Average degrees of edge-chromatic critical graphs

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1 Average degrees of edge-chromatic critical graphs Yan Cao a,guantao Chen a, Suyun Jiang b,c, Huiqing Liu d, Fuliang Lu e a Department of Mathematics and Statistics, Georgia State University, Atlanta, GA 30303, United States b Institute for Interdisciplinary Research, Jianghan University, Wuhan, Hubei , China c School of Mathematics, Shandong University, Jinan, Shandong 50100, China d Hubei Key Laboratory of Applied Mathematics, Faculty of Mathematics and Statistic, Hubei University, Wuhan, Hubei 43006, China e School of Mathematics and Statistics, Linyi University, Linyi, Shandong 76000, China Abstract Let G be a simple graph, and let (G), d(g) and χ (G) denote the maximum degree, the average degree and the chromatic index of G, respectively. We called G edge- -critical if χ (G) = (G) + 1 and χ (H) (G) for every proper subgraph H of G. Vizing in 1968 conjectured that if G is an edge- -critical graph of order n, then d(g) (G) n. We prove that for any edge- -critical graph G, { d(g) min 3, 3 (G) +1 4 d(g) }, that is, { 3 4 (G) if (G) 75; (G) if (G) This result improves the best known bound 3 ( (G) + ) obtained by Woodall in 007 for (G) 41. Keywords: edge-k-coloring; edge-critical graphs; Vizing s Adjacency Lemma 1 Introduction All graphs in this paper, unless otherwise stated, are simple graphs. Let G be a graph with vertex set V (G) and edge set E(G). Denote by (G) the maximum degree of G. An Partially supported by NSFC of China (Nos , , , , ). Corresponding author. addresses: ycao17@gsu.edu (Y. Cao), gchen@gsu.edu (G. Chen), jiang.suyun@163.com (S. Jiang), hql 008@163.com (H. Liu), flianglu@163.com (F. Lu). 1

2 edge-k-coloring of a graph G is a mapping ϕ : E(G) {1,,..., k} such that ϕ(e) ϕ(f) for any two adjacent edges e and f. We call {1,,..., k} the color set of ϕ. Denote by C k (G) the set of all edge-k-colorings of G. The chromatic index χ (G) is the least integer k 0 such that C k (G). We call G class one if χ (G) = (G). Otherwise, Vizing [13] proved χ (G) = (G) + 1 and G is said to be of class two. An edge e is called critical if χ (G e) < χ (G), where G e is the subgraph obtained from G by removing the edge e. A graph G is called (edge-) -critical if χ (G) = (G) + 1 and χ (H) (G) for any proper subgraph H of G. Clearly, if G is -critical, then G is connected and χ (G e) = (G) for any e E(G). Let d(g) denote the average degree of a graph G. Vizing [15] made the following conjecture in Conjecture 1. [Vizing s Average Degree Conjecture] If G is a -critical graph of n vertices, then d(g) (G) n. The conjecture has been verified for graphs with (G) 6, see [4, 6, 7, 9]. In general, there are a few results on the lower bound for d(g). Let G be a -critical graph with maximum degree. Fiorini [3] showed, for, d(g) { 1 1 Haile [5] improved the bounds as follows. d(g) ( + 1) if is odd; ( + ) if is even. ( + ) if = 9, 11, 13; +4 if 10, is even; if = 15; if 17, is odd. Sanders and Zhao [10] showed d(g) 1 ( + 1) for. Woodall [17] improved the bound to d(g) t( +t 1), where t = /. Improving Vizing s Adjacency Lemma, t 1 Woodall [16] improved the coefficient of from 1 to as follows. 3 d(g) ( + 1) if ; + 1 if 8; ( + ) if 15. In the same paper, Woodall provided an example demonstrating that the above result cannot be improved by the use of his new adjacency Lemmas (see Lemma and Lemma 3)

3 and Vizing s Adjacency Lemma alone. By proving a few stronger properties of -critical graphs, we get the following theorem. Theorem 1. If G is a -critical graph, then d(g) min{ 3, 3 (G) }, that +1 4 is, d(g) { 3 4 (G) if (G) 75; (G) if (G) 76. We will prove a few technical lemmas in Section and give the proof of Theorem 1 in Section 3. We will use the following terminology and notation. Let G be a graph and x be a vertex of G. Denote by N(x) the neighborhood, and by d(x) the degree of x, i.e., d(x) = N(x). For any nonnegative integer m, we call a vertex x an m-vertex if d(x) = m, a (< m)-vertex if d(x) < m, and a (> m)-vertex if d(x) > m. Similarly, we call a neighbor y of x an m-neighbor, a (< m)-neighbor and a (> m)-neighbor if d(y) = m, < m and > m, respectively. Let G be a graph, F E(G) be an edge set, and let ϕ C k (G F ) be a coloring for some integer k 0. For a vertex v V (G), define the two color sets ϕ(v) = {ϕ(e) : e is incident with v and e / F } and ϕ(v) = {1,,..., k} \ ϕ(v). We call ϕ(v) the set of colors seen by v and ϕ(v) the set of colors missing at v. A set X V (G) is called elementary with respect to ϕ if ϕ(u) ϕ(v) = for every two distinct vertices u, v X. For any color α, let E α denote the set of edges assigned color α. Clearly, E α is a matching of G. For any two colors α and β, the components of the spanning subgraph of G with edge set E α E β, named (α, β)-chains, are even cycles and paths with alternating color α and β. For a vertex v of G, we denote by P v (α, β, ϕ) the unique (α, β)-chain that contains the vertex v. Let ϕ/p v (α, β, ϕ) be the coloring obtained from ϕ by switching colors α and β on the edges on P v (α, β, ϕ). This operation is called a Kempe change. If v is not incident with any edge of color α or β, then P v (α, β, ϕ) = {v} (a path of length 0), and ϕ/p v (α, β, ϕ) = ϕ. Lemmas Let G be a graph and q be a positive number. For each edge xy E(G), let σ q (x, y) = {z N(y) \ {x} : d(z) q}, the number of neighbors of y (except x) with degree at least q. Vizing studied the case q = (G) and obtained the following result. Lemma 1. [Vizing s Adjacency Lemma [14]] If G is a -critical graph, then σ (G) (x, y) (G) d(x) + 1 for every xy E(G). 3

4 Woodall [16] studied σ q (x, y) for the case q = (G) d(x) d(y)+ and obtained the following two results. For convenience, we let σ(x, y) = σ q (x, y) when q = (G) d(x) d(y) +. Let xy E(G). To study the difference between σ(x, y) and (G) d(x) + 1, Woodall defined the following two parameters. p min (x) := min σ(x, y) (G) + d(x) 1, y N(x) (1) p(x) := d(x) min{ p min (x), 1 }. () Lemma. [Woodall [16]] Let xy be an edge in a -critical graph G. Then there are at least (G) σ(x, y) (G) d(y) + 1 vertices z N(x) \ {y} such that σ(x, z) (G) d(x) σ(x, y). Lemma 3. [Woodall [16]] Every vertex x in a -critical graph has at least d(x) p(x) 1 neighbors y for which σ(x, y) (G) p(x) 1. Inspired by Woodall s parameters (1) and (), for any positive number q, we define the following two parameters. p min (x, q) := min q(x, y) (G) + d(x) 1 y N(x) p(x, q) := d(x) min{ p min (x, q), 3 }. and The following lemma is a generalization of Lemma, which serves as a key result. Lemma 4. Let xy be an edge in a -critical graph G and q be a positive number. If (G)/ < q (G) d(x)/, then x has at least (G) σ q (x, y) vertices z N(x) \ {y} such that σ q (x, z) (G) d(x) σ q (x, y) 4. Due to its length, the proof of Lemma 4 will be placed at the end of this section. The following is a consequence of it. Lemma 5. Let G be a -critical graph, x V (G) and q be a positive number. (G)/ < q (G) d(x)/, then x has at least d(x) p(x, q) 3 neighbors y for which σ q (x, y) (G) p(x, q) 5. Proof. Let y N(x) such that p min (x, q) = σ q (x, y) (G) + d(x) 1, and = (G). If p(x, q) = p min (x, q), by Lemma 4, x has at least σ q (x, y) = d(x) p min (x, q) 3 vertices z N(x) \ {y} such that σ q (x, z) d(x) σ q (x, y) 4 = p min (x, q) 5. 4 If

5 d(x) d(x) If p(x, q) = 3 3 < p min (x, q), then for every y N(x), σ q (x, y) > d(x) = p(x, q) 6. So σ q (x, y) p(x, q) 5. d(x) Our approach is inspired by the recent development of the Tashkinov tree technique for multigraphs. Let G be a multigraph without loops, e 1 be an edge of G with endvertices y 0 and y 1 and ϕ C k (G e 1 ). A Tashkinov tree T with respect to G, e 1, ϕ is an alternating sequence T = (y 0, e 1, y 1,..., e p, y p ) with p 1 consisting of edges e 1, e,..., e p and vertices y 0, y 1,..., y p such that the following two conditions hold. (1) The vertices y 0, y 1,..., y p are distinct and e i = y r y i for each 1 i p, where r < i; () for every edge e i with i p, there is a vertex y h with 0 h < i such that ϕ(e i ) ϕ(y h ). Tashkinov [1] introduced this concept in his work on the well-known Goldberg s Conjecture. In the above definition, if we replace condition (1) with the edges e 1, e,..., e p are distinct and e i = y 0 y i for every i, then T is a multi-fan, as defined in [11]; if, in addition, ϕ(e i ) ϕ(y i 1 ) for every i, then T is a Vizing fan. Stiebitz et al. [11] showed that all multi-fans are elementary. In the definition of Tashkinov tree, if e i = y i 1 y i for every i, i.e., T is a path with end-vertices y 0 and y p, then T is a Kierstead path, which was introduced by Kierstead [8] in studying edge-colorings of multigraphs. For simple graphs, following Kierstead s proof, Zhang [18] noticed that for a Kierstead path P the set V (P ) is elementary if G is -critical and d(y i ) < for every i with i p. It is not difficult to see that every Kierstead path P with three vertices is a Vizing fan, so V (P ) is elementary if G is -critical. Kostochka and Stiebitz studied Kierstead paths with four vertices and obtained the following result. Lemma 6. [Kostochka and Stiebitz [11]] Let G be a graph with maximum degree and χ (G) = + 1. Let e 1 E(G) be a critical edge and ϕ C (G e 1 ). If K = (y 0, e 1, y 1, e, y, e 3, y 3 ) is a Kierstead path with respect to e 1 and ϕ, then the following statements hold: 1. ϕ(y 0 ) ϕ(y 1 ) = ;. if d(y ) <, then V (K) is elementary with respect to ϕ; 3. if d(y 1 ) <, then V (K) is elementary with respect to ϕ; 4. if Γ = ϕ(y 0 ) ϕ(y 1 ), then ϕ(y 3 ) Γ 1. 5

6 In the definition of Tashkinov tree T = (y 0, e 1, y 1, e, y,..., y p ), we call T a broom if e = y 1 y and for each i 3, e i = y y i, i.e., y is one of the end-vertices of e i for each i 3. Moreover, we call T a simple broom if ϕ(e i ) ϕ(y 0 ) ϕ(y 1 ) for each i 3, i.e., (y 0, e 1, y 1, e, y, e i, y i ) is a Kierstead path. Lemma 7. [Chen, Chen, Zhao []] Let G be a graph with maximum degree and χ (G) = + 1. Let e 1 = y 0 y 1 E(G) be a critical edge and ϕ C (G e 1 ), and let B = {y 0, e 1, y 1, e, y,..., e p, y p } be a simple broom. min{d(y 1 ), d(y )} <, then V (B) is elementary with respect to ϕ. If ϕ(y 0 ) ϕ(y 1 ) 4 and Lemma 8. Let G be a graph with maximum degree and χ (G) = + 1, xy E(G) be a critical edge and ϕ C (G xy). Let q be a positive number with d(x) < q 1 and Z = {z N(x)\{y} : d(z) > q, ϕ(xz) ϕ(y)}. Then d(x) + d(y) Z d(y) + 1, (3) q (d(z) q) ( q)( d(y) + 1) d(x) d(y) + +, (4) z Z and for every z Z, d(x) + d(y) + d(z) σ q (x, z) d(x) d(y). (5) q Proof. Since G is not -colorable, ϕ(x) ϕ(y) =. Let Z y := {z N(x) \ {y} : ϕ(xz) ϕ(y)}. Clearly, Z Z y and Z y = d(y) + 1. Since {y, x} Z y forms a multi-fan with center x, it is elementary, so ϕ(x) + ϕ(y) + z Z y ϕ(z). Since ϕ(x) = d(x) + 1 and ϕ(y) = d(y) + 1, we have z Z y ϕ(z) ϕ(x) ϕ(y) d(x) + d(y). (6) Since d(z) q for all z Z y Z, z Z y ϕ(z) ( Z y Z )( q). Combining this with (6), we get Z Z y d(x)+d(y) = d(y) + 1 d(x)+d(y). So (3) q holds. Since ϕ(z) = d(z) for each z Z y, by (6), we get z Z y d(z) Z y (d(x) + d(y) ). 6 q

7 Since d(z) q for every z Z y Z, we have (d(z) q) (d(z) q) Z y (d(x) + d(y) ) Z y q. z Z y z Z Substituting Z y = d(y) + 1 in the above inequality, we get (4). For each z Z, let U z = {u N(z) \ {x, y} : ϕ(zu) ϕ(x) ϕ(y)} and U z = {u U z : d(u) > q}. Since ϕ(xz) ϕ(y), U z d(x) d(y) and {y, x, z} U z forms a simple broom. Since d(x) < q 1, we have d(x). Thus ϕ(x) ϕ(y) 4 and min{d(x), d(z)} = d(x) <. By Lemma 7, {y, x, z} Uz is elementary with respect to ϕ. So u U ϕ(u) + ϕ(x) + ϕ(y) + ϕ(z), which in turn gives z u U ϕ(u) d(x) + d(y) + d(z). Since d(u) q for every u U z z U z, u U ϕ(u) ( U z z U z )( q). So, ( U z U z )( q) d(x) + d(y) + d(z). Since U z d(x) d(y), we get d(x) + d(y) + d(z) σ q (x, z) U z d(x) d(y). q So the inequality (5) holds..1 Proof of Lemma 4 Lemma 4. Let xy be an edge in a -critical graph G and q be a positive number. If (G)/ < q (G) d(x)/, then x has at least (G) σ q (x, y) vertices z N(x) \ {y} such that σ q (x, z) (G) d(x) σ q (x, y) 4. Proof. Let graph G, edge xy E(G) and q be defined as in Lemma 4. Let = (G). A vertex z N(x) \ {y} is called feasible if there exists a coloring ϕ C (G xy) such that ϕ(xz) ϕ(y), and such a coloring ϕ is called z-feasible. Denote by C z the set of all z-feasible colorings. For each feasible vertex z and each z-feasible coloring ϕ C z, let Z(ϕ) = {v N(z) \ {x} : ϕ(vz) ϕ(x) ϕ(y)}, C z (ϕ) = {ϕ(vz) : v Z(ϕ) and d(v) < q} ϕ(x) ϕ(y), Y (ϕ) = {v N(y) \ {x} : ϕ(vy) ϕ(x) ϕ(z)}, and C y (ϕ) = {ϕ(vy) : v Y (ϕ) and d(v) < q} ϕ(x) ϕ(z). 7

8 Note that Z(ϕ) and Y (ϕ) are vertex sets while C z (ϕ) and C y (ϕ) are color sets. each color α ϕ(z), let z α N(z) such that ϕ(zz α ) = α. For each color β ϕ(y), let y β N(y) such that ϕ(yy β ) = β. Let T (ϕ) = {α ϕ(x) ϕ(y) ϕ(z) : d(y α ) < q and d(z α ) < q}. Since (ϕ(x) ϕ(y)) ( ϕ(x) ϕ(y)) = and (ϕ(x) ϕ(z)) ( ϕ(x) ϕ(z)) =, we obtain that T (ϕ) (C z (ϕ) C y (ϕ)) =. Since G is -critical and ϕ is z-feasible, {x, y, z} is elementary with respect to ϕ. So ϕ(x), ϕ(y), ϕ(z) and ϕ(x) ϕ(y) ϕ(z) are mutually exclusive. It is not difficult to see that Z(ϕ) = ϕ(x) ϕ(y) 1 and Y (ϕ) = ϕ(x) ϕ(z). (7) For Also, ϕ(x) ϕ(y) ϕ(z) (ϕ(x) ϕ(y) ϕ(z)) = {1,,..., }. (8) Recall that σ q (x, y) and σ q (x, z) are the numbers of vertices with degree at least q in N(y) \ {x} and N(z) \ {x}, respectively. So, by equations (7) and (8), we have σ q (x, y) + σ q (x, z) Y (ϕ) C y (ϕ) + Z(ϕ) C z (ϕ) + ϕ(x) ϕ(y) ϕ(z) T (ϕ) = ϕ(x) ϕ(z) + ϕ(x) ϕ(y) 1 + ϕ(x) ϕ(y) ϕ(z) C y (ϕ) C z (ϕ) T (ϕ) = + ϕ(x) 1 C y (ϕ) C z (ϕ) T (ϕ) = d(x) C y (ϕ) C z (ϕ) T (ϕ). So, Lemma 4 follows from the three statements below. I. For any ϕ C z, C z (ϕ) 1 and C y (ϕ) 1; II. there exists a ϕ C z such that T (ϕ) ; and III. there are at least σ q (x, y) feasible vertices z N(x) \ {y}. For every z-feasible coloring ϕ C (G xy), let ϕ d C (G xz) be obtained from ϕ by assigning ϕ d (xy) = ϕ(xz) and keeping all colors on other edges unchanged. Clearly, ϕ d is a y-feasible coloring and Z(ϕ d ) = Z(ϕ), Y (ϕ d ) = Y (ϕ), C z (ϕ d ) = C z (ϕ) and C y (ϕ d ) = C y (ϕ). We call ϕ d the dual coloring of ϕ. Considering dual colorings, we see that some properties for vertex z also hold for vertex y. Since q d(x)/, we have 8

9 ( q) + ( d(x) + 1) 5. (9) So, for any ϕ C (G xy) and any elementary set X with x X, X \ {x} contains at most one vertex with degree at most q. (10) Let z N(x) \ {y} be a feasible vertex and ϕ C z. By the definition of Z(ϕ), {y, x, z} Z(ϕ) is the vertex-set of a simple broom, and so this set is elementary with respect to ϕ. Thus, C z (ϕ) 1 by (10). By considering its dual ϕ d, we have C y (ϕ) = C y (ϕ d ) 1. Hence, I holds. The proofs of II and III are much more complicated. Let R(ϕ) = C z (ϕ) C y (ϕ). In the remainder of the proof, we let Z = Z(ϕ), Y = Y (ϕ), C z = C z (ϕ), C y = C y (ϕ), R = R(ϕ) and T = T (ϕ) if the coloring ϕ is clear. Note that T R =, ϕ(xz) / T R and R. A coloring ϕ C z is called optimal if C z (ϕ) + C y (ϕ) is maximum over all z-feasible colorings. Note that a z-feasible coloring ϕ is optimal if and only if its dual coloring ϕ d is an optimal y-feasible coloring..1.1 Proof of Statement II. Suppose on the contrary that T (ϕ) 3 for every ϕ C z. Let ϕ be an optimal z-feasible coloring and assume, without loss of generality, ϕ(xz) = 1. Since χ (G) > and 1 ϕ(y), it follows that for every color i ϕ(x), P x (1, i, ϕ) = P y (1, i, ϕ). (11) Otherwise, an edge- -coloring of G can be gotten by ϕ/p x (1, i, ϕ) and then coloring xy with 1, a contradiction. Claim A. For each i ϕ(x) \ R and k T (ϕ), P x (i, k, ϕ) contains both y and z. Proof. We first show that z V (P x (i, k, ϕ)). Otherwise, P z (i, k, ϕ) is disjoint from P x (i, k, ϕ). Let ϕ = ϕ/p z (i, k, ϕ). Since 1 / {i, k}, ϕ is also z-feasible. Since colors in R are unchanged and d(z k ) < q, C z (ϕ ) = C z (ϕ) {i} and C y (ϕ ) C y (ϕ), giving a contradiction to the maximality of C y (ϕ) + C z (ϕ). By considering the dual coloring ϕ d, we can verify that y V (P x (i, k, ϕ)). Note that T (ϕ) 3. For each 3-element subset S(ϕ) of T (ϕ), we may assume 9

10 S(ϕ) = {k 1, k, k 3 }, let V (S(ϕ)) = {z k1, z k, z k3 } {y k1, y k, y k3 }, W (S(ϕ)) = {u V (S(ϕ)) : ϕ(u) ϕ(x) \ R = }, M(S(ϕ)) = {u V (S(ϕ)) : ϕ(u) ϕ(x) \ R }, E(S(ϕ)) = {zz k1, zz k, zz k3, yy k1, yy k, yy k3 }, E W (S(ϕ)) = {e E(S(ϕ)) : e is incident to a vertex in W (S(ϕ))}, and E M (S(ϕ)) = {e E(S(ϕ)) : e is incident to a vertex in M(S(ϕ))}. Clearly, V (S(ϕ)) = W (S(ϕ)) M(S(ϕ)) and E(S(ϕ)) = E W (S(ϕ)) E M (S(ϕ)), where denotes disjoint union. For convenience, we let W = W (S(ϕ)), M = M(S(ϕ)), E W = E W (S(ϕ)) and E M = E M (S(ϕ)) if S(ϕ) is clear. Note that one of y and z has degree at least q by (10), thus {y, z} V (S(ϕ)). We assume that E W (S(ϕ)) is minimum over all optimal z-feasible colorings ϕ and all 3-element subsets S(ϕ) of T (ϕ). For each v M, pick a color α v ϕ(v) ϕ(x) \ R. Let C M = {α v have Since R, we have : v M}. Clearly, C M M. By (9) and the condition q > /, we ϕ(x) = d(x) + 1 ( q) + 5 > 5. (1) ϕ(x) \ (R C M ) if C M 3. (13) Note that {z k1, z k, z k3 } {y k1, y k, y k3 } may be not empty, E W W E W and E M M E M. Claim B. Let u and v be two vertices of V (S(ϕ)). If ϕ(u) ϕ(v) ϕ(x) \ R, then the following statements hold. (i) If u, v W, then ϕ(x) \ (R C M ) =. (ii) There exists an optimal z-feasible coloring ϕ with V (S(ϕ )) = V (S(ϕ)) such that E W (S(ϕ )) E W and {u, v} M(S(ϕ )). Proof. Noting that ϕ(u) ϕ(v) ϕ(x) \ R, we assume α ϕ(u) ϕ(v) ϕ(x) \ R. If {u, v} M, we are done (with ϕ = ϕ). So suppose u, v W. Let β be an arbitrary color in ϕ(x) \ (R C M ) if this set is nonempty; otherwise, let β be a color in ϕ(x) \ R, which is nonempty by (1) since R. Since u, v W, we have β ϕ(u) ϕ(v). So, both u and v are endvertices of (α, β)-chains. Assume without loss of generality that P u (α, β, ϕ) is disjoint from P x (α, β, ϕ). Let ϕ = ϕ/p u (α, β, ϕ). 10

11 First, we note that T (ϕ ) T (ϕ). This is obvious if α / T (ϕ), and if α T (ϕ) it holds because in this case {z, y, z α, y α } V (P x (α, β, ϕ)) by Claim A and so {z, y, z α, y α } V (P u (α, β, ϕ)) =. So we can choose S(ϕ ) = S(ϕ), and then V (S(ϕ )) = V (S(ϕ)). Next, we claim that ϕ is an optimal z-feasible coloring. Note that ϕ (xz) = 1 ϕ (y). This is obvious if α 1 (since β 1), and if α = 1 it holds because in this case P x (α, β, ϕ) = P y (α, β, ϕ) by (11) and so V (P u (α, β, ϕ)) {x, y, z} =. So ϕ is z-feasible. Since α, β / R = C y C z, it follows that C y C y (ϕ ) and C z C z (ϕ ). Since ϕ is optimal, we have C y (ϕ ) = C y, C z (ϕ ) = C z, R(ϕ ) = R and so ϕ is optimal. Since α ϕ(u), it follows that β ϕ (u) ϕ (x) \ R(ϕ ), so u M(S(ϕ )) (that is, u has moved from W to M(S(ϕ ))). To avoid the contradiction that E W (S(ϕ )) < E W, it is necessary that P u (α, β, ϕ), which starts with an edge of color β at u, must end with an edge of color α at a vertex w M such that β is the unique color in ϕ(w) ϕ(x) \ R, thus ϕ (w) ϕ (x)\r(ϕ ) = (that is, w has moved from M to W (S(ϕ ))). But then we must have chosen α w = β, so β C M. By the choice of β, we have ϕ(x) \ (R C M ) =. Thus (i) holds. By (13), we have C M 4. Since u, v W, V (S(ϕ)) 6 and C M M, we have V (S(ϕ)) = 6, W = and M = 4. Thus E W (S(ϕ )) = E W. Hence, (ii) holds. Claim C. There exist a color k {k 1, k, k 3 }, three distinct colors i, j, l and an optimal z-feasible coloring ϕ with ϕ (xz) = 1 such that i ϕ (z k ) ϕ (x) \ R(ϕ ), j ϕ (y k ) ϕ (x) \ R(ϕ ) and l ( ϕ (z k ) ϕ (y k )) ( ϕ (x) {1} \ R(ϕ )). Proof. First we show that there exists a color k {k 1, k, k 3 } such that ϕ(z k ) ϕ(x) \ R and ϕ(y k ) ϕ(x) \ R. (14) If E M 4, by the definition of M and E M, it is not difficult to see that the above statement holds. Thus we may assume that E M 3. So E W 3 and then we have W as W E W /. Let u, v W. Since C M M E M 3, we have ϕ(x) \ (R C M ) by (13). By Claim B (i), we have ϕ(u) ϕ(v) ϕ(x) \ R =. (15) On the other hand, since u, v W V (S(ϕ)) and S(ϕ) T (ϕ), we note that d(u) < q and d(v) < q. Since R, by (9), we have ϕ(u) \ R + ϕ(v) \ R + ϕ(x) > ( q ) + d(x) + 1 >. Moreover, by the definition of W and u, v W, we have ( ϕ(u) \ R) ϕ(x) = ( ϕ(v) \ R) ϕ(x) =. This implies that ( ϕ(u) \ R) ( ϕ(v) \ R) ϕ(x), which contradicts (15). 11

12 This contradiction shows that there exists a color k {k 1, k, k 3 } such that (14) holds. Now we claim that ϕ(z k ) ϕ(y k ) ϕ(x) \ R =. (16) For otherwise, we assume i ϕ(z k ) ϕ(y k ) ϕ(x)\r, then by Claim A, the path P x (i, k, ϕ) contains three endvertices x, z k and y k, a contradiction. By (14) and (16), we can find two distinct colors i, j such that i ϕ(z k ) ϕ(x)\r and j ϕ(y k ) ϕ(x)\r. If there still exists another color l ( ϕ(z k ) ϕ(y k )) ( ϕ(x) {1}\R), then Claim C holds with ϕ = ϕ. Thus color l does not exist, that is, ϕ(z k ) ϕ(x) \ R = {i}, ϕ(y k ) ϕ(x) \ R = {j} and 1 / ϕ(z k ) ϕ(y k ), which implies that (( ϕ(z k ) \ (R {i})) ( ϕ(y k ) \ (R {j}))) ( ϕ(x) R {1}) =. (17) By the definition of T (ϕ), we have d(z k ) < q and d(y k ) < q. Since R, by (9), we have ϕ(z k )\(R {i}) + ϕ(y k )\(R {j}) + ϕ(x) (R {1}) > ( q R 1)+ d(x)+. So there exists a color α in the intersection of sets ϕ(z k ) \ (R {i}) and ϕ(y k ) \ (R {j}) by (17). Thus α ϕ(z k ) ϕ(y k ) ϕ(x) \ (R {i, j, 1}). Since R, by (1), there exists a color β ϕ(x) \ (R {i, j}). By (17), we have β ϕ(z k ) ϕ(y k ). We may assume that P zk (α, β, ϕ) is disjoint from P x (α, β, ϕ). Let ϕ = ϕ/p zk (α, β, ϕ). Similar to the proof of Claim B, ϕ is an optimal z-feasible coloring with R(ϕ ) = R and we can choose V (S(ϕ )) = V (S(ϕ)). Note that ϕ (x) = ϕ(x). Since α ϕ(z k ), we have β ϕ (z k ) ϕ (x) \ (R(ϕ ) {i, j}). Moreover, we have i ϕ (z k ) ϕ (x) \ R(ϕ ) and j ϕ (y k ) ϕ (x) \ R(ϕ ) as α, β / {i, j}. Thus i, j, β are the required colors and then Claim C holds. Let k, i, j, l and ϕ be as stated in Claim C. If l 1, we consider the coloring obtained from ϕ by interchanging colors 1 and l for edges not on the path P x (1, l, ϕ ), and rename it as ϕ. This is valid, by the argument in the proof of Claim B, since 1, l / R(ϕ ) T (ϕ ), 1 ϕ (x) and l ϕ (x). So we may assume 1 ϕ (y k ) ϕ (z k ). We first consider the case of 1 ϕ (y k ). Note that i, j ϕ (x) \ R(ϕ ). By Claim A, the paths P x (i, k, ϕ ) and P x (j, k, ϕ ) both contain y, z. Since ϕ (yy k ) = ϕ (zz k ) = k, these two paths also contain y k, z k. Since i ϕ (z k ), it follows that x and z k are the two endvertices of P x (i, k, ϕ ). So, i ϕ (y) ϕ (z) ϕ (y k ). Similarly, we have j ϕ (y) ϕ (z) ϕ (z k ). We now consider the following sequence of colorings of G xy. 1

13 Let ϕ 1 be obtained from ϕ by assigning ϕ 1 (yy k ) = 1. Since 1 was missing at both y and y k, ϕ 1 is an edge- -coloring of G xy. Now k is missing at y and y k, and i is still missing at x and z k. Since G is not -colorable, P x (i, k, ϕ 1 ) = P y (i, k, ϕ 1 ); otherwise ϕ 1 /P y (i, k, ϕ 1 ) can be extended to an edge- -coloring of G by coloring xy with i, giving a contradiction. Furthermore, z k, y k / V (P x (i, k, ϕ 1 )) since either i or k is missing at these two vertices, which in turn shows that z / V (P x (i, k, ϕ 1 )) since ϕ 1 (zz k ) = k. Let ϕ = ϕ 1 /P x (i, k, ϕ 1 ). We have k ϕ (x), i ϕ (y) ϕ (z k ) and j ϕ (x) ϕ (y k ). Since G is not edge- -colorable, we have P x (i, j, ϕ ) = P y (i, j, ϕ ), which contains neither y k nor z k. Let ϕ 3 = ϕ /P x (i, j, ϕ ). Then k ϕ 3 (x) and j ϕ 3 (y) ϕ 3 (y k ). Let ϕ 4 be obtained from ϕ 3 by recoloring yy k by j. Then 1 ϕ 4 (y), ϕ 4 (xz) = 1, k ϕ 4 (x), ϕ 4 (zz k ) = k. Since ϕ 4 (xz) = 1 ϕ 4 (y), ϕ 4 is z-feasible. Since i, j, k / R = C y C z, the colors in R are unchanged during this sequence of re-colorings, so C y (ϕ 4 ) C y and C z (ϕ 4 ) C z. Since ϕ 4 (zz k ) = k ϕ 4 (x) and d(z k ) < q, we have k = ϕ 4 (zz k ) C z (ϕ 4 ). So, C z (ϕ 4 ) C z {k}. We therefore have C y (ϕ 4 ) + C z (ϕ 4 ) C y + C z + 1, giving a contradiction. For the case of 1 ϕ (z k ), we consider the dual coloring ϕ d of G xz obtained from ϕ by uncoloring xz and coloring xy with color 1. Following the exact same argument above, we can reach a contradiction to the maximum of C y + C z. This completes the proof of Statement II..1. Proof of Statement III. For a coloring ϕ C (G xy), let X(ϕ) = {z N(x) \ {y} : ϕ(xz) ϕ(y)} and S(ϕ) = {z N(x) \ (X(ϕ) {y}) : d(y ϕ(xz) ) < q}, where y j N(y) with ϕ(yy j ) = j for any color j. We call vertices in S(ϕ) semi-feasible vertices of ϕ. Note that the vertices in X(ϕ) are feasible. Statement III clearly follows from the following claim. Claim.1. For any coloring ϕ C (G xy), the following two statements hold. a. X(ϕ) S(ϕ) σ q (x, y) 1; b. With one possible exception, for each z S(ϕ) there exists a coloring ϕ C (G xy) such that ϕ (xz) ϕ (y). Proof. Let ϕ C (G xy). Since G is -critical, it is easy to see that ϕ(y) ϕ(x) and ϕ(x) ϕ(y). To prove a, we divide ϕ(y) into two subsets: ϕ(y, q) = {i ϕ(y) : d(y i ) q} and ϕ(y, < q) = {i ϕ(y) : d(y i ) < q}. 13

14 Clearly, σ q (x, y) = ϕ(y, q) and ϕ(y) + ϕ(y, < q) = σ q (x, y). Also, X(ϕ) = ϕ(y) ϕ(x) = ϕ(y) since ϕ(y) ϕ(x), and S(ϕ) = ϕ(y, < q) ϕ(x). Since edge xy and the edges incident to y with colors in ϕ(x) form a multi-fan F, the vertex set V (F ) is elementary with respect to ϕ. By (10), V (F ) \ {x} contains at most one vertex with degree less than q. Thus ϕ(x) ϕ(y, < q) 1 and so S(ϕ) ϕ(y, < q) 1, and this proves a. To prove b, we show that for any two distinct vertices z k, z l S(ϕ), there is a coloring ϕ C (G xy) such that at least one of ϕ (xz k ) and ϕ (xz l ) is in ϕ (y). We assume ϕ(xz k ) = k and ϕ(xz l ) = l. Let y k, y l N(y) \ {x} such that ϕ(yy k ) = k and ϕ(yy l ) = l. By the definition of S(ϕ), we have d(y k ) < q and d(y l ) < q. It follows from (9) that ϕ(x) + ϕ(y k ) + ϕ(y l ) >. (18) Let z be an arbitrary vertex in X(ϕ), which exists since X(ϕ) = ϕ(y) > 0, and assume ϕ(xz) = 1, so that (11) holds. We claim that there exists a coloring ϕ C (G xy) such that ϕ (xz) = 1 ϕ (y), ϕ (xz k ) = ϕ (yy k ) = k, ϕ (xz l ) = ϕ (yy l ) = l, and ϕ (x) ( ϕ (y k ) ϕ (y l )). (19) Suppose (19) does not hold with ϕ = ϕ. Then ϕ(y k ) ϕ(y l ) ϕ(x), and so by (18) there exists a color r ϕ(x) ϕ(y k ) ϕ(y l ). Choose a color i ϕ(x). Note that {i, r} {1, k, l} = unless r = 1. Since at least one of colors i and r is missing at each of x, y k and y l, we may assume P yk (i, r, ϕ) is disjoint from P x (i, r, ϕ), and also from P y (i, r, ϕ) if r = 1, by (11). Then (19) holds with ϕ = ϕ/p yk (i, r, ϕ), since in this coloring i is missing at both x and y k. By (19), we may assume that there exists a color i ϕ (x) ϕ (y k ). Applying (11) to ϕ, we have P x (1, i, ϕ ) = P y (1, i, ϕ ), and so P yk (1, i, ϕ ) is disjoint from P x (1, i, ϕ ) and P y (1, i, ϕ ). Thus in the coloring ϕ = ϕ /P yk (1, i, ϕ ), color 1 is missing at both y and y k. Form ϕ from ϕ by changing the color of yy k from k to 1. Then ϕ (xz k ) = k ϕ (y). This proves b, and so completes the proofs of both Claim.1 and III. 3 Proof of Theorem 1 Let G be a -critical graph with maximum degree. Let n = V (G) and m = E(G). Clearly, d(g) = m/n. Let q be a positive number less than. Note that if < 1. By Woodall s result [16], Theorem 1 holds if < 1. In this section, we 14

15 assume 1. We initially assign to each vertex x of G a charge M(x, q) = d(x) and redistribute the charge according to the following -step discharging rule: Step 1: each (> q)-vertex y distributes its surplus charge of d(y) q equally among all (< q)-neighbors of y. Assume each vertex x now have charge M 1 (x, q), and call x deficient if M 1 (x, q) < q. Step : For every vertex x such that d(x) 1 and every neighbor of x has degree at least q, if x is at distance from a -vertex or a deficient 3-vertex: x sends 1 to each 4 such -vertex and 1 to each such deficient 3-vertex. 8 Denote by M (x, q) the resulting charge on each vertex x. Clearly, x V (G) M (x, q) = x V (G) M 1(x, q) = x V (G) M(x, q) = m, and for each x with d(x) q 1 we have M (x, q 1 ) M (x, q ) if q 1 q. We show that M (x, q) q for all vertices x V (G) if q = min{ 3, 3 }, which gives d(g) min{ , 3 }. Denote by +1 4 d <q (y) the number of (< q)-neighbors of y. Claim 4. in [1] shows that if 4 d(x), then M 4 1(x, 3 8) 3 8. Actually, 4 4 in the proof, it showed that M 1 (x, 3 1) 3 1. By using Step, Claim 4.3 and 4 4 Claim 4.4 in [1] showed that M (x, 3 1) 3 1 if x is a -vertex, a deficient 3-vertex, 4 4 or a ( 1) vertex. In the remainder of this proof, let q = min{ 3, 3 } If 4 d(x) < 1, we only use Step 1, it is easy to check that M (x, q) = M 1 (x, q) and M (x, q) = M 1 (x, q) q if q d(x) < 1. We show that for all vertices x with < d(x) < q, M 4 (x, q) = M 1 (x, q) q. Claim 3.1. If 4 < d(x) q + 1, then M (x, q) = M 1 (x, q) q. Proof. Since q > +1, d(x) q + 1 < q. Let y be an arbitrary neighbor of x. Since d(x) d(y) + d(x) + > q, we have σ q (x, y) σ(x, y). We will use lower bounds of σ(x, y) to estimate σ q (x, y). By (1) and (), we have 1 d <q (y) d(y) σ(x, y) d(y) ( d(x) + p(x) + 1). (0) By Lemma 3, x has at least d(x) p(x) 1 neighbors y for which σ(x, y) p(x) 1, so for these neighbors y we have 1 d <q (y) d(y) σ(x, y) d(y) ( p(x) 1). (1) By the hypothesis of Claim 3.1 and (0), we have q d(x)+1 d(x)+p(x)+ 1 < d(y). Since d(y) a d(y) b with a b is a decreasing function of d(y), for each y N(x), x 15

16 receives at least d(y) q d(y) ( d(x) + p(x) + 1) q d(x) p(x) 1. And there are at least d(x) p(x) 1 neighbors y of x giving x at least d(y) q d(y) ( p(x) 1) q p(x) + 1, where the inequality holds because q d(x) + 1 p(x) 1 as p(x) d(x) 1. Thus x receives at least (d(x) p(x) 1) q p(x) + 1 q + (p(x) + 1) d(x) p(x) 1 = (θ + θ 1 )( q) ( q), where θ = d(x) p(x) 1. It follows that M p(x)+1 1 (x, q) M(x, q)+( q) +( q) > q. 4 Claim 3.. If q + 1 < d(x) < q, then M (x, q) = M 1 (x, q) q. Proof. Let y be a neighbor of x. Then there exists a coloring ϕ C (G xy) as G is -critical. Let Z q = {z N(x) : d(z) > q}, Z y = {z N(x)\{y} : ϕ(xz) ϕ(y)} and Z y = Z q Z y. Note that Z y was called Z in Lemma 8, thus (3), (4) and (5) in Lemma 8 also hold for Z y. Moreover, by the definition of Z q and the discharging rule, for each z Z q, x receives charge at least d(z) q d(z) σ q(x,z). Thus M 1(x, q) d(x) + z Z q d(z) q d(z) σ q(x,z). Since 1, we have q 5 and so q + 1 ( q) 4. We consider the following three cases to complete the proof. Case 1. q + 1 < d(x) < q and x has a neighbor y such that d(y) q. It is easy to see that, for any vertex z, d(x) + d(y) + d(z) < q < ( q), since q < 3 4. It follows from this and (5) in Lemma 8 that, for each vertex z Z y, σ q (x, z) d(x) d(y) 1, and so d(z) σ q (x, z) d(x) + d(y) + 1. Also, by (4) in Lemma 8, z Z (d(z) q) ( q)( d(y) + 1) + 3 (d(x) + d(y) + 1). y 16

17 So, using d(y) q in the second line, we have M 1 (x, q) d(x) + z Z y > d(x) + d(z) q d(z) σ q (x, z) ( q)( d(y) + 1) d(x) + d(y) d(x) + q ( q)( q + 1) + q > 3( q) > q, ( q)( q + 1) d(x) + q q where in the penultimate line we used a + b a b for all a, b > 0, and in the final line that 3( q) q + 8 since q 3 4. Case. ( q) 4 < d(x) < q and d(y) > q for every y N(x). Let y N(x) such that the degree of y is minimum in N(x). Let u be an arbitrary vertex in N(x). Then d(u) d(y) by the choice of y. By Lemma 1, we have σ q (x, u) σ (x, u) d(x) + 1. Since d(u) d(y) and q > d(x) + 1, we have d(u) q d(u) σ q (x, u) where the inequality holds because d(u) a d(u) b d(u) q d(u) ( d(x) + 1) d(y) q d(y) ( d(x) + 1), with a > b > 0 is a increasing function of d(u). Note that N(x) \ Z y N(x) \ Z y = d(x) ( d(y) + 1). Combining this with the above inequality, we have u N(x)\Z y d(u) q d(u) σ q (x, u) For any vertex z, since d(y), d(z) and q < 3 4, d(y) q. () d(x) + d(y) + d(z) < d(x) < q < 3( q). For each z Z y, by (5) in Lemma 8, d(x) + d(y) + d(z) σ q (x, z) d(x) d(y) q d(x) d(y), 17

18 and so d(z) σ q (x, z) d(x) + d(y) +. Combining this with (4) in Lemma 8, () and using d(y) + 1 = d(x) + 3 (d(x) + d(y) + ) in the next line, gives M 1 (x, q) d(x) + z Z y d(x) + d(z) q d(z) σ q (x, y) + u N(x)\Z y ( q)( d(y) + 1) + 4 d(x) + d(y) + = d(x) + d(y) + + ( q)(d(x) + 3) > ( q 1 ) 3 q. d(u) q d(u) σ q (x, u) 1 + d(y) q ( q)(d(x) + 3) + 4 d(x) + d(y) + 3 where in the penultimate line we used a + b a b for all a, b > 0, and in the final line we used ( q)(d(x) + 3) + 4 > ( q 1 ) since d(x) > ( q) 4 and we also used q Case 3. q + 1 < d(x) ( q) 4 and d(y) > q for every y N(x). In this case, we let p := p(x, q) since it will be used heavily here. So, p = min{ p min (x, q), d(x) 3 }, where p min(x, q) = min y N(x) σ q (x, y) +d(x) 1. So σ q (x, y) d(x)+ p + 1 for every y N(x). Since d(x) ( q) 4 and p d(x) d(x) 3, we have q p 5. By Lemma 5, x has at least d(x) p 3 neighbors y for which σ q (x, y) p 5, so for these neighbors y, d(y) q d(y) σ q (x, y) d(y) q d(y) ( p 5) q p + 5. (3) 18

19 If q > d(x) + p + 1, i.e., p < d(x) + q 1, then since q 3 4. M 1 (x, q) d(x) + (d(x) p 3) q p + 5 d(x) + ( q )( q) d(x) + q + 4 = (d(x) + q + 4) + ( q)( q ) d(x) + q + 4 ( q)( q ) + q 4 3( q) 8 q, (q + 4) We now consider the case q d(x) + p + 1. Since σ q (x, y) d(x) + p + 1 for every y N(x), d(y) q d(y) σ q (x, y) d(y) q d(y) ( d(x) + p + 1) q d(x) p 1. (4) By (3), (4), the fact that d(x) > q +1 and the inequality θ+θ 1 (θ = d(x) p 3 p +5 ), we have M 1 (x, q) d(x) + (d(x) p 3) q p q (p + 3) d(x) p 1 > q ( q) ( d(x) p 3 p p + 3 ) d(x) p 1 q ( q) ( ( p + 5 d(x) p 3 p + 3 )). d(x) p 1 Since d(x) p 1 d(x) + and d(x) p 3 d(x) > q+1, we get p + 5 d(x) p 3 p + 3 d(x) p 1 = d(x) + 4 (d(x) p 3)(d(x) p 1) 8(d(x) + ) ( q + 1)(d(x) + 4) < 8 q. Thus, M 1 (x, q) > q ( q)( 8 ) = 3( q) 7 > q. q 19

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