On total coloring of 1-planar graphs without 4-cycles

Wei-Ping Ni1, Wen-yao Song1
1School of Mathematics and Statistics, Zaozhuang University, Shandong, 277160, China

Abstract

A graph is 1-planar if it can be drawn on the plane so that each edge is crossed by at most one other edge. In this paper, we confirm the total-coloring conjecture for 1-planar graphs without 4-cycles with maximum degree Δ10.

Keywords: planar graphs, graph coloring, planar graph theory, edge crossings

1. Introduction

All graphs considered are finite, simple and undirected. Let G be a graph. We use V(G), E(G), Δ(G) and δ(G) to denote its vertex set, edge set, maximum degree and minimum degree, respectively. For a vertex vV(G), NG(v) denotes the set of vertices that are adjacent to v in G. By d(v):=|NG(v)| denotes the degree of v in G. For planar graphs G, F(G) denotes its face set, the degree of a face f, denoted by d(f), is the length of a boundary walk around f in G. We call v a k-vertex, or a k+-vertex, or a k-vertex if d(v)=k, or d(v)k, or d(v)k respectively and call f a k-face, or a k+-face, or a k-face if d(f)=k, or d(f)k, or d(f)k respectively. Any undefined notation follows that of Bondy and Murty [2]. A totalkcoloring of a graph G is a coloring of V(G)E(G) using k colors such that no two adjacent or incident elements receive the same color. The totalchromatic number χ(G) of G is the smallest integer k such that G has a total-k-coloring. It is clearly that χ(G)Δ(G)+1. Behzad and Vizing [1,6] posed independently the conjecture, χ(G)Δ(G)+2 for any graph G, which is known as the total coloring conjecture.

A graph is 1-planar if it can be drawn on the plane so that each edge is crossed by at most one other edge. The notion of a 1-planar graph was introduced by Ringel [4] in connection with the problem of simultaneous coloring of adjacent/incident vertices and faces of plane graphs. In [10], Zhang et al. proved that every 1-planar graph with maximum degree Δ(G)16 is totally (Δ(G)+2)-choosable, which implies that the total-coloring conjecture holds for 1-planar graphs with maximum degree at least 16. Later, Czap [3] proved (Without discharging method) that for every 1-planar graph G with Δ(G)10 it holds χ(G)Δ(G)+3. Moreover, if χ(G)4, then χ(G)Δ(G)+2. In the same paper, the author also verified that for every 1-planar graph G without adjacent triangles and with Δ(G)10 it holds χ(G)Δ(G)+3. Moreover, if χ(G)4, then χ(G)Δ(G)+2. Zhang and Hou [7] showed the following theorem which improve the lower bound for the maximum degree in the corollary of [10] to 13.

Recently, Sun and Wu [5] verified the total coloring χ(G)r+2, for every 1-planar graph G if Δ(G)9 and g(G)4 where Δ(G) is the maximum degree of G and g(G) is the girth of G

Theorem 1.1. Let G be a 1-planar graph with maximum degree Δ(G) and let r be an integer. If Δr and r13, then χ(G)r+2.

In this paper, we shall prove the following results:

Theorem 1.2. Let G be a 1-planar graph without 4-cycles, with maximum degree Δ(G)10. Then χ(G)Δ(G)+2.

2. Preliminaries

Let G in this paper has been embedded on a plane such that every edge is crossed by at most one other edge and the number of crossings is as small as possible. The associated plane graph G× of G is obtained by turning all crossings of G into new 4-vertices on a plane. For convenience, a vertex in G× is called false if it is not a vertex of G and real otherwise. A false face means a face f in G× that is incident with at least one false vertex; otherwise, f is a normal face. For a vertex vV(G×), we use fk(v) to denote the number of k-faces which are incident with v, ni(v) to denote the number of i-vertices which are adjacent to v, and nc(v) to denote the number of false vertices which are adjacent to v.

For convenience, we use v1,v2,,vd to denote the neighbors of a d-vertex v in G× that occur around it in a clockwise order. By fi denote the face incident with vvi and vvi+1 in G×, where the addition on subscripts are taken modulo d.

Let G be a counterexample with |E(G)| as small as possible to Theorem 1.2. By minimality of G we can assume that it is connected and that it has no total (Δ(G) + 2)-colorings. First we investigate some structural of properties of G. Here, we give some known lemmas.

Lemma 2.1. .[10] Let uv be an edge in G. If min{dG(u),dG(v)}Δ+12, then dG(u)+dG(v)Δ+3.

From this lemma, we deduce that δ(G)3.

Lemma 2.2. [7] Let Vi be the set of i-vertices in G. We have |VΔ|>2|V3|.

Lemma 2.3. Let G be a 1-planar graph without 4-cycles and let G× be its associated plane graph. Then for every 5+ vertex vV(G), v is incident with at most 45dG(v) 3-faces in G×.

The proof is just similar to the one in [8], with only quite a little minor changes. So we omit it here and refer the reader to Lemma 4 of [8].

Lemma 2.4. [9] Let G be a 1-plane graph and let G× be its associated plane graph.Then the following hold:

1) For any two false vertices u and v in G×, u and v are not adjacent in G×.

2) If dG(u)=3 and v is a false vertex in G×, then either u and v are not adjacent in G×, or uv is not incident with two 3-faces.

3) Let v be a 3-vertex in G. If v is incident with two false 3-faces vv1v2 and vv1v3 in G×, then v2 and v3 are both false and v is incident with a 5+-face in G×.

Lemma 2.5. Let G be a 1-plane graph and let G× be its associated plane graph. Then, every 5-face in G× is incident with at most four 5-vertices.

The proof is just similar to the one in [7], with only quite a little minor changes. So we omit it here and refer the reader to Lemma 9 of [7].

Lemma 2.6. [10] For each integer 3k5, let Xk={xV(G)|dG(x)k}, Yk=xXkNG(x). If Xk, then there exists a bipartite subgraph Mk=(Xk,Yk) of G such that dMk(x)=1 for any xXk and dMk(y)k2 for any yYk. We call y the k-master of x if xyMk and xXk.

By this lemma, we deduce that each k-vertex (3k5) has a j-master (kj5).

Lemma 2.7. [7] Let G be a 1-plane graph and let v be a vertex of G. If dG(v)=3, then, v cannot be contained in a triangle in G. If dG(v)=4 with NG(v)=v1,v2,v3,v4, then, for any i,(1i4), the edge vvi can not be contained in two triangles.

Lemma 2.8. Let G be a 1-plane graph without 4-cycles and G× be its associated plane graph. Let v be a vertex of G, then, there are no five consecutive 3-faces that are incident with v in G×. If v is incident with i consecutive 3-faces f1,f2,,fi, (3i4) in G×, then, there is at most a real small vertex among the neighbors of v on these consecutive 3-faces. Moreover, if v is incident with 4 consecutive 3-faces f1,f2,f3,f4, then v1,v3,v5 are false vertices, v2,v4 are real vertices.

The proof is just similar to the one in [8], with only quite a little minor changes. So we omit it here and refer the reader to Lemma 4 of [8].

3. The proof of Theorem 1.2

Then, we begin to prove the main result of the paper.

A vertex v in G is small if d(v)5 and is big if d(v)6. Note that the degree of a false vertex in G× is four, so every false vertex is small.

In the following, we apply the discharging method on associated 1-planar graph G× of G and complete the proof by a contradiction. Since G× is a plane graph, we have vV(G×)(d(v)6)+fF(G×)(2d(f)6)=12, by the well-known Euler’s formular. Now we define the initial charge function ch(x) of xV(G×)F(G×). Let ch(v)=d(v)6  if xV(G×) and ch(f)=2d(f)6 if xF(G×). And we define suitable discharging rules below to change the initial charge function ch(x) to the final charge function ch(x) on V(G×)F(G×). Then we still have xV(G×)F(G×)ch(x)=xV(G×)F(G×)ch(x)=12, since any discharging procedure preserves the total charge of G×.

Our discharging rules are defined as follows.

R1. Each f in G× where d(f)4 sends 2d(f)6t(f) to each small vertex incident with it, where t(f) is the number of small vertices incident with the face f.

R2. Each 3-vertex in G receives 29 from its i-master (3i5).

R3. Each 4-vertex in G receives 625 from its i-master (4i5).

R4. Each Δ-vertex gives 12 to a common pot from which each 3-vertex receives 1, if |V3|>0.

R5. Let w be a false vertex and w is incident with a 3-face f in G×, then each 8+-neighbor of w on f sends 1350 to w.

R6. Let w be a real 4-vertex and w is incident with a normal 3-face f in G×, then each 8+-neighbor of w on f sends 1350 to w.

R7. Let u be a real 4 -vertex, v is a false vertex in G×, uvE(G×) and uv is incident with two 3-faces in G×, then v sends 1325 to u.

R8. If a false vertex v in G× is incident with four 4+-faces in G×, then v sends 512 to each 4-neighbor of v.

R9. If a false vertex v in G× is incident with exactly one 3-face f in G×, then v sends 13 to its 3-neighbor on f.

R10. Let v be a 3-vertex and v is not incident with any 3-face in G×, then v sends 16 to each false vertex which is adjacent to v.

R11. If a real 4-vertex v in G× is incident with four 4+-faces in G×, then v sends 1175 to each false vertex which is adjacent to v.

R12. If a false vertex u in G× is adjacent to a 5-vertex v in G×, and uv is incident with 4+-faces f1 and f2 which are adjacent in G×, then v sends 16 to u.

In the following, we check that the final charge ch(x) on each vertex and face is nonnegative, and we also show the final charge of the common pot is nonnegative. This implies that xV(G×)F(G×)ch(x)0 for all xV(G×)F(G×), a contradiction. This completes the proof of Theorem 1.2.

First of all, by R4, the final charge of the common pot is at least 12|VΔ||V3|>0 since |VΔ|>2|V3| by Lemma 2.2. One can also check that the final charge of every face in F(G×) is nonnegative by R1. Thus in the following we consider the vertices in G×.

Case 1. d=3. By R2 and R4, v receives 1 from the common pot and 29×3=23 from its i-masters, where 3i5. Since G is a 1-planar graph without 4-cycles, v is incident with at most two 3-faces in G× by Lemma 2.4 and Lemma 2.7. Now, we consider three subcases.

Case 1.1. If v is not incident with any 3-face in G×, then f1,f2,f3 are all 4+-faces.

First, assume that v is incident with at least one 5+-face, without loss of generality, assume that f1, then v would receive at least 1 from f1, and 24×2=1 from f2,f3, by Lemma 2.5 and R1. By R10, v sends at most 16×3=12 to false vertices which are adjacent to v. Thus, ch(v)3+1+23+1+112>0.

Second, assume that f1,f2,f3 are all 4-faces. If v is adjacent to at least one real vertex in G×, say v1, then d(v1)10, thus f1,f3 sends at least 23×2=43 to v, and f2 sends at least 24=12 to v by R1. By R10, v sends at most 16×3=12 to false vertices which are adjacent to v. Thus,ch(v)3+1+23+43+1212=0. Otherwise, v1,v2,v3 are all false vertices. Let xi be the fourth(undefined) vertices of the 4-faces fi (i=1,2,3). It is easy to check that x1x2,x2x3,x3x1E(G) by the drawing of G. Since f1,f2,f3 are all 4-faces, there are at least two big vertices among x1,x2,x3 by Lemma 2.1, without loss of generality, assume that x1,x2, thus, f1,f2 send at least 23×2=43 to v, and f3 sends at least 24=12 to v by R1. By R10, v sends at most 16×3=12 to false vertices which are adjacent to v. Thus, ch(v)3+1+23+43+1212=0.

Case 1.2. If v is incident with exactly one 3-face in G×, then without loss of generality assume that f3 is a 3-face. Since no two false vertices are adjacent in G× by Lemma 2.4, there is a real vertex, among v1 and v3, say v1, then d(v1)10.

Assume that v2 is also a real vertex, then d(v2)10. Thus, f1 sends at least 1 to v, f2 sends at least 241=23 to v by R1, Thus,ch(v)3+1+23+1+23>0. Otherwise, v2 is a false vertex. Let xi be the second neighbors of v2 on fi (i=1,2), it is easy to check that x1x2E(G) by the drawing of G. Thus, at least one of x1 and x2 is big by Lemma 2.1. This implies that v receives at least min{1+12,23×2}=43, from f1 and f2 by R1. Therefore, ch(v)3+1+23+43=0.

Case 1.3. If v is incident with exactly two 3-faces in G×, then without loss of generality assume that f2 and f3 are 3-faces. By Lemma 2.4 and Lemma 2.7, v3 must be a real vertex, v1 and v2 are false vertices, and f1 is a 5+-face. Thus, f1 sends at least 451=1 to v by Lemma 2.5 and R1. Since G is a 1-planar graph without 4-cycles, so, there is at least a vertex among v1 and v2 which is incident with exactly one 3-face, say v2. Then, v2 sends 13 to v by R9. Thus, ch(v)3+1+23+1+13=0.

Case 2. d=4 and v is a real vertex, then v has one 4-master and one 5-master. So v receives totally 625×2=1225 from its masters by R3. Since G is a 1-planar graph without 4-cycles, v is incident with at most three 3-faces in G× by Lemma 2.7.

If v is incident with exactly one 3-face in G×, say f1, then there is at most one false vertex among v1 and v2 by Lemma 2.4. Suppose that v1 is a false vertex, then, d(v2)9 by Lemma 2.1, thus, v receives at least 241=23 from f2, receives 12×2=1 from f3 and f4 by R1. Therefore, ch(v)2+1225+23+1=1175.

If v is not incident with any 3-face, then v is incident with four 4+-faces.

First, assume that v is incident with at least one 5+-face, say f1, then, v receives at least 1 from f1 by Lemma 2.5 and R1, receives 12×3=32 from f2, f3 and f4 by R1. v sends at most 1175×4=4475 to false vertices that are adjacent to v by R11. Therefore, ch(v)2+1225+1+324475>0.

Second, assume that v is incident with four 4-faces, if the neighbors of v are all false vertices, then, let xi be the fourth(undefined) vertices of the 4-faces fi (i=1,2,3,4). It is easy to check that x1x2,x3x4E(G) by the drawing of G. Thus, at least one of x1 and x2 is big, similarly to x3 and x4 by Lemma 2.1. This implies that v receives at least 12×2+23×2=73 from f1, f2,f3 and f4 by R1. v sends at most 1175×4=4475 to false vertices that are adjacent to v by R11. Therefore, ch(v)2+1225+734475=1775. Otherwise, v is adjacent to at least one real vertex. Thus, v receives at least 12×4=2 from f1, f2,f3 and f4 by R1. v sends at most 1175×3=3375 to false vertices that are adjacent to v by R11. Therefore, ch(v)2+1225+23375=375.

If v is incident with exactly three 3-faces, then without loss of generality assume that f1 f2 and f4 are 3-faces. Since G is a 1-planar graph without 4-cycles, so, v is adjacent to exactly two false vertices by Lemma 2.4 and Lemma 2.7 in G×, and moreover f3 is a 5+-face. First, assume that two false vertices are not adjacent, say v1 and v3, then d(v2)9, d(v4)9 by Lemma 2.1. Then, f3 sends at least 1 to v by R1, v1 sends 1325 to v by R7. Thus, ch(v)2+1225+1+1325=0.

Second, assume that two false vertices are v3 and v4, then, d(v1)9, d(v2)9 by Lemma 2.1. Thus, f3 sends at least 1 to v by R1, v1 and v2 send 1350×2=1325 to v by R6. This implies that ch(v)2+1225+1+1325=0.

If v is incident with exactly two 3-faces in G×, we consider four subcases.

Case 2.1. If v is not adjacent to any false vertex, then vi9 (i=1,2,3,4) by Lemma 2.1, and the two 3-faces that are incident with v have no common edge by Lemma 7, without loss of generality assume that f2 and f4 are 3-faces. Then, v receives a total of 242×2=2 from f1 and f3, thus, ch(v)2+1225+2>0.

Case 2.2. If v is adjacent to exactly one false vertex, without loss of generality assume that v1, then d(vi)9 (i=2,3,4) by Lemma 2.1. First, assume that the two 3-faces that are incident with v have no common edge, say f2 and f4, then, v receives at least 242=1 from f3, receives at least 241=23 from f1 by R1, and v receives 1350×2 from v2 and v3 by R6. Thus, ch(v)2+1225+1+23+1350×2=23. Second, assume that the two 3-faces that are incident with v have one common edge, since G has no 4-cycles, then, v1 is incident with at least one 3-face. If v1 is incident with exactly one 3-face, without loss of generality assume that f1, then f2 is a real 3-face in G×. By R6, v receives 1350×2 from v2 and v3, v receives at least 242=1 from f3 and receives at least 241=23 from f4 by R1. Thus, ch(v)2+1225+1+23+1350×2=23. If v1 is incident with two 3-faces, say f1 and f4, then, v receives at least 242×2=2 from f2 and f3. Therefore, ch(v)2+1225+2=1225.

Case 2.3. If v is adjacent to exactly two false vertices.

First, assume that two faces which are incident with v are not adjacent, say f2 and f4 are both 3-faces, then, f1 and f3 are both 4+-faces. If two false vertices that are adjacent to v are incident with the same 4+-face, say f1, then, v1 and v2 are both false vertices, v3 and v4 are both big vertices. Since G has no 4-cycles, then, f1 is a 5+-face. It implies that f1 sends at least 1 to v and f3 sends at least 242=1 to v by Lemma 2.5 and R1. Thus, ch(v)2+1225+1+1=1225. Otherwise, two false vertices that are adjacent to v are incident with different 4+-faces, say f1 and f3, since G has no 4-cycles, then, f1 and f3 are 5+-faces. Thus, v receives at least 451×2=2 from f1 and f3 by Lemma 2.5 and R1. Therefore, ch(v)2+1225+2=1225.

Second, assume that two faces which are incident with v are adjacent, say f1 and f2 are both 3-faces, then, f3 and f4 are both 4+-faces. If two false vertices that are adjacent to v are incident with the same 4+-face, without loss of generality assume that v1 and v4 are both false vertices, then, d(vi)9(i=2,3) by Lemma 2.1. So, v receives 1350×2=1325 from v2 and v3 by R6, v receives at least 24×2=1 from f3 and f4 by R1, therefore, ch(v)2+1225+1325+1=0. If two false vertices that are adjacent to v are incident with different 4+-faces, say f3 and f4, then, v1 and v3 are both false vertices, and d(vi)9(i=2,4) by Lemma 2.1. Since G has no 4-cycles, then, f3 and f4 are all 5+-faces. So, v receives at least 451×2=2 from f3 and f4 by Lemma 2.5 and R1, Therefore, ch(v)2+1225+2=1225. If two false vertices that are adjacent to v are v2 and v4, since G has no 4-cycles, there is at least one 5+-face among f3 and f4, say f3. Thus, f3 sends at least 451=1 to v, f4 sends at least 241=23 to v by Lemma 2.5 and R1. Therefore,ch(v)2+1225+1+23=1175.

Case 2.4. If v is adjacent to exactly three false vertices, say v1, v2 and v3, since G has exactly two 3-faces, so, f3 and f4 are all 3-faces by Lemma 2.4, f1 and f2 are all 4+-faces. Since G has not 4-cycles, so, f1 and f2 are either all 4-faces, or all 5+-faces, or there is at least one 6+-face. First assume that there is one 6+-face among f1 and f2, say f1. Let xi (i=1,2) be the second(undefined) neighbors of v2 on fi, it is easy to check that x1x2E(G) by the drawing of G. Thus, at least one of x1 and x2 is big by Lemma 2.1. Then, v receives at least min{661+12,66+241} =53 from f1 and f2 by R1.

Therefore, ch(v)2+1225+53=1175. Second, assume that f1 and f2 are all 5+-faces, then, v receives at least 451×2=2 from f1 and f2 by Lemma 2.5 and R1, thus, ch(v)2+1225+2=1225. Third, assume that f1 and f2 are all 4-faces, then, v2 is incident with four 4-faces, because otherwise, G has 4-cycles. By R8, v receives 512 from v2. Let xi (i=1,2) be the fourth(undefined) vertices of the 4-faces fi, it is easy to check that x1x2E(G) by the drawing of G. Thus, at least one of x1 and x2 is big by Lemma 2.1. This implies that v receives at least 12+23=76 from f1 and f2 by R1. Therefore, ch(v)2+1225+512+76=19300

Case 3. d=4 and v is a false vertex, then, the neighbors of v are real vertices, and v is adjacent to at most two small vertices in G by Lemma 2.1. Since G has no 4-cycles, so, v is incident with at most two 3-faces, we consider three subcases.

Case 3.1. If v is not incident with any 3-face in G×, then v is incident with four 4+-faces in G×.

Assume first that v has at least one 4-neighbor, say v1, then, d(v3)9, moreover, there is at least one big among v2 and v4 by Lemma 2.1, say v2, thus, v would receive at least 242=1 from f2, at least 241×2=43 from f1 and f3, at least 24=12 from f4 by R1, v sends at most 512×2=56 to its 4-neighbors by R8. Therefore, ch(v)2+1+43+1256=0. Otherwise, v does not have any 4-neighbors, then, v sends out nothing by R8, v would receive at least 24×4=2, Thus,ch(v)2+2=0.

Case 3.2. If v is incident with exactly one 3-face in G×, then without loss of generality assume that f1 is the 3-face. There is at least one big among v1 and v2, say v2.

Assume first that v1 is a 3-vertex, then, both v2 and v3 are 10+-vertices by Lemma 2.1. Thus, v would receive at least 242=1 from f2, at least 241=23 from f3, at least 24=12 from f4 by R1, v would receive 1350 from v2 by R5, v sends at most 13 to v1 by R9. Thus, ch(v)2+1+23+12+135013=775. Second, assume that 4d(v1)7, then, both v2 and v3 are 6+-vertices by Lemma 2.1. So, v would receive at least 242=1 from f2, at least 241=23 from f3, at least 24=12 from f4 by R1, v sends out nothing by R9. Therefore,ch(v)2+1+23+12=16. Third, assume that v1 is a 8+-vertex, then, v1 sends 1350 to v by R5. There is at least a big vertex among v2 and v4, so, f2 and f4 send min{242+12,241×2} =43 to v, f3 sends 24=12 to v by R1, v sends out nothing by R9. Therefore,ch(v)2+1350+43+12=775.

Case 3.3. If v is incident with two 3-faces in G×, since G has no 4-cycles, then, the two 3-faces have a common edge, without loss of generality assume that f3 and f4 are 3-faces. There is a big vertex among v1 and v3, say v1.

First assume that 6d(v4)7 , then, both v2 and v3 are big by Lemma 2.1, moreover, f1 and f2 send at least 242×2=2 to v by R1, v sends out nothing by R7. Thus, ch(v)2+2=0. Second, assume that d(v4)5, then, vi (i=1,2,3) is 8+-vertex by Lemma 2.1, moreover, f1 and f2 send at least 242×2=2 to v by R1, v1 and v3 send 1350×2=1325 to v by R5, v sends at most 1325 to v4 by R7. Thus,ch(v)2+2+13251325=0.

Third assume that v4 is a 8+-vertex. If there is at least one 5+-face among f1 and f2, say f1, then, f1 sends at least 451=1 to v, f2 sends at least 24=12 to v by Lemma 2.5 and R1, v4 sends 1350×2=1325 to v by R5. Thus, ch(v)2+1+12+1325=150. Otherwise, f1 and f2 are all 4-faces. Let xi(i=1,2) be the second(undefined) neighbors of v2 on fi, since G has no 4-cycles, then, both x1 and x2 are false vertices. Suppose that v2 is a 6+-vertex, then, f1 sends at least 242=1 to v, f2 sends at least 241=23 to v by R1, v4 sends 1350×2=1325 to v by R5. Thus, ch(v)2+1+23+1325=1475. Suppose that v2 is a 3-vertex or a 4-vertex, since G has no 4-cycles, then, v2 is not incident with any 3-faces. By R10 and R11, v2 sends at least 1175 to v. Suppose that v2 is a 5-vertex, by R12,v2 sends 16 to v. Thus, when v2 is a 5-vertex, this implies that v2 sends at least 1175 to v. And moreover, we consider the degree of v3. If v3 is a 5-vertex, then, v1 is a 8+-vertex by Lemma 2.1, v1 sends 1350 to v, v4 sends 1350×2=1325 to v by R5, f1 sends at least 241=23 to v, f2 sends at least 24=12 to v by R1. Therefore, ch(v)2+1175+1350+1325+23+12=775. If v3 is a 6+-vertex, since v1 is big, then, each of f1 and f2 sends at least 241=23 to v by R1, v4 sends 1350×2=1325 to v by R5, therefore, ch(v)2+1175+23×2+1325=0.

Case 4. d=5. v is incident with at most four 3-faces in G× by Lemma 2.3. If v would send charges to a false vertex which adjacent to v by R12, then v is incident with at most three 3-faces in G×. First assume that v is incident with exactly four 3-faces in G×, say f1, f2, f3 and f4, then v1,v3, and v5 are false vertices, v2 and v4 are real vertices by Lemma 2.8. Since G has no 4-cycles, then, f5 is a 6+-face. Thus, f5 sends 1 to v by R1, therefore, ch(v)1+1=0. Second assume that v is incident with exactly three 3-faces in G×, then, v is incident with two 4+-faces. If the two 4+-faces are not adjacent, then, v sends out nothing by R12, v would receive at least 24×2=1 from two 4+-faces which are incident with v. Thus, ch(v)1+1=0. If two 4+-faces are adjacent, without loss of generality, assume that f1 and f2 are 4+-faces. Moreover, if v2 is a real vertex, then, v sends out nothing by R12, v would receive at least 24×2=1 from f1 and f2 by R1. Thus, ch(v)1+1=0. Otherwise, v2 is a false vertex, then, v sends at most 16 to v2 by R12. Let xi be the second(undefined) neighbors of v2 on fi (i=1,2), it is easy to check that x1x2E(G) by the drawing of G. Thus, at least one of x1 and x2 is big by Lemma 2.1. This implies v would receive at least 24+241=76 from f1 and f2 by R1, therefore, ch(v)1+7616=0. Third assume that v is incident with at most two 3-faces in G×, then, v is incident with at least three 4+-faces. v sends at most 16×3=12 to false vertices which are adjacent to v by R12. v would receive at least 24×3=32 from 4+-faces which are incident with v, thus, ch(v)1+3212=0.

Case 5. 6d7. Then it is trivial that ch(v)=ch(v)0.

Case 6. 8dΔ(G)2. By Lemma 2.3, v is incident with at most 45d 3-faces in G×, then v sends at most 45d×1350 to false vertices and real 4-vertices which are adjacent to v on 3-faces by R5 and R6. Thus, ch(v)d645d×135099d750125>0, since d8.

Case 7. d=Δ(G)1. By Lemma 2.3, v is incident with at most 45d 3-faces in G×. And by Lemma 2.1, we have d(u)4 if uvE(G). So v can be a 4-master vertex of at most two vertices and a 5-master vertex of at most three vertices by Lemma 2.6.

Let Δ(G)=10, Then, d=9. By Lemma 2.3, v is incident with at most seven 3-faces in G×. If v is incident with exactly seven 3-faces in G×, then, by Lemma 2.8, there are four consecutive 3-faces and another three consecutive 3-faces which are incident with v, and v is adjacent to at most two real small vertices. Thus, v sends at most 7×1350 to false vertices and real 4-vertices which are adjacent to v by R5 and R6. v sends at most 625×2=1225 by R3. Therefore, ch(v)967×13501225=710. If v is incident with at most six 3-faces in G×, then, v sends at most 6×1350 to false vertices and real 4-vertices which are adjacent to v by R5 and R6, v sends at most 625×2+625×3=65 to real small vertices which are adjacent to it by R3. Thus, ch(v)966×135065=625.

Let Δ(G)11, then, d10. By Lemma 2.3, v is incident with at most 45d 3-faces in G×. v sends at most 45d×1350 to false vertices and real 4-vertices which are adjacent to v by R5 and R6, v sends out at most 625×2+625×3=65 by R3. Thus ch(v)Δ(G)1645(Δ(G)1)×13506599Δ(G)9991250, since Δ(G)11.

Case 8. d=Δ(G). By Lemma 2.3, v is incident with at most 45d 3-faces in G×. And by Lemma 2.1, we have d(u)3 if uvE(G). So v can be a 3-master vertex of at most one vertex, a 4-master vertex of at most two vertices and a 5-master vertex of at most three vertices by Lemma 2.6.

If d=10, then v is incident with at most eight 3-faces. When v is incident with exactly eight 3-faces, there are two groups four consecutive 3-faces that are incident with v in G×, so, v is adjacent to at most two real small vertices by Lemma 2.8. Thus, v sends at most 625×2=1225 to real small vertices that are adjacent to it by R2 and R3, v sends at most 8×1350 to false vertices and real 4-vertices that are adjacent to v by R5 and R6, and 12 to the common pot by R4. Therefore, ch(v)1068×1350122512=47500. When v is incident with at most seven 3-faces, v sends at most 7×1350 to false vertices and real 4-vertices that are adjacent to v by R5 and R6, v sends at most 625×5+29=6445 to real small vertices that are adjacent to it by R2 and R3, v sends 12 to the common pot by R4. Therefore, ch(v)1067×1350644512=582250.

If d11, v sends at most 45d×1350 to false vertices and real 4-vertices that are adjacent to v by R5 and R6, v sends at most 625×5+29=6445 to real small vertices adjacent to it by R2, R3 and 12 to the common pot by R4. Thus ch(v)Δ(G)645Δ(G)×13506445121782Δ(G)1782522500, since Δ(G)11.

Therefore, we complete the proof of the Theorem.

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