Lichiardopol's conjecture on disjoint cycles in tournaments

In 2010, N. Lichiardopol conjectured for $q \geq 3$ and $k \geq 1$ that any tournament with minimum out-degree at least $(q-1)k-1$ contains $k$ disjoint cycles of length $q$. We prove this conjecture for $q \geq 5$. Since it is already known to hold for $q\le4$, this completes the proof of the conjecture.


Introduction
We consider cycles in digraphs (directed graphs); a cycle is a strongly connected digraph in which every vertex has indegree 1 and outdegree 1. The length of a cycle is the number of edges; a q-cycle is a cycle of length q. By "k disjoint cycles" we always mean k pairwise vertex disjoint cycles. A tournament is a digraph obtained from a complete graph by assigning a direction to each edge.
A famous conjecture of Bermond and Thomassen [5] for arbitrary digraphs asserts that large minimum outdegree guarantees many disjoint cycles. Conjecture 1.1 ( [5]). If a digraph D has minimum outdegree at least 2k−1, then D contains k disjoint cycles. since a tournament on qk − 1 vertices in which every vertex has outdegree qk/2 − 1 does not have enough vertices to have k disjoint q-cycles. Given that the needed inequalities are easier to satisfy when q is large, we ask whether there is a positive constant ε such that minimum outdegree (1 − ε)qk suffices when q is sufficiently large.
Motivated by this problem on tournaments, one may wonder whether large outdegree can guarantee disjoint cycles of the same length in general digraphs, even without constraining which length it is. Our result guarantees this for tournaments. Thomassen [22] conjectured such a relationship for general digraphs, but Alon [1] showed that it cannot hold. Theorem 1.7 ( [1]). For all r ∈ N, some digraph with minimum outdegree r has no two edgedisjoint cycles of the same length (and hence also no disjoint cycles of the same length).
Other papers on disjoint cycles in digraphs include [2] and [9]. Disjoint cycles have also been studied in undirected graphs, where the results are more plentiful. The Bermond-Thomassen Conjecture is in fact the directed analogue of the Corrádi-Hajnal Theorem [12]. Theorem 1.8 ([12]). For n, k ∈ N with n ≥ 3k, every n-vertex undirected graph with minimum degree at least 2k contains k disjoint cycles.
There are many extensions and variations on this result. Those most similar to our work consider the lengths of disjoint cycles guaranteed by a threshold on the minimum degree. Let δ(G) denote the minimum degree of a graph G; also, the order of a graph or digraph is the number of vertices.
Thomassen [23] proved for k ≥ 2 that every graph G with δ(G) ≥ 3k + 1 and order at least some constant c k contains k disjoint cycles of the same length. Thomassen conjectured that minimum degree 2k suffices, which had earlier been conjectured for k = 2 by Häggkvist. Egawa [13] proved Thomassen's conjecture for k ≥ 3 with a threshold of |V (G)| ≥ 17k +o(k). Verstraëte [24] later proved Thomassen's conjecture in full. Verstraëte also conjectured that order at least 4k is enough to guarantee k disjoint cycles of the same length when δ(G) ≥ 2k.
Chiba, Fujita, Kawarabayashi, and Sakuma [10] guaranteed for k ∈ N a constant c k such that every graph with order at least c k and minimum degree at least 2k contains k disjoint even cycles, with special exceptions. Other degree conditions for disjoint cycles in undirected graphs can be found in [4] and in the survey [11].

Structure of the Proof
To prove Theorem 1.6, we prove a theorem that was mainly inspired by the proof of the Bermond-Thomassen Conjecture for tournaments by Bang-Jensen, Bessy, and Thomassé [3].
Theorem 2.1. Fix k, q ∈ N with q ≥ 5 and k ≥ 2, and let T be a tournament with minimum outdegree at least (q − 1)k − 1. For any family F of k − 1 disjoint q-cycles in T , there is a family of k disjoint q-cycles in T using at most 3q − 6 vertices outside the cycles in F .
We show first that this suffices. For a path with vertices v 1 , . . . , v q in order, we use the notation v 1 , . . . , v q (each edge in v 1 , . . . , v q is oriented from v i to v i+1 ); we also call this a v 1 , v q -path. For a cycle with vertices v 1 , . . . , v q in order, we use the notation [v 1 , . . . , v q ]. The outdegree of a vertex v is d + (v), and the minimum outdegree in a digraph D is δ + (D).
It is well-known that every tournament contains a spanning path (Rédei's Theorem [21]), and that a tournament is strongly connected if and only if it contains a spanning cycle (Camion's Theorem [8]), where a digraph is strongly connected or strong if it contains a u, vpath for any two vertices u and v. Moreover, a strong touranment (or the subtournament induced by any cycle) is pancyclic, meaning that it contains cycles of all lengths from 3 through through the number of vertices. When invoking this, we say "by pancyclicity". Finally, Moon [20] showed that every strong tournament with at least three vertices is vertex pancyclic, meaning that through any vertex there are cycles of all lengths. Proof. Assuming that Theorem 2.1 holds, we prove Theorem 1.6 by induction on k. When k = 1, we are given a tournament T with δ + (T ) ≥ q − 2. Since every tournament has a spanning path, we may let v 1 , . . . , v n be a spanning path of T . Since d + (v n ) ≥ q − 2 and v n−1 is not an outneighbor of v n , the edge to the earliest outneighbor of v n along P completes a cycle of length at least q in T . By pancyclicity, T contains a q-cycle.
For the induction step, suppose k > 1. We have δ By the induction hypothesis, T contains k − 1 disjoint q-cycles. From this family F of k − 1 disjoint q-cycles, Theorem 2.1 produces a family of k disjoint q-cycles.
We will prove Theorem 2.1 by considering two cases. In Theorem 4.1, we will prove that the conclusion holds when q ≥ 5 and k ≤ q. In Theorem 5.1, we will prove by induction on k that the conclusion holds when q ≥ 5 and k > q, using Theorem 4.1 as a basis. The restriction to using 3q − 6 vertices outside F will be helpful in proving Theorem 5.1.

Cycles and Paths in Tournaments
Here we prove some structural lemmas about tournaments that will be useful in the proof of Theorem 4.1. We use T [S] to denote the subtournament of T induced by the vertex set S. We also let d + (X, Y ) denote the number of edges with tail in X and head in Y , where X and Y may be a set of vertices or a single vertex. When uv is an edge, we say that v is a successor of u and u is a predecessor of v.
contains a cycle C ′ of length m − 1 such that the omitted vertex u of C has at least two predecessors in V (C ′ ).
Proof. Since every strong tournament is pancyclic, T [V (C)] contains a cycle C * of length m − 1. Let u ′ be the vertex of C not in C * , and let v be the predecessor of u ′ on C. If u ′ has another predecessor on C * , then let u = u ′ and C ′ = C * . Otherwise, let v, w, x be the portion of C * leaving v (this exists since m ≥ 4). Since u ′ has only one predecessor in V (C), both u ′ w and u ′ x are edges. Now form C ′ by replacing w with u ′ in C * , and let u = w. The omitted vertex u now has predecessors v and u ′ on C ′ .
We believe in general that for m ≥ 2r, a strong tournament with m vertices contains a cycle of length m − 1 such that the omitted vertex has at least r predecessors in C ′ . We proved this for r = 3, but the proof is considerably longer than for r = 2, and we only need the result for r = 2. The threshold on m is sharp; a tournament on 2r − 1 vertices in which every vertex has r − 1 predecessors and r − 1 successors has no vertex with r predecessors.
contains a cycle C ′′ of length m − 2 omitting vertices x and y with xy ∈ E(T ) such that y has at least one predecessor in V (C ′′ ).
Proof. We use Lemma 3.1 twice. First, it gives us a cycle C ′ in T [V (C)], with length m − 1, such that the remaining vertex u has at least two predecessors in V (C ′ ). We then apply Lemma 3.1 using this C ′ as C; it gives us a cycle C ′′ in T [V (C ′ )], with length m − 2, such that the remaining vertex v has at least two predecessors in V (C ′′ ). From C, the vertices u and v have been omitted. Each has at least one predecessor in V (C ′′ ). Hence we let x be the tail and y be the head of the edge joining u and v.
We say that a vertex set X dominates a vertex set Y in a tournament if every edge joining X and Y is oriented from X to Y . Proof. If the claim fails, then T [S] is a 3-cycle. Now failure of the claim requires S to dominate V (C) − S, contradicting that C is a cycle. Lemma 3.4. Let T be a tournament with minimum outdegree at least (q − 1)k − 1, where 2 ≤ k ≤ q and q ≥ 5. Let F be a family of k − 1 disjoint q-cycles in T , with vertex sets V 1 , . . . , V k−1 inducing cycles C 1 , . . . , C k−1 , and let P be a path through the remaining vertices. If S 1 and S 2 partition V (P ) with S 2 dominating S 1 in T and |S 1 | ≤ q, then (d) If S 1 dominates z ∈ V * , and z has at least r predecessors in V * , then d + (z, S 2 ) ≥ r.
Proof. of (a): Since S 2 dominates S 1 , By the pigeonhole principle, there exists Using part (a), we obtain With s = |S 1 |, the inequality can be rewritten as 2q − 2 − s(2q − 3 − s) ≥ 0. However, with 2 ≤ s ≤ q, the left side of this is negative when q ≥ 5.
Proof of (c):

Proof of (d):
Since z is dominated by S 1 and has at least r predecessors in V * , Since k ≤ q, we obtain d + (z, S 2 ) ≥ r.

The Case of Small k
In this section we prove Theorem 2.1 for k ≤ q, stated as Theorem 4.1. Essentially, we provide an algorithm to produce the desired family F * of k disjoint q-cycles by iteratively increasing the length of a cycle found outside the k − 1 disjoint q-cycles. The subtournament T ′ induced by the vertices not in the given cycles has a spanning path P ; let v be its last vertex. If v lies in a cycle of length at least q in T ′ , then by pancyclicity there is a q-cycle in T ′ , and we are done. Hence our approach, given a longest cycle through v in T ′ (in the first step the length may be 0), is to rearrange F to find a new family F ′ of k − 1 disjoint q-cycles so that the vertex at the end of the resulting remaining path P ′ lies in a longer cycle. Some of the claims in this argument are not valid when q = 4. Nevertheless, the same framework applies when q = 4, with additional more detailed reasoning.
Theorem 4.1. Given k, q ∈ N with q ≥ 5 and k ≤ q, let T be a tournament with δ + (T ) ≥ (q − 1)k − 1. For any family F of k − 1 disjoint q-cycles in T , there is a family F * of k disjoint q-cycles in T whose union has at most 3q − 6 vertices outside the cycles in F .
Proof. These hypotheses are the same as those of Lemma 3.4 once we obtain a partition (S 2 , S 1 ) of the vertices outside F such that S 2 dominates S 1 . Given such a partition, let C * with vertex set V * be the cycle in F guaranteed by Lemma 3.4(a). Let P be a spanning path through the subtournament T ′ of vertices not used by F , with last vertex v.
When T ′ has a cycle through v, let l be the maximum length of such a cycle; otherwise l = 0. If l ≥ q, then pancyclicity of the subtournament spanned by this cycle provides a q-cycle to complete F * . Otherwise, we obtain a new family F ′ where l is larger, generally by replacing C * with a new cycle C and defining a new path P ′ through the vertices outside F ′ . We will use at most two new vertices in C at each step that increases l, except that the steps to reach l ≥ 4 will use at most three new vertices. In addition, when a sufficiently long cycle appears, it uses at most q − 1 new vertices, because the subtournament that was outside the k − 1 given q-cycles entering that step did not contain a cycle of length at least q. Thus in total at most 3q − 6 new vertices are used.
Case 1: l = 0, so T ′ has no cycle through v. We seek C and P ′ so that there is a cycle outside F ′ through the last vertex of P ′ . Let u be the predecessor of v on P .
Case 1a. T ′ has no cycle through u (see Figure 1) . Choose three vertices in V (C ′′ ). By Lemma 3.3, among them is a vertex z with at least two predecessors in V * . Since also z is dominated by S 1 , Lemma 3.4(d) guarantees d + (z, S 2 ) ≥ 2. Let w be a successor of z in S 2 , and let z ′ be the successor of z on C ′′ . Replacing zz ′ in C ′′ with z, w, u, z ′ yields a q-cycle C with two vertices outside F . Replace C * with C to form F ′ from F . Since S 2 dominates S 1 , we can form the P ′ outside F ′ by appending v, x, y in order to a spanning path of T [S 2 ] − w. Since S 1 dominates V * , and y has at least x and a vertex of C ′′ as predecessors in V * , Lemma 3.4(d) yields d + (y, S 2 ) ≥ 2. Thus y has a successor in S 2 other than w, so there is a cycle through y using vertices of P ′ . Case 1b. T ′ has a cycle containing u. Let B be a longest cycle containing u in T ′ . Let S 1 = V (B) ∪ {v} and S 2 = V (T ′ ) − S 1 . Any edge from V (B) to S 2 yields a larger strong tournament in T ′ containing V (B), which contains a larger cycle containing u. Hence S 2 dominates S 1 . Also |V (B)| ≤ q − 1, since otherwise we have the kth q-cycle. Hence 4 ≤ |S 1 | ≤ q, so Lemma 3.4 applies. Choose C * with vertex set V * as given by Lemma 3.4.
We will use the following "degree fact". If F ′ is a family of k − 1 disjoint q-cycles, and the last vertex v ′ in a spanning path P ′ through the set S of remaining vertices has at least two predecessors used by F ′ , then v ′ has a successor in S. The reason, using k ≤ q, is There is then a cycle in S through v ′ , as desired.
The degree fact implies d + (V * , v) ≤ 1; otherwise already T ′ has a cycle through v. First suppose d + (V * , v) = 1 (see Figure 2). Let z be the predecessor of v in V * , and let y be the successor of z on C * . Form F ′ by replacing C * with the cycle C obtained from C * by replacing y with v (the successor of y on C * is a successor of v). Since 4 ≤ |S 1 | ≤ q, by Lemma 3.4(b) y has at least two predecessors in S 1 , at least one in B (call this vertex w). Since S 2 dominates S 1 , we can form P ′ by following P through S 2 , then B from the successor of w on B to w, and finally the edge wy. Since v and z are predecessors of y, the degree fact yields a cycle outside F ′ through y. The only vertex used by F ′ and not by F is v. Therefore, we may assume d + (V * , v) = 0, so v dominates V * (see Figures 3 and 4). In the remaining case, every vertex of B dominates or is dominated by 2q−2 |S 1 | ≤ q 2 +q 2q−2 requires q < 5, we conclude that V (B) dominates V ′ . Since also d + (V * , v) = 0, now S 1 dominates V ′ (see Figure 4). The remaining argument is similar to Case 1a. Since q − 1 ≥ 4, by Lemma 3.1 there is a cycle C ′′ of length q − 2 in T [V ′ ] such that the vertex y ′ of V ′ − V (C ′′ ) has at least two predecessors in V (C ′′ ). Now y and y ′ each have at at least two predecessors in V * . Let zz ′ be the edge joining y and y ′ .
Since S 1 dominates V ′ , any vertex of C ′′ has a successor in S 2 , by Lemma 3.4(d); let w be one such successor. Now T [V (C ′′ ) ∪ {w, v}] is strong and has a spanning cycle C of length q. Form F ′ from F by replacing C * with C (note that F ′ uses only w and v outside F ).
Using Lemma 3.4(b), d + (V (B), z) ≥ 1; let x be a predecessor of z in V (B). Since S 2 dominates S 1 , we can build a path P ′ that starts with all of S 2 − {w} (in some order), then visits all of V (B) ending with x, and finally follows x, z, z ′ . Since v dominates V * , vertex z ′ has at least two predecessors in C, and the degree fact applies. Case 2: l > 0, so T ′ has a cycle through v, the longest having length l. Let C be such a cycle of length l. We find a new family F ′ and path P ′ outside it with a longer cycle through the last vertex of P ′ . We may assume l < q, since otherwise pancyclicity yields the desired q-cycle. Let S 1 = V (C) and S 2 = V (T ′ ) − S 1 . If there is an edge from S 1 to S 2 , then S 1 and part of P induce a strong tournament, which has a longer spanning cycle (containing v). Hence S 2 dominates S 1 , and Lemma 3.4 applies.
Let C * with vertex set V * be the cycle in F guaranteed by Lemma 3.4. Since q ≥ 5, by Lemma 3.2 T [V * ] contains a cycle C ′′ of length q−2 and an edge xy with {x, y} = V * −V (C ′′ ) such that y has at least one predecessor in V (C ′′ ). Let V ′′ = V (C ′′ ). By Lemma 3.4(a), Since q/2 < q − 2 when q ≥ 5, in V ′′ there is a vertex z dominated by S 1 . By Lemma 3.4(d), z has a successor w in S 2 .
Let the cycle C through S 1 be [u ′ , x ′ , u], so x ′ x ∈ E(T ). Note that T [V ′′ ∪ {w, u}] is strong, with a spanning cycle C. Form F ′ from F by replacing C * with C.
Let P ′ follow a spanning path through S 2 − {w} and then u ′ , x ′ , x, y . If yu ′ ∈ E(T ), then we have the cycle [y, u ′ , x ′ , x] through y. Otherwise d + (y, S 1 ) = 0, and by Lemma 3.4(d) y has at least two successors in S 2 , which means it has one other than w. In this case y lies on a cycle of length more than l in T [V (P ′ )]. Case 2b. |S 1 | ≥ 4 (see Figure 6). By pancyclicity, T [S 1 ] contains a cycle B omitting one vertex u of S 1 . Since l < q, Lemma 3.4(c) implies d + (u, V * ) ≥ 3, yielding an edge uz ′ with z ′ ∈ V ′′ . Using the edge zw from V ′′ to S 2 (obtained earlier), the path z, w, u, z ′ guarantees that T [V ′′ ∪ {w, u}] is strong and hence has a spanning q-cycle C. Form F ′ from F by replacing C * with C.
By Lemma 3.4(b), x has a predecessor x ′ in V (B). Build the path P ′ outside F ′ by visiting all of S 2 − {w} (in some order), then all of V (B) ending with x ′ , and finally x ′ , x, y . Since y has at least two predecessors in V ′′ ∪ {x} and hence at most q − 3 successors in V * , If y has a successor outside {w, u} in S 1 ∪ S 2 , then with V (B) ∪ {x, y} it induces a strong tournament of order at least l + 1, yielding the desired cycle through the last vertex of P ′ . Hence w and u must be the only successors of y in S 1 ∪ S 2 . This forces d + (V ′′ , y) = 1. If any vertex of V ′′ has a successor in S 2 other than w, then we obtain F ′ as above using that vertex instead of w. Hence w must be the only successor of vertices in V ′′ , so d + (V ′′ , S 2 ) ≤ q − 2. We already computed d + (V * , S 1 ) ≤ q/2, and one of the edges counted is yu.
For any α ∈ V ′′ , using k ≤ q we have When we sum over all α ∈ V ′′ , the last term counts all edges in T [V ′′ ], possibly q − 2 edges from V ′′ to x, and one edge from V ′′ to y. Hence The upper and lower bounds on d + (V ′′ , S 1 ∪ S 2 ) require (q − 2)(q − 1)/2 − 1 ≤ 3q/2 − 3, but this inequality requires q < 5. Hence we obtain the desired improvement F ′ .
In all cases we have improved the family F as desired.

The Case of Large k
For the proof of Theorem 5.1, we will use Theorem 4.1 as a basis for induction on k. The two theorems have the same conclusion.
For any family F of k − 1 disjoint q-cycles in T , there is a family of k disjoint q-cycles in T whose union has at most 3q − 6 vertices outside the cycles in F .
We first discuss the basic set-up for the argument, defining notation to be used throughout the proof. We call the desired family an extension of F ; finding it is extending F . By Theorem 4.1, we may assume k ≥ q + 1. In the tournament T , consider a family F of k − 1 disjoint q-cycles in T . We may assume that the tournament T ′ given by deleting the vertices covered by F contains no cycle with length at least q; otherwise we have the desired extension.
We aim to find a value t ∈ {1, 2} such that we can replace t cycles in F with t + 1 cycles of length q using at most 3q − 6 vertices of T ′ . This will complete the proof.
Let X and Y be two disjoint sets of vertices in T . We say that there is an r-matching from X to Y if the set of edges with tail in X and head in Y contains r edges with no common endpoints. In order to guarantee the existence of desired matchings, we will use the famous König-Egerváry Theorem (König [15], Egerváry [14]), phrased for bipartite digraphs with all edges directed from one part to the other. Lemma 5.2 ([14,15]). If there is no r-matching from X to Y , then X ∪ Y contains a set of at most r − 1 vertices whose deletion eliminates all edges from X to Y .
For the proof of Theorem 5.1, we need a number of additional lemmas. The first is a standard application of the König-Egerváry Theorem, which we will apply with various values of the parameters. Lemma 5.3. Let X and Y be disjoint vertex sets in T , with s = min{|X|, |Y |} and t = max{|X|, |Y |}. If d + (X, Y ) > (r − 1)t, where 1 ≤ r ≤ s, then T contains an r-matching from X to Y .
Proof. Since r −1 vertices cover at most (r −1)t edges, the König-Egerváry Theorem implies that the desired matching exists.

each having at least two successors in C, then there is an extension of F .
Proof. Let W ⊆ U 2 be such a set of 3q − 6 successors of v. Since T [V (C)] is strong, by Moon's Theorem it has a (q − 1)-cycle C ′ containing v, omitting one vertex u of C. Since each vertex w ∈ W has a successor in C other than u, the subtournament T [V (C ′ ) ∪ {w}] is strong and has a spanning q-cycle (see Figure 7).
and F 0 is a family of k − 2 cycles of length q in T ′ . Using the induction hypothesis, we can extend F 0 to a family F of k − 1 cycles of length q in T ′ using at most 3q − 6 new vertices. Since |W ∪ {u}| = 3q − 5, some vertex in W ∪ {u} is not used by F.
If u is not used, then adding C to F completes the desired extension. If u is used, then at most 3q − 7 vertices not in F are used in F. In this case, some vertex w ∈ W is not used, and a spanning cycle in T [V (C ′ ) ∪ {w}] completes the extension F ′ using a total of at most 3q − 6 vertices not in F . We use Lemma 5.4 to prove the next two lemmas.
Lemma 5.5. Let C be a q-cycle in F . Suppose that F has no extension.
Proof. (i) If there is no such 2-matching, then by Lemma 5.2 one vertex covers all the edges from V (C) to U 2 . Such a vertex v can only be in C, and there is no other edge from V (C) to U 2 . Since q ≥ 3, each of the 3q − 6 successors of v in U 2 has at least two successors in V (C). By Lemma 5.4, we have an extension of F . (ii) If there is no such 3-matching, then by Lemma 5.2 two vertices u and v cover all the edges from V (C) to U 2 , of which there are at least 6q − 13. If u and v are both in V (C), then one has at least 3q − 6 successors in U 2 , each of which has at least two successors in C, since q ≥ 4. Otherwise, name u and v with u ∈ U 2 and v ∈ V (C); now v has at least 5q − 14 successors in U 2 other than u. These vertices have no predecessors in V (C) other than v; hence they have at least two successors in V (C). Since 5q − 14 ≥ 3q − 6 when q ≥ 4, in either case Lemma 5.4 applies to guarantee an extension of F . Lemma 5.6. Let C be a q-cycle in F . If T contains (i) a q-matching from U 1 to V (C) and a 2-matching from V (C) to U 2 , or (ii) a (q − 1)-matching from U 1 to V (C) and a 3-matching from V (C) to U 2 , then there is an extension of F .
Proof. We will obtain two disjoint cycles of length at least q in T [V (C) ∪V (P )], where P is a spanning path through the vertices outside F . By pancyclicity, the subtournament induced by the vertices of each such cycle contains a q-cycle. Since T ′ induces no cycle of length at least q, each of the two new cycles replacing C contains at most q − 1 new vertices.
Since q ≥ 5, we have 3q − 11 ≥ 2q − 6. Hence |S| ≥ 2q − 6. Since V (P ) induces no cycle of length at least q, any edge joining two vertices of P with at least q − 2 vertices between them on P is directed from the earlier to the later vertex.
Let M and M ′ be the given matchings from U 1 to V (C) and from V (C) to U 2 , respectively. We prove (i) and (ii) together. In either case, let u be the last vertex of P matched into C from U 1 by M; note that u ∈ {u 2 , u 1 }. After following the edge from u to V (C), let v be the vertex matched into U 2 by M ′ that is reached first when continuing along C, and let vw be this edge of M ′ . Let Q be the path thus followed, from u via M, along C, ending with vw. Now choose yz ∈ M ′ − {vw} so that y is also the head of an edge in M (under (ii), more than one edge of M ′ remains, but then at most one vertex of C is not covered by M.) Say that a vertex of U 1 leads to y if it is matched by M into the path along C that starts with the successor of v on C and ends at y. If z is closer to S than w along P , then let x be the highest-indexed vertex of U 1 (closest to S) that leads to y. Otherwise, let x be the lowest-indexed vertex of of U 1 that leads to y. Let R be the path leaving x via M, then along C to y, ending with yz. We want to form two cycles of length at least q in T [V (C) ∪ V (P )] by adding vertices of P to Q or R. Let P [a, b] or C[a, b] denote the a, b-path along P or C.
Case 1: z is closer to S than w along P , so x is the highest-indexed vertex of U 1 leading to y (see Figure 8). First consider x = u q . Let one cycle be R ∪ P [z, x]. Since |S| ≥ 2q − 6 and this cycle contains S ∪{x, y, z}, it has length at least 2q −3, which exceeds q. Meanwhile, P has at least 2q − 4 vertices between w and u q−1 . Since 2q − 4 ≥ q − 2, the edge joining them is oriented as wu q−1 . Hence Q ∪ wu q−1 ∪ P [u q−1 , u] is a cycle with at least q vertices.
When x = u q , the edge joining u 2q−2 and x has the desired orientation, because along P it skips u q and q − 3 vertices of S. Hence R ∪ P [z, u 2q−2 ] ∪ u 2q−2 x is a cycle. Since |S| ≥ 2q − 6, it has at least q − 3 vertices in S plus {x, y, z}. The other cycle is Q ∪ wu 2q−3 ∪ P [u 2q−3 , u q+1 ] ∪ u q+1 u. Since w is earlier than z along P , there are at least q − 2 vertices between w and u 2q−3 along P ; the same is true of u q+1 and u. Hence this is a cycle, and {u 2q−3 , . . . , u q+1 } ∪ {u, v, w} has at least q vertices. Figure 8: Case 1 of Lemma 5.6.
Case 2: z is farther from S than w along P , so x is the lowest indexed vertex of U 1 leading to y. Let x = u t , and define the paths Q and R as in Case 1. Note that t ≤ q. We want the two cycles to be R ∪ zu t+q−3 ∪ P [u t+q−3 , x] and [Q ∪ P [w, u t+q−2 ] ∪ u t+q−2 u t−1 ∪ P [u t−1 , u]. The jumps along P must skip at least q − 2 vertices. This is explicit for u t+q−2 u t−1 . Since (z, w) = (u j , u i ) with j > i > 4q − 11, and j − (t + q − 3) ≥ 3q − 6 − t ≥ 2q − 6, the other construction is also a cycle if 2q − 6 ≥ q − 1, which holds when q ≥ 5.
For length at least q, the first cycle has q − 3 vertices along P plus at least {x, y, z}, and the second adds to {u, v, w} all of S ∪ U 1 except the q − 2 vertices used by the first cycle and u 1 and maybe u 2 . Since 4q − 11 − q ≥ q − 3 when q ≥ 4, both cycles are long enough. Figure 9: Case 2 of Lemma 5.6.
Lemma 5.7. Let C and C ′ be two members of F , with W = V (C) and W ′ = V (C ′ ). If T contains a q-matching from U 1 to W and a 3-matching from W ′ to U 2 , and d + (W, W ′ )) ≥ q(q − 1) + 3, then there is an extension of F . Proof. Again use the same notation. We may assume that T [V (P )] contains no q-cycle. We will extend F by replacing C and C ′ in F with three q-cycles (except in one case). We must ensure that they introduce at most 3q − 6 new vertices. The other members of F remain.
Let the edges of the given 3-matching from W ′ to U 2 be y 1 z 1 , y 2 z 2 , y 3 z 3 , indexed so that z 3 , z 1 , z 2 occur in that order along P (z 2 is closest to S). Since each vertex in We complete these three paths to disjoint cycles by adding vertices along the path P . Recall that P contains the path u 4q−11 , . . . , u q+1 through S between U 2 and U 1 . Along this path of 3q − 11 vertices define three disjoint paths, each having q − 4 vertices (one vertex of S is not needed); call them Q 2 , Q 3 , Q 1 in order along P .
Let B i be the cycle formed by combining R i and Q i ; add the edges from the end of each of R i and Q i to the beginning of the other, except that between z 2 and Q 2 in B 2 we follow P to the end of U 2 , and between Q 1 and w 1 in B 1 we follow P through the beginning of U 1 . The edges from z 1 to Q 1 , from z 3 to Q 3 , from Q 3 to w 3 , and from Q 2 to w 2 , are oriented in the desired direction because they skip at least q − 2 vertices along P and hence would complete cycles of length at least q if oriented in the other direction.
Note that B 3 has exactly q vertices. Possibly B 1 or B 2 has more vertices due to picking up extras at the beginning of U 1 or the end of U 2 . However, we can shorten the cycles to length q by omitting vertices at the beginning of Q 1 and/or the end of Q 2 . This only makes the jumps along P longer, so the edges make the jumps still have the desired orientation. The resulting cycle B ′ i is a q-cycle using exactly two vertices used in F (x i and y i ). Hence {B ′ 1 , B ′ 2 , B 3 } replaces {C, C ′ } to yield an extension using 3q − 6 vertices not used by F . Finally we are ready to complete the main proof.
Proof of Theorem 5.1. We will obtain bounds on the sizes of various sets of edges under the assumption that F has no extension. These will lead to a contradiction. Let L consist of those q-cycles in F receiving at least q(q − 1) + 1 edges from U 1 (thereby guaranteeing a q-matching from U 1 , by Lemma 5.3). Let M consist of those q-cycles in F sending at least 6q − 13 edges to U 2 (thereby guaranteeing a 3-matching to U 2 , by Lemma 5.5(ii)). Let R = F − (L ∪ M). Let l, m, and r, respectively, denote the sizes of L, M and R. By Lemma 5.6, L ∩ M = ∅. Hence {L, M, R} is a partition of F , and l + m + r = k − 1. ( Now consider d + (U 1 , S ∪ U 2 ). If this is nonzero, then let u q+s be the highest-indexed (earliest) vertex of P having a predecessor in U 1 , and let u q+1−t be the lowest-indexed vertex of P having a successor in S ∪ U 2 . Since P gives a path from u q+s to u q+1−t , the tournament induced by these s + t vertices is strong and has a spanning cycle. Hence s + t < q. Also Next we obtain upper and lower bounds on d + (U 1 , F ) in order to obtain an inequality involving l and m. Using the computation above, To avoid obtaining an extension of F via Lemma 5.6, each cycle in M must avoid a (q − 1)-matching from U 1 and hence receives at most q(q − 2) edges from U 1 . By definition, each cycle in L or R receives at most q 2 or q(q − 1) edges from U 1 , respectively (the latter because otherwise it would be in L). Thus d + (U 1 , F ) ≤ q 2 l + q(q − 2)m + q(q − 1)r.
Combining (5) and (6) and collecting the terms involving q(l + r) yields q(l + r) (q − 1)k − 1 − q(l + r) − 1 2 − mq − 3q + 7 + 7 q ≤ −ml(q − 2) + (3q − 7)r − q(m + q − 2) + α. (7) Using (2), we simplify the last factor on the left: On the right side of (7), we compute α − q(q − 2) = −(q−1)(q−3) 4 + 1. On the left side, we replace l + r with k − 1 − m, and on the right we replace r with k − 1 − m − l. The inequality is now The coefficients of l in its only appearances are negative. Hence for given q, k, m, the inequality can hold only if it holds when l takes its smallest allowed numerical value. By (4), we have l ≥ m − 1 + (q/4), which yields l ≥ m + 1 when q ≥ 5 since l ∈ N. Setting l = m + 1, we now have a quadratic inequality for m in terms of k and q: We first collect terms to write this as a quadratic inequality for m: where c depends only on k and q. The inequality l ≥ m+ 1 also yields k −1 = l + m+ r ≥ 2m+ 1, and hence m ≤ ⌊k/2⌋ −1. We thus want to show that (10) cannot be satisfied when k ≥ q+1 ≥ 6 and 0 ≤ m ≤ ⌊k/2⌋−1.
In order to obtain the desired contradiction, it suffices to show that the left side of (10) is positive at its lowest allowed point. Since the coefficient of the quadratic term is positive, the quadratic polynomial is minimized where its derivative is 0. This occurs when (q 2 + 2q − 4)m = (q 2 − q)k − (3q 2 + 3 2 q − 23).
The analysis simplifies if the lowest value of the polynomial in (10) among allowed values for m occurs at the highest allowed value, ⌊k/2⌋ − 1. Since the graph of a quadratic polynomial is symmetric around the minimum, when k is even this holds if the minimizing point is at least k/2 − 3/2. When k is odd this also suffices, due to the floor function.