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Planar graphs without intersecting 5 -cycles are 4 -choosable
Discrete Mathematics 340 (2017) 1788–1792
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Discrete Mathematics journal homepage: www.elsevier.com/locate/disc
Planar graphs without intersecting 5-cycles are 4-choosable Dai-Qiang Hu a , Jian-Liang Wu b, * a b
Department of Mathematics, Jinan University, Guang Zhou, 510632, PR China School of Mathematics, Shandong University, Jinan, 250100, PR China
article
info
Article history: Received 10 March 2016 Received in revised form 11 March 2017 Accepted 11 March 2017
a b s t r a c t A graph G is k-choosable if it can be properly colored whenever every vertex has a list of at least k available colors. In the paper, it is proved that all planar graphs without intersecting 5-cycles are 4-choosable. © 2017 Elsevier B.V. All rights reserved.
Keywords: planar graph Cycle List coloring Choosable
1. Introduction All graphs considered in this paper are simple, finite and undirected, and we follow [2] for the terminologies and notations not defined here. Let G be a graph with the vertex set V (G) and the edge set E(G). For a vertex v ∈ V , let N(v ) denote the set of vertices adjacent to v , and let d(v ) = |N(v )| denote the degree of v . A k-vertex, a k+ -vertex or a k− -vertex is a vertex of degree k , at least k or at most k respectively. We use δ (G) to denote the minimum degree of G. A k-cycle is a cycle of length k, and a 3-cycle is usually called a triangle. Two cycles are adjacent( or intersecting) if they share at least one edge (or vertex, respectively). A proper k-coloring of a graph G is a mapping φ from V (G) to the color set [k] = {1, 2, . . . , k} such that φ (x) ̸ = φ (y) for every two adjacent vertices x and y of G. We say that L is an assignment for the graph G if it assigns a list L(v ) of possible colors to each vertex v of G. If G has a proper coloring φ such that φ (v ) ∈ L(v ) for any vertex v , then we say that G is L-colorable or φ is an L-coloring of G. The graph G is k-choosable if it is L-colorable for every assignment L satisfying |L(v )| ≥ k for any vertex v . The concept of a list coloring was introduced by Vizing [10] and Erdős, Rubin and Taylor [4], respectively. Thomassen [9] showed that every planar graph is 5-choosable. Examples of plane graphs which are not 4-choosable and plane graphs of girth 4 which are not 3-choosable were given by Voigt [11,12]. Since every planar graph G without 3-cycles has a vertex of degree at most 3, it is 4-choosable. Wang and Lih [13] extended the result to all planar graphs without intersecting 3-cycles. Lam, Xu and Liu [8] proved that all planar graphs without 4-cycles are 4-choosable. Lam, Shiu and Xu [7](Wang and Lih [14], respectively) proved that all planar graphs without 5-cycles are 4-choosable. Fijavž, Juvan, Mohar, Škrekovski [6] proved that all planar graphs without 6-cycles are 4-choosable. Farzad [5] proved that all planar graphs without 7-cycles are 4-choosable. Cheng, Chen and Wang [3] proved that planar graphs without 4-cycles adjacent to triangles are 4-choosable. In the paper, we prove that all planar graphs without intersecting 5-cycles are 4-choosable.
*
Corresponding author. E-mail address: [email protected] (J. Wu).
http://dx.doi.org/10.1016/j.disc.2017.03.012 0012-365X/© 2017 Elsevier B.V. All rights reserved.
D. Hu, J. Wu / Discrete Mathematics 340 (2017) 1788–1792
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Fig. 1. Three subgraphs of G.
Fig. 2. An orientation of the configurations in Figure 1.
2. Main result and its proof First, we introduce some notations and definitions used in this section. Let G be a plane graph. We use F or F (G) to denote the face set of G. For f ∈ F (G), we use V (f ) to denote the set of vertices on the boundary of f and write f = [u1 u2 · · · un ] if u1 , u2 , . . . , un are the boundary vertices of f in a cyclic order. A face of G is said to be incident with all edges and vertices in its boundary. The degree of a face f , denoted by dG (f ), is the number of edges incident with it, where a cut edge is counted twice. A k-face, k− -face or a k+ -face is a face of degree k, at most k or at least k, respectively. For convenience, a k-face f = [v1 v2 · · · vk ] is often said to be a (d(v1 ), d(v2 ), . . . , d(vk ))-face. For a face f , let ni (f ) and ni+ (f ) denote the number of i-vertices and i+ -vertices incident with f , respectively. Denote by fd (v ) and fd+ (v ) the number of d-faces and d+ -faces incident with v , respectively. Theorem 1. All planar graphs without intersecting 5-cycles are 4-choosable. Proof. Suppose, to the contrary, that Theorem 1 is false. Let G be a counterexample to Theorem 1 with the fewest vertices. Then δ (G) ≥ 4 (see [8]). First, we introduce a well-known theorem proved by Alon and Tarsi [1]. This intricate theorem reveals the connection between the list coloring of a graph G and its orientations. A digraph (directed graph) D is an ordered pair (V (D), A(D)) consisting of the vertex set V (D) and arc set A(D). For any arc a = ⟨u, v⟩, we say that u is the tail of a and v its head. The + indegree d− D (v ) of a vertex v in D is the number of arcs with head v , and the outdegree dD (v ) of v is the number of arcs with tail v . A directed cycle is denoted by a cyclic sequence u1 u2 · · · uk u1 in which each vertex dominates its successor. A subdigraph H of a directed graph D is called Eulerian if the indegree d− H (v ) of every vertex v of H is equal to its outdegree d+ ( v ). Note that we do not assume that H is connected. H is e v en if it has an even number of edges. Otherwise it is odd. Let H EE(D) and EO(D) denote the numbers of even and odd Eulerian subgraphs of D, respectively. (For convenience we agree that the empty subgraph is an even Eulerian subgraph). + Theorem 2. [1] Let D be a digraph. For each vertex v ∈ V (D), let f (v ) be a set of d+ D (v ) + 1 distinct integers, where dD (v ) is the outdegree of v . If EE(D) ̸ = EO(D), then there is a proper coloring c : V (D) → Z such that c(v ) is in f (v ) for every vertex v . That is, if L is a list assignment such that |L(v )| = d+ D (v ) + 1 for all vertices v in D, then D is L-colorable.
Lemma 3. G contains no subgraph isomorphic to one of the configurations in Fig. 1, where the vertices marked by • have degree of 4 in G and those vertices marked by ◦ have degree of 4 or 5 in G. Proof. Let L be an arbitrary list assignment of G such that each vertex is assigned precisely 4 colors. Suppose that G contains a subgraph H isomorphic to one of the configurations in Fig. 1. By the minimality of G, the graph G − V (H) has an L-coloring ϕ . An orientation of H and available colors of every vertex are shown in Fig. 2. Since EE(G1 ) = 1 < EO(G1 ) = 2, EE(G2 ) = 1 < EO(G2 ) = 2 and EE(G3 ) = 5 > EO(G3 ) = 4, any Gi (1 ≤ i ≤ 3) satisfies Theorem 2. So ϕ can be extended to G. ■
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In the following, we assume that G has been embedded in the plane. Thus G is a plane graph. By the Euler’s formula
|V (G)| − |E(G)| + |F (G)| = 2, we have ∑ ∑ (2d(v ) − 6) + (d(f ) − 6) = −6(|V (G)| − |E(G)| + |F (G)|) = −12 < 0. v∈V (G)
f ∈F (G)
Now we define the initial∑ charge function ch(x) of x ∈ V (G) ∪ F (G) to be ch(v ) = 2d(v ) − 6 if v ∈ V (G) and ch(f ) = d(f ) − 6 if f ∈ F (G). It follows that x∈V (G)∪F (G) ch(x) < 0. Now we design appropriate discharging rules and redistribute weights accordingly. Note that any discharging procedure preserves the total charge of G. If we can define suitable discharging rules to charge the initial charge function ch to the final charge function ch′ on V (G) ∪ F (G), such that ch′ (x) ≥ 0 for all x ∈ V (G) ∪ F (G), then we get an obvious contradiction. A 3-face is called special if it is adjacent to a (4, 4, 4, 4, 4)-face. A vertex v is called hard if v is a 4-vertex with f3 (v ) = 3 or f3 (v ) = 2 and f4 (v ) = 1. It is obvious that two hard 4-vertices are not adjacent. For x, y ∈ V ∪ F , let τ (x → y) denote the amount of weights transferred from x to y. Now we define the discharging rules as follows. R1. Let f be a 3-face [v1 v2 v3 ]. If d(v1 ) = d(v2 ) = d(v3 ) = 4 or every vi (1 ≤ i ≤ 3) is not hard, then τ (vi → f ) = 1 for all i = 1, 2, 3. Otherwise, assume that v1 is a hard vertex and d(v3 ) ≥ 5. Then
⎧ 2 ⎪ ⎪ , if f3 (v ) = 3 and no (4, 4, 4)-face is incident with v1 , ⎪ ⎪ 3 ⎪ ⎪ ⎨ 3 τ (v 1 → f ) = , if f3 (v ) = 2 and no (4, 4, 4)-face is incident with v1 , ⎪ 4 ⎪ ⎪ ⎪ ⎪1 ⎪ ⎩ , otherwise. ⎧2 if d(v2 ) = 4, ⎨1, τ (v2 → f ) = 3 − τ (v1 → f ) ⎩ , otherwise. 2 ⎧ ⎨2 − τ (v1 → f ), if d(v2 ) = 4, τ (v3 → f ) = 3 − τ (v1 → f ) ⎩ , otherwise. 2
R2. Let f be a 4-face [v1 v2 v3 v4 ]. Then τ (vi → f ) = 12 for i = 1, 2, 3, 4. R3. Let f be a 5-face. If n5+ (f ) > 0, then for every 5+ -vertex v incident with f , τ (v → f ) = (4, 4, 4, 4, 4)-face. Let v be a 4-vertex incident with f . If f3 (v ) ≤ 1, then τ (v → f ) = such that uv is incident with f . Then τ (w → f ) = 41 .
1 . 2
1 . n5+ (f )
Otherwise, f is a
Otherwise, let [uvw] be a 3-face
Next we show that ch′ (x) ≥ 0 for all x ∈ V ∪ F . Let f be a face of G. If d(f ) = 3, then it is obvious by R1 that ch′ (f ) = 0. If d(f ) = 4, then ch′ (f ) = −2 + 21 × 4 = 0 by R2. If d(f ) ≥ 6, then ch′ (f ) = ch(f ) ≥ 0. Let f be a 5-face of G. If n5+ (f ) ̸ = 0, then ch′ (f ) ≥ −1 + n 1 (f ) × n5+ (f ) = 0 by R3. Otherwise, f is a (4, 4, 4, 4, 4)-face 5+ [v1 v2 · · · v5 ]. If f3 (vi ) ≥ 2 for all i = 1, 2, 3, 4, 5, then f is adjacent to at least four special 3-faces since G has no intersecting 5-cycles and it follows that ch′ (f ) ≥ −1 + 41 × 4 = 0 by R3. Otherwise, without loss of generality, assume that f3 (v1 ) ≤ 1. If there is some vi (2 ≤ i ≤ 5) having f3 (vi ) ≤ 1, then ch′ (f ) ≥ −1 + 21 × 2 = 0. Otherwise, f is adjacent to at least two 3-faces and it follows that ch′ (f ) ≥ −1 + 12 + 14 × 2 = 0. Let v be a 4-vertex of G. Then f3 (v ) ≤ 3. If f3 (v ) = 0, then ch′ (v ) ≥ 2 − 12 × 4 = 0 by R2–R3. Otherwise, f6+ (v ) ≥ 1. Note that if v is incident with a 3-face [uvw] such that d(u) = d(w ) = 4, then uw is not incident with a (4, 4, 4, 4, 4)-face by Lemma 3(2). If f3 (v ) = 1, then ch′ (v ) ≥ 2 − 1 − 2 × 21 = 0. If f3 (v ) = 2 and f4 (v ) = 0, then ch′ (v ) ≥ 2 − 1 × 2 = 0 by R1. Finally, we assume that f3 (v ) = 3 or f3 (v ) = 2 and f4 (v ) = 1, that is, v is a hard vertex. By Lemma 3(1), v is incident with at most one (4, 4, 4)-face. If f3 (v ) = 3, then ch′ (v ) ≥ 2 − max{1 + 21 × 2, 23 × 3} = 0 by R1. Otherwise, ch′ (v ) ≥ 2 − max{1 + 2 × 12 , 43 × 2 + 12 } = 0. Let v be a 5+ -vertex of G. Note that any 4-vertex incident with a special 3-face is not hard. Suppose that f = [uvw] is a special 3-face such that d(u) = d(w ) = 4 and uw is incident with a (4, 4, 4, 4, 4)-face f ′ . Then τ (v → f ) = 1 by R1 and τ (v → f ′ ) = 14 by R3. In the following, sometimes we replace the case by τ (v → f ) = 54 . Lemma 4. Let f be a 3-face incident with 5+ -vertex v . Then 1 ≤ τ (v → f ) ≤
3 . 2
Proof. Let f = [uvw]. Since d(v ) ≥ 5, v is not a hard vertex. If both u and w are not hard, then 1 ≤ τ (v → f ) ≤ 54 by R1 and the above argument. Otherwise, assume that u is a hard vertex. Then w is not a hard vertex and uw is not incident with a (4, 4, 4, 4, 4)-face. So 21 ≤ τ (u → f ) ≤ 34 by R1. If d(w ) = 4, then 45 ≤ τ (v → f ) ≤ 2 − 12 = 23 . Otherwise 9 = (3 − 34 )/2 ≤ τ (v → f ) ≤ (3 − 12 )/2 = 45 . So 1 ≤ τ (v → f ) ≤ 23 . ■ 8
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Case 1. d(v ) = 5. Then ch(v ) = 2 × 5 − 6 = 4. Let v1 , v2 , . . . , v5 be the neighbors of v and f1 , f2 , . . . , f5 be the faces incident with v in clockwise order, where v1 = v6 and fi is incident with vi and vi+1 , i = 1, 2, . . . , 5. Since d(v ) = 5, f3 (v ) ≤ 3. We consider the following subcases. Subcase 1.1. f3 (v ) ≤ 1. If f3 (v ) = 0, then ch′ (v ) ≥ 4 − 21 × 4 − 1 × 1 = 1 > 0 by R2 and R3. If f3 (v ) = 1, then f6+ (v ) ≥ 1 and it follows that ′ ch (v ) ≥ 4 − 23 × 1 − 12 × 2 − 1 × 1 = 12 > 0. Subcase 1.2. f3 (v ) = 2. If the two 3-faces incident with v are not adjacent, then f6+ (v ) ≥ 2 and it follows that ch′ (v ) ≥ 4 − 23 × 2 − 1 × 1 = 0. Otherwise, f6+ (v ) ≥ 1. If f5 (v ) = 0 or f4 (v ) = 0, then ch′ (v ) ≥ 4 − 32 × 2 − max{ 21 × 2, 1} = 0. Otherwise f5 (v ) = f4 (v ) = 1, all 4-neighbors of v are not hard and v is incident with at most one special 3-face. Hence ch′ (v ) ≥ 4 − 1 × 2 − 1 − 12 − 14 = 14 > 0. Subcase 1.3. f3 (v ) = 3. Then f4 (v ) = 0 and f6+ (v ) ≥ 1. Without loss of generality, we assume that f1 , f2 are 3-faces and f5 is a 6+ -face. If f3 is a 3-face, then f4 must be a 6+ -face, fi (1 ≤ i ≤ 3) is not special and it follows that τ (v → fi ) = 1(i = 1, 2, 3) and ch′ (v ) ≥ 4 − 1 × 3 = 1 > 0. Otherwise, f4 is a 3-face, then d(f3 ) ≥ 5. Suppose that d(f3 ) = 5. Then f2 and f4 are not special, and all 4-neighbors of v are not hard. By R1, τ (v → f2 ) = τ (v → f4 ) = 1. So if f1 is special, then d(v3 ) ≥ 5 by Lemma 3(1) and ch′ (v ) ≥ 4 − 41 − 1 × 3 − 12 = 41 > 0. Otherwise, we have ch′ (v ) ≥ 4 − 1 × 3 − 1 = 0. Suppose that d(f3 ) ≥ 6. Then τ (v → f4 ) = max{1 + 41 , 32 } = 32 , v1 and v3 are not hard. If v2 is not hard, then ′ ch (v ) ≥ 4 − 14 − 1 × 2 − 32 = 14 > 0. Otherwise, by symmetry, we assume that the face, say f1′ , adjacent to f1 and incident with v1 v2 is a 4− -face. If f1′ is a 4-face, then τ (v2 → f1 ) = τ (v2 → f2 ) = 34 , τ (v → f1 ) ≤ 54 , τ (v → f2 ) ≤ 45 and we have ch′ (v ) ≥ 4 − 54 × 2 − 32 = 0. Otherwise, denote f1′ = [v1 v2 u]. If d(v1 ) ≥ 5 or d(u) ≥ 5, then τ (v2 → f1 ) = τ (v2 → f2 ) = 32 by R1. Since d(v1 ) + d(v3 ) ≥ 9 by Lemma 3(1), τ (v → f1 ) + τ (v → f2 ) ≤ 67 + 43 , and it follows that ch′ (v ) ≥ 4 − 76 − 34 − 32 = 0. So in the following, we assume that d(v1 ) = d(u) = 4, that is, f1′ is a (4, 4, 4)-face. Since f1′ is a (4, 4, 4)-face, τ (v2 → f1 ) = τ (v2 → f2 ) = 12 and τ (v → f1 ) = 23 by R1. Since d(v1 ) = 4, d(v3 ) ≥ 5 and it follows that τ (v → f2 ) = (3 − 21 )/2 = 45 by R1. Till now, we must come back to consider f4 . If τ (v → f4 ) ≤ 54 , then ch′ (v ) ≥ 4 − 23 − 2 × 54 = 0. Otherwise, by the similar argument, we have that v4 or v5 is hard, and v4 v5 is incident with a (4, 4, 4)-face. Combining (4, 4, 4)-face f1′ , it is contradicted to Lemma 3(3), so it is impossible. Case 2. d(v ) = 6. Since G contains no intersecting 5-faces, f3 (v ) ≤ 4. If f3 (v ) ≤ 2, then ch′ (v ) ≥ 6 − 2 × 32 − 1 − 2 × 12 > 0. If f3 (v ) = 3, then f4 (v ) + f5 (v ) ≤ 2 and it follows that ch′ (v ) ≥ 6 − 3 × 32 − 1 − 21 = 0. Suppose that f3 (v ) = 4. Then f4 (v ) = 0, and if f5 (v ) = 1, then v is incident with at most two special 3-faces and adjacent to no hard 4-vertex. So ch′ (v ) ≥ 6 − max{4 × 23 , 1 + 2 × 1 + 2 × 54 } = 0. Case 3. d(v ) = 7. Then f3 (v ) ≤ 5. If f3 (v ) = 5, then f6+ (v ) = 2 and it follows that ch′ (v ) ≥ 8 − 5 × f5 (v ) ≤ 1 and it follows that ch′ (v ) ≥ 8 − 4 × 23 − 1 − 2 × 12 = 0.
3 2
> 0 by Lemma 4. Otherwise f3 (v ) ≤ 4,
Case 4. d(v ) ≥ 8. 2d(v )+1 Then f3 (v ) ≤ ⌊ 3 ⌋, f4 (v ) + f5 (v ) ≤ d(v ) − f3 (v ) and ch′ (v ) ≥ 2d(v ) − 6 − f3 (v ) × 23 − f4 (v ) × 12 − f5 (v ) × 1. If d(v ) = 8, then f3 (v ) ≤ 5 and it follows that ch′ (v ) ≥ 10 − 5 × 32 − 2 × 12 − 1 > 0. If d(v ) ≥ 9, then ch′ (v ) ≥ 2d(v )+1 2d(v ) − 6 − × 32 − (d(v ) − 2d(v3)+1 − 1) × 12 − 1 = 5d(v6)−41 > 0. 3 Hence we complete the proof of the theorem. ■ Acknowledgment This work is supported by NSFC (11271006, 11631014) of China. References [1] [2] [3] [4] [5] [6] [7] [8] [9] [10]
N. Alon, M. Tarsi, Colorings and orientations of graphs, Combinatorica 12 (1992) 125–134. J.A. Bondy, U.S.R. Murty, Graph Theory with Applications, MacMillan, London, 1976. P.P. Cheng, M. Chen, Y.Q. Wang, Planar graphs without 4-cycles adjacent to triangles are 4-choosable, Discrete Math. 339 (2016) 3052–3057. P. Erdös, A.L. Rubin, H. Taylor, Choosability in graphs, Congr. Numer. 26 (1979) 125–157. B. Farzad, Planar graphs without 7-cycles are 4-choosable, SIAM J. Discrete Math. 23 (2009) 1179–1199. G. Fijavž, M. Juvan, B. Mohar, R. Škrekovski, Planar graphs without cycles of specific lengths, European J. Combin. 23 (2002) 377–388. P.C.B. Lam, W.C. Shiu, B.G. Xu, On structure of some plane graphs with applications to choosability, J. Combin. Theory Ser. B 82 (2001) 285–296. P.C.B. Lam, B.G. Xu, J.Z. Liu, The 4-choosability of plane graphs without 4-cycles, J. Combin. Theory Ser. B 76 (1999) 117–126. C. Thomassen, Every planar graph is 5-choosable, J. Combin. Theory Ser. B 62 (1994) 180–181. V.G. Vizing, Vertex colorings with given colors, Metody Diskret. Analiz, Novosibirsk 29 (1976) 3–10 (in Russian).
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D. Hu, J. Wu / Discrete Mathematics 340 (2017) 1788–1792 M. Voigt, List colourings of planar graphs, Discrete Math. 120 (1993) 215–219. M. Voigt, A not 3-choosable planar graph without 3-cycles, Discrete Math. 146 (1995) 325–328. W.F. Wang, K.W. Lih, Choosability and edge choosability of planar graphs without five cycles, Appl. Math. Lett. 15 (2002) 561–565. W.F. Wang, K.W. Lih, Choosability and edge choosability of planar graphs without intersecting triangles, SIAM J. Discrete Math. 15 (2002) 538–545.
Contents lists available at ScienceDirect
Discrete Mathematics journal homepage: www.elsevier.com/locate/disc
Planar graphs without intersecting 5-cycles are 4-choosable Dai-Qiang Hu a , Jian-Liang Wu b, * a b
Department of Mathematics, Jinan University, Guang Zhou, 510632, PR China School of Mathematics, Shandong University, Jinan, 250100, PR China
article
info
Article history: Received 10 March 2016 Received in revised form 11 March 2017 Accepted 11 March 2017
a b s t r a c t A graph G is k-choosable if it can be properly colored whenever every vertex has a list of at least k available colors. In the paper, it is proved that all planar graphs without intersecting 5-cycles are 4-choosable. © 2017 Elsevier B.V. All rights reserved.
Keywords: planar graph Cycle List coloring Choosable
1. Introduction All graphs considered in this paper are simple, finite and undirected, and we follow [2] for the terminologies and notations not defined here. Let G be a graph with the vertex set V (G) and the edge set E(G). For a vertex v ∈ V , let N(v ) denote the set of vertices adjacent to v , and let d(v ) = |N(v )| denote the degree of v . A k-vertex, a k+ -vertex or a k− -vertex is a vertex of degree k , at least k or at most k respectively. We use δ (G) to denote the minimum degree of G. A k-cycle is a cycle of length k, and a 3-cycle is usually called a triangle. Two cycles are adjacent( or intersecting) if they share at least one edge (or vertex, respectively). A proper k-coloring of a graph G is a mapping φ from V (G) to the color set [k] = {1, 2, . . . , k} such that φ (x) ̸ = φ (y) for every two adjacent vertices x and y of G. We say that L is an assignment for the graph G if it assigns a list L(v ) of possible colors to each vertex v of G. If G has a proper coloring φ such that φ (v ) ∈ L(v ) for any vertex v , then we say that G is L-colorable or φ is an L-coloring of G. The graph G is k-choosable if it is L-colorable for every assignment L satisfying |L(v )| ≥ k for any vertex v . The concept of a list coloring was introduced by Vizing [10] and Erdős, Rubin and Taylor [4], respectively. Thomassen [9] showed that every planar graph is 5-choosable. Examples of plane graphs which are not 4-choosable and plane graphs of girth 4 which are not 3-choosable were given by Voigt [11,12]. Since every planar graph G without 3-cycles has a vertex of degree at most 3, it is 4-choosable. Wang and Lih [13] extended the result to all planar graphs without intersecting 3-cycles. Lam, Xu and Liu [8] proved that all planar graphs without 4-cycles are 4-choosable. Lam, Shiu and Xu [7](Wang and Lih [14], respectively) proved that all planar graphs without 5-cycles are 4-choosable. Fijavž, Juvan, Mohar, Škrekovski [6] proved that all planar graphs without 6-cycles are 4-choosable. Farzad [5] proved that all planar graphs without 7-cycles are 4-choosable. Cheng, Chen and Wang [3] proved that planar graphs without 4-cycles adjacent to triangles are 4-choosable. In the paper, we prove that all planar graphs without intersecting 5-cycles are 4-choosable.
*
Corresponding author. E-mail address: [email protected] (J. Wu).
http://dx.doi.org/10.1016/j.disc.2017.03.012 0012-365X/© 2017 Elsevier B.V. All rights reserved.
D. Hu, J. Wu / Discrete Mathematics 340 (2017) 1788–1792
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Fig. 1. Three subgraphs of G.
Fig. 2. An orientation of the configurations in Figure 1.
2. Main result and its proof First, we introduce some notations and definitions used in this section. Let G be a plane graph. We use F or F (G) to denote the face set of G. For f ∈ F (G), we use V (f ) to denote the set of vertices on the boundary of f and write f = [u1 u2 · · · un ] if u1 , u2 , . . . , un are the boundary vertices of f in a cyclic order. A face of G is said to be incident with all edges and vertices in its boundary. The degree of a face f , denoted by dG (f ), is the number of edges incident with it, where a cut edge is counted twice. A k-face, k− -face or a k+ -face is a face of degree k, at most k or at least k, respectively. For convenience, a k-face f = [v1 v2 · · · vk ] is often said to be a (d(v1 ), d(v2 ), . . . , d(vk ))-face. For a face f , let ni (f ) and ni+ (f ) denote the number of i-vertices and i+ -vertices incident with f , respectively. Denote by fd (v ) and fd+ (v ) the number of d-faces and d+ -faces incident with v , respectively. Theorem 1. All planar graphs without intersecting 5-cycles are 4-choosable. Proof. Suppose, to the contrary, that Theorem 1 is false. Let G be a counterexample to Theorem 1 with the fewest vertices. Then δ (G) ≥ 4 (see [8]). First, we introduce a well-known theorem proved by Alon and Tarsi [1]. This intricate theorem reveals the connection between the list coloring of a graph G and its orientations. A digraph (directed graph) D is an ordered pair (V (D), A(D)) consisting of the vertex set V (D) and arc set A(D). For any arc a = ⟨u, v⟩, we say that u is the tail of a and v its head. The + indegree d− D (v ) of a vertex v in D is the number of arcs with head v , and the outdegree dD (v ) of v is the number of arcs with tail v . A directed cycle is denoted by a cyclic sequence u1 u2 · · · uk u1 in which each vertex dominates its successor. A subdigraph H of a directed graph D is called Eulerian if the indegree d− H (v ) of every vertex v of H is equal to its outdegree d+ ( v ). Note that we do not assume that H is connected. H is e v en if it has an even number of edges. Otherwise it is odd. Let H EE(D) and EO(D) denote the numbers of even and odd Eulerian subgraphs of D, respectively. (For convenience we agree that the empty subgraph is an even Eulerian subgraph). + Theorem 2. [1] Let D be a digraph. For each vertex v ∈ V (D), let f (v ) be a set of d+ D (v ) + 1 distinct integers, where dD (v ) is the outdegree of v . If EE(D) ̸ = EO(D), then there is a proper coloring c : V (D) → Z such that c(v ) is in f (v ) for every vertex v . That is, if L is a list assignment such that |L(v )| = d+ D (v ) + 1 for all vertices v in D, then D is L-colorable.
Lemma 3. G contains no subgraph isomorphic to one of the configurations in Fig. 1, where the vertices marked by • have degree of 4 in G and those vertices marked by ◦ have degree of 4 or 5 in G. Proof. Let L be an arbitrary list assignment of G such that each vertex is assigned precisely 4 colors. Suppose that G contains a subgraph H isomorphic to one of the configurations in Fig. 1. By the minimality of G, the graph G − V (H) has an L-coloring ϕ . An orientation of H and available colors of every vertex are shown in Fig. 2. Since EE(G1 ) = 1 < EO(G1 ) = 2, EE(G2 ) = 1 < EO(G2 ) = 2 and EE(G3 ) = 5 > EO(G3 ) = 4, any Gi (1 ≤ i ≤ 3) satisfies Theorem 2. So ϕ can be extended to G. ■
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In the following, we assume that G has been embedded in the plane. Thus G is a plane graph. By the Euler’s formula
|V (G)| − |E(G)| + |F (G)| = 2, we have ∑ ∑ (2d(v ) − 6) + (d(f ) − 6) = −6(|V (G)| − |E(G)| + |F (G)|) = −12 < 0. v∈V (G)
f ∈F (G)
Now we define the initial∑ charge function ch(x) of x ∈ V (G) ∪ F (G) to be ch(v ) = 2d(v ) − 6 if v ∈ V (G) and ch(f ) = d(f ) − 6 if f ∈ F (G). It follows that x∈V (G)∪F (G) ch(x) < 0. Now we design appropriate discharging rules and redistribute weights accordingly. Note that any discharging procedure preserves the total charge of G. If we can define suitable discharging rules to charge the initial charge function ch to the final charge function ch′ on V (G) ∪ F (G), such that ch′ (x) ≥ 0 for all x ∈ V (G) ∪ F (G), then we get an obvious contradiction. A 3-face is called special if it is adjacent to a (4, 4, 4, 4, 4)-face. A vertex v is called hard if v is a 4-vertex with f3 (v ) = 3 or f3 (v ) = 2 and f4 (v ) = 1. It is obvious that two hard 4-vertices are not adjacent. For x, y ∈ V ∪ F , let τ (x → y) denote the amount of weights transferred from x to y. Now we define the discharging rules as follows. R1. Let f be a 3-face [v1 v2 v3 ]. If d(v1 ) = d(v2 ) = d(v3 ) = 4 or every vi (1 ≤ i ≤ 3) is not hard, then τ (vi → f ) = 1 for all i = 1, 2, 3. Otherwise, assume that v1 is a hard vertex and d(v3 ) ≥ 5. Then
⎧ 2 ⎪ ⎪ , if f3 (v ) = 3 and no (4, 4, 4)-face is incident with v1 , ⎪ ⎪ 3 ⎪ ⎪ ⎨ 3 τ (v 1 → f ) = , if f3 (v ) = 2 and no (4, 4, 4)-face is incident with v1 , ⎪ 4 ⎪ ⎪ ⎪ ⎪1 ⎪ ⎩ , otherwise. ⎧2 if d(v2 ) = 4, ⎨1, τ (v2 → f ) = 3 − τ (v1 → f ) ⎩ , otherwise. 2 ⎧ ⎨2 − τ (v1 → f ), if d(v2 ) = 4, τ (v3 → f ) = 3 − τ (v1 → f ) ⎩ , otherwise. 2
R2. Let f be a 4-face [v1 v2 v3 v4 ]. Then τ (vi → f ) = 12 for i = 1, 2, 3, 4. R3. Let f be a 5-face. If n5+ (f ) > 0, then for every 5+ -vertex v incident with f , τ (v → f ) = (4, 4, 4, 4, 4)-face. Let v be a 4-vertex incident with f . If f3 (v ) ≤ 1, then τ (v → f ) = such that uv is incident with f . Then τ (w → f ) = 41 .
1 . 2
1 . n5+ (f )
Otherwise, f is a
Otherwise, let [uvw] be a 3-face
Next we show that ch′ (x) ≥ 0 for all x ∈ V ∪ F . Let f be a face of G. If d(f ) = 3, then it is obvious by R1 that ch′ (f ) = 0. If d(f ) = 4, then ch′ (f ) = −2 + 21 × 4 = 0 by R2. If d(f ) ≥ 6, then ch′ (f ) = ch(f ) ≥ 0. Let f be a 5-face of G. If n5+ (f ) ̸ = 0, then ch′ (f ) ≥ −1 + n 1 (f ) × n5+ (f ) = 0 by R3. Otherwise, f is a (4, 4, 4, 4, 4)-face 5+ [v1 v2 · · · v5 ]. If f3 (vi ) ≥ 2 for all i = 1, 2, 3, 4, 5, then f is adjacent to at least four special 3-faces since G has no intersecting 5-cycles and it follows that ch′ (f ) ≥ −1 + 41 × 4 = 0 by R3. Otherwise, without loss of generality, assume that f3 (v1 ) ≤ 1. If there is some vi (2 ≤ i ≤ 5) having f3 (vi ) ≤ 1, then ch′ (f ) ≥ −1 + 21 × 2 = 0. Otherwise, f is adjacent to at least two 3-faces and it follows that ch′ (f ) ≥ −1 + 12 + 14 × 2 = 0. Let v be a 4-vertex of G. Then f3 (v ) ≤ 3. If f3 (v ) = 0, then ch′ (v ) ≥ 2 − 12 × 4 = 0 by R2–R3. Otherwise, f6+ (v ) ≥ 1. Note that if v is incident with a 3-face [uvw] such that d(u) = d(w ) = 4, then uw is not incident with a (4, 4, 4, 4, 4)-face by Lemma 3(2). If f3 (v ) = 1, then ch′ (v ) ≥ 2 − 1 − 2 × 21 = 0. If f3 (v ) = 2 and f4 (v ) = 0, then ch′ (v ) ≥ 2 − 1 × 2 = 0 by R1. Finally, we assume that f3 (v ) = 3 or f3 (v ) = 2 and f4 (v ) = 1, that is, v is a hard vertex. By Lemma 3(1), v is incident with at most one (4, 4, 4)-face. If f3 (v ) = 3, then ch′ (v ) ≥ 2 − max{1 + 21 × 2, 23 × 3} = 0 by R1. Otherwise, ch′ (v ) ≥ 2 − max{1 + 2 × 12 , 43 × 2 + 12 } = 0. Let v be a 5+ -vertex of G. Note that any 4-vertex incident with a special 3-face is not hard. Suppose that f = [uvw] is a special 3-face such that d(u) = d(w ) = 4 and uw is incident with a (4, 4, 4, 4, 4)-face f ′ . Then τ (v → f ) = 1 by R1 and τ (v → f ′ ) = 14 by R3. In the following, sometimes we replace the case by τ (v → f ) = 54 . Lemma 4. Let f be a 3-face incident with 5+ -vertex v . Then 1 ≤ τ (v → f ) ≤
3 . 2
Proof. Let f = [uvw]. Since d(v ) ≥ 5, v is not a hard vertex. If both u and w are not hard, then 1 ≤ τ (v → f ) ≤ 54 by R1 and the above argument. Otherwise, assume that u is a hard vertex. Then w is not a hard vertex and uw is not incident with a (4, 4, 4, 4, 4)-face. So 21 ≤ τ (u → f ) ≤ 34 by R1. If d(w ) = 4, then 45 ≤ τ (v → f ) ≤ 2 − 12 = 23 . Otherwise 9 = (3 − 34 )/2 ≤ τ (v → f ) ≤ (3 − 12 )/2 = 45 . So 1 ≤ τ (v → f ) ≤ 23 . ■ 8
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Case 1. d(v ) = 5. Then ch(v ) = 2 × 5 − 6 = 4. Let v1 , v2 , . . . , v5 be the neighbors of v and f1 , f2 , . . . , f5 be the faces incident with v in clockwise order, where v1 = v6 and fi is incident with vi and vi+1 , i = 1, 2, . . . , 5. Since d(v ) = 5, f3 (v ) ≤ 3. We consider the following subcases. Subcase 1.1. f3 (v ) ≤ 1. If f3 (v ) = 0, then ch′ (v ) ≥ 4 − 21 × 4 − 1 × 1 = 1 > 0 by R2 and R3. If f3 (v ) = 1, then f6+ (v ) ≥ 1 and it follows that ′ ch (v ) ≥ 4 − 23 × 1 − 12 × 2 − 1 × 1 = 12 > 0. Subcase 1.2. f3 (v ) = 2. If the two 3-faces incident with v are not adjacent, then f6+ (v ) ≥ 2 and it follows that ch′ (v ) ≥ 4 − 23 × 2 − 1 × 1 = 0. Otherwise, f6+ (v ) ≥ 1. If f5 (v ) = 0 or f4 (v ) = 0, then ch′ (v ) ≥ 4 − 32 × 2 − max{ 21 × 2, 1} = 0. Otherwise f5 (v ) = f4 (v ) = 1, all 4-neighbors of v are not hard and v is incident with at most one special 3-face. Hence ch′ (v ) ≥ 4 − 1 × 2 − 1 − 12 − 14 = 14 > 0. Subcase 1.3. f3 (v ) = 3. Then f4 (v ) = 0 and f6+ (v ) ≥ 1. Without loss of generality, we assume that f1 , f2 are 3-faces and f5 is a 6+ -face. If f3 is a 3-face, then f4 must be a 6+ -face, fi (1 ≤ i ≤ 3) is not special and it follows that τ (v → fi ) = 1(i = 1, 2, 3) and ch′ (v ) ≥ 4 − 1 × 3 = 1 > 0. Otherwise, f4 is a 3-face, then d(f3 ) ≥ 5. Suppose that d(f3 ) = 5. Then f2 and f4 are not special, and all 4-neighbors of v are not hard. By R1, τ (v → f2 ) = τ (v → f4 ) = 1. So if f1 is special, then d(v3 ) ≥ 5 by Lemma 3(1) and ch′ (v ) ≥ 4 − 41 − 1 × 3 − 12 = 41 > 0. Otherwise, we have ch′ (v ) ≥ 4 − 1 × 3 − 1 = 0. Suppose that d(f3 ) ≥ 6. Then τ (v → f4 ) = max{1 + 41 , 32 } = 32 , v1 and v3 are not hard. If v2 is not hard, then ′ ch (v ) ≥ 4 − 14 − 1 × 2 − 32 = 14 > 0. Otherwise, by symmetry, we assume that the face, say f1′ , adjacent to f1 and incident with v1 v2 is a 4− -face. If f1′ is a 4-face, then τ (v2 → f1 ) = τ (v2 → f2 ) = 34 , τ (v → f1 ) ≤ 54 , τ (v → f2 ) ≤ 45 and we have ch′ (v ) ≥ 4 − 54 × 2 − 32 = 0. Otherwise, denote f1′ = [v1 v2 u]. If d(v1 ) ≥ 5 or d(u) ≥ 5, then τ (v2 → f1 ) = τ (v2 → f2 ) = 32 by R1. Since d(v1 ) + d(v3 ) ≥ 9 by Lemma 3(1), τ (v → f1 ) + τ (v → f2 ) ≤ 67 + 43 , and it follows that ch′ (v ) ≥ 4 − 76 − 34 − 32 = 0. So in the following, we assume that d(v1 ) = d(u) = 4, that is, f1′ is a (4, 4, 4)-face. Since f1′ is a (4, 4, 4)-face, τ (v2 → f1 ) = τ (v2 → f2 ) = 12 and τ (v → f1 ) = 23 by R1. Since d(v1 ) = 4, d(v3 ) ≥ 5 and it follows that τ (v → f2 ) = (3 − 21 )/2 = 45 by R1. Till now, we must come back to consider f4 . If τ (v → f4 ) ≤ 54 , then ch′ (v ) ≥ 4 − 23 − 2 × 54 = 0. Otherwise, by the similar argument, we have that v4 or v5 is hard, and v4 v5 is incident with a (4, 4, 4)-face. Combining (4, 4, 4)-face f1′ , it is contradicted to Lemma 3(3), so it is impossible. Case 2. d(v ) = 6. Since G contains no intersecting 5-faces, f3 (v ) ≤ 4. If f3 (v ) ≤ 2, then ch′ (v ) ≥ 6 − 2 × 32 − 1 − 2 × 12 > 0. If f3 (v ) = 3, then f4 (v ) + f5 (v ) ≤ 2 and it follows that ch′ (v ) ≥ 6 − 3 × 32 − 1 − 21 = 0. Suppose that f3 (v ) = 4. Then f4 (v ) = 0, and if f5 (v ) = 1, then v is incident with at most two special 3-faces and adjacent to no hard 4-vertex. So ch′ (v ) ≥ 6 − max{4 × 23 , 1 + 2 × 1 + 2 × 54 } = 0. Case 3. d(v ) = 7. Then f3 (v ) ≤ 5. If f3 (v ) = 5, then f6+ (v ) = 2 and it follows that ch′ (v ) ≥ 8 − 5 × f5 (v ) ≤ 1 and it follows that ch′ (v ) ≥ 8 − 4 × 23 − 1 − 2 × 12 = 0.
3 2
> 0 by Lemma 4. Otherwise f3 (v ) ≤ 4,
Case 4. d(v ) ≥ 8. 2d(v )+1 Then f3 (v ) ≤ ⌊ 3 ⌋, f4 (v ) + f5 (v ) ≤ d(v ) − f3 (v ) and ch′ (v ) ≥ 2d(v ) − 6 − f3 (v ) × 23 − f4 (v ) × 12 − f5 (v ) × 1. If d(v ) = 8, then f3 (v ) ≤ 5 and it follows that ch′ (v ) ≥ 10 − 5 × 32 − 2 × 12 − 1 > 0. If d(v ) ≥ 9, then ch′ (v ) ≥ 2d(v )+1 2d(v ) − 6 − × 32 − (d(v ) − 2d(v3)+1 − 1) × 12 − 1 = 5d(v6)−41 > 0. 3 Hence we complete the proof of the theorem. ■ Acknowledgment This work is supported by NSFC (11271006, 11631014) of China. References [1] [2] [3] [4] [5] [6] [7] [8] [9] [10]
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