Acyclic Chromatic Index of 1-Planar Graphs

Abstract

The acyclic chromatic index ${\chi }_a^{\prime }(G)$ of a graph G is the smallest k for which G is a proper edge colorable using k colors. A 1-planar graph is a graph that can be drawn in plane such that every edge is crossed by at most one other edge. In this paper, we prove that every 1-planar graph G has ${\chi }_a^{\prime }(G)\le \Delta +36$, where $\Delta$ denotes the maximum degree of G. This strengthens a result that if G is a triangle-free 1-planar graph, then ${\chi }_a^{\prime }(G)\le \Delta +16$.

1. Introduction

Suppose that G is a simple graph with vertex set $V(G)$, edge set $E(G)$, minimum degree $\delta (G)$, and maximum degree $\Delta (G)$ (for short, $\Delta$). A proper edge-k-coloring of G is a function $\varphi$ from $E(G)$ to the color set $\{1,2,\dots ,k\}$ such that any two adjacent edges receive distinct colors. We say that $\varphi$ is acyclic if there do not exist bichromatic cycles in G. Let ${\chi }^{\prime }(G)$ and ${\chi }_a^{\prime }(G)$ denote the chromatic index and acyclic chromatic index of G, respectively, which are defined as the smallest k such that G is edge-k-colorable and acyclic edge-k-colorable.

On the one hand, it is evident that ${\chi }_a^{\prime }(G)\ge {\chi }^{\prime }(G)\ge \Delta$ for any graph G. On the other hand, Alon et al. [1] raised the following conjecture.

Conjecture 1.

For any graph G, ${\chi }_a^{\prime }(G)\le \Delta +2$.

Alon, McDiarmid, and Reed [2] used probabilistic analysis to show that ${\chi }_a^{\prime }(G)\le 64\Delta$ for any graph G. After then, this bound has been improved to that ${\chi }_a^{\prime }(G)\le 16\Delta$ in [3], that ${\chi }_a^{\prime }(G)\le \lceil 9.62(\Delta -1)\rceil$ in [4], that ${\chi }_a^{\prime }(G)\le 4\Delta -4$ in [5], and that ${\chi }_a^{\prime }(G)\le \lceil 3.74(\Delta -1)\rceil +1$ in [6]. The acyclic edge coloring of some classical classes of graphs has been extensively investigated, including subcubic graphs [7,8,9], outerplanar graphs [10,11], and 2-degenerate graphs [12].

Suppose that G is a planar graph. Let $C_n$ denote a cycle of length n. In 2011, Basavaraju et al. [13] showed that ${\chi }_a^{\prime }(G)\le \Delta +12$. This bound was later improved to that ${\chi }_a^{\prime }(G)\le \Delta +7$ by Wang, Shu, and Wang [14] and furthermore to that ${\chi }_a^{\prime }(G)\le \Delta +6$ by Wang and Zhang [15].

A graph G is called 1-planar if it admits a drawing in the plane such that every edge has at most one crossing. Ringle [16] first introduced this concept in the study of the problem of simultaneously coloring the vertices and faces of plane graphs. Zhang, Liu, and Wu [17] proved that if G is a 1-planar graph, then ${\chi }_a^{\prime }(G)\le max\{2\Delta -2,\Delta +83\}$, which implies a upper bound that ${\chi }_a^{\prime }(G)\le 2\Delta -2$ for any 1-planar graph G with $\Delta \ge 85$. Song and Miao [18] showed that every $C_3$-free 1-planar graph G has ${\chi }_a^{\prime }(G)\le \Delta +22$. Chen, Wang, and Zhang [19] improved this result by replacing the constant 22 with 16.

In this paper, we will prove that every 1-planar graph G has ${\chi }_a^{\prime }(G)\le \Delta +36$, which improved the previously stated results on the acyclic edge coloring of 1-planar graphs.

2. Structure of 1-Planar Graphs

Given a plane graph, we denote its face set by $F(G)$. For a face $f\in F(G)$, we use $\partial (f)$ to denote the boundary of f, and write $f=[u_1{u}_2\cdots u_n]$ if $u_1,u_2,\dots ,u_n$ are the vertices of $\partial (f)$. Note that $\partial (f)$ forms a cycle once G is 2-connected. A vertex v of degree k (at most k, at least k, respectively) is called a k-vertex ($k^{-}$-vertex, $k^{+}$-vertex, respectively). The similar definition can be given for a face f. A cycle C in a plane graph G is separating if there exists at least one vertex in its interior and exterior, respectively. Let $V_{\mathrm{int}}(C)$ denote the set of vertices in G that lie interior to C.

To investigate the $L(p,q)$-labeling of planar graphs, Heuvel and McGuinness [20] established the following structural theorem on planar graphs.

Theorem 1

(Heuvel and McGuinness [20]). Every planar graph G contains a k-vertex v with neighbors $v_1,v_2,\dots ,v_k$ with $d_G(v_1)\le d_G(v_2)\le \cdots \le d_G(v_k)$ such that one of the following holds:

(A1) $k\le 2$;

(A2) $k=3$ with $d_G(v_1)\le 11$;

(A3) $k=4$ with $d_G(v_1)\le 7$ and $d_G(v_2)\le 11$;

(A4) $k=5$ with $d_G(v_1)\le 6$, $d_G(v_2)\le 7$, and $d_G(v_3)\le 11$.

Let G be a 1-planar graph. Now we shall define a new graph $G^{\times }$ as follows. Let $Z(G)$ denote the set of crossings in G. For each crossing $z\in Z(G)$, there is a unique pair of crossing edges $x_1{x}_2,y_1{y}_2\in E(G)$. Let

$V(G^{\times })=V(G)\cup Z(G)$,     $E(G^{\times })=E_0(G)\cup E_1(G),$

where $E_0(G)$ is the set of non-crossed edges in G and

$E_1(G)=\{xz,zy|xy\in E(G)\setminus E_0(G)$ and z is a crossing on $xy\}.$

When discussing the graph $G^{\times }$, we will call the vertices in $V(G)$ true vertices and the vertices in $Z(G)$false vertices. A k-vertex which is true will be called a true k-vertex. Note that $d_{G^{\times }}(v)=d_G(v)$ for each $v\in V(G)$, and $d_{G^{\times }}(v)=4$ for each $v\in Z(G)$. Let $n_{\times }(v)$ be the number of false vertices adjacent to a vertex v.

Let $G_T^{\times }$ be a triangulation of $G^{\times }$, which can be obtained by performing the following steps.

• For each $z\in Z(G)$ label the unique pair of crossing edges $x_1{x}_2$ and $y_1{y}_2$. For each $i\in \{1,2\}$ and $j\in \{1,2\}$, if $z,x_i,y_j$ does not form a 3-face in $G^{\times }$, then add an edge $x_i{y}_j$ so that $\left[z{x}_i{y}_j\right]$ is a 3-face in the new graph.
• If needed, add additional edges so that the new graph is a triangulation.

Note that the newly added edges in step 1 and existing edges may form 2-cycles but that no 2-face is created. It is easy to see that $G_T^{\times }$ is a plane triangulation without loops and 2-faces, but it may contain a cycle of length 2. However, every 2-cycle is a separating cycle of $G_T^{\times }$. A face f of $G_T^{\times }$ will be called false if f is incident to at least one false vertex, and normal otherwise.

The following theorem is a generalization of Theorem 1 for 1-planar graphs. This theorem provides additional insight into the detailed structure of 1-planar graphs.

Theorem 2.

Every 1-planar graph G contains a k-vertex v with neighbors $v_1,v_2,\dots ,v_k$ with $d_G(v_1)\le d_G(v_2)\le \cdots \le d_G(v_k)$ such that one of the following conditions holds:

(C1) $k\le 2$;

(C2) $k=3$ with $d_G(v_1)\le 23$;

(C3) $k=4$ with $d_G(v_1)\le 14$ and $d_G(v_2)\le 23$;

(C4) $k=5$ with $d_G(v_1)\le 10$, $d_G(v_2)\le 14$, and $d_G(v_3)\le 23$;

(C5) $k=6$ with $d_G(v_1)\le 9$, $d_G(v_2)\le 10$, $d_G(v_3)\le 14$, and $d_G(v_4)\le 23$;

(C6) $k=7$ with $d_G(v_1)\le 7$, $d_G(v_2)\le 9$, $d_G(v_3)\le 10$, $d_G(v_4)\le 14$, and $d_G(v_5)\le 23$.

Proof. 

Assume that Theorem 2 is false. Let G be a connected counterexample. Obviously, if G is a 1-planar counterexample graph and $u,v$ are a pair of non-adjacent vertices of G and $G^{\prime }=G+uv$ is still a 1-planar graph, then $G^{\prime }$ is still a counterexample of Theorem 2. This implies that we may assume G is 2-connected. Let $G_T^{\times }$ be a plane triangulation of G obtained by the previous operations. It is possible that there are separating 2-cycles in $G_T^{\times }$. If $G_T^{\times }$ has no 2-cycles, let $H=G_T^{\times }$. Otherwise let C be a separating 2-cycle of $G_T^{\times }$ such that the number of interior vertices of C, $|V_{\mathrm{int}}(C)|$, is minimum. Let $H=G[V(C)\cup V_{\mathrm{int}}(C)]$. Clearly, H does not contain any 2-cycles other than C. If H contains a 2-cycle, let $f_0$ denote the outer face of H, and write $F^0(H)=F(H)\setminus \left\{f_0\right\}$ and $V^0(H)=V(H)\setminus V(C)$. Otherwise let $F^0(H)=F(H)$ and $V^0(H)=V(H)$. Vertices in $V^0(H)$ will be called interior vertices and faces in $F^0(H)$ will be called interior faces. For $v\in V^0(H)$, it holds obviously that $d_H(v)=d_{G_T^{\times }}(v)$. We have the following observation. □

Observation 1.

(1) No two false vertices are adjacent in H.

(2) Let $v\in V(H)$ be a true k-vertex. Then $n_{\times }(v)=0$ if $k=3$; $n_{\times }(v)\le 1$ if $k=4$, and $n_{\times }(v)\le \lfloor \frac{k}{2}\rfloor$ if $k\ge 5$.

Two true vertices which are adjacent in H will be referred to as direct neighbors. Suppose that $x\in V^0(H)$ is a false vertex which is incident to four 3-faces $\left[xuy\right],\left[xyv\right],\left[xvz\right]$, and $\left[xzu\right]$ in clockwise order. We say that u is an indirect neighbor of v and both y and z are type 1 direct neighbors of v. Moreover, if w is a true neighbor of v incident to two normal 3-faces, then w is called a type 2 direct neighbor of v. The related configuration for a vertex v is depicted in Figure 1, where $f_1$ and $f_2$ are normal 3-faces of H. In all figures, we use • to denote a vertex that has no edges incident to it other than those shown in the graph, ∘ to denote a vertex that may have edges connected to other vertices that are not in the graph, and ⊗ to denote false vertices.

It is not hard to see that $n_{\times }(v)$ is equal to the number of indirect neighbors of v in H. A vertex v of H is called small if $d_H(v)\le 7$, and big otherwise. The following identity is deduced from applying Euler’s formula on H.

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

(1)

We now define a weight function on H. Let $w(x)=d_H(x)-4$ for each $x\in V(H)\cup F(H)$. Next, we shall redistribute the weight among vertices and faces of H and keep the sum of all weights unchanged. We will show that the resultant weight function $w^{\prime }$ satisfies the following two conditions.

(I) $w^{\prime }(x)\ge 0$ for all $x\in V^0(H)\cup F^0(H)$;

(II) if H contains a 2-cycle, then $w^{\prime }(f_0)+{\sum }_{x\in V(C)}{w}^{\prime }(x)\ge -7$.

This leads to the following inequalities.

$$-7\le \sum _{x\in V(H)\cup F(H)}{w}^{\prime }(x)=\sum _{x\in V(H)\cup F(H)}w(x)=-8$$

(2)

This contradiction implies there is no such counterexample exists and so the theorem holds.

The discharging rules are defined as follows. Note that the rules (R0) and (R1.2.1) only apply if H contains a 2-cycle C. For a vertex $v\in V^0(H)$, we use $v_1,v_2,\dots ,v_{k_v}$ to denote the internal (direct or indirect) neighbors of v with $d_H(v_1)\le d_H(v_2)\le \cdots \le d_H(v_{k_v})$, where $k_v\le d_H(v)$.

(R0) Every vertex $v\in V(C)$ sends $\frac{1}{2}$ to each internal (direct or indirect) neighbor and to each incident 3-face.

(R1) Let $f=\left[u_1{u}_2{u}_3\right]$ be a 3-face.

(R1.1) If f is false, then f gets $\frac{1}{2}$ from each of its incident internal true vertices.

(R1.2) If f is normal, then we carry out the following two sub-rules:

(R1.2.1) If $|V(f)\cup V(C)|=1$ and let $u_1\in V(C)$, then at least one of $u_2$ and $u_3$, say $u_2$, is big, then f gets $\frac{1}{2}$ from $u_2$. Otherwise, f gets $\frac{1}{4}$ from each of $u_2$ and $u_3$.

(R1.2.2) If $|V(f)\cup V(C)|=0$, then we may assume that $d_H(u_1)\le d_H(u_2)\le$ $d_H(u_3)$.

• If $d_H(u_1)\ge 8$, or $d_H(u_3)\le 7$, then f gets $\frac{1}{3}$ from each of $u_1,u_2,u_3$.
• If $d_H(u_1)\le 7$ and $d_H(u_2)\ge 8$, then f gets $\frac{1}{2}$ from each of $u_2,u_3$.
• If $d_H(u_2)\le 7$ and $d_H(u_3)\ge 8$, then f gets $\frac{1}{2}$ from $u_3$ and $\frac{1}{4}$ from each of $u_1,u_2$.

(R2) If $d_H(v)=3$, then v gets $\frac{1}{3}$ from each of $v_1,v_2,\dots ,v_{k_v}$.

(R3)$d_H(v)=4$.

(R3.1) If $d_H(v_1)\ge 15$, then v gets $\frac{1}{4}$ from each of $v_1,v_2,\dots ,v_{k_v}$.

(R3.2) Suppose $d_H(v_1)\le 14$ and $d_H(v_2)\ge 24$. If $d_H(v_1)\le 7$, $n_{\times }(v)=1$, and $v_1$ is a type 2 direct neighbor of v, then v gets $\frac{1}{2}$ from each of $v_2,v_3,\dots ,v_{k_v}$. Otherwise v gets $\frac{5}{12}$ from each of $v_2,v_3,\dots ,v_{k_v}$, see Figure 2a.

(R4)$d_H(v)=5$.

(R4.1) If $d_H(v_1)\ge 11$, then v gets $\frac{1}{5}$ from each of $v_1,v_2,\dots ,v_{k_v}$.

(R4.2) If $d_H(v_1)\le 10$ and $d_H(v_2)\ge 15$, v gets $\frac{5}{16}$ from each of $v_2,v_3,\dots ,v_{k_v}$ when $n_{\times }(v)=2$. Otherwise v gets $\frac{1}{8}$ from each of $v_2,v_3,\dots ,v_{k_v}$.

(R4.3)$d_H(v_1)\le 10$, $d_H(v_2)\le 14$, and $d_H(v_3)\ge 24$.

(R4.3.1) If $n_{\times }(v)\le 1$, then v gets $\frac{5}{18}$ from each of $v_3,\dots ,v_{k_v}$.

(R4.3.2) If $n_{\times }(v)=2$, then v gets $\frac{4}{9}$ from each of $v_3,\dots ,v_{k_v}$ when $d_H(v_1),d_H(v_2)\le 7$ and $\left[v{v}_1{v}_2\right]$ is a normal 3-face. Otherwise v gets $\frac{5}{12}$ from each of $v_3,\dots ,v_{k_v}$, see Figure 2b.

(R5)$d_H(v)=6$.

(R5.1) If $d_H(v_1)\ge 10$, then v gets $\frac{1}{6}$ from each of $v_1,v_2,\dots ,v_{k_v}$.

(R5.2) If $d_H(v_1)\le 9$ and $d_H(v_2)\ge 11$, then v gets $\frac{1}{5}$ from each of $v_2,v_3,\dots ,v_{k_v}$.

(R5.3) If $d_H(v_1)\le 9$, $d_H(v_2)\le 10$, and $d_H(v_3)\ge 15$, then v gets $\frac{1}{4}$ from each of $v_3,v_4,\dots ,v_{k_v}$ when $n_{\times }(v)=3$. Otherwise v gets $\frac{7}{48}$ from each of $v_3,v_4,\dots ,v_{k_v}$.

(R5.4) If $d_H(v_1)\le 9$, $d_H(v_2)\le 10$, $d_H(v_3)\le 14$ and $d_G(v_4)\ge 24$, then v gets $\frac{1}{3}$ from each of $v_4,\dots ,v_{k_v}$ when $n_{\times }(v)=3$. Otherwise v gets $\frac{2}{9}$ from each of $v_4,\dots ,v_{k_v}$.

(R6)$d_H(v)=7$. Then v gets

$\frac{1}{24}$ from each of $v_2,v_3,\dots ,v_{k_v}$ if $d_H(v_2)\ge 10$;

$\frac{1}{15}$ from each of $v_3,v_4,\dots ,v_{k_v}$ if $d_H(v_2)\le 9$ and $d_H(v_3)\ge 11$;

$\frac{1}{12}$ from each of $v_4,v_5,\dots ,v_{k_v}$ if $d_H(v_2)\le 9$, $d_H(v_3)\ge 10$ and $d_H(v_4)\ge 15$;

$\frac{1}{9}$ from each of $v_5,v_6,\dots ,v_{k_v}$ if $d_H(v_2)\le 9$, $d_H(v_3)\ge 10$, $d_H(v_4)\le 14$ and $d_H(v_5)\ge 24$.

Let $w^{\prime }$ denote the resultant weight function after (R0)–(R6) are performed on H. First, we prove that $w^{\prime }(x)\ge 0$ for all $x\in V^0(H)\cup F^0(H)$.

First we consider $x=f\in F^0(H)$. Let $f=\left[x_1{x}_2{x}_3\right]$ be a 3-face. Note that if H has a 2-cycle C, the vertices in $V(C)$ are true vertices. If f is false, then f is incident to exactly two true vertices. By (R0) and (R1.1), $w^{\prime }(f)\ge -1+2\times \frac{1}{2}=0$. We may assume that f is normal. If $|V(f)\cap V(C)|=2$, then $w^{\prime }(f)\ge -1+2\times \frac{1}{2}=0$ by (R0). If $|V(f)\cap V(C)|=1$, say $x_1\in V(C)$. If at least one of $x_2,x_3$ is big, then $w^{\prime }(f)\ge -1+2\times \frac{1}{2}=0$ by (R0) and (R1.2.1). Otherwise, $w^{\prime }(f)\ge -1+\frac{1}{2}+2\times \frac{1}{4}=0$ by (R0) and (R1.2.1). The only case to be considered is $|V(f)\cap V(C)|=0$. Without loss of generality, we may assume that $d_H(x_1)\le d_H(x_2)\le d_H(x_3)$. If $d_H(x_1)\ge 8$, or $d_H(x_3)\le 7$, then $w^{\prime }(f)\ge -1+3\times \frac{1}{3}=0$ by (R1.2.2). If $d_H(x_2)\ge 8$, then $w^{\prime }(f)\ge -1+2\times \frac{1}{2}=0$ by (R1.2.2). If $d_H(x_3)\ge 8$, then $w^{\prime }(f)\ge -1+\frac{1}{2}+2\times \frac{1}{4}=0$ by (R1.2.2).

Secondly, we shall consider the case that $x=v$ is an interior vertex of H. Let $v_1,v_2,\dots ,v_{k_v}$ denote the interior (direct or indirect) neighbors of v with $d_H(v_1)\le d_H(v_2)\le \cdots \le d_H(v_{k_v})$, where $d_H(v)-2\le k_v\le d_H(v)$. Since (C1) is not satisfied, $d_H(v)\ge 3$. The proof is divided into the following cases based on $d_H(v)$.

Case 1 $d_H(v)=3$.

Then $w(v)=-1$. By Observation 1(2), $n_{\times }(v)=0$. Since (C2) is not satisfied, $v_1,v_2,\dots ,v_{k_v}$ are true and big vertices. By (R0) and (R2), $w^{\prime }(v)\ge -1+3\times \frac{1}{3}=0$.

Case 2 $d_H(v)=4$.

Clearly, $w(v)=0$. If v is false, then $w^{\prime }(v)=w(v)=0$. If v is true, then $n_{\times }(v)\le 1$ by Observation 1(2). Hence v is incident to at most two false 3-faces. If $d_H(v_1)\ge 15$, then $w^{\prime }(v)\ge 0-2\times \frac{1}{2}+4\times \frac{1}{4}=0$ by (R0), (R1), and (R3.1). We may assume that $d_H(v_1)\le 14$. Since (C3) is not satisfied, $d_H(v_2)\ge 24$. If $n_{\times }(v)=1$, $d_H(v_1)\le 7$, and $v_1$ is a type 2 direct neighbor of v, then $w^{\prime }(v)\ge 0-2\times \frac{1}{2}-2\times \frac{1}{4}+3\times \frac{1}{2}=0$ by (R0), (R1), and (R3.2). Otherwise, $w^{\prime }(v)\ge 0-2\times \frac{1}{2}-\frac{1}{4}+3\times \frac{5}{12}=0$ by (R0), (R1), and (R3.2).

Case 3 $d_H(v)=5$.

Clearly, $w(v)=1$. By Observation 1(2), $n_{\times }(v)\le 2$ and it follows that v is incident to at most four false 3-faces. If $d_H(v_1)\ge 11$, then $w^{\prime }(v)\ge 1-4\times \frac{1}{2}+5\times \frac{1}{5}=0$ by (R0), (R1), and (R4.1). Otherwise, we may assume that $d_H(v_1)\le 10$.

Suppose that $d_H(v_2)\ge 15$. If $n_{\times }(v)=2$, then $w^{\prime }(v)\ge 1-4\times \frac{1}{2}-\frac{1}{4}+4\times \frac{5}{16}=0$ by (R0), (R1), and (R4.2). Otherwise, $w^{\prime }(v)\ge 1-2\times \frac{1}{2}-2\times \frac{1}{4}+4\times \frac{1}{8}=0$ by (R0), (R1), and (R4.2). Suppose that $d_H(v_2)\le 14$. Since (C4) is not satisfied, we must have $d_H(v_3)\ge 24$. If $n_{\times }(v)\le 1$, then $w^{\prime }(v)\ge 1-2\times \frac{1}{2}-\frac{1}{3}-2\times \frac{1}{4}+3\times \frac{5}{18}=0$ by (R0), (R1), and (R4.3.1). Now consider the case where $n_{\times }(v)=2$. If $d_H(v_1),d_H(v_2)\le 7$ and $\left[v{v}_1{v}_2\right]$ is a normal 3-face, then $w^{\prime }(v)\ge 1-4\times \frac{1}{2}-\frac{1}{3}+3\times \frac{4}{9}=0$ by (R0), (R1), and (R4.3.2). Otherwise, $w^{\prime }(v)\ge w(v)-4\times \frac{1}{2}-\frac{1}{4}+3\times \frac{5}{12}=0$ by (R0), (R1), and (R4.3.2).

Case 4 $d_H(v)=6$.

Clearly, $w(v)=2$. By Observation 1(2), $n_{\times }(v)\le 3$. It follows that v is incident to at most six false 3-faces. If $d_H(v_1)\ge 10$, then $w^{\prime }(v)\ge 2-6\times \frac{1}{2}+6\times \frac{1}{6}=0$ by (R0), (R1), and (R5.1). So we may assume that $d_H(v_1)\le 9$. If $d_H(v_2)\ge 11$, then $w^{\prime }(v)\ge 2-6\times \frac{1}{2}+5\times \frac{1}{5}=0$ by (R0), (R1), and (R5.2). Otherwise, $d_H(v_2)\le 10$.

Suppose that $d_H(v_3)\ge 15$. If $n_{\times }(v)=3$, then $w^{\prime }(v)\ge 2-6\times \frac{1}{2}+4\times \frac{1}{4}=0$ by (R0), (R1), and (R5.3). Otherwise, we assume that $n_{\times }(v)\le 2$. It follows that v is incident to at most four false 3-faces. Consequently, $w^{\prime }(v)\ge 2-4\times \frac{1}{2}-\frac{1}{3}-\frac{1}{4}+4\times \frac{7}{48}=0$ by (R0), (R1), and (R5.3).

Suppose that $d_H(v_3)\le 14$. Since (C5) is not satisfied, we have $d_H(v_4)\ge 24$. If $n_{\times }(v)=3$, then $w^{\prime }(v)\ge 2-6\times \frac{1}{2}+3\times \frac{1}{3}=0$ by (R0), (R1), and (R5.4). Now we consider the case that $n_{\times }(v)\le 2$. It follows that v is incident to at most four false 3-faces. Consequently, $w^{\prime }(v)\ge 2-4\times \frac{1}{2}-2\times \frac{1}{3}+3\times \frac{2}{9}=0$ by (R0), (R1), and (R5.4).

Case 5 $d_H(v)=7$.

Clearly, $w(v)=3$. By Observation 1(2), $n_{\times }(v)\le 3$. It follows that v is incident to at most six false 3-faces. If $d_H(v_1)\ge 8$, then $w^{\prime }(v)\ge 3-6\times \frac{1}{2}=0$ by (R0), (R1), and (R6). Suppose that $d_H(v_1)\le 7$. If $d_H(v_2)\ge 10$, then $w^{\prime }(v)\ge 3-6\times \frac{1}{2}-\frac{1}{4}+6\times \frac{1}{24}=0$ by (R0), (R1), and (R6). Now we may assume that $d_H(v_2)\le 9$. If $d_H(v_3)\ge 11$, then $w^{\prime }(v)\ge 3-6\times \frac{1}{2}-\frac{1}{3}+5\times \frac{1}{15}=0$ by (R0), (R1), and (R6). Suppose that $d_H(v_3)\le 10$. If $d_H(v_4)\ge 15$, then $w^{\prime }(v)\ge 3-6\times \frac{1}{2}-\frac{1}{3}+4\times \frac{1}{12}=0$ by (R0), (R1), and (R6). Suppose that $d_H(v_4)\le 14$. Since (C6) is not satisfied, $d_H(v_5)\ge 24$. Consequently, $w^{\prime }(v)\ge 3-6\times \frac{1}{2}-\frac{1}{3}+3\times \frac{1}{9}=0$ by (R0), (R1), and (R6).

Case 6 $8\le d_H(v)\le 9$.

Notice that v sends nothing to any small vertex. Thus, $w^{\prime }(v)\ge w(v)-\frac{1}{2}{d}_H(v)=d_H(v)-4-\frac{1}{2}{d}_H(v)=\frac{1}{2}{d}_H(v)-4\ge 0$.

From now on, we assume that $d_H(v)\ge 10$. Let $y_0,x_0^1,x_0^2,$$\dots ,x_0^{m_0},y_1,x_1^1,x_1^2,$$\dots ,x_1^{m_1},y_2,$$\dots ,$ $y_{k-1},x_{k-1}^1$, $x_{k-1}^2,\dots ,x_{k-1}^{m_{k-1}}$ denote the (direct or indirect) neighbors of v in H in clockwise order, where for every $i=0,1,\dots ,k-1$ and $j=1,2,\dots ,m_i$, $x_i^j$ is a small vertex, and for each $i=0,1,\dots ,k-1$ either $y_i\in V(C)$ or $y_i$ is an interior big vertex. Define

$X_i=\left\{x_i^j\right|0\le j\le m_i\}$,

$X={\bigcup }_{i=1}^{k-1}{X}_i$,

$Y=\left\{y_i\right|0\le i\le k-1\}$.

By (R1)–(R6), each v transfers at most $\frac{1}{2}$ to every (direct or indirect) neighbor and to every incident 3-face. If $\left|Y\right|\ge 8$, then $w^{\prime }(v)\ge w(v)-\frac{1}{2}{d}_H(v)-\frac{1}{2}(d_H(v)-\left|Y\right|)=\frac{1}{2}(|Y|-8)\ge 0$. Hence, we will assume that $0\le \left|Y\right|\le 7$ in the remaining cases.

For $x,y\in V(H)$, let $\tau (x\to y)$ denote the weight transferring from x to y according to (R0) and (R2)–(R6). Let $\sigma (v)$ denote the total weight that an internal vertex v sends to all its (direct or indirect) neighboring vertices according to rules (R2)–(R6). Moreover, let ${\eta }_0(v)$ denote the number of vertices in X which receive no charge from v.

Case 7 $d_H(v)=10$.

It is easy to see from (R1)–(R6) that v only transfers $\frac{1}{6}$ to each adjacent 6-vertex or 7-vertex. Hence $w^{\prime }(v)\ge 10-4-\frac{1}{2}\times 10-\frac{1}{6}\times (10-|Y|-{\eta }_0(v))$ by (R1), (R5), and (R6). It is sufficient to show $\left|Y\right|+{\eta }_0(v)\ge 4$. Noting that for $x\in X$ if $|N(x)\cap X|\ge 2$, then by (R5) and (R6), v transfers no charge to x. It follows that ${\eta }_0(v)\ge 10-3\left|Y\right|$. Therefore $\left|Y\right|+{\eta }_0(v)\ge 4$, a contradiction.

Case 8 $11\le d_H(v)\le 14$.

By inspecting our discharging rules, we find that v needs only to transfer at most $\frac{1}{5}$ to each adjacent 5-vertex, 6-vertex, or 7-vertex. Thus,

$$\begin{array}{ccc}\hfill w^{\prime }(v)& \ge & d_H(v)-4-\frac{1}{2}{d}_H(v)-\frac{1}{5}(d_H(v)-\left|Y\right|)\hfill \\ & =& \frac{1}{10}(3d_H(v)+2\left|Y\right|-40).\hfill \end{array}$$

If $\left|Y\right|\ge 4$, we can immediately conclude that $w^{\prime }(v)\ge 0$. Since (C2)–(C4) are not satisfied, for each $i=0,1,\dots ,k-1$, every vertex $x\in X_i\setminus \{x_i^1,x_i^{m_i}\}$ is a 6-vertex or a 7-vertex, therefore $\tau (v\to x)\le \frac{1}{15}$ by (R5) and (R6). Thus,

$$\begin{array}{ccc}\hfill w^{\prime }(v)& \ge & d_H(v)-4-\frac{1}{2}{d}_H(v)-\frac{1}{5}(d_H(v)-\left|Y\right|-(d_H(v)-3\left|Y\right|))-\frac{1}{15}(d_H(v)-3\left|Y\right|)\hfill \\ & =& \frac{1}{30}(13d_H(v)-6\left|Y\right|-120)\hfill \\ & \ge & \frac{1}{30}(13d_H(v)-138)>0.\hfill \end{array}$$

Case 9 $15\le d_H(v)\le 23$.

Since (C2)–(C4) are not satisfied, for each $i=0,1,\dots ,k-1$, every vertex $x\in X_i\setminus \{x_i^1,x_i^{m_i}\}$ is a 6-vertex or a 7-vertex, therefore $\tau (v\to x)\le \frac{1}{4}$ by (R5) and (R6). Moreover, $\tau (v\to x_i^1)\le \frac{5}{16}$ and $\tau (v\to x_i^{m_i})\le \frac{5}{16}$ by (R3)–(R6). Thus,

$$\begin{array}{ccc}\hfill w^{\prime }(v)& \ge & d_H(v)-4-\frac{1}{2}{d}_H(v)-\frac{5}{16}(2|Y|)-\frac{1}{4}(d_H(v)-3\left|Y\right|)\hfill \\ & =& \frac{1}{4}{d}_H(v)-4+\frac{1}{8}\left|Y\right|.\hfill \end{array}$$

If $d_H(v)\ge 16$, or $d_H(v)=15$ and $\left|Y\right|\ge 2$, then it holds obviously that $w^{\prime }(v)\ge 0$. Suppose that $d_H(v)=15$ and $0\le \left|Y\right|\le 1$.

Claim 1.If $\tau (v\to x_0^j)=\frac{1}{4}$ for some index j with $3\le j\le m_0-2$, then $\tau (v\to x_0^{j-1})\le \frac{7}{48}$ and $\tau (v\to x_0^{j+1})\le \frac{7}{48}$.

Proof. 

Since $\tau (v\to x_0^j)=\frac{1}{4}$, it follows from (R5) and (R6) that $d_H(x_0^j)=6$ and $n_{\times }(x_0^j)=3$. By symmetry, it will be sufficient to show that $\tau (v\to x_0^{j-1})\le \frac{7}{48}$. Suppose $\tau (v\to x_0^{j-1})>\frac{7}{48}$. It follows from (R5) and (R6) that $d_H(x_0^{j-1})=6$ and $n_{\times }(x_0^{j-1})=3$.

If $x_0^j$ is a direct neighbor of v in H. Since $n_{\times }(x_0^j)=3$ and false vertices can not be adjacent, it must be the case that $x_0^j{x}_0^{j-2}$ and $x_0^j{x}_0^{j+2}$ are two crossing edges of G. So $x_0^j$ is adjacent to four small vertices $x_0^{j-2},x_0^{j-1},x_0^{j+1},x_0^{j+2}$, which implies that $x_0^j$ satisfies (C5), a contradiction. Otherwise, $x_0^j$ is an indirect neighbor of v in H and $n_{\times }(x_0^{j-1})=3$, it follows that $x_0^{j-1}$ must be a direct neighbor of v. Similar to the previous argument, $x_0^{j-1}$ is adjacent to at least three small vertices and v, contradicting (C5). This completes the proof of Claim 1. □

If $\left|Y\right|=0$, then Claim 1 implies that $w^{\prime }(v)\ge 15-4-\frac{1}{2}\times 15-\frac{1}{4}\times 7-\frac{7}{48}\times 8=\frac{7}{12}$.

If $\left|Y\right|=1$, then by Claim 1 and (R4), $w^{\prime }(v)\ge 15-4-\frac{1}{2}\times 15-\frac{5}{16}\times 2-\frac{1}{4}\times 6-\frac{7}{48}\times 6=\frac{1}{2}$.

Case 10 $d_H(v)\ge 24$.

We divide the proof into the following two subcases.

Subcase 10.1 $\left|Y\right|=0$.

Then X consists of all (direct or indirect) neighbors of v in H. Let $x\in X$. Then $3\le d_H(x)\le 7$. Since x is adjacent to at least two other vertices in X and neither (C2) nor (C3) holds, we must have $d_H(x)\ne 3,4$. Hence $5\le d_H(x)\le 7$.

Claim 2.Let $x_0,x_1,x_2,x_3,x_4$ be five consecutive vertices in X in clockwise order with $5\le d_H(x_i)\le 7$ for $i=1,3$. If $\tau (v\to x_2)\in \{\frac{4}{9},\frac{5}{12}\}$, then $\tau (v\to x_i)\le \frac{2}{9}$ for $i=1,3$.

Proof. 

Note that $x_2$ is (directly or indirectly) adjacent to $v,x_1,x_3$ in H. Since $\tau (v\to x_2)=\frac{4}{9}$ or $\tau (v\to x_2)=\frac{5}{12}$, then it is easy to verify that $d_H(x_2)=5$, $n_{\times }(x_2)=2$, by (R4)-(R6). Let $u_1,u_2$ be the neighbors of $x_2$ other than $v,x_1,x_3$ such that $v{x}_1{u}_1{u}_2{x}_{3}v$ is a 5-cycle in G. Since $x_1,x_3$ are small and (C4) is not satisfied, it follows that $d_H(u_1),d_H(u_2)\ge 8$ and thus $u_1,u_2\notin X$. This implies that $u_2\ne x_4$ and $u_1\ne x_0$.

If $\tau (v\to x_2)=\frac{4}{9}$, then (R4.3.2) shows that $\left[v{x}_1{x}_3\right]$ is a normal 3-face of H, but this is impossible because of the existence of $x_2$. Thus, $\tau (v\to x_2)=\frac{5}{12}$. By (R4.3.2), v is an indirect neighbor of $x_2$, and exactly one of $u_1,u_2$ is an indirect neighbor of $x_2$, say $u_2$. Thus, $\left[x_1{x}_2{u}_1\right]$ is a normal 3-face of H.

It is easy to see that $d_H(x_3)\ge 6$. If $d_H(x_3)=7$, then $\tau (v\to x_3)\le \frac{1}{9}$ by (R6). Suppose that $d_H(x_3)=6$. If $n_{\times }(x_3)\le 2$, then $\tau (v\to x_3)\le \frac{2}{9}$ by (R5) and we are done. Otherwise, $n_{\times }(x_3)=3$. This implies that there is a crossing edge between $u_2$ and v in G, therefore $u_2$ is an indirect neighbor of v. This is impossible because $u_2\notin X$. Thus, we always have that $\tau (v\to x_3)\le \frac{2}{9}$.

Now we consider the vertex $x_1$. If $d_H(x_1)=7$, then $\tau (v\to x_1)\le \frac{1}{9}$ by (R6). If $d_H(x_1)=6$, then $n_{\times }(x_1)\le 2$ since $x_1$ is incident to a formal 3-face $\left[x_1{x}_2{u}_1\right]$. By (R5), $\tau (v\to x_1)\le \frac{2}{9}$. If $d_H(x_1)=5$, then $x_1$ is adjacent to three small vertices $x_0,x_2,x_3$, thus $x_1$ satisfies (C4), a contradiction. Thus, we have $\tau (v\to x_1)\le \frac{2}{9}$. This completes the proof. □

Claim 3.Let $x_1,x_2$ be consecutive vertices in X. Then $\tau (v\to x_1)+\tau (v\to x_2)\le \frac{2}{3}$.

Proof. 

Since (C2) and (C3) are not satisfied, $5\le d_H(x_i)\le 7$ for $i=1,2$. If $\tau (v\to x_1)\in \{\frac{4}{9},\frac{5}{12}\}$, then $\tau (v\to x_2)\le \frac{2}{9}$ by Claim 2. Thus, $\tau (v\to x_1)+\tau (v\to x_2)\le \frac{4}{9}+\frac{2}{9}=\frac{2}{3}$. If $\tau (v\to x_2)\in \{\frac{4}{9},\frac{5}{12}\}$, a similar argument holds. This implies that v gives at most $\frac{1}{3}$ to each of $x_1$ and $x_2$ by (R4)-(R6), so that $\tau (v\to x_1)+\tau (v\to x_2)\le \frac{1}{3}+\frac{1}{3}=\frac{2}{3}$. □

Claim 4.$\sigma (v)\le \frac{1}{3}\left|X\right|$.

Proof. 

Let $p=\left|X\right|=d_H(v)$. If p is even, then X can be divided into $\frac{p}{2}$ 2-sets $\{x_0^1,x_0^2\},$$\{x_0^3,x_0^4\},\dots ,\{x_0^{p-1},x_0^p\}$. By Claim 3, $\sigma (v)\le \frac{2}{3}\times \frac{p}{2}=\frac{1}{3}p=\frac{1}{3}\left|X\right|$. If p is odd, then we may choose $x_0^p$ satisfying $\tau (v\to x_0^p)\le \frac{1}{3}$ by Claim 3. Now we can divide X into $\frac{p-1}{2}$ 2-sets $\{x_0^1,x_0^2\},\{x_0^3,x_0^4\},\dots ,\{x_0^{p-2},x_0^{p-1}\}$ and one single set $\left\{x_0^p\right\}$. By Claim 3, $\sigma (v)\le \frac{2}{3}\times \frac{p-1}{2}+\frac{1}{3}=\frac{1}{3}p=\frac{1}{3}\left|X\right|$. This proves Claim 4. □

Using (R2) and Claim 4, we have the following:

$$\begin{array}{ccc}\hfill w^{\prime }(v)& \ge & w(v)-\frac{1}{2}{d}_H(v)-\sigma (v)\hfill \\ & \ge & d_H(v)-4-\frac{1}{2}{d}_H(v)-\frac{1}{3}\left|X\right|\hfill \\ & =& \frac{1}{2}{d}_H(v)-4-\frac{1}{3}{d}_H(v)\hfill \\ & =& \frac{1}{6}(d_H(v)-24)\ge 0.\hfill \end{array}$$

Subcase 10.2 $\left|Y\right|\ge 1$.

For $i=0,1,\dots ,k-1$, we use $\sigma (y_i,y_{i+1})$ to denote the total weight that v sends to all the vertices in $X_i$, where $k=\left|Y\right|$ and indices are taken as modulo k.

Claim 5.$\sigma (y_i,y_{i+1})\le \frac{1}{3}(|X_i|+1)$ for $i=0,1,\dots ,k-1$.

Proof. 

By symmetry, it suffices to consider the case $i=0$, i.e., we are going to show that $\sigma (y_0,y_1)\le \frac{1}{3}(|X_0|+1).$ Note that if $\left|Y\right|=1$, then $y_0$ is identical to $y_1$.

If $|X_0|=0$, then $\sigma (y_0,y_1)=0<\frac{1}{3}$. If $|X_0|=1$, then $\sigma (y_0,y_1)\le \frac{1}{2}<\frac{2}{3}=\frac{1}{3}(1+1)$ by (R2)-(R6). If $|X_0|=2$, then $\sigma (y_0,y_1)\le 2\times \frac{1}{2}=\frac{1}{3}(2+1)$ by (R2)-(R6).

If $|X_0|=3$, let $X_0=\{x_0^1,x_0^2,x_0^3\}$. Since $d_H(v)\ge 24$, $y_1\ne y_0$, so $\left|Y\right|\ge 2$. If every vertex in $X_0$ gets at most $\frac{4}{9}$ from v, then $\sigma (y_0,y_1)\le 3\times \frac{4}{9}=\frac{1}{3}(|X_0|+1)$. Otherwise, (R2)-(R6) imply that at least one of $x_0^1$ and $x_0^3$ gets $\frac{1}{2}$ from v, say $\tau (v\to x_0^1)=\frac{1}{2}$. Now it follows easily that $y_0$ is an indirect neighbor of v, and $x_0^2$ is a type 2 direct neighbor of $x_0^1$. So $x_0^1{x}_0^2$ is a common edge of two normal 3-faces in H. Note that $d_H(x_0^2)\ge 5$ since (C2) and (C3) are not satisfied. If $6\le d_H(x_0^2)\le 7$, then $\tau (v\to x_0^2)\le \frac{1}{3}$ by (R5) and (R6). Otherwise, we may assume that $d_H(x_0^2)=5$. Since $x_0^2$ is incident to at least two normal 3-faces, it follows that $n_{\times }(x_0^2)\le 1$, and hence $\tau (v\to x_0^2)\le \frac{5}{18}$ by (R4). Thus, $\tau (v\to x_0^2)\le \frac{1}{3}$ holds. Consequently, $\sigma (y_0,y_1)\le 2\times \frac{1}{2}+\frac{1}{3}=\frac{4}{3}=\frac{1}{3}(|X_0|+1)$.

Now we may assume that $|X_0|\ge 4$. We will prove the following claim first. □

Claim 6.$\tau (v\to x_0^1)+\tau (v\to x_0^2)\le \frac{5}{6}$.

Proof. 

If each of $x_0^1$ and $x_0^2$ gets from v at most $\frac{5}{12}$, the result holds. Otherwise, by (R3) and (R4), we need only to discuss the following cases.

•$d_H(x_0^1)=4$ such that $\tau (v\to x_0^1)=\frac{1}{2}$. According to the previous analysis, we can similarly derive that $\tau (v\to x_0^2)\le \frac{1}{3}$. It yields that $\tau (v\to x_0^1)+\tau (v\to x_0^2)\le \frac{5}{6}$.

•$d_H(x_0^1)=5$ such that $\tau (v\to x_0^1)=\frac{4}{9}$. Note that $d_H(x_0^2)\ge 5$ by (C2) and (C3). If $\tau (v\to x_0^2)\le \frac{5}{6}-\frac{4}{9}=\frac{7}{18}$, we are done. Otherwise, (R4)-(R6) implies that $d_H(x_0^2)=5$, $n_{\times }(x_0^2)=2$, and $\tau (v\to x_0^2)\in \{\frac{5}{12},\frac{4}{9}\}$. Let $w_1$ and $w_2$ be the other (direct or indirect) neighbors of $x_0^1$ other than $v,y_0,x_0^2$ in H so that $v{y}_0{w}_1{w}_2{x}_0^{2}v$ forms a 5-cycle in G. Since $\tau (v\to x_0^1)=\frac{4}{9}$, we can conclude that $n_{\times }(x_0^1)=2$; $v,w_1$ are indirect neighbors of $x_0^1$, $d_H(w_2)\le 7$, and $\left[x_0^1{x}_0^2{w}_2\right]$ is a normal 3-face of H. If $w_2\ne x_0^3$, then $x_0^2$ will be adjacent to three small vertices $x_0^1,x_0^3,w_2$. If $w_2=x_0^3$, then $x_0^2$ is adjacent to three small vertices $x_0^1,x_0^3,x_0^4$, this contradicts with (C4).

•$d_H(x_0^2)=5$ and $\tau (v\to x_0^2)=\frac{4}{9}$. Then $n_{\times }(x_0^2)=2$, $y_0,x_0^4$ are indirect neighbors of $x_0^2$ in H, and $\left[x_0^1{x}_0^2{x}_0^3\right]$ is a normal 3-face of H. This implies that $x_0^2$ is adjacent to three small vertices, a contradiction with (C4). This completes the proof. □

Now we continue the proof of Claim 5. Recall that $m_0=\left|X_0\right|$.

First, suppose that $m_0$ is even. We partition $X_0$ into $\frac{m_0}{2}$ 2-sets $\{x_0^1,x_0^2\},\{x_0^3,x_0^4\},\dots ,$$\{x_0^{m_0-3},x_0^{m_0-2}\},\{x_0^{m_0-1},x_0^{m_0}\}$. It is easy to check that Claims 2 and 3 in Subase 10.1 still hold for 2-sets $\{x_0^3,x_0^4\},\{x_0^5,x_0^6\},\dots ,$$\{x_0^{m_0-3},x_0^{m_0-2}\}$. By Claim 6, $\sigma (y_0,y_1)\le \frac{5}{6}+\frac{2}{3}\times \frac{m_0-4}{2}+\frac{5}{6}=\frac{1}{3}(m_0+1)=\frac{1}{3}(|X_0|+1)$.

Next, suppose that $m_0$ is odd. We partition $X_0$ into $\frac{m_0-3}{2}$ 2-sets $\{x_0^1,x_0^2\},\{x_0^3,x_0^4\},\dots ,$$\{x_0^{m_0-6},x_0^{m_0-5}\},\{x_0^{m_0-4},x_0^{m_0-3}\}$ and one 3-set $\{x_0^{m_0-2},x_0^{m_0-1},x_0^{m_0}\}$.

If $\tau (v\to x_0^{m_0-2})\le \frac{1}{3}$, then by Claim 6 and Claim 3, $\sigma (y_0,y_1)\le \frac{5}{6}+\frac{2}{3}\times \frac{m_0-5}{2}++\frac{1}{3}+\frac{5}{6}=\frac{1}{3}(m_0+1)=\frac{1}{3}(|X_0|+1)$.

If $\tau (v\to x_0^{m_0-2})>\frac{1}{3}$, then $x_0^{m_0-2}$ is a 5-vertex with $n_{\times }(x_0^{m_0-2})=2$ and $\tau (v\to x_0^{m_0-2})\in \{\frac{4}{9},\frac{5}{12}\}$. Similarly to the proof of Claim 2, we obtain that $\tau (v\to x_0^{m_0-1})\le \frac{2}{9}$, and hence $\tau (v\to x_0^{m_0-2})+\tau (v\to x_0^{m_0-1})\le \frac{4}{9}+\frac{2}{9}=\frac{2}{3}$. By Claim 6 and Claim 2, $\sigma (y_0,y_1)\le \frac{5}{6}+\frac{2}{3}\times \frac{m_0-5}{2}+\frac{2}{3}+\frac{1}{2}=\frac{1}{3}(m_0+1)=\frac{1}{3}(|X_0|+1)$. This completes the proof of Claim 5.

Using Claim 5 and noting that $d_H(v)=\left|Y\right|+\sum _{i=0}^{k-1}\left|X_i\right|$, we can establish the following:

$$\begin{array}{ccc}\hfill w^{\prime }(v)& \ge & w(v)-\frac{1}{2}{d}_H(v)-\sum _{i=0}^{k-1}\sigma (y_i,y_{i+1})\hfill \\ & \ge & \frac{1}{2}{d}_H(v)-4-\sum _{i=0}^{k-1}\frac{1}{3}(|X_i|+1)\hfill \\ & =& \frac{1}{2}{d}_H(v)-4-\frac{1}{3}\left|Y\right|-\frac{1}{3}\sum _{i=0}^{k-1}\left|X_i\right|\hfill \\ & =& \frac{1}{2}{d}_H(v)-4-\frac{1}{3}\left|Y\right|-\frac{1}{3}(d_H(v)-\left|Y\right|)\hfill \\ & =& \frac{1}{6}(d_H(v)-24)\ge 0.\hfill \end{array}$$

Up to now, we have proved the statement (I). That is, $w^{\prime }(x)\ge 0$ for all $x\in V^0(H)\cup F^0(H)$. To show statement (II), we first notice that $w^{\prime }(f_0)=w(f_0)=d_H(f_0)-4=2-4=-2$. Let $v\in V(C)$. Since H is a plane triangulation, $d_H(v)\ge 3$. Note that v has exactly $d_H(v)-2$ internal (direct or indirect) neighbors and has $d_H(v)-1$ internal faces incident with it. By (R0), $w^{\prime }(v)\ge d_H(v)-4-\frac{1}{2}(d_H(v)-2)-\frac{1}{2}(d_H(v)-1)=-\frac{5}{2}$. Consequently, $w^{\prime }(f_0)+\sum _{x\in V(C)}{w}^{\prime }(x)\ge -2-\frac{2}{5}\times 2=-7$. This completes the proof of Theorem 2.

Given a graph G, for a vertex $v\in V(G)$, let $n_2(v)$ denote the number of 2-vertices that are adjacent to v.

Theorem 3.

Every nonempty 1-planar graph G contains a k-vertex v with neighbors $v_1,v_2,$$\dots ,v_k$ with $d_G(v_1)\le d_G(v_2)\le \cdots \le d_G(v_k)$ such that one of the following conditions holds:

(B1) $k=1$;

(B2) $k=2$ with $n_2(v_1)\ge d_G(v_1)-23$ or $n_2(v_2)\ge d_G(v_2)-23$;

(B3) $k=3$ with $d_G(v_1)\le 23$;

(B4) $k=4$ with $d_G(v_1)\le 14$ and $d_G(v_2)\le 23$;

(B5) $k=5$ with $d_G(v_1)\le 10$, $d_G(v_2)\le 14$, and $d_G(v_3)\le 23$;

(B6) $k=6$ with $d_G(v_1)\le 9$, $d_G(v_2)\le 10$, $d_G(v_3)\le 14$, and $d_G(v_4)\le 23$;

(B7) $k=7$ with $d_G(v_1)\le 7$, $d_G(v_2)\le 9$, $d_G(v_3)\le 10$, $d_G(v_4)\le 14$, and $d_G(v_5)\le 23$.

Proof. 

Suppose that Theorem 3 is not true. Let G be a counterexample, which is a 1-planar graph and contains no vertex v satisfying (B1)–(B7). Then $\delta (G)\ge 2$ since (B1) is not satisfied. Theorem 2 implies that at least one of (C1)–(C6) is satisfied. Note that the only difference between these two theorems is the case where $\delta (G)=2$. Let H be the graph obtained by removing all 2-vertices from G. Clearly, H remains to be a 1-planar graph.

If $\delta (H)=0$, then (B2) is obviously true for G. Now we consider the case $\delta (H)=1$. Let $x\in V(H)$ with $d_H(x)=1$. Since $2=\delta (G)\le d_G(x)=d_H(x)+n_2(x)=1+n_2(x)$, it follows that $n_2(x)\ge 1$. Let $x_1,x_2,\dots ,x_m,y$ be the neighbors of x in G, where $d_G(x_i)=2$ for $i=1,2,\dots ,m(\ge 1)$, and $d_G(y)\ge 2$. So, $x_1$ is a 2-vertex of G with x as its neighbor satisfying $n_2(x)=d_G(x)-d_H(x)=d_G(x)-1\ge d_G(x)-23$, a contradiction with (B2).

Suppose that $\delta (H)=2$. Let $u\in V(H)$ with $d_H(u)=2$. We claim that $d_G(u)\ge 3$. Suppose, on contrary, if $d_G(u)=2$, then u will be removed from G by the construction of H. Thus, u is adjacent to at least one 2-vertex, say $u_1$, in G. Since $n_2(u)=d_G(u)-d_H(u)=d_G(u)-2\ge d_G(u)-23$, a contradiction with (B2).

Finally if $\delta (H)\ge 3$, then H contains a k-vertex y, $3\le k\le 7$ by Theorem 2 and it has neighbors $y_1,y_2,\dots ,y_k$ where $d_H(y_1)\le d_H(y_2)\le \cdots \le d_H(y_k)$ such that this vertex v satisfies at least one of (C2)-(C6). Note that $d_H(y_{k-2})\le 23$ by the definition of (C2)–(C6). Since (B2) is not satisfied, no vertex $x\in V(H)$ with $d_H(x)\le 23$ is adjacent to a 2-vertex in G. It follows that $d_G(y)=d_H(y)=k$ and $d_G(y_i)=d_H(y_i)$ for $i=1,2,\dots ,k-2$. This shows that y is a vertex of G satisfying at least one of (B3)–(B7), a contradiction. □

3. Acyclic Edge Coloring

A multiset S is a generalized set in which an element can appear many times. Let ${\mathrm{mul}}_S(x)$ denote the number of times that an element $x\in S$ appears in S. The union, denoted by $S\uplus S^{\prime }$, of two multisets S and $S^{\prime }$ is a multiset that consists of all elements of S and $S^{\prime }$. Moreover, we define mul${}_{S\uplus S^{\prime }}(x)={\mathrm{mul}}_S(x)+{\mathrm{mul}}_{S^{\prime }}(x)$ for any $x\in S\uplus S^{\prime }$.

Suppose that G is a 1-planar graph with an acyclic edge-k-coloring $\varphi$ using the color set $C=\{1,2,\dots ,k\}$. For a vertex $v\in V(G)$, let $C_{\varphi }(v)$ denote the subset of colors in C that are assigned to the edges incident to v. For an edge $e=uv$, let $C_{\varphi }(v;e)=C_{\varphi }(v)\setminus \left\{\varphi (e)\right\}$. Given a partial acyclic edge coloring $\pi$ of G, an $(i,j)$-maximal bichromatic path on $\pi$ is a maximal path consisting of edges colored with i and j alternately. An $(i,j;u,v)$-critical bichromatic path of $\pi$ is an $(i,j)$-maximal bichromatic path which starts at u with i and ends at v with i.

Lemma 1

(Basavaraju and Chandran, [8]). Given any two colors i and j with respect to a partial acyclic edge-coloring of a graph G, then there exists at most one $(i,j)$-maximal bichromatic path containing a particular vertex v.

Lemma 2

(Esperet and Parreau, [5]). Every graph G with $\Delta \ge 2$ has ${\chi }_a^{\prime }(G)\le 4\Delta -4$.

Theorem 4.

If G is a 1-planar graph with maximum degree Δ, then ${\chi }_a^{\prime }(G)\le \Delta +36$.

Proof. 

The proof is proceeded by induction on the edge number $|E(G)|$ of G. If $|E(G)|\le \Delta +36$, then it is trivial that G is acyclically edge-$(\Delta +36)$-colorable. We may assume that G is a connected 1-planar graph with $|E(G)|\ge \Delta +37$. If $\Delta \le 13$, then Lemma 2 confirms that ${\chi }_a^{\prime }(G)\le 4\Delta -4\le \Delta +36$. So we may assume that $\Delta \ge 14$.

If $\delta (G)=1$, let $x\in V(G)$ with $d_G(x)=1$ and let y be the unique neighbor of x in G. Consider the graph $H=G-x$. Then H is a 1-planar graph with $|E(H)|=|E(G)|-1$ and $\Delta (H)\le \Delta$. By the induction hypothesis, H admits an acyclic edge-$(\Delta +36)$-coloring $\varphi$ using the color set $C=\{1,2,\dots ,\Delta +36\}$. Based on $\varphi$, we can color $xy$ with a color in $C\setminus C_{\varphi }(y)$ to extend $\varphi$ to G.

Assume that $\delta (G)\ge 2$. By Theorem 3, there exists a k-vertex $v\in V(G)$, $2\le k\le 7$ with neighbors $v_1,v_2,\dots ,v_k$ where $d_G(v_1)\le d_G(v_2)\le \cdots \le d_G(v_k)$ such that at least one of the conditions (B2)–(B7) holds.

First, suppose that (B2) holds. That is, there exists a 2-vertex v with neighbors $v_1,v_2$ such that $n_2(v_i)\ge d_G(v_i)-23$ for some $i=1,2$, say $i=1$. Let $y_1,y_2,\dots ,y_m,y_{m+1},\dots ,y_t$ be the neighbors of $v_1$, where $t=d_G(v_1)-1$, such that $d_G(y_i)\ge 3$ for $i=1,2,\dots ,m$, and $d_G(y_i)=2$ for $i=m+1,m+2,\dots ,t$. Since $n_2(v_1)\ge d_G(v_1)-23$, we derive that $m=d_G(v_1)-n_2(v)\le 23$. Let $H=G-v$. Then H is a 1-planar graph with $|E(H)|=|E(G)|-2$ and $\Delta (H)\le \Delta$. By the induction hypothesis, H admits an acyclic edge-$(\Delta +36)$-coloring $\varphi$ using the color set $C=\{1,2,\dots ,\Delta +36\}$. Set $S_1=\{\varphi (v_1{y}_1),\varphi (v_1{y}_2),\dots ,\varphi (v_1{y}_m)\}$ and $S_2=\{\varphi (v_1{y}_{m+1}),\varphi (v_1{y}_{m+2}),\dots ,\varphi (v_1{y}_t)\}$. To extend $\varphi$ to G, we first color the edge $v{v}_2$ with a color $a\in C\setminus (C_{\varphi }(v_2)\cup S_1)$. Since $|C_{\varphi }(v_2)\cup S_1|\le |C_{\varphi }(v_2)|+|S_1|\le d_G(v_2)-1+m\le \Delta -1+m\le \Delta +22$. Since $\left|C\right|=\Delta +36$, there is the color which can be assigned to $v{v}_2$. If $a\notin S_2$, then we color $v{v}_1$ with any color in $C\setminus (C_{\varphi }(v_1)\cup \left\{a\right\})$. Otherwise, we may assume that $a=\varphi (v_1{y}_{m+1})$. Let z be the neighbor of $y_{m+1}$ other than $v_1$. We color $v{v}_1$ with a color in $C\setminus (C_{\varphi }(v_1)\cup \left\{\varphi (z{y}_{m+1})\right\})$. It is easy to check that any newly created cycle by adding $v{v}_1$ and $v{v}_2$ has at least three colors and, in turn, the extended coloring is an acyclic edge-$(\Delta +36)$-coloring of G.

Next assume that one of (B3)-(B7) holds. Let $H=G-v{v}_1$. Then H is a 1-planar graph with $|E(H)|=|E(G)|-1$ and $\Delta (H)\le \Delta$. By the induction hypothesis, H admits an acyclic edge-$(\Delta +36)$-coloring $\varphi$ using the color set $C=\{1,2,\dots ,\Delta +36\}$. Let $C^{*}=C\setminus (C_{\varphi }(v)\cup C_{\varphi }(v_1))$, and $r=|C_{\varphi }(v)\cap C_{\varphi }(v_1)|$. Of all the possible acyclic edge-$(\Delta +36)$-colorings of H choose the coloring $\varphi$ which minimizes r. The following inequality holds.

$$\begin{array}{ccc}\hfill |C^{*}|& =& \left|C\right|-|C_{\varphi }(v)|-|C_{\varphi }(v_1)|+|C_{\varphi }(v)\cap C_{\varphi }(v_1)|\hfill \\ & \ge & (\Delta +36)-(d_G(v)-1)-(d_G(v_1)-1)+r\hfill \\ & \ge & \Delta +36-(7-1)-(23-1)+r=\Delta +r+8.\hfill \end{array}$$

If $r=0$, then we can color $v{v}_1$ with any color in $C^{*}$ such that no bichromatic cycle is produced in G. Otherwise, assume that $r\ge 1$. Set $d_i=d_G(v_i)$ for $i=1,2,\dots ,k$. Let $u_1,u_2,\dots ,u_{d_1-1}$ denote the neighbors of $v_1$ other than v, and assume that $\varphi (v_1{u}_i)=i$ for $i\in \{1,2,\cdots ,d_1-1\}$. Define a multiset $S_v$ as follows:

$$S_v=C_{\varphi }(v;v{v}_2)\uplus C_{\varphi }(v;v{v}_3)\uplus \cdots \uplus C_{\varphi }(v;v{v}_k)$$

First we show that $\sum _{i=1}^{k-2}{d}_i\le 77-k-d_1$. Let ${\theta }_k=\sum _{i=1}^{k-2}{d}_i+d_1+k$. It remains to show that ${\theta }_k\le 77$ for any $k\in \{3,4,5,6,7\}$. If $k=3$, then $d_1\le 23$ and hence ${\theta }_3=d_1+d_1+k\le 23+23+3=49$. If $k=4$, then $d_1\le 14$, $d_2\le 23$, and hence ${\theta }_4=2d_1+d_2+k\le 2\times 14+23+4=55$. If $k=5$, then $d_1\le 10$, $d_2\le 14$, $d_3\le 23$, and hence ${\theta }_5=2d_1+d_2+d_3+k\le 2\times 10+14+23+5=62$. If $k=6$, then $d_1\le 9$, $d_2\le 10$, $d_3\le 14$, $d_4\le 23$ and hence ${\theta }_6=2d_1+d_2+d_3+d_4+k\le 2\times 9+10+14+23+6=71$. If $k=7$, then $d_1\le 7$, $d_2\le 9$, $d_3\le 10$, $d_4\le 14$, $d_5\le 23$ and hence ${\theta }_7=2d_1+d_2+d_3+d_4++d_5+k\le 2\times 7+9+10+14+23+7=77$.

If $r=1$, then we may assume that $\varphi (v{v}_2)=\varphi (v_1{u}_1)=1$. It follows that we can color $v{v}_1$ with a color $a\in C^{*}$ missing the neighbors of $v_2$ such that no bichromatic cycle is created in G. This is possible because $d_G(v)=7$ and $|C^{*}|\ge \Delta +9$.

Now suppose that $r\ge 2$. Let $\varphi (v{v}_{i+1})=\varphi (v_1{u}_i)=i$ for $i=1,2,\dots ,r$. Clearly, $r\le k-1$. If there exists a color $a\in C^{*}$ such that assigning a to $v{v}_1$ does not result in a bichromatic cycle in G, we are done. Otherwise, for any $a\in C^{*}$, there exists a $(i,a;v_1,v_{i+1})$-critical bichromatic path for every $i\in \{1,2,\dots ,r\}$. Note that $|C^{*}|\ge (\Delta +36)-(d_G(v)-1)-(d_G(v_1)-1)+2=\Delta +40-k-d_1$. The fact of $\sum _{i=1}^{k-2}{d}_i\le 77-k-d_1$ and $d_{k-1},d_k\le \Delta$ implies that

$$\begin{array}{ccc}\hfill \sum _{i=2}^k(d_i-1)& =& \sum _{i=1}^{k-2}{d}_i-(k-1)+d_{k-1}+d_k-d_1\hfill \\ & \le & 77-k-d_1-k+1+2\Delta -d_1\hfill \\ & =& 2\Delta +78-2d_1-2k\hfill \\ & <& 2\Delta +80-2d_1-2k\hfill \\ & =& 2(\Delta +40-k-d_1)\hfill \\ & \le & 2|C^{*}|.\hfill \end{array}$$

The above inequality implies that there exists a color $\beta \in C^{*}$ such that it appears at most once in $S_v$, say $\beta \in C_{\varphi }(v_2;v{v}_2)$. Since there exists a $(1,\beta ;v,v_1)$-critical bichromatic path through v, there is no $(1,\beta ;v,v_3)$-critical bichromatic path through $v_2$ by Lemma 1. We can recolor $v{v}_3$ with $\beta$ to yield a new acyclic edge-$(\Delta +36)$-coloring ${\varphi }^{\prime }$ of H. This implies that $|C_{{\varphi }^{\prime }}(v)\cap C_{{\varphi }^{\prime }}(v_1)|<|C_{\varphi }(v)\cap C_{\varphi }(v_1)|$, a contradiction with the choice of $\varphi$. This completes the proof of the theorem. □

4. Concluding Remarks

In this paper, we established the structural theorem on 1-planar graphs using the discharging method in Section 2, then show that every 1-planar graph G has ${\chi }_a^{\prime }(G)\le \Delta +36$ in Section 3 by the structural theorem 3. We strengthen the following two results:

(1) G is a 1-planar graph, then ${\chi }_a^{\prime }(G)\le max\{2\Delta -2,\Delta +83\}$ in [17];

(2) G is a triangle-free 1-planar graph, then ${\chi }_a^{\prime }(G)\le \Delta +16$ in [19].

Conjecture 1 seems difficult. As yet, it has been verified only for several restricted classes of planar graphs. A little work on 1-planar graphs has been done, there is a certain gap between our research results and the conjectured results ($\Delta +2$); however, this result is the best research so far. We believe that this bound is not tight, and this bound can be reduced to $\Delta +22$ or less by a good method.