ′ for some z ′ ∈ V2− . Thus, z ′ ̸ = w and abyxy′ wv z ′ a is a 8-cycle through ab. Assume V2− → v . If V2+ − {b} → v , then (V1 − {a, y}) ∪ (V2 − {b}) → v . Note that d− (v ) = r. We have |V1 − {a, y}| = 1 and r = 3, which contradicts r ≥ 4. So we have V2+ − {b} ↛ v and there exists a vertex u′ ∈ V2+ − {b} such that v → u′ . Recalling V2+ − {b} → {y, y′ , x, c }, we have u′ → {y, y′ , x, c } and v ̸ = c. Thus, abyxy′ v u′ ca is a 8-cycle through ab. Subcase 2 V3+ → y. Let x ∈ V3+ be arbitrary. Subcase 2.1 V2+ → y. In this case, we have V ′ → y → V ′′ . Note that {b, y} → c → a and a → x → {b, y}. Then there exists a vertex u ∈ (V1 −{a, y}) ∪ (V2 −{b}) such that c → u → x (as otherwise, we have d+ (c) < d+ (x), this is impossible.). Let v ∈ V3− −{c }. If u ∈ (V1 − {a, y}) ∪ (V2+ − {b}), then abcuxywv a (when w → v ) or abcuxyvw a (when v → w ) is a 8-cycle through ab for any vertex w ∈ V2− . If u ∈ V2− and |V2− | ≥ 2, then abcuxywv a (when w → v ) or abcuxyvw a (when v → w ) is a desired 8-cycle for any vertex w ∈ V2− − {u}. Assume that u ∈ V2− , |V2− | = 1 and x dominates the outset of c in (V1 − {a, y}) ∪ (V2+ − {b}). Now, we have |V2+ | = r − 1. Note that {b, y} → c. We have (V1 − {a, y}) ∪ (V2+ − {b}) ↛ c. Otherwise, d− (c) ≥ 2r − 2 > r, a contradiction. Therefore, there exists an arc cu1 for some u1 ∈ (V1 − {a, y}) ∪ (V2+ − {b}). By the assumption above, we have x → u1 . If u1 ∈ V2+ − {b}, then u1 → y and abcuxu1 yv a is a desired 8-cycle through ab. If u1 ∈ V1 − {a, y}, then by {x, c } → u1 and Lemma 2.2, there is a vertex w ∈ V2 − {b} such that u1 → w . When w = u, we have that abcu1 uxyv a is a desired 8-cycle. When w ̸ = u, we have w ∈ V2+ − {b} and w → y. Now, abcuxu1 w yv a is a 9-cycle through ab. Subcase 2.2 V2+ ↛ y. By the hypothesis, there exists a vertex u ∈ V2+ − {b} such that y → u. Suppose first that V1 −{a, y} → u. Then V1 → u → V3 . If V1 −{a, y} ↛ c, then there is an arc cy1 for some y1 ∈ V1 −{a, y}. By V3+ ∪ {b} → y and Claim 1, there is a vertex v ∈ V ′′ − {c } such that y → v . Thus, abcy1 uxyv a is a 8-cycle through ab. Assume V1 − {a, y} → c. Then N − (c) = (V1 − {a, y}) ∪ {b, u} and N + (c) = {a, y} ∪ (V2 − {b, u}). Let y′ ∈ V1 − {a, y} and w ∈ V2− . Then y′ → c → w. If x → y′ , then abyuxy′ c wa is a desired 8-cycle. If y′ → x, then y′ → {x, c }. Lemma 2.2 implies that there is a vertex u′ ∈ V2 − {b} such that u′ → y′ . Obviously, u′ ̸ = u and c → u′ . By V3+ ∪ {b} → y and Claim 1, there exists a vertex c ′ ∈ V ′′ − {u′ } such that y → c ′ . Since c → y, we have that c ′ ̸ = c and abcu′ y′ xyc ′ a is a 8-cycle through ab. Suppose now that V1 − {a, y} ↛ u. Then there exists a vertex y′ ∈ V1 − {a, y} such that u → y′ . Subcase 2.2.1 V3+ ↛ y′ . By the hypothesis, there is an arc y′ x′ for some x′ ∈ V3+ (x′ may be equal to x). Recalling that V3+ → y, we get x′ → y. If V1 − {y} ↛ x′ , then there exists a vertex y1 ∈ V1 − {a, y, y′ } such that x′ → y1 . By x′ → y1 and Claim 1, there is a vertex w ∈ V ′′ such that y1 → w . Thus, abyuy′ x′ y1 w a is a 8-cycle through ab. Assume V1 − {y} → x′ . If V2 − {b, u} ↛ x′ , then there exists an arc x′ u′ for some u′ ∈ V2 − {b, u}. When V3− ↛ u′ , then there is an arc u′ w ′ for some w ′ ∈ V3− . Thus, abyuy′ x′ u′ w ′ a is a 8-cycle through ab. When V3− → u′ , by V3− ∪ {x′ } → u′ and Lemma 2.2, there exists a vertex y1 ∈ V1 − {a, y} (y1 may be equal to y′ ) such that u′ → y1 . Then y1 → x′ . If |V3+ | ≥ 2, then by V3+ ∪ {b} → y and Claim 1, there exists a vertex v ′ ∈ V ′′ − {u′ , c } such that y → v ′ . Now, abcu′ y1 x′ yv ′ a is a 8-cycle through ab. If |V3+ | = 1, then
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V3+ = {x′ }, V3 → u′ → V1 and so u′ → y. By V3+ ∪ {b} → y and Claim 1, there exists a vertex v ′ ∈ V ′′ − {c } such that y → v ′ . Obviously, v ′ ̸ = u′ and abcu′ y1 x′ yv ′ a is a 8-cycle through ab. If V2 − {b, u} → x′ , then |V2 − {b, u}| = 1 since V1 − {y} → x′ and d− (x′ ) = r. Let V2 − {b, u} = {w}. Then r = 3 and {a, y′ , w} → x′ → {y, b, u}. So {a, y, x′ } → u → {y′ } ∪ V3− . By {b, u} → c, we get c → y′ or c → w. Let v ′ ∈ V3− − {c }. Then u → v ′ and abcy′ x′ yuv ′ a or abc w x′ yuv ′ a is a 8-cycle through ab. Subcase 2.2.2 V3+ → y′ . In this case, we have x → y′ . Suppose first that V2+ − {b} ↛ y′ . Then there exists a vertex u′ ∈ V2+ − {b, u} such that y′ → u′ . If V1 − {y} ↛ u′ , then there exists a vertex y1 ∈ V1 − {a, y, y′ } such that u′ → y1 . By u′ → y1 and Claim 1, there is a vertex w1 ∈ V ′′ such that y1 → w1 and abyuy′ u′ y1 w1 a is a 8-cycle through ab. Assume V1 − {y} → u′ . By Lemma 2.2, we have |NV+ (u′ )| ≥ r − 1. Note 3 that |V3− | ≥ 2. Then there exists a vertex v ′ ∈ V3− such that u′ → v ′ . If u → x, then ab is contained in a 8-cycle abyuxy′ u′ v ′ a. Assume x → u. Then x → {y, y′ , b, u} and r ≥ 4. Let w ∈ V2− . If w ↛ NV+ (u′ ), then there exists an arc v0 w for some v0 ∈ NV+ (u′ ). Now, abyuy′ u′ v0 w a is a desired 8-cycle. Assume 3 3 w → NV+3 (u′ ). Note that w → a and |NV+3 (u′ )| ≥ r − 1. We have N + (w) = NV+3 (u′ ) ∪ {a}, |NV+3 (u′ )| = r − 1 and V1 − {a} → w. Let y1 ∈ V1 − {a, y, y′ }. Then y1 → {u′ , w}. By Lemma 2.2, there exist two vertices v1 , v2 in V3 such that {v1 , v2 } → y1 . Since |NV+ (u′ )| = r − 1, there is at least one vertex in {v1 , v2 } which belongs to NV+ (u′ ). Without loss of generality, assume 3 3 v1 ∈ NV+3 (u′ ). Then abyuy′ u′ v1 y1 wa is a desired 9-cycle. Suppose now that V2+ − {b} → y′ . Note that V3+ → y′ . We have V ′ − {b} → y′ . If V2− → y′ , then it is easy to see that |V2− | = |V3+ | = 1 and N − (y′ ) = (V ′ − {b}) ∪ V2− . So |V3− | = r − 1, V3+ = {x} and N + (y′ ) = {b} ∪ V3− . Furthermore, we have V2− ↛ V3− (as otherwise, V2− → {a, y′ } ∪ V3− , this is impossible.). Then there exists an arc z w from V3− to V2− (z may be equal to c). If V3− ↛ w , then there exists an arc w z ′ for some z ′ ∈ V3− − {z }, and abyuy′ z w z ′ a is a 8-cycle through ab. Assume V3− → w . Let v ∈ V3− − {c }. If x → w , then V3 → w → V1 and abc w yuy′ v a is a desired 8-cycle. If w → x, then abc w xyuy′ v a is a desired 9-cycle. If V2− ↛ y′ , then there is an arc y′ w for some w ∈ V2− . By u → y′ and Lemma 2.2, there exists a vertex v ∈ V3 such that y′ → v . Note that V3+ → y′ . We have v ∈ V3− . If u → x, then abyuxy′ wv a (when w → v ) or abyuxy′ vw a (when v → w ) is a desired 8-cycle. Assume x → u. Now, x → {y, y′ , b, u} and r ≥ 4. + ′ Let X = (V1 ∪ V2 ) − {a, y, y′ , b, u}. When NX+ (c) ⊆ NX+ (x), we have d+ X (c) ≤ dX (x). Note that x → {y, y , b, u}. Then c → {a, y, y′ , u} (as otherwise, d+ (c) < d+ (x), this is impossible.). Since y′ → v , we get that v ̸ = c and abcyuy′ wv a (when w → v ) or abcyuy′ vwa (when v → w) is a desired 8-cycle. When NX+ (c) ⊊ NX+ (x), then there exists a vertex u0 ∈ X such that c → u0 → x. If u0 ̸ = w , then abcu0 xyuy′ w a is a 9-cycle through ab. Assume u0 = w . If V ′′ − {w, c } ↛ y′ , then there is an arc y′ v ′ for some v ′ ∈ V ′′ −{w, c } and abc w xyuy′ v ′ a is a desired 9-cycle. If V ′′ −{w, c } → y′ , then N − (y′ ) = (V ′ −{b})∪(V ′′ −{w, c }) and |V ′′ − {w, c }| = 1. Now, r = 3 which contradicts with r ≥ 4. □ The proof of Theorem 2.2 is complete. ■ Combining Theorem 2.1 with Theorem 2.2, we have the following theorem. Theorem 2.3. If D is an r-regular 3-partite tournament with r ≥ 3, then each arc of D is contained in a 8- or 9-cycle. Acknowledgments We would like to thank the referees for their comments and suggestions which improved the presentation. References [1] [2] [3] [4] [5]
Q. Guo, L. Cui, W. Meng, On cycles through an arc in regular 3-partite tournaments, Discrete Math. 339 (2016) 17–20. I. Stella, L. Volkmann, Each regular 3-partite tournaments has a 9-cycle, An. Univ. Bucuresti Mat. 56 (2007) 91–120. L. Volkmann, On cycles in regular 3-partite tournaments, Discrete Math. 306 (2006) 1198–1206. L. Volkmann, Multipartite tournaments: a survey, Discrete Math. 307 (2007) 3097–3129. G. Xu, S. Li, Q. Guo, H. Li, Notes on cycles through a vertex or an arc in regular 3-partite tournaments, Appl. Math. Lett. 25 (2012) 662–664.