2-distance 20-coloring of planar graphs with maximum degree 6

Kengo Aoki Department of Informatics, Graduate School of Integrated Science and Technology, Shizuoka University, 3-5-1 Johoku, Chuo-ku, Hamamatsu-shi, 432-8011, Shizuoka, Japan, aoki.kengo.19@shizuoka.ac.jp

Abstract

A 2-distance $k$ -coloring of a graph $G$ is a proper $k$ -coloring such that any two vertices at distance two or less get different colors. The 2-distance chromatic number of $G$ is the minimum $k$ such that $G$ has a 2-distance $k$ -coloring, denoted by $\chi_{2}(G)$ . In this paper, we show that $\chi_{2}(G)\leq 20$ for every planar graph $G$ with maximum degree at most six, which improves a former bound $\chi_{2}(G)\leq 21$ .

1 Introduction

All graphs considered in this paper are simple, finite, and planar. For a graph $G$ , we denote the set of vertices, the set of edges, and the set of faces by $V(G)$ , $E(G)$ , and $F(G)$ , respectively. For a graph $G$ and a vertex $v$ in $G$ , let $G-v$ denote the graph obtained from $G$ by deleting the vertex $v$ and all edges incident with $v$ . The set of neighbours of a vertex $v$ in a graph $G$ is denoted by $N_{G}(v)$ . The degree of a vertex $v$ in a graph $G$ , denoted by $d_{G}(v)$ , is the number of edges of $G$ incident with $v$ . The maximum degree and minimum degree of a graph $G$ are denoted by $\Delta(G)$ and $\delta(G)$ ( $\Delta$ and $\delta$ for short). A vertex of degree $k$ (respectively, at least $k$ , at most $k$ ) is said to be a $k$ -vertex (respectively, $k^{+}$ -vertex, $k^{-}$ -vertex). For $X\subseteq V(G)$ , let $G[X]$ denote the subgraph of $G$ induced by $X$ . A face is said to be incident with the vertices and edges in its boundary, and two faces are adjacent if their boundaries have an edge in common. The degree of a face $f$ in a graph $G$ , denoted by $d(f)$ , is the number of edges in its boundary. A face of degree $k$ (respectively, at least $k$ , at most $k$ ) is said to be a $k$ -face (respectively, $k^{+}$ -face, $k^{-}$ -face). A $[v_{1}v_{2}\ldots v_{k}]$ is a $k$ -face with vertices $v_{1},v_{2},\ldots,v_{k}$ on its boundary. For a vertex $v$ in a graph $G$ , let $m_{k}(v)$ denote the number of $k$ -faces incident with $v$ , and let $n_{k}(v)$ denote the number of $k$ -vertices adjacent to $v$ .

Let $\phi$ be a partial coloring of a graph $G$ . For a vertex $v$ in a graph $G$ , let $C_{\phi}(v)$ denote the set of colors assigned to the vertices within distance two from $v$ . A 2-distance $k$ -coloring of a graph $G$ is a mapping $\phi:V(G)\rightarrow\{1,2,\ldots,k\}$ such that $\phi(v_{1})\neq\phi(v_{2})$ for any two vertices $v_{1},v_{2}\in V(G)$ with $d_{G}(v_{1},v_{2})\leq 2$ , where $d_{G}(v_{1},v_{2})$ is the distance between the two vertices $v_{1}$ and $v_{2}$ . The 2-distance chromatic number of $G$ is the minimum $k$ such that $G$ has a 2-distance $k$ -coloring, denoted by $\chi_{2}(G)$ . Let $d_{2}(v)$ denote the number of vertices within distance two from a vertex $v$ . Any definitions and notations not explicitly stated in this paper conform to those in [1].

The study of 2-distance coloring originated from the research on square coloring, which was first introduced by Kramer and Kramer [12, 11]. The square of a graph $G$ , denoted by $G^{2}$ , is obtained by adding edges between all pairs of vertices that have a common neighbour in $G$ . In 1977, Wegner made the following conjecture.

Conjecture 1.1.

[15] If $G$ is a planar graph, then $\chi_{2}(G)\leq 7$ if $\Delta=3$ , $\chi_{2}(G)\leq\Delta+5$ if $4\leq\Delta\leq 7$ , and $\chi_{2}(G)\leq\lfloor\frac{3\Delta}{2}\rfloor+1$ if $\Delta\geq 8$ .

The case of $\Delta=3$ was independently proven by Thomassen [14] and by Hartke et al. [7]. Havet et al. [8, 9] proved that the conjecture holds asymptotically. Bousquet et al. [3] proved that $\chi_{2}(G)\leq 12$ for $\Delta\leq 4$ using an automatic discharging method. Deniz [5] proved that $\chi_{2}(G)\leq 16$ for $\Delta\leq 5$ . For a comprehensive overview of 2-distance coloring and related research, we refer the reader to [4].

The upper bound on $\chi_{2}(G)$ for $\Delta=6$ has been gradually improving. Zhu and Bu [16] proved that $\chi_{2}(G)\leq 5\Delta-7$ for $\Delta\geq 6$ , which was improved by Krzyziński et al. [13] to $\chi_{2}(G)\leq 3\Delta+4$ for $\Delta\geq 6$ . In this paper, we show that $\chi_{2}(G)\leq 20$ for every planar graph $G$ with $\Delta\leq 6$ , which improves the result of $\chi_{2}(G)\leq 21$ proved by Bousquet et al. [2]. We prove the following theorem.

Theorem 1.2.

If $G$ is a planar graph with maximum degree $\Delta\leq 6$ , then $\chi_{2}(G)\leq 20$ .

Remark.

Recently, Deniz [6] posted the first version of the proof on arXiv on March 18, 2024, showing that $\chi_{2}(G)\leq 2\Delta+7$ for planar graphs. The proof in this version covers the cases $\Delta=6$ , $7$ , and $8$ . According to this inequality, when $\Delta=6$ , it follows that $\chi_{2}(G)\leq 19$ . However, the proof for the case $\Delta=6$ in this initial version is incomplete and requires further elaboration.

2 Reducible configurations

Let $G$ be a minimum counterexample to Theorem 1.2 with minimum $|V(G)|+|E(G)|$ . That is $G$ is a planar graph with $\chi_{2}(G)>20$ , such that for any planar subgraph $G^{\prime}$ with $\Delta(G^{\prime})\leq\Delta(G)$ and $|V(G^{\prime})|+|E(G^{\prime})|<|V(G)|+|E(G)|$ , we have $\chi_{2}(G^{\prime})\leq 20$ . Obviously, $G$ is a connected graph.

Let $C=\{1,2,\ldots,20\}$ be a set of colors. We call a graph $G^{\prime}$ proper with respect to $G$ if $G^{\prime}$ is obtained from $G$ by deleting some edges or vertices and adding some edges such that for any two vertices $v_{1}$ , $v_{2}\in V(G)\cap V(G^{\prime})$ with $d_{G}(v_{1},v_{2})\leq 2$ , we have $d_{G^{\prime}}(v_{1},v_{2})\leq 2$ . This definition of proper is the same as the one used in [5, 10]. In this section, we present some reducible configurations of $G$ . The proofs of the lemmas generally follow a similar pattern: We construct a graph $G^{\prime}$ that is proper with respect to $G$ by deleting a vertex $v$ from $G$ and adding some edges. By the minimality of $G$ , there exists a 2-distance 20-coloring $\phi^{\prime}$ of $G^{\prime}$ . Let $\phi$ be a coloring of $G$ such that every vertex in $V(G)$ , except for the deleted vertex $v$ , is colored using $\phi^{\prime}$ . If $|C|-|C_{\phi}(v)|\geq 1$ , then a safe color exists for $v$ . By coloring $v$ with the safe color, $\phi^{\prime}$ can be extended to a 2-distance 20-coloring $\phi$ of $G$ . This implies that $\chi_{2}(G)\leq 20$ , which is a contradiction. The essence of the proof is to construct a proper $G^{\prime}$ such that $|C_{\phi}(v)|\leq d_{2}(v)\leq 19$ .

Lemma 2.1.

We have $\delta(G)\geq 3$ .

Proof.

Assume that $G$ contains a 1-vertex $v$ . It is clear that $G^{\prime}=G-v$ is proper with respect to $G$ . By the minimality of $G$ , $G^{\prime}$ has a 2-distance 20-coloring $\phi^{\prime}$ . Let $\phi$ be a coloring of $G$ such that every vertex in $V(G)$ , except for $v$ , is colored using $\phi^{\prime}$ . Since $\Delta\leq 6$ , it follows that $|C_{\phi}(v)|\leq 6$ and $|C|-|C_{\phi}(v)|\geq 14$ . Therefore, there exists a safe color for $v$ . By coloring $v$ with the safe color, $\phi$ becomes a 2-distance 20-coloring of $G$ , a contradiction. Next, we assume that $G$ has a 2-vertex $v$ with $N_{G}(v)=\{x,y\}$ . Let $G^{\prime}=G-v+\{xy\}$ . The graph $G^{\prime}$ is proper with respect to $G$ . By the minimality of $G$ , $G^{\prime}$ has a 2-distance 20-coloring $\phi^{\prime}$ . Let $\phi$ be a coloring of $G$ such that every vertex in $V(G)$ , except for $v$ , is colored using $\phi^{\prime}$ . Since $\Delta\leq 6$ , it follows that $|C_{\phi}(v)|\leq 12$ and $|C|-|C_{\phi}(v)|\geq 8$ . Therefore, we can color $v$ with a safe color, a contradiction. ∎

Lemma 2.2.

Let $v$ be a 3-vertex. Then,

(1)

$n_{5^{-}}(v)=0$ ,
(2)

$m_{3}(v)=0$ , and
(3)

$m_{4}(v)\leq 1$ .

Proof.

Let $v_{1}$ , $v_{2}$ , and $v_{3}$ be the neighbours of $v$ . (1) Assume that $v$ is adjacent to a $5^{-}$ -vertex. Without loss of generality, let $v_{1}$ be a $5^{-}$ -vertex. Let $G^{\prime}=G-v+\{v_{1}v_{2},v_{1}v_{3}\}$ . The graph $G^{\prime}$ is proper with respect to $G$ . By the minimality of $G$ , $G^{\prime}$ has a 2-distance 20-coloring $\phi^{\prime}$ . Let $\phi$ be a coloring of $G$ such that every vertex in $V(G)$ , except for $v$ , is colored using $\phi^{\prime}$ . Since $\Delta\leq 6$ , it follows that $|C_{\phi}(v)|\leq 17$ and $|C|-|C_{\phi}(v)|\geq 3$ . Therefore, there exists a safe color for $v$ . By coloring $v$ with the safe color, $\phi$ becomes a 2-distance 20-coloring of $G$ , a contradiction.

(2) Assume that $v$ is incident to a 3-face $[vv_{1}v_{2}]$ . Let $G^{\prime}=G-v+\{v_{1}v_{3}\}$ . The graph $G^{\prime}$ is proper with respect to $G$ . By the minimality of $G$ , $G^{\prime}$ has a 2-distance 20-coloring $\phi^{\prime}$ . Let $\phi$ be a coloring of $G$ such that every vertex in $V(G)$ , except for $v$ , is colored using $\phi^{\prime}$ . Since $|C_{\phi}(v)|\leq 16$ , we can color $v$ with a safe color, a contradiction.

(3) Assume that $v$ is incident to two 4-faces $[vv_{1}xv_{2}]$ and $[vv_{2}yv_{3}]$ . It is clear that $G^{\prime}=G-v+\{v_{1}v_{3}\}$ is proper with respect to $G$ . By the minimality of $G$ , $G^{\prime}$ has a 2-distance 20-coloring $\phi^{\prime}$ . Let $\phi$ be a coloring of $G$ such that every vertex in $V(G)$ , except for $v$ , is colored using $\phi^{\prime}$ . Since $|C_{\phi}(v)|\leq 16$ , we can color $v$ with a safe color, a contradiction. ∎

Lemma 2.3.

Let $v$ be a 4-vertex. Then $m_{3}(v)\leq 2$ . In particular, if $m_{3}(v)=2$ , then $m_{4}(v)=0$ , $n_{6}(v)=4$ , and $m_{3}(w)\leq 4$ for any 6-vertex $w$ adjacent to $v$ .

Proof.

Let $v_{1}$ , $v_{2}$ , $v_{3}$ , and $v_{4}$ be the neighbours of $v$ in clockwise order. First, we show that $m_{3}(v)\leq 2$ . Assume that $v$ is incident to three 3-faces $[vv_{1}v_{2}]$ , $[vv_{2}v_{3}]$ , and $[vv_{3}v_{4}]$ . Let $G^{\prime}=G-v+\{v_{1}v_{4}\}$ . The graph $G^{\prime}$ is proper with respect to $G$ . By the minimality of $G$ , $G^{\prime}$ has a 2-distance 20-coloring $\phi^{\prime}$ . Let $\phi$ be a coloring of $G$ such that every vertex in $V(G)$ , except for $v$ , is colored using $\phi^{\prime}$ . Since $\Delta\leq 6$ , it follows that $|C_{\phi}(v)|\leq 18$ and $|C|-|C_{\phi}(v)|\geq 2$ . Therefore, there exists a safe color for $v$ . By coloring $v$ with the safe color, $\phi$ becomes a 2-distance 20-coloring of $G$ , a contradiction.

Now, we consider the case $m_{3}(v)=2$ . Let $f_{1}$ and $f_{2}$ be two 3-faces incident to $v$ . First, we show that $m_{4}(v)=0$ . Assume that $v$ is incident to a 4-face $[vv_{1}xv_{2}]$ . If $f_{1}$ and $f_{2}$ are adjacent, say $f_{1}=[vv_{2}v_{3}]$ and $f_{2}=[vv_{3}v_{4}]$ , then let $G^{\prime}=G-v+\{v_{1}v_{4}\}$ . If $f_{1}$ and $f_{2}$ are not adjacent, say $f_{1}=[vv_{2}v_{3}]$ and $f_{2}=[vv_{4}v_{1}]$ , then let $G^{\prime}=G-v+\{v_{3}v_{4}\}$ . In both cases, $G^{\prime}$ is proper with respect to $G$ and by the minimality of $G$ , $G^{\prime}$ has a 2-distance 20-coloring $\phi^{\prime}$ . Let $\phi$ be a coloring of $G$ such that every vertex in $V(G)$ , except for $v$ , is colored using $\phi^{\prime}$ . Since $|C_{\phi}(v)|\leq 19$ in each case, we can color $v$ with a safe color, a contradiction.

Next, we prove that $n_{6}(v)=4$ . Assume that $v$ is incident to a $5^{-}$ -vertex. Without loss of generality, let $v_{1}$ be a $5^{-}$ -vertex. If $f_{1}$ and $f_{2}$ are adjacent, say $f_{1}=[vv_{1}v_{2}]$ and $f_{2}=[vv_{2}v_{3}]$ , then let $G^{\prime}=G-v+\{v_{2}v_{4}\}$ . If $f_{1}$ and $f_{2}$ are not adjacent, say $f_{1}=[vv_{1}v_{2}]$ and $f_{2}=[vv_{3}v_{4}]$ , then let $G^{\prime}=G-v+\{v_{2}v_{3},v_{4}v_{1}\}$ . In both cases, $G^{\prime}$ is proper with respect to $G$ and $d_{2}(v)\leq 19$ , a contradiction.

Finally, we prove that $m_{3}(w)\leq 4$ for any 6-vertex $w$ adjacent to $v$ . To show that $w$ cannot be incident to more than five 3-faces, it suffices to prove that no edge in $G[N_{G}(v)]$ is contained in two 3-faces of $G$ . Let $f_{1}=[vv_{1}v_{2}]$ . Assume that the edge $v_{1}v_{2}$ is contained in two 3-faces of $G$ . This implies that there exists a vertex $x$ such that $x$ is a common neighbour of $v_{1}$ and $v_{2}$ . If $f_{1}$ and $f_{2}$ are adjacent, say $f_{2}=[vv_{2}v_{3}]$ , then let $G^{\prime}=G-v+\{v_{2}v_{4}\}$ . If $f_{1}$ and $f_{2}$ are not adjacent, say $f_{2}=[vv_{3}v_{4}]$ , then let $G^{\prime}=G-v+\{v_{2}v_{3},v_{4}v_{1}\}$ . In both cases, $G^{\prime}$ is proper with respect to $G$ and $d_{2}(v)\leq 19$ , a contradiction. ∎

Lemma 2.4.

Let $v$ be a 4-vertex with $m_{3}(v)=1$ . Then $m_{4}(v)\leq 2$ . In particular, if $1\leq m_{4}(v)\leq 2$ , then $n_{4}(v)=0$ , and $n_{5}(v)\leq 1$ .

Proof.

Let $v_{1}$ , $v_{2}$ , $v_{3}$ , and $v_{4}$ be the neighbours of $v$ in clockwise order and let $[vv_{1}v_{2}]$ be a 3-face incident to $v$ . First, we show that $m_{4}(v)\leq 2$ . Assume that $v$ is incident to three 4-faces $[vv_{2}xv_{3}]$ , $[vv_{3}yv_{4}]$ , and $[vv_{4}zv_{1}]$ . Let $G^{\prime}=G-v+\{v_{2}v_{3},v_{1}v_{4}\}$ . The graph $G^{\prime}$ is proper with respect to $G$ . By the minimality of $G$ , $G^{\prime}$ has a 2-distance 20-coloring $\phi^{\prime}$ . Let $\phi$ be a coloring of $G$ such that every vertex in $V(G)$ , except for $v$ , is colored using $\phi^{\prime}$ . Since $\Delta\leq 6$ , it follows that $|C_{\phi}(v)|\leq 19$ and $|C|-|C_{\phi}(v)|\geq 1$ . Therefore, there exists a safe color for $v$ . By coloring $v$ with the safe color, $\phi$ becomes a 2-distance 20-coloring of $G$ , a contradiction.

Now, we suppose that $m_{4}(v)=2$ . Let $f_{1}$ and $f_{2}$ be two 4-faces incident to $v$ . First, we prove that $n_{4}(v)=0$ . Assume that $v$ is adjacent to a 4-vertex. Let $v_{1}$ be a 4-vertex. Regardless of whether $f_{1}$ and $f_{2}$ are adjacent, let $G^{\prime}=G-v+\{v_{1}v_{3},v_{1}v_{4}\}$ . (If a vertex $v_{i}\in N_{G}(v)$ other than $v_{1}$ is a 4-vertex, then we construct $G^{\prime}$ by deleting $v$ and adding edges from $v_{i}$ to each neighbour $v_{j}$ of $v$ with $v_{i}v_{j}\notin E(G)$ .) The graph $G^{\prime}$ is proper with respect to $G$ . By the minimality of $G$ , $G^{\prime}$ has a 2-distance 20 coloring $\phi^{\prime}$ . Let $\phi$ be a coloring of $G$ such that every vertex in $V(G)$ , except for $v$ , is colored using $\phi^{\prime}$ . Since $|C_{\phi}(v)|\leq 18$ , there exists a safe color for $v$ , a contradiction. Second, we show that $n_{5}(v)\leq 1$ . Assume that $v$ is adjacent to two 5-vertices. There are six possible arrangements of two 5-vertices among the four neighbors of $v$ . Regardless of whether $f_{1}$ and $f_{2}$ are adjacent, the construction of $G^{\prime}$ depends on which neighbours of $v$ are the two 5-vertices. If $v_{1}$ is one of the two 5-vertices, then we construct $G^{\prime}=G-v+\{v_{1}v_{3},v_{1}v_{4}\}$ . Similarly, if $v_{2}$ is one of the two 5-vertices, then we construct $G^{\prime}=G-v+\{v_{2}v_{3},v_{2}v_{4}\}$ . Otherwise, if $v_{3}$ and $v_{4}$ are two 5-vertices, then we construct $G^{\prime}=G-v+\{v_{2}v_{3},v_{3}v_{4},v_{4}v_{1}\}$ . In all cases, $G^{\prime}$ is proper with respect to $G$ . By the minimality of $G$ , $G^{\prime}$ has a 2-distance 20-coloring $\phi^{\prime}$ . Let $\phi$ be a coloring of $G$ such that every vertex in $V(G)$ , except for $v$ , is colored using $\phi^{\prime}$ . Since $|C_{\phi}(v)|\leq 18$ , we can color $v$ with a safe color, a contradiction.

Next, we suppose that $m_{4}(v)=1$ . Let $f_{1}$ be a 4-face incident to $v$ . The proof of $n_{4}(v)=0$ is similar to the proof when we supposed that $m_{4}(v)=2$ . Assume that $v$ is adjacent to a 4-vertex and let $v_{1}$ be a 4-vertex. We construct $G^{\prime}$ in the same way as before, regardless of the position of $f_{1}$ : $G^{\prime}=G-v+\{v_{1}v_{3},v_{1}v_{4}\}$ . The graph $G^{\prime}$ is proper with respect to $G$ . By the minimality of $G$ , $G^{\prime}$ has a 2-distance 20-coloring $\phi^{\prime}$ . Let $\phi$ be a coloring of $G$ such that every vertex in $V(G)$ , except for $v$ , is colored using $\phi^{\prime}$ . The only difference is that, in this case, $|C_{\phi}(v)|\leq 19$ . We can color $v$ with a safe color, a contradiction. Finally, we prove that $n_{5}(v)\leq 1$ . Assume that $v$ is adjacent to two 5-vertices. There are six possible arrangements of two 5-vertices among the four neighbors of $v$ . We consider two cases based on the position of $f_{1}$ . Case 1: $f_{1}=[vv_{2}xv_{3}]$ . If $v_{1}$ is one of the two 5-vertices, then we construct $G^{\prime}=G-v+\{v_{1}v_{3},v_{1}v_{4}\}$ . Similarly, if $v_{2}$ is one of the two 5-vertices, then we construct $G^{\prime}=G-v+\{v_{2}v_{3},v_{2}v_{4}\}$ . Otherwise, if $v_{3}$ and $v_{4}$ are the two 5-vertices, then we construct $G^{\prime}=G-v+\{v_{2}v_{3},v_{3}v_{4},v_{4}v_{1}\}$ . Case 2: $f_{1}=[vv_{3}xv_{4}]$ . In this case, we construct $G^{\prime}=G-v+\{v_{2}v_{3},v_{4}v_{1}\}$ , regardless of which neighbours of $v$ are the two 5-vertices. In all cases, $G^{\prime}$ is proper with respect to $G$ . Since $d_{2}(v)\leq 19$ , there exists a safe color for $v$ , a contradiction. ∎

Now, we discuss the properties of a 5-vertex in $G$ . Let $v$ be a 5-vertex and let $v_{1},v_{2},\ldots,v_{5}$ be the neighbours of $v$ in clockwise order.

Lemma 2.5.

Let $v$ be a 5-vertex. If $m_{3}(v)=5$ , then $n_{6}(v)=5$ and $m_{3}(w)\leq 4$ for any 6-vertex $w$ adjacent to $v$ .

Proof.

Suppose that $v$ is incident to five 3-faces. First, we show that $n_{6}(v)=5$ . Assume that $v$ is adjacent to a $5^{-}$ -vertex. Without loss of generality, let $v_{1}$ be a $5^{-}$ -vertex. It is clear that $G^{\prime}=G-v$ is proper with respect to $G$ . By the minimality of $G$ , $G^{\prime}$ has a 2-distance 20-coloring $\phi^{\prime}$ . Let $\phi$ be a coloring of $G$ such that every vertex in $V(G)$ , except for $v$ , is colored using $\phi^{\prime}$ . Since $\Delta\leq 6$ , it follows that $|C_{\phi}(v)|\leq 19$ and $|C|-|C_{\phi}(v)|\geq 1$ . Therefore, there exists a safe color for $v$ . By coloring $v$ with the safe color, $\phi$ becomes a 2-distance 20-coloring of $G$ , a contradiction.

Next, we prove that $m_{3}(w)\leq 4$ for any 6-vertex $w$ adjacent to $v$ . To show that $w$ cannot be incident to more than five 3-faces, it suffices to prove that no edge in $G[N_{G}(v)]$ is contained in two 3-faces of $G$ . Without loss of generality, we assume that the edge $v_{1}v_{2}$ is contained in two 3-faces. This implies that there exists a vertex $x$ such that $x$ is a common neighbour of $v_{1}$ and $v_{2}$ . It is clear that $G^{\prime}=G-v$ is proper with respect to $G$ . Since $d_{2}(v)\leq 19$ , there exists a safe color for $v$ , a contradiction. ∎

Lemma 2.6.

Let $v$ be a 5-vertex with $m_{3}(v)=4$ and $m_{4}(v)$ = 1. Then $n_{4^{-}}(v)=0$ , $n_{5}(v)\leq 1$ , and $m_{3}(w)\leq 5$ for any 6-vertex $w$ adjacent to $v$ .

Proof.

Suppose that $v$ is incident to four 3-faces $[vv_{1}v_{2}]$ , $[vv_{2}v_{3}]$ , $[vv_{3}v_{4}]$ , $[vv_{4}v_{5}]$ , and one 4-face $[vv_{5}xv_{1}]$ . First, we show that $n_{4^{-}}(v)=0$ . By Lemma 2.2(1), $v$ is not adjacent to any 3-vertex. Thus it suffices to show that $v$ is not adjacent to any 4-vertex. Assume that $v$ is adjacent to a 4-vertex. Let $G^{\prime}=G-v+\{v_{5}v_{1}\}$ . The graph $G^{\prime}$ is proper with respect to $G$ . By the minimality of $G$ , $G^{\prime}$ has a 2-distance 20-coloring $\phi^{\prime}$ . Let $\phi$ be a coloring of $G$ such that every vertex in $V(G)$ , except for $v$ , is colored using $\phi^{\prime}$ . Since $\Delta\leq 6$ , it follows that $|C_{\phi}(v)|\leq 19$ and $|C|-|C_{\phi}(v)|\geq 1$ . Therefore, there exists a safe color for $v$ . By coloring $v$ with the safe color, $\phi$ becomes a 2-distance 20-coloring of $G$ , a contradiction.

Next, we assume that $v$ is incident to two 5-vertices. It is clear that $G^{\prime}=G-v+\{v_{5}v_{1}\}$ is proper with respect to $G$ . By the minimality of $G$ , $G^{\prime}$ has a 2-distance 20-coloring $\phi^{\prime}$ . Let $\phi$ be a coloring of $G$ such that every vertex in $V(G)$ , except for $v$ , is colored using $\phi^{\prime}$ . Since $|C_{\phi}(v)|\leq 19$ , we can color $v$ with a safe color, a contradiction. Thus $n_{5}(v)\leq 1$ holds. This implies that $n_{6}(v)\geq 4$ .

Finally, we prove that $m_{3}(w)\leq 5$ for any 6-vertex $w$ adjacent to $v$ . Assume that $v_{1}$ is a 6-vertex. Since $v_{1}$ is incident to a 4-face, $v_{1}$ can be incident to at most five 3-faces. By symmetry, the same holds if we assume that $v_{5}$ is a 6-vertex. Next, we prove that if $v_{2}$ , $v_{3}$ , or $v_{4}$ is a 6-vertex, then it can be incident to at most five 3-faces. Without loss of generality, we assume that $v_{2}$ is a 6-vertex and is incident to six 3-faces. In this case, each of the edges $v_{1}v_{2}$ and $v_{2}v_{3}$ is contained in two 3-faces. Let $G^{\prime}=G-v+\{v_{1}v_{5}\}$ . The graph $G^{\prime}$ is proper with respect to $G$ . Since $d_{2}(v)\leq 19$ , there exists a safe color for $v$ , a contradiction. ∎

Lemma 2.7.

Let $v$ be a 5-vertex with $m_{3}(v)=4$ and $m_{5^{+}}(v)$ = 1. Then $n_{3}(v)=0$ , $n_{4}(v)\leq 1$ , and $n_{5}(v)\leq 2$ . In particular, if $n_{4}(v)=1$ , then $n_{5}(v)=0$ .

Proof.

Suppose that $v$ is incident to four 3-faces $[vv_{1}v_{2}]$ , $[vv_{2}v_{3}]$ , $[vv_{3}v_{4}]$ , $[vv_{4}v_{5}]$ , and one $5^{+}$ -face that contains $v_{1}$ and $v_{5}$ . Obviously, we have $n_{3}(v)=0$ by Lemma 2.2(1). Now, we show that $n_{4}(v)\leq 1$ . Assume that $v$ is adjacent to two 4-vertices. Let $G^{\prime}=G-v+\{v_{1}v_{5}\}$ . The graph $G^{\prime}$ is proper with respect to $G$ . By the minimality of $G$ , $G^{\prime}$ has a 2-distance 20-coloring $\phi^{\prime}$ . Let $\phi$ be a coloring of $G$ such that every vertex in $V(G)$ , except for $v$ , is colored using $\phi^{\prime}$ . Since $\Delta\leq 6$ , it follows that $|C_{\phi}(v)|\leq 18$ and $|C|-|C_{\phi}(v)|\geq 2$ . Therefore, there exists a safe color for $v$ . By coloring $v$ with the safe color, $\phi$ becomes a 2-distance 20-coloring of $G$ , a contradiction.

Next, we prove that $n_{5}(v)\leq 2$ . Assume that $v$ is adjacent to three 5-vertices. It is clear that $G^{\prime}=G-v+\{v_{1}v_{5}\}$ is proper with respect to $G$ . By the minimality of $G$ , $G^{\prime}$ has a 2-distance 20-coloring $\phi^{\prime}$ . Let $\phi$ be a coloring of $G$ such that every vertex in $V(G)$ , except for $v$ , is colored using $\phi^{\prime}$ . Since $|C_{\phi}(v)|\leq 19$ , we can color $v$ with a safe color, a contradiction.

Finally, we consider the case $n_{4}(v)=1$ . Assume that $v$ is adjacent to a 5-vertex. Obviously, $G^{\prime}=G-v+\{v_{1}v_{5}\}$ is proper with respect to $G$ . Since $d_{2}(v)\leq 19$ , there exists a safe color for $v$ , a contradiction. ∎

Lemma 2.8.

Let $v$ be a 5-vertex with $m_{3}(v)=4$ and $m_{5^{+}}(v)$ = 1. Then the number of 4-vertices, 5-vertices, and 6-vertices adjacent to $v$ must be one of the following:

(a)

( $n_{4}(v)$ , $n_{5}(v)$ , $n_{6}(v)$ ) = (1, 0, 4),
(b)

( $n_{4}(v)$ , $n_{5}(v)$ , $n_{6}(v)$ ) = (0, 2, 3),
(c)

( $n_{4}(v)$ , $n_{5}(v)$ , $n_{6}(v)$ ) = (0, 1, 4), or
(d)

( $n_{4}(v)$ , $n_{5}(v)$ , $n_{6}(v)$ ) = (0, 0, 5).

Moreover, let $w$ be any 6-vertex adjacent to $v$ . Then the following hold:

(1)

If $v$ is in case (a), then $m_{3}(w)\leq 4$ .
(2)

If $v$ is in case (b), then $m_{3}(w)\leq 4$ .
(3)

If $v$ is in case (c), then there exists at least one 6-vertex $w$ with $m_{3}(w)\leq 5$ .
(4)

If $v$ is in case (d), then there exist at least two 6-vertices $w_{1}$ , $w_{2}$ with $m_{3}(w_{1})\leq 5$ and $m_{3}(w_{2})\leq 5$ .

Proof.

The first statement of the lemma follows directly from Lemma 2.7. We now prove the remaining statements, from (1) to (4). Suppose that $v$ is incident to four 3-faces $[vv_{1}v_{2}]$ , $[vv_{2}v_{3}]$ , $[vv_{3}v_{4}]$ , $[vv_{4}v_{5}]$ , and one $5^{+}$ -face that contains $v_{1}$ and $v_{5}$ .

(1) To show that $w$ cannot be incident to more than five 3-faces, it suffices to prove that no edge in $G[N_{G}(v)]$ is contained in two 3-faces of $G$ . Assume that the edge $v_{1}v_{2}$ is contained in two 3-faces. It is clear that $G^{\prime}=G-v+\{v_{5}v_{1}\}$ is proper with respect to $G$ . By the minimality of $G$ , $G^{\prime}$ has a 2-distance 20-coloring $\phi^{\prime}$ . Let $\phi$ be a coloring of $G$ such that every vertex in $V(G)$ , except for $v$ , is colored using $\phi^{\prime}$ . Since $n_{4}(v)=1$ and $n_{6}(v)=4$ , it follows that $|C_{\phi}(v)|\leq 19$ and $|C|-|C_{\phi}(v)|\geq 1$ . Therefore, there exists a safe color for $v$ . By coloring $v$ with the safe color, $\phi$ becomes a 2-distance 20-coloring of $G$ , a contradiction.

(2) The proof is similar to that of (1). To show that $w$ cannot be incident to more than five 3-faces, it suffices to prove that no edge in $G[N_{G}(v)]$ is contained in two 3-faces of $G$ . Assume that the edge $v_{1}v_{2}$ is contained in two 3-faces. As in (1), let $G^{\prime}=G-v+\{v_{5}v_{1}\}$ , which is proper with respect to $G$ . By the minimality of $G$ , $G^{\prime}$ has a 2-distance 20-coloring $\phi^{\prime}$ . Let $\phi$ be a coloring of $G$ such that every vertex in $V(G)$ , except for $v$ , is colored using $\phi^{\prime}$ . In case (b), we have $n_{5}(v)=2$ and $n_{6}(v)=3$ , which leads to $|C_{\phi}(v)|\leq 19$ . Therefore, there exists a safe color for $v$ . By coloring $v$ with the safe color, $\phi$ becomes a 2-distance 20-coloring of $G$ , a contradiction.

(3) In case (c), we have $n_{5}(v)=1$ and $n_{6}(v)=4$ . It follows that at least one of $v_{1}$ and $v_{5}$ must be a 6-vertex. Let $w$ be such a 6-vertex. Since $w$ is incident to one $5^{+}$ -face, it can be incident to at most five 3-faces. Therefore, (3) holds.

(4) In case (d), all neighbours of $v$ are 6-vertices. It follows that both $v_{1}$ and $v_{5}$ are 6-vertices. Let $w_{1}=v_{1}$ and $w_{2}=v_{5}$ . Since each of $w_{1}$ and $w_{2}$ is incident to one $5^{+}$ -face, it can be incident to at most five 3-faces. Therefore, (4) holds. ∎

Next, we examine the properties of a 6-vertex in $G$ . Let $v$ be a 6-vertex and let $v_{1},v_{2},\ldots,v_{6}$ be the neighbours of $v$ in clockwise order.

Refer to caption — Figure 1: Illustrations of Lemma 2.9(4.3).

Lemma 2.9.

Let $v$ be a 6-vertex with $m_{3}(v)=5$ and $m_{4}(v)=1$ and let $u$ be any 5-vertex adjacent to $v$ . Then the following hold:

(1)

$n_{4}(v)\leq 2$ .
(2)

If $n_{4}(v)=2$ , then $n_{5}(v)\leq 1$ .
(3)

If $n_{4}(v)=1$ , then $n_{5}(v)\leq 3$ . Moreover, if $n_{5}(v)=3$ , then $m_{3}(u)\leq 3$ .
(4)
If $n_{4}(v)=0$ , then $n_{5}(v)\leq 5$ . Moreover, the following hold:
1. (4.1)
  
  If $n_{5}(v)=5$ , then $m_{3}(u)\leq 3$ .
2. (4.2)
  
  If $n_{5}(v)=4$ , then there exist at most two 5-vertices $u$ with $m_{3}(u)\geq 4$ .
3. (4.3)
  
  If $n_{5}(v)=3$ , then there exist at most two 5-vertices $u$ with $m_{3}(u)\geq 4$ .

Proof.

Suppose that $v$ is incident to five 3-faces $[vv_{1}v_{2}]$ , $[vv_{2}v_{3}]$ , $[vv_{3}v_{4}]$ , $[vv_{4}v_{5}]$ , $[vv_{5}v_{6}]$ , and one 4-face $[vv_{6}xx_{1}]$ . By Lemma 2.2(2), $v$ is not adjacent to any 3-vertex.

(1) In the proof of Lemma 2.3, we showed that if $v$ is a 4-vertex with $m_{3}(v)=2$ , then no edge in $G[N_{G}(v)]$ is contained in two 3-faces. Thus $v_{2}$ , $v_{3}$ , $v_{4}$ , and $v_{5}$ cannot be 4-vertices. Among the neighbours of $v$ , at most two vertices, namely, $v_{1}$ and $v_{6}$ , can be 4-vertices.

(2) We suppose that $v_{1}$ and $v_{6}$ are 4-vertices. Assume that $v$ is adjacent to two 5-vertices. Let $G^{\prime}=G-v+\{v_{1}v_{3},v_{1}v_{5},v_{1}v_{6}\}$ . The graph $G^{\prime}$ is proper with respect to $G$ . By the minimality of $G$ , $G^{\prime}$ has a 2-distance 20-coloring $\phi^{\prime}$ . Let $\phi$ be a coloring of $G$ such that every vertex in $V(G)$ , except for $v$ , is colored using $\phi^{\prime}$ . Since $\Delta\leq 6$ , it follows that $|C_{\phi}(v)|\leq 19$ and $|C|-|C_{\phi}(v)|\geq 1$ . Therefore, there exists a safe color for $v$ . By coloring $v$ with the safe color, $\phi$ becomes a 2-distance 20-coloring of $G$ , a contradiction.

(3) Without loss of generality, we suppose that $v_{1}$ is a 4-vertex. Assume that $v$ is adjacent to four 5-vertices. The graph $G^{\prime}=G-v+\{v_{1}v_{3},v_{1}v_{5},v_{1}v_{6}\}$ is proper with respect to $G$ . By the minimality of $G$ , $G^{\prime}$ has a 2-distance 20-coloring $\phi^{\prime}$ . Let $\phi$ be a coloring of $G$ such that every vertex in $V(G)$ , except for $v$ , is colored using $\phi^{\prime}$ . Since $|C_{\phi}(v)|\leq 19$ , we can color $v$ with a safe color, a contradiction.

Now, we consider the case $n_{5}(v)$ = 3. To show that any 5-vertex $u$ adjacent to $v$ cannot be incident to more than four 3-faces, it suffices to prove that no edge in $G[N_{G}(v)]$ is contained in two 3-faces of $G$ . Assume that the edge $v_{1}v_{2}$ is contained in two 3-faces. Let $G^{\prime}=G-v+\{v_{1}v_{3},v_{1}v_{5},v_{1}v_{6}\}$ . The graph $G^{\prime}$ is proper with respect to $G$ . Since $d_{2}(v)\leq 19$ , there exists a safe color for $v$ , a contradiction.

(4) We suppose that $v$ is not adjacent to any 4-vertex. Assume that all neighbours of $v$ are 5-vertices. Let $G^{\prime}=G-v+\{v_{6}v_{1},v_{3}v_{1},v_{3}v_{5}\}$ . The graph $G^{\prime}$ is proper with respect to $G$ . By the minimality of $G$ , $G^{\prime}$ has a 2-distance 20-coloring $\phi^{\prime}$ . Since $d_{2}(v)\leq 19$ , there exists a safe color for $v$ , a contradiction.

(4.1) Suppose that $v$ is adjacent to five 5-vertices. To show that any 5-vertex $u$ adjacent to $v$ cannot be incident to more than four 3-faces, it suffices to prove that no edge in $G[N_{G}(v)]$ is contained in two 3-faces of $G$ . Assume that the edge $v_{1}v_{2}$ is contained in two 3-faces. There are six possibilities for which neighbours of $v$ is a 6-vertex. Due to symmetry, it suffices to consider the cases where $v_{1}$ , $v_{2}$ , or $v_{3}$ is a 6-vertex. In each of these cases, let $G^{\prime}=G-v+\{v_{6}v_{1},v_{4}v_{2},v_{4}v_{6}\}$ . Then $G^{\prime}$ is proper with respect to $G$ . Since $d_{2}(v)\leq 19$ , there exists a safe color for $v$ , a contradiction.

(4.2) Suppose that $v$ is adjacent to four 5-vertices. We consider two cases based on whether both $v_{1}$ and $v_{6}$ are 6-vertices or not. First, we show that if $v_{1}$ and $v_{6}$ are both 6-vertices, then $m_{3}(v_{3})\leq 3$ and $m_{3}(v_{4})\leq 3$ . By symmetry, we only need to consider $v_{3}$ . Let $v_{7}$ and $v_{8}$ be the neighbours of $v_{3}$ other than $v$ , $v_{2}$ , and $v_{4}$ . Assume that $v_{3}$ is incident to four 3-faces. We have two cases: Case 1: The four 3-faces are $[vv_{2}v_{3}]$ , $[vv_{3}v_{4}]$ , $[v_{2}v_{7}v_{3}]$ , and $[v_{3}v_{8}v_{4}]$ . Case 2: The four 3-faces are $[vv_{2}v_{3}]$ , $[vv_{3}v_{4}]$ , $[v_{2}v_{7}v_{3}]$ , and $[v_{3}v_{7}v_{8}]$ . In Case 1, let $G^{\prime}=G-v_{3}+\{v_{7}v_{8}\}$ . In Case 2, let $G^{\prime}=G-v_{3}+\{v_{8}v_{4}\}$ . In each case, $G^{\prime}$ is proper with respect to $G$ . By the minimality of $G$ , $G^{\prime}$ has a 2-distance 20-coloring $\phi^{\prime}$ . Let $\phi$ be a coloring of $G$ such that every vertex in $V(G)$ , except for $v_{3}$ , is colored using $\phi^{\prime}$ . Since $|C_{\phi}(v_{3})|\leq 18$ , we can color $v_{3}$ with a safe color, a contradiction. Therefore, if $v_{1}$ and $v_{6}$ are 6-vertices, then there are at most two 5-vertices $u$ adjacent to $v$ with $m_{3}(u)\geq 4$ , namely $v_{2}$ and $v_{5}$ .

Next, we discuss the case where $v_{1}$ is not a 6-vertex or $v_{6}$ is not a 6-vertex. To show that there exist at most two 5-vertices $u$ adjacent to $v$ with $m_{3}(u)\geq 4$ , it suffices to prove that at most one edge in $G[N_{G}(v)]$ is contained in two 3-faces of $G$ . Assume that two edges $v_{i}v_{i+1}$ and $v_{j}v_{j+1}$ for $i,j\in\{1,2,3,4,5\}$ with $i\neq j$ are contained in two 3-faces. If $v_{1}$ and $v_{2}$ are 6-vertices, then we construct $G^{\prime}=G-v+\{v_{2}v_{4},v_{4}v_{6},v_{6}v_{1}\}$ . (Otherwise, we construct $G^{\prime}$ as follows: remove $v$ , add the edge $v_{6}v_{1}$ , and choose one 5-vertex $v_{i}$ in the neighbourhood of $v$ other than $v_{1}$ and $v_{6}$ , and connect $v_{i}$ to the two vertices in the neighbourhood of $v$ that are at distance two from $v_{i}$ .) The graph $G^{\prime}$ is proper with respect to $G$ . Since $d_{2}(v)\leq 19$ , there exists a safe color for $v$ , a contradiction.

(4.3) Suppose that $v$ is adjacent to three 5-vertices. There are twenty possible combinations of three 5-vertices. However, by symmetry, we only discuss ten cases: (see Figure 1.) Case 1: The three 5-vertices are $v_{4}$ , $v_{5}$ , and $v_{6}$ . Case 2: The three 5-vertices are $v_{3}$ , $v_{5}$ , and $v_{6}$ . Case 3: The three 5-vertices are $v_{3}$ , $v_{4}$ , and $v_{6}$ . Case 4: The three 5-vertices are $v_{3}$ , $v_{4}$ , and $v_{5}$ . Case 5: The three 5-vertices are $v_{2}$ , $v_{5}$ , and $v_{6}$ . Case 6: The three 5-vertices are $v_{2}$ , $v_{4}$ , and $v_{6}$ . Case 7: The three 5-vertices are $v_{2}$ , $v_{4}$ , and $v_{5}$ . Case 8: The three 5-vertices are $v_{2}$ , $v_{3}$ , and $v_{6}$ . Case 9: The three 5-vertices are $v_{1}$ , $v_{5}$ , and $v_{6}$ . Case 10: The three 5-vertices are $v_{1}$ , $v_{4}$ , and $v_{6}$ .

First, we consider Cases 3, 4, and 7. We show that $m_{3}(v_{4})\leq 3$ in these cases. Assume that $v_{4}$ is incident to four 3-faces. Let $v_{7}$ and $v_{8}$ be the neighbours of $v_{4}$ other than $v$ , $v_{3}$ , and $v_{5}$ . Since $v_{4}$ is already incident to two 3-faces, namely $[vv_{3}v_{4}]$ and $[vv_{4}v_{5}]$ , the remaining two 3-faces must be one of the following: (i) $[v_{3}v_{7}v_{4}]$ and $[v_{4}v_{8}v_{5}]$ , or (ii) $[v_{4}v_{7}v_{8}]$ and $[v_{4}v_{8}v_{5}]$ . In case (i), we construct $G^{\prime}=G-v_{4}+\{v_{7}v_{8}\}$ . In case (ii), we construct $G^{\prime}=G-v_{4}+\{v_{3}v_{7}\}$ . The graph $G^{\prime}$ is proper with respect to $G$ . By the minimality of $G$ , $G^{\prime}$ has a 2-distance 20-coloring $\phi^{\prime}$ . Let $\phi$ be a coloring of $G$ such that every vertex in $V(G)$ , except for $v_{4}$ , is colored using $\phi^{\prime}$ . Since $|C_{\phi}(v_{4})|\leq 19$ in each case, we can color $v_{4}$ with a safe color, a contradiction.

Next, we consider Cases 1, 2, 5, and 9. We prove that $m_{3}(v_{6})\leq 3$ in these cases. Assume that $v_{6}$ is incident to four 3-faces. It is clear that $G^{\prime}=G-v_{6}+\{vx\}$ is proper with respect to $G$ . Since $d_{2}(v_{6})\leq 19$ in each case, there exists a safe color for $v_{6}$ , a contradiction.

We can similarly show that $m_{3}(v_{3})\leq 3$ in Case 8. Assume that $v_{3}$ is incident to four 3-faces. Let $v_{7}$ and $v_{8}$ be the neighbours of $v_{3}$ other than $v$ , $v_{2}$ , and $v_{4}$ . Since $v_{3}$ is already incident to two 3-faces, namely $[vv_{2}v_{3}]$ and $[vv_{3}v_{4}]$ , the remaining two 3-faces must be one of the following: (i) $[v_{2}v_{7}v_{3}]$ and $[v_{3}v_{8}v_{4}]$ , or (ii) $[v_{3}v_{7}v_{8}]$ and $[v_{3}v_{8}v_{4}]$ . In case (i), we construct $G^{\prime}=G-v_{3}+\{v_{7}v_{8}\}$ . In case (ii), we construct $G^{\prime}=G-v_{3}+\{v_{2}v_{7}\}$ . The graph $G^{\prime}$ is proper with respect to $G$ . Since $d_{2}(v_{3})\leq 19$ in each case, there exists a safe color for $v_{3}$ , a contradiction.

Finally, we discuss Case 6 and Case 10. To show that there exist at most two 5-vertices $u$ adjacent to $v$ with $m_{3}(v)\geq 4$ , it suffices to prove that at most two edges in $G[N_{G}(v)]$ are contained in two 3-faces of $G$ . Assume that three edges in $G[N_{G}(v)]$ are contained in two 3-faces. In each case, we construct $G^{\prime}=G-v+\{v_{2}v_{4},v_{4}v_{6},v_{6}v_{1}\}$ . The graph $G^{\prime}$ is proper with respect to $G$ . Since $d_{2}(v)\leq 19$ , there exists a safe color for $v$ , a contradiction. From the above, there are at most two 5-vertices $u$ adjacent to $v$ with $m_{3}(u)\geq 4$ in Case 1 through Case 10. ∎

Lemma 2.10.

Let $v$ be a 6-vertex with $m_{3}(v)=5$ and $m_{5^{+}}(v)=1$ and let $u$ be any 5-vertex adjacent to $v$ . Then the following hold:

(1)

$n_{4}(v)\leq 2$ .
(2)

If $n_{4}(v)=2$ , then $n_{5}(v)\leq 2$ .
(3)

If $n_{4}(v)=1$ , then $n_{5}(v)\leq 4$ . Moreover, if $n_{5}(v)=4$ , then $m_{3}(u)\leq 3$ .
(4)

If $n_{4}(v)=0$ and $n_{5}(v)=6$ , then $m_{3}(u)\leq 3$ .
(5)

If $n_{4}(v)=0$ and $n_{5}(v)=5$ , then there exist at most two 5-vertices $u$ with $m_{3}(u)\geq 4$ .

Proof.

Suppose that $v$ is incident to five 3-faces $[vv_{1}v_{2}]$ , $[vv_{2}v_{3}]$ , $[vv_{3}v_{4}]$ , $[vv_{4}v_{5}]$ , $[vv_{5}v_{6}]$ , and one $5^{+}$ -face that contains $v_{1}$ and $v_{6}$ . By Lemma 2.2(2), $v$ is not adjacent to any 3-vertex.

(1) The proof is the same as that of Lemma 2.9(1). The vertices $v_{1}$ and $v_{6}$ can be 4-vertices.

(2) We suppose that $v_{1}$ and $v_{6}$ are 4-vertices. Assume that $v$ is adjacent to three 5-vertices. Regardless of which neighbours of $v$ other than $v_{1}$ and $v_{6}$ are the three 5-vertices, we construct $G^{\prime}=G-v+\{v_{1}v_{3},v_{1}v_{5},v_{1}v_{6}\}$ . The graph $G^{\prime}$ is proper with respect to $G$ . By the minimality of $G$ , $G^{\prime}$ has a 2-distance 20-coloring $\phi^{\prime}$ . Let $\phi$ be a coloring of $G$ such that every vertex in $V(G)$ , except for $v$ , is colored using $\phi^{\prime}$ . Since $\Delta\leq 6$ , it follows that $|C_{\phi}(v)|\leq 19$ and $|C|-|C_{\phi}(v)|\geq 1$ . Therefore, there exists a safe color for $v$ . By coloring $v$ with the safe color, $\phi$ becomes a 2-distance 20-coloring of $G$ , a contradiction.

(3) Without loss of generality, we suppose that $v_{1}$ is a 4-vertex. Assume that $v$ is adjacent to five 5-vertices. It is clear that $G^{\prime}=G-v+\{v_{1}v_{3},v_{1}v_{5},v_{1}v_{6}\}$ is proper with respect to $G$ . By the minimality of $G$ , $G^{\prime}$ has a 2-distance 20-coloring $\phi^{\prime}$ . Let $\phi$ be a coloring of $G$ such that every vertex in $V(G)$ , except for $v$ , is colored using $\phi^{\prime}$ . Since $|C_{\phi}(v)|\leq 19$ , we can color $v$ with a safe color, a contradiction. Now, we suppose that $v$ is adjacent to four 5-vertices. To show that any 5-vertex $u$ adjacent to $v$ cannot be incident to more than four 3-faces, it suffices to prove that no edge in $G[N_{G}(v)]$ is contained in two 3-faces of $G$ . Assume that the edge $v_{1}v_{2}$ is contained in two 3-faces. Let $G^{\prime}=G-v+\{v_{1}v_{3},v_{1}v_{5},v_{1}v_{6}\}$ . Then $G^{\prime}$ is proper with respect to $G$ . Since $d_{2}(v)\leq 19$ , there exists a safe color for $v$ , a contradiction.

(4) Suppose that all neighbours of $v$ are 5-vertices. To show that any 5-vertex $u$ adjacent to $v$ is incident to at most three 3-faces, it suffices to prove that no edge in $G[N_{G}(v)]$ is contained in two 3-faces of $G$ . Assume that the edge $v_{1}v_{2}$ is contained in two 3-faces. Let $G^{\prime}=G-v+\{v_{4}v_{2},v_{4}v_{6},v_{1}v_{6}\}$ . Then $G^{\prime}$ is proper with respect to $G$ . Since $d_{2}(v)\leq 19$ , there exists a safe color for $v$ , a contradiction.

(5) Suppose that $v$ is adjacent to five 5-vertices and one 6-vertex. To show that there exist at most two 5-vertices $u$ adjacent to $v$ with $m_{3}(u)\geq 4$ , it suffices to prove that at most one edge in $G[N_{G}(v)]$ is contained in two 3-faces of $G$ . Assume that two edges $v_{i}v_{i+1}$ and $v_{j}v_{j+1}$ for $i,j\in\{1,2,3,4,5\}$ with $i\neq j$ are contained in two 3-faces. Due to symmetry, it suffices to consider the cases where $v_{1}$ , $v_{2}$ , or $v_{3}$ is a 6-vertex. In each of these cases, let $G^{\prime}=G-v+\{v_{4}v_{2},v_{4}v_{6},v_{1}v_{6}\}$ . Then $G^{\prime}$ is proper with respect to $G$ . Since $d_{2}(v)\leq 19$ , there exists a safe color for $v$ , a contradiction. ∎

From Lemma 2.11 to Lemma 2.13, let $f_{i}=[vv_{i}v_{i+1}]$ for $i\in\{1,2,\ldots,5\}$ and $f_{6}=[vv_{6}v_{1}]$ be the 3-faces incident to $v$ .

Lemma 2.11.

Let $v$ be a 6-vertex with $m_{3}(v)=4$ and $m_{4}(v)=2$ . Then the following hold:

(1)

$n_{3}(v)=0$ .
(2)

If $n_{4}(v)=1$ , then $n_{5}(v)\leq 4$ .
(3)

If $n_{5}(v)=6$ , then $m_{3}(u)\leq 3$ for any 5-vertex $u$ adjacent to $v$ .

Proof.

We have three cases where $v$ is incident to four 3-faces and two 4-faces: Case 1: The 4-faces are $[vv_{1}xv_{2}]$ and $[vv_{2}yv_{3}]$ , and the 3-faces are $f_{3}$ , $f_{4}$ , $f_{5}$ , and $f_{6}$ . Case 2: The 4-faces are $[vv_{1}xv_{2}]$ and $[vv_{3}yv_{4}]$ , and the 3-faces are $f_{2}$ , $f_{4}$ , $f_{5}$ , and $f_{6}$ . Case 3: The 4-faces are $[vv_{1}xv_{2}]$ and $[vv_{4}yv_{5}]$ , and the 3-faces are $f_{2}$ , $f_{3}$ , $f_{5}$ , and $f_{6}$ .

(1) By Lemma 2.2(2), a 3-vertex is not incident to any 3-face, and by Lemma 2.2(3), a 3-vertex is incident to at most one 4-face. Thus $v$ is not adjacent to any 3-vertex in each case.

(2) Suppose that $v$ is adjacent to one 4-vertex. Assume that all other neighbours of $v$ are 5-vertices. First, we consider Case 1. In the proof of Lemma 2.3, we showed that if $v$ is a 4-vertex with $m_{3}(v)=2$ , then no edge in $G[N_{G}(v)]$ is contained in two 3-faces. Thus only $v_{1}$ , $v_{2}$ , or $v_{3}$ can be a 4-vertex. In each case, let $G^{\prime}=G-v+\{v_{1}v_{2},v_{2}v_{3},v_{3}v_{5},v_{5}v_{1}\}$ . Then $G^{\prime}$ is proper with respect to $G$ . By the minimality of $G$ , $G^{\prime}$ has a 2-distance 20-coloring $\phi^{\prime}$ . Let $\phi$ be a coloring of $G$ such that every vertex in $V(G)$ , except for $v$ , is colored using $\phi^{\prime}$ . Since $|C_{\phi}(v)|\leq 19$ and $|C|-|C_{\phi}(v)|\geq 1$ , there exists a safe color for $v$ . By coloring $v$ with the safe color, $\phi$ becomes a 2-distance 20-coloring of $G$ , a contradiction.

Next, we consider Case 2. For the same reason, only $v_{1}$ , $v_{2}$ , $v_{3}$ , or $v_{4}$ can be a 4-vertex. In each case, let $G^{\prime}=G-v+\{v_{1}v_{2},v_{3}v_{4},v_{3}v_{5},v_{5}v_{1}\}$ . Then $G^{\prime}$ is proper with respect to $G$ . Since $d_{2}(v)\leq 19$ , there exists a safe color for $v$ , a contradiction.

Finally, we consider Case 3. Each neighbour of $v$ can be a 4-vertex. By symmetry, it suffices to consider the cases where $v_{1}$ or $v_{3}$ is a 4-vertex. In each case, let $G^{\prime}=G-v+\{v_{1}v_{2},v_{4}v_{5},v_{1}v_{3},v_{3}v_{5}\}$ . Then $G^{\prime}$ is proper with respect to $G$ . Since $d_{2}(v)\leq 19$ , there exists a safe color for $v$ , a contradiction.

(3) Suppose that all neighbours of $v$ are 5-vertices. To show that any 5-vertex $u$ adjacent to $v$ cannot be incident to more than four 3-faces, it suffices to prove that no edge in $G[N_{G}(v)]$ is contained in two 3-faces of $G$ . Assume that the edge $v_{6}v_{1}$ is contained in two 3-faces. In Case 1, let $G^{\prime}=G-v+\{v_{1}v_{2},v_{2}v_{3},v_{3}v_{5},v_{5}v_{1}\}$ . In Case 2, let $G^{\prime}=G-v+\{v_{1}v_{2},v_{3}v_{4},v_{3}v_{5},v_{5}v_{1}\}$ . In Case 3, let $G^{\prime}=G-v+\{v_{1}v_{2},v_{4}v_{5},v_{1}v_{3},v_{3}v_{5}\}$ . In each case, $G^{\prime}$ is proper with respect to $G$ . Since $d_{2}(v)\leq 19$ , there exists a safe color for $v$ , a contradiction. ∎

Lemma 2.12.

Let $v$ be a 6-vertex with $m_{3}(v)=4$ , $m_{4}(v)=1$ , and $m_{5^{+}}(v)=1$ . Then $n_{3}(v)\leq 1$ . In particular, if $n_{3}(v)=1$ , then $n_{4}(v)=0$ .

Proof.

We have three cases where $v$ is incident to four 3-faces, one 4-face, and one $5^{+}$ -face: Case 1: The 4-face is $[vv_{1}xv_{2}]$ and the $5^{+}$ -face is $[vv_{2}y\ldots zv_{3}]$ , and the 3-faces are $f_{3}$ , $f_{4}$ , $f_{5}$ , and $f_{6}$ . Case 2: The 4-face is $[vv_{1}xv_{2}]$ and the $5^{+}$ -face is $[vv_{3}y\ldots zv_{4}]$ , and the 3-faces are $f_{2}$ , $f_{4}$ , $f_{5}$ , and $f_{6}$ . Case 3: The 4-face is $[vv_{1}xv_{2}]$ and the $5^{+}$ -face is $[vv_{4}y\ldots zv_{5}]$ , and the 3-faces are $f_{2}$ , $f_{3}$ , $f_{5}$ , and $f_{6}$ .

First, we show that $v$ is adjacent to at most one 3-vertex. By Lemma 2.2(2), a 3-vertex is not incident to any 3-face. Thus $v$ is not adjacent to any 3-vertex in Case 2 and Case 3. In Case 1, only $v_{2}$ can be a 3-vertex. Thus $n_{3}(v)\leq 1$ holds.

Now we consider Case 1 and suppose that $v_{2}$ is a 3-vertex. Assume that $v$ is adjacent to a 4-vertex. In the proof of Lemma 2.3, we showed that if $v$ is a 4-vertex with $m_{3}(v)=2$ , then no edge in $G[N_{G}(v)]$ is contained in two 3-faces. Hence only $v_{1}$ or $v_{3}$ can be a 4-vertex. If $v_{1}$ is a 4-vertex, then we construct $G^{\prime}=G-v_{2}+\{v_{1}y\}$ . Otherwise, we construct $G^{\prime}=G-v_{2}+\{v_{3}x,v_{3}y\}$ . In both cases, $G^{\prime}$ is proper with respect to $G$ . By the minimality of $G$ , $G^{\prime}$ has a 2-distance 20-coloring $\phi^{\prime}$ . Let $\phi$ be a coloring of $G$ such that every vertex in $V(G)$ , except for $v_{2}$ , is colored using $\phi^{\prime}$ . Since $\Delta\leq 6$ , it follows that $|C_{\phi}(v_{2})|\leq 17$ and $|C|-|C_{\phi}(v_{2})|\geq 3$ . Therefore, there exists a safe color for $v_{2}$ . By coloring $v_{2}$ with the safe color, $\phi$ becomes a 2-distance 20-coloring of $G$ , a contradiction. ∎

Lemma 2.13.

Let $v$ be a 6-vertex with $m_{3}(v)=4$ and $m_{5^{+}}(v)=2$ . Then $n_{3}(v)\leq 1$ . In particular, if $n_{3}(v)=1$ , then $n_{4}(v)=0$ .

Proof.

We have three cases where $v$ is incident to four 3-faces and two $5^{+}$ -faces: Case 1: The $5^{+}$ -faces are $[vv_{1}x\ldots yv_{2}]$ , $[vv_{2}z\ldots wv_{3}]$ , and the 3-faces are $f_{3}$ , $f_{4}$ , $f_{5}$ , and $f_{6}$ . Case 2: The $5^{+}$ -faces are $[vv_{1}x\ldots yv_{2}]$ , $[vv_{3}z\ldots wv_{4}]$ , and the 3-faces are $f_{2}$ , $f_{4}$ , $f_{5}$ , and $f_{6}$ . Case 3: The $5^{+}$ -faces are $[vv_{1}x\ldots yv_{2}]$ , $[vv_{4}z\ldots wv_{5}]$ , and the 3-faces are $f_{2}$ , $f_{3}$ , $f_{5}$ , and $f_{6}$ .

The proof is similar to that of Lemma 2.12. Only $v_{2}$ can be a 3-vertex in Case 1. Thus $n_{3}(v)\leq 1$ holds. We suppose that $v_{2}$ is a 3-vertex and assume that $v$ is adjacent to a 4-vertex. Only $v_{1}$ or $v_{3}$ can be a 4-vertex. If $v_{1}$ is a 4-vertex, then we construct $G^{\prime}=G-v_{2}+\{v_{1}y,v_{1}z\}$ . Otherwise, we construct $G^{\prime}=G-v_{2}+\{v_{3}y,v_{3}z\}$ . In both cases, $G^{\prime}$ is proper with respect to $G$ . By the minimality of $G$ , $G^{\prime}$ has a 2-distance 20-coloring $\phi^{\prime}$ . Let $\phi$ be a coloring of $G$ such that every vertex in $V(G)$ , except for $v_{2}$ , is colored using $\phi^{\prime}$ . Since $|C_{\phi}(v_{2})|\leq 18$ , we can color $v_{2}$ with a safe color, a contradiction. ∎

Lemma 2.14.

Let $f$ be a 5-face of $G$ . Then there is at most one 3-vertex incident to $f$ . In particular, if $f$ is incident to one 3-vertex, then $f$ is not incident to any 4-vertex.

Proof.

Let $[v_{1}v_{2}v_{3}v_{4}v_{5}]$ be a 5-face. By Lemma 2.2(1), a 3-vertex is not adjacent to any $5^{-}$ -vertex. Assume that $v_{1}$ and $v_{4}$ are 3-vertices with $N_{G}(v_{1})=\{v_{2},v_{5},v_{6}\}$ and $N_{G}(v_{4})=\{v_{3},v_{5},v_{7}\}$ . Let $G^{\prime}=G-v_{1}+\{v_{2}v_{4},v_{4}v_{6}\}$ . The graph $G^{\prime}$ is proper with respect to $G$ . By the minimality of $G$ , $G^{\prime}$ has a 2-distance 20-coloring $\phi^{\prime}$ . Let $\phi$ be a coloring of $G$ such that every vertex in $V(G)$ , except for $v_{1}$ , is colored using $\phi^{\prime}$ . Since $\Delta\leq 6$ , it follows that $|C_{\phi}(v_{1})|\leq 18$ and $|C|-|C_{\phi}(v_{1})|\geq 2$ . Therefore, there exists a safe color for $v_{1}$ . By coloring $v_{1}$ with the safe color, $\phi$ becomes a 2-distance 20-coloring of $G$ , a contradiction.

Now, suppose that $v_{1}$ is a 3-vertex with $N_{G}(v_{1})=\{v_{2},v_{5},v_{6}\}$ . Assume that $v_{4}$ is a 4-vertex. It is clear that $G^{\prime}=G-v_{1}+\{v_{2}v_{4},v_{4}v_{6}\}$ is proper with respect to $G$ . Let $\phi$ be a coloring of $G$ such that every vertex in $V(G)$ , except for $v_{1}$ , is colored using $\phi^{\prime}$ . Since $|C_{\phi}(v_{1})|\leq 18$ , we can color $v_{1}$ with a safe color, a contradiction. ∎

We obtain the following corollary from Lemma 2.2.

Corollary 2.15.

A $6^{+}$ -face $f$ is incident to at most $\lfloor\frac{d(f)}{2}\rfloor$ 3-vertices.

3 Discharging

In this section, we design discharging rules and complete the proof of Theorem 1.2. We can derive the following equation by Euler’s formula $|V(G)|-|E(G)|+|F(G)|=2$ .

\sum_{v\in V(G)}(d_{G}(v)-4)+\sum_{f\in F(G)}(d(f)-4)=-8.

We assign an initial charge $\mu(v)=d_{G}(v)-4$ to each vertex and $\mu(f)=d(f)-4$ to each face. We design appropriate discharging rules and redistribute the charges of the vertices and faces according to those rules. Let $\mu^{\prime}(v)$ and $\mu^{\prime}(f)$ denote the final charges of the vertices and faces, respectively, after the discharging process. During the process, the sum of charges remains constant. If $\mu^{\prime}(v)\geq 0$ and $\mu^{\prime}(f)\geq 0$ , the following contradiction arises.

0\leq\sum_{x\in V(G)\cup F(G)}\mu^{\prime}(x)=\sum_{x\in V(G)\cup F(G)}\mu(x)=-8<0.

We design the following discharging rules, which are based on the rules in [5].

R1

Every 3-face receives $\frac{1}{3}$ from each of its incident vertices.
R2

Every 3-vertex receives $\frac{1}{9}$ from each of its adjacent 6-vertices $w$ with $m_{3}(w)\leq 5$ .
R3

Every 3-vertex receives $\frac{1}{3}$ from each of its incident $5^{+}$ -faces.
R4

Every 4-vertex receives $\frac{1}{5}$ from each of its incident $5^{+}$ -faces.
R5

Every 4-vertex receives $\frac{1}{15}$ from each of its adjacent 6-vertices $w$ with $m_{3}(w)\leq 5$ .
R6

Every 5-vertex receives $\frac{1}{5}$ from each of its incident $5^{+}$ -faces.
R7

Every 5-vertex $u$ with $m_{3}(u)\geq 4$ receives $\frac{2}{15}$ from each of its adjacent 6-vertices $w$ with $m_{3}(w)\leq 5$ .
R8

Every 6-vertex $v$ receives $\frac{1}{5}$ from each of its incident $5^{+}$ -faces $f$ , if $f$ does not contain any 3-vertex adjacent to $v$ .
R9

Every 6-vertex $v$ receives $\frac{1}{9}$ from each of its incident $5^{+}$ -faces $f$ , if $f$ contains a 3-vertex adjacent to $v$ .

First, we prove that $\mu^{\prime}(f)\geq 0$ for each $f\in F(G)$ .

Case 1. $d(f)=3.$: The initial charge is $\mu(f)=d(f)-4=-1$ . By Lemma 2.2(2), $f$ is not incident to any 3-vertex. By R1, $f$ receives $\frac{1}{3}$ from each $4^{+}$ -vertex incident to $f$ . Thus $\mu^{\prime}(f)=-1+3\times\frac{1}{3}=0$ .
Case 2. $d(f)=4.$: By the discharging rules, there is no transfer of charge. Thus $\mu(f)=\mu^{\prime}(f)=0$ .
Case 3. $d(f)=5.$: The initial charge is $\mu(f)=d(f)-4=1$ . By Lemma 2.14, $f$ is incident to at most one 3-vertex, and if $f$ is incident to one 3-vertex, then $f$ is not incident to any 4-vertex. By Lemma 2.2(1), all neighbours of a 3-vertex are 6-vertices. If $f$ is incident to a 3-vertex, then $\mu^{\prime}(f)=1-\frac{1}{3}-2\times\frac{1}{5}-2\times\frac{1}{9}=\frac{2}{45}$ by R3, R6, R8, and R9. Otherwise, $\mu^{\prime}(f)=1-5\times\frac{1}{5}=0$ by R4, R6, and R8.
Case 4. $d(f)=6^{+}.$: The initial charge is $\mu(f)=d(f)-4\geq 2$ . We have $\mu^{\prime}(f)\geq 0$ by By Corollary 2.15.

Next, we prove that $\mu^{\prime}(v)\geq 0$ for each $v\in V(G)$ . By Lemma 2.1 and $\Delta\leq 6$ , we only consider the cases where $3\leq d_{G}(v)\leq 6$ .

Case 1. $d_{G}(v)=3.$

The initial charge is $\mu(v)=d_{G}(v)-4=-1$ . By Lemma 2.2(1), all neighbours of $v$ are 6-vertices. By Lemma 2.2(2) and Lemma 2.2(3), $v$ is incident to either one 4-face and two $5^{+}$ -faces or three $5^{+}$ -faces. In each case, $m_{3}(w)\leq 5$ holds for any 6-vertex $w$ adjacent to $v$ . If $v$ is incident to one 4-face and two $5^{+}$ -faces, then $\mu^{\prime}(v)=-1+3\times\frac{1}{9}+2\times\frac{1}{3}=0$ by R2 and R3. Otherwise, $\mu^{\prime}(v)=-1+3\times\frac{1}{9}+3\times\frac{1}{3}=\frac{1}{3}$ by R2 and R3.

Case 2. $d_{G}(v)=4.$

The initial charge is $\mu(v)=d_{G}(v)-4=0$ . By Lemma 2.2(1), $v$ is not adjacent to any 3-vertex. By Lemma 2.3, we have $m_{3}(v)\leq 2$ . Thus we divide the case based on the value of $m_{3}(v)$ .

Case 2.1. $m_{3}(v)=2.$

By Lemma 2.3, we have $m_{4}(v)=0$ , $n_{6}(v)=4$ , and $m_{3}(w)\leq 4$ for any 6-vertex $w$ adjacent to $v$ . This implies that $v$ is incident to two 3-faces and two $5^{+}$ -faces, all neighbours of $v$ are 6-vertices, and R5 can be applied to $v$ . By R1, R4, and R5, $\mu^{\prime}(v)=0-2\times\frac{1}{3}+2\times\frac{1}{5}+4\times\frac{1}{15}=0$ .

Case 2.2. $m_{3}(v)=1.$

By Lemma 2.4, we have $m_{4}(v)\leq 2$ . In particular, if $1\leq m_{4}(v)\leq 2$ , then $n_{4}(v)=0$ and $n_{5}(v)\leq 1$ . If $m_{4}(v)=2$ , then the remaining face incident to $v$ is a $5^{+}$ -face, and $v$ is adjacent to at least three 6-vertices. By R1, R4, and R5, $\mu^{\prime}(v)\geq 0-1\times\frac{1}{3}+1\times\frac{1}{5}+3\times\frac{1}{15}=\frac{1}{15}$ . If $m_{4}(v)=1$ , then the remaining two faces incident to $v$ are $5^{+}$ -faces, and $v$ is adjacent to at least three 6-vertices. By R1, R4, and R5, $\mu^{\prime}(v)\geq 0-1\times\frac{1}{3}+2\times\frac{1}{5}+3\times\frac{1}{15}=\frac{4}{15}$ . If $m_{4}(v)=0$ , then the remaining three faces incident to $v$ are $5^{+}$ -faces. Regardless of the number of 6-vertices adjacent to $v$ , we have $\mu^{\prime}(v)\geq 0-1\times\frac{1}{3}+3\times\frac{1}{5}=\frac{4}{15}$ by R1 and R4.

Case 2.3. $m_{3}(v)=0.$

In this case, $v$ is not incident to any 3-face, which implies that R1 cannot be applied. Thus $\mu^{\prime}(v)\geq\mu(v)=0$ .

Case 3. $d_{G}(v)=5.$

The initial charge is $\mu(v)=d_{G}(v)-4=1$ . By Lemma 2.2(1), $v$ is not adjacent to any 3-vertex. We divide the case based on the value of $m_{3}(v)$ .

Case 3.1. $m_{3}(v)=5.$

By Lemma 2.5, we have $n_{6}(v)=5$ and $m_{3}(w)\leq 4$ for any 6-vertex $w$ adjacent to $v$ . Thus R7 can be applied to $v$ . By R1 and R7, $\mu^{\prime}(v)=1-5\times\frac{1}{3}+5\times\frac{2}{15}=0$ .

Case 3.2. $m_{3}(v)=4.$

The remaining face incident to $v$ is either one 4-face or one $5^{+}$ -face. First, we consider the case where the remaining face is a 4-face. By Lemma 2.6, we have $n_{4^{-}}(v)=0$ , $n_{5}(v)\leq 1$ , and $m_{3}(w)\leq 5$ for any 6-vertex $w$ adjacent to $v$ . Thus R7 can be applied to $v$ . If $v$ is adjacent to one 5-vertex and four 6-vertices, then $\mu^{\prime}(v)\geq 1-4\times\frac{1}{3}+4\times\frac{2}{15}=\frac{1}{5}$ by R1 and R7. If $v$ is adjacent to five 6-vertices, then $\mu^{\prime}(v)\geq 1-4\times\frac{1}{3}+5\times\frac{2}{15}=\frac{1}{3}$ by R1 and R7. Next, we consider the case where the remaining face is a $5^{+}$ -face. By Lemma 2.8, the pattern of the degrees of the vertices adjacent to $v$ must be one of the cases (a) through (d). In each case, we show that $\mu^{\prime}(v)\geq 0$ .

(a).: ( $n_{4}(v)$ , $n_{5}(v)$ , $n_{6}(v)$ ) = (1, 0, 4).
By Lemma 2.8(1), we have $m_{3}(w)\leq 4$ for any 6-vertex $w$ adjacent to $v$ . Thus R7 can be applied to $v$ . By R1, R6, and R7, $\mu^{\prime}(v)=1-4\times\frac{1}{3}+1\times\frac{1}{5}+4\times\frac{2}{15}=\frac{2}{5}$ .
(b).: ( $n_{4}(v)$ , $n_{5}(v)$ , $n_{6}(v)$ ) = (0, 2, 3).
By Lemma 2.8(2), we have $m_{3}(w)\leq 4$ for any 6-vertex $w$ adjacent to $v$ . Thus R7 can be applied to $v$ . By R1, R6, and R7, $\mu^{\prime}(v)=1-4\times\frac{1}{3}+1\times\frac{1}{5}+3\times\frac{2}{15}=\frac{4}{15}$ .
(c).: ( $n_{4}(v)$ , $n_{5}(v)$ , $n_{6}(v)$ ) = (0, 1, 4).
By Lemma 2.8(3), there exists at least one 6-vertex $w$ adjacent to $v$ with $m_{3}(w)\leq 5$ . Thus $v$ receives at least $\frac{2}{15}$ from such a 6-vertex by R7. By R1, R6, and R7, $\mu^{\prime}(v)\geq 1-4\times\frac{1}{3}+1\times\frac{1}{5}+1\times\frac{2}{15}=0$ .
(d).: ( $n_{4}(v)$ , $n_{5}(v)$ , $n_{6}(v)$ ) = (0, 0, 5).
By Lemma 2.8(4), there exist at least two 6-vertices $w_{1}$ , $w_{2}$ adjacent to $v$ with $m_{3}(w_{1})\leq 5$ and $m_{3}(w_{2})\leq 5$ . Thus $v$ receives at least $2\times\frac{2}{15}$ from such 6-vertices by R7. By R1, R6, and R7, $\mu^{\prime}(v)\geq 1-4\times\frac{1}{3}+1\times\frac{1}{5}+2\times\frac{2}{15}=\frac{2}{15}$ .

Case 3.3. $m_{3}(v)\leq 3.$

The only rule by which $v$ loses charge is R1. By R1 and $m_{3}(v)\leq 3$ , we have $\mu^{\prime}(v)\geq 1-3\times\frac{1}{3}=0$ .

Case 4. $d_{G}(v)=6.$

The initial charge is $\mu(v)=d_{G}(v)-4=2$ . We divide the case based on the value of $m_{3}(v)$ .

Case 4.1. $m_{3}(v)=6.$

Since $m_{3}(v)=6$ , R2, R5, and R7 cannot be applied to $v$ . The only rule by which $v$ loses charge is R1. Thus $\mu^{\prime}(v)=2-6\times\frac{1}{3}=0$ .

Case 4.2. $m_{3}(v)=5.$

By R1, $v$ sends $\frac{1}{3}$ to each of its incident 3-faces. Since $m_{3}(v)=5$ , $v$ loses $5\times\frac{1}{3}=\frac{5}{3}$ charge. By Lemma 2.2(2), $v$ is not adjacent to any 3-vertex. The remaining face incident to $v$ is either one 4-face or one $5^{+}$ -face. First, we consider the case where the remaining face is a 4-face. By Lemma 2.9(1), we have $n_{4}(v)\leq 2$ . Thus we further divide the case based on the value of $n_{4}(v)$ .

Case 4.2.1. $m_{4}(v)=1$ , $n_{4}(v)=2.$: By Lemma 2.9(2), we have $n_{5}(v)\leq 1$ . In the worst situation, $v$ is adjacent to one 5-vertex $u$ with $m_{3}(u)\geq 4$ . By R1, R5, and R7, $\mu^{\prime}(v)\geq 2-\frac{5}{3}-2\times\frac{1}{15}-1\times\frac{2}{15}=\frac{1}{15}$ .
Case 4.2.2. $m_{4}(v)=1$ , $n_{4}(v)=1.$: By Lemma 2.9(3), we have $n_{5}(v)\leq 3$ , and if $n_{5}(v)=3$ , then $m_{3}(u)\leq 3$ for any 5-vertex $u$ adjacent to $v$ . Thus if $n_{5}(v)=3$ , then $v$ does not lose charge by R7. The vertex $v$ loses the most charge when $v$ is adjacent to two 5-vertices to which R7 applies. By R1, R5, and R7, $\mu^{\prime}(v)\geq 2-\frac{5}{3}-1\times\frac{1}{15}-2\times\frac{2}{15}=0$ .
Case 4.2.3. $m_{4}(v)=1$ , $n_{4}(v)=0.$: By Lemma 2.9(4) and Lemma 2.9(4.1), we have $n_{5}(v)\leq 5$ , and if $n_{5}(v)=5$ , then $m_{3}(u)\leq 3$ for any 5-vertex $u$ adjacent to $v$ . Thus if $n_{5}(v)=5$ , then $v$ does not lose charge by R7. If $n_{5}(v)\leq 4$ , then there exist at most two 5-vertices $u$ with $m_{3}(u)\geq 4$ by Lemma 2.9(4.2) and Lemma 2.9(4.3). This implies that $v$ loses at most $2\times\frac{2}{15}$ charge by R7. By R1 and R7, $\mu^{\prime}(v)\geq 2-\frac{5}{3}-2\times\frac{2}{15}=\frac{1}{15}$ .

Next, we consider the case where the remaining face is a $5^{+}$ -face. By Lemma 2.10(1), we have $n_{4}(v)\leq 2$ . Thus we further divide the case based on the value of $n_{4}(v)$ .

Case 4.2.4. $m_{5^{+}}(v)=1$ , $n_{4}(v)=2.$: By Lemma 2.10(2), we have $n_{5}(v)\leq 2$ . This implies that $v$ loses at most $2\times\frac{2}{15}$ charge by R7. By R1, R5, R7, and R8, $\mu^{\prime}(v)\geq 2-\frac{5}{3}-2\times\frac{1}{15}-2\times\frac{2}{15}+\frac{1}{5}=\frac{2}{15}$ .
Case 4.2.5. $m_{5^{+}}(v)=1$ , $n_{4}(v)=1.$: By Lemma 2.10(3), we have $n_{5}(v)\leq 4$ , and if $n_{5}(v)=4$ , then $m_{3}(u)\leq 3$ for any 5-vertex $u$ adjacent to $v$ . Thus if $n_{5}(v)=4$ , then $v$ does not lose charge by R7. The vertex $v$ loses the most charge when $v$ is adjacent to three 5-vertices to which R7 applies. By R1, R5, R7, and R8, $\mu^{\prime}(v)\geq 2-\frac{5}{3}-1\times\frac{1}{15}-3\times\frac{2}{15}+\frac{1}{5}=\frac{1}{15}$ .
Case 4.2.6. $m_{5^{+}}(v)=1$ , $n_{4}(v)=0.$: By Lemma 2.10(4), if $n_{5}(v)=6$ , then $m_{3}(u)\leq 3$ for any 5-vertex $u$ adjacent to $v$ . Hence if $n_{5}(v)=6$ , then $v$ does not lose charge by R7. By Lemma 2.10(5), if $n_{5}(v)=5$ , then there exist at most two 5-vertices $u$ with $m_{3}(u)\geq 4$ . Thus $v$ loses the most charge when $v$ is adjacent to four 5-vertices to which R7 applies. By R1, R7, and R8, $\mu^{\prime}(v)\geq 2-\frac{5}{3}-4\times\frac{2}{15}+\frac{1}{5}=0$ .

Case 4.3. $m_{3}(v)=4.$

By R1, $v$ sends $\frac{1}{3}$ to each of its incident 3-faces. Since $m_{3}(v)=4$ , $v$ loses $4\times\frac{1}{3}=\frac{4}{3}$ charge. We further divide the case based on the faces incident to $v$ , which can be either two 4-faces, one 4-face and one $5^{+}$ -face, or two $5^{+}$ -faces.

Case 4.3.1. $m_{4}(v)=2.$: By Lemma 2.11(1), $v$ is not adjacent to any 3-vertex. The rule by which $v$ loses the most charge, except for R1, is R7. Thus if all neighbours of $v$ are 5-vertices to which R7 applies, then $v$ loses $6\times\frac{2}{15}=\frac{4}{5}$ by R7. This discussion implies that $\mu^{\prime}(v)=2-\frac{4}{3}-\frac{4}{5}=-\frac{2}{15}<0$ . However, by Lemma 2.11(3), the situation does not arise. The next situation in which $v$ loses the most charge is when $v$ is adjacent to five 5-vertices $u$ with $m_{3}(u)\geq 4$ and one 4-vertex, but by Lemma 2.11(2), this situation cannot occur. In the possible cases, $v$ loses the most charge when $v$ is adjacent to either four 5-vertices $u$ with $m_{3}(u)\geq 4$ and two 4-vertices, or five 5-vertices $u$ with $m_{3}(u)\geq 4$ and one 6-vertex. In the former case, $\mu^{\prime}(v)=2-\frac{4}{3}-2\times\frac{1}{15}-4\times\frac{2}{15}=0$ by R1, R5, and R7. In the latter case, $\mu^{\prime}(v)=2-\frac{4}{3}-5\times\frac{2}{15}=0$ by R1 and R7.
Case 4.3.2. $m_{4}(v)=1,m_{5^{+}}(v)=1.$: By Lemma 2.12, $v$ is adjacent to at most one 3-vertex, and if $v$ is adjacent to one 3-vertex, then $v$ is not adjacent to any 4-vertex. Since $v$ is incident to one $5^{+}$ -face, R8 or R9 can be applied to $v$ . If $v$ is adjacent to one 3-vertex, then $v$ loses the most charge when $v$ is adjacent to five 5-vertices to which R7 applies. By R1, R2, R7, and R9, $\mu^{\prime}(v)=2-\frac{4}{3}-1\times\frac{1}{9}-5\times\frac{2}{15}+1\times\frac{1}{9}=0$ . If $v$ is not adjacent to a 3-vertex, then $v$ loses the most charge when $v$ is adjacent to six 5-vertices to which R7 applies. By R1, R7, and R8, $\mu^{\prime}(v)=2-\frac{4}{3}-6\times\frac{2}{15}+1\times\frac{1}{5}=\frac{1}{15}$ .
Case 4.3.3. $m_{5^{+}}(v)=2.$: By Lemma 2.13, $v$ is adjacent to at most one 3-vertex, and if $v$ is adjacent to one 3-vertex, then $v$ is not adjacent to any 4-vertex. Since $v$ is incident to two $5^{+}$ -faces, $v$ receives at least $2\times\frac{1}{9}$ charge by R9. If $v$ is adjacent to one 3-vertex, then $v$ loses the most charge when $v$ is adjacent to five 5-vertices to which R7 applies. By R1, R2, R7, and R9, $\mu^{\prime}(v)=2-\frac{4}{3}-1\times\frac{1}{9}-5\times\frac{2}{15}+2\times\frac{1}{9}=\frac{1}{9}$ . If $v$ is not adjacent to a 3-vertex, then $v$ loses the most charge when $v$ is adjacent to six 5-vertices to which R7 applies. By R1, R7, and R8, $\mu^{\prime}(v)=2-\frac{4}{3}-6\times\frac{2}{15}+2\times\frac{1}{5}=\frac{4}{15}$ .

Case 4.4. $m_{3}(v)\leq 3.$

By R1, $v$ loses at most $3\times\frac{1}{3}=1$ charge. The rule by which $v$ loses the most charge, except for R1, is R7. Thus if all neighbours of $v$ are 5-vertices $u$ with $m_{3}(u)\geq 4$ , then $v$ loses $6\times\frac{2}{15}=\frac{4}{5}$ charge by R7. The final charge is $\mu^{\prime}(v)\geq 2-1-\frac{4}{5}=\frac{1}{5}>0$ , which implies that $\mu^{\prime}(v)\geq 0$ holds when $m_{3}(v)\leq 3$ .

Now, we have confirmed $\mu^{\prime}(x)\geq 0$ for all $x\in V(G)\cup F(G)$ , which is a contradiction. Therefore, Theorem 1.2 holds.

References

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