Embedding Trees into Dense Graphs and the Tree-Star Ramsey Number

Tree Embeddings and Tree-Star Ramsey Numbers

Zilong Yan

^∗

Yuejian Peng

^†

January 3, 2022

Abstract

We say that a graphFcan be embedded into a graphGifGcontains an isomorphic copy ofF as a subgraph. Guo and Volkmann [6] conjectured that ifGis a connected graph with at leastnvertices and minimum degree at leastn−3, then any tree withnvertices and maximum degree at mostn−4 can be embedded intoG. In this paper, we give a result slightly stronger than this conjecture and obtain a sufficient and necessary condition that a tree withnvertices and maximum degree at most n−3 can be embedded into a connected graph G with at leastn vertices and minimum degree at leastn−3. Our result implies that the conjecture of Guo and Volkmann is true with one exception.

We also give an application to the Ramsey number of a tree versus a star.

Keywords: Tree, Embedding, Ramsey number, Tree-Star Ramsey number.

1 Introduction

All graphs considered throughout the paper are simple graphs, i.e. without loops and multiple edges. Let V(G) denote the vertex set ofGand letE(G) denote the edge set ofG. Forv∈V(G), letN(v) ={u∈ V(G)|uv∈E(G)},N[v] =N(v)∪{v}, andd(v) =|N(v)|. ForS⊆V(G), denoteN(S) =∪_v∈SN(v) and N[S] =N(S)∪S. Let δ(G) =min{d(v)|v∈V(G)}and ∆(G) =max{d(v)|v∈V(G)}. Let Kn denote the complete graph on n vertices andK_1,m denote the star withm+ 1 vertices. For graphs GandH, theRamsey NumberR(G, H) is the smallest integerN such that any red-blue-coloring ofE(KN) yields a redGor a blueH. For graphsF andG, we say that an injectionφ:V(F)→V(G) is anembedding ofF intoGif for any edgexyinF,φ(x)φ(y) is an edge inG. We say thatF can be embedded intoGif there is an embedding ofF intoG.

Degree conditions for tree embedding have been studied actively. A well-known conjecture of Erd˝os- S´os states that any tree with n vertices can be embedded into a simple graph with average degree exceedingn−2. In [7], Havet, Reed, Stein and Wood studied the conditions on the maximum degree and the minimum degree of a graph to embed any tree withn vertices, they proposed an interesting conjecture that any tree with n vertices can be embedded into a simple graph with maximum degree more thann+ 1 and minimum degree at leastb²⁽ⁿ⁺¹⁾₃ c. The Loebl-Koml´os-S´os conjecture [5] states that any tree with nvertices can be embedded into a simple graph with median degree at least n+ 1. We

∗School of Mathematics, Hunan University, Changsha 410082, P.R. China. Email: zilongyan@hnu.edu.cn.

†Corresponding author. School of Mathematics, Hunan University, Changsha, 410082, P.R. China. Email:

ypeng1@hnu.edu.cn. Supported in part by National Natural Science Foundation of China (No. 11931002).

arXiv:2112.14955v1 [math.CO] 30 Dec 2021

study minimum degree conditions to embed trees, and it has natural applications on tree-star Ramsey numbers.

In [8], Parson determined the path-star Ramsey numbers. A key ingredient in the proof is thatPn, a path withnvertices, can be embedded into a graph with minimum degreen−1. This is generalized to the following well-known result (see [3]).

Lemma 1.1 If Gis a graph with minimum degreen−1, then any tree withnvertices can be embedded intoG.

This lemma implies the Ramsey result of Burr [2]: R(Tn, K1,m)≤m+n−1, whereTn is a tree with nvertices and K1,m is the star withm+ 1 vertices. Applying Lemma 1.1, Cockayne [3] improved the upper bound ofR(Tn, K1,m) tom+n−2 for a special class of trees withnvertices and some values ofm andn. Further, Guo and Volkmann [6] generalized the result of Cockayne to any tree withnvertices. A key making the generalization possible is that Guo and Volkmann showed that any tree withnvertices other thanK_1,n−1can be embedded into a connected graph with at leastnvertices and minimum degree n−2. They remarked that there are connected graphsGwith at leastnvertices and minimum degree n−3, and trees with n vertices and maximum degreen−3 which cannot be embedded into G. And they proposed the following conjecture.

Conjecture 1 ([6])IfGis a connected graph with at leastnvertices and minimum degree at leastn−3, then any tree withnvertices and maximum degree at mostn−4 can be embedded into G.

We show a result slightly stronger than this conjecture and obtain a sufficient and necessary condition that trees withn vertices and maximum degree at most n−3 can be embedded into a graphG with minimum degree at leastn−3.

Letpandqbe positive integers. LetT(p, q) be a tree with a longest pathv1v2v3v4v5satisfying that d(v3) = 2, andv2andv4havepandqleaves respectively (see Figure 1). The following is the main result

Figure 1: T(p, q) in this paper.

Theorem 1.2 Let G be a connected graph with at least nvertices and minimum degree at least n−3.

Let Tn be a tree withn vertices and∆(Tn)≤n−3. ThenTn can be embedded intoGif and only if the following situations do not happen.

(i)G=Kn−3,n−3 andTn=T(p, n−3−p), wherep≥1.

(ii) G is a balanced complete (k+ 1)-partite graph with a vertices in each part and ka = n−3, and Tn=T(1, n−4), wherek≥2 is a positive integer.

This result implies that Conjecture 1 holds except in situation (i) in Theorem 1.2. A non-negative integer Nis called a linear combination of two positive integerspandqif there exist non-negative integerskand l such that N =kp+lq. Applying Theorem 1.2, we are going to prove the following tree-star Ramsey numbers in Section 4.

Theorem 1.3 Let Tn be a tree withn vertices,∆(Tn)≤n−3, andTn6=T(p, n−3−p). Ifm+n−3 is not a linear combination ofn−1andn−2, thenR(Tn, K1,m)≤m+n−3.

Theorem 1.4 LetT_nbe a tree withnvertices and∆(T_n)≤n−3. Letm=k(n−1)+3and0≤k≤n−5.

(i) If Tn 6=T(p, n−3−p), thenR(Tn, K1,m) =m+n−3.

(ii) If m+n−3 is not a linear combination of n−1, n−2, n−3 +a₁, ..., n−3 +a_d, where a1, a2, ..., ad are positive divisors of n−3 no less than 3, then R(T(1, n−4), K1,m) = m+n−3.

OtherwiseR(T(1, n−4), K_1,m) =m+n−2.

(iii) Letp≥2. Ifm+n−3 is not a linear combination of2n−6,n−1 andn−2, thenR(T(p, n−3− p), K_1,m) =m+n−3. Otherwise R(T(p, n−3−p), K_1,m) =m+n−2.

The conjecture of Erd˝os-S´os states that any tree withnvertices can be embedded into a graph with average degree exceedingn−2. It is easy to show that a connected graph with average degree greater thann−2 contains a connected subgraph with minimum degree greater than ¹₂(n−2). So if one can characterize what kind of trees withnvertices can be embedded into a connected graphGwith at least nvertices andδ(G)≥ ¹₂(n−1), then these trees will satisfy the conjecture. In general, it is interesting to study what kind ofT_ncan be embedded into a connected graphGwith at leastnvertices andδ(G)≥t.

In Section 2, we give some crucial lemmas in the proof of Theorem 1.2. The proof of Theorem 1.2 will be given in Section 3. In Section 4, we prove Theorems 1.3 to 1.4.

2 Preparations

Definition 2.1 A labelling {v1, v2, ..., vn} of vertices of a tree with n vertices is called a conventional labelling if for each j ∈[2, n],|N(vj)∩ {v1, ..., v_j−1}|= 1. To simplify the notation, we always denote the unique vertex inN(vj)∩ {v1, ..., vj−1}by vj⁰ for a conventional labelling.

Note that a treeT always has a conventional labelling. We may take any vertex inV(T) as v1 and order the other vertices asv₂, ..., v_n in increasing order of the distances tov₁.

For given graphs F andG, we say that a subset U ⊆V(F) has been embedded into G(or φ is an embedding ofU to G) if an embeddingφ of F[U] into Ghas been established. If U ⊆V(F) has been embedded toG via an embeddingφ, we say thatφcan be extended to a vertex v∈V(F)\U if there existsv⁰ ∈V(G) such that the extension function ˆφ:U∪ {v} →V(G) defined by ˆφ(u) =φ(u) foru∈U and ˆφ(v) =v⁰ is an embedding ofF[U∪ {v}] into G.

Lemma 2.2 Let G be a connected graph with at least n vertices. Let {v1, v2, ..., vn} be a conventional labelling ofTn. Let j∈ {2, ..., n}. Letφbe an embedding of {v1, ..., vj−1}intoG.

(i) If

|N(φ(vj⁰))\ {φ(v1), ..., φ(vj−1)}| ≥1, (1)

thenφcan be extended to vj.

(ii) Ifδ(G)≥n−1, then φcan be extended tovj.

(iii) Let δ(G) ≥ n−2. If 2 ≤ j ≤ n−1, then φ can be extended to vj. If j = n and there exists k∈ {1,2, ..., n−1} \ {n⁰} such thatφ(vk)∈/ N(φ(vn⁰)), thenφcan be extended to vj.

(iv) Letδ(G)≥n−3. If2≤j ≤n−2, thenφcan be extended tovj. Ifj=n−1and|{φ(v1), ..., φ(vn−2)}\

N(φ(v_(n−1)0))| ≥2, then φ can be extended tovj. Ifj =n and |{φ(v1), ..., φ(v_n−1)} \N(φ(vn⁰))| ≥3, thenφcan be extended to vj.

Proof of Lemma 2.2. The condition in (i) guarantees that there is a free vertex inN(φ(vj⁰)) to embed vj, therefore vj can be embedded to a vertex in N(φ(vj⁰)), and this proves (i). (ii), (iii) and (iv) are

direct implications of (i). 2

Lemma 2.3 ([4]) Let G be a connected graph with minimum degreeδ(G). Then G contains a path of lengthmin{n,2δ(G) + 1}.

Lemma 2.4 Let Gbe a connected graph with minimum degreeδ. LetS0⊆V(G)and let S1be a proper subset of S₀ such thatN(S₁)⊆S₀. If there existsα∈S₀ such that N(α)6⊂S₀, then there exists a path of length at least|S1|+δ− |S0|+ 1starting from a neighbor αoutsideS0.

Proof of Lemma 2.4. Let α₁ ∈ N(α)\S₀. Let α₁α₂...α_g be a maximum path outside S₀. Since N(S1) ⊆ S0, N(αg) ⊆ S0∪ {α1, α2, ..., α_g−1} \S1. Since |N(ag)| ≥ δ, |S0|+g−1− |S1| ≥ δ. So

g≥ |S₁|+δ− |S₀|+ 1. 2

Lemma 2.5 Let G be a non-complete and connected graph, then there exist α, β, γ ∈V(G) such that αβ, αγ∈E(G)andβγ /∈E(G).

Proof of Lemma 2.5. Take a shortest path between two non-adjacent vertices, then three consecutive

vertices in this path satisfy the condition. 2

3 Embedding trees

We give the proof of Theorem 1.2 below.

Proof of Theorem 1.2. Let Tn be a tree with n vertices. Let P be a longest path of Tn. Let |P| be the number of vertices of path P and i = |P| −2. Since |P| = 3 implies that T_n = K_1,n−1, it contradicts ∆(Tn) ≤ n−3. So|P| ≥ 4. For |P| ≥ 5, we label the two end vertices as v_n−1 and vn, all other vertices inP as v₁, v₂, ..., v_i such thatP =v_n−1v₁v₂...v_iv_n, i.e. v_n−1 is connected tov₁,v₁ is connected tov2, · · ·, v_i−1 is connected to vi, andvi is connected to vn. We order the vertices outside P as {vi+1, ..., v_n−2} in non-decreasing order of the distances to P. For example, v_i+1 is connected to a vertex inP. Note that{v1, v2, ..., v_n−1, vn}is a conventional labelling. We will show that there is no embeddingφ:V(T_n)→V(G) if and only if situation (i) or (ii) in Theorem 1.2 happens.

By Lemma 1.1, we can easily embed{v1, ..., v_n−2} intoG. So what we need to work on is to embed v1 andviproperly so that there is a free vertex for us to embedvn−1 in the neighborhood of the vertex v1 embedded to, and there is a free vertex for us to embed vn in the neighborhood of the vertex vi

embedded to.

By Lemma 2.5, there existu1, u2 and u3 in Gsatisfying thatu2∈N(u1) and u3 ∈N(u2)\N(u1).

Letφ(vj) =uj forj ∈[3]. By Lemma 2.2 (iv),φ can be extended to{v4, ..., v_i−1}. Denote φ(vj) =uj

forj∈[i−1]. DenoteKi={φ(v1), ..., φ(vi−1)}.

Case 1. |P| ≥6.

Case 1.1. There exists a vertex, denoted byui, inN(ui−1)\Kisuch that |Ki\N(ui)| ≥2.

In this case, let φ(vi) = ui. Since δ(G) ≥ n−3, u1, u3 ∈/ N(u1), by Lemma 2.2 (iv), φ can be extended to{v1, ..., vn−1}. Further |Ki\N(ui)| ≥2 implies that |{φ(v1), ..., φ(vn−1)} \N(ui)| ≥3, by Lemma 2.2 (iv), there exists an embedding fromTn intoG.

Case 1.2. There exists a vertex, denoted by ui, in N(ui−1)\Ki such that |Ki\N(ui)| = 1, and

|K_i\N(u)| ≤1 for anyu∈N(u_i−1)\K_i.

Note that|Ki\N(ui)|= 1 implies that there exists a vertexuxin Ki such that uxui∈/E(G).

If d(u_i) ≥ n−2, let φ(v_i) = u_i. Since u₃ ∈/ N(u₁), by Lemma 2.2 (iv), φ can be extended to {vi+1, ..., v_n−1}. Since ux, ui ∈/ N(ui), |N(ui)\ {φ(v1), ..., φ(v_n−1)}| ≥ 1 and there is at least one free vertex in N(u_i) to embedv_n. So it is sufficient to consider d(u_i) =n−3. Let K =K_i∪N[u_i], then

|K|=n−1.

Claim 3.1 N(u₁)⊆K if T_n cannot be embedded into G.

Proof of Claim 3.1. Suppose that there exists a vertex denoted byun−1inN(u1)\K. Sinceu3∈/N(u1), by Lemma 2.2 (iv),φcan be extended to{v₁, ..., v_n−1}. Furthermore, we can guarantee that some vertex va∈Tn\ {vn}is embedded tou_n−1, wherea∈[i+ 1, n−1] (This is always possible since we can always embedv_n−1 tou_n−1ifu_n−1is not previously used). Nowu_x, u_i, u_n−1∈/ N(u_i), then there exists a free vertex inN(ui)\ {φ(v1), ..., φ(v_n−2), φ(v_n−1)} to embedvn andTn can be embedded intoG. 2 Claim 3.2 N(u₂)⊆K if T_n cannot be embedded into G.

Proof of Claim 3.2. If there exists a vertexu⁰₁∈N(u2)\K, then reassignφ(v1) =u⁰₁ and letφ(vi) =ui. By the choice ofu⁰₁, we know thatu_i∈/ N(φ(v₁)). By Lemma 2.2 (iv),φcan be extended to{v₁, ..., v_n−1}.

Recall that there exists a vertex ux ∈ Ki such that uxui ∈/ E(G). We claim that ux 6= u1. Since d(u₁)≥n−3,u₁, u₃∈/ N(u₁),N(u₁)⊆K and|K|=n−1,u₁u_i ∈E(G). Sinceu_x, u⁰₁∈/ N(u_i), there is at least one free vertex inN(ui)\ {φ(v1), ..., φ(v_n−1)}to embedvn. A contradiction. 2 Claim 3.3 N(u_t)⊆K for1≤t≤i−1 ifT_n cannot be embedded into G.

Proof of Claim 3.3. Use induction on t. Fort= 1 or 2, it is guaranteed by Claim 3.2. Let 3≤t≤i−1.

We assume thatN(u_s)⊆Kfor 1≤s≤t−1. IfN(u_t)6⊂K, then there exists a vertexu⁰_t−1∈N(u_t)\K.

By Lemma 2.4, there exists a pathu⁰_t−1u⁰_t−2...u⁰₁outside K. Reassign φ(vα) =u⁰_α for 1≤α≤t−1 and letφ(v_i) =u_i. Sinceu⁰_α∈/N(u_i) for 1≤α≤t−1 andt≥3, by Lemma 2.2 (iv),φcan be extended to

{vi+1, ..., vn}. 2

Claim 3.4 N(K)⊆K ifT_n cannot be embedded into G.

Proof of Claim 3.4. If there exists a vertexu⁰_i−1 ∈N(ui)\Ki such that there exists a vertexu⁰_i−2 ∈ N(u⁰_i−1)\ K, by Lemma 2.4, there exists a path u⁰_i−2u⁰_i−3...u⁰₁ outside K. Reassign φ(vα) = u⁰_α for

1≤α≤i−1 and letφ(vi) =ui. Sincei≥4 guarantees thatu⁰₁, u⁰₂∈/N(ui), by Lemma 2.2 (iv),φcan

be extended to{vi+1, ..., vn}. 2

By Claim 3.4, if Tn cannot be embedded into G, then K is a component with n−1 vertices, contradicting thatG is a connected graph with at leastn vertices. SoTn can be embedded intoGin this case.

Case 1.3. For each vertexuin N(u_i−1)\Ki,|Ki\N(u)|= 0.

Take a vertex, denote byui, inN(ui−1)\Ki, and letφ(vi) =ui.

If d(ui) ≥ n−1, since u1, u3 ∈/ N(u1), by Lemma 2.2 (iv), φ can be extended to an embedding on {v1, ..., vn−1}. Since d(ui) ≥ n−1 and φ(vi) = ui ∈/ N(ui), there is at least one free vertex in N(ui)\ {φ(v1), ..., φ(v_n−1)}to embedvn.

Ifd(ui) =n−2, then|N[ui]|=n−1. We claim that∪_u∈N[ui]N[u] =N[ui] ifTn cannot be embedded into G. If there exists a vertex, denote byu_n−1, inN(u₁)\N[u_i], letφ(v_i) = u_i and extend φ to an embedding on {v1, ..., vn−1} such that we embed some va ∈ {vi+1, ..., vn−1} to un−1 (This is always possible since we can always embedv_n−1tou_n−1ifu_n−1is not previously used). Nowu_i, u_n−1∈/ N(u_i), therefore there exists one free vertex in N(ui) to embedded vn and Tn can be embedded into G. So we have shown that N[u₁] ⊆ N[u_i]. Assume that ∪^t_i=1N[u_i] ⊆ N[u_i] for 1 ≤ t ≤ i−1, we show that N[ut+1] ⊆ N[ui]. If there exists a vertex u⁰_t ∈ N[ut+1]\N[ui], by Lemma 2.4, there exists a pathu⁰_tu⁰_t−1...u⁰₁ outsideN[u_i]. Reassign φ(v_α) =u⁰_α for 1≤α≤t. Since u_i ∈/ N(φ(v₁)), by Lemma 2.2 (iv), φ can be extended to an embedding on {v1, ..., v_n−1}. Since d(ui) = n−2 and φ(vi) = ui, φ(v_a)∈/N(u_i) for 1≤a≤t, there is at least one free vertex inN(u_i)\ {φ(v1), ..., φ(v_n−1)}to embedv_n. Therefore we have shown thatN[Ki]⊆N[ui]. If there exist verticesu⁰, u⁰⁰ such that u⁰ ∈N(ui)\Ki

and u⁰⁰ ∈N[u⁰]\N[u_i], since d(u₃)≥n−3, |N[ui]| =n−1 and u₁, u₃ ∈/ N(u₃) andN(u₃)⊆N[u_i], u3u⁰ ∈ E(G). Reassign φ(v2) = u⁰ and φ(v1) = u⁰⁰. Since ui ∈/ N(φ(v1)), by Lemma 2.2 (iv), φ can be extended to {v1, v₂, ..., v_n−1}. Since u⁰⁰u_i ∈/ E(G) and d(u_i) = n−2, there exists a free vertex in N(ui)\ {φ(v1), ..., φ(v_n−1)} to embed vn and Tn can be embedded into G. So we have shown that

∪_u∈N[ui]N[u] = N[u_i] if T_n cannot be embedded into G. Therefore N[u_i] is a component of n−1 vertices, contradicting thatGis a connected graph with at leastnvertices.

Now we discussd(u_i) =n−3. Since u₁u₃∈/E(G) andd(u₃)≥n−3, there exists a vertex, denoted by u⁰₂, in N(u3)\N[ui]. If there exists a vertex u⁰₁ ∈ N(u⁰₂)\N[ui], then reassign φ(v2) = u⁰₂ and φ(v1) =u⁰₁. Sinceφ(v1), φ(v2)∈/N(φ(vi)), by Lemma 2.2 (iv),φcan be extended to{v1, ..., vn}andTn

can be embedded intoG. So N(u⁰₂)⊆N[ui]. Sinceu⁰₂ui ∈/ E(G) andd(u⁰₂)≥n−3,N(u⁰₂) =N(ui).

Since u⁰₂u1, u⁰₂u3 ∈ E(G), we can reassign φ(v2) = u⁰₂. Now ui has a non-neighbor in Ki. Note that u3∈/N(φ(v1)) still holds. This situation is exactlyCase 1.2we have proved.

Case 2. |P|= 5.

In this case, P =v_n−1v1v2v3vn. IfdT_n(v2)≥3, then there exists a vertex v4 connected tov2 (see Figure 2). If dT_n(v2) = 2, then NT_n(v2) = {v1, v3}. If Tn = P, since |V(G)| ≥ 5 andδ(G) ≥ 2, by Lemma 2.3, P = T_n can be embedded in G. So if d_Tn(v₂) = 2, we may assume that d_Tn(v₁)≥ 3 or dT_n(v3)≥3 (see Figure 3).

Letu⁰₃be a vertex with maximum degree inV(G). Ifd(u⁰₃)≥n−2, if there exists a vertexu⁰₂∈N(u⁰₃) such that there exists a vertex u⁰₁ ∈N(u⁰₂)\N[u⁰₃], then let φ(v1) = u⁰₁, φ(v2) = u⁰₂ and φ(v3) =u⁰₃. Since u⁰₁, u⁰₃ ∈/ N(u⁰₁) and δ(G) ≥ n−3 and d(u⁰₃) ≥ n−2, by Lemma 2.2 (iv), φ can be extended to {v4, ..., vn}. If there is no such vertex u⁰₂ in N(u⁰₃), then N[u⁰₃] is a component, so d(u⁰₃) ≥ n−1.

Figure 2: Cased_Tn(v₂)≥3 Figure 3: Cased_Tn(v₂) = 2

If there exists a vertex u⁰₁ ∈ N(u⁰₃) such that d(u⁰₁) ≥ n−2, let φ(v1) = u⁰₁ and φ(v3) = u⁰₃ and φ(v2) ∈ N(u⁰₁), then by Lemma 2.2 (iv), φ can be extended to{v4, ..., vn}. So we may assume that d(ux) =n−3 for all ux ∈N(u⁰₃). Since|N[u⁰₃]| ≥n, for any vertexu⁰⁰₃ ∈N(u⁰₃), we can find vertices u⁰⁰₁, u⁰⁰₄ ∈ N(u⁰₃)\N[u⁰⁰₃]. Let φ(v₁) = u⁰⁰₁ and φ(v₂) = u⁰₃ and φ(v₃) = u⁰⁰₃ and φ(v₄) = u⁰⁰₄. Since φ(v1), φ(v3)∈/ N(φ(v1)),d(φ(v1))≥n−3 andφ(v1), φ(v3), φ(v4)∈/N(φ(v3)), by Lemma 2.2 (iv), φcan be extended to{v₄, ..., v_n}and T_n can be embedded intoG. Therefore, it is sufficient to consider that Gis an (n−3)-regular connected graph.

Case 2.1. G is an (n−3)-regular connected graph with at least n vertices and d_Tn(v₂) ≥ 3 (see Figure 2).

Claim 3.5 LetGbe (n−3)-regular andd_Tn(v₂)≥3(see Figure 2). If there exist verticesu_x, u_y, u_a, u_b∈ V(G) such thatuxuy ∈E(G) andua, ub∈N(uy)\N[ux](see Figure 4), then Tn can be embedded into G.

Proof. Letφ(v1) =ua,φ(v2) =uy,φ(v3) =uxandφ(v4) =ub. SinceGis (n−3)-regular,φ(v1), φ(v2)∈/ N(φ(v₁)), andφ(v₁), φ(v₄), φ(v₃)∈/ N(φ(v₃)), by Lemma 2.2 (iv),φcan be extended to{v₅, ..., v_n}. 2

Figure 4: Figure for Claim 3.5 Figure 5: Figure for Case 2.1

Take a vertexu⁰₃∈V(G). Since|N[u⁰₃]|=n−2,N[u⁰₃] is not a component. So there exists vertices u⁰₁, u⁰₂such thatu⁰₂∈N(u⁰₃) andu⁰₁∈N(u⁰₂)\N[u⁰₃] (see Figure 5). By Claim 3.5,N[u⁰₂]⊆N[u⁰₃]∪ {u⁰₁}.

Ifu⁰₁ has two neighbors outside N[u⁰₃], then these two neighbors are outside N[u⁰₂]. By Claim 3.5, Tn

can be embedded intoG. If N(u⁰₁)⊆N[u⁰₃], then N(u⁰₁) = N(u⁰₃). IfTn cannot be embedded intoG, Claim 3.5 implies that∪_u∈N[u⁰

3]N(u)∪N(u⁰₁) =N[u⁰₃]∪ {u⁰₁}. Otherwise there exists a vertexu∈N(u⁰₃) such thatuhas a neighbor woutsideN[u⁰₃]∪ {u⁰₁}, thenuhas two neighborsw andu⁰₁ outsideN[u⁰₃], contradicting Claim 3.5. ThusN[u⁰₃]∪ {u⁰₁} is a component onn−1 vertices, contradicting thatGis a connected graph with at leastnvertices. Therefore, there exists exactly one vertexua ∈N(u⁰₁)\N[u⁰₃].

By Claim 3.5,N(u⁰₁) andN(ua) differ by at most one vertex. SoN(u⁰₁)⊆N(u⁰₃)∪ {ua}. If there exists a vertexu∈N(u⁰₃)∩N(u⁰₁) connected to ua, then bothua and u⁰₁ are inN(u)\N(u⁰₃), contradicting Claim 3.5. SoN(u⁰₁) andN(ua) have no common neighbor inN[u⁰₃], contradicting Claim 3.5 again. So Tn can be embedded intoG.

Case 2.2. Gis (n−3)-regular andd_Tn(v₂) = 2 (see Figure 3), i.e. T_n=T(p, n−3−p) for somep.

Claim 3.6 Let G be an (n-3)-regular connected graph on at leastnvertices. If there exists u⁰₁, u⁰₂, u⁰₃∈ V(G) such that u⁰₁u⁰₂, u⁰₂u⁰₃ ∈ E(G) andu⁰₁u⁰₃ ∈/ E(G). Then T_n can be embedded into G or N(u⁰₁) = N(u⁰₃).

Proof. Suppose that N(u⁰₁)6=N(u⁰₃). SinceGis regular, there exists a vertexu4∈N(u⁰₁)\N[u⁰₃]. Let φ(v1) = u⁰₁, φ(v2) = u⁰₂, φ(v3) = u⁰₃ and φ(v4) = u4. Since u⁰₁u⁰₃ ∈/ E(G), and u⁰₁, u⁰₃, u4 ∈/ N(u⁰₃), by

Lemma 2.2 (iv),φcan be extended to an embedding ofT_n. 2

Assume thatTn cannot be embedded intoG. Let A1 be a maximal set inGsatisfying that it is an independent set and its any two vertices have a common neighbor. By Claim 3.6, all vertices inA₁have the same set of neighbors, sayB. Then|B|=n−3. We claim thatV(G) =A1∪B. Otherwise, there exists a vertexu_x∈V(G)\(A₁∪B) such thatu_xis connected to a vertex inB and none of the vertices inA1is adjacent toux. By Claim 3.6,N(ux) =B. This contradicts the maximality ofA1. Since it holds for all such maximal sets, we can divideB intoksetsB₁, B₂, ..., B_k such thatB_j is such a maximal set for eachj ∈ [k]. SinceGis regular, |A1|=|B1|=...=|Bk|=a,n−3 = akand |V(G)|=n−3 +a (See Figure 6). Since|V(G)| ≥n,a≥3.

Figure 6: Ka,a,...,a

If k = 1, since G is (n−3)-regular, G = K_n−3,n−3, and in this case, T(p, n−3−p) cannot be embedded into K_n−3,n−3. This is because that T(p, n−3−p) is a bipartite graph with 2 and n−2 vertices in each of the two parts. This is situation (i) in Theorem 1.2.

Ifk≥2, then let φ(v₁)∈B₁,φ(v₂)∈B₂andφ(v₃)∈A₁.

Figure 7: T(p, n−3−p)(2≤p≤n−5) Figure 8: T(1, n−4)

Ifd(v1)≥3 andd(v3)≥3 (see Figure 7), letφ(vk), φ(vj)∈A1 andφ(vs), φ(vt)∈B1, then the other leaves can be embedded inA∪B sinceGis a balanced complete (k+ 1)-partite graph, andTn can be embedded intoG. Now considerd(v1) = 2 ord(v3) = 2, i.e. Tn=T(1, n−4). We claim thatT(1, n−4) cannot be embedded intoG. Without loss of generality, assume thatd(v₁) = 2 (see Figure 8). Suppose thatT(1, n−4) can be embedded intoGandv3is embedded intoA1 (without loss of generality). Then all n−3 vertices connected to v₃ should be embedded outside A₁. Since |V(G)\A₁| =n−3,φ(v₁) must be inA1. Since all vertices ofT(1, n−4)\ {v1, v3} are connected tov1 orv3, then every vertex in T(1, n−4)\ {v₁, v₃} must be embedded outside A₁, i.e. there aren−2 vertices should be embedded outsideA1, but|V(G)\A1|=n−3. This is impossible. So in this caseT(1, n−4) cannot be embedded intoG. This is situation (ii) in Theorem 1.2.

Case 3. |P|= 4.

In this case we label the vertices of T_n as in Figure 9. Note that {v₁, v₂, ..., v_n} is a conventional labelling. Sinced(v1) +d(v2) =nand ∆(Tn)≤n−3,d(v1), d(v2)≥3.

Figure 9: T_n has n-2 leaves Case 3.1. ∆(G)≥n−1.

Let u, v ∈ V(G) such that d(u) = ∆(G) and uv ∈ E(G). Embed φ(v2) = u and φ(v1) = v. By Lemma 2.2 (iv),φcan be extended to{v1, ..., v_n−2}. Recall thatv_n−1andv_n have the same parent (see Figure 9). Sinced(u)≥n−1, there are two free vertices inN(u)\ {φ(v1), ..., φ(v_n−2)} to embedv_n−1 andvn. SoTn can be embedded inG.

Case 3.2. ∆(G) =n−2.

Let φ(v2) = u⁰₂ such that d(u⁰₂) = n−2. We claim that there exists a vertex u⁰₁ ∈ N(u⁰₂) such that there exists a vertexu⁰₃∈N(u⁰₁)\N[u⁰₂]. Otherwise ∪_u∈N[u⁰

2] =N[u⁰₂]. So N[u⁰₂] is a component with n−1 vertices, contradicting that Gis a connected graph on at least nvertices. Let φ(v1) = u⁰₁ and φ(v3) = u⁰₃. By Lemma 2.2 (iv), φ can be extended to{v1, ..., v_n−2}. Since d(u⁰₂) = n−2 and u⁰₃∈/N(u⁰₂), there are two free vertices inN(u⁰₂)\ {φ(v1), ..., φ(vn−2)}to embedvn−1andvn. SoTncan be embedded intoG.

Case 3.3. Gis (n−3)-regular.

Claim 3.7 If there exists a vertex u⁰₂ ∈ V(G) such that there exists a vertex u⁰₁ ∈ N(u⁰₂) with two neighborsu_x, u_y∈N(u⁰₁)\N[u⁰₂], thenT_n can be embedded intoG.

Proof. Let φ(v₁) = u⁰₁, φ(v₂) = u⁰₂, φ(v₃) = u_x and φ(v₄) = u_y. By Lemma 2.2 (iv), φ can be extended to{v1, ..., v_n−2}. Sinceu⁰₂, ux, uy ∈/ N(u⁰₂) and d(u⁰₂) = n−3, there are two free vertices in N(u⁰₂)\ {φ(v1), ..., φ(vn−2)} to embedvn−1 andvn. SoTn can be embedded intoG. 2 Claim 3.8 If Tn cannot be embedded into G, then for any vertex ux ∈ V(G), there exists a vertex uy ∈N(ux)such that uy has exactly one neighbor outside N[ux].

Proof. By Claim 3.7, any vertex inN(ux) has at most one neighbor outsideN[ux]. If∪_u∈N[ux]N[u]⊆ N[u_x], thenN[u_x] is a component onn−2 vertices, a contradiction. 2 If T_n cannot be embedded into G, let u_x ∈V(G) and u_y ∈ N(u_x) such that u_z ∈N(u_y)\N[u_x].

SinceG is (n-3)-regular and uz ∈N(uy)\N[ux], there is exactly one vertex ua ∈N(ux)\N(uy) and N(u_y) =N[u_x]∪{u_z}\{u_y, u_a}. By Claim 3.7,N[u_y]\{u_x} ⊆N[u_z]. Sinceu_yu_a∈/E(G),|d(u_a)|=n−3 and|N[ux]|=n−2, there is exactly one vertexub∈N(ua)\N(ux). Ifub=uz, thenN(uz) =N(ux).

For any vertex u ∈ N(u_x), u_z ∈ N(u)\N(u_x). Then by Claim 3.7, N[u_x]∪ {uz} is a component on n−1 vertices, a contradiction. So ub 6= uz. Since ua has exactly one neighbor outside N(ux), N[u_a] =N[u_x]∪ {ub} \ {uy}. Since all vertices inN(u_x) have at most one neighbor outsideN(u_x) and uzis a neighbor of all vertices inN(ux)\ {ua}=N(ua)\ {ub, ux} ∪ {uy}, by Claim 3.7,ubis not adjacent to any of these vertices, i.e, u_b has no neighbor inN(u_a)\ {ub}, contradicting Claim 3.7 again. SoT_n

can be embedded intoGin this case. 2

4 Tree-star Ramsey number

In this section, we apply Theorem 1.2 to obtain the Ramsey number of a tree versus a star for some cases.

Proof of Theorem 1.3. For any red-blue-coloring of E(K_m+n−3), let G_R and G_B be the red graph and the blue graph respectively. If ∆(GB) ≥m, then there is a blue K1,m. So we may assume that

∆(G_B) ≤ m−1. Hence δ(G_R) ≥ (m+n−4)−(m−1) = n−3. Since m+n−3 is not a linear combination ofn−1 andn−2, thenGR contains a component with order at leastn. By Theorem 1.2,

T_n can be embedded intoG_R. 2

Fact 4.1 For positive integersmandn,m+n−4is a linear combination ofn−1andn−2andm+n−3 is not a linear combination ofn−1andn−2if and only if m=k(n−1) + 3 and0≤k≤n−5.

Proof of Fact 4.1. Necessity: Ifm+n−4 =k(n−1) +l(n−2),k≥0 andl≥1, then m+n−3 =k(n−1) +l(n−2) + (n−1)−(n−2) = (k+ 1)(n−1) + (l−1)(n−2),

contradicting that m+n−3 is not a linear combination of n−1 and n−2. Therefore l = 0 and m= (k−1)(n−1) + 3, wherek≥1. Let us rewritem=k(n−1) + 3, where k≥0. Ifk≥n−4, then

m+n−3 =k(n−1) +n= (k−n+ 4)(n−1) + (n−2)(n−2), andm+n−3 is a linear combination of n-1 and n-2. So 0≤k≤n−5.

Sufficiency: If m = k(n−1) + 3 and 0 ≤ k ≤ n−5, then m+n−4 = (k+ 1)(n−1) is a linear combination of n−1 and n−2. If m+n−3 is a linear combination of n−1 and n−2, then m+n−3 =k(n−1) +n=k⁰(n−1) +l⁰(n−2) (k⁰, l⁰≥0). So (k+ 2−k⁰)(n−1) = (l⁰+ 1)(n−2). Since n−1 is relatively prime ton−2,k+ 2−k⁰=c(n−2) forc≥1. Thereforek≥(n−2)−2 +k⁰ ≥n−4,

a contradiction. 2

For a positive integer a dividing n−3, let B(n−3 +a) be the balanced complete ^n−3+a_a -partite graph onn−3 +avertices.

Proof of Theorem 1.4. Since m+n−4 = (k+ 1)(n−1), we can partition V(K_m+n−4) intok+ 1 disjoint equal parts withn−1 vertices in each part. Color edges in each part red and other edges blue.

Then ∆(GB)≤m+n−4−(n−1) =m−3. So this coloring yields neither a redTn nor a blueK1,m. ThereforeR(Tn, K1,m)≥m+n−3.

(i). By Fact 4.1 and Theorem 1.3,R(Tn, K1,m)≤m+n−3. ThereforeR(Tn, K1,m) =m+n−3.

(ii). For any red-blue-coloring ofE(Km+n−3), letGR andGB be the red graph and the blue graph respectively. If ∆(GB) ≥ m, then there is a blue K1,m. So we may assume that ∆(GB) ≤ m−1.

Hence δ(GR)≥(m+n−4)−(m−1) =n−3. By Fact 4.1, GR contains a component with at least n vertices. By Theorem 1.2, ifm+n−3 is not a linear combination of n−1, n−2,n−3 +a1, ..., n−3 +ad, thenT(1, n−4) can be embedded intoGR. SoR(T(1, n−4), K1,m)≤m+n−3. Therefore R(T(1, n−4), K1,m) =m+n−3.

If m+n−3 =k1(n−1) +l1(n−2) +s1(n−3 +a1) +...+sd(n−3 +ad), for red-blue-coloring ofE(K_m+n−3), letG_R andG_B be the red graph and the blue graph respectively. Let G_R=k₁K_n−1∪ l1Kn−2∪s1B(n−3 +a1)∪...∪sdB(n−3 +ad) andGB=G^c_R. By Theorem 1.2,T(1, n−4) cannot be embedded into G_R. Sinceδ(G_R)≥n−3, then ∆(G_B)≤(m+n−4)−(n−3) = m−1. So there is neither a redTn nor a blueK1,m. ThereforeR(T(1, n−4), K1,m)≥m+n−2. The same lower bound was given in [6]. SoR(T(1, n−4), K_1,m) =m+n−2.

(iii). For any red-blue-coloring of E(K_m+n−3), letGR andGB be the red graph and the blue graph respectively. If ∆(G_B)≥m, then there is a blueK_1,m. So we may assume that ∆(G_B)≤m−1. Hence δ(GR)≥n−3. By Fact 4.1 andm+n−3 is not a linear combination of 2n−6,n−1 and n−2, there exists a componentG⁰such thatn≤ |V(G⁰)| 6= 2n−5. By Theorem 1.2,T(p, n−3−p) can be embedded intoG⁰. ThereforeR(T(p, n−3−p), K1,m) =m+n−3. Ifm+n−3 =r(2n−6) +s(n−1) +t(n−2), then we can partition K_m+n−3 into r+s+t disjoint parts such that r parts have order 2n−6 and s parts have order n−1 and t parts have order n−2. Let GR = rK_n−3,n−3∪sK_n−1∪tK_n−2 and G_B = G^c_R. Then ∆(G_B) ≤m+n−4−(n−3) = m−1 and T(p, n−3−p) cannot be embedded into G. Therefore R(T(p, n−3−p), K1,m)≥m+n−2. The same lower bound was given in [6]. So

R(T(1, n−4), K1,m) =m+n−2. 2 Acknowledgement The research is supported in part by National Natural Science Foundation of China (No. 11931002).

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