Plane graphs without 4- and 5-cycles and without ext-triangular 7-cycles are 3-colorable

TL;DR

This paper proves that certain classes of planar graphs, specifically those without 4- and 5-cycles and with restrictions on 7- and 8-cycles, are 3-colorable, advancing understanding related to Steinberg's conjecture.

Contribution

It establishes 3-colorability for plane graphs without 4- and 5-cycles under the absence of ext-triangular 7-cycles, extending known results.

Findings

Plane graphs without 4- and 5-cycles are 3-colorable if they lack ext-triangular 7-cycles.

Planar graphs without 4-, 5-, 7-cycles are 3-colorable.

Planar graphs without 4-, 5-, 8-cycles are 3-colorable.

Abstract

Listed as No. 53 among the one hundred famous unsolved problems in [J. A. Bondy, U. S. R. Murty, Graph Theory, Springer, Berlin, 2008] is Steinberg's conjecture, which states that every planar graph without 4- and 5-cycles is 3-colorable. In this paper, we show that plane graphs without 4- and 5-cycles are 3-colorable if they have no ext-triangular 7-cycles. This implies that (1) planar graphs without 4-, 5-, 7-cycles are 3-colorable, and (2) planar graphs without 4-, 5-, 8-cycles are 3-colorable, which cover a number of known results in the literature motivated by Steinberg's conjecture.

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1 Introduction

In the field of 3-colorings of planar graphs, one of the most active topics is about a conjecture proposed by Steinberg in 1976: every planar graph without cycles of length 4 and 5 is 3-colorable. There had been no progress on this conjecture for a long time, until Erdös [14] suggested a relaxation of it: does there exist a constant $k$ such that every planar graph without cycles of length from 4 to $k$ is 3-colorable? Abbott and Zhou [1] confirmed that such $k$ exists and $k\leq 11$. This result was later on improved to $k\leq 9$ by Borodin [2] and, independently, by Sanders and Zhao [13], and to $k\leq 7$ by Borodin, Glebov, Raspaud and Salavatipour [3].

Theorem 1.1** ([3]).**

Planar graphs without cycles of length from 4 to 7 are 3-colorable.

We remark that Steinberg’s conjecture was recently shown to be false in [6], by constructing a counterexample to the conjecture. The question whether every planar graph without cycles of length from 3 to 5 is 3-colorable is still open.

A more general problem than Steinberg’s Conjecture was formulated in [11, 9]:

Problem 1.2**.**

What is $\cal{A}$, a set of integers between 5 and 9, such that for $i\in\cal{A}$, every planar graph with cycles of length neither 4 nor $i$ is 3-colorable?

Thus, Steinberg’s Conjecture states that $5\in\cal{A}$. Since so far no element of $\cal{A}$ has been confirmed, it seems reasonable to consider a relaxation of Problem 1.2 where more integers are forbidden to be the length of a cycle in planar graphs. Due to a famous theorem of Grötzsch that planar graphs without triangles are 3-colorable, triangles are always allowed in further sufficient conditions. Several papers together contribute to the result below:

Theorem 1.3**.**

For any three integers $i,j,k$ with $5\leq i<j<k\leq 9$, it holds true that planar graphs having no cycles of length $4,i,j,k$ are 3-colorable.

Later on, the sufficient conditions, concerning three integers forbidden to be the length of a cycle, were considered. The corresponding problem can be formulated as follows:

Problem 1.4**.**

What is $\cal{B}$, a set of pairs of integers $(i,j)$ with $5\leq i<j\leq 9$, such that planar graphs without cycles of length $4,i,j$ are 3-colorable?

It has been proved by Borodin et al. [4] and independently by Xu [17] that every planar graph having neither 5- and 7-cycles nor adjacent 3-cycles is 3-colorable. Hence, $(5,7)\in\cal{B}$, which improves on Theorem 1.1. More elements of $B$ have been confirmed: $(6,8)\in\cal{B}$ by Wang and Chen [15], $(7,9)\in\cal{B}$ by Lu et al. [11], and $(6,9)\in\cal{B}$ by Jin et al. [9]. The result $(6,7)\in\cal{B}$ is implied in the following theorem, which reconfirms the results $(5,7)\in\cal{B}$ and $(6,8)\in\cal{B}$.

Theorem 1.5** ([5]).**

Planar graphs without triangles adjacent to cycles of length from 4 to 7 are 3-colorable.

In this paper, we show that $(5,8)\in\cal{B}$, which leaves four pairs of integers $(5,6),(5,9),(7,8),$ $(8,9)$ unconfirmed as elements of $\cal{B}$.

Recently, Mondal gave a proof of the result $(5,8)\in\cal{B}$ in [12]. Here we exhibit two couterexamples to the theorem proved in that paper which yields the result $(5,8)\in\cal{B}$. We restated this theorem as follows. Let $C$ be a cycle of length at most 12 in a plane graph without 4-, 5- and 8-cycles. $C$ is bad if it is of length 9 or 12 and the subgraph inside $C$ has a partition into 3- and 6-cycles; otherwise, $C$ is good.

Theorem 1.6** (Theorem 2 in [12]).**

Let $G$ be a graph without 4-, 5-, and 8-cycles. If $D$ is a good cycle of $G$, then every proper 3-coloring of $D$ can be extended to a proper 3-coloring of the whole graph $G$.

Counterexamples to Theorem 1.6. A plane graph $G_{1}$ consisting of a cycle $C$ of length 12, say $C:=[v_{1}\ldots v_{12}]$, and a vertex $u$ inside $C$ connected to all of $v_{1},v_{2},v_{6}$. The graph $G_{1}$ contradicts Theorem 1.6, since any proper 3-coloring of $C$ where $v_{1},v_{2},v_{6}$ receive pairwise distinct colors can not be extended to $G_{1}$. Also, a plane graph $G_{2}$ consisting of a cycle $C$ of length 12 and a triangle $T$ inside $C$, say $C:=[v_{1}\ldots v_{12}]$ and $T:=[u_{1}u_{2}u_{3}]$, and three more edges $u_{1}v_{1},u_{2}v_{4},u_{3}v_{7}$. The graph $G_{2}$ contradicts Theorem 1.6, since any proper 3-coloring of $C$ where $v_{1},v_{4},v_{7}$ receive the same color can not be extended to $G_{2}$ (see Figure 1).

1.1 Notations and formulation of the main theorem

The graphs considered in this paper are finite and simple. A graph is planar if it is embeddable into the Euclidean plane. A plane graph $(G,\Sigma)$ is a planar graph $G$ together with an embedding $\Sigma$ of $G$ into the Euclidean plane, that is, $(G,\Sigma)$ is a particular drawing of $G$ in the Euclidean plane. In what follows, we will always say a plane graph $G$ instead of $(G,\Sigma)$, which causes no confusion since no two embeddings of the same graph $G$ will be involved in.

Let $G$ be a plane graph and $C$ be a cycle of $G$. By $Int(C)$ (or $Ext(C)$) we denote the subgraph of $G$ induced by the vertices lying inside (or outside) $C$. The cycle $C$ is separating if neither $Int(C)$ nor $Ext(C)$ is empty. By $\overline{Int}(C)$ (or $\overline{Ext}(C)$) we denote the subgraph of $G$ consisting of $C$ and its interior (or exterior). The cycle $C$ is triangular if it is adjacent to a triangle, and $C$ is ext-triangular if it is adjacent to a triangle of $\overline{Ext}(C)$.

The following theorem is the main result of this paper.

Theorem 1.7**.**

Plane graphs with neither 4- and 5-cycles nor ext-triangular 7-cycles are 3-colorable.

As a consequence of Theorem 1.7, the following corollary holds true.

Corollary 1.8**.**

Planar graphs without cycles of length 4, 5, 8 are 3-colorable, that is, $(5,8)\in\cal{B}$.

We remark that Theorem 1.7 implies the known result that $(5,7)\in\cal{B}$ as well.

Denote by $d(v)$ the degree of a vertex $v$, by $|P|$ the number of edges of a path $P$, by $|C|$ the length of a cycle $C$ and by $d(f)$ the size of a face $f$. A $k$-vertex (or $k^{+}$-vertex, or $k^{-}$-vertex) is a vertex $v$ with $d(v)=k$ (or $d(v)\geq k$, or $d(v)\leq k$). Similar notations are used for paths, cycles, faces with $|P|,|C|,d(f)$ instead of $d(v)$, respectively.

Let $G[S]$ denote the subgraph of $G$ induced by $S$ with either $S\subseteq V(G)$ or $S\subseteq E(G)$. A chord of $C$ is an edge of $\overline{Int}(C)$ that connects two nonconsecutive vertices on $C$. If $Int(C)$ has a vertex $v$ with three neighbors $v_{1},v_{2},v_{3}$ on $C$, then $G[\{vv_{1},vv_{2},vv_{3}\}]$ is called a claw of $C$. If $Int(C)$ has two adjacent vertices $u$ and $v$ such that $u$ has two neighbors $u_{1},u_{2}$ on $C$ and $v$ has two neighbors $v_{1},v_{2}$ on $C$, then $G[\{uv,uu_{1},uu_{2},vv_{1},vv_{2}\}]$ is called a biclaw of $C$. If $Int(C)$ has three pairwise adjacent vertices $u,v,w$ which has a neighbor $u^{\prime},v^{\prime},w^{\prime}$ on $C$ respectively, then $G[\{uv,vw,uw,uu^{\prime},vv^{\prime},ww^{\prime}\}]$ is called a triclaw of $C$. If $G$ has four vertices $x,u,v,w$ inside $C$ and four vertices $x_{1},x_{2},v_{1},w_{1}$ on $C$ such that $S=\{uv,vw,wu,ux,xx_{1},xx_{2},vv_{1},ww_{1}\}\subseteq E(G)$, then $G[S]$ is called a combclaw of $C$ (see Figure 2).

A good cycle is an 11--cycle that has none of claws, biclaws, triclaws and combclaws. A bad cycle is an 11--cycle that is not good.

Instead of Theorem 1.7, it is easier for us to prove the following stronger one:

Theorem 1.9**.**

Let $G$ be a connected plane graph with neither 4- and 5-cycles nor ext-triangular 7-cycles. If $D$, the boundary of the exterior face of $G$, is a good cycle, then every proper 3-coloring of $G[V(D)]$ can be extended to a proper 3-coloring of $G$.

The proof of Theorem 1.9 will be proceeded by using discharging method and is given in the next section. For more information on the discharging method, we refer readers to [7]. The rest of this section contributes to other needed notations.

Let $C$ be a cycle and $T$ be one of chords, claws, biclaws, triclaws and combclaws of $C$. We call the graph $H$ consisting of $C$ and $T$ a bad partition of $C$. The boundary of any one of the parts, into which $C$ is divided by $H$, is called a cell of $H$. Clearly, every cell is a cycle. In case of confusion, let us always order the cells $c_{1},\cdots,c_{t}$ of $H$ in the way as shown in Figure 2. Let $k_{i}$ be the length of $c_{i}$. Then $T$ is further called a $(k_{1},k_{2})$-chord, a $(k_{1},k_{2},k_{3})$-claw, a $(k_{1},k_{2},k_{3},k_{4})$-biclaw, a $(k_{1},k_{2},k_{3},k_{4})$-triclaw and a $(k_{1},k_{2},k_{3},k_{4},k_{5})$-combclaw, respectively.

A vertex is external if it lies on the exterior face; internal otherwise. A vertex (or an edge) is triangular if it is incident with a triangle. We say a vertex is bad if it is an internal triangular 3-vertex; good otherwise. A path is a splitting path of a cycle $C$ if it has the two end-vertices on $C$ and all other vertices inside $C$. A $k$-cycle with vertices $v_{1},\ldots,v_{k}$ in cyclic order is denoted by $[v_{1}\ldots v_{k}]$.

Let $uvw$ be a path on the boundary of a face $f$ of $G$ with $v$ internal. The vertex $v$ is $f$-heavy if both $uv$ and $vw$ are triangular and $d(v)\geq 5$, and is $f$-Mlight if both $uv$ and $vw$ are triangular and $d(v)=4$, and $f$-Vlight if neither $uv$ nor $vw$ is triangular and $v$ is triangular and of degree 4. A vertex is $f$-light if it is either $f$-Mlight or $f$-Vlight.

Denote by $\cal{G}$ the class of connected plane graphs with neither 4- and 5-cycles nor ext-triangular 7-cycles.

2 The proof of Theorem 1.9

Suppose to the contrary that Theorem 1.9 is false. From now on, let $G$ be a counterexample to Theorem 1.9 with fewest vertices. Thus, we may assume that the boundary $D$ of the exterior face of $G$ is a good cycle, and there exists a proper 3-coloring $\phi$ of $G[V(D)]$ which cannot be extended to a proper 3-coloring of $G$. By the minimality of $G$, we deduce that $D$ has no chord.

2.1 Structural properties of the minimal counterexample $G$

Lemma 2.1**.**

Every internal vertex of $G$ has degree at least 3.

Proof.

Suppose to the contrary that $G$ has an internal vertex $v$ with $d(v)\leq 2$. We can extend $\phi$ to $G-v$ by the minimality of $G$, and then to $G$ by coloring $v$ different from its neighbors. ∎

Lemma 2.2**.**

$G$* is 2-connected and therefore, the boundary of each face of $G$ is a cycle.*

Proof.

Otherwise, we can assume that $G$ has a pendant block $B$ with cut vertex $v$ such that $B-v$ does not intersect with $D$. We first extend $\phi$ to $G-(B-v)$, and then 3-color $B$ so that the color assigned to $v$ is unchanged. ∎

Lemma 2.3**.**

$G$* has no separating good cycle.*

Proof.

Suppose to the contrary that $G$ has a separating good cycle $C$. We extend $\phi$ to $G-Int(C)$. Furthermore, since $C$ is a good cycle, the color of $C$ can be extended to its interior. ∎

By the definition of a bad cycle, one can easily conclude the lemma as follows.

Lemma 2.4**.**

If $C$ is a bad cycle of a graph in $\cal{G}$, then $C$ has length either 9 or 11. Furthermore, if $|C|=9$, then $C$ has a (3,6,6)-claw or a (3,6,6,6)-triclaw; if $|C|=11$, then $C$ has a (3,6,8)-claw, or a (3,6,6,6)- or (6,3,6,6)-biclaw, or a (3,6,6,8)-triclaw, or a (3,6,6,6,6)-combclaw.

Notice that all 3- and 6- and 8-cycles of $G$ are facial, thus the following statement is a consequence of the previous lemma together with the fact that $G\in\cal{G}$.

Lemma 2.5**.**

$G$* has neither bad cycle with a chord nor ext-triangular bad 9-cycle.*

Lemma 2.6**.**

Let $P$ be a splitting path of $D$ which divides $D$ into two cycles $D^{\prime}$ and $D^{\prime\prime}$. The following four statements hold true.

1. (1)

If $|P|=2$, then there is a triangle between $D^{\prime}$ and $D^{\prime\prime}$. 2. (2)

$|P|\neq 3$. 3. (3)

If $|P|=4$, then there is a 6- or 7-cycle between $D^{\prime}$ and $D^{\prime\prime}$. 4. (4)

If $|P|=5$, then there is a $9^{-}$-cycle between $D^{\prime}$ and $D^{\prime\prime}$.

Proof.

Since $D$ has length at most 11, we have $|D^{\prime}|+|D^{\prime\prime}|=|D|+2|P|\leq 11+2|P|$.

(1)  Let $P=xyz$. Suppose to the contrary that $|D^{\prime}|,|D^{\prime\prime}|\geq 6$. By Lemma 2.1, $y$ has a neighbor other than $x$ and $z$, say $y^{\prime}$. It follows that $y^{\prime}$ is internal since otherwise $D$ is a bad cycle with a claw. Without loss of generality, let $y^{\prime}$ lie inside $D^{\prime}$. Now $D^{\prime}$ is a separating cycle. By Lemma 2.3, $D^{\prime}$ is not good, i.e., either $D^{\prime}$ is bad or $|D^{\prime}|\geq 12$. Since every bad cycle has length either 9 or 11 by Lemma 2.4, we have $|D^{\prime}|\geq 9$. Recall that $|D^{\prime}|+|D^{\prime\prime}|\leq 15$, thus $|D^{\prime}|=9$ and $|D^{\prime\prime}|=6$. Now $D^{\prime}$ has either a (3,6,6)-claw or a (3,6,6,6)-triclaw by Lemma 2.4, which implies that $D$ has a biclaw or a combclaw respectively, a contradiction.

(2) Suppose to the contrary that $|P|=3$. Let $P=wxyz$. Clearly $|D^{\prime}|,|D^{\prime\prime}|\geq 6$. Let $x^{\prime}$ and $y^{\prime}$ be a neighbor of $x$ and $y$ not on $P$, respectively. If both $x^{\prime}$ and $y^{\prime}$ are external, then $D$ has a biclaw. Hence, we may assume $x^{\prime}$ lies inside $D^{\prime}$. By Lemmas 2.3 and 2.4 and the inequality $|D^{\prime}|+|D^{\prime\prime}|\leq 17$, we deduce that $D^{\prime}$ is a bad cycle and $D^{\prime\prime}$ is a good $8^{-}$-cycle. If $y^{\prime}$ is internal, then $y^{\prime}$ lies inside $D^{\prime}$. It follows with the specific interior of a bad cycle that $x^{\prime}=y^{\prime}$ and $D^{\prime}$ has either a claw or a biclaw, which implies that $D$ has either a triclaw or a combclaw respectively, a contradiction. Hence, $y^{\prime}$ is external. Since every bad cycle as well as every $6^{-}$- or 8-cycle contains no chord by Lemma 2.5, we deduce that $yy^{\prime}$ is a (3,6)-chord of $D^{\prime\prime}$. It follows that $D^{\prime}$ is a bad and ext-triangular 9-cycle, contradicting Lemma 2.5.

(3)  Let $P=vwxyz$. Suppose to the contrary that $|D^{\prime}|,|D^{\prime\prime}|\geq 8$. Since $|D^{\prime}|+|D^{\prime\prime}|\leq 19$, we have $|D^{\prime}|,|D^{\prime\prime}|\leq 11$. Since $G$ has no 4- and 5-cycles, if $G$ has an edge $e$ connecting two nonconsecutive vertices on $P$, then the cycle formed by $e$ and $P$ has to be a triangle, yielding a splitting 3-path of $D$, contradicting the statement (2). Therefore, no pair of nonconsecutive vertices on $P$ are adjacent.

Let $w^{\prime},x^{\prime},y^{\prime}$ be a neighbor of $w,x,y$ not on $P$, respectively. The statement (2) implies that $x^{\prime}$ is internal. Let $x^{\prime}$ lie inside of $D^{\prime}$. Thus $D^{\prime}$ is a bad 9- or 11-cycle. If $D^{\prime}$ is a bad 11-cycle, then $D^{\prime\prime}$ is a facial 8-cycle, and thus both $w^{\prime}$ and $y^{\prime}$ lie in $\overline{Int}(D^{\prime})$, which is impossible by the interior of a bad cycle. Hence, $D^{\prime}$ is a bad 9-cycle. By the statement (1), if $w^{\prime}\in V(D^{\prime\prime})$, then $G$ has the triangle $[vww^{\prime}]$, which makes $D^{\prime}$ ext-triangular, a contradiction. Hence, $w^{\prime}\notin V(D^{\prime\prime})$. Furthermore, as a bad cycle, $D^{\prime}$ has no chord by Lemma 2.5, thus $w^{\prime}$ is internal. If $w^{\prime}$ lies inside $D^{\prime}$, then it gives the interior of $D^{\prime}$ no other choices but $w^{\prime}=x^{\prime}$ and $D^{\prime}$ has a $(3,6,6)$-claw, in which case this claw contains a splitting 3-path of $D$, a contradiction. Hence, $w^{\prime}$ lies inside $D^{\prime\prime}$. Similarly, we can deduce that $y^{\prime}$ lies inside $D^{\prime\prime}$ as well. Note that $|D^{\prime\prime}|\in\{8,9,10\}$, thus $D^{\prime\prime}$ is a bad 9-cycle but has to contain both $w^{\prime}$ and $y^{\prime}$ inside, which is impossible.

(4)  Let $P=uvwxyz$. Suppose to the contrary that $|D^{\prime}|,|D^{\prime\prime}|\geq 10$. Since $|D^{\prime}|+|D^{\prime\prime}|\leq 21$, we have $|D^{\prime}|,|D^{\prime\prime}|\leq 11$. By similar argument as in the proof of the statement (3), one can conclude that $G$ has no edge connecting two nonconsecutive vertices on $P$. Let $v^{\prime},w^{\prime},x^{\prime},y^{\prime}$ be a neighbor of $v,w,x,y$ not on $P$, respectively.

The statement (2) implies that both $w^{\prime}$ and $x^{\prime}$ are internal. Let $w^{\prime}$ lie inside $D^{\prime}$. It follows that $D^{\prime}$ is a bad 11-cycle and $D^{\prime\prime}$ is a 10-cycle. Thus $x^{\prime}$ also lies inside $D^{\prime}$ and furthermore, $x^{\prime}=w^{\prime}$ and $D^{\prime}$ is a bad cycle with either a (3,6,8)-claw or a (3,6,6,6)-biclaw. It follows that $v^{\prime},y^{\prime}\in V(D^{\prime\prime})$. By the statement (1), $G$ has two triangles $[uvv^{\prime}]$ and $[yy^{\prime}z]$, at least one of them is adjacent to a 7-cycle of $\overline{Int}(D^{\prime})$, a contradiction. ∎

Lemma 2.7**.**

Let $G^{\prime}$ be a connected plane graph obtained from $G$ by deleting a set of internal vertices and identifying two other vertices so that at most one pair of edges are merged. If we

1. ($a$)

identify no two vertices of $D$, and create no edge connecting two vertices of $D$, and 2. ($b$)

create no $6^{-}$-cycle and ext-triangular 7-cycle,

then $\phi$ can be extended to $G^{\prime}$.

Proof.

The item $(a)$ guarantees that $D$ is unchanged and bounds $G^{\prime}$, and $\phi$ is a proper 3-coloring of $G^{\prime}[V(D)]$. By item $(b)$, the graph $G^{\prime}$ is simple and $G^{\prime}\in\cal{G}$. Hence, to extend $\phi$ to $G^{\prime}$ by the minimality of $G$, it remains to show that $D$ is a good cycle of $G^{\prime}$.

Suppose to the contrary that $D$ has a bad partition $H$ in $G^{\prime}$. Clearly, $H$ has a 6-cell $C^{\prime}$ such that the intersection between $D$ and $C^{\prime}$ is a path $v_{1}\ldots v_{k}$ of length $k-1$ with $k\in\{4,5\}$. Since we create no 6-cycles, $C^{\prime}$ corresponds to a 6-cycle $C$ of the original graph $G$. Recall that at most one pair of edges are merged during the process from $G$ to $G^{\prime}$, we deduce that the intersection between $D$ and $C$ is a path $P$ of one of the forms $v_{1}\ldots v_{k},v_{1}\ldots v_{k-1},v_{2}\ldots v_{k}$. Thus, $|P|\in\{3,4,5\}$. If $|P|\in\{4,5\}$, then $C$ contains a splitting 3- or 2-path of $D$ in $G$, yielding a contradiction by Lemma 2.6. Hence, $|P|=3$ and so $k=4$. By the choice of the 6-cell $C^{\prime}$, we may assume that the bad partition $H$ has either a (3,6,6,6)- or (3,6,6,8)-triclaw. Now $H$ contains three splitting 3-paths of $D$, at least one of them does not contain the identified vertex of $G^{\prime}$ no matter where it is, yielding the existence of a splitting 3-path of $D$ in $G$, contradicting Lemma 2.6. ∎

Lemma 2.8**.**

$G$* has no edge $uv$ incident with a 6-face and a 3-face such that both $u$ and $v$ are internal 3-vertices and therefore, every bad cycle of $G$ has either a (3,6,6)- or (3,6,8)-claw or a (3,6,6,6)-biclaw.*

Proof.

Suppose to the contrary that such an edge $uv$ exists. Denote by $[uvwxyz]$ and $[uvt]$ the 6-face and 3-face, respectively. Lemma 2.6 implies that not both of $w$ and $z$ are external vertices. Without loss of generality, we may assume that $w$ is internal. Let $G^{\prime}$ be the graph obtained from $G$ by deleting $u$ and $v$, and identifying $w$ with $y$ so that $wx$ and $yx$ are merged. Clearly, $G^{\prime}$ is a plane graph on fewer vertices than $G$. We will show that both the items in Lemma 2.7 are satisfied.

Since $w$ is internal, we identify no two vertices on $D$. If we create an edge connecting two vertices on $D$, then $w$ has a neighbor $w_{1}$ not adjacent to $y$ and both $y$ and $w_{1}$ are external. But now, Lemma 2.6 implies that $x$ is external and thus, $[ww_{1}x]$ is a triangle which makes the 7-cycle $[utvwxyz]$ ext-triangular. Hence, the item $(a)$ holds.

Suppose we create a $6^{-}$-cycle or an ext-triangular 7-cycle $C^{\prime}$. Thus $G$ has a $7^{-}$-path $P$ between $w$ and $y$ corresponding to $C^{\prime}$. If $x\in V(P)$, then neither $wx$ nor $xy$ are on $P$ since otherwise, $C^{\prime}$ already exists in $G$. Hence, the paths $wxy$ and $P$ form two cycles, both of them has length at least 6. It follows that $|P|\geq 10$, a contradiction. Hence, we may assume that $x\notin V(P)$. The paths $P$ and $wxy$ form a $9^{-}$-cycle, say $C$. By Lemma 2.1, we may let $x_{1}$ be a neighbor of $x$ other than $y$ and $w$. We have $x_{1}\notin V(P)$, since otherwise $P$ has length at least 8. Now $C$ has to contain either $u$ and $v$ or $x_{1}$ inside, which implies that $C$ is a bad 9-cycle. By Lemma 2.5, $C$ is not ext-triangular. Thus $C^{\prime}$ is a 7-cycle that is not ext-triangular, contradicting the supposition. Hence, the item $(b)$ holds.

By Lemma 2.7, the pre-coloring $\phi$ can be extended to $G^{\prime}$. Since $z$ and $w$ receive different colors, we can properly color $v$ and $u$, extending $\phi$ further to $G$. ∎

We follow the notations of $M$-face and $MM$-face in [3], and define weak tetrads. An $M$-face is an 8-face $f$ containing no external vertices with boundary $[v_{1}\ldots v_{8}]$ such that the vertices $v_{1},v_{2},v_{3},v_{5},v_{6},v_{7}$ are of degree 3 and the edges $v_{1}v_{2},v_{3}v_{4},v_{4}v_{5},v_{6}v_{7}$ are triangular. An $MM$-face is an 8-face $f$ containing no external vertices with boundary $[v_{1}\ldots v_{8}]$ such that $v_{2}$ and $v_{7}$ are of degree 4 and other six vertices on $f$ are of degree 3, and the edges $v_{1}v_{2},v_{2}v_{3},v_{4}v_{5},v_{6}v_{7},v_{7}v_{8}$ are triangular. A weak tetrad is a path $v_{1}\ldots v_{5}$ on the boundary of a face $f$ such that both the edges $v_{1}v_{2}$ and $v_{3}v_{4}$ are triangular, all of $v_{1},v_{2},v_{3},v_{4}$ are internal 3-vertices, and $v_{5}$ is either of degree 3 or $f$-light.

Lemma 2.9**.**

$G$* has no weak tetrad and therefore, every face of $G$ contains no five consecutive bad vertices.*

Proof.

Suppose to the contrary that $G$ has a weak tetrad $T$ following the notation used in the definition. Denote by $v_{0}$ the neighbor of $v_{1}$ on $f$ with $v_{0}\neq v_{2}$. Denote by $x$ the common neighbor of $v_{1}$ and $v_{2}$, and $y$ the common neighbor of $v_{3}$ and $v_{4}$. If $x=v_{0}$, then $v_{1}$ is an internal 2-vertex, contradicting Lemma 2.1. Hence, $x\neq v_{0}$ and similarly, $x\neq v_{3}$. Since $G$ has no 4- or 5-cycles, $x\notin\{v_{4},v_{5}\}$. Concluding above, $x\notin v_{0}\cup V(T)$. Similarly, $y\notin v_{0}\cup V(T)$. Moreover, $x\neq y$ since otherwise $[v_{1}v_{2}v_{3}x]$ is a 4-cycle. We delete $v_{1},\ldots,v_{4}$ and identity $v_{0}$ with $y$, obtaining a plane graph $G^{\prime}$ on fewer vertices than $G$. We will show that both the items in Lemma 2.7 are satisfied.

Suppose that we create a $6^{-}$-cycle or an ext-triangular 7-cycle $C^{\prime}$. Thus $G$ has a $7^{-}$-path $P$ between $v_{0}$ and $y$ corresponding to $C^{\prime}$. If $x\in V(P)$, then the cycle formed by $P$ and $v_{0}v_{1}x$ has length at least 6 and the one formed by $P$ and $xv_{2}v_{3}y$ has length at least 8, which gives $|P|\geq 9$, a contradiction. Hence, $x\notin V(P)$. The paths $P$ and $v_{0}v_{1}v_{2}v_{3}y$ form a $11^{-}$-cycle, say $C$. Now $C$ contains either $x$ or $v_{4}$ inside. Thus, $C$ is a bad cycle. By Lemma 2.8, $C$ has either a (3,6,6)- or (3,6,8)-claw or a (3,6,6,6)-biclaw. Note that both the two faces incident with $v_{2}v_{3}$ has length at least 8, thus $C$ has a bad partition owning an 8-cell no matter which one of $x$ and $v_{4}$ lies inside $C$. It follows that $C$ has a (3,6,8)-claw. If $x$ lies inside $C$, then the 6-cell is adjacent to the triangle $[xv_{1}v_{2}]$ with $d(v_{1})=d(x)=3$, contradicting Lemma 2.8. Hence, $v_{4}$ lies inside $C$. Note that $v_{4}v_{5}$ is incident with the 6-cell and the 8-cell, we deduce that $v_{5}$ is not $f$-light. By the assumption of $T$ as a weak tetrad, we may assume that $d(v_{5})=3$. We delete $v_{5}$ together with other vertices of $T$ and repeat the argument above, yielding a contradiction. Therefore, the item $(b)$ holds.

Suppose we identify two vertices on $D$ or create an edge connecting two vertices on $D$. Thus there is a splitting 4- or 5-path $Q$ of $D$ containing the path $v_{0}v_{1}v_{2}v_{3}y$. By Lemma 2.6, $Q$ together with $D$ forms a $9^{-}$-cycle which corresponds to a $5^{-}$-cycle in $G^{\prime}$. Since we create no $6^{-}$-cycle, a contradiction follows. Hence, the item $(a)$ holds.

By Lemma 2.7, the pre-coloring $\phi$ can be extended to $G^{\prime}$. We first properly color $v_{5}$ (if needed), $v_{4},v_{3}$ in turn. Since $v_{0}$ and $v_{3}$ receive different colors, we can properly color $v_{1}$ and $v_{2}$, extending $\phi$ further to $G$. ∎

Lemma 2.10**.**

$G$* has no $M$-face.*

Proof.

Suppose to the contrary that $G$ has an $M$-face $f$ following the notation used in the definition. For $(i,j)\in\{(1,2),(3,4),(4,5),(6,7)\}$, denote by $t_{ij}$ the common neighbor of $v_{i}$ and $v_{j}$. By similar argument as in the proof of previous lemma, we deduce that the vertices $t_{12},t_{34},t_{45},t_{67}$ are pairwise distinct and not incident with $f$. We delete $v_{1},v_{2},v_{3},v_{5},v_{6},v_{7}$ and identity $v_{4}$ with $v_{8}$, obtaining a plane graph $G^{\prime}$ on fewer vertices than $G$. We will show that both the items in Lemma 2.7 are satisfied.

Suppose that we create a $6^{-}$-cycle or an ext-triangular 7-cycle $C^{\prime}$. Thus $G$ has a $7^{-}$-path $P$ between $v_{4}$ and $v_{8}$ corresponding to $C^{\prime}$. By the symmetry of an $M$-face, we may assume that $P$ together with the path $v_{4}\ldots v_{8}$ forms a $11^{-}$-cycle $C$ containing $v_{1},v_{2},v_{3}$ inside. It follows with Lemma 2.8 that $C$ is a bad cycle with a $(3,6,6,6)$-claw. But now $\overline{Int}(C)$ contains $f$ that is an 8-face, a contradiction. Therefore, the item $(b)$ holds.

The satisfaction of the item $(a)$ can be proved in a similar way as in the proof of previous lemma.

By Lemma 2.7, the pre-coloring $\phi$ can be extended to $G^{\prime}$. Since we first color $v_{3}$ different from $v_{8}$, both $v_{1}$ and $v_{2}$ can be properly colored. Finally, color $v_{5},v_{6},v_{7}$ in the same way, extending $\phi$ further to $G$. ∎

Lemma 2.11**.**

$G$* has no $MM$-face.*

Proof.

Suppose to the contrary that $G$ has an $MM$-face $f$ following the notation used in the definition. For $(i,j)\in\{(1,2),(2,3),(4,5),(6,7),(7,8)\}$, denote by $t_{ij}$ the common neighbor of $v_{i}$ and $v_{j}$. Similarly, we deduce that the vertices $t_{12},t_{23},t_{45},t_{67},t_{78}$ are pairwise distinct and not incident with $f$. We delete all the vertices of $f$ and identity $t_{12}$ with $t_{67}$, obtaining a plane graph $G^{\prime}$ on fewer vertices than $G$. To extend $\phi$ to $G^{\prime}$, it suffices to fulfill the item $(a)$ of Lemma 2.7, as what we did in previous lemma.

Suppose that we create a $6^{-}$-cycle or an ext-triangular 7-cycle $C^{\prime}$. Thus $G$ has a $7^{-}$-path $P$ between $t_{12}$ and $t_{67}$ corresponding to $C^{\prime}$. If $t_{78}\in V(P)$, then both the cycles formed by $P$ and $t_{12}v_{1}v_{8}t_{78}$ and by $P$ and $t_{78}v_{7}t_{67}$ have length at least 8, which gives $|P|\geq 11$, a contradiction. Hence, $t_{78}\notin V(P)$. The paths $P$ and $t_{12}v_{1}v_{8}v_{7}t_{67}$ form a $11^{-}$-cycle, say $C$. It follows that $C$ is a bad cycle containing either $t_{78}$ or $v_{2},\ldots,v_{6}$ inside, that is, either $C$ has a bad partition owning two $8^{+}$-cell or $C$ contains five vertices inside, a contradiction in any case.

We further extend $\phi$ from $G^{\prime}$ to $G$ as follows. Let $\alpha,\beta$ and $\gamma$ be the three colors used in $\phi$. First regardless the edge $v_{1}v_{8}$, we can properly color $v_{2},v_{1},v_{3}$ and $v_{7},v_{8},v_{6}$. If $v_{1}$ and $v_{8}$ receive different colors and so do $v_{3}$ and $v_{6}$, then $v_{4}$ and $v_{5}$ can be properly colored, we are done. Hence, we may assume without loss of generality that $v_{1}$ and $v_{8}$ receive the same color, say $\beta$. Let $\alpha$ be the color assigned to $t_{12}$ and $t_{67}$. Thus $v_{2}$ and $v_{7}$ are colored with $\gamma$ and $t_{78}$ is colored with $\alpha$. We recolor $v_{8},v_{7},v_{6}$ with $\gamma,\beta,\gamma$ respectively. Now $v_{1}$ and $v_{8}$ receive different colors and so do $v_{3}$ and $v_{6}$. Again $v_{4}$ and $v_{5}$ can be properly colored, we are also done. ∎

2.2 Discharging in $G$

Let $V=V(G)$, $E=E(G)$, and $F$ be the set of faces of $G$. Denote by $f_{0}$ the exterior face of $G$. Give initial charge $ch(x)$ to each element $x$ of $V\cup F$, where $ch(f_{0})=d(f_{0})+4$, $ch(v)=d(v)-4$ for $v\in V$, and $ch(f)=d(f)-4$ for $f\in F\setminus\{f_{0}\}$. Discharge the elements of $V\cup F$ according to the following rules:

1. $R1.$

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

Every internal $6^{+}$-face sends $\frac{2}{3}$ to each incident 2-vertex. 3. $R3.$

Every internal $6^{+}$-face sends each incident 3-vertex $v$ charge $\frac{2}{3}$ if $v$ is triangular, and charge $\frac{1}{3}$ otherwise. 4. $R4.$

Every internal $6^{+}$-face $f$ sends $\frac{1}{3}$ to each $f$-light vertex, and receives $\frac{1}{3}$ from each $f$-heavy vertex. 5. $R5.$

Every internal $6^{+}$-face receives $\frac{1}{3}$ from each incident external $4^{+}$-vertex. 6. $R6.$

The exterior face $f_{0}$ sends $\frac{4}{3}$ to each incident vertex.

Let $ch^{*}(x)$ denote the final charge of each element $x$ of $V\cup F$ after discharging. On one hand, by Euler’s formula we deduce $\sum\limits_{x\in V\cup F}ch(x)=0.$ Since the sum of charges over all elements of $V\cup F$ is unchanged, it follows that $\sum\limits_{x\in V\cup F}ch^{*}(x)=0.$ On the other hand, we show that $ch^{*}(x)\geq 0$ for $x\in V\cup F\setminus\{f_{0}\}$ and $ch^{*}(f_{0})>0$. Hence, this obvious contradiction completes the proof of Theorem 1.9. It remains to show that $ch^{*}(x)\geq 0$ for $x\in V\cup F\setminus\{f_{0}\}$ and $ch^{*}(f_{0})>0$.

We remark that the discharging rules can be tracked back to the one used in [3].

Lemma 2.12**.**

$ch^{*}(v)\geq 0$* for $v\in V$.*

Proof.

First suppose that $v$ is external. Since $D$ is a cycle, $d(v)\geq 2$. If $d(v)=2$, then since $D$ has no chord, the internal face incident with $v$ is not a triangle and sends $\frac{2}{3}$ to $v$ by $R2$. Moreover, $v$ receives $\frac{4}{3}$ from $f_{0}$ by $R6$, which gives $ch^{*}(v)=d(v)-4+\frac{2}{3}+\frac{4}{3}=0$. If $d(v)=3$, then $v$ sends charge to at most one 3-face by $R1$ and thus $ch^{*}(v)\geq d(v)-4-\frac{1}{3}+\frac{4}{3}=0$. If $d(v)\geq 4$, then $v$ sends at most $\frac{1}{3}$ to each incident internal face by $R1$ and $R5$, yielding $ch^{*}(v)\geq d(v)-4-\frac{1}{3}(d(v)-1)+\frac{4}{3}>0$. Hence, we are done in any case.

It remains to suppose that $v$ is internal. By Lemma 2.1, $d(v)\geq 3.$ If $d(v)=3$, then we have $ch^{*}(v)=d(v)-4-\frac{1}{3}+\frac{2}{3}\times 2=0$ by $R1$ and $R3$ when $v$ is triangular, and $ch^{*}(v)=d(v)-4+\frac{1}{3}\times 3=0$ by $R3$ when $v$ not. If $d(v)=4$, then $v$ is incident with $k$ 3-faces with $k\leq 2$. By $R1$ and $R4$, we have $ch^{*}(v)=d(v)-4-\frac{1}{3}\times 2+\frac{1}{3}\times 2=0$ when $k=2$, $ch^{*}(v)=d(v)-4-\frac{1}{3}+\frac{1}{3}=0$ when $k=1$, and $ch^{*}(v)=d(v)-4=0$ when $k=0$. If $d(v)=5$, then $v$ sends charge to at most two 3-faces by $R1$ and to at most one $6^{+}$-face by $R4$, which gives $ch^{*}(v)\geq d(v)-4-\frac{1}{3}\times 2-\frac{1}{3}=0$. Hence, we may next assume that $d(v)\geq 6.$ Since $v$ sends at most $\frac{1}{3}$ to each incident face by our rules, we get $ch^{*}(v)\geq d(v)-4-\frac{1}{3}d(v)\geq 0.$ ∎

Lemma 2.13**.**

$ch^{*}(f_{0})>0$.

Proof.

Recall that $ch(f_{0})=d(f_{0})+4$ and $d(f_{0})\leq 11$. We have $ch^{*}(f_{0})\geq d(f_{0})+4-\frac{4}{3}d(f_{0})>0$ by $R6$. ∎

Lemma 2.14**.**

$ch^{*}(f)\geq 0$* for $f\in F\setminus\{f_{0}\}$.*

Proof.

We distinguish cases according to the size of $f$. Since $G$ has no 4- and 5-cycle, $d(f)\notin\{4,5\}$.

If $d(f)=3$, then $f$ receives $\frac{1}{3}$ from each incident vertices by $R1$, which gives $ch^{*}(f)=d(f)-4+\frac{1}{3}\times 3=0.$

Let $d(f)=6$. For any incident vertex $v$, by the rules, $f$ sends to $v$ charge $\frac{2}{3}$ if $v$ is either of degree 2 or bad, and charge at most $\frac{1}{3}$ otherwise. Since $G$ has no ext-triangular 7-cycles, $f$ is adjacent to at most one 3-face. Furthermore, by Lemma 2.8, $f$ contains at most one bad vertex. If $f$ contains a 2-vertex, say $u$, we can deduce with Lemma 2.6 that $u$ is the unique 2-vertex of $f$ and the two neighbors of $u$ on $f$ are external $3^{+}$-vertices which receive nothing from $f$. It follows that $ch^{*}(f)\geq d(f)-4-\frac{2}{3}-\frac{2}{3}-\frac{1}{3}\times 2=0.$ Hence, we may assume that $f$ contains no 2-vertices. If $f$ has no bad vertices, then $f$ sends each incident vertex at most $\frac{1}{3}$, which gives $ch^{*}(f)\geq d(f)-4-\frac{1}{3}d(f)=0$. Hence, we may let $x$ be a bad vertex of $f$. Denote by $y$ the other common vertex between $f$ and the triangle adjacent to $f$. By Lemma 2.8 again, $y$ is not a bad vertex, i.e., $y$ is either an internal $4^{+}$-vertex or an external $3^{+}$-vertex. By our rules, $f$ sends nothing to $y$, yielding $ch^{*}(f)\geq d(f)-4-\frac{2}{3}-\frac{1}{3}\times 4=0.$

Let $d(f)=7$. Since $G$ has no ext-triangular 7-cycles, $f$ contains no bad vertices. Moreover, by Lemma 2.6, we deduce that $f$ has at most two 2-vertices. Thus, $ch^{*}(f)\geq d(f)-4-\frac{2}{3}\times 2-\frac{1}{3}\times 5=0.$

Let $d(f)\geq 8$. On the hand, if $f$ contains precisely one external vertex, say $w$, then $d(w)\geq 4$ and so $f$ receives $\frac{1}{3}$ from $w$ by $R5$. Furthermore, since $f$ contains no weak tetrad by Lemma 2.9, $f$ has a good vertex other than $w$ and sends at most $\frac{1}{3}$ to it. Hence, $ch^{*}(f)\geq d(f)-4+\frac{1}{3}-\frac{1}{3}-\frac{2}{3}(d(f)-2)\geq 0.$ On the other hand, if $f$ contains at least two external vertices, then at least two of them are of degree more than 2. Since $f$ sends nothing to external $3^{+}$-vertices, we have $ch^{*}(f)\geq d(f)-4-\frac{2}{3}(d(f)-2)\geq 0$. By the two hands above, we may assume that all the vertices of $f$ are internal. We distinguish two cases.

Case 1: assume that $d(f)=8$. Denote by $r$ the number of bad vertices of $f$. We have $ch^{*}(f)\geq d(f)-4-\frac{2}{3}r-\frac{1}{3}(d(f)-r)=\frac{4-r}{3}\geq 0$, provided by $r\leq 4$. Since $f$ contains no weak tetrad, $r\leq 6$. Hence, we may assume that $r\in\{5,6\}$. For $r=5$, we claim that $f$ has a vertex failing to take charge from $f$, which gives $ch^{*}(f)\geq d(f)-4-\frac{2}{3}\times 5-\frac{1}{3}\times 2=0.$ Suppose to the contrary that no such vertex exists. Thus, the bad vertices of $f$ can be paired so that any good vertex of the path of $f$ between each pair is $f$-Mlight, contradicting the parity of $r$. For $r=6$, since again $f$ contains no five consecutive bad vertices, these six bad vertices of $f$ are divided by the two good ones into cyclically either 3+3 or 2+4. We may assume that $f$ has a good vertex that is either $f$-light or of degree 3, since otherwise we are done with $ch^{*}(f)\geq d(f)-4-\frac{2}{3}\times 6=0$. Denote by $u$ such a good vertex and by $v$ the other one. By the drawing of $u$ and of the 3-faces adjacent to $f$, we deduce that, for the case 3+3, $f$ is an $M$-face, contradicting Lemma 2.10, and for the case 2+4, if $u$ is $f$-Mlight then either $f$ is an $MM$-face or $v$ is $f$-heavy; otherwise $f$ contains a weak tetrad. It follows with Lemmas 2.11 and 2.9 that $v$ is $f$-heavy, which is the only possible case. Hence, $f$ receives $\frac{1}{3}$ from $v$ by $R4$, yielding $ch^{*}(f)\geq ch(f)-4-\frac{2}{3}\times 6+\frac{1}{3}-\frac{1}{3}=0$.

Case 2: assume that $d(f)\geq 9$. By Lemma 2.9, we deduce that $f$ contains at least two good vertices, each of them receives at most $\frac{1}{3}$ from $f$. Thus, $ch^{*}(f)\geq d(f)-4-\frac{2}{3}(d(f)-2)-\frac{1}{3}\times 2=\frac{d(f)-10}{3}\geq 0$, provided by $d(f)\geq 10$. It remains to suppose $d(f)=9$. If $f$ has at most six bad vertices, then $ch^{*}(f)\geq d(f)-4-\frac{2}{3}\times 6-\frac{1}{3}\times 3=0.$ Hence, we may assume that $f$ has precisely seven bad vertices. By the same argument as for the case $d(f)=8$ and $f$ has five bad vertices above, $f$ has a vertex failing to take charge from $f$, which gives $ch^{*}(f)\geq d(f)-4-\frac{2}{3}\times 7-\frac{1}{3}=0.$ ∎

By the previous three lemmas, the proof of Theorem 1.9 is completed.

3 Acknowledgement

The first author is supported by Deutsche Forschungsgemeinschaft (DFG) grant STE 792/2-1. The fourth author is supported by National Natural Science Foundation of China (NSFC) 11271335.