On Erdős-Sós conjecture for trees of large size

Erdős and Sós conjectured that every graph G of average degree greater than k−1 contains every tree of size k. Several results based upon the number of vertices in G have been proved including the special cases where G has exactly k+1 vertices (Zhou), k + 2 vertices (Slater, Teo and Yap), k + 3 vertices (Woźniak) and k + 4 vertices (Tiner). We further explore this direction. Given an arbitrary integer c > 1, we prove Erdős-Sós conjecture in the case when G has k + c vertices provided that k > k0(c) (here k0(c) = c12polylog(c)). We also derive a corollary related to the Tree Packing Conjecture.


Introduction
A set of (simple) graphs G 1 , G 2 , . . ., G q are said to pack into a complete graph K n (in short pack) if G 1 , G 2 , . . ., G q can be found as pairwise edge-disjoint subgraphs in K n .Many classical problems in Graph Theory can be stated as packing problems.In particular, H is a subgraph of G if and only if H and the complement of G pack.
Erdős and Sós conjectured that every graph G with average degree greater than k − 1 contains every tree with k edges.This conjecture has been restated by Woźniak [16] as follows.
Conjecture 1. Suppose that G is a graph with n vertices and T is any tree with k edges.
, then G and T pack (into the complete graph K n ).
Ajtai, Komlós, Simonovits and Szemerédi have announced a proof of Conjecture 1 for sufficiently large k.There are many partial results concerning this conjecture.They have been obtained either for some special families of graphs [2,5,6,15] or for some special families of trees [7,11,12] or else for certain values of the parameters k and n.In particular, the cases where n is equal to k + 1, k + 2, k + 3, or k + 4 were proved by Zhou [17], by Slater, Teo, and Yap [13], by Woźniak [16], and by Tiner [14], respectively.We extend these results to n = k + c for any c, provided k is sufficiently large.Theorem 2. Let c be a positive integer and let k 0 (c) = γc 12 ln 4 c where γ is some universal sufficiently large constant.Then for every t = 1, . . ., c and for every integer k k 0 (c) the following holds.If T is a tree with k edges and G is a graph on k + t vertices with |E(G)| < t(k+t) 2 , then T and G pack into K k+t .Another famous tree packing conjecture (TPC) posed by Gyárfás [9] states that any set of n − 1 trees T n , T n−1 , . . ., T 2 such that T i has i vertices pack into K n .In [8] Bollobás suggested the following weakening of TPC Conjecture 3.For every c 1 there is an n(c) such that if n n(c), then any set of c trees T 1 , T 2 , . . ., T c such that T i has n − (i − 1) vertices pack into K n .
Bourgeois, Hobbs and Kasiraj [4] showed that any three trees T n , T n−1 , T n−2 pack into K n .Recently, Balogh and Palmer [3] proved that any set of t = 1 10 n 1/4 trees T 1 , . . ., T t such that no tree is a star and T i has n − i + 1 vertices pack into K n .We obtain the following corollary of Theorem 2: Corollary 4. Let c be a positive integer and let n 0 (c) = γc 12 ln 4 c where γ is some universal sufficiently large constant.If n n 0 (c), then any set of c trees T 1 , T 2 , . . ., T c , such that Proof.The proof is by induction on c.For c = 1 the statement is obvious.So fix some c > 1 and assume that the statement is true for c − 1.Let T 1 , T 2 , . . ., T c be any set of c trees such that T i has n − 2(i − 1) vertices.By the induction hypothesis T 1 , T 2 , . . ., T c−1 pack into K n .Let G be a graph with Furthermore, T c has n − (2c − 1) edges.Thus, by Theorem 2, G and T c pack, which completes the proof of the corollary.
The notation is standard.In particular denotes the degree of a vertex v in G, δ(G) and ∆(G) denote the minimum and the maximum degree of G, respectively.N G (v) denotes the set of neighbors of v and, for a subset of vertices the electronic journal of combinatorics 23(1) (2016), #P1.52

Preliminaries
In the proof we refine the approach of Alon and Yuster from [1].However, we apply it in a slightly different way as we choose random subsets B i (to be defined later) in a denser graph.
We write Bin(p, n) for the binomial distribution with n trials and success probability p.Let X ∈ Bin(n, p).We will use the following two versions of the Chernoff bound which follows from formulas (2.5) and (2.6) from [10] by taking t = 2µ − np and t = np − µ/2, respectively. If On the other hand, if µ E[X] = np then Proposition 5. Let G be a graph with n vertices and at most m edges.Let Proof.The proposition is true because The following technical lemma is the main tool in the proof.A version of it appeared in [1].Lemma 6.Let G be a graph with n vertices and at most m edges.Let . ., n, be any subsets of V (G) with the additional requirement that if u ∈ A i then d(u) < a.For i = 1, . . ., n let B i be a random subset of A i where each vertex of A i is independently selected to B i with probability p < 1/a.Let the electronic journal of combinatorics 23(1) (2016), #P1.52 Proof.Fix some vertex Observe that |C i | is a sum of d(v i ) independent indicator random variables each of which has success probability at most 2mp d(v i ) .Thus, the expectation of |C i | is at most 2mp.Therefore, by (1), the probability of |C i | being larger than 4mp satisfies Consider now the second part of the lemma.Observe that for u ∈ A i , the probability that u ∈ B i is p.On the other hand, for any j, the probability that u ∈ N [B j ] is at least 1 − ap.Indeed, u ∈ N [B j ] if and only if u ∈ B j or one of its neighbors belongs to B j .Since u ∈ A i , it has at most a − 1 neighbors.Hence, the probability that u ∈ N [B j ] is at most ap.Therefore, as long as i 1/(ap), Observe that |D i | is a sum of |A i | independent indicator random variables, each having success probability at least p e .Therefore the expectation of 3 Proof of Theorem 2 The proof is by induction on t.By Zhou's result the theorem holds for t = 1.So fix some t, 2 t c, and assume that the statement is true for t − 1.Let G be a (bipartite) graph that arises from T by adding a set Clearly, G and G pack if and only if T and G pack.
the electronic journal of combinatorics 23(1) (2016), #P1.52 Thus, by the induction hypothesis, G − v and T pack.Therefore, G and T pack as well.Hence, we may assume that non-neighbors.Suppose that α vertices of G have degree greater than or equal to 5c.Thus and so α n 5 .Therefore Now, we divide the proof into two cases depending whether ∆(T ) < 60cn 3/4 or ∆(T ) 60cn 3/4 .

Case ∆(T ) < 60cn 3/4
Recall that S i ⊂ V (G) \ N [v i ] with the assumption that if u ∈ S i then d G (u) < 5c.
For i = 1, . . ., n let B i be a random subset of S i where each vertex of S i is independently selected to B i with probability Let Claim 8.The following conditions hold simultaneously with positive probability: the electronic journal of combinatorics 23(1) (2016), #P1.52 Proof.Recall that |E(G)| < tn 2 cn 2 .Thus, by Lemma 6, Furthermore, by Claim 7, Hence, by Lemma 6 (with a = 5c and Thus, by the union bound, each part of the lemma holds with probability greater than 1/2.Hence both hold with positive probability.Therefore, we may fix sets B 1 , . . ., B n satisfying all the conditions of Claim 8 with respect to the cardinalities of the sets C i and D i .We construct a packing f : V (G) → V (G ) in three stages.At each point of the construction, some vertices of V (G) are matched to some vertices of V (G ), while the other vertices of V (G) and V (G ) are yet unmatched.Initially, all vertices are unmatched.We always maintain the packing property, that is for any In Stage 1 we match certain number of vertices of G that have the largest degrees.After this stage, by the assumption that ∆(G ) 60cn 3/4 , both G and G do not have unmatched vertices of high degree (vertices of high degree are the main obstacle in packing).This fact enables us to complete the packing in Stages 2 and 3.
Stage 1 Let x be the largest integer such that d G (v x ) n 1/4 300c .Thus, by Proposition 5, This stage is done repeatedly for i = 1, . . ., x and throughout it we maintain the following two invariants 1.At iteration i we match v i with some vertex f 2. Furthermore, we match all yet unmatched neighbors of f (v i ) to some vertices of B i (this way all neighbors of f (v i ) in G are matched to vertices of i j=1 B j ).To see that this is possible, consider the i'th iteration of Stage 1 where v i is some yet unmatched vertex of G. Let Q be the set of all yet unmatched vertices of G having degree less than or equal to 3. Note that, by Proposition 5, the number of vertices of degree less than or equal to 3 in G is at least n/2.Hence, the electronic journal of combinatorics 23(1) (2016), #P1.52 Let X be the set of already matched neighbors of v i and let Y = N G (f (X)).Thus, the valid choice for f (v i ) would be a vertex of Q \ Y .To see that such a choice is possible, it is enough to show that Thus, by the first invariant of Stage 1, and by (6) and Claim 8 In order to maintain the second invariant it remains to match the yet unmatched neighbors of f (v i ) with vertices from B i .Let R be the set of neighbors of f (v i ) in G that are still unmatched.Recall that |R | 3. We have to match vertices of R with some vertices of B i .Since , a valid choice of such vertices is by taking an |R |-subset of D i .By Claim 8 and by (6), Thus, all vertices of D i are still unmatched.Hence, such a choice is possible.
Stage 2 Let M 1 and M 1 be the set of matched vertices of G and G after Stage 1, respectively.Clearly In Stage 2 we match the vertices from V (G ) \ (M 1 ∪ J ), one by one, with some What is more, by the second invariant of Stage 1, the neighbors of each f (v i ), i = 1, . . ., x, are already matched.Hence, Thus, by the definition of x, for each u ∈ X we have the electronic journal of combinatorics 23(1) (2016), #P1.52 Therefore, Stage 3 Let M 2 and M 2 be the sets of matched vertices of G and G after Stage 2, respectively.In order to complete a packing of G and G , it remains to match the vertices of V (G) \ M 2 with the vertices of J .Consider a bipartite graph B whose sides are V (G) \ M 2 and J .For two vertices u ∈ V (G) \ M 2 and v ∈ J , we place an edge uv ∈ E(B) if and only if it is possible to match u with v (by this we mean that mapping u to v will not violate the packing property).Thus u is not allowed to be matched to at most d G (u)∆(G ) vertices of J .Hence Now we will evaluate d B (v ).We define X and Y in the same way as in Stage 2. Then (7) holds again.Hence, v is not allowed to be matched to at most ∆(G ) Therefore, by Hall's Theorem there is a perfect matching in B, and so a packing of G and G .

Case ∆(T ) 60cn 3/4
In this case we will follow the ideas from the previous subsection.However, the key difference is that now both G and G may have vertices of high degrees.Because of this obstacle, a packing has two more stages at the beginning.After a preparatory Stage 1, in Stage 2 we match the vertices of G that have high degrees with vertices of G that have small degrees.Then in Stage 3, we match the vertices of G having high degree.This stage is very similar to Stage 1 from the previous subsection, but with the change of the role of G and G .Finally, we complete the packing in Stages 4 and 5, which are analogous to Stages 2 and 3 from the previous subsection.Let Let P ⊆ N G (v 1 ) be the set of neighbors of v 1 such that each vertex in P has degree at most q in G , and every neighbor different from v 1 of every vertex from P has degree at most q in G .Proof.Note that every vertex v ∈ N G (v 1 ) \ P has the property that d G (v ) > q or v has a neighbor w = v 1 such that d G (w ) > q.Therefore, and the statement follows.
We construct a packing f : V (G) → V (G ) in five stages.At each point of the construction, some vertices of V (G) are matched to some vertices of V (G ), while the other vertices of V (G) and V (G ) are yet unmatched.Initially, all vertices are unmatched.
Stage 1.We first match v n with v 1 , i.e. f (v n ) = v 1 .Next we match the neighbors of v n with d G (v n ) vertices from I .This is possible since, by (3) Moreover, since I is a set of isolated vertices, this maping does not violate the packing property.
Stage 2. Let z be the largest integer such that d G (v z ) n 1/4 .Since |E(G)| < cn/2, by Proposition 5 This stage is done repeatedly for i = 1, . . ., z and throughout it we maintain the following invariants: 2. Furthermore, we also make sure that all neighbors of f (v i ) in G are matched to vertices of S i ∪ {v n }.
Note that because G is acyclic and since there are no edges (in G) between v i and S i ∪{v n } for those v i that are non-neighbors of v n , such a mapping does not violate the packing property.
To see that this mapping is possible, consider the i'th iteration of Stage 2, where v i is a vertex of G with d G (v i ) n 1/4 5c.In particular v i ∈ i−1 j=1 S j .Thus, if v i is already matched, then it was matched in Stage 1 and so f (v i ) ∈ I .Then, the second invariant of Stage 2 is automatically preserved because f (v i ) is isolated.
Therefore, we may assume that v i is yet unmatched.In this case we may take f (v i ) to be any vertex of P .Indeed, note that |P | z and before iteration i, the number of already matched vertices of P was at most i − 1.
Furthermore, observe that since v 1 is a common neighbor of all f (v j ), j = 1, . . ., i, at iteration i the overall number of matched vertices is at most Let R be the set of neighbors of f (v i ) in G that are still unmatched.Note that R contains all neighbors of f (v i ) apart from v 1 .Thus, in order to maintain the second invariant, it suffices to match vertices of R with some vertices of S i .Note that by the choice of P and since v 1 is already matched, |R | q − 1.Let Q be the set of yet unmatched vertices of S i .By Claim 7 and formula (10), |Q| n/4 + t − (t + n/59) n 1/4 59c > q − 1.
Hence, this is possible.Before we describe Stage 3, we need some preparations.Let M 2 be the set of all vertices of G that were matched in Stage 1 or 2. Similarly, let M 2 be the set of all vertices of G that were matched in Stage 1 or 2. Recall that Let For each 1 i r we define a set S i ⊆ V (H ) \ N H [w i ] to be a largest independent set of vertices but with the additional requirement that each w ∈ S i has d H (w ) < 180.
set of all yet unmatched vertices of G. Clearly, |Q| |J | 4n/9 since the vertices of J remain unmatched in every step of Stage 2. Let X be the set of already matched neighbors of v .Let Y = N G (f −1 (X )).Thus, the valid choice for f −1 (v ) would be a vertex of Q \ Y .To see that such a choice is possible we will prove that |Q \ Y | > 0. Recall that |X | 60cn 3/4 .