STRONG EDGE-COLORING FOR PLANAR GRAPHS WITH LARGE

LILY CHEN1, KECAI DENG1, GEXIN YU2,3, AND XIANGQIAN ZHOU1,4

1 School of Mathematical Sciences, Huaqiao University, China 2 School of Mathematics and Statistics, Central China Normal University, Wuhan, Hubei, China. 3 Department of Mathematics, College of William and Mary, Williamsburg, VA, USA. 4Department of Mathematics and Statistics, Wright State University, Dayton, Ohio, 45435

Abstract. A strong edge-coloring of a graph G = (V,E) is a partition of its edge set E into induced matchings. Let G be a with girth k ≥ 26 and maximum ∆. We show that either G is isomorphic to a subgraph of a very special ∆- 12(∆−2) with girth k, or G has a strong edge-coloring using at most 2∆ + d k e colors.

1. introduction Graphs in this article are assumed to be simple and undirected. Let G be a graph. A proper edge-coloring of G is an assignment of colors to the edges such that no two adjacent edges receive the same color. Clearly, every coloring class is a of G. However, these matchings may not be induced. A strong edge-coloring of a graph G, first introduced by Fouquet and Jolivet [6], is a proper edge-coloring such that every two edges joined by another edge are colored differently. In a strong edge-coloring, every color class is an . The minimum number of colors required in a strong edge-coloring of G is called 0 the strong chromatic index and is denoted by χs(G). Let e and e0 be two edges of G. We say that e sees e0 if e and e0 are adjacent or share a common adjacent edge. So, equivalently, a strong edge-coloring is an assignment of colors to all edges such that every two edges that can see each other receive distinct colors. Let ∆ be the maximum degree of G, and for u ∈ V (G), let dG(u) denote the degree of u in G. In this paper, we study strong edge-coloring of planar graphs. Faudree et al [5] conjectured that every subcubic planar graph has a strong edge-coloring using at most 9 colors. This conjecture was recently confirmed by Kostochka et al [8]. For an arbitrary planar graph G,

E-mail address: [email protected], [email protected]. Date: November 8, 2017. 1991 Mathematics Subject Classification. 05C15. Key words and phrases. planar graph, strong edge-coloring, girth. Both the first and the second author are supported by NSFC (No.11501223 and No.11701195), and Science Foundation of the Fujian Province, China (No.2016J05009); the first author is also supported by Scientific Research Funds of Huaqiao University (No.14BS311); and the second author is also supported by Research Funds of Huaqiao University (No.16BS808) and NSFC (No. 11471273). The third author is supported in part by the NSA grant: H98230-16-1-0316 and Natural Science Foundation of China (11728102). The fourth author is partially supported by the Minjiang Scholar Program hosted by Huaqiao University, Quanzhou, Fujian, China. 1 0 0 0 Faudree et al [5] proved that χs(G) ≤ 4χ (G), where χ (G) is the chromatic index of G. Now 0 it follows from Vizing’s Theorem [10] that χs(G) ≤ 4∆ + 4. Faudree et al [5] also showed 0 that there exists a planar graph G with χs(G) = 4∆ − 4 for every integer value of ∆ ≥ 2. 0 Better bounds for χs(G) were obtained when G is required to have a large girth. The next result of Hud´aket al [7] is a good example of such results. Theorem 1.1. (Hud´aket al 2014) 0 1. If G is a planar graph with girth at least 7, then χs(G) ≤ 3∆(G). 0 2. If G is a planar graph with girth at least 6, then χs(G) ≤ 3∆(G) + 5. For planar graph with girth 6, the bound given in Theorem 1.1 was improved to 3∆ + 1 by Bensmail et al [1], and independently by Ruksasakchai and Wang [9], in 2014. 0 Note that, when a graph G has two adjacent degree-∆ vertices, then χs(G) ≥ 2∆ − 1. Let G be a planar graph with ∆ ≥ 3 and girth g. Borodin and Ivanova [2] proved in 2013 that ∆ 0 if g ≥ 40b 2 c + 1, then χs(G) ≤ 2∆ − 1. The bound on the girth of G was later improved to 10∆ + 46 by Chang et al [4]. The best bound on the girth was established by Wang and Zhao [11] in the following result. Theorem 1.2. (Wang and Zhao 2015) If G is a planar graph with ∆ ≥ 4 and girth at least 0 10∆ − 4, then χs(G) ≤ 2∆ − 1. Chang and Duh [3] introduced another parameter σ on a graph G defined as follows.

σ(G) = max {dG(x) + dG(y) − 1}. xy∈E(G) 0 It is clear that for any graph G, χs(G) ≥ σ(G). Chang and Duh [3] proved the following theorem. Theorem 1.3. (Chang and Duh 2015) Let G be a planar graph with σ(G) ≥ 5 and σ(G) ≥ 0 ∆(G) + 2. If the girth of G is at least 5σ(G) + 16, then χs(G) = σ(G). Note that, the bounds on the girth given in Theorem 1.2 and 1.3 are both larger than a linear function of ∆, and so they become really big for graphs with large ∆ values. It would be nice if one can improve the lower bound on the girth down to a constant instead of a linear function of ∆. However, Hud´aket al [7] showed that there exists a planar graph G 0 ∆ with girth k and with χs(G) > 2∆ by constructing the following examples: Let Gk be the graph obtained from a Ck of length k by adding ∆ − 2 pendent edges at each vertex ˘ of Ck. Hud´ak,Lu˘zar,Sot´ak,and Skrekovski [7] proved the following result. Proposition 1.4. (Hud´akat al 2014) For every odd integer k ≥ 3 and every integer ∆ ≥ 3, the following inequality holds.

2k(∆ − 1) 2k(∆ − 2) ≤ χ0 (G∆) ≤ + 5. k − 1 s k k − 1 Motivated by these constructions, Hud´aket al [7] made the following conjecture. Conjecture 1.5. (Hud´akat al 2014) There exists a constant c such that for every planar graph G of girth k ≥ 5, 2(∆ − 1) χ0 (G) ≤ 2∆ + + c. s k − 1 2 In the paper, we prove the following theorem, which is a weaker result than Conjecture 1.5. Theorem 1.6. Let G be a planar graph with ∆(G) = ∆ ≥ 4 and girth g = k ≥ 26. Then ∆ either G is isomorphic to a subgraph of Gk , or 12(∆ − 2) χ0 (G) ≤ 2∆ + . s k We would like to point out that, with the exception of some small values of ∆, the upper 0 ∆ bound on χs(Gk ) given by Hud´aket al [7] in Proposition 1.4 is better than our bound given ∆ in Theorem 1.6. That is the reason we keep subgraphs of Gk as one of the outcomes in Theorem 1.6. The paper is organized as follows: we present the proof for Theorem 1.6 in Section 2 and then in Section 3, we talk about some possible extensions of our result.

2. proof of theorem 1.6 First we will introduce some notions to be used in our proof. A k-vertex is a vertex with degree exactly k. A vertex is called a k+-vertex (resp. k−-vertex) if it has degree at least k (resp. at most k). The notions of k-neighbor, k+-neighbor, and k−-neighbor of a vertex v are defined in a similar manner. A vertex v in a graph G is called a poor vertex if dG(v) ≥ 2 and v has exactly dG(v) − 2 1-neighbors. A vertex v is called a rich vertex if dG(v) ≥ 3 and v has at least three 2+-neighbors. To prove Theorem 1.6, we choose G to be a minimal counterexample, that is, G is a planar graph with girth g = k ≥ 26 and ∆(G) = ∆ such that 1) G does not have a strong 12(∆−2) edge-coloring using at most 2∆ + d k e colors; and 2) |V (G)| is as small as possible. We will first prove some structural results of G. 2.0.1. A 1-vertex must be adjacent to a 4+-vertex. Proof. Suppose u is a 1-vertex and whose neighbor v is a 3−-vertex. Then the edge uv sees at most 2∆ edges. Since we have more than 2∆ + 1 colors, a strong edge-coloring of G\u can be easily extended to a strong edge-coloring of G, a contradiction.  It follows from 2.0.1 that every 2-vertex is a poor vertex. 2.0.2. Every 2+-vertex of G has at least two 2+-neighbors. Proof. Let v be a 2+-vertex of G. If v has no 2+-neighbor, then since G is connected, G is isomorphic to a , which has a strong using ∆ colors. Next assume that v + has exactly one 2 -neighbor. Since dG(v) ≥ 2, v has at least one 1-neighbor. Let u be a 1-neighbor of v. Since G is a minimal counterexample, G\u has a strong edge-coloring using 12(∆−2) at most 2∆ + d k e colors. The edge uv sees at most 2∆ − 2 edges in G, so a coloring of G\u can be easily extended to a coloring of G, a contradiction.  Recall that a rich vertex v is a 3+-vertex that has at least three 2+-neighbors. By 2.0.2, every 3+-vertex is either a poor vertex or a rich vertex. 2.0.3. If v is a poor vertex of G, then v is incident to exactly two faces in every planar embedding of G. 3 Proof. Let v be a poor vertex of G. Then by the definition of a poor vertex, v is incident to at most two faces in every planar embedding of G. Suppose that v is incident to exactly one face. Then G\v has exactly two non-trivial connected components G1 and G2. Let u1 and + u2 be the two 2 -neighbors of v. Assume that u1 ∈ V (G1) and u2 ∈ V (G2). For i ∈ {1, 2}, ¯ let Gi be the graph obtained from Gi by adding the vertex v and the edge vui. Since G ¯ ¯ is a minimum counterexample, each of G1 and G2 has a strong edge-coloring using at most 12(∆−2) 2∆ + d k e colors. By switching colors if necessary, we can assume that 1) there exists a color set K of size at most ∆ − 1 such that the edges of G1 incident to u1 and the edges of G2 incident to u2 both used colors from K; and 2) the edge vu1 and the edge vu2 receive ¯ ¯ distinct colors. Now it is easy to extend the colorings of G1 and G2 to a coloring a G, a contradiction. 

Let Gp be the graph spanned by all poor vertices of G. Then by the definition of a poor vertex, ∆(Gp) ≤ 2. So Gp is a union of cycles and paths. If Gp contains a cycle, then since ∆ G is connected, it is not hard to see that the graph G is a subgraph of the graph Gk . Recall 0 ∆ that Hudak et al [7] gave a lower bound and an upper bound for χs(Gk ) and these two bounds differ by at most 5; see Proposition 1.4. From now on we assume that Gp contains no cycle.

k 2.0.4. Let P be a path of Gp and let l be the number of vertices on P . Then l ≤ d 6 e − 1.

Proof. Suppose that P = v1v2 ··· vl is a path in Gp. Then each of v1, v2, ··· , vl is a poor + + vertex of G. Let v0 be the 3 -neighbor of v1 other than v2 and let vl+1 be the 3 -neighbor of 0 vl other than vl−1. Let G be the graph obtained from G by deleting all 1-vertices adjacent to a vertex on P . Since G is a minimal counterexample, G0 has a strong edge-coloring using at most 2∆ + 12(∆−2) d k e colors. Now we will construct a strong edge-coloring of G. For 1 ≤ i ≤ l, let Ei be the set of pendent edges at vi and let Ki be the set of colors that will be assigned to edges in Ei. First assign colors to E1; this is possible since |E1| ≤ ∆ − 2 and E1 sees at most ∆ + 2 colored edges. There are two cases depending on the parity of l. First assume that l is even. In this case, we will color edges in El (i.e., pendent edges at vl) in such a way that |Kl ∩ K1| is as small as possible and then we color edges in El−1 such that |K1 ∩ Kl−1| is as large as possible. Each of E1 and El sees at most ∆ + 2 colored edges. In the worst case, E1 and El forbid the same set of ∆+2 colors. Denote this set of forbidden colors by K0. So K0 = ∆ + 2 ≥ 6. By switching colors if necessary, we may assume that every edge on the path P is colored by a color from K0. 12(∆−2) Note that there are at least ∆ − 2 + d k e colors available for E1 and El; so we can 12(∆−2) color El such that Kl\K1 ≥ d k e. Next we color El−1 such that |K1 ∩ Kl−1| is as large as possible. Since colors in Kl are forbidden for El−1; we can choose Kl−1 such that 12(∆−2) Kl−1 ∩K1 = K1\Kl. Therefore, we can guarantee that |K1 ∩Kl−1| ≥ d k e. By repeating the same coloring procedure, we get that

12(∆ − 2) l − 2 12(∆ − 2) |K ∩ K | ≥ 2 × , and |K ∩ K | ≥ × . 1 l−3 k 1 3 2 k

So the number of forbidden colors for E2 is at most 4 l − 2 12(∆ − 2) 2∆ − × , 2 k

l 12(∆−2) and therefore, the number of available colors for E2 is at least 2 × d k e. We may assume that l 12(∆ − 2) × < ∆ − 2, 2 k for otherwise, E2 can be colored. Let 12(∆ − 2) = km + t with 0 ≤ t < k. Then we have l t km + t (m + d e) < ∆ − 2 = . 2 k 12 It follows that t k(m + k ) k l < t ≤ . 6(m + d k e) 6 k Since l is an integer, we have l ≤ d 6 e − 1.  2.0.5. Every cycle of G contains at least six rich vertices. Proof. Let C be a cycle of G. Then |V (C)| ≥ k since the girth of G is k. Now if C contains at most five rich vertices. Then the graph spanned by all poor vertices on C has at most five connected components, and each component is a path of Gp. By the Pigeonhole Principle, k−5 at least one of these paths contains at least d 5 e vertices. Since k ≥ 26, it is routine to k−5 k k check that d 5 e = d 5 e − 1 > d 6 e − 1, contrary to 2.0.4.  We will consider a planar embedding of G and assign initial charge to every vertex and k−2 every face as follows: every v ∈ V (G) gets an initial charge of µ(v) = 2 dG(v) − k; and every face F gets an initial charge of µ(F ) = dG(F ) − k. By the Euler formula, X k − 2 X ( d (v) − k) + (d(F ) − k) = −2k. 2 G v F Recall that every poor vertex of G is incident to exactly two faces by 2.0.3. Now we will redistribute the charge among vertices and faces by the following discharging rules: R1) Every face sends 2 to every incident 1-vertex. k−2 R2) Every 1-vertex receives 2 from its neighbor. R3) Every poor vertex receive 1 from each of its incident face. k−6 R4) Each rich vertex sends 6 to each of its incident face. Let µ∗(v) (resp. µ∗(F )) be the new charge of a vertex v (resp. a face F ). First note that if v is a 1-vertex, then k − 2 k − 2 µ∗(v) = − k + 2 + = 0. 2 2 If v is a poor vertex, then k − 2 k − 2 µ∗(v) = d (v) × − k + 2 − (d (v) − 2) × = 0. G 2 G 2 5 Next assume that v is a rich vertex. Let t1 be the number of 1-neighbors of v. Then the vertex v is incident to at most dG(v) − t1 different faces, therefore k − 2 k − 2 k − 6 µ∗(v) ≥ d (v) × − k − t − (d (v) − t ) × G 2 2 1 G 1 6 k k = (d (v) − t ) × − k ≥ 3 × − k = 0 G 1 3 3 Remark: in the red line, we assume that v has only three neighbors that are not poor. In fact, if v has more rich neighbors, then v has more remaining charges. On the other hand, if v is incident to exactly three threads of poor vertices and each thread has about k/6 poor vertices, then it is very likely to be reducible. Can we use this to improve our result?? Finally for a face F , let s1 be the number of 1-vertices that lies in F , let sp be the number of poor vertices on the boundary of F , and sr be the number of rich vertices on the boundary of F . Then by 2.0.5, sr ≥ 6. Also note that d(F ) − 2s1 = sp + sr. Therefore the new charge of the face F is the following. k − 6 µ∗(F ) = d(F ) − k − 2s − s + s × 1 p r 6 ≥ d(F ) − k − 2s1 − sp + (k − 6) = (d(F ) − 2s1) − sp − 6

= sr + sp − sp − 6 = sr − 6 ≥ 0 Therefore, we have a contradiction since X X X X −2k = µ(v) + µ(F ) = µ∗(v) + µ∗(F ) ≥ 0. v F v F

3. Possible extensions We would like to point out that our main result, i.e. Theorem 1.6 is weaker than Conjec- 12(∆−2) ture 1.5 in two aspects: first of all, we require 2∆ + d k e colors, while Conjecture 1.5 2∆ says that 2∆ + k−1 + c colors would suffices. There exist planar graphs that require more than 2∆ colors in a strong edge-coloring. Conjecture 1.5 basically says that by using a small 2∆ number of extra colors (i.e. k−1 + c), one can color all planar graphs with girth at least 5. 12(∆−2) The number of extra colors required by our result (i.e. d k e) is roughly six times of the 2∆ number given in Conjecture 1.5 (i.e. k−1 + c). Secondly, our lower bound on the girth is 26, which is much larger than the conjectured bound of 5, and we do not see a quick way to improve that bound without increasing the number of colors being used. We realized that the major difficulty was how to deal with graphs with a large number of poor vertices of degree ∆. If we restrict ourselves to the class of planar graphs without any poor vertices of degree larger than a fixed integer (say 4 for example), then it is straight- forward to show that the strong chromatic index of every such graph is bounded above by 2∆ + c for some integer c. If the graph has many poor vertices of large degree, then on the one hand, it is not easy to find reducible configurations as every pendent edge at a poor vertex may see up to 3∆ − 3 other edges; on the other hand, these poor vertices are also problematic in the discharging process since they have to give away a lot of charges to its 1-neighbors (many of them). Our way to overcome the aforementioned difficulty was to use more extra colors; this way, we can bound the length of a path formed by poor vertices. To improve our result, it 6 seems inevitable to study more carefully on local structures that have a large number of poor vertices. Hopefully one can find better ways to color pendent edges at these poor vertices. We conclude by stating that, despite the difficulty, we strongly believe that Conjecture 1.5 is true.

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