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On 3-choosability of planar graphs without certain cycles
Information Processing Letters 107 (2008) 102–106 www.elsevier.com/locate/ipl
On 3-choosability of planar graphs without certain cycles ✩ Haihui Zhang a,b,∗ , Zhiren Sun a a School of Mathematics and Computer Science, Nanjing Normal University, 122, Ninghai Road, Nanjing 210097, PR China b Mathematics Department, Huaiyin Teachers College, 223001 Huaian, PR China
Received 12 January 2008; received in revised form 28 January 2008 Available online 14 February 2008 Communicated by L. Boasson
Abstract A graph G = (V , E) is L-colorable if for a given list assignment L = {L(v): v ∈ V (G)}, there exists a proper coloring c of G such that c(v) ∈ L(v) for all v ∈ V . If G is L-colorable for every list assignment L with |L(v)| k for all v ∈ V , then G is said to be k-choosable. In this paper, we prove that every planar graph with neither 5-, 6-, and 7-cycles nor triangles of distance less than 3, or with neither 5-, 6-, and 8-cycles nor triangles of distance less than 2 is 3-choosable. © 2008 Elsevier B.V. All rights reserved. Keywords: Combinatorial problems; Choosability; Planar graph; Cycle
1. Introduction All graphs considered in this paper are finite, simple planar graphs. A graph G is planar if G can be drawn on the plane so that its edges meet only at the vertices of the graph. A plane graph is such a particular drawing of a planar graph. Let G = (V , E, F ) denote a planar graph, with V , E and F being the set of vertices, edges and faces of G, respectively. We use b(f ) to denote the boundary walk of a face f and write b(f ) = [v1 v2 v3 . . . vn ] if v1 , v2 , v3 , . . . , vn are the vertices of b(f ) in a cyclic order. A face f is incident with all vertices and edges on b(f ). The degree of a face f of G, denoted ✩
Research is partially supported by the NSFC 10371055.
* Corresponding author at: School of Mathematics and Computer
Science, Nanjing Normal University, 122, Ninghai Road, Nanjing 210097, PR China. E-mail address: [email protected] (H. Zhang). 0020-0190/$ – see front matter © 2008 Elsevier B.V. All rights reserved. doi:10.1016/j.ipl.2008.01.008
by dG (f ), is the number of edges incident with it, where cut edges are counted twice. A vertex (face) of degree k is called a k-vertex (k-face). If r k or 1 k r, then a k-vertex (k-face) is called an r + - or r − -vertex (r + - or r − -face), respectively. A k-cycle is a cycle with k edges. The vertex set of a cycle C will also be denoted by C. A graph G = (V , E) is L-colorable if for a given list assignment L = {L(v): v ∈ V (G)}, there exists a proper coloring c of G such that c(v) ∈ L(v) for all v ∈ V . If G is L-colorable for every list assignment L with |L(v)| k for all v ∈ V , then G is said to be k-choosable. The choice number of G, denoted by χl (G) or ch(G), is the minimum k such that G is k-choosable. All 2-choosable graphs were characterized completely in [2]. Thomassen proved that every planar graph is 5-choosable [11]. Examples of planar graphs which are not 4-choosable were given by Voigt [13], and by Mirzakhani [8] independently. Voigt and Wirth [14], and Gutner [4] independently, presented some planar graphs of girth 4 which are not 3-choosable. Thus, it re-
H. Zhang, Z. Sun / Information Processing Letters 107 (2008) 102–106
mains to determine whether a given planar graph is 3or 4-choosable. In [4], Gutner proved that these problems are NP-complete. Therefore, many authors tried to find sufficient conditions for a planar graph to be 3- or 4-choosable. Alon and Tarsi [1] proved that every planar bipartite graph is 3-choosable. Thomassen [12] proved that every planar graph of girth 5 is 3-choosable. Lam et al. [5,6] proved that plane graphs without i-cycles are 4-choosable, for i = 3, 4, 5 or 6. Independently, Wang and Lih [15] proved that plane graphs without 5-cycles are 4-choosable, Fijavž et al. [3] proved that plane graphs without 6-cycles are 4-choosable. Xu [18], and Wang and Lih [16] independently, proved that plane graphs without two triangles sharing a common vertex are 4-choosable. More recently, some new sufficient conditions for a planar graph to be 3-choosable have been given. These include: having no 3-, 5-, and 6-cycles [7]; having no 3-, 8-, and 9-cycles [19]. Together with other known parallel results, Wang et al. [17] obtained a theorem on 3choosability of planar graphs: planar graphs without cycles of length 4, i, j , 9 with i < j and i, j ∈ {5, 6, 7, 8} are 3-choosable. The distance between two vertices x and y, denoted by dist(x, y), is the length of a shortest path connecting them in G. The distance between two triangles T and T is defined to be the value min{dist(x, y) | x ∈ V (T ) and y ∈ V (T )}. In [9], Montassier et al. proved that every planar graph either without 4- and 5-cycles and without triangles at distance less than 4, or without 4-, 5-, and 6-cycles and without triangles at distance less than 3 is 3-choosable. In [10] Montassier proposed a conjecture that every planar graph without cycles of length 4, 5, 6, is 3-choosable. In this paper, we prove that: Theorem 1. Every planar graph without 5-, 6-, and 7cycles, and without triangles of distance less than 3 is 3-choosable. Theorem 2. Every planar graph without 5-, 6-, and 8cycles, and without triangles of distance less than 2 is 3-choosable. For x ∈ V (G) ∪ F (G), we use Fk (x) to denote the set of all k-faces that are incident with or adjacent to x, and Vk (x) to denote the set of all k-vertices that are incident with or adjacent to x. Let v be a k-vertex. If |F3 (v)| = 1 and each other face incident with v is of degree at least 8, then v is light. If |F4 (v)| = 1 and each other face incident with v is of degree at least 8, then v is sublight. If v is only incident with 8+ -faces, then v is
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good. For convenience, we use N (f ) and V (f ) to denote the set of faces adjacent to the face f and vertices incident with f , respectively. 2. Preliminary lemmas Following lemmas will be needed for proving the main theorem. A graph G is minimal non-3-choosable if G itself is not 3-choosable, but G − v is 3-choosable for each vertex v of G. Lemma 1. (See [2].) Every cycle of even length is 2choosable. Lemma 2. (See [19].) Let G be a minimal non-3choosable graph. Then G does not contain any vertices of degree less than 3. That is δ(G) 3. Lemma 3. Let G be a minimal non-3-choosable graph. Then any 2n-cycle C with no chord in G contains at least one 4+ -vertex. Proof. Suppose dG (v) = 3 for all v ∈ C by Lemma 2, and L is a color-list of G with |L(v)| = 3 for all v ∈ V (G). By assumption, there exists an L0 -coloring φ0 of G0 = G − C, where L0 is the restriction of L to V (G0 ). Let L = {L (vi ): 1 i 2n} where L (vi ) = L(vi ) \ {φ0 (u): u ∈ NG (vi ) \ C}. It is clear that |L (vi )| 2. Clearly, there exists an L -coloring φ on C by Lemma 1. An L-coloring of G immediately follows by combining φ0 and φ . This contradiction implies that C contains at least one 4+ -vertex. 2 3. Proof of main theorems Proof of Theorem 1. By contradiction. Suppose that G is a minimal counterexample to the theorem. That is, G is a minimal non-3-choosable graph. Let L be a list assignment of G with |L(v)| = 3 for all v ∈ V (G) such that G is not L-colorable. Thus, G is connected, without 5-, 6-, and 7-faces, and we have the following claims: (I) (II) (III) (IV)
There are no triangles of distance less than 3. Neither 3-face nor 4-face is adjacent to a 4-face. |F3 (f )| d(f )/4 for any face f ∈ F (G). δ(G) 3 and every 2n-cycle contains at least one 4+ -vertex.
Statement (I), (II) and (III) directly follows from G is a graph without 5- and 6-cycles and without triangles of distance less than 3, and statement (IV) follows from Lemmas 2 and 3.
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Fig. 1.
We define a weight ω on V ∪ F by letting ω(x) = dG (x) − 6 if x ∈ V and ω(x) = 2dG (x) − 6 if x ∈ F . By Euler’s formula for planar graphs, we have x∈V ∪F ω(x) = −12. If we obtain a new nonnegative weight ω∗ (x) for all x ∈ V ∪ F by transferring weightsfrom one element to another, then we have −12 = x∈V ∪F ω(x) = x∈V ∪F ω∗ (x) 0. This contradiction will complete the proof of Theorem 1. Our transferring rules are as follows, in which, f is a l-face (l 4) and v is a k-vertex on b(f ). (R1 ) If k = 3, l 8. Then (a) f transfers 1 to each incident good v. (b) f transfers 32 to each incident light v. (c) f transfers 76 to each incident sublight v. (R2 ) If k = 4, l 8. Then f transfers 12 to each incident v in the following two cases, 1 to v otherwise. (a) v is good. (b) v is light or sublight and f is adjacent to the 3-face in F (v). (R3 ) If k = 5, l 8. Then f transfers 13 to v. (R4 ) If k = 3, l = 4. Then f transfers 23 to each incident sublight v. The rules are illustrated in Fig. 1. Note that our discharging rules are designed to insure ω∗ (v) 0 for all v ∈ V (G). Now let f be a face with d(f ) = h. Then h ∈ {3, 4, 8+ }. If h = 3, then ω∗ (f ) = ω(f ) = 0 since no charge is discharged from or to f . If h = 4, then ω(f ) = 2. By (VI), f is incident with at least one 4+ -vertex. Because f transfers weight only to the incident sublight 3-vertex, we derive that ω∗ (f ) ω(f ) − 3 × 23 = 0 by (R4 ). If h = 8, then ω(f ) = 10. By (VI), V (f ) contains at least one 4+ -vertex. For convenience, we denote by p the number of light 3-vertices, by q the number of sub-
light 3-vertices and by r the number of rest vertices on b(f ), respectively. Because whatever happens, the weight 13 , which transferred from a 8-f to the incident 5-vertices, is less than the weight which transferred from a 8-f to 4− vertices. So we only consider the 4− -vertices on b(f ). That is, f must be incident with a 4-vertex by (VI). So r 1, p 4 by |F3 (f )| h/4 and q 7 for r 1. Because the weight transferred from a 8-face to light 3-vertices is 32 by (R1 )(a), which is more than 76 transferred from a 8-face to sublight 3-vertices by (R1 )(b) and is more than weight transferred from a 8-face to the vertices in other cases. So we consider the case that p is as larger as possible, and assume f transfers 1 to every incident 4-vertex. We will consider the following cases depending on the number of light 3-vertices on b(f ): Case 1. p = 0. By (R1 )(b), ω∗ (f ) ω(f ) − 76 × p − 1 × r. The worst possible case is that q = 7 and r = 1. So ω∗ (f ) 10 − 76 × 7 − 1 × 1 = 56 > 0. Case 2. p = 1, then f must be adjacent to a 3-face f and one of the two vertices on V (f ) ∩ V (f ) is a 4-vertex. By (R1 )(a) and (R1 )(b), ω∗ (f ) ω(f ) − 32 × p − 76 × q − 1 × r = 10 − 32 × 1 − 76 × q − 1 × r 10 − 32 × 1 − 76 × 6 − 1 × 1 = 36 > 0. Case 3. p = 2. By (R1 )(a) and (R1 )(b), ω∗ (f ) ω(f ) − 32 × p − 76 × q − 1 × r 10 − 32 × 2 − 76 × q − 1 × r 10 − 32 × 2 − 76 × 5 − 1 × 1 = 16 > 0. Case 4. p = 3, then f must be adjacent to two 3faces f , f and one of the four vertices on (V (f ) ∪ V (f )) ∩ V (f ) is a 4-vertex. In this case, we will show that the only 4-vertex, denoted by v = v2 , as shown in Fig. 2, should be received 12 from f . Let f be such a 8-face with b(f ) = v1 v2 v3 v4 v5 v6 v7 v8 , that is, f is adjacent to two 3-faces f and f on edges v1 v2 , v5 v6 and
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(V (B) ∪ F (B)) \ {v0 }. It follows that {w (x) | x ∈ (V (B) ∪ F (B)) \ {v0 }} 0, a contradiction. This contradiction completes the proof of Theorem 1. 2
Fig. 2.
incident with three light 3-vertices and four sublight 3vertices, then the face incident with v and adjacent to f and f1 must be a 8+ -face by (II) and (III). So the weight transferred from 8-f to the vertex v is 12 by (R2 ), then ω∗ (f ) ω(f ) − 32 × 3 − 76 × 4 − 12 × 1 = 13 > 0. Case 5. p = 4, then f is adjacent to two 3-faces f , f and all the four vertices on (V (f ) ∪ V (f )) ∩ V (f ) are light 3-vertices. And the only 4-vertex must incident with one of the adjacent 4-faces. Without loss of generality, we can choose one of the four vertices to consider, applying the same method as used in Case 4 and (II), (III). We also can show that the weight from f to the only 4-vertex is 12 , So, by (R1 )(a) and (R1 )(b), ω∗ (f ) ω(f ) − 32 × p − 76 × q − 12 × r = 10 − 32 × 4 − 76 × 3 − 12 × 1 = 0. If h 9, by (III), we have |F3 (f )| h/4. Even f is incident with 2 × h/4 light 3-vertices and transfers 7 ∗ 6 to other sublight 3-vertices on b(f ), then ω (f ) 3 7 ω(f ) − 2 × 2 × h/4 − 6 × (h − 2 × h/4) = 5h 6 − 2 6 − 3 × h/4. Consider d(f )/4 − 1 < d(f )/4 d(f )/4, so ω∗ (f ) 5h/6 − 6 − 23 × h/4 5h/6 − 6 − 23 × h/4 = (4h − 36)/6 0 while h 9. Now, we get that ω∗ (x) 0 for each x ∈ V (G) ∪ F (G). If follows that −12 = x∈V ∪F ω(x) = ∗ (x) 0. ω x∈V ∪F Suppose now that G is not 2-connected. We consider that G has cut-vertices. Let B be a block containing a unique cut vertex v0 of G. Thus, dB (v0 ) 2 and dB (v) 3 for every v ∈ V (B) \ {v0 } by (IV). For the block B, we define the weight function w on V (B) ∪ F (B) by w(v) = dB (v) − 6 for v ∈ V (B) and f ∈ F (B). Since w(v0 ) = w(f ) = 2dB (f ) − 6 for dB (v0 ) − 6 −4, we have x∈(V (B)∪F (B))\{v0 } w(x) = −12 − w(v0 ) −8. Let w denote the new weight after discharging the weights by the same rules as in the preceding proof, except that v0 receives nothing from its incident faces. We have x∈(V (B)∪F (B))\{v0 } w (x) −8 since the total weight is kept fixed as we have seen before. However, by the similar argument as for the 2-connected case, we obtain w (x) 0 for all x ∈
Proof of Theorem 2. Preliminary assumptions are identical to those of Theorem 1. But in this case, we have the following claims: (I) (II) (III) (IV) (V)
There are no two triangles at distance less than 2. Neither 3-face nor 4-face is adjacent to a 4-face. A 3-face is not adjacent to a 7-face. |F3 (f )| d(f )/3. δ(G) 3 and every 2n-cycle contains at least one 4+ -vertex.
Statement (III) is proved as follows. Let f be a 3-face with b(f ) = v1 v2 v3 and f be a 7-face with b(f ) = v1 v2 u1 u2 u3 u4 u5 . Suppose to the contrary that f is adjacent to f , then they must share at least three vertices on their boundary, otherwise it is a contradiction for G contains no 8-cycles. But sharing at least three vertices on their boundary will lead to 5- or 6-cycles. Our transferring rules are as follows. (R1 ) Charge to a 3-vertex v. (a) Every 7+ -face f transfers 1 and 76 to each incident good and sublight v, respectively. (b) Every 9+ -face f transfers 32 to each incident light v. (c) Every 4-face f transfers 23 to each incident sublight v. (R2 ) Every 7+ -face transfers 1 to each incident 4- or 5-vertex. Note that our discharging rules are designed to insure ω∗ (v) 0 for all v ∈ V (G). Now let f be a face with d(f ) = h. Then h ∈ {3, 4, 7, 9+ }. If h = 3, then ω∗ (f ) = ω(f ) = 0 since no charge is discharged from or to f . If h = 4, then ω(f ) = 2. By (V), f is incident with at least one 4+ -vertex. Because f transfers weight only to the incident sublight 3-vertex, we derive that ω∗ (f ) ω(f ) − 3 × 23 = 0 by (R1 )(c). If h = 7, then f is not adjacent to any 3-faces by (III) and f is incident with at least one vertex which is neither light 3-vertex nor sublight 3-vertex by (II). So ω∗ (f ) ω(f ) − 76 × 6 − 1 × 1 = 0. If h = 9, then there are at most three 3-faces adjacent to f by (IV). And while the number of the adjacent 3faces is exactly 6 and all of them are light 3-vertex, then
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the leaving three vertices must not be light or sublight, so ω∗ (f ) ω(f ) − 32 × 6 − 1 × 3 = 0. If h 10, by (IV), we have |F3 (f )| h/3. Even f is incident with 2 × h/3 light 3-vertices and transfers 7 ∗ 6 to other sublight 4-vertices on b(f ), then ω (f ) ω(f ) − 32 × 2 × h/3 − 76 × (h − 2 × h/3) = 5h/6 − 6 − 23 × h/3. Consider d(f )/3 − 1 < d(f )/3 d(f )/3, so ω∗ (f ) 5h/6 − 6 − 23 × h/3 5h/6 − 6 − 23 × h/3 = (11h − 108)/18 > 0 while h 10. Now, we get that ω∗ (x) 0 for each x ∈ V (G) ∪ F (G). If follows that −12 = x∈V ∪F ω(x) = ∗ (x) 0. Applying the same method as used ω x∈V ∪F in the proof of Theorem 1 for the case G has cut-vertices will complete the proof of Theorem 2. 2 Acknowledgement We would like to express our gratitude to the referees for their careful reading and valuable comments and suggestions concerning this paper. We are also grateful to Professor Xu Baogang for fruitful discussions about the problem and Professor Sun Zhihong for his help in the writing of this paper. References [1] N. Alon, M. Tarsi, Colorings and orientations of graphs, Combinatorica 12 (2) (1992) 125–134. [2] P. Erdös, A.L. Rubin, H. Taylor, Choosability in graphs, Congres. Numer. 26 (1979) 125–157. [3] G. Fijavž, M. Juvan, B. Mohar, R. Škrekovski, Planar graphs without cycles of specific lengths, European J. Combin. 23 (2002) 377–388.
[4] S. Gutner, The complexity of planar graph choosability, Discrete Math. 159 (1996) 119–130. [5] P.C.B. Lam, B. Xu, J. Liu, The 4-choosability of plane graphs without 4-cycles, J. Combin. Theory Ser. B 76 (1999) 117–126. [6] P.C.B. Lam, W.C. Shiu, B. Xu, On structure of some plane graphs with application to choosability, J. Combin. Theory Ser. B 82 (2001) 285–297. [7] P.C.B. Lam, W.C. Shiu, Z.M. Song, The 3-choosability of plane graphs of girth 4, Discrete Math. 294 (2005) 297–301. [8] M. Mirzakhani, A small non-4-choosable planar graph, Bull. Inst. Combin. Appl. 17 (1996) 15–18. [9] M. Montassier, A. Raspaud, W. Wang, Bordeaux 3-color conjecture and 3-choosability, Discrete Math. 306 (2006) 573–579. [10] M. Montassier, A note on the not 3-choosability of some families of planar graphs, Discrete Math. 99 (2006) 68–71. [11] C. Thomassen, Every planar graph is 5-choosable, J. Combin. Theory Ser. B 62 (1994) 180–181. [12] C. Thomassen, 3-list-coloring planar graph of girth 5, J. Combin. Theory Ser. B 64 (1995) 101–107. [13] M. Voigt, List colourings of planar graphs, Discrete Math. 120 (1993) 215–219. [14] M. Voigt, B. Wirth, A not 3-choosable planar graph without 3cycles, Discrete Math. 146 (1995) 325–328. [15] W.F. Wang, K.W. Lih, Choosability and edge choosability of planar graphs without five cycles, Appl. Math. Lett. 15 (2002) 561–565. [16] W.F. Wang, K.W. Lih, Choosability and edge choosability of planar graph without intersecting triangles, SIAM J. Discrete Math. 15 (2002) 538–545. [17] Y. Wang, H. Lu, M. Chen, A note on 3-choosability of planar graphs, Inform. Process. Lett. 105 (2008) 206–211. [18] B. Xu, (4m, m)-choosability of plane graphs, J. Systems Sci. Complexity 14 (2001) 174–178. [19] H. Zhang, B. Xu, Z. Sun, Every plane graph with girth at least 4 without 8- and 9-circuits is 3-choosable, Ars Combin. 80 (2006) 247–257.
On 3-choosability of planar graphs without certain cycles ✩ Haihui Zhang a,b,∗ , Zhiren Sun a a School of Mathematics and Computer Science, Nanjing Normal University, 122, Ninghai Road, Nanjing 210097, PR China b Mathematics Department, Huaiyin Teachers College, 223001 Huaian, PR China
Received 12 January 2008; received in revised form 28 January 2008 Available online 14 February 2008 Communicated by L. Boasson
Abstract A graph G = (V , E) is L-colorable if for a given list assignment L = {L(v): v ∈ V (G)}, there exists a proper coloring c of G such that c(v) ∈ L(v) for all v ∈ V . If G is L-colorable for every list assignment L with |L(v)| k for all v ∈ V , then G is said to be k-choosable. In this paper, we prove that every planar graph with neither 5-, 6-, and 7-cycles nor triangles of distance less than 3, or with neither 5-, 6-, and 8-cycles nor triangles of distance less than 2 is 3-choosable. © 2008 Elsevier B.V. All rights reserved. Keywords: Combinatorial problems; Choosability; Planar graph; Cycle
1. Introduction All graphs considered in this paper are finite, simple planar graphs. A graph G is planar if G can be drawn on the plane so that its edges meet only at the vertices of the graph. A plane graph is such a particular drawing of a planar graph. Let G = (V , E, F ) denote a planar graph, with V , E and F being the set of vertices, edges and faces of G, respectively. We use b(f ) to denote the boundary walk of a face f and write b(f ) = [v1 v2 v3 . . . vn ] if v1 , v2 , v3 , . . . , vn are the vertices of b(f ) in a cyclic order. A face f is incident with all vertices and edges on b(f ). The degree of a face f of G, denoted ✩
Research is partially supported by the NSFC 10371055.
* Corresponding author at: School of Mathematics and Computer
Science, Nanjing Normal University, 122, Ninghai Road, Nanjing 210097, PR China. E-mail address: [email protected] (H. Zhang). 0020-0190/$ – see front matter © 2008 Elsevier B.V. All rights reserved. doi:10.1016/j.ipl.2008.01.008
by dG (f ), is the number of edges incident with it, where cut edges are counted twice. A vertex (face) of degree k is called a k-vertex (k-face). If r k or 1 k r, then a k-vertex (k-face) is called an r + - or r − -vertex (r + - or r − -face), respectively. A k-cycle is a cycle with k edges. The vertex set of a cycle C will also be denoted by C. A graph G = (V , E) is L-colorable if for a given list assignment L = {L(v): v ∈ V (G)}, there exists a proper coloring c of G such that c(v) ∈ L(v) for all v ∈ V . If G is L-colorable for every list assignment L with |L(v)| k for all v ∈ V , then G is said to be k-choosable. The choice number of G, denoted by χl (G) or ch(G), is the minimum k such that G is k-choosable. All 2-choosable graphs were characterized completely in [2]. Thomassen proved that every planar graph is 5-choosable [11]. Examples of planar graphs which are not 4-choosable were given by Voigt [13], and by Mirzakhani [8] independently. Voigt and Wirth [14], and Gutner [4] independently, presented some planar graphs of girth 4 which are not 3-choosable. Thus, it re-
H. Zhang, Z. Sun / Information Processing Letters 107 (2008) 102–106
mains to determine whether a given planar graph is 3or 4-choosable. In [4], Gutner proved that these problems are NP-complete. Therefore, many authors tried to find sufficient conditions for a planar graph to be 3- or 4-choosable. Alon and Tarsi [1] proved that every planar bipartite graph is 3-choosable. Thomassen [12] proved that every planar graph of girth 5 is 3-choosable. Lam et al. [5,6] proved that plane graphs without i-cycles are 4-choosable, for i = 3, 4, 5 or 6. Independently, Wang and Lih [15] proved that plane graphs without 5-cycles are 4-choosable, Fijavž et al. [3] proved that plane graphs without 6-cycles are 4-choosable. Xu [18], and Wang and Lih [16] independently, proved that plane graphs without two triangles sharing a common vertex are 4-choosable. More recently, some new sufficient conditions for a planar graph to be 3-choosable have been given. These include: having no 3-, 5-, and 6-cycles [7]; having no 3-, 8-, and 9-cycles [19]. Together with other known parallel results, Wang et al. [17] obtained a theorem on 3choosability of planar graphs: planar graphs without cycles of length 4, i, j , 9 with i < j and i, j ∈ {5, 6, 7, 8} are 3-choosable. The distance between two vertices x and y, denoted by dist(x, y), is the length of a shortest path connecting them in G. The distance between two triangles T and T is defined to be the value min{dist(x, y) | x ∈ V (T ) and y ∈ V (T )}. In [9], Montassier et al. proved that every planar graph either without 4- and 5-cycles and without triangles at distance less than 4, or without 4-, 5-, and 6-cycles and without triangles at distance less than 3 is 3-choosable. In [10] Montassier proposed a conjecture that every planar graph without cycles of length 4, 5, 6, is 3-choosable. In this paper, we prove that: Theorem 1. Every planar graph without 5-, 6-, and 7cycles, and without triangles of distance less than 3 is 3-choosable. Theorem 2. Every planar graph without 5-, 6-, and 8cycles, and without triangles of distance less than 2 is 3-choosable. For x ∈ V (G) ∪ F (G), we use Fk (x) to denote the set of all k-faces that are incident with or adjacent to x, and Vk (x) to denote the set of all k-vertices that are incident with or adjacent to x. Let v be a k-vertex. If |F3 (v)| = 1 and each other face incident with v is of degree at least 8, then v is light. If |F4 (v)| = 1 and each other face incident with v is of degree at least 8, then v is sublight. If v is only incident with 8+ -faces, then v is
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good. For convenience, we use N (f ) and V (f ) to denote the set of faces adjacent to the face f and vertices incident with f , respectively. 2. Preliminary lemmas Following lemmas will be needed for proving the main theorem. A graph G is minimal non-3-choosable if G itself is not 3-choosable, but G − v is 3-choosable for each vertex v of G. Lemma 1. (See [2].) Every cycle of even length is 2choosable. Lemma 2. (See [19].) Let G be a minimal non-3choosable graph. Then G does not contain any vertices of degree less than 3. That is δ(G) 3. Lemma 3. Let G be a minimal non-3-choosable graph. Then any 2n-cycle C with no chord in G contains at least one 4+ -vertex. Proof. Suppose dG (v) = 3 for all v ∈ C by Lemma 2, and L is a color-list of G with |L(v)| = 3 for all v ∈ V (G). By assumption, there exists an L0 -coloring φ0 of G0 = G − C, where L0 is the restriction of L to V (G0 ). Let L = {L (vi ): 1 i 2n} where L (vi ) = L(vi ) \ {φ0 (u): u ∈ NG (vi ) \ C}. It is clear that |L (vi )| 2. Clearly, there exists an L -coloring φ on C by Lemma 1. An L-coloring of G immediately follows by combining φ0 and φ . This contradiction implies that C contains at least one 4+ -vertex. 2 3. Proof of main theorems Proof of Theorem 1. By contradiction. Suppose that G is a minimal counterexample to the theorem. That is, G is a minimal non-3-choosable graph. Let L be a list assignment of G with |L(v)| = 3 for all v ∈ V (G) such that G is not L-colorable. Thus, G is connected, without 5-, 6-, and 7-faces, and we have the following claims: (I) (II) (III) (IV)
There are no triangles of distance less than 3. Neither 3-face nor 4-face is adjacent to a 4-face. |F3 (f )| d(f )/4 for any face f ∈ F (G). δ(G) 3 and every 2n-cycle contains at least one 4+ -vertex.
Statement (I), (II) and (III) directly follows from G is a graph without 5- and 6-cycles and without triangles of distance less than 3, and statement (IV) follows from Lemmas 2 and 3.
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Fig. 1.
We define a weight ω on V ∪ F by letting ω(x) = dG (x) − 6 if x ∈ V and ω(x) = 2dG (x) − 6 if x ∈ F . By Euler’s formula for planar graphs, we have x∈V ∪F ω(x) = −12. If we obtain a new nonnegative weight ω∗ (x) for all x ∈ V ∪ F by transferring weightsfrom one element to another, then we have −12 = x∈V ∪F ω(x) = x∈V ∪F ω∗ (x) 0. This contradiction will complete the proof of Theorem 1. Our transferring rules are as follows, in which, f is a l-face (l 4) and v is a k-vertex on b(f ). (R1 ) If k = 3, l 8. Then (a) f transfers 1 to each incident good v. (b) f transfers 32 to each incident light v. (c) f transfers 76 to each incident sublight v. (R2 ) If k = 4, l 8. Then f transfers 12 to each incident v in the following two cases, 1 to v otherwise. (a) v is good. (b) v is light or sublight and f is adjacent to the 3-face in F (v). (R3 ) If k = 5, l 8. Then f transfers 13 to v. (R4 ) If k = 3, l = 4. Then f transfers 23 to each incident sublight v. The rules are illustrated in Fig. 1. Note that our discharging rules are designed to insure ω∗ (v) 0 for all v ∈ V (G). Now let f be a face with d(f ) = h. Then h ∈ {3, 4, 8+ }. If h = 3, then ω∗ (f ) = ω(f ) = 0 since no charge is discharged from or to f . If h = 4, then ω(f ) = 2. By (VI), f is incident with at least one 4+ -vertex. Because f transfers weight only to the incident sublight 3-vertex, we derive that ω∗ (f ) ω(f ) − 3 × 23 = 0 by (R4 ). If h = 8, then ω(f ) = 10. By (VI), V (f ) contains at least one 4+ -vertex. For convenience, we denote by p the number of light 3-vertices, by q the number of sub-
light 3-vertices and by r the number of rest vertices on b(f ), respectively. Because whatever happens, the weight 13 , which transferred from a 8-f to the incident 5-vertices, is less than the weight which transferred from a 8-f to 4− vertices. So we only consider the 4− -vertices on b(f ). That is, f must be incident with a 4-vertex by (VI). So r 1, p 4 by |F3 (f )| h/4 and q 7 for r 1. Because the weight transferred from a 8-face to light 3-vertices is 32 by (R1 )(a), which is more than 76 transferred from a 8-face to sublight 3-vertices by (R1 )(b) and is more than weight transferred from a 8-face to the vertices in other cases. So we consider the case that p is as larger as possible, and assume f transfers 1 to every incident 4-vertex. We will consider the following cases depending on the number of light 3-vertices on b(f ): Case 1. p = 0. By (R1 )(b), ω∗ (f ) ω(f ) − 76 × p − 1 × r. The worst possible case is that q = 7 and r = 1. So ω∗ (f ) 10 − 76 × 7 − 1 × 1 = 56 > 0. Case 2. p = 1, then f must be adjacent to a 3-face f and one of the two vertices on V (f ) ∩ V (f ) is a 4-vertex. By (R1 )(a) and (R1 )(b), ω∗ (f ) ω(f ) − 32 × p − 76 × q − 1 × r = 10 − 32 × 1 − 76 × q − 1 × r 10 − 32 × 1 − 76 × 6 − 1 × 1 = 36 > 0. Case 3. p = 2. By (R1 )(a) and (R1 )(b), ω∗ (f ) ω(f ) − 32 × p − 76 × q − 1 × r 10 − 32 × 2 − 76 × q − 1 × r 10 − 32 × 2 − 76 × 5 − 1 × 1 = 16 > 0. Case 4. p = 3, then f must be adjacent to two 3faces f , f and one of the four vertices on (V (f ) ∪ V (f )) ∩ V (f ) is a 4-vertex. In this case, we will show that the only 4-vertex, denoted by v = v2 , as shown in Fig. 2, should be received 12 from f . Let f be such a 8-face with b(f ) = v1 v2 v3 v4 v5 v6 v7 v8 , that is, f is adjacent to two 3-faces f and f on edges v1 v2 , v5 v6 and
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(V (B) ∪ F (B)) \ {v0 }. It follows that {w (x) | x ∈ (V (B) ∪ F (B)) \ {v0 }} 0, a contradiction. This contradiction completes the proof of Theorem 1. 2
Fig. 2.
incident with three light 3-vertices and four sublight 3vertices, then the face incident with v and adjacent to f and f1 must be a 8+ -face by (II) and (III). So the weight transferred from 8-f to the vertex v is 12 by (R2 ), then ω∗ (f ) ω(f ) − 32 × 3 − 76 × 4 − 12 × 1 = 13 > 0. Case 5. p = 4, then f is adjacent to two 3-faces f , f and all the four vertices on (V (f ) ∪ V (f )) ∩ V (f ) are light 3-vertices. And the only 4-vertex must incident with one of the adjacent 4-faces. Without loss of generality, we can choose one of the four vertices to consider, applying the same method as used in Case 4 and (II), (III). We also can show that the weight from f to the only 4-vertex is 12 , So, by (R1 )(a) and (R1 )(b), ω∗ (f ) ω(f ) − 32 × p − 76 × q − 12 × r = 10 − 32 × 4 − 76 × 3 − 12 × 1 = 0. If h 9, by (III), we have |F3 (f )| h/4. Even f is incident with 2 × h/4 light 3-vertices and transfers 7 ∗ 6 to other sublight 3-vertices on b(f ), then ω (f ) 3 7 ω(f ) − 2 × 2 × h/4 − 6 × (h − 2 × h/4) = 5h 6 − 2 6 − 3 × h/4. Consider d(f )/4 − 1 < d(f )/4 d(f )/4, so ω∗ (f ) 5h/6 − 6 − 23 × h/4 5h/6 − 6 − 23 × h/4 = (4h − 36)/6 0 while h 9. Now, we get that ω∗ (x) 0 for each x ∈ V (G) ∪ F (G). If follows that −12 = x∈V ∪F ω(x) = ∗ (x) 0. ω x∈V ∪F Suppose now that G is not 2-connected. We consider that G has cut-vertices. Let B be a block containing a unique cut vertex v0 of G. Thus, dB (v0 ) 2 and dB (v) 3 for every v ∈ V (B) \ {v0 } by (IV). For the block B, we define the weight function w on V (B) ∪ F (B) by w(v) = dB (v) − 6 for v ∈ V (B) and f ∈ F (B). Since w(v0 ) = w(f ) = 2dB (f ) − 6 for dB (v0 ) − 6 −4, we have x∈(V (B)∪F (B))\{v0 } w(x) = −12 − w(v0 ) −8. Let w denote the new weight after discharging the weights by the same rules as in the preceding proof, except that v0 receives nothing from its incident faces. We have x∈(V (B)∪F (B))\{v0 } w (x) −8 since the total weight is kept fixed as we have seen before. However, by the similar argument as for the 2-connected case, we obtain w (x) 0 for all x ∈
Proof of Theorem 2. Preliminary assumptions are identical to those of Theorem 1. But in this case, we have the following claims: (I) (II) (III) (IV) (V)
There are no two triangles at distance less than 2. Neither 3-face nor 4-face is adjacent to a 4-face. A 3-face is not adjacent to a 7-face. |F3 (f )| d(f )/3. δ(G) 3 and every 2n-cycle contains at least one 4+ -vertex.
Statement (III) is proved as follows. Let f be a 3-face with b(f ) = v1 v2 v3 and f be a 7-face with b(f ) = v1 v2 u1 u2 u3 u4 u5 . Suppose to the contrary that f is adjacent to f , then they must share at least three vertices on their boundary, otherwise it is a contradiction for G contains no 8-cycles. But sharing at least three vertices on their boundary will lead to 5- or 6-cycles. Our transferring rules are as follows. (R1 ) Charge to a 3-vertex v. (a) Every 7+ -face f transfers 1 and 76 to each incident good and sublight v, respectively. (b) Every 9+ -face f transfers 32 to each incident light v. (c) Every 4-face f transfers 23 to each incident sublight v. (R2 ) Every 7+ -face transfers 1 to each incident 4- or 5-vertex. Note that our discharging rules are designed to insure ω∗ (v) 0 for all v ∈ V (G). Now let f be a face with d(f ) = h. Then h ∈ {3, 4, 7, 9+ }. If h = 3, then ω∗ (f ) = ω(f ) = 0 since no charge is discharged from or to f . If h = 4, then ω(f ) = 2. By (V), f is incident with at least one 4+ -vertex. Because f transfers weight only to the incident sublight 3-vertex, we derive that ω∗ (f ) ω(f ) − 3 × 23 = 0 by (R1 )(c). If h = 7, then f is not adjacent to any 3-faces by (III) and f is incident with at least one vertex which is neither light 3-vertex nor sublight 3-vertex by (II). So ω∗ (f ) ω(f ) − 76 × 6 − 1 × 1 = 0. If h = 9, then there are at most three 3-faces adjacent to f by (IV). And while the number of the adjacent 3faces is exactly 6 and all of them are light 3-vertex, then
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the leaving three vertices must not be light or sublight, so ω∗ (f ) ω(f ) − 32 × 6 − 1 × 3 = 0. If h 10, by (IV), we have |F3 (f )| h/3. Even f is incident with 2 × h/3 light 3-vertices and transfers 7 ∗ 6 to other sublight 4-vertices on b(f ), then ω (f ) ω(f ) − 32 × 2 × h/3 − 76 × (h − 2 × h/3) = 5h/6 − 6 − 23 × h/3. Consider d(f )/3 − 1 < d(f )/3 d(f )/3, so ω∗ (f ) 5h/6 − 6 − 23 × h/3 5h/6 − 6 − 23 × h/3 = (11h − 108)/18 > 0 while h 10. Now, we get that ω∗ (x) 0 for each x ∈ V (G) ∪ F (G). If follows that −12 = x∈V ∪F ω(x) = ∗ (x) 0. Applying the same method as used ω x∈V ∪F in the proof of Theorem 1 for the case G has cut-vertices will complete the proof of Theorem 2. 2 Acknowledgement We would like to express our gratitude to the referees for their careful reading and valuable comments and suggestions concerning this paper. We are also grateful to Professor Xu Baogang for fruitful discussions about the problem and Professor Sun Zhihong for his help in the writing of this paper. References [1] N. Alon, M. Tarsi, Colorings and orientations of graphs, Combinatorica 12 (2) (1992) 125–134. [2] P. Erdös, A.L. Rubin, H. Taylor, Choosability in graphs, Congres. Numer. 26 (1979) 125–157. [3] G. Fijavž, M. Juvan, B. Mohar, R. Škrekovski, Planar graphs without cycles of specific lengths, European J. Combin. 23 (2002) 377–388.
[4] S. Gutner, The complexity of planar graph choosability, Discrete Math. 159 (1996) 119–130. [5] P.C.B. Lam, B. Xu, J. Liu, The 4-choosability of plane graphs without 4-cycles, J. Combin. Theory Ser. B 76 (1999) 117–126. [6] P.C.B. Lam, W.C. Shiu, B. Xu, On structure of some plane graphs with application to choosability, J. Combin. Theory Ser. B 82 (2001) 285–297. [7] P.C.B. Lam, W.C. Shiu, Z.M. Song, The 3-choosability of plane graphs of girth 4, Discrete Math. 294 (2005) 297–301. [8] M. Mirzakhani, A small non-4-choosable planar graph, Bull. Inst. Combin. Appl. 17 (1996) 15–18. [9] M. Montassier, A. Raspaud, W. Wang, Bordeaux 3-color conjecture and 3-choosability, Discrete Math. 306 (2006) 573–579. [10] M. Montassier, A note on the not 3-choosability of some families of planar graphs, Discrete Math. 99 (2006) 68–71. [11] C. Thomassen, Every planar graph is 5-choosable, J. Combin. Theory Ser. B 62 (1994) 180–181. [12] C. Thomassen, 3-list-coloring planar graph of girth 5, J. Combin. Theory Ser. B 64 (1995) 101–107. [13] M. Voigt, List colourings of planar graphs, Discrete Math. 120 (1993) 215–219. [14] M. Voigt, B. Wirth, A not 3-choosable planar graph without 3cycles, Discrete Math. 146 (1995) 325–328. [15] W.F. Wang, K.W. Lih, Choosability and edge choosability of planar graphs without five cycles, Appl. Math. Lett. 15 (2002) 561–565. [16] W.F. Wang, K.W. Lih, Choosability and edge choosability of planar graph without intersecting triangles, SIAM J. Discrete Math. 15 (2002) 538–545. [17] Y. Wang, H. Lu, M. Chen, A note on 3-choosability of planar graphs, Inform. Process. Lett. 105 (2008) 206–211. [18] B. Xu, (4m, m)-choosability of plane graphs, J. Systems Sci. Complexity 14 (2001) 174–178. [19] H. Zhang, B. Xu, Z. Sun, Every plane graph with girth at least 4 without 8- and 9-circuits is 3-choosable, Ars Combin. 80 (2006) 247–257.