Gallai-Ramsey number of even cycles with chords

TL;DR

This paper determines the Gallai-Ramsey number for even cycles with chords, providing a unified formula for all such cycles with at least four vertices under Gallai colorings.

Contribution

It establishes an exact formula for the Gallai-Ramsey number of even cycles with chords, extending understanding beyond traditional Ramsey numbers.

Findings

Gallai-Ramsey number for n-cycles with chords is (n-1)k + n + 1 for all k  and n .

Unified proof for the Gallai-Ramsey number of all even cycles with at least four vertices.

Results generalize previous bounds and provide exact values for a broad class of cycles.

Abstract

For a graph $H$ and an integer $k\ge1$, the $k$-color Ramsey number $R_k(H)$ is the least integer $N$ such that every $k$-coloring of the edges of the complete graph $K_N$ contains a monochromatic copy of $H$. Let $C_m$ denote the cycle on $m\ge4$ vertices and let $\Theta_m$ denote the family of graphs obtained from $C_m$ by adding an additional edge joining two non-consecutive vertices. Unlike Ramsey number of odd cycles, little is known about the general behavior of $R_k(C_{2n})$ except that $R_k(C_{2n})\ge (n-1)k+n+k-1$ for all $k\ge2$ and $n\ge2$. 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 $k\geq 1$, the Gallai-Ramsey number $GR_k(H)$ of a graph $H$ is the least positive integer $N$ such that every Gallai $k$-coloring of the complete graph $K_N$ contains a monochromatic copy of $H$. We prove that $GR_k(\Theta_{2n})=(n-1)k+n+1$ for all $k\geq 2$ and $n\geq 3$. This implies that $GR_k(C_{2n})=(n-1)k+n+1$ all $k\geq 2$ and $n\geq 3$. Our result yields a unified proof for the Gallai-Ramsey number of all even cycles on at least four vertices.

Equations

$$R(C_{2\lfloor {\alpha }_{1}n\rfloor },C_{2\lfloor {\alpha }_{3}n\rfloor },C_{2\lfloor {\alpha }_{3}n\rfloor })=({\alpha }_1+{\alpha }_2+{\alpha }_3+\max \{{\alpha }_1,{\alpha }_2,{\alpha }_3\})n+o(n),$$

$$R_k(C_{2n})\ge \{\begin{array}{ll}(n-1)k+n+k& \text{if }k\text{ is odd}\\ (n-1)k+n+k-1& \text{if }k\text{ is even.}\end{array}$$

$$G{R}_k(K_t)=\{\begin{array}{ll}(R_2(K_t)-1{)}^{k/2}+1& \text{if }k\text{ is even}\\ (t-1)(R_2(K_t)-1{)}^{(k-1)/2}+1& \text{if }k\text{ is odd.}\end{array}$$

$$G{R}_k(C_{2n})\ge \{\begin{array}{ll}(n-1)k+n+1& \text{if }k\ge 2\text{ and }n\ge 3\\ (n-1)k+n+2& \text{if }k\ge 2\text{ and }n=2.\end{array}$$

$$|V(G)|\ge \{\begin{array}{ll}(n-1)\cdot q_{\tau }(G,n)+n+1& \text{if }n\ge 3\\ (n-1)\cdot q_{\tau }(G,n)+n+2& \text{if }n=2,\end{array}$$

$$G{R}_k(C_{2n})=\{\begin{array}{ll}(n-1)k+n+1& \text{if }n\ge 3\\ (n-1)k+n+2& \text{if }n=2.\end{array}$$

$$G{R}_k(P_m)=\{\begin{array}{ll}(n-1)k+n+1& \text{if }m=2n\\ (n-1)k+n+2& \text{if }m=2n+1.\end{array}$$

$$M^{\ast }:=\{a_i\in \{a_1,\dots ,a_{\delta -1}\}\mid b_i\in N(x),\text{ where }i\in [\delta -1]\}.$$

$$|M^{\ast }|+|N(x)\cap \overline{N}|=d(x)>|\overline{N}|=|N(x)\cap \overline{N}|+|\overline{N}\setminus N(x)|,$$

$$\begin{array}{rl}{\mathcal{V}}_r& :=\{V_i\in \{V_1,\dots ,V_{p-1}\}\mid V_i\text{ is red-complete to }{V}_p\text{ in }(G\setminus X,\tau )\}\text{ and}\\ {\mathcal{V}}_b& :=\{V_i\in \{V_1,\dots ,V_{p-1}\}\mid V_i\text{ is blue-complete to }{V}_p\text{ in }(G\setminus X,\tau )\}.\end{array}$$

$$|G|=|V_p\cup R\cup X_1|+|X_2|+\cdots +|X_q|\le (2n-1)+(n-1)(q-1)=(n-1)q+n,$$

$$|G\setminus X_q|\ge (n-1)q+n+1-(n-1)=(n-1)\cdot q_{\tau }(G\setminus X_q,n)+n+1.$$

$$|H|=|G|-3\ge 3n-4\ge ((n-1)-1)\cdot q_{{\tau }^{\prime }}(H,n-1)+(n-1)+1.$$

$$|H|=|G|-3\ge 3n-4\ge ((n-1)-1)\cdot q_{\sigma }(H,n-1)+(n-1)+1.$$

$$\begin{array}{rl}{N}_b(u)& :=\{v\in V(C)\mid uv\text{ is colored blue in }(G,\sigma )\}\text{ and}\\ N_r(u)& :=\{v\in V(C)\mid uv\text{ is colored red in }(G,\sigma )\}.\end{array}$$

$$\begin{array}{rl}|S\cup T\cup (L\setminus X^{\ast })|& =|G|-|X^{\ast }\cup (V(C)\setminus S)|\\ & \ge \{\begin{array}{ll}(3n-1)-(n+1)=2n-2& \text{if }|X^{\ast }|=1\\ (3n-1)-(n+3)=2n-4& \text{if }|X^{\ast }|=2\end{array}\\ & =2n-2|X^{\ast }|.\end{array}$$

Full text

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 $P_{m}$, $C_{m}$ and $K_{m}$ to denote the path, cycle and complete graph on $m$ vertices, respectively. For $m\geq 4$, let $\Theta_{m}$ denote the family of graphs obtained from $C_{m}$ by adding an additional edge joining two non-consecutive vertices. For any positive integer $k$, we write $[k]$ for the set $\{1,\ldots,k\}$. Given an integer $k\geq 1$ and graphs $H_{1},\ldots,H_{k}$, the classical Ramsey number $R(H_{1},\ldots,H_{k})$ is the least integer $N$ such that every $k$-coloring of the edges of $K_{N}$ contains a monochromatic copy of $H_{i}$ in color $i$ for some $i\in[k]$. When $H=H_{1}=\dots=H_{k}$, we simply write $R_{k}(H)$ 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 $(G,\tau)$ denote a $k$-colored complete graph, where $G$ is a complete graph and $\tau:E(G)\rightarrow[k]$. We say $(G,\tau)$ is $\mathcal{F}$-free if $G$ does not contain a monochromatic copy of a graph in a given family $\mathcal{F}$ under the $k$-coloring $\tau$; when $\mathcal{F}=\{F\}$, we simply say $(G,\tau)$ is $F$-free. By abusing notation, we say $(G,\tau)$ contains a monochromatic copy of $\mathcal{F}$ if $G$ contains a monochromatic copy of a graph in $\mathcal{F}$ under $\tau$. One of the earliest and well-known problems is that of determining the Ramsey number $R_{k}(C_{m})$. When $k=2$, the behavior of the Ramsey number $R(C_{\ell},C_{m})$ 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 $R_{k}(C_{2n})$ in general. Figaj and Łuczak [16] proved that for $\alpha_{1},\alpha_{2},\alpha_{3}>0$,

[TABLE]

as $n\rightarrow\infty$. Following the ideas of Gyárfás, Ruszinkó, Sárközy, and Szemerédi [23] in determining the value of $R(P_{m},P_{m},P_{m})$, Benevides and Skokan [1] proved that $R(C_{2n},C_{2n},C_{2n})=4n$ for sufficiently large $n$. For general $k$, Dzido, Nowik and Szuca [11] showed that

Theorem 1.1** **([11])

For all $k\geq 1$ and $n\geq 2$,

[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 $k\geq 1$ and graphs $H_{1},\ldots,H_{k}$, Gallai-Ramsey number $GR(H_{1},\ldots,H_{k})$ is defined to be the least integer $N$ such that every Gallai $k$-coloring of $K_{N}$ contains a monochromatic copy of $H_{i}$ in color $i$ for some $i\in[k]$. When $H=H_{1}=\dots=H_{k}$, we simply write $GR_{k}(H)$. Clearly, $GR_{k}(H)\leq R_{k}(H)$ for all $k\geq 1$ and $GR(H_{1},H_{2})=R(H_{1},H_{2})$. Theorem 1.2 below is a result of Gyárfás, Sárközy, Sebő and Selkow [25] which characterizes the general behavior of $GR_{k}(H)$.

Theorem 1.2** **([25])

Let $H$ be a fixed graph with no isolated vertices and let $k\geq 1$ be an integer. Then $GR_{k}(H)$ 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 $H=K_{3}$), $GR_{k}(H)$ behaves nicely, while the order of magnitude of $R_{k}(H)$ seems hopelessly difficult to determine. We will utilize the following important structural result of Gallai [20].

Theorem 1.3** **([20])

Let $(G,\tau)$ be a Gallai $k$-colored complete graph with $|V(G)|\geq 2$. Then $V(G)$ can be partitioned into nonempty sets $V_{1},\dots,V_{p}$ with $p\geq 2$ so that at most two colors are used on the edges in $E(G)\setminus(E(V_{1})\cup\cdots\cup E(V_{p}))$ and only one color is used on the edges between any fixed pair $(V_{i},V_{j})$ under $\tau$, where $E(V_{i})$ denotes the set of edges with both ends in $V_{i}$ for all $i\in[p]$.

The partition $\{V_{1},\dots,V_{p}\}$ given in Theorem 1.3 is a Gallai partition of $(G,\tau)$. Let $(\mathcal{R},\tau)$ be obtained from $(G,\tau)$ by first contracting each $V_{i}$ into a single vertex $v_{i}$ and then coloring $v_{i}v_{j}$ by the unique color on the edges between $V_{i}$ and $V_{j}$ in $(G,\tau)$. We say $(\mathcal{R},\tau)$ is the reduced graph of $(G,\tau)$ corresponding to the Gallai partition $\{V_{1},\dots,V_{p}\}$. Note that $\mathcal{R}=K_{p}$. By Theorem 1.3, all edges in $\mathcal{R}$ are colored by at most two colors under $\tau$. One can see that any monochromatic copy of $H$ in $(\mathcal{R},\tau)$ will result in a monochromatic copy of $H$ in $(G,\tau)$. It is not surprising that the $2$-color Ramsey number $R_{2}(H)$ plays an important role in determining the value of $GR_{k}(H)$ when $H$ is a complete graph. Fox, Grinshpun and Pach [17] posed the following conjecture.

Conjecture 1.4** **([17])

For all $k\geq 1$ and $t\geq 3$,

[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 $(G,\tau)$ to denote a Gallai $k$-colored complete graph, where $G$ is a complete graph and $\tau:E(G)\rightarrow[k]$ is a Gallai $k$-coloring. For each $(G,\tau)$, let $G^{\tau}_{i}$ denote the spanning subgraph of $G$ with $E(G^{\tau}_{i}):=\{e\in E(G)\mid\tau(e)=i\}$ for all $i\in[k]$. We simply write $G^{\tau}_{r}$ if the color $i$ is red; $G^{\tau}_{b}$ if the color $i$ is blue. For every $n\in\mathbb{N}$, let $q_{\tau}(G,n)$ denote the number of colors $i\in[k]$ such that $G^{\tau}_{i}$ has a component of order at least $n$. Then $q_{\tau}(G,n)\leq k$. 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 $(G,\tau)$ be a Gallai $k$-colored complete graph with $|V(G)|\geq n\geq 2$. Then the following hold.

1. (i)

$q_{\tau}(G,n)\geq 1$. 2. (ii)

Let $\{V_{1},\ldots,V_{p}\}$ be a Gallai partition of $(G,\tau)$ with $p\geq 2$ as small as possible. Then for any color $i$ on the edges in $(\mathcal{R},\tau)$, $\mathcal{R}^{\tau}_{i}$ is connected.

Proof.  Let $\{V_{1},\ldots,V_{p}\}$ be any Gallai partition of $(G,\tau)$ and let $(\mathcal{R},\tau)$ be the corresponding reduced graph of $(G,\tau)$. By Theorem 1.3, we may further assume that every edge of $\mathcal{R}$ is colored red or blue. Note that for any $i\in\{r,b\}$, $G_{i}^{\tau}$ is connected if $\mathcal{R}^{\tau}_{i}$ is connected. Hence the statement is trivially true if $p=2$ or $(\mathcal{R},\tau)$ is monochromatic. We may then assume that $p\geq 3$ and $(\mathcal{R},\tau)$ contains both red and blue edges. If both $\mathcal{R}^{\tau}_{r}$ and $\mathcal{R}^{\tau}_{b}$ are connected, then we are done, so we may assume that $\mathcal{R}^{\tau}_{r}$ is disconnected. Let $D$ be a component of $\mathcal{R}^{\tau}_{r}$. Then all edges between $V(D)$ and $V(\mathcal{R})\setminus V(D)$ in $(\mathcal{R},\tau)$ are blue, and so $\mathcal{R}^{\tau}_{b}$ must be connected. Hence $q_{\tau}(G,n)\geq 1$. Next assume that $p\geq 3$ is chosen as small as possible. We may assume that $V(D)=\{v_{1},\ldots,v_{d}\}$, where $d<p$ and $v_{1},\ldots,v_{d}$ are obtained from $V_{1},\ldots,V_{d}$ by contracting each $V_{i}$ into $v_{i}$. Let $V_{1}^{\prime}:=\cup_{j=1}^{d}V_{j}$ and $V_{2}^{\prime}:=V(G)\setminus V_{1}^{\prime}$. Then $\{V_{1}^{\prime},V_{2}^{\prime}\}$ yields a Gallai partition of $(G,\tau)$ with two parts, contrary to the minimality of $p$. Consequently, if $p$ is chosen as small as possible, then $\mathcal{R}^{\tau}_{i}$ is connected for any color $i$ on the edges in $(\mathcal{R},\tau)$. $\blacksquare$

In this paper, we focus on determining the exact values of $GR_{k}(\Theta_{2n})$ for all $k\geq 2$ and $n\geq 3$, and $GR_{k}(C_{2n})$ for all $k\geq 2$ and $n\geq 2$. Note that $GR_{k}(\Theta_{2n})\geq GR_{k}(C_{2n})$ for all $k\geq 2$ and $n\geq 2$. Using the construction of Erdős, Faudree, Rousseau and Schelp (see Section 2 in [12]) for $R_{k}(C_{2n})$, we have

[TABLE]

Indeed when $k=2$, simply take $(G,\tau)$ to be a 2-colored $K_{5}$ with each color being a monochromatic $C_{5}$ when $n=2$, and a 2-colored $K_{3n-2}$ obtained from 1-colored $K_{2n-1}$ by first adding $n-1$ vertices and then coloring all the new edges with the second color when $n\geq 3$; for $k\geq 3$, let $(G,\tau)$ 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 $GR_{k}(C_{2n})$ were first studied in [19], later improved in [27]. More recently, the exact value of $GR_{k}(C_{2n})$ for $2\leq n\leq 6$ 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 $\Theta_{2n}$ in Gallai $k$-colored complete graphs. We include $C_{4}$ 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 $(G,\tau)$ be a Gallai $k$-colored complete graph with $|V(G)|\geq n\geq 2$. If

[TABLE]

then $(G,\tau)$ has a monochromatic $C_{4}$ when $n=2$ and a monochromatic $\Theta_{2n}$ for all $n\geq 3$.

Theorem 1.6 turns out to be very powerful. Since $q_{\tau}(G,n)\leq k$, we see that Theorem 1.6 yields the exact values of $GR_{k}(\Theta_{2n})$ for all $k\geq 2$ and $n\geq 3$; exact values of $GR_{k}(C_{2n})$ for all $k\geq 2$ and $n\geq 2$. Furthermore, Theorem 1.6 also provides a unified proof for the Gallai-Ramsey number of $C_{2n}$ for all $k\geq 2$ and $n\geq 2$.

Theorem 1.7

For all $k\geq 2$ and $n\geq 3$, we have $GR_{k}(\Theta_{2n})=(n-1)k+n+1$.

Theorem 1.8

For all $k\geq 2$ and $n\geq 2$, we have

[TABLE]

Using a similar argument in [32, Proposition 1.14], Theorem 1.8 further yields the exact value of $GR_{k}(P_{m})$ for all $k\geq 2$ and $m\geq 6$ (see [32, Proposition 1.12] for the lower bound construction for $GR_{k}(P_{m})$).

Theorem 1.9

For all $k\geq 2$ and $m\geq 6$,

[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 $S\subseteq V(G)$ and $F\subseteq E(G)$, we use $|G|$ to denote the number of vertices of $G$, $G\setminus S$ the subgraph obtained from $G$ by deleting all vertices in $S$, $G\setminus F$ the subgraph obtained from $G$ by deleting all edges in $F$, and $G[S]$ the subgraph obtained from $G$ by deleting all vertices in $V(G)\setminus S$. We simply write $G\setminus v$ when $S=\{v\}$, and $G\setminus uv$ when $F=\{uv\}$. For two disjoint sets $A,B\subseteq V(G)$, $A$ 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 $(G,\tau)$ be a Gallai $k$-colored complete graph. For two disjoint sets $A,B\subseteq V(G)$, $A$ is mc-complete to $B$ if all the edges between $A$ and $B$ in $(G,\tau)$ are colored the same color. We simply say $A$ is $j$-complete to $B$ if all the edges between $A$ and $B$ in $(G,\tau)$ are colored by some color $j\in[k]$, and $A$ is blue-complete to $B$ if all the edges between $A$ and $B$ in $(G,\tau)$ are colored blue. We say a vertex $x\in V(G)$ is blue-adjacent to a vertex $y\in V(G)$ if the edge $xy$ is colored blue in $(G,\tau)$, and $x$ is blue-complete to an edge $yz\in E(G)$ if $x$ is blue-complete to $\{y,z\}$ in $(G,\tau)$. Similar definitions hold when blue is replaced by another color. For convenience, we use $A\setminus B$ to denote $A-B$; and $A\setminus b$ to denote $A\setminus\{b\}$ 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 $(G,\tau)$, $n$ and $k$ be as given in the statement of Theorem 1.6. Suppose the statement is false. Choose $n\geq 2$ as small as possible, and subject to the choice of $n$, choose $(G,\tau)$ so that $q_{\tau}(G,n)$ is minimum. Let $q:=q_{\tau}(G,n)\geq 1$. We may assume that for each color $i\in[q]$, $G^{\tau}_{i}$ has a component of order at least $n$. Let $X_{1},\ldots,X_{q}$ be disjoint subsets of $V(G)$ such that for each $i\in[q]$, $X_{i}$ (possible empty) is mc-complete in color $i$ to $V(G)\setminus\bigcup_{i=1}^{q}X_{i}$. Choose $X_{1},\ldots,X_{q}$ so that $|G|-\sum_{i=1}^{q}|X_{i}|\geq n$ and $\sum_{i=1}^{q}|X_{i}|$ is as large as possible. Let $X:=\bigcup_{i=1}^{q}X_{i}$ and let $\{V_{1},\ldots,V_{p}\}$ be a Gallai partition of $(G\setminus X,\tau)$ with $p\geq 2$ as small as possible and $|V_{1}|\leq\cdots\leq|V_{p}|$. Then $(G,\tau)$ is $C_{2n}$-free and the reduced graph $(\mathcal{R},\tau)$ of $(G\setminus X,\tau)$ are colored by at most two colors in $[q]$, say red and blue. Let $R$ be set of all vertices $v\in V(G)\setminus(X\cup V_{p})$ such that $v$ is red-complete to $V_{p}$, and $B$ be set of all vertices $v\in V(G)\setminus(X\cup V_{p})$ such that $v$ is blue-complete to $V_{p}$. With some effort it can be shown that $R\neq\emptyset$, $B\neq\emptyset$, $q=2$, $n\geq 3$ and $|X_{i}|\leq n-1$ for all $i\in[2]$. We may assume $X_{1}$ is red-complete to $V(G)\setminus X$ and $X_{2}$ is blue-complete to $V(G)\setminus X$. We then prove two crucial claims (5) and (7) that $|V_{p}|\leq n-2$ and either $X_{1}=\emptyset$ or $X_{2}=\emptyset$. These allow us to reduce $\tau$ to be a Gallai $3$-coloring as follows: let $\sigma$ be obtained from $\tau$ by recoloring all the edges of $G[X_{1}]$ blue if $X=X_{1}$ and $G[X_{2}]$ red if $X=X_{2}$; and all the edges of $G[V_{i}]$ green for all $i\in[p]$. Then $(G,\sigma)$ is Gallai $3$-colored with no monochromatic copy of $C_{2n}$, $q_{\sigma}(G,n)=q=2$ and $\{V_{1},\ldots,V_{p}\}$ a Gallai partition of $(G\setminus X,\sigma)$. Let $x\in V_{p}$, $y\in R\cup X_{1}$ and $z\in B\cup X_{2}$ be such that $y\in X_{1}$ if $X_{1}\neq\emptyset$ and $z\in X_{2}$ if $X_{2}\neq\emptyset$. Let $H:=G\setminus\{x,y,z\}$. Then $q_{\sigma}(H,n-1)\leq q_{\tau}(H,n-1)=2$. By minimality of $n$, $(H,\sigma)$ contains a red or blue copy of $C:=C_{2n-2}$, say blue. We then define three pairwise disjoint sets $S,T,L$ according to the green edges between $V(C)$ and $V(G)\setminus V(C)$ in $(G,\sigma)$: $S$ consists of all (special) vertices $v\in V(C)$ such that $v$ is green-adjacent to some vertex in $V(G)\setminus V(C)$; $T$ consists of all vertices $u\in(R\cup B)\setminus V(C)$ such that $u$ is green-adjacent to some vertex in $S$; finally, let $L:=V(G)\setminus(V(C)\cup T)$, all leftover vertices. In the next step we prove that $|L|\leq n$, $|T|\geq 1$ and $1\leq|S|\leq n-4$ by either finding a red $C_{2n}$ using edges between $V(C)$ and $T\cup L$ or extending $C$ to be a blue $C_{2n}$. The key claim (15) states that: there exists a subgraph $J$ of $C\setminus S$ with $|J|\geq 2n-2-2|S|\geq 6$ such that each component of $J$ is a path of odd length, and for all $u\in T\cup L$ 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 $T\cup L$ and a subset $W^{*}$ of $(T\cup L)\setminus W$ with $|W|=n-1-s$ and $|W^{*}|=s$. With some efforts it can be shown that $(G[W^{*}\cup S],\sigma)$ has a red cycle $C^{*}$ on $2s$ vertices, and $(G[W\cup V(J)],\sigma)$ 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 $C_{2n}$ 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 $n\geq 3$, $R(C_{2n},C_{2n})=3n-1$.

Theorem 3.2** **([2])

Let $G$ be a graph on $n\geq 3$ vertices. If $\delta(G)\geq n/{2}$, then either $G$ has a cycle of length $\ell$ for every $\ell$ satisfying $3\leq\ell\leq n$ or $n$ is even and $G$ is isomorphic to $K_{n/2,n/2}$.

For a bipartite graph $G$ with bipartition $\{M,N\}$, let $\delta(M):=\min\{d_{G}(x):x\in M\}$ and $\Delta(M):=\max\{d_{G}(x):x\in M\}$. 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 $|M|\geq 2$, $|N|\geq 4$ and $\delta(M)\geq(|N|+1)/{2}$. Then $G$ has a cycle of length $2\ell$ for any $\ell$ satisfying $2\leq\ell\leq\min\{|M|,\delta(M)-1\}$, or $\Delta(M)=(|N|+1)/2$, $M=M_{1}\cup M_{2}$ and $N=N_{1}\cup N_{2}\cup N_{3}$, where $M_{1},M_{2},N_{1},N_{2},N_{3}$ are non-empty, pairwise disjoint sets in $G$, $|N_{3}|=1$, $|N_{1}|=|N_{2}|$, and $M_{i}$ is complete to $N_{i}\cup N_{3}$ but anti-complete to $N_{3-i}$ for all $i\in[2]$.

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 $|M|\geq 2$, $|N|\geq 3$ and $\delta(M)\geq|N|/{2}$. Let $\ell=\min\{|M|,\delta(M)\}$. Then $G$ has a path on $2\ell-1$ vertices with both ends in $M$ or $M=M_{1}\cup M_{2}$ and $N=N_{1}\cup N_{2}$, where $M_{1},M_{2},N_{1},N_{2}$ are non-empty, pairwise disjoint sets, $|N_{1}|=|N_{2}|$, and $M_{i}$ is complete to $N_{i}$ but anti-complete to $N_{3-i}$ for all $i\in[2]$. Moreover, if $|M|\geq 2\ell-3$ and $|N|=2\ell$, then $G$ has a path on $2\ell-3$ 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 $\{M,N\cup\{x\}\}$. For any vertex $v\in M$, $d_{H}(v)\geq d_{G}(v)+1\geq(|N\cup\{x\}|+1)/2$. By Lemma 3.3, $H$ contains a cycle $C_{2\ell}$ (and thus $G$ has a desired path on $2\ell-1$ vertices with both ends in $M$ and a desired path on $2\ell-3$ vertices with both ends in $N$) or $M=M_{1}\cup M_{2}$ and $N=N_{1}\cup N_{2}$, where $M_{1},M_{2},N_{1},N_{2}$ are non-empty, pairwise disjoint sets, $|N_{1}|=|N_{2}|=\ell$, and $M_{i}$ is complete to $N_{i}$ but anti-complete to $N_{3-i}$ for all $i\in[2]$. Moreover, if $|M|\geq 2\ell-3$ and $|N|=2\ell$, then $|M_{1}|+|M_{2}|=|M|\geq 2\ell-3$, we see that $G$ contains a path on $2\ell-3$ vertices with both ends in $N$, as desired. $\blacksquare$

Corollary 3.5

Let $G$ be a bipartite graph with bipartition $\{M,N\}$ such that $|N|\geq|M|\geq 3$ and $\delta(M)\geq|N|-1$. 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 $M^{\prime}$ be the set of all vertices $v\in M$ such that $v$ is complete to $N$. Then $|M^{\prime}|\geq 2$. Since $\delta(M)\geq|N|-1$, the statement is trivially true if $|M\setminus M^{\prime}|\leq 1$. So we may assume that $|M\setminus M^{\prime}|\geq 2$. By Corollary 3.4 applied to $G\setminus M^{\prime}$ and the fact $|N|-1>|N|/2$, $G\setminus M^{\prime}$ has an $(x,y)$-path $P$ on $2|M\setminus M^{\prime}|-1$ vertices with $x,y\in M\setminus M^{\prime}$. Then $|N\setminus V(P)|\geq|M^{\prime}|+1\geq 3$. Let $u,w\in N\setminus V(P)$ be distinct such that $xu,yw\in E(G)$ and let $Q$ be a $(u,w)$-path using edges between $M^{\prime}$ and $N\setminus V(P)$. Then $G$ has a cycle on $2|M|$ vertices with edge set $E(P)\cup E(Q)\cup\{xu,yw\}$. $\blacksquare$

Corollary 3.6

Let $G$ be a bipartite graph with bipartition $\{M,N\}$ such that $|M|\geq 2$ and $|N|\geq 4$. If $\delta(M)>(|N|+1)/{2}$ or $\Delta(M)>\delta(M)=(|N|+1)/{2}$, then $G$ has a cycle of length ${2\ell}$ for any $\ell$ satisfying $2\leq\ell\leq\min\{|M|,\delta(M)\}$.

Proof.  Let $\delta:=\delta(M)$ and let $x$ be a vertex in $M$ with $d(x)=\Delta(M)$. Assume first that $\delta>|M|\geq 2$. By our assumption and Lemma 3.3, $G$ has a cycle of length $2\ell$ for any $\ell$ satisfying $2\leq\ell\leq|M|$. Assume next that $|M|\geq\delta$. By assumption, $|M|\geq\delta\geq 3$ and $|M\setminus x|\geq\delta-1\geq 2$. By Lemma 3.3, $G$ has a cycle of length $2\ell$ for any $\ell$ satisfying $2\leq\ell\leq\delta-1$, and $G\setminus x$ has a cycle $C:=C_{2\delta-2}$, say with vertices $a_{1},b_{1},a_{2},b_{2},\ldots,a_{\delta-1},b_{\delta-1}$ in order, where $a_{1},\ldots,a_{\delta-1}\in M$ and $b_{1},\ldots,b_{\delta-1}\in N$. We next show that $G$ has a cycle of length $2\delta$. Let $\overline{N}:=N\setminus V(C)$. By the choice of $x$ and $C$, we see that $x\in M\setminus V(C)$, $N(x)\cap\overline{N}\neq\emptyset$ and $|\overline{N}|=|N|-(\delta-1)<d(x)$. Then $N(x)\cap V(C)\neq\emptyset$. Let

[TABLE]

Assume first that there exists a vertex, say $b\in N(x)\cap\overline{N}$, such that $b$ is adjacent to some $a_{i}\in M^{*}$. Then $G$ contains a cycle of length $2\delta$ with edge set $E(C\setminus a_{i}b_{i})\cup\{a_{i}b,bx,xb_{i}\}$, as desired. So we may assume that $M^{*}$ is anti-complete to $N(x)\cap\overline{N}$. Since

[TABLE]

we have $|M^{*}|>|\overline{N}\setminus N(x)|$. Note that for each $a_{i}\in M^{*}$, $|N(a_{i})\cap\overline{N}|=d(a_{i})-(\delta-1)\geq 1$. It follows that $\overline{N}\setminus N(x)\neq\emptyset$, $|M^{*}|\geq 2$, and there must exist a vertex, say $b\in\overline{N}\setminus N(x)$, such that $b$ is complete to $\{a_{i},a_{j}\}$, where $a_{i},a_{j}\in M^{*}$ with $i<j$. Then $G$ contains a cycle of length $2\delta$ with edge set $E(C\setminus\{a_{i}b_{i},a_{j}b_{j}\})\cup\{a_{i}b,ba_{j},b_{i}x,xb_{j}\}$, as desired.

This completes the proof of Corollary 3.6 $\blacksquare$

4 Proof of Theorem 1.6

Let $(G,\tau)$, $n$ and $k$ be as given in the statement. Suppose $(G,\tau)$ is $C_{4}$-free when $n=2$ and $\Theta_{2n}$-free when $n\geq 3$. Choose $n\geq 2$ as small as possible, and subject to the choice of $n$, choose $(G,\tau)$ so that $q_{\tau}(G,n)$ is minimum. Let $q:=q_{\tau}(G,n)$. By Lemma 1.5(i), $q\geq 1$. We may assume that for each color $i\in[q]$, $G^{\tau}_{i}$ has a component of order at least $n$. Let $X_{1},\ldots,X_{q}$ be disjoint subsets of $V(G)$ such that for each $i\in[q]$, $X_{i}$ (possiblely empty) is mc-complete in color $i$ to $V(G)\setminus\bigcup_{i=1}^{q}X_{i}$. Choose $X_{1},\ldots,X_{q}$ so that $|G|-\sum_{i=1}^{q}|X_{i}|\geq n$ and $\sum_{i=1}^{q}|X_{i}|$ is as large as possible. Let $X:=\bigcup_{i=1}^{q}X_{i}$. Then $|G\setminus X|\geq n\geq 2$. Since $(G,\tau)$ has no rainbow triangle, we see that for $i,j\in[q]$ with $i\neq j$, each edge between $X_{i}$ and $X_{j}$ is colored $i$ or $j$. We next prove a series of claims.

(1) $(G,\tau)$ is $C_{2n}$-free for all $n\geq 2$.

Proof.  Suppose $(G,\tau)$ contains a monochromatic copy of $C:=C_{2n}$, say with vertices $x_{1},x_{2},\ldots,x_{2n}$ in order. Then $n\geq 3$. We may assume that all edges of $C$ are colored blue. Then no chord of $C$ is colored blue because $(G,\tau)$ is $\Theta_{2n}$-free. We may further assume that $x_{1}x_{3}$ is colored red. Then $x_{1}x_{j}$ is colored red for all $j\in\{3,4,\ldots,2n-1\}$ because $(G,\tau)$ 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 $C_{2n}$. It can be easily checked that $H$ contains a chorded $C_{2n}$ because $n\geq 3$. Thus $(G,\tau)$ contains a red copy of $\Theta_{2n}$, a contradiction. $\blacksquare$

(2) For all $i\in[q]$, $|X_{i}|\leq n-1$.

Proof.  Suppose $|X_{i}|\geq n$ for some color $i\in[q]$, say blue. By the choice of $X_{1},\ldots,X_{q}$, $X_{i}$ is blue-complete to $V(G)\setminus X$. It follows that $(G,\tau)$ has a blue $C_{2n}$ using edges between $X_{i}$ and $V(G)\setminus X$, contrary to (4). $\blacksquare$

Let $\{V_{1},\ldots,V_{p}\}$ be a Gallai partition of $(G\setminus X,\tau)$ with $p\geq 2$ as small as possible. We may assume that $|V_{1}|\leq\cdots\leq|V_{p}|$. By Theorem 1.3 and Lemma 1.5(ii), all edges of the reduced graph of $(G\setminus X,\tau)$ are colored by at most two colors in $[q]$, say red and blue. Then for all $i\in[p-1]$, $V_{i}$ is either red- or blue-complete to $V_{p}$ in $(G\setminus X,\tau)$. Let

[TABLE]

Let $R=\bigcup_{V_{j}\in\mathcal{V}_{r}}V_{j}$ and $B=\bigcup_{V_{j}\in\mathcal{V}_{b}}V_{j}$. Then $R\cup B=V(G)\setminus(X\cup V_{p})$, and $R$ and $B$ are disjoint. We may further assume that $X_{1}$ is red-complete to $V(G)\setminus X$ and $X_{2}$ is blue-complete to $V(G)\setminus X$.

(3) $R\neq\emptyset$ and $B\neq\emptyset$.

Proof.  Suppose $R=\emptyset$ or $B=\emptyset$, say the latter. Since $p\geq 2$, we see that $|R|\geq 1$ and $R$ is red-complete to $V_{p}$. Then $|V_{p}|\leq n-1$, else, let $X^{\prime}_{1}:=X_{1}\cup R$ and $X^{\prime}_{i}:=X_{i}$ for all $i\in\{2,\ldots,q\}$. But then $|X^{\prime}_{1}|+\cdots+|X_{q}^{\prime}|=|X\cup R|>|X|$, contrary to the choice of $X_{1},\ldots,X_{q}$. Similarly, $|R|\leq n-1$. By (4), $|X_{i}|\leq n-1$ for all $i\in[q]$. If $|V_{p}\cup R\cup X_{1}|\leq 2n-1$, then

[TABLE]

contrary to the assumption that $|G|\geq(n-1)q+n+1$. Thus $|V_{p}\cup R\cup X_{1}|\geq 2n$. Let $H$ be the subgraph of $G$ with $V(H)=V_{p}\cup R\cup X_{1}$ and $E(H)$ consisting of all red edges in $(G[V_{p}\cup R\cup X_{1}],\tau)$. Then $\delta(H)\geq|H|/2$. By Theorem 3.2, $H$ has a red $C_{2n}$, which yields a red $C_{2n}$ in $(G,\tau)$, contrary to (4). Thus $R\neq\emptyset$ and $B\neq\emptyset$. $\blacksquare$

By (4) and Lemma 1.5(ii), both $G_{r}^{\tau}\setminus X$ and $G_{b}^{\tau}\setminus X$ are connected. Thus $q\geq 2$. By minimality of $p$, $R$ is neither red- nor blue-complete to $B$ in $(G\setminus X,\tau)$. Thus $p\geq 4$ and so $|G\setminus X|\geq p\geq 4$.

(4) $|V_{p}|\leq n-1$ and so $X_{i}\neq\emptyset$ for every color $i\in[q]$ that is neither red nor blue.

Proof.  Suppose $|V_{p}|\geq n$. Then every vertex in $R\cup B$ is either red- or blue-complete to $V_{p}$. Let $X^{\prime}_{1}:=X_{1}\cup R$, $X^{\prime}_{2}:=X_{2}\cup B$, and $X^{\prime}_{i}:=X_{i}$ for all $i\in\{3,\ldots,q\}$. But then $|X^{\prime}_{1}|+\cdots+|X_{q}^{\prime}|=|X\cup R\cup B|>|X|$, contrary to the choice of $X_{1},\ldots,X_{q}$. This proves that $|V_{p}|\leq n-1$. Next, suppose there exists a color $j\in[q]$ such that $j$ is neither red nor blue but $X_{j}=\emptyset$. Then no edges between pairs of $X_{1},\ldots,X_{q}$ are colored by color $j$ in $(G,\tau)$. But then $G^{\tau}_{j}$ has no component of order at least $n$, because $|V_{\ell}|\leq|V_{p}|\leq n-1$ for all $\ell\in[p-1]$, and $|X_{i}|\leq n-1$ for all $i\in[q]$ by (4), a contradiction. $\blacksquare$

(5) $n\geq 3$ and $q=2$.

Proof.  Suppose first $n=2$. By (4) and (4), $|V_{p}|=1$ and $|X_{i}|\leq 1$ for all $i\in[q]$. Since $(G,\tau)$ is $C_{4}$-free, we see that $q\geq 2$. Then $|G\setminus\bigcup_{i=3}^{q}X_{i}|\geq q+4-(q-2)=6$. Thus $(G,\tau)$ contains a red or blue $C_{4}$ because $R_{2}(C_{4})=6$ [8], contrary to (4). Suppose next $n\geq 3$ and $q\neq 2$. Then $q\geq 3$ and the color $q$ is neither red nor blue. By (4), $X_{q}\neq\emptyset$ and $|V_{p}|\leq n-1$. Thus $q_{\tau}(G\setminus X_{q},n)=q_{\tau}(G,n)-1=q-1$. By (4), $|X_{q}|\leq n-1$, and so

[TABLE]

By minimality of $q$ and the fact that $n\geq 3$, $(G\setminus X_{q},\tau)$ has a monochromatic copy of $C_{2n}$, contrary to (4). This proves that $q=2$. $\blacksquare$

By (4), $n\geq 3$ and $q=2$ and so $|G|\geq 2(n-1)+n+1=3n-1$. By Theorem 3.1, $k\geq 3$ and edges of $(G,\tau)$ must be colored by at least three colors. We may further assume that the third color on $E(G)$ is green under $\tau$.

(6) $|V_{p}|\leq n-2$.

Proof.  Suppose $|V_{p}|\geq n-1$. By (4), $|V_{p}|=n-1$. We may assume that $|R\cup X_{1}|\geq|B\cup X_{2}|$. Then $|R\cup X_{1}|\geq n$ because $|R\cup X_{1}|+|B\cup X_{2}|\geq 2n$. Then for any two distinct vertices $a,b\in R\cup X_{1}$, there exists a red $(a,b)$-path on $2n-1$ vertices using edges between $R\cup X_{1}$ and $V_{p}$. Since $(G,\tau)$ contains no red $C_{2n}$, we see that no vertex in $B$ is red-adjacent to two vertices in $R\cup X_{1}$. It follows that $|X_{1}|=0$ because $R$ is not blue-complete to $B$ in $(G,\tau)$. Suppose $|B\cup X_{2}|\geq n$. By a similar argument, $|X_{2}|=0$. 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 $V_{p}$, yields a blue $C_{2n}$ in $(G,\tau)$. Thus $|B\cup X_{2}|\leq n-1$ and so $|R|\geq n+1$. Then $(G[R],\tau)$ contains at most one red edge, else we obtain a red $C_{2n}$ using two red edges in $G[R]$ and edges between $R$ and $V_{p}$; and $(G[R\cup B],\tau)$ has no red $P_{3}$ with both ends in $R$. Furthermore, for any $V_{\ell}\in\mathcal{V}_{r}$ with $|V_{\ell}|\geq 2$, $V_{\ell}$ is blue-complete to $(R\setminus V_{\ell})\cup B\cup X_{2}$, else $(G[R\cup B],\tau)$ contains a red $P_{3}$ with both ends in $R$. Suppose $|B\cup X_{2}|=n-1\geq 2$. Let $x,y\in B\cup X_{2}$ be two distinct vertices. Let $P$ be a blue $(x,y)$-path on $2n-3$ vertices using edges between $B\cup X_{2}$ and $V_{p}$. If there exists a $V_{\ell}\in\mathcal{V}_{r}$ with $|V_{\ell}|\geq 2$, then $V_{\ell}$ is blue-complete to $(R\setminus V_{\ell})\cup B\cup X_{2}$. Let $u,v\in V_{\ell}$ and $z\in R\setminus V_{\ell}$. Then we obtain a blue $C_{2n}$ in $(G,\tau)$ with edge set $\{zu,zv,ux,vy\}\cup E(P)$. Thus $|\mathcal{V}_{r}|=|R|\geq n+1\geq 4$. Then $(G[R],\tau)$ contains at most one red edge and all other edges are colored blue. It can be easily checked that $(G[R],\tau)$ contains a blue $P_{3}$ with vertices $u,v,w$ in order such that $ux$ and $wy$ are colored blue in $(G,\tau)$. But then $(G,\tau)$ contains a blue $C_{2n}$ with edge set $\{uv,vw,ux,wy\}\cup E(P)$. This proves that $|B\cup X_{2}|\leq n-2$. Then $|R|\geq n+2\geq|B\cup X_{2}|+4$. Let $H$ be the subgraph of $G$ with $V(H)=R\cup B\cup X_{2}$ and $E(H)$ being the set of all blue edges in $(G[R\cup B\cup X_{2}],\tau)$. Since $G[R\cup B]$ contains no red $P_{3}$ with both ends in $R$, and for any $V_{\ell}\in\mathcal{V}_{r}$ with $|V_{\ell}|\geq 2$, $V_{\ell}$ is blue-complete to $(R\setminus V_{\ell})\cup B\cup X_{2}$, it follows that for all $v\in B\cup X_{2}$, $d_{H}(v)\geq|R|-1\geq(|H|+2)/2$; and for all $v\in R$, $d_{H}(v)\geq|H|/2$. By Theorem 3.2, $H$ contains a cycle of length $2n$, which yields a blue $C_{2n}$ in $(G,\tau)$. $\blacksquare$

(7) $n\geq 4$.

Proof.  Suppose $n=3$. Then $|G|\geq 8$. By (4), $|V_{p}|=1$. Then every edge of $(G\setminus X,\tau)$ is colored red or blue. Since at least one edge of $(G,\tau)$ is colored green, we may assume that $G[X_{1}]$ contains a green edge, say $xy$. Note that $R$ is not blue-complete to $B$ in $(G,\tau)$ and $|R\cup B|\geq 3$. Let $u,v,w\in R\cup B$ be all distinct such that $u\in R$ and $v\in B$ with $uv$ colored red under $\tau$. Then we obtain a red copy of $C_{6}$ in $(G,\tau)$ with vertices $x,w,y,v,u,z$ in order, where $\{z\}=V_{p}$, contrary to (4). $\blacksquare$

(8) $X_{1}=\emptyset$ or $X_{2}=\emptyset$.

Proof.  Suppose $X_{1}\neq\emptyset$ and $X_{2}\neq\emptyset$. Let $H:=G\setminus\{v_{p},x_{1},x_{2}\}$, where $v_{p}\in V_{p}$, $x_{1}\in X_{1}$ and $x_{2}\in X_{2}$. Let $\tau^{\prime}$ be obtained from $\tau$ by recoloring all the edges of $G[V_{i}]$ green for all $i\in[p]$. Then $(H,\tau^{\prime})$ is Gallai $k$-colored with no monochromatic copy of $C_{2n}$. Since $|V_{p}|\leq n-2$, we see that $q_{\tau^{\prime}}(H,n-1)\leq q_{\tau}(H,n-1)=2$. Then

[TABLE]

By (4), $n-1\geq 3$. By minimality of $n$, $(H,\tau^{\prime})$ contains a monochromatic, say red, copy of $C:=C_{2n-2}$. We first claim that $C$ contains no vertex in $R\cup V_{p}$. Suppose there exists a vertex $u\in R\cup V_{p}$ such that $u$ lies on $C$. Let $v$ be one neighbor of $u$ on the cycle $C$. Then $uv$ is colored red under $\tau^{\prime}$. Since all the edges in $G[V_{i}]$ are colored green in $(G,\tau^{\prime})$ for all $i\in[p]$, we see that either $u\notin V_{p}$ or $v\notin V_{p}$, say the latter. By the choice of $u$, $v\in R\cup B\cup X_{1}$. Then we obtain a red $C_{2n}$ in $(G,\tau)$ from the cycle $C$ by replacing $uv$ with the path having vertices $u,x_{1},v_{p},v$ in order if $v\in R\cup X_{1}$; and with the path having vertices $u,v_{p},x_{1},v$ in order if $v\in B$ (and thus $u\in R$), contrary to (4). We next claim that $C$ contains no vertex in $X_{1}$. Suppose $V(C)\cap X_{1}\neq\emptyset$. Since $X_{2}$ is blue-complete to $B$ and $|X_{i}\setminus x_{i}|\leq n-2$ for all $i\in[2]$, there must exist $u\in X_{1}\setminus x_{1}$ and $v\in B$ such that $uv\in E(C)$. But then we obtain a red $C_{2n}$ in $(G,\tau)$ from the cycle $C$ by replacing $uv$ with the path having vertices $u,v_{p},x_{1},v$ in order. This proves that $C$ contains no vertex in $R\cup V_{p}\cup X_{1}$. Thus $V(C)\subseteq B$. By the choice of $p$ and Lemma 1.5(ii), $G_{r}^{\tau}\setminus X$ is connected. Thus there exist $u\in R\cup(B\setminus V(C))$ and $v\in V(C)$ such that $uv$ is colored red under $\tau$. Let $w$ be one neighbor of $v$ on the cycle $C$. We obtain a red $C_{2n}$ in $(G,\tau)$ from the cycle $C$ by replacing $vw$ with the path having vertices $v,u,x_{1},w$ in order, contrary to (4). $\blacksquare$

By (4), $X=X_{1}$ or $X=X_{2}$. For the remainder of the proof, let $\sigma$ be obtained from $\tau$ by recoloring all the edges of $G[X_{1}]$ blue if $X=X_{1}$, all the edges of $G[X_{2}]$ red if $X=X_{2}$, and all the edges of $G[V_{i}]$ green for all $i\in[p]$. Then $(G,\sigma)$ is Gallai $3$-colored with no monochromatic copy of $C_{2n}$. Note that $q_{\sigma}(G,n)=q=2$ and $\{V_{1},\ldots,V_{p}\}$ is a Gallai partition of $(G\setminus X,\sigma)$. By (4), $(G,\sigma)$ contains no green cycle on more than $n-2$ vertices.

(9) $|V_{p}|\geq 2$.

Proof.  Suppose $|V_{p}|=1$. Then $(G,\sigma)$ is Gallai $2$-colored. By Theorem 3.1, $(G,\sigma)$ contains a red or blue cycle $C:=C_{2n}$. By (4), we may assume that $X_{2}=\emptyset$. By (4) and the choice of $\sigma$, the cycle $C$ contains no edge of $G[X_{1}]$. It follows that $C$ is a red or blue $C_{2n}$ in $(G,\tau)$, contrary to (4). $\blacksquare$

Let $x\in V_{p}$, $y\in R\cup X_{1}$ and $z\in B\cup X_{2}$ be such that $y\in X_{1}$ if $X_{1}\neq\emptyset$ and $z\in X_{2}$ if $X_{2}\neq\emptyset$. Let $H:=G\setminus\{x,y,z\}$. By (4) and (4), $q_{\sigma}(H,n-1)\leq q_{\tau}(H,n-1)=2$. Then

[TABLE]

By (4), $n-1\geq 3$. By minimality of $n$, $(H,\sigma)$ contains a red or blue copy of $C:=C_{2n-2}$ with vertices, say $a_{1},a_{2},\ldots,a_{2n-2}$ in order. We may further assume that the cycle $C$ is blue. By (4), $C$ contains no vertex in $X_{1}$. By the choice of $\sigma$, the cycle $C$ contains no edge in $(G[X_{2}],\sigma)$. We claim that $V(C)\subseteq V_{p}\cup R\cup B$. Suppose the cycle $C$ contains a vertex in $X_{2}$, say $a_{1}\in X_{2}$. By (4), $X_{1}=\emptyset$. By the choice of $\sigma$, $a_{2}\in R\cup B\cup V_{p}$. By the choice of $z$, we see that $z\in X_{2}\setminus a_{1}$. But then we obtain a blue $C_{2n}$ in $(G,\tau)$ with vertices $a_{1},y,z,a_{2},\ldots,a_{2n-2}$ in order, contrary to (4). This proves that $V(C)\subseteq V_{p}\cup R\cup B$. Then $V_{p}\cap V(C)=\emptyset$, else say $a_{1}\in V_{p}\setminus x$, then $a_{2}x$ is blue because $a_{2}a_{1}$ is blue, and thus $(G,\tau)$ has a blue $C_{2n}$ with vertices $a_{1},z,x,a_{2},\ldots,a_{2n-2}$ in order, contrary to (4). It follows that $V(C)\subseteq R\cup B$.

For the remainder of the proof, we say a vertex $v$ on the cycle $C$ is special if there exist a vertex $u\in(R\cup B)\setminus V(C)$ and $V_{j}\in\{V_{1},\ldots,V_{p-1}\}$ such that $u,v\in V_{j}$, that is, $uv$ is colored green under $\sigma$. Let $S$ be the set of all special vertices on the cycle $C$ and let $T$ be the set of all vertices $u\in(R\cup B)\setminus V(C)$ such that $u$ is green-adjacent to some $v\in S$. Finally, let $L:=V(G)\setminus(V(C)\cup T)$. Then no vertex in $L$ is green-adjacent to any vertex in $V(C)\cup T$. Moreover, $V_{p}\subseteq L$ because $V(C)\subseteq R\cup B$. By the choice of $\sigma$ and the fact that $(G,\sigma)$ has neither red nor blue $C_{2n}$, we see that no two consecutive vertices on $C$ are both special vertices. Let $s:=|S|$. Then $s\leq n-1$. It is worth noting that for each $V_{j}\in\{V_{1},\ldots,V_{p-1}\}$, if $V_{j}\cap S\neq\emptyset$, then $V_{j}\subseteq S\cup T$ and $|V_{j}|\geq 2$. Since $|V_{p}|\geq 2$, we see that the cycle $C$ does not contain two edges in $(G[B],\sigma)$, else we obtain a blue $C_{2n}$ in $(G,\sigma)$.

(10) For each $a_{i}\in V(C)$, if $a_{i}\in R$, then either $a_{i+1}\in R$ or $a_{i-1}\in R$, where all arithmetic on indices here and henceforth is done modulo $2n-2$. Moreover, $|V(C)\cap R|\geq n$ and $B\setminus V(C)\subseteq L$.

Proof.  Suppose there exists a vertex $a_{i}\in V(C)$ such that $a_{i}\in R$ but $a_{i-1},a_{i+1}\in B$. Let $x’\in V_{p}\setminus x$. Note that $V_{p}\cap V(C)=\emptyset$. But then we obtain a blue $C_{2n}$ in $(G,\sigma)$ with vertices $a_{1},\ldots,a_{i-1},x,z,x^{\prime},a_{i+1},\ldots,a_{2n-2}$ in order, contrary to (4). It follows that $|V(C)\cap R|\geq n$, because $C$ does not contain two edges in $(G[B],\sigma)$. Next, suppose there exists a vertex $u\in B\setminus V(C)$ such that $u\in T$. We may assume that $a_{1},u\in V_{j}$ for some $j\in[p]$. Then we obtain a blue $C_{2n}$ in $(G,\sigma)$ with vertices $a_{1},x,u,a_{2},\ldots,a_{2n-2}$ in order, contrary to (4). Thus $B\setminus V(C)\subseteq L$. $\blacksquare$

(11) Every vertex in $L$ is red-complete to $S\cup T$ in $(G,\sigma)$, and for all $i,j\in[p]$ with $i\neq j$, if $V_{i}\subseteq S\cup T$ and $V_{j}\subseteq S\cup T$, then $V_{i}$ is red-complete to $V_{j}$ in $(G,\sigma)$.

Proof.  Suppose there exists a vertex $v\in L$ such that $v$ is not red-complete to $S\cup T$ in $(G,\sigma)$. By the choice of $S$ and $T$, $v$ is not green-adjacent to any vertex in $S\cup T$ and thus there must exist some $V_{\ell}\subseteq S\cup T$ such that $v$ is blue-complete to $V_{\ell}$. We may assume that $a_{j},u\in V_{\ell}$ for some $j\in[2n-2]$ and $u\in T$. But then $a_{j-1}u$ is colored blue and we obtain a blue $C_{2n}$ in $(G,\sigma)$ with vertices $a_{1},a_{2},\ldots,a_{j-1},u,v,a_{j},\ldots,a_{2n-2}$ in order, contrary to (4). This proves that every vertex in $L$ is red-complete to $S\cup T$ in $(G,\sigma)$. Next, suppose there exist $V_{i}\subseteq S\cup T$ and $V_{j}\subseteq S\cup T$ such that $V_{i}$ is blue-complete to $V_{j}$ in $(G,\sigma)$. We may assume that $\{a_{1},u_{1}\}\subseteq V_{i}$ and $\{a_{\ell},u_{2}\}\subseteq V_{j}$ for some $\ell\in\{2,\ldots,2n-2\}$ and $u_{1},u_{2}\in T$. But then $u_{2}a_{\ell+1}$ is colored blue and we obtain a blue $C_{2n}$ in $(G,\sigma)$ with vertices $a_{1},\ldots,a_{\ell},u_{1},u_{2},a_{\ell+1},\ldots,a_{2n-2}$ in order, contrary to (4). $\blacksquare$

(12) If a vertex $v\in T\cup L$ is blue-complete to an edge, say $a_{1}a_{2}$, on the cycle $C$, then no vertex in $(T\cup L)\setminus v$ is blue-complete to any edge on the cycle with vertices $a_{1},v,a_{2},\ldots,a_{2n-2}$ in order.

Proof.  It follows from the fact that $(G,\sigma)$ has no blue $C_{2n}$. $\blacksquare$

For each vertex $u\in T\cup L$, let

[TABLE]

Since no vertex in $L$ is green-adjacent to any vertex on the cycle $C$, we see that $|N_{r}(u)|+|N_{b}(u)|=2n-2$ for all $u\in L$. By (4), $|N_{r}(u)|=|V(C)\cap R|\geq n$ for all $u\in V_{p}$.

(13) $|L|\leq n$. Consequently, $|T|\geq 1$ and so $s\geq 1$.

Proof.  Suppose for a contradiction that $|L|\geq n+1$. Let $L^{*}$ be the set of all vertices $u\in L$ with $|N_{r}(u)|\geq n$. 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 $(G,\sigma)$. By the choice of $\sigma$ and the fact that $V(C)\subseteq R\cup B$, we see that $H^{*}$ contains no cycle of length $2n$. By Corollary 3.6 applied to $H^{*}$ with $M=L^{*}$, $N=V(C)$ and $\delta(M)\geq n>(|N|+1)/2$, we have $|L^{*}|\leq n-1$. It follows that $|L\setminus L^{*}|\geq 2$ and for any $v\in L\setminus L^{*}$, $|N_{r}(v)|\leq n-1$ and so $|N_{b}(v)|\geq n-1$. Then $(L\setminus L^{*})\cap V_{p}=\emptyset$ by (4). We next claim that there exist two distinct vertices $w_{1},w_{2}\in L\setminus L^{*}$ with $N_{r}(w_{1})=N_{r}(w_{2})$ such that $|N_{r}(w_{1})|=|N_{r}(w_{2})|=n-1$ and every vertex in $L\setminus\{w_{1},w_{2}\}$ is red-adjacent to at least $n-2$ vertices in $N_{r}(w_{1})$.

Suppose first that no vertex in $L\setminus L^{*}$ is blue-complete to any edge on the cycle $C$. Then for any $v\in L\setminus L^{*}$, $|N_{b}(v)|=n-1$ and so $|N_{r}(v)|=n-1$. Let $w_{1},w_{2}$ be two distinct vertices in $L\setminus L^{*}$. We may assume that $w_{1}$ is blue-complete to $\{a_{1},a_{3},\ldots,a_{2n-3}\}$ in $(G,\sigma)$. Then $L\setminus L^{*}$ must be blue-complete to $\{a_{1},a_{3},\ldots,a_{2n-3}\}$ and no vertex in $L^{*}$ is blue-adjacent to two vertices in $\{a_{2},a_{4},\ldots,a_{2n-2}\}$, else in each case we obtain a blue $C_{2n}$ in $(G,\sigma)$. Then $N_{r}(w_{1})=N_{r}(w_{2})=\{a_{2},a_{4},\ldots,a_{2n-2}\}$ and every vertex in $L\setminus\{w_{1},w_{2}\}$ is red-adjacent to at least $n-2$ vertices in $N_{r}(w_{1})$, as claimed. Suppose next that some vertex in $L\setminus L^{*}$ is blue-complete to an edge, say $a_{1}a_{2}$, on the cycle $C$. Let $w_{1}$ be such a vertex, and let $C^{*}$ be the blue cycle on $2n-1$ vertices with edge set $\{w_{1}a_{1},w_{1}a_{2}\}\cup E(C\setminus a_{1}a_{2})$. Suppose first $\{w_{1}\}\in\{V_{1},\ldots,V_{p-1}\}$. By (4), every vertex in $L\setminus w_{1}$ is red-adjacent to at least $n$ vertices on $C^{*}$. If each vertex in $V_{p}$ is red-adjacent to at least $n+1$ vertices on $C^{*}$, let $H^{*}$ be the bipartite subgraph of $G$ with bipartition $\{L\setminus w_{1},V(C^{*})\}$ and $E(H^{*})$ consisting of all red edges between $L\setminus w_{1}$ and $V(C^{*})$ in $(G,\sigma)$. By Corollary 3.6 applied to $H^{*}$ with $M=L\setminus w_{1}$, $N=V(C^{*})$, $\Delta(M)\geq n+1$ and $\delta(M)\geq n=(|N|+1)/2$, $H^{*}$ has a cycle of length $2n$, which yields a red $C_{2n}$ in $(G,\sigma)$, contrary to (4). Thus each vertex in $V_{p}$ is red-adjacent to at most $n$ vertices on $C^{*}$. Since $|N_{r}(u)|\geq n$ for each $u\in V_{p}$, we see that $w_{1}$ is blue-complete to $V_{p}$ in $(G,\sigma)$ and $V_{p}$ is red-adjacent to exactly $n$ vertices on $C$. Then $V_{p}$ must be red-complete to $\{a_{1},a_{2}\}$ in $(G,\sigma)$ and no vertex in $V_{p}$ is blue-complete to any edge on $C$, else we obtain a blue $C_{2n}$ in $(G,\sigma)$. Thus one of $a_{3}$ and $a_{2n-2}$, say $a_{3}$, must be blue-complete to $V_{p}$ under $\sigma$. But then $(G,\sigma)$ contains a blue $C_{2n}$ with edge set $E(C\setminus a_{2}a_{3})\cup\{a_{2}w_{1},w_{1}x,xa_{3}\}$, contrary to (4). This proves that $\{w_{1}\}\notin\{V_{1},\ldots,V_{p-1}\}$. Thus there exists $V_{\ell}\in\{V_{1},\ldots,V_{p-1}\}$ such that $w_{1}\in V_{\ell}$ and $|V_{\ell}|\geq 2$. Let $w_{2}\in V_{\ell}$ with $w_{2}\neq w_{1}$. By (4), no vertex in $V_{\ell}$ is blue-complete to any edge on $C\setminus a_{1}a_{2}$. This, together with fact that $|N_{b}(w_{i})|\geq n-1$, implies that $|N_{b}(w_{i})|=n-1$ for all $i\in[2]$, and no vertex in $L\setminus\{w_{1},w_{2}\}$ is blue-adjacent to two vertices in $N_{r}(w_{1})$, otherwise we obtain a blue $C_{2n}$ in $(G,\sigma)$. Then every vertex in $L\setminus\{w_{1},w_{2}\}$ is red-adjacent to at least $n-2$ vertices in $N_{r}(w_{1})$, as claimed.

Let $L^{\prime}$ be a subset of $L$ such that $|L^{\prime}|=n-1$, $w_{1},w_{2}\in L^{\prime}$ and $(V_{p}\setminus x)\cap L^{\prime}\neq\emptyset$. Then $|L^{\prime}|+|N_{r}(w_{1})|=(n-1)+(n-1)=2n-2$. Let $H^{*}$ be the bipartite subgraph of $G$ with bipartition $\{L^{\prime},N_{r}(w_{1})\}$ and $E(H^{*})$ consisting of all red edges between $L^{\prime}$ and $N_{r}(w_{1})$ in $(G,\sigma)$. Then by the choice of $w_{1},w_{2}$, we see that $\{w_{1},w_{2}\}$ is red-complete to $N_{r}(w_{1})$. By Corollary 3.5 applied to $H^{*}$ with $M=L^{\prime}$ and $N=N_{r}(w_{1})$, $H^{*}$ has a cycle $C^{*}$ on $2n-2$ vertices. By the choice of $L^{\prime}$, let $x_{1}x_{2}$ be an edge of $C^{*}$ with $x_{1}\in V_{p}\setminus x$. Let $u\in V(C)\cap R$ with $u\notin V(C^{*})$. But then $(G,\sigma)$ has a red $C_{2n}$ with edge set $E(C^{*}\setminus x_{1}x_{2})\cup\{x_{1}u,ux,xx_{2}\}$, contrary to (4). This proves that $|L|\leq n$. Consequently, $|T|=|G|-|V(C)\cup L|\geq 3n-1-(2n-2+n)\geq 1$ and so $s=|S|\geq 1$. $\blacksquare$

By (4) and (4), $X_{2}=\emptyset$. By (4), $z\in L\setminus V_{p}$. We next claim that

(14) $s\leq n-4$.

Proof.  Suppose that $s\geq n-3$. By (4) and the fact that $V_{p}\subseteq L$ and $|V_{p}|\leq n-2$, we may assume that $S\cup T=V_{1}\cup\cdots\cup V_{\ell}$ for some $\ell\in[p-1]$. By (4) again, $L$ is red-complete to $V_{1}\cup\cdots\cup V_{\ell}$ and $V_{i}$ is red-complete to $V_{j}$ for all $i,j\in[\ell]$ with $i\neq j$ when $\ell\geq 2$. By (4), $|V(C)\cap R|\geq n$ and so $(V(C)\setminus S)\cap R\neq\emptyset$. Let $X^{*}:=\{x\}$ when $s\geq n-2$ and $X^{*}:=\{x,x^{\prime}\}$ when $s=n-3$, where $x^{\prime}\in V_{p}\setminus x$. Then

[TABLE]

We define $L^{*}$ to be a subset of $S\cup T\cup(L\setminus X^{*})$ with $|L^{*}|=2n-2|X^{*}|$ such that when $|X^{*}|=1$, $L^{*}\cap V_{p}\neq\emptyset$; and when $|X^{*}|=2$, $z\in L^{*}$ and $L^{*}\cap T\neq\emptyset$. Let $H^{*}$ be the subgraph of $G$ with $V(H^{*})=L^{*}$ and $E(H^{*})$ consisting of all red edges in $(G,\sigma)$ between $L^{*}\cap(L\setminus X^{*})$ and $L^{*}\cap(V_{1}\cup\cdots\cup V_{\ell})$, and all edges between each pair $L^{*}\cap V_{i}$ and $L^{*}\cap V_{j}$ for all $i,j\in[\ell]$ with $i\neq j$. Since $|V_{i}|\leq n-2$ for all $i\in[\ell]$, $|L\setminus X^{*}|\leq n-|X^{*}|$, we see that $\delta(H^{*})\geq|H^{*}|/2$. When $X^{*}=\{x\}$, let $u\in(V(C)\cap R)\setminus S$. By Theorem 3.2, $H^{*}$ contains a cycle $C^{*}:=C_{2n-2}$, say with vertices $v_{1},v_{2},\ldots,v_{2n-2}$ in order. We may assume that $v_{1}\in V_{p}$. Then $\{u,v_{2}\}$ is red-complete to $\{x,v_{1}\}\subseteq V_{p}$ and $(G,\sigma)$ contains a red $C_{2n}$ with vertices $v_{1},u,x,v_{2},\ldots,v_{2n-2}$ in order, contrary to (4). Thus $X^{*}=\{x,x^{\prime}\}$ and $s=n-3$. By Theorem 3.2 again, $H^{*}$ contains a cycle $C^{*}:=C_{2n-4}$, say with vertices $v_{1},v_{2},\ldots,v_{2n-4}$ in order. We may further assume that $v_{1}=z$. By the choice of $E(H^{*})$, $v_{2}\in S\cup T$. Note that $v_{1}\in B$ by the choice of $z$.

Since $C\setminus S$ contains at most $s-1$ isolated vertices and $2n-2-(s+s-1)=2n-1-2(n-3)=5$, it follows that $C\setminus S$ contains at least three edges. By (4), $V_{p}$ is not blue-complete to any two edges on the cycle $C$. Let $ab,cd\in E(C\setminus S)$ such that $a,c\in R$. Then $b,d\in R$, else, say $b\in B$, then $(G,\sigma)$ contains a blue $C_{2n}$ with edge set $E(C\setminus ab)\cup\{av_{1},v_{1}x,xb\}$ if $av_{1}$ is blue; and a red $C_{2n}$ with vertices $v_{1},a,x,c,x^{\prime},v_{2},\ldots,v_{2n-4}$ in order if $av_{1}$ is red, contrary to (4). Thus $v_{1}$ is blue-complete to $\{a,b,c,d\}$, else, say $v_{1}a$ is red, then $(G,\sigma)$ contains a red $C_{2n}$ with vertices $v_{1},a,x,c,x^{\prime},v_{2},\ldots,v_{2n-4}$ in order, contrary to (4). By (4), no vertex in $T\cup(L\setminus z)$ is blue-complete to $\{a,b\}$ or $\{c,d\}$ in $(G,\sigma)$. By the choice of $S\cup T$, we see that no vertex in $S$ is blue-complete to $\{a,b\}$ or $\{c,d\}$ in $(G,\sigma)$. We may assume that $v_{3}a$ is colored red because $v_{3}\in S\cup T\cup(L\setminus z)$. But then $(G,\sigma)$ contains a red $C_{2n}$ with vertices $v_{1},v_{2},x,b,x^{\prime},a,v_{3},\ldots,v_{2n-4}$ in order, contrary to (4). $\blacksquare$

(15) There exists a subgraph $J$ of $C\setminus S$ with $|J|\geq 2n-2-2s\geq 6$ such that each component of $J$ is a path of even order, and for all $u\in T\cup L$ 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 $|N_{r}(u)\cap V(J)|\geq|J|/2$.

Proof.  By (4), $s\geq 1$ and $|T|\geq 1$. By (4), $|V(C)\setminus S|\geq n+2$. Since $S$ is an independent set of the cycle $C$, we see that $C\setminus S$ has exactly $s$ components such that each component is a path. Assume first that no vertex in $T\cup L$ is blue-complete to an edge on the cycle $C$. Let $J$ be obtained from $C\setminus S$ by deleting one end of each odd component of $C\setminus S$. Then $|J|\geq 2n-2-2s$ and for all $u\in T\cup L$, $u$ is not green-adjacent to any vertex in $V(J)$, and so $|N_{r}(u)\cap V(J)|\geq|J|/2$ because $u$ is not blue-complete to any edge on the cycle $C$. By (4), $|J|\geq 2n-2-2s\geq 6$. Assume next that there exists a vertex $w\in T\cup L$ such that $w$ is blue-complete to an edge, say $a_{1}a_{2}$, on the cycle $C$. By (4), no vertex in $(T\cup L)\setminus w$ is blue-complete to any edge on the path $P:=C\setminus a_{1}a_{2}$. Let $J$ be obtained from $P\setminus S$ by deleting one end of each odd component of $P\setminus S$. We claim that $P\setminus S$ has at most $s$ odd components. Suppose $P\setminus S$ has at least $s+1$ odd components. Then $P\setminus S$ 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 $P\setminus S$ has at most $s$ odd components, as claimed. Then $|J|\geq 2n-2-2s\geq 6$ and for all $u\in T\cup L$ with $u\neq w$, $u$ is not green-adjacent to any vertex in $V(J)$, and so $|N_{r}(u)\cap V(J)|\geq|J|/2$ because $u$ is not blue-complete to any edge in $J$. $\blacksquare$

By (4), let $w$ be the possible vertex in $T\cup L$ such that $w$ is blue-complete to some edge in $J$. Let $T^{\prime}:=T\setminus w$ and $L^{\prime}:=L\setminus w$. Then $|T^{\prime}\cup L^{\prime}|=|(T\cup L)\setminus w|\geq n$. We next show that $T^{\prime}\neq\emptyset$. Suppose $T^{\prime}=\emptyset$. Then $T=\{w\}$ and $L=L^{\prime}$ with $|L|\geq n$. By (4), for any $u\in L$, $|N_{r}(u)\cap V(J)|\geq|J|/2\geq n-1-s$ and $u$ is not green-adjacent to any vertex in $V(J)$. By (4), $L$ is red-complete to $S\cup T$. Let $H^{*}$ be the bipartite graph of $G$ with bipartition $\{L,V(C)\cup\{w\}\}$ and $E(H^{*})$ consisting of all red edges between $L$ and $V(C)\cup\{w\}$ in $(G,\sigma)$. Then each vertex in $L\setminus V_{p}$ is red-adjacent to at least $|N_{r}(u)\cap V(J)|+|S\cup T|\geq(n-1-s)+s+1=n$ vertices in $V(C)\cup\{w\}$, and each vertex in $V_{p}$ is red-adjacent to at least $|T|+|R\cap V(C)|\geq n+1$ vertices in $V(C)\cup\{w\}$. By Corollary 3.6 applied to $H^{*}$ with $M=L$, $N=V(C)\cup\{w\}$, $\delta(M)\geq n$ and $\Delta(M)\geq n+1$, $H^{*}$ has a cycle of length $2n$, which yields a red $C_{2n}$ in $(G,\sigma)$, contrary to (4). Thus $T^{\prime}\neq\emptyset$.

Let $x_{1}\in V_{p}$ with $x_{1}\neq x$. Note that $|T^{\prime}\cup L^{\prime}|\geq n=(n-s-1)+(s+1)\geq(n-s-1)+2$. Let $W$ be a subset of $T^{\prime}\cup L^{\prime}$ with $|W|=n-s-1$ such that $x,x_{1}\notin W$, $|W\cap T^{\prime}|$ is as large as possible and $|W\cap V_{p}|$ is as small as possible. Let $W^{*}$ be a subset of $(T^{\prime}\cup L^{\prime})\setminus W$ with $|W^{*}|=s$ such that $x_{1}\in W^{*}$, $x\notin W^{*}$ and $|W^{*}\cap V_{p}|$ is as large as possible. Let $H^{*}$ be the subgraph of $G$ with $V(H^{*})=W^{*}\cup S$ and $E(H^{*})$ consisting of all red edges in $(G[W^{*}\cup S],\sigma)$. We claim that $\delta(H^{*})\geq|H^{*}|/2$. Assume first $|V_{p}|\geq s+1$. By the choice of $W$ and $W^{*}$, $W^{*}\subseteq V_{p}$. Then $W^{*}$ is red-complete to $S$ and so $\delta(H^{*})\geq|H^{*}|/2$. Assume next $|V_{p}|\leq s$. By (4), every vertex in $W^{*}\cap L^{\prime}$ is red-complete to $S$; for every vertex $u\in W^{*}\cap T^{\prime}$, we may assume that $u\in V_{\ell}$ for some $\ell\in[p-1]$. Then $u$ is red-complete to $V(H^{*})\setminus V_{\ell}$, and so $\delta(H^{*})\geq|H^{*}|/2$ because $|V_{\ell}|\leq|V_{p}|\leq s$. Hence in both cases, $\delta(H^{*})\geq|H^{*}|/2$. 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 $H^{\prime}$ be the bipartite subgraph of $G$ with bipartition $\{W,V(J)\}$ and $E(H^{\prime})$ consisting of all red edges between $W$ and $V(J)$ in $(G,\sigma)$. 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 $M_{1}\neq\emptyset$ and $M_{2}\neq\emptyset$ and $V(J)$ is a disjoint union of $N_{1}$ and $N_{2}$ such that $|N_{1}|=|N_{2}|=|V(J)|/2\geq n-1-s\geq 3$ and $M_{i}$ is red-complete to $N_{i}$ but blue-complete to $N_{3-i}$ for all $i\in[2]$. By (4) and the choice of $W$, neither $J[N_{1}]$ nor $J[N_{2}]$ contains an edge and $J$ contains at least three independent edges, say $a_{i}a_{i+1},a_{j}a_{j+1},a_{\ell}a_{\ell+1}$ with $i<j<\ell$. We may further assume $a_{i},a_{j}\in N_{1}$. Then $a_{i+1},a_{j+1}\in N_{2}$ and $(G,\sigma)$ has a blue $C_{2n}$ with edge set $E(C\setminus\{a_{i}a_{i+1},a_{j}a_{j+1}\})\cup\{a_{i}b_{2},b_{2}a_{j},a_{i+1}b_{1},b_{1}a_{j+1}\}$, where $b_{1}\in M_{1}$ and $b_{2}\in M_{2}$, contrary to (4). This proves that $W$ and $V(J)$ have no such partition. By Corollary 3.4 applied to $H^{\prime}$ with $M=W$, $N=V(J)$, $\delta(M)\geq|V(J)|/2$, $H^{\prime}$ has an $(a,b)$-path $P$ on $2(n-1-s)-1=2n-(2s+3)$ vertices with $a,b\in W$. Note that $|(V(P)\cup V(C^{*}))\cap V(C)|=(n-2-s)+s=n-2$.

(16) If $(G,\sigma)$ contains a red cycle $C_{*}$ on $2n-2$ vertices such that $x_{1}\in V(C_{*})$ and $x\notin V(C_{*})$, then $|V(C_{*})\cap V(C)|\geq n$.

Proof.  Suppose $|V(C_{*})\cap V(C)|\leq n-1$. By (4), $|R\setminus V(C_{*})|\geq|(R\cap V(C))\setminus V(C_{*})|\geq 1$. Let $u\in R\setminus V(C_{*})$ and $x_{2}$ be a neighbor of $x_{1}$ on $C_{*}$. Then $xx_{2}$ is red because $x,x_{1}\in V_{p}$. But then $(G,\sigma)$ has a red $C_{2n}$ with edge set $E(C_{*}\setminus x_{1}x_{2})\cup\{x_{1}u,ux,xx_{2}\}$, contrary to (4). $\blacksquare$

Let $\ell\in[p-1]$ such that $V_{\ell}\cap T^{\prime}\neq\emptyset$. This is possible because $T^{\prime}\neq\emptyset$. By the choice of $T$, $S\cap V_{\ell}\neq\emptyset$. For the remainder of the proof, let $b_{1}\in S\cap V_{\ell}$. By the choice of $H^{*}$, $b_{1},x_{1}\in V(C^{*})$. Let $b_{2}$ be a neighbor of $b_{1}$ on the cycle $C^{*}$; let $E_{0}=\{b_{1}b_{2}\}$ when $s\geq 2$ and $E_{0}=\emptyset$ when $s=1$. Since $|V(P)\cap V(J)|=n-2-s$ and every minimum vertex cover of $J$ has at least $n-1-s$ vertices, we see that $J\setminus V(P)$ has at least one edge, say $d_{1}d_{2}$. We next claim that we can choose $P$ so that either $a\in T^{\prime}$ or $b\in T^{\prime}$. Suppose not. Then $H^{\prime}$ has no cycle on $2(n-1-s)$ vertices because $T^{\prime}\cap W\neq\emptyset$. By the choice of $W$, $a,b\in L^{\prime}$. By (4), $ab_{1}$ and $bb_{1}$ are red because $b_{1}\in S$. Then neither $d_{1}$ nor $d_{2}$ is red-complete to $\{a,b\}$ because $H^{\prime}$ has no cycle on $2(n-1-s)$ vertices. By (4) and the choice of $b_{1}$ and $J$, no vertex in $T^{\prime}\cup L^{\prime}$ is blue-complete to $\{d_{1},d_{2}\}$. Thus we may assume that $ad_{1},bd_{2},b_{1}d_{1}$ are red. If $b_{2}\in S$, then $bb_{2}$ is red. But then $(G,\sigma)$ has a red $C_{*}$ on $2n-2$ vertices with edge set $E(P)\cup(E(C^{*})\setminus E_{0})\cup\{b_{1}d_{1},d_{1}a,bb_{2}\}$ such that $|V(C_{*})\cap V(C)|\leq|(V(P)\cup V(C^{*}))\cap V(C)|+1=n-1$, contrary to (4). Thus $b_{2}\notin S$ and so $b_{2}$ is not blue-complete to $\{d_{1},d_{2}\}$. We may assume that $b_{2}d_{2}$ is red. But then $(G,\sigma)$ has a red $C_{*}$ on $2n-2$ vertices with edge set $E(P)\cup(E(C^{*})\setminus E_{0})\cup\{b_{1}a,bd_{2},d_{2}b_{2}\}$ such that $|V(C_{*})\cap V(C)|\leq|(V(P)\cup V(C^{*}))\cap V(C)|+1=n-1$, contrary to (4). This proves that we can choose $P$ so that either $a\in T^{\prime}$ or $b\in T^{\prime}$, say the former.

By the choice of $T$ and $b_{1}$, we may further assume that $a\in V_{\ell}$ and so $ab_{1}$ is colored green. Then $ab_{2}$ is red. Recall that $a\in T^{\prime}$. By (4), $a$ is not blue-complete to $\{d_{1},d_{2}\}$ and neither $ad_{1}$ nor $ad_{2}$ is green. We may assume that $ad_{1}$ is red. Then $b_{1}d_{1}$ is red. It follows that $bb_{2}$ is not colored red, otherwise $(G,\sigma)$ has a red $C_{*}$ on $2n-2$ vertices with edge set $E(P)\cup(E(C^{*})\setminus E_{0})\cup\{b_{1}d_{1},d_{1}a,bb_{2}\}$ such that $|V(C_{*})\cap V(C)|\leq|(V(P)\cup V(C^{*}))\cap V(C)|+1=n-1$, contrary to (4). Similarly, $bd_{1}$ is blue because $ab_{2}$ is red. Note that $ad_{1}$ is red and $bd_{1}$ is blue, we see that $b\notin V_{\ell}$. Then $b$ is red-complete to $\{a,b_{1}\}$ by (4) and $bd_{2}$ is red by (4). If $b_{2}d_{2}$ is blue, then $bb_{2}$ is not colored green because $bd_{2}$ is red. Thus $bb_{2}$ is blue because $bb_{2}$ is not red. But then $(G,\sigma)$ has a blue $C_{2n}$ with edge set $E(C\setminus d_{1}d_{2})\cup\{d_{1}b,bb_{2},b_{2}d_{2}\}$, contrary to (4). Thus $b_{2}d_{2}$ is red. Since $a\in T^{\prime}$ and $b$ is red-complete to $\{a,b_{1}\}$, by the choice of $H^{\prime}$, there must exist three consecutive vertices, say $v_{1},v_{2},v_{3}$, on $P$ with $v_{1},v_{3}\in W$ and $v_{2}\in V(J)$ such that either $\{v_{1},v_{3}\}\cap T\neq\emptyset$ and $\{v_{1},v_{3}\}\cap L\neq\emptyset$, or $v_{1}\in V_{\ell}^{\prime}\cap T$ and $v_{3}\in V_{\ell^{\prime\prime}}\cap T$ for some $\ell^{\prime},\ell^{\prime\prime}\in[p-1]$ with $\ell^{\prime}\neq\ell^{\prime\prime}$. By (4), $v_{1}v_{3}$ is colored red. But then $(G,\sigma)$ has a red $C_{*}$ on $2n-2$ vertices with edge set $E(P\setminus v_{2})\cup(E(C^{*})\setminus E_{0})\cup\{v_{1}v_{3},b_{1}d_{1},d_{1}a,bd_{2},d_{2}b_{2}\}$ such that $|V(C_{*})\cap V(C)|\leq|(V(P\setminus v_{2})\cup V(C^{*}))\cap V(C)|+2=n-1$, contrary to (4).

This completes the proof of Theorem 1.6. $\blacksquare$

Acknowledgements

The authors would like to thank Christian Bosse and Jingmei Zhang for their helpful discussion.