Proof of a conjecture of Thomassen on Hamilton cycles in highly connected tournaments

A conjecture of Thomassen from 1982 states that for every k there is an f(k) so that every strongly f(k)-connected tournament contains k edge-disjoint Hamilton cycles. A classical theorem of Camion, that every strongly connected tournament contains a Hamilton cycle, implies that f(1)=1. So far, even the existence of f(2) was open. In this paper, we prove Thomassen's conjecture by showing that f(k)=O(k^2*log^2(k)). This is best possible up to the logarithmic factor. As a tool, we show that every strongly 10^4*k*log(k)-connected tournament is k-linked (which improves a previous exponential bound). The proof of the latter is based on a fundamental result of Ajtai, Koml\'os and Szemer\'edi on asymptotically optimal sorting networks.


and Szemeréd
on asymptotically optimal sorting networks.

1. Introduction 1.1.Main result.A tournament is an orientation of a complete graph and a Hamilton cycle in a tournament is a (consistently oriented) cycle which contains all the vertices of the tournament.Hamilton cycles in tournaments have a long and rich history.For instance, one of the most basic results about tournaments is Camion's theorem, which states that every strongly connected tournament has a Hamilton cycle [10].This is strengthened by Moon's theorem [19], which implies that such a tournament is even pancyclic, i.e. contains cycles of all possible lengths.Many related results have been proved; the monograph by Bang-Jensen 1. Introduction 1.1.Main result.A tournament is an orientation of a complete graph and a Hamilton cycle in a tournament is a (consistently oriented) cycle which contains all the vertices of the tournament.Hamilton cycles in tournaments have a long and rich history.For instance, one of the most basic results about tournaments is Camion's theorem, which states that every strongly connected tournament has a Hamilton cycle [10].This is strengthened by Moon's theorem [19], which implies that such a tournament is even pancyclic, i.e. contains cycles of all possible lengths.Many related results have been proved; the monograph by Bang-Jensen and Gutin [5] gives an overview which also includes many recent results.
and Gutin [5] gives an overview which also includes many recent results.

In 1982, Thomassen [22] made a very natural conjecture on how to guarantee not just one Hamilton cycle, but many edge-disjoint ones: he conjectured that for every k there is an f (k) so that every strongly f (k)-connected tournament contains k edge-disjoint Hamilton cycles (see also the recent surveys [4,16]).This turned out to be surprisingly difficult: not even the existence of f (2) was known so far.Our main result shows that f (k) = O(k 2 log 2 k).

Date: February 6, 2014.D. Kühn, D. Osthus and V. Patel were supported by the EPSRC, g In 1982, Thomassen [22] made a very natural conjecture on how to guarantee not just one Hamilton cycle, but many edge-disjoint ones: he conjectured that for every k there is an f (k) so that every strongly f (k)-connected tournament contains k edge-disjoint Hamilton cycles (see also the recent surveys [4,16]).This turned out to be surprisingly difficult: not even the existence of f (2) was known so far.Our main result shows that f (k) = O(k 2 log 2 k).
Date: February 6, 2014.D. Kühn, D. Osthus and V. Patel were supported by the EPSRC, grant no.EP/J008087/1.D. Kühn was also supported by the ERC, grant no.258345.D. Osthus was also supported by the ERC, grant no.306349.Theorem 1.1.There exists C > 0 such that for all k ∈ N with k ≥ 2 every strongly Ck 2 log 2 k-connected tournament contains k edge-disjoint Hamilton cycles.
ant no.EP/J008087/1.D. Kühn was also supported by the ERC, grant no.258345.D. Osthus was also supported by the ERC, grant no.306349.Theorem 1.1.There exists C > 0 such that for all k ∈ N with k ≥ 2 every strongly Ck 2 log 2 k-connected tournament contains k edge-disjoint Hamilton cycles.

In Proposition 5.1, we describe an example which shows that f (k) ≥ (k − 1 In Proposition 5.1, we describe an example which shows that f (k) ≥ (k − 1) 2 /4, i.e. our bound on the connectivity is asymptotically close to best possible.Thomassen [22] observed that f (2) > 2 and conjectured that f (2) = 3.He also observed that one cannot weaken the assumption in Theorem 1.1 by replacing strong connectivity with strong edge-connectivity.
2 /4, i.e. our bound on the connectivity is asymptotically close to best possible.Thomassen [22] observed that f (2) > 2 and conjectured that f (2) = 3.He also observed that one cannot weaken the assumption in Theorem 1.1 by replacing strong connectivity with strong edge-connectivity.

To simplify the presentation, we have made no attempt to optimize the value of the constant C. Our exposition shows that one can take C := 10 12 for k ≥ 20.Rather than proving Theorem 1.1 directly, we deduce it as an im To simplify the presentation, we have made no attempt to optimize the value of the constant C. Our exposition shows that one can take C := 10 12 for k ≥ 20.Rather than proving Theorem 1.1 directly, we deduce it as an immediate consequence of two further results, which are both of independent interest: we show that every sufficiently highly connected tournament is highly linked (see Theorem 1.3) and show that every highly linked tournament contains many edge-disjoint Hamilton cycles (see Theorem 1.2).

diate consequence of two fu
ther results, which are both of independent interest: we show that every sufficiently highly connected tournament is highly linked (see Theorem 1.3) and show that every highly linked tournament contains many edge-disjoint Hamilton cycles (see Theorem 1.2).

1.2.Linkedness in tournaments.Given sets A, B of size k in a strongly k-connected digraph D, Menger's theorem implies that D contains k vertexdisjoint paths from 1.2.Linkedness in tournaments.Given sets A, B of size k in a strongly k-connected digraph D, Menger's theorem implies that D contains k vertexdisjoint paths from A to B. In a k-linked digraph, we can even specify the initial and final vertex of each such path (see Section 2 for the precise definition).
A to B. In a k-linked digraph, we can even specify the initial and final vertex of each such path (see Section 2 for the precise definition).

Theorem 1.2.There exists C > 0 such that for all k ∈ N with k ≥ 2 every C k 2 log k-linke Theorem 1.2.There exists C > 0 such that for all k ∈ N with k ≥ 2 every C k 2 log k-linked tournament contains k edge-disjoint Hamilton cycles.
tournament contains k edge-disjoint Hamilton cycles.

The bound in Theorem 1.2 is asymptotically close to best possible, as we shall discuss below.We will show that C := 10 7 works for all k ≥ 20.(As mentioned earlier, we have made no attempt to optimise the value of this constant.)

It is not clear from the definition that every (very) highly connected tournament is also highly linked.In fact, for general digraphs this is far from true: Thomassen [24] showed that for all k there are strongly k-connected digraphs which are not even 2-linked.On the other hand, he showed that there is an (exponential) function g(k) so that every strongly g(k)-connected tournament is k-linked [23].The next result sho The bound in Theorem 1.2 is asymptotically close to best possible, as we shall discuss below.We will show that C := 10 7 works for all k ≥ 20.(As mentioned earlier, we have made no attempt to optimise the value of this constant.) It is not clear from the definition that every (very) highly connected tournament is also highly linked.In fact, for general digraphs this is far from true: Thomassen [24] showed that for all k there are strongly k-connected digraphs which are not even 2-linked.On the other hand, he showed that there is an (exponential) function g(k) so that every strongly g(k)-connected tournament is k-linked [23].The next result shows that we can take g(k) to be almost linear in k.Note that this result together with Proposition 5.1 shows that Theorem 1.2 is asymptotically best possible up to logarithmic terms.Theorem 1.3.For all k ∈ N with k ≥ 2 every strongly 10 4 k log k-connected tournament is k-linked.
s that we can take g(k) to be almost linear in k.Note that this result together with Proposition 5.1 shows that Theorem 1.2 is asymptotically best possible up to logarithmic terms.Theorem 1.3.For all k ∈ N with k ≥ 2 every strongly 10 4 k log k-connected tournament is k-linked.

For small k, the constant 10 4 can easily be improved (see Theorem 4.5).The proof of Theorem 1.3 is based on a fundamental result of Ajtai, Komlós and Szemerédi [1,2] on the existence of asymptotically optimal sorting networks.Though their result is asymptotically optimal, it is not clear whether this is the case for Theorem 1.3.In fact, for the case of (undirected) graphs, a deep result of Bollobás and Thomason [8] states that every 22k-con For small k, the constant 10 4 can easily be improved (see Theorem 4.5).The proof of Theorem 1.3 is based on a fundamental result of Ajtai, Komlós and Szemerédi [1,2] on the existence of asymptotically optimal sorting networks.Though their result is asymptotically optimal, it is not clear whether this is the case for Theorem 1.3.In fact, for the case of (undirected) graphs, a deep result of Bollobás and Thomason [8] states that every 22k-connected graph is k-linked (this was improved to 10k by Thomas and Wollan [21]).Thus one might believe that a similar relation also holds in the case of tournaments: ected graph is k-linked (this was improved to 10k by Thomas and Wollan [21]).Thus one might believe that a sim lar relation also holds in the case of tournaments:

Similarly, we believe that the logarithmic terms can also be removed in Theorems 1.1 and 1.2: Conjecture 1.5.

(i) There exists C > 0 such that for all k ∈ N every C k 2 -linked tournament contains Similarly, we believe that the logarithmic terms can also be removed in Theorems 1.1 and 1.2: Conjecture 1.5.
(i) There exists C > 0 such that for all k ∈ N every C k 2 -linked tournament contains k edge-disjoint Hamilton cycles.(ii) There exists C > 0 such that for all k ∈ N every strongly C k 2connected tournament contains k edge-disjoint Hamilton cycles.
edge-disjoint Hamilton cycles.(ii) There exists C > 0 such that for all k ∈ N every strongly C k 2connected tournament contains k edge-disjoint Hamilton cycles.

Note that Conjectures 1.4 and 1.5(i) together imply Conjecture 1.5(ii).

1.3.Algorithmic aspects.Both Hamiltonicity and linkedness in tournaments have also been studied from an algorithmic perspective.Camion's theorem implies that the Hamilton cycle problem (though NP-complete in general) is solvable in polynomial time for tournaments.Chudnovsky, Scott and Seymour [11] solved a long-standing problem of Bang-Jensen and Thomassen [6] by showing that the linkedness problem is also solvable in polynomial time for tournaments.More precisely, for a given tournament on n vertices, one can determine in time polynomial in n whether it is k-linked and if yes, one can produce a correspon Note that Conjectures 1.4 and 1.5(i) together imply Conjecture 1.5(ii).
1.3.Algorithmic aspects.Both Hamiltonicity and linkedness in tournaments have also been studied from an algorithmic perspective.Camion's theorem implies that the Hamilton cycle problem (though NP-complete in general) is solvable in polynomial time for tournaments.Chudnovsky, Scott and Seymour [11] solved a long-standing problem of Bang-Jensen and Thomassen [6] by showing that the linkedness problem is also solvable in polynomial time for tournaments.More precisely, for a given tournament on n vertices, one can determine in time polynomial in n whether it is k-linked and if yes, one can produce a corresponding set of k paths (also in polynomial time).Fortune, Hopcroft and Wyllie [13] showed that for general digraphs, the problem is NP-complete even for k = 2.
ing set of k paths (also in polynomial time).Fortune, Hopcroft and Wyllie [13] showed that for general digraphs, the problem is NP-complete even for k = 2.

We can use the result in [11] to obtain an algorithmic version of Theorem 1.2.More precisely, given a C k 2 log k-linked tournament on n vertices, one can find k edge-disjoint Hamilton cycles in time po We can use the result in [11] to obtain an algorithmic version of Theorem 1.2.More precisely, given a C k 2 log k-linked tournament on n vertices, one can find k edge-disjoint Hamilton cycles in time polynomial in n (where k is fixed).We discuss this in more detail in Section 9. Note that this immediately results in an algorithmic version of Theorem 1.1.

nomia
in n (where k is fixed).We discuss this in more detail in Section 9. Note that this immediately results in an algorithmic version of Theorem 1.1.

1.4.Related results and spanning regular subgraphs.Proposition 5.1 actually suggests that the 'bottleneck' to finding k edge-disjoint Hamilton cycles is the existence of a k-regular subdigraph: it states that if the connectivity of a tournament T is significantly lower than in Theorem 1.1, then T may not even contain a spanning k-regular subdigraph.There are other results which exhibit this phenomenon: if T is itself regular, then Kelly's conjecture from 1968 states that T itself has a Hamilton decomposition.Kelly's conjecture was proved very recently (for large tournaments) by Kühn and Osthus [17].Erdős raised a 'probabilistic' version of Kelly's conjecture: for a tournament T , let δ 0 (T ) denote the minimum of the minimum out-degree and the minimum indegree.He conjectured that for almost all tournaments T , the maximum number of edge-disjoint Hamilton cycles in T is exactly δ 0 (T ).In particular, this would imply that with high probability, δ 0 (T ) is also the degree of a densest span 1.4.Related results and spanning regular subgraphs.Proposition 5.1 actually suggests that the 'bottleneck' to finding k edge-disjoint Hamilton cycles is the existence of a k-regular subdigraph: it states that if the connectivity of a tournament T is significantly lower than in Theorem 1.1, then T may not even contain a spanning k-regular subdigraph.There are other results which exhibit this phenomenon: if T is itself regular, then Kelly's conjecture from 1968 states that T itself has a Hamilton decomposition.Kelly's conjecture was proved very recently (for large tournaments) by Kühn and Osthus [17].Erdős raised a 'probabilistic' version of Kelly's conjecture: for a tournament T , let δ 0 (T ) denote the minimum of the minimum out-degree and the minimum indegree.He conjectured that for almost all tournaments T , the maximum number of edge-disjoint Hamilton cycles in T is exactly δ 0 (T ).In particular, this would imply that with high probability, δ 0 (T ) is also the degree of a densest spanning regular subdigraph in a random tournament T .This conjecture of Erdős was proved by Kühn and Osthus [18], based on the main result in [17].
ing regular subdigraph in a random tournament T .This conjecture of Erdős was proved by Kühn and Osthus [18], based on the main result in [17].

It would be interesting to obtain further conditions which relate the degree of the densest spanning regular subdigraph of a tournament T to the number of edge-disjoint Hamilton cycles in T .For undirected graphs, one such conjecture was made in [15]: it states that for any graph G satisfying the conditions of Dirac's theorem, the number of edge-disjoint Hamilton cycles in G is exactly half the degree of a densest spanning even-regular subgraph It would be interesting to obtain further conditions which relate the degree of the densest spanning regular subdigraph of a tournament T to the number of edge-disjoint Hamilton cycles in T .For undirected graphs, one such conjecture was made in [15]: it states that for any graph G satisfying the conditions of Dirac's theorem, the number of edge-disjoint Hamilton cycles in G is exactly half the degree of a densest spanning even-regular subgraph of G.An approximate version of this conjecture was proved by Ferber, Krivelevich and Sudakov [12], see e.g.[15,18] for some related results.1.5.Organization of the paper.The methods used in the current paper are quite different from those used e.g. in the papers mentioned in Section 1.4.A crucial ingredient is the construction of highly structured dominating sets (see Section 3 for an informal description).We believe that this approach will have further applications.
of G.An approximate version of this conjecture was proved by Ferber, Krivelevich and Sudakov [12], see e.g.[15,18] for some related results.1.5.Organization of the paper.The methods used in the current paper are quite different from those used e.g. in the papers mentioned in Section 1.4.A crucial ingredient is the construction of high y structured dominating sets (see Section 3 for an informal description).We believe that this approach will have further applications.

In the next section, we introduce the notation that will be used for the remainder of the paper.In Section 3, we give an overview of the proof of Theorem 1.2.In Sections 4 and 5, we give the relatively short proofs of Theorem 1.3 and Proposition 5.1.In Section 6, we show that given a 'linked domination structure' (as introduced in the proof sketch), we can find a single Hamilton cycle (Lemma 6.7).In Section 7, we show that given several suitable linked domination structures, we can repeatedly apply Lemma 6.7 to find k edge-disjoint Hamilton cycles.In Section 8 we show that any In the next section, we introduce the notation that will be used for the remainder of the paper.In Section 3, we give an overview of the proof of Theorem 1.2.In Sections 4 and 5, we give the relatively short proofs of Theorem 1.3 and Proposition 5.1.In Section 6, we show that given a 'linked domination structure' (as introduced in the proof sketch), we can find a single Hamilton cycle (Lemma 6.7).In Section 7, we show that given several suitable linked domination structures, we can repeatedly apply Lemma 6.7 to find k edge-disjoint Hamilton cycles.In Section 8 we show that any highly linked tournament contains such suitable linked domination structures.Finally, Section 9 contains some concluding remarks.

ghly link
d tournament contains such suitable linked domination structures.Finally, Section 9 contains some concluding remarks.


Notation

The digraphs considered in this paper do not have loops and we allow up to two edges between any pair of x, y of distinct vertices, at most one in each direction.A digraph is an oriented g

Notation
The digraphs considered in this paper do not have loops and we allow up to two edges between any pair of x, y of distinct vertices, at most one in each direction.A digraph is an oriented graph if there is at most one edge between any pair x, y of distinct vertices, i.e. if it does not contain a cycle of length two.
aph if there is at most one edge between any pair x, y of distinct vertices, i.e. if it does not contain a cycle of length two.

Given a digraph D, we write V (D) for its vertex set, E(D) for its edge set, e(D) := |E(D)| for the number of its edges and |D| for its order, i.e. for the number of its vertices.We write H ⊆ D to mean that H is a subdigraph of D, i.e.V (H) ⊆ V (D) and E(H) ⊆ E(D).Given X ⊆ V (D), we write D − X for the digraph obtained from D by deleting all vertices in X, and D[X] for the subdigraph of D induced by X.Given F ⊆ E(D), we write D − F for the digraph obtained from D by deleting all edges i Given a digraph D, we write V (D) for its vertex set, E(D) for its edge set, e(D) := |E(D)| for the number of its edges and |D| for its order, i.e. for the number of its vertices.We write H ⊆ D to mean that H is a subdigraph of D, i.e.V (H) ⊆ V (D) and E(H) ⊆ E(D).Given X ⊆ V (D), we write D − X for the digraph obtained from D by deleting all vertices in X, and D[X] for the subdigraph of D induced by X.Given F ⊆ E(D), we write D − F for the digraph obtained from D by deleting all edges in F .We write V (F ) for the set of all endvertices of edges in F .We write V (F ) for the set of all endvertices of edges in
F . If H is a subdigraph of D, we write D − H for D − E(H).
We write xy for an edge directed from x to y.Unless stated otherwise, when we refer to paths and cycles in digraphs, we mean directed paths and cycles, i.e. the edges on these paths and cycles are oriented consistently.Given a path P = x . . .y from x to y and a vertex z outside P which sends an edge to x, we write zxP for the path obtained from P by appending the edge zx.The length of a path or cycle is the number of its edges.We call the terminal vertex of a path P the head of P and denote it by h(P ).Similarly, we call the initial vertex of a path P the tail of P and denote it by t(P ).The interior Int(P ) of a path P is the subpath obtained by deleting t(P ) and h(P ).Thus Int(P ) = ∅ if P has length at most one.Two paths P and P are internally disjoint if P = P and V (Int(P )) ∩ V (Int(P )) = ∅.A path system P is a collection of vertex-disjoint paths.We write V (P) for the set of all vertices lying on paths in P and E(P) for the set of all edges lying on paths in P. We write h(P) for the set consisting of the heads of all paths in P and t(P) for the set consisting of the tails of all paths in P. If v ∈ V (P), we write v + and v − for the successor and predecessor of v on the path in P containing v. A path system P is a path cover of a directed graph D if every path in P lies in D and together the paths in P cover all th We write xy for an edge directed from x to y.Unless stated otherwise, when we refer to paths and cycles in digraphs, we mean directed paths and cycles, i.e. the edges on these paths and cycles are oriented consistently.Given a path P = x . . .y from x to y and a vertex z outside P which sends an edge to x, we write zxP for the path obtained from P by appending the edge zx.The length of a path or cycle is the number of its edges.We call the terminal vertex of a path P the head of P and denote it by h(P ).Similarly, we call the initial vertex of a path P the tail of P and denote it by t(P ).The interior Int(P ) of a path P is the subpath obtained by deleting t(P ) and h(P ).Thus Int(P ) = ∅ if P has length at most one.Two paths P and P are internally disjoint if P = P and V (Int(P )) ∩ V (Int(P )) = ∅.A path system P is a collection of vertex-disjoint paths.We write V (P) for the set of all vertices lying on paths in P and E(P) for the set of all edges lying on paths in P. We write h(P) for the set consisting of the heads of all paths in P and t(P) for the set consisting of the tails of all paths in P. If v ∈ V (P), we write v + and v − for the successor and predecessor of v on the path in P containing v. A path system P is a path cover of a directed graph D if every path in P lies in D and together the paths in P cover all the vertices of D. If X ⊆ V (D) and P is a path cover of D[X], we sometimes also say that P is a path cover of X.
vertices of D. If X ⊆ V (D) and P is a path cover of D[X], we sometimes also say that P is a path cover of X.

If A digraph D is strongly connected if for all x, y ∈ V (D), there is a directed path in D from x to y.Given k ∈ N, we say a digraph is strongly k-connected If A digraph D is strongly connected if for all x, y ∈ V (D), there is a directed path in D from x to y.Given k ∈ N, we say a digraph is strongly k-connected if |D| > k and for every S ⊆ V (D) of size at most k−1, D−S is strongly connected.We say a digraph D is k-linked if |D| ≥ 2k and whenever x 1 , . . ., x k , y 1 , . . ., y k are 2k distinct vertices of D, there exist vertex-disjoint paths P 1 , . . ., P k such that P i is a path from x i to y i .
Given a digraph D and sets X, Y ⊆ V (D), we say that X in-dominates Y if each vertex in Y is an in-neighbour of some vertex in X.Similarly, we say that X out-dominates Y if each vertex in Y is an out-neighbour of some vertex in X.
vertex in X.

A tournament T is transitive if there exists an ordering v 1 , . . ., v n of its vertices such that v i v j ∈ E(T ) if and only if i < j.In this case, we often say that v 1 is the tail of T and v n is the head of T .

Given k ∈ N, we write [k] := {1, . . ., k}.We write log for the binary logarithm and l A tournament T is transitive if there exists an ordering v 1 , . . ., v n of its vertices such that v i v j ∈ E(T ) if and only if i < j.In this case, we often say that v 1 is the tail of T and v n is the head of T .
g 2 n := (log n) 2 .


Sketch of the proof of Theorem 1.2

In this section, we give an outline of the proof of Theorem 1.2.An important idea is the notion of a 'covering edge'.Given a small (pre-determined) set S of vertices in a tournament T , this will mean that it will suffice to find a cycle covering all vertices of T − S.More precisely, let T be a tournament, let x ∈ V (T ), and suppose C is a cycle in T covering T − x.If yz ∈ E(C) and yx, xz ∈ E(T ), then we can replace yz by yxz in C to turn C into a Hamilton cycle.We call yz a covering edge for x.More generally, if S ⊆ V (T ) and C is a cycle in T spanning V (T ) − S such that C contains a covering edge for each x ∈ S, then

Sketch of the proof of Theorem 1.2
In this section, we give an outline of the proof of Theorem 1.2.An important idea is the notion of a 'covering edge'.Given a small (pre-determined) set S of vertices in a tournament T , this will mean that it will suffice to find a cycle covering all vertices of T − S.More precisely, let T be a tournament, let x ∈ V (T ), and suppose C is a cycle in T covering T − x.If yz ∈ E(C) and yx, xz ∈ E(T ), then we can replace yz by yxz in C to turn C into a Hamilton cycle.We call yz a covering edge for x.More generally, if S ⊆ V (T ) and C is a cycle in T spanning V (T ) − S such that C contains a covering edge for each x ∈ S, then we can turn C into a Hamilton cycle by using all these covering edges.Note that this idea still works if C covers some part of S. On the other hand, note that S needs to be fixed at the beginning -this is different than in the recently popularized 'absorbing method'.
we can turn C into a Hamilton cycle by using all these covering edges.Note that this idea still works if C covers some part of S. On the other hand, note that S needs to be fixed at the beginning -this is different than in the recently popularized 'absorbing method'.

Another important tool will be the following consequence of t Another important tool will be the following consequence of the Gallai-Milgram theorem: suppose that G is an oriented graph on n vertices with δ(G) ≥ n − .Then the vertices of G can be covered with vertex-disjoint paths.We use this as follows: suppose we are given a highly linked tournament T and have already found i edge-disjoint Hamilton cycles in T .Then the Gallai-Milgram theorem implies that we can cover the vertices of the remaining oriented graph by a set of 2i vertex-disjoint paths.Very roughly, the aim is to link together these paths using the high linkedness of the original tournament T .
e Gallai-Milgram theorem: suppose that G is an oriented graph on n vertices with δ(G) ≥ n − .Then the vertices of G can be covered with vertex-disjoint paths.We use this as follows: suppose we are given a highly linked tournament T and have already found i edge-disjoint Hamilton cycles in T .Then the Gallai-Milgram theorem implies that we can cover the vertices of the remaining oriented graph by a set of 2i vertex-disjoint paths.Very roughly, the aim is to link together these paths using the high linkedness of the original tournament T .

To achieve this aim, we introduce and use the idea of 'transitive dominating sets'.Here a transitive out-dominating set A has the following properties:
• A out-dominates V (T ) \ A , i.e. every vertex of V (T ) \ A receives an edge from A . • A induces a transitive tournament in T .
Transitive in-dominating sets B are defined similarly.

Now suppose that we have already found i edge-disjoint Hamilt To achieve this aim, we introduce and use the idea of 'transitive dominating sets'.Here a transitive out-dominating set A has the following properties: Transitive in-dominating sets B are defined similarly.
Now suppose that we have already found i edge-disjoint Hamilton cycles in a highly linked tournament T .Let T be the oriented subgraph of T obtained by removing the edges of these Hamilton cycles.Suppose that we also have the following 'linked dominating structure' in T , which consists of: n cycles in a highly linked tournament T .Let T be the oriented subgraph of T obtained by removing the edges of these Hamilton cycles.Suppose that we also have the following 'linked dominating structure' in T , which consists of:

• small disjoint transitive out-dominating sets A 1 , . . ., A t , where t := 2i+1; • small disjoint transitive in-dominating sets B 1 , . . ., B t ; • a set of short vertex-disjoint paths P 1 , . . ., P t , where each P is a path from the head b of B to the tail a of A .

Recall that the head of a transitive tournament is the vertex of out-degree zero and the tail is defined • small disjoint transitive out-dominating sets A 1 , . . ., A t , where t := 2i+1; • small disjoint transitive in-dominating sets B 1 , . . ., B t ; • a set of short vertex-disjoint paths P 1 , . . ., P t , where each P is a path from the head b of B to the tail a of A .
Recall that the head of a transitive tournament is the vertex of out-degree zero and the tail is defined analogously.The paths P are found at the outset of the proof, using the assumption that the original tournament T is highly linked.(Note that T need not be highly linked.)Let A * denote the union of the A i and let B * denote the union of the B i .Note that δ(T − A * ∪ B * ) ≥ n − 1 − 2i = n − t.So the Gallai-Milgram theorem implies that we can cover the vertices of T −A * ∪B * with t vertex-disjoint paths Q 1 , . . ., Q t .Now we can link up successive paths using the above dominating sets as follows.The final vertex of Q 1 sends an edge to some vertex b in B 2 (since B 2 is in-dominating).Either b is equal to the head b 2 of B 2 or there is an edge in T [B 2 ] from b to b 2 (since T [B 2 ] is a transitive tournament).Now follow the path P 2 from b 2 to the tail a 2 of A 2 .Using the fact that T [A 2 ] is transitive and that A 2 is out-dominating, we can similarly find a path of length at most  two from a 2 to the initial vertex of Q 2 .Continuing in this way, we can link up all the paths Q and P into a single cycle C which covers all vertices outside A * ∪ B * (and some of the vertices inside A * ∪ B * ).The idea is illustrated in Figure 1.
nalogously.The paths P are found at the outset of the proof, using the assumption that the original tournament T is highly linked.(Note that T need not b 2i = n − t.So the Gallai-Milgram theorem implies that we can cover the vertices of T −A * ∪B * with t vertex-disjoint paths Q 1 , . . ., Q t .Now we can link up successive paths using the above dominating sets as follows.The final vertex of Q 1 sends an edge to some vertex b in B 2 (since B 2 is in-dominating).Either b is equal to the head b 2 of 2 or there is an edge in T [B 2 ] from b to b 2 (since T [B 2 ] is a transitive tournament).Now follow the path P 2 from b 2 to the tail a 2 of A 2 .Using the fact that T [A 2 ] is transitive and that A 2 is out-dominating, we can similarly find a path of length at most  two from a 2 to the initial vertex of Q 2 .Continuing in this way, we can link up all the paths Q and P into a single cycle C which covers all vertices outside A * ∪ B * (and some of the vertices inside A * ∪ B * ).The idea is illustrated in igure 1.

In our construction, we will ensure that the paths P contain a set of covering edges for A * ∪ B * .So C also contains covering edges for A * ∪ B * , and so we can transform C into a Hamilton cycle as discussed earlier.

A major obstacle to the above strategy is that in order to guarantee the P in T − A * ∪ B * , we would need the linkedness of T to be significantly larger than |A * ∪ B * | (and thus larger than |A |).However, there are many tournaments where any in-or out-dominating set contains Ω(log n) vertices In our construction, we will ensure that the paths P contain a set of covering edges for A * ∪ B * .So C also contains covering edges for A * ∪ B * , and so we can transform C into a Hamilton cycle as discussed earlier.
A major obstacle to the above strategy is that in order to guarantee the P in T − A * ∪ B * , we would need the linkedness of T to be significantly larger than |A * ∪ B * | (and thus larger than |A |).However, there are many tournaments where any in-or out-dominating set contains Ω(log n) vertices (consider a random tournament).This leads to a linkage requirement on T which depends on n (and not just on k, as required in Theorem 1.2).
a linkage requirement on T which depends on n (and not just on k, as required in Theorem 1.2).

We overcome this problem by considering 'almost dominating sets': instead of out-dominating all vertices outside A , the A will out-dominate almost all vertices outside A .(Analogous comments apply to the in-dominating sets B .)This means that we have a small 'exceptional set' E of vertices which are not out-dominated by all of the A .The problem with allowing an excep We overcome this problem by considering 'almost dominating sets': instead of out-dominating all vertices outside A , the A will out-dominate almost all vertices outside A .(Analogous comments apply to the in-dominating sets B .)This means that we have a small 'exceptional set' E of vertices which are not out-dominated by all of the A .The problem with allowing an exceptional set is that if the tail of a path Q in our cover is in the exceptional set E, we cannot extend it directly into the out-dominating set A as in the above description.However, if we make sure that the A include the vertices of smallest in-degree of T , we can deal with this issue.Indeed, in this case we can show that every vertex v ∈ E has in-degree d − (v) > 2|E| say, so we can always extend the tail of a path out of the exceptional set if necessary (and then into an almost outdominating set A as before).Unfortunately, we may 'break' one of the paths P in the process.However, if we are careful about the place where we break it and construct some 'spare' paths at the outset, it turns out that the above strategy can be made to work.
ional set is that if the tail of a path Q in our cover is in the exceptional set E, we cannot extend it directly into the out-dominating set A as in the above description.However, if we make sure that the A include the ve tices of smallest in-degree of T , we can deal with this issue.Indeed, in this case we can show that every vertex v ∈ E has in-degree d − (v) > 2|E| say, so we can always extend the tail of a path out of the exceptional set if necessary (and then into an almost outdominating set A as before).Unfortunately, we may 'break' one of the paths P in the process.However, if we are careful about the place where we break it and construct some 'spa e' paths at the outset, it turns out that the above strategy can be made to work.


Connectivity and linkedness in tournaments

In this section we give the proof of Theorem 1.3.We will also collect some simple properties of highly linked directed graphs which we will use later on.The proof of Theorem 1.3 is based on an important result of Ajtai, Komlós and Szemerédi [1,2] on sorting networks.Roughly speaking, the proof idea of Theorem 1.3 is as follows.Suppose that we are given a highly connected tournament T and we want to link an ordered set X of k vertices to a set Y of the same size.Then we construct the equivalent of a sorting network D inside T − Y with 'initial vertices' in X and 'final vertices' in a set Z. The high connectivity of T guarantees an 'unsorted' set of k ZY -paths which avoid the vertices in D − Z.One can then extend these paths via D to the appropriate vertices in X.In this way, we obtain paths linking the vertices in X to the appropriate ones in Y .An example is shown in Figure 2.

We now introduce the necessary background on non-adaptive sorting algorithms and sorting networks; see [1

Connectivity and linkedness in tournaments
In this section we give the proof of Theorem 1.3.We will also collect some simple properties of highly linked directed graphs which we will use later on.The proof of Theorem 1.3 is based on an important result of Ajtai, Komlós and Szemerédi [1,2] on sorting networks.Roughly speaking, the proof idea of Theorem 1.3 is as follows.Suppose that we are given a highly connected tournament T and we want to link an ordered set X of k vertices to a set Y of the same size.Then we construct the equivalent of a sorting network D inside T − Y with 'initial vertices' in X and 'final vertices' in a set Z. The high connectivity of T guarantees an 'unsorted' set of k ZY -paths which avoid the vertices in D − Z.One can then extend these paths via D to the appropriate vertices in X.In this way, we obtain paths linking the vertices in X to the appropriate ones in Y .An example is shown in Figure 2.
We now introduce the necessary background on non-adaptive sorting algorithms and sorting networks; see [14] for a more detailed treatment.In a sorting problem, we are given k registers R 1 , . . ., R k , and each register R i is assigned a distinct element from [k], which we call the value of R i ; thus there is some permutation π of [k] such that value i has been assigned to register R π(i) .Our task is to sort the values into their corresponding registers (so that value i is assigned to R i ) by making a sequence of comparisons: a comparison entails taking two registers and reassigning their values so that the higher value is assigned to the higher register and the lower value to the lower register.A non-adaptive sorting algorithm is a sequence of comparisons specified in advance such that for any initial assignment of k values to k registers, applying the prescribed sequence of comparisons results in every value being assigned to its corresponding register.

for a more detailed treatment.In a sorting
problem, we are given k registers R 1 , . . ., R k , and each register R i is assigned a distinct element from [k], which we call the value of R i ; thus there is some permutation π of [k] such that value i has been assigned to register R π(i) .Our task is to sort the values into their corresponding registers (so that value i is assigned to R i ) by making a sequence of comparisons: a comparison entails taking two registers and reassigning their values so that the higher value is assigned to the higher register and the lower value to the lower register.A non-adaptive sorting algorithm is a sequence of comparisons specified in advance such that for any initial assignment of k values to k registers, applying the prescribed sequence of comparisons results in every value being assigned to its corresponding register.

Ajtai, Komlós and Szemerédi [1,2] proved, via the construction of sorting Ajtai, Komlós and Szemerédi [1,2] proved, via the construction of sorting networks, that there exists an absolute constant C and a non-adaptive sorting algorithm (for k registers and values) that requires C k log k comparisons, and this is asymptotically best possible.It is known that we can take C := 3050 [20] (results of this type are often stated in terms of the depth of a sorting network rather than the number of comparisons).
etworks, that there exists an absolute constant C and a non-adaptive sorting algorithm (for k registers and values) that requires C k log k comparisons, and this is asymptotically best possible.It is known that we can take C := 3050 [20] (results of this type are often stated in terms of the depth of a sorting network rather than the number of comparisons).

The next theorem is a consequence of the above.Before we can state it, we first need to introduce some notation.A comparison c, which is part of some non-adaptive sorting algorithm for k registers, will be denoted by c = (s; t), where 1 ≤ s < t ≤ k, to indicate that c is a comparison in which the values of registers R s and R t are compared (and sorted so the higher value is assigned to the higher register).We now show how to obtain a structure within a highly connected tournament that simulates the function of a non-adaptive sorting algorithm.Each comparison in the sorting algorithm will be simulated by a 'swi The next theorem is a consequence of the above.Before we can state it, we first need to introduce some notation.A comparison c, which is part of some non-adaptive sorting algorithm for k registers, will be denoted by c = (s; t), where 1 ≤ s < t ≤ k, to indicate that c is a comparison in which the values of registers R s and R t are compared (and sorted so the higher value is assigned to the higher register).We now show how to obtain a structure within a highly connected tournament that simulates the function of a non-adaptive sorting algorithm.Each comparison in the sorting algorithm will be simulated by a 'switch', which we now define.
ch', which we now define.
An (a 1 , a 2 )-switch is a digraph D on 5 distinct vertices a 1 , a 2 , b, b 1 , b 2 , where ei- ther E(D) = {a 1 b, bb 1 , bb 2 , a 2 b 1 , a 2 b 2 } or E(D) = {a 2 b, bb 1 , bb 2 , a 1 b 1 , a 1 b 2 }.
We call b 1 and b 2 the terminal vertices of the (a 1 , a 2 )-switch.Note that for any permutation π of {1, 2}, there exist vertex-disjoint paths P 1 , P 2 of D such that Given k ∈ N, we write S k for the We call b 1 and b 2 the terminal vertices of the (a 1 , a 2 )-switch.Note that for any permutation π of {1, 2}, there exist vertex-disjoint paths P 1 , P 2 of D such that Given k ∈ N, we write S k for the set of permutations of [k] and id k for the identity permutation of [k].The following structural lemma for tournaments is at the heart of the proof of Theorem 1.3.It constructs the equivalent of a sorting network in a tournament of high minimum outdegree.Lemma 4.3.Let C := 3050 and k ∈ N be such that k ≥ 2. Let T be a tournament with δ + (T ) ≥ (3C + 5)k log k, and let x 1 , . . ., x k ∈ V (T ) be distinct vertices.Then there exists a digraph D ⊆ T and distinct vertices z 1 , . . ., z k ∈ V (D) with the following properties: et of permutations of [k] and id k for the identity permutation of [k].The following structural lemma for tournaments is at the heart of the proof of Theorem 1.3.It constructs the equivalent of a sorting network in a tournament of high minimum outdegree.Lemma 4.3.Let C := 3050 and k ∈ N be such that k ≥ 2. Let T be a tournament with δ + (T ) ≥ (3C + 5)k log k, and let x 1 , . . ., x k ∈ V (T ) be distinct verti es.Then there exists a digraph D ⊆ T and distinct vertices z 1 , . . ., z k ∈ V (D) with the following properties:
P i joins a i to b π(i) for i = 1, 2. Proposition 4.2. Let T be a tournament. Given distinct vertices a 1 , a 2 ∈ V (T ), if d + T (a 1 ), d + T (a 2 ) ≥ 7, then T contains an (a 1 , a 2 )-switch. Proof. We may choos disjoint sets A 1 ⊆ N + T (a 1 ) \ {a 2 } and A 2 ⊆ N + T (a 2 ) \ {a 1 } with |A 1 | = |A 2 | = 3. Consider the bipartite digraph H induced by T between A 1 and A 2 . It is easy to check that there exists b ∈ A 1 ∪ A 2 with d + H (b) (iii) For any π ∈ S k , we can find vertex-disjoint paths P 1 , . . ., P k such that Proof.Consider the sorting problem for k registers, and apply Theorem 4.1 to obtain a sequence c 1 , . . ., c r of r ≤ C k log k comparisons such that for any π ∈ S k , if value i is initially assigned to register R π(i) , then applying the comparisons c 1 , . . ., c r results in every value being assigned to its corresponding register.Given π ∈ S k , we write π q ∈ S k for the permutation such that after applying the first q comparisons c 1 , . . ., c q , value i is assigned to register R πq(i) for all i; thus π r = id k .Let D 0 be the digraph with vertex set {x 1 , . . ., x k } and empty edge set.We inductively construct digraphs D 0 ⊆ D 1 ⊆ • • • ⊆ D r ⊆ T and for each D q we maintain a set Z q = {z q 1 , . . ., z q k } of k distinct final vertices such that the following holds: D 0 ⊆ D 1 ⊆ • • • ⊆ D r ⊆ T and for each D q we maintain a set Z q = {z q 1 , . . ., z q k } of k distinct final vertices such that the following holds:
(a) |D q | = 3q + k. (b) Whenever π ∈ S k
is a permutation, there exist vertex-disjoint paths P q 1 , . . ., P q k in D q such that P q i joins x π(i) to z q πq(i) for all i ∈ [k].Assuming the above statement holds for q = 0, . . ., r, then taking
D := D r with z i := z r i for all i ∈ [k] proves the lemma. Indeed |D r | = 3r+k ≤ 3C k log k+ k ≤ (3C + 1)k log k and π r = id k .
Having already defined D 0 , let us describe the inductive step of our construction.Suppose that for some q ∈ [r] we have constructed D q−1 ⊆ T and a corre-
sponding set Z q−1 = {z q−1 1 , . . . , z q−1 k } of final vertices. Let s is a permutation, there exist vertex-disjoint paths P q 1 , . . ., P q k in D q such that P q i joins x π(i) to z q πq(i) for all i ∈ [k].Assuming the above statement holds for q = 0, . . ., r, then taking Having already defined D 0 , let us describe the inductive step of our construction.Suppose that for some q ∈ [r] we have constructed D q−1 ⊆ T and a corre- tournament T := T − (V (D q−1 ) \ {z q−1 s , z q−1 t }).
Then T has minimum out-degree Then T has minimum out-degree at least C + 5)k log k − 3r − k ≥ 5k log k − k ≥ 7,
and so in particular d + T (z q−1 s ), d + T (z q−1 t ) ≥ 7. Thus we may apply Proposition 4.2 to obtain a (z q−1 s , z q−1 t

)-switch σ in T .Write b 1 , b 2 for the terminal vertices of σ.Now D q is constructed from D q−1 by adding the vertices and edges of σ to D q−1 ; note that z q−1 s and z q−1 t are precisely the common vertices of D q−1 and σ.We define the set Z q = {z q 1 , . . ., z q k } by setting z q i := z q−1 i for all i = s, t and z q s := b 1 as well as z q t := b 2 .Note that z q 1 , . . ., z q k are distinct.Finally we check that conditions (a) and (b) hold for D q .Condition (a) holds since D q has exactly 3 more vertices than D q−1 .For (b), by induction we may assume that there are vertex-disjoint paths
P q−1 1 , . . . , P q−1 k and so in particular d + T (z q−1 s ), d + T (z q−1 t ) ≥ 7. Thus we may apply Proposition 4.2 to obtain a (z q−1 s , z q−1 t )-switch σ in T .Write b 1 , b 2 for the terminal vertices of σ.Now D q is constructed from D q−1 by adding the vertices and edges of σ to D q−1 ; note that z q−1 s and z q−1 t are precisely the common vertices of D q−1 and σ.We define the set Z q = {z q 1 , . . ., z q k } by setting z q i := z q−1 i for all i = s, t and z q s := b 1 as well as z q t := b 2 .Note that z q 1 , . . ., z q k are distinct.Finally we check that conditions (a) and (b) hold for D q .Condition (a) holds since D q has exactly 3 more vertices than D q−1 .For (b), by induction we may assume that there are vertex-disjoint paths then Q s joins z q−1 s to z q t and Q t joins z q−1 t to z q s ; • if c q d s and Q t joins z q−1 t to z q t .Now exactly two of the paths from P q−1 1 , . . ., P q−1 k end at z q−1 s and z q−1 t , namely those indexed by π −1 q−1 (s) and π −1 q−1 (t).We extend these two paths using Q s and Q t , and leave all others unchanged to obtain paths P q 1 , . . ., P q k .It is straightforward to check that these paths are vertex-disjoint and that P i joins x π(i) to z q πq(i) for all i ∈ [k].
the paths from P q−1 1 , . . ., P q−1 k end at z q−1 s and z q−1 t , namely those indexed by π −1 q−1 (s) and π −1 q−1 (t).We extend these two paths using Q s r all i ∈ [k].

It is now an easy step to It is now an easy step to prove Theorem 1.3.We will use the following directed version of Menger's Theorem.Proof of Theorem 1.3.Set C := 3050 and C := 3C + 6 < 10 4 .We must show that, given a strongly Ck log k-connected tournament T and distinct vertices x 1 , . . ., x k , y 1 , . . ., 10 4 .We must show that, given a strongly Ck log k-connected tournament T and distinct vertices x 1 , . . ., x k , y 1 , . . .,
y k ∈ V (T ), we can find vertex-disjoint paths R 1 , . . . , R k such that R i joins x i to y i for all i ∈ [k].
Let X := {x 1 , . . ., x k }, Y := {y 1 , . . ., y k } and T := T − Y .Note that T is strongly (3C + 5)k log k-connected, and in particular δ + (T ) ≥ (3C + 5)k log k.Thus we can apply Lemma 4.3 to T and x 1 , . . ., x k to obtain a digraph D ⊆ T and vertices z
1 , . . . , z k ∈ V (D) satisfying properties (i)-(iii) of Lemma 4.3. Let Z := {z 1 , . . . , z k }. Since |D| ≤ (3C + 1)k log k, the tournament T := T − (V (D) \ Z) is strongly k-con Let X := {x 1 , . . ., x k }, Y := {y 1 , . . ., y k } and T := T − Y .Note that T is strongly (3C + 5)k log k-connected, and in particular δ + (T ) ≥ (3C + 5)k log k.Thus we can apply Lemma 4.3 to T and x 1 , . . ., x k to obtain a digraph D ⊆ T and vertices z Therefore, by Theorem 4.4, there exist k vertex-disjoint paths, with each path starting in Z and ending in Y .For each i ∈ [k], let us assume that P π(i) is the path that joins z i to y π(i) , where π is some permutation of [k].By Lemma 4.3, we can find vertex-disjoint paths Q 1 , . . ., Q k in D such that Q i joins x π(i) to z i .Then the path R i := Q π −1 (i) P π −1 (i) joins x i to y i and these paths are vertex-disjoint.
at P π(i) is the path that joins z i to y π(i) , where π is some permutation of [k].By Lemma 4.3, we can find vertex-disjoint paths Q 1 , . . ., Q k in D such that Q i joins x π(i) to z i .Then the path R i := Q π −1 (i) P π −1 (i) joins x i to y i and these paths are vertex-disjoint.

Batcher [7] (see also [14]) gave a construction of sorting networks which is asymptotically not optimal but which gives better values for small k.More precisely, it uses at most 2k log 2 k comparisons for k ≥ 3.If we use these as a building block in the proof of Lemma 4.3 instead of the asymptoti Batcher [7] (see also [14]) gave a construction of sorting networks which is asymptotically not optimal but which gives better values for small k.More precisely, it uses at most 2k log 2 k comparisons for k ≥ 3.If we use these as a building block in the proof of Lemma 4.3 instead of the asymptotically optimal ones leading to Theorem 4.1, we immediately obtain the following result which improves Theorem 1.3 for small values of k.Theorem 4.5.For all k ∈ N with k ≥ 3, every strongly 12k log 2 k-connected tournament is k-linked.
4.1, we immediately obtain the following result which improves Theorem 1.3 for small values of k.Theorem 4.5.For all k ∈ N with k ≥ 3, every strongly 12k log 2 k-connected tournament is k-linked.

For k = 2, the best bound is obtained by a result of Bang- For k = 2, the best bound is obtained by a result of Bang-Jensen [3], who showed that every strongly 5-connected semi-complete digraph is 2-linked.
We will now collect some simple properties of highly linked directed graphs which we will use later on.The first two follow straightforwardly from the definition of linkedness.Proposition 4.6.Let k ∈ N. Then a digraph D is k-linked if and only if |D| ≥ 2k and whenever (x 1 , y 1 ), . . ., (x k , y k ) are ordered pairs of (not necessarily distinct) vertices of D, there exist internally disjoint paths P 1 , . . ., P k such that P i joins x i to y i .
rdly from the definition of linkedness.Proposition 4.6.Let k ∈ N. Then a digraph D is k-linked if and only if |D| ≥ 2k and whenever (x 1 , y 1 ), . . ., (x k , y k ) are ordered pairs of (not D) be such that |X| + 2|F | ≤ 2 . Then D − X − F is (k − )-linked.
The next lemma shows that in a sufficiently highly linked digraph we can link given pairs of vertices by vertex-disjoint paths which together do not contain too many vertices.Lemma 4.8.Let k, s ∈ N, and let D be a 2ks-linked digraph.Let (x 1 , y 1 ), . . ., (x k , y k ) be ordered pairs of (not necessarily distinct) vertices in D. Then there exist internally disjoint paths P 1 , . . ., P k The next lemma shows that in a sufficiently highly linked digraph we can link given pairs of vertices by vertex-disjoint paths which together do not contain too many vertices.Lemma 4.8.Let k, s ∈ N, and let D be a 2ks-linked digraph.Let (x 1 , y 1 ), . . ., (x k , y k ) be ordered pairs of (not necessarily distinct) vertices in D. Then there exist internally disjoint paths P 1 , . . ., P k such that P i joins x i to y i for all i ∈ [k] and such that P i joins x i to y i for all i ∈ [k] and
|P 1 ∪ • • • ∪ P k | ≤ |D|/s.

Proof.

By Proposition 4.6 there exist internally disjoint paths P 1 1 , . . ., P 2s k such that P j i joins x i to y i for all i ∈ [k] and all j ∈ [2s].For any j, the interiors of P j 1 , . . ., P j k contain at least
|P j 1 ∪ • • • ∪ P j k | − 2k vertices. So the disjointness of the paths implies that there is a j ∈ [2s] with |P j 1 ∪ • • • ∪ P j k

Proof.
By Proposition 4.6 there exist internally disjoint paths P 1 1 , . . ., P 2s k such that P j i joins x i to y i for all i ∈ [k] and all j ∈ [2s].For any j, the interiors of P j 1 , . . ., P j k contain at least . The result now follows by setting P i := P j i and noting that 2k ≤ |D|/2s.

Nearly extremal example
ly extremal example

The aim of this section is to prove the following proposition, which shows that the bound on the connectivity in Theorem 1.1 is The aim of this section is to prove the following proposition, which shows that the bound on the connectivity in Theorem 1.1 is close to best possible.Proposition 5.1.Fix n, k ∈ N with k ≥ 2 and n > k 2 + k + 2. There exists a strongly k 2 /4 -connected tournament T of order n such that if D ⊆ T is a spanning r-regular subdigraph, then r ≤ k.In particular, T contains at most k edge-disjoint Hamilton cycles.
close to best possible.Proposition 5.1.Fix n, k ∈ N with k ≥ 2 and n > k 2 + k + 2. There exists a strongly k 2 /4 -connected tournament T of order n such that if D ⊆ T is a span ing r-regular subdigraph, then r ≤ k.In particular, T contains at most k edge-disjoint Hamilton cycles.

It is easy to see that the above tournament T is also Ω(k 2 )-linked.This shows that the bound in Theorem 1.2 has to be at least quadratic in k.

Proof.Let ∈ N. We will first d It is easy to see that the above tournament T is also Ω(k 2 )-linked.This shows that the bound in Theorem 1.2 has to be at least quadratic in k.
Proof.Let ∈ N. We will first describe a tournament T = (V , E ) of order 2 + 1 which is strongly -connected.We then use T as a building block to construct a tournament as desired in the proposition.
ition.

Let V := {v 0 , . . ., v 2 } and let E consist of the edges v i v i+t for all i = 0, . . ., 2 and all t ∈ [ ], where indices are understood to be modulo 2 + 1.One may t Let V := {v 0 , . . ., v 2 } and let E consist of the edges v i v i+t for all i = 0, . . ., 2 and all t ∈ [ ], where indices are understood to be modulo 2 + 1.One may think of T as the tournament with vertices v 0 , . . ., v 2 placed in order, clockwise, around a circle, where the out-neighbours of each v i are the closest vertices to v i in the clockwise direction, and the in-neighbours are the closest vertices in the anticlockwise direction.Note that T is regular.Note also that, for any distinct x, y ∈ V , we can find a path in T from x to y by traversing vertices from x to y in clockwise order; this remains true even if we delete any − 1 vertices from T .
ink of T as the tournament with vertices v 0 , . . ., v 2 placed in order, clockwise, around a circle, where the out-neighbours of each v i are the closest vertices to v i in the clockwise direction, and the in-neighbours are the closest vertices in the anticlockwise direction.Note that T is regular.No

also t
at, for any distinct x, y ∈ V , we can find a path in T from x to y by travers emains true even if we delete any − 1 vertices from T .

Next we construct a tournament T m, = (V m, , E m, ) as follows.We take V m, to be the disjoint union of sets A := {a 0 , . . . is a transitive tournament which respects the given order of the vertices in C m (i.e.c i c j is an edge if and only if i < j).
Next we construct a tournament T m, = (V m, , E m, ) as follows.We take V m, to be the disjoint union of sets A := {a 0 , . . . is a transitive tournament which respects the given order of the vertices in C m (i.e.c i c j is an edge if and only if i < j).Each vertex in A is an in-neighbour of all vertices in C m , and each vertex in B is an out-neighbour of all vertices in C m .Finally, a vertex a i ∈ A is an in-neighbour of a vertex b j ∈ B if and

ch vertex in A is an in-
eighbour of all vertices in C m , and each vertex in B is an out-neighbour of all vertices in C m .Finally, a vertex a i ∈ A is an in-neighbour of a vertex b j ∈ B if and
only if i = j. Note that |T m, | = m + 4 + 2. Claim 1. The tournament T m, is strongly -connected. To see that T m, is strongly -connected, we check that if S ⊆ V m, with |S| ≤ − 1, then T m, − S is, . . . , c r of C m . Since N − D (c i ) ⊆ N − T m, (c i ) = A ∪ {c 1 , . . . , c i−1 } and |N − D (c i )| = r, we have |N − D (c i ) ∩ A | ≥ r − i + 1, so that e D (A , {c i }) ≥ r − i + 1.
l ∈ N.This completes the proof of Claim 2.

To prove the proposition, we set := k 2 /4 and m := n − 4 − 2, and take T to be T m, .
Thus |T | = |T m, | = m + 4 + 2 = n. By Claim 1, T is strongly k 2 /4 -connected. Since n > k 2 + k + 2 ≥ 4 + √ 4 + 2, we have m > √ 4 , so Claim 2 implies that if D ⊆ T = T m, is a spanning r-regular subdigraph, then r ≤ √ 4 ≤ k.

Finding a single Hamilton cycle in suitable oriented graphs

We first state two simple, well-known facts concerning the degree sequences of tournaments.Proposition 6.1.Let T be a tournament on n vertices.Then T contains at least one vertex of in-degree at most n/2, and at least one vertex of out-degree at mos To prove the proposition, we set := k 2 /4 and m := n − 4 − 2, and take T to be T m, .

Finding a single Hamilton cycle in suitable oriented graphs
We first state two simple, well-known facts concerning the degree sequences of tournaments.Proposition 6.1.Let T be a tournament on n vertices.Then T contains at least one vertex of in-degree at most n/2, and at least one vertex of out-degree at most n/2.Proposition 6.2.Let T be a tournament on n vertices and let d ≥ 0. Then T has at most 2d + 1 vertices of in-degree at most d, and at most 2d + 1 vertices of out-degree at most d.
n/2.Proposition 6.2.Let T be a tournament on n vertices and let d ≥ 0. Then T has at most 2d + 1 vertices of in-degree at most d, and at most 2d + 1 vertices of out-degree at most d.

We will also use the following well-known result due to Gallai and Milgram (see for example [9]).(The independence number of a digraph T is the maximal size of a set X ⊆ V (T ) such that T [X] contains no edges.)Theorem 6.3.Let T be a digraph with independence number at most k.Then T has a path cove We will also use the following well-known result due to Gallai and Milgram (see for example [9]).(The independence number of a digraph T is the maximal size of a set X ⊆ V (T ) such that T [X] contains no edges.)Theorem 6.3.Let T be a digraph with independence number at most k.Then T has a path cover consisting of at most k paths.
The following corollary is an immediate consequence of Theorem 6.3.Corollary 6.4.Let T be an oriented graph on n vertices with δ(T ) ≥ n − k.Then T has a path cover consisting of at most k paths.
a path cover consisting of at most k paths.

Given a digraph T , we define a covering edge for a vertex v to be an edge xy of T such that xv, vy ∈ E(T ).We call xv and vy the activating edges of xy.Note that if xy is a covering edge for v and C is a cycle in T containing xy but not v, we can form a new cycle C with V (C ) = V (C) ∪ {v} by replacing xy with xvy in C. We will see in Section 8 that covering edges are easy t Given a digraph T , we define a covering edge for a vertex v to be an edge xy of T such that xv, vy ∈ E(T ).We call xv and vy the activating edges of xy.Note that if xy is a covering edge for v and C is a cycle in T containing xy but not v, we can form a new cycle C with V (C ) = V (C) ∪ {v} by replacing xy with xvy in C. We will see in Section 8 that covering edges are easy to find in strongly 2-connected tournaments.
Recall that, given a path system P, we write h(P) for the set of heads of paths in P and t(P) for the set of tails of paths in P. If v ∈ V (P), we write v + and v − respectively for the successor and predecessor of v on the path in P containing v.
The following lemma allows us to take a path cover P of a digraph and modify it into a path cover P with no heads in some "bad" set I, without adding any heads or tails in I ∪ J for some other "bad" set J.Moreover, we can do this without losing any edges in some "good" set F ⊆ E(P), and without altering too many paths in P. In our applications, F will consist of covering edges.We require that every vertex in I has high out-degree.Lemma 6.5.Let T be a digraph.Let I, J ⊆ V (T ) be disjoint.Let P = P 1 ∪P 2 be a path cover of T satisfying h(P 2 ) ∩ I = ∅.Let F ⊆ E(P).Suppose d + (v) > 3(|I| + |J|) + 2|F | for all v ∈ I. Then there exists a path cover P of T satisfying the following properties: plications, F will consist of covering edges.We require that every vertex in I has high out-degree.Lemma 6.5.Let T be a digraph.Let I, J ⊆ V (T ) be disjo nt.Let P = P 1 ∪P 2 be a path cover of T satisfying h(P 2 ) ∩ I = ∅.Let F ⊆ E(P).Suppose d + (v) > 3(|I| + |J|) + 2|F | for all v ∈ I. Then there exists a path cover P of T satisfying the following properties:
(i) h(P ) ∩ I = ∅. (ii) h(P ) ∩ J = h(P) ∩ J. (iii) t(P ) ∩ (I ∪ J) = t(P) ∩ (I ∪ J). (iv) F ⊆ E(P ). (v) |P

≤ |P| + |P 1 |. (vi) |P ∩ P 2 | ≥ |P 2 | − |P 1 |. If in ad
ition d + (v) > 3(|I| + |J|) + 2|F | + |V (P 2 )
| for all v ∈ I, then we may strengthen (vi) to P 2 ⊆ P .

Proof.We will use the degree condition on the vertices in I in the hypothesis to repeatedly extend paths with heads in I out of I, breaking other paths in P as a result.We must ensure that we do not create new paths with endpoints in I ∪ J in the process.Let r := |P 1 | and P 0 := P. We shall find path covers P 1 , . . ., P r of | for all v ∈ I, then we may strengthen (vi) to P 2 ⊆ P .
Proof.We will use the degree condition on the vertices in I in the hypothesis to repeatedly extend paths with heads in I out of I, breaking other paths in P as a result.We must ensure that we do not create new paths with endpoints in I ∪ J in the process.Let r := |P 1 | and P 0 := P. We shall find path covers P 1 , . . ., P r of T such that the following properties hold for all 0 ≤ i ≤ r: such that the following properties hold for all 0 ≤ i ≤ r:
(P1) |h(P i ) ∩ I| ≤ r − i. (P2) h(P i ) ∩ J = h(P) ∩ J. (P3) t(P i ) ∩ (I ∪ J) = t(P) ∩ (I ∪ J). (P4) F ⊆ E(P i ). (P5) |P i | ≤ |P| + i. (P6) |P i ∩ P 2 | ≥ |P 2 | − i.
If this is possible, we may then take P := P r .

By hypothesis, P 0 satisfies (P1)-(P6).So suppose we hav If this is possible, we may then take P := P r .
By hypothesis, P 0 satisfies (P1)-(P6).So suppose we have found P 0 , . . ., P i−1 for some i ∈ [r].We then form P i as follows.If |h(P i−1 ) ∩ I| ≤ r − i, we simply let P i := P i−1 .Otherwise, let P ∈ P i−1 be a path with head v ∈ I.We will form P i by extending the head v of P and breaking the path in P i−1 which P now intersects into two subpaths.Define found P 0 , . . ., P i−1 for some i ∈ [r].We then form P i as follows.If |h(P i−1 ) ∩ I| ≤ r − i, we simply let P i := P i−1 .Otherwise, let P ∈ P i−1 be a path with head v ∈ I.We will form P i by e tending the head v of P and breaking the path in P i−1 which P now intersects into two subpaths.Define
X := {x ∈ V (T ) : {x + , x, x − } ∩ (I ∪ J) = ∅}.
We have
d + (v) > 3(|I| + |J|) + 2|F | ≥ |X| + |V (F )| ≥ |X ∪ V (F )|,
and so there exists w ∈ N + (v) \ (X ∪ V (F )).Let Q be the path in P i−1 containing w (note that we may have Q = P ).Split Q into (at most) two paths and an isolated vertex by removing any of the ed We have and so there exists w ∈ N + (v) \ (X ∪ V (F )).Let Q be the path in P i−1 containing w (note that we may have Q = P ).Split Q into (at most) two paths and an isolated vertex by removing any of the edges w − w, ww + that exist, and let P * be the set of paths obtained from P i−1 in this way.Let P * be the path in P * containing v. (Note that P * = P unless w ∈ V (P ).)We then form P i by replacing P * by P * vw in P * .
es w − w, ww + that exist, and let P * be the set of paths obtained from P i−1 in this way.Let P * be the path in P * containing v. (Note that P * = P unless w ∈ V (P ).)We then form P i by replacing P * by P * vw in P * .

First suppose w ∈ Int(Q) First suppose w ∈ Int(Q).Then P i is a path cover of T such that h(P i ) = (h(P i−1 ) \ {v}) ∪ {w, w − } and t(P i ) = t(P i ) ∪ {w + }.
Then P i is a path cover of T such that h(P i ) = (h(P i−1 ) \ {v}) ∪ {w, w − } and t(P i ) = t(P i ) ∪ {w + }.

Since w / ∈ X, we have w, w − / ∈ I and hence
|h(P i ) ∩ I| = |h(P i−1 ) ∩ I| − 1 ≤ r − i.
Thus (P1) holds.Similarly,
h(P i ) ∩ J = h(P i−1 ) ∩ J = h(P) ∩ J, t(P i ) ∩ (I ∪ J) = t(P i−1 ) ∩ (I ∪ J) = t(P) ∩ (I ∪ J),
and so (P2) and (P3) hold.By similar arguments, (P1)-(P3) also hold if w is an endpoint of Q.Since w / ∈ V (F ) and F ⊆ E(P i−1 ) we have F ⊆ E(P i ) and (P4) holds.(P5) holds too since |P i | ≤ |P i−1 | + 1.Finally, we have altered at most two paths in P i−1 .One of these had its head in I, so we have altered at most one path in P i−1 ∩ P 2 .Thus (P6) holds.

If in addition w Since w / ∈ X, we have w, w − / ∈ I and hence Thus (P1) holds.Similarly, and so (P2) and (P3) hold.By similar arguments, (P1)-(P3) also hold if w is an endpoint of Q.Since w / ∈ V (F ) and F ⊆ E(P i−1 ) we have F ⊆ E(P i ) and (P4) holds.(P5) holds too since |P i | ≤ |P i−1 | + 1.Finally, we have altered at most two paths in P i−1 .One of these had its head in I, so we have altered at most one path in P i−1 ∩ P 2 .Thus (P6) holds.
If in addition we have then we may use almost exactly the same argument to prove the strengthened version of the result.Instead of choosing w ∈ N + (v)\(X ∪V (F )), we may choose \ (X ∪ V (F ) ∪ V (P 2 )
).We also strengthen (P6) to the ).We also strengthen (P6) to the requirement that P 2 ⊆ P i .The strengthened (P6) must hold in each step since we now have that w / ∈ V (P 2 ).
equirement that P 2 ⊆ P i .The strengthened (P6) must hold in each step since we now have that w / ∈ V (P 2 ).

The following analogue of Lemma 6.5 for tails can be obtained by reversing the orientation of each edge of T .Lemma 6.6.Let T be a digraph.Let I, J ⊆ V (T ) be disjoint.Let P = P 1 ∪P 2 be a path cover of T satisfying t(P 2 ) ∩ I = ∅.Let F ⊆ E(P).Suppose d − (v) > 3(|I| + |J|) + 2|F The following analogue of Lemma 6.5 for tails can be obtained by reversing the orientation of each edge of T .Lemma 6.6.Let T be a digraph.Let I, J ⊆ V (T ) be disjoint.Let P = P 1 ∪P 2 be a path cover of T satisfying t(P 2 ) ∩ I = ∅.Let F ⊆ E(P).Suppose d − (v) > 3(|I| + |J|) + 2|F | for all v ∈ I. Then there exists a path cover P of T satisfying the following properties: ∪ J). (iv) F ⊆ E(P ). (v) |P | ≤ |P| + |P 1 |. ( i) |P ∩ P 2 | ≥ |P 2 | − |P 1 |. If in addition d − (v) > 3(|I| + |J|) + 2|F | + |V (P 2 )
| for all v ∈ I, then we may strengthen (vi) to P 2 ⊆ P .

The following lemma is the main building block of the proof of Theorem 1.2.It will be applied repeatedly to find the required edge-disjoint Hamilton cycles.Roughly speaking, the lemma guarantees a Hamilton cycle provi | for all v ∈ I, then we may strengthen (vi) to P 2 ⊆ P .
The following lemma is the main building block of the proof of Theorem 1.2.It will be applied repeatedly to find the required edge-disjoint Hamilton cycles.Roughly speaking, the lemma guarantees a Hamilton cycle provided that we have well-chosen disjoint (almost) dominating sets A i and B i which are linked by short paths containing covering edges for all vertices in these dominating sets.(This is the linked dominating structure described in Sections 1 and 3.) An additional assumption is that we have not removed too many edges of our tournament T already.In general, the statement and proof roughly follow the sketch in Section 3, with the addition of a set X ⊆ V (T ).
nating s ing edges for all vertices in these dominating sets.(This is the linked dominating structure described in Sections 1 and 3.) An additional assumption is that we have not removed too many edges of our tournament T already.In general, the statement and proof roughly follow the sketch in Section 3, with the addition of a set X ⊆ V (T ).

The role of X is as follows.The sets A i and B i in the lemma dominate only almost all ver The role of X is as follows.The sets A i and B i in the lemma dominate only almost all vertices of T , so we have some small exceptional sets E A and E B of vertices which are not dominated.We will use Lemmas 6.5 and 6.6 to extend a certain path system out of these exceptional sets E A and E B .For this we need that the vertices in E A ∪ E B have relatively high in-and out-degree.But T may have vertices which do not satisfy this degree condition.When we apply Lemma 6.7, these problematic vertices will be the elements of X. Lemma 6.7.Let C := 10 6 , k ≥ 20, t := 164k, and c := log 50t + 1 .Suppose that T is an oriented graph of order n satisfying δ(T ) > n−4k and δ 0 (T ) ≥ Ck 2 .Suppose moreover that T contains disjoint sets of vertices A 1 , . . ., A t , B 1 , . . ., B t and X, a matching F , and vertex-disjoint paths P 1 , . . ., P t such that the following conditions hold, where ices of T , so we have some small exceptional sets E A and E B of d a certain path system out of these exception e vertices in E A ∪ E B have ion.When we apply Lemma 6.7, these problematic vertices will be the elements of X. Lemma 6.7.Let C := 10 6 , k ≥ 20, t := 164k, and c := log 50t + 1 .Suppose that T is an oriented graph of order n satisfying δ(T ) > n−4k and δ 0 (T ) ≥ Ck 2 .Suppose moreover that T contains disjoint sets of vertices A 1 , . . ., A t , B 1 , . . ., B t and X, a matching F , and verte -disjoint paths P 1 , .
hold, where
A * := A 1 ∪ • • • ∪ A t and B * := B 1 ∪ • • • ∪ B t : (i) 2 ≤ |A i | ≤ c for all i ∈ [t]. Moreover, T [A i ] is a transitive tournament
whose head has out-degree at least n/3 in T .(ii) There exists a set (iv) There exists a set
E A ⊆ V (T ) \ (A * ∪ B * ), such that each A i out- dominates V (T ) \ (A * ∪ B * ∪ E A ). Moreo whose head has out-degree at least n/3 in T .(ii) There exists a set (iv) There exists a set er, |E A | ≤ d − /40, where d − := min{d − T (v) : v ∈ E A \ X}. (iii) 2 ≤ |B i | ≤ c for all i ∈ [t]. Moreover, T [B i ]E B ⊆ V (T ) \ (A * ∪ B * ), such that each B i in- dominates V (T ) \ (A * ∪ B * ∪ E B ). Moreover, |E B | ≤ d + /40, where d + := min{d + T (v) : v ∈ E B \ X}. (v) For all i ∈ [t], P i is a path from the head of T [B i ] to the tail of T [A i ] which is inter * }
, where e v is a covering edge for v and e v = e v whenever v = v .In particular,
|F | = |A * ∪ B * | ≤ 2ct. (v , where e v is a covering edge for v and e v = e v whenever v = v .In particular, i) We have X ⊆ V (P 1 ∪ • • • ∪ P t ), X ∩ (A * ∪ B * ) = ∅ and |X| ≤ 2kt.
Then T contains a Hamilton cycle.

Proof.Without loss of generality, suppose that d − ≤ d + .(Otherwise, reverse the orientation of every edge in T .)Write a i for the head of T [A i ] and a i for its tail.Similarly, write b i for the head of T [B i ] and b i for its tail.Let By Corollary 6.4, there exists a path cover P 1 of N \V (P 2 ) with |P 1 | ≤ 4k.Then Q 1 := P 1 ∪P 2 is a path cover of T .The situation is illustrated in Figure 3.

Claim.There exists an oriented graph T with T ⊆ T ⊆ T [V (T ) ∪ A ∪ B ] and a path cover Q of T such that the following properties hold:
(Q1) F ⊆ E(Q). (Q2) t(Q) Then T contains a Hamilton cycle.
Proof.Without loss of generality, suppose that d − ≤ d + .(Otherwise, reverse the orientation of every edge in T .)Write a i for the head of T [A i ] and a i for its tail.Similarly, write b i for the head of T [B i ] and b i for its tail.Let By Corollary 6.4, there exists a path cover P 1 of N \V (P 2 ) with |P 1 | ≤ 4k.Then Q 1 := P 1 ∪P 2 is a path cover of T .The situation is illustrated in Figure 3.
Claim.There exists an oriented graph T with T ⊆ T ⊆ T [V (T ) ∪ A ∪ B ] and a path cover Q of T such that the following properties hold: E A = ∅. (Q3) h(Q) ∩ E B = ∅. (Q4) |Q ∩ P 2 | ≥ |Q 1 | − 20k. (Q5) If a i or b i is in V (Q), then P i / ∈ Q. (Q6) |Q| ≤ |Q 1 | + 124k. (Q7) No paths in Q \ P 2 have endpoints in A * ∪ B * .
We will prove the claim by applying Lemmas 6.5 and 6.6 repeatedly to improve our current path cover.More precisely, we will construct path covers Q 2 , . . ., Q 6 such that eventually Q 6 satisfies (Q1)-(Q7).So we can take Q := Q 6 .

In order to be able to apply Lemmas 6.5 and 6.6, we must first bound the degrees of the vertices in T from below.For all v ∈ V (T ), we have
d + T (v) ≥ d + T (v) − |A * ∪ B * | (i),(iii) ≥ d + T (v) − 2ct ≥ d + T (v) − δ 0 (T ) 5 ≥ 4 5 d + T (v).(1)
Similarly,
d − T (v) ≥ 4 5 d − T (v)(2)
for all v ∈ V (T ).

We will first extend the tails of paths in Q 1 out of E A .We do this by applying Lemma 6.6 to T and Q 1 = P 1 ∪P 2 with I := E A \ X, J := X ∪ A ∪ B to form a new path cover Q 2 of T which will satisfy (Q1) and (Q2).By con We will prove the claim by applying Lemmas 6.5 and 6.6 repeatedly to improve our current path cover.More precisely, we will construct path covers Q 2 , . . ., Q 6 such that eventually Q 6 satisfies (Q1)-(Q7).So we can take Q := Q 6 .
In order to be able to apply Lemmas 6.5 and 6.6, we must first bound the degrees of the vertices in T from below.For all v ∈ V (T ), we have Similarly, for all v ∈ V (T ).
We will first extend the tails of paths in Q 1 out of E A .We do this by applying Lemma 6.6 to T and Q 1 = P 1 ∪P 2 with I := E A \ X, J := X ∪ A ∪ B to form a new path cover Q 2 of T which will satisfy (Q1) and (Q2).By conditions (ii) and (v), no paths in P 2 have endpoints in I.By condition (vi), F ⊆ E(Q 1 ).Moreover, 2 have endpoints in I.By condition (vi), F ⊆ E(Q 1 ).Moreover,
3(|I| + |J|) + 2|F | ≤ 3|E A | In the final inequality we used the fact that d − ≥ δ 0 (T ) ≥ Ck 2 .Thus for all v ∈ I we have d − (3) > 3(|I| + |J|) + 2|F |.
Thus the requirements of Lemma 6.6 are satisfied, and we can apply the lemma to obta Thus the requirements of Lemma 6.6 are satisfied, and we can apply the lemma to obtain a path cover Q 2 of T .Lemma 6.6(iv) implies that Q 2 satisfies (Q1).Moreover, Lemma 6.6(v),(vi) imply that as well as We will now extend the heads of paths in Q 2 out of E B .We do this by applying Lemma 6.5 to T , (Q extend the heads of paths in Q 2 out of E B .We do this by applying Lemma 6.5 to T , (Q
2 \ P 2 ) ∪(Q 2 ∩ P 2 ) with I := E B \ X, J := (E A \ E B ) ∪ X ∪ A ∪ B to form
a new path cover Q 3 of T which will satisfy (Q1)-(Q4).As before, no paths in a new path cover Q 3 of T which will satisfy (Q1)-(Q4).As before, no paths in P 2 ⊇ Q 2 ∩ P 2 have endpoints in I, and F ⊆ E(Q 2 ) by (Q1) for Q 2 .Moreover, similarly as in (3) we obtain
a 2 ) and L ly that no path in Q 3 has its head in (E B \ X) ∪ X ⊇ E B and so Q 3 satisfies (Q3).Finally, (a 1 ) and Lemma 6.5(ii),(iii) together imply that (b 1 ) no paths in Q 3 have tails in A or heads in B.

We will now extend the paths in Q 3 \ P 2 so that their endpoints lie in A ∪ B rather than A ∪ B.More precisely, if P ∈ Q 3 \ P 2 has head a i ∈ A We will now extend the paths in Q 3 \ P 2 so that their endpoints lie in A ∪ B rather than A ∪ B.More precisely, if P ∈ Q 3 \ P 2 has head a i ∈ A , then we replace P by P a i a i (recall that a i a i ∈ E(T ) by condition (i) and nition of N ). If P ∈ Q 3 \P 2 has tail b i ∈ B, we replace P by b i b i P (recall that b i b i ∈ e path system thus obtained from Q 3 . Let T := T [V (Q 4 )]. Then T ⊆ T ⊆ T [V (T ) ∪ A ∪ B ].
and Q 4 is a path cover of T satisfying (Q1)-(Q4) and such that ( 7)
|Q 4 | = |Q 3 | and Q 4 ∩ P 2 = and Q 4 is a path cover of T satisfying (Q1)-(Q4) and such that ( 7) h(Q 4 \ P Together with (b 1 ) this implies that (c 1 ) no paths in Q 4 \ P 2 have endpoints in A ∪ B.Moreover, by construction of Q 4 , every vertex a i ∈ V (Q 4 ) ∩ A is a head of some path P ∈ Q 4 \P 2 and this path P also contains a i (so in particular P i / ∈ Q 4 ∩P 2 ).Similarly, every vertex in b i ∈ V (Q 4 ) ∩ B is a tail of some path P ∈ Q 4 \ P 2 and this path P also contains b i (in particular P i / ∈ Q 4 ∩ P 2 ).Thus (Q5) as well as the following assertion hold: er with ( hat (c 1 ) no paths in Q 4 \ P 2 have endpoints in A ∪ B.Moreover, by construction of Q 4 , every vertex a i ∈ V (Q 4 ) ∩ A is a head of some path P ∈ Q 4 \P 2 and this path P also contains a i (so in particular P i / ∈ Q 4 ∩P 2 ).Similarly, every vertex in b i ∈ V (Q 4 ) ∩ B is a tail of some path P ∈ Q 4 \ P 2 and this path P also contains b i (in particular P i / ∈ Q 4 (c 2 ) no paths in Q 4 have heads in B or tails in A.
2 ) no paths in Q 4 have heads in B or tails in A.

We will now extend the tails of paths in Q 4 \ P 2 out of A * ∪ B * .We do this by applying the strengthened form of Lemma 6.6 to T , (Q  3), it follows that
4 \ P 2 ) ∪(Q 4 ∩ P 2 ) with I := B , J := E A ∪E B ∪A3(|I| + |J|) + 2|F | + |V (Q 4 ∩ P 2 )| ≤ 3(|A | + |A| + |B | + |B| + |E A | + |E B |) + 2|F | + |V (P 2 ) We will now extend the tails of paths in Q 4 \ P 2 out of A * ∪ B * .We do this by applying the strengthened form of Lemma 6.6 to T , (Q  3), it follows that Thus the requirements of the strengthened form of Lemma 6.6 are satisfied, and we can apply the lemma to obtain a path cover Q 5 of T such that Q 5 ∩ P 2 ⊇ Q 4 ∩ P 2 .Note that Lemma 6.6(ii),(iii) imply that the endpoints of Q 5 \ (P 2 ∩ Q 4 ) in J are the same as those of Q 4 \ P 2 .Together with (c 1 ) this implies that no paths in Q 5 \ (P 2 ∩ Q 4 ) have endpoints in A ∪ B. In particular, this means that Q 5 ∩ P 2 = Q 4 ∩ P 2 and so (d 1 ) no paths in Q 5 \ P 2 have endpoints in A ∪ B. Thus (Q5) for Q 4 implies that Q 5 satisfies (Q5) as well.Lemma 6.6(ii)-(iv), (vi) (strengthened) and (Q1)-(Q4) for Q 4 together imply that Q 5 satisfies (Q1)-(Q4).Moreover, Lemma 6.6(v) implies that .6 are satisfied, and we can apply the lemma to obtain a path cover Q 5 of T such that Q 5 ∩ P 2 ⊇ Q 4 ∩ P 2 .Note that Lemma 6.6(ii),(iii) imply that the endpoints of Q 5 \ (P 2 ∩ Q 4 ) in J are the same as those of Q 4 \ P 2 .Together with (c 1 ) this implies that no paths in Q 5 \ (P 2 ∩ Q 4 ) have endpoints in A ∪ B. In particula 2 have endpoints in A ∪ B. Thus (Q5) (vi) (strengthened) and (Q1)-(Q4) for Q 4 together imply that Q 5 satisfies (Q1)-(Q4).Moreover, Lemma 6.6(v) implies that
|Q 5 | ≤ |Q 4 | + |Q 4 \ P 2 | (7) = |Q 3 | + |Q 3 \ P 2 | = 2|Q 3 | − |Q 3 ∩ P 2 |
(5),( 6)
≤ |Q 1 | + 52k. (8)
By Lemma 6.6(i),(ii) and (c 2 ), we can also strengthen (d 1 ) to (d 2 ) no paths in Q 5 \ P 2 have endpoints (5),( 6) By Lemma 6.6(i),(ii) and (c 2 ), we can also strengthen (d 1 ) to (d 2 ) no paths in Q 5 \ P 2 have endpoints in A ∪ B ∪ B and no paths in Q 5 have tails in A. Finally, we will extend the heads of paths in Q 5 \ P 2 out of A * ∪ B * .We do this by applying the strengthened form of Lemma 6.5 to T , n A ∪ B ∪ B and no paths in Q 5 have tails in A. Finally, we will extend the heads of paths in Q 5 \ P 2 out of A * ∪ B * .We do this by applying the strengthened form of Lemma 6.5 to T ,
(Q 5 \P 2 ) ∪(Q 5 ∩P 2 ) with I := A, J := E A ∪ E B ∪ A ∪ B ∪ B to form a new path cover Q 6 of
T which will satisfy (Q1)-(Q7).Clearly no paths T which will satisfy (Q1)-(Q7).Clearly no paths in P 2 ⊇ Q 5 ∩ P 2 have heads in I, and F ⊆ E(Q 5 ) by (Q1).Similarly as before, condition (i) and ( 1) together imply that i) and ( 1) together imply that
3(|I| + |J|) + 2|F | + |V (Q 5 ) ∩ P 2 | < n 4 ≤ Thus the requirements of the strengthened form of Lemma 6.5 are satisfied, and we can apply the lemma to obtain a path cover Q 6 of T such that Q 6 ∩ P 2 = Q 5 ∩ P 2 .(The fact that we have equality follows using a similar argument as in (d 1 ) above.)Thus (Q5) for Q 5 implies that Q 6 satisfies (Q5) as well.Lemma 6.5(ii)-(iv), (vi) (strengthened) and (Q1)-(Q4) for Q 5 together imply that Q 6 satisfies (Q1)-(Q4).Also, by Lemma 6.5(v) we have s the requirements of the strengthened form of Lemma 6.5 are sat ath cover Q 6 of T such that Q 6 ∩ P 2 = Q 5 ∩ P 2 .(The fact that we have equality follows using a similar argument as in (d 1 ) above.)Thus (Q5) for Q 5 implies that Q 6 satisfies (Q5) as well.Lemma 6.5(ii)-(iv), (vi) (strengthened) and (Q1)-(Q4) for Q 5 together imply that Q 6 satisfies (Q1)-(Q4).Also, by Lemma 6.5(v) we have
|Q 6 | ≤ |Q 5 | + |Q 5 \ P 2 | = 2|Q 5 | − |Q 5 ∩ P 2 | (Q4),(8) ≤ |Q 1 | + 124k.
So (Q6) holds.Moreover, by Lemma 6.5(i)-(iii), (d 2 ) and the fact that
Q 6 ∩P 2 = Q 5 ∩ P 2 , no paths in Q 6 \ P 2 h So (Q6) holds.Moreover, by Lemma 6.5(i)-(iii), (d 2 ) and the fact that ve endpoints in A ∪ A ∪ B ∪ B. Since no vertex in (A * ∪ B * ) \ (A ∪ A ∪ B ∪ B) lies in V (T ) = V (Q 6 )
, this in turn implies (Q7).So the path system Q := Q 6 is as required in the claim.

We will now use the fact that each A i and each B i is an almost dominating set in order to extend the paths in Q \ P 2 into those A i and B i which contain the endpoints of paths in Q ∩ P 2 .We then use the paths in Q ∩ P 2 to join these extended paths into a long c , this in turn implies (Q7).So the path system Q := Q 6 is as required in the claim.
We will now use the fact that each A i and each B i is an almost dominating set in order to extend the paths in Q \ P 2 into those A i and B i which contain the endpoints of paths in Q ∩ P 2 .We then use the paths in Q ∩ P 2 to join these extended paths into a long cycle C covering (at least) N , and with F ⊆ E(C).Finally, we will deploy whatever covering edges we need from F in order to absorb any vertices in A * ∪ B * not already covered into C.
vertices in A * ∪ B * not already covered into C.

Let R := Q \ P 2 and S := Q ∩ P 2 .In o Let R := Q \ P 2 and S := Q ∩ P 2 .In order to carry out the steps above, we would like to have |R| = |S| to avoid having any paths in S left over.So we first split the paths in R until we have exactly |S| of them.In this process, we wish to preserve (Q1)-(Q3), (Q5) and (Q7).To show that this can be done, first note that by (Q4) and (Q6), we have hem.In this process, we wish to preserve (Q1)-(Q3), (Q5) and (Q7).To show that this can be done, first note that by (Q4) and (Q6), we have
|R| = |Q \ P 2 | ≤ 144k = t − 20k ≤ |Q 1 | − 20k ≤ |Q ∩ P 2 | = |S|.
The number of edges in R which are incident to vertices in E A ∪ E B ∪ A * ∪ B * , or which belong to F , is bounded above by
2(|E A | + |E B | + |A * | + |B * |) + |F | ≤ d + 10 + 6ct ≤ n 4 .
On the other hand,
|E(R)| = |V (R)| − |R| ≥ (n − |A * ∪ B * | − |V (P 2 )|) − 144k ≥ n − 2ct − n 20 − 144k ≥ n 2 . Hence |E(R)| − 2(|E A | + |E B | + |A * | + |B * |) − |F | ≥ n 4 > t ≥ |S|.
We may therefore form a path cover R of T [V (R)] with |R | = |S| by greedily removing edges of paths in R The number of edges in R which are incident to vertices in E A ∪ E B ∪ A * ∪ B * , or which belong to F , is bounded above by On the other hand, We may therefore form a path cover R of T [V (R)] with |R | = |S| by greedily removing edges of paths in R which are neither incident to Q5) and (Q7).Next, we extend the paths in R into A * ∪ B * and join them with the paths in S to form a long cycle C.By relabeling the P i if necessary, we may assume that S = {P 1 , . . ., P }.Let R 1 , . . ., R denote the paths in R and for each j ∈ [ ] let x j be the tail of R j and y j the head of R j .Recall from (Q2) and (Q7) that x j / ∈ A * ∪ B * ∪ E A .Hence by condition (ii) there exists x j ∈ A j−1 with x j x j ∈ E(T ), where the indices are understood to be modulo .Similarly y j / ∈ A * ∪ B * ∪ E B by (Q3) and (Q7), so by condition (iv) there exists y j ∈ B j with y j y j ∈ E(T ).Let R j := x j x j R j y j y j .If x j = a j−1 , then we extend R j by adding the edge a j−1 x j .Similarly, if y j = b j we extend R j by adding the edge y j b j .In all cases, we still denote the resulting path from a j−1 to b j by R j .
e paths in S to form a long cycle C.By relabeling the P i if necessary, we may assume that S = {P 1 , . . ., P }.Let R 1 , . . ., R denote the paths in R and for each j ∈ [ ] let x j be the tail of R j and y j the head of R j .Recall from (Q2) and (Q7) that x j / ∈ A * ∪ B * ∪ E A .Hence by conditio indices are understood to be modulo .Similarly y j / ∈ A * ∪ B * ∪ E B by (Q3) and (Q7), so by condition (iv) there exists y j ∈ B j with y j y j ∈ E(T ).Let R j := x j x j R j y j y j .If x j = a j−1 , th extend R j by adding the edge y j b j .In all cases, we still denote the resulting path from a j−1 to b j by R j .
A * ∪ B * ∪ E A ∪ E B nor elements of F . Then R ∪ S satisfies (Q1)-(Q3), (
Recall that P j is a path from b j to a j for all j ∈ [ ].Moreover, we have
x j , y j / ∈ V (Q \ P 2 ) = V (R ) for all j ∈ [ ].
(Indeed, if x j = a j this follows since for the oriented graph T defined in the claim we have V (T ) ∩ A i ⊆ {a i , a i }.If x j Recall that P j is a path from b j to a j for all j ∈ [ ].Moreover, we have (Indeed, if x j = a j this follows since for the oriented graph T defined in the claim we have V (T ) ∩ A i ⊆ {a i , a i }.If x j = a j , this follows since P j ∈ Q and so (Q5) implies that a j / ∈ V (Q).The argument for y j is similar.)Thus R 1 , . . ., R are pairwise vertex-disjoint and internally disjoint from the paths in S.So we can define a cycle C by ment for y j is similar.)Thus R 1 , . . ., R are pairwise vertex-disjoint Note that N ⊆ V (C) since R ∪ S is a path cover of T , and F ⊆ E(C) by (Q1).Recall from condition (vi) that F consists of covering edges e v for all v ∈ A * ∪ B * and that these e v are pairwise distinct.Thus each e v lies on C and so neither of the two activating edges of e v can lie on C. Writing e v = x v y v , it follows from these observations that we may form a new cycle C by replacing cover of T , and F ⊆ E(C) by (Q1).Recall from condition (vi) that F consists of cover ng edges e v for all v ∈ A * ∪ B * and that these e v are pairwise distinct.Thus each e v lies on C and so neither of the two activating edges of e v can lie on C. Writing e v = x v y v , it follows from these observations that we may form a new cycle C by replacing
x v y v by x v vy v in C for all v ∈ (A * ∪ B * ) \ V (C).
Then C is a Hamilton cycle of T , as desired.


Finding many edge-disjoint Hamilton cycles in a good tournament

In the proof of Then C is a Hamilton cycle of T , as desired.

Finding many edge-disjoint Hamilton cycles in a good tournament
In the proof of Theorem 1.2, we will find the edge-disjoint Hamilton cycles in a given highly-linked tournament by repeatedly applying Lemma 6.7.In each application, we will need to set up all the dominating sets and paths required by Lemma 6.7.The following definition encapsulates this idea.(Recall that Int(P ) denotes the interior of a path P .)Definition 7.1.We say that a tournament T is (C, k, t, c)-good if it contains vertex sets A 1 1 , . . ., A t k , B 1 1 , . . ., B t k , E A,1 , . . ., E A,k , E B,1 , . . ., E B,k , edge sets F 1 , . . ., F k , and paths P 1 1 , . . ., P t k such that the following statements hold, where Theorem 1.2, we will find the edge-disjoint Hamilton cycles in a given highly-linked tournament by repeatedly applying Lemma 6.7.In each application, we will need to set up all the dominating sets and paths required by Lemma 6.7.The following definition encapsulates this idea.(Recall that Int(P ) denotes the interior of a path P .)Definition 7.1.We sa A 1 1 , . . ., A t k , B 1 1 , . . ., B t k , E A,1 , . . ., E A,k , E B,1 , . . ., E B,k , edge sets F 1 , . . ., F k , and where
A * i := A 1 * 1 ∪ ≤ |A i | ≤ c for all i ∈ [k] and ∈ [t]. Moreover, each T [A i ] is a transitive tournament whose head has out-degree at least 2n/5 in T . Write A := {h(T [A i ]) : i ∈ [k], ∈ [t]}. (G2) The sets B 1
1 , . . ., B t k are disjoint from each other and from A * , and
2 ≤ |B i | ≤ c for all i ∈ [k] and ∈ [t]. Moreover, each T [B i ] is a transitive tournament whose tail has in-degree at least 2n/5 in T . Write B := {t(T [B i ]) : i ∈ [k], ∈ [t]}. (G3) Write d − := min{d − (v) : v ∈ V (T ) \ (A ∪ B )}. Each A i out-dominates V (T )\(A * ∪B * ∪E A,i ). Moreover, |E A,i | ≤ d − /50 and E A,i ∩(A * i ∪B * i ) = ∅ for all i ∈ [k]. (G4) Write d + := min{d + (v) : v ∈ V (T ) \ (A ∪ B )}. Each B i in-dominates V (T )\(A * ∪B * ∪E B,i ). Moreover, |E B,i | ≤ d + /50 and E B,i ∩(A * i ∪B * i ) = ∅ for all i ∈ [k]. (G5) Each P i is a path from the head of T [B i ] to the tail of T [A i ]. For each i ∈ [k], the paths P 1 i , . . . , P t i are vertex-disjoint and |P 1 1 ∪• • •∪P t k | ≤ n/20
. For all i = j and all , m ∈ [t], P i and P m j are edge-disjoint and
V 1 , . . ., B t k are disjoint from each other and from A * , and . For all i = j and all , m ∈ [t], P i and P m j are edge-disjoint and (Int(P i )) ∩ (A * ∪ B * ) ⊆ (A ∪ B ) \ (A * i ∪ B * i ). (G6) F i ⊆ E(P t i l i ∈ [k]. (G7) The set F 1 ∪ • • • ∪ F k is a matching in T − (A * ∪ B * ). For all i ∈ [k]
we have
F i = {e v : v ∈ A * i ∪ B * i }
, where e v is a covering edge for v and e v = e v whenever v = v .Moreover, for each i ∈ [k], let F act i be the set of activating edges corresponding to the covering edges in F i .Then
F act i ∩ E(P j ) = ∅ for all i, j ∈ [k] and we have , where e v is a covering edge for v and e v = e v whenever v = v .Moreover, for each i ∈ [k], let F act i be the set of activating edges corresponding to the covering edges in F i .Then 2 log k.
For convenience, we collect the various disjointness conditions of Definition 7.1 into a single statement.


(G9)

• The sets
A 1 1 , . . . , A t k , B 1 1 , . . . , B t k are disjoint. • (E A,i ∪ E B,i ) ∩ (A * i ∪ B * i ) = ∅ for all i ∈ [k]. • F 1 ∪ • • • ∪ F k is a matching in T − (A * ∪ B * ).
• For each i ∈ [k], the paths P 1 i , . . ., P t i are vertex-disjoint.• For all i = j For convenience, we collect the various disjointness conditions of Definition 7.1 into a single statement.

(G9)
• The sets • For each i ∈ [k], the paths P 1 i , . . ., P t i are vertex-disjoint.• For all i = j and all , m ∈ [t], P i and P m j are edge-disjoint and Int(P i )) ∩ (A * ∪ B * ) ⊆ (A ∪ B ) \ (A * i ∪

* i ).
In particular, P 1 i , . . ., P t i are internally disjoi
In particular, P 1 i , . . ., P t i are internally disjoint from A * i ∪ B * i .The next lemma shows that for suitable parameters C, t = t(k) and c = c(k), every (C, k, t, c)-good tournament contains k edge-disjoint Hamilton cycles.In the next section we then show that there exists a constant C > 0 such that any C k 2 log k-linked tournament is (C, k, t, c)-good (see Lemma 8.7).These two results together immediately imply Theorem 1.2.
t from A * i ∪ B * i .The next lemma shows that for suitable parameters C, t = t(k) and c = c(k), every (C, k, t, c)-good tournament contains k edge-disjoint Hamilton cycles.In the next section we then show that there exists a constant C > 0 such that any C k 2 log k-linked tournament is (C, k, t, c)-good (see Lemma 8.7).These two results together immediately imply Theorem 1.2.

As mentioned at the beginning of this section, in order to prove Lemma 7.2 we will apply Lemma 6.7 k times.In the notation for Definition 7.1, our convention is that the sets with subscript i will be used in the ith application of Lemma 6.7 to find the ith Hamilton cycle.


Proof.

Let T be a (C, k, t, c)-good tournament, and let n : as in Definition 7.1.)Our aim is to appl As mentioned at the beginning of this section, in order to prove Lemma 7.2 we will apply Lemma 6.7 k times.In the notation for Definition 7.1, our convention is that the sets with subscript i will be used in the ith application of Lemma 6.7 to find the ith Hamilton cycle.

Proof.
Let T be a (C, k, t, c)-good tournament, and let n : as in Definition 7.1.)Our aim is to apply Lemma 6.7 repeatedly to find k edge-disjoint Hamilton cycles.So suppose that for some i ∈ [k] we have already found edge-disjoint Hamilton cycles C 1 , . . ., C i−1 such that the following conditions hold: Intuitively, these conditions guarantee that none of the edges we will need in order to find C i , . . ., C k are contained in C 1 , . . ., C i−1 .We have to show that Define P j (A * i ) ∪ A * ∪ B *   \ (A * i ∪ B * i )   , E B,i := E B,i ∪ Then it suffices to find a Hamilton cycle C i of T i .We will do so by applying Lemma 6.7 to T i , A 1 i , . . ., A t i , B 1 i , . . ., B t i , P 1 i , . . ., P t i , E A,i , E B,i , F i and X i .It therefore suffices to verify that the conditions of Lemma 6.7 hold.
n it suffices to find a Hamilton cycle C i of T i .We will do so by applying Lemma 6.7 to T i , A 1 i , . . ., A t i , B 1 i , . . ., B t i , P 1 i , . . ., P t i , E A,i , E B,i , F i and X i .It therefore suffices to verify that the conditions of Lemma 6.7 hold.

We claim that for each v ∈ V (T i ), we have
(9) d + T i (v) ≥ d + T (v) − (i − 1) − (k − i) − 1 − We claim that for each v ∈ V (T i ), we have > d + T (v) − 2k. Indeed, it is immediate that d + C 1 ∪•••∪C i−1 (v) = i − 1.
Since by (G9) for each j > i the paths P 1 j , . . ., P t j are vertex-disjoint, v is covered by at most k−i of the paths P 1 i+1 , . . ., P t k and hence d +
P 1 i+1 ∪•••∪P t k (v) ≤ k−i. Recal Since by (G9) for each j > i the paths P 1 j , . . ., P t j are vertex-disjoint, v is covered by at most k−i of the paths P 1 i+1 , . . ., P t k and hence d + from (G7) that F 1 ∪• • •∪F k consists of one covering edge e v for each v ∈ A * ∪ B * . Moreover, by (G9) the set F 1 ∪ • • • ∪ F k is a matchi
We have δ 0 (T ) > Ck 2 by (G8), and hence δ 0 (T i ) > 10 6 k 2 as required by Lemma 6.7.The disjointness conditions of Lemma 6.7 are satisfied by (G9) and the definition of X i .Since V (T i ) = V (T ), it is immediate that A 1 i , . . ., A t i , B 1 i , . . ., B t i , X i ⊆ V (T i ).We claim that P 1 i , . . ., P t i ⊆ T i .Indeed, by (a) and (G5), each P i is edge-disjoint from C 1 ∪ • • • ∪ C i−1 and from P m j for all j > i and all m ∈ [t].By (G7), each . . ., P t i ⊆ T i .Indeed, by (a) and (G5), each P i is edge-disjoint from C 1 ∪ • • • ∪ C i−1 and from P isjoint from F act 1 ∪ • • • ∪ F act k . Moreover, by (G5), each P i is edge-disjoint from T [A m j ] ∪ T [B m j ]
for all j > i and all m ∈ [t].Altogether this implies that P 1 i , . . ., P t i ⊆ T i .We have
F i ⊆ E(P t i ) ⊆ E(T i
) by (G6).It therefore suffices to prove that c for all j > i and all m ∈ [t].Altogether this implies that P 1 i , . . ., P t i ⊆ T i .We have ) by (G6).It therefore suffices to prove that conditions (i)-(vii) of Lemma 6.7 hold. ndition Condition (v) follows from (G5).Condition (vi) follows from (G6) and (G7).(Note that (G7) implies that F act i ∩ F act j = ∅ for all i = j.So (G7), (b) and the definition of T i imply that F act i ⊆ T i .)By (G6) we have X i ⊆ V (P t i ) and by (G1) and (G2) we have |X i | ≤ |A ∪ B | = 2kt, so condition (vii) holds too.
oo.

It therefore remains to verify conditions (i)-(iv).We first check (i).We have
2 ≤ |A i | It therefore remains to verify conditions (i)-(iv).We first check (i).We have Indeed, to see this, note that C 1 , . . ., C i−1 are edge-disjoint from T [A i ] by (a); by (G9) for all j > i and all m ∈ [t] each path P m j and each , . . ., C i−1 are edge-disjoint from T [A i ] by (a); by (G9) for all j > i and all m ∈ [t] each path P m j and eac ; by (G7) all edges in F act j for j > i are incident to a vertex in A * j ∪B * j , and hence by (G9) none of these edges belongs to j , and hence by (G9) none of these edges belongs to
T [A i ]. Thus T i [A i ] = T [A i ]
is a transitive tournament by (G1).Finally, by (G1) the head of each T is a transitive tournament by (G1).Finally, by (G1) the head of each T [A i ] has out-degree at least 2n/5 in T , and so by (9) out-degree at least n/3 in T i .Hence condition (i) of Lemma 6.7 is satisfied.A similar argument shows that condition (iii) of Lemma 6.7 is also satisfied.
e at least n/3 in T i .Hence condition (i) of Lemma 6.7 is satisfied.A similar argument shows that condition (iii) of Lemma 6.7 is also satisfied.

We will next verify that condition (ii) of Lemma 6.7 holds too.(G9) and the definition of E A,i together imply that
E A,i ∩ (A * i ∪ B * i ) = ∅. By (G3), each A i out-dominates V (T ) \ (A * ∪ B * ∪ E A,i ) in T , and hence out-dominates V (T i ) \ (A * ∪ B * ∪ E A,i ∪ N + T −T i (A * i )) in T i .
However, i We will next verify that condition (ii) of Lemma 6.7 holds too.(G9) and the definition of E A,i together imply that However, it follows from (G9) that for all j > i and all , m ∈ [t], no edge in F act j has an endpoint in A i and that hat
A i ∩ A m j = A i ∩ B m j = ∅.
Hence by (G9) we have that
N + T −T i (A * i ) = j Hence by (G9) we have that P j (A * i ).
It therefore follows from the definitions of It therefore follows from the definitions of E A,i and T i that A i out-dominates E A,i and T i that A i out-dominates
V (T i ) \ (A * i ∪ B * i ∪ E A,i ) in T i for all ∈ [t]
. So in order to check that condition (ii) of Lemma 6.7 holds, it remains only to . So in order to check that condition (ii) of Lemma 6.7 holds, it remains only to bound |E A,i | from above.To do this, first note that by (G9), each vertex in A * i is contained in at most k −i of the paths P 1 i+1 , . . ., P t k .Moreover, |E A,i | ≤ d − /50 by (G3).It therefore follows from the definition of E A,i , (G1) and (G2) that the paths P 1 i+1 , . . ., P t k .Moreover, |E A,i | ≤ d − /50 by (G3).It therefore follows from the definition of E A,i , (G1) and (G2) that
|E A,i | ≤ |E A,i | + j<i N + C i (A * i ) + j>i, ∈[t] N + P j (A * i ) + |A * | + |B * | ≤ d − 50 + (i − 1)|A * i | + (k − i)|A * i | + 2kct ≤ d − 50 + kct + 2kct ≤ d − 45 .
The last inequality follows since
d − ≥ δ 0 (T ) ≥ Ck 2 log k by (G8). Since E A,i is dis The last inequality follows since E A,i \ X i = E A,i \ (A ∪ B ). Hence for all v ∈ E A,i \ X i we have d − T i (v)(10)≥ d − T (v) − 2k (G3) ≥ d − − 2k ≥ 19 20 d − and so |E A,i | ≤ d − 45 ≤ 1 40 min{d − T i (v) : v ∈ E A,i \ X i }.
This shows that condition (ii) of Lemma 6.7 is satisfied.The argument that (iv) holds is similar.We may therefore apply Lemma 6.7 to find a Hamilton cycle C i in T i as desired.


Highly-linked tournaments This shows that condition (ii) of Lemma 6.7 is satisfied.The argument that (iv) holds is similar.We may therefore apply Lemma 6.7 to find a Hamilton cycle C i in T i as desired.

Highly-linked tournaments are good
are good

The aim of this section is to prove that any sufficiently highly-linked tournament is (C, k, t, c)-good.We first show that it is very easy to find covering edges for any given vertex -we will use the following lemma to find matchings F 1 , . . ., F k consisting of covering edges a The aim of this section is to prove that any sufficiently highly-linked tournament is (C, k, t, c)-good.We first show that it is very easy to find covering edges for any given vertex -we will use the following lemma to find matchings F 1 , . . ., F k consisting of covering edges as in Definition 7.1.Lemma 8.1.Suppose that T is a strongly 2-connected tournament, and v ∈ V (T ).Then there exists a covering edge for v.
that T is a strongly 2-connected tournament, and v ∈ V (T ).Then there exists a covering e Proof.Since T is strongly connected and |T | > 1, we have N + (v), N − (v) = ∅.Since T − v is strongly connected, there is an edge xy from N − (v) to N + (v).But then xv, vy ∈ E(T ), so xy is a covering edge for v, as desired.
ing edge for v, as desired.

The next lemma will be used to obtain paths P 1 1 , . . ., P t k as in Definition 7.1.Reca The next lemma will be used to obtain paths P 1 1 , . . ., P t k as in Definition 7.1.Recall that we require F i ⊆ E(P t i ) and i ) and
(A ∪ B ) \ (A * i ∪ We will ensure the latter requirement by first covering (A∪B )\(A * i ∪B * i ) with few paths and then linking these paths together -hence the form of the lemma.
ent by first covering (A∪B )\(A * i ∪B * i ) with few paths and Lemma 8.2.Let s ∈ N, and let T be a digraph.Let x 1 , . . ., x k , y 1 , . . ., y k be distinct vertices of T , and let Q 1 , . . ., Q k be (possibly empty) path systems in T − {x 1 , . . ., x k , y 1 , . . ., y k } with E(Q i ) ∩ E(Q j ) = ∅ whenever i = j.Write (11) m Let s ∈ N, and let T be a digraph.Let x 1 , . . ., x k , y 1 , . . ., y k be distinct vertices of T , and let Q 1 , . . ., Q k be (possibly empty) path systems in T − {x 1 , . . ., x k , y 1 , . . ., y k } with E(Q i ) ∩ E(Q j ) = ∅ whenever i = j.Write (11) m
:= k + k i=1 |Q i | + k i=1 V (Q i ) ,
and suppose that T is 2sm-linked.Then there and suppose that T is 2sm-linked.Then there exist edge-disjoint paths P 1 , . . ., P k ⊆ T satisfying the following properties: [k]. (iii) V (P i ) ∩ V (P j ) ⊆ Proof.For all i ∈ [k], let a 1 i . . .b 1 i , . . ., a t i i . . .b t i i denote the paths in Q i .Let F ⊆ E(T ) denote the set of all those edges which form a path of length one in ∪ • • • ∪ Q k . Let T := T V (T ) \ k i=1 V (Q i ) ∪ k i=1 t i j=1 {a j i , b j i } − F. Note that E(T ) ∩ (E(Q 1 ) ∪ • • • ∪ E(Q k )) = ∅.
Define sets X 1 , . . ., X k of ordered Define sets X 1 , . . ., X k of ordered pairs of vertices of T by

irs of vertices of T by
X i := {(x
, a 1 i ), (b 1 i , a 2 i ), . . . , (b t i −1 i , a t i i ), (b t i i , y i )}, if Q i = ∅, {(x i , y i )} if Q i = ∅,
and let
X := X 1 ∪ • • • ∪ X k . Let := 2sm − 2s|X|. Since |V (T ) \ V (T )| + |F | ≤ |V (Q 1 ) ∪ • • • ∪ V (Q k )| and |X| = k + k i=1 |Q i |, it follows that 2 = 4s(m − |X|)(11)
= 4s
k i=1 V (Q i ) ≥ |V (T ) \ V (T )| + 2|F |.
Thus and let = 4s Thus by Proposition 4.7, T is 2s|X|-linked.We may therefore apply Lemma 4.8 to X in order to obtain, for each i ∈ [k], a path system P i whose paths link the pairs in X i and such that whenever i = j, we have E(P i ) ∩ E(P j ) = ∅ and V (P i ) ∩ V (P j ) consists exactly the vertices that lie in a pair in both X i and X j .Let P i be the path obtained from the union of all paths in P i and all paths in Q i .Then P 1 , . . ., P k are edge-disjoint paths satisfying (i)-(iv).
k], a path system P i whose paths link the pairs in X i and such that whenever i = j, we have E(P i ) ∩ E(P j ) = ∅ and V (P i ) ∩ V (P j ) consists exactly t ll paths in P i and all paths in Q i .Then P 1 , . . ., P k are edge-disjoint paths satisfying (i)-(iv).

The next lemma shows that given a vertex v in a tournament The next lemma shows that given a vertex v in a tournament T , we can find a small transitive subtournament whose head is v and which out-dominates almost all vertices of T .Lemma 8.3.Let T be a tournament on n vertices, let v ∈ V (T ), and suppose that c ∈ N satisfies 2 ≤ c ≤ log d − (v) − 1.Then there exist disjoint sets A, E ⊆ V (T ) such that the following properties hold: T , we can find a small transitive subtournament whose head is v and which out-dominates almost all vertices of T .Lemma 8.3.Let T be a tournament on n vertices, let v ∈ V (T ), and suppose that c ∈ N satisfies 2 ≤ c ≤ log d − (v) − 1.Then there exist disjoint sets A, E ⊆ V (T ) d:
(i) 2 ≤ |A| ≤ c and T [A] is a transitive tournament with head v. (ii) A out-dominates V (T ) \ (A ∪ E). (iii) |E| ≤ (1/2) c−1 The fact that the bound in (iii) depends on d − (v) is crucial: for instance, we can apply Lemma 8.3 with v being the vertex of lowest in-degree.Then (iii) implies that the 'exceptional set' |E| is much smaller than d − (v) ≤ d − (w) for any w ∈ E. So while w is not dominated by A directly, it is dominated by many vertices outside E. This will make it possible to cover E by paths whose endpoints lie outside E. (More formally, the lemma is used to ensure (G3), which in turn is used for (Q2) in the proof of Lemma 6.7).
his will make it possible to cover E by paths whose endpoints lie outside E. (More formally, he proof of Lemma 6.7).

Proof.Let v 1 := v.We will find A by repeatedly choosing vertices v 1 , . . ., v i such that Proof.Let v 1 := v.We will find A by repeatedly choosing vertices v 1 , . . ., v i such that the size of their common in-neighbourhood is minimised at each step.More precisely, let A 1 := {v 1 }.Suppose that for some i < c we have already found a set A i = {v 1 , . . ., v i } such that T [A i ] is a transitive tournament with head v 1 , and such that the common in-neighbourhood E i of v 1 , . . ., v i satisfies have already found a set A i = {v 1 , Note that these conditions are satisfied for i = 1.Moreover, note that E i is the set of all those vertices in T − A i which are not out-dominated by A i .If conditions are satisfied for i = 1.Moreover, note that E i is the set of all those vertices in T − A i which are not out-dominated by A i .If
|E i | < 4, then we have (12) |E i | < 4 = 1 2 log d − (v)−2 d − (v) ≤ 1 2 c−1 d − (v),
and so A i satisfies (i)-(iii).(Note that |A i | ≥ 2 since the assumptions imply that d − (v) ≥ 8.) Thus in this case we can take A := A i and E := E i .So suppose next that |E i | ≥ 4. In this case we will extend A i to A i+1 by adding a suitable vertex and so A i satisfies (i)-(iii).(Note that |A i | ≥ 2 since the assumptions imply that d − (v) ≥ 8.) Thus in this case we can take A := A i and E := E i .So suppose next that |E i | ≥ 4. In this case we will extend A i to A i+1 by adding a suitable vertex v i+1 .By Proposition 6.1, E i contains a vertex v i+1 of in-degree at most v i+1 .By Proposition 6.1, E i contains a vertex v i+1 of in-degree at most
|E i |/2 in T [E i ]. Let A i+1 {v 1 , . . . , v i+1 } and let E i+1 be the common in-neighbourhood th head v 1 and |E i+1 | ≤ 1 2 |E i | ≤ 1 2 i d − (v)
. By repeating this construction, either we will find |E i | < 4 for some i < c (and therefore take A := A i and E := E i ) or we will obtain sets A c and E c satisfying (i)-(iii).

We will also need the following analogue of Lemma 8.3 for in-dominating sets.It immediately follows from Lemma 8.3 by reversing the orientations of all edges.We will now apply Lemma 8.3 repeatedly to obtain many pairwise disjoint small almost-out-dominating sets.We will also prove an ana . By repeating this construction, either we will find |E i | < 4 for some i < c (and therefore take A := A i and E := E i ) or we will obtain sets A c and E c satisfying (i)-(iii).
We will also need the following analogue of Lemma 8.3 for in-dominating sets.It immediately follows from Lemma 8.3 by reversing the orientations of all edges.We will now apply Lemma 8.3 repeatedly to obtain many pairwise disjoint small almost-out-dominating sets.We will also prove an analogue for in-dominating sets.These lemmas will be used in order to obtain sets A 1 1 , . . ., A t k , B 1 1 , . . ., B t k , E A,1 , . . ., E A,k and E B,1 , . . ., E B,k as in Definition 7.1.Lemma 8.5.Let T be a tournament on n vertices, U ⊆ V (T ) and c ∈ N with c ≥ 2. Suppose that δ − (T ) ≥ 2 c+1 +c|U |.Then there exist families {A v : v ∈ U } and {E v : v ∈ U } of subsets of V (T ) such that the following properties hold: ogue for in-dominating sets.These lemmas will be used in order to obtain sets A 1 1 , . . ., A t k , B 1 1 , . . ., B t k , E A,1 , . . ., E A,k and E B,1 , . . ., E B,k as in Definition 7.1.Lemma 8.5.Let T be a tournament on n vertices, U ⊆ V (T ) and c ∈ N with c ≥ 2. Suppose that δ − (T ) ≥ 2 c+1 +c|U |.Then there exist families {A v : v ∈ U } and {E v : v ∈ U } of subsets of V (T ) such that the following propertie tes V (T ) \ (E v ∪ u∈U A u ) for all v ∈ U . (ii) T [A v ] is a transitive tournament with head v for all v ∈ U . (iii) |E v | ≤ (1/2) c−1 d − (v) for all v ∈ U . (iv) 2 ≤ |A v | ≤ c f A v = ∅ for all u = v.
Proof.We repeatedly apply Lemma 8.3.Suppose that for some U ⊆ U with U = U we have already found {A u : u ∈ U } and {E u : u ∈ U } satisfying (ii)-(vi) (with U playing the role of U and E u playing the role of E u ) such that (a)
A v out-dominates V (T ) \ ( u∈U Proof.We repeatedly apply Lemma 8.3.Suppose that for some U ⊆ U with U = U we have already found {A u : u ∈ U } and {E u : u ∈ U } satisfying (ii)-(vi) (with U playing the role of U and E u playing the role of E u ) such that (a) in A and B suitably, we may assume that A ∩ B = ∅.(Since n ≥ δ 0 (T ) ≥ 2kt, this is indeed possible.)Define Our first aim is to choose the sets A 1 1 , . . ., A t k using Lemma 8.5.Partition arbitrarily into sets A 1 , . . ., A k of size t, and write A i =: {a 1 i , . . ., a t i }.Since |B | = kt ≤ δ 0 (T )/2, we have ose the sets A 1 1 , . . ., A t k using Lemma 8 , . . ., A k of size t, and write A i =: {a 1 i , . . ., a t i }.Since |B | = kt ≤ δ 0 (T )/2, we have
d − := min{d − (v) : v ∈ V (T ) \ (A ∪ B )}, d + := min{d + (v) : v ∈ V (T ) (A ∪ B )}.2 c+1 + c|A| ≤ 400t + ckt ≤ C 2 k 2 log k ≤ δ − (T ) − |B | ≤ δ − (T − B ).
Thus we can apply Lemma 8.5 to T − B , A and c in order to obtain almost out-dominating sets A i a i and corresponding exceptional sets E A,i as in the statement of Lemma 8.5 (for all i ∈ [k] and all
∈ [t]). Write A * i := A 1 i ∪ • • • ∪ A t i and A * := A * 1 ∪ • • • ∪ A * k .
Let us now verify (G1).By Lemma 8.5(ii), (iv) and (vi), each T [A i ] is a transitive tournament with head a i , 2 ≤ |A i Thus we can apply Lemma 8.5 to T − B , A and c in order to obtain almost out-dominating sets A i a i and corresponding exceptional sets E A,i as in the statement of Lemma 8.5 (for all i ∈ [k] and all Let us now verify (G1).By Lemma 8.5(ii), (iv) and (vi), each T [A i ] is a transitive tournament with head a i , 2 ≤ |A i | ≤ c, and the sets A 1 1 , . . ., A t k are all disjoint.In particular, A = {h(A i ) : i ∈ [k], ∈ [t]}.We claim in addition that d + (a i ) ≥ 2n/5.Indeed, Proposition 6.2 implies that T has at most 4n/5+1 vertices of out-degree at most 2n/5, and hence at least n/5 − 1 vertices of outdegree at least 2n/5.Moreover, ≤ c, and the sets A 1 1 , . . ., A t k are all disjoint.In particular, A = {h(A i ) : i ∈ [k], ∈ [t]}.We claim in addition that d + (a i ) ≥ 2n/5.Indeed, Proposition 6.2 implies that T has at most 4n/5+1 vertices of out-degree at most 2n/5, So since the vertices of A were chosen to have minimal in-degree in T , it follows that d + (a i ) ≥ 2n/5 for all i ∈ [k] and all ∈ [t].Thus (G1) holds.We will next apply Lemma 8.6 in order to obtain the sets B 1 1 , . . ., B t k .To do this, we first partition B arbitrarily into sets B 1 , . . ., B k of size t, and write 8.6 in order to obtain the sets B 1 1 , . . ., B t k .To do this, we first partition B arbitrarily into sets B 1 , . . ., B k of size t, and write
B i =: {b 1 i , . . . , b t i }. Since |A * | ≤ ktc ≤ δ 0 (T )/2, we have 2 c+1 i and B * := B * 1 ∪ • • • ∪ B * k .
Similarly as befor Similarly as before one can show that (G2) holds.We now define the exceptional sets E A,i and E B,i .For all i ∈ ,i := (E 1 B,i ∪ • • • ∪ E t B,i ).
Recall from Lemmas 8.5(v) and 8.6
(v) that E A,i ∩A * = ∅ and E B,i ∩(A * ∪B * ) = ∅ for all i ∈ [k] and all ∈ [t]. Thus E A,i ∩ (A * i ∪ B * i ) = ∅ and E B,i ∩ (A * i ∪ B * i ) = ∅ for all i ∈ [k]
. By Lemma 8.5(i), each A i out-dominates V (T ) \ (A * ∪ B Recall from Lemmas 8.5(v) and 8.6 . By Lemma 8.5(i), each A i out-dominates V (T ) \ (A * ∪ B * ∪ E A,i ).
.
F i := {e v : v ∈ A * i ∪ B * i }.
Then the first part of (G7) holds.It remains to choose the paths P 1 1 , . . ., P t k .Recall from (G6) that we need to ensure that (A ∪ B ) \ (A * i ∪ B * i ) ⊆ V (P t i ) for all i ∈ [k].We could achieve this by incorporating each of these vertices using the high linkedness of T .However, since |A ∪ B | = 2kt, Then the first part of (G7) holds.It remains to choose the paths P 1 1 , . . ., P t k .Recall from (G6) that we need to ensure that (A ∪ B ) \ (A * i ∪ B * i ) ⊆ V (P t i ) for all i ∈ [k].We could achieve this by incorporating each of these vertices using the high linkedness of T .However, since |A ∪ B | = 2kt, a direct application of linkedness would require T to be Θ(k 3 )-linked.For each i ∈ [k], we will therefore first choose a path cover Q i of T [(A ∪ B ) \ (A * i ∪ B * i )] consisting of few paths and then use Lemma 8.2 (and thereby the high linkedness of T ) to incorporate these paths into P t i .This has the advantage that we will only need T to be Θ(k 2 log k)-linked.
nto P t i .This has the advantage that we will only need T to be Θ(k 2 log k)-linked.

Let us first choose the path covers
Q i of T [(A ∪ B ) \ (A * i ∪ B * i )]
. Suppose that for some Let us first choose the path covers . Suppose that for some j ∈ [k] we have already found path systems Q 1 , . . ., Q j−1 such that, for each i < j, Q i is a path cover of T [(A ∪ B ) \ (A * i ∪ B * i )] with |Q i | ≤ 2k, and such that for all i < i < j the paths in Q i are edge-disjoint from paths in Q i .To choose Q j , apply Corollary 6.For all i ∈ [k] and all ∈ [t − 1] let Q i := ∅.For all i ∈ [k] let Q t i be the path system consisting of all the edges in F i (each viewed as a path of length one) and all the paths in Q i .Let T := T − ((A * ∪ B * ) \ (A ∪ A ∪ B ∪ B )).Our aim is to apply Lemma 8.2 with s := 30 to T , the vertices b 1 1 , . . ., b t k , a 1 1 , . . ., a t k , and the path systems Q 1 1 , . . ., Q t k .To verify that T is sufficiently highly linked, let m be as defined in (11) and note that m = kt + 3 j ∈ [k] we have already found path systems Q 1 , . . ., Q j−1 such that, for each i < j, Q i is a path cover of T [(A ∪ B ) \ (A * i ∪ B * i )] with |Q i | ≤ 2k, and such that for all i < i < j the paths in Q i are edge-disjoint from paths in Q i .To choose Q j , apply Corollary 6.For all i ∈ [k] and all ∈ [t − 1] let Q i := ∅.For all i ∈ [k] let Q t i be the path system consisting of all the edges in F i (each viewed as a path of length one) and all the paths in Q i .Let T := T − ((A * ∪ B * ) \ (A ∪ A ∪ B ∪ B )).O b t k , a 1 1 , . . ., a t k , and the path systems Q 1 1 , . . ., Q t k .To verify that T is sufficiently highly linked, let m be as defined in (11) and note that m = kt + 3
k i=1 |F i | + k i=1 |Q i | + k i=1 V (Q i ) ≤ kt + 6ckt + 2kP i for all Q ∈ Q i . (iii) V (P i ) ∩ V (P m j ) ⊆ V (Q i ) ∩ V (Q m j
) for all (i, ) = (j, m).(iv) We have that
|P 1 1 ∪ • • • ∪ P t k | ≤ n 30 + 2 k i=1 |F i | + k i=1 V (Q i ) = n 30 + 2|A * ∪ B * | + |A ∪ B | ≤ n 30 + 4ckt + 2kt ≤ n 20 .
Condition (ii) implies that F i ⊆ P t i and (A∪B )\(
A * i ∪B * i ) = V (Q i ) ⊆ V (Q t i ) ⊆ V (P t i ) for all i ∈ [k]
. Thus (G6) holds.We now prove that (G5) holds.From (iii) and the fact that that V (Q i ) ∩ V (Q m i ) = ∅ for all i ∈ [k] ) for all (i, ) = (j, m).(iv) We have that Condition (ii) implies that F i ⊆ P t i and (A∪B )\( . Thus (G6) holds.We now prove that (G5) holds.From (iii) and the fact that that V (Q i ) ∩ V (Q m i ) = ∅ for all i ∈ [k], = m, it follows that P 1 i , . . ., P t i are vertex-disjoint for all i ∈ [k].Together with (i) and (iv) this implies that in order to check (G5), it remains to show that  Fix i ∈ [k], ∈ [t] and take j ∈ [k]\{i}.We have (A∪B )∩(A * i ∪B * i )∩V (Q i ) = ∅, and by (G6) we have (A ∪ B ) ∩ (A * i ∪ B * i ) ⊆ (A ∪ B ) \ (A * j ∪ B * j ) ⊆ V (P t j ).Applying (iii) to P i and P t j , it therefore follows that ( 17) t i are vertex-disjoint for all i ∈ [k].Together with (i) and (iv) this implies that in order to check (G5), it remains to show that  Fix i ∈ [k], ∈ [t] and take j ∈ [k]\{i}.We have (A∪B )∩(A * i ∪B * i )∩V (Q i ) = ∅, and by (G6) we have (A ∪ B ) ∩ (A * i ∪ B * i ) ⊆ (A ∪ B ) \ (A * j ∪ B * j ) ⊆ V (P t j ).Applying (iii) to P i and P t j , it therefore follows that ( 17)
V (P i ) ∩ (A ∪ B ) ∩ (A * (15)-( 17) now imply (14).Thus (G5) holds.
ds.

So it remains to check that the last part of (G7) holds too, i.e. that F act i ∩ E(P j ) = ∅ for all i, j ∈ [k] and all ∈ [t].Consider any covering edge e v = x v y v ∈ F i .Then (G6) implies that x v and y v are contained in P t i .Moreover, (iii) implies that V (P t i ) ∩ V (P j ) ⊆ V (Q t i ) ∩ V (Q j ) ⊆ A ∪ B whenever (i, t) = (j, ).Since x v , y v / ∈ A ∪ B , this shows that x v v, vy v / ∈ E(P j ) whenever (i, t) = (j, ).But since e v ∈ E(P t i ), we also have x v v, vy v / ∈ E(P t i ).This completes the proof that T is (C, k, t, c)-good.9. Concluding remarks 9.1.Eliminating the logarithmic factor.A natural approach to improve the bound in Theorem 1.2 would be to reduce the param So it remains to check that the last part of (G7) holds too, i.e. that F act i ∩ E(P j ) = ∅ for all i, j ∈ [k] and all ∈ [t].Consider any covering edge e v = x v y v ∈ F i .Then (G6) implies that x v and y v are contained in P t i .Moreover, (iii) implies that V (P t i ) ∩ V (P j ) ⊆ V (Q t i ) ∩ V (Q j ) ⊆ A ∪ B whenever (i, t) = (j, ).Since x v , y v / ∈ A ∪ B , this shows that x v v, vy v / ∈ E(P j ) whenever (i, t) = (j, ).But since e v ∈ E(P t i ), we also have x v v, vy v / ∈ E(P t i ).This completes the proof that T is (C, k, t, c)-good.9. Concluding remarks 9.1.Eliminating the logarithmic factor.A natural approach to improve the bound in Theorem 1.2 would be to reduce the parameter c, i.e. to consider smaller 'almost dominating' sets.In particular, if we could choose c independent of k, then we would obtain the (conjectured) optimal bound of Θ(k 2 ) for the linkedness.The obstacle to this in our argument is given by ( 13), which requires that c has a logarithmic dependence on k. 9.2.Algorithmic aspects.As remarked in the introduction, the proof of Theorem 1.2 is algorithmic.Indeed, when we apply the assumption of high linkedness to find appropriate paths in the proof of Lemma 8.7 (via Lemma 8.2), we can make use of the main result of [11] that these can be found in polynomial time.Moreover, the proof of the Gallai-Milgram theorem (Theorem 6.3) is also algorithmic (see [9]).These are the only tools we need in the proof, and the proof itself immediately translates into a polynomial time algorithm.
we would obtain the (conjectured) optimal bound of Θ(k 2 ) for the linkedness.The obstacle to this in our argument is given by ( 13), which requires that c has a logarithmic dependence on k. 9.2.Algorithmic aspects.As remarked in the introducti paths in the proof of Lemma 8.7 (via Lemm itself immediately translates into a polynomial time algorithm.



x is a vertex of a digraph D, then N + D (x) denotes the out-neighbourhood of x, i.e. the set of all those vertices y for which xy ∈ E(D).Similarly, N − D (x) denotes the in-neighbourhood of x, i.e. the set of all those vertices y f x is a vertex of a digraph D, then N + D (x) denotes the out-neighbourhood of x, i.e. the set of all those vertices y for which xy ∈ E(D).Similarly, N − D (x) denotes the in-neighbourhood of x, i.e. the set of all those vertices y for which yx ∈ E(D).We write d + D (x) := |N + D (x)| for the out-degree of x and d − D (x) := |N − D (x)| for its in-degree.We denote the minimum out-degree of D by δ + (D) := min{d + D (x) : x ∈ V (D)} and the maximum out-degree of D by ∆ + (D) := max{d + D (x) : x ∈ V (D)}.We define the minimum in-degree δ − (D) and the maximum in-degree ∆ − (D) similarly.The minimum degree of D is defined by δ(D) := min{d + D (x) + d − D (x) : x ∈ V (D)} and its minimum semi-degree by δ 0 (D) := min{δ + (D), δ − (D)}.Whenever X, Y ⊆ V (D) are disjoint, we write e D (X) for the number of edges of D having both endvertices in X, and e D (X, Y ) for the number of edges of D with tail in X and head in Y .We write N + D (X) := x∈X N + D (x) and define N − D (X) similarly.In all these definitions we often omit the subscript D if the digraph D is clear from the context.
x) := |N : x ∈ V (D)} and the maximum out-degree of D by ∆ + (D) := max{d + D (x) : x ∈ V (D)}.We define the minimum in-degree δ − (D) and the maximum in-degree ∆ − (D) similarly.The minimum degree of D is defined by δ(D) := min{d + D (x) + d − D (x) : x ∈ V (D)} and its mini ver X, Y ⊆ V (D) are disjoint, we write e D (X) for the number of edges of D having both endvertices in X and e D (X, Y ) for the number of edges of D with tail in X and head in Y .We write N + D (X) := x∈X N + D (x) and define N − D (X) similarly.In all these definitions we often omit the subscript D if the digraph D is clear from the context.


Figure 1 .
1
Figure 1.Illustrating the paths Qi and Pi as well as the edges linking them up via the linked domination structure.


Figure 2 .
2
Figure 2. Illustrating our construction of a digraph D which corresponds to a sorting network for k = 4. D is used to link xi to yi.In the notation of the proof of Theore

Figure 1 .
Figure 1.Illustrating the paths Qi and Pi as well as the edges linking them up via the linked domination structure.

Figure 2 .
Figure 2. Illustrating our construction of a digraph D which corresponds to a sorting network for k = 4. D is used to link xi to yi.In the notation of the proof of Theorem 1.3, we have π(3) = 1.
1.3, we have π(3) = 1.


Theorem 4 . 1 .
41
Let C := 3050 and k ∈ N be such that k ≥ 2. Then there exist r ≤ C k log k and a sequence of comparisons c 1 , . . ., c r satisfying the following property: for any initial assignment of k values to k registers, applying the comparisons in sequence results in register R i being assigned the value i for all i ∈ [k].




b 1 and b 2 be two out-neighbours of b in H. Now the vertices a 1 , a 2 , b, b 1 , b 2 wi

Theorem 4 . 1 .
Let C := 3050 and k ∈ N be such that k ≥ 2. Then there exist r ≤ C k log k and a sequence of comparisons c 1 , . . ., c r satisfying the following property: for any initial assignment of k values to k registers, applying the comparisons in sequence results in register R i being assigned the value i for all i ∈ [k].
b 1 and b 2 be two out-neighbours of b in H. Now the vertices a 1 , a 2 , b, b 1 , b 2 with suitably chosen edges from T form an (a 1 , a 2 )-switch (with terminal vertices b 1 and b 2 ).
h suitably chosen edges from T form an (a 1 , a 2 )-switch (with terminal vertices b 1 and b 2 ).


Theorem 4 . 4 (
44
Menger's Theorem).Suppose D is a strongly k-connected digraph with A, B ⊆ V (D) and |A|, |B| ≥ k.Then there exist k vertex-disjoint paths in D each starting in A and ending in B.


Proposition 4 . 7 .
47
Let k, ∈ N with < k, and let D be a




, a 2 }, B := {b 0 , . . ., b 2 }, and C m := {c 1 , . . ., c m }.The edges of T m, are defined as follows: T m, [A ] and T m, [B ] are isomorphic to T (with the natural

Theorem 4 . 4 (
Menger's Theorem).Suppose D is a strongly k-connected digraph with A, B ⊆ V (D) and |A|, |B| ≥ k.Then there exist k vertex-disjoint paths in D each starting in A and ending in B.

Proposition 4 . 7 .
Let k, ∈ N with < k, and let D be a , a 2 }, B := {b 0 , . . ., b 2 }, and C m := {c 1 , . . ., c m }.The edges of T m, are defined as follows: T m, [A ] and T m, [B ] are isomorphic to T (with the natural labelling of vertices), and T [C m ]

4 .
strongly connected.Write A , B and C m respectively for A \ S, B \ S, and C m \ S. Note that there is at least one edge of T m, − S from B to A , which we may assume by symmetry to be b 0 a 0 .Ordering the vertices of T m, as a 0 , . . ., a 2 , c 1 , . . ., c m , b 1 , . . ., b 2 , b 0 and removing the vertices of S from this ordering gives a Hamilton cycle in T m, − S. Thus T m, − S must be strongly connected.This completes the proof of Claim 1. Claim 2. Let m, ∈ N be such that m > √ 4 .Then for every r-regular spanning subdigraph D ⊆ T m, we have r ≤ √ Suppose for a contradiction that D ⊆ T m, is an r-regular spanning subdigraph with r := √ 4 +1 > √ 4 .Since D is regular, we have e D (A , Ā ) = e D ( Ā , A ), where Ā := V (D)\A .Noting that r ≤ m, consider the first r vertices c 1

Figure 3 .
Figure 3.Our linked domination structure and path cover at the beginning of the proof of Lemma 7.2.

a t }, A :
∪A∪B to form a new path cover Q 5 of T which still satisfies (Q1)-(Q5), and such that no path in Q 5 \P 2 has endpoints in A ∪B ∪B.Clearly no paths in P 2 ⊇ Q 4 ∩ P 2 have tails in I, and F ⊆ E(Q 4 ) by (Q1).By condition (iii) we have d − T (v) ≥ n/3 for all v ∈ I. Together with (2) this implies that d − T (v) ≥ d − T (v) ≥ n/4 for all v ∈ I.Note also that |V (P 2 )| ≤ n/20 by condition (v).So similarly as in (

1
contains a Hamilton cycle C i which satisfies (a) and (b) (with i replaced by i + 1).
Note that d − (a) ≤ d − for all a ∈ A and d + (b) ≤ d + for all b ∈ B .

Lemma 8 . 5 (
iii) and the fact that a i ∈ A together imply that1 d − (a i ) ≤ t 2 c−1 d − ≤ d − 50, so (G3) holds.Similarly, by Lemma 8.6(i), eachB i in-dominates V (T ) \ (A * ∪ B * ∪ E B,i), and as in(13) one can show that |E B,i | ≤ d + /50.Thus (G4) holds.We now use Lemma 8.1 in order to define the sets F 1 , . . ., F k of covering Recall from (G7) that we requireF 1 ∪ • • • ∪ F k to be a matching in T − (A * ∪ B * ).Suppose that for some (possibly empty) subset V A * ∪ B * we have defined a set {e v : v ∈ V } of independent edges in T − (A * ∪ B * ) such that e v is a covering edge for v and e v = e v whenever v = v .Pick any vertex v ∈ (A * ∪ B * ) \ V .We will next define e v .Let T be the tournament obtained from T by deleting (A * ∪ B * ) \ {v} as well as the endvertices of the covering edges e v for all v ∈ V .Then|V (T ) \ V (T )| ≤ |A * ∪ B * | + 2|A * ∪ B * | ≤ 3ktc ≤ C 2 k 2 log k,so by Proposition 4.7, T is still (Ck 2 log k/2)-linked and hence strongly 2connected.We may therefore apply Lemma 8.1 to find a covering edge e v for v in T .Continue in this way until we have chosen e v for each v ∈ A * ∪ B * and let 4 to the oriented graph T obtained fromT [(A ∪ B ) \ (A * j ∪ B * j )] by deleting the edges of all the paths in Q 1 , . . ., Q j−1 .Since δ(T ) ≥ |T |−1−2(j−1) ≥ |T |−2k, Corollary 6.4 ensures that |Q j | ≤ 2k.We will now choose P 1 1 , . . ., P t k .For each i ∈ [k] and each∈ [t], let a i denote the tail of T [A i ] and b i the head of T [B i ].Let A := {a i : i ∈ [k], ∈ [t]} and B := {b i : i ∈ [k], ∈ [t]}.

that
Thus we can apply Lemma 8.6 to T − A * , B and c in order to obtain almost in-dominating sets B i b i and corresponding exceptional sets E B,i as in the statement of Lemma 8.6 (for all i ∈ [k] and all ∈ [t]).Write B * i 2 + |A ∪ B | Together with the fact that |T | − |T | ≤ 2ckt and Proposition 4.7 this implies that T is 2 • 30m-linked.So we indeed apply Lemma 8.2 to find edgedisjoint paths P i in T (for all i ∈ [k] and all ∈ [t]) satisfying the following properties: (i) P i is a path from b i to a i .(ii) Q ⊆