Edge Colorings of Planar Graphs Without 6-Cycles with Three Chords

Abstract

A graph G is of class 1 if its edges can be colored with k colors such that adjacent edges receive different colors, where k is the maximum degree of G. It is proved here that every planar graph is of class 1 if its maximum degree is at least 6 and any 6-cycle contains at most two chords.

1 Introduction

All graphs considered here are finite and simple. Let G be a graph with the vertex set V(G) and edge set E(G). We denote the maximum degree of G by $\Delta (G)$. If $v\in V(G)$, then its neighbor set $N_G(v)$ (or simply N(v)) is the set of the vertices in G adjacent to v and the degree d(v) of v is $|N_G(v)|$. For $V'\subseteq V(G)$, denote $N(V')=\cup _{u\in V'}N(u)$. A k-vertex, $k^{-}$-vertex, or a $k^{+}$-vertex is a vertex of degree k, at most k or at least k, respectively. A k (or $k^{+}$)-vertex adjacent to a vertex x is called a k (or $k^{+}$)-neighbor of x. Let $d_{k}(x)$, $d_{k^{+}}(x)$ denote the number of k-neighbors, $k^{+}$- neighbors of x. A k-cycle is a cycle of length k. Two cycles sharing a common edge are said to be adjacent. Given a cycle C of length k in G, an edge $xy\in E(G)\backslash E(C)$ is called a chord of C if $x,y\in V(C)$. Such a cycle C is also called a chordal-k-cycle.

Let G be a plane graph, F(G) be the face set of G. A face of an graph is said to be incident with all edges and vertices in its boundary. Two faces sharing an edge e are said to be adjacent at e. The degree of a face f, denoted by $d_G(f)$ is the number of edges incident with f where each cut edge is counted twice. A k-, $k^{+}$-face is a face of degree k, at least k. A k-face of G is called an $(i_1, i_2,\ldots ,i_k)$-face if the vertices in its boundary in clockwise order are of degrees $i_1, i_2,\ldots ,i_k$ respectively. A 3-face incident with distinct vertices x, y, z is denoted by (x, y, z), moreover, by [x, y, z] if $d(x)\le d(y)\le d(z)$. A 4-face incident with distinct vertices w, v, x, y is denoted by (w, v, x, y), moreover, by [w, v, x, y] if $d(x)=2$ and v, x, y form a 3-face, we call this 4-face special. For a vertex $v\in V(G)$, we denote by $f_k(v)$ the number of k-faces incident with v.

A graph has an edge k-coloring if its edges can be colored with color set $\{1,2,\ldots ,k\}$ such that adjacent edges receive different colors. A graph is k-edge-colorable if it has an edge k-coloring. The edge chromatic number of a graph G, denoted by $\chi '(G)$, is the smallest integer k such that G is k-edge-colorable. In 1964, Vizing proved that for any simple graph G, $\Delta (G) \le \chi '(G) \le \Delta (G)+1$. A graph G is said to be of class 1 if $\chi '(G)=\Delta (G)$, and of class 2 if $\chi '(G)=\Delta (G)+1$. A graph G is critical if it is connected and of class 2, and $\chi '(G-e)<\chi '(G)$ for any edge e of G. A critical graph with the maximum degree $\Delta$ is called a $\Delta$-critical graph. It is clear that every critical graph is 2-connected.

For planar graphs, Vizing [2] noted that if $\Delta \in \{2,3,4,5\}$, there exist $\Delta$-critical planar graphs, and proved that every planar graph with $\Delta \ge 8$ is of class 1 and then conjectured that every planar graph with maximum degree 6 or 7 is of class 1 (There are more general results, see [3] and [5]). The case $\Delta =7$ for the conjecture has been verified by Zhang [11] and, independently, by Sanders and Zhao [7]. The case $\Delta =6$ remains open, but some partial results are obtained. Theorem 16.3 in [2] stated that a planar graph with the maximum degree $\Delta$ and the girth g is of class 1 if $\Delta \ge 3$ and $g\ge 8$, or $\Delta \ge 4$ and $g\ge 5$, or $\Delta \ge 5$ and $g\ge 4$. Lam, Liu, Shiu, and Wu [4] proved that a planar graph G is of class 1 if $\Delta \ge 6$ and no two 3-cycles of G sharing a common vertex. Zhou [12] obtained that every planar graph with $\Delta \ge 6$ and without 4 or 5-cycles is of class 1. Bu and Wang [1] proved that every planar graph with $\Delta \ge 6$ and without chordal 5-cycles and chordal 6-cycles is of class 1. Wu and Xue [9] extended the result that every planar graph with $\Delta \ge 6$ and without 5-cycles with two chord is of class 1. Ni [6] proved that every planar graph with $\Delta \ge 6$ and without chordal 6-cycles is of class 1. Recently, Xue and Wu [10] extended the result that every planar graph with $\Delta \ge 6$ and without 6-cycles with two chords is of class 1. In the paper, we shall improve the above result to planar graphs with $\Delta =6$ and without 6-cycles with three chords.

2 The Main Result and its Proof

Firstly, we introduce some known lemmas.

Lemma 1

[7],[11] Every planar graph with maximum degree at least 7 is of class 1.

Lemma 2

(Vizing’s Adjacency Lemma [2]) Let G be a $\Delta$-critical graph, and let u and v be adjacent vertices of G with $d(v)=k$.

• (a) If $k<\Delta$, then u is adjacent to at least $\Delta -k+1$ vertices of degree $\Delta$;

• (b) If $k=\Delta$, then u is adjacent to at least two vertices of degree $\Delta$.

From the above Lemma, it is easy to get the following corollary.

Corollary 3

Let G be a $\Delta$-critical graph. Then

• (a) every vertex is adjacent to at most one 2-vertex and at least two $\Delta$-vertices; 

• (b) the sum of the degree of any two adjacent vertices is at least $\Delta +2;$

(c) if $uv\in E(G)$ and $d(u)+d(v)=\Delta +2$, then every vertex of $N(\{u,v\})\setminus \{u,v\}$ is a $\Delta$-vertex.

Lemma 4

[11] Suppose that G is a $\Delta$-critical graph, $uv\in E(G)$ and $d(u)+d(v)=\Delta +2$. Then

• (a) every vertex of $N(N(\{u,v\}))\setminus \{u,v\}$ is of degree at least $\Delta -1$;

• (b) if $d(u),d(v)<\Delta$, then every vertex of $N(N(\{u,v\}))\setminus \{u,v\}$ is a $\Delta$-vertex.

Lemma 5

[7] No $\Delta$-critical graph has distinct vertices x, y, z such that x is adjacent to y and z, $d(z)<2\Delta -d(x)-d(y)+2$, and xz is in at least $d(x)+d(y)-\Delta -2$ triangles not containing y.

Lemma 6

[8] Let G be a $\Delta$-critical graph with $\Delta (G) \ge 6$ and let x be a 4-vertex. Then the following hold:

• (a) If x is adjacent to a $(\Delta -2)$-vertex, say y, then every vertex of $N_G(N_G(x))\setminus \{x, y\}$ is a $\Delta$-vertex;

• (b) Suppose that x is not adjacent to any $(\Delta -2)$-vertex and y is one neighbor of x. If y is adjacent to $d_G(y)-(\Delta -3)$ $(\Delta -2)^{-}$-vertices, then each of the other three neighbors of x is adjacent to only one $(\Delta -2)^{-}$-vertex, which is x;

• (c) If x is adjacent to a $(\Delta -1)$-vertex, then there are at least two $\Delta$-vertices in $N_G(x)$ which are adjacent to at most two $(\Delta - 2)^{-}$-vertices. Moreover, if x is adjacent to two $(\Delta -1)$-vertices, then each of the two $\Delta$-neighbors of x is adjacent to exactly one $(\Delta -2)^{-}$-vertex, which is x.

Let the edges of a graph be colored with colors from $C = \{1,\ldots ,k\}$ and let $u \in V$. If an edge incident with u is colored i, we say u sees i. Otherwise, we say u misses i. If a vertex u sees a color i, we use (u, i) to denote the edge incident with u colored i. Given two colors $i, j \in \{1,\ldots , k\}$, an (i, j)-chain is a path whose edges are colored alternatively i and j, and we use $(u, i)\sim (v, j)$ to denote that there is a (i, j)-chain containing (u, i) and (v, j). Let $L_{i,j}(u)$ denote the longest (i, j)-chain passing through u.

The following is the key fact when dealing with a $\Delta$-critical graph G.

Fact 7

Let G be a $\Delta$-critical graph and $xy\in E(G)$. Giving any edge $\Delta$-coloring of $G-xy$, if x misses j and y misses k, then x sees k, y sees j, and $(x,k)\sim (y,j)$.

Proof

If x does not see k, then we can color xy with k to obtain an edge $\Delta$-coloring of G, a contradiction. By the same argument, y sees j. If $(x,k)\not \sim (y,j)$, then we can swap colors on $L_{k,j}(x)$ and color xy with k to obtain an edge $\Delta$-coloring of G, a contradiction, too. $\square$

Lemma 8

No 6-critical graph has distinct vertices v, w, x, y, z such that $d(x)=d(w)=4, d(y)= 5$, and vwz and xyz are triangles (see Fig. 1).

Fig. 1

Black vertices do not have neighbors other than presented in the picture, while white vertices can be adjacent to each other as well as to some other vertices

Proof

Suppose, to be contrary, that a 6-critical graph G contains such vertices v, w, x, y, z. Since G is 6-critical, $G-xy$ has an edge 6-coloring $\phi$. By Fact 7, we can assume that $\phi (xz)=1$, x sees 2, 3 and y sees 4, 5, 6. We consider the following cases.

Case 1 $\phi (yz)\in \{4,5,6\}$, With out loss of generality (WLOG), assume that $\phi (yz)=4$.

Subcase 1.1 y misses 1.

Since $d(y)=5$, y must miss 2 or 3. WLOG, assume that y misses 2. Then y sees 3. By Fact 7, we have $(x,i)\sim (y,j)$, where $i\in \{1,2\}$ and $j\in \{4,5,6\}$.

Subcase 1.1.1 $\phi (wz)=2$.

Since $(x,2)\sim (y,4)$, w sees 4. If w misses 1, then we can obtain an edge 6-coloring of G by recoloring wz with 1, zx with 4, yz with 2, and coloring xy with 1, a contradiction. So w sees 1. Since $d(w)=4$, w must miss 5 or 6. WLOG, assume that w misses 5. Since $(x, 1)\sim (y, 5)$, $L_{5, 1}(w)$ does not pass x and y. So we swap colors on $L_{1,5}(w)$, recolor wz with 1, zx with 4, yz with 2, and coloring xy with 1 to obtain an edge 6-coloring of G, a contradiction.

Subcase 1.1.2 $\phi (wz)=3$ and $\phi (vz)=2$.

Suppose that $\phi (vw)=1$. Then w sees 2, for otherwise, we can obtain an edge 6-coloring of G by recoloring zv with 1, yz and vw with 2, zx with 4, and coloring xy with 1, a contradiction. If w misses $i\in \{4, 5, 6\}$, then we can swap colors on $L_{2,i}(w)$ to satisfy that w misses 2. So w sees 4, 5, and 6. It is impossible. If $\phi (vw)=4$, then w sees 2 according to $(x,2)\sim (y,4)$, and it follows that w sees 5, 6, it is also impossible. Suppose that $\phi (vw)=5$. Then w sees 1, for otherwise we swap colors on $L_{5,1}(w)$ to go back to the previous case that $\phi (vw)=1$. By w seeing 1, we can induce that w also sees 6. Since $d(w)=4$ and w sees 1, 3, 5, 6, w misses 4. Thus we swap colors on $L_{4,1}(w)$ to satisfy that w misses 1 to go back to the above case. It is similar to settle the case $\phi (vw)=6$.

Subcase 1.1.3 $\phi (wz)=3$ and $\phi (vz)\in \{5,6\}$. WLOG, assume that $\phi (vz)=5$.

Suppose that $\phi (vw)=1$. Since $(x,1)\sim (y,5)$, w sees 5. Consecutively, it is easy to check that w sees 2, 4, and 6, it is impossible. If $\phi (vw)=4$, Then w sees 1, for otherwise, we just need to swap colors on $L_{4,1}(w)$ to go back the above case. Consecutively, w sees 2, 5, and 6, it is also impossible. Suppose that $\phi (vw)=2$. Then w sees 5, for otherwise, we just need to swap colors on $L_{2,5}(w)$ to go back to Subcase 1.1.2. Consecutively, w sees 1 and 6, it is also impossible. It is similar to settle the case that $\phi (vw)=6$.

Subcase 1.1.4 $\phi (wz)\in \{5, 6\}$. WLOG, assume that $\phi (wz)=5$.

Since $(x,1)\sim (y,5)$, w sees 1. If w misses 4, then we can obtain an edge 6-coloring of G by recoloring wz with 4, zx with 5, yz with 1, and coloring xy with 4, a contradiction. So w sees 4. w also sees 6, for otherwise, we swap colors on $L_{1,6}(w)$ to obtain the case that w misses 1. Hence w sees 1, 4, 5, 6. It follows from $d(w)=4$ that w misses 2, and we swap colors on $L_{4,2}(w)$ to obtain that w misses 4, a contradiction.

Subcase 1.2 y sees 1.

Since $d(y)=5$, y misses 2 and 3. By Fact 7, we have $(x,i)\sim (y,j)$, where $i\in \{2, 3\}$ and $j\in \{4,5,6\}$. Suppose that $\phi (wz)\in \{2, 3\}$. WLOG, assume that $\phi (wz)=2$. Since $(x, 2)\sim (y,4)$, w sees 4. If w misses 3, then we can swap colors on $L_{4,3}(w)$ to obtain a contradiction. So w sees 3. Also w sees 5, for otherwise, we swap colors on $L_{3,5}(w)$ to get a contradiction. It is similar to check that w sees 6. That is impossible. Suppose that $\phi (wz)\in \{5, 6\}$. WLOG, assume that $\phi (wz)=5$. If w misses 2, then we swap colors on $L_{2,5}(w)$ and go back to the above case. So w sees 2. In the same way, w sees 3. Consecutively, we have that w sees 4 and 6, contrary to that $d(w)=4$.

Case 2 $\phi (yz)\in \{2, 3\}$, WLOG, assume that $\phi (yz)=3$.

By Fact 7, we have $(x,i)\sim (y,j)$, where $i\in \{1,2\}$ and $j\in \{4,5,6\}$. Suppose that $\phi (wz)\in \{4, 5, 6\}$. WLOG, assume that $\phi (wz)=4$. Since $(x,1)\sim (y,4)$, w sees 1. By the similar argument, we have that w sees 5 and 6. Since $d(w)=4$, w misses 2. After swapping colors on $L_{6,2}(w)$, w misses 6, a contradiction. Suppose that $\phi (wz)=2$. Then w sees 4, 5, and 6. Since $d(w)=4$, w misses 1. After swapping colors on $L_{6,1}(w)$, w misses 6, a contradiction. $\square$

Now, we begin to prove our main result.

Theorem 9

Let G be a planar graph with $\Delta \ge 6$. If any 6-cycle contains at most two chords, then G is of class 1.

Proof

Suppose that G is a counterexample to our theorem with the minimum number of edges and suppose that G is embedded in the plane. Then G is a 6-critical graph by Lemma 1, and it is 2-connected. By Euler’s formula $|V(G)|-|E(G)|+|F(G)|=2$, we have

$$\begin{aligned} \sum _{x \in V(G)}(d(x)-4)+\sum _{x \in F(G)}(d(x)-4)=-8<0. \end{aligned}$$

We define ch to be the initial charge. Let $ch(x)=d(x)-4$ for each $x\in V\cup F$. So $\sum _{x\in V\cup F}ch(x)<0$. In the following, we will reassign a new charge denoted by $ch^{'}(x)$ to each $x\in V \cup F$ according to the discharging rules, since our rules only move charges around, and do not affect the sum. If we show that $ch^{'}(x)\ge 0$ for each $x\in V \cup F$, then we get an obvious contradiction $0\le \sum _{x\in V\cup F}ch^{'}(x)=\sum _{x\in V\cup F}ch(x)<0$, which completes our proof.

• R1 Let v be a 2-vertex. If v is incident with a $5^+$-face f, then v receives 1 from f, $\frac{1}{2}$ from each adjacent vertex; If v is incident with a special 4-face f, then v receives $\frac{1}{3}$ from f, $\frac{5}{6}$ from each adjacent vertex; Otherwise v receives 1 from each adjacent vertex.

• R2 Every 3-vertex receives $\frac{1}{3}$ from each adjacent vertex.

• R3 Let f be a 3-face [x, y, z] with $d(x)\le d(y)\le d(z)$. If $d(x)\le 4$ and $d(y)\ge 5$, then f receives $\frac{1}{2}$ from y, $\frac{1}{2}$ from z; If $d(x)=d(y)=4$ and $d(z)=6$, then f receives 1 from z; If $d(x)\ge 5$, then f receives $\frac{1}{3}$ from x, y, z, respectively.

• R4 Let v be a 5-vertex.

• R4.1 If v is adjacent to a 3-vertex u and $N(u)=\{v,w,x\}$, then v receives $\frac{1}{3}$ from w and $\frac{1}{3}$ from x;

• R4.2 If v is incident with a (4, 5, 5)-face [u, v, w] and $N(u)=\{v,w,x,y\}$, then v receives $\frac{1}{6}$ from x, and $\frac{1}{6}$ from y;

• R4.3 If v is incident with a (4, 5, 6)-face [u, v, w], then v receives $\frac{1}{6}$ from w.

• R5 Let v be a 6-vertex.

• R5.1 If v is incident with a special 4-face $f=[w,v,x,y]$ such that $d(y)=2$, then v sends $\frac{1}{3}$ to f;

• R5.2 If v is incident with two adjacent 3-faces (u, v, x), (v, x, y), and $d(u)=d(x)=4$, then v receives $\frac{1}{6}$ from y.

Now, let’s began to check $ch'(x)\ge 0$ for all $x\in V\cup F$. Let $f\in F(G)$. If $d(f)\ge 5$, then f is incident with at most $d(f)-4$ 2-vertices by Corollary 3(c), so $ch'(f)\ge ch(f)-(d(f)-4)=0$ by R1. Suppose $d(f)=4$. If f is special, then $ch'(f)=0+\frac{1}{3}-\frac{1}{3}=0$ by R1 and R5.1; Otherwise, $ch'(f)=ch(f)=0$. Suppose $d(f)=3$. Since $\Delta =6$, f must be the $(2^{+},6,6)$-face, $(3,5^+,6)$-face, (4, 4, 6)-face, or $(4^{+},5^{+},5^{+})$-face by Lemma 2. Hence $ch'(f)=ch(f)+\min \{2\times \frac{1}{2}, 1, 3\times \frac{1}{3}\}=0$ by R3.

Let $w\in V(G)$. Then $d(w)\ge 2$. If $d(w)=2$, then w is adjacent to two 6-vertices by Corollary 3(a), so $ch'(w)=ch(w)+\min \{1+2\times \frac{1}{2}, \frac{1}{3}+2\times \frac{5}{6}, 2\times 1\}=0$ by R1. If $d(w)=3$, then w is adjacent to three $5^{+}$-vertices by Corollary 3(b), and it follows that $ch'(w)=-1+3 \times \frac{1}{3}=0$ by R2. If $d(w)=4$, then $ch'(w)=ch(w)=0$.

Suppose that $d(w)=5$. Then $ch(w)=1$, $\min \{d(u)|u\in N(w)\} \ge 3$, $d_{3}(w) \le 1$ ,and $d_{6}(w) \ge 2$ by Corollary 3, and $f_3(w)\le 3$ since all 6-cycles in G contain at most two chords. If all neighbors of w are $5^+$-vertices, then $ch'(w)\ge 1- 3\times \frac{1}{3}=0$ by R3. If w is adjacent to a 3-vertex, say $w_1$, then w receives $\frac{1}{3}$ from each of neighbors of $w_1$ except w by R4.1, and it follows that $ch'(w)\ge 1+2\times \frac{1}{3}-\frac{1}{3}- 2\times \frac{1}{2}-\frac{1}{3}=0$ by R2 and R3. Suppose that w is adjacent to a 4-vertex. Then $d_6(w)\ge 3$ by Lemma 2a. If $f_3(w)\le 2$, then $ch'(w)\ge 1- 2\times \frac{1}{2}=0$ by R3; Otherwise, $f_3(w)=3$, w is incident with at most one $(4,5,5^{+})$-face by Lemma 5. So $ch'(w)\ge 1- \max \{2\times \frac{1}{2}+ \frac{1}{3}-2\times \frac{1}{6}, \frac{1}{2}+2\times \frac{1}{3}-\frac{1}{6}\}=0$ by R3 and R4.

Suppose that $d(w)=6$. Then $ch(w)=2$, $d_{2}(w) \le 1$, and $d_{6}(w) \ge 2$ by Lemma 2, and $f_3(w)\le 4$ since all 6-cycles in G contain at most two chords. Note that every special 4-face is adjacent to at most two 3-faces since all 6-cycles contain at most two chords. We denote by $f_{s4}(w)$ the number of special 4-faces incident with w and $f_{34}(w)=f_3(w)+f_{s4}(w)$. It is easy to check that $f_{s4}(w)\le 3$ and $f_{34}(w)\le 4$.

Case 1 w sends some charge to a 5-vertex v (see R4).

Suppose that v is adjacent to a 3-vertex u. Then w sends $\frac{1}{3}$ to v by R4.1. By Lemma 4b, w is adjacent to five 6-vertices, that is, $d_6(w)=5$. Since $f_{34}(w)\le 4$, $ch'(w)\ge 2-\frac{1}{3}-2\times \frac{1}{2}-2\times \frac{1}{3}=0$ by R3 and R4.1.

Suppose that v is incident with a (4, 5, 5)-face [u, v, x] such that $N(u)=\{v,w,x,y\}$. Then v receives $\frac{1}{6}$ from w by R4.2. By Lemma 6c, w is adjacent to one $4^{-}$-vertex, which is u. Since $f_{34}(w)\le 4$, $ch'(w)=2-2\times \frac{1}{6}-2\times \frac{1}{2}-2\times \frac{1}{3}=0$.

Suppose that v is incident with a (4, 5, 6)-face [u, v, w]. Then v receives $\frac{1}{6}$ from w by R4.3. If $d_4(w)=1$, then $ch'(w)\ge 2-2\times \frac{1}{2}-2\times \frac{1}{3}-2\times \frac{1}{6}=0$; otherwise $d_4(w)=2$. By Lemma 8, w is incident a (4, 5, 6)-face and at most one (4, 6, 6)-face, then $ch'(w)\ge 2-2\times \frac{1}{2}-2\times \frac{1}{3}-\frac{1}{6}=\frac{1}{6}>0$ by R3 and R4.3.

Case 2 w sends $\frac{1}{6}$ to some 6-vertex v (see R5).

Suppose v is incident with two 3-faces [w, v, x] and [v, x, y] such that $d(x)=d(y)=4$. Then $f_{34}(w)\le 4$ and $d_6(w)=5$ by Lemma 4b. So $ch'(w)=2-\frac{1}{6}-2\times \frac{1}{2}-2\times \frac{1}{3}>0$.

Case 3 w sends no charge to its $5^+$-vertices.

Let $k=\min \{d(u)|u\in N(w)\}$. If $k\ge 5$, then $ch'(w)\ge 2-4\times \frac{1}{3}>0$. Suppose that $k=4$. Then $d_6(w)\ge 3$ by Lemma 2a. If w is incident with two 3-faces [u, w, x] and [w, x, y] such that $d(x)=d(y)=4$, then $d_6(w)=4$. If wy is incident with a $4^+$-face, then w receives $\frac{1}{6}$ from u, and it follows that $ch'(w)\ge 2+\frac{1}{6}-1-\frac{1}{2}-2\times \frac{1}{3}=0$ since $f_{34}(w)\le 4$; otherwise wy is incident with another 3-face [w, y, z], then w receives $\frac{1}{6}$ from each of u, z, and it follows that $ch'(w)\ge 2+2\times \frac{1}{6}-1-2\times \frac{1}{2}-\frac{1}{3}=0$ since $f_{34}(w)\le 4$; otherwise, $ch'(w)\ge 2-\max \{1+3\times \frac{1}{3}, 4\times \frac{1}{2}\}=0$.

Suppose that $k=3$. Then $d_6(w)\ge 4$ by Lemma 2a. If $d_{3}(w)=1$ and $d_{5^+}(w)\ge 5$, then $ch'(w)\ge 2-\frac{1}{3}-2\times (\frac{1}{2}+\frac{1}{3})=0$; otherwise, w is incident with two $4^-$-vertices u, v, then u and v are incident with at most one 3-face by Lemma 5 since $d(u)+d(v)+d(w)\le 3+4+6<14$. So $f_{34}(w)\le 4$, and it follows that $ch'(w)\ge 2-\frac{1}{3}-\max \{\frac{1}{2}+2\times \frac{1}{3}, \frac{1}{3}+3\times \frac{1}{3}\}>0$ by R1 and R3.

Suppose that $k=2$, that is, w is adjacent to a 2-vertex v. Then $d_6(w)=5$ by Lemma 2a. If v is incident with a special 4-face $f=[u,v,w,x]$, then $f_3(v)\le 3$ and w sends $\frac{5}{6}$ to v, and it follows that $ch'(w)\ge 2-\frac{5}{6}-\frac{1}{2}-2\times \frac{1}{3}=0$; otherwise, v is incident with a $5^+$-face or two 4-faces. If v is incident with a $5^+$-face, then w sends $\frac{1}{2}$ to v, and it follows that $ch'(w)\ge 2-\frac{1}{2}-(\frac{1}{2}+3\times \frac{1}{3})=0$. If v is incident with two 4-faces, then $f_{34}\le 3$, and it follows that $ch'(w)\ge 2-1-3\times \frac{1}{3}=0$. $\square$