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Relaxed equitable colorings of planar graphs with girth at least 8
Discrete Mathematics 343 (2020) 111790
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Relaxed equitable colorings of planar graphs with girth at least 8✩ Ming Li, Xia Zhang
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School of Mathematics and Statistics, Shandong Normal University, Jinan 250358, PR China
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Article history: Received 9 May 2018 Received in revised form 8 December 2019 Accepted 17 December 2019 Available online xxxx Keywords: Equitable coloring Relaxed coloring Discharging Planar graph
a b s t r a c t An RE-m-coloring of a graph G is a vertex m-coloring of G, which is relaxed (every vertex shares the same color with at most one neighbor) and equitable (the sizes of all color classes differ by at most one). In this article, we prove that every planar graph with minimum degree at least 2 and girth at least 8 has an RE-m-coloring for each integer m ≥ 4. We use the discharging method and Hall’s Theorem to simply the structures of counterexamples. © 2019 Elsevier B.V. All rights reserved.
1. Introduction Given a graph G = (V , E), a k-coloring of G is a mapping f : V (G) → [k] and a k-coloring of G is proper if f (u) ̸ = f (v ) for each uv ∈ E(G), where [k] = {1, 2, . . . , k}. That is to say, each subset of vertices colored with a same color is an independent (or conflict-free) set. A proper coloring is equitable if the sizes of all color classes differ by at most one. The equitable chromatic number of G, denoted by χeq (G), is the smallest integer k such that G is equitable k-colorable. The ∗ equitable chromatic threshold of G, denoted by χeq (G), is the smallest integer m such that G is equitable k-colorable for all ∗ k ≥ m. Note that χeq (G) ≤ χeq (G). In fact, the gap between these two parameters can be arbitrarily large. For example, ∗ χeq (K2t +1,2t +1 ) = 2 while χeq (K2t +1,2t +1 ) = 2t + 2 for each t ≥ 1. The study of equitable colorings began with a conjecture due to Erdős [8] which was proved by Hajnal and Szemerédi [11]: Every graph G with maximum degree at most ∆ has an equitable k-coloring for every k ≥ ∆ + 1. It is an NP-hard problem to decide χeq (G) even for planar graphs. There are a series of results on equitable colorings of graphs (see [13]). Zhang and Yap [20] proved that a planar graph is equitably m-colorable if m ≥ ∆ ≥ 13. Luo, Sereni, Stephens and Yu [15] considered equitable colorings for planar graphs with some restrictions on girth (the girth of a graph G is the minimum length of the cycles in G and denoted by g(G)). Theorem 1 ([15]). Let G be a planar graph with δ (G) ≥ 2. ∗ (a) If g(G) ≥ 14, then χeq (G) ≤ 3. ∗ (b) If g(G) ≥ 10, then χeq (G) ≤ 4.
✩ This work is supported by the Shandong Provincial Natural Science Foundation, China (Grant No. ZR2019MA032), the National Natural Science Foundation of China (Grant No. 11701342) of China. ∗ Corresponding author. E-mail addresses: [email protected], [email protected] (X. Zhang). https://doi.org/10.1016/j.disc.2019.111790 0012-365X/© 2019 Elsevier B.V. All rights reserved.
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M. Li and X. Zhang / Discrete Mathematics 343 (2020) 111790
A d-relaxed k-coloring, also known as a d-defectiv e k-coloring, of G is a k-coloring such that every vertex shares the same color with at most d neighbors. An early result of Lovász [14] says that: For any graph G, if d1 + d2 +· · ·+ dk = ∆(G) − k + 1, then there exists a k-coloring of V (G) such that ∆(G[Vi ]) ≤ di for every i ∈ [k], where Vi denotes the set of vertices colored with color i. The d-relaxed coloring of graphs have been widely studied (see [1,2,4–7,10,12,19]). In this paper, a 1-relaxed coloring is called a relaxed one for short. In some applications, ‘‘equitable" colorings are more important than the ‘‘conflict-free" ones. Also, it could be less costly to have an ‘‘equitable" coloring in which all color classes are not necessarily independent. Such a coloring is called a d-relaxed equitable coloring. Concretely, a d-relaxed equitable k-coloring of a graph G is a k-coloring of G, which is d-relaxed and the sizes of all color classes differ by at most one. When d = 1, it is called an RE-k-coloring for short. The relaxed ∗ equitable chromatic threshold of G, denoted by χre (G), is the smallest integer m such that G is RE-k-colorable for all k ≥ m. Williams, Vandenbussche and Yu [18] firstly studied RE-colorings of graphs and obtained the following result. ∗ Theorem 2 ([18]). A planar graph G with δ (G) ≥ 2 and g(G) ≥ 10 is RE-m-colorable for all m ≥ 3, that is χre (G) ≤ 3.
This result extends Theorem 1 on equitable colorings of planar graphs. Fan, Kierstead, Liu, Molla, Wu and Zhang [9] proved that a graph with maximum degree ∆ can have a d-relaxed equitable k-coloring for every k ≥ ∆ + 1 − d, which ∗ ∗ implies that χre (G) ≤ ∆. Molla [16] in his dissertation proved that every graph with maximum degree ∆ has χre (G) ≤ 56 ∆. Cummuang and Nakprasit [3] presented the d-relaxed equitable chromatic numbers of paths, cycles, complete graphs, hypercubes, stars, and wheels for any positive integer d. Nakprasit and Cummuang [17] gave the d-relaxed equitable chromatic numbers of complete bipartite graphs for d = 1, 2. ∗ The planar graph K2,n has girth 4 but χre (K2,n ) is not bounded by any constant. Williams, Vandenbussche and Yu [18] asked that: ∗ Question 1. What is the smallest girth g for which χre (G) can be bounded by a constant in planar graphs G? Is g = 5?
In this paper, our following result gives a step towards the question by extending the girth condition to g(G) ≥ 8. Theorem 3.
χre∗ (G) ≤ 4.
A planar graph G with minimum degree δ (G) ≥ 2 and girth g(G) ≥ 8 is RE-m-colorable for all m ≥ 4, that is,
Note that the condition δ (G) ≥ 2 above is necessary, because, for any fixed integer k, the graph K1,n is not RE-k-coloring for a sufficient large integer n. It is possible that the girth condition g(G) ≥ 8 may be relaxed or the condition m ≥ 4 may be replaced by m ≥ 3, but that should be difficult. For a graph G, we denote |V (G)| by n(G), the neighborhood of a vertex v in G by NG (v ). An m-coloring is ascending equitable if |V1 | ≤ |V2 | ≤ · · · ≤ |Vm | ≤ |V1 | + 1 and descending equitable if |V1 | ≥ |V2 | ≥ · · · ≥ |Vm | ≥ |V1 | − 1. Note that an ascending (or descending) equitable coloring may be non-relaxed. For convenience, let m (mod m) = m for any positive integer m. Let G be a counterexample to Theorem 3 with a smallest order. We construct an auxiliary bipartite graph B(H) = (V (H), [n(H)]) in such a way that u ∈ V (H) is adjacent to i ∈ [n(H)] if and only if the color i (mod m) is not used on NG (u) ∩ V (G − H) in the coloring c of G − H. We first show that B(H) exists a perfect matching by way of Hall’s Theorem, then modify it to obtain an RE-m-coloring of G. Our method is similar to the one in [18]. The proof uses a standard discharging argument, but the proofs for reducible structures employ the Hall’s matching theorem to avoid lengthy cases analysis. 2. The structure of minimal counterexamples Let G be a counterexample to Theorem 3 with smallest order. Before discussing the structure of G, we first clarify some necessary definitions and notations. The degree of a vertex v in G is denoted by dG (v ). (When there is no confusion, the subscript G is omitted usually.) A k-v ertex is a vertex of degree k, and a k+ -v ertex is a vertex whose degree is at least k. Let u1 , u2 , . . . , us (s ≥ 2) be distinct vertices and ui are colored with c(ui ), i ∈ [s]. A (u1 , u2 , . . . , us )-sw ap is an operation that recoloring ui+1 with c(ui ), for i ∈ [s − 1], and u1 with c(us ). Let t be a nonnegative integer. A t-thread in G is a path u0 , u1 , . . . , ut , ut +1 with d(u0 ), d(ut +1 ) ≥ 3 and d(ui ) = 2 for i ∈ [t ]. Two distinct vertices that are the endpoints of a thread are called pseudo-adjacent. For each 3+ -vertex u, we denote by t(u) the number of 2-vertices in its incident threads, and by ai (u) the number of incident i-threads. Lemma 4.
If G has a t-thread, then t ≤ 2.
Proof. Supposed that G has a t-thread u0 , u1 , . . . . ut , ut +1 with t ≥ 3. Let H = {u1 , u2 , . . . , ut −1 , ut }. Let m ≥ 4. Case 1. G − H has an RE-m-coloring c. We may assume that the coloring c is ascending equitable. Extend c to G by coloring ui with i (mod m) for every vertex in H. If c(u0 )̸ =c(u1 ) and c(ut )̸ =c(ut +1 ), then we obtain an RE-m-coloring of G. If c(u0 ) = c(u1 ) or c(ut ) = c(ut +1 ), we do a (u1 , u2 )-swap, or (ut −1 , ut )-swap accordingly. We can always make c(u0 )
M. Li and X. Zhang / Discrete Mathematics 343 (2020) 111790
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and c(ut +1 ) different from the colors of its adjacent vertices in H because t ≥ 3. In other words, we gain a resultful RE-m-coloring of G. (Especially, if t = 3, c(u0 ) = c(u1 ) and c(u3 ) = c(u4 ), do a (u1 , u3 )-swap; if u0 = u4 , then c(u1 )̸ =c(u3 ) by m ≥ 4. ) Case 2. G − H has no RE-m-coloring. Then by the minimality of G, δ (G − H) ≤ 1. That means u0 = ut +1 and d(u0 ) = 3. If G − H − {u0 } has minimum degree at least 2, we can extend an RE-m-coloring of graph G − H − {u0 } to G as above. Otherwise, let x be the pseudo-neighbor of u0 along a path from u0 in G − H. Let x, v1 , . . . , vt = ut be the path from x to ut containing every ui . Clearly, G − {v1 , . . . , vt } has minimum degree at least 2. Again by the minimality of G, the graph G − {v1 , . . . , vt } has an RE-m-coloring. We can also extend the coloring to G as before. In conclusion, we can obtain an RE-m-coloring of G. By contradiction, t ≤ 2. ■ Lemma 5.
The minimal counterexample G satisfies all of the following:
(P1) (P2) (P3) (P4) (P5) (P6)
Every 3-vertex u has t(u) ≤ 1 or t(u) = 2 with a1 (u) = 2. Every 4-vertex u has t(u) ≤ 4 with a2 (u) ≤ 1. Every 5-vertex u has a2 (u) ≤ 2. Every 6-vertex u has a2 (u) ≤ 4. Every 7-vertex u has a2 (u) ≤ 6. Let u be a 3-vertex with a1 (u) = 2, and p′ is adjacent to u by a 0-thread of u. Then: (i) d(p′ ) = 3 with t(p′ ) = 1, or (ii) d(p′ ) = 4 with t(p′ ) ≤ 1 or t(p′ ) = 2 with a2 (p′ ) = 0, or (iii) d(p′ ) ≥ 5. (P7) Let u be a 3-vertex with a1 (u) = 2, p′ is adjacent to u by a 0-thread of u and d(p′ ) = 3 with t(p′ ) = 1. Let p′′ is the endpoint of another 0-thread of p′ . Then: (i) d(p′′ ) = 4 with t(p′′ ) = 3 and a2 (p′′ ) = 0, or (ii) d(p′′ ) ≥ 5.
Proof. Consider the earliest property to fail in G. When (Pi) fails, i ∈ [5], let H1 be the graph induced by u and the 2-vertices in its incident threads. When (P6) fails, let H2 be the graph induced by u, p′ and the 2-vertices in their incident threads. When (P7) fails, let H3 be the graph induced by u, p′ , p′′ and the 2-vertices in their incident threads. The below are some basic facts on Hi , i ∈ [3]. (B1) n(H1 ) = t(u) + 1, 3 ≤ n(H1 ) ≤ 13 or n(H1 ) = 15; n(H2 ) = t(p′ ) + 4, 6 ≤ n(H2 ) ≤ 8 or n(H2 ) = 4; n(H3 ) = t(p′′ ) + 6, 6 ≤ n(H3 ) ≤ 10. (B2) For H2 , there is d(p′ ) ≤ 4. When d(p′ ) = 3, t(p′ ) = 0 or 2 by (P1). When d(p′ ) = 4, t(p′ ) = 2 and a2 (p′ ) = 1, or t(p′ ) = 3 or 4 by (P2). (B3) For H3 , there is d(p′′ ) ≤ 4. When d(p′′ ) = 3, t(p′′ ) ≤ 2 with a2 (u) = 0 by (P1). When d(p′′ ) = 4, t(p′′ ) ≤ 4 with a2 (p′′ ) ≤ 1 by (P2). In particular, a2 (p′′ ) = 1 when t(p′′ ) = 3. Let H ∈ {H1 , H2 , H3 }. Since g(G) ≥ 8 and the diameter of H is at most 5, each H is a tree and δ (G − H) ≥ 2. A vertex in H is called free if it has no neighbor in G − H. For a non-free vertex v , the neighbors of v in G − H are called outer neighbors of v . Observation 1. (1) The vertices in H except u (in H1 ), p′ (in H2 ) and p′′ (in H3 ) have no more than one outer neighbor. (2) u in H1 , p′ in H2 , p′′ in H3 have at most 2, 2, 3 outer neighbors, respectively. (3) u, p′ , p′′ is the only possible non-free vertex with degree more than 2 in H1 , H2 , H3 , respectively. Due to the minimality of G, the graph G − H has an RE-m-coloring for an arbitrary integer m ≥ 4. Let c : V (G − H)→ [m] be an ascending equitable RE-m-coloring. Claim. The coloring c can be extended to an ‘‘equitable" coloring (but not necessarily an RE-coloring), i.e. the sizes of all color classes differ by at most one, of G so that every non-free vertex in H gets a different color from its outer neighbors. Proof. Construct an auxiliary bipartite graph B(H) = (V (H), [n(H)]) so that u∈V (H) is adjacent to i∈[n(H)] if and only if the color i (mod m) is not used on its outer neighbors in the coloring c of G − H. Here, we denote the degree, the neighborhood, of a vertex v in B(H) (B(Hi )) by dB (v ), NB (v ) (dBi (v ), NBi (v )), respectively. For graph B(H), we obtain conclusions (F1 ) − (F3 ) as follows. (F1) By m ≥ 4 and Observation 1, each vertex v ∈ V (H) has degree dB (v ) ≥ n(H) − ⌈ 4 ⌉ except the following three n(H) n(H) n(H) vertices: dB1 (u) ≥ n(H) − 2⌈ 4 ⌉, dB2 (p′ ) ≥ n(H) − 2⌈ 4 ⌉ and dB3 (p′′ ) ≥ n(H) − 3⌈ 4 ⌉. Especially, there must be ′ ′′ dB1 (u) ≥ 1, dB2 (p ) ≥ 2 and dB3 (p ) ≥ 1 by n(H1 ) ≥ 3, n(H2 ) ≥ 4, n(H3 ) ≥ 6 (see (B1)) and Observation 1 (2). n(H) (F2) Let s(H) be the number of vertices that are not free in H. If s(H) ≤ n(H) − ⌈ 4 ⌉, then B(H) has a perfect matching. n(H)
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M. Li and X. Zhang / Discrete Mathematics 343 (2020) 111790
Table 1 n(H) The relationship between s(H) and n(H) − ⌈ 4 ⌉. n(H) s(H) n(H) n(H) − ⌈ 4 ⌉
3 2 2
4 3 3
5 3 3
6 3, 4 4
7 3, 4, 5 5
8 4, 5, 6 6
9 4, 5, 6 6
10 5, 6 7
11 5, 6 8
12 6 9
13 6 9
15 7 11
(F3) For a perfect matching M in B(H), we give a coloring c ′ of H by coloring v ∈ V (H) with i (mod m) when v i ∈ M. Then c ′ satisfies that (a) each non-free vertex in H gets a different color from its outer neighbors. (c) c ′ is descending equitable. Above conclusions are easy to verify except (F2). Now we give a proof for (F2). By Hall’s Theorem, if B(H) contains no perfect matching, then there exists a set S0 ⊂ V (H) such that |NB (S0 )| < |S0 |. Then S0 contains no free vertex, because each free vertex x in H has |NB (x)| = n(H). Furthermore, S0 contains at least one non-free 2-vertex of H. (Otherwise, by Observation 1, S0 contains the unique vertex: u for H1 , or p′ for H2 , or p′′ for H3 . n(H) By (F1), NBi (S0 ) ≥ |S0 | = 1, a contradiction.) By (F1), s(H) ≥ |S0 | > |NB (S0 )| ≥ n(H) − ⌈ 4 ⌉. n(H) Next, we show that s(H) ≤ n(H) − ⌈ 4 ⌉. Clearly, there are
{
d(u), d(u) − 1,
if u has at most one outer neighbor, if u has two outer neighbors.
{
d(p′ ) + 1, d(p′ ),
if p′ has at most one outer neighbor, if p′ has two outer neighbors.
{
d(p′′ ) + 2, d(p′′ ) + 1, d(p′′ ),
if p′′ has at most one outer neighbor, if p′′ has two outer neighbors, if p′′ has three outer neighbors.
s(H1 ) =
s(H2 ) =
s(H3 ) =
According to the above facts, we give a summary table (see Table 1). Note that, when n(H) = 3, there must be H = H1 and t = 2 by (B1). Actually, it is the case that (P1) fails, i.e. d(u) = 3 and a2 (u) = 1. This means that s(H) = 2 when n(H) = 3. By Table 1 and (F2), B(H) contains a perfect matching. Thus we can get a coloring c ′ of H satisfying (F3). Combining coloring c with coloring c ′ , we obtain a required coloring of G as described earlier. The proof is done. ■ The coloring c ′ of H described above is not relaxed only if the graph H contains a monochromatic subtree with order b ≥ 3. Such a subtree is called a bad tree. When H contains a bad tree with order b, there should be n(H)
n(H)
⌉ ≥ b ≥ 3. (1) 4 m Choose a perfect matching of B(H) that minimizes the maximum order of the monochromatic subtrees. Denote by L a largest bad tree in H. Then |L| = b ≥ 3. By Inequality (1) and 3 ≤ n(H) ≤ 15, there must be b ∈ {3, 4}. We consider the cases H = H1 , H2 , H3 respectively. Case 1. H = H1 . By Inequality (1) and n(H1 ) ≤ 15, b = 3 or 4. Observe that u must be in L. Hence L is the only bad tree in H1 . The color of the bad tree L is denoted by α . And when H1 has a bad tree, the order of H1 is at least 9 by m ≥ 4 and c ′ is descending equitable. So we can only consider the cases with 9 ≤ n(H1 ) ≤ 15, which implies that |Vi (H1 )| ≤ 4, i ∈ [m]. A 2-vertex in H1 is called an inside v ertex if it is adjacent to u, the other 2-vertices in H1 are called outside v ertices. Subcase 1.1. b = 3. First, we have the following: Observation 2. Since 9 ≤ n(H1 ) ≤ 15, there must be 8 ≤ t(u) ≤ 14 and then d(u) ∈ {4, 5, 6, 7}. Noting that t(u) ≥ 8, here a2 (u) ≥ 3 when (Pi) fails, 2 ≤ i ≤ 5. Also, for i ∈ [m], either 2 ≤ |Vi (H1 )| ≤ 3 or 3 ≤ |Vi (H1 )| ≤ 4. Next, we consider two cases. (I) u is not the center of the bad tree L. Since a2 (u) ≥ 3, u has two 2-threads whose 2-vertices do not belong to L, denoted by ux1 y1 z1 and ux2 y2 z2 . Since each |Vi | ≤ 4, at least one of the vertices y1 and y2 , say y1 , has a different color from α . Denote the center of L by x∗ . Since all inside vertices except x∗ are not colored with α , after an (x1 , x∗ )-swap, the bad tree L is eliminated and no new bad tree is created. (II) u is the center of the bad tree L. First, consider the case that u is free. Since d(u) ∈ {4, 5, 6, 7}, u has at most five inside vertices which do not belong to L. Also, by m ≥ 4, there must exist a color β which colors at most one inside vertex. If color β appears exactly on one inside vertex, say x, and x is in a 2-thread, then do an (x, u)-swap. Clearly, this operation eliminates the bad tree L and does not create a new bad tree. If x is in a 1-thread, denote its outer neighbor by z. If c(z) ̸ = α , then do an (x, u)-swap. If c(z) = α , noting that a2 (u) ≥ 3, there is a 2-thread ux∗ y∗ z ∗ whose 2-vertices do / {α, β}. Do an (x, u, x∗ )-swap. not belong to L. That is c(x∗ ) ∈ ⌈
⌉≥⌈
M. Li and X. Zhang / Discrete Mathematics 343 (2020) 111790
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Fig. 1. A bad tree L with order 4.
Fig. 2. (1)H3 with order 9. (2) H3 with order 10.
If color β does not appear on inside vertices, then β only appears on outside vertices. When the vertices colored with β are all adjacent to the vertices of L, the number of such vertices is exactly two, because c ′ is descending equitable. Denote the two vertices colored with β by y′ , y′′ . Since a2 (u) ≥ 3, there must exist a 2-thread ux1 y1 z1 whose 2-vertices do not belong to L and c(x1 ) ∈ / {α, β}. If the color of the outer neighbor of y′ (or y′′ ) is not α , do a (y′ , u)-swap (or (y′′ , u)-swap). Otherwise, do a (u, x1 , y′ )-swap. When one of the vertices colored with β , say y∗ , is not adjacent to L, we denote the 2-thread by ux∗ y∗ z ∗ . So c(x∗ ) ∈ / {α, β}. If c(z ∗ ) ̸= α , do a (u, y∗ )-swap. If c(z ∗ ) = α , do a (u, x∗ , y∗ )-swap. In either case, the bad tree L is eliminated and no new bad tree is created. Next, consider the case that u is not free. Let z be an outer neighbor of u. Note that u is not free only if d(u) = 5 with t(u) = 8, or d(u) = 6 with t(u) = 10 by Observation 2. So z is the unique outer neighbor of u. When d(u) = 6 (d(u) = 5), then u has three (two) inside vertices which do not belong to L. Also m ≥ 4, so there must exist two colors β and γ , which are different to α and color at most one inside vertex. If c(z) = β , consider the color γ and let xγ be a vertex colored with γ such that d(u, xr ) is smallest, where d(u, xr ) is the distance between u and xr . Clearly, 1 ≤ d(u, xr ) ≤ 2. If color γ appears exactly once on inside vertices, do a (u, xγ )-swap. Otherwise, color γ only appears on outside vertices. For an arbitrary inside vertex x∗ ̸ ∈ L, c(x∗ ) ∈ / {α, γ }. When the outer neighbor of xγ is not colored with α , do a (u, xγ )-swap. When the outer neighbor of xγ is colored with α , do a (u, x∗ , xγ )-swap. In either case, the bad tree L is eliminated and no new bad tree is created. Analogously, if c(z) = γ , consider the color β ; if c(z) ∈ / {β, γ }, consider β or γ . Subcase 1.2. b = 4. n(H ) Since ⌈ 4 1 ⌉ ≥ b = 4, n(H1 ) = 13, 15 and t(u) ≥ 12. That is d(u) = 6 with t(u) = 12, or d(u) = 7 with t(u) = 14. So u must be free. Furthermore, 3 ≤ |Vi (H1 )| ≤ 4, i ∈ [m], and then no other vertex colored with α in H1 except the ones in L. Let ux1 y1 z1 be a 2-thread whose 2-vertices do not belong to L. If L contains two 2-vertices in a 2-thread, denote the inside vertex in the 2-thread of L by x∗ . (See Fig. 1, in which the bad tree L is marked with thick lines and the dotted lines just appear in the case d(u) = 7.) Do an (x1 , x∗ )-swap. Since c(y1 ) ̸ = α , the operation eliminates the bad tree L and creates a new bad tree L′ with |L′ | = 3. This is a contradiction to the choice of largest bad tree in H. Now assume that L only contains u and three inside vertices. Denote the set of all inside vertices not in L by U. If a unique color, say β , appears on the vertices in U, then there exists an outer vertex, say y1 , such that c(y1 ) ∈ / {α, β} because |Vi (H1 )| ≤ 4, i ∈ [m]. If c(z1 ) ̸ = α , do a (u, y1 )-swap. If c(z1 ) = α , do a (u, x1 , y1 )-swap. In either case, the bad tree L is eliminated and no new bad tree is created. If at least two colors appear on the vertices of U, then there must exist a color, say β , which appears at most two times in U. W.l.o.g., we assume that c(x1 ) = β . Do a (u, x1 )-swap. This operation eliminates the bad tree L and may create a new bad tree. Furthermore, a new bad tree appears only if u is exactly adjacent to two inside vertices colored with β , x1 and x2 , and c(y2 ) = β , where ux2 y2 z2 is a 2-thread different to ux1 y1 z1 . Obviously, the new bad tree has order 3 in the case. This is a contradiction to the choice of largest bad tree in H. n(H ) n(H ) Case 2. H = H2 . Recall that n(H2 ) = 4, 6, 7, 8. Then ⌈ m2 ⌉ ≤ ⌈ 4 2 ⌉ ≤ 2. There is no bad tree in the case. n(H3 ) n(H3 ) Case 3. H = H3 . Recall that 6 ≤ n(H3 ) ≤ 10 and ⌈ m ⌉ ≤ ⌈ 4 ⌉ ≤ 3. So when H3 contains a bad tree, b = 3 and n(H3 ) ≥ 9. There are only two cases in which graph H3 contains a bad tree. (I) d(p′′ ) = 4 with t(p′′ ) = 3 and a2 (p′′ ) = 1. In this case n(H3 ) = 9. Assume the vertices in the bad tree L are colored with α . Then |Vα (H3 )| = 3 and |Vi (H3 )| = 2 for each i ̸ = α . Let w be a vertex that is adjacent to p′′ by a 2-thread. (See Fig. 2(1).) Let z be the center of L. Clearly z ∈ {u, p′ , p′′ , w} and u, p′ , w are free. If z = w , do a (w, u)-swap. If z ∈ {u, p′ }, do a (z , w )-swap. If z = p′′ , then L must contain at least one vertex in {w, p′ }. When p′ in L, do a (u, p′ )-swap. Otherwise, do a (u, w )-swap. Clearly, it will not create a new bad tree when we eliminate the original bad tree L.
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M. Li and X. Zhang / Discrete Mathematics 343 (2020) 111790
(II) d(p′′ ) = 4 with t(p′′ ) = 4. By (P2), p′′ has a2 (p′′ ) = 1. In this case, n(H3 ) = 10. Since c ′ is descending equitable and there is a bad tree with order 3, there must be m = 4. Then there are two colors, say α, β , either of which appears 3 times in L. If there is only one bad tree L, w.l.o.g. assume the vertices in L are colored with α . Let z be the center of L. Clearly z ∈ F = {u, p′ , p′′ , w} and every vertex in F is free. Then following the rules recoloring the vertices in H3 will not create a new bad tree while we eliminate the original bad tree L.
• z = w . Do a (w, u)-swap. • z = u. If p′ ∈ V (L), do a (p′ , p′′ )-swap; if p′ ∈ / V (L), do a (u, p′ )-swap. ′ • z =p. If u, p′′ ∈ V (L), do a (w, p′ )-swap; if p′′ ∈ / V (L), do a (p′ , p′′ )-swap; if u ∈ / V (L), do a (u, p′ )-swap. • z = p′′ . If p′ ∈ V (L), do a (u, p′ )-swap; if p′ ∈ / V (L), w ∈ V (L), do a (p′ , p′′ )-swap; if p′ , w ∈ / V (L) (see Fig. 2(2), in which the bad tree L is marked with thick lines.), do a (p′ , p′′ )-swap when all of ′ w, p , u are colored with β , otherwise, do a (t , p′′ )-swap, where vertex t ∈ {w, p′ , u} and c(t) ̸= β . If there are two bad trees L1 and L2 , then swap the colors of the centers of L1 and L2 . In any case, we can always obtain a descending equitable RE-coloring of H, a contradiction with the choice of the coloring c ′ of H. ■ 3. Discharging rules By Euler’s formula, a planar graph G with girth g satisfies mad(G) <
2g , g −2
in which
mad(G) = max 2e(H)/n(H). H ⊆G
Consider a minimal counterexample G to Theorem 3. Since g(G) ≥ 8, we have mad(G) < 38 , then v∈V (G) (d(v ) − 83 ) < 0. Let µ(v ) = d(v ) − 38 be the initial charge, and µ∗ (v ) be the final charge after distributing charge among vertices by the following rules:
∑
(R1) Every 3+ -vertex gives 13 to each 2-vertex in its incident threads. (R2) We call a 3-vertex u with a1 (u) = 2 as a bad v ertex. Every 3+ -vertex gives 13 to each adjacent bad vertex. (R3) Every 4+ -vertex gives 31 to each adjacent 3-vertex u, if t(u) = 1 and u is adjacent to a bad vertex. First, if d(v ) = 2, µ(v ) = 2 − 83 = − 32 . By (R1), the vertex v can receive 13 from each endpoint of the thread containing v and it will not give charge to other vertices. So µ∗ (v ) = 0. If d(v ) ≥ 5, the vertex v gives 31 · t(v ) to all 2-vertices by (R1) and gives at most 13 · a0 (v ) to all adjacent 3-vertices by (R2) and (R3). So the vertex v gives at most 31 [t(v ) + a0 (v )] = 31 [d(v ) + a2 (v )]. Then by (P3)-(P5) and a2 (v ) ≤ d(v ), 1 [d(v ) + a2 (v )] ≤ d(v ) − 83 . Therefore µ∗ (v ) ≥ d(v ) − 38 − 13 [d(v ) + a2 (v )] ≥ 0. 3 Next, if d(v ) = 3, µ(v ) = 3 − 83 = 13 . Then by above rules, v gives charge to each 2-vertex in its incident threads and bad vertices. By (P1), a 3-vertex v has t(v ) ≤ 1 or t(v ) = 2 with a1 (v ) = 2.
• If t(v ) = 0, v does not receive any charge by (R2) and (R3). Due to (P6), a 3-vertex v that is adjacent to a bad vertex must have t(v ) = 1. So v does not give charge to the other vertices by (R2). Then µ∗ (v ) = 3 − 38 = 31 > 0. • If t(v ) = 1, v may receive charge from adjacent 4+ -vertices by (R3) and v also gives charge to the adjacent 2-vertex and bad vertices. If v does not adjacent to any bad vertex, then µ∗ (v ) = 3 − 83 − 13 = 0. If v is adjacent to bad vertices, then the number of bad vertices is one, because, if v is adjacent to a bad vertex, another 3+ -neighbor of v must have degree at least four by (P7). In this case, v can receive 13 from this 4+ -vertex by (R3). So µ∗ (v ) = 3− 38 −( 13 + 13 )+ 31 = 0. • If t(v ) = 2 with a1 (v ) = 2, v receives 31 from the adjacent 3+ -vertex. And there is no bad vertices that are adjacent to v by (P6)(i). So v only gives 13 to each adjacent 2-vertex. In other words, µ∗ (v ) = 3 − 38 − 23 + 31 = 0. Finally, if d(v ) = 4, then t(v ) ≤ 4 and a2 (v ) ≤ 1 by (P2). Also, by (P6)(ii) and (P7)(i), a 4-vertex that gives charge to 3-vertex must have a2 (v ) = 0. Thus v gives charge at most 34 . Then µ∗ (v ) ≥ 4 − 38∑ − 43 ≥ 0. ∑ ∗ ∗ In conclusion, for a vertex in G, we always have µ (v ) ≥ 0. Therefore 0 ≤ v∈V (G) µ (v ) = v∈V (G) µ(v ) < 0, a contradiction. Declaration of competing interest The authors declare that there is no conflict of interest in this paper.
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Acknowledgments The authors would like to thank the referees for careful reading and valuable comments. References [1] L. Cowen, R. Cowen, D. Woodall, Defective colorings of graphs in surfaces: Partitions into subgraphs of bounded valency, J. Graph Theory 10 (2) (1986) 187–195. [2] L. Cowen, W. Goddard, C. Jesurum, Defective coloring revisited, J. Graph Theory 24 (3) (1997) 205–219. [3] S. Cummuang, K. Nakprasit, Equitable defective colorings of some classes of graphs, KKU Sci. J. 39 (1) (2011) 687–694. [4] W. Cusing, H. Kierstead, Planar graphs are 1-relaxed, 4-choosable, European J. Combin. 31 (2010) 1385–1397. [5] W. Deuber, X. Zhu, Relaxed coloring of a graph, Graphs Combin. 14 (1998) 121–130 (English summary). [6] P. Dorbec, T. Kaiser, M. Montassier, A. Raspaud, Limits of near-coloring of sparse graphs, J. Graph Theory 75 (2) (2014) 191–202. [7] N. Eaton, T. Hull, Defective list colorings of planar graphs, Bull. Inst. Combin. Appl. 25 (1999) 79–87. [8] P. Erdős, Theory of Graphs and Its Applications, 159, Czech. Acad. Sci. Publ., Prague, 1964, Problem 9. [9] H. Fan, H.A. Kierstead, G. Liu, T. Molla, J. Wu, X. Zhang, A note on relaxed equitable coloring of graphs, Inform. Process. Lett. 111 (2011) 1062–1066. [10] M. Frick, F. Henning, Extremal results on defective colorings of graphs, Discrete Math. 126 (1994) 151–158. [11] A. Hajnal, E. Szemerédi, Proof of a conjecture of P. Erdős, in: P. Erdős, A. Rényi, V.T. Sós (Eds.), Combinatorial Theory and Its Applications, in: Colloquia Mathematica Societatis János Bolyai, North-Holland, Amsterdam, 1970, pp. 601–623. [12] J. Kim, A. Kostochka, X. Zhu, Improper coloring of sparse graphs with a given girth, I: (0, 1)-colorings of triangle-free graphs, European J. Combin. 42 (2014) 26–48. [13] K. Lih, Equitable coloring of graphs, in: P.M. Pardalos, D.Z. Du, R.L. Graham (Eds.), Handbook of Combinatorial Optimization, second ed., Springer, New York, 2013, pp. 1199–1248. [14] L. Lovász, On decomposition of graphs, Stud. Sci. Math. Hung. 1 (1966) 237–238. [15] R. Luo, J. Sereni, C. Stephens, G. Yu, Equiable coloring sparse planar graphs, SIAM J. Discrete Math. 24 (4) (2010) 1572–1583. [16] T. Molla, Equitable and relaxed graph coloring (Dissertation Prospectus), 2011. [17] K. Nakprasit, S. Cummuang, Equitable defective colorings of complete bipartite graphs, Int. J. Pure Appl. Math. 92 (1) (2014) 73–86. [18] L. Williams, J. Vandenbussche, G. Yu, Equitable defective coloring of sparse planar graphs, Discrete Math. 312 (5) (2012) 957–962. [19] B. Xu, On (3,1)∗ -coloring of plane graphs, SIAM J. Discrete Math. 23 (2008) 205–220. [20] Y. Zhang, H. Yap, Equitable colorings of planar graphs, J. Combin. Math. Combin. Comput. 27 (1998) 97–105.
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Discrete Mathematics journal homepage: www.elsevier.com/locate/disc
Relaxed equitable colorings of planar graphs with girth at least 8✩ Ming Li, Xia Zhang
∗
School of Mathematics and Statistics, Shandong Normal University, Jinan 250358, PR China
article
info
Article history: Received 9 May 2018 Received in revised form 8 December 2019 Accepted 17 December 2019 Available online xxxx Keywords: Equitable coloring Relaxed coloring Discharging Planar graph
a b s t r a c t An RE-m-coloring of a graph G is a vertex m-coloring of G, which is relaxed (every vertex shares the same color with at most one neighbor) and equitable (the sizes of all color classes differ by at most one). In this article, we prove that every planar graph with minimum degree at least 2 and girth at least 8 has an RE-m-coloring for each integer m ≥ 4. We use the discharging method and Hall’s Theorem to simply the structures of counterexamples. © 2019 Elsevier B.V. All rights reserved.
1. Introduction Given a graph G = (V , E), a k-coloring of G is a mapping f : V (G) → [k] and a k-coloring of G is proper if f (u) ̸ = f (v ) for each uv ∈ E(G), where [k] = {1, 2, . . . , k}. That is to say, each subset of vertices colored with a same color is an independent (or conflict-free) set. A proper coloring is equitable if the sizes of all color classes differ by at most one. The equitable chromatic number of G, denoted by χeq (G), is the smallest integer k such that G is equitable k-colorable. The ∗ equitable chromatic threshold of G, denoted by χeq (G), is the smallest integer m such that G is equitable k-colorable for all ∗ k ≥ m. Note that χeq (G) ≤ χeq (G). In fact, the gap between these two parameters can be arbitrarily large. For example, ∗ χeq (K2t +1,2t +1 ) = 2 while χeq (K2t +1,2t +1 ) = 2t + 2 for each t ≥ 1. The study of equitable colorings began with a conjecture due to Erdős [8] which was proved by Hajnal and Szemerédi [11]: Every graph G with maximum degree at most ∆ has an equitable k-coloring for every k ≥ ∆ + 1. It is an NP-hard problem to decide χeq (G) even for planar graphs. There are a series of results on equitable colorings of graphs (see [13]). Zhang and Yap [20] proved that a planar graph is equitably m-colorable if m ≥ ∆ ≥ 13. Luo, Sereni, Stephens and Yu [15] considered equitable colorings for planar graphs with some restrictions on girth (the girth of a graph G is the minimum length of the cycles in G and denoted by g(G)). Theorem 1 ([15]). Let G be a planar graph with δ (G) ≥ 2. ∗ (a) If g(G) ≥ 14, then χeq (G) ≤ 3. ∗ (b) If g(G) ≥ 10, then χeq (G) ≤ 4.
✩ This work is supported by the Shandong Provincial Natural Science Foundation, China (Grant No. ZR2019MA032), the National Natural Science Foundation of China (Grant No. 11701342) of China. ∗ Corresponding author. E-mail addresses: [email protected], [email protected] (X. Zhang). https://doi.org/10.1016/j.disc.2019.111790 0012-365X/© 2019 Elsevier B.V. All rights reserved.
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M. Li and X. Zhang / Discrete Mathematics 343 (2020) 111790
A d-relaxed k-coloring, also known as a d-defectiv e k-coloring, of G is a k-coloring such that every vertex shares the same color with at most d neighbors. An early result of Lovász [14] says that: For any graph G, if d1 + d2 +· · ·+ dk = ∆(G) − k + 1, then there exists a k-coloring of V (G) such that ∆(G[Vi ]) ≤ di for every i ∈ [k], where Vi denotes the set of vertices colored with color i. The d-relaxed coloring of graphs have been widely studied (see [1,2,4–7,10,12,19]). In this paper, a 1-relaxed coloring is called a relaxed one for short. In some applications, ‘‘equitable" colorings are more important than the ‘‘conflict-free" ones. Also, it could be less costly to have an ‘‘equitable" coloring in which all color classes are not necessarily independent. Such a coloring is called a d-relaxed equitable coloring. Concretely, a d-relaxed equitable k-coloring of a graph G is a k-coloring of G, which is d-relaxed and the sizes of all color classes differ by at most one. When d = 1, it is called an RE-k-coloring for short. The relaxed ∗ equitable chromatic threshold of G, denoted by χre (G), is the smallest integer m such that G is RE-k-colorable for all k ≥ m. Williams, Vandenbussche and Yu [18] firstly studied RE-colorings of graphs and obtained the following result. ∗ Theorem 2 ([18]). A planar graph G with δ (G) ≥ 2 and g(G) ≥ 10 is RE-m-colorable for all m ≥ 3, that is χre (G) ≤ 3.
This result extends Theorem 1 on equitable colorings of planar graphs. Fan, Kierstead, Liu, Molla, Wu and Zhang [9] proved that a graph with maximum degree ∆ can have a d-relaxed equitable k-coloring for every k ≥ ∆ + 1 − d, which ∗ ∗ implies that χre (G) ≤ ∆. Molla [16] in his dissertation proved that every graph with maximum degree ∆ has χre (G) ≤ 56 ∆. Cummuang and Nakprasit [3] presented the d-relaxed equitable chromatic numbers of paths, cycles, complete graphs, hypercubes, stars, and wheels for any positive integer d. Nakprasit and Cummuang [17] gave the d-relaxed equitable chromatic numbers of complete bipartite graphs for d = 1, 2. ∗ The planar graph K2,n has girth 4 but χre (K2,n ) is not bounded by any constant. Williams, Vandenbussche and Yu [18] asked that: ∗ Question 1. What is the smallest girth g for which χre (G) can be bounded by a constant in planar graphs G? Is g = 5?
In this paper, our following result gives a step towards the question by extending the girth condition to g(G) ≥ 8. Theorem 3.
χre∗ (G) ≤ 4.
A planar graph G with minimum degree δ (G) ≥ 2 and girth g(G) ≥ 8 is RE-m-colorable for all m ≥ 4, that is,
Note that the condition δ (G) ≥ 2 above is necessary, because, for any fixed integer k, the graph K1,n is not RE-k-coloring for a sufficient large integer n. It is possible that the girth condition g(G) ≥ 8 may be relaxed or the condition m ≥ 4 may be replaced by m ≥ 3, but that should be difficult. For a graph G, we denote |V (G)| by n(G), the neighborhood of a vertex v in G by NG (v ). An m-coloring is ascending equitable if |V1 | ≤ |V2 | ≤ · · · ≤ |Vm | ≤ |V1 | + 1 and descending equitable if |V1 | ≥ |V2 | ≥ · · · ≥ |Vm | ≥ |V1 | − 1. Note that an ascending (or descending) equitable coloring may be non-relaxed. For convenience, let m (mod m) = m for any positive integer m. Let G be a counterexample to Theorem 3 with a smallest order. We construct an auxiliary bipartite graph B(H) = (V (H), [n(H)]) in such a way that u ∈ V (H) is adjacent to i ∈ [n(H)] if and only if the color i (mod m) is not used on NG (u) ∩ V (G − H) in the coloring c of G − H. We first show that B(H) exists a perfect matching by way of Hall’s Theorem, then modify it to obtain an RE-m-coloring of G. Our method is similar to the one in [18]. The proof uses a standard discharging argument, but the proofs for reducible structures employ the Hall’s matching theorem to avoid lengthy cases analysis. 2. The structure of minimal counterexamples Let G be a counterexample to Theorem 3 with smallest order. Before discussing the structure of G, we first clarify some necessary definitions and notations. The degree of a vertex v in G is denoted by dG (v ). (When there is no confusion, the subscript G is omitted usually.) A k-v ertex is a vertex of degree k, and a k+ -v ertex is a vertex whose degree is at least k. Let u1 , u2 , . . . , us (s ≥ 2) be distinct vertices and ui are colored with c(ui ), i ∈ [s]. A (u1 , u2 , . . . , us )-sw ap is an operation that recoloring ui+1 with c(ui ), for i ∈ [s − 1], and u1 with c(us ). Let t be a nonnegative integer. A t-thread in G is a path u0 , u1 , . . . , ut , ut +1 with d(u0 ), d(ut +1 ) ≥ 3 and d(ui ) = 2 for i ∈ [t ]. Two distinct vertices that are the endpoints of a thread are called pseudo-adjacent. For each 3+ -vertex u, we denote by t(u) the number of 2-vertices in its incident threads, and by ai (u) the number of incident i-threads. Lemma 4.
If G has a t-thread, then t ≤ 2.
Proof. Supposed that G has a t-thread u0 , u1 , . . . . ut , ut +1 with t ≥ 3. Let H = {u1 , u2 , . . . , ut −1 , ut }. Let m ≥ 4. Case 1. G − H has an RE-m-coloring c. We may assume that the coloring c is ascending equitable. Extend c to G by coloring ui with i (mod m) for every vertex in H. If c(u0 )̸ =c(u1 ) and c(ut )̸ =c(ut +1 ), then we obtain an RE-m-coloring of G. If c(u0 ) = c(u1 ) or c(ut ) = c(ut +1 ), we do a (u1 , u2 )-swap, or (ut −1 , ut )-swap accordingly. We can always make c(u0 )
M. Li and X. Zhang / Discrete Mathematics 343 (2020) 111790
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and c(ut +1 ) different from the colors of its adjacent vertices in H because t ≥ 3. In other words, we gain a resultful RE-m-coloring of G. (Especially, if t = 3, c(u0 ) = c(u1 ) and c(u3 ) = c(u4 ), do a (u1 , u3 )-swap; if u0 = u4 , then c(u1 )̸ =c(u3 ) by m ≥ 4. ) Case 2. G − H has no RE-m-coloring. Then by the minimality of G, δ (G − H) ≤ 1. That means u0 = ut +1 and d(u0 ) = 3. If G − H − {u0 } has minimum degree at least 2, we can extend an RE-m-coloring of graph G − H − {u0 } to G as above. Otherwise, let x be the pseudo-neighbor of u0 along a path from u0 in G − H. Let x, v1 , . . . , vt = ut be the path from x to ut containing every ui . Clearly, G − {v1 , . . . , vt } has minimum degree at least 2. Again by the minimality of G, the graph G − {v1 , . . . , vt } has an RE-m-coloring. We can also extend the coloring to G as before. In conclusion, we can obtain an RE-m-coloring of G. By contradiction, t ≤ 2. ■ Lemma 5.
The minimal counterexample G satisfies all of the following:
(P1) (P2) (P3) (P4) (P5) (P6)
Every 3-vertex u has t(u) ≤ 1 or t(u) = 2 with a1 (u) = 2. Every 4-vertex u has t(u) ≤ 4 with a2 (u) ≤ 1. Every 5-vertex u has a2 (u) ≤ 2. Every 6-vertex u has a2 (u) ≤ 4. Every 7-vertex u has a2 (u) ≤ 6. Let u be a 3-vertex with a1 (u) = 2, and p′ is adjacent to u by a 0-thread of u. Then: (i) d(p′ ) = 3 with t(p′ ) = 1, or (ii) d(p′ ) = 4 with t(p′ ) ≤ 1 or t(p′ ) = 2 with a2 (p′ ) = 0, or (iii) d(p′ ) ≥ 5. (P7) Let u be a 3-vertex with a1 (u) = 2, p′ is adjacent to u by a 0-thread of u and d(p′ ) = 3 with t(p′ ) = 1. Let p′′ is the endpoint of another 0-thread of p′ . Then: (i) d(p′′ ) = 4 with t(p′′ ) = 3 and a2 (p′′ ) = 0, or (ii) d(p′′ ) ≥ 5.
Proof. Consider the earliest property to fail in G. When (Pi) fails, i ∈ [5], let H1 be the graph induced by u and the 2-vertices in its incident threads. When (P6) fails, let H2 be the graph induced by u, p′ and the 2-vertices in their incident threads. When (P7) fails, let H3 be the graph induced by u, p′ , p′′ and the 2-vertices in their incident threads. The below are some basic facts on Hi , i ∈ [3]. (B1) n(H1 ) = t(u) + 1, 3 ≤ n(H1 ) ≤ 13 or n(H1 ) = 15; n(H2 ) = t(p′ ) + 4, 6 ≤ n(H2 ) ≤ 8 or n(H2 ) = 4; n(H3 ) = t(p′′ ) + 6, 6 ≤ n(H3 ) ≤ 10. (B2) For H2 , there is d(p′ ) ≤ 4. When d(p′ ) = 3, t(p′ ) = 0 or 2 by (P1). When d(p′ ) = 4, t(p′ ) = 2 and a2 (p′ ) = 1, or t(p′ ) = 3 or 4 by (P2). (B3) For H3 , there is d(p′′ ) ≤ 4. When d(p′′ ) = 3, t(p′′ ) ≤ 2 with a2 (u) = 0 by (P1). When d(p′′ ) = 4, t(p′′ ) ≤ 4 with a2 (p′′ ) ≤ 1 by (P2). In particular, a2 (p′′ ) = 1 when t(p′′ ) = 3. Let H ∈ {H1 , H2 , H3 }. Since g(G) ≥ 8 and the diameter of H is at most 5, each H is a tree and δ (G − H) ≥ 2. A vertex in H is called free if it has no neighbor in G − H. For a non-free vertex v , the neighbors of v in G − H are called outer neighbors of v . Observation 1. (1) The vertices in H except u (in H1 ), p′ (in H2 ) and p′′ (in H3 ) have no more than one outer neighbor. (2) u in H1 , p′ in H2 , p′′ in H3 have at most 2, 2, 3 outer neighbors, respectively. (3) u, p′ , p′′ is the only possible non-free vertex with degree more than 2 in H1 , H2 , H3 , respectively. Due to the minimality of G, the graph G − H has an RE-m-coloring for an arbitrary integer m ≥ 4. Let c : V (G − H)→ [m] be an ascending equitable RE-m-coloring. Claim. The coloring c can be extended to an ‘‘equitable" coloring (but not necessarily an RE-coloring), i.e. the sizes of all color classes differ by at most one, of G so that every non-free vertex in H gets a different color from its outer neighbors. Proof. Construct an auxiliary bipartite graph B(H) = (V (H), [n(H)]) so that u∈V (H) is adjacent to i∈[n(H)] if and only if the color i (mod m) is not used on its outer neighbors in the coloring c of G − H. Here, we denote the degree, the neighborhood, of a vertex v in B(H) (B(Hi )) by dB (v ), NB (v ) (dBi (v ), NBi (v )), respectively. For graph B(H), we obtain conclusions (F1 ) − (F3 ) as follows. (F1) By m ≥ 4 and Observation 1, each vertex v ∈ V (H) has degree dB (v ) ≥ n(H) − ⌈ 4 ⌉ except the following three n(H) n(H) n(H) vertices: dB1 (u) ≥ n(H) − 2⌈ 4 ⌉, dB2 (p′ ) ≥ n(H) − 2⌈ 4 ⌉ and dB3 (p′′ ) ≥ n(H) − 3⌈ 4 ⌉. Especially, there must be ′ ′′ dB1 (u) ≥ 1, dB2 (p ) ≥ 2 and dB3 (p ) ≥ 1 by n(H1 ) ≥ 3, n(H2 ) ≥ 4, n(H3 ) ≥ 6 (see (B1)) and Observation 1 (2). n(H) (F2) Let s(H) be the number of vertices that are not free in H. If s(H) ≤ n(H) − ⌈ 4 ⌉, then B(H) has a perfect matching. n(H)
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M. Li and X. Zhang / Discrete Mathematics 343 (2020) 111790
Table 1 n(H) The relationship between s(H) and n(H) − ⌈ 4 ⌉. n(H) s(H) n(H) n(H) − ⌈ 4 ⌉
3 2 2
4 3 3
5 3 3
6 3, 4 4
7 3, 4, 5 5
8 4, 5, 6 6
9 4, 5, 6 6
10 5, 6 7
11 5, 6 8
12 6 9
13 6 9
15 7 11
(F3) For a perfect matching M in B(H), we give a coloring c ′ of H by coloring v ∈ V (H) with i (mod m) when v i ∈ M. Then c ′ satisfies that (a) each non-free vertex in H gets a different color from its outer neighbors. (c) c ′ is descending equitable. Above conclusions are easy to verify except (F2). Now we give a proof for (F2). By Hall’s Theorem, if B(H) contains no perfect matching, then there exists a set S0 ⊂ V (H) such that |NB (S0 )| < |S0 |. Then S0 contains no free vertex, because each free vertex x in H has |NB (x)| = n(H). Furthermore, S0 contains at least one non-free 2-vertex of H. (Otherwise, by Observation 1, S0 contains the unique vertex: u for H1 , or p′ for H2 , or p′′ for H3 . n(H) By (F1), NBi (S0 ) ≥ |S0 | = 1, a contradiction.) By (F1), s(H) ≥ |S0 | > |NB (S0 )| ≥ n(H) − ⌈ 4 ⌉. n(H) Next, we show that s(H) ≤ n(H) − ⌈ 4 ⌉. Clearly, there are
{
d(u), d(u) − 1,
if u has at most one outer neighbor, if u has two outer neighbors.
{
d(p′ ) + 1, d(p′ ),
if p′ has at most one outer neighbor, if p′ has two outer neighbors.
{
d(p′′ ) + 2, d(p′′ ) + 1, d(p′′ ),
if p′′ has at most one outer neighbor, if p′′ has two outer neighbors, if p′′ has three outer neighbors.
s(H1 ) =
s(H2 ) =
s(H3 ) =
According to the above facts, we give a summary table (see Table 1). Note that, when n(H) = 3, there must be H = H1 and t = 2 by (B1). Actually, it is the case that (P1) fails, i.e. d(u) = 3 and a2 (u) = 1. This means that s(H) = 2 when n(H) = 3. By Table 1 and (F2), B(H) contains a perfect matching. Thus we can get a coloring c ′ of H satisfying (F3). Combining coloring c with coloring c ′ , we obtain a required coloring of G as described earlier. The proof is done. ■ The coloring c ′ of H described above is not relaxed only if the graph H contains a monochromatic subtree with order b ≥ 3. Such a subtree is called a bad tree. When H contains a bad tree with order b, there should be n(H)
n(H)
⌉ ≥ b ≥ 3. (1) 4 m Choose a perfect matching of B(H) that minimizes the maximum order of the monochromatic subtrees. Denote by L a largest bad tree in H. Then |L| = b ≥ 3. By Inequality (1) and 3 ≤ n(H) ≤ 15, there must be b ∈ {3, 4}. We consider the cases H = H1 , H2 , H3 respectively. Case 1. H = H1 . By Inequality (1) and n(H1 ) ≤ 15, b = 3 or 4. Observe that u must be in L. Hence L is the only bad tree in H1 . The color of the bad tree L is denoted by α . And when H1 has a bad tree, the order of H1 is at least 9 by m ≥ 4 and c ′ is descending equitable. So we can only consider the cases with 9 ≤ n(H1 ) ≤ 15, which implies that |Vi (H1 )| ≤ 4, i ∈ [m]. A 2-vertex in H1 is called an inside v ertex if it is adjacent to u, the other 2-vertices in H1 are called outside v ertices. Subcase 1.1. b = 3. First, we have the following: Observation 2. Since 9 ≤ n(H1 ) ≤ 15, there must be 8 ≤ t(u) ≤ 14 and then d(u) ∈ {4, 5, 6, 7}. Noting that t(u) ≥ 8, here a2 (u) ≥ 3 when (Pi) fails, 2 ≤ i ≤ 5. Also, for i ∈ [m], either 2 ≤ |Vi (H1 )| ≤ 3 or 3 ≤ |Vi (H1 )| ≤ 4. Next, we consider two cases. (I) u is not the center of the bad tree L. Since a2 (u) ≥ 3, u has two 2-threads whose 2-vertices do not belong to L, denoted by ux1 y1 z1 and ux2 y2 z2 . Since each |Vi | ≤ 4, at least one of the vertices y1 and y2 , say y1 , has a different color from α . Denote the center of L by x∗ . Since all inside vertices except x∗ are not colored with α , after an (x1 , x∗ )-swap, the bad tree L is eliminated and no new bad tree is created. (II) u is the center of the bad tree L. First, consider the case that u is free. Since d(u) ∈ {4, 5, 6, 7}, u has at most five inside vertices which do not belong to L. Also, by m ≥ 4, there must exist a color β which colors at most one inside vertex. If color β appears exactly on one inside vertex, say x, and x is in a 2-thread, then do an (x, u)-swap. Clearly, this operation eliminates the bad tree L and does not create a new bad tree. If x is in a 1-thread, denote its outer neighbor by z. If c(z) ̸ = α , then do an (x, u)-swap. If c(z) = α , noting that a2 (u) ≥ 3, there is a 2-thread ux∗ y∗ z ∗ whose 2-vertices do / {α, β}. Do an (x, u, x∗ )-swap. not belong to L. That is c(x∗ ) ∈ ⌈
⌉≥⌈
M. Li and X. Zhang / Discrete Mathematics 343 (2020) 111790
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Fig. 1. A bad tree L with order 4.
Fig. 2. (1)H3 with order 9. (2) H3 with order 10.
If color β does not appear on inside vertices, then β only appears on outside vertices. When the vertices colored with β are all adjacent to the vertices of L, the number of such vertices is exactly two, because c ′ is descending equitable. Denote the two vertices colored with β by y′ , y′′ . Since a2 (u) ≥ 3, there must exist a 2-thread ux1 y1 z1 whose 2-vertices do not belong to L and c(x1 ) ∈ / {α, β}. If the color of the outer neighbor of y′ (or y′′ ) is not α , do a (y′ , u)-swap (or (y′′ , u)-swap). Otherwise, do a (u, x1 , y′ )-swap. When one of the vertices colored with β , say y∗ , is not adjacent to L, we denote the 2-thread by ux∗ y∗ z ∗ . So c(x∗ ) ∈ / {α, β}. If c(z ∗ ) ̸= α , do a (u, y∗ )-swap. If c(z ∗ ) = α , do a (u, x∗ , y∗ )-swap. In either case, the bad tree L is eliminated and no new bad tree is created. Next, consider the case that u is not free. Let z be an outer neighbor of u. Note that u is not free only if d(u) = 5 with t(u) = 8, or d(u) = 6 with t(u) = 10 by Observation 2. So z is the unique outer neighbor of u. When d(u) = 6 (d(u) = 5), then u has three (two) inside vertices which do not belong to L. Also m ≥ 4, so there must exist two colors β and γ , which are different to α and color at most one inside vertex. If c(z) = β , consider the color γ and let xγ be a vertex colored with γ such that d(u, xr ) is smallest, where d(u, xr ) is the distance between u and xr . Clearly, 1 ≤ d(u, xr ) ≤ 2. If color γ appears exactly once on inside vertices, do a (u, xγ )-swap. Otherwise, color γ only appears on outside vertices. For an arbitrary inside vertex x∗ ̸ ∈ L, c(x∗ ) ∈ / {α, γ }. When the outer neighbor of xγ is not colored with α , do a (u, xγ )-swap. When the outer neighbor of xγ is colored with α , do a (u, x∗ , xγ )-swap. In either case, the bad tree L is eliminated and no new bad tree is created. Analogously, if c(z) = γ , consider the color β ; if c(z) ∈ / {β, γ }, consider β or γ . Subcase 1.2. b = 4. n(H ) Since ⌈ 4 1 ⌉ ≥ b = 4, n(H1 ) = 13, 15 and t(u) ≥ 12. That is d(u) = 6 with t(u) = 12, or d(u) = 7 with t(u) = 14. So u must be free. Furthermore, 3 ≤ |Vi (H1 )| ≤ 4, i ∈ [m], and then no other vertex colored with α in H1 except the ones in L. Let ux1 y1 z1 be a 2-thread whose 2-vertices do not belong to L. If L contains two 2-vertices in a 2-thread, denote the inside vertex in the 2-thread of L by x∗ . (See Fig. 1, in which the bad tree L is marked with thick lines and the dotted lines just appear in the case d(u) = 7.) Do an (x1 , x∗ )-swap. Since c(y1 ) ̸ = α , the operation eliminates the bad tree L and creates a new bad tree L′ with |L′ | = 3. This is a contradiction to the choice of largest bad tree in H. Now assume that L only contains u and three inside vertices. Denote the set of all inside vertices not in L by U. If a unique color, say β , appears on the vertices in U, then there exists an outer vertex, say y1 , such that c(y1 ) ∈ / {α, β} because |Vi (H1 )| ≤ 4, i ∈ [m]. If c(z1 ) ̸ = α , do a (u, y1 )-swap. If c(z1 ) = α , do a (u, x1 , y1 )-swap. In either case, the bad tree L is eliminated and no new bad tree is created. If at least two colors appear on the vertices of U, then there must exist a color, say β , which appears at most two times in U. W.l.o.g., we assume that c(x1 ) = β . Do a (u, x1 )-swap. This operation eliminates the bad tree L and may create a new bad tree. Furthermore, a new bad tree appears only if u is exactly adjacent to two inside vertices colored with β , x1 and x2 , and c(y2 ) = β , where ux2 y2 z2 is a 2-thread different to ux1 y1 z1 . Obviously, the new bad tree has order 3 in the case. This is a contradiction to the choice of largest bad tree in H. n(H ) n(H ) Case 2. H = H2 . Recall that n(H2 ) = 4, 6, 7, 8. Then ⌈ m2 ⌉ ≤ ⌈ 4 2 ⌉ ≤ 2. There is no bad tree in the case. n(H3 ) n(H3 ) Case 3. H = H3 . Recall that 6 ≤ n(H3 ) ≤ 10 and ⌈ m ⌉ ≤ ⌈ 4 ⌉ ≤ 3. So when H3 contains a bad tree, b = 3 and n(H3 ) ≥ 9. There are only two cases in which graph H3 contains a bad tree. (I) d(p′′ ) = 4 with t(p′′ ) = 3 and a2 (p′′ ) = 1. In this case n(H3 ) = 9. Assume the vertices in the bad tree L are colored with α . Then |Vα (H3 )| = 3 and |Vi (H3 )| = 2 for each i ̸ = α . Let w be a vertex that is adjacent to p′′ by a 2-thread. (See Fig. 2(1).) Let z be the center of L. Clearly z ∈ {u, p′ , p′′ , w} and u, p′ , w are free. If z = w , do a (w, u)-swap. If z ∈ {u, p′ }, do a (z , w )-swap. If z = p′′ , then L must contain at least one vertex in {w, p′ }. When p′ in L, do a (u, p′ )-swap. Otherwise, do a (u, w )-swap. Clearly, it will not create a new bad tree when we eliminate the original bad tree L.
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M. Li and X. Zhang / Discrete Mathematics 343 (2020) 111790
(II) d(p′′ ) = 4 with t(p′′ ) = 4. By (P2), p′′ has a2 (p′′ ) = 1. In this case, n(H3 ) = 10. Since c ′ is descending equitable and there is a bad tree with order 3, there must be m = 4. Then there are two colors, say α, β , either of which appears 3 times in L. If there is only one bad tree L, w.l.o.g. assume the vertices in L are colored with α . Let z be the center of L. Clearly z ∈ F = {u, p′ , p′′ , w} and every vertex in F is free. Then following the rules recoloring the vertices in H3 will not create a new bad tree while we eliminate the original bad tree L.
• z = w . Do a (w, u)-swap. • z = u. If p′ ∈ V (L), do a (p′ , p′′ )-swap; if p′ ∈ / V (L), do a (u, p′ )-swap. ′ • z =p. If u, p′′ ∈ V (L), do a (w, p′ )-swap; if p′′ ∈ / V (L), do a (p′ , p′′ )-swap; if u ∈ / V (L), do a (u, p′ )-swap. • z = p′′ . If p′ ∈ V (L), do a (u, p′ )-swap; if p′ ∈ / V (L), w ∈ V (L), do a (p′ , p′′ )-swap; if p′ , w ∈ / V (L) (see Fig. 2(2), in which the bad tree L is marked with thick lines.), do a (p′ , p′′ )-swap when all of ′ w, p , u are colored with β , otherwise, do a (t , p′′ )-swap, where vertex t ∈ {w, p′ , u} and c(t) ̸= β . If there are two bad trees L1 and L2 , then swap the colors of the centers of L1 and L2 . In any case, we can always obtain a descending equitable RE-coloring of H, a contradiction with the choice of the coloring c ′ of H. ■ 3. Discharging rules By Euler’s formula, a planar graph G with girth g satisfies mad(G) <
2g , g −2
in which
mad(G) = max 2e(H)/n(H). H ⊆G
Consider a minimal counterexample G to Theorem 3. Since g(G) ≥ 8, we have mad(G) < 38 , then v∈V (G) (d(v ) − 83 ) < 0. Let µ(v ) = d(v ) − 38 be the initial charge, and µ∗ (v ) be the final charge after distributing charge among vertices by the following rules:
∑
(R1) Every 3+ -vertex gives 13 to each 2-vertex in its incident threads. (R2) We call a 3-vertex u with a1 (u) = 2 as a bad v ertex. Every 3+ -vertex gives 13 to each adjacent bad vertex. (R3) Every 4+ -vertex gives 31 to each adjacent 3-vertex u, if t(u) = 1 and u is adjacent to a bad vertex. First, if d(v ) = 2, µ(v ) = 2 − 83 = − 32 . By (R1), the vertex v can receive 13 from each endpoint of the thread containing v and it will not give charge to other vertices. So µ∗ (v ) = 0. If d(v ) ≥ 5, the vertex v gives 31 · t(v ) to all 2-vertices by (R1) and gives at most 13 · a0 (v ) to all adjacent 3-vertices by (R2) and (R3). So the vertex v gives at most 31 [t(v ) + a0 (v )] = 31 [d(v ) + a2 (v )]. Then by (P3)-(P5) and a2 (v ) ≤ d(v ), 1 [d(v ) + a2 (v )] ≤ d(v ) − 83 . Therefore µ∗ (v ) ≥ d(v ) − 38 − 13 [d(v ) + a2 (v )] ≥ 0. 3 Next, if d(v ) = 3, µ(v ) = 3 − 83 = 13 . Then by above rules, v gives charge to each 2-vertex in its incident threads and bad vertices. By (P1), a 3-vertex v has t(v ) ≤ 1 or t(v ) = 2 with a1 (v ) = 2.
• If t(v ) = 0, v does not receive any charge by (R2) and (R3). Due to (P6), a 3-vertex v that is adjacent to a bad vertex must have t(v ) = 1. So v does not give charge to the other vertices by (R2). Then µ∗ (v ) = 3 − 38 = 31 > 0. • If t(v ) = 1, v may receive charge from adjacent 4+ -vertices by (R3) and v also gives charge to the adjacent 2-vertex and bad vertices. If v does not adjacent to any bad vertex, then µ∗ (v ) = 3 − 83 − 13 = 0. If v is adjacent to bad vertices, then the number of bad vertices is one, because, if v is adjacent to a bad vertex, another 3+ -neighbor of v must have degree at least four by (P7). In this case, v can receive 13 from this 4+ -vertex by (R3). So µ∗ (v ) = 3− 38 −( 13 + 13 )+ 31 = 0. • If t(v ) = 2 with a1 (v ) = 2, v receives 31 from the adjacent 3+ -vertex. And there is no bad vertices that are adjacent to v by (P6)(i). So v only gives 13 to each adjacent 2-vertex. In other words, µ∗ (v ) = 3 − 38 − 23 + 31 = 0. Finally, if d(v ) = 4, then t(v ) ≤ 4 and a2 (v ) ≤ 1 by (P2). Also, by (P6)(ii) and (P7)(i), a 4-vertex that gives charge to 3-vertex must have a2 (v ) = 0. Thus v gives charge at most 34 . Then µ∗ (v ) ≥ 4 − 38∑ − 43 ≥ 0. ∑ ∗ ∗ In conclusion, for a vertex in G, we always have µ (v ) ≥ 0. Therefore 0 ≤ v∈V (G) µ (v ) = v∈V (G) µ(v ) < 0, a contradiction. Declaration of competing interest The authors declare that there is no conflict of interest in this paper.
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