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Some sufficient conditions for 1-planar graphs to be Class 1
Theoretical Computer Science 566 (2015) 50–58
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Theoretical Computer Science www.elsevier.com/locate/tcs
Some sufficient conditions for 1-planar graphs to be Class 1 ✩ Wenwen Zhang, Jian-Liang Wu ∗ School of Mathematics, Shandong University, Jinan, 250100, China
a r t i c l e
i n f o
Article history: Received 5 June 2013 Received in revised form 15 October 2014 Accepted 29 November 2014 Available online 5 December 2014 Communicated by G. Ausiello
a b s t r a c t 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. Let G be a 1-planar graph with maximum degree Δ(G ). In this paper, it is shown that χ (G ) = Δ(G ) if (1) Δ(G ) ≥ 9 and G contains no adjacent chordal 5-cycles, or (2) Δ(G ) ≥ 8 and G contains no adjacent 4-cycles or no 5-cycles. © 2014 Elsevier B.V. All rights reserved.
Keywords: 1-Planar graphs Edge coloring Cycle
1. Introduction All graphs considered in this paper are finite, simple and undirected. Any undefined notation follows that of Bondy and Murty [1]. For a real number x, let x be the greatest integer not larger than x. Let G be a graph. We use V (G ) and E (G ) to denote its vertex set and its edge set, respectively. Let v ∈ V (G ). If uv ∈ E (G ), then the vertex u is said to be a neighbor of v in G. We denote by N G ( v ) the set of neighbors of v in G and by d G ( v ) = | N G ( v )| the degree of v in G. We use δ(G ) and Δ(G ) to denote the minimum degree and the maximum degree of G, respectively. For u , v ∈ V (G ), N G (u , v ) = N G (u ) ∪ N G ( v ) . If G is a planar graph, we assume that G has always been embedded in the plane. Let G be a planar graph. We denote by F (G ) the face set of G. The degree of a face f in G, denoted by d G ( f ), is the number of edges incident with it, where each cut-edge is counted twice. Throughout this paper, a k-, k+ - and k− -vertex (or face) in a planar graph is a vertex (or face) of degree k, at least k and at most k, respectively. A k-cycle is a cycle of length k. Two cycles sharing at least a common edge are said to be adjacent. Given a cycle C of length k in G, an edge uv ∈ E (G )\ E (C ) is called a chord of C if u , v ∈ V (C ). Such a cycle C is also called a chordal-k-cycle. A graph is k edge-colorable if its edges can be colored with k colors in such a way that adjacent edges receive different colors. The edge chromatic number (or chromatic index) of G, denoted by χ (G ), is the smallest integer k such that G is k edge-colorable. For edge coloring, Vizing’s Theorem states that for any graph G, Δ(G ) ≤ χ (G ) ≤ Δ(G ) + 1. A graph G is said to be of Class 1 if χ (G ) = Δ(G ), and of Class 2 if χ (G ) = Δ(G ) + 1. A graph G is critical if G is a connected graph of Class 2 and χ (G − e ) < χ (G ) for every edge e of G. If a graph G is critical and its maximum degree is Δ, we say that G is Δ-critical. For planar graphs, more is known. In [2], Vizing presented examples of planar graphs of Class 2 with maximum degree Δ for each Δ ∈ {2, 3, 4, 5}. He proved that every planar graph with Δ ≥ 8 is of Class 1 and then conjectured that every planar graph with maximum degree 6 or 7 is of Class 1. The case Δ = 7 for the conjecture has been verified by ✩
*
This work is supported by a research grant NSFC (11271006) of China. Corresponding author. E-mail address: [email protected] (J.-L. Wu).
http://dx.doi.org/10.1016/j.tcs.2014.11.031 0304-3975/© 2014 Elsevier B.V. All rights reserved.
W. Zhang, J.-L. Wu / Theoretical Computer Science 566 (2015) 50–58
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Zhang [5] and, independently, by Sanders and Zhao [4]. The case Δ = 6 remains open, but some partial results are obtained, such as [6,7] and [8]. A graph G 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 1-planar graphs was introduced by Ringel [3] in connection with the problem of simultaneous coloring of adjacent/incidence vertices and faces of plane graphs. Compared to the family of planar graphs, 1-planar graphs have not been extensively studied in the literature. The first result concerning the edge colorings of 1-planar graphs is due to Zhang and Wu [10], who proved that every 1-planar graph with maximum degree Δ ≥ 10 is of Class 1. Recently, Zhang and Liu [11] proved that every 1-planar graph without adjacent triangles and with maximum degree Δ ≥ 8 is of Class 1. They [12] also proved that every 1-planar graph without chordal 5-cycles and with maximum degree Δ ≥ 9 is of Class 1, and constructed Class 2 1-planar graphs with maximum degree 6 or 7, so they posed the following conjecture: Conjecture 1. Every 1-planar graph with maximum degree at least 8 is of Class 1. In this paper, we get the following theorems. Theorem 2. Let G be a 1-planar graph without adjacent 4-cycles or without 5-cycles. If Δ(G ) ≥ 8, then χ (G ) = Δ(G ). Theorem 3. Let G be a 1-planar graph without adjacent chordal 5-cycles. If Δ(G ) ≥ 9, then χ (G ) = Δ(G ). In the following, we always assume that G is a 1-planar graph and 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 a 1-plane graph G is the plane graph that is obtained from G by turning all crossings of G into new 4-vertices. A vertex in G × is called false if it is not a vertex of G and true otherwise. We call a 3-face in G × false or true according to whether it is incident with a false vertex or not. Let f k ( v ) denote the number of k-faces incident with v in G × . In [10], the authors showed some basic properties on a 1-planar graph G and its associated plane graph G × , see Lemma 4. Lemma 4. (See [10].) Let G be a 1-planar graph and G × be its associated plane graph. Then the following results hold.
/ E ( G × ); For any two false vertices u and v in G × , uv ∈ If there is a 3-face uv wu in G × such that d G ( v ) = 2, then u and w are not false vertices; / E (G × ) or uv is not incident with two 3-faces; If d G (u ) = 3 and v is a false vertex in G × , then either uv ∈ If a 3-vertex v in G is incident with two 3-faces and adjacent to two false vertices in G × , then v must also be incident with a 5+ -face; (5) For any 4-vertex v in G, v is incident with at most three false 3-faces.
(1) (2) (3) (4)
Here, we give some known results on edge-coloring critical graphs. Lemma 5 (Vizing’s Adjacency Lemma). (See [9].) Let G be a Δ-critical graph and let u , v be adjacent vertices of G with d G ( v ) = k. Then (1) if k < Δ, then u is adjacent to at least (Δ − k + 1) Δ-vertices; (2) if k = Δ, then u is adjacent to at least two Δ-vertices. Lemma 6. (See [4].) No Δ-critical graph has distinct vertices x, y, z such that x is adjacent to y and z, d G ( z) < 2Δ − d G (x) − d G ( y ) + 2, and xz is in at least d G (x) + d G ( y ) − Δ − 2 triangles not containing y. Lemma 7. (See [8].) Let G be a Δ-critical graph with Δ(G ) ≥ 6 and let x be a 4-vertex. Then the following hold: (1) If x is adjacent to a (Δ − 2)-vertex, say y, then every vertex of N G ( N G (x)) \ {x, y } is a Δ-vertex; (2) If x is not adjacent to any (Δ − 2)-vertex and if one of the neighbors y of x is adjacent to d G ( y ) − (Δ − 3) (Δ − 2)− -vertices, then each of the other three neighbors of x is adjacent to only one (Δ − 2)-vertex, which is x; (3) If x is adjacent to a (Δ − 1)-vertex, then there are at least two Δ-vertices in N G (x) which are adjacent to at most two (Δ − 2)− -vertices. Moreover, if x is adjacent to two (Δ − 1)-vertices, then each of the two Δ-neighbors of adjacent to exactly one (Δ − 2)− -vertex, which is x. Let x ∈ V (G ), we denote δi (x) as the degree of ith small neighbor of x in G. In the following, we use Δ instead of Δ(G ) for simplicity and say u is a k-neighbor of v if uv ∈ E (G ) and d G (u ) = k.
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2. The proof of Theorem 2 Lemma 8. Let G be a 1-planar graph without adjacent 4-cycles and let G × be its associated plane graph. Let v be a k-vertex of G. Then the following results hold. (1) If 5 ≤ k ≤ 7, then f 3 ( v ) ≤ k − 1 in G × ; (2) If 8 ≤ k ≤ 9, then f 3 ( v ) ≤ k − 2 in G × ; (3) Suppose that 6 ≤ k ≤ 7. If f 3 ( v ) = k − 1, then v is incident with a 5+ -face. Moreover, if d G ( v ) = 7 and f 3 ( v ) = 6, then there are at least two false vertices on this 5+ -face. Proof. Let v 1 , v 2 , . . . , v k be neighbors of v of G × in an anticlockwise order. Let f i of G × be face incident with v, v i and v i +1 , for all i such that i ∈ {1, 2, . . . , k}. Note that all subscripts in the paper are taken modulo k. If f i , f i +1 , f i +2 are true 3-faces for some 1 ≤ i ≤ k, then v v i v i +1 v i +2 v and v v i +1 v i +2 v i +3 v are two adjacent 4-cycles. If f i and f i +1 are two false 3-faces, then v v i v i +2 v forms a 3-cycle. (1) If f 3 ( v ) = k, then v must be incident with three 3-cycles C 1 , C 2 , C 3 such that C 2 is adjacent to C 1 and C 3 , and it follows that there are two adjacent 4-cycles in G, a contradiction. So v is incident with at most k − 1 3-faces in G × . (2) It is similar to be proved as (1), we omit here. (3) Suppose f 3 ( v ) = k − 1. If k = 7, then all 3-faces incident with v must be false. Let f be the 4+ -face incident with v. If d G × ( f ) = 4, then f is incident with a false vertex, and there is a 3-cycle incident with a 3-cycle and a 4-cycle, and it follows that there are two adjacent 4-cycles, a contradiction. So d G × ( f ) ≥ 5. If k = 6, then the number of false vertices must be three, otherwise there are two adjacent 4-cycles. By symmetry, there are two cases. If v 1 , v 3 and v 5 are false vertices, then v v 2 v 4 v 6 v forms a 4-cycle. If d G × ( f 6 ) = 4, then there are two adjacent 4-cycles, a contradiction. So d G × ( f ) ≥ 5. If v 1 , v 3 and v 6 are false vertices, then v v 2 v 4 v 5 v forms a 4-cycle. If d G × ( f 6 ) = 4, then there are two adjacent 4-cycles, a contradiction. So d G × ( f ) ≥ 5. Suppose f 3 ( v ) = 6 and such six 3-faces are v v i v i +1 in G × , where 1 ≤ i ≤ 6. Then v is incident with four false vertex since G has no adjacent 4-cycles. Without loss of generality, we assume that v 1 , v 3 , v 5 and v 7 are false vertices. So there are at least two false vertices on this 5+ -face. 2 Lemma 9. Let G be a 1-planar graph without 5-cycles and let G × be its associated plane graph. Let v be a k-vertex of G. Then the following results hold. (1) If k = 7, then f 3 ( v ) ≤ 6 in G × . Moreover, if f 3 ( v ) = 6, then v is incident with a 6+ -face which is incident with at least two false vertices; (2) If k = 8, then f 3 ( v ) ≤ 6 in G × . Moreover, if f 3 ( v ) = 6, then v are incident with two 6+ -faces; (3) If k = 9, then f 3 ( v ) ≤ 7 in G × . Proof. Let v 1 , v 2 , . . . , v k be neighbors of v of G × in an anticlockwise order. Let f i of G × be face incident with v, v i and v i +1 , for all i such that i ∈ {1, 2, . . . , k}. Note that all subscripts in the paper are taken modulo k. If f i , f i +1 , f i +2 are true 3-faces for some 1 ≤ i ≤ k, then v v i v i +1 v i +2 v i +3 v is a 5-cycle. If f i and f i +1 are two false 3-faces and v i +1 is the false vertex, then v v i v i +2 v forms a 3-cycle. (1) If f 3 ( v ) = 7, then v is incident with three false vertices since G has no 5-cycles. So v are incident with four 3-cycles C 1 , C 2 , C 3 and C 4 such that C 2 , C 4 are adjacent to C 1 and C 3 , and it follows that there are 5-cycles in G, a contradiction. So f 3 ( v ) ≤ 6 in G × . Suppose f 3 ( v ) = 6 and such six 3-faces are v v i v i +1 in G × , where 1 ≤ i ≤ 6. Then v is incident with four false vertices since G has no 5-cycles. Without loss of generality, we assume that v 1 , v 3 , v 5 and v 7 are false vertices. Let the remaining face be f . If d G × ( f ) = 4, 5, then v is incident with 5-cycles. So d G × ( f ) ≥ 6 and there are at least two false vertices on this face. (2) Suppose f 3 ( v ) = 7 and such seven 3-faces are v v i v i +1 in G × , where 1 ≤ i ≤ 7. Since the number of false vertices containing in v 1 , v 2 , . . . , v 8 cannot exceed four by Lemma 4, there are at least four true vertices. Without loss of generality, we assume that v 1 , v 3 , v 5 , v 7 are false vertices and v 2 , v 4 , v 6 , v 8 are true vertices. Obviously, there is one 5-cycle incident with v, a contradiction. So f 3 ( v ) ≤ 6. If f 3 ( v ) = 6, then G have subgraphs isomorphic to Fig. 1, obviously, the remaining two faces are 6+ -faces. (3) It is similar to be proved as (2), we omit here. 2 Now we prove Theorem 2. Proof. Since every 1-planar graph with maximum degree Δ ≥ 10 has chromatic index Δ (see [10]), we assume that 8 ≤ Δ ≤ 9 in the following proof. Suppose that G is a counterexample to the theorem with the smallest number of edges. Then
W. Zhang, J.-L. Wu / Theoretical Computer Science 566 (2015) 50–58
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Fig. 1. Black vertices do not have neighbors other than presented in the picture, white vertices can be adjacent to some other vertices, while gray vertices are false vertices.
G is a Δ-critical 1-planar graph. By Vizing’s Adjacency Lemma (VAL for short), we have δ(G ) ≥ 2. In the following, we apply the discharging method on the associated planar graph G × and complete the proof by a contradiction. Since G × is a plane graph, we have | V (G × )| − | E (G × )| + | F (G × )| = 2. It follows that
dG × ( v ) − 4 +
v ∈ V (G × )
d G × ( f ) − 4 = −8.
f ∈ F (G × )
× × × × Now we define ch(x) to be the initial charge of x ∈ V (G ) ∪ F (G ). Let ch(x) = d G × (x) − 4 for each x ∈ V (G ) ∪ F (G ). It follows that x∈ V (G × )∪ F (G × ) ch (x) = −8. We now redistribute the initial charge ch (x) and form a new charge ch (x) for each
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W. Zhang, J.-L. Wu / Theoretical Computer Science 566 (2015) 50–58
x ∈ V (G × ) ∪ F (G × ) by discharging method. Since our rules only move charge around, and do not affect the sum, we have show that ch (x) ≥ 0 for each x ∈ V (G × ) ∪ F (G × ), then we x∈ V (G × )∪ F (G × ) ch (x) = x∈ V (G × )∪ F (G × ) ch (x) = −8. If we can get an obvious contradiction to 0 ≤ x∈ V (G × )∪ F (G × ) ch (x) = x∈ V (G × )∪ F (G × ) ch(x) = −8, which completes our proof. Case 1. Suppose that G is a 1-planar graph without adjacent 4-cycles. In the case, we define the discharging rules as follows. R1. Let f be a false 3-face in G × , then f receives 1 2
1 2
from each of its incident true vertices; let f be a true 3-face in G × ,
from each of its incident 5+ -vertices.
then f receives R2. Let f be a 5+ -face. Each 3-vertex incident with f receives d G × ( f )−4− 2t m
1 2
from f , each of the remaining true vertex incident with
f receives from f , where t denotes the number of 3-vertices and m denotes the number of the remaining true vertices incident with f . R3. Let v be a 2-vertex in G. Then v receives 1 from each of its neighbors in G. R4. Let v be a 3-vertex in G. Then v receives 12 from each of its neighbors in G. 5 12
R5. Let v be a 4-vertex in G. If Δ = 9, then v receives 1 2
from each of its 7-neighbors,
from each of its 9-neighbors in G; otherwise, Δ = 8.
R5.1 If δ1 ( v ) = 6, then v receives
1 2
from 6-vertex,
1 3
from each of its 8-neighbors and
from each of its 8-neighbors in G;
5 R5.2 If δ1 ( v ) = 7 and δ2 ( v ) = 7, then v receives from each of its 8-neighbors and 12 from each of its 7-neighbors in G; R5.3 If δ1 ( v ) = 7, δ2 ( v ) = 8 and some 8-neighbor y of v is adjacent to three 6− -vertices, then v receives 31 from y and 2 3
1 3
1 3
from each of its another two 8-neighbors in G;
5 R5.4 If δ1 ( v ) = 7, δ2 ( v ) = 8 and every 8-neighbor of v is adjacent to at most two 6− -vertices, then v receives 12 from each of its 7+ -neighbors in G; R5.5 If δ1 ( v ) = 8 and some 8-neighbor y of v is adjacent to three 6− -vertices, then v receives 23 from each of its 8-neighbors in G except y; 5 R5.6 If δ1 ( v ) = 8 and every 8-neighbor of v is adjacent to at most two 6− -vertices, then v receives 12 from each of its 8-neighbors in G. R6. Let v be a 5-vertex in G. If f 3 ( v ) ≤ 3, then v receives 81 from each of its (Δ − 2)+ -neighbors in G; otherwise v receives 1 4
from each of its (Δ − 2)+ -neighbors in G.
R7. Let v be a 6-vertex in G. Then v receives R8. Let v be a 7-vertex in G. Then v receives
1 8 1 6
from each of its 7-neighbors and
1 5
from each of its Δ-neighbors in G.
from each of its Δ-neighbors in G.
Next we show that ch (x) ≥ 0 for all x ∈ V (G × ) ∪ F (G × ). Let f be a face of G × . Suppose d G × ( f ) = 3. If f is a false 3-face in G × , then f is incident with two true vertices by Lemma 4, so we have ch ( f ) ≥ −1 + 2 × 12 = −1 + 1 = 0 by R1. Otherwise, f is a true 3-face in G × , every true 3-face in G × is incident with at least two 5+ -vertices since any two 4− -vertices are not adjacent by VAL. So we have ch ( f ) ≥ −1 + min{2 × 12 , 3 × 12 } = −1 + 1 = 0 by R1. If d G × ( f ) = 4, then ch ( f ) = ch( f ) = 0. If d × ( f )−4− t
2 d G × ( f ) ≥ 5, then ch ( f ) = d G × ( f ) − 4 − 2t − G m × m = 0 by R2. Let v be a vertex of G. If d G ( v ) = 2, then v is adjacent to two Δ-vertices in G by VAL. By Lemma 4, v cannot be incident with a false 3-face in G × . It follows that ch ( v ) = −2 + 2 × 1 = 0 by R3. Suppose that d G ( v ) = 3. By Lemma 4, if v is incident with two false 3-faces in G × , then v shall also be incident with a + 5 -face f , from which v receives 12 by R2 and R4. So we have ch ( v ) ≥ −1 − max{ 12 , 2 × 12 − 12 } + 3 × 12 = 0. Suppose that d G ( v ) = 4. If v is a false vertex, then ch ( v ) = ch( v ) = 0; otherwise, v is a true vertex. Suppose d G ( v ) = 4. 5 By Lemma 4, v is incident with at most three false 3-faces. If Δ = 9, then ch ( v ) ≥ 0 − 3 × 12 + min{3 × 12 + 12 , 2 × 12 +
2 × 13 , 3 ×
1 2
+ 13 , 4 × 12 } > 0 by R1 and R5. Suppose Δ = 8. If δ1 ( v ) = 6, another three neighbors of v must be 8-vertices 1 + 12 + 3 × 13 = 0 by R1 and R5.1. Suppose δ1 ( v ) = 7. Then v is adjacent to at least 2 5 two 8-vertices in G. If δ2 ( v ) = 7, then ch ( v ) ≥ 0 − 3 × 12 + 2 × 12 + 2 × 13 = 0 by R1 and R5.2; otherwise, δ2 ( v ) = 8, then 1 2 1 5 1 ch ( v ) ≥ 0 − 3 × 2 + min{2 × 3 + 3 , 4 × 12 } = 6 > 0 by R1, R5.3 and R5.4. If δ1 ( v ) = 8, then ch ( v ) ≥ 0 − 3 × 12 + min{3 × 23 , 5 4 × 12 } = 16 > 0 by R1, R5.5 and R5.6. Suppose that d G ( v ) = 5. By Lemma 8 and VAL, v is incident with at most four 3-faces in G × and is adjacent to at least four (Δ − 2)+ -vertices in G. If f 3 ( v ) ≤ 3, then v receives 18 from each of its (Δ − 2)+ -neighbors in G by R6. So ch ( v ) ≥ 1 − 3 × 12 + 4 × 18 = 0 by R1; otherwise f 3 ( v ) = 4, then v receives 41 from each of its (Δ − 2)+ -neighbors in G. So ch ( v ) ≥ 1 − 4 × 12 + 4 × 14 = 0 by R1 and R6. Suppose that d G ( v ) = 6. By Lemma 8, v is incident with at most five 3-faces in G × . Suppose f 3 ( v ) ≤ 4. If δ1 ( v ) ≤ 5, then v is adjacent to at least four Δ-vertices in G by VAL, so ch ( v ) ≥ 2 − 4 × 12 − max{ 12 , 2 × 14 } + 4 × 15 > 0 by R1, R5, R6 and R7. If δ1 ( v ) ≥ 6, then ch ( v ) ≥ 2 − 4 × 12 = 0 by R1. So f 3 ( v ) = 5. If δ1 ( v ) = 4, then by VAL, v is adjacent to five Δ-vertices in G. So ch ( v ) ≥ 2 − 5 × 12 + 5 × 15 − 12 = 0 by R1, R5.1 and R7. If δ1 ( v ) = 5 (denote the vertex as x), then v is adjacent to by VAL. So we have ch ( v ) ≥ 0 − 3 ×
W. Zhang, J.-L. Wu / Theoretical Computer Science 566 (2015) 50–58
four Δ-vertices in G by VAL. If δ2 ( v ) ≥ 6, then ch ( v ) ≥ 2 − 5 ×
1 2
+4×
1 5
−
1 4
55
> 0 by R1, R6 and R7; otherwise δ2 ( v ) = 5
(denote the vertex as y). By Lemma 6, wx and w y are not incident with any 3-faces, so f 3 (x), f 3 ( y ) ≤ 3 and w sends
1 2
1 5
1 8
1 8
to
x and y, respectively by R6. It follows that ch ( v ) ≥ 2 − 5 × + 4 × − 2 × > 0 by R1, R6 and R7. Suppose that d G ( v ) = 7. By Lemma 8 and VAL, v is incident with at most six 3-faces in G × and at least two Δ-vertices. 5 If f 3 ( v ) ≤ 5, then ch ( v ) ≥ 3 − 5 × 12 − max{ 12 , 2 × 12 , 3 × 14 , 4 × 18 } + 2 × 16 = 0 by R1, R4, R5.2 and R6–R8. So f 3 ( v ) = 6, + + the remaining face is 5 -face by Lemma 8. Let the 5 -face be f . If δ1 ( v ) = 3, then v is adjacent to at least six Δ-vertices in G by VAL. So ch ( v ) ≥ 3 − 6 × 12 + 6 × 16 − 12 = 12 > 0 by R1, R4 and R8. If δ1 ( v ) = 4, then by VAL, v is adjacent to at least 5 five Δ-vertices in G. So ch ( v ) ≥ 3 − 6 × 12 + 5 × 16 − 2 × 12 = 0 by R1, R5 and R8. Suppose δ1 ( v ) = 5. Then v is adjacent to at least four Δ-vertices in G by VAL and the remaining face is 5+ -face and there are at least two false vertices on this face by Lemma 8. If d G × ( f ) = 5, then there is only one 3-vertex on this face since δ1 ( v ) = 5. So v receives at least 14 from
f by R2; otherwise d G × ( f ) ≥ 6, then there are at most
d G × ( f )−3 2
3-vertices on this face since any two 4− -vertices are not
adjacent by VAL and there are at least two false vertices on this face by Lemma 8. So v receives at least So ch ( v ) ≥ 3 − 6 ×
1 2
from f by R2.
− 3 × 14 + 14 = 16 > 0 by R1, R2, R6 and R8. If δ1 ( v ) = 6, then v is adjacent to at least three Δ-vertices in G by VAL. So ch ( v ) ≥ 3 − 6 × 12 + 3 × 16 − 4 × 18 = 0 by R1, R7 and R8. If δ1 ( v ) ≥ 7, then v is adjacent to at least two Δ-vertices in G by VAL. So ch ( v ) ≥ 3 − 6 × 12 + 2 × 16 = 13 > 0 by R1 and R8. Suppose that d G ( v ) = 8. Note that a 8-vertex v may be Δ-vertex here. We assume that Δ = 8 here. By Lemma 8, v is incident with at most six 3-faces in G × . If δ1 ( v ) = 2, then by VAL, v is adjacent to seven Δ-vertices in G. So ch ( v ) ≥ 4 − 6 × 12 − 1 ≥ 0 by R1 and R3. If δ1 ( v ) ≥ 5, then ch ( v ) ≥ 4 − 6 × 12 − max{4 × 14 , 5 × 15 , 6 × 16 } = 0 by R1 and R6–R8. Suppose δ1 ( v ) = 3. By VAL, v is adjacent to at least six Δ-vertices in G. If δ2 ( v ) = 3 or δ2 ( v ) ≥ 5, then ch ( v ) ≥ 4 − 6 × 1 5 − max{2 × 12 , 12 + 14 } ≥ 0 by R1, R5 and R7. So δ2 ( v ) = 4. By R5, v sends at most 12 to the 4-vertex adjacent to v. So 2 1 1 5 1 ch ( v ) ≥ 4 − 6 × 2 − 2 − 12 = 12 > 0. Suppose δ1 ( v ) = 4. By VAL, v is adjacent to at least five Δ-vertices in G. Let u be a 4-vertex that is adjacent to v in G. If δ1 (u ) = 6, then every vertex of N G (u , v ) \ {u , v } is a 8-vertex by Lemma 7. So ch ( v ) ≥ 5 4 − 6 × 12 − 13 = 23 > 0 by R5.1. If δ1 (u ) = δ2 (u ) = 7, then ch ( v ) ≥ 4 − 6 × 12 − max{3 × 13 , 2 × 13 + 13 , 13 + 2 × 13 , 13 + 12 + 16 } = 0 1 1 5 1 by R5.2, R5.3 and R5.4. If δ1 (u ) = 7 and δ2 (u ) = 8, then ch ( v ) ≥ 4 − 6 × 2 − max{3 × 3 , 2 × 12 + 6 } = 0 by R5.3 and R5.4. 5 If δ1 (u ) = 8, then ch ( v ) ≥ 4 − 6 × 12 − max{ 23 + 2 × 16 , 2 × 12 } = 16 > 0 by (2) of Lemma 7 and by R5.5 and R5.6. Suppose d G ( v ) = 9. By Lemma 8, v is incident with at most seven 3-faces in G × . If δ1 ( v ) = 2, then v is adjacent to eight 9-vertices in G by VAL. So ch ( v ) ≥ 5 − 7 × 12 − 1 = 12 ≥ 0 by R1 and R3. If δ1 ( v ) = 3, then v is adjacent to at least seven Δ-vertices in G by VAL. So ch ( v ) ≥ 5 − 7 × 12 − 2 × 12 > 0 by R1, R4 and R5. If δ1 ( v ) = 4. By VAL, v is adjacent to at least five 9-vertices in G. So ch ( v ) ≥ 5 − 7 × 12 − 3 × 12 = 0 by R1 and R5. If δ1 ( v ) ≥ 5, then ch ( v ) ≥ 5 − 7 × 12 − max{4 × 14 , 4 × 16 , 5 × 15 } = 12 > 0 by R1 and R6–R8. +4×
1 6
1 2
Case 2. Suppose that G is a 1-planar graph without 5-cycles. In the case, we define the discharging rules as follows. R1. Let f be a false 3-face in G × , then f receives 1 2
1 2
from each of its incident true vertices; let f be a true 3-face in G × ,
from each of its incident 5+ -vertices.
then f receives R2. Let f be a 6+ -face. If the number of false vertices incident with f is at least two, then each true vertex incident with f receives f receives
d G × ( f )−4 d G × ( f )−2 d G × ( f )−4 d G × ( f )−1
from f ; if the number of false vertices incident with f is one, then each true vertex incident with from f ; otherwise, each true vertex incident with f receives
R3. Let v be a 2-vertex in G. Then v receives 1 from each of its neighbors in G. R4. Let v be a 3-vertex in G. Then v receives 23 from each of its neighbors in G. R5. Let v be a 4-vertex in G. Then v receives
1 2
each of its 6-neighbors and
3 8
from each of its 8+ -neighbors; otherwise, v receives
from each of its 7+ -neighbors in G.
R7. Let v be a 6-vertex in G. If f 3 ( v ) = 6 and δ1 ( v ) = δ2 ( v ) = δ3 ( v ) = 6, then v receives 6-neighbors, receives 3 10
3 10
from f .
from each of its neighbors in G.
R6. Let v be a 5-vertex in G. If f 3 ( v ) ≤ 3, then v receives 3 8
d G × ( f )−4 dG × ( f )
1 10
3 16
from
from one of these three
5 from each of its 8+ -neighbors in G; otherwise, v receives 16 from each of its 7-neighbors and
from each of its 8+ -neighbors in G.
R8. Let v be a 7-vertex in G. Then v receives
1 4
from each of its Δ-neighbors in G.
Next we show that ch (x) ≥ 0 for all x ∈ V (G × ) ∪ F (G × ). Let f be a face of G × . Suppose d G × ( f ) = 3. If f is a false 3-face in G × , then f is incident with two true vertices by Lemma 4, so we have ch ( f ) ≥ −1 + 2 × 12 = −1 + 1 = 0 by R1. Otherwise, f is a true 3-face in G × , every true 3-face in G × is incident with at least two 5+ -vertices since any two 4− -vertices are not adjacent by VAL. So we have ch ( f ) ≥ −1 + min{2 × 12 , 3 × 12 } = −1 + 1 = 0 by R1. If 4 ≤ d G × ( f ) ≤ 5, then ch ( f ) = ch( f ) = 0. If d G × ( f ) ≥ 6, obviously, ch ( f ) ≥ 0 by R2. Let v be a vertex of G. If d G ( v ) = 2, then v is adjacent to two Δ-vertices in G by VAL. By Lemma 4, v cannot be incident with a false 3-face in G × . It follows that ch ( v ) = −2 + 2 × 1 = 0 by R3.
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If d G ( v ) = 3, then v is adjacent to three (Δ − 1)+ -vertices in G and v is incident with at most two false 3-faces in G × by VAL and Lemma 4. So we have ch ( v ) ≥ −1 − max{ 12 , 2 × 12 } + 3 × 23 = 0 by R1 and R4. Suppose that d( v ) = 4. If v is a false vertex, then ch ( v ) = ch( v ) = 0; otherwise, v is a true vertex. Suppose d G ( v ) = 4, then v is adjacent to four (Δ − 2)+ -vertices in G and v is incident with at most three false 3-faces in G × by VAL and Lemma 4. So we have ch ( v ) ≥ 0 − 3 × 12 + 4 × 12 = 12 > 0 by R1 and R5. Suppose that d G ( v ) = 5. Then v is adjacent to at least two Δ-vertices in G by VAL. If f 3 ( v ) ≤ 3, then ch ( v ) ≥ 1 − 3 × 1 + 2 × 38 = 14 > 0 by R1 and R6; otherwise, f 3 ( v ) ≥ 4. If δ1 ( v ) = 5, then v is adjacent to four Δ-vertices in G by VAL, so 2 ch ( v ) ≥ 1 − 5 ×
+ 4 × 38 = 0 by R1 and R6. If δ1 ( v ) = 6, then by VAL, v is adjacent to at least three Δ-vertices in G. So 3 ch ( v ) ≥ 1 − 5 × + 2 × 16 + 3 × 38 = 0 by R1 and R6. If δ1 ( v ) ≥ 7, then v is adjacent to at least two Δ-vertices in G by 1 VAL. So ch ( v ) ≥ 1 − 5 × 2 + 5 × 38 = 38 > 0 by R1 and R6. 1 Suppose that d G ( v ) = 6. There are two subcases: (1): v sends 10 to some adjacent 6-vertex w (Ref. R7). Note that one 1 6-vertex v sends charge 10 to its 6-neighbor w if and only if f 3 ( w ) = 6 and δ1 ( w ) = δ2 ( w ) = δ3 ( w ) = 6. Then f 6+ ( v ) ≥ 2 3 3 1 1 since G contains no 5-cycles. So ch ( v ) ≥ 2 − 4 × 12 + 2 × 13 + 2 × 10 − max{ 16 + 10 , 3 × 10 } > 0 by R1, R2, R6 and R7. (2): v sends no charge to some adjacent 6-vertex w. If δ1 ( v ) = 4, then by VAL, v is adjacent to five Δ-vertices in G. So 3 ch ( v ) ≥ 2 − 6 × 12 + 5 × 10 − 12 = 0 by R1, R5 and R7. If δ1 ( v ) = 5, then v is adjacent to at least four Δ-vertices in G 3 3 1 by VAL. If δ2 ( v ) ≥ 6, then ch ( v ) ≥ 2 − 6 × 12 + 4 × 10 − 16 = 80 > 0 by R1, R6 and R7. So δ2 ( v ) = 5. If f 3 ( v ) ≤ 5, then 1 3 3 13 ch ( v ) ≥ 2 − 5 × 2 + 4 × 10 − 2 × 16 = 40 > 0 by R1, R6 and R8; otherwise, f 3 ( v ) = 6, then there are at least two 6+ -face 3 incidents with each of these two 5-neighbors since G contains no 5-cycles, so ch ( v ) ≥ 2 − 6 × 12 + 4 × 10 = 15 > 0 by R1, R6 and R7. If δ1 ( v ) = 6, then v is adjacent to at least three Δ-vertices in G by VAL. If δ1 ( v ) = δ2 ( v ) = δ3 ( v ) = 6, then there are at least two 6+ -face incidents with one of these three 6-neighbors since G contains no 5-cycles, so ch ( v ) ≥ 3 1 3 5 5 2 − 6 × 12 + 3 × 10 + 10 = 0 by R1, R6 and R7; otherwise, ch ( v ) ≥ 2 − 6 × 12 + 3 × 10 + min{2 × 16 , 16 } > 0. If δ1 ( v ) ≥ 7, 1 3 5 then v is adjacent to at least two Δ-vertices in G by VAL. So ch ( v ) ≥ 2 − 6 × 2 + 2 × 10 + 4 × 16 > 0 by R1 and R7. Suppose that d G ( v ) = 7. By Lemma 9 and VAL, v is incident with at most six 3-faces in G × . Suppose f 3 ( v ) ≤ 5. If δ1 ( v ) = 3, then v is adjacent to six Δ-vertices in G by VAL. So ch ( v ) ≥ 3 − 5 × 12 + 6 × 14 − 23 > 0 by R1, R4 and R8. If δ1 ( v ) = 4, then by VAL, v is adjacent to at least five Δ-vertices in G. So ch ( v ) ≥ 3 − 5 × 12 + 5 × 14 − 2 × 12 > 0 by R1, R5 and R8. If δ1 ( v ) = 5, then by VAL, v is adjacent to at least four Δ-vertices in G. So ch ( v ) ≥ 3 − 5 × 12 + 4 × 14 − 3 × 38 > 0 by R1, 5 R6 and R8. If δ1 ( v ) = 6, then v is adjacent to at least three Δ-vertices in G by VAL. So ch ( v ) ≥ 3 − 5 × 12 + 3 × 14 − 4 × 16 =0 1 by R1, R7 and R8. If δ1 ( v ) ≥ 7, then v is adjacent to at least two Δ-vertices in G by VAL. So ch ( v ) ≥ 3 − 5 × 2 + 2 × 14 > 0 by R1. So f 3 ( v ) = 6, the remaining face is 6+ -face by Lemma 9 and v receives at least 12 from f by R2. If δ1 ( v ) = 3, then v is adjacent to at least six Δ-vertices in G by VAL. So ch ( v ) ≥ 3 − 6 × 12 + 12 + 6 × 14 − 23 > 0 by R1, R4 and R8. If δ1 ( v ) = 4, then by VAL, v is adjacent to at least five Δ-vertices in G. So ch ( v ) ≥ 3 − 6 × 12 + 12 + 5 × 14 − 2 × 12 > 0 by R1, R5 and R8. If δ1 ( v ) = 5, then by VAL, v is adjacent to at least four Δ-vertices in G. So ch ( v ) ≥ 3 − 6 × 12 + 12 + 4 × 14 − 3 × 38 > 0 by R1, R6 5 and R8. If δ1 ( v ) = 6, then v is adjacent to at least three Δ-vertices in G by VAL. So ch ( v ) ≥ 3 − 6 × 12 + 12 + 3 × 14 − 4 × 16 =0 1 1 by R1, R7 and R8. If δ1 ( v ) ≥ 7, then v is adjacent to at least two Δ-vertices in G by VAL. So ch ( v ) ≥ 3 − 6 × 2 + 2 + 2 × 14 > 0
1 2 1 2
by R1 and R8. Suppose that d G ( v ) = 8. Note that a 8-vertex v may be Δ-vertex here. We assume that Δ = 8 here. By Lemma 9, v is 3 incident with at most six 3-faces in G × . If f 3 ( v ) ≤ 5, then ch ( v ) ≥ 4 − 5 × 12 − max{1, 2 × 23 , 3 × 12 , 4 × 38 , 4 × 10 , 6 × 14 } = 0. So f 3 ( v ) = 6, the remaining faces are 6+ -face by Lemma 9 and v receives at least
ch ( v ) ≥ 4 − 6 ×
1 2
+2×
1 3
− max{1, 2 ×
2 3 , 3 × 12 , 4 × 38 , 4 × 10 , 6 × 14 } 3
1 3
from each of two 6+ -faces by R2. So
> 0.
Suppose that d G ( v ) = 9. By Lemma 9, v is incident with at most seven 3-faces in G × . So ch ( v ) ≥ 5 − 7 × 12 − max{1, 2 × 23 ,
3 3 × 12 , 4 × 38 , 4 × 10 , 6 × 14 } = 0. The proof of Theorem 2 is completed.
2
3. The proof of Theorem 3 Lemma 10. Let G be a 1-planar graph without adjacent chordal 5-cycles and let G × be its associated plane graph. Let v be a k-vertex of G. Then the following results hold. (1) If k = 7, then f 3 ( v ) ≤ 6 in G × ; (2) If k = 8, then f 3 ( v ) ≤ 7 in G × . Moreover, if f 3 ( v ) = 7, then v is incident with a 5+ -face which is incident with at least one false vertex; (3) If k = 9, then f 3 ( v ) ≤ 8 in G × . Moreover, if f 3 ( v ) = 8, then v is incident with a 5+ -face which is incident with at least two false vertices on this face. Proof. Let v 1 , v 2 , . . . , v k be neighbors of v of G × in an anticlockwise order. Let f i of G × be face incident with v, v i and v i +1 , for all i such that i ∈ {1, 2, . . . , k}. Note that all subscripts in the paper are taken modulo k. If f i , f i +1 , f i +2 are true
W. Zhang, J.-L. Wu / Theoretical Computer Science 566 (2015) 50–58
57
3-faces for some 1 ≤ i ≤ k, then v v i v i +1 v i +2 v i +3 v is a chordal 5-cycle. If f i and f i +1 are two false 3-faces and v i +1 is the false vertex, then v v i v i +2 v forms a 3-cycle. (1) Suppose f 3 ( v ) = 7. Then v is at most incident with three false vertices. If the number of false vertices is no more than two, obviously there are two adjacent chordal 5-cycles in G, a contradiction. If the number of false vertices is three, then v are incident with four 3-cycles C 1 , C 2 , C 3 and C 4 such that C 2 , C 4 are adjacent to C 1 and C 3 , and it follows that there are two adjacent chordal 5-cycles in G, a contradiction. So f 3 ( v ) ≤ 6. (2) If f 3 ( v ) = 8, then the proof of this case is similar to be proved as (1), we omit here. So f 3 ( v ) ≤ 7. Suppose f 3 ( v ) = 7 and such seven 3-faces are v v i v i +1 in G × , where 1 ≤ i ≤ 7. Then v is incident with four false vertices since G has no adjacent chordal 5-cycles. Suppose that the remaining face is f . Without loss of generality, we assume that v 1 , v 3 , v 5 and v 7 are false vertices and there is at least one false vertex on f . If d G × ( f ) = 4, then v is incident with adjacent chordal 5-cycles. So d G × ( f ) ≥ 5 and there is at least one false vertex on f . (3) If f 3 ( v ) = 9, then the proof of this case is similar to be proved as (1), we omit here. So f 3 ( v ) ≤ 8. Suppose f 3 ( v ) = 8 and such seven 3-faces are v v i v i +1 in G × , where 1 ≤ i ≤ 8. Then v is incident with five false vertices since G has no adjacent chordal 5-cycles. Suppose that the remaining face is f . Without loss of generality, we assume that v 1 , v 3 , v 5 , v 7 and v 9 are false vertices and there are at least two false vertices on f . If d G × ( f ) = 4, then v is incident with adjacent chordal 5-cycles. So d G × ( f ) ≥ 5 and there are at least two false vertices on f . 2 Now we prove Theorem 3. Proof. Since every 1-planar graph with maximum degree Δ ≥ 10 has chromatic index Δ (see [10]), it suffices to prove the following result. (*) Let G be a 1-planar graph without adjacent chordal 5-cycles. If Δ(G ) = 9, then
χ (G ) = 9.
Suppose that G is a counterexample to (*) with the smallest number of edges. Then G is an 9-critical 1-plane graph. By VAL, we have δ(G ) ≥ 2. In the following, we apply the discharging method on the associated planar graph G × and complete the proof by a contradiction. Since G × is a plane graph, we have | V (G × )| − | E (G × )| + | F (G × )| = 2. It follows that
dG × ( v ) − 4 +
v ∈ V (G × )
d G × ( f ) − 4 = −8.
f ∈ F (G × )
× × × × Now we define ch(x) to be the initial charge of x ∈ V (G ) ∪ F (G ). Let ch(x) = d G × (x) − 4 for each x ∈ V (G ) ∪ F (G ). It follows that x∈ V (G )∪ F (G × ) ch (x) = −8. Now we redistribute the initial charge ch (x) and form a new charge ch (x) for each x ∈ V (G × ) ∪ F (G × ) by discharging method. Since our rules only move charge around, and do not affect the sum, we have × × x∈ V (G × )∪ F (G × ) ch (x) = x∈ V (G × )∪ F (G × ) ch (x) = −8. If we can show that ch (x) ≥ 0 for each x ∈ V ( G ) ∪ F ( G ), then we
get an obvious contradiction to 0 ≤ x∈ V (G × )∪ F (G × ) ch (x) = Now we define the discharging rules as follows. R1. Let f be a false 3-face in G × , then f receives 1 2
1 2
x∈ V (G × )∪ F (G × ) ch (x)
= −8, which completes our proof.
from each of its incident true vertices; let f be a true 3-face in G × ,
from each of its incident 5+ -vertices.
then f receives R2. Let f be a 5+ -face. If the number of false vertices incident with f is one, then each true vertex incident with f receives d G × ( f )−4 d G × ( f )−1
receives
from f ; if the number of false vertices incident with f is at least two, then each true vertex incident with f d G × ( f )−4 d G × ( f )−2
from f .
R3. Let v be a 2-vertex in G. Then v receives 1 from each of its neighbors in G. R4. Let v be a 3-vertex in G. Then v receives 23 from each of its neighbors in G. R5. Let v be a 4-vertex in G. Then v receives R6. Let v be a 5-vertex in G. Then v receives R7. Let v be a 6-vertex in G. Then v receives R8. Let v be a 7-vertex in G. Then v receives R9. Let v be a 8-vertex in G. Then v receives
3 from each of its 7+ -neighbors in G. 8 1 from each of its 6+ -neighbors in G. 3 4 from each of its 7+ -neighbors in G. 15 2 from each of its 8+ -neighbors in G. 9 4 from each of its 9-neighbors in G. 21
Next we show that ch (x) ≥ 0 for all x ∈ V (G × ) ∪ F (G × ). Let f be a face of G × . If d G × ( f ) = 3, then every false 3-face in G × is incident with two true vertices and every true 3-face in G × is incident with at least two 5+ -vertices by Lemma 4 and VAL. So we have ch ( f ) ≥ ch( f ) + 2 × 12 = 3 − 4 + 1 = 0 by R1. If d G × ( f ) = 4, then ch ( f ) = ch( f ) ≥ 0. If d G × ( f ) ≥ 5, obviously, ch ( f ) ≥ 0 by R2. Let v be a vertex of G. If d G ( v ) = 2, then v is adjacent to two 9-vertices in G by VAL. By Lemma 4, v cannot be incident with a false 3-face in G × . It follows that ch ( v ) = ch( v ) + 2 × 1 = 0 by R1 and R3. If d G ( v ) = 3, then v is adjacent
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W. Zhang, J.-L. Wu / Theoretical Computer Science 566 (2015) 50–58
to three 8+ -vertices in G and v is incident with at most two false 3-faces in G × by VAL and Lemma 4. So we have ch ( v ) ≥ −1 − max{ 12 , 2 × 12 } + 3 × 23 = 0 by R1 and R3. Suppose that d( v ) = 4. If v is a false vertex, then ch ( v ) = ch( v ) = 0; otherwise, v is a true vertex. Suppose d G ( v ) = 4, then v is adjacent to four 7+ -vertices in G and v is incident with at most three false 3-faces in G × by VAL and Lemma 4. So we have ch ( v ) ≥ 0 − 3 × 12 + 4 × 38 = 0 by R1 and R5. If d G ( v ) = 5, then v is adjacent to five 6+ -vertices in G by VAL. So we have ch ( v ) ≥ 5 − 4 − 5 ×
1 2
+ 5 × 13 = 16 > 0 by R1 and R6. 4 Suppose that d G ( v ) = 6. If δ1 ( v ) = 5, then v is adjacent to at least five 9-vertices by VAL. So ch ( v ) ≥ 2 − 6 × 12 + 5 × 15 − 1 1 4 1 = 0 by R1, R6 and R7. If δ ( v ) = 6, then v is adjacent to four 9-vertices by VAL. So ch ( v ) ≥ 2 − 6 × + 4 × = > 0 1 3 2 15 15 4 by R1 and R7. If δ1 ( v ) ≥ 7, then v is adjacent to at least two 9-vertices by VAL. So ch ( v ) ≥ 2 − 6 × 12 + 6 × 15 = 35 > 0 by
R1 and R8. Suppose that d G ( v ) = 7. If δ1 ( v ) = 4, then v is adjacent to six 9-vertices by VAL. So ch ( v ) ≥ 3 − 6 × 12 + 6 × 29 − 38 = by R1, R5 and by R1, R6 and
23 24 R8. If δ1 ( v ) = 5, then v is adjacent to five 9-vertices by VAL. So ch ( v ) ≥ 3 − 6 × + 5 × − 2 × = 49 4 4 R8. If δ1 ( v ) = 6, then v is adjacent to four 9-vertices by VAL. So ch ( v ) ≥ 3 − 6 × 12 + 4 × 29 − 3 × 15 = 45 R8. If δ1 ( v ) ≥ 7, then v is adjacent to at least two 9-vertices by VAL. So ch ( v ) ≥ 3 − 6 × 12 + 2 × 29 = 49
1 2
2 9
1 3
>0 >0 >0 >0
by R1, R7 and by R1 and R8. Suppose that d G ( v ) = 8. By Lemma 10 and VAL, v is incident with at most seven 3-faces in G × and adjacent to at least 4 4 two 9-vertices. If f 3 ( v ) ≤ 6, then ch ( v ) ≥ 4 − 6 × 12 + 2 × 21 − max{ 23 , 2 × 38 , 3 × 13 , 4 × 15 , 5 × 29 } > 0 by R1 and R4–R7; otherwise f 3 ( v ) = 7. The remaining face is 5+ -face by Lemma 10. Let the 5+ -face be f . By R2, v receives at least
1 4
from f .
4 4 So ch ( v ) ≥ 4 − 7 × 12 + 14 + 2 × 21 − max{ 23 , 2 × 38 , 3 × 13 , 4 × 15 , 5 × 29 } > 0 by R1, R2 and R4–R7. Suppose that d G ( v ) = 9. By Lemma 10 and VAL, v is incident with at most eight 3-faces in G × and adjacent to at least 4 4 two 9-vertices. If f 3 ( v ) ≤ 7, then ch ( v ) ≥ 5 − 7 × 12 − max{1, 2 × 23 , 3 × 38 , 4 × 13 , 5 × 15 , 6 × 29 , 7 × 21 } = 0 by R1 and R3–R8; otherwise f 3 ( v ) = 8. The remaining face is 5+ -face and there are at least two false vertices on this face by Lemma 10. Let 4 the 5+ -face be f . By R2, v receives at least 13 from f . So ch ( v ) ≥ 5 − 8 × 12 + 13 − max{1, 2 × 23 , 3 × 38 , 4 × 13 , 5 × 15 ,6 × 2 4 , 7 × 21 } 9
= 0 by R1, R2 and R3–R8. The proof of Theorem 3 is completed.
2
References [1] [2] [3] [4] [5] [6] [7] [8] [9] [10] [11] [12]
J.A. Bondy, U.S.R. Murty, Graph Theory with Applications, North-Holland, New York, 1976. V.G. Vizing, Critical graphs with given chromatic class, Diskretn. Anal. 5 (1965) 9–17. G. Ringel, Ein Sechsfarbenproblem auf der Kugel, Abh. Math. Semin. Univ. Hambg. 29 (1965) 107–117. D.P. Sanders, Y. Zhao, Planar graphs of maximum degree seven are Class I, J. Combin. Theory Ser. B 83 (2) (2002) 348–360. L. Zhang, Every planar graph with maximum degree 7 is of class 1, Graphs Combin. 16 (2000) 467–495. Y.H. Bu, W.F. Wang, Some sufficient conditions for a planar graph of maximum degree six to be class 1, Discrete Math. 306 (13) (2006) 1440–1445. W.F. Wang, Y.Z. Chen, A sufficient condition for a planar graph to be class 1, Theoret. Comput. Sci. 385 (1–3) (2007) 71–77. R. Luo, L. Miao, Y. Zhao, The size of edge chromatic critical graphs with maximum degree 6, J. Graph Theory 60 (2009) 149–171. H.P. Yap, Some Topics in Graph Theory, Cambridge University Press, New York, 1986. X. Zhang, J.-L. Wu, On edge colorings of 1-planar graphs, Inform. Process. Lett. 111 (3) (2011) 124–128. X. Zhang, G. Liu, On edge colorings of 1-planar graphs without adjacent triangles, Inform. Process. Lett. 111 (3) (2012) 138–142. X. Zhang, G. Liu, On edge colorings of 1-planar graphs without chordal 5-cycles, Ars Combin. 104 (2012) 431–436.
Contents lists available at ScienceDirect
Theoretical Computer Science www.elsevier.com/locate/tcs
Some sufficient conditions for 1-planar graphs to be Class 1 ✩ Wenwen Zhang, Jian-Liang Wu ∗ School of Mathematics, Shandong University, Jinan, 250100, China
a r t i c l e
i n f o
Article history: Received 5 June 2013 Received in revised form 15 October 2014 Accepted 29 November 2014 Available online 5 December 2014 Communicated by G. Ausiello
a b s t r a c t 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. Let G be a 1-planar graph with maximum degree Δ(G ). In this paper, it is shown that χ (G ) = Δ(G ) if (1) Δ(G ) ≥ 9 and G contains no adjacent chordal 5-cycles, or (2) Δ(G ) ≥ 8 and G contains no adjacent 4-cycles or no 5-cycles. © 2014 Elsevier B.V. All rights reserved.
Keywords: 1-Planar graphs Edge coloring Cycle
1. Introduction All graphs considered in this paper are finite, simple and undirected. Any undefined notation follows that of Bondy and Murty [1]. For a real number x, let x be the greatest integer not larger than x. Let G be a graph. We use V (G ) and E (G ) to denote its vertex set and its edge set, respectively. Let v ∈ V (G ). If uv ∈ E (G ), then the vertex u is said to be a neighbor of v in G. We denote by N G ( v ) the set of neighbors of v in G and by d G ( v ) = | N G ( v )| the degree of v in G. We use δ(G ) and Δ(G ) to denote the minimum degree and the maximum degree of G, respectively. For u , v ∈ V (G ), N G (u , v ) = N G (u ) ∪ N G ( v ) . If G is a planar graph, we assume that G has always been embedded in the plane. Let G be a planar graph. We denote by F (G ) the face set of G. The degree of a face f in G, denoted by d G ( f ), is the number of edges incident with it, where each cut-edge is counted twice. Throughout this paper, a k-, k+ - and k− -vertex (or face) in a planar graph is a vertex (or face) of degree k, at least k and at most k, respectively. A k-cycle is a cycle of length k. Two cycles sharing at least a common edge are said to be adjacent. Given a cycle C of length k in G, an edge uv ∈ E (G )\ E (C ) is called a chord of C if u , v ∈ V (C ). Such a cycle C is also called a chordal-k-cycle. A graph is k edge-colorable if its edges can be colored with k colors in such a way that adjacent edges receive different colors. The edge chromatic number (or chromatic index) of G, denoted by χ (G ), is the smallest integer k such that G is k edge-colorable. For edge coloring, Vizing’s Theorem states that for any graph G, Δ(G ) ≤ χ (G ) ≤ Δ(G ) + 1. A graph G is said to be of Class 1 if χ (G ) = Δ(G ), and of Class 2 if χ (G ) = Δ(G ) + 1. A graph G is critical if G is a connected graph of Class 2 and χ (G − e ) < χ (G ) for every edge e of G. If a graph G is critical and its maximum degree is Δ, we say that G is Δ-critical. For planar graphs, more is known. In [2], Vizing presented examples of planar graphs of Class 2 with maximum degree Δ for each Δ ∈ {2, 3, 4, 5}. He proved that every planar graph with Δ ≥ 8 is of Class 1 and then conjectured that every planar graph with maximum degree 6 or 7 is of Class 1. The case Δ = 7 for the conjecture has been verified by ✩
*
This work is supported by a research grant NSFC (11271006) of China. Corresponding author. E-mail address: [email protected] (J.-L. Wu).
http://dx.doi.org/10.1016/j.tcs.2014.11.031 0304-3975/© 2014 Elsevier B.V. All rights reserved.
W. Zhang, J.-L. Wu / Theoretical Computer Science 566 (2015) 50–58
51
Zhang [5] and, independently, by Sanders and Zhao [4]. The case Δ = 6 remains open, but some partial results are obtained, such as [6,7] and [8]. A graph G 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 1-planar graphs was introduced by Ringel [3] in connection with the problem of simultaneous coloring of adjacent/incidence vertices and faces of plane graphs. Compared to the family of planar graphs, 1-planar graphs have not been extensively studied in the literature. The first result concerning the edge colorings of 1-planar graphs is due to Zhang and Wu [10], who proved that every 1-planar graph with maximum degree Δ ≥ 10 is of Class 1. Recently, Zhang and Liu [11] proved that every 1-planar graph without adjacent triangles and with maximum degree Δ ≥ 8 is of Class 1. They [12] also proved that every 1-planar graph without chordal 5-cycles and with maximum degree Δ ≥ 9 is of Class 1, and constructed Class 2 1-planar graphs with maximum degree 6 or 7, so they posed the following conjecture: Conjecture 1. Every 1-planar graph with maximum degree at least 8 is of Class 1. In this paper, we get the following theorems. Theorem 2. Let G be a 1-planar graph without adjacent 4-cycles or without 5-cycles. If Δ(G ) ≥ 8, then χ (G ) = Δ(G ). Theorem 3. Let G be a 1-planar graph without adjacent chordal 5-cycles. If Δ(G ) ≥ 9, then χ (G ) = Δ(G ). In the following, we always assume that G is a 1-planar graph and 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 a 1-plane graph G is the plane graph that is obtained from G by turning all crossings of G into new 4-vertices. A vertex in G × is called false if it is not a vertex of G and true otherwise. We call a 3-face in G × false or true according to whether it is incident with a false vertex or not. Let f k ( v ) denote the number of k-faces incident with v in G × . In [10], the authors showed some basic properties on a 1-planar graph G and its associated plane graph G × , see Lemma 4. Lemma 4. (See [10].) Let G be a 1-planar graph and G × be its associated plane graph. Then the following results hold.
/ E ( G × ); For any two false vertices u and v in G × , uv ∈ If there is a 3-face uv wu in G × such that d G ( v ) = 2, then u and w are not false vertices; / E (G × ) or uv is not incident with two 3-faces; If d G (u ) = 3 and v is a false vertex in G × , then either uv ∈ If a 3-vertex v in G is incident with two 3-faces and adjacent to two false vertices in G × , then v must also be incident with a 5+ -face; (5) For any 4-vertex v in G, v is incident with at most three false 3-faces.
(1) (2) (3) (4)
Here, we give some known results on edge-coloring critical graphs. Lemma 5 (Vizing’s Adjacency Lemma). (See [9].) Let G be a Δ-critical graph and let u , v be adjacent vertices of G with d G ( v ) = k. Then (1) if k < Δ, then u is adjacent to at least (Δ − k + 1) Δ-vertices; (2) if k = Δ, then u is adjacent to at least two Δ-vertices. Lemma 6. (See [4].) No Δ-critical graph has distinct vertices x, y, z such that x is adjacent to y and z, d G ( z) < 2Δ − d G (x) − d G ( y ) + 2, and xz is in at least d G (x) + d G ( y ) − Δ − 2 triangles not containing y. Lemma 7. (See [8].) Let G be a Δ-critical graph with Δ(G ) ≥ 6 and let x be a 4-vertex. Then the following hold: (1) If x is adjacent to a (Δ − 2)-vertex, say y, then every vertex of N G ( N G (x)) \ {x, y } is a Δ-vertex; (2) If x is not adjacent to any (Δ − 2)-vertex and if one of the neighbors y of x is adjacent to d G ( y ) − (Δ − 3) (Δ − 2)− -vertices, then each of the other three neighbors of x is adjacent to only one (Δ − 2)-vertex, which is x; (3) If x is adjacent to a (Δ − 1)-vertex, then there are at least two Δ-vertices in N G (x) which are adjacent to at most two (Δ − 2)− -vertices. Moreover, if x is adjacent to two (Δ − 1)-vertices, then each of the two Δ-neighbors of adjacent to exactly one (Δ − 2)− -vertex, which is x. Let x ∈ V (G ), we denote δi (x) as the degree of ith small neighbor of x in G. In the following, we use Δ instead of Δ(G ) for simplicity and say u is a k-neighbor of v if uv ∈ E (G ) and d G (u ) = k.
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2. The proof of Theorem 2 Lemma 8. Let G be a 1-planar graph without adjacent 4-cycles and let G × be its associated plane graph. Let v be a k-vertex of G. Then the following results hold. (1) If 5 ≤ k ≤ 7, then f 3 ( v ) ≤ k − 1 in G × ; (2) If 8 ≤ k ≤ 9, then f 3 ( v ) ≤ k − 2 in G × ; (3) Suppose that 6 ≤ k ≤ 7. If f 3 ( v ) = k − 1, then v is incident with a 5+ -face. Moreover, if d G ( v ) = 7 and f 3 ( v ) = 6, then there are at least two false vertices on this 5+ -face. Proof. Let v 1 , v 2 , . . . , v k be neighbors of v of G × in an anticlockwise order. Let f i of G × be face incident with v, v i and v i +1 , for all i such that i ∈ {1, 2, . . . , k}. Note that all subscripts in the paper are taken modulo k. If f i , f i +1 , f i +2 are true 3-faces for some 1 ≤ i ≤ k, then v v i v i +1 v i +2 v and v v i +1 v i +2 v i +3 v are two adjacent 4-cycles. If f i and f i +1 are two false 3-faces, then v v i v i +2 v forms a 3-cycle. (1) If f 3 ( v ) = k, then v must be incident with three 3-cycles C 1 , C 2 , C 3 such that C 2 is adjacent to C 1 and C 3 , and it follows that there are two adjacent 4-cycles in G, a contradiction. So v is incident with at most k − 1 3-faces in G × . (2) It is similar to be proved as (1), we omit here. (3) Suppose f 3 ( v ) = k − 1. If k = 7, then all 3-faces incident with v must be false. Let f be the 4+ -face incident with v. If d G × ( f ) = 4, then f is incident with a false vertex, and there is a 3-cycle incident with a 3-cycle and a 4-cycle, and it follows that there are two adjacent 4-cycles, a contradiction. So d G × ( f ) ≥ 5. If k = 6, then the number of false vertices must be three, otherwise there are two adjacent 4-cycles. By symmetry, there are two cases. If v 1 , v 3 and v 5 are false vertices, then v v 2 v 4 v 6 v forms a 4-cycle. If d G × ( f 6 ) = 4, then there are two adjacent 4-cycles, a contradiction. So d G × ( f ) ≥ 5. If v 1 , v 3 and v 6 are false vertices, then v v 2 v 4 v 5 v forms a 4-cycle. If d G × ( f 6 ) = 4, then there are two adjacent 4-cycles, a contradiction. So d G × ( f ) ≥ 5. Suppose f 3 ( v ) = 6 and such six 3-faces are v v i v i +1 in G × , where 1 ≤ i ≤ 6. Then v is incident with four false vertex since G has no adjacent 4-cycles. Without loss of generality, we assume that v 1 , v 3 , v 5 and v 7 are false vertices. So there are at least two false vertices on this 5+ -face. 2 Lemma 9. Let G be a 1-planar graph without 5-cycles and let G × be its associated plane graph. Let v be a k-vertex of G. Then the following results hold. (1) If k = 7, then f 3 ( v ) ≤ 6 in G × . Moreover, if f 3 ( v ) = 6, then v is incident with a 6+ -face which is incident with at least two false vertices; (2) If k = 8, then f 3 ( v ) ≤ 6 in G × . Moreover, if f 3 ( v ) = 6, then v are incident with two 6+ -faces; (3) If k = 9, then f 3 ( v ) ≤ 7 in G × . Proof. Let v 1 , v 2 , . . . , v k be neighbors of v of G × in an anticlockwise order. Let f i of G × be face incident with v, v i and v i +1 , for all i such that i ∈ {1, 2, . . . , k}. Note that all subscripts in the paper are taken modulo k. If f i , f i +1 , f i +2 are true 3-faces for some 1 ≤ i ≤ k, then v v i v i +1 v i +2 v i +3 v is a 5-cycle. If f i and f i +1 are two false 3-faces and v i +1 is the false vertex, then v v i v i +2 v forms a 3-cycle. (1) If f 3 ( v ) = 7, then v is incident with three false vertices since G has no 5-cycles. So v are incident with four 3-cycles C 1 , C 2 , C 3 and C 4 such that C 2 , C 4 are adjacent to C 1 and C 3 , and it follows that there are 5-cycles in G, a contradiction. So f 3 ( v ) ≤ 6 in G × . Suppose f 3 ( v ) = 6 and such six 3-faces are v v i v i +1 in G × , where 1 ≤ i ≤ 6. Then v is incident with four false vertices since G has no 5-cycles. Without loss of generality, we assume that v 1 , v 3 , v 5 and v 7 are false vertices. Let the remaining face be f . If d G × ( f ) = 4, 5, then v is incident with 5-cycles. So d G × ( f ) ≥ 6 and there are at least two false vertices on this face. (2) Suppose f 3 ( v ) = 7 and such seven 3-faces are v v i v i +1 in G × , where 1 ≤ i ≤ 7. Since the number of false vertices containing in v 1 , v 2 , . . . , v 8 cannot exceed four by Lemma 4, there are at least four true vertices. Without loss of generality, we assume that v 1 , v 3 , v 5 , v 7 are false vertices and v 2 , v 4 , v 6 , v 8 are true vertices. Obviously, there is one 5-cycle incident with v, a contradiction. So f 3 ( v ) ≤ 6. If f 3 ( v ) = 6, then G have subgraphs isomorphic to Fig. 1, obviously, the remaining two faces are 6+ -faces. (3) It is similar to be proved as (2), we omit here. 2 Now we prove Theorem 2. Proof. Since every 1-planar graph with maximum degree Δ ≥ 10 has chromatic index Δ (see [10]), we assume that 8 ≤ Δ ≤ 9 in the following proof. Suppose that G is a counterexample to the theorem with the smallest number of edges. Then
W. Zhang, J.-L. Wu / Theoretical Computer Science 566 (2015) 50–58
53
Fig. 1. Black vertices do not have neighbors other than presented in the picture, white vertices can be adjacent to some other vertices, while gray vertices are false vertices.
G is a Δ-critical 1-planar graph. By Vizing’s Adjacency Lemma (VAL for short), we have δ(G ) ≥ 2. In the following, we apply the discharging method on the associated planar graph G × and complete the proof by a contradiction. Since G × is a plane graph, we have | V (G × )| − | E (G × )| + | F (G × )| = 2. It follows that
dG × ( v ) − 4 +
v ∈ V (G × )
d G × ( f ) − 4 = −8.
f ∈ F (G × )
× × × × Now we define ch(x) to be the initial charge of x ∈ V (G ) ∪ F (G ). Let ch(x) = d G × (x) − 4 for each x ∈ V (G ) ∪ F (G ). It follows that x∈ V (G × )∪ F (G × ) ch (x) = −8. We now redistribute the initial charge ch (x) and form a new charge ch (x) for each
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x ∈ V (G × ) ∪ F (G × ) by discharging method. Since our rules only move charge around, and do not affect the sum, we have show that ch (x) ≥ 0 for each x ∈ V (G × ) ∪ F (G × ), then we x∈ V (G × )∪ F (G × ) ch (x) = x∈ V (G × )∪ F (G × ) ch (x) = −8. If we can get an obvious contradiction to 0 ≤ x∈ V (G × )∪ F (G × ) ch (x) = x∈ V (G × )∪ F (G × ) ch(x) = −8, which completes our proof. Case 1. Suppose that G is a 1-planar graph without adjacent 4-cycles. In the case, we define the discharging rules as follows. R1. Let f be a false 3-face in G × , then f receives 1 2
1 2
from each of its incident true vertices; let f be a true 3-face in G × ,
from each of its incident 5+ -vertices.
then f receives R2. Let f be a 5+ -face. Each 3-vertex incident with f receives d G × ( f )−4− 2t m
1 2
from f , each of the remaining true vertex incident with
f receives from f , where t denotes the number of 3-vertices and m denotes the number of the remaining true vertices incident with f . R3. Let v be a 2-vertex in G. Then v receives 1 from each of its neighbors in G. R4. Let v be a 3-vertex in G. Then v receives 12 from each of its neighbors in G. 5 12
R5. Let v be a 4-vertex in G. If Δ = 9, then v receives 1 2
from each of its 7-neighbors,
from each of its 9-neighbors in G; otherwise, Δ = 8.
R5.1 If δ1 ( v ) = 6, then v receives
1 2
from 6-vertex,
1 3
from each of its 8-neighbors and
from each of its 8-neighbors in G;
5 R5.2 If δ1 ( v ) = 7 and δ2 ( v ) = 7, then v receives from each of its 8-neighbors and 12 from each of its 7-neighbors in G; R5.3 If δ1 ( v ) = 7, δ2 ( v ) = 8 and some 8-neighbor y of v is adjacent to three 6− -vertices, then v receives 31 from y and 2 3
1 3
1 3
from each of its another two 8-neighbors in G;
5 R5.4 If δ1 ( v ) = 7, δ2 ( v ) = 8 and every 8-neighbor of v is adjacent to at most two 6− -vertices, then v receives 12 from each of its 7+ -neighbors in G; R5.5 If δ1 ( v ) = 8 and some 8-neighbor y of v is adjacent to three 6− -vertices, then v receives 23 from each of its 8-neighbors in G except y; 5 R5.6 If δ1 ( v ) = 8 and every 8-neighbor of v is adjacent to at most two 6− -vertices, then v receives 12 from each of its 8-neighbors in G. R6. Let v be a 5-vertex in G. If f 3 ( v ) ≤ 3, then v receives 81 from each of its (Δ − 2)+ -neighbors in G; otherwise v receives 1 4
from each of its (Δ − 2)+ -neighbors in G.
R7. Let v be a 6-vertex in G. Then v receives R8. Let v be a 7-vertex in G. Then v receives
1 8 1 6
from each of its 7-neighbors and
1 5
from each of its Δ-neighbors in G.
from each of its Δ-neighbors in G.
Next we show that ch (x) ≥ 0 for all x ∈ V (G × ) ∪ F (G × ). Let f be a face of G × . Suppose d G × ( f ) = 3. If f is a false 3-face in G × , then f is incident with two true vertices by Lemma 4, so we have ch ( f ) ≥ −1 + 2 × 12 = −1 + 1 = 0 by R1. Otherwise, f is a true 3-face in G × , every true 3-face in G × is incident with at least two 5+ -vertices since any two 4− -vertices are not adjacent by VAL. So we have ch ( f ) ≥ −1 + min{2 × 12 , 3 × 12 } = −1 + 1 = 0 by R1. If d G × ( f ) = 4, then ch ( f ) = ch( f ) = 0. If d × ( f )−4− t
2 d G × ( f ) ≥ 5, then ch ( f ) = d G × ( f ) − 4 − 2t − G m × m = 0 by R2. Let v be a vertex of G. If d G ( v ) = 2, then v is adjacent to two Δ-vertices in G by VAL. By Lemma 4, v cannot be incident with a false 3-face in G × . It follows that ch ( v ) = −2 + 2 × 1 = 0 by R3. Suppose that d G ( v ) = 3. By Lemma 4, if v is incident with two false 3-faces in G × , then v shall also be incident with a + 5 -face f , from which v receives 12 by R2 and R4. So we have ch ( v ) ≥ −1 − max{ 12 , 2 × 12 − 12 } + 3 × 12 = 0. Suppose that d G ( v ) = 4. If v is a false vertex, then ch ( v ) = ch( v ) = 0; otherwise, v is a true vertex. Suppose d G ( v ) = 4. 5 By Lemma 4, v is incident with at most three false 3-faces. If Δ = 9, then ch ( v ) ≥ 0 − 3 × 12 + min{3 × 12 + 12 , 2 × 12 +
2 × 13 , 3 ×
1 2
+ 13 , 4 × 12 } > 0 by R1 and R5. Suppose Δ = 8. If δ1 ( v ) = 6, another three neighbors of v must be 8-vertices 1 + 12 + 3 × 13 = 0 by R1 and R5.1. Suppose δ1 ( v ) = 7. Then v is adjacent to at least 2 5 two 8-vertices in G. If δ2 ( v ) = 7, then ch ( v ) ≥ 0 − 3 × 12 + 2 × 12 + 2 × 13 = 0 by R1 and R5.2; otherwise, δ2 ( v ) = 8, then 1 2 1 5 1 ch ( v ) ≥ 0 − 3 × 2 + min{2 × 3 + 3 , 4 × 12 } = 6 > 0 by R1, R5.3 and R5.4. If δ1 ( v ) = 8, then ch ( v ) ≥ 0 − 3 × 12 + min{3 × 23 , 5 4 × 12 } = 16 > 0 by R1, R5.5 and R5.6. Suppose that d G ( v ) = 5. By Lemma 8 and VAL, v is incident with at most four 3-faces in G × and is adjacent to at least four (Δ − 2)+ -vertices in G. If f 3 ( v ) ≤ 3, then v receives 18 from each of its (Δ − 2)+ -neighbors in G by R6. So ch ( v ) ≥ 1 − 3 × 12 + 4 × 18 = 0 by R1; otherwise f 3 ( v ) = 4, then v receives 41 from each of its (Δ − 2)+ -neighbors in G. So ch ( v ) ≥ 1 − 4 × 12 + 4 × 14 = 0 by R1 and R6. Suppose that d G ( v ) = 6. By Lemma 8, v is incident with at most five 3-faces in G × . Suppose f 3 ( v ) ≤ 4. If δ1 ( v ) ≤ 5, then v is adjacent to at least four Δ-vertices in G by VAL, so ch ( v ) ≥ 2 − 4 × 12 − max{ 12 , 2 × 14 } + 4 × 15 > 0 by R1, R5, R6 and R7. If δ1 ( v ) ≥ 6, then ch ( v ) ≥ 2 − 4 × 12 = 0 by R1. So f 3 ( v ) = 5. If δ1 ( v ) = 4, then by VAL, v is adjacent to five Δ-vertices in G. So ch ( v ) ≥ 2 − 5 × 12 + 5 × 15 − 12 = 0 by R1, R5.1 and R7. If δ1 ( v ) = 5 (denote the vertex as x), then v is adjacent to by VAL. So we have ch ( v ) ≥ 0 − 3 ×
W. Zhang, J.-L. Wu / Theoretical Computer Science 566 (2015) 50–58
four Δ-vertices in G by VAL. If δ2 ( v ) ≥ 6, then ch ( v ) ≥ 2 − 5 ×
1 2
+4×
1 5
−
1 4
55
> 0 by R1, R6 and R7; otherwise δ2 ( v ) = 5
(denote the vertex as y). By Lemma 6, wx and w y are not incident with any 3-faces, so f 3 (x), f 3 ( y ) ≤ 3 and w sends
1 2
1 5
1 8
1 8
to
x and y, respectively by R6. It follows that ch ( v ) ≥ 2 − 5 × + 4 × − 2 × > 0 by R1, R6 and R7. Suppose that d G ( v ) = 7. By Lemma 8 and VAL, v is incident with at most six 3-faces in G × and at least two Δ-vertices. 5 If f 3 ( v ) ≤ 5, then ch ( v ) ≥ 3 − 5 × 12 − max{ 12 , 2 × 12 , 3 × 14 , 4 × 18 } + 2 × 16 = 0 by R1, R4, R5.2 and R6–R8. So f 3 ( v ) = 6, + + the remaining face is 5 -face by Lemma 8. Let the 5 -face be f . If δ1 ( v ) = 3, then v is adjacent to at least six Δ-vertices in G by VAL. So ch ( v ) ≥ 3 − 6 × 12 + 6 × 16 − 12 = 12 > 0 by R1, R4 and R8. If δ1 ( v ) = 4, then by VAL, v is adjacent to at least 5 five Δ-vertices in G. So ch ( v ) ≥ 3 − 6 × 12 + 5 × 16 − 2 × 12 = 0 by R1, R5 and R8. Suppose δ1 ( v ) = 5. Then v is adjacent to at least four Δ-vertices in G by VAL and the remaining face is 5+ -face and there are at least two false vertices on this face by Lemma 8. If d G × ( f ) = 5, then there is only one 3-vertex on this face since δ1 ( v ) = 5. So v receives at least 14 from
f by R2; otherwise d G × ( f ) ≥ 6, then there are at most
d G × ( f )−3 2
3-vertices on this face since any two 4− -vertices are not
adjacent by VAL and there are at least two false vertices on this face by Lemma 8. So v receives at least So ch ( v ) ≥ 3 − 6 ×
1 2
from f by R2.
− 3 × 14 + 14 = 16 > 0 by R1, R2, R6 and R8. If δ1 ( v ) = 6, then v is adjacent to at least three Δ-vertices in G by VAL. So ch ( v ) ≥ 3 − 6 × 12 + 3 × 16 − 4 × 18 = 0 by R1, R7 and R8. If δ1 ( v ) ≥ 7, then v is adjacent to at least two Δ-vertices in G by VAL. So ch ( v ) ≥ 3 − 6 × 12 + 2 × 16 = 13 > 0 by R1 and R8. Suppose that d G ( v ) = 8. Note that a 8-vertex v may be Δ-vertex here. We assume that Δ = 8 here. By Lemma 8, v is incident with at most six 3-faces in G × . If δ1 ( v ) = 2, then by VAL, v is adjacent to seven Δ-vertices in G. So ch ( v ) ≥ 4 − 6 × 12 − 1 ≥ 0 by R1 and R3. If δ1 ( v ) ≥ 5, then ch ( v ) ≥ 4 − 6 × 12 − max{4 × 14 , 5 × 15 , 6 × 16 } = 0 by R1 and R6–R8. Suppose δ1 ( v ) = 3. By VAL, v is adjacent to at least six Δ-vertices in G. If δ2 ( v ) = 3 or δ2 ( v ) ≥ 5, then ch ( v ) ≥ 4 − 6 × 1 5 − max{2 × 12 , 12 + 14 } ≥ 0 by R1, R5 and R7. So δ2 ( v ) = 4. By R5, v sends at most 12 to the 4-vertex adjacent to v. So 2 1 1 5 1 ch ( v ) ≥ 4 − 6 × 2 − 2 − 12 = 12 > 0. Suppose δ1 ( v ) = 4. By VAL, v is adjacent to at least five Δ-vertices in G. Let u be a 4-vertex that is adjacent to v in G. If δ1 (u ) = 6, then every vertex of N G (u , v ) \ {u , v } is a 8-vertex by Lemma 7. So ch ( v ) ≥ 5 4 − 6 × 12 − 13 = 23 > 0 by R5.1. If δ1 (u ) = δ2 (u ) = 7, then ch ( v ) ≥ 4 − 6 × 12 − max{3 × 13 , 2 × 13 + 13 , 13 + 2 × 13 , 13 + 12 + 16 } = 0 1 1 5 1 by R5.2, R5.3 and R5.4. If δ1 (u ) = 7 and δ2 (u ) = 8, then ch ( v ) ≥ 4 − 6 × 2 − max{3 × 3 , 2 × 12 + 6 } = 0 by R5.3 and R5.4. 5 If δ1 (u ) = 8, then ch ( v ) ≥ 4 − 6 × 12 − max{ 23 + 2 × 16 , 2 × 12 } = 16 > 0 by (2) of Lemma 7 and by R5.5 and R5.6. Suppose d G ( v ) = 9. By Lemma 8, v is incident with at most seven 3-faces in G × . If δ1 ( v ) = 2, then v is adjacent to eight 9-vertices in G by VAL. So ch ( v ) ≥ 5 − 7 × 12 − 1 = 12 ≥ 0 by R1 and R3. If δ1 ( v ) = 3, then v is adjacent to at least seven Δ-vertices in G by VAL. So ch ( v ) ≥ 5 − 7 × 12 − 2 × 12 > 0 by R1, R4 and R5. If δ1 ( v ) = 4. By VAL, v is adjacent to at least five 9-vertices in G. So ch ( v ) ≥ 5 − 7 × 12 − 3 × 12 = 0 by R1 and R5. If δ1 ( v ) ≥ 5, then ch ( v ) ≥ 5 − 7 × 12 − max{4 × 14 , 4 × 16 , 5 × 15 } = 12 > 0 by R1 and R6–R8. +4×
1 6
1 2
Case 2. Suppose that G is a 1-planar graph without 5-cycles. In the case, we define the discharging rules as follows. R1. Let f be a false 3-face in G × , then f receives 1 2
1 2
from each of its incident true vertices; let f be a true 3-face in G × ,
from each of its incident 5+ -vertices.
then f receives R2. Let f be a 6+ -face. If the number of false vertices incident with f is at least two, then each true vertex incident with f receives f receives
d G × ( f )−4 d G × ( f )−2 d G × ( f )−4 d G × ( f )−1
from f ; if the number of false vertices incident with f is one, then each true vertex incident with from f ; otherwise, each true vertex incident with f receives
R3. Let v be a 2-vertex in G. Then v receives 1 from each of its neighbors in G. R4. Let v be a 3-vertex in G. Then v receives 23 from each of its neighbors in G. R5. Let v be a 4-vertex in G. Then v receives
1 2
each of its 6-neighbors and
3 8
from each of its 8+ -neighbors; otherwise, v receives
from each of its 7+ -neighbors in G.
R7. Let v be a 6-vertex in G. If f 3 ( v ) = 6 and δ1 ( v ) = δ2 ( v ) = δ3 ( v ) = 6, then v receives 6-neighbors, receives 3 10
3 10
from f .
from each of its neighbors in G.
R6. Let v be a 5-vertex in G. If f 3 ( v ) ≤ 3, then v receives 3 8
d G × ( f )−4 dG × ( f )
1 10
3 16
from
from one of these three
5 from each of its 8+ -neighbors in G; otherwise, v receives 16 from each of its 7-neighbors and
from each of its 8+ -neighbors in G.
R8. Let v be a 7-vertex in G. Then v receives
1 4
from each of its Δ-neighbors in G.
Next we show that ch (x) ≥ 0 for all x ∈ V (G × ) ∪ F (G × ). Let f be a face of G × . Suppose d G × ( f ) = 3. If f is a false 3-face in G × , then f is incident with two true vertices by Lemma 4, so we have ch ( f ) ≥ −1 + 2 × 12 = −1 + 1 = 0 by R1. Otherwise, f is a true 3-face in G × , every true 3-face in G × is incident with at least two 5+ -vertices since any two 4− -vertices are not adjacent by VAL. So we have ch ( f ) ≥ −1 + min{2 × 12 , 3 × 12 } = −1 + 1 = 0 by R1. If 4 ≤ d G × ( f ) ≤ 5, then ch ( f ) = ch( f ) = 0. If d G × ( f ) ≥ 6, obviously, ch ( f ) ≥ 0 by R2. Let v be a vertex of G. If d G ( v ) = 2, then v is adjacent to two Δ-vertices in G by VAL. By Lemma 4, v cannot be incident with a false 3-face in G × . It follows that ch ( v ) = −2 + 2 × 1 = 0 by R3.
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If d G ( v ) = 3, then v is adjacent to three (Δ − 1)+ -vertices in G and v is incident with at most two false 3-faces in G × by VAL and Lemma 4. So we have ch ( v ) ≥ −1 − max{ 12 , 2 × 12 } + 3 × 23 = 0 by R1 and R4. Suppose that d( v ) = 4. If v is a false vertex, then ch ( v ) = ch( v ) = 0; otherwise, v is a true vertex. Suppose d G ( v ) = 4, then v is adjacent to four (Δ − 2)+ -vertices in G and v is incident with at most three false 3-faces in G × by VAL and Lemma 4. So we have ch ( v ) ≥ 0 − 3 × 12 + 4 × 12 = 12 > 0 by R1 and R5. Suppose that d G ( v ) = 5. Then v is adjacent to at least two Δ-vertices in G by VAL. If f 3 ( v ) ≤ 3, then ch ( v ) ≥ 1 − 3 × 1 + 2 × 38 = 14 > 0 by R1 and R6; otherwise, f 3 ( v ) ≥ 4. If δ1 ( v ) = 5, then v is adjacent to four Δ-vertices in G by VAL, so 2 ch ( v ) ≥ 1 − 5 ×
+ 4 × 38 = 0 by R1 and R6. If δ1 ( v ) = 6, then by VAL, v is adjacent to at least three Δ-vertices in G. So 3 ch ( v ) ≥ 1 − 5 × + 2 × 16 + 3 × 38 = 0 by R1 and R6. If δ1 ( v ) ≥ 7, then v is adjacent to at least two Δ-vertices in G by 1 VAL. So ch ( v ) ≥ 1 − 5 × 2 + 5 × 38 = 38 > 0 by R1 and R6. 1 Suppose that d G ( v ) = 6. There are two subcases: (1): v sends 10 to some adjacent 6-vertex w (Ref. R7). Note that one 1 6-vertex v sends charge 10 to its 6-neighbor w if and only if f 3 ( w ) = 6 and δ1 ( w ) = δ2 ( w ) = δ3 ( w ) = 6. Then f 6+ ( v ) ≥ 2 3 3 1 1 since G contains no 5-cycles. So ch ( v ) ≥ 2 − 4 × 12 + 2 × 13 + 2 × 10 − max{ 16 + 10 , 3 × 10 } > 0 by R1, R2, R6 and R7. (2): v sends no charge to some adjacent 6-vertex w. If δ1 ( v ) = 4, then by VAL, v is adjacent to five Δ-vertices in G. So 3 ch ( v ) ≥ 2 − 6 × 12 + 5 × 10 − 12 = 0 by R1, R5 and R7. If δ1 ( v ) = 5, then v is adjacent to at least four Δ-vertices in G 3 3 1 by VAL. If δ2 ( v ) ≥ 6, then ch ( v ) ≥ 2 − 6 × 12 + 4 × 10 − 16 = 80 > 0 by R1, R6 and R7. So δ2 ( v ) = 5. If f 3 ( v ) ≤ 5, then 1 3 3 13 ch ( v ) ≥ 2 − 5 × 2 + 4 × 10 − 2 × 16 = 40 > 0 by R1, R6 and R8; otherwise, f 3 ( v ) = 6, then there are at least two 6+ -face 3 incidents with each of these two 5-neighbors since G contains no 5-cycles, so ch ( v ) ≥ 2 − 6 × 12 + 4 × 10 = 15 > 0 by R1, R6 and R7. If δ1 ( v ) = 6, then v is adjacent to at least three Δ-vertices in G by VAL. If δ1 ( v ) = δ2 ( v ) = δ3 ( v ) = 6, then there are at least two 6+ -face incidents with one of these three 6-neighbors since G contains no 5-cycles, so ch ( v ) ≥ 3 1 3 5 5 2 − 6 × 12 + 3 × 10 + 10 = 0 by R1, R6 and R7; otherwise, ch ( v ) ≥ 2 − 6 × 12 + 3 × 10 + min{2 × 16 , 16 } > 0. If δ1 ( v ) ≥ 7, 1 3 5 then v is adjacent to at least two Δ-vertices in G by VAL. So ch ( v ) ≥ 2 − 6 × 2 + 2 × 10 + 4 × 16 > 0 by R1 and R7. Suppose that d G ( v ) = 7. By Lemma 9 and VAL, v is incident with at most six 3-faces in G × . Suppose f 3 ( v ) ≤ 5. If δ1 ( v ) = 3, then v is adjacent to six Δ-vertices in G by VAL. So ch ( v ) ≥ 3 − 5 × 12 + 6 × 14 − 23 > 0 by R1, R4 and R8. If δ1 ( v ) = 4, then by VAL, v is adjacent to at least five Δ-vertices in G. So ch ( v ) ≥ 3 − 5 × 12 + 5 × 14 − 2 × 12 > 0 by R1, R5 and R8. If δ1 ( v ) = 5, then by VAL, v is adjacent to at least four Δ-vertices in G. So ch ( v ) ≥ 3 − 5 × 12 + 4 × 14 − 3 × 38 > 0 by R1, 5 R6 and R8. If δ1 ( v ) = 6, then v is adjacent to at least three Δ-vertices in G by VAL. So ch ( v ) ≥ 3 − 5 × 12 + 3 × 14 − 4 × 16 =0 1 by R1, R7 and R8. If δ1 ( v ) ≥ 7, then v is adjacent to at least two Δ-vertices in G by VAL. So ch ( v ) ≥ 3 − 5 × 2 + 2 × 14 > 0 by R1. So f 3 ( v ) = 6, the remaining face is 6+ -face by Lemma 9 and v receives at least 12 from f by R2. If δ1 ( v ) = 3, then v is adjacent to at least six Δ-vertices in G by VAL. So ch ( v ) ≥ 3 − 6 × 12 + 12 + 6 × 14 − 23 > 0 by R1, R4 and R8. If δ1 ( v ) = 4, then by VAL, v is adjacent to at least five Δ-vertices in G. So ch ( v ) ≥ 3 − 6 × 12 + 12 + 5 × 14 − 2 × 12 > 0 by R1, R5 and R8. If δ1 ( v ) = 5, then by VAL, v is adjacent to at least four Δ-vertices in G. So ch ( v ) ≥ 3 − 6 × 12 + 12 + 4 × 14 − 3 × 38 > 0 by R1, R6 5 and R8. If δ1 ( v ) = 6, then v is adjacent to at least three Δ-vertices in G by VAL. So ch ( v ) ≥ 3 − 6 × 12 + 12 + 3 × 14 − 4 × 16 =0 1 1 by R1, R7 and R8. If δ1 ( v ) ≥ 7, then v is adjacent to at least two Δ-vertices in G by VAL. So ch ( v ) ≥ 3 − 6 × 2 + 2 + 2 × 14 > 0
1 2 1 2
by R1 and R8. Suppose that d G ( v ) = 8. Note that a 8-vertex v may be Δ-vertex here. We assume that Δ = 8 here. By Lemma 9, v is 3 incident with at most six 3-faces in G × . If f 3 ( v ) ≤ 5, then ch ( v ) ≥ 4 − 5 × 12 − max{1, 2 × 23 , 3 × 12 , 4 × 38 , 4 × 10 , 6 × 14 } = 0. So f 3 ( v ) = 6, the remaining faces are 6+ -face by Lemma 9 and v receives at least
ch ( v ) ≥ 4 − 6 ×
1 2
+2×
1 3
− max{1, 2 ×
2 3 , 3 × 12 , 4 × 38 , 4 × 10 , 6 × 14 } 3
1 3
from each of two 6+ -faces by R2. So
> 0.
Suppose that d G ( v ) = 9. By Lemma 9, v is incident with at most seven 3-faces in G × . So ch ( v ) ≥ 5 − 7 × 12 − max{1, 2 × 23 ,
3 3 × 12 , 4 × 38 , 4 × 10 , 6 × 14 } = 0. The proof of Theorem 2 is completed.
2
3. The proof of Theorem 3 Lemma 10. Let G be a 1-planar graph without adjacent chordal 5-cycles and let G × be its associated plane graph. Let v be a k-vertex of G. Then the following results hold. (1) If k = 7, then f 3 ( v ) ≤ 6 in G × ; (2) If k = 8, then f 3 ( v ) ≤ 7 in G × . Moreover, if f 3 ( v ) = 7, then v is incident with a 5+ -face which is incident with at least one false vertex; (3) If k = 9, then f 3 ( v ) ≤ 8 in G × . Moreover, if f 3 ( v ) = 8, then v is incident with a 5+ -face which is incident with at least two false vertices on this face. Proof. Let v 1 , v 2 , . . . , v k be neighbors of v of G × in an anticlockwise order. Let f i of G × be face incident with v, v i and v i +1 , for all i such that i ∈ {1, 2, . . . , k}. Note that all subscripts in the paper are taken modulo k. If f i , f i +1 , f i +2 are true
W. Zhang, J.-L. Wu / Theoretical Computer Science 566 (2015) 50–58
57
3-faces for some 1 ≤ i ≤ k, then v v i v i +1 v i +2 v i +3 v is a chordal 5-cycle. If f i and f i +1 are two false 3-faces and v i +1 is the false vertex, then v v i v i +2 v forms a 3-cycle. (1) Suppose f 3 ( v ) = 7. Then v is at most incident with three false vertices. If the number of false vertices is no more than two, obviously there are two adjacent chordal 5-cycles in G, a contradiction. If the number of false vertices is three, then v are incident with four 3-cycles C 1 , C 2 , C 3 and C 4 such that C 2 , C 4 are adjacent to C 1 and C 3 , and it follows that there are two adjacent chordal 5-cycles in G, a contradiction. So f 3 ( v ) ≤ 6. (2) If f 3 ( v ) = 8, then the proof of this case is similar to be proved as (1), we omit here. So f 3 ( v ) ≤ 7. Suppose f 3 ( v ) = 7 and such seven 3-faces are v v i v i +1 in G × , where 1 ≤ i ≤ 7. Then v is incident with four false vertices since G has no adjacent chordal 5-cycles. Suppose that the remaining face is f . Without loss of generality, we assume that v 1 , v 3 , v 5 and v 7 are false vertices and there is at least one false vertex on f . If d G × ( f ) = 4, then v is incident with adjacent chordal 5-cycles. So d G × ( f ) ≥ 5 and there is at least one false vertex on f . (3) If f 3 ( v ) = 9, then the proof of this case is similar to be proved as (1), we omit here. So f 3 ( v ) ≤ 8. Suppose f 3 ( v ) = 8 and such seven 3-faces are v v i v i +1 in G × , where 1 ≤ i ≤ 8. Then v is incident with five false vertices since G has no adjacent chordal 5-cycles. Suppose that the remaining face is f . Without loss of generality, we assume that v 1 , v 3 , v 5 , v 7 and v 9 are false vertices and there are at least two false vertices on f . If d G × ( f ) = 4, then v is incident with adjacent chordal 5-cycles. So d G × ( f ) ≥ 5 and there are at least two false vertices on f . 2 Now we prove Theorem 3. Proof. Since every 1-planar graph with maximum degree Δ ≥ 10 has chromatic index Δ (see [10]), it suffices to prove the following result. (*) Let G be a 1-planar graph without adjacent chordal 5-cycles. If Δ(G ) = 9, then
χ (G ) = 9.
Suppose that G is a counterexample to (*) with the smallest number of edges. Then G is an 9-critical 1-plane graph. By VAL, we have δ(G ) ≥ 2. In the following, we apply the discharging method on the associated planar graph G × and complete the proof by a contradiction. Since G × is a plane graph, we have | V (G × )| − | E (G × )| + | F (G × )| = 2. It follows that
dG × ( v ) − 4 +
v ∈ V (G × )
d G × ( f ) − 4 = −8.
f ∈ F (G × )
× × × × Now we define ch(x) to be the initial charge of x ∈ V (G ) ∪ F (G ). Let ch(x) = d G × (x) − 4 for each x ∈ V (G ) ∪ F (G ). It follows that x∈ V (G )∪ F (G × ) ch (x) = −8. Now we redistribute the initial charge ch (x) and form a new charge ch (x) for each x ∈ V (G × ) ∪ F (G × ) by discharging method. Since our rules only move charge around, and do not affect the sum, we have × × x∈ V (G × )∪ F (G × ) ch (x) = x∈ V (G × )∪ F (G × ) ch (x) = −8. If we can show that ch (x) ≥ 0 for each x ∈ V ( G ) ∪ F ( G ), then we
get an obvious contradiction to 0 ≤ x∈ V (G × )∪ F (G × ) ch (x) = Now we define the discharging rules as follows. R1. Let f be a false 3-face in G × , then f receives 1 2
1 2
x∈ V (G × )∪ F (G × ) ch (x)
= −8, which completes our proof.
from each of its incident true vertices; let f be a true 3-face in G × ,
from each of its incident 5+ -vertices.
then f receives R2. Let f be a 5+ -face. If the number of false vertices incident with f is one, then each true vertex incident with f receives d G × ( f )−4 d G × ( f )−1
receives
from f ; if the number of false vertices incident with f is at least two, then each true vertex incident with f d G × ( f )−4 d G × ( f )−2
from f .
R3. Let v be a 2-vertex in G. Then v receives 1 from each of its neighbors in G. R4. Let v be a 3-vertex in G. Then v receives 23 from each of its neighbors in G. R5. Let v be a 4-vertex in G. Then v receives R6. Let v be a 5-vertex in G. Then v receives R7. Let v be a 6-vertex in G. Then v receives R8. Let v be a 7-vertex in G. Then v receives R9. Let v be a 8-vertex in G. Then v receives
3 from each of its 7+ -neighbors in G. 8 1 from each of its 6+ -neighbors in G. 3 4 from each of its 7+ -neighbors in G. 15 2 from each of its 8+ -neighbors in G. 9 4 from each of its 9-neighbors in G. 21
Next we show that ch (x) ≥ 0 for all x ∈ V (G × ) ∪ F (G × ). Let f be a face of G × . If d G × ( f ) = 3, then every false 3-face in G × is incident with two true vertices and every true 3-face in G × is incident with at least two 5+ -vertices by Lemma 4 and VAL. So we have ch ( f ) ≥ ch( f ) + 2 × 12 = 3 − 4 + 1 = 0 by R1. If d G × ( f ) = 4, then ch ( f ) = ch( f ) ≥ 0. If d G × ( f ) ≥ 5, obviously, ch ( f ) ≥ 0 by R2. Let v be a vertex of G. If d G ( v ) = 2, then v is adjacent to two 9-vertices in G by VAL. By Lemma 4, v cannot be incident with a false 3-face in G × . It follows that ch ( v ) = ch( v ) + 2 × 1 = 0 by R1 and R3. If d G ( v ) = 3, then v is adjacent
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to three 8+ -vertices in G and v is incident with at most two false 3-faces in G × by VAL and Lemma 4. So we have ch ( v ) ≥ −1 − max{ 12 , 2 × 12 } + 3 × 23 = 0 by R1 and R3. Suppose that d( v ) = 4. If v is a false vertex, then ch ( v ) = ch( v ) = 0; otherwise, v is a true vertex. Suppose d G ( v ) = 4, then v is adjacent to four 7+ -vertices in G and v is incident with at most three false 3-faces in G × by VAL and Lemma 4. So we have ch ( v ) ≥ 0 − 3 × 12 + 4 × 38 = 0 by R1 and R5. If d G ( v ) = 5, then v is adjacent to five 6+ -vertices in G by VAL. So we have ch ( v ) ≥ 5 − 4 − 5 ×
1 2
+ 5 × 13 = 16 > 0 by R1 and R6. 4 Suppose that d G ( v ) = 6. If δ1 ( v ) = 5, then v is adjacent to at least five 9-vertices by VAL. So ch ( v ) ≥ 2 − 6 × 12 + 5 × 15 − 1 1 4 1 = 0 by R1, R6 and R7. If δ ( v ) = 6, then v is adjacent to four 9-vertices by VAL. So ch ( v ) ≥ 2 − 6 × + 4 × = > 0 1 3 2 15 15 4 by R1 and R7. If δ1 ( v ) ≥ 7, then v is adjacent to at least two 9-vertices by VAL. So ch ( v ) ≥ 2 − 6 × 12 + 6 × 15 = 35 > 0 by
R1 and R8. Suppose that d G ( v ) = 7. If δ1 ( v ) = 4, then v is adjacent to six 9-vertices by VAL. So ch ( v ) ≥ 3 − 6 × 12 + 6 × 29 − 38 = by R1, R5 and by R1, R6 and
23 24 R8. If δ1 ( v ) = 5, then v is adjacent to five 9-vertices by VAL. So ch ( v ) ≥ 3 − 6 × + 5 × − 2 × = 49 4 4 R8. If δ1 ( v ) = 6, then v is adjacent to four 9-vertices by VAL. So ch ( v ) ≥ 3 − 6 × 12 + 4 × 29 − 3 × 15 = 45 R8. If δ1 ( v ) ≥ 7, then v is adjacent to at least two 9-vertices by VAL. So ch ( v ) ≥ 3 − 6 × 12 + 2 × 29 = 49
1 2
2 9
1 3
>0 >0 >0 >0
by R1, R7 and by R1 and R8. Suppose that d G ( v ) = 8. By Lemma 10 and VAL, v is incident with at most seven 3-faces in G × and adjacent to at least 4 4 two 9-vertices. If f 3 ( v ) ≤ 6, then ch ( v ) ≥ 4 − 6 × 12 + 2 × 21 − max{ 23 , 2 × 38 , 3 × 13 , 4 × 15 , 5 × 29 } > 0 by R1 and R4–R7; otherwise f 3 ( v ) = 7. The remaining face is 5+ -face by Lemma 10. Let the 5+ -face be f . By R2, v receives at least
1 4
from f .
4 4 So ch ( v ) ≥ 4 − 7 × 12 + 14 + 2 × 21 − max{ 23 , 2 × 38 , 3 × 13 , 4 × 15 , 5 × 29 } > 0 by R1, R2 and R4–R7. Suppose that d G ( v ) = 9. By Lemma 10 and VAL, v is incident with at most eight 3-faces in G × and adjacent to at least 4 4 two 9-vertices. If f 3 ( v ) ≤ 7, then ch ( v ) ≥ 5 − 7 × 12 − max{1, 2 × 23 , 3 × 38 , 4 × 13 , 5 × 15 , 6 × 29 , 7 × 21 } = 0 by R1 and R3–R8; otherwise f 3 ( v ) = 8. The remaining face is 5+ -face and there are at least two false vertices on this face by Lemma 10. Let 4 the 5+ -face be f . By R2, v receives at least 13 from f . So ch ( v ) ≥ 5 − 8 × 12 + 13 − max{1, 2 × 23 , 3 × 38 , 4 × 13 , 5 × 15 ,6 × 2 4 , 7 × 21 } 9
= 0 by R1, R2 and R3–R8. The proof of Theorem 3 is completed.
2
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