Theoretical Computer Science 522 (2014) 54–61

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Theoretical Computer Science

www.elsevier.com/locate/tcs

Total coloring of planar graphs with maximum degree 8

Huijuan Wang a, Lidong Wu b, Jianliang Wu a,∗a School of Mathematics, Shandong University, Jinan, 250100, Chinab Department of Computer Science, University of Texas at Dallas, Richardson, TX 75080, USA

a r t i c l e i n f o a b s t r a c t

Article history:Received 17 October 2013Accepted 10 December 2013Communicated by D.-Z. Du

Keywords:Planar graphTotal coloringCycle

Let G be a planar graph with Δ � 8 and without adjacent cycles of size i and j, for some3 � i � j � 5. In this paper, it is proved that G is (Δ + 1)-total-colorable.

© 2013 Elsevier B.V. All rights reserved.

1. Introduction

All graphs considered in this paper are simple, finite and undirected, and we follow [2] for terminologies and notationsnot defined here. Let G be a graph. We use V , E , Δ and δ to denote the vertex set, the edge set, the maximum degree andthe minimum degree of G , respectively. A k-vertex, k−-vertex or k+-vertex is a vertex of degree k, at most k or at least k,respectively. Similarly, we define a k-face, k−-face and k+-face. We say that two cycles are adjacent if they share at least oneedge. We use (v1, v2, . . . , vn) to denote a cycle whose vertices are consecutively v1, v2, . . . , vn .

A k-total-coloring of a graph G = (V , E) is a coloring of V ∪ E using k colors such that no two adjacent or incident ele-ments receive the same color. A graph G is k-total-colorable if it admits a k-total-coloring. The total chromatic number χ ′′(G)

of G is the smallest integer k such that G is k-total-colorable. Clearly, χ ′′(G) � Δ + 1. Behzad, and Vizing independently,posed the famous conjecture, known as the Total Coloring Conjecture (TCC).

Conjecture 1. For any graph G, χ ′′(G) �Δ + 2.

Conjecture 1 was confirmed for graphs with Δ � 5 [9]. In recent years, the study of total colorings of planar graphs hasattracted considerable attention. For planar graphs, the only open case of Conjecture 1 is that of Δ = 6 [9,13]. For graphsembedded in a surface Σ of Euler characteristic χ(Σ)� 0, the only open case of Conjecture 1 is also that of Δ = 6 [16,20].Furthermore, the total chromatic number of planar graphs with higher maximum degree can be determined. More precisely,it is known that χ ′′(G) = Δ + 1 if G is a planar graph with Δ � 9 [10]. Some related results can be found in [3,4,6,7,11,12,17,19]. For a planar graph G with Δ � 8, it is known that χ ′′(G) = Δ + 1 if G does not contain 5-cycles or 6-cycles [8], orintersecting triangles [14,18], or adjacent triangles [5], or adjacent 4-cycles [15]. In this paper, we generalize some of theformer results and get the following theorem.

* Corresponding author.E-mail address: jlwu@sdu.edu.cn (J.L. Wu).

0304-3975/$ – see front matter © 2013 Elsevier B.V. All rights reserved.http://dx.doi.org/10.1016/j.tcs.2013.12.006

H.J. Wang et al. / Theoretical Computer Science 522 (2014) 54–61 55

Fig. 1. Reducible configurations of Lemma 2, where d(v) = 7 in (1).

Theorem 1. Suppose that G is a planar graph with Δ � 8 and without adjacent cycles of size i and j, for some 3 � i � j � 5. Thenχ ′′(G) = Δ + 1.

2. Reducible configurations

According to [10], planar graphs with Δ � 9 are (Δ + 1)-total-colorable. Therefore, to prove Theorem 1, it suffices toprove the case of Δ = 8. Let G = (V , E, F ) be a minimal counterexample to Theorem 1, in the sense that the quantity|V | + |E| is minimum. By [13], every planar graph with Δ � 7 has a 9-total-coloring, so every proper subgraph of G has a9-total-coloring ϕ using the color set C = {1,2, . . . ,9}. For a vertex v , let C(v) be the set of colors assigned to the edgesincident with v , and let C ′(v) = C\(C(v) ∪ {ϕ(v)}). This section is devoted to investigating some structural information,which shows that certain configurations are reducible, that is, they cannot occur in G .

Lemma 1. (See [1].)

(1) G is 2-connected.(2) If uv is an edge of G with d(u) � 4, then d(u) + d(v) �Δ + 2 = 10.(3) The subgraph G28 of G induced by all edges joining 2-vertices to 8-vertices is a forest.

In each component T of G28, there is a matching M in T which pairs off all the 2-vertices with some of the 8-vertices:in the graph T , choose an 8-vertex u as the root of T , and math each 2-vertex v with the 8-vertex w adjacent to v whichis farther than its another neighbor z from u (note that z is also an 8-vertex and the leaves of T are all 8-vertices). In thiscase, the vertex w is called the child of v , and the vertex z is called the parent of v . So every 2-vertex has exactly oneparent and exactly one child, which are all 8-vertices. Moreover, if an 8-vertex is adjacent to at least two 2-vertices, thenthis 8-vertex is a child of at most one 2-vertex and the parent of the remaining 2-vertices adjacent to it.

Lemma 2. (See [5,10].) G has no configurations depicted in Fig. 1(1)–(9), where the vertices marked by • have no other neighbors inG, and d(v) = 7 in Fig. 1(1).

Lemma 3. Suppose that v is an 8-vertex, u is a neighbor of v with d(u) = 2, and v1, v2, . . . , vk are consecutive neighbors of v withd(vi) � 3 for 1 � i � k, where k ∈ {4,5, . . . ,7}. If v is incident with two 3-cycles (v1, v, v2) and (vk−1, v, vk), and incident with4-cycles (v, vi, xi, vi+1) for 2 � i � k − 2, then at least one vertex in {v2, v3, . . . , vk−1} is a 4+-vertex.

Proof. Assume to the contrary that d(vi) = 3 for i = 2,3, . . . ,k − 1 (see Fig. 2(1)). Let w be the neighbor of u differentfrom v in G . By the minimality of G , G ′ = G − uv has a 9-total-coloring ϕ . Erase the colors on u, v2, v3, . . . , vk−1 andwithout loss of generality, we assume that ϕ(uw) = 1. Then 1 /∈ C(v) ∪ {ϕ(v)}, since otherwise, vu touches nine colorsand can be colored properly. Without loss of generality, let ϕ(v v1) = 2 and ϕ(v vk) = 3. It is easy to see that 1 ∈ C(vi) fori = 2,3, . . . ,k − 1, since otherwise, recolor v vi with 1, color vu with ϕ(v vi) and color u, v2, v3, . . . , vk−1 properly, also acontradiction. We also have 1 ∈ {ϕ(v1 v2)∪ϕ(vk−1 vk)}, since otherwise, there is a vertex xt (2 � t � k − 2) on which color 1appears twice.

First, suppose ϕ(v1 v2) = ϕ(vk−1 vk) = 1. Then we have ϕ(v2x2) = 2, since otherwise, exchange the colors on v v1 andv1 v2, color vu with 2 and color u, v2, v3, . . . , vk−1 properly, which yields a proper coloring of G . Similarly, ϕ(vk−1xk−2) = 3.Suppose that ϕ(v3x2) �= 1. Now, ϕ(v3x3) = ϕ(v4x4) = · · · = ϕ(vk−2xk−2) = 1. Thus ϕ(vk−2xk−3) = 3, since otherwise, wemay get a contradiction by exchanging the colors, respectively, on v vk and vk−1 vk , and on vk−1xk−2 and vk−2xk−2, col-oring vu with 3 and coloring u, v2, v3, . . . , vk−1 properly. Similarly, ϕ(vk−3xk−4) = ϕ(vk−4xk−5) = · · · = ϕ(v3x2) = 3. Then

56 H.J. Wang et al. / Theoretical Computer Science 522 (2014) 54–61

Fig. 2. Reducible configurations of Lemma 3.

Fig. 3. Reducible configuration of Lemma 4.

Fig. 4. Reducible configuration of Lemma 5.

exchange the colors, respectively, on v v1 and v1 v2, and on v2x2 and v3x2, and color vu with 2, also a contradiction. Soϕ(v3x2) = 1. Similarly, ϕ(v4x3) = ϕ(v5x4) = · · · = ϕ(vk−2xk−3) = 1 and ϕ(v3x3) = ϕ(v4x4) = · · · = ϕ(vk−2xk−2) = 2 (seeFig. 2(1)). Then, the coloring as shown in Fig. 2(2) is a nice coloring of G , if we color u, v2, v3, . . . , vk−1 properly.

Second, suppose exactly one of ϕ(v1 v2) and ϕ(vk−1 vk) is 1. Without loss of generality, let ϕ(v1 v2) = ϕ(v3x2) =ϕ(v4x3) = · · · = ϕ(vk−1xk−2) = 1. Then we have ϕ(v2x2) = ϕ(v3x3) = · · · = ϕ(vk−2xk−2) = ϕ(vk−1 vk) = 2. By exchangingcolors, respectively, on v v1 and v1 v2, on v2x2 and v3x2, on v3x3 and x3 v4, . . . , on vk−2xk−2 and vk−1xk−2, on vk−1 vk andv vk , coloring vu with 3 and coloring u, v2, v3, . . . , vk−1 properly, we get a proper coloring of G, a contradiction. �Lemma 4. Suppose that v is an 8-vertex and v1, v2, . . . , vk are consecutive neighbors of v with d(v2) = 2, d(v1) � 3 and d(vi) � 3for 3 � i � k, where k ∈ {4,5, . . . ,8}. If v is incident with a 3-cycle (v, vk−1, vk), and incident with 4-cycles (v, vi, xi, vi+1) for1 � i � k − 2, then at least one vertex in {v1, v3, v4, . . . , vk−1} is a 4+-vertex.

Proof. Assume to the contrary that d(vi) = 3 for i = 1,3,4, . . . ,k − 1 (see Fig. 3). Note that x1 = x2. Denote by x0 thevertex adjacent to v1 different from v and x2. By the minimality of G , G ′ = G − v v2 has a 9-total-coloring ϕ . Erase thecolors on v1, v2, . . . , vk−1 and assume that ϕ(x2 v2) = 1. Then color 1 does not appear at v . So without loss of generality,assume ϕ(v vk) = 2. By a similar argument as in the proof of Lemma 3, we have ϕ(x0 v1) = ϕ(x3 v3) = · · · = ϕ(xk−2 vk−2) =ϕ(vk−1 vk) = 1 and ϕ(vk−1xk−2) = ϕ(vk−2xk−3) = · · · = ϕ(v4x3) = ϕ(v3x2) = 2 (for otherwise, we can get a contradictioneasily). Now, we first exchange colors, respectively, on v vk and vk−1 vk , on xk−2 vk−1 and xk−2 vk−2, on xk−3 vk−2 andxk−3 vk−3, . . . , on x3 v4 and x3 v3, on x2 v3 and x2 v2. Next, exchange the colors on x2 v2 and x2 v1. Finally, we color v v2with 2 and color v1, v2, . . . , vk−1 properly. Thus we get a 9-total-coloring of G , a contradiction. �Lemma 5. Suppose that v is an 8-vertex and v2, v3, . . . , vk are consecutive neighbors of v with d(v2) = 2 and d(vi) � 3 for 3 �i � k, where k ∈ {4,5, . . . ,7}. If v is incident with a 3-cycle (v, vk−1, vk), incident with a 3-cycle (v, vs, vt) where d(vs) = 3 and{s, t} �= {k − 1,k}, and incident with 4-cycles (v, vi, xi, vi+1) for 2 � i � k − 2, then at least one vertex in {v3, v4, . . . , vk−1} is a4+-vertex.

Proof. Assume to the contrary that d(vi) = 3 for i = 3,4, . . . ,k − 1 (see Fig. 4). By the minimality of G , G ′ = G − v v2 hasa 9-total-coloring ϕ . Erase the colors on v2, v3, . . . , vk−1, vs and assume that ϕ(x2 v2) = 1. Then color 1 does not appearat v . So without loss of generality, assume ϕ(v vk) = 2, ϕ(v vs) = 3 and ϕ(v vt) = 4. By a similar argument as in the proof

H.J. Wang et al. / Theoretical Computer Science 522 (2014) 54–61 57

of Lemma 3, we have ϕ(x3 v3) = ϕ(x4 v4) = · · · = ϕ(xk−2 vk−2) = ϕ(vk−1 vk) = 1 and ϕ(vk−1xk−2) = ϕ(vk−2xk−3) = · · · =ϕ(v4x3) = ϕ(v3x2) = 2. Moreover, we also have 1 ∈ C(vs). Suppose ϕ(vs vt) = 1. Then ϕ(xs vs) = 4, since otherwise, wecan exchange the colors on v vt and vs vt , color v v2 with 4 and color v2, v3, . . . , vk−1, vs properly, a contradiction. Next,exchange the colors, respectively, on v vk and vk−1 vk , on xk−2 vk−1 and xk−2 vk−2, on xk−3 vk−2 and xk−3 vk−3, . . . , on x3 v4and x3 v3, on x2 v3 and x2 v2. Finally, we recolor v vs with 2, color v v2 with 3 and color v2, v3, . . . , vk−1, vs properly. Thuswe get a 9-total-coloring of G , a contradiction. Suppose ϕ(xs vs) = 1. Then ϕ(vs vt) = 2, since otherwise, we can exchangethe colors, respectively, on v vk and vk−1 vk , on xk−2 vk−1 and xk−2 vk−2, on xk−3 vk−2 and xk−3 vk−3, . . . , on x3 v4 and x3 v3,on x2 v3 and x2 v2, recolor v vs with 2, color v v2 with 3 and color v2, v3, . . . , vk−1, vs properly, also a contradiction. Next,exchange the colors, respectively, on v vk and vk−1 vk , on xk−2 vk−1 and xk−2 vk−2, on xk−3 vk−2 and xk−3 vk−3, . . . , on x3 v4and x3 v3, on x2 v3 and x2 v2, on v vt and vs vt , color v v2 with 4 and color v2, v3, . . . , vk−1, vs properly. Thus we get a9-total-coloring of G , a contradiction. �3. Discharging

We shall complete the proof of Theorem 1 by using the discharging method in order to obtain a contradiction. Denoteby nd(v) the number of d-vertices adjacent to the vertex v , by nd( f ) the number of d-vertices incident with the face f , andby fd(v) the number of d-faces incident with the vertex v . Let (V , E, F ) be a plane embedding of G .

First, by Euler’s formula |V | − |E| + |F | = 2, we have∑

v∈V

(2d(v) − 6

) +∑

f ∈F

(d( f ) − 6

) = −6(|V | − |E| + |F |) = −12 < 0.

Next, we define c(x) to be the initial charge of x ∈ V ∪ F . Let c(v) = 2d(v) − 6 for each v ∈ V and c( f ) = d( f ) − 6 foreach f ∈ F . So

∑x∈V ∪F c(x) = −12 < 0. Then we apply the following rules to redistribute the initial charge that leads to a

new charge c′(x).

(R1) To each 2-vertex from its child, transfer 32 , and from its parent, transfer 1

2 .(R2) From each 4-vertex to each of the k-faces incident with it, where 3 � k � 5, transfer 1

2 .(R3) From each 5-vertex to each of the k-faces incident with it, transfer

1, if k = 3;12 , if k = 4 or 5.

(R4) From each 6-vertex to each of the k-faces f incident with it, transfer

54 , if k = 3 and n4−( f ) = 1;1, if k = 3 and n4−( f ) = 0;23 , if k = 4 and n3−( f ) = 1;12 , if k = 4 and n3−( f ) = 0;13 , if k = 5.

(R5) From each 7+-vertex to each of the k-faces f incident with it, transfer

32 , if k = 3 and n3−( f ) = 1;54 , if k = 3, n3−( f ) = 0 and n4( f ) = 1;1, if k = 3 and n4−( f ) = 0;1, if k = 4 and n3−( f ) = 2;34 , if k = 4 and n3−( f ) = 1,n5−( f ) = 2;23 , if k = 4 and n3−( f ) = 1,n6+( f ) = 3;12 , if k = 4 and n3−( f ) = 0;13 , if k = 5 and n3−( f ) = 2;14 , if k = 5 and n3−( f ) � 1.

The rest of this paper is to check that c′(x) � 0 for all x ∈ V ∪ F which will be the desired contradiction.

58 H.J. Wang et al. / Theoretical Computer Science 522 (2014) 54–61

3.1. Final charge of faces

Let f ∈ F be a face of G . Clearly, c′( f ) = c( f )� 0 if d( f )� 6. First, suppose that d( f ) = 3. If n3− ( f ) = 1, then n7+ ( f ) = 2by Lemma 1(2). So c′( f ) = c( f ) + 3

2 × 2 = 0 by (R5). If n4( f ) = 1, then n6+ ( f ) = 2 by Lemma 1(2), and c′( f ) � c( f ) +12 + 5

4 × 2 = 0 by (R2), (R4) and (R5). If n5+ ( f ) � 3, then c′( f ) � c( f ) + 1 × 3 = 0 by (R3)–(R5). Second, suppose d( f ) = 4.If n3− ( f ) = 2, then n7+ ( f ) = 2 by Lemma 1(2), and c′( f ) = c( f ) + 1 × 2 = 0 by (R5). If n3− ( f ) = 1 and n5− ( f ) = 2, thenc′( f )� c( f )+ 1

2 + 34 × 2 = 0. If n3− ( f ) = 1 and n6+ ( f ) = 3, then it follows that c′( f ) � c( f )+ 2

3 × 3 = 0. If n3− ( f ) = 0, then

c′( f )� c( f )+ 12 ×4 = 0. Third, suppose d( f ) = 5. If n3− ( f ) = 2, then n7+ ( f ) = 3 by Lemma 1(2), so c′( f ) � c( f )+ 1

3 ×3 = 0.

Otherwise, c′( f )� c( f ) + 14 × 4 = 0. Hence, c′( f )� 0 for all faces.

For convenience, we first give a claim.

Claim 1. Let v be a vertex of G. By the condition of Theorem 1, v is incident with at most three consecutive 3-faces. Moreover, if v isincident with three consecutive 3-faces, then these three 3-faces are adjacent with two 6+-faces incident with v.

3.2. Final charge of vertices

Let v be a vertex of G . If d(v) = 2, then c′(v) = −2 + 32 + 1

2 = 0 by (R1). If d(v) = 3, then clearly c′(v) = c(v) = 0. Ifd(v) = 4, then c′(v) � c(v) − 1

2 × 4 = 0 by (R2). If d(v) = 5, then f3(v) � 3 by Claim 1. By (R3), c′(v) � c(v) − 1 × 3 − 12 ×

2 = 0.Suppose d(v) = 6. Then c(v) = 6, f3(v) � 4 by Claim 1 and all the neighbors of v are 4+-vertices by Lemma 1(2). If

f3(v) � 3, then c′(v) � 6 − ( 54 × 3 + 2

3 × 3) = 14 > 0 by (R4). Otherwise, f3(v) = 4 and f5+ (v) = 2 by the condition of

Theorem 1. Then we have c′(v) � 6 − ( 54 × 4 + 1

3 × 2) = 13 > 0 by (R4).

Suppose d(v) = 7. Then c(v) = 8, f3(v) � 5 by Claim 1 and all the neighbors of v are 3+-vertices by Lemma 1(2).By Lemma 2(1), v is incident with at most two 3-faces each of which is incident with a 3-vertex, that is, there are atmost two 3-faces each of which receives 3

2 from v . If f3(v) = 5, then clearly f6+ (v) = 2 by the condition of Theorem 1and Claim 1. So c′(v) � 8 − ( 3

2 × 2 + 54 × 3) = 5

4 > 0 by (R5). Suppose f3(v) = 4. If v is incident with a 3-face which isincident with a 3-vertex u, then v is not adjacent to 3-vertex except u by Lemma 2(1). So each of the 4+-faces incidentwith v receives at most 3

4 from v by (R5), and c′(v) � 8 − ( 32 × 2 + 5

4 × 2 + 34 × 3) = 1

4 > 0. Otherwise, each of the 3-facesincident with v receives at most 5

4 from v , and c′(v) � 8 − ( 54 × 4 + 1 × 3) = 0. Suppose f3(v) = 3. If f5+ (v) � 1, then

c′(v) � 8 − ( 32 × 2 + 5

4 + 1 × 3 + 13 ) = 5

12 > 0. Otherwise, f4(v) = 4 and v is not incident with adjacent 3-faces by thecondition of Theorem 1. So v is incident with at least one 4-face which is incident with at least three 4+-vertices. Hence,c′(v) � 8 − ( 3

2 × 2 + 54 + 1 × 3 + 3

4 ) = 0. Suppose f3(v) � 2. Then c′(v) � 8 − ( 32 × 2 + 1 × 5) = 0.

Suppose d(v) = 8. Then c(v) = 10. Now, suppose n2(v) = 0. Then v does not send any charge to 2-vertices and f3(v) � 6by Claim 1. Suppose f3(v) = 6. Then v is not incident with 4-face and 5-face by the condition of Theorem 1. Hence,f6+ (v) = 2 and it follows that c′(v) � 10 − 3

2 × 6 = 1 > 0. Suppose f3(v) = 5. If f5+ (v) � 1, then c′(v) � 10 − ( 32 × 5 +

1 × 2 + 13 ) = 1

6 > 0. Otherwise, f4(v) = 3. Clearly, v is incident with adjacent cycles of size i and j, for 3 � i � j � 5,

a contradiction to Theorem 1. Suppose f3(v) � 4. Then we have c′(v) � 10 − ( 32 × 4 + 1 × 4) = 0. So we assume in the

following that n2(v) � 1. Then by Lemma 2(8), v is not incident with two adjacent 3-faces which share a common 3-vertex.By earlier arguments, note that v is child of at most one 2-vertex and parent of the remaining 2-vertices incident with v .So v sends a total of at most n2(v)+2

2 to the 2-vertices adjacent to it by (R1). In the following, we will discuss how muchcharge is transferred from v to its incident 5−-faces.

Case 1. n2(v) � 5. Then f3(v) � 2 by Lemma 2(3) and f6+ (v) � 2 by Lemma 2(2). If f6+ (v) � 4, then c′(v) �10 − (

n2(v)+22 + 3

2 × 2 + 1 × 2) � 0. Otherwise, f6+ (v) � 3 and hence by Lemma 2(2), n2(v) = 5. If f6+ (v) = 3, then byLemma 2(2, 3) we have f3(v) � 1 and c′(v) � 10 − ( 5+2

2 + 32 + 1 × 4) = 1 > 0. If f6+ (v) = 2, then by Lemma 2(2, 3), it

follows that f3(v) = 0 and c′(v) � 10 − ( 5+22 + 1 × 6) = 1

2 > 0.

Case 2. n2(v) = 4. Then f3(v) � 3 by Lemma 2(3) and c(v) − 4+22 = 10 − 3 = 7. Suppose f6+ (v) � 3. Then c′(v) �

7−( 32 ×3+1×2) = 1

2 > 0. Suppose f6+ (v) = 2. Then by Lemma 2(2, 3) we have f3(v) � 2 and c′(v) � 7−( 32 ×2+1×4) = 0.

Suppose f6+ (v) = 1. Then f3(v) � 1 by Lemma 2(2, 3). So we consider the number of 3-faces which are incident with v .Suppose f3(v) = 1. If f5(v) � 1, then c′(v) � 7 − ( 3

2 + 1 × 5 + 13 ) = 1

6 > 0. Otherwise, f4(v) = 6. If the 3-face incidentwith v is not incident with a 3-vertex, then v is incident with at least two 4-faces each of which is incident with at leastthree 4+-vertices and c′(v) � 7 − ( 5

4 + 1 × 4 + 34 × 2) = 1

4 > 0. Otherwise, this 3-face is incident with a 3-vertex, denotedby y. Then the other face which is incident with v and y is a 5-face by Lemma 2(5), a contradiction to f4(v) = 6. Supposef3(v) = 0. Then c′(v) � 7 − 1 × 7 = 0. Suppose f6+ (v) = 0. Then clearly we have f3(v) = 0 by Lemma 2(2, 3). If there is a3-vertex adjacent to v , say u, then u is incident with at least one 5+-face which is incident with v by Lemma 2(6) and vsends at most 1

3 + 1 to the two faces which are incident with uv . Otherwise, we can assume that v has four neighbors of

degree at least 4. For every such neighbor u, v sends at most 34 × 2 to the two faces which are incident with uv . Hence,

c′(v) � 7 − 3 × 8 = 1 > 0.

4

H.J. Wang et al. / Theoretical Computer Science 522 (2014) 54–61 59

Case 3. n2(v) = 3. Then f3(v) � 3 by Claim 1 and Lemma 2(3), and c(v) − 3+22 = 10 − 5

2 = 152 . Suppose f6+ (v) � 2. Then

c′(v) � 152 − ( 3

2 × 3 + 1 × 3) = 0. So we assume f6+ (v) � 1. In the following we consider two cases.

Case 3.1. f6+ (v) = 1. Suppose f3(v) � 1. Then c′(v) � 152 − ( 3

2 + 1 × 6) = 0. Suppose f3(v) = 2. If f5(v) � 1, thenc′(v) � 15

2 − ( 32 × 2 + 1 × 4 + 1

3 ) = 16 > 0. Otherwise, f4(v) = 5. Then v is incident with at least two 4-faces each of which

is incident with at least three 4+-vertices by Lemma 2(5, 6, 8), and c′(v) � 152 − ( 3

2 × 2 + 1 × 3 + 34 × 2) = 0. Suppose

f3(v) = 3. If none of these three 3-faces is incident with a 3-vertex, then c′(v) � 152 − ( 5

4 × 3 + 1 × 2 + 34 × 2) = 1

4 > 0. Ifv is incident with exactly one 3-face which is incident with a 3-vertex, then f5(v) � 1 by Lemma 2(2, 5, 8) and c′(v) �152 − ( 3

2 + 54 ×2+1×2+ 3

4 + 13 ) = 5

12 > 0. If v is incident with exactly two 3-faces each of which is incident with a 3-vertex,

then f5(v) � 2 by Lemma 2(2, 5, 8) and c′(v) � 152 − ( 3

2 × 2 + 54 + 1 × 2 + 1

3 × 2) = 712 > 0. If each of these three 3-faces

incident with v is incident with a 3-vertex, then f5(v) � 3 by Lemma 2(2, 5, 8) and c′(v) � 152 − ( 3

2 ×3+1+ 13 ×3) = 1 > 0.

Case 3.2. f6+ (v) = 0. Then f3(v) � 2 by Lemma 2(2, 3). Suppose f3(v) = 0. If f5(v) � 1, then c′(v) � 152 − (1 × 7 +

13 ) = 1

6 > 0. Otherwise, f4(v) = 8. Then v is adjacent to at least one 4+-vertex by Lemma 2(2, 6), and c′(v) � 152 − (1 ×

6 + 34 × 2) = 0. Suppose f3(v) = 1. If f5(v) � 2, then c′(v) � 15

2 − ( 32 + 1 × 5 + 1

3 × 2) = 13 > 0. If f5(v) = 1, then by

Lemma 2(2, 5, 6), v is incident with at least two 4-faces each of which is incident with at least three 4+-vertices, andc′(v) � 15

2 − ( 32 + 1 × 4 + 3

4 × 2 + 13 ) = 1

6 > 0. If f5(v) = 0, then by Lemma 2(2, 5, 6, 7, 9), v is incident with at least four

4-faces each of which is incident with at least three 4+-vertices, and c′(v) � 152 − ( 3

2 +1×3+ 34 ×4) = 0. Suppose f3(v) = 2.

If none of these two 3-faces is incident with a 3-vertex, then v is incident with at most two 4-faces each of which is incidentwith two 3−-vertices by Lemma 2(2, 5, 6). Hence, we have c′(v) � 15

2 − ( 54 × 2 + 1 × 2 + 3

4 × 4) = 0. If v is incident withexactly one 3-face which is incident with a 3-vertex, then f5(v) � 1 by Lemma 2(2, 5, 8), and v is incident with at most two4-faces each of which is incident with two 3−-vertices by Lemma 2(2, 6, 8). So c′(v) � 15

2 − ( 32 + 5

4 + 1 × 2 + 34 × 3 + 1

3 ) =16 > 0. If each of the two 3-faces incident with v is incident with a 3-vertex, then f5(v) � 2 by Lemma 2(2, 5), and v isincident with at most two 4-faces each of which is incident with two 3−-vertices by Lemma 2(2, 6, 8). Thus, it follows thatc′(v) � 15

2 − ( 32 × 2 + 1 × 2 + 3

4 × 2 + 13 × 2) = 1

3 > 0.

Case 4. n2(v) = 2. Then f3(v) � 4 by Lemma 2(3) and Claim 1, and c(v) − 2+22 = 10 − 2 = 8. If f6+ (v) � 2, then none of

these 6+-faces receives charge from v , hence c′(v) � 8 − ( 32 × 4 + 1 × 2) = 0. Thus we assume f6+ (v) � 1. Next, we consider

the following two cases.

Case 4.1. f6+ (v) = 1. Then we consider the number of 3-faces incident with v . Suppose f3(v) � 2. Then clearly c′(v) �8 − ( 3

2 × 2 + 1 × 5) = 0. Suppose f3(v) = 3. If f5(v) � 1, then c′(v) � 8 − ( 32 × 3 + 1 × 3 + 1

3 ) = 16 > 0. Otherwise, f4(v) = 4.

If there are two adjacent 3-faces incident with v , then v is incident with adjacent cycles of size i and j, for 3 � i � j � 5, acontradiction. Otherwise, v is not incident with adjacent 3-faces. Then each of the 4-faces incident with v is incident withat least three 4+-vertices by Lemma 2(5) and Lemma 3. Then c′(v) � 8 − ( 3

2 × 3 + 34 × 4) = 1

2 > 0. Suppose f3(v) = 4. Iff5(v) � 2, then c′(v) � 8 − ( 3

2 × 4 + 1 + 13 × 2) = 1

3 > 0. Otherwise, f5(v) � 1. If f5(v) = 0, then v is incident with adjacentcycles of size i and j, for 3 � i � j � 5, also a contradiction. If f5(v) = 1, then v is incident with at least two 3-faces none ofwhich is incident with a 3−-vertex by Lemma 2(2, 5, 8) and Lemma 3. Hence, c′(v) � 8− ( 3

2 ×2+ 54 ×2+1×2+ 1

3 ) = 16 > 0.

Case 4.2. f6+ (v) = 0. Then we consider the number of 5-faces incident with v . Suppose f5(v) � 3. Then clearly c′(v) �8 − ( 3

2 × 4 + 1 + 13 × 3) = 0. Suppose f5(v) = 2. If f3(v) � 2, then c′(v) � 8 − ( 3

2 × 2 + 1 × 4 + 13 × 2) = 1

3 > 0. If f3(v) � 4,then v is incident with adjacent cycles of size i and j, for 3 � i � j � 5, a contradiction. So f3(v) = 3. If there are atmost two 3-faces each of which is incident with v and a 3-vertex, then c′(v) � 8 − ( 3

2 × 2 + 54 + 1 × 3 + 1

3 × 2) = 112 > 0.

Otherwise, each of 3-faces incident with v is incident with a 3-vertex. So v is not incident with three consecutive 3-faces byLemma 2(2, 8). Then v is incident with at least one 4-face which is incident with at least three 4+-vertices by Lemma 2(5, 8)and Lemma 3. Hence it follows that c′(v) � 8 − ( 3

2 × 3 + 1 × 2 + 34 + 1

3 × 2) = 112 > 0. Suppose f5(v) = 1. If f3(v) � 1, then

c′(v) � 8 − ( 32 + 1 × 6 + 1

3 ) = 16 > 0. If f3(v) � 4, then v is incident with adjacent cycles of size i and j, for 3 � i � j � 5,

a contradiction. If f3(v) = 3, then v is not incident with two adjacent 3-faces by the condition of Theorem 1. So eachof the 4-faces incident with v is incident with at least three 4+-vertices by Lemma 2(2, 5) and Lemma 3. Then c′(v) �8 − ( 3

2 × 3 + 34 × 4 + 1

3 ) = 16 > 0. So we may assume f3(v) = 2. Then the same arguments as above show that v is not

incident with two adjacent 3-faces, and v is incident with at least two 4-faces each of which is incident with at leastthree 4+-vertices. Hence, c′(v) � 8 − ( 3

2 × 2 + 1 × 3 + 34 × 2 + 1

3 ) = 16 > 0. Suppose f5(v) = 0. Then we also consider the

number of 3-faces incident with v . Clearly, by the condition of Theorem 1, f3(v) � 3. Suppose f3(v) = 3. Then v is notincident with two adjacent 3-faces by the condition of Theorem 1, and v is incident with at most one 3-face which isincident with a 3-vertex by Lemma 2(2, 5) and Lemma 3, and by the same argument, each of the 4-faces incident with vis incident with at least three 4+-vertices. Hence, c′(v) � 8 − ( 3

2 + 54 × 2 + 3

4 × 5) = 14 > 0. Suppose f3(v) = 2. Then the

3-faces incident with v are not adjacent by the condition of Theorem 1. If none of the 3-faces incident with v is incidentwith a 3-vertex, then v is incident with at least three 4-faces each of which is incident with at least three 4+-vertices.So c′(v) � 8 − ( 5

4 × 2 + 1 × 3 + 34 × 3) = 1

4 > 0. If v is incident with exactly one 3-face which is incident with a 3-vertex,then v is incident with at least three 4-faces each of which is incident with at least three 4+-vertices Lemma 2(2, 5, 6). So

60 H.J. Wang et al. / Theoretical Computer Science 522 (2014) 54–61

c′(v) � 8 − ( 32 + 5

4 + 1 × 3 + 34 × 3) = 0. If each of the 3-faces incident with v is incident with a 3-vertex, then v is incident

with at least four 4-faces each of which is incident with at least three 4+-vertices Lemma 2(2, 5, 6, 9) and Lemma 3. Soc′(v) � 8 − ( 3

2 × 2 + 1 × 2 + 34 × 4) = 0. Suppose f3(v) = 1. Then by Lemma 2(2, 5, 6, 7, 9) and Lemma 4, v is incident with

at most five 4-faces each of which is incident with two 3−-vertices. Hence c′(v) � 8 − ( 32 + 1 × 5 + 3

4 × 2) = 0. Supposef3(v) = 0. Then it follows that c′(v) � 8 − (1 × 8) = 0.

Case 5. n2(v) = 1. Then c(v) − 1+22 = 10 − 3

2 = 172 . Let u be the 2-vertex adjacent to v . Next, we consider the following

two cases.

Case 5.1. u is incident with a 3-cycle. Then by Claim 1, f3(v) � 6 and by Lemma 2(4), none of the 3-faces incidentwith v is incident with a 3-vertex. If f3(v) = 6, then clearly f6+ (v) = 2 by the condition of Theorem 1, and c′(v) �172 − ( 3

2 + 54 × 5) = 3

4 > 0. If f3(v) = 5, then by the condition of Theorem 1, f5+ (v) � 2 or f6+ (v) � 1. Hence, c′(v) �172 − ( 3

2 + 54 × 4 + 1 × 2) = 0. Suppose f3(v) = 4. If the 3-faces incident with v are not adjacent, then each of the 4-faces

is incident with at least three 4+-vertices by Lemma 2(4), and c′(v) � 172 − ( 3

2 + 54 × 3 + 3

4 × 4) = 14 > 0. Otherwise,

the same arguments above show that f5+ (v) � 2 or f6+ (v) � 1, then c′(v) � 172 − ( 3

2 + 54 × 3 + 1 × 3) = 1

4 > 0. Supposef3(v) = 3. If f5+ (v) � 1, then c′(v) � 17

2 − ( 32 + 5

4 × 2 + 1 × 4 + 13 ) = 1

6 > 0. Otherwise, f5+ (v) = 0. Then, similarly asbefore, v is not incident with adjacent 3-faces, and v is incident with at least two 4-faces each of which is incidentwith at least three 4+-vertices. Hence, we have c′(v) � 17

2 − ( 32 + 5

4 × 2 + 34 × 2 + 1 × 3) = 0. Suppose f3(v) � 2. Then

c′(v) � 172 − ( 3

2 + 54 + 1 × 5 + 3

4 ) = 0.

Case 5.2. u is not incident with a 3-cycle. Then f3(v) � 5 by Claim 1. Suppose f3(v) = 5. Then by the condition ofTheorem 1 and Claim 1, f6+ (v) � 2 and f4(v) � 1. So c′(v) � 17

2 − ( 32 × 5 + 1) = 0. Suppose f3(v) = 4. First, suppose

f5+ (v) � 3 or f6+ (v) � 2 or f6+ (v) � 1 and f5(v) � 1. Then c′(v) � 172 − ( 3

2 × 4 + 1 × 2 + 13 ) = 1

6 > 0. Second, supposef6+ (v) = 1 and f5(v) = 0. Then f4(v) = 3 and v is not incident with adjacent 3-faces by the condition of Theorem 1. Butit contradicts to u is not incident with a 3-cycle. Third, suppose f6+ (v) = 0. Then v is not incident with three consecutive3-faces by Claim 1. Suppose f5(v) � 1. Then it is a contradiction to Theorem 1. Suppose f5(v) = 2 and f4(v) = 2. If v isincident with at most three 3-faces each of which is incident with a 3-vertex, then c′(v) � 17

2 − ( 32 ×3+ 5

4 +1×2+ 13 ×2) =

112 > 0. Otherwise, v is incident with at least one 4-face which is incident with at least three 4+-vertices by Lemma 3 andLemma 5. So c′(v) � 17

2 − ( 32 × 4 + 1 + 3

4 + 13 × 2) = 1

12 > 0. Suppose f3(v) = 3. If f5+ (v) � 2 or f6+ (v) � 1, then c′(v) �172 − ( 3

2 × 3 + 1 × 4) = 0. Otherwise, f4(v) = 5 or f4(v) = 4 and f5(v) = 1. And the 3-faces incident with v are not adjacentto each other by the condition of Theorem 1. Suppose f4(v) = 4. Then v is incident with at least two 4-faces each of whichis incident with at least three 4+-vertices by Lemma 3 and Lemma 4. Hence, c′(v) � 17

2 − ( 32 ×3+1×2+ 3

4 ×2+ 13 ) = 1

6 > 0.Suppose f4(v) = 5. Then v is incident with at least two 4-faces each of which is incident with at least three 4+-vertices. Ifv is incident with at most one 3-face which is incident with a 3-vertex, then c′(v) � 17

2 − ( 32 + 5

4 × 2 + 1 × 3 + 34 × 2) = 0.

Otherwise, v is incident with at least four 4-faces each of which is incident with at least three 4+-vertices by Lemma 3 andLemma 5. So c′(v) � 17

2 − ( 32 × 3 + 1 + 3

4 × 4) = 0. Suppose f3(v) = 2. If f5+ (v) � 1, then c′(v) � 172 − ( 3

2 × 2 + 1 × 5 + 13 ) =

16 > 0. Otherwise, f4(v) = 6 and by the condition of Theorem 1, the two 3-faces incident with v are not adjacent. Thenby Lemma 4, v is incident with at least two 4-faces each of which is incident with at least three 4+-vertices. Hence,c′(v) � 17

2 − ( 32 × 2 + 1 × 4 + 3

4 × 2) = 0. Suppose f3(v) � 1. Then c′(v) � 172 − max{ 3

2 + 1 × 6 + 34 , 1 × 8} = 1

4 > 0.

In all cases we have shown that c′(x) � 0 for every x ∈ V ∪ F , and therefore,∑

x∈V ∪F c(x) = ∑x∈V ∪F c′(x) � 0, a contra-

diction. This completes the proof of Theorem 1.

Acknowledgements

This work was supported in part by the National Natural Science Foundation of China under grants 11201440, 11271006,11301410, Natural Science Basic Research Plan in Shaanxi Province of China under grant 2013JQ1002, and ScientificResearch Foundation for the Excellent Young and Middle-Aged Scientists of Shandong Province of China under grantBS2013DX002. This work was also supported in part by the National Science Foundation of USA under grants CNS-0831579and CCF-0728851.

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