List vertex arboricity of planar graphs without 5-cycles intersecting with 6-cycles

1.   Introduction

All graphs considered in this paper are finite, simple and undirected. For a graph $G$, we use $V(G), E(G),$ $\Delta(G)$, and $\delta(G)$ to denote its vertex set, edge set, maximum degree, and minimum degree, respectively. A plane graph is a planar graph with a given planar drawing on the Euclidean plane. If $G$ is a plane graph, let $F(G)$ denote the set of faces in $G$. We say that two cycles (or faces) are adjacent if they share at least one edge. In particular, when they share exactly one edge and two vertices, they are said to be normally adjacent. Two cycles (or faces) are intersecting if they share at least one vertex.

The vertex arboricity, denoted by $a(G)$, of a graph $G$ is the minimum number of subsets into which $V(G)$ can be partitioned so that each subset induces a forest. Obviously, $a(G) = 1$ if and only if $G$ itself is a forest. In 1968, Chartrand, Kronk and Wall ^[2] first introduced the vertex arboricity of a graph and proved that $a(G)\leq \lceil \frac {\Delta(G)+1}{2}\rceil$ for any graph $G$ and $a(G)\leq 3$ for any planar graph $G$. It is known that there exist infinitely many planar graphs $G$ such that $a(G) = 3$. In 1989, Hakimi and Schmeichel ^[6] provided a characterization by showing that a plane graph $G$ has $a(G) = 2$ if and only if $G^*$, the dual of $G$, contains a connected Eulerian spanning subgraph. In 2008, Raspaud and Wang ^[10] conjectured that every planar graph $G$ without adjacent 3-cycles has $a(G)\le 2$. To attack this conjecture, Chen, Raspaud and Wang ^[3] confirmed a weak version, i.e., every planar graph $G$ without intersecting 3-cycles has $a(G)\le 2$. Some sufficient conditions for a planar graph $G$ to have $a(G)\le 2$ have been obtained in ^[5,7,10,13].

A graph $G$ is said to be $L$-forested-colorable if for any color list $L = \{L(v) \mid v \in V(G)\}$, one can choose a color for each vertex $v$ from its list $L(v)$ so that the subgraph induced by every color class is a forest. The list vertex arboricity $a_l(G)$ is the minimum number of integer $k$ such that $G$ is $L$-forested-colorable with $|L(v)|\ge k$ for each $v\in V(G)$. Obviously, $a(G)\le a_l(G)$ for any graph $G$. In 2009, Borodin and Ivanova ^[1] proved that $a_l(G)\le 2$ if $G$ is a planar graph without 4-cycles adjacent to 3-cycles. This result has been recently extended by Chen, Huang and Wang ^[4] to a toroidal graph without 4-cycles adjacent to 3-cycles. In 2020, Wang, Huang and Chen ^[9] proved that every planar graph $G$ without intersecting $5$-cycles has $a_l(G)\le 2$. The list vertex arboricity of toroidal graphs has also been extensively investigated, see ^[8,11,14].

In this paper, we prove the following result:

Theorem 1. If $G$ is a planar graph without 5-cycles intersecting with 6-cycles, then $a_l(G)\le 2$.

We first introduce a few concepts and terminology used in the paper. Let $G$ be a plane graph. For $x\in V(G)\cup F(G)$, let $d_G(x)$, or simply $d(x)$, denote the degree of $x$ in $G$. A vertex of degree $k$ (resp., at least $k$, at most $k$) is called a $k$-vertex (resp., $k^+$-vertex, $k^-$-vertex). Similarly, we can define a $k$-face, $k^+$-face and $k^-$-face. For a $k$-vertex $v\in V(G)$, we usually use $v_0, v_1, \ldots, v_{k-1}$ to denote the neighbors of $v$ in $G$ in a clockwise order, and let $f_i$ denote the incident face of $v$ that contains $vv_i, vv_{i+1}$ as two boundary edges for $i = 0, 1, \ldots, k-1$, where indices are taken modulo $k$. For a face $f\in F(G)$, we use $b(f)$ to denote the boundary walk of $f$ and write $f = [v_1v_2\cdots v_n]$ if $v_1, v_2, \ldots, v_n$ are the vertices of $b(f)$ in a clockwise order. Let $V(f) = V(b(f))$. Moreover, the face adjacent to $f$ with $e = v_iv_{i+1}$ as a common boundary edge is simply denoted by $f^{v_iv_{i+1}}$. For $v\in V(G)$, let $F_{i}(v)$ (or $F_{i^+}(v)$) denote the set of $i$-faces (or $i^+$-faces) incident to $v$. Moreover, let $m_i(v) = |F_i(v)|$ and $m_{i^+}(v) = |F_{i^+}(v)|$. A $k$-vertex $v$ is called a $(d_0, d_1, \ldots, d_{k-1})$- vertex if $d(f_i) = d_i$ for $i = 0, 2, \ldots, k-1$. A cycle $C$ in a plane graph $G$ is called separating if both its interior and exterior contain at least one vertex of $G$.

This paper is organized as follows: In section 2 we give the proof of Theorem 1. Initially, we explore structural properties of a minimal counterexample. Then we use the discharging technique to contradict the existence of such a graph. Finally, in Section 3 we conclude with conjectures.

2.   Proof of Theorem 1

We prove Theorem 1 by contradiction. Suppose that $G$ is a minimal counterexample to the Theorem 1, i.e., $G$ is a planar graph satisfying the following conditions:

(ⅰ) without 5-cycles intersecting with 6-cycles;

(ⅱ) $a_l(G) > 2$; and

(ⅲ) having as few as possible vertices.

Let $L$ be a list assignment of $V(G)$ such that every $v\in V(G)$ has $|L(v)| = 2$. If $G$ contains a vertex $v$ of degree at most 3, let $H = G-v$. By the minimality of $G$, $H$ admits an $L$-forested-coloring $\pi$. Based on $\pi$, we may color $v$ with a color in $L(v)$ which appears at most once in its neighbors such that $\pi$ is extended to $G$, which is a contradiction. Hence suppose that $\delta(G)\ge 4$.

2.1.   Structural properties of $G$

As given in ^[9], the following lemma still holds for the current graph $G$:

Lemma 1. $G$ contains no a $4$-cycle $C = x_1x_2x_3x_4x_1$ with a chord $x_1x_3$ such that $d_G(x_1)\le 5$ and $d_G(x_i) = 4$ for $i = 2, 3, 4$, as shown in the configuration $C_1$ of Figure 1.

For $S\subset V(G)$, let $G[S]$ be a subgraph of $G$ induced on $S$ and let $\pi$ be an $L$-forested-coloring of $G[S]$. Note that by definition every color class of $\pi$ induces a forest in $G$. We call $\pi$ a partial $L$-forested-coloring of $G$. For a vertex $u\in V(G)\setminus S$ and a color $c\in L(u)$, we use $\tau_L(c, u)$ to denote the number of times that $c$ appears in the colored neighbors of $u$ in $G$. Set

Vertex $u$ is said to be free with respect to $(L, \pi)$ if $\pi$ can be extended into a partial $L$-forested-coloring of $G$ by coloring $u$ with a color $c\in L(u)$ such that $\tau_L(c, u) = \tau_L(u)$.

Lemma 2. (^[12]) Let $u\in V(G)\setminus S$ be a $4$-vertex. If at least one of the following conditions holds, then $u$ is free.

$\rm (1)$ At least one of neighbors of $u$ is uncolored$;$

$\rm (2)$ At least three colors appear in the neighbors of $u$$;$

$\rm (3)$ Some color appears at least three times in the neighbors of $u$.

Lemma 3. $G$ does not contain a internal $4$-vertex $v$ incident to four $3$-faces such that $v_1, v_2$ are internal 4-vertices and $v_0, v_3$ are internal $5$-vertices, as shown in the configuration $C_2$ of Figure 1.

Proof. Suppose that $G$ contains such a 4-vertex $v$. By symmetry, we have to discuss the following cases.

Case 1. $v_0v_2\notin G$.

Let $v'_0, v''_0$ be the neighbors of $v_0$ other than $v, v_1, v_3$; $v_3', v_3''$ be the neighbors of $v_3$ other that $v, v_0, v_2$; $v_1'$ be the neighbor of $v_1$ other than $v, v_0, v_2$; $v_2'$ be the neighbor of $v_2$ other than $v, v_1, v_3$. Consider the graph $H = G-\{v, v_0, v_1, v_2, v_3\}$. By the minimality of $G$, $H$ has an $L$-forested-coloring $\pi$. Define a sublist $L'$ of colors as follows:

$L'(v_i) = L(v_i)\setminus \{\pi(v'_i), \pi(v''_i)\}$ for $i = 0, 3$.

It is easy to see that $|L'(v_i)|\ge 0$ for $i = 0, 3$. Moreover, if $\pi(v'_i) = \pi(v''_i)$, or $\pi(v'_i)\ne \pi(v''_i)$ and $L(v_i)\ne \{\pi(v'_i), \pi(v''_i)\}$, then $|L'(v_i)|\ge 1$. Let $S = \{x \in \{v_0, v_3\} \ |\ |L'(x)|\ge 1\}$. Obviously, $0\le |S|\le 2$. We are going to extend $\pi$ to $G$, which leads to a contradiction.

Firstly, we give each $v_i$ a color, $i = 0, 1, 2, 3$. For $i = 1, 2$, since $|L(v_i)|\geq 2$ and $v_i$ has only one neighbor which is colored, there exists a color $a_i\in L(v_i)\setminus\pi(v_i')$ that we use to color $v_i$ with $\alpha_i$. Suppose that $i\in\{0, 3\}$. If $v_i\in S$, then we color the vertex $v_i$ with a color $b_i\in L'(v_i)$. Otherwise, color $v_i$ with the color $\pi(v_i')$. We have to discuss the following cases.

Case 1.1 $|S|\geq 1$.

By symmetry, assume that $|L'(v_0)|\ge 1$ and $a_0 \in L'(v_0)$. Let $\pi(v_1) = a_1$, $\pi(v_2) = a_2$, and $\pi(v_3) = a_3$.

Case 1.1.1. $a_0\neq a_3$.

Suppose that $\{a_1, a_2\}\neq\{a_0, a_3\}$. Then $v$ is free by Lemma 2 and $\pi$ can be extended to $G$, a contradiction. Otherwise, $\{a_1, a_2\} = \{a_0, a_3\}$. If $a_1 = a_0$ and $a_2 = a_3$, then we recolor $v_2$ with a color in $L(v_2)\backslash \{a_2\}$ and color $v$ with a color in $L(v)\backslash \{a_0\}$. If $a_1 = a_3$ and $a_2 = a_0$, then we color $v$ with a color in $L(v)$.

Case 1.1.2. $a_0 = a_3$.

If $a_1\neq a_2$, then $v$ is free by Lemma 2 and $\pi$ can be extended to $G$, a contradiction. If $a_1 = a_2 = a_0$, then recolor $v_1$ with a color in $L(v_1)\backslash \{a_0\}$, and color $v$ with a color in $L(v)\backslash \{a_0\}$. Otherwise, $a_1 = a_2\neq a_0$, then we recolor $v_2$ with a color in $L(v_2)\backslash \{a_2\}$ and color $v$ with a color in $L(v)\backslash \{a_0\}$. Now if $G$ does not contain monochromatic cycle, then we get an $L$-forested-coloring of $G$, a contradiction. Otherwise, $G$ contains a monochromatic cycle $v_2v_3\cdots v_2'v_2$, then $|L'(v_3)| = 0$, $L(v_3) = \{\pi(v_3'), \pi(v_3'')\}$ and $\pi(v_3')\neq \pi(v_3'')$. Recoloring $v_3$ with a color in $L(v_3)\backslash \{a_3\}$, we extend $\pi$ to $G$, a contradiction.

Case 1.2. $|S| = 0$.

Without loss of generality, let $\pi(v_i) = a_i$, $i = 0, 1, 2, 3$.

Case 1.2.1. $a_0\neq a_3$.

Suppose that $\{a_1, a_2\}\neq\{a_0, a_3\}$. Then $v$ is free by Lemma 2 and $\pi$ can be extended to $G$, a contradiction. Otherwise, $\{a_1, a_2\} = \{a_0, a_3\}$. If $a_1 = a_0$ and $a_2 = a_3$, then we recolor $v_0$ with the color $\pi(v_0'')$, and color $v$ with a color in $L(v)\setminus \{a_3\}$. If there is no monochromatic cycle in $G$, then $\pi$ is extended to $G$, a contradiction. Otherwise, if $G$ contains a monochromatic cycle $v_0v_3\cdots v_0''v_0$, then color $v_3$ with the color $\pi(v_3'')$. If $a_1 = a_3$ and $a_2 = a_0$, then we color $v$ with a color in $L(v)$.

Case 1.2.2. $a_0 = a_3$.

Recolor $v_0$ with the color $\pi(v_0'')$, and then the proof can be given as in Case 1.2.1.

Case 2. $v_0v_2\in G$.

Let $v'_0$ be the neighbors of $v_0$ other than $v, v_1, v_2, v_3$; $v_3', v_3''$ be the neighbors of $v_3$ other that $v, v_0, v_2$; $v_1'$ be the neighbor of $v_1$ other than $v, v_0, v_2$.

Consider the graph $H = G-\{v, v_0, v_1, v_2, v_3\}$. By the minimality of $G$, $H$ has an $L$-forested-coloring $\pi$. Define a sublist $L'(v_3) = L(v_3)\setminus \{\pi(v'_3), \pi(v''_3)\}$.

Firstly, we give each $v_i$ a color, $i = 0, 1, 2, 3$. For $i = 0, 1$, since $|L(v_i)|\geq 2$ and $v_i$ has only one neighbor which is colored, there exists a color $a_i\in L(v_i)\setminus\{\pi(v_i')\}$ that we use to color $v_i$ with $a_i$. We color the vertex $v_2$ with a color $a_2\in L(v_2)\setminus\{a_0\}$. If $L'(v_3)\ge 1$, then we color the vertex $v_3$ with a color $a_3\in L'(v_3)$. Otherwise, color $v_3$ with the color $\pi(v_3')$.

Case 2.1. $a_0\neq a_3$.

Suppose that $\{a_1, a_2\}\neq\{a_0, a_3\}$. Then $v$ is free by Lemma 2 and $\pi$ can be extended to $G$, a contradiction. Otherwise, $\{a_1, a_2\} = \{a_0, a_3\}$. Thus, $a_0 = a_1$ and $a_2 = a_3$. We recolor $v_1$ with a color in $L(v_1)\backslash \{a_0\}$ and color $v$ with a color in $L(v)\backslash \{a_3\}$.

Case 2.2. $a_0 = a_3$.

If $a_1\neq a_2$, then $v$ is free by Lemma 2 and $\pi$ can be extended to $G$, a contradiction. If $a_1 = a_2$, then recolor $v_1$ with a color in $L(v_1)\backslash \{a_2\}$, and color $v$ with a color in $L(v)\backslash \{a_0\}$.

This completes the proof of Lemma $3$.

2.2.   The discharging technique

To obtain a contradiction by applying discharging technique, we define a new graph $H$ from $G$ as follows: If there is no separating 3-cycle in $G$, let $H = G$; otherwise, choose a separating 3-cycle $T$ with the least internal vertices, and let $H = G[V^0(T)\cup V(T)]$, where $V^0(T)$ is the set of internal vertices of $T$. Let $V^0(H) = V^0(T)$, which are called the set of internal vertices in $H$. Let $f^\Delta$ denote the outer face of $H$, and let the set of interior faces of $H$ be $F^0(H) = F(H)\setminus \{f^\Delta\}$. For each vertex $v\in V^0(H)$, it holds obviously that $d_H(v) = d_G(v)$. Since $G$ is connected, so is $H$. In the following, we will contradict the existence of subgraph $H$ and therefore of $G$, by using the discharging technique. Using Euler's formula $|V(H)|-|E(H)|+|F(H)| = 2$, we can deduce the following identity.

Define an initial weight $w$ on vertices and faces of $H$ by $w(v) = 2d_H(v)-6$ if $v\in V(H)$ and $w(f) = d_H(f)-6$ if $f\in F(H)$. It follows from (2.1) that the total sum of weights is equal to $-12$. In what follows, we will design some discharging rules and redistribute weights accordingly. Once the discharging is finished, a new weight function $w'$ is produced. However, the total sum of weights is kept fixed during the discharging process. Nevertheless, we can show that the sum of $w'(x)$ is at least $-11.6$ for all $x\in V(H)\cup F(H)$. This leads to the following obvious contradiction

and hence the theorem follows.

For a 3-face $f\in F(G)$, we say that it is bad if $V(f)$ contains three internal $4$-vertices; weak if $V(f)$ contains two internal $4$-vertices and one $5^+$-vertex, or two internal $4$-vertices with $|V(f)\cap V(T)| = 1$; and good otherwise.

Observation 1. Let $f', f''\in F^0(H)$ with $d_H(f') = 3$ and $d_H(f'') = 4$. If $f'$ is normally adjacent to $f''$, then $b(f')\cup b(f'')$ contains a 5-cycle.

Let $C_5 \bowtie C_6$ denote a configuration that a vertex lies in a 5-cycle and a 6-cycle at the same time.

Observation 2. Let $f', f''\in F^0(H)$ with $d_H(f') = 3$ and $d_H(f'') = 5$. If $f'$ is normally adjacent to $f''$, then $b(f')\cup b(f'')$ contains a configuration $C_5 \bowtie C_6$.

Proof. Suppose that $f'' = [x_1x_2x_3x_4x_5]$ and $f' = [x_1x_2x_6]$. Since $f''$ is a 5-face, $x_1, x_2, \ldots, x_5$ are mutually distinct. Since $f'$ and $f''$ are normally adjacent, it follows that $x_6\notin V(f'')$ and hence a 6-cycle $x_1x_6x_2x_3x_4x_5x_1$ is found. Thus, $b(f')\cup b(f'')$ has a configuration $C_5 \bowtie C_6$ at the vertex $x_1$.

Claim 1. If $v\in V^0(H)$ is a $5^+$-vertex, then $m_3(v)\leq \lfloor \frac {3d_H(v)}{4} \rfloor$.

Proof. Since $v$ is an internal vertex, all the faces incident to $v$ are internal faces. It is easy to check that if $v$ is incident to four consecutive 3-faces, then $G$ will contain a configuration $C_5 \bowtie C_6$. This shows that $m_3(v)\leq \lfloor \frac {3d_H(v)}{4} \rfloor$.

Claim 2. If $v\in V^0(H)$ is incident to two $3$-faces $f_i, f_{i+1}$, then $d_H(f_j)\ne 4, 5$ for each $j\in \{i-1, i+2\}$.

Proof. Without loss of generality, assume that $i = 1$.

$\bullet$ $d_H(f_{3}) = 4$.

Suppose that $f_3 = [vv_3uv_4]$. If $u, v_1, v_2$ are mutually distinct, then a 5-cycle $vv_2v_3uv_4v$ and a 6-cycle $vv_1v_2v_3uv_4v$ are found. Thus, $b(f_1)\cup b(f_2) \cup b(f_3)$ has a configuration $C_5 \bowtie C_6$ at the vertex $v$, a contradiction. If $v_2 = u$, then a separating 3-cycle $vv_2v_4v$ with fewer internal vertices than $T$ appears, contradicting the choice of $T$. If $v_1 = u$, then a separating 3-cycle $vv_1v_3v$ with fewer internal vertices than $T$ appears, contradicting the choice of $T$.

$\bullet$ $d_H(f_{3}) = 5$.

Suppose that $f_3 = [vv_3u_1u_2v_4]$. By Observation 2, $f_2, f_3$ are not normally adjacent, thus $v_2 = u_1$ or $v_2 = u_2$. If $v_2 = u_1$, then $v_3$ is a $2$-vertex and $v$ is a vertex on $T$, a contradiction. If $v_2 = u_2$, then a separating 3-cycle $vv_2v_4v$ with fewer internal vertices than $T$ appears, contradicting the choice of $T$.

Claim 3. Let $v\in V^0(H)$ with $d_H(v)\ge 4$. If $d_H(f_i) = d_H(f_{i+1}) = d_H(f_{i+2}) = 3$, then $d_H(f_{i-1}), d_H(f_{i+3})$ $\ge 7$.

Proof. Without loss of generality, assume that $i = 1$. By Claim 1, $d_H(f_{0})\neq 3$. By Claim 2, $d_H(f_{0})\neq 4, 5$. If $d_H(f_{0}) = 6$, a 5-cycle $vv_1v_2v_3v_4v$ and a 6-cycle $f_0$ are found. Thus, $b(f_0)\cup b(f_1) \cup b(f_2)\cup b(f_3)$ has a configuration $C_5 \bowtie C_6$ at the vertex $v$, a contradiction.

Claim 4. If $v\in V^0(H)$ is incident to two $3$-faces $f_i, f_{i+2}$, then $d_H(f_{i+1})\ne 4, 5$.

Proof. Without loss of generality, assume that $i = 1$.

$\bullet$ $d_H(f_{2}) = 4$.

Suppose that $f_2 = [vv_2uv_3]$. If $u, v_1, v_4$ are mutually distinct, then a 5-cycle $vv_1v_2uv_3v$ and a 6-cycle $vv_1v_2uv_3v_4v$ are found. Thus, $b(f_1)\cup b(f_2) \cup b(f_3)$ has a configuration $C_5 \bowtie C_6$ at the vertex $v$, a contradiction. If $v_1 = u$, then a separating 3-cycle $vv_1v_3v$ with fewer internal vertices than $T$ appears, contradicting the choice of $T$. If $v_4 = u$, then a separating 3-cycle $vv_2v_4v$ with fewer internal vertices than $T$ appears, contradicting the choice of $T$.

$\bullet$ $d_H(f_{2}) = 5$.

Suppose that $f_2 = [vv_2u_1u_2v_3]$. By Observation 2, $f_1, f_2$ are not normally adjacent and $f_2, f_3$ are not normally adjacent, thus $v_1 = u_1$ and $v_4 = u_2$. In this case, $v_2$ and $v_3$ are both 2-vertices and they are stand on different 3-faces, a contradiction.

Our discharging rules are defined as follows:

(R0) Let $v\in V(T)$. We carry out the following three subrules (R0.1), (R0.2) and (R0.3):

(R0.1) $v$ sends $0.5$ to each incident 4-face, and $0.2$ each incident 5-face.

(R0.2) If $d_H(v)\geq5$, then $v$ sends $2$ to each incident internal $3$-face. Otherwise, assume that $3\leq d_H(v)\leq4$. If $m_3(v) = d_H(v)$, then $v$ sends $1.5$ to each incident internal $3$-face. If $m_3(v)\leq d_H(v)-1$, then $v$ sends $2$ to each incident internal $3$-face.

(R0.3) If $f = [vx_1x_2x_3]$ is a 4-face, then $v$ sends $0.1$ to $x_2$ through the face $f$.

(R1) Every 4-face gets $0.5$ from each of its incident internal vertices.

(R2) Every 5-face gets $0.2$ from each of its incident internal vertices.

(R3) Let $v\in V^0(H)$ with $m_3(v)\ge 1$.

(R3.1) Suppose that $d_H(v) = 4$.

$\bullet$ Assume that $m_3(v) = 4$. If $v$ is incident with a bad 3-face and a good 3-face, then $v$ sends $1$ to bad 3-face, 0 to good 3-face and $0.5$ to each of its incident weak 3-faces. Otherwise, $v$ sends $0.5$ to each of its incident 3-faces.

$\bullet$ Assume that $m_3(v) = 3$. If $v$ is incident with a bad 3-face, then $v$ sends $1$ to bad 3-face, and $0.5$ to each of its other incident 3-faces. If $v$ is incident with at least one good 3-face, then $v$ sends $0.5$ to good 3-face, and $0.75$ to each of its incident weak 3-faces. Otherwise, $v$ sends $0.5$ to each of its incident weak 3-faces.

$\bullet$ If $m_3(v)\le 2$, then $v$ sends $1$ to each of its incident 3-faces.

(R3.2) Suppose that $d_H(v) = 5$.

$\bullet$ If $d_H(f_i) = 3$ for $i = 0, 1, 2$, and $d_H(f_3), d_H(f_4)\ge7$, then $v$ sends $1.25$ to each of $f_0, f_1, f_2$.

$\bullet$ If $d_H(f_i) = 3$ for $i = 0, 1, 3$, and $d_H(f_2), d_H(f_4)\ge6$, then $v$ sends $1.25$ to each of $f_0, f_1$ and $1.5$ to $f_3$.

$\bullet$ For other cases, $v$ sends $1.5$ to each of its incident 3-faces.

(R3.3) If $d_H(v)\ge 6$, then $v$ sends $1.5$ to each of its incident 3-faces.

(R4) Assume that $v\in V^0(H)$ is a $(3, 4, 5, 4)$-vertex with $f_1 = [vv_1x_1v_2]$ and $f_3 = [vv_3x_2v_0]$, as shown in the configuration $G_1$ of Figure 2. Then each of $x_1$ and $x_2$ send $0.1$ to $v$.

(R5) Assume that $v\in V^0(H)$ is an internal $(3, 3, 3, 3)$-vertex and $v_0$ is an internal $(3, 3, 7^+, 3, 7^+)$-vertex. Let $f^*$, other than $f_3$, denote the incident face of the edge $v_0v_3$. Such configuration $G_2$ is depicted Figure 2.

(1) If $v_1, v_2$ are internal $4$-vertices and $v_3$ is an internal $6^+$-vertex, then $f^*$ sends $0.25$ to $f_3$.

(2) If $v_3$ is an internal $4$-vertex, then $f^*$ send $0.25$ to $f_3$.

If $v$ is a $k$-vertex incident with $m$ 3-faces, then $v$ is called a $k_m$-vertex. If a $k$-vertex $v$ sends the weight $a$ to an incident 3-face $f$, then $v$ is called a $k^a$-vertex for $f$. A $k_m^a$-vertex for a 3-face $f$ is a $k_m$-vertex that sends the weight $a$ to $f$.

For $x, y \in V (H)\cup F(H)$, let $\tau (x \rightarrow y)$ denote the amount of weights transferred from $x$ to $y$ according to the rules (R0)-(R5).

Observation 3. Let $v$ be an internal $4^+$-vertex incident to an internal 3-face $f$. Then

$\rm (1)$ $\tau(v\to f)\in \{1, 0.75, 0.5, 0\}$ if $d_H(v) = 4$$;$

$\rm (2)$ $\tau(v\to f)\in \{1.25, 1.5\}$ if $d_H(v) = 5$$;$

$\rm (3)$ $\tau(v\to f) = 1.5$ if $d_H(v)\ge 6$.

Claim 5. Suppose that $f = [x_1x_2x_3]$ is an internal $3$-face. Let $s$ denote the total number of $4^{0.75}$-vertices, $4^{0.5}$-vertices and $4^0$-vertices which are adjacent to $f$. Then $s \leq 1$.

Proof. Suppose that $s \geq2$. By the discharging rules, if $x_i$ is a $4^{0.75}$-vertex, $4^{0.5}$-vertex or $4^0$-vertex, then $x_i$ is an internal $4_3$-vertex or $4_4$-vertex. We consider the following possibilities by symmetry.

$\bullet$ At least one of $x_1$, $x_2$ is a $4_3$-vertex in $V^0(H)$.

If $f^{x_1x_2}$ and $f^{x_1x_3}$ are 3-faces, then neither $x_2$ nor $x_3$ is an internal $4_3$-vertex or $4_4$-vertex, for otherwise $H$ will contain a separating 3-cycle with fewer internal vertices than $T$ or a configuration $C_5 \bowtie C_6$ at the vertex $x_1$, a contradiction. The case in which $f^{x_1x_2} = [x_1x_2w]$ and $f^{x_1w} = [x_1wy]$ are 3-faces can be similarly discussed, where $y$ and $w$ are the neighbors of $x_1$ other than $x_2$ and $x_3$.

$\bullet$ Both $x_1$ and $x_2$ are internal $4_4$-vertices.

In this case, $H$ contains a separating 3-cycle with fewer internal vertices than $T$ or a configuration $C_5 \bowtie C_6$ at the vertex $x_1$, also a contradiction.

The above discussion implies that $s \leq 1$.

Lemma 4. $w'(f)\ge 0$ for each $f\in F^0(H)$.

Proof. Let $f\in F^0(H)$. Clearly, $d_H(f)\ge 3$.

Case 1. Suppose that $d_H(f) = 3$ and $f = [x_1x_2x_3]$.

Then $w(f) = -3$. Since $f$ is an internal face of $H$, we can derive that $|V(f)\cap V(T)|\leq2$.

Case 1.1. $|V(f)\cap V(T)| = 2$.

Then $f$ is a good 3-face. By (R0), $f$ receives $1.5$ or 2 from each of its two vertices of $T$. Hence $w'(f)\ge -3+1.5\times 2 = 0$.

Case 1.2. $|V(f)\cap V(T)| = 1$.

Without loss of generality, let $x_1\in V(T)$. We distinguish cases depending on whether $x_2$ and $x_3$ are internal $4$-vertices or internal $5^+$-vertices. Assume that both $x_2$ and $x_3$ are internal $4$-vertices. Then $f$ is a weak $3$-face. Rule (R3.1) assures that $4$-vertices always send non-zero weight to weak faces. Hence by Observation 3, $x_2$ and $x_3$ send either $0.5$ or $0.75$ or $1$ to $f$. Now, by Claim 5, at most one of $x_2$ and $x_3$ is an $4^{0.5}$-vertex or $4^{0.75}$-vertex. Equivalently at least one of them is a $4^1$-vertex. Thus, $w'(f)\ge -3+1.5+1+0.5 = 0$ by (R0) and (R3). Assume that $x_2$ is a internal $4$-vertex and $x_3$ is a internal $5^+$-vertex. If $x_2$ is not a $4^0$-vertex, then $w'(f)\ge -3+1.5+1.25+0.5 = 0.25$ by (R0) and (R3). Otherwise, $x_2$ is a $4^0$-vertex. By (R3.1), $x_2$ is a $4_4$-vertex. If $d_H(x_1) = 3$, then $|V(f)\cap V(T)| = 2$, a contradiction. If $d_H(x_1)\ge4$, then $f^{x_1x_3}$ is a $4^+$-face and $m_3(x_1)\le d_H(x_1)-1$, otherwise it will produce a $3$-vertex in $v^0(H)$, or a separating $3$-cycle with fewer internal vertices than $T$ or a configuration $C_5 \bowtie C_6$ at the vertex $x_1$, a contradiction. Thus $f$ gets $2$ from $x_1$. Therefore, $w'(f)\ge -3+2+1.25 = 0.25$ by (R0) and (R3). Assume that both $x_2$ and $x_3$ are $5^+$-vertices, then $w'(f)\ge -3+1.5+1.25\times2 = 1$ by (R0) and (R3).

Case 1.3. $|V(f)\cap V(T)| = 0$.

Let $d_H(x_1) \leq d_H(x_2) \leq d_H(x_3)$.

$\bullet$ Assume that $d_H(x_1)\geq5$, then $f$ gets at least $1.25$ from each of $x_1, x_2, x_3$ by Observation 3. Thus $w'(f)\geq -3+1.25\times 3 = 0.75$.

$\bullet$ Assume that $d_H(x_1) = 4, d_H(x_2)\geq5$, then $f$ gets at least $1.25$ from each of $x_2, x_3$ by (R3.2) and (R3.3). If $x_1$ is not a $4^0$-vertex, then $f$ gets at least $0.5$ from $x_1$ by (R3.1). Hence, $w'(f)\geq -3+0.5+1.25\times 2 = 0$. Otherwise, $x_1$ is a $4^0$-vertex, then $x_1$ is a internal $4_4$-vertex and $x_1$ is incident with a bad 3-face by (R3.1). By Lemma 1 and Lemma 3, $x_2$ or $x_3$ is a internal $6^+$-vertex, say $x_2$, and $f$ gets $1.5$ from $x_2$ by (R3.3). If $x_3$ is a internal $6^+$-vertex, or a internal 5-vertex with $m_3(x_3) = 2$, then $f$ gets $1.5$ from $x_3$ by (R3.2) and (R3.3). Thus, $w'(f)\geq -3+1.5\times 2 = 0$. Otherwise, $x_3$ is a internal 5-vertex with $m_3(x_3) = 3$. If $x_3$ is a internal $(3, 3, 3, 6^+, 6^+)$-vertex, a $(3, 3, 6, 3, 6^+)$-vertex, or a $(3, 3, 6^+, 3, 6)$-vertex, it will contain a configuration $C_5 \bowtie C_6$ at the vertex $x_3$, a contradiction. If $x_3$ is a internal $(3, 3, 7^+, 3, 7^+)$-vertex, it is the construction $G_2$ and $f$ gets $0.25$ from the $7^+$-face $f^{x_2x_3}$ by (R5). Thus, $w'(f)\geq -3+1.5+1.25+0.25 = 0$ by (R3.2) and (R3.3).

$\bullet$ Assume that $d_H(x_1) = d_H(x_2) = 4$ and $d_H(x_3)\geq5$. By definition, $f$ is a weak 3-face and $f$ gets at least $0.5$ from each of $x_1, x_2$ by (R3.1). By Claim 5, at least one of $x_1$ and $x_2$ is a $4^1$-vertex. Without loss of generality, assume that $x_2$ is a $4^1$-vertex. If $x_1$ is a $4^1$-vertex or a $4^{0.75}$-vertex, then $w'(f)\geq -3+1+0.75+1.25 = 0$ by (R3). Otherwise, $x_1$ is a $4^{0.5}$-vertex. If $x_3$ is a internal $6^+$-vertex or a $5^{1.5}$-vertex, then $w'(f)\geq -3+1+1.5+0.5 = 0$ by (R3). Therefore, $x_3$ is a internal 5-vertex with $m_3(x_3) = 3$. By Lemma 1 and the definition of good 3-face, we can derive that $x_1$ is incident with a good 3-face. If $x_1$ is a internal $4_3$-vertex, then $f$ gets $0.75$ from $x_1$ by (R3.1), a contradiction. Otherwise, $x_1$ is a internal $4_4$-vertex. If $x_3$ is a $(3, 3, 3, 6^+, 6^+)$-vertex, $(3, 3, 6, 3, 6^+)$-vertex or $(3, 3, 6^+, 3, 6)$-vertex, it will contain a configuration $C_5 \bowtie C_6$ at the vertex $x_3$, which is a contradiction. If $x_3$ is a $(3, 3, 7^+, 3, 7^+)$-vertex, it is the construction $G_2$ and $f$ gets $0.25$ from the $7^+$-face $f^{x_2x_3}$ by (R5). Thus, $w'(f)\geq -3+1+0.5+1.25+0.25 = 0$ by (R3).

$\bullet$ Assume that $d_H(x_1) = d_H(x_2) = d_H(x_3) = 4$.

By definition, $f$ is a bad 3-face and $f$ gets $1$ from each of $x_1, x_2, x_3$ by (R3). Thus, $w'(f)\geq -3+1\times3 = 0$.

Case 2. $4\leq d_H(f)\leq 6$.

(1) If $d_H(f) = 4$, then $w'(f) = -2+0.5\times4 = 0$ by (R0) and (R1).

(2) If $d_H(f) = 5$, then $w'(f) = -1+0.2\times5 = 0$ by (R0) and (R2).

(3) If $d_H(f) = 6$, then $w'(f) = 0$.

Next, if $f = [v_1v_2v_3 \cdots v_n]$ is a $7^+$-face, $v_iv_{i+1}$ is in the configuration $G_2$, $[v_iv_{i+1}x]$ is a 3-face and $x$ is a $4_4$-vertex, then we say that $f$ is adjacent to a configuration $G_2$ by $v_iv_{i+1}$. Let $t$ denote the number of the configuration $G_2$ which $f$ is adjacent to.

Claim 6. Suppose that $f$ is a $7^+$-face in $H$, then $t \leq \lfloor\frac{2d_H(f)}{3}\rfloor$.

Proof. Let $f = [v_1v_2v_3 \cdots v_n]$ be a $7^+$-face in $H$.

If $f$ is adjacent to a configuration $G_2$ by $v_iv_{i+1}$, then $v_i$ or $v_{i+1}$ is a internal $(3, 3, 7^+, 3, 7^+)$-vertex. By symmetry, let $v_i$ be a internal $(3, 3, 7^+, 3, 7^+)$-vertex. If $f$ is adjacent to another $G_2$ by edge $v_{i-1}v_i$, then $v_i$ would be a $(3, 3, 3, 3, 7^+)$-vertex, contradicting the definition of $G_2$. Hence, if $f$ is adjacent to a $G_2$ by edge $v_iv_{i+1}$ at least one of $v_{i-1}v_i$ or $v_{i+1}v_{i+2}$ is not adjacent to $G_2$. It means that the three consecutive edges on $f$ are adjacent to at most two $G_2$ configurations. Consequently, $t\leq \lfloor\frac{2d_H(f)}{3}\rfloor$.

Case 3. $d_H(f)\geq7$.

Suppose that $d_H(f)$ = 7. By Claim 6, $t\leq \lfloor\frac{2d_H(f)}{3}\rfloor = 4$, then $w'(f)\ge 7-6-0.25\times 4 = 0$ by (R5). Suppose that $d_H(f)\geq8$, then $w'(f)\ge d_H(f)-6-0.25\times \lfloor\frac{2d_H(f)}{3}\rfloor\geq\frac{5}{6}d_H(f)-6\geq0$ by (R5).

This completes the proof of Lemma $4$.

Lemma 5. $w'_H(v)\ge 0$ for each $v\in V^0(H)$.

Proof. Since $v$ is a vertex of $G$ and $\delta(G)\geq 4$, it follows that $d_H(v) = d_G(v)\ge 4$. We have six cases to consider.

Case 1. $d_H(v) = 4$.

Case 1.1. Suppose that $v$ is $x_1$ (or $x_2$) in the configuration $G_1$.

In the configuration $G_1$, we can derive that $m_5^+(v)\ge2$ and $m_3(v)\le1$, otherwise it will produce a separating 3-cycle with fewer internal vertices than $T$ or a configuration $C_5 \bowtie C_6$, a contradiction. Hence, $w'(v)\geq 2-0.5-0.1-0.2\times 2-1 = 0$.

Case 1.2. Suppose that $v$ is not $x_1$ (or $x_2$) in the construction $G_1$.

$\bullet$ Assume that $m_3(v) = 0$, then $w'(v)\geq 2-0.5\times 4 = 0$ by (R1) and (R2).

$\bullet$ Assume that $m_3(v) = 1$. If $m_4(v)+m_5(v)\leq 2$, then $w'(v)\geq 2-1-0.5\times2 = 0$ by (R1), (R2) and (R3.1). Suppose that $m_4(v)+m_5(v) = 3$. If $m_5(v)\geq2$, then $w'(v)\geq 2-1-0.5-0.2\times2 = 0.1$ by (R1), (R2) and (R3.1). Suppose that $m_5(v) = 1$ and $m_4(v) = 2$. If $v$ is a $(4, 3, 5, 4)$-vertex, then it will produce a separating 3-cycle with fewer internal vertices than $T$ or a configuration $C_5 \bowtie C_6$ at the vertex $v$, a contradiction. If $v$ is a $(4, 3, 4, 5)$-vertex, then it is the configuration $G_1$. Thus, $w'(v)\geq 2-1-0.5\times2-0.2+0.1\times2 = 0$ by (R1), (R2), (R3.1) and (R4). If $m_4(v) = 3$, then it will produce a separating 3-cycle or a configuration $C_5 \bowtie C_6$, a contradiction.

$\bullet$ Assume that $m_3(v) = 2$. If $d_H(f_0) = d_H(f_1) = 3$, then $d_H(f_2)\geq 6$ and $d_H(f_3)\geq6$ by Claim 2. Thus, $w'(v)\geq 2-1\times2 = 0$ by (R3.1). If $d_H(f_0) = d_H(f_2) = 3$, then $d_H(f_1)\geq 6$ and $d_H(f_3)\geq6$ by Claim 4. Thus, $w'(v)\geq 2-1\times2 = 0$ by (R3.1).

$\bullet$ Assume that $m_3(v) = 3$. Suppose that $d_H(f_0) = d_H(f_1) = d_H(f_2) = 3$, then $d_H(f_3)\geq 7$ by Claim 3. By Lemma 1, at least one of $v_0, v_1, v_2, v_3$ is an internal $5^+$-vertex or belongs to $V(T)$. If $v_0$ or $v_3$ is an internal $5^+$-vertex or belongs to $V(T)$, say $v_0$, then one of $v_1, v_2, v_3$ is an internal $5^+$-vertex or belongs to $V(T)$ by Lemma 1 and $v$ is incident with at most one bad 3-face. If $v_1$ or $v_2$ is an internal $5^+$-vertex or belongs to $V(T)$, then $v$ is incident with at most one bad 3-face. Thus, $w'(v)\geq 2-1-0.5\times2 = 0$, or $w'(v)\geq 2-0.5-0.75\times2 = 0$, or $w'(v)\geq 2- 0.5\times3 = 0.5$ by (R3.1).

$\bullet$ Assume that $m_3(v) = 4$. If $v$ is not incident with any bad 3-face, then $w'(v)\geq 2-0.5\times4 = 0$ (R3.1). Otherwise, $v$ is incident with bad 3-faces. By Lemma 1, at least two of $v_0, v_1, v_2, v_3$ are internal $5^+$-vertices or belong to $V(T)$, it means that $v$ is incident with one bad 3-face, one good 3-face and two weak 3-faces. Thus, $w'(v)\geq 2-1-0.5\times2 = 0$ by (R3.1).

Case 2. $d_H(v) = 5$.

Note that $w(v) = 4$. By Claim 1, $m_3(v)\le\lfloor\frac 34d_H(v)\rfloor = 3$. If $m_{6^+}(v) \geq 3$, then $w'(v)\geq 4-1.5\times2 = 0$. Assume that $m_{6^+}(v)\leq2$. If $m_3(v) = 0$, then $w'(v)\geq 4-0.5\times5-0.1\times5 = 1$ by (R1)-(R4). If $m_3(v) = 1$, then $w'(v)\geq 4-1.5-0.5\times 4-0.1\times 4 = 0.1$ by (R1)-(R4). Suppose that $m_3(v) = 2$. If $d_H(f_0) = d_H(f_1) = 3$, then $d_H(f_2)\geq 6$ and $d_H(f_4)\geq6$ by Claim 2. Thus, $w'(v)\geq 4-1.5\times2-0.5-0.1 = 0.4$ by (R1)-(R4). If $d_H(f_0) = d_H(f_2) = 3$, then $d_H(f_1)\geq 6$ by Claim 4. If $d_H(f_3) = d_H(f_4) = 4$, it will produce a separating 3-cycle with fewer internal vertices than $T$ or a configuration $C_5 \bowtie C_6$ at the vertex $v$, a contradiction. Thus $w'(v)\geq 4-1.5\times2-0.5-0.1-0.2 = 0.2$ by (R1)-(R4).

Suppose that $m_3(v) = 3$. If $d_H(f_0) = d_H(f_1) = d_H(f_2) = 3$, then $d(f_3)\geq 7$ and $d(f_4)\geq7$ by Claim $3$. Thus $w'(v)\geq 4-1.25\times3 = 0.25$ by (R3.2). If $d_H(f_0) = d_H(f_1) = d_H(f_3) = 3$, then $d_H(f_2)\geq 6$ and $d_H(f_4)\geq6$ by Claim $2$. Thus, $w'(v)\geq 4-1.25\times2-1.5 = 0$ by (R3.2).

Case 3. $d_H(v) = 6$.

Note that $w(v) = 6$. By Claim 1, $m_3(v)\le\lfloor\frac 34d_H(v)\rfloor = 4$. If $m_{6^+}(v) \geq 2$, then $w'(v)\geq 6-1.5\times4 = 0$ by (R3.3).

Suppose that $m_{6^+}(v) \leq 1$. If $m_3(v)\leq 2$, then $w'(v)\geq 6-1.5 \times 2-0.5\times4-0.1\times4 = 0.6$ by (R1)-(R4). Assume that $m_3(v) = 3$. If $d_H(f_0) = d_H(f_1) = d_H(f_2) = 3$, then $d_H(f_3)\geq 7$ and $d_H(f_5)\geq7$ by Claim 3, a contradiction. If $d_H(f_0) = d_H(f_1) = d_H(f_3) = 3$, then $d_H(f_2)\geq 6$ and $d_H(f_5)\geq6$ by Claim 2, a contradiction. If $d_H(f_0) = d_H(f_2) = d_H(f_4) = 3$, then $d_H(f_1)\geq 6$, $d_H(f_3)\geq6$ and $d_H(f_5)\geq6$ by Claim 4. Assume that $m_3(v) = 4$. If $d_H(f_0) = d_H(f_1) = d_H(f_2) = d_H(f_4) = 3$, then $d_H(f_3)\geq 7$ and $d_H(f_5)\geq7$ by Claim 3, a contradiction. If $d_H(f_0) = d_H(f_1) = d_H(f_3) = d_H(f_4) = 3$, then $d_H(f_2)\geq 6$ and $d_H(f_5)\geq6$ by Claim 2, a contradiction.

Case 4. $d_H(v) = 7$.

Note that $w(v) = 8$. By Claim 1, $m_3(v)\le\lfloor\frac 34d_H(v)\rfloor = 5$. If $m_{6^+}(v) \geq 2$, then $w'(v)\geq 8-1.5\times5 = 0.5$ by (R3.3). Suppose that $m_{6^+}(v) \leq 1$. If $m_3(v)\leq 4$, then $w'(v)\geq 8-1.5 \times 4-0.5\times3-0.1\times3 = 0.2$ by (R1)-(R4). Assume that $m_3(v) = 5$. If $d_H(f_0) = d_H(f_1) = d_H(f_2) = 3$, then $d_H(f_3)\geq 7$ and $d_H(f_6)\geq7$ by Claim 3, a contradiction. If $d_H(f_0) = d_H(f_1) = 3$, then $d_H(f_2)\geq 6$ and $d_H(f_6)\geq6$ by Claim 2, a contradiction.

Case 5. $d_H(v) = 8$.

Note that $w(v) = 10$. By Claim 1, $m_3(v)\le\lfloor\frac 34d_H(v)\rfloor = 6$. If $m_{6^+}(v) \geq 2$, then $w'(v)\geq 10-1.5\times6 = 1$ by (R3.3). Suppose that $m_{6^+}(v) \leq 1$. If $m_3(v)\leq 5$, then $w'(v)\geq 10-1.5 \times 5-0.5\times3-0.1\times3 = 0.7$ by (R1)-(R4). Assume that $m_3(v) = 6$. If $d_H(f_0) = d_H(f_1) = d_H(f_2) = 3$, then $d_H(f_3)\geq 7$ and $d_H(f_7)\geq7$ by Claim 3, a contradiction. If $d_H(f_0) = d_H(f_1) = 3$, then $d_H(f_2)\geq 6$ and $d_H(f_7)\geq6$ by Claim 2, a contradiction.

Case 6. $d_H(v)\geq 9$.

We have the following estimation:

This completes the proof of Lemma 5.

Lemma 6. $w'(f^\Delta)+\Sigma_{x\in V(T)}w'(x)\geq -11.6$.

Proof. By (R0), $w'(f^\Delta)\geq 3-6 = -3$. Let $v \in V(T)$. First assume that $d_H(v) = 2$. Note that $v$ is not be incident with any $3$-face or $4$-face for otherwise we can find another separating $3$-cycle with fewer internal vertices than $T$, which is impossible. Consequently, $w'(v)\geq 2\times2-6-0.2 = -2.2$ by (R0). Next assume that $d_H(v) = 3$. If $m_3(v) = 3$, then $w'(v)\geq 2\times3-6-1.5\times2 = -3$ by (R0). Otherwise, $w'(v)\geq 2\times3-6-2-0.6 = -2.6$ by (R0). Now assume that $d_H(v) = 4$. If $m_3(v) = 4$, then $w'(v)\geq 2\times4-6-1.5\times3 = -2.5$ by (R0). Otherwise, $w'(v)\geq 2\times4-6-2\times2-0.6 = -2.6$ by (R0). Finally assume that $d_H(v)\geq 5$. It follows that $m_3(v)\leq d_H(v)-1$, otherwise a configuration $C_5 \bowtie C_6$ at the vertex $v$ will be found. Thus, $w'(v)\geq 2\times d_H(v)-6-2\times (d_H(v)-2)-0.6 = -2.6$ by (R0).

If $f^\Delta$ is incident with a vertex that is not a $3_3$-vertex, then $w'(f^\Delta)+\Sigma_{x\in V(T)}w'(x)\geq -3-3-3-2.6 = -11.6$. Otherwise there is an internal $3$-vertex, which is impossible.

Recall that the sum of the initial weights of vertices and faces of $H$ equals $-12$. From Lemma 4-6, the corresponding sum of the new weights satisfies $\Sigma_{x\in V^0(H)\cup F^0(H)}w'(x) + w'(f^\Delta)+\Sigma_{x\in V(T)}w'(x)\geq -11.6$. Since the total weight of $H$ has been preserved during the discharging process, this is a contradiction to our initial assumptions that $G$, the minimal counterexample to Theorem 1, exists. This completes the proof of Theorem 1.

3.   Conjectures

In this paper, we show that the list vertex arboricity of every planar graph without a 5-cycle intersecting with a 6-cycle is at most 2. This implies directly that if $G$ is a planar graph without 5-cycles or without 6-cycles, then $a_l(G)\le 2$. We like to conclude this paper by raising the following conjectures:

Conjecture 1. Every planar graph $G$ without intersecting $4$-cycles has $a_l(G)\le 2$.

Conjecture 2. If $G$ is a planar graph without a vertex lying on all cycles of length from $3$ to $7$, then $a_l(G)\le 2$.

Acknowledgments

The authors would like to thank the anonymous referees for their valuable comments and suggestions, which led to considerable improvement of the article.

The research is supported by the Natural Science Foundation of China (Grant No. 12031018).

Conflict of interest

The authors declare that they have no competing interests.