Blow-ups and extensions of trees in tournaments
Abstract
A class of acyclic digraphs is linearly unavoidable if there exists a constant such that every digraph is contained in all tournaments of order . The class of all acyclic digraphs is not linearly avoidable, and Fox, He, and Widgerson recently showed that this is not even the case for acyclic digraphs with bounded maximum degree. On the positive side, Thomason and Häggkvist proved that the class of oriented trees is linearly unavoidable. In this work, we generalize this result to acyclic digraphs obtained from an oriented tree by adding at most vertices, and -blow-ups of oriented trees, for every fixed integer .
More precisely, we show that if is obtained from an oriented tree of order by adding universal vertices, then is contained in every tournament of order ; and if is obtained from by replacing each vertex by an independent set of size and every arc by all possible arcs from to , then is contained in every tournament of order .
1 Introduction
A digraph is -unavoidable, for an integer , if it is contained in every tournament of order . For a digraph , the unavoidability of , denoted by , is the smallest integer such that is -unavoidable, if such a exists. It is easy and well-known that a digraph is -unavoidable for some integer if and only if it is acyclic. Indeed, every digraph containing a directed cycle is not contained in any transitive tournament, and so is not unavoidable. Moreover, as observed by Erdős and Moser [8], an immediate induction and Proposition 1 shows that the transitive tournament of order is -unavoidable and thus every acyclic digraph of order is -unavoidable.
Proposition 1.
Let be an acyclic digraph. For every source or sink of ,
This is somehow tight in the sense that any upper bound on the unavoidability of acyclic digraphs as a function of their order must be exponential. Indeed, considering a random tournament and using the First Moment Method, Erdős and Moser [8] established . Since this seminal paper, only little progress has been made on the unavoidability of transitive tournaments. Using the Local Lemma, one can slightly improve on the lower bound and show that when is large enough. The unavoidability of small transitive tournaments have been established. Clearly , , (because of the directed -cycle), and (because of the Paley tournament on vertices). Reid and Parker [22] showed , , and Sanchez-Flores [25] proved . This result and Proposition 1 imply for all .
The average degree of a nonempty digraph is the average degree of its underlying graph, that is . The maximum average degree of a digraph is the maximum average degree over all its nonempty subdigraphs. Similarly to Erdős and Moser [8], one can show that the unavoidability of an acyclic digraph is exponential in its maximum average degree.
Proposition 2.
Let be an acyclic digraph with maximum average degree . Then .
Proof.
Consider a random tournament on vertices. Let be a labelling of the vertices of a subdigraph of with average degree . For each ordered -uple of distinct vertices in , the probability that contains a labelled copy of (i.e. such that is an arc in whenever is an arc in ) is . Thus the expected number of labelled copies of is . ∎
Proposition 2 gives a non-trivial lower bound on the unavoidability of dense acyclic digraphs, that is with maximum average degree larger than logarithmic in their order. However, smaller bounds on the unavoidability are expected for sparse digraphs. In particular, one can ask which families of digraphs are linearly unavoidable, that is such that there exists a constant such that for every digraph in , and which are polynomially unavoidable, that is such that there exists a polynomial such that for every digraph in .
Note that by Proposition 2, the digraphs of a linearly unavoidable family must have maximum average degree at most . For every integer with , Linial, Saks and Sós [19] constructed an -unavoidable digraph with vertices and arcs for some fixed constant . Those digraphs form a linearly unavoidable family with logarithmic maximum average degree.
Several other classes of acyclic digraphs have been proved to be linearly unavoidable. Much attention has been devoted to oriented paths, oriented trees, and oriented forests which are orientations of paths, trees, and forests respectively. It started with Rédei’s theorem [21] which states that the unavoidability of , the directed path on vertices, is exactly .
Theorem 3 (Rédei [21]).
Every tournament has a Hamiltonian directed path.
In 1971, Rosenfeld [24] conjectured that there is an integer such that for every oriented path of order at least . Several papers gave partial answers to this conjecture [11, 1, 9, 26] until Rosenfeld’s conjecture was verified by Thomason, who proved in [27] Rosenfeld’s conjecture for . Finally, Havet and Thomassé [15] showed that for every oriented path except the antidirected paths of order , , and .
Regarding oriented trees, Sumner (see [23]) made the following celebrated conjecture.
Conjecture 4 (Sumner).
Every oriented tree of order is -unavoidable.
Since every forest is the subdigraph of a tree, this conjecture is equivalent to the analogue for oriented forests. If true, Sumner’s conjecture would be tight. Indeed, the out-star , which is the digraph on vertices consisting of a vertex dominating the others, is not contained in the regular tournaments of order . The first linear upper bound was given by Häggkvist and Thomason [12]. Following improvements of Havet [13], Havet and Thomassé [14], and El Sahili [7], Dross and Havet [6] used the notion of median order to prove that every oriented tree of order is -unavoidable. Kühn, Mycroft and Osthus [16] proved that Sumner’s conjecture is true for all sufficiently large .
Theorem 5 (Kühn, Mycroft, and Osthus [16]).
There is an integer such that for every , every oriented tree of order is -unavoidable.
Motivated by those results and a work of Lee [18], Bucić, Letzter and Sudakov [3] asked whether the digraphs with bounded maximum average degree are linearly unavoidable. Draganić et al. [5] proved that this is the case for -th powers of directed paths. The -th power of a digraph is the digraph, denoted by , with vertex set in which in an arc if and only if . Fox, He, and Widgerson [10] showed that this is not the case in general: they proved that for any and any sufficiently large , there exists an acyclic digraph with vertices and maximum degree such that
This result raises the two following questions. Firstly, are the digraphs with bounded maximum average degree polynomially unavoidable?
Problem 6.
Let be a positive real number. Does there exist a polynomial such that for every digraph with maximum average degree at most ?
Secondly, which families of acyclic digraphs (with bounded maximum average degree) are linearly unavoidable? In this direction, Fox, He, and Widgerson [10] showed that for every acyclic digraph of maximum degree , if there is a partition of such that for every arc of , there exists such that and , then .
Our results
In this paper, we give some more partial answers to the above second question by proving some new families of acyclic digraphs to be linearly unavoidable. We are in particular interested in operations to form linearly unavoidable families from others.
Let be a positive integer. The -blow-up of a digraph , denoted by , is obtained by blowing-up each vertex into an independent set of size . Formally, is defined by
We believe that if a family is linearly unavoidable, then the family of its -blow-ups is also linearly unavoidable.
Conjecture 7.
If is linearly unavoidable and has bounded maximum average degree, then is also linearly unavoidable.
Observe first that the bounded maximum average degree condition in the above conjecture is necessary. Indeed, for every , let be the digraph of order which is the disjoint union of a transitive tournament of order and isolated vertices. Each is clearly -unavoidable, but has maximum average degree and so by Proposition 2.
Observe that the -blow-up of the directed path of order is contained in the -th power of the directed path of order . Thus, the result of Draganić et al. [5] implies Conjecture 7 for directed paths. As a more general evidence to this conjecture, we show in Section 3 that it holds for oriented trees (and thus more generally oriented forests). Precisely, we prove in Theorem 18 that the -blow-up of an oriented tree of order is -unavoidable. This upper bound is certainly not tight. However, the maximum average degree of the -blow-up of a tree of order is . Therefore, by Proposition 2, the optimal bound is of the form . Hence we are left with the following question.
Problem 8.
What is the infimum of all the constants such that for every large enough integer , the -blow-up of every oriented tree of order is -unavoidable?
Generalizing Proposition 1, we believe that adding a vertex cannot affect too much the unavoidability, in a sense that it can at most multiply it by a constant factor.
Conjecture 9.
There exists a constant such that for every acyclic digraph and for every vertex of .
Let be an acyclic digraph and let be a positive integer. An acyclic digraph is a -extension of if there exists a set of vertices such that . The set is the added set and its vertices are the added vertices. A vertex is universal in a digraph if . A -extension of is full, if its added vertices are universal in . A -extension of is a -twin extension, if all added vertices have the same in- and out-neighbourhood in . Conjecture 9, together with Conjecture 4, implies the following.
Conjecture 10.
There is an absolute constant such that for every integer , every -extension of an oriented tree of order is -unavoidable.
More generally, Conjecture 9 would imply the following.
Conjecture 11.
If is linearly unavoidable, then the family of -extensions of digraphs in is also linearly unavoidable.
In Section 4, we prove this last conjecture for the family of oriented trees (and thus also of oriented forest). More precisely; we show in Corollary 25 that every -extension of a sufficiently large tree is -unavoidable.
This upper bound on the unavoidability of extension of forests is certainly not tight. In Proposition 28, we show that the unavoidability of some -extensions of the arcless digraph on vertices is as . Hence we are left with the following analogue to Problem 8.
Problem 12.
What is the minimum function such that every -extension of every forest of order is -unavoidable?
The above-mentioned results show that the function is at least linear and at most quadratic. The same question applies to any subfamily, starting with that of empty digraphs. Generally, determining the precise value or good approximations of the unavoidability of -extensions of oriented trees would be interesting. In Section 5, we show an almost tight upper bound on the unavoidability of full -twin extensions and the exact unavoidability of the full -extensions of directed paths.
2 Preliminaries
2.1 Basic properties of tournaments
We will make use of the following folklore properties of tournaments.
Lemma 13 ([2, Proposition 2.2.2]).
Let be a positive integer. Every tournament has at most vertices of out-degree less than .
The out-section of a vertex in a tournament , denoted by or simply , is the set of vertices that can be reached from by a directed path. Similarly, the in-section of a vertex in a tournament , denoted by or simply is the set of vertices from which can be reached by a directed path. We set and . An out-generator (resp. in-generator) of is a vertex such that (resp. ).
Lemma 14 ([2, Proposition 2.7.6]).
Let be a tournament and a vertex of . In , is the initial (resp. terminal) vertex of a directed path of order (resp. ). In particular, if is an out-generator (resp. in-generator) of , then is the initial vertex (resp. terminal vertex) of a directed Hamiltonian path.
Let be a tournament, and and be two disjoint sets of vertices of . If every vertex of dominates every vertex of , then we write .
Lemma 15.
Let be a tournament and let be the set of vertices that are initial vertices of a directed path of order in . Then .
Proof.
Suppose for contradiction that there exists and such that . Since , there is a path of order in with initial vertex . Hence and so is the initial vertex of a directed path of order by Lemma 14. ∎
2.2 Median order
The notion of median order has been introduced in [14], see also [2, Chapter 2] and [5]. A median order of a digraph is a linear order of its vertex set such that , the number of forward arcs (i.e. directed from left to right), is as large as possible. Let us first note some basic properties of median orders of tournaments.
Lemma 16.
Let be a tournament and a median order of . Then, for any two indices with :
-
(M1)
the suborder is a median order of the induced subtournament ;
-
(M2)
dominates at least half of the vertices , and is dominated by at least half of the vertices , in particular dominates ;
-
(M3)
if , then when , and when ;
-
(M4)
if , then there exists such that ; moreover, if , then both and are median orders of ;
-
(M5)
if is a median order of , then is also a median order of .
Proof.
(M3) Assume for a contradiction that but there is an arc from to . By (M2) there is exactly one such arc and its tail belongs to . If , then has two out-neighbours in , contradicting (M2). If , then has less forward arcs than , a contradiction to (M1). By directional duality, we also have .
(M4) Suppose that . By (M2), dominates at least half of , and so more than half of . Symmetrically, is dominated by more than half of . Hence there exists such that . Now suppose that this is unique. Hence dominates exactly vertices among . It follows that the number of forward arcs in is , which is maximum since is a median order. This proves that is a median order. Symmetrically, is a median order too. ∎
2.3 A Kővári-Sós-Turán-like lemma
The following lemma is inspired by the classical result by Kővári, Sós, and Turán [17], and a similar one was recently used in [5].
Lemma 17.
Let be two positive reals and a positive integer. Let be a bipartite graph with and , such that for every . Then there exist and such that
-
1.
,
-
2.
, and
-
3.
is complete to .
Proof.
Suppose for contradiction that the result does not hold. We will count by two different methods the number of pairs with of size and such that . On the first hand, for every of size , there are less than vertices in such that . We thus have
On the other hand, by Jensen’s inequality, we have
This contradiction proves the lemma. ∎
Note that satisfies the hypothesis of Lemma 17. Indeed, simple computations show that is a non-increasing function of , and as we have
3 Blow-ups of oriented trees
In this section, we show that for every fixed integer , the class of -blow-ups of oriented trees is linearly avoidable.
Theorem 18.
Let be an oriented tree of order and a positive integer. Every tournament on at least vertices contains .
Proof.
We root on an arbitrary vertex . Let , and . Observe that we have , , and . Let (using and ) and let be a tournament of order . We will show that has a subdigraph isomorphic to . Let be a median order of . We partition into intervals for . We will find a copy of in iteratively, by adding at each step the out-children (i.e. its children which are out-neighbours) or the in-children (i.e. its children which are in-neighbours) of a node in . Let be a sequence of subtrees of such that
-
is the singleton ,
-
for every , is obtained from by adding all the out-children in of a vertex without any out-children in ,
-
for every , is obtained from by adding all the in-children in of a vertex without any in-children in , and
-
.
Observe that such a sequence exists with . Moreover, we say that a vertex in is fully processed if all its children in are also in , half processed if all its out-children in are in . Otherwise we say that is not processed. We prove the following property by induction:
For every , there is an injection and for every there exists a set such that
-
if is not processed,
-
if is half processed,
-
if is fully processed,
-
if is an arc in , then every vertex in dominates , and
-
for every a fraction of at most of the elements of the interval (respectively ) is in . More precisely,
Observe that, for , setting and taking for any set in of size satisfies the invariant. Note also that, for , the invariant implies the theorem. Now suppose that the property holds for some . Let and witnessing that the property holds at step . Let be the vertex in such that is obtained from by adding the out-children of in , or the in-children of in . Our goal is to find a new value for , and values for for every out-children (respectively in-children) of .
- Case 1:
-
is obtained from by adding the out-children of in .
Let and . Note that . We first justify that every vertex in dominates at least vertices in . To this purpose, let be any vertex in , so . Let be the set of vertices , which is disjoint from ( is empty if ). Observe that
Since , and because is a median order, by (M2) we have . Therefore, we obtain
This shows that every vertex in dominates at least vertices in , and so does every vertex in . By Lemma 17 applied for , there exist and such that every vertex in dominates , , and . We set .
Let be the set of indices such that and . In what follows we show that is large enough, which will enable us to find for every out-child of in .
Claim 1.
Proof of claim. Suppose that this does not hold, so . We define and as follows:
Observe that by definition. By the induction hypothesis, . Since by assumption, we obtain . This leads to the following inequalities, in which we also use
This is a contradiction since .
We can now extend to the out-children of by taking values in , and then we can arbitrarily take for every out-child of . Recall that . The invariant still holds because we add vertices at the end and at the beginning of the median order:
-
For every , .
-
For every , .
-
For every , .
-
For every , .
-
For every , .
-
- Case 2:
-
is obtained from by adding the in-children of in .
Let and . Since is half processed, we have . Moreover and every vertex in is dominated by at least vertices in . Using Lemma 17, we deduce that there are sets with , and . We set . We now find for every in-children of in . With a proof almost identical to the one of Claim 1 we deduce the following claim.
Claim 2.
There is a set of size at least such that for every and for every .
Then we build for every in-children of as in the first case. This concludes the proof of the theorem.∎
4 Extensions of oriented trees
In this section, we show that for every fixed integer , the class of -extensions of oriented trees is linearly unavoidable. Our method is based on the fact that the number of copies of in any -vertex tournaments is in . We start with the case , which is a good warm-up for the general case.
Lemma 19 (Folklore).
Every tournament of order contains at least copies of .
Proof.
Let be a tournament of order . Let be the number of directed triangles in and let be the number of in . The number of copies of (the directed path of order ) in is . But the number of is also . We deduce that and so . ∎
We will also need the following well-known lemma. Recall that a hypergraph is a pair where is a finite set and is a subset of . We call the elements of the vertices of , and the elements of the hyperedges of . An hypergraph is a subhypergraph of , and we write , if and .
Lemma 20 (Folklore).
Let be a hypergraph on vertices and with hyperedges. There is a hypergraph such that every vertex of belongs to at least hyperedges of .
Proof.
We proceed by induction on . If every vertex in is in at least hyperedges we are done. Otherwise there is a vertex in less than hyperedges, then has hyperedges and vertices. Hence, by the induction hypothesis, has a subhypergraph with minimum degree at least as wanted. ∎
Given an acyclic digraph , we denote by the maximum of over all the subdigraphs of .
Theorem 21.
Let be a -extension of an oriented tree on vertices, then .
By applying Theorem 5 to bound , we obtain the following.
Corollary 22.
There exists a constant such that the following holds. Let be a -extension of an oriented tree on at least vertices, then .
An embedding of a digraph in a digraph is an injective function such that for every . Observe that contains a copy of if and only if there is an embedding of in .
Proof of Theorem 21.
Let be a -extension of with added vertex . We assume that is a full -extension of , for otherwise we just add missing arcs. Let be a tournament on vertices. By Lemma 19, the number of in is at least . We deduce that there is a vertex which is the middle vertex (i.e. the vertex with in- and out-degree ) of at least copies of in . In other words there are at least arcs from to . Consider the bipartite graph where . By Lemma 20, has a nonempty subgraph with minimum degree at least . Let . Since is bipartite, note that in particular .
We will now inductively build and embedding such that
-
•
for every , , and
-
•
for every , .
Extending such an embedding of by setting yields the desired embedding of in .
If is empty, then the result is clear. Suppose now that is not empty and consider the tree obtained from by contracting every connected component of and into a single vertex . Consider a leaf of , and without loss of generality, assume that . If , then , and we can embed in since . Otherwise, assume by induction that we have an embedding with and . Let be the arc joining to in . Observe that we have and , for otherwise is a directed triangle in , a contradiction to being acyclic.
As has minimum degree at least , and because , the size of is at least , hence there are at least vertices in . Thus we can extend to the vertices in in , and this gives the desired embedding of . See Figure 1 for an illustration. ∎
We now move to the general case. We first prove an analogue of Lemma 19 for larger values of .
Lemma 23.
Let be a positive integer and let be a tournament on vertices. Then the number of distinct copies of in is at least .
Proof.
We proceed by induction on . For the result is clear. Now suppose and that the result holds for smaller values of . By Proposition 13, in there are at least vertices of out-degree at least . For every such vertex , there are by induction at least copies of in the subtournament induced by . Together with , this gives as many copies of in . We conclude that the number of copies of in is at least . ∎
This bound could be improved to , see [4].
Theorem 24.
Let be integers with and let be an oriented tree of order . Let be a -extension of , then .
Again, applying Theorem 5 to bound , we deduce the following corollary.
Corollary 25.
There is a constant such that the following holds. Let be a -extension of an oriented tree on at least vertices, then .
Proof of Theorem 24.
Let be a tournament on vertices.
Let be a copy of with acyclic ordering . By Lemma 23, contains at least copies of . By the Pigeon-hole Principle, there are vertices in such that there are at least embeddings of in with mapped to for every . From now on, let us say that an embedding is valid if for every .
We consider the hypergraph with vertex set which contains as a hyperedge for every valid embedding . By Lemma 20 applied to , there is a set of vertices such that every vertex in has degree at least in . Let be the partition of where contains all the vertices such that for some valid embedding .
For all with and for every vertex , we claim that has at least out-neighbours in . Assume that this is not the case, then the degree of in , which is exactly the number of valid embedding with the extra conditions and for every , is at most
a contradiction to the choice of . Similarly, if then has at least in-neighbours in . Since , this implies in particular that for every .
Let be a full -extension of . As is acyclic, let be a topological ordering of . Let be such that and the ordering of with respect to . We let be the partition of where contains every vertex between and in , for every , contains every vertex before in , and contains every vertex after in . By construction, every arc between and in is from to if . We will now construct by induction on an embedding such that for every . By setting for every , we will get the desired embedding of in .
If is empty, then the result is clear. Now assume . Consider the tree obtained from by contracting every connected component of into a single vertex , for every . Let be a leaf of and be the index such that . If , then, as , there is an embedding of into and we are done. Otherwise let be the unique neighbour in of a vertex in which does not belong to . Assume that is the tail of the (unique) arc between and , the other case being symmetric. Let be the index such that . Recall that, by construction, we have . By induction hypothesis, there is an embedding such that for every . We know that has at least in-neighbours in . Moreover and so (by definition of ). It follows that can be embedded in . This gives the desired embedding of and concludes the proof of the theorem. ∎
5 Particular extensions of some particular forests
5.1 Twin extensions of arcless digraphs
For every positive integers , let be full -twin-extension of the arcless digraph of order in which the added vertices have the same in-neighbourhood of size and the same out-neighbourhood of size (with a partition of ). Note that is simply the oriented star made of a vertex dominated by vertices and dominating vertices.
Proposition 26.
For every positive integers with , each of the following holds
-
1.
, and
-
2.
.
Proof.
The fact that and follows from Lemma 13. For the lower bounds, consider first any -regular tournament . Then has vertices and does not contain . Thus . Similarly, if is an -regular tournament and an -regular tournament, then is a tournament of order which does not contain , implying that . ∎
To prove a lower bound on , we will use a probabilistic argument relying on Chernoff’s bound.
Proposition 27 (Chernoff’s Bound).
If is a random variable following a binomial law with parameters and , then for every
Proposition 28.
For every fixed , as .
Proof.
The added vertices of are sources, so by Proposition 1, .
To prove the lower bound, let and let be a tournament of order taken uniformly at random. For every of size , let be the number of common out-neighbours of the vertices in , that is . Then follows a binomial law with parameters and . By Chernoff’s bound (Proposition 27),
It follows by the Union Bound that for large enough. This proves that for large enough. ∎
We will now prove a similar upper bound for . To do so, we will use Ramsey’s theorem, that we first recall here.
Theorem 29 (Ramsey’s theorem [20]).
For every positive integers , and , there exists such that for every -colouring of the -subsets of , either there is a set of size such that every -subset of is coloured , or there is a set of size such that every -subset of is coloured .
Theorem 30.
Let and let be a positive integer. There is a constant depending only on and such that .
Proof.
For fixed constants , simple computations show the existence of constants such that for every integer . We let be the maximum between and .
We first assume that . Let be a tournament on vertices. Consider a median order of and let be the first vertices in this order, be the last vertices, and be set of the remaining vertices in the middle.
By (M2), every vertex is dominated by at least vertices in (using ). Let be any set of size . By Lemma 17, there is a subset of of size such that has at least common in-neighbours in . Since this holds for every subset of size , and by choice of , by Ramsey’s Theorem there is a subset of of size such that every subset of vertices in have at least common in-neighbours in . By repeating the same reasoning, because every vertex in dominates at least vertices in , we deduce that contains a set of vertices with at least common out-neighbours in . Since every -subset of has at least common in-neighbours in , this gives the desired copy of .
We now consider the case or . We apply the previous case for replaced respectively by and . This shows that for . ∎
While Proposition 28 shows that this upper is tight when or is in , we do not know the precise asymptotic of when .
5.2 -extensions of directed paths
In this section, we give the exact value of when is a full -extension of a directed path.
Theorem 31.
Let be a full -extension of a directed path with added vertex . If is a source or a sink, or if is , then . Otherwise, .
Proof.
Observe that every orientation of contains , and that the directed triangle does not contain , so . We now assume that is not .
Set , and let be a vertex of such that is a directed path . Let and and set and . Observe that . By Proposition 26 we obtain , and if is a sink or a source.
If or , then is a source or a sink. Thus, by Proposition 1, we have .
It remains to show that , assuming and , and . By directional duality, assume without loss of generality that . Let be a tournament of order , and let be a median order of . Set and . Let and . Observe that, by (M2), and . We first consider the case .
Claim 3.
If , then contains a copy of with .
Proof of claim. Let be the set of in-generators of and be the set of vertices which are initial vertices of a directed path of order in . Note that and . Moreover, by Lemma 15, and . Let be the vertex in with largest index and be the vertex in with smallest index. By Lemma 14, is the terminus of a directed path of order in and is the initial vertex of a directed path of order in . Hence we may assume , for otherwise the union of , , , the arcs between those paths and , and is a copy of in .
Assume first that . Then, since , has exactly out-neighbour and in-neighbours in . Consequently is also a median order of . By (M4) for , since , there is a vertex in such that . Observe that by maximality of and because . Thus , and similarly . In particular, . If , remove the end-vertex from , otherwise remove the initial vertex of . In both cases, the resulting union of , , and the arcs between those paths and is a copy of in (with ) as desired.
Henceforth, we may assume that . In particular it implies : to see this, take any Hamiltonian directed path of (which exists by Theorem 3), then the two first vertices of this directed path are initial vertices of directed paths of order in . If there is a vertex in such that , then as above there is a copy of in (with ), so assume that there is no such . Hence, by (M4) and (M5), is a median order of . Let be the vertex of smallest index in . Let be a directed path of order in with initial vertex . If , then the union of , , , the arcs between those paths and , and is a copy of in . Henceforth we may assume . By (M4) for , there is a vertex with such that . Note that because , and as above . As above, by removing either the end-vertex of or the initial vertex of , we get that the union of , , and the arcs between those paths and is a copy of in .
Since Claim 3 settles the case , and because , we now assume that . Then by (M4) and (M5) applied on , we have that is a median order of . In particular this implies and . Observe that is then a Hamiltonian directed cycle of . Not also that if we can apply Claim 3 with playing the roles of respectively. By repeating this process, either we eventually find a copy of in with added vertex or we ensure that . Henceforth we assume that .
The end of the proof is similar to the proof of Claim 3, with playing the role of . Let and . We have and, by (M2) on , . Since is a directed cycle, in particular both and are the terminal vertices of a directed path of order in . Let be the set of vertices which are initial vertex of a directed path of order in . Note that because and, by Lemma 15, . Let be the vertex in with smallest index. Let be the directed path and be a directed path of order in starting on .
We may assume that , for otherwise , , , the arcs between those paths and , and is a copy of in . If there is a vertex in such that , then by removing either the end-vertex of or the initial vertex of , we get that the union of , , and the arcs between those paths and is a copy of in . This holds because by choice of and is a directed path in . Henceforth, we assume that there is no such . Therefore, by (M4) and (M5) applied on , is a median order of . In particular, . Let be . We thus obtain that the union of , , and the arcs between those paths and is a copy of in . ∎
References
- [1] B. Alspach and M. Rosenfeld. Realization of certain generalized paths in tournaments. Discrete Mathematics, 34(2):199–202, 1981.
- [2] J. Bang-Jensen and G. Gutin, editors. Classes of directed graphs. Springer monographs in mathematics. Springer International Publishing, Cham, Switzerland, 1st edition, July 2018.
- [3] M. Bucić, S. Letzter, and B. Sudakov. Directed ramsey number for trees. Journal of Combinatorial Theory, Series B, 137:145–177, 2019. arXiv:1708.04504.
- [4] L. N. Coregliano and A. A. Razborov. On the density of transitive tournaments. Journal of Graph Theory, 85(1):12--21, 2017. arXiv:1501.04074.
- [5] N. Draganić, F. Dross, J. Fox, A. Girão, F. Havet, D. Korándi, W. Lochet, D. M. Correia, A. Scott, and B. Sudakov. Powers of paths in tournaments. Combinatorics, Probability and Computing, 30(6):894--898, 2021. arXiv:2010.05735.
- [6] F. Dross and F. Havet. On the unavoidability of oriented trees. Journal of Combinatorial Theory, Series B, 151:83--110, 2021. arXiv:1812.05167.
- [7] A. El Sahili. Trees in tournaments. Journal of Combinatorial Theory, Series B, 92(1):183--187, 2004.
- [8] P. Erdős and L. Moser. On the representation of directed graphs as unions of orderings. Magyar Tud. Akad. Mat. Kutató Int. Közl., 9:125--132, 1964.
- [9] R. Forcade. Parity of paths and circuits in tournaments. Discrete Mathematics, 6(2):115--118, 1973.
- [10] J. Fox, X. He, and Y. Wigderson. Ramsey numbers of sparse digraphs. Israel Journal of Mathematics, 2024. arXiv:2105.02383.
- [11] B. Grünbaum. Antidirected hamiltonian paths in tournaments. Journal of Combinatorial Theory, Series B, 11(3):249--257, 1971.
- [12] R. Häggkvist and A. Thomason. Trees in tournaments. Combinatorica, 11(2):123–130, June 1991.
- [13] F. Havet. Trees in tournaments. Discrete Mathematics, 243(1):121--134, 2002.
- [14] F. Havet and S. Thomassé. Median orders of tournaments: a tool for the second neighborhood problem and Sumner’s conjecture. Journal of Graph Theory, 35(4):244--256, 2000.
- [15] F. Havet and S. Thomassé. Oriented Hamiltonian paths in tournaments: a proof of Rosenfeld’s conjecture. Journal of Combinatorial Theory, Series B, 78(2):243--273, 2000.
- [16] D. Kühn, R. Mycroft, and D. Osthus. An approximate version of Sumner’s universal tournament conjecture. Journal of Combinatorial Theory, Series B, 101(6):415--447, 2011. arXiv:1010.4429.
- [17] T. Kóvari, V. T. Sós, and P. Turán. On a problem of K. Zarankiewicz. Colloquium Mathematicum, 3(1):50–57, 1954.
- [18] C. Lee. Ramsey numbers of degenerate graphs. Annals of Mathematics, 185(3), 2017. arXiv:1505.04773.
- [19] N. Linial, M. Saks, and V. T. Sós. Largest digraphs contained in alln-tournaments. Combinatorica, 3(1):101--104, 1983.
- [20] F. P. Ramsey. On a problem of formal logic. In Classic Papers in Combinatorics, pages 1--24. Springer, 1987.
- [21] L. Rédei. Ein kombinatorischer Satz. Acta Litteraria Szeged, 7:39--43, 1934.
- [22] K. Reid and E. Parker. Disproof of a conjecture of Erdös and Moser on tournaments. Journal of Combinatorial Theory, 9(3):225 -- 238, 1970.
- [23] K. B. Reid and N. C. Wormald. Embedding oriented -trees in tournaments. Studia Scientiarum Mathematicarum Hungarica, 18(2-4):377--387, 1983.
- [24] M. Rosenfeld. Antidirected Hamiltonian paths in tournaments. Journal of Combinatorial Theory, Series B, 12(1):93–99, Feb. 1972.
- [25] A. Sanchez-Flores. On tournaments free of large transitive subtournaments. Graphs and Combinatorics, 14(2):181--200, Jun 1998.
- [26] H. J. Straight. The existence of certain type of semi-walks in tournaments. Congressus Numerantium, 29:901--908, 1980.
- [27] A. Thomason. Paths and cycles in tournaments. Transactions of the American Mathematical Society, 296(1):167--180, 1986.