Adjacent Vertex Distinguishing Edge Coloring of Planar Graphs Without 4-Cycles

Abstract

The adjacent vertex distinguishing edge coloring of a graph G is a proper edge coloring of G such that the edge coloring set on any pair of adjacent vertices is distinct. The minimum number of colors required for an adjacent vertex distinguishing edge coloring of G is denoted by $\chi _{a}'(G)$. It is observed that $\chi _a'(G)\ge \Delta (G)+1$ when G contains two adjacent vertices of degree $\Delta (G)$. In this paper, we prove that if G is a planar graph without 4-cycles, then $\chi _a'(G)\le \max \{9,\Delta (G)+1\}$.

1 Introduction

All graphs considered in this paper are finite and simple. A plane graph is a particular drawing of a planar graph on the Euclidean plane so that any two edges intersect only at their ends. For a plane graph G, we denote its vertex set, edge set, and face set by V(G), E(G), and F(G), respectively. For $x\in V(G)\cup F(G)$, let $d_G(x)$ denote the degree of x. Let $N_G(v)$ be the set of neighbors of v. Then, $d_G(v)=|N_G(v)|$. Let $\Delta (G)=\max \{d_G(v)|v\in V(G)\}$ be the maximum degree of G and $\delta (G)=\min \{d_G(v)|v\in V(G)\}$ be the minimum degree of G. In G, a vertex of degree k, at least k, or at most k is called a k-vertex, $k^{+}$-vertex, or $k^{-}$-vertex, respectively. We use $d_k^G(v)$, $d_{k^+}^G(v)$, and $d_{k^-}^G(v)$ to denote the number of k-vertices, $k^{+}$-vertices, or $k^{-}$-vertices adjacent to v, respectively.

A proper edge k-coloring of G is a mapping  $\phi$: $E(G)\rightarrow \{1, 2,\ldots , k\}$ such that $\phi (e)\ne \phi (e')$ for any two adjacent edges e and $e'$. The chromatic index, denoted by $\chi '(G)$, is the smallest integer k such that G has a proper edge k-coloring. Let $C_\phi (v)= \{\phi (xv)|xv\in E(G)\}$ denote the set of colors assigned to edges incident with vertex v. We call two adjacent vertices u and vconflict under an edge coloring $\phi$ if $C_\phi (u)=C_\phi (v)$. A proper edge k-coloring $\phi$ of G is adjacent vertex distinguishing, or a k-avd-coloring in short, if any two adjacent vertices do not conflict. That is $C_{\phi }(u)\ne C_{\phi }(v)$ for any $uv\in E(G)$. The adjacent vertex distinguishing chromatic index, denoted by $\chi _a'(G)$, is the smallest integer k such that G has a k-avd-coloring.

Note that if G contains an isolated edge, then G has no avd-coloring. A graph G is normal if it has no isolated edges. The adjacent vertex distinguishing coloring is variously known as adjacent strong edge coloring (Zhang et al. [11]) and 1-strong edge colorings (Akbari et al. [1]). It follows from the definition that $\chi _a'(G)\ge \chi '(G)\ge \Delta (G)$. The well-known Vizing’s theorem (1964) asserts that $\chi '(G)\le \Delta (G)+{1}$ for any simple graph G. However, there exists infinitely many graphs G such that $\chi _a'(G)>\Delta (G)+{1}$. For example, for a cycle $C_n$ of length n, $\chi _a'(C_n)=4=\Delta (C_n)+{2}$ if $n\not \equiv 0(\text{ mod }\ 3)$ and $n\ne 5$, and $\chi _a'(C_5)=5=\Delta (C_5)+{3}$.

Zhang et al. [11] completely determined $\chi _a'(G)$ for some special graphs. Based on these results, they proposed the following conjecture.

Conjecture 1

Let G be a normal connected graph and $G\ne C_5$. Then, $\chi _a'(G)\le \Delta (G)+{2}$.

Balister et al. [2] confirmed Conjecture 1 for all bipartite graphs and graphs with $\Delta (G)\le 3$ and proved that $\chi _a'(G)\le \Delta (G)+O(\log k)$, where k is the vertex chromatic number of the normal graph G. Hatami [4] showed that every normal graph G with $\Delta (G)\ge 10^{20}$ has $\chi _a'(G)\le \Delta (G)+300$ by the probabilistic method. Wang and Wang [8] characterized $\chi _a'(G)$ for $K_4$-minor free graphs G with $\Delta (G)\ge 5$. For more results about the adjacent vertex distinguishing edge coloring with the small maximum degree, the readers may refer to [5, 7, 9].

Bu et al. [3] confirmed Conjecture 1 for planar graphs with girth at least 6. Yan et al. [10] showed that if G is a normal planar graph with girth $g(G)\ge 5$ and $G\ne C_5$, then $\chi _a'(G)\le \Delta (G)+2$. Huang et al. [6] showed that if G is a connected planar graph without 3-cycles and maximum degree at least 12, then $\chi _a'(G)\le \Delta (G)+1$. In this paper, we will prove the following result:

Theorem 1

Let G be a normal planar graph without 4-cycles. Then, $\chi _a'(G)\le \max \{9,\Delta (G)+1\}$.

To obtain our main result, we need to introduce some notations. Let G be a plane graph. For $k=2$ or 3, a k-vertex v is called bad in G if $d_k^{G}(v)=1$; it is called good in G if $d_k^{G}(v)=0$. We use $d_{kb}^{G}(v)$ or $d_{kg}^{G}(v)$ to denote the number of bad k-vertices or good k-vertices in G adjacent to a vertex v, respectively. Let $t_G(v)$ denote the number of 3-faces incident with a vertex v. An edge uv is called light k-edge if $d_G(u)=d_G(v)=k$ with $k=2$ or 3. For $f\in F(G)$, we use b(f) to denote the boundary walk of f, and $f=[v_0v_1\cdots v_{k-1}]$ if $v_0$, $v_1$,..., $v_{k-1}$ are the boundary vertices of f in cyclic order. Let $d_k^G(f)$ be the number of k-vertices incident with f. Similarly, we can define $d_{k^+}^G(f)$, $d_{k^-}^G(f)$, $d_{kg}^G(f)$, and $d_{kb}^G(f)$. A 3-face $[v_0v_1v_2]$ is called a $(d_0,d_1,d_2)$-face if $d_G(v_i)=d_i$ for $i=0,1,2$. A 5-face f is called bad if $d_{2b}^G(f)=2$. A 5-face f is called special if $d_{2g}^G(f)=2$. A 6-face f is called special if $d_2^G(f)=3$ and $d_{2b}^G(f)=2$. A 7-face is called special if $d_{2b}^G(f)=4$.

2 Proof of Theorem 1

Theorem 1 will be proved by contradiction. For any graph G, set $n_i(G)=|\{v\in V(G)|d_G(v)=i\}|$ for $i=1,2,\ldots , \Delta (G)$. We denote that a graph H is “smaller” than G if $(|E(H)|,n_t(H),n_{t-1}(H),\ldots ,n_2(H),n_1(H))$ precedes $(|E(G)|,n_t(G),n_{t-1}(G),\ldots ,n_2(G),n_1(G))$ with respect to the standard lexicographic order, where $t=\max \{\Delta (G),\Delta (H)\}$. Let G be the “smallest” counterexample of Theorem 1. Obviously, G is connected. First, we shall prove the following claims on the characterizations of G. Then, we will derive a contradiction using the discharging method.

For two positive integers p, q with $p<q$, we use [p, q] to denote the set of all integers between p and q (including p and q). Let $T(G)=\max \{9,\Delta (G)+1\}$. Then, $T(G)\ge 9$. We may assume that $\Delta (G)\ge 4$ by the result in [2].

Claim 1

There is no edge uv with $d_G(u)=1$ and $d_G(v)\le 5$.

Proof

Assume to the contrary that G contains an edge uv with $d_G(u)=1$ and $d_G(v)\le 5$. Let $H=G-u$. Then, H is connected and hence normal as $\Delta (G)\ge 4$. By the smallest of G, H has a T(G)-avd-coloring $\phi$ with the color set C. Since $|C|\ge 9$ and $d_G(v)\le 5$, we can color uv with a color in $C\setminus C_{\phi }(v)$, such that v does not conflict with its neighbors. $\square$

Remark 1

Claim 1 implies that if e is an arbitrary edge of G, then $H=G-e$ is a normal subgraph, and therefore, $\chi '_a(H)\le T(H)\le T(G)$ by the smallest of G.

Claim 2

Let v be a k-vertex with $2\le k\le 4$. Then, $d_k^G(v)\le 1$.

Proof

We only need to proof the case that $k=4$. (For $k=2$ and 3, the proof is similar and simple.) Assume to the contrary that $d_4^G(v)\ge 2$. Let $v_1, v_2,v_3, v_4$ be the neighbors of v with $d_G(v_1)=d_G(v_2)=4$. Let $u_i\ (i=1,2,3)$ be the neighbors of $v_1$ other than v; $w_i\ (i=1,2,3)$ be the neighbors of $v_2$ other than v. Then, $G-vv_1$ has a T(G)-avd-coloring $\phi$ by Remark 1. Let $\phi (vv_i)=i-1$ for $i\in [2,4]$, and $\phi (v_1u_i)=a_i$ for $i=1,2,3$. The proof of this claim is divided into four cases, depending on the value of $s=|\{a_1,a_2,a_3\}\setminus \{1,2,3\}|$.

• (i) Suppose that $s=1$, say $a_1=1$, $a_2=2$, and $a_3=4$. Suppose that $vv_1$ cannot be colored without creating a conflict. That is, colors $\{5,6,\ldots ,9\}$ must be occurred on the vertices $\{v_2,v_3,v_4,u_1,u_2,u_3\}$. Without loss of generality, we may assume that $C_\phi (v_i)=\{1,2,3,i+3\}$ for $i=2,3,4$, $C_\phi (u_1)=\{1,2,4,8\}$, and $\{1,2,4,9\}\in \{C_{\phi }(u_2), C_{\phi }(u_3)\}$. Note that $vv_2$ can be recolored with a color $\alpha \in \{6,7,8,9\}$ such that $v_2$ does not conflict with its neighbors. Let $\beta \in \{6,7\}\setminus \{\alpha \}$. If $\{1,2,4,\beta \}\notin \{C_{\phi }(u_2), C_{\phi }(u_3)\}$, then we can recolor $vv_2$ with $\alpha$ and color $vv_1$ with $\beta$. If $\{1,2,4,\beta \}\in \{C_{\phi }(u_2), C_{\phi }(u_3)\}$, then we can recolor $v_1u_1$ with a color $\alpha _1\in \{3,5,6,7\}$ such that $u_1$ does not conflict with its neighbors, and color $vv_1$ with 9.

• (ii) Suppose that $s=0$. If $vv_2$ can be recolored without creating a conflict, turn to the case that $s=1$. So we may assume that $vv_2$ cannot be recolored without creating a conflict, say $\phi (v_2w_i)=i+3$ and $C_{\phi }(w_i)=\{4,5,6,i+6\}$ for $i=1,2,3$. We can recolor $v_2w_1$ with a color $\alpha \in \{2,3,8,9\}$ such that $w_1$ does not conflict with its neighbors, and recolor $vv_2$ with a color in $\{8,9\}\setminus \{\alpha \}$, turn to the case that $s=1$.

• (iii) Suppose that $s=2$, say $a_1\in [1,3]$, $a_2=4$, and $a_3=5$. Suppose that $vv_1$ cannot be colored without creating a conflict. That is, $C_\phi (v_i)=\{1,2,3,i+4\}$ for $i=2,3,4$, and $\{a_1,4,5,9\}\in \{C_{\phi }(u_i)| i=1,2,3\}$; or $C_\phi (v_i)=\{1,2,3,i+4\}$ for $i=2,3$, and $\{a_1,4,5,8\},\{a_1,4,5,9\}\in \{C_{\phi }(u_i)| i=1,2,3\}$.

• Suppose that $C_\phi (v_i)=\{1,2,3,i+4\}$ for $i=2,3,4$, and $\{a_1,4,5,9\}\in \{C_{\phi }(u_i)| i=1,2,3\}$. If $\{a_1,4,5,7\}\notin \{C_{\phi }(u_i)|i=1,2,3\}$, then we can recolor $vv_2$ with a color $\alpha \in \{4,5,8,9\}$ such that $v_2$ does not conflict with its neighbors, and color $vv_1$ with 7. So we may assume that $\{a_1,4,5,7\}\in \{C_{\phi }(u_i)|i=1,2,3\}$. Similarly, we can show that $\{a_1,4,5,8\}\in \{C_{\phi }(u_i)|i=1,2,3\}$. Now, we can recolor $vv_3$ with a color $\beta \in \{4,5,8,9\}$ such that $v_3$ does not conflict with its neighbors, and color $vv_1$ with 6.

• Suppose that $C_\phi (v_i)=\{1,2,3,i+4\}$ for $i=2,3$, and $\{a_1,4,5,8\},\{a_1,4,5,9\}\in \{C_{\phi }(u_i)| i=1,2,3\}$. Note that $\{a_1,4,5,6\}$ and $\{a_1,4,5,7\}$ is not the color set of $u_1,u_2$, and $u_3$; otherwise, the proof is similar as the above arguments. It is observed that $vv_2$ can be recolored with a color $\alpha \in \{4,5,8,9\}$ such that $v_2$ does not conflict with its neighbors. If $\{2,3,\alpha ,7\}\ne C_{\phi }(v_4)$, then we can recolor $vv_2$ with $\alpha$ and color $vv_1$ with 7. So we may assume that $\{2,3,\alpha ,7\}=C_{\phi }(v_4)$. Then, we can recolor $vv_3$ with a color $\beta \in \{4,5,8,9\}$ such that $v_3$ does not conflict with its neighbors, and color $vv_1$ with 6.

• (iv) Suppose that $s=3$, say $a_1=4$, $a_2=5$, and $a_3=6$. Suppose that $vv_1$ cannot be colored without creating a conflict. That is, $C_\phi (v_i)=\{1,2,3,i+5\}$ for $i=2,3$, and $C_{\phi }(v_4)=\{1,2,3,9\}$ or $\{4,5,6,9\}\in \{C_{\phi }(u_i)|i=1,2,3\}$.

• Suppose that $C_{\phi }(v_4)=\{1,2,3,9\}$. If $\{4,5,6,9\}\notin \{C_{\phi }(u_i)|i=1,2,3\}$, then we can recolor $vv_2$ with a color in $\{4,5,6,8\}$ such that $v_2$ does not conflict with its neighbors, and color $vv_1$ with 9. So we may assume that $\{4,5,6,9\}=\{C_{\phi }(u_1)\}$. Similarly, we can show that $\{4,5,6,7\},\{4,5,6,8\}\in \{C_{\phi }(u_i)|i=2,3\}$. We can recolor $vv_2$ with a color in $\{4,5,6,9\}$ such that $v_2$ does not conflict with its neighbors, recolor $v_1u_1$ with a color in $\{1,2,3,7\}$ such that $u_1$ does not conflict with its neighbors, and color $vv_1$ with 8.

• Suppose that $\{4,5,6,9\}\in \{C_{\phi }(u_i)|i=1,2,3\}$, say $\{4,5,6,9\}=C_{\phi }(u_1)$. Note that $\{\{4,5,6,7\},\{4,5,6,8\}\}\ne \{C_{\phi }(u_2), C_{\phi }(u_3)\}$; otherwise, the proof is similar as the above arguments. It is observed that $vv_i$ can be recolored with a color $\alpha _i\in \{4,5,6,9\}$ such that $v_i$ does not conflict with its neighbors for $i=2,3$. If $\{4,5,6,8\}$ or $\{2,3,\alpha _2,8\}$ is not the color set of $u_2,u_3$, and $v_4$, then we can recolor $vv_2$ with $\alpha _2$ and color $vv_1$ with 8. If $\{4,5,6,7\}$ or $\{1,3,\alpha _3,7\}$ is not the color set of $u_2,u_3$, and $v_4$, then we can recolor $vv_3$ with $\alpha _3$ and color $vv_1$ with 7. Hence, by the above notation, we may assume that $C_{\phi }(u_2)=\{4,5,6,8\}$ and $C_{\phi }(v_4)=\{1,3,\alpha _3,7\}$. Then, $C_{\phi }(u_3)\ne \{4,5,6,7\}$. We can recolor $vv_4$ with a color in $\{4,5,6,8,9\}\setminus \{\alpha _3\}$ such that $v_4$ does not conflict with its neighbors, and color $vv_1$ with 7.

$\square$

Claim 3

Let v be a k-vertex of G with $k\le 4$. Then, $d_i^G(v)=0$, where $i<k$.

Proof

Assume to the contrary that G contains a k-vertex v adjacent to a i-vertex u with $i<k\le 4$. Then, $G-vu$ has a T(G)-avd-coloring $\phi$ using the color set $C=[1,T(G)]$ by Remark 1. By Claim 2, $d_k^G(v)\le 1$ and $d_i^G(u)\le 1$. Note that $|C_{\phi }(v)|\le k-1\le 3$ and $|C_{\phi }(u)|\le i-1\le 2$. Since $T(G)\ge 9$, we can color uv with a color in $[1,T(G)]\setminus (C_{\phi }(v)\cup C_{\phi }(u))$ such that neither v nor u conflicts with its neighbors. $\square$

Claim 4

For each vertex $v\in V(G)$, if $d_1^G(v)\ge i$, then $d_{i+1}^G(v)=0$ for $i=1,2$.

Proof

Let $v_1, v_2,\ldots , v_k$ be the neighbors of v with $d_G(v_1)=1$.

• (i) First, we show that $d_2^G(v)=0$ when $d_1^G(v)\ge 1$. Assume to the contrary that $d_2^G(v)\ge 1$, say $d_G(v_2)=2$ with $N_G(v_2)=\{u,v\}$. Let H be the graph obtained from $G-v_2u$ by adding a new vertex $v_2'$ and a new edge $uv_2'$. Then, $|E(H)|=|E(G)|$ and $n_2(H)<n_2(G)$, i.e., H is smaller than G. Hence, H has a T(G)-avd-coloring $\phi$ by the smallest of G. Now, we need to find the T(G)-avd-coloring $\psi$ of G. Let $\psi (e)=\phi (e)$ for $e\in (E(H)\cap E(G))\setminus \{vv_1,vv_2\}$.

Suppose that $d_G(u)\ne 2$. Let $\psi (v_2u)=\phi (v_2'u)$, $\psi (vv_2)\in \{\phi (vv_1),\phi (vv_2)\}\setminus \{\psi (v_2u)\}$, and $\psi (vv_1)\in \{\phi (vv_1),\phi (vv_2)\}\setminus \{\psi (vv_2)\}$. Suppose that $d_G(u)=2$ with $N_G(u)=\{v_2,u_1\}$. Let $\psi (vv_2)\in \{\phi (vv_1),\phi (vv_2)\}\setminus \{\psi (uu_1)\}$, and $\psi (vv_1)\in \{\phi (vv_1),\phi (vv_2)\}\setminus \{\psi (vv_2)\}$. Then, we can color $v_2u$ properly.

• (ii) Next, we show that $d_3^G(v)=0$ when $d_1^G(v)\ge 2$. Suppose that $d_G(v_2)=1$. Assume to the contrary that $d_3^G(v)\ge 1$, say $d_G(v_3)=3$ with $N_G(v_3)=\{u_1,u_2,v\}$. Let H be the graph obtained from $G-\{v_3u_1,v_3u_2\}$ by adding a new vertex $v_3'$ and two new edges $v_3'u_1$ and $v_3'u_2$. Then, $|E(H)|=|E(G)|$ and $n_3(H)<n_3(G)$, i.e., H is smaller than G. Hence, H has a T(G)-avd-coloring $\phi$ by the smallest of G. Now, we need to find the T(G)-avd-coloring $\psi$ of G. Let $\psi (e)=\phi (e)$ for $e\in (E(H)\cap E(G))\setminus \{vv_1,vv_2,vv_3\}$.

Suppose that $d_G(u_1)\ne 3$ and $d_G(u_2)\ne 3$. Let $\psi (v_3u_i)=\phi (v_3'u_i)$ for $i=1,2$, $\psi (vv_3)\in \{\phi (vv_1),\phi (vv_2),\phi (vv_3)\}\setminus \{\psi (v_3u_1),\psi (v_3u_2)\}$, $\{\psi (vv_1),\psi (vv_2)\}=\{\phi (vv_1),\phi (vv_2),\phi (vv_3)\}\setminus \{\psi (vv_3)\}$. By Claim 2, $d_3^G(v_3)\le 1$. Suppose that $d_G(u_1)=3$ and $d_G(u_2)\ne 3$. We remove the color of $v_3'u_1$ and set $\alpha \in C_{\phi }(u_1)\setminus \{\phi (v_3'u_2)\}$. Let $\psi (v_3u_2)=\phi (v_3'u_2)$, $\psi (vv_3)\in \{\phi (vv_1),\phi (vv_2),\phi (vv_3)\}\setminus \{\psi (v_3u_2), \alpha \}$, $\{\psi (vv_1),\psi (vv_2)\}=\{\phi (vv_1),\phi (vv_2),\phi (vv_3)\}\setminus \{\psi (vv_3)\}$. Then, we can color $v_3u_1$ properly. $\square$

Claim 5

There is no 5-cycle incident with three 2-vertices in G; and there is no 5-cycle incident with two bad 2-vertices and a bad 3-vertex in G simultaneously.

Proof

Let $C=v_0v_1v_2v_3v_4v_0$ be the 5-cycle of G with $d_G(v_0)=d_G(v_1)=2$. Then, $G-v_0v_1$ has a T(G)-avd-coloring $\phi$.

First, we show that G does not contain any 5-cycle incident with three 2-vertices. Assume to the contrary that C is incident with three 2-vertices. By Claim 2, we may assume that $d_G(v_3)=2$. If $\phi (v_0v_4)\ne \phi (v_1v_2)$, we can color $v_0v_1$ properly to obtain a T(G)-avd-coloring of G, a contradiction. Assume that $\phi (v_0v_4)=\phi (v_1v_2)$. Then, $\phi (v_3v_4)\ne \phi (v_1v_2)$. We can exchange the colors of $v_1v_2$ and $v_2v_3$ and color $v_0v_1$ properly.

Next, we show that G does not contain any 5-cycle incident with two bad 2-vertices and a bad 3-vertex simultaneously. Assume to the contrary that C is incident with bad 3-vertex. By Claim 3, $d_G(v_3)=3$ with $d_3^G(v_3)=1$. Let x be the neighbor of $v_3$ other than $v_2$ and $v_4$. Then, $d_G(x)=3$. Let $x_1$ and $x_2$ be the neighbors of x other than $v_3$. If $\phi (v_0v_4)\ne \phi (v_1v_2)$, we can color $v_0v_1$ properly to obtain a T(G)-avd-coloring of G, a contradiction. So we may assume that $\phi (v_0v_4)=\phi (v_1v_2)=1$. Then, we can assume that $\phi (v_2v_3)=2$ and $\phi (v_3v_4)=3$. We remove the color of $v_3x$. Note that $\{\{1,2\},\{1,3\}\}\setminus \{\phi (xx_1),\phi (xx_2)\}\ne \emptyset$, say $\{\phi (xx_1),\phi (xx_2)\}\ne \{1,2\}$. Then, we can exchange the colors of $v_0v_4$ and $v_3v_4$ and color $v_0v_1$ and $v_3x$ properly. $\square$

Claim 6

G does not contain any configuration as shown in Fig. 1.

Proof

Assume to the contrary that G contains $G_i$ as a subgraph as shown in Fig. 1, where $i\in [1,7]$. We will use the labels of graphs as shown in Fig. 1. Note that we first remove the colors of all light 2-edges and light 3-edges. After recoloring some edges of $G_i$ for $i\in [1,7]$ so that the color sets of each pair of end vertices of light edges are distinct, then we can color them properly to deduce a T(G)-avd-coloring of G.

Case 1G contains $G_1$ as a subgraph.

Let H be the graph obtained from $G-vv_1$ by adding a new vertex $v_1'$ and a new edge $vv_1'$. Then, $|E(H)|=|E(G)|$ and $n_2(H)<n_2(G)$, i.e., H is smaller than G. Thus, H has a T(G)-avd-coloring $\phi$. Now, we need to construct a T(G)-avd-coloring $\psi$ of G. First, let $\psi (e)=\phi (e)$ for $e\in E(H)\cap E(G)$.

Let $\alpha =\phi (v_1u_1)$ when $d_G(u_1)\ne 2$. When $d_G(u_1)=2$, let x be the second neighbor of $u_1$ other than $v_1$. Then, $d_G(x)\ne 2$ by Claim 2. We remove the color of $v_1u_1$ and let $\alpha =\phi (u_1x)$. If $\phi (vv_1')\ne \alpha$, then set $\psi (vv_1)=\phi (vv_1')$. So we may assume that $\phi (vv_1')= \alpha$. If $\phi (v_2u_2)\ne \alpha$, then we can recolor $vv_2$ with $\alpha$ and color $vv_1$ with $\phi (vv_2)$. If $\phi (v_2u_2)=\alpha$, then we can exchange the color of $vu_2$ and $v_2u_2$ and color $vv_1$ with $\psi (vu_2)$.

Fig. 1

The configurations used in Claim 6. In this figure solid vertices have no other vertices adjacent to them other than those showed. In $G_2$, we further assume that $d_G(v_3)\le 2$

Case 2G contains $G_2$ or $G_3$ as a subgraph.

Then, $G-v_1v_2$ admits a T(G)-avd-coloring $\phi$ by Remark 1. Suppose that $v_1v_2$ cannot be colored without creating a conflict. That is, $\{\phi (vv_1),\phi (v_1u_1)\}=\{\phi (vv_2),\phi (v_2u_2)\}$.

• (i) Suppose that G contains $G_2$ as a subgraph. If $d_G(v_3)=1$ or $v_3$ is a good 2-vertex in G, then we can recolor $vv_3$ with a color in $\{\phi (vv_1),\phi (vv_2)\}\setminus \{\phi (v_3u_3)\}$, say $\phi (vv_1)$, and color $vv_1$ with $\phi (vv_3)$. So we may assume that $v_3$ is a bad 2-vertex in G. Let $N_G(v_3)=\{v,u_3\}$ and $N_G(u_3)=\{v_3,u_3'\}$. Then, $d_G(u_3')\ne 2$ by Claim 2. We can recolor $vv_3$ with a color in $\{\phi (vv_1),\phi (vv_2)\}\setminus \{\phi (u_3u_3')\}$, say $\phi (vv_1)$, and color $vv_1$ with $\phi (vv_3)$.

• (ii) Suppose that G contains $G_3$ as a subgraph. Note that $\{\phi (vv_3),\phi (v_3u_3)\}\ne \{\phi (vv_4),\phi (v_4u_4)\}$, say $\phi (v_3u_3)\ne \phi (vv_4)$. We can recolor $vv_3$ with a color in $\{\phi (vv_1),\phi (vv_2)\}\setminus \{\phi (v_3u_3)\}$, say $\phi (vv_1)$. Then, we can recolor $vv_1$ with $\phi (vv_3)$ and color $v_1v_2$, $v_3v_4$ properly.

Case 3G contains $G_i$ as a subgraph, where $i\in [4,7]$.

Then, $G-v_1v_2$ admits a T(G)-avd-coloring $\phi$ by Remark 1. Suppose that $v_1v_2$ cannot be colored without creating a conflict. Without loss of generality, we may assume $\phi (v_2v_3)= \phi (v_0v_1)=1$.

• (i) G contains $G_4$ as a subgraph. We can exchange the color of $v_2v_3$ and $v_3v_4$ and color two light edges $v_1v_2$ and $v_4v_5$ properly.

• (ii) G contains $G_5$ as a subgraph. If $\phi (v_4v_5)\ne 1$, then we can exchange the colors of $v_0v_1$ and $v_0v_5$. So we may assume that $\phi (v_4v_5)=1$. If $\phi (v_3v_4)\ne \phi (v_0v_5)$, then we can recolor $v_2v_3$, $v_3v_4$, and $v_4v_5$ with $\phi (v_3v_4)$, 1, $\phi (v_3v_4)$, respectively. Hence, assume that $\phi (v_3v_4)=\phi (v_0v_5)=2$. Let $\alpha =\phi (x_1y_1)$ when $d_G(y_1)\ne 2$. When $d_G(y_1)=2$, let $y_1'$ be the second neighbor of $y_1$ other than $x_1$. Then, $d_G(y_1')\ne 2$ by Claim 2. We remove the color of $x_1y_1$ and let $\alpha =\phi (y_1y_1')$. If $\alpha \ne 1$, then we can recolor $v_3x_1$, $v_2v_3$ with 1, $\phi (v_3x_1)$, respectively. If $\alpha =1$, then we can recolor $v_3v_2$, $v_3v_4$, $v_4v_5$, $v_5v_0$, $v_0v_1$, $v_3x_1$ with $\phi (v_3x_1)$, 1, 2, 1, 2, 2, respectively.

• (iii) G contains $G_6$ as a subgraph. We remove the color of $v_4v_5$. If $\phi (v_5v_6)\ne 1$, then we can exchange the colors of $v_2v_3$ and $v_3v_4$. So we may assume that $\phi (v_5v_6)=1$. If $\phi (v_3v_4)\ne \phi (v_0v_6)$, then we can recolor $v_0v_1$, $v_0v_6$, and $v_5v_6$ with $\phi (v_0v_6)$, 1, $\phi (v_0v_6)$, respectively. Hence, assume that $\phi (v_3v_4)=\phi (v_0v_6)=2$. Let $\alpha =\phi (x_1y_1)$ when $d_G(y_1)\ne 2$. When $d_G(y_1)=2$, let $y_1'$ be the second neighbor of $y_1$ other than $x_1$. Then, $d_G(y_1')\ne 2$ by Claim 2. We remove the color of $x_1y_1$ and let $\alpha =\phi (y_1y_1')$. If $\alpha \ne 1$, then we can recolor $v_3x_1$, $v_2v_3$ with 1, $\phi (v_3x_1)$, respectively. Therefore, $\alpha =1$. We can recolor $v_3v_4$, $v_5v_6$, $v_6v_0$, $v_0v_1$, $v_3x_1$, with $\phi (v_3x_1)$, 2, 1, 2, 2, respectively.

• (iv) G contains $G_7$ as a subgraph. We remove the color of $v_5v_6$. If $\phi (v_4v_5)\ne 1$, then we can exchange the colors of $v_0v_1$ and $v_0v_6$. So we may assume that $\phi (v_4v_5)=1$. If $\phi (v_3v_4)\ne \phi (v_0v_6)$, then we can recolor $v_2v_3$, $v_3v_4$, and $v_4v_5$ with $\phi (v_3v_4)$, 1, $\phi (v_3v_4)$, respectively. Hence, assume that $\phi (v_3v_4)=\phi (v_0v_6)=2$. Let $\alpha =\phi (x_1y_1)$ when $d_G(y_1)\ne 2$. When $d_G(y_1)=2$, let $y_1'$ be the second neighbor of $y_1$ other than $x_1$. Then, $d_G(y_1')\ne 2$ by Claim 2. We remove the color of $x_1y_1$ and let $\alpha =\phi (y_1y_1')$. If $\alpha \ne 1$, then we can recolor $v_3x_1$, $v_2v_3$ with 1, $\phi (v_3x_1)$, respectively. Therefore, $\alpha =1$. We can recolor $v_2v_3$, $v_3v_4$, $v_4v_5$, $v_6v_0$, $v_0v_1$, $v_3x_1$ with $\phi (v_3x_1)$, 1, 2, 1, 2, 2, respectively.

$\square$

Claim 7

Every light 2-edge must be incident with some 5-cycle.

Proof

Assume to the contrary that there exists some light 2-edge uv which is not incident with any 5-cycle. Let $N_G(u)=\{v,u'\}$ and $N_G(v)=\{u,v'\}$. If $u'=v'$, then $G-uv$ admits a T(G)-avd-coloring $\phi$. Note that $\phi (u'u) \ne \phi (v'v)$. We can color uv properly to obtain a T(G)-avd-coloring of G, a contradiction. If $u'\ne v'$, then it must be incident with two $6^+$-cycle. Let H be the graph obtained by contract edge uv and denote the new vertex by w. Then, H is a planar graph without 4-cycles. Thus, H has a T(G)-avd-coloring $\phi$. Now, we can construct the T(G)-avd-color $\psi$ of G: $\psi (e)=\phi (e)$ for $e\in E(H)\cap E(G)$, $\psi (uu')=\phi (wu')$, $\psi (vv')=\phi (wv')$, and color uv properly. $\square$

By the proof of Claim 7, we can deduce that every light 2-edge is not incident with 3-cycle.

Claim 8

Let v be a 5-vertex of G.

• (1) $d_{3^-}^G(v)\le 1$;

• (2) v is not adjacent to bad 2-vertex;

• (3) If $d_{3^-}^G(v)=1$, then v is not incident with any 3-cycle.

Proof

Let $v_1,v_2,\ldots ,v_5$ be the neighbors of v with $d_G(v_1)=k\le 3$. Let $u_1,u_2,\ldots ,v_{k-1}$ be the neighbors of $v_1$ other than v. Then, $G-vv_1$ has a T(G)-avd-coloring $\phi$ with color set $C=[1,T(G)]$ by Remark 1. Note that $T(G)\ge 9$. Suppose that $\phi (vv_i)=i-1$ for $i\in [2,5]$.

• (1) Assume to the contrary that $d_{3^-}^G(v)\ge 2$. Then, $d_5^G(v)\le 3$. Note that $k\ge 2$ by Claim 1.

Suppose that $k=2$. Let $\alpha =\phi (v_1u_1)$ if $d_G(u_1)\ne 2$. When $d_G(u_1)=2$, $N_G(u_1)=\{y,v_1\}$. We can recolor $v_1u_1$ with a color in $[1,4]\setminus C_{\phi }(u_1)$, and let $\alpha =\phi (u_1y)$. Then, we can color $vv_1$ with a color in $[5,9]\setminus \{\alpha \}$ such that v does not conflict with its neighbors.

Suppose that $k=3$. By Claim 2, $d_3^G(v_1)\le 1$. Let $\alpha _i=\phi (v_1u_i)$ if $d_3^G(v_1)=0$. When $d_3^G(v_1)=1$, let $d_G(u_1)=3$. We can recolor $v_1u_1$ with a color in $[1,4]\setminus C_{\phi }(u_1)$ and set $\alpha _1\in C_{\phi }(u_1)\setminus C_{\phi }(v_1)$, $\alpha _2=\phi (v_1u_2)$. Without loss of generality, assume that $\alpha _1$, $\alpha _2\in [1,6]$. Firstly, $vv_1$ can be colored with any color in $\{7,8,9\}$. Secondly, we can recolor $vv_2$ with a color $\beta \in [5,9]\setminus C_\phi (v_2)$ such that $v_2$ does not conflict its neighbors, and color $vv_1$ with a color in $[7,9]\backslash \{\beta \}$. Hence, we have at least 4 different ways to recolor or color the edges incident to v. Note that v has at most 3 conflict vertices. So we can extend $\phi$ to G, a contradiction.

• (2) Assume to the contrary that v is adjacent to a bad 2-vertex $v_1$.

Let $N_G(v_1)=\{u_1,v\}$ and $N_G(u_1)=\{v_1, x\}$. By Claim 7, $v_1u_1$ is incident with some 5-cycle, say $xv_2\in E(G)$. Let $\alpha =\phi (u_1x)$. If $\alpha \in [1,4]$, then we can color $vv_1$ with a color in [5, 9] such that v does not conflict with its neighbors. So we may assume that $\alpha =5$. Suppose that $vv_1$ cannot be colored with a color in [6,9] so that $C_{\phi }(v)\ne C_{\phi }(v_i)$ for $i=2,3,4,5$, say $C_\phi (v_i)=\{1,2,3,4,i+4\}$. If $\{2,3,4,6,a\}\notin \{C_{\phi }(y)| yv_2\in E(G)\ \mathrm{and} ~y\ne x\}$ for some $a\in [7,9]$, then we can recolor $vv_2$ with a and color $vv_1$ with a color in $[7,9]\setminus \{a\}$. So we may assume that $\{\{2,3,4,6,a\}|a=7,8,9\}=\{C_{\phi }(y)| yv_2\in E(G)\ \mathrm{and} ~y\ne x\}$. We exchange the colors of $u_1x$ and $v_2x$ and recolor $vv_2$ with a color $b\in [7,9]$ such that $v_2$ does not conflict with x. Then, we can color $vv_1$ with a color in $[7,9]\setminus \{b\}$.

• (3) Assume to the contrary that v is incident with some 3-cycle when $d_{3^-}^G(v)=1$.

Case 1 Suppose that $v_1$ is incident with some 3-cycle, say $v_1v_2\in E(G)$.

• (i) Suppose that $d_G(v_1)=2$. Let $\phi (v_1v_2)=\alpha$. If $\alpha \subseteq [1,4]$, then we can color $vv_1$ with a color in [5, 9] such that v does not conflict with its neighbors. So we may assume that $\alpha =5$. Then, we can color $vv_1$ with a color in [6, 9] such that v does not conflict with its neighbors.

• (ii) Suppose that $d_G(v_1)=3$ with $N_G(v_1)=\{u_1,v,v_2\}$. Recall that $d_3^G(v_1)\le 1$ by Claim 2. When $d_3^G(v_1)=0$, let $\phi (v_1u_1)=a_1$ and $\phi (v_1v_2)=a_2$. When $d_3^G(v_1)=1$, $d_G(u_1)=3$ by Claim 8 (1). We can recolor $v_1u_1$ with a color in $[1,4]\setminus C_{\phi }(u_1)$, and let $a_1\in C_{\phi }(u_1)\setminus \{\phi (v_1v_2)\}$ and $a_2=\phi (v_1v_2)$.

• Suppose that $a_1$, $a_2\in [1,4]$. Then, we can color $vv_1$ with a color in [5,9] such that v does not conflict with its neighbors.

• Suppose that $a_1\in [1,4]$, $a_2=5$. Then, we can color $vv_1$ with a color in [6,9] such that $C_{\phi }(v)\ne C_{\phi }(v_i)$ for $i=3,4,5$. Note that $a_2=5$, then $C_{\phi }(v)\ne C_{\phi }(v_2)$.

• Suppose that $a_1=5$ and $a_2\in [2,4]$, say $a_2=2$. Suppose that $vv_1$ cannot be colored without creating a conflict. That is, $C_\phi (v_i)=\{1,2,3,4,4+i\}$ for $i=2,3,4,5$. We can recolor $vv_2$ with a color in $\alpha \in \{5,7,8,9\}$ such that $v_2$ does not conflict with its neighbors. Then, we can color $vv_1$ with a color in $\{7,8,9\}\backslash \{\alpha \}$.

• Suppose that $a_1=5, a_2=6$. Suppose that $vv_1$ cannot be colored without creating a conflict. That is, $C_\phi (v_i)=\{1,2,3,4,4+i\}$ for $i=3,4,5$. Then, we can exchange the colors of $vv_2$ and $v_1v_2$ and color $vv_1$ with a color in $\{7,8,9\}$ such that $C_{\phi }(v)\ne C_{\phi }(v_2)$.

Case 2 Suppose that $v_1$ is not incident with any 3-cycle, say $v_2v_3\in E(G)$.

Subcase 2.1$d_G(v_1)=2$.

Let $\alpha =\phi (v_1u_1)$ if $d_G(u_1)\ne 2$. When $d_G(u_1)=2$, $N_G(u_1)=\{y,v_1\}$. We can recolor $v_1u_1$ with a color in $[1,4]\setminus C_{\phi }(u_1)$, and let $\alpha =\phi (u_1y)$. If $\alpha \in [1,4]$, then we can color $vv_1$ with a color in [5,9] such that v does not conflict with its neighbors. So we may assume that $\alpha =5$. Suppose that $vv_1$ cannot be colored without creating a conflict. That is, $C_\phi (v_i)=\{1,2,3,4,i+4\}$ for $i=2,3,4,5$. We can recolor $vv_2$ with color a color $\beta \in \{5,7,8,9\}$ such that $v_2$ does not conflict with its neighbors. Then, we can color $vv_1$ with a color in $\{7,8,9\}\backslash \{\beta \}$.

Subcase 2.2$d_G(v_1)=3$ with $N_G(v_1)=\{u_1,u_2,v\}$.

By Claim 2, $d_3^G(v_1)\le 1$. When $d_3^G(v_1)=0$, let $\phi (v_1u_1)=a_1$ and $\phi (v_1u_2)=a_2$. When $d_3^G(v_1)=1$, say $d_G(u_1)=3$. We can recolor $v_1u_1$ with a color in $[1,4]\setminus C_{\phi }(u_1)$, and let $a_1\in C_{\phi }(u_1)\setminus \{\phi (v_1u_2)\}$ and $a_2=\phi (v_1u_2)$. Without loss of generality, assume that $a_1,a_2\in [1,6]$. Suppose that $vv_1$ cannot be colored without creating a conflict. That is, $C_\phi (v_2)=\{1,2,3,4,7\}$, $\{1,2,3,4,8\},\{1,2,3,4,9\}\in \{C_{\phi }(v_i)|i=3,4,5\}$. Then, $d_G(v_2)=5$.

• (i) Suppose that $C_{\phi }(v_3)\in \{\{1,2,3,4,8\},\{1,2,3,4,9\}\}$, say $C_{\phi }(v_3)=\{1,2,3,4,8\}$. We can recolor $vv_2$ with a color in $\beta \in \{5,6,8,9\}$ such that $v_2$ does not conflict with its neighbors. If $\beta \in \{5,6\}$, then we can color $vv_1$ with a color in $\{8,9\}$ such that v does not conflict with its neighbors. Suppose that $\beta \in \{8,9\}$. If $C_{\phi }(v_5)\ne \{2,3,4,8,9\}$, then we can color $vv_1$ with a color in $\{8,9\}\setminus \{\beta \}$. So we may suppose that $C_\phi (v_5)=\{2,3,4,8,9\}$. Then, we can recolor $vv_3$ with a color in $\gamma \in \{5,6,7,9\}$ such that $v_3$ does not conflict with its neighbors, and color $vv_1$ with a color in $\{7,9\}\setminus \{\gamma \}$.

• (ii) Suppose that $C_{\phi }(v_4)=\{1,2,3,4,8\}$ and $C_{\phi }(v_5)=\{1,2,3,4,9\}$.

• Suppose that $vv_2$ can be recolored with a color $\beta \in \{5,6,8,9\}$ such that $v_2$ does not conflict with its neighbors. If $\beta \in \{5,6\}$, then we can color $vv_1$ with a color in $\{8,9\}$ such that v does not conflict with $v_3$. Suppose that $\beta \in \{8,9\}$. If $C_{\phi }(v_3)\ne \{2,3,4,8,9\}$, then we can color $vv_1$ with a color in $\{8,9\}\setminus \{\beta \}$. So we may suppose that $C_\phi (v_3)=\{2,3,4,8,9\}$. Then, we can recolor $vv_2$ with $\beta$, recolor $vv_3$ with a color in $\gamma \in \{1,5,6,7\}$ such that $v_3$ does not conflict with its neighbors, and color $vv_1$ with a color in $\{1,5,6,7\}\setminus \{a_1,a_2,\gamma \}$.

• Suppose that $vv_2$ cannot be recolored with a color $\beta \in \{5,6,8,9\}$ such that $v_2$ does not conflict with its neighbors. That is, $\{\{2,3,4,7,i\}|i=5,6,8,9\}=\{C_{\phi }(x)|x\in N_G(v_2)\}$. First, assume that $C_{\phi }(v_3)\in \{\{2,3,4,7,i\}|i=5,6\}$, say $C_{\phi }(v_3)=\{2,3,4,5,7\}$. We can recolor $vv_2$ with 5, recolor $vv_3$ with a color $\beta \in \{1,6,8,9\}$ such that $v_3$ does not conflict with its neighbors, and color $vv_1$ with a color in $\{8,9\}\setminus \{\beta \}$. Next, assume that $C_{\phi }(v_3)\in \{\{2,3,4,7,i\}|i=8,9\}$, say $C_{\phi }(v_3)=\{2,3,4,7,8\}$. We can recolor $vv_2$ with 8, recolor $vv_3$ with a color $\beta \in \{1,5,6,9\}$ such that $v_3$ does not conflict with its neighbors, and color $vv_1$ with a color in $\{1,5,6,9\}\setminus \{a_1,a_2,\beta \}$. $\square$

Claim 9

Let v be a 6-vertex of G.

• (1) $d_{3^-}^G(v)\le 2$;

• (2) If $d_{2^-}^G(v)\ge 1$, then $d_{3^-}^G(v)\le 1$;

• (3) If vertex v is adjacent to a $(2,6,k_1)$-cycle, then $d_4^G(v)=0$;

• (4) If v is incident with a $(3,6,k_1)$-cycle, then $d_3^G(v)=1$. Hence, v is not incident with (3, 3, 6)-cycle.

Proof

Let $v_1, v_2,\ldots , v_6$ be the neighbors of v with $d_G(v_1)=k\le 3$. Let $u_1, u_2,\ldots ,u_{k-1}$ be the neighbors of $v_1$ other than v. Then, $G-vv_1$ admits a T(G)-avd-coloring $\phi$ with the color set $C=[1,T(G)]$ by Remark 1. Suppose that $\phi (vv_i)=i-1$ for $i\in [2,6]$.

• (1) Assume to the contrary that $d_{3^-}^G(v)\ge 3$. Then, $d_G(v_2)$, $d_G(v_3)\le 3$. Recall that $k\le 3$. By Claim 2, $d_k^G(v_1)\le 1$. When $d_k^G(v_1)=0$, let $\phi (v_1u_1)=a_1$ and $\phi (v_1u_{k-1})=a_2$. When $d_k^G(v_1)=1$ with $k=2$, say $d_G(u_1)=2$, let $C_{\phi }(u_1)=\{a_1,a_2\}$. When $d_k^G(v_1)=1$ with $k=3$, say $d_G(u_1)=3$. We recolor $v_1u_1$ with a color in $[1,5]\setminus C_{\phi }(u_1)$, and let $a_1\in C_{\phi }(u_1)\setminus \{\phi (v_1u_2)\}$ and $a_2=\phi (v_1u_2)$. Without loss of generality, assume that $a_1,a_2\in [1,7]$.

Firstly, $vv_1$ can be colored with any color in $\{8,9\}$. Next, for each $i\in \{2,3\}$, $v_i$ has at most one possible conflict vertex by Claim 2. So $vv_i$ can be recolored with $b_i\in \{6,7,8,9\}\setminus C_\phi (v_i)$ such that $v_i$ does not conflict with its neighbors. Then, we can color $vv_1$ with a color in $\{8,9\}\backslash \{b_i\}$. Hence, we have at least $2+1\times 2=4$ different ways to recolor or color the edges incident to v. Since v has at most three conflict vertices, we can extend $\phi$ to G, a contradiction.

• (2) Assume to the contrary that $d_{3^-}^G(v)\ge 2$ when $d_{2^-}^G(v)\ge 1$. Then, $k\le 2$ and $d_G(v_2)\le 3$. When $d_G(v_1)=2$ and $d_G(u_1)\ne 2$, let $\alpha =\phi (v_1u_1)$. When $d_G(v_1)=2$ and $d_G(u_1)=2$, $N_G(u_1)=\{y,v_1\}$. By Claim 2, $d(y)\ne 2$. We can recolor $v_1u_1$ with a color in $[1,5]\backslash C_\phi (u_1)$, and let $\alpha =\phi (u_1y)$. Without loss of generality, assume that $\alpha \in [1,6]$.

Firstly, $vv_1$ can be colored with any color in $\{7,8,9\}$. Next, we can recolor $vv_2$ with a color $b_1\in \{6,7,8,9\}\backslash C_\phi (v_2)$ such that $v_2$ does not conflict with its neighbors and color $vv_1$ with a color in $\{7,8,9\}\backslash \{b_1\}$. Hence, we have at least 3+2=5 different ways to recolor or color the edges incident to $v_1$. Since v has at most 4 conflict vertices, we can extend $\phi$ to G, a contradiction.

• (3) Assume to the contrary that $d_4^G(v)\ge 1$ when v is adjacent to a $(2,6,k_1)$-cycle. Then, $k=2$ and $v_1v_2\in E(G)$. Then, $d_G(v_2)\ge 5$ by Claims 9 (2) and 3. Let $\phi (v_1v_2)=a\in [1,6]$. Assume that $d_G(v_3)=4$ with $N_G(v_3)=\{x_1,x_2,x_3,v\}$. Firstly, $vv_1$ can be colored with any color in $\{7,8,9\}$. Next, when $d_4^G(v_3)=0$, we can recolor $vv_3$ with a color $b\in [6,9]\setminus C_{\phi }(v_3)$ and color $vv_1$ with a color in $[7,9]\setminus \{b\}$. When $d_4^G(v_3)=1$, say $d_G(x_1)=4$, we remove the color of $v_3x_1$. Then, we can recolor $vv_3$ with a color $b\in \{6,7,8,9\}\backslash C_\phi (v_3)$ such that $v_3$ does not conflict with $x_1$, color $v_3x_1$ properly and color $vv_1$ with a color in $\{7,8,9\}\backslash \{b\}$. Hence, we have at least 3+2=5 different ways to recolor or color the edges incident to $v_1$. Since v has at most 4 conflict vertices, we can extend $\phi$ to G, a contradiction.

• (4) Assume to the contrary that $d_3^G(v)\ge 2$ when v is incident with some $(3,6,k_1)$-cycle C. Then, $k=3$ and $N_G(v_1)=\{u_1,v_2,v\}$, say $v_1v_2\in E(G)$.

Case 1$d_G(v_2)=3$ with $N_G(v_2)=\{x,v_1,v\}$.

We can recolor $v_1v_2$ with a color in $[2,5]\setminus \{\phi (v_1u_1),\phi (v_2x)\}$. Let $\phi (v_1u_1)=a_1$ and$\phi (v_2x)=a_2$.

• (i) Suppose that at most one of $a_1$ and $a_2$ belongs to [1, 5], say $a_1,a_2\in [1,6]$. We can color $vv_1$ with a color in [7,9], or we can recolor $vv_2$ with 7 and color $vv_1$ with a color in $\{8,9\}$. Hence, we have at least 3+2=5 different ways to recolor or color the edges incident to v. Since v has at most 4 conflict vertices, we can extend $\phi$ to G, a contradiction.

• (ii) Suppose that neither $a_1$ nor $a_2$ belongs to [1,5], say $a_1=6$ and $a_2=7$. We can color $vv_1$ with a color in [7,9], or we can recolor $vv_2$ with 9 and color $vv_1$ with a color in $\{7,8\}$. So we have at least 3+2=5 different ways to recolor or color the edges incident to v. While v has at most 4 conflict vertices, we can extend $\phi$ to G, a contradiction.

Case 2$d_G(v_2)\ne 3$.

Suppose that $d_G(v_3)=3$ with $N_G(v_3)=\{x_1,x_2\}$. Recall that $d_3^G(v_3)\le 1$ by Claim 2. When $d_G(u_1)\ne 3$, let $\phi (v_1u_1)=a_1$ and $\phi (v_1v_2)=a_2$. When $d_G(u_1)=3$, we can recolor $v_1u_1$ with a color in $[1,5]\setminus (\{\phi (v_1v_2\}\cup C_{\phi }(u_1))$. Let $a_2=\phi (v_1v_2)$ and $a_1\in C_{\phi }(u_1)\setminus \{a_2\}$.

• (i) Suppose that at most one of $a_1$ and $a_2$ belongs to [1,5], say $a_1,a_2\in [1,6]$. Firstly, $vv_1$ can be colored with any color in $\{7,8,9\}$. Next, we can recolor $vv_3$ with a color $\alpha \in \{6,7,8,9\}\backslash \{C_\phi (v_3)\}$ such that $v_3$ does not conflict with its neighbors, and color $vv_1$ with a color in $\{7,8,9\}\backslash \{\alpha \}$. Thus, we have at least 3+2=5 different ways to recolor the edges incident to v. While v has at most 4 conflict vertices, we can extend $\phi$ to G, a contradiction.

• (ii) Suppose that neither $a_1$ nor $a_2$ belongs to [1,5], say $a_1=6$, $a_2=7$. Firstly, $vv_1$ can be recolored with 8 or 9. Secondly, we can exchange the colors of $v_1v_2$ and $vv_2$ and color $vv_1$ with 8 or 9. Finally, we can recolor $vv_3$ with a color $\alpha \in \{6,7,8,9\}\backslash \{C_\phi (v_3)\}$ such that $v_3$ does not conflict with its neighbors, and color $vv_1$ with a color in $\{8,9\}\backslash \{\alpha \}$. Thus, we have at least 2+2+1=5 different ways to recolor the edges incident to v. While v has at most 4 conflict vertices, we can extend $\phi$ to G, a contradiction. $\square$

Claim 10

Let v be a 7-vertex of G.

• (1) If $d_1^G(v)\ge 1$, then $d_{3^-}^G(v)\le 2$.

• (2) If $d_{2^-}^G(v)\ge 1$, then $d_{3^-}^G(v)\le 3$.

• (3) Suppose that v is incident with a (3, 3, 7)-cycle. Then, $d_{3^-}^G(v)\le 2$.

Proof

Let $v_1, v_2,\ldots , v_7$ be the neighbors of v with $d(v_1)=k\le 3$. Let $u_1, u_2,\ldots ,u_{k-1}$ be the neighbors of $v_1$ other than v. Then, $G-vv_1$ admits a T(G)-avd-coloring $\phi$ with the color set $C=[1,T(G)]$ by Remark 1. Suppose that $\phi (vv_i)=i-1$ for $i\in [2,7]$. $\square$

Remark 2

Suppose that $v_i$ is a $3^-$-vertex. If $d_G(v_i)\le 2$ or $d_G(v_i)=3$ with $d_3^G(v_i)=0$, then we can recolor $vv_i$ with a color $b_i\in \{7,8,9\}\setminus C_\phi (v_i)$ such that $v_i$ does not conflict with its neighbors. If $d_G(v_i)=3$ with $d_3^G(v_i)=1$, let $x_i$ be the 3-vertex adjacent to $v_i$. We can recolor $v_ix_i$ with a color in $[1,6]\setminus (C_\phi (v_i)\cup C_\phi (x_i))$ and recolor $vv_i$ with a color $b_i\in \{7,8,9\}\setminus C_\phi (v_i)$ so that $v_i$ does not conflict with $x_i$. Hence, we can recolor $vv_i$ with some color $b_i\in \{7,8,9\}$ without creating a conflict.

• (1) Assume to the contrary that $d_{3^-}^G(v)\ge 3$ when $d_1^G(v)\ge 1$. Then, $d_G(v_1)=1$, and $d_G(v_2)$, $d_G(v_3)\le 3$. Firstly, $vv_1$ can be colored with any color in $\{7,8,9\}$. Secondly, for each $i\in [2,3]$, $vv_i$ can be recolored with $b_i\in \{7,8,9\}$ without creating a conflict by Remark 2. Then, we can color $vv_1$ with a color in $\{7,8,9\}\backslash \{b_i\}$. Hence, we have at least $3+1\times 2=5$ different ways to recolor the edges incident to v. While v has at most 4 conflict vertices, we can extend $\phi$ to G, a contradiction.

• (2) Assume to the contrary that $d_{3^-}^G(v)\ge 4$ when $d_{2^-}^G(v)\ge 1$. Then, by Claim 10 (1), $d_G(v_1)=2$, and $d_G(v_2)$, $d_G(v_3)$, $d_G(v_4)\le 3$. Let $u_1$ be the neighbor of $v_1$ other than v. Let $\phi (v_1u_1)=\alpha$ when $d_G(u_1)\ne 2$. When $d_G(u_1)=2$, let y be the neighbor of $u_1$ other than $v_1$. We can recolor $v_1u_1$ with a color in $[1,6]\setminus \{C_{\phi }(u_1y)\}$, and let $\phi (u_1y)=\alpha$. Without loss of generality, assume that $\alpha \in [1,7]$. Firstly, $vv_1$ can be colored with 8 or 9. Secondly, for each $i\in [2,4]$, $vv_i$ can be recolored with $b_i\in \{7,8,9\}$ without creating a conflict by Remark 2. Then, we can color $vv_1$ with a color in $\{8,9\}\backslash \{b_i\}$. Hence, we have at least $2+1\times 3=5$ different ways to recolor or color the edges incident to v. While v has at most 3 conflict vertices, we can extend $\phi$ to G, a contradiction.

• (3) Assume to the contrary that $d_{3^-}^G(v)\ge 3$ when $k_1=3$. Let $d_G(v_2)=3$ with $N_G(v_2)=\{x,v_1,v\}$ and $d_G(v_3)\le 3$. We recolor $v_1v_2$ with a color in $[1,6]\setminus \{\phi (v_1u_1),\phi (v_2x)\}$. Let $\phi (v_1u_1)=a_1$ and $\phi (v_2x)=a_2$.

• Suppose that $a_1,a_2\in [1,6]$. We can color $vv_1$ with a color in [7,9], or we can recolor $vv_2$ with 7 and color $vv_1$ with a color in $\{8,9\}$. Hence, we have at least 3+2=5 different ways to recolor or color the edges incident to v. Since v has at most 4 conflict vertices, we can extend $\phi$ to G, a contradiction.

• Suppose that $a_1\in [1,6]$ and $a_2=7$. Firstly, $vv_1$ can be recolored with 8 or 9. Secondly, $vv_2$ can be recolored with 8, and $vv_1$ can be colored with 7 or 9. Finally, $vv_2$ can be recolored with 9, and $vv_1$ can be colored with 7. Hence, we have at least 2+2+1=5 different ways to recolor or color the edges incident to v. Since v has at most 4 conflict vertices, we can extend $\phi$ to G, a contradiction.

• Suppose that $a_2\in [1,6]$ and $a_1=7$. Firstly, $vv_1$ can be recolored with 8 or 9. Secondly, $vv_1$ can be colored with 8, and $vv_2$ can be recolored with 7 or 9. Finally, $vv_1$ can be colored with 9, and $vv_2$ can be recolored with 7. Hence, we have at least 2+2+1=5 different ways to recolor or color the edges incident to v. Since v has at most 4 conflict vertices, we can extend $\phi$ to G, a contradiction.

• Suppose that $a_1=7$ and $a_2=8$. Firstly, $vv_1$ can be recolored with 8 or 9. Secondly, $vv_2$ can be recolored with 7, and $vv_1$ can be colored with 1 or 9. Finally, $vv_1$ can be colored with 8, and $vv_2$ can be recolored with 9. Hence, we have at least 2+2+1=5 different ways to recolor or color the edges incident to v. Since v has at most 4 conflict vertices, we can extend $\phi$ to G, a contradiction. $\square$

Claim 11

Let v be a k-vertex of G with $k\ge 8$.

• (1) If $d_1^G(v)\ge i$ for $i=1,2$, then $d_k^G(v)\ge d_{(i+1)^-}^G(v)+1$.

• (2) If $d_2^G(v)\ge 1$, then $d_k^G(v)\ge 1$.

Proof

• (1) We will show the case that $i=2$. (The proof of $i=1$ is similar.) Let $t=d_{3^-}^G(v)$. Assume to the contrary that $d_k^G(v)\le t$. Let $v_1,v_2,\ldots ,v_k$ be the neighbors of v with $d_G(v_1)=d_G(v_2)=1$, $d_G(v_3)\le \cdots \le d_G(v_t)\le 3$. Then, $H=G-vv_1$ has a T(H)-avd-coloring $\phi$ using the color set $C=[1,T(G)]$ by Remark 1. Let $\phi (vv_i)=i-1$ for $2\le i\le k$.

Firstly, $vv_1$ can be colored with k or $k+1$. Secondly, $vv_2$ can be recolored with k and $vv_1$ can be colored with $k+1$. Finally, for $3\le i\le t$, $vv_i$ can be recolored with a color $b_i \in \{1,k,k+1\}$ without creating a conflict by Remark 2. For $b_i\in \{k,k+1\}$, we can color $vv_1$ with a color in $\{k,k+1\}\backslash \{b_i\}$. For $b_i=1$, we can recolor $vv_2$ with k and color $vv_1$ with $k+1$. This implies that there are at least $2+1+(t-2)=t+1$ different ways to recolor or color the edges incident with v. While v has at most t conflict vertices, we can extend $\phi$ to G, a contradiction.

• (2) Similarly, it is easy to show (2) is true. $\square$

Claim 12

Let v be a k-vertex of G and $v_1,v_2,\ldots ,v_k$ be the neighbors of v in the clockwise order. If $d_2^G(v)\ge 3$, then there are no two consecutive 2-vertices with at least one of them is bad, where $v_k$ and $v_1$ are considered to be consecutive.

Proof

Assume to the contrary that there exist some $i(1\le i\le k)$, say $i=1$, such that $d_G(v_1)=d_G(v_2)=2$ and $v_1$ is a bad 2-vertex in G. Let $d_G(v_j)=2$ by $d_2^G(v)\ge 3$. Let $u_l$ be the second neighbor of $v_l$ other than v for $l=1,2,j$, and $y_1$ be the second neighbor of $u_1$ other than $v_1$. Note that by Claim 5, no 5-cycle is incident with both $v_1$ and $v_2$. Recall that by the proof of Claim 7, $y_1\ne v$. Now, the proof is split into the following two cases.

Case 1 $v_2$ is a good 2-vertex in G.

Suppose that $v_1$ and $v_2$ are incident with some 6-cycle $vv_1u_1y_1u_2v_2v$. Then, $H=G-v_1u_1$ admits a T(G)-avd-coloring $\phi$ using the color set $C=[1,T(G)]$ by Remark 1. Assume that $\phi (vv_i)=i$ for $i\in [1,k]$. If $\phi (u_1y_1)\ne 1$, then $v_1u_1$ can be colored properly. So we may suppose that $\phi (u_1y_1)=1$. If $\phi (v_2u_2)\ne 1$, then we can exchange the colors of $vv_1$ and $vv_2$ and color $v_1u_1$ properly. So $\phi (v_2u_2)=1$. If $\phi (u_2y_1)\ne 2$, then we can recolor $u_1y_1$, $u_2y_1$, $v_2u_2$ with $\phi (u_2y_1)$, 1, $\phi (u_2y_1)$ and color $u_1v_1$ properly. Suppose that $\phi (u_2y_1)=2$. When $d(u_j)\ne 2$, set $\alpha =\phi (v_ju_j)$. When $d(u_j)=2$, let $y_j$ be the second neighbor of $u_j$ other than $v_j$. Recolor $v_ju_j$ with a color in $[3,k]\backslash C_\phi (u_j)$, and set $\alpha =\phi (u_jy_j)$. Recolor $vv_j$ with a color $\beta \in \{1,2\}\backslash \{\alpha \}$. If $\beta =1$, then we can recolor $vv_1$ with j and color $v_1u_1$ properly. If $\beta =2$, then we can recolor $vv_2$, $v_2u_2$, $u_2y_1$, $y_1u_1$ with j, 2, 1, 2 and color $v_1u_1$ properly.

Suppose that no 6-cycle is incident with both $v_1$ and $v_2$. Let H be the graph obtained by adding a new vertex w to $G-\{v_1u_1,v_2u_2\}$ and joining w to both $u_1$ and $u_2$. Then, H contains no 4-cycles, $|E(H)|=|E(G)|$, $\Delta (H)=\Delta (G)$, and $n_2(H)<n_2(G)$. By the minimality of G, H has a T(G)-avd-coloring $\phi$ using the color set $C=[1,T(G)]$. Assume that $\phi (vv_i)=i$ for $i\in [1,k]$, $\phi (y_1u_1)=a$, $\phi (wu_2)=b$. Note that $a\ne b$, Since $\phi$ is the avd-coloring. Now, we need to find the T(G)-avd-coloring $\psi$ of G. First, let $\psi (e)=\phi (e)$ for $e\in E(H)\cap E(G)$. If $a\ne 1$ and $b\ne 2$, then we can color $v_1u_1$ properly and color $v_2u_2$ with b. If $a=1$ or $b=2$, then $b\ne 1$ or $a\ne 2$, respectively. We can recolor or color $v_2u_2$, $vv_1$ and $vv_2$ with b, 2, 1, respectively, and color $v_1u_1$ properly.

Case 2$v_2$ is a bad 2-vertex, say $d_G(u_2)=2$ and $N_G(u_2)=\{v_2,y_2\}$.

By Claim 6, no 6-cycle and 7-cycle are incident with both $v_1$ and $v_2$. Let $H=(G-\{v_1u_1,v_2u_2\})\cup \{u_1u_2\}$. Then, H contains no 4-cycles, $|E(H)|<|E(G)|$, and $\Delta (H)=\Delta (G)$. Since H is smaller than G, H has a T(G)-avd-coloring $\phi$ using the color set $C=[1,T(G)]$. Assume that $\phi (vv_i)=i$ for $i\in [1,k]$, $\phi (u_1y_1)=a$, $\phi (u_2y_2)=b$. Note that $a\ne b$, since $\phi$ is the avd-coloring. Now, we need to find the T(G)-avd-coloring $\psi$ of G. First, let $\psi (e)=\phi (e)$ for $e\in E(H)\cap E(G)$. If $a\ne 1$ and $b\ne 2$, then we can color $v_1u_1$ and $v_2u_2$ properly. If $a=1$ or $b=2$, then $b\ne 1$ or $a\ne 2$, respectively. We can recolor $vv_1$ and $vv_2$ with 2, 1, respectively, and color $v_1u_1$, $v_2u_2$ properly. $\square$

3 Discharging Methods

Let H be the graph obtained from G by removing all 1-vertices. Then, H is a planar graph without 4-cycles. We show the relationship between $d_G(v)$ and $d_H(v)$ for $v\in V(H)$ by claims 1, 9-11 in Table 1. Hence, we can deduce that $\delta (H)\ge 2$. Therefore, H is normal. Other properties of H are collected in the following.

Claim 13

• (1) Let v be a k-vertex of H with $2\le k\le 4$. Then, $d_k^H(v)\le 1$ and $\sum \limits _{2\le i\le 4, i\ne k}d_i^H(v)=0$.

• (2) Let v be a 5-vertex of H. Then, $d_{2b}^H(v)=0$ and $d_{3^-}^H(v)\le 1$. Furthermore, if $d_{3^-}^H(v)=1$, then $t_H(v)=0$.

• (3) Let v be a 6-vertex of H. Then,

  • (3.1) $d_{3^-}^H(v)\le 2$.

  • (3.2) If $d_2^H(v)\ge 1$, then $d_{3^-}^H(v)\le 1$.

  • (3.3) If v is incident with $(2,6,k_1)$-cycle, then $d_{4}^H(v)=0$.

  • (3.4) If v is incident with $(3,6,k_1)$-cycle, then $d_{3}^H(v)=1$.

• (4) Let v be a 7-vertex of H. Then,

  • (4.1) If $d_2^H(v)\ge 1$, then $d_{3^-}^H(v)\le 3$.

  • (4.2) If v is incident with (3, 3, 7)-cycle, then $d_{3}^H(v)\le 2$.

• (5) Let v be a k-vertex of H with $d_2^H(v)\ge 1$ and $k\ge 8$, then $d_{5^+}^H(v)\ge 1$.

• (6) Each 3-face in H must be a $(2,6^+,6^+)$-face, $(3,6^+,6^+)$-face, $(3,3,7^+)$-face, $(4,4,7^+)$-face, or $(4^+,5^+,5^+)$-face.

• (7) Each f of H is incident with at most $\lfloor \frac{2}{3}d_H(f)\rfloor$$4^-$-vertices.

• (8) For each vertex v of H, $t_H(v)\le \lfloor \frac{d_H(v)}{2}\rfloor$.

Table 1 The relation between $d_G(v)$ and $d_H(v)$

Proof

By Table 1 and Claims 1, 2, 10, the statements (1) and (5) are obviously true.

• (2) Suppose that $d_H(v)=5$. When $d_G(v)=5$, statement (2) is true by Claim 8. When $d_G(v)\ge 7$, $d_1^G(v)\ge 2$. By Claim 4, we have $d_2^G(v)+d_{3}^G(v)=0$. Assume that $d_G(v)=6$. Then, $d_1^G(v)=1$. By Claim 9, $d_2^G(v)+d_{3}^G(v)=0$. Thus, by Table 1, $d_{3^-}^H(v)=0$ when $d_G(v)\ge 6$. Hence, statement (2) is also true for $d_G(v)\ge 6$ by Table 1.

• (3) Suppose that $d_H(v)=6$. When $d_G(v)=6$, statement (3) is true by Claim 9. When $d_G(v)=7$, $d_1^G(v)=1$. By Claims 4 and  10, $d_2^G(v)=0$ and $d_{3^-}^G(v)\le 2$, which implies that $d_2^H(v)=0$ and $d_{3}^H(v)\le 1$ by Table 1. Thus, statement (3) holds. When $d_G(v)\ge 8$, $d_1^G(v)\ge 2$. By Claim 4, we have $d_2^G(v)+d_{3}^G(v)=0$, which implies that $d_{3^-}^H(v)=0$ by Table 1. Hence, statement (3) holds.

• (4) Suppose that $d_H(v)=7$. When $d_G(v)=7$, statement (4) holds by Claim 10. When $d_G(v)\ge 8$, then $d_1^G(v)\ge 1$. By Claim 4, we have $d_2^G(v)=0$, which implies that $d_2^H(v)=0$ by Table 1. Hence, statement (4.1) holds. By Claim 6, G does not contain $G_2$ as subgraph, which implies that v is not incident with any (3, 3, 7)-cycle. So statement (4.2) holds.

• (6) Let $f=[v_0v_1v_2]$ be a $(d_0,d_1,d_2)$-face in H with $d_0\le d_1\le d_2$. Suppose that $d_0=2$, then by the proof of Claim 7, $d_1\ne 2$. By Claim 13 (1) and (2), $d_1\ge 6$, say f is a $(2,6^+,6^+)$-face.

• (i) Suppose that $d_0=3$. If $d_1=3$, then $d_2\ge 7$ by Claim 13 (1), (2), and (3.4). So f is a $(3,3,7^+)$-face. Suppose that $d_1\ne 3$. By Claim 13 (1), (2), we have $d_1\ge 6$, say f is a $(3,6^+,6^+)$-face.

• (ii) Suppose that $d_0\ge 4$. If f is neither the $(4^+,5^+,5^+)$-face nor the $(4,4,7^+)$-face, then f must be the (4, 4, 5)-face or (4, 4, 6)-face by Claim 13 (1). Now, we need to show that H does not contain (4, 4, 5)-face and (4, 4, 6)-face. Assume to the contrary that H contains a (4, 4, 5)-face or a (4, 4, 6)-face. Then, $d_G(v_0)=d_G(v_1)=4$. Then, $H'=G-v_0v_1$ admits a T(G)-avd-coloring $\phi$ using the color set $C=[1,T(G)]$ by Remark 1. Note that $T(G)\ge 9$. If $C_{\phi }(v_0)\ne C_{\phi }(v_1)$, then we can color $v_0v_1$ properly to obtain a T(G)-avd-coloring of G, a contradiction. So we may assume that $C_{\phi }(v_0)=C_{\phi }(v_1)=\{1,2,3\}$, $\phi (v_2v_0)=1$, and $\phi (v_2v_1)=2$. If $d_1^G(v_2)\ge 2$, then let $u_1$ and $u_2$ be the 1-vertices adjacent to $v_2$. Note that $\{\phi (v_2u_1),\phi (v_2u_2)\}\setminus \{3\}\ne \emptyset$, say $\phi (v_2u_1)\ne 3$. We can exchange the color of $v_2u_1$ and $v_2v_0$ and color $v_0v_1$ properly to achieve a T(G)-avd-coloring of G, a contradiction. So $d_1^G(v_2)\le 1$, say $d_G(v_2)=5$, 6, or 7.

• Suppose that $d_G(v_2)=5$. First, we can recolor $v_2v_0$ with any color in $[4,9]\setminus C_{\phi }(v_2)$. Next, we can recolor $v_2v_1$ with any color in $[4,9]\setminus C_{\phi }(v_2)$. Hence, there are at least 3+3=6 different ways to recolor the edges incident with $v_2$. While $v_2$ has at most 3 conflict vertices, we can extend $\phi$ to G, a contradiction.

• Suppose that $d_G(v_2)=6$. Note that color 3 may not belong to $C_{\phi }(v_2)$. Without loss of generality, assume that $C_{\phi }(v_2)\subseteq [1,7]$. Firstly, $v_2v_0$ can be recolored with 8 or 9. Secondly, $v_2v_1$ can be recolored with 8 or 9. Finally, $v_2v_0$ can be recolored with 8, and $v_2v_1$ can be recolored with 9. Hence, there are at least 2+2+1=5 different ways to recolor the edges incident with $v_2$. While $v_2$ has at most 4 conflict vertices, we can extend $\phi$ to G, a contradiction.

• Suppose that $d_G(v_2)=7$. Then, $d_1^G(v_2)=1$ by the previous discussion. Let $u_1$ be the 1-vertex adjacent to $v_2$. If $\phi (v_2u_1)\ne 3$, then we can exchange the color of $v_2u_1$ and $v_2v_0$ and color $v_0v_1$ properly to achieve a T(G)-avd-coloring of G, a contradiction. Suppose that $\phi (v_2u_1)=3$. Without loss of generality, assume that $C_{\phi }(v_2)=[1,7]$. Firstly, $v_2v_0$ can be recolored with 8 or 9. Secondly, $v_2v_1$ can be recolored with 8 or 9. Finally, $v_2v_0$ can be recolored with 8, and $v_2v_1$ can be recolored with 9. Hence, there are at least 2+2+1=5 different ways to recolor the edges incident with $v_2$. While $v_2$ has at most 4 conflict vertices, we can extend $\phi$ to G, a contradiction.

• (7) By Claim 13 (1), it is easy to check that statement (7) holds.

• (8) By the fact that H contains no 4-cycles, statement (8) holds.

$\square$

Using Euler’s formula $|V(H)|-|E(H)|+|F(H)|=2$, we have :

$$\begin{aligned} \sum \limits _{v\in V(H)}(d_H(v)-4)+\sum \limits _{f\in F(H)}(d_H(f)-4)=-8 \end{aligned}$$

In order to complete the proof, we make use of discharging method. First, we define an initial charge function $w(x)=d_H(x)-4$ for every $x\in V(H)\cup F(H)$. Next, we design some discharging rules and redistribute weights accordingly. Once the discharging is finished, a new charge function $w'$ is produced. However, the sum of all charge is kept fixed when the discharging is in progress. Nevertheless, we can show that $w'(x)\ge 0$ for all $x\in V(H)\cup F(H)$. This leads to the following obvious contradiction:

$$\begin{aligned} 0\le \sum \limits _{x\in V(H)\cup F(H)}w'(x)=\sum \limits _{x\in V(H)\cup F(H)}w(x)=-8<0 \end{aligned}$$

and hence demonstrates that the counterexample cannot exist.

We define the following discharging rules:

• R1 Let f be a 3-face incident with $5^{+}$-vertex v. If f is $(4,4,7^+)$-face, then v sends $\frac{1}{2}$ to f; otherwise, v sends $\frac{1}{d_{5^+}^H(f)}$ to f.

• R2 Every 5-vertex sends $\frac{1}{2}$ to its adjacent $3^-$-vertex.

• R3 Let v be a $6^+$-vertex adjacent to a $3^-$-vertex u.

  • R3.1 If u is bad 2-vertex, then v sends $\frac{5}{6}$ to u.

  • R2.2 If u is good 2-vertex with $t_H(u)=1$, then v sends $\frac{3}{4}$ to u.

  • R3.3 If u is good 2-vertex with $t_H(u)=0$, then v sends $\frac{1}{2}$ to u.

  • R3.4 If $d_H(u)=3$, then v sends $\frac{1}{3}$ to u.

• R4 Every $5^{+}$-face sends $\frac{1}{2}$ to each its incident good 2-vertex, $\frac{7}{12}$ to each incident bad 2-vertex, $\frac{1}{6}$ to each incident bad 3-vertex, and $\frac{1}{6}$ to each adjacent $(4,4,7^+)$-face.

• R5 Every $6^{+}$-face sends $\frac{1}{3}$ to each incident bad 5-face through every light 2-edge.

• R6 Let f be a $5^{+}$-face, and f is not a special k-face for $k=5,6,7$.

  • R6.1 If $d_H(f)\ge 6$, then f sends $\frac{1}{12}$ to each incident $5^+$-vertex.

  • R6.2 Suppose that $d_H(f)=5$ with $d_2^H(f)\le 1$. Then, f sends $\frac{1}{12}$ to each incident $5^+$-vertex.

  • R6.3 Suppose that $d_H(f)=5$ with $d_{2b}^H(f)=2$, say $u_1$ be the bad 2-vertex incident with f. Let u be the $5^+$-vertex adjacent to $u_1$. If f is not adjacent to $(4,4,7^+)$-face, then f sends $\frac{1}{12}$ to u.

Remark 3

By Claim 13 (1), $d_4^H(v)=1$ for any 4-vertex v. So a $5^+$-face f sends at most $\frac{1}{6}d_4^H(f)$ to adjacent $(4,4,7^+)$-faces in total.

Now, we will show that $w'(x)\ge 0$ for all $x\in V(H)\cup F(H)$ in the following two lemmas.

Lemma 1

$w'(f)\ge 0$ for all $f\in F(H)$.

Proof

• (1) Suppose that $d_H(f)=3$. Then, $w(f)=-1$. Recall that by Claim 13 (6), $d_{5^+}^H(f)\ge 1$. If f is a $(4,4,7^+)$-face, then $w'(f)=-1+\frac{1}{2}+\frac{1}{6}\times 3=0$ by R1 and R4. Otherwise, $w'(f)=-1+d_{5^+}^H(f)\times \frac{1}{d_{5^+}^H(f)}=0$ by R1.

• (2) Suppose that $d_H(f)=5$. Then, $w(f)=1$. By Claim 5, $d_2^H(f)=d_2^G(f)\le 2$. Suppose that $d_2^H(f)\le 1$. Note that $d_{4^-}^H(f)\le 3$ by Claim 13 (7). So $d_3^H(f)+d_4^H(f)\le 2$. Thus, $w'(f)\ge 1-\frac{1}{2}d_2^H(f)-\frac{1}{6}d_3^H(f)- \frac{1}{6}d_4^H(f)-\frac{1}{12}d_{5^+}^H(f)\ge 1-\frac{1}{2}- \frac{1}{6}\times 2-\frac{1}{12}\times 2=0$ by R4 and R6.2. Suppose that $d_2^H(f)=2$, say $d_{2g}^H(f)=2$ or $d_{2b}^H(f)=2$. If $d_{2g}^H(f)=2$, then f is special and $d_3^H(f)+d_4^H(f)=0$ by Claim 13 (1). Thus, $w'(f)=1-\frac{1}{2}\times 2=0$ by R4. If $d_{2b}^H(f)=2$, then f must be adjacent to a $6^+$-face by the fact that H contains no 4-cycles. By Claims 5 and  13 (1), $d_4^H(f)\le 1$ and f is not incident with bad 3-vertex. If f is not adjacent to $(4,4,7^+)$-face, then $w'(f)=1-\frac{7}{12}\times 2+\frac{1}{3}-\frac{1}{12}\times 2=0$ by R4 and R6.3. If f is adjacent to $(4,4,7^+)$-face, then $w'(f)=1-\frac{7}{12}\times 2+\frac{1}{3}-\frac{1}{6}=0$ by R4 and R6.3.

• (3) Suppose that $d_H(f)=k\ge 6$. By Claim 13 (7), we have $d_{4^-}^H(f)\le \lfloor \frac{2}{3}k\rfloor$. Thus, by R4–R6 and Remark 3, we have

$$\begin{aligned} w'(f)\ge & {} (k-4)-\frac{1}{2}d_{2g}^H(f)-\frac{7}{12}(d_2^H(f)-d_{2g}^H(f))- \frac{1}{6}d_{3b}^H(f)\nonumber \\&-\frac{1}{6}d_{4}^H(f)-\frac{1}{3}\times \frac{1}{2}(d_2^H(f)-d_{2g}^H(f))-\frac{1}{12}d_{5^+}^H(f)\nonumber \\\ge & {} \frac{11}{12}k-4-\frac{8}{12}d_{2}^H(f)-\frac{1}{12}d_3^H(f)- \frac{1}{12}d_4^H(f)+\frac{1}{4}d_{2g}^H(f) \end{aligned}$$

(1)

$$\begin{aligned}\ge & {} \frac{11}{12}k-4-\frac{8}{12}d_{4^-}^H(f) \end{aligned}$$

(2)

• Suppose that $k\ge 8$. Then, $w'(f)\ge 0$ by (2) and $d_{4^-}^H(f)\le \lfloor \frac{2}{3}k\rfloor$.

• Suppose that $k=7$. If $d_{4^-}^H(f)\le 3$, then $w'(f)\ge \frac{11}{12}\times 7-4-\frac{8}{12}\times 3=\frac{5}{12}$ by (2). Suppose that $d_{4^-}^H(f)=4$. If $d_{2b}^H(f)\le 2$, then by (1), we have $w'(f)\ge \frac{11}{12}\times 7-4-\frac{8}{12}d_{2}^H(f)-\frac{1}{12}d_{3}^H(f)-\frac{1}{12}d_{4}^H(f)+\frac{1}{4}d_{2g}^H(f)\ge \frac{29}{12}-\frac{8}{12} d_{2b}^H(f)-\frac{5}{12}(d_{2g}^H(f)+d_3^H(f)+d_4^H(f))\ge \frac{29}{12}-\frac{1}{4}d_{2b}^H(f)-\frac{5}{12}d_{4^-}^H(f)\ge \frac{29}{12}-\frac{1}{4}\times 2-\frac{5}{12}\times 4=\frac{1}{4}$. If $d_{2b}^H(f)=4$, then f is special. Thus, $w'(f)\ge (7-4)-\frac{7}{12}\times 4-\frac{1}{3}\times 2=0$ by R4–R6.

• Suppose that $k=6$. By Claim 6, $d_2^H(f)\le 3$. If $d_{2}^H(f)=3$ and $d_{2b}^H(f)=2$, then f is special. Thus, $w'(f)\ge (6-4)-\frac{7}{12}\times 2-\frac{1}{2}-\frac{1}{3}\times 1=0$ by R4–R6. If $d_{2g}^H(f)=3$, then $d_3^H(f)+d_4^H(f)=0$ by Claim 13 (1). Thus, $w'(f)\ge (6-4)-\frac{1}{2}\times 3-\frac{1}{12}\times 3=\frac{1}{4}$ by R4 and R6. Suppose that $d_{2}^H(f)=2$. By Claim 13 (7), $d_3^H(f)+d_4^H(f)\le 2$. Thus, $w'(f)\ge (6-4)-\frac{7}{12}\times 2-\frac{1}{6}\times 2-\frac{1}{3}\times 1-\frac{1}{12}\times 2=0$ by R4–R6 and Remark 3. Suppose that $d_{2}^H(f)\le 1$. By Claim 13 (7), we have $d_{4^-}^H(f)\le 4$. Thus, $w'(f)\ge (6-4)-\frac{1}{2}d_2^H(f)-\frac{1}{6}d_{3}^H(f)-\frac{1}{6} d_4^H(f)-\frac{1}{12}d_{5^+}^H(f)\ge 2-\frac{1}{2}-\frac{1}{6}\times 3-\frac{1}{12}\times 2=\frac{5}{6}$ by R4–R6 and Remark 3.

$\square$

Lemma 2

$w'(v)\ge 0$ for all $v\in V(H)$.

Proof

Let v be a k-vertex in H. Let $v_0,v_1,\ldots ,v_{k-1}$ be the neighbors of v in the clockwise order. For each $i\in [0,k-1]$, let $f_i$ be the face with $vv_i$ and $vv_{i+1}$ as its boundary edges, where the indices are taken modular k.

• (1) Suppose that $d_H(v)$=2. Then, $w(v)=-2$. If $d_2^H(v)=1$, by Claim 13 (1), (2), and (6), $d_{6^+}^H(v)=1$ and v is incident with two $5^+$-faces. Then, $w'(v)=-2+\frac{5}{6}+\frac{7}{12}\times 2=0$ by R3 and R4. Suppose that $d_2^H(v)=0$. We have $d_{5^+}^H(v)=2$ by Claim 13 (1). If $t_H(v)=0$, then $w'(v)=-2+\frac{1}{2}\times 2+\frac{1}{2}\times 2=0$ by R2–R4. If $t_H(v)=1$, then $w'(v)=-2+\frac{3}{4}\times 2+\frac{1}{2}=0$ by R2–R4.

• (2) Suppose that $d_H(v)=3$. Then, $w(v)=-1$. If v is good, by Claim 13 (1), we have $d_{5^+}^H(v)=3$. Then, $w'(v)=-1+\frac{1}{3}\times 3=0$ by R2 and R3. If v is bad, by Claim 13 (1), we have $d_{5^+}^H(v)=2$. Note that v is incident with at least two $5^+$-face. Thus, $w'(v)=-1+\frac{1}{3}\times 2+\frac{1}{6}\times 2=0$ by R2–R4.

• (3) Suppose that $d_H(v)=4$. Then, $w'(v)=w(v)=0$

• (4) Suppose that $d_H(v)=5$. Then, $w(v)=1$. Note that v sends at most $\frac{1}{2}$ to each incident 3-face by R1 and Claim 13 (6). If $d_{3^-}^H(v)=1$, then $t_H(v)=0$ and $d_{2b}^H(v)=0$ by Claim 13 (2). Thus, $w'(v)=1-\frac{1}{2}=\frac{1}{2}$ by R2. If $d_{3^-}^H(v)=0$, then $t_H(v)\le 2$ by Claim 13 (8). Thus, $w'(v)=1-\frac{1}{2}\times 2=0$ by R1.

• (5) Suppose that $d_H(v)=6$. Then, $w(v)=2$. Note that $t_H(v)\le 3$ by Claim 13 (8), and v sends at most $\frac{1}{2}$ to each incident 3-face by R1 and Claim 13 (6). First, assume that $d_2^H(v)=0$. Then, $d_3^H(v)\le 2$ by Claim 13 (3). If v is incident with a $(3,6,k_1)$-face, then $d_3^H(v)=1$ by Claim 13 (3). Thus, $w'(v)=2-\frac{1}{3}-\frac{1}{2}\times 3=\frac{1}{6}$ by R1 and R3. If v is not incident with any $(3,6,k_1)$-face, then $t_H(v)\le 2$ when $d_3^H(v)\ge 1$. Thus, $w'(v)=2-\max \{\frac{1}{3}\times 2+\frac{1}{2}\times 2,\frac{1}{2}\times 3\}=\frac{1}{3}$ by R1 and R3. Next, assume that v is adjacent to a 2-vertex u. Then, $d_2^H(v)=1$ and $d_3^H(v)=0$ by Claim 13 (3). If $d_2^H(u)=0$ and $t_H(u)=0$, then $w'(v)\ge 2-\frac{1}{2}-\frac{1}{2}\times 3=0$ by R1 and R3. If $d_2^H(u)=0$ and $t_H(u)=1$, then v is incident with a $(2,6,k_1)$-face $f'$. Hence, $d_4^H(v)=0$ by Claim 13 (3). So v sends $\frac{1}{3}$ to each incident 3-face $f\ne f'$. Thus, $w'(v)\ge 2-\frac{3}{4}-\frac{1}{2}-\frac{1}{3}\times 2=\frac{1}{12}$ by R1 and R3. If $d_2^H(u)=1$, then $t_H(v)\le 2$ since u is not incident with 3-face. Thus, $w'(v)\ge 2-\frac{5}{6}-\frac{1}{2}\times 2=\frac{1}{6}$ by R1 and R3.

• (6) Suppose that $d_H(v)=7$. Then, $w(v)=3$ and $t_H(v)\le 3$ by Claim 13 (8). By Claim 13 (4), $d_{3^-}^H(v)\le 3$ when $d_2^H(v)\ge 1$.

• Assume that $d_2^H(v)=0$. If v is incident with a (3,3,7)-face, then $d_3^H(v)\le 2$ by Claim 13 (4). Thus, $w'(x)=3-\frac{1}{3}\times 2-1-\frac{1}{2}\times 2=\frac{1}{3}$ by R1 and R3. So suppose that v is not incident with (3,3,7)-face. Then, $d_3^H(v)\le 7-t_H(v)$. Recall that v sends at most $\frac{1}{2}$ to each incident 3-face by R1 and Claim 13 (6). Thus, $w'(x)\ge 3-\frac{1}{3}(7-t_H(v))-\frac{1}{2}t_H(v)=\frac{2}{3}-\frac{1}{6}t_H(v)\ge \frac{2}{3}-\frac{1}{6}\times 3=\frac{1}{6}$ by R1 and R3.

• Assume that $d_2^H(v)\ge 1$, then $d_3^H(v)\le 2$. By Claim 6, v is not incident with (3,3,7)-face. So v sends at most $\frac{1}{2}$ to each incident 3-face by R1 and Claim 13 (6). If $d_2^H(v)=1$, then $w'(v)\ge 3-\frac{5}{6}-\frac{1}{3}\times 2-\frac{1}{2}\times 3=0$ by R1 and R3. So suppose that $d_2^H(v)\ge 2$. By Claim 6, v is not incident with $(2,7,k_1)$-face. Hence, v sends $\frac{1}{2}$ to each adjacent good 2-vertex. If $d_{2g}^H(v)\ge 2$, then $t_H(v)\le 2$. Thus, $w'(v)=3-\frac{1}{2}\times 2-\frac{5}{6}-\frac{1}{2}\times 2=\frac{1}{6}$ by R1 and R3. If $d_{2g}^H(v)=1$, then $t_H(v)\le 2$ when $d_2^H(v)=2$, and $t_H(v)\le 1$ when $d_2^H(v)=3$. Thus, $w'(v)\ge 3-\frac{1}{2}-\max \{\frac{5}{6}+\frac{1}{3}+\frac{1}{2}\times 2, \frac{5}{6}\times 2+\frac{1}{2}\}=\frac{1}{3}$ by R1 and R3. If $d_{2g}^H(v)=0$, then $t_H(v)\le 2$ when $d_2^H(v)=2$, and $t_H(v)\le 1$ when $d_2^H(v)=3$. Thus, $w'(v)\ge 3-\max \{\frac{5}{6}\times 2+\frac{1}{3}+\frac{1}{2}\times 2, \frac{5}{6}\times 3+\frac{1}{2}\}=0$ by R1 and R3.

• (7) Suppose that $d_H(v)=k\ge 8$.

Case 1$d_2^H(v)=0$.

If v is incident with (3, 3, k)-face, say $f_2$, then by Claim 6, v is incident with exactly one (3, 3, k)-face. Thus, $d_3^H(v)\le k+1-t_H(v)$. For $i=1,3$, it is easy to check that $d_H(f_i)\ge 5$ and $f_i$ is not a special face. Further, when $d_H(f_i)=5$, then $d_2^H(f_i)\le 1$ by Claim 13 (1) and $d_2^H(v)=0$. So $f_i$ sends $\frac{1}{12}$ to v by R6 for $i=1,3$. Hence, $w'(v)\ge (k-4)-\frac{1}{3}d_3^H(v)-1-\frac{1}{2}(t_H(v)-1)+\frac{1}{12}\times 2\ge \frac{2}{3}k-\frac{1}{6}t_H(v)-\frac{14}{3} \ge \frac{2}{3}k-\frac{1}{6}\lfloor \frac{k}{2}\rfloor -\frac{14}{3}\ge 0$ by R1 and R3. If v is not incident with (3, 3, k)-face, then $d_3^H(v)\le k-t_H(v)$. Thus, $w'(v)\ge (k-4)-\frac{1}{3}d_3^H(v)-\frac{1}{2}t_H(v) \ge \frac{2}{3}k-\frac{1}{6}t_H(v)-4\ge \frac{2}{3}k-\frac{1}{6}\lfloor \frac{k}{2}\rfloor -4\ge \frac{2}{3}$.

Case 2$d_2^H(v)\ge 1$.

Then, $d_{5^+}^H(v)\ge 1$ by Claim 13 (5). By Claim 6, v is not incident with any (3, 3, k)-face. Hence, $d_3^H(v)\le k-d_2^H(v)-t_H(v)$. If v is incident with some $(2,k,k_1)$-face, then $d_2^H(v)=1$ by Claim 6. Thus, $w'(v)\ge (k-4)-\frac{3}{4}-\frac{1}{3}d_3^H(v)-\frac{1}{2}t_H(v)\ge k-4-\frac{3}{4}-\frac{1}{3}(k-1-t_H(v))-\frac{1}{2}t_H(v)=\frac{2}{3}k-\frac{1}{6}t_H(v)-\frac{53}{12}\ge \frac{1}{4}$ by R1 and R3. So assume that v is not incident with any $(2,k,k_1)$-face. Then, $d_2^H(v)+t_H(v)\le k$. If $d_{2b}^H(v)=0$, then $w'(v)\ge (k-4)-\frac{1}{2}d_2^H(v)-\frac{1}{3}d_3^H(v)-\frac{1}{2}t_H(v)=\frac{2}{3}k-\frac{1}{6}d_2^H(v)-\frac{1}{6}t_H(v)-4\ge \frac{2}{3}k-\frac{1}{6}k-4\ge 0$ by R1 and R3. Hence, assume that $d_{2b}^H(v)\ge 1$. Now, the proof is split into following two cases, depending on the value of $d_2^H(v)$.

Subcase 2.1$d_2^H(v)\ge 3$.

Remark 4

By Claim 12, if $v_i$ is a bad 2-vertex, then $d_H(v_{i+1})\ge 3$ and $d_H(v_{i-1})\ge 3$, where the indices are taken modular k. So $d_{2b}^H(v)\le \lfloor \frac{k}{2}\rfloor$. Further, if $d_{2b}^H(v)<\frac{k}{2}$, then $d_{3^+}^H(v)\ge d_{2b}^H(v)+1$; if $d_{2b}^H(v)=\frac{k}{2}$, then $d_{3^+}^H(v)\ge d_{2b}^H(v)$. This implies that $d_2^H(v)=k-d_{3^+}^H(v)\le k-d_{2b}^H(v)$. By Claim 6 and Claim 12, $f_i$ and $f_{i-1}$ are not special when $v_i$ is a bad 2-vertex. By the fact that G contains no 4-cycles, we have $\max \{d_H(f_{i-1}),d_H(f_i)\}\ge 6$. So at least one of $f_i$ and $f_{i-1}$ sends $\frac{1}{12}$ to v by R6.

Recall that $d_3^H(v)\le k-d_2^H(v)-t_H(v)$. Since v is not incident with any $(2,k,k_1)$-face, we have $t_H(v)\le \lfloor \frac{k-d_2^H(v)}{2}\rfloor$. By R1, R3, and Remark 4, we have

$$\begin{aligned} w'(v)= & {} (k-4)-\frac{5}{6}d_{2b}^H(v)-\frac{1}{2}(d_2^H(v)-d_{2b}^H(v))-\frac{1}{3}d_3^H(v)-\frac{1}{2}t_H(v)+\frac{1}{12}d_{2b}^H(v)\nonumber \\\ge & {} \frac{2}{3}k-\frac{1}{4}d_{2b}^H(v)-\frac{1}{6}d_2^H(v)-\frac{1}{6}t_H(v)-4 \nonumber \\\ge & {} \frac{2}{3}k-\frac{1}{4}d_{2b}^H(v)-\frac{1}{6}d_2^H(v)-\frac{1}{6}\times \frac{k-d_2^H(v)}{2}-4 \nonumber \\\ge & {} \frac{7}{12}k-\frac{1}{4}d_{2b}^H(v)-\frac{1}{12}(k-d_{2b}^H(v))-4 \nonumber \\\ge & {} \frac{1}{2}k-\frac{1}{6}d_{2b}^H(v)-4 \end{aligned}$$

(3)

$$\begin{aligned}\ge & {} \frac{1}{2}k-\frac{1}{6}\times \lfloor \frac{k}{2}\rfloor -4 \end{aligned}$$

(4)

• (i) Suppose that $k\ge 10$. Then, $w'(v)\ge \frac{1}{6}$ by (4).

• (ii) Suppose that $k=9$. If $d_{2b}^H(v)\le 3$, then $w'(v)\ge 0$ by (3). So suppose that $d_{2b}^H(v)=4$. Then, $d_{3^+}^H(v)=5$, $d_{5^+}^H(v)\ge 1$, and $t_H(v)\le 1$ by Claim 13 (5) and Remark 4. Thus, by R1, R3, and Remark 4, we have $w'(v)\ge (9-4)-\frac{5}{6}\times 4-\frac{1}{3}\times 4-\frac{1}{2}+\frac{1}{12}\times 4=\frac{1}{6}$.

• (iii) Suppose that $k=8$. Then, $w(v)=4$, $d_{2b}^H(v)\le 4$ and $d_{5^+}^H(v)\ge 1$.

• Suppose that $d_{2b}^H(v)=4$, then $d_3^H(v)\le 3$ and $t_H(v)=0$. Thus, $w'(v)\ge 4-\frac{5}{6}\times 4-\frac{1}{3}d_3^H(v)+\frac{1}{12}\times 4\ge 0$ by R1, R3, and Remark 4.

• Suppose that $d_{2b}^H(v)=3$. If $v_0$, $v_2$, and $v_4$ are bad 2-vertices, then $d_H(v_i)\ge 3$ for $i=1,3,5,7$. Moreover, if $d_H(v_6)=2$, then $t_H(v)=0$; if $d_H(v_6)\ge 3$, then $t_H(v)\le 1$. It is easy to check that both $f_1$ and $f_2$ send $\frac{1}{12}$ to v by R6, since $d_H(f_0)\ne 3$ and $d_H(f_3)\ne 3$. Hence, $w'(v)\ge 4-\frac{5}{6}\times 3-\frac{1}{3}\times 4-\frac{1}{2}+\frac{1}{12}\times 4=0$ by R1, R3, and Remark 4. So assume that $v_0$, $v_2$, and $v_5$ are bad 2-vertices. Then, $d_H(v_i)\ge 3$ for $i=1,3,4,6,7$ and $d_H(f_j)\ne 3$ for $j=0,1,2,4,5,7$. If at least one of $f_3$ and $f_6$ is not a 3-face, say $d_H(f_3)\ne 3$, then it is easy to check that both $f_1$ and $f_2$ send $\frac{1}{12}$ to v by R6. Hence, $w'(v)\ge 4-\frac{5}{6}\times 3-\frac{1}{3}\times 4-\frac{1}{2}+\frac{1}{12}\times 4=0$ by R1, R3, and Remark 4. So assume that $d_H(f_3)=d_H(f_6)=3$. If at least one of $v_3$, $v_4$, $v_6$, and $v_7$ is a 4-vertex, then $d_3^H(v)\le 2$ by Claim 13 (1). Hence, $w'(v)\ge 4-\frac{5}{6}\times 3-\frac{1}{3}\times 2-\frac{1}{2}\times 2+\frac{1}{12}\times 3=\frac{1}{12}$ by R1, R3, and Remark 4. Therefore, none of $v_3$, $v_4$, $v_6$, and $v_7$ is a 4-vertex. So for $j=0,1,2,4,5,7$, $f_j$ is not special, and $f_j$ is not adjacent to $(4,4,7^+)$-face when $d_H(f_j)=5$. Then, $f_j$ sends $\frac{1}{12}$ to v by R6. Hence, $w'(v)\ge 4-\frac{5}{6}\times 3-\frac{1}{3}\times 3-\frac{1}{2}\times 2+\frac{1}{12}\times 6=0$ by R1 and R3.

• Suppose that $d_{2b}^H(v)=2$. Recall that v is not incident with any $(2,k,k_1)$-face. Then, it is easy to check that $d_{2g}^H(v)+t_H(v)\le 3$ by Remark 4. Thus, $w'(v)\ge 4-\frac{5}{6}\times 2-\frac{1}{2}d_{2g}^H(v)-\frac{1}{3}d_3^H(v)-\frac{1}{2}t_H(v)+\frac{1}{12}\times 2 \ge \frac{5}{2}-\frac{1}{2}d_{2g}^H(v)-\frac{1}{3}(8-2-d_{2g}^H(v)-t_H(v))-\frac{1}{2}t_H(v)=\frac{1}{2}-\frac{1}{6}(d_{2g}^H(v)+t_H(v))\ge \frac{1}{2}-\frac{1}{6}\times 3=0$ by R1, R3, and Remark 4.

• Suppose that $d_{2b}^H(v)=1$. Then, it is easy to check that $d_{2g}^H(v)+t_H(v)\le 5$ by Remark 4. Thus, $w'(v)\ge 4-\frac{5}{6}-\frac{1}{2}d_{2g}^H(v)-\frac{1}{3}d_3^H(v)-\frac{1}{2}t_H(v)+\frac{1}{12} \ge \frac{13}{4}-\frac{1}{2}d_{2g}^H(v)-\frac{1}{3}(8-1-d_{2g}^H(v)-t_H(v))-\frac{1}{2}t_H(v)=\frac{11}{12}-\frac{1}{6}(d_{2g}^H(v)+t_H(v))\ge \frac{11}{12}-\frac{1}{6}\times 5=\frac{1}{12}$ by R1, R3, and Remark 4.

Subcase 2.2$1\le d_2^H(v)\le 2$.

Then, $d_{2b}^H(v)\le d_2^H(v)\le 2$. Recall that $d_3^H(v)\le k-d_2^H(v)-t_H(v)$, since v is not incident with (3, 3, k)-face by Claim 6. Since v is not incident with any $(2,k,k_1)$-face, we have $t_H(v)\le \lfloor \frac{k-d_2^H(v)}{2}\rfloor$. By R1 and R3, we have

$$\begin{aligned} w'(v)= & {} (k-4)-\frac{5}{6}d_{2b}^H(v)-\frac{1}{2}(d_2^H(v)-d_{2b}^H(v))-\frac{1}{3}d_3^H(v)-\frac{1}{2}t_H(v) \nonumber \\\ge & {} \frac{2}{3}k-\frac{1}{3}d_{2b}^H(v)-\frac{1}{6}d_2^H(v)-\frac{1}{6}t_H(v)-4 \nonumber \\\ge & {} \frac{2}{3}k-\frac{1}{3}d_{2b}^H(v)-\frac{1}{6}d_2^H(v)-\frac{1}{6}\times \frac{k-d_2^H(v)}{2}-4 \nonumber \\\ge & {} \frac{7}{12}k-\frac{1}{3}d_{2b}^H(v)-\frac{1}{12}d_2^H(v)-4 \end{aligned}$$

(5)

$$\begin{aligned}\ge & {} \frac{7}{12}k-\frac{29}{6} \end{aligned}$$

(6)

• (i) Suppose that $k\ge 9$. Then, $w'(v)\ge \frac{5}{12}$ by (6).

• (ii) Suppose that $k=8$. If $d_{2b}^H(v)\le 1$, then $w'(v)\ge \frac{7}{12}\times 8-\frac{1}{3}\times 1-\frac{1}{12}\times 2-4=\frac{1}{6}$ by (5). So suppose that $d_{2b}^H(v)=2$. Then, $d_{2g}^H(v)=0$ and $t_H(v)\le 3$. Without loss of generality, assume that $v_0$ and $v_i$ are a bad 2-vertex for $i\in [1,7]$. Then, $\max \{d_H(f_0),d_H(f_7)\}\ge 6$ and $\max \{d_H(f_{i-1}),d_H(f_i)\}\ge 6$ by the fact that G contains no 4-cycles. If $d_H(v_1)\ge 3$ and $d_H(v_7)\ge 3$, then $f_0$ or $f_7$ is a non-special $6^+$-face by Claim 6. Similarly, $f_{i-1}$ or $f_i$ is a non-special $6^+$-face. Thus, $w'(v)\ge (8-4)-\frac{5}{6}\times 2-\frac{1}{3}d_3^H(v)-\frac{1}{2}t_H(v)+\frac{1}{12}\times 2\ge \frac{5}{2}-\frac{1}{3}(8-2-t_H(v))-\frac{1}{2}t_H(v)=\frac{1}{2}-\frac{1}{6}t_H(v)\ge 0$ by R1, R3, and R6. So suppose that $d_H(v_1)=2$ or $d_H(v_7)=2$, say $d_H(v_1)=2$. Then, $d_H(v_7)\ge 3$ and $d_H(v_2)\ge 3$. If $t_H(v)\le 2$, then $w'(v)\ge (8-4)-\frac{5}{6}\times 2-\frac{1}{3}d_3^H(v)-\frac{1}{2}t_H(v)\ge \frac{7}{3}-\frac{1}{3}(8-2-t_H(v))-\frac{1}{2}t_H(v)=\frac{1}{3}-\frac{1}{6}t_H(v)\ge 0$ by R1 and R3. Thus, $t_H(v)=3$, say $d_H(f_j)=3$ for $j=2,4,6$. If at least one of $v_2$, $v_3,\cdots ,v_7$ is a 4-vertex, then $d_3^H(v)\le 2$ by Claim 13 (1). Hence, $w'(v)\ge (8-4)-\frac{5}{6}\times 2-\frac{1}{3}\times 2-\frac{1}{2}\times 3=\frac{1}{6}$ by R1 and R3. Therefore, none of $v_2$, $v_3,\cdots ,v_7$ is a 4-vertex. So for $j=1,7$, $f_j$ is not special, and $f_j$ is not adjacent to $(4,4,7^+)$-face when $d_H(f_j)=5$. Then, $f_j$ sends $\frac{1}{12}$ to v by R6. Hence, $w'(v)\ge (8-4)-\frac{5}{6}\times 2-\frac{1}{3}\times 3-\frac{1}{2}\times 3+\frac{1}{12}\times 2=0$ by R1 and R3. $\square$