Planar graphs without normally adjacent short cycles

TL;DR

This paper proves that certain classes of planar graphs with specific cycle restrictions are 3-choosable, using a stronger form of DP-coloring, and also shows their vertices can be partitioned into an independent set and a forest.

Contribution

It introduces a new proof technique for 3-choosability of graphs with cycle restrictions and extends previous results, also providing vertex partition results.

Findings

Graphs in class  are 3-choosable.

Planar graphs without certain cycles are 3-choosable.

Vertices can be partitioned into an independent set and a forest.

Abstract

Let $\mathscr{G}$ be the class of plane graphs without triangles normally adjacent to $8^{-}$-cycles, without $4$-cycles normally adjacent to $6^{-}$-cycles, and without normally adjacent $5$-cycles. In this paper, it is shown that every graph in $\mathscr{G}$ is $3$-choosable. Instead of proving this result, we directly prove a stronger result in the form of ``weakly'' DP-$3$-coloring. The main theorem improves the results in [J. Combin. Theory Ser. B 129 (2018) 38--54; European J. Combin. 82 (2019) 102995]. Consequently, every planar graph without $4$-, $6$-, $8$-cycles is $3$-choosable, and every planar graph without $4$-, $5$-, $7$-, $8$-cycles is $3$-choosable. In the third section, using almost the same technique, we prove that the vertex set of every graph in $\mathscr{G}$ can be partitioned into an independent set and a set that induces a forest, which strengthens the result in [Discrete Appl. Math. 284 (2020) 626--630]. In the final section, tightness is discussed.

Equations

$$|V(G)|+|E(G)|-|S|\text{ is minimized.}$$

$$3$$

$$\sum_{v\,\in\,V(G)}\big{(}\deg(v)-4\big{)}+\sum_{f\,\in\,F(G)\setminus D}\big{(}\deg(f)-4\big{)}+(\deg(D)+4)=0.$$

$${\mu }^{\prime }(f)\ge k-4-(k-2)\times \frac{2}{3}=\frac{1}{3}(k-8)>0.$$

Full text

Keywords: DP-coloring; Near-bipartite; IF-coloring; List coloring

1 Introduction

All graphs in this paper are finite, undirected and simple. For a graph $G$, a list-assignment $L$ assigns to each vertex $v$ a set $L(v)$ of colors available at $v$. An $L$-coloring of $G$ is a proper coloring $\phi$ of $G$ such that $\phi(v)\in L(v)$ for all $v\in V(G)$. A list $k$-assignment $L$ is a list-assignment such that $|L(v)|\geq k$ for all $v\in V(G)$. A graph $G$ is $k$-choosable or list $k$-colorable if it has an $L$-coloring for any list $k$-assignment $L$. The list chromatic number or choice number $\chi_{\ell}(G)$ is the least integer $k$ such that $G$ is $k$-choosable.

The Four Color Theorem states that every planar graph is $4$-colorable. Grötzsch [6] showed that every planar graph without triangles is $3$-colorable. Much more sufficient conditions for $3$-colorability and $3$-choosability are extensively studied. Thomassen [21] showed that every planar graph with girth at least five is $3$-choosable. Borodin [1] conjectured that every planar graph without cycles of length $4$ to $8$ is $3$-choosable.

A widely used technique in ordinary vertex coloring is the identification of vertices, but this is not feasible in general for list-coloring because different vertices may have different lists. To overcome this difficulty, Dvořák and Postle [5] introduced DP-coloring, also called correspondence coloring, as a generalization of list-coloring.

Definition 1**.**

Let $G$ be a simple graph and $L$ be a list-assignment for $G$. For each vertex $v\in V(G)$, let $L_{v}=\{v\}\times L(v)$; for each edge $uv\in E(G)$, let $\mathscr{M}_{uv}$ be a matching between the sets $L_{u}$ and $L_{v}$, and let $\mathscr{M}:=\bigcup_{uv\in E(G)}\mathscr{M}_{uv}$. We call $\mathscr{M}$ a matching assignment. The matching assignment is called a $k$-matching assignment if $L(v)=[k]$ for each $v\in V(G)$. A cover of $G$ is a graph $H_{L,\mathscr{M}}$ (simply write $H$) satisfying the following two conditions:

1. (C1)

the vertex set of $H$ is the disjoint union of $L_{v}$ for all $v\in V(G)$; 2. (C2)

the edge set of $H$ is the matching assignment $\mathscr{M}$.

Note that the matching $\mathscr{M}_{uv}$ is not required to be a perfect matching between the sets $L_{u}$ and $L_{v}$, and possibly it is empty. The induced subgraph $H[L_{v}]$ is an independent set for each vertex $v\in V(G)$.

Definition 2**.**

Let $G$ be a simple graph and $H$ be a cover of $G$. An $\mathscr{M}$-coloring of $G$ is an independent set $\mathcal{I}$ in $H$ such that $|\mathcal{I}\cap L_{v}|=1$ for each vertex $v\in V(G)$. The graph $G$ is DP-$k$-colorable if for any list-assignment $L(v)\supseteq[k]$ and any matching assignment $\mathscr{M}$, it has an $\mathscr{M}$-coloring. The DP-chromatic number $\chi_{\mathrm{DP}}(G)$ of $G$ is the least integer $k$ such that $G$ is DP-$k$-colorable.

DP-coloring is quite different from list-coloring, for example each even cycle is $2$-choosable but it is not DP-$2$-colorable. Dvořák and Postle gave a relation between DP-coloring and list-coloring.

Let $W=w_{1}w_{2}\dots w_{m}$ with $w_{m}=w_{1}$ be a closed walk of length $m-1$ in $G$, a matching assignment is inconsistent on $W$, if there exist $c_{1},\dots,c_{m}$ such that $c_{i}\in L(w_{i})$ for $i\in[m]$ and $(w_{i},c_{i})(w_{i+1},c_{i+1})$ is an edge in $\mathscr{M}_{w_{i}w_{i+1}}$ for $i\in[m-1]$ and $c_{1}\neq c_{m}$. Otherwise, the matching assignment is consistent on $W$. We say that a matching assignment is consistent if it is consistent on every closed walk in $G$.

Theorem 1.1** (Dvořák and Postle [5]).**

A graph $G$ is $k$-choosable if and only if $G$ is $\mathscr{M}$-colorable for every consistent $k$-matching assignment $\mathscr{M}$.

With the aid of DP-coloring, Dvořák and Postle [5] solved the longstanding conjecture by Borodin.

Theorem 1.2** (Dvořák and Postle [5]).**

Every planar graph without cycles of length $4$ to $8$ is $3$-choosable.

An edge $uv$ in $G$ is straight in a $k$-matching assignment $\mathscr{M}$ if $(u,c_{1})(v,c_{2})\in\mathscr{M}_{uv}$ satisfies $c_{1}=c_{2}$. An edge $uv$ in $G$ is full in a $k$-matching assignment $\mathscr{M}$ if $\mathscr{M}_{uv}$ is a perfect matching.

Lemma 1.1** (Dvořák and Postle [5]).**

Let $G$ be a graph with a $k$-matching assignment $\mathscr{M}$, and let $K$ be a subgraph of $G$. If for every cycle $\mathcal{Q}$ in $K$, the assignment $\mathscr{M}$ is consistent on $\mathcal{Q}$ and all edges of $\mathcal{Q}$ are full, then we may rename $L(u)$ for $u\in V(K)$ to obtain a $k$-matching assignment $\mathscr{M}^{\prime}$ for $G$ such that all edges of $K$ are straight in $\mathscr{M}^{\prime}$.

In order to prove Theorem 1.2, they showed a stronger result as the following.

Theorem 1.3** (Dvořák and Postle [5]).**

Let $G$ be a plane graph without cycles of length $4$ to $8$. Let $S$ be a set of vertices of $G$ such that $|S|\leq 1$ or $S$ consists of all vertices on a face of $G$. Let $\mathscr{M}$ be a $3$-matching assignment for $G$ such that $\mathscr{M}$ is consistent on every closed walk of length three in $G$. If $|S|\leq 12$, then every $\mathscr{M}$-coloring $\phi$ of $G[S]$ can be extended to an $\mathscr{M}$-coloring $\varphi$ of $G$.

Two cycles are adjacent if they have at least one common edge. An $\ell_{1}$-cycle and an $\ell_{2}$-cycle are normally adjacent if they form an $(\ell_{1}+\ell_{2}-2)$-cycle with exactly one chord. In other words, two cycles are normally adjacent if their intersection is $K_{2}$. Recently, Liu and Li [13] improved Theorem 1.3 to the following result by allowing cycles of length $4$ to $8$ but forbidding adjacent cycles of length at most $8$.

Theorem 1.4** (Liu and Li [13]).**

Let $G$ be a plane graph without adjacent cycles of length at most $8$. Let $S$ be a set of vertices of $G$ such that $|S|\leq 1$ or $S$ consists of all vertices on a face of $G$. Let $\mathscr{M}$ be a $3$-matching assignment for $G$ such that $\mathscr{M}$ is consistent on every closed walk of length three in $G$. If $|S|\leq 12$, then every $\mathscr{M}$-coloring $\phi$ of $G[S]$ can be extended to an $\mathscr{M}$-coloring $\varphi$ of $G$.

This implies the $3$-choosability of planar graphs without adjacent cycles of length at most $8$.

Theorem 1.5** (Liu and Li [13]).**

Every planar graph without adjacent cycles of length at most $8$ is $3$-choosable.

The first goal of this paper is to further improve Theorem 1.4 to the following result by allowing adjacent cycles of length $6$ to $8$ and changing the condition on precolored vertices from faces to cycles. But before we state the main theorem, it’s necessary to give a new concept. A cycle is abnormal if it is the $11$- or $12$-cycle in a subgraph isomorphic to a configuration in Fig. 1. A cycle is normal if it is not an abnormal cycle. A $d$-vertex, $d^{+}$-vertex or $d^{-}$-vertex is a vertex of degree $d$, at least $d$, or at most $d$ respectively. Similar definitions can be applied to faces and cycles.

Let $\mathscr{G}$ be the class of plane graphs without triangles normally adjacent to $8^{-}$-cycles, without $4$-cycles normally adjacent to $6^{-}$-cycles, and without normally adjacent $5$-cycles.

Theorem 1.6**.**

Let $G$ be a graph in $\mathscr{G}$. Let $S$ be a set of vertices of $G$ such that $|S|\leq 1$ or $S$ consists of all vertices on a normal cycle of $G$. Let $\mathscr{M}$ be a $3$-matching assignment for $G$ such that $\mathscr{M}$ is consistent on every closed walk of length three in $G$. If $|S|\leq 12$, then every $\mathscr{M}$-coloring $\phi$ of $G[S]$ can be extended to an $\mathscr{M}$-coloring of $G$.

Remark 1**.**

The graphs in Fig. 1 are in the class $\mathscr{G}$. It is observed that not every $\mathscr{M}$-coloring of the $11$- or $12$-cycle can be extended to the whole graph. Thus, we require the condition that $S$ consists of all vertices on a “normal” cycle. On the other hand, we find all the non-extendable $12^{-}$-cycles.

1. Observation:

Every $10^{-}$-cycle is normal. For a normal $12^{-}$-cycle $\mathcal{O}$, every vertex not on $\mathcal{O}$ has at most two neighbors on $\mathcal{O}$. Every edge on an abnormal $12^{-}$-cycle is contained in a $4$-, $5$-, $6$- or $7$-cycle.

The following result is a direct consequence of Theorem 1.6, and it extends Theorem 1.5.

Theorem 1.7**.**

Every graph in $\mathscr{G}$ is $3$-choosable.

The following three results are immediate consequences of Theorem 1.7. The first one generalizes the $3$-colorability of such graphs described by Luo, Chen and Wang [19], and the second one generalizes the $3$-colorability of such graphs described by Wang and Chen [22].

Corollary 1.8**.**

Every planar graph without $4$-, $6$-, $8$-cycles is $3$-choosable.

Corollary 1.9**.**

Every planar graph without $4$-, $5$-, $7$-, $8$-cycles is $3$-choosable.

Remark 2**.**

Theorem 1.2, Theorem 1.5 and Theorem 1.7 are only for $3$-choosable, but not for DP-$3$-colorable. Since we require the “consistency” on every closed walk of length three, the current arguments cannot guarantee the DP-3-colorability of the graphs in Theorem 1.3, Theorem 1.4 and Theorem 1.6. So it is interesting to know whether such graphs are DP-$3$-colorable.

Note that we only require the “consistency” on every closed walk of length three, if triangles are forbidden in a graph, then we can obtain the following results on DP-3-coloring.

Corollary 1.10** (Liu et al. [16]).**

Every planar graph without $3$-, $5$-, $6$-cycles is DP-3-colorable.

Corollary 1.11** (Liu et al. [16]).**

Every planar graph without $3$-, $6$-, $7$-, $8$-cycles is DP-3-colorable.

There are some other sufficient conditions for planar graphs to be DP-$3$-colorable which extend the $3$-choosability of such graphs. We refer the reader to [15, 16, 24]. DP-$4$-colorable planar or toroidal graphs can be found in [14, 12, 11, 2]. Thomassen [20] showed that every planar graph is $5$-choosable. Dvořák and Postle [5] observed that every planar graph is DP-$5$-colorable. Recently, Li and Wang [10] extended these results to $K_{5}$-minor-free graphs.

In addition, we prove a result on the vertex partition using almost the same technique as that in Theorem 1.6. An IF-coloring of a graph $G$ is a mapping $\phi:V(G)\rightarrow\{I,F\}$ such that the subgraph induced by all the vertices $\phi^{\mbox{-}1}(I)$ is an independent set, and the subgraph induced by all the vertices $\phi^{\mbox{-}1}(F)$ is a forest. In other words, an IF-coloring of a graph $G$ is a partition of $V(G)$ into two parts $I$ and $F$, such that $G[I]$ is an independent set and $G[F]$ is a forest.

Kawarabayashi and Thomassen [8] proved that every planar graph with girth at least five has an IF-coloring. Wang and Chen [23] showed that every planar graph without $4$-, $6$- and $8$-cycles is $3$-colorable. Very recently, Liu and Yu [17] proved that every planar graph without $4$-, $6$- and $8$-cycles has an IF-coloring. We prove that every graph in $\mathscr{G}$ has an IF-coloring.

Theorem 1.12**.**

The vertex set of every graph in $\mathscr{G}$ can be partitioned into an independent set and a set that induces a forest.

This paper is organized as follows. In the remainder of this section, we introduce some notations and results utilized in the proof of Theorem 1.6. In section 2, we present a proof of Theorem 1.6. In section 3, we prove a slightly stronger result than Theorem 1.12. Finally, we conclude with some open questions in section 4.

Let $G$ be a plane graph. The edges and vertices divide the plane into a number of faces. The unbounded face is called the outer face, and the other faces are called inner faces. An internal vertex is a vertex that is not incident with the outer face. An internal face is a face having no common vertices with the outer cycle. Let $\mathcal{O}$ be a cycle of a plane graph $G$, the cycle $\mathcal{O}$ divides the plane into two regions, the subgraph induced by all the vertices in the unbounded region is denoted by $\mathrm{ext}(\mathcal{O})$, and the subgraph induced by all the vertices in the other region is denoted by $\mathrm{int}(\mathcal{O})$. If both $\mathrm{int}(\mathcal{O})$ and $\mathrm{ext}(\mathcal{O})$ contain at least one vertex, then we call the cycle $\mathcal{O}$ a separating cycle of $G$. The subgraph obtained from $G$ by deleting all the vertices in $\mathrm{ext}(\mathcal{O})$ is denoted by $\mathrm{Int}(\mathcal{O})$, and the subgraph obtained from $G$ by deleting all the vertices in $\mathrm{int}(\mathcal{O})$ is denoted by $\mathrm{Ext}(\mathcal{O})$. Let $\mathcal{N}$ be the set of inner faces having at least one common vertex with the outer face.

We need the following two special covers of cycles.

• •

The circular ladder $\Gamma_{n}$ is the Cartesian product of the cycle $C_{n}$ and an independent set with two vertices.

• •

The Möbius ladder $M_{n}$ is the graph with vertex set \big{\{}\,(i,j)\mid i\in[n],j\in[2]\,\big{\}}, in which two vertices $(i,j)$ and $(i^{\prime},j^{\prime})$ are adjacent if and only if either

1. —

$i^{\prime}=i+1$ and $j=j^{\prime}$ for $1\leq i\leq n-1$, or 2. —

$i=n$, $i^{\prime}=1$ and $j\neq j^{\prime}$.

The following result can be derived from a theorem in [18, 9].

Theorem 1.13**.**

Let $C$ be a cycle, and let $H$ be a cover with a $2$-list assignment. If the cover is not a circular ladder or a Möbius ladder, then $H$ has an $\mathscr{M}$-coloring.

2 Proof of Theorem 1.6

In this section, we prove the following main result. See 1.6

• Proof.

Suppose that $G$ is a minimal counterexample to Theorem 1.6. That is, there exists an $\mathscr{M}$-coloring of $G[S]$ that cannot be extended to an $\mathscr{M}$-coloring of $G$ such that

[TABLE]

Subject to (1),

[TABLE]

By the condition (2), each edge that is not in a triangle is full in the matching assignment $\mathscr{M}$. By the structure of $G$, we immediately have the following result on $8^{-}$-cycles.

Lemma 2.1**.**

Every $8^{-}$-cycle has no chords.

Next, we give some structural results on $G$. Some of the lemmas are almost the same as the ones in [5] and [13], but for completeness we give detailed proofs here.

Lemma 2.2**.**

**

1. (a)

$S\neq V(G)$. 2. (b)

$G$* is $2$-connected, and the boundary of every face is a cycle.* 3. (c)

Each vertex not in $S$ has degree at least three. 4. (d)

Either $|S|=1$ or $G[S]$ is an induced cycle of $G$. 5. (e)

There are no separating normal $k$-cycles for $3\leq k\leq 12$. Thus, every edge on an abnormal cycle is incident with a $4$-, $5$-, $6$- or $7$-face. 6. (f)

$G[S]$* is an induced cycle of $G$. For convenience, we can redraw the graph $G$ such that $G[S]$ is the outer cycle $\mathcal{C}$ of $G$. Let $D$ be the outer face which is bounded by $G[S]$.* 7. (g)

Every $5$-face ($\neq$* outer face) is incident with at most one $2$-vertex.*

• Proof.

(a) Suppose to the contrary that $S=V(G)$. Every $\mathscr{M}$-coloring of $G[S]$ is an $\mathscr{M}$-coloring of $G$, a contradiction.

(b) By the condition (1), $G$ is connected. Suppose to the contrary that $G$ has a cut-vertex $w$. We may assume that $G=G_{1}\cup G_{2}$ and $G_{1}\cap G_{2}=\{w\}$. By the assumption of the set $S$, either $S\subseteq V(G_{1})$ or $S\subseteq V(G_{2})$. We may assume that $S\subseteq V(G_{1})$. By the condition (1), the $\mathscr{M}$-coloring $\phi$ of $G[S]$ can be extended to an $\mathscr{M}$-coloring $\phi_{1}$ of $G_{1}$, and $\phi_{1}(w)$ can be extended to an $\mathscr{M}$-coloring $\phi_{2}$ of $G_{2}$. These two colorings $\phi_{1}$ and $\phi_{2}$ together give an $\mathscr{M}$-coloring of $G$ whose restriction on $G[S]$ is $\phi$, a contradiction.

(c) Suppose that there exists a vertex $w$ not in $S$ having degree two. By the condition (1), the $\mathscr{M}$-coloring of $G[S]$ can be extended to an $\mathscr{M}$-coloring of $G-w$. Since $w$ has degree two, there are at most two forbidden colors for $w$, thus we can extend the $\mathscr{M}$-coloring of $G-w$ to an $\mathscr{M}$-coloring of $G$, a contradiction.

(d) If $S=\emptyset$, then we put any vertex into $S$ to make $|S|=1$. Suppose that $S=V(\mathcal{Q})$ and $\mathcal{Q}$ is a cycle with a chord $uv$. It is observed that the $\mathscr{M}$-coloring of $G[S]$ is also an $\mathscr{M}$-coloring of the induced subgraph in $G-uv$. By the condition (1), the $\mathscr{M}$-coloring $\phi$ of $G[S]$ can be extended to an $\mathscr{M}$-coloring of $G-uv$, and hence it is also an $\mathscr{M}$-coloring of $G$, a contradiction.

(e) We first show that $G[S]$ cannot be a separating cycle. Suppose to the contrary that $G[S]$ is a separating (normal) cycle $\mathcal{O}$. By the condition (1), the $\mathscr{M}$-coloring $\phi$ of $\mathcal{O}$ can be extended to an $\mathscr{M}$-coloring $\phi_{1}$ of $\mathrm{Int}(\mathcal{O})$, and another $\mathscr{M}$-coloring $\phi_{2}$ of $\mathrm{Ext}(\mathcal{O})$. These two colorings $\phi_{1}$ and $\phi_{2}$ together give an $\mathscr{M}$-coloring of $G$ whose restriction on $G[S]$ is $\phi$, a contradiction.

Thus, either $|S|=1$ or $S$ consists of all vertices on a face of $G$. Let $\mathcal{Q}$ be a separating normal $k$-cycle with $3\leq k\leq 12$. Thus, we may assume that $S\subseteq\mathrm{Ext}(\mathcal{Q})$. By the condition (1), the $\mathscr{M}$-coloring $\phi$ of $G[S]$ can be extended to an $\mathscr{M}$-coloring $\varphi_{1}$ of $\mathrm{Ext}(\mathcal{Q})$. Similarly, the restriction of $\varphi_{1}$ on $Q$ can be extended to an $\mathscr{M}$-coloring $\varphi_{2}$ of $\mathrm{Int}(\mathcal{Q})$. These two colorings $\varphi_{1}$ and $\varphi_{2}$ together give an $\mathscr{M}$-coloring of $G$ whose restriction on $G[S]$ is $\phi$, a contradiction.

(f) According to (d), suppose to the contrary that $S=\{w\}$. We first assume that $w$ is on a $10^{-}$-cycle $\mathcal{Q}$. Without loss of generality, we may assume that $\mathcal{Q}$ is a shortest cycle containing $w$. Then $\mathcal{Q}$ is an induced cycle. By (e), we may assume that $\mathrm{ext}(\mathcal{Q})=\emptyset$ and $\mathcal{Q}$ is the outer cycle. By (c) and $\mathcal{Q}$ is an induced cycle, every vertex on $\mathcal{Q}$ other than $w$ has a neighbor in $\mathrm{int}(\mathcal{Q})$, which implies that $\mathrm{int}(\mathcal{Q})\neq\emptyset$. By the condition (1), the $\mathscr{M}$-coloring $\phi$ of $\{w\}$ can be extended to an $\mathscr{M}$-coloring $\phi_{1}$ of $\mathcal{Q}$. By the condition (1), the $\mathscr{M}$-coloring $\phi_{1}$ of $\mathcal{Q}$ can be further extended to an $\mathscr{M}$-coloring of $G$, a contradiction.

So we may assume that every cycle containing $w$ has length at least $11$. Let $w$ be incident with a face $[w_{1}ww_{2}\dots w_{1}]$. Let $G^{\prime}$ be obtained from $G$ by adding a chord $w_{1}w_{2}$ in the face, let $S^{\prime}=\{w,w_{1},w_{2}\}$ and let the $3$-matching assignment $\mathscr{M}^{\prime}$ for $G^{\prime}$ be obtained from $\mathscr{M}$ by setting the matching corresponding to $w_{1}w_{2}$ is edgeless. We can easily check that $G^{\prime}$ is a plane graph satisfying the assumption of Theorem 1.6. By the condition (1), the $\mathscr{M}$-coloring $\phi$ of $\{w\}$ can be extended to an $\mathscr{M^{\prime}}$-coloring $\phi_{1}$ of $G^{\prime}[S^{\prime}]$. By the condition (1), the $\mathscr{M^{\prime}}$-coloring $\phi_{1}$ of $G^{\prime}[S^{\prime}]$ can be further extended to an $\mathscr{M^{\prime}}$-coloring $\varphi$ of $G^{\prime}$. It is observed that $\varphi$ is an $\mathscr{M}$-coloring of $G$, a contradiction.

(g) Note that every $2$-vertex and its two neighbors are all on the outer cycle. Suppose to the contrary that $f=[x_{1}x_{2}x_{3}x_{4}x_{5}]$ is a $5$-face which is incident with two $2$-vertices. Note that the two $2$-vertices must be adjacent on the outer cycle, say $x_{2}$ and $x_{3}$. It follows that $x_{1}$ and $x_{4}$ are on the outer cycle $\mathcal{C}$. If $x_{5}$ has three neighbors on the outer cycle $\mathcal{C}$, then $\mathcal{C}$ is abnormal (see 1(a) and 1(b)), a contradiction. Thus, $x_{5}$ has a neighbor not on the outer cycle $\mathcal{C}$, and $\mathcal{C}^{\prime}=(\mathcal{C}-\{x_{2},x_{3}\})\cup\{x_{1}x_{5},x_{4}x_{5}\}$ is a separating $11^{-}$-cycle. By (e), $\mathcal{C}^{\prime}$ is an abnormal $11$-cycle (see 1(a)). It follows that $\mathcal{C}$ is an abnormal $12$-cycle (see 1(d)), a contradiction. ∎

Lemma 2.3**.**

There are no $3$-faces adjacent to $8^{-}$-faces, no $4$-faces adjacent to $6^{-}$-faces, and no adjacent $5$-faces.

• Proof.

Recall that every face is bounded by a cycle. Assume that $f$ is an $8^{-}$-face and it is adjacent to a $5^{-}$-face $g$. By 2.1, it suffices to consider that $g$ is a $4$- or $5$-face.

Suppose that $f=[w_{1}w_{2}\dots w_{k}]$ is a $6^{-}$-face and $g=[uvw_{3}w_{2}]$ is a $4$-face. Since there are no $4$-cycles normally adjacent to $6^{-}$-cycles, we have that either $u$ or $v$ is on $f$. By symmetry, we may assume that $u$ is on $f$. Recall that every $6^{-}$-face is bounded by a cycle and this cycle has no chord, so $u=w_{1}$ and $v$ is not on $f$. It is observed that $w_{2}$ is a $2$-vertex and it must be on the outer cycle $\mathcal{C}$. It follows that either $f$ or $g$ is the outer face. If $f$ is the outer face, then $v$ is an internal vertex and it has a neighbor not on the outer cycle $\mathcal{C}$ (since $[w_{1}vw_{3}\dots w_{k}]$ has no chords by 2.1), thus there is a separating $6^{-}$-cycle $[w_{1}vw_{3}\dots w_{k}]$, a contradiction. Similarly, if $g$ is the outer face, then there is an internal vertex on $f$ having a neighbor not on the outer cycle $\mathcal{C}$, thus there is a separating $6^{-}$-cycle containing $w_{1}vw_{3}$, a contradiction.

Suppose that $f$ and $g$ are $5$-faces. Since every $8^{-}$-cycle has no chord, there are only four cases (up to symmetry) for the local structures (see Fig. 2). Since there are no normally adjacent $5$-cycles, the first case will not appear. For the other three cases, we first assume that $x$ is an internal vertex. Since every internal vertex has degree at least three, $x$ has a neighbor $x^{\prime}$ other than $w_{2}$ and $y$. It is observed that $x$ is on a $6^{-}$-cycle $\mathcal{O}_{x}$ not containing $w_{3}$. If $x^{\prime}$ is on $\mathcal{O}_{x}$, then $xx^{\prime}$ is a chord of $\mathcal{O}_{x}$, but this contradicts 2.1; if $x^{\prime}$ is not on $\mathcal{O}_{x}$, then $\mathcal{O}_{x}$ is a separating $6^{-}$-cycle, this contradicts 2.2(e). So we may assume that $x$ is on the outer cycle. In the second and third cases, $w_{3}$ is a $2$-vertex, so it is on the outer cycle, and $g$ must be the outer face. In the fourth case, by the symmetry of $x$ and $z$, we have that $z$ is on the outer cycle, and $g$ is the outer face. Therefore, $g$ is the outer face in the last three cases, and there is an internal vertex on $f$ having a neighbor not on $g$, and then there is a separating $6^{-}$-cycle containing $w_{2}xy$, a contradiction. ∎

Lemma 2.4**.**

If $[x_{1}x_{2}x_{3}]$ is a triangle and $x_{2},x_{3}$ are internal $3$-vertices, then all the edges in the triangle are full.

• Proof.

Suppose to the contrary that at least one of $x_{1}x_{3},x_{2}x_{3}$ and $x_{1}x_{2}$ is not full. By applying 1.1 to $\{x_{1}x_{3},x_{2}x_{3}\}$, we may assume that $x_{1}x_{3}$ and $x_{2}x_{3}$ are straight in $\mathscr{M}$. Let $\mathscr{M}^{\prime}$ be a new $3$-matching assignment for $G$ by setting $\mathscr{M}_{e}^{\prime}=\mathscr{M}_{e}$ for each $e\notin\{x_{1}x_{3},x_{2}x_{3},x_{1}x_{2}\}$ and all edges in $\{x_{1}x_{3},x_{2}x_{3},x_{1}x_{2}\}$ are straight and full. Note that $x_{1}x_{3}$ and $x_{2}x_{3}$ are straight in $\mathscr{M}$, thus $\mathscr{M}_{x_{1}x_{3}}\subseteq\mathscr{M}_{x_{1}x_{3}}^{\prime}$ and $\mathscr{M}_{x_{2}x_{3}}\subseteq\mathscr{M}_{x_{2}x_{3}}^{\prime}$. Since all the edges in $\{x_{1}x_{3},x_{2}x_{3},x_{1}x_{2}\}$ are full in $\mathscr{M}^{\prime}$ but not in $\mathscr{M}$, the number of edges in $\mathscr{M}^{\prime}$ is greater than that in $\mathscr{M}$. Since there are no adjacent triangles, every closed walk of length three is consistent in $\mathscr{M}^{\prime}$. By the condition (2), the $\mathscr{M}$-coloring $\phi$ (also $\mathscr{M}^{\prime}$-coloring) of the outer cycle $\mathcal{C}$ can be extended to an $\mathscr{M}^{\prime}$-coloring $\phi^{\prime}$ of $G$, but $\phi^{\prime}$ is not an $\mathscr{M}$-coloring of $G$ by our assumption. Note that $\mathscr{M}_{e}\subseteq\mathscr{M}_{e}^{\prime}$ for any $e\neq x_{1}x_{2}$, so we may assume that $\phi^{\prime}(x_{1})=1$, $\phi^{\prime}(x_{2})=2$ and $(x_{1},1)(x_{2},2)\in\mathscr{M}_{x_{1}x_{2}}$. If $(x_{1},1)$ has an incident edge in $\mathscr{M}_{x_{1}x_{3}}$ and $(x_{2},2)$ has an incident edge in $\mathscr{M}_{x_{2}x_{3}}$, then the closed walk $x_{3}x_{1}x_{2}$ is not consistent in $\mathscr{M}$, a contradiction. If $(x_{1},1)$ has no incident edge in $\mathscr{M}_{x_{1}x_{3}}$, then we can modify $\phi^{\prime}$ to obtain an $\mathscr{M}$-coloring of $G$ by recoloring $x_{2}$ and $x_{3}$ in order, a contradiction. So we may assume that $(x_{1},1)$ has an incident edge in $\mathscr{M}_{x_{1}x_{3}}$ and $(x_{2},2)$ has no incident edge in $\mathscr{M}_{x_{2}x_{3}}$. Since $x_{1}x_{3}$ is straight in $\mathscr{M}$, we have that $(x_{1},1)(x_{3},1)\in\mathscr{M}_{x_{1}x_{3}}$. Furthermore, we may assume that $(x_{3},1)$ has no incident edge in $\mathscr{M}_{x_{2}x_{3}}$, otherwise the closed walk $x_{2}x_{3}x_{1}$ is not consistent in $\mathscr{M}$. Now, we can obtain a new $3$-matching assignment $\mathscr{M}^{*}$ for $G$ by adding an edge $(x_{3},1)(x_{2},2)$ to $\mathscr{M}$. By the adjacency of cycles, $x_{2}x_{3}$ is only contained in an unique triangle $x_{1}x_{2}x_{3}$, so the addition of $(x_{3},1)(x_{2},2)$ does not make $\mathscr{M}^{*}$ inconsistent on closed $3$-walk, but this contradicts the condition (2). ∎

Lemma 2.5**.**

Let $w_{0},w_{1},w_{2},w_{3}$ and $w_{4}$ be five consecutive vertices on a $5^{+}$-face. If $w_{1},w_{2},w_{3}$ and $w_{4}$ are all $3$-vertices, and $w_{1}w_{2}$ is incident with a $3$-face $ww_{1}w_{2}$, then at least one vertex in $\{w_{1},w_{2},w_{3},w_{4}\}$ is on the outer cycle $\mathcal{C}$.

• Proof.

Suppose to the contrary that none of $\{w_{1},w_{2},w_{3},w_{4}\}$ is on the outer cycle $\mathcal{C}$. Let $w^{\prime}$ be the neighbor of $w_{3}$ other than $w_{2},w_{4}$, and let $G^{*}=G-\{w_{1},w_{2},w_{3},w_{4}\}$. It is observed that $w_{0},w_{1},w_{2},w_{3},w_{4},w$ and $w^{\prime}$ are seven distinct vertices. We claim that the distance between $w_{0}$ and $w^{\prime}$ is at least nine in $G^{*}$. Let $P$ be a shortest path between $w_{0}$ and $w^{\prime}$ in $G^{*}$. It is observed that $\mathcal{Q}=P\cup w_{0}w_{1}w_{2}w_{3}w^{\prime}$ is a cycle. If $w$ is on the path $P$, then $P[w_{0},w]\cup w_{0}w_{1}w$ and $P[w,w^{\prime}]\cup ww_{2}w_{3}w^{\prime}$ are all cycles (see Fig. 3), which implies that these two cycles have length at least nine and $|E(P)|\geq(9-2)+(9-3)=13$. If $w$ is not on the path $P$, then $\mathcal{Q}$ is a separating normal cycle (note that $w$ and $w_{4}$ are in different sides of the cycle $\mathcal{Q}$) and it has length at least $13$, which implies that $|E(P)|=|\mathcal{Q}|-4\geq 13-4=9$. Therefore, the distance between $w_{0}$ and $w^{\prime}$ is at least nine in $G^{*}$.

By 2.4 and 1.1, we may assume that all the edges incident with the vertices in $\{w_{1},w_{2},w_{3}\}$ are straight. Let $G^{\prime}$ be the graph obtained from $G^{*}$ by identifying $w_{0}$ and $w^{\prime}$, and let $\mathscr{M^{\prime}}$ be the restriction of $\mathscr{M}$ on $E(G^{\prime})$. Since the distance between $w_{0}$ and $w^{\prime}$ is at least nine in $G^{*}$, the graph $G^{\prime}$ has no loops, no multiple edges and no new $8^{-}$-cycles, thus $G^{\prime}$ is a simple plane graph satisfying the assumption of Theorem 1.6, and $\mathscr{M}^{\prime}$ is consistent on every closed walk of length three in $G^{\prime}$. Moreover, $\mathcal{C}$ is also a normal cycle of $G^{\prime}$ and it has no chords in $G^{\prime}$. This implies that $\phi$ is an $\mathscr{M}^{\prime}$-coloring of $G^{\prime}[S]$. Since $|V(G^{\prime})|<|V(G)|$, the $\mathscr{M}^{\prime}$-coloring $\phi$ of $G^{\prime}[S]$ can be extended to an $\mathscr{M}^{\prime}$-coloring $\varphi$ of $G^{\prime}$. Since $w_{3}$ and $w_{4}$ are all $3$-vertices, we can extend $\varphi$ to $w_{4}$ and $w_{3}$ in order. Recall that all the edges incident with vertices in $\{w_{1},w_{2},w_{3}\}$ are straight, thus we may assume that $w_{0}$ and $w_{3}$ have distinct colors, and then we can further extend the coloring to $w_{2}$ and $w_{1}$, a contradiction. ∎

Let $w$ be a vertex on the outer cycle $\mathcal{C}$, and let $w_{1},w_{2},\dots,w_{k}$ be consecutive neighbors in a cyclic order. If $f$ is a face in $\mathcal{N}$ incident with $ww_{i}$ and $ww_{i+1}$, but neither $ww_{i}$ nor $ww_{i+1}$ is an edge of $\mathcal{C}$, then we call $f$ a special face (at $w$). A $4$-face is a $4^{*}$-face if it has three common vertices with $\mathcal{C}$. An internal $3$-vertex is bad if it is incident with a $3$-face which is not special, light if it is incident with an internal $4$-face or a $4^{*}$-face or a special $3$-face, good if it is neither bad nor light. According to 2.5, we have the following result on bad vertices.

Lemma 2.6**.**

There are no five consecutive bad vertices on the boundary of a $5^{+}$-face.

Lemma 2.7**.**

If a $4$-face in $\mathcal{N}$ has exactly two common vertices with $\mathcal{C}$, then these two vertices are consecutive on the $4$-face.

• Proof.

Suppose that $f$ is a $4$-face in $\mathcal{N}$ that has exactly two common vertices with $\mathcal{C}$. If these two vertices are not consecutive on the $4$-face, then there exists a separating normal $8^{-}$-cycle, a contradiction. ∎

Assume that $f=[v_{1}v_{2}\dots v_{l}]$ is an internal $(3,3,3,3)$-face or an internal $(3,3,3,3,3)$-face. Let $u_{i}$ be the third neighbor of $v_{i}$ for $1\leq i\leq l$. Note that every $8^{-}$-cycle has no chords, and there are no separating $4$- or $5$-cycles. It is observed that $\{u_{i},v_{i}\mid 1\leq i\leq l\}$ contains $2l$ distinct vertices. Let $\Gamma$ be the graph $G\setminus(V(f)\setminus\{v_{3}\})$, and let $G^{*}$ be the graph obtained from $\Gamma$ by identifying $u_{1}$ and $v_{3}$ into a new vertex $z$.

Lemma 2.8**.**

The graph $G^{*}$ is in the class $\mathscr{G}$, and $\mathcal{C}$ is an induced cycle of $G^{*}$.

• Proof.

Let $P$ be an arbitrary path between $u_{1}$ and $v_{3}$ in $\Gamma$. It is observed that $\mathcal{Q}=P\cup u_{1}v_{1}v_{2}v_{3}$ is a cycle of length $|E(P)|+3$.

(i) Assume that $u_{2}$ is not on the path $P$. It is easy to check that $|E(P)|\leq 8$ only if $|E(P)|=8$ and $\mathcal{Q}$ is a separating abnormal $11$-cycle (note that $u_{2}$ and $v_{l}$ are in different sides of the cycle $\mathcal{Q}$). Note that there are no $3$-cycles normally adjacent to abnormal cycles. Even though the identification may make the path $P$ becomes an $8$-cycle, but the new cycle is not normally adjacent to any $3$-cycle in $G^{*}$.

(ii) Assume that $u_{2}$ is on the path $P$. Then $Q_{1}=P[u_{1},u_{2}]\cup u_{1}v_{1}v_{2}u_{2}$ and $Q_{2}=P[u_{2},v_{3}]\cup u_{2}v_{2}v_{3}$ are all cycles, which implies that each of these two cycles has length at least six and $|E(P)|\geq(6-2)+(6-3)=7$. It is easy to check that $|E(P)|=7$ only if $|Q_{1}|=|Q_{2}|=6$, and $|E(P)|=8$ only if $|Q_{i}|=6$ and $|Q_{3-i}|=7$. Recall that there are no $3$-cycles normally adjacent to $8^{-}$-cycles, and there are no $4$-cycles normally adjacent to $6^{-}$-cycles. Therefore, even though the identification may make the path $P$ become a $7$- or $8$-cycle, but the new cycle is not normally adjacent to any $3$-cycle in $G^{*}$.

According to the above arguments on the two cases, $G^{*}$ satisfies the requirement for adjacency in Theorem 1.6, and the distance of $u_{1}$ and $v_{3}$ in $\Gamma$ is at least seven. Note that $v_{3}$ is an internal vertex, $\mathcal{C}$ is still a $12^{-}$-cycle in $G^{*}$. It is easy to check that $\mathcal{C}$ is also a normal $12^{-}$-cycle in $G^{*}$, for otherwise, two new $7$-cycles are created to make $\mathcal{C}$ abnormal in $G^{*}$, but there is a separating normal $10$-cycle in $G$, a contradiction.

Suppose that the identification creates a chord for $\mathcal{C}$. Since $v_{3}$ is an internal vertex, the vertex $u_{1}$ is on the outer cycle $\mathcal{C}$. Note that $u_{3}$ is also on the outer cycle, the two vertices $u_{1}$ and $u_{3}$ divide the cycle $\mathcal{C}$ into two paths $P^{\prime}$ and $P^{\prime\prime}$. Since the distance of $u_{1}$ and $v_{3}$ in $\Gamma$ is at least seven, we have that $|E(P^{\prime})|=|E(P^{\prime\prime})|=6$. Thus, $P^{\prime}\cup u_{1}v_{1}v_{2}v_{3}u_{3}$ or $P^{\prime\prime}\cup u_{1}v_{1}v_{2}v_{3}u_{3}$ is a separating normal $10$-cycle in $G$, a contradiction. Hence, the identification does not create chords of $\mathcal{C}$. ∎

Lemma 2.9**.**

There are no internal $(3,3,3,3)$-faces or $(3,3,3,3,3)$-faces.

• Proof.

Suppose to the contrary that $f=[v_{1}v_{2}\dots v_{l}]$ is an internal $(3,3,3,3)$-face or an internal $(3,3,3,3,3)$-face. All the related vertices are labeled as the above. Noting that none of $u_{1}v_{1},v_{1}v_{2},v_{2}v_{3},v_{3}u_{3}$ is in a triangle, it follows that these four edges are full. So we may assume that each of these four edges is straight. Let $H^{*}$ be the cover obtained from $H-\cup L_{v}$ by identifying $(u_{1},j)$ and $(v_{3},j)$ into a new vertex $(z,j)$, where $1\leq j\leq 3$, and the union is taken over $v\in V(f)\setminus\{v_{3}\}$. By the minimality of $G$, the precoloring $\phi$ can be extended to a coloring $T^{*}$ of the cover $H^{*}$. Without loss of generality, we may assume that $(z,1)\in T$. This coloring naturally gives a coloring $T$ of $H-\cup_{x\in V(f)}L_{x}$. For each $x\in V(f)$, let L^{\prime}_{x}=L_{x}\setminus\{(x,j_{x})\mid\text{(x,j_{x})$is the neighbor of a vertex in$T}\}. Let $H^{\prime}$ be the subgraph of $H$ induced by $\cup_{x\in V(f)}L^{\prime}_{x}$. Note that $(v_{1},1)\notin L^{\prime}_{v_{1}}$ and $(v_{3},1)\in L^{\prime}_{v_{3}}$. Recall that $v_{1}v_{2}$ and $v_{2}v_{3}$ are full and straight, thus $H^{\prime}$ is neither a circular ladder nor a Möbius ladder. By Theorem 1.13, the cover $H^{\prime}$ has a coloring $T^{\prime}$. Therefore, $T\cup T^{\prime}$ is a coloring of the cover $H$, a contradiction. ∎

We give the initial charge $\mu(v)=\deg(v)-4$ for any $v\in V(G)$, $\mu(f)=\deg(f)-4$ for any face $f\in F(G)$ other than outer face $D$, and $\mu(D)=\deg(D)+4$ for the outer face $D$. By the Euler formula, the sum of the initial charges is zero. That is,

[TABLE]

Next, we give the discharging rules to redistribute the charges, preserving the sum, such that the final charge of every element in $V(G)\cup F(G)$ is nonnegative, and at least one element in $V(G)\cup F(G)$ has positive final charge. This leads a contradiction to complete the proof.

A small face is a $5^{-}$-face. We say that a face is a negative face if it is a non-special $3$-face, or an internal $(3,3,3,3,4)$-face, or an internal $4$-face.

The followings are the discharging rules.

1. R1.

Each non-special $3$-face receives $\frac{1}{3}$ from each incident internal vertex. 2. R2.

Let $w$ be an internal $3$-vertex.

1. (a)

If $w$ is a bad vertex, then it receives $\frac{2}{3}$ from each incident $9^{+}$-face. 2. (b)

If $w$ is incident with a special $3$-face or a $4^{*}$-face, then it receives $\frac{1}{2}$ from each incident $7^{+}$-face. 3. (c)

If $w$ is incident with an internal $4$-face, then it receives $\frac{1}{9}$ from the incident $4$-face, and $\frac{4}{9}$ from each incident $7^{+}$-face. 4. (d)

If $w$ is a good vertex, then it receives $\frac{1}{3}$ from each incident face. 3. R3.

Let $w$ be an internal $4$-vertex. Then $w$ sends $\frac{1}{3}$ to each incident internal $(3,3,3,3,4)$-face and internal $4$-face.

1. (a)

If $w$ is incident with two negative faces, then it receives $\frac{1}{3}$ from each of the other two incident faces. 2. (b)

If $w$ is incident with exactly one negative face $f$, and the opposite face at $w$ is a $9^{+}$-face, then $w$ receives $\frac{1}{3}$ from the opposite face. 3. (c)

If $w$ is incident with exactly one negative face $f$, and the opposite face at $w$ is an $8^{-}$-face, then $w$ receives $\frac{1}{6}$ from each face at $w$ that is not the opposite face. 4. R4.

Each internal $5^{+}$-vertex sends $\frac{1}{3}$ to each incident $5$-face. 5. R5.

Each $2$-vertex on the outer cycle $\mathcal{C}$ receives $\frac{2}{3}$ from the incident face in $\mathcal{N}$ and $\frac{4}{3}$ from the outer face. 6. R6.

Each $3$-vertex on the outer cycle $\mathcal{C}$ receives $\frac{4}{3}$ from the outer face, and sends $\frac{1}{3}$ to each incident $4^{-}$-face in $\mathcal{N}$ and $\frac{1}{6}$ to each incident $5$- or $6$-face in $\mathcal{N}$. 7. R7.

Each $4$-vertex on the outer cycle $\mathcal{C}$ receives $1$ from the outer face, and sends $1$ to each incident special $4^{-}$-face, and $\frac{1}{3}$ to each of the other incident $6^{-}$-face in $\mathcal{N}$. 8. R8.

Each $5^{+}$-vertex on the outer cycle $\mathcal{C}$ receives $1$ from the outer face, and sends $1$ to each incident special $4^{-}$-face, and $\frac{1}{2}$ to each of the other incident $6^{-}$-face in $\mathcal{N}$.

Lemma 2.10**.**

Every face other than $D$ has nonnegative final charge.

• Proof.

According to the discharging rules, inner $3$-faces never give charges. If $f$ is a special $3$-face, then $\mu^{\prime}(f)=3-4+1=0$ by R7 and R8. If $f$ is a non-special $3$-face having no vertex on the outer cycle $\mathcal{C}$, then $\mu^{\prime}(f)=3-4+3\times\frac{1}{3}=0$ by R1. If $f$ is a non-special $3$-face having two vertices on the outer cycle $\mathcal{C}$, then $f$ has a common edge with the outer face by 2.2(f), and then $\mu^{\prime}(f)\geq 3-4+3\times\frac{1}{3}=0$ by R1, R6, R7 and R8. Note that no inner $3$-faces have three common vertices with $\mathcal{C}$.

Let $f$ be a $4$-face. If $f$ is an internal face, then it is incident with at least one $4^{+}$-vertex by 2.9, and then $\mu^{\prime}(f)\geq 4-4+\frac{1}{3}-3\times\frac{1}{9}=0$ by R2c and R3. So we may assume that $f$ is a face in $\mathcal{N}$. If $f$ has exactly one common vertex with $\mathcal{C}$, then it receives $1$ from the vertex on the outer cycle $\mathcal{C}$, and then $\mu^{\prime}(f)\geq 4-4+1-3\times\frac{1}{3}=0$ by R2d, R7 and R8. If $f$ has exactly two common vertices with the outer cycle $\mathcal{C}$, then these two vertices are consecutive on $f$ by 2.7, and then $\mu^{\prime}(f)\geq 4-4+2\times\frac{1}{3}-2\times\frac{1}{3}=0$ by R2d, R6, R7 and R8. If $f$ has exactly three common vertices with the outer cycle $\mathcal{C}$, i.e., $f$ is a $4^{*}$-face, then $\mu^{\prime}(f)\geq 4-4+2\times\frac{1}{3}-\frac{2}{3}=0$ by R5, R7 and R8.

Let $f$ be a $5$-face. By 2.9, $f$ cannot be incident with five internal $3$-vertices. If $f$ is an internal $(3,3,3,3,4^{+})$-face, then $f$ receives $\frac{1}{3}$ from the incident $4^{+}$-vertex and sends $\frac{1}{3}$ to each incident $3$-vertex, thus $\mu^{\prime}(f)=5-4+\frac{1}{3}-4\times\frac{1}{3}=0$ by R2d, R3 and R4. If $f$ is an internal face incident with at least two $4^{+}$-vertices, then $\mu^{\prime}(f)\geq 5-4-3\times\frac{1}{3}=0$ by R2d. So we may assume that $f$ is a face in $\mathcal{N}$. If $f$ has exactly one common vertex with the outer cycle $\mathcal{C}$, then it receives at least $\frac{1}{3}$ from the vertex on the outer cycle $\mathcal{C}$, and then $\mu^{\prime}(f)\geq 5-4+\frac{1}{3}-4\times\frac{1}{3}=0$ by R2d, R7 and R8. If $f$ has at least two common vertices with the outer cycle $\mathcal{C}$, and it is not incident with any $2$-vertex, then it receives at least $\frac{1}{6}$ from each incident vertex on the outer cycle $\mathcal{C}$, and then $\mu^{\prime}(f)\geq 5-4+2\times\frac{1}{6}-3\times\frac{1}{3}=\frac{1}{3}$ by R2d, R6, R7 and R8. If $f$ is incident with exactly one $2$-vertex, then $\mu^{\prime}(f)\geq 5-4+2\times\frac{1}{6}-\frac{2}{3}-2\times\frac{1}{3}=0$ by R2d, R6, R7 and R8. By 2.2(g), $f$ cannot be incident with more than one $2$-vertex.

Let $f$ be an internal $6$-face. Note that there are no $4^{-}$-faces adjacent to $6$-faces, it follows that $f$ sends at most $\frac{1}{3}$ to each incident vertex, thus $\mu^{\prime}(f)\geq 6-4-6\times\frac{1}{3}=0$.

Let $f=[x_{1}x_{2}\dots x_{6}]$ be a $6$-face in $\mathcal{N}$. (i) Suppose that $f$ is incident with three $2$-vertices and an internal vertex $x_{6}$. If $x_{6}$ has at least three neighbors on the outer cycle $\mathcal{C}$, then $\mathcal{C}$ is an abnormal cycle (see 1(a), 1(b) and 1(c)), a contradiction. So $x_{6}$ has a neighbor not on $\mathcal{C}$, and $\mathcal{C}^{\prime}=(\mathcal{C}-\{x_{2},x_{3},x_{4}\})\cup\{x_{1}x_{6},x_{5}x_{6}\}$ is a separating $10^{-}$-cycle, a contradiction. (ii) Suppose that $f$ is incident with one or two $2$-vertices. Thus, $f$ receives at least $\frac{1}{6}$ from each incident $3^{+}$-vertex on the outer cycle $\mathcal{C}$, and sends at most $\frac{1}{3}$ to each incident internal vertex, which implies that $\mu^{\prime}(f)\geq 6-4+2\times\frac{1}{6}-2\times\frac{2}{3}-2\times\frac{1}{3}=\frac{1}{3}$. (iii) If $f$ is not incident with any $2$-vertex, then it sends at most $\frac{1}{3}$ to each incident internal vertex, but does not send any charge to the incident vertices on the outer cycle $\mathcal{C}$, thus $\mu^{\prime}(f)\geq 6-4-5\times\frac{1}{3}>0$.

Let $f=[x_{1}x_{2}\dots x_{7}]$ be a $7$-face. Suppose that $f$ is an internal face. Since there are no $3$-faces adjacent to $7$-faces, we have that $f$ sends at most $\frac{4}{9}$ to each incident internal vertex. Note that seven is an odd number, $f$ is incident with at most six $3$-vertices which are incident with internal $4$-faces, thus $\mu^{\prime}(f)\geq 7-4-6\times\frac{4}{9}-\frac{1}{3}=0$. Now, assume that $f$ is a face in $\mathcal{N}$. If $f$ is not incident with any $2$-vertex, then $\mu^{\prime}(f)\geq 7-4-6\times\frac{1}{2}=0$. If $f$ is incident with one or two $2$-vertices, then $\mu^{\prime}(f)\geq 7-4-2\times\frac{2}{3}-3\times\frac{1}{2}>0$. Consider the case $f$ is incident with exactly three $2$-vertices. It is observed that the three $2$-vertices are consecutive on the outer cycle, say $x_{2},x_{3}$ and $x_{4}$. By the discharging rules, $f$ sends at most $\frac{1}{2}$ to each of $x_{6}$ and $x_{7}$. It follows that $\mu^{\prime}(f)\geq 7-4-3\times\frac{2}{3}-2\times\frac{1}{2}=0$. Furthermore, $\mu^{\prime}(f)=0$ only if both $x_{6}$ and $x_{7}$ are $3$-vertices, and each of which is incident with a $4^{*}$-face, which implies that $\mathcal{C}$ is abnormal (see 1(f)), a contradiction. The final case: $f$ is incident with four $2$-vertices, say $x_{2},x_{3},x_{4}$ and $x_{5}$. Then it is incident with exactly one internal vertex $x_{7}$. Note that $x_{7}$ has degree at least three. If $x_{7}$ has another neighbor on the outer cycle other than $x_{1}$ and $x_{6}$, then $\mathcal{C}$ is abnormal (see 1(b) and 1(e)), a contradiction. If $x_{7}$ has a neighbor not on the outer cycle, then $(\mathcal{C}\setminus\{x_{2},x_{3},x_{4},x_{5}\})\cup\{x_{1}x_{7},x_{6}x_{7}\}$ is a separating normal $9^{-}$-cycle, a contradiction. In summary, every $7$-face has nonnegative final charge, and each $7$-face adjacent to $D$ has positive final charge.

Let $f$ be an $8$-face. By 2.3, there are no $3$-faces adjacent to $8$-faces, so $f$ sends at most $\frac{1}{2}$ to each incident internal vertex. If $f$ is an internal face, then $\mu^{\prime}(f)\geq 8-4-8\times\frac{1}{2}=0$. So we may assume that $f$ is an $8$-face in $\mathcal{N}$. If $f$ is not incident with any $2$-vertex, then $\mu^{\prime}(f)\geq 8-4-7\times\frac{1}{2}=\frac{1}{2}$. Suppose that $f$ is incident with at least one $2$-vertex. It follows that $f$ has at least two common $3^{+}$-vertices with the outer cycle $\mathcal{C}$, and it sends nothing to these vertices. Thus, $f$ sends $\frac{2}{3}$ to each incident $2$-vertex, and sends at most $\frac{1}{2}$ to each incident internal vertex, which implies that $\mu^{\prime}(f)\geq 8-4-5\times\frac{2}{3}-\frac{1}{2}>0$.

Let $f$ be a $k$-face with $k\geq 9$. By the discharging rules, it is easy to show the following fact.

1. Fact-1.

The face $f$ sends nothing to the $3^{+}$-vertices on the outer cycle $\mathcal{C}$, and sends at most $\frac{1}{3}$ to each incident internal $4$-vertex, and sends at most $\frac{2}{3}$ to any vertex.

If $f$ is incident with some $2$-vertices, then $f$ is incident with at least two $3^{+}$-vertices on the outer cycle $\mathcal{C}$ and it sends nothing to these vertices, which implies that

[TABLE]

So we may assume that $f$ is not incident with any $2$-vertex.

Let $\alpha$ be the number of incident bad vertices, $\beta$ be the number of incident light vertices, and let $\rho$ be the number of incident internal $3$-vertices. It is observed that $\alpha+\beta\leq\rho$. By 2.5, we can easily show the following fact on the parameters $\alpha$ and $\beta$.

1. Fact-2.

If $\alpha\geq 3$, then $\alpha+\beta\leq\rho\leq k-2$.

If $\alpha\leq 3$, then $\mu^{\prime}(f)\geq k-4-3\times\frac{2}{3}-(k-3)\times\frac{1}{2}=\frac{k-9}{2}\geq 0$. If $\alpha\geq 4$ and $k\geq 10$, then $\mu^{\prime}(f)\geq k-4-(k-2)\times\frac{2}{3}-2\times\frac{1}{3}=\frac{1}{3}(k-10)\geq 0$. It remains to assume that $\alpha\geq 4$, $k=9$ and $f=[w_{1}w_{2}\dots w_{9}]$.

Suppose that $\alpha=7$. By 2.6, the two non-bad vertices divide the bad vertices on $f$ into two parts, one consisting of three consecutive bad vertices and the other consisting of four consecutive bad vertices. Without loss of generality, we may assume that none of $w_{1}$ and $w_{6}$ is a bad vertex. By 2.5, $w_{1}w_{2}$, $w_{3}w_{4}$ and $w_{5}w_{6}$ are incident with $3$-faces. By symmetry, we may further assume that $w_{6}w_{7}$ and $w_{8}w_{9}$ are incident with $3$-faces. By 2.5, $w_{1}$ cannot be an internal $3$-vertex. If $w_{1}$ is an internal $5^{+}$-vertex or on the outer cycle $\mathcal{C}$, then $f$ sends nothing to $w_{1}$. If $w_{1}$ is an internal $4$-vertex, then $w_{1}$ is incident with three $9^{+}$-faces, and it receives nothing from $f$ by R3b. Thus, $f$ sends nothing to $w_{1}$ in all cases, which implies that $\mu^{\prime}(f)\geq 9-4-7\times\frac{2}{3}-\frac{1}{3}=0$.

If $4\leq\alpha\leq 5$, then $\mu^{\prime}(f)\geq 9-4-\alpha\times\frac{2}{3}-(\rho-\alpha)\times\frac{1}{2}-(9-\rho)\times\frac{1}{3}=2-\frac{\alpha+\rho}{6}\geq 2-\frac{5+7}{6}=0$. It remains to assume that $\alpha=6$. Recall that $f$ is not incident with any $2$-vertex. If $f$ has a common vertex with the outer cycle $\mathcal{C}$, then $f$ sends nothing to this vertex and $\mu^{\prime}(f)\geq 9-4-6\times\frac{2}{3}-2\times\frac{1}{2}=0$. So we may assume that $f$ is an internal $9$-face. If $f$ is not incident with any light vertex, then $\mu^{\prime}(f)\geq 9-4-6\times\frac{2}{3}-3\times\frac{1}{3}=0$. So we may assume that there is a light vertex on $f$. By the definitions of bad vertices and light vertices, a light vertex cannot be adjacent to two bad vertices on $f$, thus a light vertex must be adjacent to a non-bad vertex on $f$, which implies that the bad vertices on $f$ are divided into two parts by 2.6. Moreover, each part has at most four consecutive bad vertices, thus each part has at least two consecutive bad vertices. Without loss of generality, we may assume that $w_{1}$ is a light vertex and $w_{9}$ is a non-bad vertex. Then $w_{2}$ and $w_{8}$ must be bad vertices. Since bad vertex and light vertex cannot be incident with the same $3$-face, $w_{2}w_{3}$ is incident with a $3$-face and $w_{1}w_{9}$ is incident with a $4^{-}$-face. By 2.5 and 2.6, $w_{3}$ is also a bad vertex and $w_{9}$ is not an internal $3$-vertex. Hence, $w_{9}$ must be an internal $4^{+}$-vertex. Since $w_{8}$ is a bad vertex, either $w_{7}w_{8}$ or $w_{8}w_{9}$ is incident with a $3$-face. If $w_{7}w_{8}$ is incident with a $3$-face, then $f$ sends nothing to $w_{9}$, which implies that $\mu^{\prime}(f)\geq 9-4-6\times\frac{2}{3}-2\times\frac{1}{2}=0$. In the other case, $w_{8}w_{9}$ is incident with a $3$-face. Then $w_{6}w_{7}$ is incident with a $3$-face. By 2.5, one vertex in $\{w_{4},w_{5}\}$ is an internal $4^{+}$-vertex, and the other is a bad $3$-vertex, thus $w_{4}w_{5}$ is incident with a $3$-face. Whenever $w_{4}$ or $w_{5}$ is an internal $4^{+}$-vertex, it receives nothing from $f$ by R3b, which implies that $\mu^{\prime}(f)\geq 9-4-6\times\frac{2}{3}-\frac{1}{2}-\frac{1}{3}>0$. ∎

Lemma 2.11**.**

Every vertex has nonnegative final charge.

• Proof.

If $v$ is a $2$-vertex, then it is on the outer cycle, and it receives $\frac{2}{3}$ from the incident face in $\mathcal{N}$ and $\frac{4}{3}$ from the outer face $D$ by R5, which implies that $\mu^{\prime}(v)=2-4+\frac{2}{3}+\frac{4}{3}=0$. Let $v$ be a $3$-vertex on the outer cycle. By the adjacency of the faces, $v$ is incident with a $4^{-}$-face and a $7^{+}$-face, or it is incident with two $5^{+}$-faces. Thus, $\mu^{\prime}(v)\geq 3-4+\frac{4}{3}-\max\left\{\frac{1}{3},2\times\frac{1}{6}\right\}=0$ by R6. If $v$ is a $4$-vertex on the outer cycle, then it receives $1$ from the outer face, sends $1$ to an incident special $4^{-}$-face and at most $\frac{1}{3}$ to each of the other incident $6^{-}$-face in $\mathcal{N}$ by R7, which implies that $\mu^{\prime}(v)\geq 4-4+1-\max\{1,3\times\frac{1}{3}\}=0$. If $v$ is a $5^{+}$-vertex on the outer cycle, then it receives $1$ from the outer face, and averagely sends at most $\frac{1}{2}$ to each incident face in $\mathcal{N}$, and then $\mu^{\prime}(v)\geq\deg(v)-4+1-(\deg(v)-1)\times\frac{1}{2}=\frac{\deg(v)-5}{2}\geq 0$.

If $v$ is a bad vertex, then $\mu^{\prime}(v)=3-4+2\times\frac{2}{3}-\frac{1}{3}=0$ by R1 and R2a. If $v$ is incident with a $4^{*}$-face or a special $3$-face, then $\mu^{\prime}(v)=3-4+2\times\frac{1}{2}=0$ by R2b. If $v$ is incident with an internal $4$-face, then $\mu^{\prime}(v)=3-4+\frac{1}{9}+2\times\frac{4}{9}=0$ by R2c. If $v$ is a good vertex, then $\mu^{\prime}(v)=3-4+3\times\frac{1}{3}=0$ by R2d. If $v$ is an internal $4$-vertex which is incident with two negative faces, then $\mu^{\prime}(v)=4-4+2\times\frac{1}{3}-2\times\frac{1}{3}=0$ by R1 and R3. If $v$ is an internal $4$-vertex which is incident with exactly one negative face and a $9^{+}$-face at the opposite side, then $\mu^{\prime}(v)=4-4+\frac{1}{3}-\frac{1}{3}=0$ by R1 and R3. If $v$ is incident with exactly one negative face and an $8^{-}$-face at the opposite side, then $\mu^{\prime}(v)=4-4+2\times\frac{1}{6}-\frac{1}{3}=0$ by R1 and R3. If $v$ is an internal $4$-vertex which is not incident with any negative face, then $\mu^{\prime}(v)=4-4=0$. Note that there are no adjacent $5^{-}$-faces, thus every vertex $v$ is incident with at most $\lfloor\frac{\deg(v)}{2}\rfloor$ small faces. If $v$ is an internal $5^{+}$-vertex, then it sends at most $\frac{1}{3}$ to each incident $5^{-}$-face, which implies that $\mu^{\prime}(v)\geq\deg(v)-4-\frac{1}{3}\times\lfloor\frac{\deg(v)}{2}\rfloor>0$. ∎

Lemma 2.12**.**

The outer face $D$ has nonnegative charge, and there exists an element in $V(G)\cup F(G)$ having positive final charge.

By R5–R8, the outer face $D$ sends $\frac{4}{3}$ to each incident $3^{-}$-vertex, and sends $1$ to each incident $4^{+}$-vertex, thus $\mu^{\prime}(D)\geq|D|+4-\frac{4}{3}|D|=\frac{1}{3}(12-|D|)\geq 0$. Therefore, every element in $V(G)\cup F(G)$ has a nonnegative final charge. In particular, $\mu^{\prime}(D)=0$ holds if and only if $|D|=12$ and each vertex incident with $D$ receives $\frac{4}{3}$ from $D$. So we may assume that $|D|=12$ and each vertex on the outer cycle $\mathcal{C}$ is a $3^{-}$-vertex.

Let $f$ be an arbitrary $k$-face adjacent to $D$. By the discharging rules, $f$ sends nothing to $3^{+}$-vertices on the outer cycle, and sends at most $\frac{2}{3}$ to each of the other incident vertex. If $k\geq 9$, then $\mu^{\prime}(f)\geq k-4-(k-2)\times\frac{2}{3}>0$. Recall that every $6$-, $7$- and $8$-face adjacent to $D$ has a positive final charge, thus $D$ is not adjacent to any $6^{+}$-face. Therefore, there exists a $3$-vertex on the outer cycle such that it is incident with two adjacent $5^{-}$-faces, a contradiction. ∎

3 IF-coloring

For IF-coloring, a vertex is colored with $I$ if the image is $I$ in a mapping, or it is in the part $I$ of an $(I,F)$-partition. An $F$-path is a path whose vertices are all colored with $F$, and an $F$-cycle is a cycle whose vertices are all colored with $F$.

Given a graph $G$ and a cycle $C$ in $G$, an IF-coloring $\phi_{C}$ of $G[V(C)]$ can be superextended to $G$ if there exists an IF-coloring $\phi_{G}$ of $G$ that extends $\phi_{C}$ with the property that there are no $F$-paths (having at least one vertex not on $C$) linking two vertices of $C$. We say that $C$ is superextendable to $G$ if every IF-coloring of $G[V(C)]$ can be superextended to $G$. For convenience, we also say that a vertex $w$ is superextendable to $G$ if every IF-coloring of $w$ can be extended to an IF-coloring of $G$.

For an IF-coloring, a vertex $v$ is $F$-reachable to a cycle if there is a path from $v$ to a vertex on the cycle such that all the vertices on the path are colored with $F$.

Instead of proving Theorem 1.12, we prove the following stronger result.

Theorem 3.1**.**

Let $G$ be a graph in $\mathscr{G}$. Let $S$ be a set of vertices of $G$ such that $|S|\leq 1$ or $S$ consists of all vertices on a normal cycle of $G$. If $|S|\leq 12$, then every IF-coloring $\phi$ of $G[S]$ can be superextended to an IF-coloring of $G$.

Remark 3**.**

Analogously, not every IF-coloring of the $11$- or $12$-cycle can be superextended to the whole graph. Thus, we require the condition that $S$ consists of all vertices on a “normal” cycle.

The following result is a direct consequence of Theorem 3.1.

Theorem 3.2**.**

Every graph in $\mathscr{G}$ is IF-colorable.

• Sketch of a proof for Theorem 3.1.

Suppose that $G$ is a minimal counterexample to Theorem 3.1. That is, there exists an IF-coloring of $G[S]$ that cannot be superextended to an IF-coloring of $G$ such that $|V(G)|+|E(G)|-|S|$ is minimized.

Lemma 3.1**.**

Every $8^{-}$-cycle has no chords.

Lemma 3.2**.**

**

1. (a)

$S\neq V(G)$. 2. (b)

$G$* is $2$-connected, and thus the boundary of every face is a cycle.* 3. (c)

Each vertex not in $S$ has degree at least three. 4. (d)

Either $|S|=1$ or $G[S]$ is an induced cycle of $G$. 5. (e)

There are no separating normal $k$-cycles for $3\leq k\leq 12$. Thus, every edge on an abnormal cycle is incident with a $4$-, $5$-, $6$- or $7$-face. 6. (f)

$G[S]$* is an induced cycle of $G$. For convenience, we can redraw the graph $G$ such that $G[S]$ is the outer cycle $\mathcal{C}$ of $G$. Let $D$ be the outer face which is bounded by $G[S]$.* 7. (g)

Every $5$-face ($\neq$* outer face) is incident with at most one $2$-vertex.*

• Proof.

The proof is similar to the one in 2.2, so the reader can do it as an exercise, or find it in the arXiv version. ∎

Lemma 3.3**.**

There are no $3$-faces adjacent to $8^{-}$-faces, no $4$-faces adjacent to $6^{-}$-faces, and no adjacent $5$-faces.

• Proof.

The proof is the same with that in 2.1. ∎

Lemma 3.4**.**

Let $w_{0},w_{1},w_{2},w_{3}$ and $w_{4}$ be five consecutive vertices on a $5^{+}$-face. If $w_{1},w_{2},w_{3}$ and $w_{4}$ are all $3$-vertices, and $w_{1}w_{2}$ is incident with a $3$-face $ww_{1}w_{2}$, then at least one vertex in $\{w_{1},w_{2},w_{3},w_{4}\}$ is on the outer cycle $\mathcal{C}$.

• Proof.

Suppose to the contrary that none of $\{w_{1},w_{2},w_{3},w_{4}\}$ is on the outer cycle $\mathcal{C}$. Let $w^{\prime}$ be the neighbor of $w_{3}$ other than $w_{2},w_{4}$, and let $G^{*}=G-\{w_{1},w_{2},w_{3},w_{4}\}$. It is observed that $w_{0},w_{1},w_{2},w_{3},w_{4},w$ and $w^{\prime}$ are seven distinct vertices. Similar to 2.5, we can prove that the distance between $w_{0}$ and $w^{\prime}$ is at least nine in $G^{*}$.

Let $G^{\prime}$ be the graph obtained from $G^{*}$ by identifying $w_{0}$ and $w^{\prime}$. Since the distance between $w_{0}$ and $w^{\prime}$ is at least nine in $G^{*}$, the graph $G^{\prime}$ has no loop, no multiple edge and no new $8^{-}$-cycle, thus $G^{\prime}$ is a simple plane graph satisfying the assumption of Theorem 3.1. Moreover, $\mathcal{C}$ is also a normal cycle of $G^{\prime}$ and it has no chords in $G^{\prime}$. This implies that $\phi$ is an IF-coloring of $G^{\prime}[S]$. Since $|V(G^{\prime})|<|V(G)|$, the IF-coloring $\phi$ of $G^{\prime}[S]$ can be superextended to an IF-coloring $\varphi$ of $G^{\prime}$. We color $w_{0}$ and $w^{\prime}$ with the same color as the new vertex in $G^{\prime}$. If one neighbor of $w_{4}$ is colored with $I$, then we color $w_{4}$ with $F$; otherwise, we color $w_{4}$ with $I$.

If $\varphi(w_{0})=\varphi(w^{\prime})=I$, then let $\varphi(w_{1})=\varphi(w_{3})=F$ and $\varphi(w_{2})\neq\varphi(w)$. So we may assume that $\varphi(w_{0})=\varphi(w^{\prime})=F$, and let $\varphi(w_{3})\neq\varphi(w_{4})$. If $\varphi(w)=I$, then let $\varphi(w_{1})=\varphi(w_{2})=F$. So we may assume that $\varphi(w)=F$. If $\varphi(w_{3})=I$, then let $\varphi(w_{1})=I$ and $\varphi(w_{2})=F$. In the final, we may assume that $\varphi(w_{0})=\varphi(w)=\varphi(w^{\prime})=\varphi(w_{3})=F$. It is observed that there are no $F$-paths in $H$ between $w^{\prime}$ and $w_{0}$. Meanwhile, at least one of $w^{\prime}$ and $w_{0}$ is not reachable to $\mathcal{C}$. When we color $w_{1}$ with $F$, it is a superextension to $G-w_{2}$, we can color $w_{2}$ with $I$. Otherwise, there is an $F$-path in $H$ between $w$ and $w_{0}$, or both $w$ and $w_{0}$ are reachable to $\mathcal{C}$. It follows that there are no $F$-paths between $w$ and $w^{\prime}$, and at least one of $w$ and $w^{\prime}$ is not reachable to $\mathcal{C}$. Then let $\varphi(w_{1})=I$ and $\varphi(w_{2})=F$.

It is easy to check that the resulting coloring is always a superextension to $G$, a contradiction. ∎

Let $w$ be a vertex on the outer cycle $\mathcal{C}$, and let $w_{1},w_{2},\dots,w_{k}$ be consecutive neighbors in a cyclic order. If $f$ is a face in $\mathcal{N}$ incident with $ww_{i}$ and $ww_{i+1}$, but neither $ww_{i}$ nor $ww_{i+1}$ is an edge of $\mathcal{C}$, then we call $f$ a special face (at $w$). A $4$-face is a $4^{*}$-face if it has three common vertices with $\mathcal{C}$. An internal $3$-vertex is bad if it is incident with a $3$-face which is not special, light if it is incident with an internal $4$-face or a $4^{*}$-face or a special $3$-face, good if it is neither bad nor light. According to 3.4, we have the following result on bad vertices.

Lemma 3.5**.**

There are no five consecutive bad vertices on the boundary of a $5^{+}$-face.

Lemma 3.6**.**

If a $4$-face in $\mathcal{N}$ has exactly two common vertices with $\mathcal{C}$, then these two vertices are consecutive on the $4$-face.

• Proof.

Suppose that $f$ is a $4$-face in $\mathcal{N}$ that has exactly two common vertices with $\mathcal{C}$. If these two vertices are not consecutive on the $4$-face, then there exists a separating normal $8^{-}$-cycle, a contradiction. ∎

Assume that $f=[v_{1}v_{2}\dots v_{l}]$ is an internal $(3,3,3,3)$-face or an internal $(3,3,3,3,3)$-face. Let $u_{i}$ be the third neighbor of $v_{i}$ for $1\leq i\leq l$. Note that every $8^{-}$-cycle has no chords, and there are no separating $4$- or $5$-cycles. It is observed that $\{u_{i},v_{i}\mid 1\leq i\leq l\}$ contains $2l$ distinct vertices. Let $\Gamma$ be the graph $G\setminus(V(f)\setminus\{v_{3}\})$, and let $G^{*}$ be the graph obtained from $\Gamma$ by identifying $u_{1}$ and $v_{3}$ into a new vertex $z$.

Lemma 3.7**.**

The graph $G^{*}$ is in the class $\mathscr{G}$, and $\mathcal{C}$ is an induced cycle of $G^{*}$.

• Proof.

The proof is the same with that in 2.8. ∎

Lemma 3.8**.**

There are no internal $(3,3,3,3)$-faces or $(3,3,3,3,3)$-face.

• Proof.

We know that $(G^{*},\mathcal{C},\phi)$ satisfies all the requirements in Theorem 3.1. By the minimality of $G$, the IF-coloring $\phi$ can be superextended to an IF-coloring $\varphi$ of $G^{*}$. We color $u_{1}$ and $v_{3}$ with the same color as the new vertex by the identification.

$\bullet$ $\bm{l=5}$.

Assume that $\varphi(u_{1})=\varphi(v_{3})=I$. Then let $\varphi(v_{1})=\varphi(v_{2})=\varphi(v_{4})=F$. Furthermore, we color $v_{5}$ such that $\varphi(v_{5})\neq\varphi(u_{5})$, this is invalid only if $u_{2}v_{2}v_{1}v_{5}v_{4}u_{4}$ is on an $F$-cycle or is on an $F$-path linking two vertices of $\mathcal{C}$. In both cases, we recolor $v_{2}$ and $v_{4}$ with $I$, and $v_{3}$ with $F$.

Assume that $\varphi(u_{1})=\varphi(v_{3})=F$. If $\varphi(u_{4})=I$, then let $\varphi(v_{1})=\varphi(v_{4})=F$, $\varphi(v_{2})\neq\varphi(u_{2})$ and $\varphi(v_{5})\neq\varphi(u_{5})$, this is invalid only if $v_{1},v_{2},v_{3},v_{4},v_{5}$ are all colored with $F$. In this case, we recolor $v_{1}$ with $I$. So we may assume that $\varphi(u_{4})=F$. If $\varphi(u_{2})=I$ or $\varphi(u_{3})=I$, then let $\varphi(v_{1})=\varphi(v_{4})=I$ and $\varphi(v_{2})=\varphi(v_{5})=F$. So we may assume that $\varphi(u_{1})=\varphi(u_{2})=\varphi(u_{3})=\varphi(u_{4})=F$. If $\varphi(u_{5})=I$, then let $\varphi(v_{2})=\varphi(v_{4})=I$, and $\varphi(v_{1})=\varphi(v_{5})=F$. Now, we may assume that all the vertices $u_{i}$ are colored with $F$. Then let $\varphi(v_{3})=\varphi(v_{5})=I$ and $\varphi(v_{1})=\varphi(v_{2})=\varphi(v_{4})=F$, this is invalid only if $u_{1}v_{1}v_{2}u_{2}$ is on an $F$-cycle or an $F$-path linking two vertices of $\mathcal{C}$. In this case, let $\varphi(v_{1})=\varphi(v_{4})=I$ and $\varphi(v_{2})=\varphi(v_{3})=\varphi(v_{5})=F$.

$\bullet$ $\bm{l=4}$.

Assume that $\varphi(u_{1})=\varphi(v_{3})=I$. Then let $\varphi(v_{1})=\varphi(v_{2})=\varphi(v_{4})=F$, this is invalid only if $u_{2}v_{2}v_{1}v_{4}u_{4}$ is on an $F$-cycle or is on an $F$-path linking two vertices of $\mathcal{C}$. In both cases, let $\varphi(v_{1})=\varphi(v_{3})=F$, $\varphi(v_{2})=\varphi(v_{4})=I$.

Assume that $\varphi(u_{1})=\varphi(v_{3})=F$. Then let $\varphi(v_{1})=F$, $\varphi(v_{2})\neq\varphi(u_{2})$ and $\varphi(v_{4})\neq\varphi(u_{4})$, this is invalid only if $\varphi(u_{2})=\varphi(u_{4})=I$. In this case, we recolor $v_{1}$ with $I$. ∎

Note that 2.4 is only prepared for 2.5, so we do not need it here. All the structural lemmas are the same in the proof processes, so the discharging part is the same with that in Theorem 3.1. ∎

4 Final discussion

If we can relax the normally adjacent $5$-cycles in Theorem 1.6, then Theorem 1.6 implies that every planar graph without $3$-, $6$-, $7$-cycles is DP-$3$-colorable. On the other hand, if we can allow the normally adjacent $4$-cycles in Theorem 1.6, then Theorem 1.6 implies that every planar graph without $3$-, $7$-, $8$-cycles is DP-$3$-colorable. But until now, we don’t know whether every planar graph without $3$-, $6$-, $7$-cycles is DP-$3$-colorable or not, and we don’t know whether every planar graph without $3$-, $7$-, $8$-cycles is DP-$3$-colorable or not. It seems that it is not easy to solve these two problems.

Dvořák, Lidický, and Škrekovski [4] proved that every planar graph without $3$-, $6$-, $7$-cycles is $3$-choosable. The same authors [3] also proved that every planar graph without $3$-, $7$-, $8$-cycles is $3$-choosable. So it is interesting to consider the following problems.

Problem 1**.**

Is every planar graph without $3$-, $6$-, $7$-cycles DP-$3$-colorable?

Problem 2**.**

Is every planar graph without $3$-, $7$-, $8$-cycles DP-$3$-colorable?

Problem 3**.**

Does every planar graph without $3$-, $6$-, $7$-cycles have an IF-coloring?

Problem 4**.**

Does every planar graph without $3$-, $7$-, $8$-cycles have an IF-coloring?

Added Note. Recently, Han et al. [7] proved that every triangle-free planar graph without 4-cycles normally adjacent to 4- and 5-cycles is DP-$3$-colorable. This solves 1.

Acknowledgments. This work was supported by the Fundamental Research Funds for Universities in Henan (YQPY20140051).

Appendix

• Proof of 3.2.

(a) Suppose to the contrary that $S=V(G)$. Every IF-coloring of $G[S]$ is an IF-coloring of $G$, a contradiction.

(b) It is observed that $G$ is connected. Suppose to the contrary that $G$ has a cut-vertex $w$. We may assume that $G=G_{1}\cup G_{2}$ and $G_{1}\cap G_{2}=\{w\}$. By the assumption of the set $S$, either $S\subseteq V(G_{1})$ or $S\subseteq V(G_{2})$. We may assume that $S\subseteq V(G_{1})$. By the minimality of $G$, the IF-coloring $\phi$ of $G[S]$ can be superextended to an IF-coloring $\phi_{1}$ of $G_{1}$, and $\phi_{1}(w)$ can be superextended to an IF-coloring $\phi_{2}$ of $G_{2}$. These two colorings $\phi_{1}$ and $\phi_{2}$ together give an IF-coloring of $G$ whose restriction on $G[S]$ is $\phi$, a contradiction.

(c) Suppose that there exists a vertex $w$ not in $S$ having degree two. By the minimality of $G$, the IF-coloring of $G[S]$ can be superextended to an IF-coloring of $G-w$. If the two neighbors of $w$ are colored with $F$, then we color $w$ with $I$; otherwise, we color $w$ with $F$.

(d) If $S=\emptyset$, then we put any vertex into $S$ to make $|S|=1$. Suppose that $S=V(\mathcal{Q})$ and $\mathcal{Q}$ is a cycle with a chord $uv$. It is observed that the IF-coloring of $G[S]$ is also an IF-coloring of the induced subgraph in $G-uv$. By the minimality of $G$, the IF-coloring $\phi$ of $G[S]$ can be superextended to an IF-coloring of $G-uv$, and hence it is also a superextension of $G$, a contradiction.

(e) We first show that $G[S]$ cannot be a separating cycle. Suppose to the contrary that $G[S]$ is a separating (normal) cycle $\mathcal{O}$. By the minimality of $G$, the IF-coloring $\phi$ of $\mathcal{O}$ can be superextended to an IF-coloring $\phi_{1}$ of $\mathrm{Int}(\mathcal{O})$, and another IF-coloring $\phi_{2}$ of $\mathrm{Ext}(\mathcal{O})$. These two colorings $\phi_{1}$ and $\phi_{2}$ together give a superextension, a contradiction.

Thus, either $|S|=1$ or $S$ consists of all vertices on a face of $G$. Let $\mathcal{Q}$ be a separating normal $k$-cycle with $3\leq k\leq 12$. Thus, we may assume that $S\subseteq\mathrm{Ext}(\mathcal{Q})$. By the minimality of $G$, the IF-coloring $\phi$ of $G[S]$ can be superextended to an IF-coloring $\varphi_{1}$ of $\mathrm{Ext}(\mathcal{Q})$. Similarly, the restriction of $\varphi_{1}$ on $\mathcal{Q}$ can be superextended to an IF-coloring $\varphi_{2}$ of $\mathrm{Int}(\mathcal{Q})$. These two colorings $\varphi_{1}$ and $\varphi_{2}$ together give a superextension of $G$, a contradiction.

(f) According to (d), suppose to the contrary that $S=\{w\}$. We first assume that $w$ is on a $10^{-}$-cycle $\mathcal{Q}$. Without loss of generality, we may assume that $\mathcal{Q}$ is a shortest cycle containing $w$. Then $\mathcal{Q}$ is an induced cycle. By (e), we may assume that $\mathrm{ext}(\mathcal{Q})=\emptyset$ and $\mathcal{Q}$ is the outer cycle. By (c) and $\mathcal{Q}$ is an induced cycle, every vertex on $\mathcal{Q}$ other than $w$ has a neighbor in $\mathrm{int}(\mathcal{Q})$, which implies that $\mathrm{int}(\mathcal{Q})\neq\emptyset$. By the minimality of $G$, the IF-coloring $\phi$ of $\{w\}$ can be superextended to an IF-coloring $\phi_{1}$ of $\mathcal{Q}$. By the minimality of $G$, the IF-coloring $\phi_{1}$ of $\mathcal{Q}$ can be further superextended to an IF-coloring of $G$, a contradiction.

So we may assume that every cycle containing $w$ has length at least $11$. Let $w$ be incident with a face $[w_{1}ww_{2}\dots w_{1}]$. Let $G^{\prime}$ be obtained from $G$ by adding a chord $w_{1}w_{2}$ in the face, let $S^{\prime}=\{w,w_{1},w_{2}\}$. We can easily check that $G^{\prime}$ is a plane graph satisfying the assumption of Theorem 3.1. By the minimality of $G$, the IF-coloring $\phi$ of $\{w\}$ can be superextended to an IF-coloring $\phi_{1}$ of $G^{\prime}[S^{\prime}]$. By the minimality of $G$, the IF-coloring $\phi_{1}$ of $G^{\prime}[S^{\prime}]$ can be further superextended to an IF-coloring $\varphi$ of $G^{\prime}$. It is observed that $\varphi$ is an IF-coloring of $G$, a contradiction.

(g) Note that every $2$-vertex and its two neighbors are all on the outer cycle. Suppose to the contrary that $f=[x_{1}x_{2}x_{3}x_{4}x_{5}]$ is a $5$-face which is incident with two $2$-vertices. Note that the two $2$-vertices must be adjacent on the outer cycle, say $x_{2}$ and $x_{3}$. It follows that $x_{1}$ and $x_{4}$ are on the outer cycle $\mathcal{C}$. If $x_{5}$ has three neighbors on $\mathcal{C}$, then $\mathcal{C}$ is abnormal (see 1(a) and 1(b)), a contradiction. Thus, $x_{5}$ has a neighbor not on the outer cycle $\mathcal{C}$, and $\mathcal{C}^{\prime}=(\mathcal{C}-\{x_{2},x_{3}\})\cup\{x_{1}x_{5},x_{4}x_{5}\}$ is a separating $11^{-}$-cycle. By 3.2(e), $\mathcal{C}^{\prime}$ is an abnormal $11$-cycle (see 1(a)). It follows that $\mathcal{C}$ is an abnormal $12$-cycle (see 1(d)), a contradiction. ∎