Results on proper conflict-free list coloring of graphs

TL;DR

This paper investigates the proper conflict-free list coloring of graphs, especially for degree-based list assignments with added constants, providing new results for graphs with maximum degree up to 4 and special classes like subdivided graphs.

Contribution

It introduces the concept of proper conflict-free degree+k-choosability, proves new results for graphs with maximum degree up to 4, and explores special graph classes like subdivided graphs.

Findings

Connected graphs with max degree ≤ 3 (except 5-cycle) are degree+2-choosable.

Graphs obtained by subdividing each edge once are degree+2-choosable.

Graphs with max degree ≤ 4 are degree+3-choosable using DP-coloring techniques.

Abstract

Given a graph $G$ and a mapping $f:V(G) \to \mathbb{N}$, an $f$-list assignment of $G$ is a function that maps each $v \in V(G)$ to a set of at least $f(v)$ colors. For an $f$-list assignment $L$ of a graph $G$, a proper conflict-free $L$-coloring of $G$ is a proper coloring $\phi$ of $G$ such that $\phi(v) \in L(v)$ for every vertex $v\in V(G)$ and $v$ has a color that appears precisely once at its neighborhood for every non-isolated vertex $v\in V(G)$. We say that $G$ is proper conflict-free $f$-choosable if for any $f$-list assignment $L$ of $G$, there exists a proper conflict-free $L$-coloring of $G$. For a non-negative integer $k$, we say that $G$ is \emph{proper conflict-free $({\rm degree}+k)$-choosable} if $G$ is proper conflict-free $f$-choosable where $f$ is a mapping with $f(v)= d_G(v)+k$ for every vertex $v\in V(G)$. Motivated by degree-choosability of graphs, we investigate the proper conflict-free $({\rm degree}+k)$-choosability of graphs, especially for cases $k=1,2,3$. As the 5-cycle is not proper conflict-free $({\rm degree}+2)$-choosable and it is the only such graph we know, it is possible that every connected graph other than the 5-cycle is proper conflict-free $({\rm degree}+2)$-choosable and thus every graph is proper conflict-free $({\rm degree}+3)$-choosable. To support these, we show that every connected graph with maximum degree at most 3 distinct from the 5-cycle is proper conflict-free $(\text{degree}+2)$-choosable, and that $S(G)$ is proper conflict-free $(\text{degree}+2)$-choosable for every graph $G$, where $S(G)$ is a graph obtained from $G$ by subdividing each edge once. Furthermore, by adapting the technique of DP-colorings, we prove that every graph with maximum degree at most $4$ is proper conflict-free $({\rm degree}+3)$-choosable.

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Full text

Motivated by degree-choosability of graphs, we investigate the proper conflict-free $({\rm degree}+k)$-choosability of graphs, especially for cases $k=1,2,3$. As the 5-cycle is not proper conflict-free $({\rm degree}+2)$-choosable and it is the only such graph we know, it is possible that every connected graph other than the 5-cycle is proper conflict-free $({\rm degree}+2)$-choosable and thus every graph is proper conflict-free $({\rm degree}+3)$-choosable. To support these, we show that every connected graph with maximum degree at most 3 distinct from the 5-cycle is proper conflict-free $(\text{degree}+2)$-choosable, and that $S(G)$ is proper conflict-free $(\text{degree}+2)$-choosable for every graph $G$, where $S(G)$ is a graph obtained from $G$ by subdividing each edge once. Furthermore, by adapting the technique of DP-colorings, we prove that every graph with maximum degree at most $4$ is proper conflict-free $({\rm degree}+3)$-choosable.

Keywords: proper conflict-free coloring, odd coloring, list coloring, maximum degree, degree choosability

1 Introduction

Throughout this paper, we only consider simple, finite and undirected graphs. We refer readers to Bondy and Murty [1] for all terminologies and notations not defined here.

A $k$-coloring of a graph $G$ is a map $\phi:V(G)\to\{1,2,\ldots,k\}$. A coloring $\phi$ is called proper if every adjacent pair of vertices receive distinct colors. The chromatic number of a graph $G$, denoted by $\chi(G)$, is the least positive integer $k$ such that $G$ admits a proper $k$-coloring.

For a graph $G$, a coloring $\phi$ is called proper conflict-free if $\phi$ is a proper coloring of $G$, and for every non-isolated vertex $v\in V(G)$, there exists a color that appears exactly once in the neighborhood of $v$. The proper conflict-free chromatic number of a graph $G$, denoted by $\chi_{\textrm{pcf}}(G)$, is the least positive integer $k$ such that $G$ admits a proper conflict-free $k$-coloring. The notion of proper conflict-free coloring of a graph was introduced by Fabrici, Lužar, Rindošová and Sotá [12], where they investigated proper conflict-free coloring of planar graphs and outerplanar graphs.

Another related concept is odd coloring of graphs, which was first introduced by Petruševski and Škrekovski [17]. A coloring $\phi$ of $G$ is called an odd coloring of $G$ if $\phi$ is a proper coloring of $G$, and for every vertex $v\in V(G)$, there exists a color that appears odd times in the neighborhood of $v$. The odd chromatic number of a graph $G$, denoted by $\chi_{o}(G)$, is the least positive integer $k$ such that $G$ admits an odd $k$-coloring. By definitions, a proper conflict-free coloring of a graph $G$ is an odd coloring of $G$ as well, hence $\chi_{o}(G)\leq\chi_{\textrm{pcf}}(G)$ for any graph $G$.

In proper colorings of graphs, there is a trivial bound of the chromatic number that $\chi(G)\leq\Delta(G)+1$ for any graph $G$. This bound does not work for proper conflict-free colorings or odd colorings of graphs since $C_{5}$, the 5-cycle, has the maximum degree 2 and $\chi_{\textrm{pcf}}(C_{5})=\chi_{o}(C_{5})=5$. Caro, Petruševski and Škrekovski [4] conjectured that $\chi_{o}(G)\leq\Delta(G)+1$ for every connected graph $G$ provided $\Delta(G)\geq 3$, and verified the case $\Delta(G)=3$. They also conjectured the following stronger form in [5].

Conjecture 1**.**

For any connected graph $G$ with $\Delta(G)\geq 3$, $\chi_{\textup{pcf}}(G)\leq\Delta(G)+1$.

Conjecture 1 for the case $\Delta(G)=3$ follows from an old result by Liu and Yu [14], where they studied the similar concept in the name of superliniar coloring, and their result also implies the list version. For general cases, the best known bound of proper conflict-free chromatic number in terms of the maximum degree is the following by Cho, Choi, Kwon and Park [6].

Theorem 2**.**

For any graph $G$, $\chi_{\textup{pcf}}(G)\leq 2\Delta(G)-1$.

In fact, they showed a result on proper $h$-conflict-free coloring of a graph, which is a coloring with more general setting. For graphs with sufficient large maximum degree, it was proved recently by Liu and Reed [16] that $\chi_{\textrm{pcf}}(G)\leq(1+o(1))\Delta(G)$ by using the probabilistic method. There are some other results which describe asymptotic behaviors of proper conflict-free chromatic numbers and odd chromatic numbers (See [7, 8, 13]).

We say a graph $G$ is degree-choosable if $G$ is $f$-choosable with $f(v)=d_{G}(v)$ for each $v\in V(G)$. It is well known that a connected graph $G$ is not degree-choosable if and only if $G$ is a Gallai-tree, i.e. each block of $G$ is either a complete graph or an odd cycle. Degree-choosable graphs have been investigated in many papers [2, 3, 11, 18, 19]. We consider the analogous concept of degree-choosability for proper conflict-free coloring. We say that a graph $G$ is proper conflict-free $({\rm degree}+k)$-choosable if $G$ is proper conflict-free $f$-choosable with $f(v)=d_{G}(v)+k$ for each $v\in V(G)$ and some constant $k$. The following conjecture states the fundamental problem in this concept.

Conjecture 3**.**

There exists an absolute constant $k$ such that every graph is proper conflict-free $({\rm degree}+k)$-choosable.

Since the 5-cycle is not proper conflict-free $4$-choosable, if Conjecture 3 has a positive answer, then the constant $k$ must be at least 3. On the other hand, the 5-cycle is the only connected graph we know which is not proper conflict-free $({\rm degree}+2)$-choosable. This naturally derives the following conjecture.

Conjecture 4**.**

Every connected graph other than the 5-cycle is proper conflict-free $({\rm degree}+2)$-choosable.

We cannot reduce $(\text{degree}+2)$ in Conjecture 4 to $(\text{degree}+1)$ as it is shown by Caro, Petruševski and Škrekovski [5] that the cycle of length $\ell$ is not proper conflict-free $3$-colorable if $\ell$ is not divided by 3. More strongly, there are many graphs which are not proper conflict-free $(\text{degree}+1)$-choosable in a sense. We show the following.

Theorem 5**.**

For any connected graph $H$, there is a connected graph $G$ such that $H$ is an induced subgraph of $G$ and $G$ is not proper conflict-free $(\text{degree}+1)$-choosable.

In particular, large maximum degree does not imply proper conflict-free $({\rm degree}+1)$-choosability of graphs. This is significantly different from the behavior of proper conflict-free $(\Delta+1)$-colorability conjectured in Conjecture 1.

To support Conjecture 4, we verify it for two classes of graphs as follows. For a graph $G$, the 1-subdivision of $G$, denoted by $S(G)$, is a graph obtained from $G$ by replacing each edge with a path of length 2.

Theorem 6**.**

If $G$ is a connected graph with maximum degree at most $3$ and $G$ is not isomorphic to $C_{5}$, then $G$ is proper conflict-free $({\rm degree}+2)$-choosable.

Theorem 7**.**

For every graph $G$, $S(G)$ is proper conflict-free $({\rm degree}+2)$-choosable.

If Conjecture 4 has a positive answer, then obviously it implies that Conjecture 3 has a positive answer by choosing $k=3$. Thus we consider proper conflict-free $({\rm degree}+3)$-choosability of graphs, and prove the following by using a technique of DP-coloring, which was introduced by Dvořák and Postle [9].

Theorem 8**.**

Every graph $G$ with maximum degree at most $4$ is $({\rm degree}+3)$-choosable. In particular, every graph with maximum degree at most $4$ is proper conflict-free $7$-choosable.

Theorem 2 implies the bound $\chi_{\rm pcf}(G)\leq 5$ for graphs with maximum degree 3, and the bound $\chi_{\rm pcf}(G)\leq 7$ for graphs with maximum degree 4. Therefore, Theorems 6 and 8 respectively extend this two results to list version.

The paper is organized as follows. We give a proof of Theorem 5 in Section 2, proofs of Theorems 6 and 7 in Section 3, and a proof of Theorem 8 in Section 4.

We end this section by introducing some notations. For a positive integer $k$, let $[k]$ denote the set of integers $\{0,1,\dots,k-1\}$. Given a graph $G$ and a subgraph $H$ of $G$, for $v\in V(G)$ (not necessary in $H$), we denote $N_{H}(v)$ the neighbors of $v$ in $V(H)$. For a set $S\subseteq V(G)$, we denote $N_{H}(S)$ all the vertices in $V(H)$ which have a neighbor in $S$. For a (partial) coloring $\phi$ of $G$ and a subset $R\subseteq V(G)$, let $\phi(R)$ be the set comprised of all colors used on $R$ of $\phi$ (not every vertex in $R$ needs to be colored in $\phi$). For a (partial) proper $h$-conflict-free coloring $\phi$ of $G$, let $\mathcal{U}_{\phi}(v,G)$ be the set of colors that appear exactly once at the neighborhood of $v$ in $G$, and let $\mathcal{U}_{\phi}(R,G)=\bigcup_{v\in R}\mathcal{U}_{\phi}(v,G)$.

2 Degree plus 1 colors

In this section, we give a proof of Theorem 5.

Let $H$ be an arbitrary connected graph and let $v_{0}$ be a vertex of $H$. Let $d_{H}(v_{0})=d$. For each $i\in\{1,2,\dots,d+1\}$, let $C_{i}=v_{0}^{i}v_{1}^{i}v_{2}^{i}v_{3}^{i}v_{0}^{i}$ be a cycle of length $4$. Let $G$ be a graph obtained from $H$ and $C_{1},C_{2},\dots,C_{d+1}$ by identifying $v_{0}$ and $v_{0}^{1},v_{0}^{2},\dots,v_{0}^{d+1}$ into a vertex $v$. Then $H$ is an induced subgraph of $G$. We define a list assignment $L$ of $G$ as follows.

• •

for each $i\in\{1,2,\dots,d+1\}$, let $L(v_{1}^{i})=L(v_{2}^{i})=L(v_{3}^{i})=\{3i-2,3i-1,3i\}$,

• •

$L(v)=\{1,2,\dots,3d+3\}$, and

• •

for each vertex $u\in V(G)\setminus\{v\}$, let $L(u)=\{1,2,\dots,d_{G}(u)\}$.

Since $d_{G}(v)=d_{H}(v_{0})+2(d+1)=d+2(d+1)=3d+2$, $L$ is a list assignment of $G$ such that $|L(u)|\geq d_{G}(u)+1$ for every vertex $u$ of $G$. We show that $G$ is not proper conflict-free $L$-colorable. Suppose that $G$ admits a proper conflict-free $L$-coloring $\phi$. For each $i\in\{1,2,\dots,d+1\}$, since $\mathcal{U}_{\phi}(v_{2}^{i},G)\neq\emptyset$, we have $\{\phi(v_{1}^{i}),\phi(v_{2}^{i}),\phi(v_{3}^{i})\}=\{3i-2,3i-1,3i\}$. In addition, since $\phi(v_{j}^{i})\neq\phi(v)$ and $\mathcal{U}_{\phi}(v_{j}^{i},G)\neq\emptyset$ for $j=1,3$, we have $\phi(v)\notin\{\phi(v_{1}^{i}),\phi(v_{2}^{i}),\phi(v_{3}^{i})\}=\{3i-2,3i-1,3i\}$. Combining these, we have $\phi(v)\notin\bigcup_{i=1}^{d+1}\{3i-2,3i-1,3i\}=\{1,2,\dots,3d-3\}=L(v)$, a contradiction.

3 Degree plus 2 colors

In this section, we give proofs of Theorems 6 and 7.

3.1 Proof of Theorem 6

Let $L$ be a list assignment of $G$ with $|L(v)|\geq d_{G}(v)+2$ for each $v\in V(G)$. The statement is trivial when $G$ is isomorphic to $K_{1}$ or $K_{2}$, so we assume that $G$ has at least three vertices.

If $G$ is a cycle of length $\ell\leq 4$, then obviously $G$ admits a proper conflict-free $L$-coloring with $\ell$ distinct colors. We assume that $G=u_{1}u_{2}\cdots u_{\ell}u_{1}$ is a cycle for some $\ell\geq 6$. If $L(u_{i})=L(u_{j})$ for any $i\neq j$, then by the result of Caro et al. [5], $G$ is $L$-colorable. Thus without loss of generality, we assume that $L(u_{1})\neq L(u_{k})$. Let $\phi(u_{1})$ be a color in $L(u_{1})\setminus L(u_{k})$. We choose colors for $\{u_{2},u_{3},\dots,u_{k-1}\}$ in the ascending order of indices such that $\phi(u_{2})\in L(u_{2})\setminus\{\phi(u_{1})\}$ and $\phi(u_{i})\in L(u_{i})\setminus\{\phi(u_{i-1}),\phi(u_{i-2})\}$. Finally, we choose a color in $L(u_{k})\setminus\{\phi(u_{2}),\phi(u_{k-2}),\phi(u_{k-1})\}$ as $\phi(u_{k-1})$ to obtain an $L$-coloring $\phi$ of $G$. It is easy to check $\phi$ is a proper conflict-free $L$-coloring of $G$.

Hence we may assume that $G$ is not a cycle, and thus $\Delta(G)=3$. Let $G_{0},G_{1},\dots,G_{k}$ be a sequence of subgraphs of $G$ such that $G_{0}=G$, $G_{k}$ is either isomorphic to $K_{2}$ or a graph with minimum degree at least 2, and $G_{i+1}=G_{i}-v_{i}$ for some $v_{i}\in V(G_{i})$ with $d_{G_{i}}(v_{i})=1$ for each $i\in\{0,1,\dots,k-1\}$. Note that $G_{i}$ is connected for every $i\in\{0,1,\dots,k\}$. Then for each $i\in\{0,1,\dots,k-1\}$, $G_{i}$ is proper conflict-free $L$-colorable if $G_{i+1}$ is proper conflict-free $L$-colorable. Indeed, suppose $\phi$ is a proper conflict-free $L$-coloring of $G_{i+1}$, and $x$ is the neighbor of $v_{i}$ in $G_{i}$. We choose a color $c_{x}\in\mathcal{U}_{\phi}(x,G_{i+1})$ arbitrarily, and choose a color in $L(v_{i})\setminus\{\phi(x),c_{x}\}$ as $\phi(v_{i})$ to extend $\phi$ to $G$. Obviously $\phi$ is a proper conflict-free $L$-coloring of $G_{i}$, as desired.

By the last paragraph, it suffices to show that $G_{i}$ is proper conflict-free $L$-colorable for some $i\in\{1,2,\dots,k\}$. If $G_{k}$ is isomorphic to $K_{2}$, then it is proper conflict-free $L$-colorable. If $G_{k}$ has a vertex of degree 3, then by the result in Liu and Yu [14], $G_{k}$ is proper conflict-free $4$-choosable, and hence proper conflict-free $L$-colorable. Thus, we assume that $G_{k}=w_{1}w_{2}\cdots w_{t}w_{1}$ is a cycle of length $t\geq 3$. Without loss of generality, we assume $v_{k-1}$ is adjacent to $w_{1}$, and thus $d_{G}(w_{1})=3$. Now we define an $L$-coloring $\phi$ of $G_{k-1}$ as follows: We choose colors $\phi(w_{2})\in L(w_{2})$, $\phi(w_{3})\in L(w_{3})\setminus\{\phi(w_{2})\}$, and choose a color $\phi(w_{j})$ for each of $\{w_{4},w_{5},\dots,w_{t}\}$ in the ascending order of indices such that $\phi(w_{j})\in L(w_{j})\setminus\{\phi(w_{j-1}),\phi(w_{j-2})\}$. Finally we choose a color in $L(w_{1})\setminus\{\phi(w_{2}),\phi(w_{3}),\phi(w_{t-1}),\phi(w_{t})\}$ as $\phi(w_{1})$ and a color in $L(v_{k-1})\setminus\{\phi(w_{1}),\phi(w_{2})\}$ as $\phi(v_{k-1})$. It is easy to check that $\phi$ is a proper conflict-free $L$-coloring of $G_{k-1}$.

3.2 Proof of Theorem 7

Let $G$ be a graph with $n$ vertices $\{v_{1},v_{2},\dots,v_{n}\}$. The proof goes by induction on $n$. If $G$ is disconnected, then each component of $S(G)$ has a desired coloring and we simply combine them to obtain a desired coloring of $S(G)$. Thus we assume $G$ is connected. If $n=1$, then obviously $S(G)$ is isomorphic to $K_{1}$, which is proper conflict-free $(\text{degree}+2)$-choosable, so we assume $n\geq 2$. Let $S(G)$ be labeled as $V(S(G))=\{v_{i}\mid 1\leq i\leq n\}\cup\{u_{e}\mid e\in E(G)\}$ and $E(S(G))=\{v_{i}u_{e}\mid v_{i}\text{ incident with }e\text{ in }G\}$. Note that $d_{S(G)}(v_{i})=d_{G}(v_{i})$ for $i\in\{1,2,\ldots,n\}$ and $d_{S(G)}(u_{e})=2$ for every $e\in E(G)$. Let $L$ be a list assignment of $G$ such that $|L(v_{i})|\geq d_{G}(v_{i})+2$ for each $i\in\{1,2,\dots,n\}$ and $|L(u_{e})|=4$ for each edge $e\in E(G)$. If $G$ is isomorphic to $K_{2}$, then we choose distinct colors $\phi(v_{1})\in L(v_{1})$ and $\phi(v_{2})\in L(v_{2})$, and choose a color in $L(u_{v_{1}v_{2}})\setminus\{\phi(v_{1}),\phi(v_{2})\}$ as $\phi(u_{v_{1}v_{2}})$ to obtain a proper conflict-free $L$-coloring $\phi$ of $G$. Now we assume that $n\geq 3$ and the statement holds for every proper subgraph of $G$.

Suppose that $G$ contains a vertex of degree 1. Without loss of generality, we assume that $d_{G}(v_{1})=1$ and let $e=v_{1}v_{2}$ be the unique edge incident with $v_{1}$ in $G$. We choose an arbitrary color $\alpha$ from $L(v_{1})$. Let $G^{\prime}=G-v_{1}$ and let $L^{\prime}$ be a list assignment of $S(G^{\prime})$ defined by $L^{\prime}(v_{2})=L(v_{2})\setminus\{\alpha\}$ and $L^{\prime}(w)=L(w)$ for every other vertex $w$. By induction hypothesis, $S(G^{\prime})$ admits a proper conflict-free $L^{\prime}$-coloring $\phi$. As $n\geq 3$ and $G$ is connected, $v_{2}$ is not an isolated vertex of $S(G^{\prime})$, so we choose a color $c_{2}\in\mathcal{U}_{\phi}(v_{2},S(G^{\prime}))$ arbitrarily. Then we let $\phi(v_{1})=\alpha$ and choose a color in $L(u_{e})\setminus\{\phi(v_{2}),c_{2},\alpha\}$ as $\phi(u_{e})$ to extend $\phi$ to $S(G)$. It is easy to check that $\phi$ is a proper $L$-coloring of $S(G)$ and $\mathcal{U}_{\phi}(w,S(G))\neq\emptyset$ for every non-isolated vertex $w\in V(S(G^{\prime}))\setminus\{v_{2}\}$. Since $\phi(u_{e})\neq c_{2}$, the color $c_{2}$ is contained in $\mathcal{U}_{\phi}(v_{2},S(G))$. Furthermore, $\alpha\in\mathcal{U}_{\phi}(u_{e},S(G))$ since $\phi(v_{2})\neq\alpha$, and $\phi(u_{e})\in\mathcal{U}_{\phi}(v_{1},S(G))$, these imply that $\phi$ is a proper conflict-free $L$-coloring. So we assume that $\delta(G)\geq 2$, and thus $\delta(S(G))\geq 2$ as well.

If every vertex of $G$ has degree 2, and $L(u)=L(v)$ for any two distinct vertices $u,v$ in $S(G)$, then $S(G)$ is a cycle of length $2n$ and it is proper conflict-free $L$-colorable since every cycle other than the 5-cycle is proper conflict-free 4-colorable. Thus without loss of generality, we may assume that either $d_{S(G)}(v_{1})=2$ and there is an edge $e$ incident to $v_{1}$ in $G$ satisfying $L(v_{1})\neq L(u_{e})$, or $d_{S(G)}(v_{1})\geq 3$. Without loss of generality, we assume that $N_{G}(v_{1})=\{v_{2},v_{3},\dots,v_{d+1}\}$, where $d$ is the degree of $v_{1}$ in $G$. Let $e_{i}=v_{1}v_{i}\in E(G)$ for each $i\in\{2,3,\dots,d+1\}$. For each color $c\in L(v_{1})$, let $U_{c}=\{u_{e_{i}}\mid 2\leq i\leq d+1,c\in L(u_{e_{i}})\}$. Then there is a color $\alpha\in L(v_{1})$ such that $|U_{\alpha}|\leq\min\{3,d-1\}$. Indeed, if $d=2$, then we may assume that $L(v_{1})\neq L(u_{e_{2}})$, and thus the color $\alpha\in L(v_{1})\setminus L(u_{e_{2}})$ satisfies $U_{\alpha}\subseteq\{u_{e_{3}}\}$. If $d=3$, then since $|L(v_{1})|=5$, a color $\alpha\in L(v_{1})\setminus L(u_{e_{2}})$ satisfies $U_{\alpha}\subseteq\{u_{e_{3}},u_{e_{4}}\}$. If $d\geq 4$, then we have $|L(v_{1})|\geq d+2$, which implies that there must be a color $\alpha\in L(v_{1})$ such that $|U_{\alpha}|\leq 3$, for otherwise $4(d+2)\leq\sum_{c\in L(v_{1})}|U_{c}|=\sum_{i=2}^{d+1}|L(u_{e_{i}})|=4d$, a contradiction. Without loss of generality, we may assume that $u_{e_{2}}\notin U_{\alpha}$.

Let $G^{\prime}=G-(\{v_{1}\}\cup\{u_{e_{i}}\mid 2\leq i\leq d+1\})$ and let $L^{\prime}$ be a list assignment of $S(G^{\prime})$ such that $L^{\prime}(v_{i})=L(v_{i})\setminus\{\alpha\}$ for each $i\in\{2,3,\dots,d+1\}$ and $L^{\prime}(w)=L(w)$ for every vertex $w\in V(S(G^{\prime}))\setminus\{v_{i}\mid 2\leq i\leq d+1\}$. By induction hypothesis, $S(G^{\prime})$ admits a proper conflict-free $L^{\prime}$-coloring $\phi$. We extend $\phi$ to a proper conflict-free $L$-coloring of $G$. Note that $G^{\prime}$ has no isolated vertices since $\delta(G)\geq 2$. For each $i\in\{2,3,\dots,d+1\}$, let $c_{i}$ be a color in $\mathcal{U}_{\phi}(v_{i},S(G^{\prime}))$. Let $\phi(v_{1})=\alpha$, and we choose $\phi(u_{e_{i}})\in L(u_{e_{i}})\setminus\{\phi(v_{i}),c_{i},\alpha\}$ arbitrarily for each $u_{e_{i}}\in U_{\alpha}$. We define a color $\beta$ as follows: If $U_{\alpha}=\emptyset$, then let $\beta$ be a color in $L(u_{e_{2}})\setminus\{\phi(v_{2}),c_{2},\alpha\}$. If $U_{\alpha}\neq\emptyset$ and $|\phi(U_{\alpha})|=1$, then since $\alpha\notin L(u_{e_{2}})$, let $\beta$ be a color in $L(u_{e_{2}})\setminus(\{\phi(v_{2}),c_{2},\alpha\}\cup\phi(U_{\alpha}))$. Otherwise, i.e. $U_{\alpha}\neq\emptyset$ and $|\phi(U_{\alpha})|\geq 2$, the fact $|U_{\alpha}|\leq 3$ implies that some color appears exactly once at $U_{\alpha}$, so let $\beta$ be the color. Finally for each uncolored vertex $u_{e_{i}}$, since $u_{e_{i}}\notin U_{\alpha}$, we can choose a color $L(u_{e_{i}})\setminus\{\phi(v_{j}),c_{j},\alpha,\beta\}$ as $\phi(u_{e_{i}})$. It is easy to verify that $\phi$ is a proper $L$-coloring of $S(G)$, and that $\mathcal{U}_{\phi}(w,S(G))=\mathcal{U}_{\phi}(w,S(G^{\prime}))\neq\emptyset$ for each $w\in V(S(G^{\prime}))\setminus\{v_{i}\mid 2\leq i\leq d+1\}$. Since $\phi(u_{e_{i}})\neq c_{i}$, $\mathcal{U}_{\phi}(v_{i},S(G))$ contains the color $c_{i}$ for each $i\in\{2,3,\dots,d+1\}$. By definition of $L^{\prime}$, we have $\phi(v_{i})\neq\alpha=\phi(v_{1})$, so $\alpha\in\mathcal{U}_{\phi}(u_{e_{i}},S(G))$ for each $j\in\{2,3,\dots,d+1\}$. Furthermore, by the choice of colors $\{\phi(u_{e_{i}})\mid 2\leq i\leq d+1\}$, we have $\beta\in\mathcal{U}_{\phi}(v_{1},S(G))$. Hence $\phi$ is a proper conflict-free $L$-coloring of $S(G)$. This completes the proof of Theorem 7.

We remark that $({\rm degree}+2)$ cannot be reduced to $({\rm degree}+1)$ in Theorems 6 and 7. Indeed, Caro, Petruševski and Škrekovski [5] observed that if $n$ is not divided by $3$, then $C_{n}$ is not proper conflict-free $3$-choosable. Clearly $S(C_{3k+1})=C_{6k+2}$ for every positive integer $k$, and thus $S(C_{3k+1})$ is not proper conflict-free $({\rm degree}+1)$-choosable.

4 Degree plus 3 colors

In this section, we give a proof of Theorem 8.

We adapt the technique of the DP-coloring, which was introduced by Dvořák and Postle [9]. Recall that $[k]$ denote the set of integers $\{0,1,\dots,k-1\}$, and in this section, the indices are taken with modulo $k$ when it is related to the label of vertices.

Suppose that the statement is false, and let $G$ be a minimum counterexample in terms of the order, i.e. there is a list assignment $L$ of $G$ satisfying $|L(v)|\geq d_{G}(v)+3$ for each $v\in V(G)$ such that $G$ is not proper conflict-free $L$-colorable. Clearly, $G$ is connected and has at least 5 vertices as every vertex has at least 4 colors.

Claim 1**.**

The minimum degree of $G$ is at least 3.

Proof.

If there is a vertex $v$ of degree $1$ in $G$ with the neighbor $u$, then by the minimality of $G$, there is a proper conflict-free $L$-coloring $\phi$ of $G-v$. We can extend $\phi$ to $G$ by setting $\phi(v)\in L(v)\setminus\{\phi(u),\alpha\}$ for $v$, where $\alpha\in\mathcal{U}_{\phi}(u,G)$. Thus have $\delta(G)\geq 2$ as $G$ is connected.

Now we assume $v\in V(G)$ is a vertex of degree $2$ with neighbors $u_{1}$ and $u_{2}$. Let $G^{\prime}=(G-v)\cup\{u_{1}u_{2}\}$. As $\Delta(G^{\prime})\leq 4$, by the minimality of $G$, $G^{\prime}$ admits a proper conflict-free $L$-coloring $\phi$. For each $i\in\{1,2\}$, we define $c_{u_{i}}$ as follows: If $|\mathcal{U}_{\phi}(u_{i},G-v)|=0$, then all the neighbors of $u_{i}$ in $G-v$ receives the same color, say $\alpha_{i}$, in $\phi$, we let $c_{u_{i}}=\alpha_{i}$. Otherwise, $|\mathcal{U}_{\phi}(u_{i},G-v)|\geq 1$, we choose $c_{u_{i}}\in\mathcal{U}_{\phi}(u_{i},G-v)$. We choose a color in $L(v)\setminus\{\phi(u_{1}),\phi(u_{2}),c_{u_{1}},c_{u_{2}}\}$ as $\phi(v)$, which extends $\phi$ to $G$, a contradiction. ∎

Let $C=v_{0}v_{1}\cdots v_{k-1}v_{0}$ be a shortest cycle of $G$. Let $G^{\prime}=G-V(C)$. For each $i\in[k]$, let $N_{G}(v_{i})\setminus V(C)=\{u_{i},w_{i}\}$ if $d_{G}(v_{i})=4$, and let $N_{G}(v_{i})\setminus V(C)=\{u_{i}\}$ if $d_{G}(v_{i})=3$.

In the rest of the proof, we will first give an $L$-coloring (not necessarily proper conflict-free) $\phi$ of $G^{\prime}$, and then we extend $\phi$ to a proper conflict-free $L$-coloring of $G$. For a coloring $\phi$ of $G^{\prime}$, a pair of colors $(\alpha,\beta)$ and a vertex $v_{i}\in V(C)$, we say $v_{i}$ blocks $(\alpha,\beta)$ with respect to $\phi$ if either $d_{G}(v_{i})=4$, $\alpha\neq\beta$ and $\phi(\{u_{i},w_{i}\})=\{\alpha,\beta\}$, or $d_{G}(v_{i})=3$ and $\phi(u_{i})=\alpha=\beta$. For example, in the middle drawing of Fig. 1, $y$ blocks $(1,4)$ and $(4,1)$, $z$ blocks $(3,3)$. If $(\alpha,\beta)$ is not blocked by $v_{i}$, then we say $(\alpha,\beta)$ is feasible for $v_{i}$. Observe that if $v_{i}$ blocks $(\alpha,\beta)$, then we can not extend $\phi$ to $G$ as $\phi(\{v_{i-1},v_{i+1}\})=\{\alpha,\beta\}$ because such an extension forces the vertex $v_{i}$ to have $\mathcal{U}_{\phi}(v_{i},G)=\emptyset$. By the definition, we have the following observation.

Observation 1**.**

Every vertex blocks at most two color pairs. If $u$ blocks both $(\alpha_{1},\beta_{1})$ and $(\alpha_{2},\beta_{2})$, then they satisfy that $\alpha_{1}=\beta_{2}\neq\alpha_{2}=\beta_{1}$.

Claim 2**.**

$k\geq 4$, i.e., $G$ contains no triangle.

Proof.

Assume $k=3$. By the minimality of $G$, $G^{\prime}$ admits a proper $L$-coloring $\phi$ such that every non-isolated vertex $v$ of $G^{\prime}$ satisfies $\mathcal{U}_{\phi}(v,G^{\prime})\neq\emptyset$. For each vertex $u\in N_{G^{\prime}}(V(C))$ which is non-isolated in $G^{\prime}$, we choose $c_{u}\in\mathcal{U}_{\phi}(u,G^{\prime})$ arbitrarily. Let $L_{C}$ be a list assignment of $C$ defined by $L_{C}(v_{i})=L(v_{i})\setminus(\{c_{u}\mid u\in N_{G^{\prime}}(v_{i})\}\cup\phi(N_{G^{\prime}}(v_{i}))$ for each $i\in[3]$, and observe that $|L_{C}(v_{i})|\geq 3$. Now it suffices to show that $\phi$ can be extended to a proper $L$-coloring $\phi$ of $G$ such that $\phi(v_{i})\in L_{C}(v_{i})$ and $\mathcal{U}_{\phi}(v_{i},G)\neq\emptyset$ for each $i\in[3]$. Indeed, if that is possible, then for each isolated vertex $x$ of $G^{\prime}$, we have $|\mathcal{U}_{\phi}(x,G)|=3$, and for each non-isolated vertex $y$ of $G^{\prime}$, we have $|\mathcal{U}_{\phi}(y,G)|>0$ because we do not use $c_{u}$ in $V(C)$ for each $u\in N_{G^{\prime}}(V(C))$.

If $L_{C}(v_{0})=L_{C}(v_{1})=L_{C}(v_{2})$, then we extend $\phi$ to $G$ by giving an arbitrary $L_{C}$-coloring of $C$. Note that for each $i\in[3]$, $\phi(v_{i-1})\neq\phi(v_{i+1})$ and $\phi(\{v_{i-1},v_{i+1}\})\cap\phi(N_{G^{\prime}}(v_{i}))=\emptyset$, so $\mathcal{U}_{\phi}(v_{i},G)\neq\emptyset$, a contradiction.

Thus without loss of generality, we may assume $L_{C}(v_{0})\neq L_{C}(v_{1})$, and that there exists four distinct colors $1,2,3,4$ such that $\{1,2\}\subsetneq L_{C}(v_{0})$ and $\{3,4\}\subsetneq L_{C}(v_{1})$. By Observation 1, $v_{2}$ blocks at most one of $(1,3),(1,4),(2,3),(2,4)$, so without loss of generality, we may assume that $(1,4),(2,3),(2,4)$ are all feasible for $v_{2}$. Let $\alpha$ be a color in $L_{C}(v_{2})\setminus\{1,4\}$, then we deduce that $\phi(\{u_{0},w_{0}\})=\{1,\alpha\}$ or $\phi(\{u_{1},w_{1}\})=\{4,\alpha\}$. For otherwise we can extend $\phi$ by setting $\phi(v_{0})=1$, $\phi(v_{1})=4$ and $\phi(v_{2})=\alpha$. Without loss of generality, we may also assume that $\phi(\{u_{1},w_{1}\})=\{1,\alpha\}$, and this implies that $\alpha\neq 3$ as $3\in L_{C}(v_{1})$. If $\phi(\{u_{0},w_{0}\})=\{4,\alpha\}$, then $\alpha\neq 2$, and thus we extend $\phi$ by setting $\phi(v_{0})=2$, $\phi(v_{1})=3$ and $\phi(v_{2})=\alpha$, a contradiction. Thus we have $\phi(\{u_{0},w_{0}\})\neq\{4,\alpha\}$. If $\alpha\neq 2$, then we can extend $\phi$ by setting $\phi(v_{0})=2$, $\phi(v_{1})=4$ and $\phi(v_{2})=\alpha$, a contradiction. Therefore, we have that $\alpha=2$. Let $\beta\in L_{C}(v_{0})\setminus\{1,2\}$. Clearly, there exists a color $\gamma\in L_{C}(v_{1})\setminus\{\beta\}$ such that $(\beta,\gamma)$ is feasible for $v_{2}$ and we know that $\gamma\neq 2$ as $2\in\phi(\{u_{1},w_{1}\})$, we then extend $\phi$ by setting $\phi(v_{0})=\beta$, $\phi(v_{1})=\gamma$, and $\phi(v_{2})=2$. See the leftmost drawing on Fig. 1 for illustration. ∎

Now we observe that $G^{\prime}$ has no isolated vertices. Indeed, if $u$ is an isolated vertex of $G^{\prime}$, then since $\delta(G)\geq 3$ by Claim 2, there exist vertices $v_{a},v_{b},v_{c}\in V(C)\cap N_{G}(u)$ with $0\leq a<b<c\leq k-1$. By Claim 2, $b-a\geq 2$ and $c-b\geq 2$. Thus $v_{a}uv_{c}v_{c+1}\cdots v_{a-1}v_{a}$ is a shorter cycle than $C$, a contradiction.

Let $G^{\prime\prime}$ be the graph obtained from $G^{\prime}$ by adding all the edges $u_{i}w_{i}$ for each $i\in[k]$ if $w_{i}$ exists, and removing the multiple edges. Clearly, $\Delta(G^{\prime\prime})\leq 4$, and by the minimality of $G$, $G^{\prime\prime}$ has a proper conflict-free $L$-coloring $\phi$ with $\phi(u_{i})\neq\phi(w_{i})$ for every $i\in[k]$ if $d_{G}(v_{i})=4$. Note that for each $u\in N_{G}(V(C))$, $\mathcal{U}_{\phi}(u,G^{\prime\prime})\neq\emptyset$, but it is possible that $\mathcal{U}_{\phi}(u,G^{\prime})=\emptyset$ which holds only if $|\phi(N_{G^{\prime}}(u))|=1$ (the vertex $x$ in the second drawing on Fig. 1 is a such example). For each $u\in N_{G}(V(C))$, we define a color $c_{u}$ as follows (see again the second drawing on Fig. 1 for illustration): If $\mathcal{U}_{\phi}(u,G^{\prime})\neq\emptyset$, then let $c_{u}$ be an arbitrarily color in it, otherwise, let $c_{u}$ be the unique color in $\phi(N_{G^{\prime}}(u))$.

Let $L_{C}$ be a list assignment of $C$ with $L_{C}(v_{i})=L(v_{i})\setminus(\{c_{u}\mid u\in N_{G^{\prime}}(v_{i})\}\cup\phi(N_{G^{\prime}}(v_{i})))$ for each $i\in[k]$. Observe that $|L_{C}(v_{i})|\geq 3$, moreover $|L_{C}(v_{i})|\geq 4$ if $d_{G}(v_{i})=3$. In the rest of the proof, we will extend $\phi$ to a proper $L$-coloring of $G$ by choosing a color from $L_{C}(v_{i})$ for each $i\in[k]$. We observe that after doing this, for each $u\in V(G)\setminus V(C)$, there will be one color that appears exactly once at the neighborhood of $v$ in $G$. Indeed, this is clearly true if $u\in V(G)\setminus(V(C)\cup N_{G^{\prime}}(C))$. Suppose $u\in N_{G^{\prime}}(v_{i})$ for some $i\in[k]$. If $\mathcal{U}_{\phi}(u,G^{\prime})=\emptyset$, then the color we chose for $v_{i}$ appears exactly once at the neighborhood of $u$, otherwise, $c_{u}$ does. Therefore, it suffices to give a proper $L_{C}$-coloring of $C$ such that in the resulting coloring of $G$, for each vertex in $C$, some color appears exactly once at its neighborhood.

Now we have the following two observations which we use.

Observation 2**.**

Let $C=xzywx$, $L^{\prime}_{C}(x)\subseteq L_{C}(x)$ and $L^{\prime}_{C}(y)\subseteq L_{C}(y)$ be two non-empty sets. If $|L^{\prime}_{C}(x)|+|L^{\prime}_{C}(y)|\geq 4$, then there exists a pair of colors in $L^{\prime}_{C}(x)\times L^{\prime}_{C}(y)$ which is feasible for both $z$ and $w$.

Proof.

If $z$ blocks two color pairs $(\alpha_{1},\beta_{1}),(\alpha_{2},\beta_{2})\in L^{\prime}_{C}(x)\times L^{\prime}_{C}(y)$, then by Observation 1, we have $\alpha_{1}=\beta_{2}\neq\alpha_{2}=\beta_{1}$, and $\{\alpha_{1},\alpha_{2}\}\subseteq L^{\prime}_{C}(x)\cap L^{\prime}_{C}(y)$. Thus $(\alpha_{1},\alpha_{1})$ or $(\alpha_{2},\alpha_{2})$ is not blocked by $w$. Hence by the symmetry of $z$ and $w$, we may assume that both $z$ and $w$ blocks at most one color pair in $L^{\prime}_{C}(x)\times L^{\prime}_{C}(y)$, but note that the condition $|L^{\prime}_{C}(x)|+|L^{\prime}_{C}(y)|\geq 4$ implies $|L^{\prime}_{C}(x)|\geq 3$, or $|L^{\prime}_{C}(y)|\geq 3$ or $\min\{|L^{\prime}_{C}(x)|,|L^{\prime}_{C}(y)|\}\geq 2$. In any case, $L^{\prime}_{C}(x)\times L^{\prime}_{C}(y)$ contains at least three pairs, thus there is a desired pair in $L^{\prime}_{C}(x)\times L^{\prime}_{C}(y)$. ∎

Observation 3**.**

Let $C=xzywx$. If there is a color pair $(\alpha,\beta)\in L_{C}(x)\times L_{C}(y)$ which is feasible for both $z$ and $w$ (possibly, $\alpha=\beta$), and a color pair in $L_{C}(z)\setminus\{\alpha,\beta\}\times L_{C}(w)\setminus\{\alpha,\beta\}$ which is feasible for both $x$ and $y$, then $\phi$ can be extended to a proper conflict-free $L$-coloring of $G$.

Claim 3**.**

$k\geq 5$, i.e., $G$ contains no 4-cycle.

Proof.

Suppose to the contrary, $k=4$. By Observation 1 and the fact that $L_{C}(v_{1})\times L_{C}(v_{3})$ contains at least six pair of distinct colors, there must be a pair of distinct colors, say $(1,2)\in L_{C}(v_{1})\times L_{C}(v_{3})$, which is feasible for both $v_{0}$ and $v_{2}$. By Observations 2, 3 and the assumption that $|L_{C}(v_{i})|\geq 3$, we have $1\leq|L_{C}(v_{i})\setminus\{1,2\}|\leq 2$ for $i\in\{0,2\}$, and at least one of $|L_{C}(v_{0})\setminus\{1,2\}|$ and $|L_{C}(v_{2})\setminus\{1,2\}|$ is equal to 1. Without loss of generality, we may assume that $1\in L_{C}(v_{0})$ and $\{1,2\}\subsetneq L_{C}(v_{2})$. We deduce that $d_{G}(v_{3})=3$ and $\phi(u_{3})=1$, for otherwise $(1,1)\in L_{C}(v_{0})\times L_{C}(v_{2})$ is feasible for both $v_{1}$ and $v_{3}$ and $|L_{C}(v_{1})\setminus\{1\}|+|L_{C}(v_{3})\setminus\{1\}|\geq 4$, a contradiction by Observation 3. Thus $1\notin L_{C}(v_{3})$. Observe that $(1,2)$ is feasible for both $v_{1}$ and $v_{3}$, this implies that $2\in L_{C}(v_{1})$ and $|L_{C}(v_{1})|=3$ again by Observation 3. So we deduce that $d_{G}(v_{0})=3$ and $\phi(u_{0})=2$ as $(2,2)\in L_{C}(v_{1})\times L_{C}(v_{3})$ must be blocked by $v_{0}$. Therefore, $2\notin L_{C}(v_{0})$, but recall that $2\in L_{C}(v_{1})$ and $|L_{C}(v_{1})|=3$, so there exists a color $\alpha\in L_{C}(v_{0})\setminus L_{C}(v_{1})$. Then $(\alpha,1)\in L_{C}(v_{0})\times L_{C}(v_{2})$ is feasible for both $v_{1}$ and $v_{3}$, and $|L_{C}(v_{1})\setminus\{\alpha,1\}|+|L_{C}(v_{3})\setminus\{\alpha,1\}|\geq 2+2\geq 4$, a contradiction by Observation 3. See the rightmost drawing in Fig. 1 for an illustration. ∎

By Claims 2 and 3, $G$ has no triangles and $4$-cycles. Recall that by the assumption, $C$ is a shortest cycle, so we have $N_{G^{\prime}}(v_{i})\cap N_{G^{\prime}}(v_{j})=\emptyset$ for any distinct $i,j\in[k]$.

We construct an auxiliary graph $H$ with $V(H)=\bigcup_{i=1}^{k}X_{i}$ where $X_{i}=\{v_{i}\}\times L_{C}(v_{i})$, and $E(H)=E_{0}\cup E_{1}\cup E_{2}$, where

[TABLE]

The edges in $E_{1}$ (resp. $E_{2}$) are called short (resp. long) edges. The edges incident with $(v_{i},\alpha)\in X_{i}$ with other endpoints in $X_{i+1}\cup X_{i+2}$ (resp. $X_{i-1}\cup X_{i-2}$) are called forward (resp. backward) edges of $(v_{i},\alpha)$. See Fig. 2 for illustration.

The following observation follows immediately from the definition of $H$.

Observation 4**.**

For each $i\in[k]$, there are at most two long edges between $X_{i-1}$ and $X_{i+1}$. Furthermore, if there are two long edges between $X_{i-1}$ and $X_{i+1}$, then they must be $(v_{i-1},\alpha)(v_{i+1},\beta)$ and $(v_{i-1},\beta)(v_{i+1},\alpha)$ for some distinct $\alpha$ and $\beta$ from $L_{C}(v_{i-1})\cap L_{C}(v_{i+1})$.

Claim 4**.**

$H$* has no independent set of size $|V(C)|=k$.*

Proof.

Suppose $I$ is an independent set of $H$ with $|I|=k$. We extend $\phi$ to $G$ so that for each $(v,\alpha)\in I$, $\phi(v)=\alpha$. As $|I|=k$ and $I$ contains at most one vertex of $X_{i}$ for each $i\in[k]$, all the vertices of $C$ are colored in this manner. Clearly, $\phi$ is proper and $\mathcal{U}_{\phi}(u,G)\neq\emptyset$ for each $u\in V(C)$. By the definition of $L_{C}$, $\mathcal{U}_{\phi}(u,G)\neq\emptyset$ for each $u\in V(G)\setminus V(C)$, which implies that $\phi$ is a proper conflict-free $L$-coloring of $G$, a contradiction. ∎

Claim 5**.**

For every $i\in[k]$ and every $\alpha\in L_{C}(v_{i})$, there are at most one forward edge and at most one backward edge from $(v_{i},\alpha)$.

Proof.

Suppose that there is a forward short edge from $(v_{i},\alpha)$, thus $\alpha\in L_{C}(v_{i+1})$. Then $\alpha\notin\phi(N_{G}(v_{i+1}))$, and hence there is no long forward edges incident with $(v_{i},\alpha)$ by definition. So there is at most one forward edge from $(v_{i},\alpha)$. By symmetry, the statement holds for backward edges. ∎

An extendable structure $(U,I)$ of $H$ is a pair of a set $U\subseteq X_{i}\cup X_{i+t}$ for some $i\in[k]$, $t\in\{1,2\}$ and an independent set $I$ of $H$ satisfying the followings:

1. (C1)

$|U\cap X_{i}|\geq 3$ and $|U\cap X_{i+t}|=2$. 2. (C2)

No backward long edges or no backward short edges incident with vertices from $U\cap X_{i+t}$. 3. (C3)

If $t=1$, then $I$ consists of one vertex in $X_{i+2}$ and one in $X_{i+3}$. And if $t=2$, then $I$ consists of a vertex in each of $X_{i+1}$, $X_{i+3}$ and $X_{i+4}$. 4. (C4)

No backward edges from vertices in $I$ to $U\cup X_{i-1}$.

Note that an extendable structure is a local structure contained in $\bigcup_{i\leq j\leq i+4}X_{j}$ for some $i\in[k]$, and we know $k\geq 5$ by Claim 3. If $H$ contains an extendable structure, then it can be extended to an independent set of $H$ with $k$ vertices as in the proof of next claim.

Claim 6**.**

$H$* contains no extendable structure.*

Proof.

Suppose to the contrary, $(U,I)$ is an extendable structure of $H$. Without loss of generality, we may assume that $U\subseteq X_{0}\cup X_{t}$ for some $t\in\{1,2\}$. Since every vertex of $H$ has at most one forward edge by Claim 5 and $|L_{C}(v_{i})|\geq 3$ for each $i\in[k]$, we can greedily add vertices to $I$ from each of $X_{t+3},X_{t+4},\ldots,X_{k-1},X_{0}$ to obtain an independent set $J^{-}$ of size $k-1$. Observe that $J^{-}\cap X_{t}=\emptyset$. By condition (C2), at least one of two vertices in $X_{t}\cap U$ has no neighbors in $I\cap(X_{t-1}\cup X_{t-2})$, so we add one such vertex to $J^{-}$ to have a set $J$ of vertices. By (C4), $J$ is still an independent set of $H$ and $|J|=k$, a contradiction to Claim 4. ∎

The following observation is frequently used in the the sequel.

Observation 5**.**

Suppose $\alpha,\beta\in L_{C}(v_{i})$ for some $i\in[k]$ and $\alpha\neq\beta$. Then there exists a color $\gamma\in L_{C}(v_{i+1})\setminus\{\alpha,\beta\}$ and a color $\tau\in L_{C}(v_{i+2})\setminus\{\gamma\}$ such that $(v_{i+2},\tau)$ is adjacent to neither $(v_{i},\alpha)$ nor $(v_{i},\beta)$. (Possibility $\tau\in\{\alpha,\beta\}$ is not excluded.)

Proof.

Suppose that one of $(v_{i},\alpha)$ and $(v_{i},\beta)$, say $(v_{i},\alpha)$, has a neighbor in $X_{i+2}$. As $|L_{C}(v_{i+2})|\geq 3$, there exists a color $\tau\in L_{C}(v_{i+2})$ such that $(v_{i+2},\tau)$ is adjacent to neither $(v_{i},\alpha)$ nor $(v_{i},\beta)$. By Claim 5, we know $\alpha\notin L_{C}(v_{i+1})$, so we can choose a color $\gamma$ from $L_{C}(v_{i+1})\setminus\{\alpha,\beta,\tau\}$, as desired. Hence we may assume that both $(v_{i},\alpha)$ and $(v_{i},\beta)$ have no neighbors in $X_{i+2}$, then we can arbitrarily choose a color $\gamma\in L_{C}(v_{i+1})\setminus\{\alpha,\beta\}$ and a color $\tau\in L_{C}(v_{i+2})\setminus\{\gamma\}$ as desired. ∎

Claim 7**.**

Every vertex of $H$ is incident with exactly one forward edge and one backward edge.

Proof.

Suppose to the contrary, $(v_{0},\alpha)$ has no backward edge for some $\alpha\in L_{C}(v_{0})$. Argument for forward edge is symmetric.

First assume that there exists a color $\beta\in L_{C}(v_{1})$ such that $\beta\neq\alpha$ and $(v_{1},\beta)$ has no neighbors in $X_{k-1}$. As $|L_{C}(v_{k-1})|\geq 3$, by Pigeonhole Principle, there exists two colors, say $1,2\in L_{C}(v_{k-1})$, such that backward long edges or backward short edges are missed from $(v_{k-1},1),(v_{k-1},2)\in X_{k-1}$. Then $U=X_{k-2}\cup\{(v_{k-1},1),(v_{k-1},2)\}$ and $I=\{(v_{0},\alpha),(v_{1},\beta)\}$ form an extendable structure with $t=1$: The conditions (C1) and (C3) are obvious. The condition (C2) follows from the choice of $(v_{k-1},1)$ and $(v_{k-1},2)$, and the condition (C4) is satisfied since $(v_{0},\alpha)$ has no backward edge and $(v_{1},\beta)$ has no backward edge to $X_{k-1}$, a contradiction.

Thus by the definition of $H$, without loss of generality, we deduce that there exist two distinct colors, say $1,2$, such that $L_{C}(v_{1})=\{\alpha,1,2\}$, $\phi(N_{G^{\prime}}(v_{1}))=\{1,2\}$ and $\{1,2\}\subsetneq L_{C}(v_{k-1})$. By Observation 5, there exists a color $\gamma\in L_{C}(v_{2})\setminus\{1,2\}$ and $\tau\in L_{C}(v_{3})\setminus\{\gamma\}$ such that $(v_{3},\tau)$ has no neighbors in $\{(v_{1},1),(v_{1},2)\}$. Then $U=X_{k-1}\cup\{(v_{1},1),(v_{1},2)\}$ and $I=\{(v_{0},\alpha),(v_{2},\gamma),(v_{3},\tau)\}$ form an extendable structure with $t=2$: The conditions (C1), (C2) and (C3) are obvious. The condition (C4) is satisfied since $(v_{0},\alpha)$ has no backward edge, and both $(v_{2},\gamma)$ and $(v_{3},\tau)$ have no backward edge to $\{(v_{1},1),(v_{1},2)\}$, a contradiction. ∎

Claim 8**.**

For every $i\in[k]$, there is at most one forward short edge incident with vertices of $X_{i}$.

Proof.

Without loss of generality assume $\{1,2,3\}\subseteq L_{C}(v_{0})$, both $(v_{0},1)$ and $(v_{0},2)$ incident with forward short edges. Let $U=\{(v_{0},1),(v_{0},2),(v_{0},3),(v_{1},1),(v_{1},2)\}$. If $(v_{0},3)$ is incident with forward short edge, then by Observation 5, there exist colors $\gamma\in L_{C}(v_{2})\setminus\{1,2\}$ and $\tau\in L_{C}(v_{3})\setminus\{\gamma\}$ such that neither $(v_{1},1)$ nor $(v_{1},2)$ is adjacent to $(v_{3},\tau)$. So $U$ and $I=\{(v_{2},\gamma),(v_{3},\tau)\}$ form an extendable structure with $t=1$, a contradiction. Hence $(v_{0},3)$ is adjacent to $(v_{2},\alpha)$ for some $\alpha\in L_{C}(v_{2})\setminus\{1,2\}$.

First assume that there exists a color $\beta\in L_{C}(v_{2})\setminus\{1,2,\alpha\}$. Clearly, $(v_{2},\beta)$ has no neighbors in $\{(v_{0},1),(v_{0},2),(v_{0},3)\}$. If there exists a color $\gamma\in L_{C}(v_{3})\setminus\{\beta\}$ such that $(v_{3},\gamma)$ has no neighbors in $\{(v_{1},1),(v_{1},2)\}$, then $U$ and $I=\{(v_{2},\beta),(v_{3},\gamma)\}$ form an extendable structure with $t=1$, a contradiction. Thus we deduce that $L_{C}(v_{3})=\{1,2,\beta\}$ and $\phi(N_{G^{\prime}}(v_{2}))=\{1,2\}$, so $1,2\notin L_{C}(v_{2})$. As there is a color $\beta^{\prime}\in L_{C}(v_{2})\setminus\{1,2,\alpha,\beta\}$, $U$ and $I=\{(v_{2},\beta^{\prime}),(v_{3},\beta)\}$ form an extendable structure with $t=1$, a contradiction.

So we have $L_{C}(v_{2})=\{1,2,\alpha\}$. Using the same argument repeatedly, we conclude that $\{1,2\}\subseteq L_{C}(v_{i})$ for every $i\in[k]$. Then $I=\{(v_{2m+1},1)\mid 0\leq m\leq\left\lfloor(k-1)/2\right\rfloor\}\cup\{(v_{2m},2)\mid 0\leq m\leq\left\lfloor(k-1)/2\right\rfloor\}\cup\{(v_{0},3)\}$ is an independent set of $H$ with $|I|=k$, a contradiction. ∎

If $|X_{i}|\geq 4$ for some $i\in[k]$, then by Claims 7 and 8, there are at least three long edges between $X_{i}$ and $X_{i+2}$, a contradiction to Observation 4. Thus $|X_{i}|=3$ for each $i\in[k]$, two of which incident with forward long edges and the rest one incident with a forward short edge.

If all the short edges form a matching of $H$, then we can easily find an independent set $I$ of $H$ with $|I|=k$ by choosing vertices incident with forward short edges, a contradiction to Claim 4. Therefore, without loss of generality, there must exist a color, say $\alpha$, which appears in $L_{C}(v_{0})\cap L_{C}(v_{1})\cap L_{C}(v_{2})$. Then there are two distinct colors in $L_{C}(v_{0})\setminus\{\alpha\}$, say 1 and 2, and there are two distinct colors in $L_{C}(v_{1})\setminus\{\alpha,1,2\}$, say 3 and 4 such that all of $\{(v_{0},1),(v_{0},2),(v_{1},3),(v_{1},4)\}$ are incident with forward long edges. By Observation 4 and the fact $|L_{C}(v_{i})|=3$ for every $i\in[k]$, we know that $L_{C}(v_{0})=L_{C}(v_{2})=\{\alpha,1,2\}$ and $L_{C}(v_{1})=\{\alpha,3,4\}$, $L_{C}(v_{3})=\{\beta,3,4\}$ for some color $\beta\notin\{3,4\}$. If $\beta\neq\alpha$, then $U=X_{0}\cup\{(v_{1},3),(v_{1},4)\}$ and $I=\{(v_{2},\alpha),(v_{3},\beta)\}$ form an extendable structure with $t=1$: the conditions (C1), (C2) and (C3) are obvious by the choice of vertices, and the condition (C4) is satisfied since $(v_{2},\alpha)$ has a backward short edge to $(v_{1},\alpha)\notin U$ and the long edges between $X_{1}$ and $X_{3}$ are $(v_{1},3)(v_{3},4)$ and $(v_{1},4)(v_{3},3)$, which are not incident to $(v_{3},\beta)$, a contradiction.

Thus $L_{C}(v_{3})=\{\alpha,3,4\}$. Using the same argument repeatedly, we conclude that $k$ is even, moreover, $L_{C}(v_{j})=\{\alpha,1,2\}$ if $j$ is even, and $L_{C}(v_{j})=\{\alpha,3,4\}$ if $j$ is odd. Then $I=\{(v_{j},1)\mid j\in[k]\text{ and }j\text{ is even}\}\cup\{(v_{j},3)\mid j\in[k]\text{ and }j\text{ is odd}\}$ is an independent set of $H$ with $|I|=k$, a contradiction. This completes the proof of Theorem 8.

Acknowledgement

This work is based on the discussion at Zhejiang Normal University, China. We appreciate Xuding Zhu for inviting the first and the second authors there. M. Kashima has been supported by Keio University SPRING scholarship JPMJSP2123 and JST ERATO JPMJER2301. R. Škrekovski has been partially supported by the Slovenian Research Agency and Innovation (ARIS) program P1-0383, project J1-3002, and the annual work program of Rudolfovo. R. Xu has been supported by National Science Foundation for Young Scientists of China, grant number: 12401472.