Spanning trees in dense directed graphs
SSpanning trees in dense directed graphs
Amarja Kathapurkar ∗ Richard Montgomery † Abstract
In 2001, Koml´os, S´ark¨ozy and Szemer´edi proved that, for each α >
0, there is some c > n such that, if n ≥ n , then every n -vertex graph with minimum degree at least(1 / α ) n contains a copy of every n -vertex tree with maximum degree at most cn/ log n .We prove the corresponding result for directed graphs. That is, for each α >
0, there is some c > n such that, if n ≥ n , then every n -vertex directed graph with minimum semi-degree at least (1 / α ) n contains a copy of every n -vertex oriented tree whose underlyingmaximum degree is at most cn/ log n .As with Koml´os, S´ark¨ozy and Szemer´edi’s theorem, this is tight up to the value of c . Ourresult improves a recent result of Mycroft and Naia, which requires the oriented trees to haveunderlying maximum degree at most ∆, for any constant ∆ ∈ N and sufficiently large n . Incontrast to the previous work on spanning trees in dense directed or undirected graphs, ourmethods do not use Szemer´edi’s regularity lemma. Given two graphs H and G , when may we expect to find a copy of H in G ? In general, thisdecision problem is NP-complete, and therefore we seek simple conditions on G which imply itcontains a copy of H . An important early result is Dirac’s theorem from 1952 that, when n ≥ n -vertex graph with minimum degree at least n/ n -vertex graph H , what is the lowest minimum degreecondition on an n -vertex graph G which guarantees it contains a copy of H ? As such a copy of H would contain every vertex in G , we say it is a spanning copy of H .This question has been studied for many different graphs H , for example when H is a K -factorfor some small fixed graph K [8, 14], the k -th power of a Hamilton cycle for any k ≥ H has bounded chromatic number and maximum degree, and sublinear bandwith [4]. Formore details on these results, and those for other graphs, see the survey by K¨uhn and Osthus [13].Here, we will concentrate on the minimum degree required to guarantee different spanning trees.Koml´os, S´ark¨ozy and Szemer´edi [10] proved in 1995 that, for each α, ∆ >
0, there is some n such that, if n ≥ n , then every n -vertex graph with minimum degree at least (1 / α ) n containsa copy of every n -vertex tree with maximum degree at most ∆, thus confirming a conjecture ofBollob´as [2]. This result is furthermore notable as one of the earliest applications of the blow-uplemma. In 2001, Koml´os, S´ark¨ozy and Szemer´edi [12] relaxed the maximum degree condition,showing that, for each α >
0, there is some c > n such that, if n ≥ n , then every ∗ University of Birmingham, Birmingham, B15 2TT, UK. Email:
[email protected] . † University of Birmingham, Birmingham, B15 2TT, UK. Email: [email protected] . Supportedby the European Research Council (ERC) under the European Union Horizon 2020 research and innovationprogramme (grant agreement No. 947978). a r X i v : . [ m a t h . C O ] F e b -vertex graph with minimum degree at least (1 / α ) n contains a copy of every n -vertex treewith maximum degree at most cn/ log n . This is tight up to the constant c . In this paper, wewill prove the corresponding version of this result for directed graphs (digraphs) .The minimum semidegree of a digraph D , denoted by δ ( D ), is the smallest in- or out-degree over the vertices in D , that is, δ ( D ) = min v ∈ V ( D ) , (cid:5)∈{ + , −} d (cid:5) ( v ). Ghouila-Houri [7] solvedthe minimum semidegree problem for the directed Hamilton cycle, showing that, if an n -vertexdigraph D has δ ( D ) ≥ n/
2, then it contains a directed Hamilton cycle. That is, an n -vertex cyclewith the edges oriented in the same direction. DeBiasio, K¨uhn, Molla, Osthus and Taylor [5]showed that, when n is sufficiently large, this holds in fact for any n -vertex cycle with anyorientations on its edges, except for when the edges change direction at every vertex around thecycle. This latter cycle, known as the anti-directed Hamilton cycle , is only guaranteed to appearif δ ( D ) ≥ n/ H is an oriented n -vertex tree, with somebound on the degree of its underlying (undirected) tree. Mycroft and Naia [16, 17] proved that,for each α, ∆ >
0, there is some n such that, if n ≥ n , then every n -vertex digraph withminimum semidegree at least (1 / α ) n contains a copy of every oriented n -vertex tree T with∆ ± ( T ) ≤ ∆. Moreover, their result holds for a slightly wider class of trees, allowing them toshow that, for each α >
0, almost every labelled oriented n -vertex tree appears in every n -vertexdigraph with minimum semidegree at least (1 / α ) n .In this paper, we introduce new methods to embed oriented trees in digraphs, relaxing themaximum degree condition to give a full directed version of Koml´os, S´ark¨ozy and Szemer´edi’sresult, as follows. Theorem 1.1.
For each α > , there exists c > and n ∈ N such that the following holds forevery n ≥ n . Any n -vertex digraph D with δ ( D ) ≥ (1 / α ) n contains a copy of every oriented n -vertex tree T with ∆ ± ( T ) ≤ cn/ log n . We note that the undirected version follows immediately from Theorem 1.1. Indeed, given any n -vertex tree T and an n -vertex graph G , we can apply Theorem 1.1 to a copy of T with each edgeoriented arbitrarily and a digraph formed from G by replacing each edge uv with an edge from u to v and an edge from v to u . This demonstrates that, as with Koml´os, S´ark¨ozy and Szemer´edi’sresult, Theorem 1.1 is tight up to the constant c . Furthermore, through Theorem 1.1 we givea new proof of the undirected result without using Szemer´edi’s regularity lemma, in contrast tothe work of both Koml´os, S´ark¨ozy and Szemer´edi [10], and Mycroft and Naia [16, 17]. Key toour result is to use a random embedding of part of the tree using ‘guide sets’ and embeddingmany leaves (and small subtrees) of the tree using ‘guide graphs’. This replaces the regularitymethods of [10, 16, 17], and is sketched in Section 2, where we also outline the rest of this paper. Let D be a digraph. We denote by V ( D ) and E ( D ) the vertex set and edge set of D , respectively,where every element of the edge set of D is an ordered pair of vertices. We let | D | = | V ( D ) | ,which we call the size of D , and let e ( D ) = | E ( D ) | . Letting u, v ∈ V ( D ), if uv ∈ E ( D ), thenwe say that u is an in-neighbour of v and v is an out-neighbour of u . Denote by N − D ( v ) and N + D ( v ), respectively, the set of all in- and out-neighbours of v . We let d − D ( v ) = (cid:12)(cid:12) N − D ( v ) (cid:12)(cid:12) and d + D ( v ) = (cid:12)(cid:12) N + D ( v ) (cid:12)(cid:12) , and we refer to these as the in- and out-degree of v , respectively. For each2 ∈ { + , −} , we let δ (cid:5) ( D ) and ∆ (cid:5) ( D ) be, respectively, the minimum and maximum (cid:5) -degree of D . For any A, B ⊆ V ( D ), and each (cid:5) ∈ { + , −} , let N (cid:5) D ( A, B ) = (cid:83) a ∈ A ( N (cid:5) D ( a ) ∩ B ), and let d (cid:5) D ( A, B ) = | N (cid:5) D ( A, B ) | . We omit the subscript when the graph is clear from context. Note that,for simplicity of notation, we use ‘ − ’ and ‘in’ interchangeably, and, similarly, we use ‘+’ and ‘out’interchangeably. We use ‘ ± ’ to represent that a property holds for both ‘ − ’ and ‘+’.Suppose that A and B are disjoint subsets of V ( D ). We write D [ A ] to mean D induced onthe set A , that is, the graph obtained from D by deleting all vertices which are not in A . For each (cid:5) ∈ { + , −} , a (cid:5) -matching from A into B is a set of vertex-disjoint edges such that every edge inthe set has one endpoint in A and one endpoint in B , and the endpoint in B is a (cid:5) -neighbour ofthe endpoint in A , that is, every edge is a (cid:5) -edge from A into B . We say this matching covers A if every vertex of A belongs to some edge in the matching, and we call this a perfect (cid:5) -matching if it covers both A and B . A bare path of length m in a tree is a path with m edges such thateach of the internal vertices have degree 2 in the tree. When P is a path in D , we let D − P denote the subgraph of D obtained by removing the internal vertices of P .For any n ∈ N , we let [ n ] := { , . . . , n } . In order to simplify notation, we use hierarchies tostate our results. That is, for a, b ∈ (0 , a (cid:28) b (or b (cid:29) a ), we mean that there exists a non-decreasing function f : (0 , → (0 ,
1] such thatthe statement holds whenever a ≤ f ( b ). We define similar expressions with multiple variablesanalogously. We say a random event occurs with high probability if the probability of the eventoccurring tends to 1 as n tends to infinity. In our proofs, when we have shown that a propertyholds with high probability, we will implicitly assume that this property holds from that pointonwards. For simplicity, we ignore floors and ceilings wherever this does not affect the argument. When 1 /n (cid:28) c (cid:28) α , we will embed any oriented n -vertex tree T with ∆ ± ( T ) ≤ cn/ log n intoany n -vertex digraph D with δ ( D ) ≥ (1 / α ) n . We embed T using the absorption method,an approach first introduced in general by R¨odl, Ruci´nski and Szemer´edi [18] which has beeneffective on a range of embedding problems for spanning graphs and digraphs (see, for example,the survey [3]). We first partially embed a subtree T (cid:48)(cid:48) of T into a set A such that, given anysubset B ⊂ V ( D ) with A ⊂ B and | B | = | T (cid:48)(cid:48) | , we can complete this embedding of T (cid:48)(cid:48) into D [ B ](see Theorem 2.1).We then use an almost-spanning embedding to embed the vertices in V ( T ) \ V ( T (cid:48)(cid:48) ) to extendthe partial embedding of T (cid:48)(cid:48) (see Theorem 2.2). We will have chosen T (cid:48)(cid:48) so that in this stagea tree, called T (cid:48) , is attached to an embedded vertex of T (cid:48)(cid:48) . Using the property of the partialembedding of T (cid:48)(cid:48) , we then complete the embedding of T (cid:48)(cid:48) with the unused vertices in D . Thedecomposition of T that we need follows from a simple proposition (Proposition 2.3).In Section 2.2.1, we state these three results, Theorem 2.1, Theorem 2.2 and Proposition 2.3,before deducing Theorem 1.1 from them. In Section 2.2.2, we then discuss in detail the proof ofTheorem 2.2, which is the major challenge overcome by this paper.In the rest of Section 2, we restate the probabilistic tools we will use, and give a basicstructural decomposition of trees and some simple results on matchings. In Section 3, we proveTheorem 2.2. In Section 4, we prove Theorem 2.1. For Theorem 1.1, we will first find a suitable subtree T (cid:48)(cid:48) ⊂ T and a set A ⊂ V ( D ) with slightlyfewer than | T (cid:48)(cid:48) | vertices, so that, given any set B of | T (cid:48)(cid:48) | vertices containing A , we can embed3 (cid:48)(cid:48) in D [ B ]. Furthermore, we will ensure that some pre-specified vertex t ∈ V ( T (cid:48)(cid:48) ) is alwaysembedded to some fixed vertex v ∈ A , as follows. Theorem 2.1.
Let /n (cid:28) c (cid:28) ε (cid:28) µ (cid:28) α . Let D be an n -vertex digraph with minimumsemidegree at least (1 / α ) n . Let T be an oriented tree with µn vertices and ∆ ± ( T ) ≤ cn/ log n ,and let t ∈ V ( T ) .Then, V ( D ) contains a vertex set A with size ( µ − ε ) n containing a vertex v ∈ A such thatthe following holds. For any set B ⊂ V ( D ) with A ⊂ B and | B | = µn , D [ B ] contains a copy of T in which t is copied to v . Theorem 2.1 is proved in Section 4 by randomly embedding most of T and taking A to bethe image of this embedding. We then show that the partial embedding of T can be extendedusing any new vertex in y ∈ V ( D ) \ A by switching y into the partial embedding in place of somevertex in A that can instead be used to embed a new vertex of T . Repeatedly doing this willallow the embedding of T to be completed using any set of | T | − | A | new vertices in V ( D ) \ A .This is sketched in more detail at the start of Section 4, before Theorem 2.1 is proved.We will embed the majority of the tree for Theorem 1.1, using the following almost-spanningembedding. Theorem 2.2.
Let /n (cid:28) c (cid:28) ε, α . Let D be an n -vertex digraph with minimum semidegree atleast (1 / α ) n and let v ∈ V ( D ) . Let T be an oriented tree with at most (1 − ε ) n vertices and ∆ ± ( T ) ≤ cn/ log n , and let t ∈ V ( T ) .Then, D contains a copy of T in which t is copied to v . Using in addition the following simple proposition (see, for example, [15, Proposition 3.22]),we can now deduce Theorem 1.1.
Proposition 2.3.
Let n, m ∈ N satisfy ≤ m ≤ n/ . Given any n -vertex tree T , there are twoedge-disjoint trees T , T ⊂ T such that E ( T ) ∪ E ( T ) = E ( T ) and m ≤ | T | ≤ m .Proof of Theorem 1.1 from Theorems 2.1 and 2.2. Let ε, µ > c (cid:28) ε (cid:28) µ (cid:28) α .Let D be an n -vertex digraph with δ ( D ) ≥ (1 / α ) n . Let T be an oriented n -vertex tree with∆ ± ( T ) ≤ cn/ log n .Using Proposition 2.3 with m = µn , find edge-disjoint trees T (cid:48) , T (cid:48)(cid:48) ⊂ T such that E ( T (cid:48) ) ∪ E ( T (cid:48)(cid:48) ) = E ( T ) and µn ≤ | T (cid:48)(cid:48) | ≤ µn . Let t be the vertex which is in both T (cid:48) and T (cid:48)(cid:48) . ByTheorem 2.1 applied with µ (cid:48) = | T (cid:48)(cid:48) | /n , there is a set A ⊂ V ( D ) such that | A | = | T (cid:48)(cid:48) | − εn , and avertex v ∈ A such that, for any set B ⊂ V ( D ) with A ⊂ B and | B | = | T (cid:48)(cid:48) | , D [ B ] contains a copyof T (cid:48)(cid:48) in which t is copied to v .Let D (cid:48) = D − ( A \ { v } ). Let n (cid:48) = | D (cid:48) | , so that (1 − µ ) n ≤ n (cid:48) ≤ n . Let α (cid:48) be such that D (cid:48) has minimum semidegree (1 / α (cid:48) ) n (cid:48) . Note that (1 / α (cid:48) ) n ≥ (1 / α (cid:48) ) n (cid:48) ≥ (1 / α − µ ) n ,so that α (cid:48) ≥ α/
2. Furthermore, n (cid:48) = n − | T (cid:48)(cid:48) | + εn + 1 = | T (cid:48) | + εn , and therefore | T (cid:48) | n (cid:48) = | T (cid:48) || T (cid:48) | + εn ≤ | T (cid:48) || T (cid:48) | (1 + ε ) ≤ − ε/ . Thus, by Theorem 2.2, we can find a copy, S (cid:48) say, of T (cid:48) in D (cid:48) in which t is copied to v . Byapplying the property of A from Theorem 2.1, we can then find a copy of T (cid:48)(cid:48) in D − ( V ( S (cid:48) ) \ { v } )in which t is copied to v . Together, these give us a copy of T .4 .2.2 Proof Sketch of Theorem 2.2 We will embed a (1 − ε ) n -vertex tree T for Theorem 2.2 by dividing most of T into a small coreforest T ⊂ T and a collection of constant-sized subtrees, which are either attached to T by asingle edge or by two short paths. It is the trees attached to T by a single edge that will be themost challenging to embed, and so we dedicate most of our attention in the proof sketch to this.More precisely, we will find a tree T (cid:48) ⊂ T , containing a core forest T ⊂ T (cid:48) and vertex-disjointtrees S , . . . , S (cid:96) ⊂ T (cid:48) − V ( T ), for some (cid:96) ∈ N , such that T (cid:48) is formed from T by, for each i ∈ [ (cid:96) ],(1) either adding S i to T using two bare paths with length 2,(2) or adding S i to T with a single edge.Furthermore, for some µ > K ∈ N , with 1 /n (cid:28) /K, µ (cid:28) α, ε , we will have that • | T | ≤ µn (i.e., T is small), • | T (cid:48) | ≥ | T | − µn (i.e., T (cid:48) is most of T ), • there are at most µn trees S i which are in Case (1), and • each tree S i has at most K vertices.In Case (1), we say S i is added to T as a path, and in Case (2) we say S i is added to T as aleaf. The crux of our method is to embed T along with the trees S i in Case (2) connected to theembedding of T by the appropriate edge. This is encapsulated in the following lemma, which isproved in Section 3.1. Lemma 2.4.
Let /n (cid:28) c (cid:28) µ (cid:28) α, ε , let c (cid:28) /K and let (cid:96) ∈ N . Suppose D is an n -vertexdigraph with δ ( D ) ≥ (1 / α ) n and v ∈ V ( D ) .Suppose that T is an oriented tree with | T | ≤ (1 − ε ) n and ∆ ± ( T ) ≤ cn/ log n . Supposethat T (cid:48) , S , . . . S (cid:96) ⊂ T are vertex-disjoint subtrees with | T (cid:48) | ≤ µn , and | S i | ≤ K for each i ∈ [ (cid:96) ] .Suppose that T is formed from T (cid:48) by attaching each S i , i ∈ [ (cid:96) ] , to T (cid:48) by an edge. Finally, let t ∈ V ( T (cid:48) ) .Then, D contains a copy of T in which t is copied to v . We will now briefly sketch how Theorem 2.2 can be proved from Lemma 2.4. Let m be thetotal number of vertices that appear in the trees S i in Case (1) above. To embed these trees,we use the fact that two random sets in D of the same (linear) size are likely to have a perfectmatching from one to the other (see Proposition 2.12). Taking p (cid:29) /n and Kp ≤
1, we can,with high probability, find pn copies of an oriented tree with K vertices in a random set of Kpn vertices in D by taking randomly K disjoint subsets within this set of size pn and findingappropriate matchings between them (see Section 2.5). Collecting isomorphic trees S i together,and applying this to each of the constantly many (depending on K ) isomorphism classes, allowsus to embed the trees S i in Case (1) with high probability in a random set with size m + εn/ εn/ V ( D ) into sets V ∪ V ∪ V ∪ V chosen uniformly at random so that | V | = n − m − εn/ | V | = m + εn/ | V | = | V | = εn/
4, with high probability, the followingoccur. • δ ± ( D [ V ]) ≥ (1 / α/ | V | , so that, applying Lemma 2.4, we can embed T along withthe trees S i in Case (2) connected to the embedding of T by the appropriate edge.5 We can embed the trees S i in Case (1) in D [ V ]. • Then, using that there are at most µn trees in Case (1), we can greedily attach them to theembedding of T using two paths with length 2 whose interior vertex is an unused vertexin V (see Section 3.2). • Finally, as | T | − | T (cid:48) | ≤ µn , we can greedily extend the resulting embedding of T (cid:48) to one of T , by adding a sequence of leaves using vertices in V (see Section 3.3).Here, the last two steps are (with high probability) possible using the semi-degree condition of D . Note that, as µ (cid:28) ε , we only embed a small proportion of vertices into V and V .We will now give a detailed proof sketch of Lemma 2.4. Proof sketch of Lemma 2.4
To simplify our discussion, let us assume that each tree S i in Lemma 2.4 consists of only a singlevertex, which is an out-neighbour in the tree T of a vertex of T , and that every vertex in T is attached to exactly one such tree. That is, T consists of T with an out-matching attached.Our embedding of T is randomised, which will allow the methods described to be used to findmatchings attached from different subsets of the image of the embedding of T to different randomsets. This will allow the embedding below for T to be used for the general case.Let us detail the example situation precisely. Suppose we have a µn -vertex tree T and choosetwo disjoint random sets V , V ⊂ V ( D ) with size p n and p n respectively, where p (cid:29) µ and p = (1 + o (1)) µ . We will randomly embed T into V , so that there is an out-matching from thevertex set of the embedding of T into V . Note that there are many spare vertices in V , possibleas in the general case we embed T once. However, as we then find (potentially) many differentmatchings, we need to do this with few spare vertices, and therefore use most of the vertices in V (as p is only a little larger than µ ).We will embed T vertex-by-vertex, say in order t , . . . , t (cid:96) , so that each new vertex is embeddedas an in- or out-leaf of the previously embedded subtree. Having chosen the random sets V , V ,and before beginning the embedding, we will find guide sets A v, (cid:5) ⊂ N (cid:5) D ( v, V ), v ∈ V and (cid:5) ∈ { + , −} , which we use to guide the random embedding. We then start the random embedding,under the rule that if, for some v ∈ V and (cid:5) ∈ { + , −} , we are attaching a (cid:5) -edge as a leaf to v ,then we choose this leaf uniformly at random from the unused vertices in A v, (cid:5) .The guide sets ensure that, with high probability, there will be a matching from the embeddingof T into V . These guide sets are found using Lemma 3.5, and they exist (with high probabilityfor the choice of V , V ) due to the semi-degree condition in D . Essentially, for some constants β, γ , we find, for each v ∈ V ( D ) and (cid:5) ∈ { + , −} , a set A v, (cid:5) ⊂ N (cid:5) D ( v, V ) with size βn and bipartitedigraphs H ◦ v, (cid:5) ⊂ D ◦ [ A v, (cid:5) , V ], ◦ ∈ { + , −} , so that in H ◦ v, (cid:5) each vertex in A v, (cid:5) has around γp n ◦ -neighbours in V , and each vertex in V has around γβn ◦ -edges leading into it. That is, H ◦ v, (cid:5) is approximately regular on each side with edge density approximately γ .The guide graphs H + v, (cid:5) can be used to find the matching from the embedding of T as follows.When a vertex t i is embedded using a guide set A v i , (cid:5) i , to some vertex s i say, we add the edgesin H + v i , (cid:5) i adjacent to s i to an auxiliary graph K – note that approximately γp n edges are addednext to s i . Note further that, as most of the vertices in A v i , (cid:5) i will be unused, each w ∈ V willhave an edge added from s i to w with probability approximately d − H + vi, (cid:5) i ( w ) | A v i , (cid:5) i | ≈ γβnβn = γ. (1)6hen this is complete, K is a bipartite digraph with vertex classes { s , . . . , s (cid:96) } and V . Eachvertex s i will have out-degree approximately γp n , and, due to the randomness of the embeddingand (1), each vertex in V will have in-degree which is approximately γ(cid:96) = γ | T | ≈ γp n .Thus, K will be a bipartite graph with the in-degrees in one vertex class approximately equalto the out-degrees in the other. Via Hall’s matching criterion, an out-matching will exist from { s , . . . , s (cid:96) } to V which covers most of the vertices in { s , . . . , s (cid:96) } . By ensuring that V is likelyto be a little larger than (cid:96) , we in fact will get with high probability that such an out-matchingcan cover { s , . . . , s (cid:96) } .Note that, in the sketch above, we do not use the graph H − v, (cid:5) . However, in practice, we findsuch guide sets and guide graphs with V = V ( D ) \ V (see Lemma 3.3), before taking randomsubsets of V . We will find out-matchings into some of these random sets, and in-matchingsinto some others. Therefore, it is important to have both guide graphs H − v, (cid:5) and H + v, (cid:5) , and,furthermore, that the same set A v, (cid:5) is used for both graphs.Finally, let us note where the condition ∆ ± ( T ) ≤ cn/ log n is used in our proof of Lemma 2.4.In the sketch above the set V will always have size which is linear in n , but we may need toattach the trees in Lemma 2.4 to few vertices in T . The maximum in- or out-degree condition on T ensures, that, if the trees S i in Lemma 2.4 together comprise linearly (in n ) many vertices in T , then they are attached to at least C log n different vertices, for some large constant C , whichgives us sufficient probability concentrations when these vertices are randomly embedded for thecorresponding versions of Hall’s criterion to hold (see the proof of Claim 3.7). Let n, m, k ∈ N be such that max { m, k } ≤ n . Let A be a set of size n , and B ⊆ A be suchthat | B | = m . Let A (cid:48) be a uniformly random subset of A of size k . Then the random variable X = | A (cid:48) ∩ B | is said to have hypergeometric distribution with parameters n, m and k , which wedenote by X ∼ Hyp( n, m, k ). We will use the following Chernoff-type bound.
Lemma 2.5 (see, for example, [9]) . Suppose X ∼ Hyp( n, m, k ) . Then for any < α < / , wehave P [ | X − E [ X ] | ≥ α E [ X ]] ≤ (cid:0) α E [ X ] / (cid:1) . A sequence of random variables ( X i ) i ≥ is a martingale if E [ X i +1 | X , . . . , X i ] = X i for each i ≥
0. We will use the following Azuma-type bound for martingales.
Lemma 2.6 (see, for example, [1]) . Let ( X i ) i ≥ be a martingale and let c i > for each i ≥ . If | X i − X i − | < c i for each i ≥ , then, for each n ≥ , P [ | X n − X | ≥ t ] ≤ (cid:18) − t (cid:80) ni =1 c i (cid:19) . We will use this bound for supermartingales and submartingales. A sequence of randomvariables ( X i ) i ≥ is a supermartingale if E [ X i +1 | X , . . . , X i ] ≤ X i for each i ≥
0, and asubmartingale if E [ X i +1 | X , . . . , X i ] ≥ X i for each i ≥
0. The bound on the upper tail inLemma 2.6 holds for supermartingales, while the bound on the lower tail holds for submartingales.
In this section we decompose undirected trees. Note that we will later apply this to directedtrees as the edge directions do not affect the decompositions. We will use the following simplebut useful lemma (see [15, Lemma 4.1]) which tells us that either a tree has many leaves, or ithas many bare paths. 7 emma 2.7.
Let t, m ≥ , and suppose that T is a tree with at most t leaves. Then there is some s and some vertex-disjoint bare paths P i , i ∈ [ s ] , in T with length m so that | T − P − · · · − P s | ≤ mt + 2 | T | / ( m + 1) . We can now prove the following key lemma, in which we decompose a tree for our embedding.
Lemma 2.8.
Let (cid:28) /n (cid:28) /K (cid:28) /k (cid:28) η . Let T be a tree on n vertices with t ∈ V ( T ) .Then, T contains forests T ⊂ T ⊂ T ⊂ T = T , such that T is a tree, and the following hold. P1 | T | ≤ ηn and t ∈ V ( T ) . P2 T is formed from T by the vertex-disjoint addition of trees, S v , v ∈ V ( T ) , so that, foreach v ∈ V ( T ) , S v − v is a forest of trees with size at most K . P3 T is formed from T by the addition of trees with size at least k and at most K attachedto T with exactly two bare paths of length 2. P4 | T | − | T | ≤ ηn .Proof. Let ε satisfy 1 /K (cid:28) ε (cid:28) /k , and let S = T . Do the following for i = 0 , , . . . as faras possible, where a set of independent leaves is a set of leaves which pairwise have no commonneighbours in the tree. If S i has at least εn independent leaves, the set L i say, then remove L i \ { t } from S i to get the tree S i +1 . Suppose this finishes with S (cid:96) , which does not have at least εn independent leaves. Note that (cid:96) ≤ /ε + 1. We will show the following claim. Claim 2.9.
To get from T to S (cid:96) , for each v ∈ V ( S (cid:96) ) , there is a tree removed from v which hasat most (cid:96) vertices.Proof of Claim 2.9. We will show by induction on i = 0 , , . . . , (cid:96) , that, to get from S (cid:96) − i to S (cid:96) ,for each v ∈ V ( S (cid:96) ), there is a tree removed from v which has at most 2 i vertices. Thus the claimfollows when i = (cid:96) . Note that this is trivially true for i = 0 and label the tree removed from v ∈ V ( S (cid:96) ) to get from S (cid:96) − i to S (cid:96) as T v,i .Now, let 0 ≤ i < (cid:96) , and assume that | T v,i | ≤ i for each v ∈ V ( S (cid:96) ). As we remove aset of independent leaves from S (cid:96) − i − to get to S (cid:96) − i , for each v ∈ V ( S (cid:96) ), we remove a set ofindependent leaves of T v,i +1 to get T v,i . Therefore, for each v ∈ V ( S (cid:96) ), | T v,i +1 | ≤ | T v,i | ≤ i +1 ,as required.Let L ( S (cid:96) ) be the set of leaves of S (cid:96) . Remove L ( S (cid:96) ) \ { t } and call the resulting tree S (cid:48) . Notethat, as S (cid:96) does not have at least εn independent leaves, S (cid:48) does not have at least εn leaves.Thus, by Lemma 2.7, for some m ≤ n/ ( k + 1), S (cid:48) contains vertex disjoint bare paths P , . . . , P m with length k such that t / ∈ V ( P i ) for each i ∈ [ k ] and | S (cid:48) − P − · · · − P m | ≤ k · εn + 2 n/ ( k + 1) + k + 1 ≤ ηn/ . (2)For each path P i , i ∈ [ m ], if possible, find within P i a path P (cid:48) i with length at least k − η k , suchthat, labelling its endvertices x i and y i the following hold.(i) Each of x i and y i had a tree with size at most ηk/ T to reach S (cid:48) .(ii) Letting Q i be the component of T − { x i , y i } containing P (cid:48) i − { x i , y i } , we have | Q i | ≤ K .8ay, with relabelling, these paths are P (cid:48) , . . . , P (cid:48) m (cid:48) . We will show that m (cid:48) ≥ m − ηn/ k . Notefirst that the number of i ∈ [ m ] with no vertices x i and y i ∈ V ( P i ) respectively within η k ofthe two endvertices of P i , so that each of x i and y i had a tree with at most ηk/ n/ ( η k · ηk/ ≤ ηn/ k . Note further that the number of i ∈ [ m ] with atleast K vertices in V ( P i ) or in a component of T − E ( P i ) containing an interior vertex of P i isat most n/K ≤ ηn/ k . Therefore, we can find such a path P (cid:48) i for all but at most ηn/ k valuesof i ∈ [ m ], so that m (cid:48) ≥ m − ηn/ k .Letting T = S (cid:48) − P (cid:48) − . . . − P (cid:48) m (cid:48) , we will now show that | T | ≤ ηn . Note that, for each i ∈ [ m (cid:48) ], | V ( P i ) \ V ( P (cid:48) i ) | ≤ η k . Therefore, as m ≤ n/k , | T | ≤ | S (cid:48) − P − . . . − P m | + k · ηn/ k + m · η k (2) ≤ ηn. Furthermore, clearly t ∈ V ( T ), and thus P1 holds.Note that, by Claim 2.9, for each i ∈ [ m (cid:48) ], | Q i | ≤ k (cid:96) ≤ K . For each v ∈ V ( T ), let R v be thetree containing v in T [( V ( T ) \ V ( T )) ∪ { v } ], without any of the x i , y i as neighbours. Now, byClaim 2.9, R v − v consists of trees with at most 2 (cid:96) ≤ K vertices. Let T = T ∪ ( (cid:83) v ∈ V ( T ) R v ).Thus, P2 holds.Let T = T and let T be T [ V ( T ) ∪ ( (cid:83) i ∈ [ m (cid:48) ] ( { x i , y i } ∪ V ( Q i )))]. Note that P3 holds byconstruction, and as | Q i | ≤ K for each i ∈ [ m (cid:48) ]. Furthermore, the only missing vertices from T are those in R v − v , for each v ∈ { x i , y i : i ∈ [ m (cid:48) ] } , and thus T is a tree. For each such v , | R v | ≤ ηk/ | T | − | T | ≤ ( n/k ) · (2 ηk/ ≤ ηn , and hence P4 holds. With high probability, any random subset of vertices in the digraph in Theorem 1.1 satisfies asimilar minimum semidegree condition, as follows.
Lemma 2.10. /n (cid:28) c, α , and suppose D is an n -vertex digraph with δ ( D ) ≥ (1 / α ) n .Let A ⊆ V ( D ) be chosen uniformly at random subject to | A | = cn . Then, with high probability,for every vertex v ∈ V ( D ) , we have (cid:12)(cid:12) N ± D ( v, A ) (cid:12)(cid:12) ≥ (1 / α/ | A | .Proof. Let v be an arbitrary vertex of D and let A ⊆ V ( D ) be a uniformly random subset with | A | = cn . For (cid:5) ∈ { + , −} , we let Z (cid:5) v be the random variable which measures | N (cid:5) ( v ) ∩ A | . Then Z (cid:5) v has hypergeometric distribution with expectation E [ Z (cid:5) v ] = | N (cid:5) ( v ) | | A | n ≥ (cid:18)
12 + α (cid:19) cn. Therefore, by Lemma 2.5, we have P (cid:20) | Z (cid:5) v − E [ Z (cid:5) x ] | > α/ / α (1 / α ) cn (cid:21) ≤ (cid:32) − (cid:18) α/ / α (cid:19) (1 / α ) cn (cid:33) = 2 exp (cid:18) − α cn α (cid:19) . Then, applying a union bound, with probability at least 1 − n exp (cid:0) − α cn/ (6 + 12 α ) (cid:1) = 1 − o (1),we have that Z (cid:5) v ≥ (1 / α/ | A | for each (cid:5) ∈ { + , −} and v ∈ V ( D ).The following digraph version of Hall’s matching criterion implies a matching exists, as followsdirectly from the same result for undirected graphs.9 emma 2.11. Let D be a bipartite digraph with vertex classes A and B , and let (cid:5) ∈ { + , −} .Suppose that for every S ⊂ A , | N (cid:5) D ( S, B ) | ≥ | S | . Then there is a (cid:5) -matching from A into B which covers A . We will refer to the condition in Lemma 2.11 as
Hall’s criterion . In combination withLemma 2.10, Lemma 2.11 shows that with high probability there is a perfect matching betweena large random pair of disjoint equal-sized vertex subsets in the digraph, as follows.
Proposition 2.12.
Let /n (cid:28) p, α , and suppose D is an n -vertex digraph with δ ( D ) ≥ (1 / α ) n . Let A, B be chosen uniformly at random from all disjoint pairs of subsets of V ( D ) , eachwith size pn , and let (cid:5) ∈ { + , −} . Then, with high probability, there is a perfect (cid:5) -matching from A into B .Proof. By Lemma 2.10, with high probability we can assume the following. For all v ∈ A , wehave | N ± ( v, B ) | ≥ (1 / α/ | B | , and, for all v ∈ B , we have | N ± ( v, A ) | ≥ (1 / α/ | A | . Wewill now show that Hall’s criterion holds.Let S ⊆ A , such that S (cid:54) = ∅ and | S | ≤ (1 / α/ pn , and let x ∈ S . Then, | N (cid:5) ( S, B ) | ≥| N (cid:5) ( x, B ) | ≥ (1 / α/ pn ≥ | S | , so Hall’s condition is trivially satisfied. Now take S ⊆ A , | S | > (1 / α/ pn , and assume for a contradiction that | N (cid:5) ( S, B ) | < | S | . Then in particular, B \ N (cid:5) ( S, B ) (cid:54) = ∅ . Take b ∈ B \ N (cid:5) ( S, B ), and let ◦ ∈ { + , −} be such that ◦ (cid:54) = (cid:5) . We have | N ◦ ( b, A ) | ≥ (1 / α/ pn . However, since b (cid:54)∈ N (cid:5) ( S, B ), we have N ◦ ( b, A ) ∩ S = ∅ . So, pn = | A | ≥ | N ◦ ( b, A ) | + | S | ≥ (1 / α/ pn + (1 / α/ pn = (1 + α ) pn > pn, giving a contradiction. Thus, Hall’s criterion is satisfied for all S ⊆ A and so, since | A | = | B | , byLemma 2.11, there is a perfect (cid:5) -matching from A into B .We use Proposition 2.12 to embed many vertex disjoint small trees, via the following twolemmas. In Lemma 2.13, we embed linearly many copies of a given constant-sized tree intospecified subsets of our digraph. In Lemma 2.13, we embed a forest of constant-sized treescovering almost all the vertices in our digraph. Lemma 2.13.
Let /n (cid:28) /K, p, α with pK ≤ . Suppose T is an oriented K -vertex treecontaining t ∈ V ( T ) . Let D be an n -vertex digraph with δ ( D ) ≥ (1 / α ) n . Let V , V be vertexdisjoint subsets of V ( D ) chosen uniformly at random subject to | V | = pn and | V | = ( K − pn .Then, with high probability, D [ V ∪ V ] contains pn vertex disjoint copies of T , in which t iscopied into V in each copy of T .Proof. Let V = U , and let U ∪ · · · ∪ U K be a partition of V chosen uniformly at random sothat | U i | = pn for each i ∈ { , . . . , K } . Note that the distribution of any pair of sets U i , U j with1 ≤ i < j ≤ K is that of two disjoint vertex sets with size pn in V ( D ), uniformly at randomdrawn from all such pairs.Label the vertices of T by t , . . . , t K so that t = t and T [ { t , . . . , t i } ] is a tree for each i ∈ { , . . . , K } . For each i ∈ { , . . . , K } , let j i ∈ { , . . . , i − } be such that t j i is the in- orout-neighbour in T [ { t , . . . , t i − } ] of the vertex t i , and let (cid:5) i ∈ { + , −} be such that t i ∈ N (cid:5) i T ( t j i ).Now by Proposition 2.12, for each i ∈ { , . . . , K } , with high probability, we can find a (cid:5) i -matching from U j i into U i . By applying a union bound, we see that, with high probability, forevery i ∈ { , . . . , K } , there is a (cid:5) i -matching, M i say, from U j i into U i .Note that the union of these matchings, (cid:83) ≤ i ≤ K M i ⊂ D [ V ∪ V ] is the disjoint union of pn copies of T , in which, for each i ∈ [ K ], the copy of t i is in V i . Thus, in each of these pn copies of T , t = t is copied into V = U , as required. 10 emma 2.14. Let /n (cid:28) /K, ε , and suppose F is a digraph with at most (1 − ε ) n verticeswhich is the disjoint union of trees with size at most K . Let D be an n -vertex digraph with δ ( D ) ≥ (1 / α ) n . Then, with high probability, D contains a copy of F .Proof. Arrange the components of F into isomorphic classes of trees R , . . . , R (cid:96) , noting that wemay take (cid:96) ≤ (2 K ) K − . For each i ∈ [ (cid:96) ], let t i = |R i | and let s i be the size of each component in R i . Uniformly at random, take, in V ( D ), disjoint subsets V i, and V i, , i ∈ [ (cid:96) ], with | V i, | = p i n and | V i, | = ( s i − p i n , where p i = t i /n + ε/(cid:96)s i , for each i ∈ [ (cid:96) ]. Note that this is possible, since (cid:96) (cid:88) i =1 s i p i n = (cid:96) (cid:88) i =1 (cid:16) s i t i + εn(cid:96) (cid:17) ≤ n. For each i ∈ [ (cid:96) ], we can apply Lemma 2.13 to show that, with high probability, there are p i n copies of the underlying tree of R i in D i = D [ V i, ∪ V i, ]. Since p i n ≥ t i , this implies that withhigh probability, we can find a copy of R i in D i for each i ∈ [ (cid:96) ]. By applying a union bound andusing that 1 /n (cid:28) /(cid:96) , we have, with high probability, that there is a copy of F in D . The key aim of this section is to prove Theorem 2.2, that is, to prove we can embed an almost-spanning tree T in our digraph. By Lemma 2.8, we can find T ⊂ T ⊂ T ⊂ T = T , satisfying P1 to P4 . In Section 3.1, we show that we can embed T . In Section 3.2, we show that we canembed T \ T , and T \ T . We conclude in Section 3.3 by combining this to obtain an embeddingof T . As sketched in Section 2.2, we will embed T randomly, leaf by leaf, using a guide set to embedeach new vertex. Each guide set has an accompanying guide graph, which we later use to finda matching. The property of the guide graph that we use to find the matching is that it is skew-bounded , as follows. Definition 3.1.
A digraph D with vertex sets A, B ⊂ V ( D ) is ( a, b, (cid:5) )-skew-bounded on ( A, B ) if d (cid:5) D ( v, B ) ≥ a for each v ∈ A and d ◦ D ( v, A ) ≤ b for each v ∈ B , where ◦ ∈ { + , −} and ◦ (cid:54) = (cid:5) . This property can imply a matching exists via Hall’s criterion, as follows.
Proposition 3.2.
Let a ≥ b and (cid:5) ∈ { + , −} . Suppose D is a digraph containing disjoint vertexsets A, B ⊂ V ( D ) , such that D is ( a, b, (cid:5) ) -skew-bounded on ( A, B ) . Then, there is a (cid:5) -matchingfrom A into B in D .Proof. Let U ⊂ A . As D is ( a, b, (cid:5) )-skew-bounded on ( A, B ), there are at least a | U | and at most b | N (cid:5) D ( U, B ) | (cid:5) -edges from U to N (cid:5) D ( U, B ). Thus, | N (cid:5) D ( U, B ) | ≥ a | U | /b ≥ | U | . Therefore, byLemma 2.11, there is a (cid:5) -matching from A into B .In the following lemmas, we find our guide sets and guide graphs. We start by finding in D ,for each v ∈ V ( D ) and (cid:5) ∈ { + , −} , a guide set A and guide graphs which are skew-bounded on( A, V ( D )). 11 emma 3.3. Let /n (cid:28) ε (cid:28) α, η ≤ and /n (cid:28) µ ≤ α / . Let D be an n -vertex digraph with δ ( D ) ≥ (1 / α ) n , let v ∈ V ( D ) and let (cid:5) ∈ { + , −} .Then, there is a set A ⊂ N (cid:5) D ( v ) with size µn and digraphs H + , H − ⊂ D such that, for each ◦ ∈ { + , −} , H ◦ is ( εn, (1 + η ) µεn, ◦ ) -skew-bounded on ( A, V ( D )) .Proof. We start by showing that we can label the vertices of V ( D ) as V ( D ) = { x , . . . , x n } = { y , . . . , y n } so that, for each i ∈ [ n ], | N − D ( x i ) ∩ N (cid:5) D ( v ) ∩ N + D ( y i ) | ≥ α n. (3)To do this, create an auxiliary graph, as follows. For each w ∈ V ( D ), create distinct newvertices w − and w + , and let V + = { w + : w ∈ V ( D ) } and V − = { w − : w ∈ V ( D ) } . Consider theauxiliary bipartite graph H with vertex set V + ∪ V − , where for each x, y ∈ V ( D ), there is anedge between x + and y − if and only if (cid:12)(cid:12) N − D ( x ) ∩ N (cid:5) D ( v ) ∩ N + D ( y ) (cid:12)(cid:12) ≥ α n . Claim 3.4. δ ( H ) ≥ (1 / α/ n .Proof of Claim 3.4. Let x ∈ V ( D ). We have | N − D ( x ) ∩ N (cid:5) D ( v ) | ≥ n − ( n − d − D ( x )) − ( n − d (cid:5) D ( v )) ≥ αn . Let B = N − D ( x ) ∩ N (cid:5) D ( v ) and Y = { y ∈ V ( D ) : | N + D ( y ) ∩ B | ≥ α n } , and note that d H ( x + ) = | Y | .For each u ∈ B , we have | N − D ( u ) | ≥ (1 / α ) n , and thus e D ( V ( D ) , B ) ≥ (1 / α ) | B | n . Bythe choice of Y , we have e D ( V ( D ) , B ) ≤ | Y || B | + α n . Therefore, as, in addition, 2 αn ≤ | B | ,we have (1 / α ) | B | n ≤ | Y || B | + α n ≤ | Y || B | + α | B | n/ . Thus, (1 / α/ | B | n ≤ | Y || B | , so that | Y | ≥ (1 / α/ n . Therefore, d H ( x + ) = | Y | ≥ (1 / α/ n .A similar argument, with the signs reversed, shows that d H ( y − ) ≥ (1 / α/ n for each y ∈ V ( D ), completing the proof of the claim.As in the proof of Proposition 2.12, Claim 3.4 easily implies that Hall’s criterion is satisfied,so that there is a matching from V + to V − in H . That is, we can label the vertices of V ( D ) as V ( D ) = { x , . . . , x n } = { y , . . . , y n } so that, for each i ∈ [ n ], (3) holds.We will now show by induction that, for each 0 ≤ i ≤ µn , there is a set A i ⊂ N (cid:5) D ( v ) with size i and graphs H + i , H − i ⊂ D such that, for each ◦ ∈ { + , −} , H ◦ i is ( εn, (1 + η ) µεn, ◦ )-skew-boundedon ( A i , V ( D )), e ( H ◦ i ) = iεn , and, for each j ∈ [ n ], d − H + i ( x j ) = d + H − i ( y j ).Note that if A = ∅ and if H +0 , H − have no edges and vertex set V ( D ), then the conditionshold, so assume that 0 ≤ i < µn and we have A i ⊂ N (cid:5) D ( v ) and H + i , H − i ⊂ D as described.Let J i ⊂ [ n ] be the set of j ∈ [ n ] for which d − H + i ( x j ) = d + H − i ( y j ) ≤ (1 + η/ µεn . Note that, as e ( H + i ) = e ( H − i ) = iεn ≤ µεn , we have( n − | J i | )(1 + η/ µεn ≤ µεn . Thus, as η ≤
1, ( n − | J i | ) ≤ n/ (1 + η/ ≤ n (1 − η/ | J i | ≥ ηn/ j ∈ J i , let W i,j = ( N − D ( x j ) ∩ N (cid:5) D ( v ) ∩ N + D ( y j )) \ A i , noting that, by (3), | W i,j | ≥ α n − i > α n − µn ≥ α n/
2. By averaging, choose some w i ∈ V ( D ) such that |{ j ∈ J i : w i ∈ W i,j }| ≥ (cid:80) j ∈ J i | W i,j | n ≥ ηn/ · α n/ n ≥ εn, using that α, η (cid:29) ε . Choose a set J (cid:48) i ⊂ { j ∈ J i : w i ∈ W i,j } with size εn . Let A i +1 = A i ∪ { w i } .Let H + i +1 be the digraph H + i with edges w i x j , j ∈ J (cid:48) i , added. Note that, as d − H + i ( x j ) ≤ (1+ η/ µεn j ∈ J (cid:48) i , H + i +1 is ( εn, (1 + η ) µεn, +)-skew-bounded on ( A i +1 , V ( D )). Furthermore, by thedefinition of W i,j , the edges added to H + i are in D , and therefore H + i +1 ⊂ D .Let H − i +1 be the digraph H − i with the edges y j w i , j ∈ J (cid:48) i , added. Note that, similarly, H − i +1 is ( εn, (1 + η ) µεn, − )-skew-bounded on ( A i +1 , V ( D )). Finally, noting that A i +1 has size i + 1,that e ( H + i +1 ) = e ( H − i +1 ) = ( i + 1) εn and that, for each j ∈ [ n ], d − H + i +1 ( x j ) = d + H − i +1 ( y j ), completesthe inductive step, and hence the proof.We now show that the guide sets and guide graphs found by Lemma 3.3 have a similarskew-bounded property when restricted to random vertex subsets, as follows. Lemma 3.5.
Let /n (cid:28) ε (cid:28) α, η ≤ and /n (cid:28) /k, p , p , . . . , p k ≤ . Let µ = α p / . Let D be an n -vertex digraph with δ ( D ) ≥ (1 / α ) n . Let V , V , . . . , V k ⊂ V ( D ) be disjoint randomsets chosen uniformly at random subject to | V i | = p i n for each i ∈ { , . . . , k } .Then, with high probability, for each v ∈ V ( D ) and (cid:5) ∈ { + , −} , there is a set A v, (cid:5) ⊂ N (cid:5) D ( v ) ∩ V with size µn and digraphs H ◦ v, (cid:5) ⊂ D , ◦ ∈ { + , −} , such that, for each ◦ ∈ { + , −} and i ∈ [ k ] , H ◦ v, (cid:5) is ( εp i n, (1 + η ) εµn, ◦ ) -skew-bounded on ( A v, (cid:5) , V i ) .Proof. By Lemma 3.3, applied with ε (cid:48) = (1 + η/ ε , η (cid:48) = η/ µ (cid:48) = (1 + η/ α /
4, for each v ∈ V ( D ) and (cid:5) ∈ { + , −} , there is a set ¯ A v, (cid:5) ⊂ N (cid:5) D ( v ) with size (1 + η/ α n/ H + v, (cid:5) , H − v, (cid:5) ⊂ D such that, for each ◦ ∈ { + , −} , H ◦ v, (cid:5) is ((1 + η/ εn, (1 + η/ εα n/ , ◦ )-skew-bounded on ( ¯ A v, (cid:5) , V ( D )).Select V , V , . . . , V k ⊂ V ( D ) according to the distribution in the lemma. Using Lemma 2.5,and a union bound, we have that, with high probability, the following hold. Q1 For each v ∈ V ( D ) and (cid:5) ∈ { + , −} , | ¯ A v, (cid:5) ∩ V | ≥ α p n/ µn . Q2 For each v ∈ V ( D ), (cid:5) , ◦ ∈ { + , −} , and w ∈ ¯ A v, (cid:5) , | N ◦ H ◦ v, (cid:5) ( w, V i ) | ≥ εp i n . Q3 For each v ∈ V ( D ), (cid:5) , ◦ ∈ { + , −} , and w ∈ V ( D ), | N ¯ ◦ H ◦ v, (cid:5) ( w, ¯ A v, (cid:5) ) ∩ V ) | ≤ (1+ η ) εα p n/ η ) εµn , where ¯ ◦ ∈ { + , −} is such that ¯ ◦ (cid:54) = ◦ .Indeed, by Lemma 2.5, as ε, η, α, p , p , . . . , p k (cid:29) /n , for any instance of v ∈ V ( D ), (cid:5) , ◦ ∈{ + , −} , and w ∈ V ( D ), the property Q1 above holds with probability 1 − exp( − Ω( n )), and thesame is true for Q2 and Q3 . Therefore, by a union bound, with high probability, the properties Q1 , Q2 and Q3 hold.Now, for each v ∈ V ( D ) and (cid:5) ∈ { + , −} , using Q1 , choose A v, (cid:5) ⊂ ¯ A v, (cid:5) ∩ V with | A v, (cid:5) | = µn .By Q2 and Q3 , we have, for each ◦ ∈ { + , −} and i ∈ [ k ], that H ◦ v, (cid:5) is ( εp i n, (1 + η ) εµn, ◦ )-skew-bounded on ( A v, (cid:5) , V i ), as required,We will now use the guide sets produced by Lemma 3.5 to randomly embed T , the smallcore of the original tree, and then use the guide graphs to find matchings from certain subsetsof the image of the embedding to other random sets, as follows. Lemma 3.6.
Let /n (cid:28) c (cid:28) β (cid:28) ε, q, α ≤ and /n (cid:28) c (cid:28) p (cid:28) /m . Let D be an n -vertexdigraph with δ ( D ) ≥ (1 / α ) n .Let T be an oriented tree with ∆ ± ( T ) ≤ cn/ log n consisting of a subtree T ⊂ T with | T | ≤ βn , such that every vertex in V ( T ) \ V ( T ) is attached as a leaf to T . Let t ∈ V ( T ) . Let U = V ( T ) and let U ∪ . . . ∪ U m be a partition of V ( T ) \ V ( T ) such that, for each i ∈ [ m ] , either e T ( V ( T ) , U i ) = 0 or e T ( U i , V ( T )) = 0 . Let V , V , . . . , V m ⊂ V ( D ) be disjoint random setschosen uniformly at random subject to | V | = qn , and, for each i ∈ [ m ] , | V i | = (cid:98) (1 + ε ) | U i |(cid:99) + pn .Then, with high probability, for each s ∈ V , there is an embedding of T into D such that t isembedded to s , and, for each i ∈ { , , . . . , m } , U i is embedded into V i . roof. Choose µ such that c, β (cid:28) µ (cid:28) ε, q, α . For each j ∈ [ m ], let p j = ( (cid:98) (1 + ε ) | U i |(cid:99) /n ) + p .Choose V , V , . . . , V m according to the distribution in the lemma. By Lemma 3.5 with p = q ,with high probability, for each v ∈ V ( D ) and (cid:5) ∈ { + , −} , there is R1 a set A v, (cid:5) ⊂ N (cid:5) D ( v ) ∩ V with size qα n/
4, and R2 digraphs H ◦ v, (cid:5) ⊂ D , ◦ ∈ { + , −} , such that, for each j ∈ [ m ] and ◦ ∈ { + , −} , H ◦ v, (cid:5) is( µp j n, (1 + ε/ µqα n/ , ◦ )-skew-bounded on ( A v, (cid:5) , V j ).We will now show that, given only R1 and R2 , we can embed T as required in the lemma foreach s ∈ V . Let then s ∈ V . We will randomly embed T into D [ V ], as follows, before showingthat, with positive probability, it can be extended into the required copy of T . Let (cid:96) = | T | andlabel V ( T ) = { t , . . . , t (cid:96) } , so that t = t and T [ { t , . . . , t i } ] is a tree for each i ∈ [ (cid:96) ]. Let s = s and embed t to s . For each i ∈ { , . . . , (cid:96) } in turn, let j i ∈ { , . . . , i − } be such that t j i isthe in- or out-neighbour of t i in T [ { t , . . . , t i } ] and let (cid:5) i ∈ { + , −} be such t i ∈ N (cid:5) i T ( t j i ), and,uniformly at random, embed t i to s i ∈ A s ji , (cid:5) i \ { s , . . . , s i − } . Claim 3.7.
For each j ∈ [ m ] , with high probability, the embedding of T can be extended to anembedding of T [ V ( T ) ∪ U j ] by embedding U j into V j . As p (cid:29) /n , and m ≤ /p , we can take a union bound over all j ∈ [ m ], to show that, withpositive probability, for each j ∈ [ m ], the embedding of T can be extended to T [ V ( T ) ∪ U j ]by embedding U j into V j , and hence T can be embedded as required in the lemma. Therefore,there is some choice of the embedding of T for which this can be done. It is left then to proveClaim 3.7. Proof of Claim 3.7.
Let j ∈ [ m ] and let ◦ j ∈ { + , −} be such that all the edges from V ( T ) to U j in T are ◦ j -edges. For each i ∈ [ (cid:96) ], let d j,i = | N ◦ j T ( t i , U j ) | . For each i ∈ [ (cid:96) ], take d j,i new verticesand call them w j,i,i (cid:48) , i (cid:48) ∈ [ d j,i ]. Let W j = { w j,i,i (cid:48) : i ∈ [ (cid:96) ] , i (cid:48) ∈ [ d j,i ] } . Let K j be the directed graphwith vertex set W j ∪ V j , containing only ◦ j -edges from W j to V j , and where, for each i ∈ [ (cid:96) ], i (cid:48) ∈ [ d j,i ] and v ∈ V j , there is a ◦ j -edge from w j,i,i (cid:48) to v in K j if, and only if, s i v ∈ E ( H ◦ j s ji , (cid:5) i ).We will show that, with high probability, K j is ( µp j n, µp j n, ◦ j )-skew-bounded on ( W j , V j ).This is enough to prove the claim, as, by Proposition 3.2, there is a perfect ◦ j -matching from W j into V j in K j . Thus, we can label distinct vertices v (cid:48) j,i,i (cid:48) , i ∈ [ (cid:96) ], i (cid:48) ∈ [ d j,i ] in V j so that w j,i,i (cid:48) v (cid:48) j,i,i (cid:48) , i ∈ [ (cid:96) ] and i (cid:48) ∈ [ d j,i ], is a matching in K j . For each i ∈ [ (cid:96) ], use the vertices v (cid:48) j,i,i (cid:48) , i (cid:48) ∈ [ d j,i ],to embed d j,i ◦ j -neighbours of t i in U j into V j . This is possible as, by the definition of K j and H ◦ j s ji , (cid:5) i , s i v (cid:48) j,i,i (cid:48) is a ◦ j -edge in D . Therefore, this extends the embedding of T to an embeddingof T ∪ T [ U j ] with U j embedded into V j , as required.Thus, it is sufficient to prove that, with high probability, K j is ( µp j n, µp j n, ◦ j )-skew-boundedon ( W j , V j ). Now, for each i ∈ [ (cid:96) ], s i ∈ A j i , (cid:5) i , and therefore s i has at least µp j n ◦ j -neighboursin V j in H ◦ j s ji , (cid:5) i by R2 . Therefore, for each i ∈ [ (cid:96) ] and i (cid:48) ∈ [ d j,i ], w j,i,i (cid:48) has at least µp i n ◦ j -neighbours in K j . That is, each v ∈ W j has at least µp j n ◦ j -neighbours in K j . Thus, letting¯ ◦ j ∈ { + , −} with ¯ ◦ j (cid:54) = ◦ j , it is sufficient to prove that, for each v ∈ V j , with probability 1 − o ( n − ), d ¯ ◦ j K j ( v, W j ) ≤ µp j n .Let then v ∈ V j . For each i ∈ [ (cid:96) ], let X j,vi = (cid:26) d j,i if s i v ∈ E ( H ◦ j s ji , (cid:5) i )0 otherwise,so that d ¯ ◦ j K j ( v, W j ) = (cid:80) i ∈ [ (cid:96) ] X j,vi . Note that, when s i ∈ A s ji , (cid:5) i \ { s , . . . , s i − } is chosen uniformlyat random, by R1 and R2 , and as β (cid:28) ε, α, q and i ≤ (cid:96) ≤ βn , if d i,j >
0, then X j,vi = d j,i with14robability at most d H ◦ jsji, (cid:5) i ( v ) | A s ji , (cid:5) i \ { s , . . . , s i − }| ≤ (1 + ε/ µqα n/ qα n/ − ( i − ≤ (1 + ε ) µ. Let γ = (1 + ε ) µ . Then, for each i ∈ [ (cid:96) ], E ( X j,vi | X j,v , . . . , X j,vi − ) ≤ γ · d j,i .Let Y j,vi = 0 and, for each i ∈ [ (cid:96) ], let Y j,vi = (cid:80) ≤ i (cid:48) ≤ i ( X j,vi (cid:48) − γd j,i ). Then, Y j,v , Y j,v , . . . , Y j,v(cid:96) is a supermartingale, since E [ Y j,vi +1 | Y , . . . , Y i ] = E [ X j,vi +1 | Y , . . . , Y i ] − γd j,i +1 + Y j,vi ≤ γd j,i +1 − γd j,i +1 + Y j,vi = Y j.vi for each i ≥
0. Note further that | Y j,vi +1 − Y j,vi | = | X j,vi +1 − γd j,i +1 | ≤ d j,i +1 for each j ∈ [ (cid:96) − d j,i ≤ cn/ log n for each i ∈ [ (cid:96) ], and (cid:80) i ∈ [ (cid:96) ] d j,i ≤ | U j | ≤ n ,we have (cid:80) i ∈ [ (cid:96) ] d j,i ≤ cn / log n . Therefore, by Azuma’s inequality (Lemma 2.6) with t = pn/ c (cid:28) p , P ( Y j,v(cid:96) ≥ pn/ ≤ − p n log n/ cn ) = o ( n − ) . Thus, with probability 1 − o ( n − ), we have Y j,v(cid:96) < pn/
3, so that d ¯ ◦ j K j ( v, W j ) = (cid:88) i ∈ [ (cid:96) ] X j,vi = γ · (cid:88) i ∈ [ (cid:96) ] d j,i + Y j,v(cid:96) = γ | U j | + Y j,v(cid:96) < γ | U j | + pn/ ≤ γp j n/ (1 + ε ) = (1 + ε ) µp j n/ (1 + ε ) = µp j n, completing the proof of the claim, and hence the lemma.Finally, by combining Lemma 3.6 and Lemma 2.13, we can prove Lemma 2.4. Proof of Lemma 2.4.
Let p satisfy 1 /n (cid:28) c (cid:28) p (cid:28) /K . For each j ∈ [ (cid:96) ], let s j be the vertex of S j with an in- or out-neighbour in V ( T (cid:48) ) in T . Let R be a maximal set of pairs ( R, r ) for which R is a directed tree with at most K edges and r ∈ V ( R ), such that the pairs ( R, r ) are unique upto isomorphism. Let m = |R| and enumerate R as ( R , r ) , . . . , ( R m , r m ). Note that p (cid:28) /m Let T (cid:48)(cid:48) = T [ V ( T (cid:48) ) ∪ N + T ( V ( T (cid:48) )) ∪ N − T ( V ( T (cid:48) ))]. For each i ∈ [ m ] and (cid:5) ∈ { + , −} , let U i, (cid:5) ⊂ V ( T (cid:48)(cid:48) ) be the set of vertices s j , j ∈ [ (cid:96) ], for which ( S j , s j ) is isomorphic to ( R i , r i ) and the edgefrom V ( T (cid:48) ) to s j in T is a (cid:5) -edge.In V ( D ), take disjoint random sets V and V i, (cid:5) ,j , i ∈ [ m ], (cid:5) ∈ { + , −} and j ∈ { , } , uniformlyat random subject to the following. • | V | = εn/ • For each i ∈ [ m ] and (cid:5) ∈ { + , −} , we have that | V i, (cid:5) , | = (cid:98) (1 + ε/ | U i, (cid:5) |(cid:99) + pn and | V i, (cid:5) , | = ( (cid:98) (1 + ε/ | U i, (cid:5) |(cid:99) + pn )( | R i | − | V | + (cid:88) i ∈ [ m ] , (cid:5)∈{ + , −} ( | V i, (cid:5) , | + | V i, (cid:5) , | ) = | V | + (cid:88) i ∈ [ m ] , (cid:5)∈{ + , −} ( (cid:98) (1 + ε/ | U i, (cid:5) |(cid:99) + pn ) | R i |≤ εn/ ε/ (cid:88) j ∈ [ (cid:96) ] | S j | + (cid:88) i ∈ [ m ] pn · | R i |≤ εn/ ε/ | T | + (2 pn ) · m · K ≤ n. ε/ v ∈ V . By Lemma 3.6, with high probability, if v ∈ V , then thereis an embedding of T (cid:48)(cid:48) into D such that t is embedded to v , V ( T (cid:48) ) ⊂ V , and, for each i ∈ [ m ]and (cid:5) ∈ { + , −} , U i, (cid:5) is embedded into V i, (cid:5) , . By Lemma 2.13, for each i ∈ [ m ] and (cid:5) ∈ { + , −} , D [ V i, (cid:5) , ∪ V i, (cid:5) , ] contains | V i, (cid:5) , | vertex disjoint copies of R i , in which r i is copied into V i, (cid:5) , . Foreach i ∈ [ m ] and (cid:5) ∈ { + , −} , add each copy of R i containing an embedded vertex of U i, (cid:5) to theembedding of T (cid:48)(cid:48) . Note that this results in a copy of T . Given our decomposition T ⊂ T ⊂ T ⊂ T = T , we have now embedded T . We now embedthe vertices from V ( T ) \ V ( T ), recalling that we obtain T from T by adding constant-sizedtrees, where each tree is attached to T by exactly two bare paths of length 2. In the followinglemma, we embed T \ T so that the vertices in V ( T ) ∩ V ( T ) are embedded to preselectedvertices (labelled a i , b i , i ∈ [ (cid:96) ]). This allows us to extend our embedding of T to one of T . Lemma 3.8.
Let /n (cid:28) /K ≤ /k (cid:28) α, ε . Suppose T is a forest formed of vertex-disjointoriented trees T i , i ∈ [ (cid:96) ] , with at most (1 − ε ) n vertices in total, and so that k ≤ | T i | ≤ K , foreach i ∈ [ (cid:96) ] , and each tree T i contains distinct vertices r i and s i which are leaves in T i whoseneighbour has total in- and out-degree 2.Suppose D is an n -vertex digraph with δ ( D ) ≥ (1 / α ) n , containing the distinct vertices a i , b i , i ∈ [ (cid:96) ] . Then, D contains a copy of T in which, for each i ∈ [ (cid:96) ] , r i is embedded to a i and s i is embedded to b i .Proof. Let β be such that 1 /k (cid:28) β (cid:28) α, ε . For each i ∈ [ (cid:96) ], let r (cid:48) i and s (cid:48) i be the neighboursin T i of r i and s i , respectively, and let T (cid:48) i = T i − { r i , r (cid:48) i , s i , s (cid:48) i } . Let T (cid:48) be the forest composedof connected components T (cid:48) i , i ∈ [ (cid:96) ], so that | T (cid:48) i | ≤ (1 − ε ) n . Let A = { a i , b i : i ∈ [ (cid:96) ] } . Then | A | = 2 (cid:96) ≤ n/k . Let B ⊂ V ( D ) \ A be a random subset of vertices with | B | = βn .Let D (cid:48) = D − A − B . As 1 /k, β (cid:28) α, ε , we have | D (cid:48) | ≥ (1 − ε/ n and δ ( D (cid:48) ) ≥ (1 / α/ | D (cid:48) | .Since (cid:12)(cid:12) T (cid:48) (cid:12)(cid:12) ≤ (1 − ε ) n ≤ (1 − ε )(1 − ε/ | D (cid:48) | ≤ (1 − ε/ (cid:12)(cid:12) D (cid:48) (cid:12)(cid:12) , we have, by Lemma 2.14, with high probability we can find a copy, S (cid:48) say, of T (cid:48) inside D (cid:48) .Let r (cid:48)(cid:48) i and s (cid:48)(cid:48) i be the neighbours in T (cid:48) of r (cid:48) i and s (cid:48) i , respectively, for each i ∈ [ (cid:96) ], and let a (cid:48)(cid:48) i and b (cid:48)(cid:48) i be the copy of r (cid:48)(cid:48) i and s (cid:48)(cid:48) i in S (cid:48) , respectively. Claim 3.9.
The following holds with high probability. For any pair of vertices u, v ∈ V ( D ) and (cid:5) , ◦ ∈ { + , −} , we have that | N (cid:5) ( u ) ∩ N ◦ ( v ) ∩ B | ≥ αβn .Proof of Claim 3.9. Let u, v ∈ V ( D ) and (cid:5) , ◦ ∈ { + , −} . Note that, by the semi-degree conditionon D , | N (cid:5) ( u ) ∩ N ◦ ( v ) | ≥ αn , and hence | N (cid:5) ( u ) ∩ N ◦ ( v ) ∩ B | has a hypergeometric distributionwith E | N (cid:5) ( u ) ∩ N ◦ ( v ) ∩ B | ≥ αβn . By Lemma 2.5, and a union bound over all pairs u, v ∈ D and (cid:5) , ◦ ∈ { + , −} , the statement in the claim thus holds with probability 1 − o (1).Thus, with high probability, we can assume the property in the claim holds. Now, for each i ∈ [ (cid:96) ], embed a i and b i to r i and s i , respectively. Let (cid:5) i , ◦ i , (cid:5) (cid:48) i , ◦ (cid:48) i ∈ { + , −} be such that r (cid:48) i ∈ N (cid:5) i ( r i ) ∩ N ◦ i ( r (cid:48)(cid:48) i ), and s (cid:48) i ∈ N (cid:5) (cid:48) i ( s i ) ∩ N ◦ (cid:48) i ( s (cid:48)(cid:48) i ). Greedily and disjointly, for each i ∈ [ r ],embed r (cid:48) i to a vertex in N (cid:5) i ( a i ) ∩ N ◦ i ( a (cid:48)(cid:48) i ) ∩ B and embed s (cid:48) i to a vertex in N (cid:5) (cid:48) i ( b i ) ∩ N ◦ (cid:48) i ( b (cid:48)(cid:48) i ) ∩ B .Note that this is possible, since, from the property in the claim we have, for each i ∈ [ r ] (cid:12)(cid:12) N (cid:5) i ( a i ) ∩ N ◦ i ( a (cid:48)(cid:48) i ) ∩ B (cid:12)(cid:12) , (cid:12)(cid:12)(cid:12) N (cid:5) (cid:48) i ( b i ) ∩ N ◦ (cid:48) i ( b (cid:48)(cid:48) i ) ∩ B (cid:12)(cid:12)(cid:12) ≥ αβn ≥ nk ≥ r. This completes the embedding of T with the property required in the lemma.16 .3 Proof of Theorem 2.2 We now combine Lemma 2.4 and Lemma 3.8 to find a copy of any almost-spanning tree.
Proof of Theorem 2.2.
Take
K, k and η so that 1 /n (cid:28) /K (cid:28) /k (cid:28) η (cid:28) ε, α . Let D be an n -vertex graph with δ ( D ) ≥ (1 / α ) n . Let T be an oriented tree on at most (1 − ε ) n verticeswith ∆ ± ( T ) ≤ cn/ log n . By Lemma 2.8, we can find forests T ⊂ T ⊂ T ⊂ T = satisfying P1 to P4 . Randomly partition V ( D ) into three parts, V ( D ) = V ∪ V ∪ V so that | V | = | T | + εn/ | V | = | T | − | T | + εn/
3, and | V | = | T | − | T | + εn/
3. Note that, with probability at least ε/ v ∈ V .By applying Lemma 2.10 with A = V , with high probability we have δ ( D [ V ]) ≥ (1 / α/ | V | . Thus, by applying Lemma 2.4 to D = D [ V ] and T = T , we can find a copy of T in V in which t is copied to v . By P3 , for some (cid:96) ∈ N , T is formed from T by the addition oftrees F i , i ∈ [ (cid:96) ], where k ≤ | F i | ≤ K , which are each attached to T by exactly two bare paths oflength 2, P i and Q i say. For each i ∈ [ (cid:96) ], let p i and q i be the endpoint of P i and Q i , respectively,which belongs to T . Let a i and b i be the embedding in V of p i and q i , respectively, and let A = { a i , b i : i ∈ [ (cid:96) ] } .By Lemma 2.10 again, we have, with high probability, δ ( D [ A ∪ V ]) ≥ (1 / α/ | A ∪ V | .Applying Lemma 3.8 to D [ A ∪ V ] with T i = F i ∪ P i ∪ Q i , r i = p i , and s i = q i , for each i ∈ [ (cid:96) ], wecan find a copy of T in D [ V ∪ V ]. Now since T is a tree, any vertex in T \ T can have at mostone neighbour in T . Note that, by Lemma 2.10, we know that with high probability every vertexin D has at least (1 / α/ | V | ≥ ηn in-neighbours in V and at least (1 / α/ | V | ≥ ηn out-neighbours in V . Let j = | T | − | T | ≤ ηn and order the vertices of T \ T by u , . . . , u j , sothat T [ V ( T ) ∪ { u , . . . , u i } ] is a tree for each i ∈ [ j ]. Embed the vertices u , . . . , u j greedily into V , to complete the copy of T in D . Noting that this embedding was successful with probabilityat least ε/ − o (1) >
0, there must always be such a copy of T . The aim of this section is to prove Theorem 2.1. The main idea is as follows. Given a small tree T , we split it into two trees T (cid:48) and T (cid:48)(cid:48) and randomly embed T (cid:48) vertex by vertex. With positiveprobability, the resulting tree is such that, given the right number of other vertices in the graph,we can embed T (cid:48)(cid:48) to extend this into a copy of T while making some small modifications to thecopy of T (cid:48) . Essentially, we show that, for each vertex y , there are many vertices in the embeddingof T (cid:48) which we can switch with y and still get a copy of T . We then embed T (cid:48)(cid:48) vertex-by-vertex,at each step switching an unused vertex into the copy of T (cid:48) in place of a vertex which we caninstead use to extend the (partial) embedding of T (cid:48)(cid:48) . Proof of Theorem 2.1.
Take λ such that ε (cid:28) λ (cid:28) µ . Using Proposition 2.3, let T = T (cid:48) ∪ T (cid:48)(cid:48) ,where t ∈ V ( T (cid:48) ) and εn < | T (cid:48)(cid:48) | ≤ εn . Let (cid:96) = | T (cid:48) | , and label V ( T (cid:48) ) as t , . . . , t (cid:96) so that t = t , T (cid:48) [ t , . . . , t i ] is a tree for each i ∈ [ (cid:96) ], and the leaves of T (cid:48) appear last in this order (except for t )and in any bare path of length 6 the middle 3 vertices appear consecutively. For each i ∈ [ (cid:96) ], let T i = T (cid:48) [ { t , . . . , t i } ].Pick an arbitrary vertex v ∈ V ( D ), and let R be the graph with only the vertex v . Foreach i = 2 , . . . , (cid:96) , do the following. Let (cid:5) i ∈ { + , −} be such that N (cid:5) i T i ( t i ) is non-empty (and thuscontains exactly one vertex. Let ◦ i ∈ { + , −} with ◦ i (cid:54) = (cid:5) i . Take R i − , which is a copy of T i − ,and let w i be the copy of the sole vertex in N (cid:5) i T i ( t i ) in R i − . Pick a vertex v i independently atrandom from N ◦ i D ( w i ) \ V ( R i − ). Embed t i to v i to get R i , a copy of T i .17ote that this process always ends with a copy of T (cid:48) , as N ◦ i D ( w i ) \ V ( R i − ) always has sizeat least d ◦ i D ( w i ) − | T | ≥ (1 / α ) n − µn and µ (cid:28) α . Let R = R (cid:96) , so that R is a copy of T (cid:48) . Wewill show that, with positive probability the following property holds. S For each distinct x, y ∈ V ( D ) and (cid:5) ∈ { + , −} , |{ i ∈ [ (cid:96) ] : v i ∈ N (cid:5) D ( x ) and N ± R ( v i ) ⊂ N ± D ( y ) }| ≥ λn. Noting | R | = | T (cid:48) | ≤ | T |−| T (cid:48)(cid:48) | +1 ≤ ( µ − ε ) n , let A ⊂ V ( D ) contain V ( R ) so that | A | = ( µ − ε ) n ,and let v be the copy of t . We will show in two claims that, with positive probability S holds,and that, if S holds, then A and v satisfy the property in the theorem. Thus, the theorem followsfrom these two claims. Claim 4.1.
With positive probability, S holds.Proof of Claim 4.1. Fix x, y ∈ V ( D ) and (cid:5) ∈ { + , −} with x (cid:54) = y . We will show that S holds for x, y and (cid:5) with probability at least 1 − / n , so that the result follows by a union bound.For convenience, let us take two cases. Either T (cid:48) has 2 µ n leaves (Case I) or µ n vertex-disjoint bare paths with length 6 (Case II). One of these cases must hold, as, suppose thatCase I does not hold and thus T (cid:48) has fewer than 2 µ n leaves. Then, by Lemma 2.7, we knowthat there is some s and some vertex-disjoint bare paths P i , i ∈ [ s ], in T (cid:48) of length 6 so that | T (cid:48) − P − · · · − P s | ≤ µ n + 2 (cid:96)/
7. Removing the internal vertices of each path P i , i ∈ [ s ], from T (cid:48) removes 5 vertices, and | T (cid:48) | = (cid:96) , so that (cid:96) − s ≤ µ n + 2 (cid:96)/
7, and therefore s ≥ ( (cid:96) − (cid:96)/ / − µ n/ ≥ (cid:96)/ − µ n ≥ ( µ − ε ) n/ − µ n ≥ µ n, where the final inequality holds since ε (cid:28) µ . Case I.
Assume that at least µ n leaves of T (cid:48) are out-leaves, where the proof whenever T (cid:48) hasat least µ n in-leaves follows similarly. Let (cid:96) (cid:48) be the smallest integer such that, for each i > (cid:96) (cid:48) , t i is a leaf of T (cid:48) . We will analyse the embedding of T (cid:48) in two stages. First, for the embeddingof t , . . . , t (cid:96) (cid:48) , we show that with high probability there will be plenty of these vertices which areadjacent to out-leaves in t (cid:96) (cid:48) +1 , . . . , t (cid:96) that are embedded to in-neighbours of y . Then, we willanalyse the embedding of t (cid:96) (cid:48) +1 , . . . , t (cid:96) , and show that plenty of these vertices whose in-neighbourin t , . . . , t (cid:96) (cid:48) was embedded to an in-neighbour of y are themselves embedded to a (cid:5) -neighbour of x . For each i ∈ [ (cid:96) (cid:48) ], let c i be the number of out-leaves of t i in T (cid:48) . For each i ∈ [ (cid:96) (cid:48) ], let X i be therandom variable which takes value c i if v i ∈ N − D ( y ), and 0 otherwise. Note that, for each i ∈ [ (cid:96) ],if c i >
0, then, when the process selects v i , having chosen v , . . . , v i − , X i = c i with probabilityat least | ( N ◦ i D ( w i ) \ V ( R i )) ∩ N − D ( y ) | n ≥ | ( N ◦ i D ( w i )) ∩ N − D ( y ) | − | R i | n ≥ αn − µnn ≥ α, (4)as α (cid:29) µ . Thus, for each i ∈ [ (cid:96) ], E [ X i | X , . . . X i − ] ≥ αc i .Note that (cid:80) i ∈ [ (cid:96) (cid:48) ] c i is the number of out-leaves of T (cid:48) , so that (cid:80) i ∈ [ (cid:96) (cid:48) ] c i ≥ µ n and, as ∆( T ) ≤ cn/ log n , (cid:80) i ∈ [ (cid:96) (cid:48) ] c i ≤ cn / log n . Let Z = 0 and, for each i ∈ [ (cid:96) (cid:48) ], let Z i = (cid:80) j ≤ i ( X j − αc j ). Then,( Z i ) i ≥ is a submartingale, since E [ Z i +1 | Z , . . . , Z i ] = Z i + E [ X i +1 − αc i +1 | X , . . . X i ] ≥ Z i for each i ∈ [ (cid:96) (cid:48) ]. Furthermore, for each i ∈ [ (cid:96) (cid:48) ], we have | Z i − Z i − | = | X i − αc i | ≤ c i . Therefore,by Azuma’s inequality (Lemma 2.6) with t = αµ n/
2, we have P (cid:88) i ∈ [ (cid:96) (cid:48) ] ( X i − αc i ) ≤ − t ≤ (cid:32) − t (cid:80) i ∈ [ (cid:96) (cid:48) ] c i (cid:33) ≤ (cid:18) − t log ncn (cid:19) ≤ n . (5)18ere, the final inequality holds because c (cid:28) µ, α . Therefore, with probability at least 1 − / n ,we have (cid:80) i ∈ [ (cid:96) (cid:48) ] X i ≥ (cid:80) i ∈ [ (cid:96) (cid:48) ] αc i − αµ n/ ≥ αµ n/ m = (cid:80) i ∈ [ (cid:96) (cid:48) ] X i ≥ αµ n/
2. Consider now the embedding of t (cid:96) (cid:48) +1 , . . . , t (cid:96) . Let j , . . . , j m ∈{ (cid:96) (cid:48) + 1 , . . . , (cid:96) } be such that t j i is an out-leaf of T (cid:48) and the image of N − T (cid:48) ( t j i ) is an in-neighbourof y for each i ∈ [ m ]. For each i ∈ [ m ], let Y i be the random variable which takes value 1 if v j i isin N (cid:5) D ( x ), and 0 otherwise. Note that, similarly to the calculation in (4), E [ Y i | Y , . . . Y i − ] ≥ α for each i ∈ [ m ]. Let Z = 0 and, for each i ∈ [ m ], let Z i = (cid:80) j ≤ i ( Y j − α ). Then, ( Z i ) i ≥ is asubmartingale, since E [ Z i +1 | Z , . . . , Z i ] = Z i + E [ Y i − α | Y , . . . , Y i − ] ≥ Z i for each i ∈ [ m ].Furthermore, | Z i − Z i − | = | Y i − α | ≤ i ∈ [ m ]. Therefore, by Azuma’s inequality(Lemma 2.6) with t = αm/
2, we see that P (cid:88) i ∈ [ m ] Y i − α < − t ≤ (cid:18) − t (1 − α ) m (cid:19) ≤ n , (6)where the final inequality holds because 1 /n (cid:28) µ, α . Hence, with probability at least 1 − / n ,we have (cid:80) i ∈ [ m ] Y i ≥ αm/
2. Note that (cid:12)(cid:12) { i ∈ [ (cid:96) ] : v i ∈ N (cid:5) D ( x ) and N ± R ( v i ) ⊂ N ± D ( y ) } (cid:12)(cid:12) ≥ (cid:80) i Y i .Thus, by taking a simple union bound over the events in (5) and (6) and using λ (cid:28) α, µ , wesee that in total, with probability at least 1 − / n , (cid:12)(cid:12) { i ∈ [ (cid:96) ] : v i ∈ N (cid:5) D ( x ) and N ± R ( v i ) ⊂ N ± D ( y ) } (cid:12)(cid:12) ≥ αm/ ≥ λn. Taking a union bound over all possible x, y ∈ V ( D ) and (cid:5) ∈ { + , −} , we see that in this case S holds with probability at least 1 / Case II.
Let m = µ n . Let P , . . . , P m be vertex disjoint paths of length 6 in T , so that, if,for each i ∈ [ m ], j i is such that t j i is the middle vertex of P i , then the vertices t j i appear in orderin t , . . . , t (cid:96) .For each i ∈ [ m ], let X i be the random variable taking value 1 if v j i ∈ N (cid:5) D ( x ) and N ± R ( v j i ) ⊂ N ± D ( y ) (7)and 0 otherwise. Note that, by virtue of the labelling of the t , . . . , t (cid:96) , the vertices that appearin N ± R ( v j i ) are exactly the vertices v j i − and v j i +1 . When we choose each of v j i − , v j i , v j i +1 , theprobability that it satisfies its condition in (7) (however the previous vertices v i (cid:48) are chosen) isat least α , in a calculation similar to (4). Therefore, we have, for each i ∈ [ m ], that E [ X i | X , . . . X i − ] ≥ α .Now, let Z = 0 and, for each i ∈ [ m ], let Z i = (cid:80) j ≤ i ( X j − α ). Then, E [ Z i +1 | Z , . . . , Z i ] = Z i + E [ X i +1 | X , . . . , X i ] − α ≥ Z i for each i ∈ [ m ], and thus ( Z i ) i ≥ is a submartingale.Furthermore, | Z i − Z i − | = (cid:12)(cid:12) X i − α (cid:12)(cid:12) ≤ i ∈ [ m ]. Thus, by Azuma’s inequality(Lemma 2.6), letting t = α m/
2, we have P [ Z m ≤ − t ] ≤ (cid:18) − t m (cid:19) = 2 exp (cid:18) − α m (cid:19) ≤ n , as 1 /n (cid:28) α, µ . Therefore, with probability at least 1 − / n , we have Z m > − t , so that, as λ (cid:28) µ, α , (cid:12)(cid:12) { i ∈ [ (cid:96) ] : v i ∈ N (cid:5) D ( x ) and N ± R ( v i ) ⊂ N ± D ( y ) } (cid:12)(cid:12) ≥ (cid:12)(cid:12) { i ∈ [ m ] : v j i ∈ N (cid:5) D ( x ) and N ± R ( v j i ) ⊂ N ± D ( y ) } (cid:12)(cid:12) = (cid:88) i ∈ [ m ] X i = Z m + α m ≥ α m − t ≥ λn. x, y ∈ V ( D ) and (cid:5) ∈ { + , −} , we see that in this case S holds with probability at least 1 / Claim 4.2. If S holds then A and v satisfy the property in the theorem.Proof of Claim 4.2. Let B ⊂ V ( D ) with A ⊂ B and | B | = µn . Let k = | T (cid:48)(cid:48) | − ≤ εn and labelthe vertices of V ( T (cid:48)(cid:48) ) \ V ( T (cid:48) ) as s , . . . , s k , so that, for each i ∈ [ k ], T (cid:48) i := T (cid:48) ∪ T (cid:48)(cid:48) [ { s , . . . , s i } ] isa tree. Note that | B \ V ( R ) | = k and label the vertices of B \ V ( R ) as y , . . . , y k .Let S = R . Now, for each i = 1 , . . . , k in turn, do the following. Let x i ∈ V ( S i − ) and (cid:5) i ∈ { + , −} be such that we need to add a (cid:5) i -neighbour to x i as a leaf to get a copy of T (cid:48) i .Choose some j (cid:48) i ∈ [ (cid:96) ] \ { , j (cid:48) , . . . , j (cid:48) i − } such that v j (cid:48) i ∈ N (cid:5) i D ( x i ) and N ± S i − ( v j (cid:48) i − ) ⊂ N ± D ( y i ) and d + S i − ( v j (cid:48) i ) + d − S i − ( v j (cid:48) i ) ≤ /λ. Replace v j (cid:48) i with y i in S i − and add v j (cid:48) i as a (cid:5) i -neighbour of x i to get S i , a copy of T (cid:48) i with vertexsets V ( S i − ) ∪ { y i } .We need only show that there is such a vertex v j (cid:48) i in each case, as if this process finds S k ,then we have a copy of T (cid:48) k = T . Fix then i ∈ [ k ]. Note that there are at most (4 /λ ) · εn ≤ λn/ v j (cid:48) , . . . , v j (cid:48) i in R i − , and therefore N + R ( v i (cid:48) ) = N + R i − ( v i (cid:48) ) and N − R ( v i (cid:48) ) = N − R i − ( v i (cid:48) ) for all but at most λn/ i (cid:48) ∈ [ (cid:96) ]. Furthermore, as (cid:80) i (cid:48) ∈ [ (cid:96) ] ( d + T ( t i (cid:48) ) + d − T ( t i (cid:48) )) ≤ n , at most λn/ i ∈ [ k ] can have d + S i − ( v j (cid:48) i ) + d − S i − ( v j (cid:48) i ) > /λ . Thus, such an j (cid:48) i willalways exist by S . References [1] N. Alon and J. H. Spencer.
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