On (p, 1)-total labelling of planar graphs

Abstract

A k-(p, 1)-total labelling of a graph G is a function f from $V(G)\cup E(G)$ to the color set $\{0, 1, \ldots , k\}$ such that $|f(u)-f(v)|\ge 1$ if $uv\in E(G), |f(e_1)-f(e_2)|\ge 1$ if $e_1$ and $e_2$ are two adjacent edges in G and $|f(u)-f(e)|\ge p$ if the vertex u is incident with the edge e. The minimum k such that G has a k-(p, 1)-total labelling, denoted by $\lambda _p^T(G)$, is called the (p, 1)-total labelling number of G. In this paper, we prove that, for any planar graph G with maximum degree $\Delta \ge 4p+4$ and $p\ge 2, \lambda _p^T(G)\le \Delta +2p-2$.

1 Introduction

Many real-world situations can conveniently be described by means of a diagram consisting of a set points together with lines joining certain pairs of these points. A mathematical abstraction of situations of this type gives rise to the concept of a graph. A graph G is an ordered triple $(V(G),E(G),\psi _G)$ consisting of a nonempty set V(G) of vertices, a set E(G), disjoint from V(G), of edges, and an incidence function $\psi _G$ that associates with each edge of G an unordered pair of vertices of G.

Motivated by the frequency channel assignment problem, Griggs and Yeh (1992) first introduced the L(2, 1)-labelling of graphs. This notion was subsequently extended to a general form, named as L(p, q)-labelling of graphs. An L(p, q)-labelling of a graph G is a mapping from the set of vertices V(G) to the set of integers $Z_k=\{0,1,\ldots ,k\}$ such that $|f(x)-f(y)|\ge p$ if x and y are adjacent and $|f(x)-f(y)|\ge q$ if x and y are at distance 2. Interested readers can refer to Calamoneti (2006) and Yeh (2006). Whittlesey et al. (1995) investigated the L(2, 1)-labelling of the incidence graph I(G) of a graph G, which is obtained from G by replacing each edge by a path of length 2. The L(2, 1)-labelling of I(G) can be easily translated to another kind of labelling, the so called (2, 1)-total labelling of G introduced by Havet and Yu (2002) and Havet (2003).

A k-(p, 1)-total labelling of a graph G is a function f from $V(G)\cup E(G)$ to the color set $\{0, 1,\ldots , k\}$ such that $|f(u)-f(v)|\ge 1$ if $uv\in E(G), |f(e_1)-f(e_2)|\ge 1$ if $e_1$ and $e_2$ are two adjacent edges in G and $|f(u)-f(e)|\ge p$ if the vertex u is incident with the edge e. The minimum k such that G has a k-(p, 1)-total labelling, denoted by $\lambda _p^T(G)$ (or simply by $\lambda _p^T$), is called the (p, 1)-total labelling number of G. When $p=1$, the (1, 1)-total labelling is the well-known total coloring of graphs. One can see that $\lambda _p^T (G) \ge \Delta (G)+p-1$ with $p\ge 1$, where $\Delta (G)$ is the maximum degree of G. On the other hand, we can construct a (p, 1)-total labelling of G by properly coloring its edges with $\chi '(G)$ integers of $[0, \chi '(G)-1]$, and its vertices with $\chi (G)$ integers of $[\chi '(G)+p-1, \chi (G)+\chi '(G)+p-2]$, where $\chi (G)$ and $\chi '(G)$ denote the vertex chromatic number and the edge chromatic number of G, respectively. Thus, we have $\Delta (G)+p-1\le \lambda _p^T (G) \le \chi (G)+\chi '(G)+p-2$. Meanwhile, it is easy to get that $\lambda _p^T (G) \le 2\Delta (G)+p-1$ by Brooks’ theorem and Vizing’s theorem. However, this upper bound for $\lambda _p^T$ seems to be not tight. As a natural extension of the total coloring conjecture, Havet and Yu (2002, 2008) conjectured the following known as the (p, 1)-total labelling conjecture.

Conjecture 1

(Havet and Yu 2002, 2008) Let G be a graph. Then $\lambda _p^T(G)\le \min \{\Delta (G)+2p-1, 2\Delta (G)+p-1\}$.

If $p=1$, this conjecture is the total coloring conjecture, which has been extensively studied in many papers, we give some (Kowalik et al. 2008; Wang and Wu 2012; Wang et al. 2014). For $p=2$, the (2, 1)-total labelling of outerplanar graphs has been discussed in Chen and Wang (2007) and Hasunuma et al. (2011). In Yu et al. (2011), Yu et al. proved that every planar graph with maximum degree $\Delta \ge 12$ satisfies that $\Delta +1\le \lambda _2^T\le \Delta +2$. The (p, 1)-total labelling conjecture in general has been considered for some other classes such as planar graphs with high girth and high maximum degree (Bazzaro et al. 2007) and graphs with a given maximum average degree (Montassier and Raspaud 2006). Particularly, Bazzaro, Montassier and Raspaud proved the following theorem for all planar graphs (Bazzaro et al. 2007).

Theorem 2

(Bazzaro et al. 2007) Let G be a planar graph with maximum degree $\Delta$. If $\Delta \ge 8p+2$ and $p\ge 2$, then $\lambda _p^T (G) \le \Delta +2p-2$.

In this paper, we obtain a result shown in Theorem 3, which is an improvement of Theorem 2.

Theorem 3

Let $p\ge 2$ be a positive integer and let G be a planar graph with maximum degree $\Delta \ge 4p+4$. Then $\lambda _p^T (G) \le \Delta +2p-2$.

For planar graphs with maximum degree $\Delta \ge 7$, $\chi '=\Delta$ (Sanders and Zhao 2001) and $\chi \le 4$. Recalling that $\lambda ^T_p\le \chi +\chi '+p-2$, we have $\lambda _p^T\le \Delta +p+2$ for planar graphs with maximum degree $\Delta \ge 4p+4$. So the upper bound for $\lambda ^T_p$ in Theorem 3 is better for small p. Particularly, we will prove the following theorem, which is only a technical strengthening of Theorem 3. Without it we would get complications when considering a subgraph $G'\subset G$ such that $\Delta (G')< \Delta (G)$. The interesting case of Theorem 4 is $M=\Delta$.

Theorem 4

Let M, p be two positive integers such that $M\ge 4p+4$ and let G be a planar graph with maximum degree $\Delta \le M$. Then $\lambda _p^T (G) \le M+2p-2$ with $p\ge 2$.

Some notations should be introduced. All graphs considered here are simple, finite and undirected. For a graph $G, \delta (G)$ and $\Delta (G)$ (or simply $\delta$ and $\Delta$) denote the minimum degree and the maximum degree of G, respectively. For a vertex $v\in V(G), N_G(v)$ denotes the set of vertices adjacent to v in $G, d_G(v)=|N_G(v)|$ denotes the degree of v. A k-, $k^+$- and $k^-$-vertex is a vertex of degree k, at least k and at most k, respectively. A vertex u is called a k-neighbor (resp. $k^-$-neighbor, $k^+$-neighbor) of a vertex v if $uv\in E(G)$ and $d_G(u)=k$(resp. $d_G(u)\le k, d_G(u)\ge k$). For a given plane graph G, F(G) denotes the face set of G. The degree of a face f, denoted by $d_G(f)$, is the number of edges incident with it, where cut edge is counted twice. A k-, $k^+$- and $k^-$-face in a plane graph G is defined analogously to counterparts of a vertex. For the terminologies and notations not defined here, we follow Bondy and Murty (1976).

2 Structural properties of the minimum counterexample

Let G be a counterexample to Theorem 4 with $|V(G)|+|E(G)|$ being minimum. The following lemmas are useful to the proof of Theorem 4.

Lemma 5

(Zhang et al. 2011) For any edge $uv\in E(G)$, if $min\{d_G(u),d_G(v)\}\le \left\lfloor {\frac{M+2p-2}{2p}}\right\rfloor$, then $d_G(u)+d_G(v)\ge M+2$, otherwise, $d_G(u)+d_G(v)\ge M-2p+3$.

Lemma 6

(Zhang et al. 2011) For any integer k satisfying $2\le k\le \left\lfloor {\frac{M+2p-2}{2p}}\right\rfloor$, let $X_k=\{x\in V(G): d_G(x)\le k\}$ and $Y_k=\cup _{x\in {X_k}}N_G(x)$. If $X_k\ne \emptyset$, then there exists a bipartite subgraph $M_k$ of G with partite sets $X_k$ and $Y_k$ such that $d_{M_k}(x)=1$ for every $x\in X_k$ and $d_{M_k}(y)\le k-1$ for every $y\in Y_k$.

Following the terms of Borodin, and Woodall in Borodin et al. (1997), in Lemma 6 we call y the k-master of x if $xy\in M_k$ and $x\in X_k$ and we call x the k-dependent of y. From Lemma 6, we can get the following useful lemma.

Lemma 7

(Zhang et al. 2011) Every i-vertex in G has a j-master, where $2\le i\le j\le \left\lfloor {\frac{M+2p-2}{2p}}\right\rfloor$, and every vertex in G has at most $k-1\,k$-dependents, $2\le k\le \left\lfloor {\frac{M+2p-2}{2p}}\right\rfloor$.

Through this paper, we do not distinct the terms color and label. For a set A, we usually denote the cardinality of A by |A|. In a partial (p, 1)-total labelling $\phi$ of G from A to the color interval $C=\{0,1,\ldots ,M+2p-2\}$, let $\phi (x)$ denote the color of element $x\in A$, satisfying all the conditions in the definition of (p, 1)-total labelling of graphs. For the sake of clarity of the presentation, we need the following notations.

$$\begin{aligned} E_{\phi }(v)= & {} \{\phi (e)|e\in E(G)\ \text{ is } \text{ incident } \text{ with } \text{ vertex } v\}\hbox { for }v\in V(G); \\ N_{\phi }(v)= & {} \{\phi (u)|u\in N_G(v)\};\\ I_{\phi }(x)= & {} \{\phi (x)-(p-1),\ldots ,\phi (x)-1,\phi (x),\phi (x)+1,\ldots ,\phi (x)+(p-1)\}\cap \\&C\hbox { for }x\in V(G)\cup E(G); \end{aligned}$$

$A_{\phi }(x)$ denotes the set of colors which are still available for labelling element $x\in V(G)\cup E(G)$ under the partial (p, 1)-total labelling $\phi$.

Firstly, we prove a simple lemma as follows.

Lemma 8

For a vertex $u\in V(G)$, if $d_G(u)\le \left\lfloor {\frac{M+2p-2}{2p}}\right\rfloor$, then a partial (p, 1)-total labelling of G with the vertex u uncolored can be extended to a (p, 1)-total labelling of G.

Proof

Let $\phi$ denote a partial (p, 1)-total labelling of G with the vertex u uncolored, and the color interval is C. Since $|A_{\phi }(u)|=|C\setminus (\cup _{w\in N_G(u)}{I_{\phi }(uw)} \cup N_{\phi }(u))|\ge |C|-(2p-1)d_G(u)-d_G(u)=|C|-2pd_G(u)\ge (M+2p-1)-2p\cdot \frac{M+2p-2}{2p}=1$. Then we can color the vertex u properly under $\phi$. $\square$

In order to make our proof much simpler, we need the following lemmas and give the proofs in detail.

Lemma 9

For $k=\left\lfloor {\frac{M+2p-2}{2p}}\right\rfloor$. Let $M=ap+b$ and a, b be two nonnegative integers with $0\le b\le 2p+2$, if one of the following conditions holds in G:

• (1) $0\le b < p+2$ and $(a,b)\in \{(2d,0),(2d,1),(2d+1,2),(2d+1,3),(s,t): s=2d\ \text{ or }\ 2d+1, 4\le t< p+2\ \text{ with }\ p\ge 3\}$;

• (2) $p+2\le b < p+4$ and $a=2d$;

• (3) $p+4\le b < 2p+2$ and a is 2d or $2d+1, p\ge 3$;

• (4) $b =2p+2$ and $a=2d+1, p\ge 3$.

Then a ($k+1$)-vertex is adjacent to $(M-2p+3-k)^+$-vertices in G.

Proof

By Lemma 5, it is enough to prove that there is no ($k+1$)-vertex is adjacent to ($M-2p+3-k-1$)-vertices. Otherwise, let $uv\in E(G), d_G(u)=k+1$ and $d_G(v)=M-2p+3-k-1$. By the minimality of $G, H=G-uv$ has a (p, 1)-total labelling with the color interval C. Erase the color of vertex u, and denote this partial (p, 1)-total labelling by $\phi$. Then $|A_{\phi }(uv)|\ge |C|-|E_{\phi }(u)|-|E_{\phi }(v)|-|I_{\phi }(v)|= (M+2p-1)-(d_G(u)-1)-(d_G(v)-1)-(2p-1)=(2p-1), |A_{\phi }(u)|\ge |C|-|\cup _{w\in N_G(u)}{I_{\phi }(uw)}|-|N_{\phi }(u)|=(M+2p-1)-k\cdot (2p-1)- (k+1)=(M+2p-1)-2pk-1$. Now, as for $|A_{\phi }(u)|$, we will divide this problem to eight cases:

Case 1. $(a,b)=(2d,0)$.

From the assumption, we have $k=d$. Then $|A_{\phi }(u)|\ge (M+2p-1)-2pk-1=2p-2\ge 2\ (p\ge 2)$. Choose $\alpha \in A_{\phi }(u)$ to color u. If $A_{\phi }(uv)\ne \{\alpha -(p-1), \cdots , \alpha -1, \alpha , \alpha +1,\ldots , \alpha +(p-1)\}$, then we choose $\beta \in A_{\phi }(uv)\setminus \{\alpha -(p-1),\ldots , \alpha -1, \alpha , \alpha +1,\ldots , \alpha +(p-1)\}$ to color edge uv. Otherwise, $A_{\phi }(uv)=\{\alpha -(p-1),\ldots , \alpha -1, \alpha , \alpha +1,\ldots , \alpha +(p-1)\}$. Then we choose $\gamma \in A_{\phi }(u)\setminus \{\alpha \}$ to color u. Since $A_{\phi }(uv)\ne \{\gamma -(p-1),\ldots , \gamma -1, \gamma , \gamma +1,\ldots , \gamma +(p-1)\}$, we can choose $\pi \in A_{\phi }(uv)\setminus \{\gamma -(p-1),\ldots , \gamma -1, \gamma , \gamma +1,\ldots , \gamma +(p-1)\}$ to color edge uv. Thus, we extend $\phi$ from subgraph H to G, a contradiction.

Case 2. $(a,b)=(2d,1)$.

From the assumption, we have $k=d$. Then $|A_{\phi }(u)|\ge (M+2p-1)-2pk-1=2p-1\ge 3\ (p\ge 2)$.

Case 3. $(a,b)=(2d+1,2)$.

From the assumption, we have $k=d+1$. Then $|A_{\phi }(u)|\ge (M+2p-1)-2pk-1=p\ge 2\ (p\ge 2)$.

Case 4. $(a,b)=(2d+1,3)$.

From the assumption, we have $k=d+1$. Then $|A_{\phi }(u)|\ge (M+2p-1)-2pk-1=p+1\ge 3\ (p\ge 2)$.

Case 5. $(a,b)\in \{(s,t):s=2d\ or\ 2d+1, 4\le t< (p+2)\ \text{ with }\ p\ge 3\}$.

From the assumption, When $a=2d, k=d+1$ and $|A_{\phi }(u)|\ge (M+2p-1)-2pk-1=b-2\ge 2\ (b\ge 4)$; when $a=2d+1, k=d+1$ and $|A_{\phi }(u)|\ge (M+2p-1)-2pk-1=p+b-2\ge 4\ (b\ge 4)$, where, $0< \frac{b-2}{2p}< \frac{1}{2}$.

Case 6. When $p+2\le b< p+4, \frac{1}{2}\le \frac{b-2}{2p}< 1$.

If $a=2d$, then $k=d+1$. Hence, $|A_{\phi }(u)|\ge (M+2p-1)-2pk-1=b-2\ge p\ge 2\ (b\ge p+2)$.

Case 7. When $p+4\le b< 2p+2, p\ge 3, \frac{1}{2}\le \frac{b-2}{2p}< 1$.

If $a=2d$, then $k=d+1$. Hence, $|A_{\phi }(u)|\ge (M+2p-1)-2pk-1=b-2\ge p+2> 2\ (b\ge p+4)$. If $a=2d+1$, then $k=d+2$. Hence, $|A_{\phi }(u)|\ge (M+2p-1)-2pk-1=b-p-2\ge 2\ (b\ge p+4)$.

Case 8. When $b=2p+2, p\ge 3$.

If $a=2d+1$, then $k=d+2$. Hence, $|A_{\phi }(u)|\ge (M+2p-1)-2pk-1=p\ge 2$.

From Case 2 to Case 8, we can always get that $|A_{\phi }(u)|\ge 2$. Analogous to Case 1, we can extend $\phi$ to a (p, 1)-total labelling of G, a contradiction. $\square$

For convenience, a cycle of length k in G is denoted by $[d_G(v_1), d_G(v_2),\ldots , d_G(v_k)]$ if $v_1, v_2,\ldots , v_k$ are the boundary vertices of the cycle in a clockwise order. The following lemma focuses on the discussion of a cycle of length 3 (triangle).

Lemma 10

Let $k=\left\lfloor {\frac{M+2p-2}{2p}}\right\rfloor \ge 2$. Then G has the following structural properties:

• (1) There is no triangle $[d_G(v_1),d_G(v_2),d_G(v_3)]$ with $d_G(v_1)=k+2$ and $d_G(v_2)=d_G(v_3)=M-2p-k+1$.

• (2) If $f=[d_G(v_1),d_G(v_2),d_G(v_3)]$ is a triangle with $d_G(v_1)=k+2, d_G(v_2)=M-2p-k+1$ and $d_G(v_3)=M-2p-k+2$, then $v_1$ has no $(M-2p-k+1)$-neighbor besides $v_2$.

Proof

(1) On the contrary, suppose that G contains a triangle $[d_G(v_1),d_G(v_2),d_G(v_3)]$ with $d_G(v_1)=k+2$ and $d_G(v_2)=d_G(v_3)=M-2p-k+1$. Assume $H=G-\{v_1v_2, v_1v_3\}$. By minimality of G, H has a (p, 1)-total labelling $\phi$ with interval C.

Case 1. $\phi (v_1)\in \cup _{u\in \{v_2,v_3\}}{E_{\phi }(u)}\bigcup \cup _{u\in \{v_2,v_3\}}{I_{\phi }(u)}$.

Without loss of generality, suppose that $\phi (v_1)\in E_{\phi }(v_2)\cup I_{\phi }(v_2)$. Then $|(E_{\phi }(v_1)\cup I_{\phi }(v_1))\bigcap (E_{\phi }(v_2)\cup I_{\phi }(v_2))|\ge 1$. Therefore, $|A_{\phi }(v_1v_2)|=M+2p-1-|(E_{\phi }(v_1)\cup I_{\phi }(v_1))\bigcup (E_{\phi }(v_2)\cup I_{\phi }(v_2))|\ge M+2p-1-[(d_G(v_1)-2+d_G(v_2)-1)+2\cdot (2p-1)-1]=(M+2p-1)-(M-2p+3)-2\cdot (2p-1)+4=2$ and $|A_{\phi }(v_1v_3)|=M+2p-1-|(E_{\phi }(v_1)\cup I_{\phi }(v_1))\bigcup (E_{\phi }(v_3)\cup I_{\phi }(v_3))|\ge M+2p-1-[(d_G(v_1)-2+d_G(v_3)-1)+2\cdot (2p-1)]=(M+2p-1)-(M-2p+3)-2\cdot (2p-1)+3=1$. Hence, we can extend $\phi$ to a (p, 1)-total labelling of G, a contradiction.

Case 2. $\phi (v_1)\notin \cup _{u\in \{v_2,v_3\}}{E_{\phi }(u)}\bigcup \cup _{u\in \{v_2,v_3\}}{I_{\phi }(u)}$.

From the above condition, we can recolor the edge $v_2v_3$ with color $\phi (v_1)$. Denote this new partial (p, 1)-total labelling by $\phi '$. Then, $|(E_{\phi '}(v_1)\cup I_{\phi '}(v_1))\bigcap (E_{\phi '}(v_2)\cup I_{\phi '}(v_2))|\ge 1$, the analysis of which is analogous to Case 1, a contradiction.

(2) On the contrary, suppose that G contains such a configuration and $v_4$ is a $(M-2p-k+1)$-neighbor of $v_1$ different from $v_2$. By the minimality of $G, H=G-\{v_1v_2,v_1v_3\}$ has a (p, 1)-total labelling $\phi$ with color interval C.

Claim 1. $\phi (v_1)\ne \phi (v_2v_3)$. Otherwise, $|(E_{\phi }(v_1)\cup I_{\phi }(v_1))\bigcap (E_{\phi }(v_2)\cup I_{\phi }(v_2))|\ge 1$ and $|(E_{\phi }(v_1)\cup I_{\phi }(v_1))\bigcap (E_{\phi }(v_3)\cup I_{\phi }(v_3))|\ge 1$, which implies that $|A_{\phi }(v_1v_2)|=M+2p-1-|(E_{\phi }(v_1)\cup I_{\phi }(v_1))\bigcup (E_{\phi }(v_2)\cup I_{\phi }(v_2))|\ge M+2p-1-[(d_G(v_1)-2+d_G(v_2)-1)+2\cdot (2p-1)-1]=(M+2p-1)-(M-2p+3)-2\cdot (2p-1)+4=2$ and $|A_{\phi }(v_1v_3)|=M+2p-1-|(E_{\phi }(v_1)\cup I_{\phi }(v_1))\bigcup (E_{\phi }(v_3)\cup I_{\phi }(v_3))|\ge M+2p-1-[(d_G(v_1)-2+d_G(v_3)-1)+2\cdot (2p-1)-1]=(M+2p-1)-(M-2p+4)-2\cdot (2p-1)+4=1$. Then we can extend the partial (p, 1)-total labelling $\phi$ to G, a contradiction.

Claim 2. $E_{\phi }(v_1)\subseteq \cup _{u\in \{v_2,v_3\}}{E_{\phi }(u)}\bigcup \cup _{u\in \{v_2,v_3\}}{I_{\phi }(u)}$. Otherwise, we can choose a color $\alpha \in E_{\phi }(v_1)\setminus (\cup _{u\in \{v_2,v_3\}}{E_{\phi }(u)}\bigcup \cup _{u\in \{v_2,v_3\}}{I_{\phi }(u)})$ to recolor edge $v_2v_3$. Denote this new coloring of H by $\phi '$. Then $|(E_{\phi '}(v_1)\cup I_{\phi '}(v_1))\bigcap (E_{\phi '}(v_2)\cup I_{\phi '}(v_2))|\ge 1$ and $|(E_{\phi '}(v_1)\cup I_{\phi '}(v_1))\bigcap (E_{\phi '}(v_3)\cup I_{\phi '}(v_3))|\ge 1$. Therefore, $|A_{\phi '}(v_1v_2)|\ge 2$ and $|A_{\phi '}(v_1v_3)|\ge 1$, a contradiction.

Claim 3. $E_{\phi }(v_1)\subseteq E_{\phi }(v_2)\cup I_{\phi }(v_2)$. Otherwise, there are two cases: the first is $E_{\phi }(v_1)\bigcap (E_{\phi }(v_2)\cup I_{\phi }(v_2)))\ne \emptyset$ and $E_{\phi }(v_1)\bigcap (E_{\phi }(v_3)\cup I_{\phi }(v_3)))\ne \emptyset$, which is impossible by the similar analysis in Claim 1; the second is $E_{\phi }(v_1)\subseteq E_{\phi }(v_3)\cup I_{\phi }(v_3)$, which implies that $|(E_{\phi }(v_1)\cup I_{\phi }(v_1))\bigcap (E_{\phi }(v_3)\cup I_{\phi }(v_3))|\ge d_G(v_1)-2$. In the second case, $|A_{\phi }(v_1v_3)|=M+2p-1-|(E_{\phi }(v_1)\cup I_{\phi }(v_1))\bigcup (E_{\phi }(v_3)\cup I_{\phi }(v_3))|\ge M+2p-1-[(d_G(v_1)-2+d_G(v_3)-1)+2\cdot (2p-1)-(d_G(v_1)-2)]= M-2p-d_G(v_3)+2=d_G(v_1)-2\ge 2\,(d_G(v_1)=k+2\ge 4)$, and $|A_{\phi }(v_1v_2)|\ge 1$. Therefore, we can extend $\phi$ to G, a contradiction.

By Claim 2 and Claim 3, we have $|(E_{\phi }(v_1)\cup I_{\phi }(v_1))\bigcap (E_{\phi }(v_2)\cup I_{\phi }(v_2))|\ge (d_G(v_1)-2)$ and $E_{\phi }(v_1)\bigcap (E_{\phi }(v_3)\cup I_{\phi }(v_3))=\emptyset$. Since $\phi (v_1v_4)\in E_{\phi }(v_1)$, then $\phi (v_1v_4)\notin E_{\phi }(v_3)\cup I_{\phi }(v_3)$. For edge $v_1v_4, |A_{\phi }(v_1v_4)|=M+2p-1-|E_{\phi }(v_1)\cup I_{\phi }(v_1))\bigcup (E_{\phi }(v_4)\cup I_{\phi }(v_4))|\ge M+2p-1-[(d_G(v_1)-3)+(d_G(v_4)-1)+2\cdot (2p-1)]=2$. Hence, there exists a color $\alpha \in A_{\phi }(v_1v_4)$ such that $\alpha \ne \phi (v_1v_4)$. We use $\alpha$ to recolor $v_1v_4$ and denote this new partial (p, 1)-total labelling by $\phi '$. Note that $E_{\phi '}(v_1)=E_{\phi }(v_1)\cup \{\alpha \}\setminus \{\phi (v_1v_4)\}, I_{\phi '}(v_1)=I_{\phi }(v_1), E_{\phi '}(v_2)\cup I_{\phi '}(v_2)=E_{\phi }(v_2)\cup I_{\phi }(v_2), E_{\phi '}(v_3)\cup I_{\phi '}(v_3)=E_{\phi }(v_3)\cup I_{\phi }(v_3)$. Hence, $\phi (v_1v_4)\notin (E_{\phi '}(v_1)\cup I_{\phi '}(v_1))\bigcup (E_{\phi '}(v_3)\cup I_{\phi '}(v_3))$, then we can color $v_1v_3$ with $\phi (v_1v_4)$. Denote this new partial (p, 1)-total labelling by $\phi ''$. For edge $v_1v_2$, we have $|A_{\phi ''}(v_1v_2)|=M+2p-1-|E_{\phi ''}(v_1)\cup I_{\phi ''}(v_1))\bigcup (E_{\phi ''}(v_2)\cup I_{\phi ''}(v_2))|\ge M+2p-1-[(d_G(v_1)-1)+(d_G(v_2)-1)+2\cdot (2p-1)-(d_G(v_1)-2)]=2$ $(d_G(v_1)=k+2\ge 4)$. Therefore, we can obtain a (p, 1)-total labelling of G, a contradiction. $\square$

Lemma 11

If a vertex v is adjacent to two vertices $v_1, v_2$ such that $2\le d_G(v_1)=d_G(v_2)=M+2-d_G(v)\le \left\lfloor {\frac{M+2p-2}{2p}}\right\rfloor$, then every face incident with $vv_1$ or $vv_2$ must be a $4^+$-face.

Proof

Suppose that there is triangle face $uvv_1$ such that $vv_2\in E(G)$ and $2\le d_G(v_1)=d_G(v_2)=M+2-d_G(v)\le \left\lfloor {\frac{M+2p-2}{2p}}\right\rfloor$. By the minimality of $G, H=G-\{vv_1,vv_2\}$ has a (p, 1)-total labelling with color interval C. Erase the colors of $v_1$ and $v_2$, and denote this partial (p, 1)-total labelling by $\phi$. Then $|A_{\phi }(vv_1)|\ge (M+2p-1)-|E_{\phi }(v_1)|-|E_{\phi }(v)|-|I_{\phi }(v)|\ge (M+2p-1)-(d_G(v_1)-1)-(d_G(v)-2)-(2p-1)=1$. Analogously, $|A_{\phi }(vv_2)|\ge 1$.

Case 1. $\max \{|A_{\phi }(vv_1)|, |A_{\phi }(vv_2)|\}\ge 2$ or $A_{\phi }(vv_1)\ne A_{\phi }(vv_2)$. Then $vv_1$ and $vv_2$ can be colored properly by choosing colors from $A_{\phi }(vv_1)$ and $A_{\phi }(vv_2)$, respectively.

Case 2. Otherwise, suppose that $A_{\phi }(vv_1)=A_{\phi }(vv_2)=\{\alpha \}$. Then $(E_{\phi }(v_1)\cup E_{\phi }(v_2))\bigcap (E_{\phi }(v)\cup I_{\phi }(v))=\emptyset$, otherwise, at least one of $vv_1$ and $vv_2$ has at least two available colors under the partial (p, 1)-total labelling $\phi$, which is similar to Case 1. Note that $E_{\phi }(v_1)= E_{\phi }(v_2)$ otherwise $A_{\phi }(vv_1)\ne A_{\phi }(vv_2)$. Hence exchange the colors of $uv_1$ and uv. Denote this new partial (p, 1)-total labelling by $\phi '$. we can see that $A_{\phi '}(vv_1)=A_{\phi }(vv_1)=\{\alpha \}$ and $|E_{\phi '}(v)\cap E_{\phi '}(v_2)|=1$. So we have $|A_{\phi '}(vv_2)|\ge (M+2p-1)-|E_{\phi '}(v_2)\cup E_{\phi '}(v)\cup I_{\phi '}(v)|\ge (M+2p-1)-(d_G(v_2)-1)-(d_G(v)-2)-(2p-1)+1=2$, which implies that we can extend $\phi '$ to a new partial (p, 1)-total labelling $\phi ''$ of G with $v_1$ and $v_2$ uncolored. Since $2\le d_G(v_1)=d_G(v_2)\le \left\lfloor {\frac{M+2p-2}{2p}}\right\rfloor$, by Lemma 8, we can obtain a (p, 1)-total labelling of G, a contradiction. $\square$

Lemma 12

Each $\Delta$-vertex can be adjacent to at most one 2-vertex.

The proof of this Lemma is similar to that of Lemma 2.6 (e) in Yu et al. (2011), so we omit it here.

3 Proof of Theorem 4

Proof of Theorem 4

Suppose that G is a minimum counterexample to the theorem in terms of $|V(G)|+|E(G)|$ with $M\ge 4p+4$. Then G is connected and $\delta (G)\ge 2$. By Lemmas 9 and 10, G has some structural properties as follows:

(C1):

A 4-vertex is adjacent to $(2P+4)^+$-vertices;

(C2):

There is no $[5, 2p+2, 2p+2]$-face;

(C3):

If $f=[d_G(v_1),d_G(v_2),d_G(v_3)]$ is a triangle with $d_G(v_1)=5, d_G(v_2)=2p+2$ and $d_G(v_3)=2p+3$, then $v_1$ has no $(2p+2)$-neighbor besides $v_2$.

By Euler’s formula, we can easily deduce that $\sum _{x\in V(G)\cup E(G)}c(x)=\sum _{v\in V(G)}(d_{G}(v)-4)+\sum _{f\in F(G)}(d_{G}(f)-4)=-8<0$. Now we assign an initial charge function c on $x\in V(G)\cup F(G)$ by letting $c(x)=d_G(x)-4$. The discharging rules are defined as follows.

R1. :

Each 2-vertex receives charge $\frac{1}{2}$ from each of its adjacent $\Delta$-vertex and receives charge 1 from its 3-master.

R2. :

Each 3-vertex receives charge 1 from its 3-master.

R3. :

Each 5-vertex transfers charge $\frac{1}{4}$ to each of its incident 3-faces $[5, 2p+2, 2p+3]$ and $\frac{1}{6}$ otherwise.

R4. :

Each k-vertex with $6\le k\le 2p+3$ transfers charge $\frac{k-4}{k}$ to each 3-face that is incident with it.

R5. :

Each $(2p+4)^+$-vertex transfer charge $\frac{1}{2}$ to each 3-face that is incident with it.

Let f be a k-face of G. If $k\ge 4$, then $c'(f)=c(f)\ge 0$ since $4^+$-faces do not participate in the rules. If $k=3$, assume that $f=[d_G(v_1), d_G(v_2), d_G(v_3)]$ with $d_G(v_1)\le d_G(v_2)\le d_G(v_3)$. Consider the following subcases.

• (1) Suppose $d_G(v_1)\le 3$. Then by Lemma 5, $\min \{d_G(v_2), d_G(v_3)\}\ge M+2-3\ge M-1\ge 4p+3$. Hence, $c'(f)=c(f)+2 \times \frac{1}{2}=0$ by R5;

• (2) Suppose $d_G(v_1)=4$. Then by $(C1), d_G(v_3)\ge d_G(v_2)\ge 2p+4$. Hence, $c'(f)=c(f)+2 \times \frac{1}{2}=0$ by R5;

• (3) Suppose $d_G(v_1)=5$. By (C2) and $(C3), d_G(v_2)= 2p+2, d_G(v_3)\ge 2p+3$ or $d_G(v_3)\ge d_G(v_2)\ge 2p+3$. If $d_G(v_2)= 2p+2, d_G(v_3)= 2p+3$, then $c'(f)\ge c(f)+\frac{1}{4}+\frac{2p-2}{2p+2}+\frac{2p-1}{2p+3}\ge -1+\frac{1}{4}+\frac{1}{3}+\frac{3}{7}>0$ by R3, R4 and $p\ge 2$. If f is a 3-face other than $[5, 2p+2, 2p+3]$, then $d_G(v_2)= 2p+2, d_G(v_3)\ge 2p+4$ or $d_G(v_3)\ge d_G(v_2)\ge 2p+3$. Therefore, $c'(f)\ge c(f)+\frac{1}{6}+\min \left\{ \frac{2p-2}{2p+2}+\frac{1}{2}, 2\times \frac{2p-1}{2p+3}, 2\times \frac{1}{2}\right\} \ge 0$ by R3, R4, R5 and $p\ge 2$;

• (4) Suppose $d_G(v_1)=t\ge 6, d_G(v_3)\ge d_G(v_2)\ge 6$. Therefore, $c'(f)\ge c(f)+3\times \min \left\{ \frac{t-4}{t}, \frac{1}{2}\right\} =0$ by R4 and R5.

Let v be a s-vertex of G. If $s=2$, then $c'(v)=c(v)+1+2\times \frac{1}{2}=-2+1+1=0$ by R1 and Lemma 7. If $s=3$, then $c'(v)=c(v)+1=-1+1=0$ by R2 and Lemma 7. If $s=4$, then $c'(v)=c(v)=0$ since 4-vertices do not participate in any rules. If $s=5$, then $c'(v)\ge c(v)-2\times \frac{1}{4}-3\times \frac{1}{6}=0$ by R3, since v is incident with at most two 3-faces $[5, 2p+2, 2p+3]$ by (C2) and (C3). If $6\le s\le 2p+3$, then $c'(v)\ge c(v)-s\frac{s-4}{s}=0$ by R4. If $2p+4\le s\le M-2$, then $c'(v)\ge c(v)-s\frac{1}{2}\ge 0$ by R5 and Lemma 5.

By Lemma 5, if $\Delta (G)=M$, then $\delta (G)\ge 2$; if $\Delta (G)=M-1$, then $\delta (G)\ge 3$; if $\Delta (G)\le M-2$, then $\delta (G)\ge 4$. Thus, when $\Delta (G)\le M-2, c'(v)\ge 0$ for all $v\in V(G)$. Consider the cases of $\Delta (G)=M-1$ and $\Delta (G)=M$.

Case 1. $\Delta (G)=M-1$. Then $\delta (G)\ge 3$. By Lemma 11, $(M-1)$-vertex is incident with at most $(M-4)$ 3-faces if it has at least two 3-neighbors. Hence, we have $c'(v)\ge \min \{c(v)-\frac{1}{2}\Delta (G)-1, c(v)-\frac{1}{2}(\Delta (G)-3)-2\}=\frac{M-1}{2}-5\ge \frac{1}{2}$ by Lemma 7, Lemma 11, R2 and R5.

Case 2. $\Delta (G)=M$. Then $\delta (G)\ge 2$. For $s=\Delta -1=M-1$, see Case 1. For $s=\Delta =M$, note that v is adjacent to at most one 2-vertex by Lemma 12. Hence, together with R1, R2, R5, Lemma 7, Lemma 11 and Lemma 12, we have $c'(v)\ge c(v)-\frac{1}{2}\Delta (G)-2=\frac{M}{2}-6\ge 0$ when v is adjacent to no 2-vertex and at least one 3-vertices; $c'(v)\ge c(v)-\frac{1}{2}(\Delta (G)-1)-2-\frac{1}{2}=\frac{M}{2}-6\ge 0$ when v is adjacent to one 2-vertex and at least one 3-vertex; $c'(v)\ge c(v)-\frac{1}{2}(\Delta (G)-1)-1-\frac{1}{2}=\frac{M}{2}-5\ge 1$ when v is adjacent to one 2-vertex and no 3-vertex; $c'(v)\ge c(v)-\frac{1}{2}\Delta (G)=\frac{M}{2}-4\ge 2$ when v is adjacent to no 2-vertex and no 3-vertex.

Therefore, we obtain that $0> \sum _{x\in V(G)\cup E(G)}c(x)=\sum _{x\in V(G)\cup E(G)}c'(x)\ge 0$. This contradiction completes our proof.