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(Δ+1) -total-colorability of plane graphs with maximum degree Δ at least 6 and without adjacent short cycles
Information Processing Letters 110 (2010) 830–834
Contents lists available at ScienceDirect
Information Processing Letters www.elsevier.com/locate/ipl
( + 1)-total-colorability of plane graphs with maximum degree at least 6 and without adjacent short cycles ✩ Jingwen Zhang, Yingqian Wang ∗ College of Mathematics, Physics and Information Engineering, Zhejiang Normal University, Jinhua, 321004, China
a r t i c l e
i n f o
a b s t r a c t
Article history: Received 6 May 2010 Received in revised form 14 June 2010 Accepted 1 July 2010 Communicated by J. Torán
By a short cycle we mean a cycle of length at most 4. In this paper, we prove that every plane graph with maximum degree at least 6 and without adjacent short cycles is ( + 1)-totally-colorable. © 2010 Elsevier B.V. All rights reserved.
Keywords: Combinatorial problems Plane graph Total coloring Maximum degree
1. Introduction All graphs considered here are finite, simple and undirected. Undefined terminology and notation in this paper can be found in [2]. A graph is planar if it can be embedded into the plane so that its edges meet only at their ends. Any such embedding of a planar graph is called a plane graph. For a plane graph G, we denote its vertex set, edge set, face set, maximum degree and minimum degree by V (G ), E (G ), F (G ), (G ) and δ(G ) (or simply V , E, F , and δ if this causes no confusion), respectively. A k-cycle is a cycle of length k. A k− -cycle or a k+ -cycle is a cycle of length at most or at least k, respectively. Call v ∈ V a k-vertex or a k+ -vertex or a k− -vertex if d( v ), the degree of v in G, is equal to or at least or at most k, respectively. A k-vertex adjacent to v is often called a k-neighbor of v. An (i , j )-edge is an edge connecting an i-vertex to a j-vertex. For a face f ∈ F , we define its degree, denoted by d( f ), ✩ Supported by the Natural Science Foundation of Zhejiang Province, China, Grant No. Y6090699, partially supported by the Natural Science Foundation of China, Grant No. 10971198, and Zhejiang Innovation Project, Grant No. T200905. Corresponding author. E-mail address: [email protected] (Y. Wang).
*
0020-0190/$ – see front matter doi:10.1016/j.ipl.2010.07.003
© 2010
Elsevier B.V. All rights reserved.
to be the length of the boundary cycle of f . The notions of a k-face, a k+ -face, or a k− -face are analogously defined for vertices. By a (d( v 1 ), d( v 2 ), . . . , d( v k ))-face we mean a face with the boundary cycle v 1 v 2 . . . v k v 1 . Two cycles (or faces) are adjacent if they share at least one edge. A forest is a graph without cycles. A k-total-coloring of a graph G is a mapping φ from V ∪ E to the set of available colors {1, 2, . . . , k} such that φ(x) = φ( y ) for every pair of adjacent or incident elements x, y ∈ V ∪ E. G is called k-totally-colorable if it admits a k-total-coloring. Clearly, at least ( + 1) colors are needed to color G totally. Vizing [22] and Behzad [1] independently conjectured that every graph is ( + 2)-totally-colorable. This conjecture is known as the Total Coloring Conjecture, or TCC, and has been extensively studied. However, even for plane graphs, TCC remains open. It is unknown whether every plane graph with maximum degree 6 is 8-totallycolorable. More precisely, with one exception = 6, it has been proved that every plane graph with maximum degree is ( + 2)-totally-colorable, see [14,21,10,3,16,11,25]. It is interesting to note that plane graphs with maximum degree 9 are ( + 1)-totally-colorable, see [4,24,12]. It is challenging to determine the ( + 1)total-colorability neatly as in [4,24,12] for plane graphs
J. Zhang, Y. Wang / Information Processing Letters 110 (2010) 830–834
with maximum degree 8. It seems wise to study the ( + 1)-total-colorability for plane graphs with maximum degree 8 under some additional constraints. For = 8, Hou et al. [9] proved that a plane graph with maximum degree 8 is 9-totally-colorable if it does not contain any 5-cycle; or any 6-cycle. This result was extended later by a series of papers. For example, Shen and Wang proved that a plane graph with maximum degree 8 is 9-totallycolorable if it does not contain any chordal 5-cycle [17]; or chordal 6-cycle [17]; or intersecting triangles [18]; or adjacent triangles [7]. Very recently, Roussel and Zhu [15] proved that a plane graph with maximum degree 8 is 9-totally-colorable if, for every vertex x in the graph, there is a kx -cycle, kx ∈ {3, 4, 5, 6, 7, 8}, being absent at x. For = 7, as far as we know, it has been proved that a plane graph with maximum degree 7 is 8-totally-colorable if it does not contain any 3-cycle [6]; 4-cycle [23] (for the list version see [8]); 5-cycle [20]; intersecting 4-cycles [13]. As for = 6, [5] first proved that a plane graph with maximum degree 5 is ( + 1)-totally-colorable if it contains neither 3- nor 4-cycle. Recently, Shen and Wang [19] proved that a plane graph with maximum degree 6 is 7-totally-colorable if it does not contain any 4-cycle, and conjectured that every plane graph with maximum degree 4 is ( + 1)-totally-colorable. This conjecture may be known as PTCC since it only concerns total colorability of plane graphs. This paper offers PTCC a new moderate evidence as follows: Theorem 1. Every plane graph with maximum degree 6 and without adjacent 4− -cycles is ( + 1)-totally-colorable. 2. Lemmas Let G = ( V , E , F ) be a counterexample to Theorem 1 with σ (G ) = | V | + | E | as small as possible. It is easy to see that G is 2-connected, see [3]. Hence, δ 2 and the boundary of every face in G is a cycle. Lemmas 1–4 below were first proved for higher maximum degree in some previous papers. Those proofs still apply for all 6. Hence, we cite them here without proofs. Lemma 1. (See [3].) Let uv ∈ E. If d(u ) , then d(u ) + 2 d( v ) + 2. As corollaries, the two neighbors of a 2-vertex are -vertices; the neighbors of a 3-vertex are ( − 1)+ -vertices. Lemma 2. (See [3].) The subgraph induced by all (2, )-edges in G is a forest. In particular, G has no (2, , 2, )-face. Let H be the forest stated in Lemma 2 and T , a maximal tree (i.e., a component) in H . It is easy to see that all leaves of T are -vertices. It is easy to prove by induction that T has a maximum matching M that matches every 2-vertex in T . Fix a maximum matching for each maximal tree in H . Let v be a 2-vertex in G. The neighbor of v which is matched to v by the prescribed matching is called the master of v. The notion of master was introduced in [5].
831
Fig. 1. Forbidden configuration.
Lemma 3. (See [24].) G has no configuration depicted in Fig. 1 where u 1 and u 4 are 2-vertices. Namely, if a -vertex has a 2neighbor on a 3-cycle, then every other neighbor of the -vertex is a 3+ -vertex. Lemma 4. (See [20].) G has no (4, 4, 4)-face. The following lemma follows from the absence of adjacent 4− -cycles in G. Lemma 5. Let f and f be two adjacent faces with d( f ) = 3. Then f is a 6+ -face if f and f share two edges; a 5+ -face, otherwise. 3. Discharging Using a discharging procedure, we derive a contradiction showing that G does not exist. Euler’s formula | V | − | E | + | F | = 2 can be easily rewritten as
d(x) − 4 = −8.
x∈ V ∪ F
We define the initial charge function ch on V ∪ F by letting ch(x) = d(x) − 4 for x ∈ V ∪ F . So, x∈ V ∪ F ch(x) = −8. Since any discharging procedure preserves the total charge of G, if we can define suitable discharging rules to change the initial charge function ch to the final charge function ch on V ∪ F such that ch (x) 0 for all x ∈ V ∪ F , then we get an obvious contradiction 0 x∈ V ∪ F ch (x) = ch ( x ) = − 8, which establishes Theorem 1. x∈ V ∪ F We define the discharging rules as follows: R1: Charge to a 2-vertex v R1.1. If v is incident with a 3-face, then v gets 1 from its incident 6+ -face and 12 from each of its two -neighbors. R1.2. If v is incident with a 4-face, then v gets 12 from its only incident 5+ -face, 1 from its master and 1 from its other -neighbor. 2 R1.3. If v is incident with two 5+ -faces, then v gets 1 from each of its incident faces and 1 from its 2 master. R2: Charge to a 3-vertex v R2.1. If v is incident with a 4− -face, then v gets 12 from each of its two incident 5+ -faces. R2.2. If v is not incident with any 4− -face, then v gets 1 from each of its three incident 5+ -faces. 3
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J. Zhang, Y. Wang / Information Processing Letters 110 (2010) 830–834
Fig. 2. Discharging rules.
R3: Charge to a 3-face f Note that f is incident with at most one 3− -vertex by Lemma 1; two 4-vertices by Lemma 4. R3.1. If f is incident with exactly one 4− -vertex, then it gets 12 from each of its two 5+ -vertices. R3.2. If f is incident with exactly two 4-vertices, then it gets 14 from each of its two 4-vertices and 12 from its remaining 5+ -vertex. R3.3. If f is not incident with any 4− -vertex, then it gets 13 from each of its three vertices. R4: Charge to a 4-vertex v R4.1. If v is incident with exactly one (4, 4, 5+ )-face f , then v gets 18 from each of its two 5+ -faces which are adjacent to f and incident with v.
R4.2. Assume that v is incident with two (4, 4, 5+ )faces. R4.2.1. If there is exactly one 4-neighbor of v on each of its remaining two 5+ -faces, then v gets 14 from each of its remaining two incident 5+ -faces. R4.2.2. Otherwise, v gets 13 from its incident 5+ face which has two 5+ -neighbors of v and from its remaining incident 5+ -face.
1 5
The rest of this paper is devoted to checking that ch (x) 0 for all x ∈ V ∪ F . The final charge of vertices. First note that R1 and R2 are just designed so that ch ( v ) 0 for all 3− -vertices v ∈ V .
J. Zhang, Y. Wang / Information Processing Letters 110 (2010) 830–834
So, we only need to check that ch ( v ) 0 for all 4+ vertices v ∈ V . Let t be the number of 3-faces incident with v. Since G has no adjacent 3-cycles, t d(2v ) . Let v be a 4-vertex. If v is not incident with any (4, 4, 5+ )-face, according to our discharging rules, no charge is sent to or got from v, then ch ( v ) = ch( v ) = 0. If v is incident with exactly one (4, 4, 5+ )-face, then ch ( v ) = ch( v ) − 14 + 2 × 18 = 0 by R3.2 and R4.1. If v is incident with exactly two (4, 4, 5+ )-faces, then v sends
1 4 1 2
to each inci-
from its incident 5+ -faces by R4.2. Thus, ch ( v ) ch( v )− 2 × 14 + 12 = 0.
dent 3-face by R3.2, and receives at least
Let v be a 5-vertex. By our rules, v gives at most 12 to each incident 3-face. Note that t 2. So, ch ( v ) ch( v ) − 2 × 12 = 0. Let v be a 6+ -vertex. If v is not adjacent to any 2vertex, then v only sends at most 12 to each incident 3-face
by R3. Since t 12 d( v ), we have ch ( v ) ch( v ) − 1 2
1 d( v ) 2
3 d( v ) − 4 4
1 2
×t
d( v ) − 4 − × = > 0 by R3. Assume that v is adjacent to at least one 2-vertex. In this case, v is a vertex by Lemma 1. If v has a 2-neighbor on a 3-face, then v is not adjacent to any other 2-vertex by Lemma 3. Note that t 12 d( v ) = 12 . We have ch ( v ) ch( v ) − 12 − 12 × t
−4−
1 2
−
1 2
× 12 = 34 −
9 2
0 by R1.1 and R3. As-
sume that no 2-neighbor of v is incident with any 3-face. Let n be the number of 2-neighbors of v each of which is incident with a 4-face, then n + t 12 d( v ) = 12 by our hypothesis. If v is not a master of any its 2-neighbor, then v sends 12 to each 2-neighbor that is incident with a 4-face and at most
1 2
to each incident 3-face. Hence,
ch ( v ) = ch( v ) − 12 (n + t ) − 4 − 12 × 12 = 34 − 4 > 0. Assume that v is a master of its some 2-neighbor, say u. If u is incident with a 4-face, then ch ( v ) = ch( v ) − 1 − 1 [(n − 1) + t ] − 4 − 1 − 12 × ( 12 − 1) = 34 − 92 0. 2 Assume that u is not incident with any 4-face. Note that 2 n + t −
. If is even, then ch ( v ) = ch( v ) − 1 − 2 1 (n + t ) 2
− 4 − 1 − 12 × ( 12 − 1) = 34 − 92 0. If is odd, then ch ( v ) = ch( v ) − 1 − 12 (n + t ) − 4 − 1 − 12 × 1 ( + − 1) = 34 − 19 > 0. 2 4
The final charge of faces. Note that R3 is just designed, according to Lemmas 1 and 4, so that ch ( f ) 0 for all 3faces f ∈ F . If f is a 4-face, then ch ( f ) = ch( f ) = 0 since no charge is sent to or got from f by our rules. Let f be a 5+ -face and nk , the number of k-vertices incident with d( f ) f . By Lemma 1, n3− = n2 + n3 2 . In what follows, n∗2 is the number of 2-vertices on f that are incident with a 3-face. According to the values of n3− , by analyzing the values of n∗2 , we can complete the final charge checking as follows. Let f be a 5-face. By Lemma 1, n3− 2. Note that n∗2 = 0 by Lemma 5. If n3− = 2, then the remaining three vertices on f are 5+ -vertices. Applying R1.2, R1.3 and R2, ch ( f ) ch( f ) − 2 × 12 = 0. Let n3− = 1. By Lemma 1, f is incident with at most two 4-vertices. If f is incident with exactly two 4-vertices, then ch ( f ) ch( f ) − 12 − 2 × 14 = 0 by R1, R2 and R4.
Otherwise, ch ( f ) ch( f ) −
1 2
−
1 3
> 0 by R1, R2 and R4.
833
Finally let n3− = 0. According to the values of n4 , we can check ch ( f ) 0 as follows. If n4 = 5, then ch ( f ) ch( f ) − 5 × 15 = 0 by R4.1 and R4.2.2. If n4 = 4, then
ch ( f ) ch( f ) − 4 ×
1 4
= 0 by R4.1 and R4.2.1. If n4 3, 1 0 by R4. 3
then ch ( f ) ch( f ) − 3 ×
Let f be a 6-face. Note that n3− 3 by Lemma 1 and n2 2 by Lemma 2. Due to the absence of adjacent 4− cycles in G, n∗2 1. Assume that n3− = 3. In this case, f is incident with three 5+ -vertices by Lemma 1. Hence, ch ( f ) = ch( f ) − 1 × n∗2 − 12 × (n3− − n∗2 ) 2 − 1 − 2 × 12 = 0 by R1 and R2. Assume that n3− = 2. Now, there are at least three 5+ vertices on f , hence n4 1. Thus, ch ( f ) ch( f )− 1 × n∗2 −
×(n3− − n∗2 )− 13 × n4 2 − 1 − 12 − 13 > 0 by R1, R2 and R4. Assume that n3− = 1. Note that there are at least two 5+ -vertices on f , namely, n4 3. Thus, ch ( f ) ch( f ) − 1 × n3− − 13 × n4 2 − 1 − 3 × 13 = 0 by R1, R2 and R4. Lastly, assume that n3− = 0. We have ch ( f ) ch( f ) − 6 × 13 = 0 by R4. 1 2
Let f be a 7-face. Note that n3− 3 by Lemma 1 and n2 3 by Lemma 2. Due to the absence of adjacent 4− -cycles in G, n∗2 2. Arguing as above can yield that ch ( f ) 0. Finally, let f be an 8+ -face. By Lemma 1, n3− 12 d( f ).
If n3− = 0, then ch ( f ) ch( f ) −
1 3
× n4 d( f ) − 4 − − 4 > 0 by R4. Assume that n3− > 0. Clearly, n5+ n3− by Lemma 1. Thus, n4 = d( f ) − n3− − n5+ d( f ) − n3− − n3− = d( f ) − 1 d( f ) 3
=
2 d( f ) − 4 3
16 3
2n3− . It follows that
1 × n3− − n∗2 − × n4 2 3 1 1 d( f ) − 4 − n∗2 − n3− + n∗2 2 2 1 − × d( f ) − 2n3− 3 1 1 1 2 = d( f ) − 4 − n∗2 − n3− − d( f ) + n3− 2 2 3 3 2 1 ∗ 1 = d ( f ) − 4 − n 2 + n 3− 3 2 6 2 1 d ( f ) − 4 − n 3− 3 3 2 1 1 d( f ) − 4 − × d( f ) 3 3 2 1 = d( f ) − 4 2 0
ch ( f ) ch( f ) − 1 × n∗2 −
1
by R1, R2 and R4. So, for each x ∈ V ∪ F , ch (x) 0. This completes the proof of Theorem 1.
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J. Zhang, Y. Wang / Information Processing Letters 110 (2010) 830–834
References [1] M. Behzad, Graphs and their chromatic numbers, Ph.D. Thesis, Michigan State University, 1965. [2] J.A. Bondy, U.S.R. Murty, Graph Theory, Springer, Berlin, 2008. [3] O.V. Borodin, On the total coloring of planar graphs, J. Reine Angew. Math. 394 (1989) 180–185. [4] O.V. Borodin, A.V. Kostochka, D.R. Woodall, Total colorings of planar graphs with large maximum degree, J. Graph Theory 26 (1997) 53– 59. [5] O.V. Borodin, A.V. Kostochka, D.R. Woodall, List edge and list total colourings of multigraphs, J. Combin. Theory Ser. B 71 (1997) 184– 204. [6] O.V. Borodin, A.V. Kostochka, D.R. Woodall, Total colourings of planar graphs with large girth, European J. Combin. 19 (1998) 19–24. [7] D. Du, L. Shen, Y. Wang, Planar graphs with maximum degree 8 and without adjacent triangles are 9-total-colorable, Discrete Appl. Math. 157 (2009) 2778–2784. [8] J. Hou, G. Liu, J. Cai, List edge and list total colorings of planar graphs without 4-cycles, Theoret. Comput. Sci. 369 (2006) 250–255. [9] J. Hou, Y. Zhu, G. Liu, J. Wu, Total coloring of planar graphs without small cycles, Graphs Combin. 24 (2008) 91–100. [10] A.V. Kostochka, The total coloring of a multigraph with maximum degree 4, Discrete Math. 17 (1977) 161–163. [11] A.V. Kostochka, The total chromatic number of any multigraph with maximum degree five is at most seven, Discrete Math. 162 (1996) 199–214. [12] L. Kowalik, J.-S. Sereni, R. Škrekovski, Total coloring of plane graphs with maximum degree nine, SIAM J. Discrete Math. 22 (2008) 1462– 1479.
[13] B. Liu, J. Hou, J. Wu, G. Liu, Total colorings and list total colorings of planar graphs without intersecting 4-cycles, Discrete Math. 309 (2009) 6035–6043. [14] M. Rosenfeld, On the total coloring of certain graphs, Israel J. Math. 9 (1971) 396–402. [15] N. Roussel, X. Zhu, Total coloring of planar graphs of maximum degree eight, Inform. Process. Lett. 110 (2010) 321–324. [16] D.P. Sanders, Y. Zhao, On total 9-colorings of maximum degree seven, J. Graph Theory 31 (1999) 67–73. [17] L. Shen, Y. Wang, Total coloring of planar graphs with maximum degree at least 8, Sci. China Ser. A 52 (2009) 1733–1742. [18] L. Shen, Y. Wang, On the 9 total colorability of planar graphs with maximum degree 8 and without intersecting triangles, Appl. Math. Lett. 22 (2009) 1369–1373. [19] L. Shen, Y. Wang, On the 7 total colorability of planar graphs with maximum degree 6 and without 4-cycles, Graphs Combin. 25 (2009) 401–407. [20] L. Shen, Y. Wang, Planar graphs with maximum degree 7 and without 5-cycles are 8-totally-colorable, Discrete Math. 310 (2010) 2372– 2379. [21] N. Vijayaditya, On total chromatic number of a graph, J. Lond. Math. Soc. 3 (1971) 405–408. [22] V.G. Vizing, Some unsolved problems in graph theory, Uspekhi Mat. Nauk 23 (1968) 117–134 (in Russian). [23] P. Wang, J. Wu, A note on total colorings of planar graphs without 4-cycles, Discuss. Math. Graph Theory 24 (2004) 125–135. [24] W. Wang, Total chromatic number of planar graphs with maximum degree ten, J. Graph Theory 54 (2007) 91–102. [25] H.P. Yap, Total Coloring of Graphs, Lecture Notes in Mathematics, vol. 1623, Springer, Berlin, 1996.
Contents lists available at ScienceDirect
Information Processing Letters www.elsevier.com/locate/ipl
( + 1)-total-colorability of plane graphs with maximum degree at least 6 and without adjacent short cycles ✩ Jingwen Zhang, Yingqian Wang ∗ College of Mathematics, Physics and Information Engineering, Zhejiang Normal University, Jinhua, 321004, China
a r t i c l e
i n f o
a b s t r a c t
Article history: Received 6 May 2010 Received in revised form 14 June 2010 Accepted 1 July 2010 Communicated by J. Torán
By a short cycle we mean a cycle of length at most 4. In this paper, we prove that every plane graph with maximum degree at least 6 and without adjacent short cycles is ( + 1)-totally-colorable. © 2010 Elsevier B.V. All rights reserved.
Keywords: Combinatorial problems Plane graph Total coloring Maximum degree
1. Introduction All graphs considered here are finite, simple and undirected. Undefined terminology and notation in this paper can be found in [2]. A graph is planar if it can be embedded into the plane so that its edges meet only at their ends. Any such embedding of a planar graph is called a plane graph. For a plane graph G, we denote its vertex set, edge set, face set, maximum degree and minimum degree by V (G ), E (G ), F (G ), (G ) and δ(G ) (or simply V , E, F , and δ if this causes no confusion), respectively. A k-cycle is a cycle of length k. A k− -cycle or a k+ -cycle is a cycle of length at most or at least k, respectively. Call v ∈ V a k-vertex or a k+ -vertex or a k− -vertex if d( v ), the degree of v in G, is equal to or at least or at most k, respectively. A k-vertex adjacent to v is often called a k-neighbor of v. An (i , j )-edge is an edge connecting an i-vertex to a j-vertex. For a face f ∈ F , we define its degree, denoted by d( f ), ✩ Supported by the Natural Science Foundation of Zhejiang Province, China, Grant No. Y6090699, partially supported by the Natural Science Foundation of China, Grant No. 10971198, and Zhejiang Innovation Project, Grant No. T200905. Corresponding author. E-mail address: [email protected] (Y. Wang).
*
0020-0190/$ – see front matter doi:10.1016/j.ipl.2010.07.003
© 2010
Elsevier B.V. All rights reserved.
to be the length of the boundary cycle of f . The notions of a k-face, a k+ -face, or a k− -face are analogously defined for vertices. By a (d( v 1 ), d( v 2 ), . . . , d( v k ))-face we mean a face with the boundary cycle v 1 v 2 . . . v k v 1 . Two cycles (or faces) are adjacent if they share at least one edge. A forest is a graph without cycles. A k-total-coloring of a graph G is a mapping φ from V ∪ E to the set of available colors {1, 2, . . . , k} such that φ(x) = φ( y ) for every pair of adjacent or incident elements x, y ∈ V ∪ E. G is called k-totally-colorable if it admits a k-total-coloring. Clearly, at least ( + 1) colors are needed to color G totally. Vizing [22] and Behzad [1] independently conjectured that every graph is ( + 2)-totally-colorable. This conjecture is known as the Total Coloring Conjecture, or TCC, and has been extensively studied. However, even for plane graphs, TCC remains open. It is unknown whether every plane graph with maximum degree 6 is 8-totallycolorable. More precisely, with one exception = 6, it has been proved that every plane graph with maximum degree is ( + 2)-totally-colorable, see [14,21,10,3,16,11,25]. It is interesting to note that plane graphs with maximum degree 9 are ( + 1)-totally-colorable, see [4,24,12]. It is challenging to determine the ( + 1)total-colorability neatly as in [4,24,12] for plane graphs
J. Zhang, Y. Wang / Information Processing Letters 110 (2010) 830–834
with maximum degree 8. It seems wise to study the ( + 1)-total-colorability for plane graphs with maximum degree 8 under some additional constraints. For = 8, Hou et al. [9] proved that a plane graph with maximum degree 8 is 9-totally-colorable if it does not contain any 5-cycle; or any 6-cycle. This result was extended later by a series of papers. For example, Shen and Wang proved that a plane graph with maximum degree 8 is 9-totallycolorable if it does not contain any chordal 5-cycle [17]; or chordal 6-cycle [17]; or intersecting triangles [18]; or adjacent triangles [7]. Very recently, Roussel and Zhu [15] proved that a plane graph with maximum degree 8 is 9-totally-colorable if, for every vertex x in the graph, there is a kx -cycle, kx ∈ {3, 4, 5, 6, 7, 8}, being absent at x. For = 7, as far as we know, it has been proved that a plane graph with maximum degree 7 is 8-totally-colorable if it does not contain any 3-cycle [6]; 4-cycle [23] (for the list version see [8]); 5-cycle [20]; intersecting 4-cycles [13]. As for = 6, [5] first proved that a plane graph with maximum degree 5 is ( + 1)-totally-colorable if it contains neither 3- nor 4-cycle. Recently, Shen and Wang [19] proved that a plane graph with maximum degree 6 is 7-totally-colorable if it does not contain any 4-cycle, and conjectured that every plane graph with maximum degree 4 is ( + 1)-totally-colorable. This conjecture may be known as PTCC since it only concerns total colorability of plane graphs. This paper offers PTCC a new moderate evidence as follows: Theorem 1. Every plane graph with maximum degree 6 and without adjacent 4− -cycles is ( + 1)-totally-colorable. 2. Lemmas Let G = ( V , E , F ) be a counterexample to Theorem 1 with σ (G ) = | V | + | E | as small as possible. It is easy to see that G is 2-connected, see [3]. Hence, δ 2 and the boundary of every face in G is a cycle. Lemmas 1–4 below were first proved for higher maximum degree in some previous papers. Those proofs still apply for all 6. Hence, we cite them here without proofs. Lemma 1. (See [3].) Let uv ∈ E. If d(u ) , then d(u ) + 2 d( v ) + 2. As corollaries, the two neighbors of a 2-vertex are -vertices; the neighbors of a 3-vertex are ( − 1)+ -vertices. Lemma 2. (See [3].) The subgraph induced by all (2, )-edges in G is a forest. In particular, G has no (2, , 2, )-face. Let H be the forest stated in Lemma 2 and T , a maximal tree (i.e., a component) in H . It is easy to see that all leaves of T are -vertices. It is easy to prove by induction that T has a maximum matching M that matches every 2-vertex in T . Fix a maximum matching for each maximal tree in H . Let v be a 2-vertex in G. The neighbor of v which is matched to v by the prescribed matching is called the master of v. The notion of master was introduced in [5].
831
Fig. 1. Forbidden configuration.
Lemma 3. (See [24].) G has no configuration depicted in Fig. 1 where u 1 and u 4 are 2-vertices. Namely, if a -vertex has a 2neighbor on a 3-cycle, then every other neighbor of the -vertex is a 3+ -vertex. Lemma 4. (See [20].) G has no (4, 4, 4)-face. The following lemma follows from the absence of adjacent 4− -cycles in G. Lemma 5. Let f and f be two adjacent faces with d( f ) = 3. Then f is a 6+ -face if f and f share two edges; a 5+ -face, otherwise. 3. Discharging Using a discharging procedure, we derive a contradiction showing that G does not exist. Euler’s formula | V | − | E | + | F | = 2 can be easily rewritten as
d(x) − 4 = −8.
x∈ V ∪ F
We define the initial charge function ch on V ∪ F by letting ch(x) = d(x) − 4 for x ∈ V ∪ F . So, x∈ V ∪ F ch(x) = −8. Since any discharging procedure preserves the total charge of G, if we can define suitable discharging rules to change the initial charge function ch to the final charge function ch on V ∪ F such that ch (x) 0 for all x ∈ V ∪ F , then we get an obvious contradiction 0 x∈ V ∪ F ch (x) = ch ( x ) = − 8, which establishes Theorem 1. x∈ V ∪ F We define the discharging rules as follows: R1: Charge to a 2-vertex v R1.1. If v is incident with a 3-face, then v gets 1 from its incident 6+ -face and 12 from each of its two -neighbors. R1.2. If v is incident with a 4-face, then v gets 12 from its only incident 5+ -face, 1 from its master and 1 from its other -neighbor. 2 R1.3. If v is incident with two 5+ -faces, then v gets 1 from each of its incident faces and 1 from its 2 master. R2: Charge to a 3-vertex v R2.1. If v is incident with a 4− -face, then v gets 12 from each of its two incident 5+ -faces. R2.2. If v is not incident with any 4− -face, then v gets 1 from each of its three incident 5+ -faces. 3
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J. Zhang, Y. Wang / Information Processing Letters 110 (2010) 830–834
Fig. 2. Discharging rules.
R3: Charge to a 3-face f Note that f is incident with at most one 3− -vertex by Lemma 1; two 4-vertices by Lemma 4. R3.1. If f is incident with exactly one 4− -vertex, then it gets 12 from each of its two 5+ -vertices. R3.2. If f is incident with exactly two 4-vertices, then it gets 14 from each of its two 4-vertices and 12 from its remaining 5+ -vertex. R3.3. If f is not incident with any 4− -vertex, then it gets 13 from each of its three vertices. R4: Charge to a 4-vertex v R4.1. If v is incident with exactly one (4, 4, 5+ )-face f , then v gets 18 from each of its two 5+ -faces which are adjacent to f and incident with v.
R4.2. Assume that v is incident with two (4, 4, 5+ )faces. R4.2.1. If there is exactly one 4-neighbor of v on each of its remaining two 5+ -faces, then v gets 14 from each of its remaining two incident 5+ -faces. R4.2.2. Otherwise, v gets 13 from its incident 5+ face which has two 5+ -neighbors of v and from its remaining incident 5+ -face.
1 5
The rest of this paper is devoted to checking that ch (x) 0 for all x ∈ V ∪ F . The final charge of vertices. First note that R1 and R2 are just designed so that ch ( v ) 0 for all 3− -vertices v ∈ V .
J. Zhang, Y. Wang / Information Processing Letters 110 (2010) 830–834
So, we only need to check that ch ( v ) 0 for all 4+ vertices v ∈ V . Let t be the number of 3-faces incident with v. Since G has no adjacent 3-cycles, t d(2v ) . Let v be a 4-vertex. If v is not incident with any (4, 4, 5+ )-face, according to our discharging rules, no charge is sent to or got from v, then ch ( v ) = ch( v ) = 0. If v is incident with exactly one (4, 4, 5+ )-face, then ch ( v ) = ch( v ) − 14 + 2 × 18 = 0 by R3.2 and R4.1. If v is incident with exactly two (4, 4, 5+ )-faces, then v sends
1 4 1 2
to each inci-
from its incident 5+ -faces by R4.2. Thus, ch ( v ) ch( v )− 2 × 14 + 12 = 0.
dent 3-face by R3.2, and receives at least
Let v be a 5-vertex. By our rules, v gives at most 12 to each incident 3-face. Note that t 2. So, ch ( v ) ch( v ) − 2 × 12 = 0. Let v be a 6+ -vertex. If v is not adjacent to any 2vertex, then v only sends at most 12 to each incident 3-face
by R3. Since t 12 d( v ), we have ch ( v ) ch( v ) − 1 2
1 d( v ) 2
3 d( v ) − 4 4
1 2
×t
d( v ) − 4 − × = > 0 by R3. Assume that v is adjacent to at least one 2-vertex. In this case, v is a vertex by Lemma 1. If v has a 2-neighbor on a 3-face, then v is not adjacent to any other 2-vertex by Lemma 3. Note that t 12 d( v ) = 12 . We have ch ( v ) ch( v ) − 12 − 12 × t
−4−
1 2
−
1 2
× 12 = 34 −
9 2
0 by R1.1 and R3. As-
sume that no 2-neighbor of v is incident with any 3-face. Let n be the number of 2-neighbors of v each of which is incident with a 4-face, then n + t 12 d( v ) = 12 by our hypothesis. If v is not a master of any its 2-neighbor, then v sends 12 to each 2-neighbor that is incident with a 4-face and at most
1 2
to each incident 3-face. Hence,
ch ( v ) = ch( v ) − 12 (n + t ) − 4 − 12 × 12 = 34 − 4 > 0. Assume that v is a master of its some 2-neighbor, say u. If u is incident with a 4-face, then ch ( v ) = ch( v ) − 1 − 1 [(n − 1) + t ] − 4 − 1 − 12 × ( 12 − 1) = 34 − 92 0. 2 Assume that u is not incident with any 4-face. Note that 2 n + t −
. If is even, then ch ( v ) = ch( v ) − 1 − 2 1 (n + t ) 2
− 4 − 1 − 12 × ( 12 − 1) = 34 − 92 0. If is odd, then ch ( v ) = ch( v ) − 1 − 12 (n + t ) − 4 − 1 − 12 × 1 ( + − 1) = 34 − 19 > 0. 2 4
The final charge of faces. Note that R3 is just designed, according to Lemmas 1 and 4, so that ch ( f ) 0 for all 3faces f ∈ F . If f is a 4-face, then ch ( f ) = ch( f ) = 0 since no charge is sent to or got from f by our rules. Let f be a 5+ -face and nk , the number of k-vertices incident with d( f ) f . By Lemma 1, n3− = n2 + n3 2 . In what follows, n∗2 is the number of 2-vertices on f that are incident with a 3-face. According to the values of n3− , by analyzing the values of n∗2 , we can complete the final charge checking as follows. Let f be a 5-face. By Lemma 1, n3− 2. Note that n∗2 = 0 by Lemma 5. If n3− = 2, then the remaining three vertices on f are 5+ -vertices. Applying R1.2, R1.3 and R2, ch ( f ) ch( f ) − 2 × 12 = 0. Let n3− = 1. By Lemma 1, f is incident with at most two 4-vertices. If f is incident with exactly two 4-vertices, then ch ( f ) ch( f ) − 12 − 2 × 14 = 0 by R1, R2 and R4.
Otherwise, ch ( f ) ch( f ) −
1 2
−
1 3
> 0 by R1, R2 and R4.
833
Finally let n3− = 0. According to the values of n4 , we can check ch ( f ) 0 as follows. If n4 = 5, then ch ( f ) ch( f ) − 5 × 15 = 0 by R4.1 and R4.2.2. If n4 = 4, then
ch ( f ) ch( f ) − 4 ×
1 4
= 0 by R4.1 and R4.2.1. If n4 3, 1 0 by R4. 3
then ch ( f ) ch( f ) − 3 ×
Let f be a 6-face. Note that n3− 3 by Lemma 1 and n2 2 by Lemma 2. Due to the absence of adjacent 4− cycles in G, n∗2 1. Assume that n3− = 3. In this case, f is incident with three 5+ -vertices by Lemma 1. Hence, ch ( f ) = ch( f ) − 1 × n∗2 − 12 × (n3− − n∗2 ) 2 − 1 − 2 × 12 = 0 by R1 and R2. Assume that n3− = 2. Now, there are at least three 5+ vertices on f , hence n4 1. Thus, ch ( f ) ch( f )− 1 × n∗2 −
×(n3− − n∗2 )− 13 × n4 2 − 1 − 12 − 13 > 0 by R1, R2 and R4. Assume that n3− = 1. Note that there are at least two 5+ -vertices on f , namely, n4 3. Thus, ch ( f ) ch( f ) − 1 × n3− − 13 × n4 2 − 1 − 3 × 13 = 0 by R1, R2 and R4. Lastly, assume that n3− = 0. We have ch ( f ) ch( f ) − 6 × 13 = 0 by R4. 1 2
Let f be a 7-face. Note that n3− 3 by Lemma 1 and n2 3 by Lemma 2. Due to the absence of adjacent 4− -cycles in G, n∗2 2. Arguing as above can yield that ch ( f ) 0. Finally, let f be an 8+ -face. By Lemma 1, n3− 12 d( f ).
If n3− = 0, then ch ( f ) ch( f ) −
1 3
× n4 d( f ) − 4 − − 4 > 0 by R4. Assume that n3− > 0. Clearly, n5+ n3− by Lemma 1. Thus, n4 = d( f ) − n3− − n5+ d( f ) − n3− − n3− = d( f ) − 1 d( f ) 3
=
2 d( f ) − 4 3
16 3
2n3− . It follows that
1 × n3− − n∗2 − × n4 2 3 1 1 d( f ) − 4 − n∗2 − n3− + n∗2 2 2 1 − × d( f ) − 2n3− 3 1 1 1 2 = d( f ) − 4 − n∗2 − n3− − d( f ) + n3− 2 2 3 3 2 1 ∗ 1 = d ( f ) − 4 − n 2 + n 3− 3 2 6 2 1 d ( f ) − 4 − n 3− 3 3 2 1 1 d( f ) − 4 − × d( f ) 3 3 2 1 = d( f ) − 4 2 0
ch ( f ) ch( f ) − 1 × n∗2 −
1
by R1, R2 and R4. So, for each x ∈ V ∪ F , ch (x) 0. This completes the proof of Theorem 1.
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