Planar graphs without pairwise adjacent 3-,4-,5-, and 6-cycle are 4-choosable

TL;DR

This paper proves that planar graphs lacking pairwise adjacent 3-, 4-, 5-, and 6-cycles are 4-choosable, extending previous results by Xu and Wu to a broader class of graphs.

Contribution

It establishes a new sufficient condition for 4-choosability in planar graphs, generalizing earlier work by removing adjacency restrictions among certain cycles.

Findings

Planar graphs without pairwise adjacent 3-, 4-, 5-, and 6-cycles are 4-choosable.

The result broadens the class of graphs known to be 4-choosable.

Improves upon previous conditions requiring non-adjacency of 5-cycles to 3- and 4-cycles.

Abstract

Xu and Wu proved that if every 5-cycle of a planar graph G is not simultaneously adjacent to 3-cycles and 4-cycles, then G is 4-choosable. In this paper, we improve this result as follows. If G is a planar graph without pairwise adjacent 3-,4-,5-, and 6-cycle, then G is 4-choosable.

Equations

$$\sum _{u\in V(G)}\mu (u)+\sum _{f\in F(G)}\mu (f)=-12.$$

$${\mu }^{\ast }(D)$$

$$=9+2\sum _{v\in C_0}(d(v)-2)-2\times 3-\frac{12}{5}{f}_3^{\prime }-2f^{\prime }$$

$$=3-\frac{2}{5}{f}_3^{\prime }+2(e(C_0,V(G)-C_0)-f_3^{\prime }-f^{\prime })$$

Full text

Planar graphs without pairwise adjacent 3-,4-,5-, and 6-cycle are 4-choosable

Pongpat Sittitrai                   Kittikorn Nakprasit

Abstract

Xu and Wu proved that if every $5$-cycle of a planar graph $G$ is not simultaneously adjacent to $3$-cycles and $4$-cycles, then $G$ is $4$-choosable. In this paper, we improve this result as follows. If $G$ is a planar graph without pairwise adjacent $3$-,$4$-,$5$-, and $6-$cycle, then $G$ is $4$-choosable.

1. Introduction

Every graph in this paper is finite, simple, and undirected. The concept of choosability was introduced by Vizing in 1976 [12] and by Erdős, Rubin, and Taylor in 1979 [5], independently. A $k$-assignment $L$ of a graph $G$ assigns a list $L(v)$ (a set of colors) with $|L(v)|=k$ to each vertex $v.$ A graph $G$ is $L$-colorable if there is a proper coloring $f$ where $f(v)\in L(v).$ If $G$ is $L$-colorable for any $k$-assignment $L,$ then we say $G$ is $k$-choosable.

It is known that every planar graphs is $4$-colorable [1, 2]. Thomassen [11] proved that every planar graph is $5$-choosable. Meanwhile, Voight [13] presented an example of non $4$-choosable planar graph. Additionally, Gutner [8] showed that determining whether a given planar graph $4$-choosable is NP-hard. Since every planar graph without $3$-cycle always has a vertex of degree at most $3,$ it is $4$-choosable. More conditions for a planar graph to be $4$-choosable are investigated. It is shown that a planar graph is $4$-choosable if it has no $4$-cycles [10], $5$-cycles [15], $6$-cycles [7], $7$-cycles [6], intersecting $3$-cycles [16], intersecting $5$-cycles [9], or $3$-cycles adjacent to $4$-cycles [3, 4]. Xu and Wu [14] proved that if every $5$-cycle of a planar graph $G$ is not simultaneously adjacent to $3$-cycles and $4$-cycles, then $G$ is $4$-choosable. In this paper, we improve this result as follows.

Theorem 1.1**.**

If $G$ is a planar graph without pairwise adjacent $3$-,$4$-,$5$-, and $6-$cycle, then $G$ is $4$-choosable.

2. Preliminaries

First, we introduce some notations and definitions.

Let $G$ be a plane graph. We use $V(G),E(G),$ and $F(G)$ for the vertex set, the edge set, and the face set respectively. We use $B(f)$ to denote a boundary of a face $f.$ A wheel $W_{n}$ is an $n$-vertex graph formed by connecting a single vertex (hub) to all vertices (external vertices) of an $(n-1)$-cycle. A $k$-vertex ($k^{+}$-vertex, $k^{-}$-vertex, respectively) is a vertex of degree $k$ (at least $k,$ at most $k,$ respectively). The same notations are applied to faces.

A $(d_{1},d_{2},\dots,d_{k})$-face $f$ is a face of degree $k$ where vertices on $f$ have degree $d_{1},d_{2},\dots,d_{k}$ in a cyclic order. A $(d_{1},d_{2},\dots,d_{k})$-vertex $v$ is a vertex of degree $k$ where faces incident to $v$ have degree $d_{1},d_{2},\dots,d_{k}$ in a cyclic order. Note that some face may appear more than one time in the order. A face is called poor, semi-rich, and rich, respectively if it is incident to no $5^{+}$-vertices, exactly one $5^{+}$-vertex, and at least two $5^{+}$-vertices, respectively. An extreme face is a bounded face that shares a vertex with the unbounded face. An inner face is a bounded face that is not an extreme face. A $(3,5,3,5^{+})$-vertex $v$ is called a flaw $4$-vertex if $v$ is incident to a poor $5$-face and all incident faces of $v$ are inner faces.

We say $xy$ is a chord in an embedding cycle $C$ if $x,y\in V(C)$ but $xy\in E(G)-E(C).$ An internal chord is a chord inside $C$ while external chord is a chord outside $C.$ A triangular chord is a chord $e$ such that two edges in $C$ and $e$ form a $3$-cycle. A graph $C(m,n)$ is obtained from a cycle $x_{1}x_{2}\ldots x_{m+n-2}$ with an internal chord $x_{1}x_{m}.$ A graph $C(l,m,n)$ is obtained from a cycle $x_{1}x_{2}\ldots x_{l+m+n-4}$ with internal chords $x_{1}x_{l}$ and $x_{1}x_{l+m-2}.$ A graph $C(m,n,p,q)$ can be defined similarly. We use $int(C)$ and $ext(C)$ to denote the graphs induced by vertices inside and outside a cycle $C$, respectively. The cycle $C$ is a separating cycle if $int(C)$ and $ext(C)$ are not empty.

Let $L$ be a list assignment of $G$ and let $H$ be an induced subgraph of $G.$ Suppose $G-H$ has an $L$-coloring $\phi$ on $G-H$ where $L$ is restricted to $G-H.$ For a vertex $v\in H,$ let $L^{\prime\prime}(v)$ be a set of colors used on the neighbors of $v$ by $\phi.$ We define a residual list assignment $L^{\prime}$ of $H$ by $L^{\prime}(v)=L(v)-L^{\prime\prime}(v).$ One can see that if $G-H$ has an an $L$-coloring $\phi$ and $H$ has an $L^{\prime}$-coloring, then $G$ has an $L$-coloring.

The following is a fact on list colorings that we use later.

Lemma 2.1**.**

[5]** Let $L$ be a $2$-assignment. A cycle $C_{n}$ is $L$-colorable if and only if $n$ is even or $L$ does not assign the same list to all vertices.

Let $\mathcal{A}$ denote the family of planar graphs without pairwise adjacent $3$-,$4$-,$5$-, and $6-$cycle.

Next, we explore some properties of graphs in $\mathcal{A}$ which are helpful in a proof of the main results.

Lemma 2.2**.**

*Every graph $G$ in $\mathcal{A}$ does not contain each of the followings:

(1) $C(3,3,4),$

(2) $C(3,3,5),$

(3) $C(3,4,4^{-}),$

(4) $C(4,3,5).$

(5) $W_{5}$ that shares exactly one edge with a $6^{-}$-cycle.*

Proof.

Let $C(l,m,n)$ be obtained from a cycle $x_{1}x_{2}\ldots x_{l+m+n-4}$ with internal chords $x_{1}x_{l}$ and $x_{1}x_{l+m-2}.$

(1) Suppose $G$ contains $C(3,3,4).$ Then we have four pairwise adjacent cycles $x_{1}x_{2}x_{3},$ $x_{1}x_{2}x_{3}x_{4},$ $x_{1}x_{3}x_{4}x_{5}x_{6},$ and $x_{1}x_{2}x_{3}x_{4}x_{5}x_{6},$ contrary to $G\in\mathcal{A}.$

(2) Suppose $G$ contains $C(3,3,5).$ Then we have four pairwise adjacent cycles $x_{1}x_{3}x_{4},$ $x_{1}x_{2}x_{3}x_{4},$ $x_{1}x_{4}x_{5}x_{6}x_{7},$ and $x_{1}x_{3}x_{4}x_{5}x_{6}x_{7},$ contrary to $G\in\mathcal{A}.$

(3) Suppose $G$ contains $C(3,4,3).$ Then we have four pairwise adjacent cycles $x_{1}x_{2}x_{3},$ $x_{1}x_{3}x_{4}x_{5},$ $x_{1}x_{2}x_{3}x_{4}x_{5},$ and $x_{1}x_{2}x_{3}x_{4}x_{5}x_{6},$ contrary to $G\in\mathcal{A}.$

Suppose $G$ contains $C(3,4,4).$ Then we have four pairwise adjacent cycles $x_{1}x_{2}x_{3},$ $x_{1}x_{3}x_{4}x_{5},$ $x_{1}x_{2}x_{3}x_{4}x_{5},$ and $x_{1}x_{3}x_{4}x_{5}x_{6}x_{7},$ contrary to $G\in\mathcal{A}.$

(4) Suppose $G$ contains $C(4,3,5).$ Then we have four pairwise adjacent cycles $x_{1}x_{4}x_{5},$ $x_{1}x_{2}x_{3}x_{4},$ $x_{1}x_{2}x_{3}x_{4}x_{5},$ and $x_{1}x_{4}x_{5}x_{6}x_{7}x_{8},$ contrary to $G\in\mathcal{A}.$

(5) Let the hub of $W_{5}$ be $q$ and let external vertices be $r,s,u,$ and $v$ in a cyclic order.

Suppose there is a cycle $uvw.$ Then we have four pairwise adjacent cycles $vwu,$ $vwuq,$ $vwusq,$ and $vwusqr,$ contrary to $G\in\mathcal{A}.$

Suppose there is a cycle $uvwx.$ Then we have four pairwise adjacent cycles $usq,$ $usqv,$ $usqrv,$ and $usqvwx,$ contrary to $G\in\mathcal{A}.$

Suppose there is a cycle $uvwxy.$ Then we have four pairwise adjacent cycles $uqv,$ $uqrv,$ $uqsrv,$ and $uqvwxy,$ contrary to $G\in\mathcal{A}.$

Suppose there is a cycle $uvwxyz.$ Then we have four pairwise adjacent cycles $uvq,$ $uvqs,$ $uvqrs,$ and $uvwxyz,$ contrary to $G\in\mathcal{A}.$

∎

Lemma 2.3**.**

If $C$ is a $6$-cycle with a triangular chord, then $C$ has exactly one chord.

Proof.

Let $C=tuvxyz$ with a chord $tv.$ Suppose to the contrary that $C$ has another chord $e.$ By symmetry, it suffices to assume that $e=ux,uy,tx,ty,$ or $xz.$

If $e=ux,$ then we have four pairwise adjacent cycles $tuv,tuxv,tvxyz,$ and $tuvxyz,$ contrary to $G\in\mathcal{A}.$

If $e=uy,$ then we have four pairwise adjacent cycles $tuv,uvxy,tvxyz,$ and $tuvxyz,$ contrary to $G\in\mathcal{A}.$

If $e=tx,$ then we have four pairwise adjacent cycles $tuv,tuvx,tvxyz,$ and $tuvxyz,$ contrary to $G\in\mathcal{A}.$

If $e=ty,$ then we have four pairwise adjacent cycles $tuv,tvxy,tvxyz,$ and $tuvxyz,$ contrary to $G\in\mathcal{A}.$

If $e=xz,$ then we have four pairwise adjacent cycles $tuv,tvxz,tvxyz,$ and $tuvxyz,$ contrary to $G\in\mathcal{A}.$

Thus $C$ has exactly one chord. ∎

3. Structure

To prove Theorem 1.1, we prove a stronger result as follows.

Theorem 3.1**.**

If $G\in\mathcal{A}$ with a $4$-assignment $L,$ then each precoloring of a $3$-cycle in $G$ can be extended to an $L$-coloring of $G$.

If $G$ does not contain a $3$-cycles, then $G$ is $4$-choosable as stated above. So we consider $(G,C_{0})$ and a $4$-assignment $L$ where $C_{0}$ is a precolored $3$-cycle as a minimal counterexample to Theorem 3.1. Embed $G$ in the plane.

Lemma 3.2**.**

$G$* has no separating $3$-cycles.*

Proof.

Suppose to the contrary that there exists $G$ contains a separating $3$-cycle $C.$ By symmetry, we assume $V(C_{0})\subseteq V(C)\cup int(C).$ By the minimality of $G,$ a precoloring of $C_{0}$ can be extended to $V(C)\cup int(C).$ After $C$ is colored, then again the coloring of $C$ can be extended to $ext(C)$. Thus we have an $L$-coloring of $G$, a contradiction. ∎

So we may assume that a minimal counterexample $(G,C_{0})$ has no separating $3$-cycles and $C_{0}$ is the boundary of the unbounded face $D$ of $G$ in the rest of this paper.

Lemma 3.3**.**

Each vertex in $int(C_{0})$ has degree at least four.

Proof.

Suppose otherwise that there exists a $3^{-}$-vertex $v$ in $int(C_{0})$. By the minimality of $(G,C_{0})$, $(G-v,C_{0})$ has an $L$-coloring. One can see that the residual list $L^{\prime}(v)$ is not empty. Thus we can color $v$ and thus extend a coloring to $G,$ a contradiction. ∎

Lemma 3.4**.**

*For faces in $G,$ each of the followings holds.

(1) The boundary of a bounded $6^{-}$-face is a cycle.

(2) If a bounded $k_{1}$-face $f$ and a bounded $k_{2}$-face $g$ are adjacent where $k_{1}+k_{2}\leq 8,$ then $B(f)\cup B(g)=C(k_{1},k_{2}).$

(3) If a bounded $4$-face $f$ and a bounded $5$-face $g$ are adjacent, then $B(f)\cup B(g)$ is $C(4,5)$ or a configuration as in Figure 1 where $tuy$ is $C_{0}.$

(4) If bounded $5$-faces $f$ and $g$ are adjacent, then $B(f)\cup B(g)$ is $C(5,5)$ or a configuration as in Figure 2.*

Proof.

(1) One can observe that a boundary of a $5^{-}$-face is always a cycle. Consider a bounded $6$-face $f.$ If $B(f)$ is not a cycle, then a boundary closed walk is in a form of $uvwxywu.$ By Lemma 3.3, $u$ or $x$ has degree at least $4.$ Consequently, $uvw$ or $xyw$ is a separating $3$-cycle, contrary to Lemma 3.2.

(2) It suffices to show that such $f$ and $g$ share exactly two vertices. Let $B(f)=uvw$ and $B(g)=vwx.$ If $u=x,$ then $f$ or $g$ is the unbounded face, a contradiction.

Let $B(f)=uvw$ and $B(g)=vwxy.$ If $u=x$ or $y,$ then $d(w)=2$ or $d(v)=2,$ contrary to Lemma 3.3.

Let $B(f)=uvw$ and $B(g)=vwxyz.$ If $u=x$ or $z,$ then $d(v)=2$ or $d(w)=2,$ contrary to Lemma 3.3. If $u=y,$ then $vyz$ or $wxy$ is a separating $3$-cycle, contrary to Lemma 3.2.

Let $B(f)=stuv$ and $B(g)=uvwx.$ If $s=w,$ then $d(v)=2,$ contrary to Lemma 3.3. If $s=x,$ then $utx$ or $vwx$ is a separating $3$-cycle, contrary to Lemma 3.2. The remaining cases are similar.

(3) Let $B(f)=stuv$ and $B(g)=uvwxy.$ It suffices to show that $V(B(f))\cap V(B(g))=\{u,v\}$ or $\{u,v,x\}$ where $x=s$ or $t.$ If $t=w,$ then $svw$ or $uvw$ is a separating $3$-cycle, contrary to Lemma 3.2. If $t=x,$ then $tuy$ is $C_{0},$ otherwise $tuy$ is a separating cycle, contrary to Lemma 3.2. The remaining cases are similar.

(4) Let $B(f)=rstuv$ and $B(g)=uvwxy.$ It suffices to show that $V(B(f))\cap V(B(g))=\{u,v\}$ or $\{u,v,x=s\}.$ If $r=w,$ then $d(v)=2,$ contrary to Lemma 3.3. If $B(f)\cap B(g)=\{u,v,r=x\},$ then $vwx,uvxy,uvwxy,$ and $stuvwx$ are four pairwise adjacent cycles, contrary to $G\in\mathcal{A}.$ If $B(f)\cap B(g)=\{u,v,r=x,s=y\},$ then $rvx,rvuz,rvuts,$ and $rstuvs$ are four pairwise adjacent cycles, contrary to $G\in\mathcal{A}.$ then $uts$ or $vwx$ is a separating $3$-cycle, contrary to Lemma 3.2. If $B(f)\cap B(g)=\{u,v,r=y\},$ then $ruv$ is a separating $3$-cycle, contrary to Lemma 3.2. If $B(f)\cap B(g)=\{u,v,s=w\},$ then $rvw,tuvw,uvwxy,$ and $rwxyuv$ are four pairwise adjacent cycles, contrary to $G\in\mathcal{A}.$ The remaining cases are similar.

∎

Lemma 3.5**.**

*If a $k$-vertex $v$ is incident to bounded faces $f_{1},\ldots,f_{k}$ in a cyclic order and $d_{i}$ is a degree of a face $f_{i}$ for each $i\in\{1,\ldots,k\},$ then each of the followings holds.

(1) $(d_{1},d_{2},d_{3})\neq(3,3,4).$

(2) $(d_{1},d_{2},d_{3})\neq(3,3,5).$

(3) $(d_{1},d_{2},d_{3})\neq(3,4,4^{-}).$

(4) $(d_{1},d_{2},d_{3})\neq(4,3,5).$

(5) Let $H$ be $W_{5}$ such that a hub and each two vertices of consecutive external vertices form a boundary of an inner $3$-face. Then $H$ is not adjacent to a boundary of a $6^{-}$-face other than these $3$-faces.*

Proof.

Let $F=B_{1}\cup B_{2}\cup B_{3}$ where $B_{i}$ denote $B(f_{i}).$

(1) Suppose $(d_{1},d_{2},d_{3})=(3,3,4).$ Let $B_{1}=rsv,$ $B_{2}=vst,$ and $B_{3}=vtxy.$ It follows from Lemma 3.4 (2) that $V(B_{1})\cap V(B_{2})=\{s,v\}$ and $V(B_{2})\cap V(B_{3})=\{t,v\}.$ If $r=x,$ then $stx$ or $vxy$ is a separating $3$-cycle, contrary to Lemma 3.2. If $r=y,$ then $d(v)=3,$ contrary to Lemma 3.3. Thus $V(B_{1})\cap V(B_{3})=\{v\}.$ Altogether we have $F=C(3,3,4),$ contrary to Lemma 2.2(1).

(2) Suppose $(d_{1},d_{2},d_{3})=(3,3,5).$ Let $B_{1}=rsv,$ $B_{2}=vst,$ and $B_{3}=vtxyz.$ It follows from Lemma 3.4 (2) that $V(B_{1})\cap V(B_{2})=\{s,v\}$ and $V(B_{2})\cap V(B_{3})=\{t,v\}.$ We have $C=stxyzv$ is a $6$-cycle with a triangular chord $tv.$ If $r\in\{x,y,z\},$ then $C$ has another chord, contrary to Lemma 2.3. Thus $V(B_{1})\cap V(B_{3})=\{v\}.$ Altogether we have $F=C(3,3,5),$ contrary to Lemma 2.2(2).

(3) Suppose $(d_{1},d_{2},d_{3})=(3,4,3).$ Let $B_{1}=rsv,$ $B_{2}=vstu,$ and $B_{3}=vuw.$ It follows from Lemma 3.4 (2) that $V(B_{1})\cap V(B_{2})=\{s,v\}$ and $V(B_{2})\cap V(B_{3})=\{u,v\}.$ If $r=w,$ then $d(v)=3,$ contrary to Lemma 3.3. Thus $V(B_{1})\cap V(B_{3})=\{v\}.$ Altogether we have $F=C(3,4,3),$ contrary to Lemma 2.2(3).

Suppose $(d_{1},d_{2},d_{3})=(3,4,4).$ Let $B_{1}=rsv,$ $B_{2}=vstu,$ and $B_{3}=uvxy.$ It follows from Lemma 3.4 (2) that $V(B_{1})\cap V(B_{2})=\{s,v\}$ and $V(B_{2})\cap V(B_{3})=\{u,v\}.$ If $r=x,$ then $d(v)=3,$ contrary to Lemma 3.3. If $r=y,$ then $vuy$ is a separating $3$-cycle, contrary to Lemma 3.2. Thus $V(B_{1})\cap V(B_{3})=\{v\}.$ Altogether we have $F=C(3,4,4),$ contrary to Lemma 2.2(3).

(4) Suppose $(d_{1},d_{2},d_{3})=(4,3,5).$ Let $B_{1}=qrsv,$ $B_{2}=vst,$ and $B_{3}=vtxyz.$ It follows from Lemma 3.4 (2) that $V(B_{1})\cap V(B_{2})=\{s,v\}$ and $V(B_{2})\cap V(B_{3})=\{t,v\}.$ We have $C=stxyzv$ is a $6$-cycle with a triangular chord $tv.$ If $\{q,r\}$ and $\{x,y,z\}$ are not disjoint, then $C$ has another chord, contrary to Lemma 2.3. Thus $V(B_{1})\cap V(B_{3})=\{v\}.$ Altogether we have $F=C(4,3,5),$ contrary to Lemma 2.2(2).

(5) Let $v$ be a hub and let $w,x,y,z$ be external vertices of $H$ in the cyclic order. Suppose to the contrary that $H$ is adjacent to a face $f$ with $B(f)=wxq,wxqr,wxqrs,$ or $wxqrst.$ Now we have $\{w,x\}\subseteq V(H)\cap V(B(f)).$ By Lemma 2.2(5), $V(H)\cap V(B(f))\neq\{w,x\}.$ If $q=y,$ then $d(x)=3,$ contrary to Lemma 3.3. If $r=y,$ then $vwxqyz$ is a $6$-cycle with four triangular chords, contrary to Lemma 2.3. If $s=y,$ then $vxw,vxwz,vxwzy,$ and $vxqryz$ are four pairwise adjacent cycles, contrary to $G\in\mathcal{A}.$ If $t=y,$ then $vxw,vxwz,vxwzy,$ and $vxqrsy$ are four pairwise adjacent cycles, contrary to $G\in\mathcal{A}.$ The remaining cases lead to similar contradictions. Thus $f$ is not a $6^{-}$-face. ∎

The following is rephrased from ([4], Lemma 4).

Lemma 3.6**.**

Let $C(m,n)$ in $int(C_{0})$ be obtained from a cycle $C=x_{1}\ldots x_{m+n-2}$ with a chord $x_{1}x_{m}$ and $d(x_{1})\leq 5.$ If $C$ has at most one additional chord $e$ and $e$ is not incident to $x_{1},$ then there exists $i\in\{2,\ldots,m+n-2\}$ with $d(x_{i})\geq 5.$

Corollary 3.7**.**

*If $v$ is a flaw vertex, then we have the followings.

(1) $v$ is incident to exactly one poor $5$-face.

(2) Each $3$-face that is incident to $v$ is a semi-rich face.*

Proof.

Let $v$ be incident to inner faces $f_{1},f_{2},f_{3},f_{4}$ in a cyclic order where $f_{1}$ and $f_{3}$ are $3$-faces, $f_{2}$ is a poor $5$-face, and $f_{4}$ is a $5^{+}$-face. By Lemma 3.4, $B(f_{1})\cup B(f_{2})$ and $B(f_{2})\cup B(f_{3})$ are $C(3,5).$ It follows from Lemma 3.6 that some vertex in $B(f_{1})\cup B(f_{2})$ and in $B(f_{2})\cup B(f_{3})$ has degree at least $5.$ Observe that some vertex in $B(f_{1})$ and in $B(f_{3})$ has degree at least $5$ since $f_{2}$ is a poor face.

(1) If $f_{4}$ is also a poor $5$-face, then $f_{1}$ is a poor face, contrary to the observation above. (2) By observation above, $f_{1}$ and $f_{3}$ are not poor $3$-faces. Since $f_{2}$ is a poor face, we obtain that $f_{1}$ and $f_{3}$ are not rich faces. ∎

Lemma 3.8**.**

If $H$ in Figure 2 is in $int(C_{0})$ and contains a $5^{-}$-vertex $v,$ then there is another vertex of $H$ with degree at least $5$ in $G.$

Proof.

First, we show that $H$ is an induced subgraph. Suppose to the contrary that there is an edge $e$ joining vertices in $V(H)$ such that $e\notin E(H).$ If $e=ty,$ then $sty$ or $tuy$ is a separating $3$-cycle. If $e=ux,$ then $stu$ is a separating $3$-cycle. If $e=sv,$ then $rsv$ is a separating $3$-cycle. If $e=rw,$ then $rvw$ is a separating $3$-cycle. All consequences contradicts Lemma 3.2. Thus $H$ is an induced subgraph.

Suppose to the contrary that $d(v)\leq 5$ but each of remaining vertices has degree at most $4.$ By minimality, $G-H$ has an $L$-coloring where $L$ is restricted to $G-H.$ Consider a residual list assignment $L^{\prime}$ on $H.$ Since $L$ is a $4$-assignment, we have $|L^{\prime}(s)|=4,$ $|L^{\prime}(u)|\geq 3,$ and $|L^{\prime}(v)|,|L^{\prime}(r)|,|L^{\prime}(t)|,|L^{\prime}(y)|,|L^{\prime}(w)|\geq 2.$ We begin by choosing a color $c$ from $L^{\prime}(u)$ such that $|L^{\prime}(y)-{c}|\geq 2.$ Then we choose colors of $v,r,w,t,s,$ and $y$ in this order, we obtain an $L^{\prime}$-coloring on $H$. Thus we can extend an $L$-coloring to $G,$ a contradiction. ∎

Corollary 3.9**.**

Let $v$ be a $k$-vertex in $int(C_{0})$ with consecutive incident faces $f_{1},\ldots,f_{k}$ where $k\leq 5.$ If $f_{1}$ and $f_{2}$ are inner $5^{-}$-faces, then there exists $w\in B(f_{1})\cup B(f_{2})$ such that $w\neq v$ and $d(w)\geq 5.$

Proof.

The result follows directly from Lemmas  3.4, 3.6, and 3.8. ∎

Corollary 3.10**.**

If $v$ is a $5$-vertex in which each incident face is an inner $5^{-}$-face, then $v$ is incident to at least three rich faces.

Proof.

Suppose to the contrary that $v$ is incident to at most two rich faces. Consequently, $v$ is incident to two adjacent semi-rich faces $f$ and $g.$ But $d(f)\leq 5$ and $d(g)\leq 5$ while all vertices in $B(f)\cup B(g)$ except $v$ has degree $4.$ This contradicts Corollary 3.9. ∎

Lemma 3.11**.**

Let $C(l_{1},\ldots,l_{k})$ in $int(C_{0})$ be obtained from a cycle $C=x_{1}\ldots x_{m}$ with $k$ internal chords sharing a common endpoint $x_{1}.$ Suppose $x_{1}$ is not incident to other chords while $x_{2}$ or $x_{m}$ is not incident to any chord. If $d(x_{1})\leq k+3,$ then there exists $i\in\{2,3,\ldots,m\}$ such that $d(x_{i})\geq 5.$

Proof.

By symmetry, we assume $x_{m}$ is not an endpoint of any chord in $C.$ Suppose to the contrary that $d(x_{i})\leq 4$ for each $i=2,3,\ldots,m.$ By the minimality of $G,$ the subgraph $G-\{x_{1},\ldots,x_{m}\}$ has an $L$-coloring where $L$ is restricted to $G-\{x_{1},\ldots,x_{m}\}$. Consider a residual list assignment $L^{\prime}$ on $x_{1},\ldots,x_{m}.$ Since $L$ is a $4$-assignment, we have $|L^{\prime}(x_{1})|\geq 3$ and $|L^{\prime}(v)|\geq 3$ for each $v\in V(C)$ with an edge $x_{1}v$ and $|L^{\prime}(x_{i})|\geq 2$ for each of the remaining vertices $x_{i}$ in $V(C).$ Since $x_{m}$ is not an endpoint of a chord in $C,$ we can choose a color $c$ from $L^{\prime}(x_{1})$ such that $|L^{\prime}(x_{m})-{c}|\geq 2.$ By choosing colors of $x_{2}$, $x_{3},\dots,x_{m}$ in this order, we obtain an $L^{\prime}$-coloring on $G^{\prime}$. Thus we can extend an $L$-coloring to $G,$ a contradiction. ∎

Corollary 3.12**.**

Let $v$ be a $6$-vertex with consecutive incident faces $f_{1},\ldots,f_{6}$ and let $F=B_{1}\cup B_{2}\cup B_{3}\cup B_{4}$ where $B_{i}$ denote $B(f_{i}).$ If $f_{1}\ldots f_{4}$ are bounded faces and $(d(f_{1}),d(f_{2}),d(f_{3}),d(f_{4}))=(5,3,5,3),$ then there exists $w\in V(F)-\{v\}$ with $d(w)\geq 5.$

Proof.

By Lemma 3.11, it suffices to show that $F=C(5,3,5,3).$ Let cycles $B_{1}=vqrst,B_{2}=vtu,$ $B_{3}=vuwxy,$ and $B_{4}=vyz.$ Using Lemma 3.4, we have that $V(B_{1})\cap V(B_{2})=\{v,t\},$ $V(B_{2})\cap V(B_{3})=\{v,u\},$ and $V(B_{3})\cap V(B_{4})=\{v,y\}.$ It suffices to show that $V(B_{1})\cap V(B_{3})=\{v\}$ $=V(B_{4})\cap(V(B_{1})\cup V(B_{2})).$

Suppose to the contrary that $V(B_{1})\cap V(B_{3})\neq\{v\}.$ Consider a $6$-cycle $vtuwxy$ with a triangular chord $uv.$ If $s=u,w,x,$ or $y,$ then $vtuwxy$ has another chord, contrary to Lemma 2.3. Thus $s\notin V(B_{1})\cap V(B_{3}).$ Similarly each of $q,w,$ and $y$ is not in $V(B_{1})\cap V(B_{3}).$ The only remaining possibility is that $r=x.$ Suppose this holds. Then $vyz,vyxq,vyxwu,$ and $vyrstu$ are four pairwise adjacent cycle, contrary to $G\in\mathcal{A}.$ Thus $V(B_{1})\cap V(B_{3})=\{v\}$ which implies $B_{1}\cup B_{2}\cup B_{3}=C(5,3,5).$ As a consequence, we have $vqrstu$ and $vtuwxy$ are $6$-cycles with a triangular chord.

If there is a vertex $b\in V(B_{4})\cap(V(B_{1})\cup V(B_{2}))$ such that $b\neq v,$ then $vqrstu$ or $vtuwxy$ has another chord, contrary to Lemma 2.3. This completes the proof. ∎

Corollary 3.13**.**

Let $v$ be a $4$-vertex incident to four inner $3$-faces. If all four neighbors of $v$ are $5^{-}$-vertices, then at least three of them are $5$-vertices.

Proof.

Let $w,x,y,z$ be neighbor of $v$ in a cyclic order. Let cycles $B_{1}=vwx$ and $B_{2}=vxy.$ Note that $w$ and $y$ are not adjacent, otherwise $vwy$ is a separating $3$-cycle, contrary to Lemma 3.2. Similarly, $x$ and $z$ are not adjacent.

Suppose to the contrary that there are at least two $4$-vertices among $w,x,y,$ and $z.$ If those two $4$-vertices are not adjacent, say $w$ and $y,$ then $B_{1}\cup B_{2}$ contradicts Lemma 3.6. Thus we assume that $w$ and $x$ are $4$-vertices.

Let $H$ be the graph induced by $v$ and its neighbors. By minimality of $G,$ the graph $G-H$ has an $L$-coloring where $L$ is restricted to $G-H.$ Consider a residual list assignment $L^{\prime}$ on $H.$ Since $L$ is a $4$-assignment, we have $|L^{\prime}(y)|,|L^{\prime}(z)|\geq 2,$ $|L^{\prime}(w)|,|L^{\prime}(x)|\geq 3,$ and $|L^{\prime}(v)|=4.$ It suffices to assume that equalities holds for these list sizes. We aim to show that $H$ has an $L^{\prime}$-coloring, and thus an $L$-coloring can be extended to $G,$ a contradiction.

CASE 1. There is a color $t$ in $L^{\prime}(v)-(L^{\prime}(y)\cup L^{\prime}(z))$.

We begin by choosing $t$ for $v.$ Each of the residual lists of $w,x,y,z$ now has sizes at least $2.$ By Lemma 2.1, an even cycle is $2$-choosable, thus $H$ has an $L^{\prime}$-coloring.

CASE 2. $L^{\prime}(v)-(L^{\prime}(y)\cup L^{\prime}(z))=\emptyset.$

This implies $L^{\prime}(y)\cap L^{\prime}(z)=\emptyset.$ Choose $t\in L^{\prime}(v)-L^{\prime}(w)$ for $v.$ If $t\in L^{\prime}(y),$ then $t\notin L^{\prime}(z)$ and we can color $y,x,z,$ and $w$ in this order, otherwise we can color $z,y,x,$ and $w$ in this order. Thus $H$ has an $L^{\prime}$-coloring. This contradiction completes the proof. ∎

4. Proof of Theorem 3.1

Let the initial charge of a vertex $u$ in $G$ be $\mu(u)=2d(u)-6$ and the initial charge of a face $f$ in $G$ be $\mu(f)=d(f)-6$. Then by Euler’s formula $|V(G)|-|E(G)|+|F(G)|=2$ and by the Handshaking lemma, we have

[TABLE]

Now we design the discharging rule transferring charge from one element to another to provide a new charge $\mu^{*}(x)$ for all $x\in V(G)\cup F(G).$ The total of new charges remains $-12$. If the final charge $\mu^{*}(x)\geq 0$ for all $x\in V(G)\cup F(G)$, then we get a contradiction and complete the proof.

Before we establish a discharging rule, some definitions are required. A graph $C(3,3,3)$ in $int(C_{0}$ is called a trio . A vertex that is not in any trio is called a good vertex. We call a vertex $v$ incident to a face $f$ in a trio $T$ a bad (worse, worst, respectively) vertex of $f$ if $v$ is incident to exactly one (two, three, respectively) $3$-face(s) in $T.$ We call a face $f$ in a trio $T$ a bad (worse,worst, respectively) face of a vertex $v$ if $v$ is a bad (worse, worst, respectively) vertex of $f$ in $T.$ For our purpose, we regard an external vertex of $W_{5}$ as a worse vertex of its incident $3$-faces in $W_{5}.$

Let $w(v\rightarrow f)$ be the charge transferred from a vertex $v$ to an incident face $f.$ From now on, a vertex $v$ is in $int(C_{0})$ unless stated otherwise. The discharging rules are as follows.

(R1) Let $f$ be an inner $3$-face that is not adjacent to another $3$-face.

(R1.1) For a $4$-vertex $v$,

$w(v\rightarrow f)=\begin{cases}\frac{9}{10},&\text{if }v\text{ is flaw},\\ 1,&\text{otherwise.}\end{cases}$

(R1.2) For a $5^{+}$-vertex $v$,

$w(v\rightarrow f)=\begin{cases}\frac{6}{5},&\text{if }f\text{ is a }(4,4,5^{+})\text{-face},\\ 1,&\text{otherwise.}\end{cases}$

(R2) Let $f$ be an inner $3$-face that is adjacent to another $3$-face.

(R2.1) For a $4$-vertex $v$,

$w(v\rightarrow f)=\begin{cases}\frac{1}{2},&\text{if }v\text{ is a hub of }W_{5},\\ 1,&\text{if }f\text{ is a good, bad, or worse face of }v,\\ \frac{2}{3},&\text{if }f\text{ is a worst face of }v.\end{cases}$

(R2.2) For a $5$-vertex $v$,

$w(v\rightarrow f)=\begin{cases}1,&\text{if }f\text{ is a good or worst face of }v,\\ \frac{5}{4},&\text{if }f\text{ is a worse face of }v,\\ \frac{3}{2},&\text{if }f\text{ is a bad face of }v.\end{cases}$

(R2.3) For a $6^{+}$-vertex $v$,

$w(v\rightarrow f)=\begin{cases}1,&\text{if }f\text{ is a good or worst face of }v,\\ \frac{3}{2},&\text{if }f\text{ is a bad or worse face of }v.\end{cases}$

(R3) Let $f$ be an inner $4$-face.

(R3.1) For a $4$-vertex $v$, let $w(v\rightarrow f)=\frac{1}{3}$.

(R3.2) For a $5^{+}$-vertex $v$,

$w(v\rightarrow f)=\begin{cases}1,&\text{if }f\text{ is a }(4,4,4,5^{+})\text{-face},\\ \frac{2}{3},&\text{if }f\text{ is rich. }\end{cases}$

(R4) Let $f$ be an inner $5$-face.

(R4.1) For a $4$-vertex $v$,

$w(v\rightarrow f)=\begin{cases}\frac{1}{5},&\text{if }v\text{ is flaw and }f\text{ is a poor }5\text{-face},\\ \frac{1}{3},&\text{if }v\text{ is incident to at most one }3\text{-face,}\\ 0,&\text{otherwise.}\end{cases}$

(R4.2) For a $5^{+}$-vertex $v$,

$w(v\rightarrow f)=\begin{cases}1,&\text{if }f\text{ is a }(4,4,4,4,5^{+})\text{-face adjacent to five }3\text{-faces},\\ \frac{2}{3},&\text{if }f\text{ is a }(4,4,4,4,5^{+})\text{-face adjacent to at least one }4^{+}\text{-face other than }f,\\ \frac{1}{t},&\text{if }f\text{ is a rich face with }t\text{ incident }5^{+}\text{-vertices.}\\ \end{cases}$

(R5) Let $f$ be an inner $3$-face incident to a $4$-vertex $v$ that is incident to four $3$-faces. If $f$ is adjacent to a $7^{+}$-face $g,$ we let $w(g\rightarrow f)=\frac{1}{8}$.

(R6) The unbounded face $D$ gets $\mu(v)$ from each incident vertex.

(R7) Let $f$ be an extreme face.

$w(x\rightarrow f)=\begin{cases}\frac{5}{2},&\text{if }f\text{ is a 3-face that }B(f)\text{ shares an edge with }C_{0}\text{ and }x=D,\\ 2,&\text{if }f\text{ is a 4- or 5-face and }x=D,\\ 2,&\text{if }f\text{ is a 3-face that }B(f)\text{ shares exactly one vertex with }C_{0}\text{ and }x=D,\\ \frac{1}{2},&\text{if }f\text{ is a 3-face and }x\text{ is a vertex in }int(C_{0}),\\ 0,&\text{otherwise.}\\ \end{cases}$

(R8) After (R1) to (R7), redistribute the total of charges of $3$-faces in the same cluster of at least three adjacent inner $3$-faces (trio or $W_{5}$) equally among its $3$-faces.

It remains to show that resulting $\mu^{*}(x)\geq 0$ for all $x\in V(G)\cup F(G)$. Let $v$ be a $k$-vertex incident to faces $f_{1},\ldots,f_{k}$ in a cyclic order. By (R6), we only consider $v$ in $int(C_{0}).$ Consider the following cases.

• (1)

$v$ is a $4$-vertex.

• (1.1)

A vertex $v$ is incident to a $3$-face that is adjacent to another $3$-face.

• (1.1.1)

$v$ is incident to at least two consecutive $3$-faces.

If $v$ is incident to four $3$-faces, then $v$ is a hub of $W_{5}$. Thus $\mu^{*}(v)\geq\mu(v)-4\times\frac{1}{2}=0$ by (R2.1) and (R7). If $v$ is incident to exactly three $3$-faces, say $f_{1},f_{2},$ and $f_{3},$ then $f_{4}$ is a $6^{+}$-face by Lemma 3.5(1) , (2). Thus $\mu^{*}(v)\geq\mu(v)-3\times\frac{2}{3}=0$ by (R2.1) and (R7). If $v$ is incident to exactly two $3$-faces, say $f_{1}$ and $f_{2},$ then $f_{3}$ and $f_{4}$ are $6^{+}$-faces by Lemma 3.5(1), (2). Thus $\mu^{*}(v)\geq\mu(v)-2\times 1=0$ by (R2.1) and (R7).

• (1.1.2)

$v$ has no adjacent incident $3$-faces.

Let $f_{1}$ be a $3$-face adjacent to another $3$-cycle. If follows from Lemma 3.5(1) and (2) that $f_{2}$ and $f_{4}$ are $6^{+}$-faces. Then $w(v\rightarrow f_{1})\leq 1$ by (R2.1) and (R7), and $w(v\rightarrow f_{3})\leq 1$ by (R2.1), (R3.1), (R4.1), and (R7). Thus $\mu^{*}(v)\geq\mu(v)-2\times 1=0.$

• (1.2)

$v$ is not incident to a $3$-face that is adjacent to another $3$-face and $v$ is adjacent to at most one $3$-face.

Using the fact that $w(v\rightarrow f_{i})\leq 1$ for a $3$-face $f_{i}$ by (R1.1) and (R7), and $w(v\rightarrow f_{i})\leq\frac{1}{3}$ for each $4^{+}$-face $f_{i}$ by (R3.1), (R4.1), and (R7), we obtain that $\mu^{*}(v)\geq\mu(v)-1-3\times\frac{1}{3}=0.$

• (1.3)

$v$ is not incident to a $3$-face that is adjacent to another $3$-face and $v$ is adjacent to at least two $3$-faces.

Consequently, $v$ is incident to exactly two $3$-faces, say $f_{1}$ and $f_{3}.$ It follows from Lemma 3.5(3) that $f_{2}$ and $f_{4}$ are $5^{+}$-faces. If $f_{2}$ is an extreme face, then $w(v\rightarrow f_{2})=0$ by (R7), $w(v\rightarrow f_{4})=0$ by (R4.1) and (R7), and $w(v\rightarrow f_{i})\leq 1$ for $i=1$ and $3$ by (R1.1) and (R7). Thus $\mu^{*}(v)\geq\mu(v)-2\times 1=0.$ Now assume that all incident faces of $v$ are inner faces. If $v$ is flaw, then $v$ is incident to exactly one poor $5$-face, say $f_{2}$ by Corollary 3.7(1). $w(v\rightarrow f_{i})=\frac{9}{10}$ for $i=1$ and $3$ by (R1.1), $w(v\rightarrow f_{2})\leq\frac{1}{5}$ and $w(v\rightarrow f_{4})=0$ by (R4.1) and (R7). Thus $\mu^{*}(v)\geq\mu(v)-2\times\frac{9}{10}-\frac{1}{5}=0.$ If $v$ is not flaw, then $w(v\rightarrow f_{i})=1$ for $i=1$ and $3$ by (R1.1) and $w(v\rightarrow f_{i})=0$ for $i=2$ and $4$ by (R4.1). Thus $\mu^{*}(v)=\mu(v)-2\times 1=0.$

• (2)

A $5$-vertex $v$ is incident to a $3$-face that is adjacent to another $3$-face.

• (2.1)

$v$ has at least two consecutive incident $3$-faces.

If $v$ is incident to four $3$-faces say $f_{1},f_{2},f_{3},$ and $f_{4}$, then one can see that $B(f_{1})\cup B(f_{2})\cup B(f_{3})\cup B(f_{4})=C(3,3,3,3).$ But $C(3,3,3,3)$ contains four pairwise adjacent cycles that contradict $G\in\mathcal{A}.$ Thus $v$ is incident to at most three consecutive $3$-faces.

If $v$ incident to consecutive three $3$-faces say $f_{1},f_{2},$ and $f_{3}$, then $f_{4}$ and $f_{5}$ are $6^{+}$-faces by Lemma 3.5(1) and (2). Thus $\mu^{*}(v)=\mu(v)-3\times 1>0$ by (R2.2) and (R7).

If $v$ incident to exactly two consecutive $3$-faces say $f_{1}$ and $f_{2}$, then $f_{3}$ and $f_{5}$ are $6^{+}$-faces by Lemma 3.5(1) and (2). Consequently, $w(v\rightarrow f_{i})\leq\frac{5}{4}$ for $i=1$ and $2,$ and $w(v\rightarrow f_{4})\leq\frac{3}{2}$ by (R1.2), (R3.2), (R4.2), and (R7). Thus $\mu^{*}(v)\geq\mu(v)-2\times\frac{5}{4}-\frac{3}{2}=0.$

• (2.2)

$v$ has no adjacent incident $3$-faces.

Let $f_{1}$ be a $3$-face adjacent to another $3$-face. If follows from Lemma 3.5(1) and (2) that $f_{2}$ and $f_{5}$ are $6^{+}$-faces. By (R2.2) and (R7), $w(v\rightarrow f_{1})\leq\frac{3}{2}.$ If neither $f_{3}$ nor $f_{4}$ are $3$-faces, then $w(v\rightarrow f_{i})\leq 1$ for $i=3$ and $4$ by (R3.2), (R4.2), and (R7). Thus $\mu^{*}(v)\geq\mu(v)-\frac{3}{2}-2\times 1>0.$

Now assume that $f_{3}$ is a $3$-face. By the condition of (2.2), $f_{4}$ is a $4^{+}$-face which implies $w(v\rightarrow f_{4})\leq 1$ by (R3.2), (R4.2), and (R7). If $f_{3}$ is adjacent to another $3$-face, then $f_{4}$ is a $6^{+}$-face by Lemma 3.5(1) and (2). Moreover, $w(v\rightarrow f_{3})\leq\frac{3}{2}$ by (R2.2) and (R7). Thus $\mu^{*}(v)\geq\mu(v)-2\times\frac{3}{2}>0.$ If $f_{3}$ is not adjacent to another $3$-face, then $w(v\rightarrow f_{3})\leq 1$ by (R2.2) and (R7). Thus $\mu^{*}(v)\geq\mu(v)-\frac{3}{2}-2\times 1>0.$

• (3)

A $5$-vertex $v$ is not incident to a $3$-face that is adjacent to another $3$-face and $v$ is incident to at least one $6^{+}$-face.

Consequently, $v$ is incident to at most two $3$-faces.

• (3.1)

$v$ is incident to at least two $6^{+}$-faces.

Recall that $w(v\rightarrow f_{i})\leq\frac{6}{5}$ for each $3$-face $f_{i}$ by (R1.2) and (R7), and $w(v\rightarrow f_{i})\leq 1$ for each $k$-face $f_{i}$ where $k=4,5$ by (R3.2), (R4.2), and (R7). If $v$ is incident to $t$ $3$-faces, then there are at most $3-t$ faces $f$ with $d(f)=4$ or $5.$ Thus $\mu^{*}(v)\geq\mu(v)-t\times\frac{6}{5}-(3-t)\times 1>0$ by $t\leq 3.$

• (3.2)

$v$ is incident to exactly one $6^{+}$-face and incident to at most one $3$-face.

If $v$ has no incident $3$-faces, then $v$ has all incident faces $f$ except one $6^{+}$-face has $d(f)=4$ or $5.$ Thus $\mu^{*}(v)\geq\mu(v)-4\times 1=0$ by (R3.2), (R4.2), and (R7).

Assume $v$ is incident to exactly one $3$-face, say $f_{1}.$ By Lemma 3.5(3), $v$ is not a $(3,4,4,4,6^{+})$- or a $(3,4,4,6^{+},4)$-face. Consequently, $v$ has at least one incident $5$-face $f_{j}.$ Moreover, $f_{j}$ is adjacent to at least one $4^{+}$-face. We have $w(v\rightarrow f_{1})\leq\frac{6}{5}$ by (R1.2) and (R7), $w(v\rightarrow f_{j})\leq\frac{2}{3}$ by (R4.2) and (R7), and $w(v\rightarrow f_{i})\leq 1$ for each remaining $k$-face $f_{i}$ where $k=4,5$ by (R3.2), (R4.2), and (R7). Thus $\mu^{*}(v)\geq\mu(v)-\frac{6}{5}-\frac{2}{3}-2\times 1>0.$

• (3.3)

$v$ is incident to exactly one $6^{+}$-face and incident to exactly two $3$-faces, say $f_{1}$ and $f_{3}.$

It follows from Lemma 3.5(3) and (4) that $v$ is either a $(3,5,3,5,6^{+})$-, $(3,5,5,3,6^{+})$- or $(3,5,4,3,6^{+})$-vertex.

Assume $v$ is a $(3,5,3,5,6^{+})$- or $(3,5,5,3,6^{+})$-vertex. Applying Corollary 3.9 to $B(f_{2})\cup B(f_{3}),$ $v$ has an incident $5$-face $f_{j}$ which is rich or extreme. Recall that $w(v\rightarrow f_{i})\leq\frac{6}{5}$ for each $3$-face $f_{i}$ by (R1.2) and (R7), $w(v\rightarrow f_{j})\leq\frac{1}{2}$ by (R4.2) and (R7), and $w(v\rightarrow f_{i})\leq 1$ for the remaining $5$-face $f_{i}$ by (R4.2) and (R7). Thus $\mu^{*}(v)\geq\mu(v)-2\times\frac{6}{5}-\frac{1}{2}-1>0.$

Assume $v$ is a $(3,5,4,3,6^{+})$-vertex. Applying Corollary 3.9 to $B(f_{1})\cup B(f_{2}),$ we obtain that $f_{1}$ or $f_{2}$ is rich or extreme. In the former case, $w(v\rightarrow f_{1})\leq 1$ by (R1.2) and (R7), and $w(v\rightarrow f_{2})\leq\frac{2}{3}$ by (R4.2) and (R7). In the latter case, $w(v\rightarrow f_{1})\leq\frac{6}{5}$ by (R1.2) and (R7), and $w(v\rightarrow f_{2})\leq\frac{1}{2}$ by (R4.2) and (R7). Combining with $w(v\rightarrow f_{3})\leq 1$ by (R3.2) and (R7) and $w(v\rightarrow f_{4})\leq\frac{6}{5}$ by (R1.2) and (R7), we have $\mu^{*}(v)\geq\mu(v)-2\times 1-\frac{2}{3}-\frac{6}{5}>0$ or $\mu^{*}(v)\geq\mu(v)-2\times\frac{6}{5}-\frac{1}{2}-1>0.$

• (4)

A $5$-vertex $v$ is not incident to a $3$-face that is adjacent to another $3$-face and $v$ is not incident to a $6^{+}$-face.

Consequently, $v$ is incident to at most two $3$-faces. Using Corollary 3.10, we have that $v$ has at least three incident faces that are rich or extreme.

• (4.1)

$v$ has no incident $3$-faces.

If $f$ has an extreme face $f_{i},$ then $w(v\rightarrow f_{i})=0$ by (R7) and $w(v\rightarrow f_{i})\leq 1$ for each remaining $f_{i}$ by (R3.2), (R4.2), and (R7). Thus $\mu^{*}(v)\geq\mu(v)-4\times 1=0.$

If $f$ has $t$ rich faces, then $\mu^{*}(v)\geq\mu(v)-t\times\frac{2}{3}-(5-t)\times 1\geq 0$ by (R3.2), (R4.2), (R7), and $t\geq 3.$

• (4.2)

$v$ is incident to exactly one $3$-face, say $f_{1}.$

It follows from Lemma 3.5(3) that $v$ has at most two incident $4$-faces.

• (4.2.1)

$v$ has no incident $4$-faces.

We have that $w(v\rightarrow f_{1})\leq\frac{6}{5}$ by (R1.2) and (R7) and $w(v\rightarrow f_{i})\leq\frac{2}{3}$ for each $5$-face $f_{i}$ by (R4.2) and (R7). Thus $\mu^{*}(v)\geq\mu(v)-\frac{6}{5}-4\times\frac{2}{3}>0.$

• (4.2.2)

$v$ has exactly one incident $4$-face.

It follows from Lemma 3.5(4) that $v$ is a $(3,5,4,5,5)$-face. Recall that $w(v\rightarrow f_{1})\leq\frac{6}{5}$ by (R1.2) and (R7), $w(v\rightarrow f_{3})\leq 1$ by (R3.2) and (R7), and $w(v\rightarrow f_{i})\leq\frac{2}{3}$ for each remaining $f_{i}$ by (R4.2) and (R7). If $f_{3}$ is rich, then $w(v\rightarrow f_{3})\leq\frac{2}{3}$ by (R3.2) and (R7). Thus $\mu^{*}(v)\geq\mu(v)-\frac{6}{5}-4\times\frac{2}{3}>0.$ If $f_{3}$ is not rich, then $f_{2}$ and $f_{4}$ are rich by Corollary 3.9. Consequently, $w(v\rightarrow f_{i})\leq\frac{1}{2}$ for $i=2$ or $4$ by (R4.2) and (R7). Thus $\mu^{*}(v)\geq\mu(v)-\frac{6}{5}-1-2\times\frac{1}{2}-\frac{2}{3}>0.$

• (4.2.3)

$v$ has exactly two incident $4$-faces.

It follows from Lemma 3.5(3) and (4) that $v$ is a $(3,4,5,5,4)$- or a $(3,5,4,4,5)$-face. Moreover, $v$ has at least three incident rich faces by Corollary 3.10. Consequently, we have (i) $f_{1}$ and at least one $4$-face $f_{i}$ are rich, (ii) $f_{1}$ and two $5^{+}$-faces are rich, (iii) a $4$-face and two $5$-faces are rich, or (iv) two $4$-faces and a $5$-face are rich.

Recall that $w(v\rightarrow f_{1})\leq\frac{6}{5}$ by (R1.2) and (R7), $w(v\rightarrow f_{i})\leq 1$ for each $4$-face $f_{i}$ by (R3.2) and (R7), and $w(v\rightarrow f_{i})\leq\frac{2}{3}$ for each $5$-face $f_{i}$ by (R4.2) and (R7). Additionally, $w(v\rightarrow f_{1})\leq 1$ if $f_{1}$ is rich by (R1.2) and (R7), $w(v\rightarrow f_{i})\leq\frac{2}{3}$ for each rich $4$-face $f_{i}$ by (R3.2) and (R7), and $w(v\rightarrow f_{i})\leq\frac{1}{2}$ for each rich $5$-face $f_{i}$ by (R4.2) and (R7).

If $f_{1}$ and a $4$-face $f_{i}$ are rich, then $\mu^{*}(v)\geq\mu(v)-2\times 1-3\times\frac{2}{3}=0.$ If $f_{1}$ and two $5^{+}$-faces are rich, then $\mu^{*}(v)\geq\mu(v)-1-2\times 1-2\times\frac{1}{2}=0.$ If a $4$-face and two $5^{+}$-faces are rich, then $\mu^{*}(v)\geq\mu(v)-\frac{6}{5}-1-\frac{2}{3}-2\times\frac{1}{2}>0.$ If two $4$-faces and a $5$-face are rich, then $\mu^{*}(v)\geq\mu(v)-\frac{6}{5}-3\times\frac{2}{3}-\frac{1}{2}>0.$

• (4.3)

$v$ is incident to exactly two $3$-faces, say $f_{1}$ and $f_{3}.$

It follows from Lemma 3.5(3) and (4) that $v$ has no incident $4$-faces. This implies $v$ is a $(3,5,3,5,5)$-vertex. Recall that $w(v\rightarrow f_{i})\leq\frac{6}{5}$ for each $3$-face $f_{i}$ by (R1.2) and (R7), and $w(v\rightarrow f_{i})\leq\frac{2}{3}$ for each $5$-face $f_{i}$ by (R4.2) and (R7). Additionally, $w(v\rightarrow f_{1})\leq 1$ if $f_{1}$ is rich by (R1.2) and (R7), and $w(v\rightarrow f_{i})\leq\frac{1}{2}$ for each rich $5$-face $f_{i}$ by (R4.2) and (R7).

If each incident $5$-face is rich, then $\mu^{*}(v)\geq\mu(v)-2\times\frac{6}{5}-3\times\frac{1}{2}>0.$ If $f_{2}$ is not rich, then $f_{1}$ and $f_{3}$ are rich by Corollary 3.9. Consequently, $f_{4}$ and $f_{5}$ are also rich. Thus $\mu^{*}(v)\geq\mu(v)-3\times 1-2\times\frac{1}{2}=0.$ If $f_{4}$ is not rich, then $f_{3}$ and $f_{5}$ are rich by Corollary 3.9. Consequently, $f_{2}$ is also rich. Thus $\mu^{*}(v)\geq\mu(v)-\frac{6}{5}-1-\frac{2}{3}-2\times\frac{1}{2}>0.$ The case that $f_{5}$ is not rich is similar.

• (5)

A $6$-vertex $v$ is incident to a $3$-face that is adjacent to another $3$-face.

• (5.1)

$v$ is incident to at least two consecutive $3$-faces.

Let $f_{1},\ldots,f_{k}$ be consecutive $3$-faces. Similar to Case (2.1), we have $k\leq 3.$ It follows from Lemma 3.5(1) and (2) that $v$ is a $(3,3,6^{+},k_{4},k_{5},6^{+})$- or $(3,3,3,6^{+},k_{5},6^{+})$-face. Since $w(v\rightarrow f_{i})\leq\frac{3}{2}$ for each $5^{-}$-face $f_{i}$ by (R2.3), (R3.2), (R4.2), and (R7), Thus $\mu^{*}(v)\geq\mu(v)-4\times\frac{3}{2}=0.$

• (5.2)

$v$ has no adjacent incident $3$-faces.

Let $f_{1}$ be a $3$-face adjacent to another $3$-face. It follows from Lemma 3.5(1) and (2) that $f_{2}$ and $f_{6}$ are $6^{+}$-faces. Similar to Case 5.1, we obtain that $\mu^{*}(v)\geq\mu(v)-4\times\frac{3}{2}=0.$

• (6)

A $6$-vertex $v$ is not incident to a $3$-face that is adjacent to another $3$-face.

Consequently, $v$ is incident to at most three $3$-faces.

• (6.1)

$v$ is incident to at least one $6^{+}$-face.

Recall that $w(v\rightarrow f_{i})\leq\frac{6}{5}$ for each $3$-face $f_{i}$ by (R1.2) and (R7), and $w(v\rightarrow f_{i})\leq\frac{3}{2}$ for each $k$-face $f_{i}$ where $k=4$ or $5$ by (R3.2), (R4.2), and (R7). Thus $\mu^{*}(v)\geq\mu(v)-t\times\frac{6}{5}-(5-t)\times 1>0$ where $t\leq 3$ is the number of incident $3$-faces.

• (6.2)

$v$ has no incident $6^{+}$-face.

• (6.2.1)

$v$ has no incident $3$-faces.

By (R3.2), (R4.2), and (R7), we have $\mu^{*}(v)\geq\mu(v)-6\times 1=0.$

• (6.2.2)

$v$ has exactly one incident $3$-face, say $f_{1}.$

It follows from Lemma 3.5(3) that $v$ is not a $(3,4,4,4,4,4)$-vertex. Consequently, $v$ has $s$ $5$-faces where $t\geq 1.$ Note that each incident face of $v$ is adjacent to another $4^{+}$-face. It follows that $w(v\rightarrow f_{i})\leq\frac{2}{3}$ for each $5$-face $f_{i}$ by (R4.2) and (R7). Recall that $w(v\rightarrow f_{1})\leq\frac{6}{5}$ by (R1.2) and (R7), and $w(v\rightarrow f_{i})\leq 1$ for each $4$-face $f.$ Thus $\mu^{*}(v)\geq\mu(v)-\frac{6}{5}-s\times\frac{2}{3}-(5-s)\times 1>0.$

• (6.2.3)

$v$ has exactly two incident $3$-faces.

Consequently, $v$ is a $(3,k_{2},3,k_{4},k_{5},k_{6})$- or $(3,k_{2},k_{3},3,k_{5},k_{6})$-vertex.

Assume $v$ is a $(3,k_{2},3,k_{4},k_{5},k_{6})$-face. Then $k_{2}=5$ by Lemma 3.5(3). This implies $k_{4}=k_{6}=5$ by Lemma 3.5(4). Since $v$ is a $(3,5,3,5,4^{+},5)$-vertex, we have $w(v\rightarrow f_{i})\leq\frac{6}{5}$ for $i=1$ and $3$ by (R1.2) and (R7), $w(v\rightarrow f_{i})\leq 1$ for $i=2$ and $5$ by (R3.2),(R4.2) and (R7), and $w(v\rightarrow f_{i})\leq\frac{2}{3}$ for $i=4$ and $6$ by (R4.2) and (R7). Thus $\mu^{*}(v)\geq\mu(v)-2\times\frac{6}{5}-2\times 1-2\times\frac{2}{3}>0.$

Assume $v$ is a $(3,k_{2},k_{3},3,k_{5},k_{6})$-vertex. It follows from Lemma 3.5(4) that $\{k_{2},k_{6}\}\neq\{4,5\}.$ If $k_{2}=k_{6}=4,$ then $k_{3}=k_{5}=5$ by Lemma 3.5(3). Consequently, we may assume that $v$ is a $(3,4,5,3,5,4)$- and $(3,5,5,3,5,5)$-vertex. Recall that $w(v\rightarrow f_{i})\leq\frac{6}{5}$ for $i=1$ and $4$ by (R1.2) and (R7), $w(v\rightarrow f_{i})\leq 1$ for each $4$-face $f_{i}$ by (R3.2) and (R7). and $w(v\rightarrow f_{i})\leq\frac{2}{3}$ for each $5$-face $f_{i}$ by (R4.2) and (R7). Thus a $(3,4,5,3,5,4)$-vertex has $\mu^{*}(v)\geq\mu(v)-2\times\frac{6}{5}-2\times 1-2\times\frac{2}{3}>0,$ and a $(3,5,5,3,5,5)$-vertex has $\mu^{*}(v)\geq\mu(v)-2\times\frac{6}{5}-4\times\frac{2}{3}-1>0.$

• (6.2.4)

$v$ has exactly three incident $3$-faces.

Consequently, $v$ is a $(3,5,3,5,3,5)$-vertex by Lemma 3.5(3). It follows from Corollary 3.12 that two incident $5$-faces of $v$, say $f_{2},$ and $f_{4},$ are rich. This implies $w(v\rightarrow f_{i})\leq\frac{6}{5}$ for $i=1,3$ and $5$ by (R1.2) and (R7), $w(v\rightarrow f_{i})\leq\frac{1}{2}$ for $i=2$ and $4,$ $w(v\rightarrow f_{6})\leq 1$ by (R4.2) and (R7). Thus $\mu^{*}(v)\geq\mu(v)-3\times\frac{6}{5}-2\times\frac{1}{2}-1>0.$

• (7)

$v$ is a $k$-vertex where $k\geq 7.$

• (7.1)

A vertex $v$ is incident to a $3$-face that is adjacent to another $3$-face.

Then $v$ is incident to at least two $6^{+}$-faces by Lemma 3.5(1) and (2). Thus $\mu^{*}(v)\geq\mu(v)-(k-2)\times\frac{3}{2}>0$ by (R2.3), (R3.2), (R4.2), and (R7).

• (7.2)

A vertex $v$ is not incident to a $3$-face that is adjacent to another $3$-face.

Consequently $v$ is incident to $t$ $3$-faces where $t\leq k/2.$ Thus $\mu^{*}(v)\geq\mu(v)-t\times\frac{6}{5}-(k-t)\times 1>0$ by (R1.2), (R3.2), (R4.2), and (R7).

• (8)

An inner $3$-face $f$ is not adjacent to another $3$-face.

If $f$ has no incident flaw $4$-vertices, then $\mu^{*}(f)\geq\mu(f)+3\times 1=0$ by (R1.1) adn (R1.2). If $f$ has an incident flaw vertex, then $f$ is a $(4,4,5^{+})$-face by Corollary 3.7(2). Recall that $w(v\rightarrow f)\geq\frac{9}{10}$ for an incident $4$-vertex $v$ by (R1.1), and $w(v\rightarrow f)\geq\frac{6}{5}$ for an incident $5^{+}$-vertex $v$ by (R1.2). Thus $\mu^{*}(f)\geq\mu(f)+2\times\frac{9}{10}+\frac{6}{5}=0.$

• (9)

An inner $3$-face $f$ is adjacent to another $3$-face.

Note that we use only (R2) to calculate a new charge.

• (9.1)

A face $f$ is not in a trio.

Then $\mu^{*}(f)\geq\mu(f)+3\times 1=0.$

• (9.2)

A face $f$ is in a trio $T$ but not in $W_{5}$ formed by four inner $3$-faces.

Let $f_{1},f_{2},$ and $f_{3}$ be $3$-faces in the same trio $T.$ Define $\mu(T):=\mu(f_{1})+\mu(f_{2})+\mu(f_{3})=-9$ and $\mu^{*}(T):=\mu^{*}(f_{1})+\mu^{*}(f_{2})+\mu^{*}(f_{3}).$ By (R8), it suffices to prove that $\mu^{*}(T)\geq 0.$

• (9.2.1)

A worst vertex is a $5^{+}$-vertex.

Then $\mu^{*}(T)\geq\mu(T)+9\times 1=0.$

• (9.2.2)

A worst vertex is a $4$-vertex and each worse vertex is a $4$-vertex.

Then two bad vertices are $5^{+}$-vertices by Corollary 3.9. Thus $\mu^{*}(T)\geq\mu(T)+3\times\frac{2}{3}+2\times\frac{3}{2}+4\times 1=0.$

• (9.2.3)

A worst vertex is a $4$-vertex and one of worse vertices is a $5$-vertex.

Then Corollary 3.9 yields that the other worse vertex or at least one bad vertex is a $5^{+}$-vertex. Thus $\mu^{*}(T)\geq\mu(T)+3\times\frac{2}{3}+4\times\frac{5}{4}+2\times 1=0$ or $\mu^{*}(T)\geq\mu(T)+3\times\frac{2}{3}+2\times\frac{5}{4}+\frac{3}{2}+3\times 1=0$, respectively.

• (9.2.3)

A worst vertex is a $4$-vertex and one of worse vertices is a $6^{+}$-vertex.

Then $\mu^{*}(T)\geq\mu(T)+3\times\frac{2}{3}+2\times\frac{3}{2}+4\times 1=0.$

• (9.3)

A face $f$ is in $W_{5}$ formed by four inner $3$-faces incident to $v.$

Let $f_{1},f_{2},f_{3},$ and $f_{4}$ be $3$-faces in the same $W_{5}.$ Define $\mu(W_{5}):=\mu(f_{1})+\mu(f_{2})+\mu(f_{3})+\mu(f_{4})=-12$ and $\mu^{*}(W_{5}):=\mu^{*}(f_{1})+\mu^{*}(f_{2})+\mu^{*}(f_{3})+\mu^{*}(f_{4}).$ By (R8), it suffices to prove that $\mu^{*}(W_{5})\geq 0.$ Note that each $3$-face in $W_{5}$ is adjacent to a $7^{+}$-face by Lemma 3.5(5). Thus $W_{5}$ always obtains $4\times\frac{1}{8}$ from four $7^{+}$-faces by (R5).

• (9.3.1)

Each vertex of $W_{5}$ is a $5^{-}$-vertex.

Then at least three of them are $5$-vertices by Corollary 3.13. Thus $\mu^{*}(W_{5})\geq\mu(W_{5})+6\times\frac{5}{4}+2\times 1+4\times\frac{1}{2}+4\times\frac{1}{8}=0.$

• (9.3.2)

Exactly one vertex of $W_{5}$ is a $6^{+}$-vertex.

Then one of the remaining vertices is a $5^{+}$-vertex by Corollary 3.9. Thus $\mu^{*}(W_{5})=\mu(W_{5})+2\times\frac{3}{2}+2\times\frac{5}{4}+4\times 1+4\times\frac{1}{2}+4\times\frac{1}{8}=0.$

• (9.3.3)

At least two vertices of $W_{5}$ are $6^{+}$-vertices.

Then $\mu^{*}(W_{5})\geq\mu(W_{5})+4\times\frac{3}{2}+4\times 1+4\times\frac{1}{2}+4\times\frac{1}{8}>0.$

• (10)

$f$ is an inner $4$-face.

We claim that $f$ is a $(4^{+},4^{+},4^{+},5^{+})$-face. Suppose to the contrary that $f$ is a $(4,4,4,4)$-face. By the minimality of $G,$ there is an $L$-coloring of $G-B(f)$ where $L$ is restricted to $G-B(f).$ After the coloring, each vertex of $B(f)$ has at least two legal colors. By Lemma 2.1, we can extend an $L$-coloring to $G,$ a contradiction.

If $f$ is a $(4,4,4,5^{+})$-face, then $\mu^{*}(f)\geq\mu(f)+3\times\frac{1}{3}+1=0$ by (R3). If $f$ is a $(4^{+},4^{+},5^{+},5^{+})$- or $(4^{+},5^{+},4^{+},5^{+})$-face, then $f$ is a rich face and thus $\mu^{*}(f)\geq\mu(f)+2\times\frac{1}{3}+2\times\frac{2}{3}=0$ by (R3).

• (11)

$f$ is an inner $5$-face.

• (11.1)

$f$ is a poor $5$-face, that is $f$ is a $(4,4,4,4,4)$-face.

It follows from Lemma 3.5(2) that each incident $4$-vertex of $f$ is incident to at most two $3$-faces.

If an incident vertex $v$ of $f$ is incident to at most one $3$-face, then $w(v\rightarrow f)=\frac{1}{3}$ by (R4.1), otherwise $v$ is a flaw vertex by Lemma 3.5(2) and (3) which implies $w(v\rightarrow f)=\frac{1}{5}$ by (R4.1). Consequently, $f$ gains at least $\frac{1}{5}$ from each incident vertex. Thus $\mu^{*}(f)\geq\mu(f)+5\times\frac{1}{5}=0.$

• (11.2)

$f$ is a $(4,4,4,4,5^{+})$-face.

• (11.2.1)

$f$ is adjacent to at least one $4^{+}$-face $g.$

It follows from (R4.2) that $w(v\rightarrow f)=\frac{2}{3}$ for an incident $5^{+}$-vertex $v$ of $f.$ Consider a $4$-vertex $u\in V(B(f))\cap V(B(g)).$ It follows from Lemma 3.5(2) that $u$ is incident to at most one $3$-face. Consequently, $w(u\rightarrow f)=\frac{1}{3}$ by (R4.1). Thus $\mu^{*}(f)\geq\mu(f)+\frac{2}{3}+\frac{1}{3}=0.$

• (11.2.2)

$f$ is adjacent to five $3$-faces.

Then $\mu^{*}(f)=\mu(f)+1=0$ by (R4.2).

• (11.3)

$f$ is a rich face with $t$ incident $5^{+}$-vertices.

Then $\mu^{*}(f)\geq\mu(f)+t\times\frac{1}{t}=0$ by (R4.2).

• (12)

$f$ is an inner $6^{+}$-face.

If $f$ is a $6$-face, then $\mu^{*}(f)=\mu(f)=0.$ If $f$ is a $k$-face where $k\geq 7,$ then $\mu^{*}(f)\geq\mu(f)-k\times\frac{1}{8}>0$ by (R5).

• (13)

$f$ is an extreme face.

• (13.1)

$f$ is a $3$-face that shares exactly one vertex, say $u,$ with $C_{0}.$

It follows from (R7) that $w(D\rightarrow f)=2$ and $w(v\rightarrow f)=\frac{1}{2}$ for each incident vertex $v$ in $int(C_{0}).$ Thus $\mu^{*}(f)=\mu(f)+2+2\times\frac{1}{2}=0.$

• (13.2)

$f$ is a $3$-face that shares an edge with $C_{0}.$

It follows from (R7) that $w(D\rightarrow f)=\frac{5}{2}$ and $w(v\rightarrow f)=\frac{1}{2}$ for an incident vertex $v$ in $int(C_{0}).$ Thus $\mu^{*}(f)=\mu(f)+\frac{5}{2}+\frac{1}{2}=0.$

• (13.3)

$f$ is a $4$- or $5$-face.

Then $\mu^{*}(f)\geq\mu(f)+2\geq 0$ by (R7).

• (14)

$D$ is the unbounded face.

Let $f_{3}^{\prime},f^{\prime}$ be the number of $3$-faces sharing exactly one edge with $D$, $3$-faces sharing exactly one vertex with $D$ or $4$-or $5$-faces sharing vertices with $D$, respectively. Let $E(C_{0},V(G)-C_{0})$ be the set of edges between $C_{0}$ and $V(G)-C_{0}$ and let $e(C_{0},V(G)-C_{0})$ be its size. Then by (R6) and (R7),

[TABLE]

So we may consider that each edge $e\in E(C_{0},V(G)-C_{0})$ gives a charge of $2$ to $D.$ Since each $5^{-}$-face is a cycle, it contains two edges in $E(C_{0},V(G)-C_{0})$. It follows that $e(C_{0},V(G)-C_{0})-f_{3}^{\prime}-f^{\prime}\geq 0$. Note that $f_{3}^{\prime}\leq 3$. Thus $\mu^{*}(D)>0.$