Gallai-Ramsey number of even cycles with chords
Fangfang Zhang1,
This work was done while the first author visited the University of Central Florida as a visiting student. The hospitality of the
hosting institution is greatly acknowledged. The visit was funded by the Chinese Scholarship Council. E-mail address: [email protected].
1Department of Mathematics, Nanjing University, Nanjing 210093, P.R. China
2Department of Mathematics, University of Central Florida, Orlando, FL 32816, USA
Zi-Xia Song2,
Partially supported by the National Science Foundation of China under Grant No. DMS-1854903. E-mail address: [email protected].
1Department of Mathematics, Nanjing University, Nanjing 210093, P.R. China
2Department of Mathematics, University of Central Florida, Orlando, FL 32816, USA
Yaojun Chen1,
Yaojun Chen and Fangfang Zhang are partially supported by the National Natural Science Foundation
of China under grant numbers 11671198 and 11871270. E-mail address: [email protected].
1Department of Mathematics, Nanjing University, Nanjing 210093, P.R. China
2Department of Mathematics, University of Central Florida, Orlando, FL 32816, USA
Abstract
For a graph H and an integer k≥1, the k-color Ramsey number Rk(H) is the least integer N such that every k-coloring of the edges of the complete graph KN contains a monochromatic copy of H. Let Cm denote the cycle on m≥4 vertices and let Θm denote the family of graphs obtained from Cm by adding an additional edge joining two non-consecutive vertices. Unlike Ramsey number of odd cycles, little is known about the general behavior of Rk(C2n) except that Rk(C2n)≥(n−1)k+n+k−1 for all k≥2 and n≥2. In this paper, we study Ramsey number of even cycles with chords under Gallai colorings, where a Gallai coloring is a coloring of the edges of a complete graph without rainbow triangles. For an integer k≥1, the Gallai-Ramsey number GRk(H) of a graph H is the least positive integer N such that every Gallai k-coloring of the complete graph KN contains a monochromatic copy of H. We prove that GRk(Θ2n)=(n−1)k+n+1 for all k≥2 and n≥3. This implies that
GRk(C2n)=(n−1)k+n+1 all k≥2 and n≥3. Our result yields a unified proof for the Gallai-Ramsey number of all even cycles on at least four vertices.
Keywords: Gallai coloring, Gallai-Ramsey, cycles, rainbow triangle
2020 Mathematics Subject Classification: 05C55; 05D10; 05C15
1 Introduction
In this paper we consider graphs that are finite, simple and undirected. We use Pm, Cm and Km to denote the path, cycle and complete graph on m vertices, respectively. For m≥4, let Θm denote the family of graphs obtained from Cm by adding an additional edge joining two non-consecutive vertices.
For any positive integer k, we write [k] for the set {1,…,k}.
Given an integer k≥1 and graphs H1,…,Hk, the classical Ramsey number R(H1,…,Hk) is the least integer N such that every k-coloring of the edges of KN contains a monochromatic copy of Hi in color i for some i∈[k]. When H=H1=⋯=Hk, we simply write Rk(H) to denote the k-color Ramsey number of H. In the seminal paper of Ramsey [34], it is shown that Ramsey numbers are finite. This was rediscovered independently by Erdős and Szekeres [13]. Since the 1970s, Ramsey theory has grown into one of the most active areas of research in combinatorics, overlapping variously with graph theory, number theory, geometry and logic.
Let (G,τ) denote a k-colored complete graph, where G is a complete graph and τ:E(G)→[k]. We say (G,τ) is F-free if G does not contain a monochromatic copy of a graph in a given family F under the k-coloring τ; when F={F}, we simply say (G,τ) is F-free. By abusing notation, we say (G,τ) contains a monochromatic copy of F if G contains a monochromatic copy of a graph in F under τ.
One of the earliest and well-known problems is that of determining the Ramsey number Rk(Cm).
When k=2, the behavior of the Ramsey number R(Cℓ,Cm) has been studied and fully determined by several authors, including Bondy and Erdős [3], Faudree and Schelp [15] and Rosta [35]. However, in the case where more than two colors are involved such results are still rather rare.
For even cycles, not much is known about the behavior of Rk(C2n) in general. Figaj and Łuczak [16] proved that for α1,α2,α3>0,
[TABLE]
as n→∞. Following the ideas of Gyárfás, Ruszinkó, Sárközy, and Szemerédi [23] in determining the value of R(Pm,Pm,Pm), Benevides and Skokan [1] proved that
R(C2n,C2n,C2n)=4n for sufficiently large n.
For general k, Dzido, Nowik and Szuca [11] showed that
Theorem 1.1** **([11])
For all k≥1 and n≥2,
[TABLE]
For further results in this direction, we refer the reader to Graham, Rothchild and Spencer [21] and Radziszowski [31] for a dynamic survey.
In this paper we investigate Ramsey numbers of even cycles and even cycles with chords under Gallai colorings,
where a Gallai coloring is a coloring of the edges of a complete graph without rainbow triangles (that is, a triangle with all its edges colored differently). Gallai colorings naturally arise in several areas including: information theory [28]; the study of partially ordered sets, as in Gallai’s original paper [20] (his result was restated in [26] in the terminology of graphs); and the study of perfect graphs [7]. There are now a variety of papers which consider Ramsey-type problems in Gallai colorings (see, e.g., [4, 5, 6, 9, 19, 24, 25, 27, 29, 30, 32, 33, 36]). More information on this topic can be found in [17, 18].
A Gallai k-coloring is a Gallai coloring that uses at most k colors.
Given an integer k≥1 and graphs H1,…,Hk, Gallai-Ramsey number GR(H1,…,Hk) is defined to be the least integer N such that every Gallai k-coloring of KN contains a monochromatic copy of Hi in color i for some i∈[k]. When H=H1=⋯=Hk, we simply write GRk(H). Clearly, GRk(H)≤Rk(H) for all k≥1 and GR(H1,H2)=R(H1,H2).
Theorem 1.2 below is a result of Gyárfás, Sárközy, Sebő and Selkow [25] which characterizes the general behavior of GRk(H).
Theorem 1.2** **([25])
Let H be a fixed graph with no isolated vertices
and let k≥1 be an integer. Then
GRk(H) is exponential in k if H is not bipartite, linear in k if H is bipartite but not a star, and constant (does not depend on k) when H is a star.
It turns out that for some graphs H (e.g., when H=K3), GRk(H) behaves nicely, while the order of magnitude of Rk(H) seems hopelessly difficult to determine.
We will utilize the following important structural result of Gallai [20].
Theorem 1.3** **([20])
Let (G,τ) be a Gallai k-colored complete graph with ∣V(G)∣≥2. Then V(G) can be partitioned into nonempty sets V1,…,Vp with p≥2 so that at most two colors are used on the edges in E(G)∖(E(V1)∪⋯∪E(Vp)) and only one color is used on the edges between any fixed pair (Vi,Vj) under τ, where E(Vi) denotes the set of edges with both ends in Vi for all i∈[p].
The partition {V1,…,Vp} given in Theorem 1.3 is a Gallai partition of (G,τ). Let (R,τ) be obtained from (G,τ) by first contracting each Vi into a single vertex vi and then coloring vivj by the unique color on the edges between Vi and Vj in (G,τ). We say (R,τ) is the reduced graph of (G,τ) corresponding to the Gallai partition {V1,…,Vp}. Note that R=Kp.
By Theorem 1.3, all edges in R are colored by at most two colors under τ. One can see that any monochromatic copy of H in (R,τ) will result in a monochromatic copy of H in (G,τ). It is not surprising that the 2-color Ramsey number R2(H) plays an important role in determining the value of GRk(H) when H is a complete graph. Fox, Grinshpun and Pach [17] posed the following conjecture.
Conjecture 1.4** **([17])
For all k≥1 and t≥3,
[TABLE]
The first case of Conjecture 1.4 follows directly from a result of Chung and Graham [10] in 1983. The proof in [10] does not rely on Theorem 1.3.
A simpler proof of the case t=3 using Theorem 1.3 can be found in [25]. The next open case, when t=4, was recently settled in [30].
For the remainder of the paper,
we use (G,τ) to denote a Gallai k-colored complete graph, where G is a complete graph and τ:E(G)→[k] is a Gallai k-coloring. For each (G,τ), let Giτ denote the spanning subgraph of G with E(Giτ):={e∈E(G)∣τ(e)=i} for all i∈[k]. We simply write Grτ if the color i is red; Gbτ if the color i is blue.
For every n∈N, let qτ(G,n) denote the number of colors i∈[k] such that Giτ has a component of order at
least n. Then qτ(G,n)≤k.
We begin with Lemma 1.5 ( which follows directly from the proof of Lemma 9 given in [27]), which we restate and prove here for completeness.
Lemma 1.5** **([27])
Let (G,τ) be a Gallai k-colored complete graph with ∣V(G)∣≥n≥2. Then the following hold.
- (i)
qτ(G,n)≥1.
2. (ii)
Let {V1,…,Vp} be a Gallai partition of (G,τ) with p≥2 as small as possible. Then for any color i on the edges in (R,τ), Riτ is connected.
Proof. Let {V1,…,Vp} be any Gallai partition of (G,τ) and let (R,τ) be the corresponding reduced graph of (G,τ). By Theorem 1.3, we may further assume that every edge of R is colored red or blue. Note that for any i∈{r,b}, Giτ is connected if Riτ is connected. Hence the statement is trivially true if p=2 or (R,τ) is monochromatic. We may then assume that p≥3 and (R,τ) contains both red and blue edges. If both Rrτ and Rbτ are connected, then we are done, so we may assume that Rrτ is disconnected. Let D be a component of Rrτ. Then all edges between V(D) and V(R)∖V(D) in (R,τ) are blue, and so Rbτ must be connected. Hence qτ(G,n)≥1. Next assume that p≥3 is chosen as small as possible. We may assume that V(D)={v1,…,vd}, where d<p and v1,…,vd are obtained from V1,…,Vd by contracting each Vi into vi. Let V1′:=∪j=1dVj and V2′:=V(G)∖V1′. Then {V1′,V2′} yields a Gallai partition of (G,τ) with two parts, contrary to the minimality of p. Consequently, if p is chosen as small as possible, then Riτ is connected for any color i on the edges in (R,τ). ■
In this paper, we focus on determining the exact values of GRk(Θ2n) for all k≥2 and n≥3, and GRk(C2n) for all k≥2 and n≥2. Note that GRk(Θ2n)≥GRk(C2n) for all k≥2 and n≥2. Using the construction of Erdős, Faudree, Rousseau and Schelp (see Section 2 in [12]) for Rk(C2n), we have
[TABLE]
Indeed when k=2, simply take (G,τ) to be a 2-colored K5 with each color being a monochromatic C5 when n=2, and a 2-colored K3n−2 obtained from 1-colored K2n−1 by first adding n−1 vertices and then coloring all the new edges with the second color when n≥3; for k≥3, let (G,τ) be obtained from the construction for k−1 by first adding n−1 vertices and then coloring all the new edges with a new color. General upper bounds for GRk(C2n) were first studied in [19], later improved in [27].
More recently, the exact value of GRk(C2n) for 2≤n≤6 has been completely settled, see [14, 19, 22, 29, 32]. Building on the ideas in [27], we establishes a sufficient condition for the existence of monochromatic copy of Θ2n in Gallai k-colored complete graphs. We include C4 in the statement of Theorem 1.6 below in order to provide a unified proof for the Gallai-Ramsey number of all even cycles on at least four vertices.
Theorem 1.6
Let (G,τ) be a Gallai k-colored complete graph with ∣V(G)∣≥n≥2. If
[TABLE]
then (G,τ) has a monochromatic C4 when n=2 and a monochromatic Θ2n for all n≥3.
Theorem 1.6 turns out to be very powerful. Since qτ(G,n)≤k, we see that Theorem 1.6 yields the exact values of GRk(Θ2n) for all k≥2 and n≥3; exact values of GRk(C2n) for all k≥2 and n≥2. Furthermore, Theorem 1.6 also provides a unified proof for the Gallai-Ramsey number of C2n for all k≥2 and n≥2.
Theorem 1.7
For all k≥2 and n≥3,
we have GRk(Θ2n)=(n−1)k+n+1.
Theorem 1.8
For all k≥2 and n≥2, we have
[TABLE]
Using a similar argument in [32, Proposition 1.14], Theorem 1.8 further yields the exact value of GRk(Pm) for all k≥2 and m≥6 (see [32, Proposition 1.12] for the lower bound construction for GRk(Pm)).
Theorem 1.9
For all k≥2 and m≥6,
[TABLE]
We prove Theorem 1.6 in Section 4. The proof of Theorem 1.6 uses recoloring method and several structural results on the existence of a cycle or path in bipartite graphs. An overview of the proof is given in Section 2.
We conclude this section by introducing more notation. Given a graph G, sets S⊆V(G) and F⊆E(G), we use ∣G∣ to denote the number
of vertices of G, G∖S the subgraph obtained from G by deleting all vertices in S, G∖F the subgraph obtained from G by deleting all edges in F, and G[S] the subgraph obtained from G by deleting all vertices in V(G)∖S. We simply write G∖v when S={v}, and G∖uv when F={uv}.
For two disjoint sets A,B⊆V(G), A is complete to B in G if each vertex in A is adjacent to all vertices in B, and anti-complete to B in G if no vertex in A is adjacent to any vertex in B.
Let (G,τ) be a Gallai k-colored complete graph.
For two disjoint sets A,B⊆V(G), A is mc-complete to B if all the edges between A and B in (G,τ) are colored the same color. We simply say A is j-complete to B if all the edges between A and B in (G,τ) are colored by some color j∈[k], and A is blue-complete to B if all the edges between A and B in (G,τ) are colored blue. We say a vertex x∈V(G) is blue-adjacent to a vertex y∈V(G) if the edge xy is colored blue in (G,τ), and x is blue-complete to an edge yz∈E(G) if x is blue-complete to {y,z} in (G,τ).
Similar definitions hold when blue is replaced by another color. For convenience, we use A∖B to denote A−B; and A∖b to denote A∖{b} when B={b}. We use the convention “S:=” to mean that S is defined to be the right-hand side of the relation.
2 An overview of the proof of Theorem 1.6
Let (G,τ), n and k be as given in the statement of Theorem 1.6. Suppose the statement is false. Choose n≥2 as small as possible, and subject to the choice of n, choose (G,τ) so that qτ(G,n) is minimum. Let q:=qτ(G,n)≥1. We may assume that for each color i∈[q], Giτ has a component of order at
least n. Let X1,…,Xq be disjoint subsets of V(G) such that for each i∈[q], Xi (possible empty) is mc-complete in color i to V(G)∖⋃i=1qXi. Choose X1,…,Xq so that ∣G∣−∑i=1q∣Xi∣≥n and ∑i=1q∣Xi∣ is as large as possible. Let X:=⋃i=1qXi and let {V1,…,Vp} be a Gallai partition of (G∖X,τ) with p≥2 as small as possible and ∣V1∣≤⋯≤∣Vp∣. Then (G,τ) is C2n-free and the reduced graph (R,τ) of
(G∖X,τ) are colored by at most two colors in [q], say red and blue. Let R be set of all vertices v∈V(G)∖(X∪Vp) such that v is red-complete to Vp, and B be set of all vertices v∈V(G)∖(X∪Vp) such that v is blue-complete to Vp. With some effort it can be shown
that R=∅, B=∅, q=2, n≥3 and ∣Xi∣≤n−1 for all i∈[2]. We may assume X1 is red-complete to V(G)∖X and X2 is blue-complete to V(G)∖X. We then prove two crucial claims (5) and (7) that ∣Vp∣≤n−2 and either X1=∅ or X2=∅. These allow us to reduce τ to be a Gallai 3-coloring as follows:
let σ be obtained from τ by recoloring all the edges of G[X1] blue if X=X1 and G[X2] red if X=X2; and all the edges of G[Vi] green for all i∈[p]. Then (G,σ) is Gallai 3-colored with no monochromatic copy of C2n, qσ(G,n)=q=2 and {V1,…,Vp} a Gallai partition of (G∖X,σ). Let x∈Vp, y∈R∪X1 and z∈B∪X2 be such that y∈X1 if X1=∅ and z∈X2 if X2=∅. Let H:=G∖{x,y,z}.
Then qσ(H,n−1)≤qτ(H,n−1)=2. By minimality of n, (H,σ) contains a red or blue copy of C:=C2n−2, say blue. We then define three pairwise disjoint sets S,T,L according to the green edges between V(C) and V(G)∖V(C) in (G,σ): S consists of all (special) vertices v∈V(C) such that v is green-adjacent to some vertex in V(G)∖V(C); T consists of all vertices u∈(R∪B)∖V(C) such that u is green-adjacent to some vertex in S; finally, let L:=V(G)∖(V(C)∪T), all leftover vertices. In the next step we prove that ∣L∣≤n, ∣T∣≥1 and 1≤∣S∣≤n−4 by either finding a red C2n using edges between V(C) and T∪L or extending C to be a blue C2n. The key claim (15) states that: there exists a subgraph J of C∖S with ∣J∣≥2n−2−2∣S∣≥6 such that each component of J is a path of odd length, and for all u∈T∪L except possiblely one vertex, u is red-adjacent to at least ∣J∣/2 many vertices in J. Finally, in the last step, we choose “wisely” a subset W of T∪L and a subset W∗ of (T∪L)∖W with ∣W∣=n−1−s and ∣W∗∣=s. With some efforts it can be shown that (G[W∗∪S],σ) has a red cycle C∗ on 2s vertices, and (G[W∪V(J)],σ) has a red path P on 2n−(2s+3) vertices with both ends in W such that P uses only edges between W and V(J). By the “wise” choice of W and W∗, we obtain a red C2n from C∗ and P by joining them through three additional vertices.
3 Preliminaries
In this section we first list some known results and then prove several new corollaries that shall be applied in the proof of our main results.
Theorem 3.1** **([35])
For all n≥3, R(C2n,C2n)=3n−1.
Theorem 3.2** **([2])
Let G be a graph on n≥3 vertices. If δ(G)≥n/2, then either G has a cycle of length ℓ for every ℓ satisfying 3≤ℓ≤n or n is even and G is isomorphic to Kn/2,n/2.
For a bipartite graph G with bipartition {M,N}, let δ(M):=min{dG(x):x∈M} and Δ(M):=max{dG(x):x∈M}. Lemma 3.3 follows from Lemma 12, Lemma 13, Lemma 14 in [27].
Lemma 3.3** **([27])
Let G be a bipartite graph with bipartition {M,N} such that ∣M∣≥2, ∣N∣≥4 and δ(M)≥(∣N∣+1)/2. Then G has a cycle of length 2ℓ for any ℓ satisfying 2≤ℓ≤min{∣M∣,δ(M)−1}, or Δ(M)=(∣N∣+1)/2, M=M1∪M2 and N=N1∪N2∪N3, where M1,M2,N1,N2,N3 are non-empty, pairwise disjoint sets in G, ∣N3∣=1, ∣N1∣=∣N2∣, and Mi is complete to Ni∪N3 but anti-complete to N3−i for all i∈[2].
We next prove several structural results on the existence of a cycle or path in bipartite graphs, which will be useful in the proof of our main results.
Corollary 3.4
Let G be a bipartite graph with bipartition {M,N} such that ∣M∣≥2, ∣N∣≥3 and δ(M)≥∣N∣/2. Let ℓ=min{∣M∣,δ(M)}.
Then G has a path on 2ℓ−1 vertices with both ends in M or M=M1∪M2 and N=N1∪N2, where M1,M2,N1,N2 are non-empty, pairwise disjoint sets, ∣N1∣=∣N2∣, and Mi is complete to Ni but anti-complete to N3−i for all i∈[2]. Moreover, if ∣M∣≥2ℓ−3 and ∣N∣=2ℓ, then G has a path on 2ℓ−3 vertices with both ends in N.
Proof. Let H be obtained from G by adding a new vertex x adjacent to all vertices in M. Then H is a bipartite graph with bipartition {M,N∪{x}}. For any vertex v∈M, dH(v)≥dG(v)+1≥(∣N∪{x}∣+1)/2. By Lemma 3.3, H contains a cycle C2ℓ (and thus G has a desired path on 2ℓ−1 vertices with both ends in M and a desired path on 2ℓ−3 vertices with both ends in N) or M=M1∪M2 and N=N1∪N2, where M1,M2,N1,N2 are non-empty, pairwise disjoint sets, ∣N1∣=∣N2∣=ℓ, and Mi is complete to Ni but anti-complete to N3−i for all i∈[2]. Moreover, if ∣M∣≥2ℓ−3 and ∣N∣=2ℓ, then ∣M1∣+∣M2∣=∣M∣≥2ℓ−3, we see that G contains a path on 2ℓ−3 vertices with both ends in N, as desired. ■
Corollary 3.5
Let G be a bipartite graph with bipartition {M,N} such that ∣N∣≥∣M∣≥3 and δ(M)≥∣N∣−1. If M has at least two vertices each complete to N, then G contains a cycle of length 2∣M∣.
Proof. Let G, M and N be as in the statement. Let M′ be the set of all vertices v∈M such that v is complete to N. Then ∣M′∣≥2. Since δ(M)≥∣N∣−1, the statement is trivially true if ∣M∖M′∣≤1. So we may assume that ∣M∖M′∣≥2. By Corollary 3.4 applied to G∖M′ and the fact ∣N∣−1>∣N∣/2, G∖M′ has an (x,y)-path P on 2∣M∖M′∣−1 vertices with x,y∈M∖M′. Then ∣N∖V(P)∣≥∣M′∣+1≥3. Let u,w∈N∖V(P) be distinct such that xu,yw∈E(G) and let Q be a (u,w)-path using edges between M′ and N∖V(P). Then G has a cycle on 2∣M∣ vertices with edge set E(P)∪E(Q)∪{xu,yw}. ■
Corollary 3.6
Let G be a bipartite graph with bipartition {M,N} such that ∣M∣≥2 and ∣N∣≥4. If δ(M)>(∣N∣+1)/2 or Δ(M)>δ(M)=(∣N∣+1)/2, then G has a cycle of length 2ℓ for any ℓ satisfying 2≤ℓ≤min{∣M∣,δ(M)}.
Proof. Let δ:=δ(M) and let x be a vertex in M with d(x)=Δ(M). Assume first that δ>∣M∣≥2. By our assumption and Lemma 3.3, G has a cycle of length 2ℓ for any ℓ satisfying 2≤ℓ≤∣M∣. Assume next that ∣M∣≥δ.
By assumption, ∣M∣≥δ≥3 and ∣M∖x∣≥δ−1≥2. By Lemma 3.3, G has a cycle of length 2ℓ for any ℓ satisfying 2≤ℓ≤δ−1, and
G∖x has a cycle C:=C2δ−2, say with vertices a1,b1,a2,b2,…,aδ−1,bδ−1 in order, where a1,…,aδ−1∈M and b1,…,bδ−1∈N. We next show that G has a cycle of length 2δ. Let N:=N∖V(C). By the choice of x and C, we see that x∈M∖V(C), N(x)∩N=∅ and
∣N∣=∣N∣−(δ−1)<d(x).
Then N(x)∩V(C)=∅. Let
[TABLE]
Assume first that there exists a vertex, say b∈N(x)∩N, such that b is adjacent to some ai∈M∗. Then G contains a cycle of length 2δ with edge set E(C∖aibi)∪{aib,bx,xbi}, as desired. So we may assume that M∗ is anti-complete to N(x)∩N. Since
[TABLE]
we have ∣M∗∣>∣N∖N(x)∣. Note that for each ai∈M∗,
∣N(ai)∩N∣=d(ai)−(δ−1)≥1. It follows that N∖N(x)=∅, ∣M∗∣≥2, and there must exist a vertex, say b∈N∖N(x), such that b is complete to {ai,aj}, where ai,aj∈M∗ with i<j. Then G contains a cycle of length 2δ with edge set E(C∖{aibi,ajbj})∪{aib,baj,bix,xbj}, as desired.
This completes the proof of Corollary 3.6
■
4 Proof of Theorem 1.6
Let (G,τ), n and k be as given in the statement. Suppose (G,τ) is C4-free when n=2 and Θ2n-free when n≥3. Choose n≥2 as small as possible, and subject to the choice of n, choose (G,τ) so that qτ(G,n) is minimum. Let q:=qτ(G,n). By Lemma 1.5(i), q≥1.
We may assume that for each color i∈[q], Giτ has a component of order at
least n. Let X1,…,Xq be disjoint subsets of V(G) such that for each i∈[q], Xi (possiblely empty) is mc-complete in color i to V(G)∖⋃i=1qXi. Choose X1,…,Xq so that ∣G∣−∑i=1q∣Xi∣≥n and ∑i=1q∣Xi∣ is as large as possible. Let X:=⋃i=1qXi. Then ∣G∖X∣≥n≥2. Since (G,τ) has no rainbow triangle, we see that for i,j∈[q] with i=j, each edge between Xi and Xj is colored i or j. We next prove a series of claims.
(1)
(G,τ) is C2n-free for all n≥2.
Proof. Suppose (G,τ) contains a monochromatic copy of C:=C2n,
say with vertices x1,x2,…,x2n in order. Then n≥3. We may assume that all edges of C are colored blue. Then no chord of C is colored blue because (G,τ) is Θ2n-free. We may further assume that x1x3 is colored red. Then x1xj is colored red for all j∈{3,4,…,2n−1} because (G,τ) has a no rainbow triangle. It follows that all chords of C are colored red. Let H be the graph with V(H)=V(C) and E(H) consisting of all chords of C. Then H is the complement of C2n. It can be easily checked that H contains a chorded C2n because n≥3. Thus (G,τ) contains a red copy of Θ2n, a contradiction. ■
(2)
For all i∈[q], ∣Xi∣≤n−1.
Proof. Suppose ∣Xi∣≥n for some color i∈[q], say blue. By the choice of X1,…,Xq, Xi is blue-complete to V(G)∖X. It follows that (G,τ) has a blue C2n using edges between Xi and V(G)∖X, contrary to (4).
■
Let {V1,…,Vp} be a Gallai partition of (G∖X,τ) with p≥2 as small as possible. We may assume that ∣V1∣≤⋯≤∣Vp∣. By Theorem 1.3 and Lemma 1.5(ii), all edges of the reduced graph of
(G∖X,τ) are colored by at most two colors in [q], say red and blue. Then for all i∈[p−1], Vi is either red- or blue-complete to Vp in (G∖X,τ). Let
[TABLE]
Let R=⋃Vj∈VrVj and B=⋃Vj∈VbVj. Then R∪B=V(G)∖(X∪Vp), and R and B are disjoint. We may further assume that X1 is red-complete to V(G)∖X and X2 is blue-complete to V(G)∖X.
(3)
R=∅ and B=∅.
Proof. Suppose R=∅ or B=∅, say the latter. Since p≥2, we see that ∣R∣≥1 and R is red-complete to Vp. Then ∣Vp∣≤n−1, else, let X1′:=X1∪R and Xi′:=Xi for all i∈{2,…,q}. But then
∣X1′∣+⋯+∣Xq′∣=∣X∪R∣>∣X∣,
contrary to the choice of X1,…,Xq. Similarly, ∣R∣≤n−1. By (4), ∣Xi∣≤n−1 for all i∈[q]. If ∣Vp∪R∪X1∣≤2n−1, then
[TABLE]
contrary to the assumption that ∣G∣≥(n−1)q+n+1. Thus ∣Vp∪R∪X1∣≥2n. Let H be the subgraph of G with V(H)=Vp∪R∪X1 and E(H) consisting of all red edges in (G[Vp∪R∪X1],τ). Then δ(H)≥∣H∣/2. By Theorem 3.2, H has a red C2n, which yields a red C2n in (G,τ), contrary to (4). Thus R=∅ and B=∅. ■
By (4) and Lemma 1.5(ii), both Grτ∖X and Gbτ∖X are connected. Thus q≥2. By minimality of p, R is neither red- nor blue-complete to B in (G∖X,τ). Thus p≥4 and so ∣G∖X∣≥p≥4.
(4)
∣Vp∣≤n−1 and so Xi=∅ for every color i∈[q] that is neither red nor blue.
Proof. Suppose ∣Vp∣≥n. Then every vertex in R∪B is either red- or blue-complete to Vp. Let X1′:=X1∪R, X2′:=X2∪B, and Xi′:=Xi for all i∈{3,…,q}. But then
∣X1′∣+⋯+∣Xq′∣=∣X∪R∪B∣>∣X∣,
contrary to the choice of X1,…,Xq. This proves that ∣Vp∣≤n−1. Next, suppose there exists a color j∈[q] such that j is neither red nor blue but Xj=∅. Then no edges between pairs of X1,…,Xq are colored by color j in (G,τ). But then Gjτ has no component of order at least n, because ∣Vℓ∣≤∣Vp∣≤n−1 for all ℓ∈[p−1], and ∣Xi∣≤n−1 for all i∈[q] by (4), a contradiction.
■
(5)
n≥3 and q=2.
Proof. Suppose first n=2. By (4) and (4), ∣Vp∣=1 and ∣Xi∣≤1 for all i∈[q]. Since (G,τ) is C4-free, we see that q≥2. Then
∣G∖⋃i=3qXi∣≥q+4−(q−2)=6. Thus (G,τ) contains a red or blue C4 because R2(C4)=6 [8], contrary to (4). Suppose next n≥3 and q=2. Then q≥3 and the color q is neither red nor blue. By (4), Xq=∅ and ∣Vp∣≤n−1. Thus qτ(G∖Xq,n)=qτ(G,n)−1=q−1. By (4), ∣Xq∣≤n−1, and so
[TABLE]
By minimality of q and the fact that n≥3, (G∖Xq,τ) has a monochromatic copy of C2n, contrary to (4). This proves that q=2.
■
By (4), n≥3 and q=2 and so ∣G∣≥2(n−1)+n+1=3n−1. By Theorem 3.1, k≥3 and edges of (G,τ) must be colored by at least three colors. We may further assume that the third color on E(G) is green under τ.
(6)
∣Vp∣≤n−2.
Proof. Suppose ∣Vp∣≥n−1. By (4), ∣Vp∣=n−1. We may assume that ∣R∪X1∣≥∣B∪X2∣. Then ∣R∪X1∣≥n because ∣R∪X1∣+∣B∪X2∣≥2n. Then for any two distinct vertices a,b∈R∪X1, there exists a red (a,b)-path on 2n−1 vertices using edges between R∪X1 and Vp. Since (G,τ) contains no red C2n, we see that no vertex in B is red-adjacent to two vertices in R∪X1. It follows that ∣X1∣=0 because R is not blue-complete to B in (G,τ).
Suppose ∣B∪X2∣≥n. By a similar argument, ∣X2∣=0. It is easy to see that there exists a vertex in R which is blue-adjacent to two vertices, say x,y, in B. This, together with a blue (x,y)-path on 2n−1 vertices using edges between B and Vp, yields a blue C2n in (G,τ). Thus ∣B∪X2∣≤n−1 and so ∣R∣≥n+1. Then (G[R],τ) contains at most one red edge, else we obtain a red C2n using two red edges in G[R] and edges between R and Vp; and (G[R∪B],τ) has no red P3 with both ends in R. Furthermore, for any Vℓ∈Vr with ∣Vℓ∣≥2, Vℓ is blue-complete to (R∖Vℓ)∪B∪X2, else (G[R∪B],τ) contains a red P3 with both ends in R.
Suppose ∣B∪X2∣=n−1≥2.
Let x,y∈B∪X2 be two distinct vertices. Let P be a blue (x,y)-path on 2n−3 vertices using edges between B∪X2 and Vp. If there exists a Vℓ∈Vr with ∣Vℓ∣≥2, then Vℓ is blue-complete to (R∖Vℓ)∪B∪X2. Let u,v∈Vℓ and z∈R∖Vℓ. Then we obtain a blue C2n in (G,τ) with edge set {zu,zv,ux,vy}∪E(P). Thus ∣Vr∣=∣R∣≥n+1≥4. Then (G[R],τ) contains at most one red edge and all other edges are colored blue. It can be easily checked that (G[R],τ) contains a blue P3 with vertices u,v,w in order such that ux and wy are colored blue in (G,τ). But then (G,τ) contains a blue C2n with edge set {uv,vw,ux,wy}∪E(P). This proves that ∣B∪X2∣≤n−2.
Then ∣R∣≥n+2≥∣B∪X2∣+4. Let H be the subgraph of G with V(H)=R∪B∪X2 and E(H) being the set of all blue edges in (G[R∪B∪X2],τ). Since G[R∪B] contains no red P3 with both ends in R, and for any Vℓ∈Vr with ∣Vℓ∣≥2, Vℓ is blue-complete to (R∖Vℓ)∪B∪X2, it follows that for all v∈B∪X2, dH(v)≥∣R∣−1≥(∣H∣+2)/2; and for all v∈R, dH(v)≥∣H∣/2. By Theorem 3.2, H contains a cycle of length 2n, which yields a blue C2n in (G,τ). ■
(7)
n≥4.
Proof. Suppose n=3. Then ∣G∣≥8. By (4), ∣Vp∣=1. Then every edge of (G∖X,τ) is colored red or blue. Since at least one edge of (G,τ) is colored green, we may assume that G[X1] contains a green edge, say xy. Note that R is not blue-complete to B in (G,τ) and ∣R∪B∣≥3. Let u,v,w∈R∪B be all distinct such that u∈R and v∈B with uv colored red under τ. Then we obtain a red copy of C6 in (G,τ) with vertices x,w,y,v,u,z in order, where {z}=Vp, contrary to (4). ■
(8)
X1=∅ or X2=∅.
Proof. Suppose X1=∅ and X2=∅. Let H:=G∖{vp,x1,x2}, where vp∈Vp, x1∈X1 and x2∈X2. Let τ′ be obtained from τ by recoloring all the edges of G[Vi] green for all i∈[p]. Then (H,τ′) is Gallai k-colored with no monochromatic copy of C2n. Since ∣Vp∣≤n−2, we see that qτ′(H,n−1)≤qτ(H,n−1)=2. Then
[TABLE]
By (4), n−1≥3. By minimality of n, (H,τ′) contains a monochromatic, say red, copy of C:=C2n−2.
We first claim that C contains no vertex in R∪Vp. Suppose there exists a vertex u∈R∪Vp such that u lies on C. Let v be one neighbor of u on the cycle C. Then uv is colored red under τ′. Since all the edges in G[Vi] are colored green in (G,τ′) for all i∈[p], we see that either u∈/Vp or v∈/Vp, say the latter. By the choice of u, v∈R∪B∪X1. Then we obtain a red C2n in (G,τ) from the cycle C by replacing uv with the path having vertices u,x1,vp,v in order if v∈R∪X1; and with the path having vertices u,vp,x1,v in order if v∈B (and thus u∈R), contrary to (4). We next claim that C contains no vertex in X1. Suppose V(C)∩X1=∅. Since X2 is blue-complete to B and ∣Xi∖xi∣≤n−2 for all i∈[2], there must exist u∈X1∖x1 and v∈B such that uv∈E(C). But then we obtain a red C2n in (G,τ) from the cycle C by replacing uv with the path having vertices u,vp,x1,v in order. This proves that C contains no vertex in R∪Vp∪X1. Thus V(C)⊆B. By the choice of p and Lemma 1.5(ii), Grτ∖X is connected. Thus there exist u∈R∪(B∖V(C)) and v∈V(C) such that uv is colored red under τ. Let w be one neighbor of v on the cycle C. We obtain a red C2n in (G,τ) from the cycle C by replacing vw with the path having vertices v,u,x1,w in order, contrary to (4). ■
By (4), X=X1 or X=X2.
For the remainder of the proof, let σ be obtained from τ by recoloring all the edges of G[X1] blue if X=X1, all the edges of G[X2] red if X=X2, and all the edges of G[Vi] green for all i∈[p]. Then (G,σ) is Gallai 3-colored with no monochromatic copy of C2n. Note that qσ(G,n)=q=2 and {V1,…,Vp} is a Gallai partition of (G∖X,σ). By (4), (G,σ) contains no green cycle on more than n−2 vertices.
(9)
∣Vp∣≥2.
Proof. Suppose ∣Vp∣=1. Then (G,σ) is Gallai 2-colored. By Theorem 3.1, (G,σ) contains a red or blue cycle C:=C2n. By (4), we may assume that X2=∅.
By (4) and the choice of σ, the cycle C contains no edge of G[X1]. It follows that C is a red or blue C2n in (G,τ), contrary to (4).
■
Let x∈Vp, y∈R∪X1 and z∈B∪X2 be such that y∈X1 if X1=∅ and z∈X2 if X2=∅. Let H:=G∖{x,y,z}. By (4) and (4), qσ(H,n−1)≤qτ(H,n−1)=2. Then
[TABLE]
By (4), n−1≥3. By minimality of n, (H,σ) contains a red or blue copy of C:=C2n−2 with vertices, say a1,a2,…,a2n−2 in order. We may further assume that the cycle C is blue. By (4), C contains no vertex in X1.
By the choice of σ, the cycle C contains no edge in (G[X2],σ). We claim that V(C)⊆Vp∪R∪B. Suppose the cycle C contains a vertex in X2, say a1∈X2. By (4), X1=∅. By the choice of σ, a2∈R∪B∪Vp. By the choice of z, we see that z∈X2∖a1. But then we obtain a blue C2n in (G,τ) with vertices a1,y,z,a2,…,a2n−2 in order, contrary to (4). This proves that V(C)⊆Vp∪R∪B. Then Vp∩V(C)=∅, else say a1∈Vp∖x, then a2x is blue because a2a1 is blue, and thus (G,τ) has a blue C2n with vertices a1,z,x,a2,…,a2n−2 in order, contrary to (4). It follows that V(C)⊆R∪B.
For the remainder of the proof, we say a vertex v on the cycle C is special if there exist a vertex u∈(R∪B)∖V(C) and Vj∈{V1,…,Vp−1} such that u,v∈Vj, that is, uv is colored green under σ. Let S be the set of all special vertices on the cycle C and let T
be the set of all vertices u∈(R∪B)∖V(C) such that u is green-adjacent to some v∈S. Finally, let L:=V(G)∖(V(C)∪T). Then no vertex in L is green-adjacent to any vertex in V(C)∪T. Moreover, Vp⊆L because V(C)⊆R∪B.
By the choice of σ and the fact that (G,σ) has neither red nor blue C2n, we see that no two consecutive vertices on C are both special vertices. Let s:=∣S∣. Then s≤n−1. It is worth noting that for each Vj∈{V1,…,Vp−1}, if Vj∩S=∅, then Vj⊆S∪T and ∣Vj∣≥2. Since ∣Vp∣≥2, we see that the cycle C does not contain two edges in (G[B],σ), else we obtain a blue C2n in (G,σ).
(10)
For each ai∈V(C), if ai∈R, then either ai+1∈R or ai−1∈R, where all arithmetic
on indices here and henceforth is done modulo 2n−2. Moreover, ∣V(C)∩R∣≥n and B∖V(C)⊆L.
Proof. Suppose
there exists a vertex ai∈V(C) such that ai∈R but ai−1,ai+1∈B. Let x’∈Vp∖x. Note that Vp∩V(C)=∅. But then we obtain a blue C2n in (G,σ) with vertices a1,…,ai−1,x,z,x′,ai+1,…,a2n−2 in order, contrary to (4). It follows that ∣V(C)∩R∣≥n, because C does not contain two edges in (G[B],σ). Next, suppose there exists a vertex
u∈B∖V(C) such that u∈T. We may assume that a1,u∈Vj for some j∈[p]. Then we obtain a blue C2n in (G,σ) with vertices a1,x,u,a2,…,a2n−2 in order, contrary to (4). Thus B∖V(C)⊆L. ■
(11)
Every vertex in L is red-complete to S∪T in (G,σ), and for all i,j∈[p] with i=j, if Vi⊆S∪T and Vj⊆S∪T, then Vi is red-complete to Vj in (G,σ).
Proof. Suppose there exists a vertex v∈L such that v is not red-complete to S∪T in (G,σ). By the choice of S and T, v is not green-adjacent to any vertex in S∪T and thus there must exist some Vℓ⊆S∪T such that v is blue-complete to Vℓ. We may assume that aj,u∈Vℓ for some j∈[2n−2] and u∈T. But then aj−1u is colored blue and we obtain a blue C2n in (G,σ) with vertices a1,a2,…,aj−1,u,v,aj,…,a2n−2 in order, contrary to (4). This proves that every vertex in L is red-complete to S∪T in (G,σ). Next,
suppose there exist Vi⊆S∪T and Vj⊆S∪T such that Vi is blue-complete to Vj in (G,σ). We may assume that {a1,u1}⊆Vi and {aℓ,u2}⊆Vj for some ℓ∈{2,…,2n−2} and u1,u2∈T. But then u2aℓ+1 is colored blue and we obtain a blue C2n in (G,σ) with vertices a1,…,aℓ,u1,u2,aℓ+1,…,a2n−2 in order,
contrary to (4). ■
(12)
If a vertex v∈T∪L is blue-complete to an edge, say a1a2, on the cycle C, then no vertex in (T∪L)∖v is blue-complete to any edge on the cycle with vertices a1,v,a2,…,a2n−2 in order.
Proof. It follows from the fact that (G,σ) has no blue C2n. ■
For each vertex u∈T∪L, let
[TABLE]
Since no vertex in L is green-adjacent to any vertex on the cycle C, we see that ∣Nr(u)∣+∣Nb(u)∣=2n−2 for all u∈L. By (4), ∣Nr(u)∣=∣V(C)∩R∣≥n for all u∈Vp.
(13)
∣L∣≤n. Consequently, ∣T∣≥1 and so s≥1.
Proof. Suppose for a contradiction that ∣L∣≥n+1. Let L∗ be the set of all vertices u∈L with ∣Nr(u)∣≥n.
Let H∗ be the bipartite subgraph of G with bipartition {L∗,V(C)} and E(H∗) consisting of all red edges between L∗ and V(C) in (G,σ). By the choice of σ and the fact that V(C)⊆R∪B, we see that H∗ contains no cycle of length 2n. By Corollary 3.6 applied to H∗ with M=L∗, N=V(C) and δ(M)≥n>(∣N∣+1)/2, we have ∣L∗∣≤n−1. It follows that ∣L∖L∗∣≥2 and for any v∈L∖L∗, ∣Nr(v)∣≤n−1 and so ∣Nb(v)∣≥n−1. Then (L∖L∗)∩Vp=∅ by (4). We next claim that there exist two distinct vertices w1,w2∈L∖L∗ with Nr(w1)=Nr(w2) such that ∣Nr(w1)∣=∣Nr(w2)∣=n−1 and every vertex in L∖{w1,w2} is red-adjacent to at least n−2 vertices in Nr(w1).
Suppose first that no vertex in L∖L∗ is blue-complete to any edge on the cycle C. Then for any v∈L∖L∗, ∣Nb(v)∣=n−1 and so ∣Nr(v)∣=n−1. Let w1,w2 be two distinct vertices in L∖L∗. We may assume that w1 is blue-complete to {a1,a3,…,a2n−3} in (G,σ). Then L∖L∗ must be blue-complete to {a1,a3,…,a2n−3} and no vertex in L∗ is blue-adjacent to two vertices in {a2,a4,…,a2n−2},
else in each case we obtain a blue C2n in (G,σ). Then Nr(w1)=Nr(w2)={a2,a4,…,a2n−2} and every vertex in L∖{w1,w2} is red-adjacent to at least n−2 vertices in Nr(w1), as claimed.
Suppose next that some vertex in L∖L∗ is blue-complete to an edge, say a1a2, on the cycle C. Let w1 be such a vertex, and let C∗ be the blue cycle on 2n−1 vertices with edge set {w1a1,w1a2}∪E(C∖a1a2). Suppose first {w1}∈{V1,…,Vp−1}. By (4), every vertex in L∖w1 is red-adjacent to at least n vertices on C∗.
If each vertex in Vp is red-adjacent to at least n+1 vertices on C∗, let H∗ be the bipartite subgraph of G with bipartition {L∖w1,V(C∗)} and E(H∗) consisting of all red edges between L∖w1 and V(C∗) in (G,σ). By Corollary 3.6 applied to H∗ with M=L∖w1, N=V(C∗), Δ(M)≥n+1 and δ(M)≥n=(∣N∣+1)/2, H∗ has a cycle of length 2n, which yields a red C2n in (G,σ), contrary to (4). Thus each vertex in Vp is red-adjacent to at most n vertices on C∗. Since ∣Nr(u)∣≥n for each u∈Vp, we see that
w1 is blue-complete to Vp in (G,σ) and Vp is red-adjacent to exactly n vertices on C. Then Vp must be red-complete to {a1,a2} in (G,σ) and no vertex in Vp is blue-complete to any edge on C, else we obtain a blue C2n in (G,σ). Thus one of a3 and a2n−2, say a3, must be blue-complete to Vp under σ. But then (G,σ) contains a blue C2n with edge set E(C∖a2a3)∪{a2w1,w1x,xa3}, contrary to (4).
This proves that {w1}∈/{V1,…,Vp−1}. Thus there exists Vℓ∈{V1,…,Vp−1} such that w1∈Vℓ and ∣Vℓ∣≥2. Let w2∈Vℓ with w2=w1.
By (4), no vertex in Vℓ is blue-complete to any edge on C∖a1a2.
This, together with fact that ∣Nb(wi)∣≥n−1, implies that ∣Nb(wi)∣=n−1 for all i∈[2],
and no vertex in L∖{w1,w2} is blue-adjacent to two vertices in Nr(w1), otherwise we obtain a blue C2n in (G,σ). Then every vertex in L∖{w1,w2} is red-adjacent to at least n−2 vertices in Nr(w1), as claimed.
Let L′ be a subset of L such that ∣L′∣=n−1, w1,w2∈L′ and (Vp∖x)∩L′=∅. Then ∣L′∣+∣Nr(w1)∣=(n−1)+(n−1)=2n−2. Let H∗ be the bipartite subgraph of G with bipartition {L′,Nr(w1)} and E(H∗) consisting of all red edges between L′ and Nr(w1) in (G,σ). Then by the choice of w1,w2, we see that {w1,w2} is red-complete to Nr(w1). By Corollary 3.5 applied to H∗ with M=L′ and N=Nr(w1), H∗ has a cycle C∗ on 2n−2 vertices. By the choice of L′, let x1x2 be an edge of C∗ with x1∈Vp∖x. Let u∈V(C)∩R with u∈/V(C∗). But then (G,σ) has a red C2n with edge set E(C∗∖x1x2)∪{x1u,ux,xx2}, contrary to (4).
This proves that ∣L∣≤n. Consequently, ∣T∣=∣G∣−∣V(C)∪L∣≥3n−1−(2n−2+n)≥1 and so s=∣S∣≥1. ■
By (4) and (4), X2=∅. By (4), z∈L∖Vp. We next claim that
(14)
s≤n−4.
Proof. Suppose that s≥n−3.
By (4) and the fact that Vp⊆L and ∣Vp∣≤n−2, we may assume that S∪T=V1∪⋯∪Vℓ for some ℓ∈[p−1]. By (4) again, L is red-complete to V1∪⋯∪Vℓ and Vi is red-complete to Vj for all i,j∈[ℓ] with i=j when ℓ≥2. By (4), ∣V(C)∩R∣≥n and so (V(C)∖S)∩R=∅. Let X∗:={x} when s≥n−2 and X∗:={x,x′} when s=n−3, where x′∈Vp∖x. Then
[TABLE]
We define L∗ to be a subset of S∪T∪(L∖X∗) with ∣L∗∣=2n−2∣X∗∣ such that when ∣X∗∣=1, L∗∩Vp=∅; and when ∣X∗∣=2, z∈L∗ and L∗∩T=∅. Let H∗ be the subgraph of G with V(H∗)=L∗ and E(H∗) consisting of all red edges in (G,σ) between L∗∩(L∖X∗) and L∗∩(V1∪⋯∪Vℓ), and all edges between each pair L∗∩Vi and L∗∩Vj for all i,j∈[ℓ] with i=j. Since ∣Vi∣≤n−2 for all i∈[ℓ], ∣L∖X∗∣≤n−∣X∗∣, we see that δ(H∗)≥∣H∗∣/2. When X∗={x}, let u∈(V(C)∩R)∖S. By Theorem 3.2, H∗ contains a cycle C∗:=C2n−2, say with vertices v1,v2,…,v2n−2 in order. We may assume that v1∈Vp. Then {u,v2} is red-complete to {x,v1}⊆Vp and (G,σ) contains a red C2n with vertices v1,u,x,v2,…,v2n−2 in order, contrary to (4). Thus X∗={x,x′} and s=n−3. By Theorem 3.2 again, H∗ contains a cycle C∗:=C2n−4, say with vertices v1,v2,…,v2n−4 in order. We may further assume that v1=z. By the choice of E(H∗), v2∈S∪T. Note that v1∈B by the choice of z.
Since C∖S contains at most s−1 isolated vertices and 2n−2−(s+s−1)=2n−1−2(n−3)=5, it follows that C∖S contains at least three edges. By (4), Vp is not blue-complete to any two edges on the cycle C. Let ab,cd∈E(C∖S) such that a,c∈R.
Then b,d∈R, else, say b∈B, then (G,σ) contains a blue C2n with edge set E(C∖ab)∪{av1,v1x,xb} if av1 is blue; and a red C2n with vertices v1,a,x,c,x′,v2,…,v2n−4 in order if av1 is red, contrary to (4). Thus v1 is blue-complete to {a,b,c,d}, else, say v1a is red, then (G,σ) contains a red C2n with vertices v1,a,x,c,x′,v2,…,v2n−4 in order, contrary to (4).
By (4), no vertex in T∪(L∖z) is blue-complete to {a,b} or {c,d} in (G,σ).
By the choice of S∪T, we see that no vertex in S is blue-complete to {a,b} or {c,d} in (G,σ). We may assume that v3a is colored red because v3∈S∪T∪(L∖z).
But then (G,σ) contains a red C2n with vertices v1,v2,x,b,x′,a,v3,…,v2n−4 in order, contrary to (4). ■
(15)
There exists a subgraph J of C∖S with ∣J∣≥2n−2−2s≥6 such that each component of J is a path of even order, and for all u∈T∪L except possiblely one vertex, u is neither blue-complete to an edge in J nor green-adjacent to a vertex in V(J), and so
∣Nr(u)∩V(J)∣≥∣J∣/2.
Proof. By (4), s≥1 and ∣T∣≥1.
By (4), ∣V(C)∖S∣≥n+2. Since S is an independent set of the cycle C,
we see that C∖S has exactly s components such that each component is a path. Assume first that no vertex in T∪L is blue-complete to an edge on the cycle C. Let J be obtained from C∖S by deleting one end of each odd component of C∖S. Then ∣J∣≥2n−2−2s and for all u∈T∪L, u is not green-adjacent to any vertex in V(J), and so ∣Nr(u)∩V(J)∣≥∣J∣/2 because u is not blue-complete to any edge on the cycle C. By (4), ∣J∣≥2n−2−2s≥6.
Assume next that there exists a vertex w∈T∪L such that w is blue-complete to an edge, say a1a2, on the cycle C. By (4), no vertex in (T∪L)∖w is blue-complete to any edge on the path P:=C∖a1a2. Let J be obtained from P∖S by deleting one end of each odd component of P∖S. We claim that P∖S has at most s odd components. Suppose P∖S has at least s+1 odd components. Then P∖S has exactly s+1 components such that each component is odd. But then P must have an odd number of vertices, contrary to ∣P∣=2n−2. Thus P∖S has at most s odd components, as claimed. Then ∣J∣≥2n−2−2s≥6 and for all u∈T∪L with u=w, u is not green-adjacent to any vertex in V(J), and so ∣Nr(u)∩V(J)∣≥∣J∣/2 because u is not blue-complete to any edge in J. ■
By (4), let w be the possible vertex in T∪L such that w is blue-complete to some edge in J. Let T′:=T∖w and L′:=L∖w. Then ∣T′∪L′∣=∣(T∪L)∖w∣≥n. We next show that T′=∅. Suppose T′=∅. Then T={w} and L=L′ with ∣L∣≥n. By (4), for any u∈L, ∣Nr(u)∩V(J)∣≥∣J∣/2≥n−1−s and u is not green-adjacent to any vertex in V(J).
By (4), L is red-complete to S∪T. Let H∗ be the bipartite graph of G with bipartition {L,V(C)∪{w}} and E(H∗) consisting of all red edges between L and V(C)∪{w} in (G,σ). Then each vertex in L∖Vp is red-adjacent to at least ∣Nr(u)∩V(J)∣+∣S∪T∣≥(n−1−s)+s+1=n vertices in V(C)∪{w}, and each vertex in Vp is red-adjacent to at least ∣T∣+∣R∩V(C)∣≥n+1 vertices in V(C)∪{w}. By Corollary 3.6 applied to H∗ with M=L, N=V(C)∪{w}, δ(M)≥n and Δ(M)≥n+1, H∗ has a cycle of length 2n, which yields a red C2n in (G,σ), contrary to (4). Thus T′=∅.
Let x1∈Vp with x1=x. Note that ∣T′∪L′∣≥n=(n−s−1)+(s+1)≥(n−s−1)+2. Let W be a subset of T′∪L′ with ∣W∣=n−s−1 such that x,x1∈/W, ∣W∩T′∣ is as large as possible and ∣W∩Vp∣ is as small as possible. Let W∗ be a subset of (T′∪L′)∖W with ∣W∗∣=s such that x1∈W∗, x∈/W∗ and ∣W∗∩Vp∣ is as large as possible.
Let H∗ be the subgraph of G with V(H∗)=W∗∪S and E(H∗) consisting of all red edges in (G[W∗∪S],σ). We claim that δ(H∗)≥∣H∗∣/2. Assume first ∣Vp∣≥s+1. By the choice of W and W∗, W∗⊆Vp. Then W∗ is red-complete to S and so δ(H∗)≥∣H∗∣/2. Assume next ∣Vp∣≤s. By (4), every vertex in W∗∩L′ is red-complete to S; for every vertex u∈W∗∩T′, we may assume that u∈Vℓ for some ℓ∈[p−1]. Then u is red-complete to V(H∗)∖Vℓ, and so δ(H∗)≥∣H∗∣/2 because ∣Vℓ∣≤∣Vp∣≤s. Hence in both cases, δ(H∗)≥∣H∗∣/2. By Theorem 3.2, H∗ has a cycle C∗ on 2s vertices (here by abusing the notation, C∗ denotes an edge on two vertices when s=1). Next,
let H′ be the bipartite subgraph of G with bipartition {W,V(J)} and E(H′) consisting of all red edges between W and V(J) in (G,σ).
By (4), every vertex in W is red-adjacent to at least ∣V(J)∣/2 vertices in V(J). Suppose first that W is a disjoint union of M1=∅ and M2=∅ and V(J) is a disjoint union of N1 and N2 such that
∣N1∣=∣N2∣=∣V(J)∣/2≥n−1−s≥3 and Mi is red-complete to Ni but blue-complete to N3−i for all i∈[2].
By (4) and the choice of W, neither J[N1] nor J[N2] contains an edge and J contains at least three independent edges, say aiai+1,ajaj+1,aℓaℓ+1 with i<j<ℓ. We may further assume ai,aj∈N1. Then ai+1,aj+1∈N2 and (G,σ) has a blue C2n with edge set E(C∖{aiai+1,ajaj+1})∪{aib2,b2aj,ai+1b1,b1aj+1}, where b1∈M1 and b2∈M2, contrary to (4). This proves that W and V(J) have no such partition.
By Corollary 3.4 applied to H′ with M=W, N=V(J), δ(M)≥∣V(J)∣/2, H′ has an (a,b)-path P on 2(n−1−s)−1=2n−(2s+3) vertices with a,b∈W. Note that ∣(V(P)∪V(C∗))∩V(C)∣=(n−2−s)+s=n−2.
(16) If (G,σ) contains a red cycle C∗ on 2n−2 vertices such that x1∈V(C∗) and x∈/V(C∗), then ∣V(C∗)∩V(C)∣≥n.
Proof. Suppose ∣V(C∗)∩V(C)∣≤n−1. By (4), ∣R∖V(C∗)∣≥∣(R∩V(C))∖V(C∗)∣≥1. Let u∈R∖V(C∗) and x2 be a neighbor of x1 on C∗. Then xx2 is red because x,x1∈Vp. But then (G,σ) has a red C2n with edge set E(C∗∖x1x2)∪{x1u,ux,xx2}, contrary to (4). ■
Let ℓ∈[p−1] such that Vℓ∩T′=∅. This is possible because T′=∅. By the choice of T, S∩Vℓ=∅. For the remainder of the proof, let b1∈S∩Vℓ. By the choice of H∗, b1,x1∈V(C∗). Let b2 be a neighbor of b1 on the cycle C∗; let E0={b1b2} when s≥2 and E0=∅ when s=1.
Since ∣V(P)∩V(J)∣=n−2−s and every minimum vertex cover of J has at least n−1−s vertices, we see that J∖V(P) has at least one edge, say d1d2. We next claim that we can choose P so that either a∈T′ or b∈T′. Suppose not. Then H′ has no cycle on 2(n−1−s) vertices because T′∩W=∅. By the choice of W, a,b∈L′. By (4), ab1 and bb1 are red because b1∈S. Then neither d1 nor d2 is red-complete to {a,b} because H′ has no cycle on 2(n−1−s) vertices. By (4) and the choice of b1 and J, no vertex in T′∪L′ is blue-complete to {d1,d2}. Thus we may assume that ad1,bd2,b1d1 are red. If b2∈S, then bb2 is red. But then (G,σ) has a red C∗ on 2n−2 vertices with edge set E(P)∪(E(C∗)∖E0)∪{b1d1,d1a,bb2} such that ∣V(C∗)∩V(C)∣≤∣(V(P)∪V(C∗))∩V(C)∣+1=n−1, contrary to (4). Thus b2∈/S and so b2 is not blue-complete to {d1,d2}. We may assume that b2d2 is red. But then (G,σ) has a red C∗ on 2n−2 vertices with edge set E(P)∪(E(C∗)∖E0)∪{b1a,bd2,d2b2} such that ∣V(C∗)∩V(C)∣≤∣(V(P)∪V(C∗))∩V(C)∣+1=n−1, contrary to (4). This proves that we can choose P so that either a∈T′ or b∈T′, say the former.
By the choice of T and b1, we may further assume that a∈Vℓ and so ab1 is colored green. Then ab2 is red. Recall that a∈T′. By (4), a is not blue-complete to {d1,d2} and neither ad1 nor ad2 is green. We may assume that ad1 is red. Then b1d1 is red.
It follows that bb2 is not colored red, otherwise (G,σ) has a red C∗ on 2n−2 vertices with edge set E(P)∪(E(C∗)∖E0)∪{b1d1,d1a,bb2} such that ∣V(C∗)∩V(C)∣≤∣(V(P)∪V(C∗))∩V(C)∣+1=n−1, contrary to (4). Similarly, bd1 is blue because ab2 is red. Note that ad1 is red and bd1 is blue, we see that b∈/Vℓ. Then b is red-complete to {a,b1} by (4) and bd2 is red by (4). If b2d2 is blue, then bb2 is not colored green because bd2 is red. Thus bb2 is blue because bb2 is not red. But then (G,σ) has a blue C2n with edge set E(C∖d1d2)∪{d1b,bb2,b2d2}, contrary to (4). Thus b2d2 is red. Since a∈T′ and b is red-complete to {a,b1}, by the choice of H′, there must exist three consecutive vertices, say v1,v2,v3, on P with v1,v3∈W and v2∈V(J) such that either {v1,v3}∩T=∅ and {v1,v3}∩L=∅, or v1∈Vℓ′∩T and v3∈Vℓ′′∩T for some ℓ′,ℓ′′∈[p−1] with ℓ′=ℓ′′.
By (4), v1v3 is colored red.
But then (G,σ) has a red C∗ on 2n−2 vertices with edge set E(P∖v2)∪(E(C∗)∖E0)∪{v1v3,b1d1,d1a,bd2,d2b2} such that ∣V(C∗)∩V(C)∣≤∣(V(P∖v2)∪V(C∗))∩V(C)∣+2=n−1, contrary to (4).
This completes the proof of Theorem 1.6.
■
Acknowledgements
The authors would like to thank Christian Bosse and Jingmei Zhang for their helpful discussion.