Average degrees of edge-chromatic critical graphs

Yan Caoa,Guantao Chena, Suyun Jiangb∗, Huiqing Liuc∗, Fuliang Lud∗

a Department of Mathematics and Statistics, Georgia State University, Atlanta, GA 30303

b School of Mathematics, Shandong University, Jinan, 250100

c Faculty of Mathematics and Statistics, Hubei University, Wuhan 430062

d School of Mathematics and Statistics, Linyi University, Linyi, Shandong 276000

Abstract

Given a graph G, denote by ∆(G), d(G) and χ′(G) the maximum degree, theaverage degree and the chromatic index ofG, respectively. A simple graphG is callededge-∆-critical if χ′(G) = ∆(G) + 1 and χ′(H) ≤ ∆(G) for every proper subgraphH of G. Vizing in 1968 conjectured that if G is edge-∆-critical, then d(G) ≥∆(G)−1+ 3

n . We show that for any edge-∆-critical graph with ∆(G) ≥ 32 and any

positive number c, d̄(G) ≥ q−3q−2∆(G)−23+f(c) , where q = min{2

√2(∆(G)−1)−2

2√

2+1, 3

4∆(G)−2}

and f(c) = max{3− c∆(G)−q , 2} ×

(5c+2)∆(G)−(6c+3)q+3c+2c∆(G) . Consequently,

d̄(G) ≥

{0.69242∆(G)− 0.1701 if ∆(G) ≥ 66,

0.68707∆(G) + 0.0136 if 32 ≤ ∆(G) ≤ 65.

This result improves the best known bound 23(∆(G) + 2) obtained by Woodall in

2007 for ∆(G) ≥ 66. Additionally, Woodall constructed an infinite family of graphsshowing his result cannot be improved by well-known Vizing’s Adjacency Lemmaand other known edge-coloring techniques. To over come the barrier, we follow therecently developed recoloring technique of Tashkinov trees to expand Vizing fanstechnique to a larger class of trees.

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. 11671232, 11571096, 61373019, 11671186).

1

edge-k-coloring of a graph G is a mapping ϕ : E(G)→ {1, 2, . . . , k} such that ϕ(e) 6= ϕ(f)

for any two adjacent edges e and f . We call {1, 2, . . . , k} the color set of ϕ. Denote by

Ck(G) the set of all edge-k-colorings of G. The chromatic index χ′(G) is the least integer

k ≥ 0 such that Ck(G) 6= ∅. We call G class one if χ′(G) = ∆(G). Otherwise, Vizing’

theorem [12] gives 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 [14] made the following conjecture in 1968.

Conjecture 1. [Vizing’s Average Degree Conjecture] If G is a ∆-critical graph of n

vertices, then d(G) ≥ ∆(G)− 1 + 3n.

The conjecture has been verified for graphs with ∆(G) ≤ 6, see [3, 5, 6, 8]. In general,

there are a few results on the lower bound for d̄(G). Let G be a ∆-critical graph with

maximum degree ∆. Fiorini [2] showed, for ∆ ≥ 2,

d̄(G) ≥

{12(∆ + 1) if ∆ is odd;

12(∆ + 2) if ∆ is even.

Haile [4] improved the bounds as follows.

d̄(G) ≥

35(∆ + 2) if ∆ = 9, 11, 13;

∆+62− 12

∆+4if ∆ ≥ 10, ∆ is even;

15+√

292

if ∆ = 15;∆+7

2− 16

∆+5if ∆ ≥ 17,∆ is odd.

Sanders and Zhao [9] showed d̄(G) ≥ 12(∆ +

√2∆− 1) for ∆ ≥ 2. Woodall [16] improved

the bound to d̄(G) ≥ t(∆+t−1)2t−1

,where t = d√

∆/2e. Improving Vizing’s Adjacency Lemma,

Woodall [15] improved the coefficient of ∆ from 12

to 23

as follows.

d(G) ≥

23(∆ + 1) if ∆ ≥ 2;

23∆ + 1 if ∆ ≥ 8;

23(∆ + 2) if ∆ ≥ 15.

In the same paper, Woodall provided the following example demonstrating that the

above result cannot be improved by the use of his new adjacency Lemmas (see Lemma 2

and Lemma 3) and Vizing’s Adjacency Lemma alone.

2

Let G be a graph comprising k vertices of degree 4, all of whose neighbors have degree

∆(G), and 2k vertices of degree ∆(G), each of which is adjacent to two vertices of degree

4 and ∆(G)− 2 vertices of degree ∆(G). Graph G can be chosen to be triangle-free, and

indeed to have arbitrarily large girth. Then G may not be ∆-critical, but it satisfies the

conclusions of all the existing lemmas at that time including two mentioned above, and it

has average degree 23(∆(G) + 2). So, using these known results, it is impossible to prove

that the example is not ∆-critical. On the other hand, we note that using our new result,

Claim 3.4 in Section 3, it is readily seen that if ∆(G) ≥ 6 then the above example is not

∆-critical. By proving a few stronger properties of ∆-critical graphs, we get the following

theorem.

Theorem 1. Let G be a ∆-critical graph with maximum degree ∆ ≥ 32, c be a positive

number, and q = min{2√

2(∆−1)−2

2√

2+1, 3

4∆− 2}. Then d̄(G) ≥ q − 3q−2∆−2

3+f(c), where

f(c) = max{3− c

∆− q, 2} × (5c+ 2)∆− (6c+ 3)q + 3c+ 2

c∆.

By taking c = 18.5, we get Corollary 1, which improve Woodall’s result for ∆ ≥ 66.

Corollary 1. If G is a ∆-critical graph with maximum degree ∆ ≥ 32, then

d̄(G) ≥

{0.69242∆− 0.1701 if ∆ ≥ 66,

0.68707∆ + 0.0136 if 32 ≤ ∆ ≤ 65.

We will prove a few technical lemmas in Section 2 and give the proof of Theorem 1

in Section 3 and give the proof of Corollary 1 in Section 4. We will use the following

terminology and notation. Let G be a graph. Denote by N(x) the neighborhood of x for

any x ∈ V (G), 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 ∈ Ck(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 ϕ̄(v) = {1, 2, . . . , k} \ ϕ(v). We call ϕ(v) the set

of colors seen by v and ϕ̄(v) the set of colors missing at v. A set X ⊆ V (G) is called

elementary with respect to ϕ if ϕ̄(u) ∩ ϕ̄(v) = ∅ for every two distinct vertices u, v ∈ X.

For any color α, let Eα denote the set of edges assigned color α. Clearly, Eα is a matching

of G. For any two colors α and β, the components of the spanning subgraph of G with

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edge set Eα∪Eβ, named (α, β)-chains, are even cycles and paths with alternating color α

and β. For a vertex v of G, we denote by Pv(α, β, ϕ) the unique (α, β)-chain that contains

the vertex v. Let ϕ/Pv(α, β, ϕ) be the coloring obtained from ϕ by switching colors α

and β on the edges on Pv(α, β, ϕ). If v is not incident with any edge of color α or β, then

Pv(α, β, ϕ) = {v} (a path of length 0), and ϕ/Pv(α, β, ϕ) = ϕ.

2 Lemmas

Let G be a graph and q := q(G) be a positive number. For each edge xy ∈ E(G), let

σq(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 [13]] If G is a ∆-critical graph, then σ∆(G)(x, y) ≥∆(G)− d(x) + 1 for every xy ∈ E(G).

Woodall [15] studied σq(x, y) for the case q = 2∆(G)− d(x)− d(y) + 2 and obtained

the following two results. For convenience, we let σ(x, y) = σq(x, y) when q = 2∆(G) −d(x)−d(y)+2. Let G be a graph and xy ∈ E(G). To study the difference between σ(x, y)

and ∆(G)− d(x) + 1, Woodall defined the following two parameters.

pmin(x) := miny∈N(x)

σ(x, y)−∆(G) + d(x)− 1, (1)

p(x) := min{ pmin(x),

⌊d(x)

2

⌋− 1 }. (2)

Lemma 2. [Woodall [15]] 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) ≥2∆(G)− d(x)− σ(x, y).

Lemma 3. [Woodall [15]] 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 (2), for any positive number q, we define

the following two parameters.

pmin(x, q) := miny∈N(x)

σq(x, y)−∆(G) + d(x)− 1 and

p(x, q) := min{ pmin(x, q),

⌊d(x)

2

⌋− 3 }.

The following lemma is a generalization of Lemma 2, which serve as a key result.

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Lemma 4. Let xy be an edge in a ∆-critical graph G and q be a positive number. If

∆(G)/2 < q ≤ ∆(G) − d(x)/2 − 2, then x has at least ∆(G) − σq(x, y) − 2 vertices

z ∈ N(x) \ {y} such that σq(x, z) ≥ 2∆(G)− d(x)− σq(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. If

∆(G)/2 < q ≤ ∆(G)− d(x)/2− 2, then x has at least d(x)− p(x, q)− 3 neighbors y for

which σq(x, y) ≥ ∆(G)− p(x, q)− 5.

Proof. Let y ∈ N(x) such that pmin(x, q) = σq(x, y)−∆(G) + d(x)− 1, and ∆ = ∆(G).

If p(x, q) = pmin(x, q), by Lemma 4, x has at least ∆−σq(x, y)−2 = d(x)−pmin(x, q)−3

vertices z ∈ N(x) \ {y} such that σq(x, z) ≥ 2∆− d(x)−σq(x, y)− 4 = ∆− pmin(x, q)− 5.

If p(x, q) =⌊d(x)

2

⌋− 3 < pmin(x, q), then for every y ∈ N(x), σq(x, y) > ∆− d(x) + 1 +⌊

d(x)2

⌋− 3 ≥ ∆−

⌊d(x)

2

⌋− 3 = ∆− p(x, q)− 6. So σq(x, y) ≥ ∆− p(x, q)− 5.

Let G be a graph, e1 = y0y1 ∈ E(G) and ϕ ∈ Ck(G − e1). A Tashkinov tree T is a

sequence T = (y0, e1, y1, . . . , ep, yp) with p ≥ 1 consisting of distinct edges e1, e2, . . . , ep

and vertices y0, y1, . . . , yp such that the following two conditions hold.

• For each i with 1 ≤ i ≤ p, ei = yryi for some r < i;

• For every edge ei with 2 ≤ i ≤ p, there is a vertex yh with 0 ≤ h < i such that

ϕ(ei) ∈ ϕ̄(yh).

Tashkinov [11] introduced Taskinov technique in his work on the well-know Goldberg’s

Conjecture. In the above definition, if ei = y0yi for every i (i.e., T is a star with y0 as

the center), then T is a multi-fan defined in [10], which is a generalization of a Vizing

fan. Stiebitz et al. [10] showed that all multi-fans are elementary. In the definition of

Tashkinov tree, if ei = yi−1yi for every i (i.e. T is a path with end-vertices y0 and yp),

then T is a Kierstead path, which was introduced by Kierstead [7] in study edge-colorings

of multigraphs. For simple graphs, following Kierstead’s proof, Zhang [17] noticed that

for a Kierstead path P the set V (P ) is elementary if G is ∆-critical and d(yi) < ∆ for

every i with 2 ≤ 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.

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Lemma 6. [Kostochka and Stiebitz [10]] Let G be a graph with maximum degree ∆

and χ′(G) = ∆ + 1. Let e1 ∈ E(G) be a critical edge and ϕ ∈ C∆(G − e1). If K =

(y0, e1, y1, e2, y2, e3, y3) is a Kierstead path with respect to e1 and ϕ, then the following

statements hold:

1. ϕ̄(y0) ∩ ϕ̄(y1) = ∅;

2. if d(y2) < ∆, then V (K) is elementary with respect to ϕ;

3. if d(y1) < ∆, then V (K) is elementary with respect to ϕ;

4. if Γ = ϕ̄(y0) ∪ ϕ̄(y1), then |ϕ̄(y3) ∩ Γ| ≤ 1.

In the definition of Tashkinov tree T = (y0, e1, y1, e2, y2, . . . , yp), we call T a broom if

e2 = y1y2 and for each i ≥ 3, ei = y2yi, i.e., y2 is one of the end-vertices of ei for each

i ≥ 3. Moreover, we call T a simple broom if ϕ(ei) ∈ ϕ̄(y0) ∪ ϕ̄(y1) for each i ≥ 3, i.e.,

(y0, e1, y1, e2, y2, ei, yi) is a Kierstead path.

Lemma 7. [Chen, Chen, Zhao [1]] Let G be a graph with maximum degree ∆ and

χ′(G) = ∆ + 1. Let e1 = y0y1 ∈ E(G) be a critical edge and ϕ ∈ C∆(G − e1), and

let B = {y0, e1, y1, e2, y2, . . . , ep, yp} be a simple broom. If |ϕ̄(y0) ∪ ϕ̄(y1)| ≥ 4 and

min{d(y1), d(y2)} < ∆, then V (B) is elementary with respect to ϕ.

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) ∈ ϕ̄(y)}. Then

|Z| ≥ ∆− d(y) + 1−⌊d(x) + d(y)−∆− 2

∆− q

⌋, (3)∑

z∈Z

(d(z)− q) ≥ (∆− q)(∆− d(y) + 1)− d(x)− d(y) + ∆ + 2, (4)

and for every z ∈ Z,

σq(x, z) ≥ 2∆− d(x)− d(y) + 1−⌊d(x) + d(y) + d(z)− 2∆− 2

∆− q

⌋. (5)

Proof. Since G is not ∆-colorable, ϕ̄(x) ∩ ϕ̄(y) = ∅. Let Zy := {z ∈ N(x) \ {y} :

ϕ(xz) ∈ ϕ̄(y)}. Clearly, Z ⊆ Zy and |Zy| = ∆ − d(y) + 1. Since {y, x} ∪ Zy forms a

6

multi-fan with center x, it is elementary, so |ϕ̄(x)| + |ϕ̄(y)| +∑

z∈Zy |ϕ̄(z)| ≤ ∆. Since

|ϕ̄(x)| = ∆− d(x) + 1 and |ϕ̄(y)| = ∆− d(y) + 1, we have∑z∈Zy

|ϕ̄(z)| ≤ ∆− |ϕ̄(x)| − |ϕ̄(y)| ≤ d(x) + d(y)−∆− 2. (6)

Since d(z) ≤ q for all z ∈ Zy − Z,∑

z∈Zy |ϕ̄(z)| ≥ (|Zy| − |Z|)(∆ − q). Combining this

with (6), we get |Z| ≥ |Zy| −⌊d(x)+d(y)−∆−2

∆−q

⌋= ∆ − d(y) + 1 −

⌊d(x)+d(y)−∆−2

∆−q

⌋. So (3)

holds.Since |ϕ̄(z)| = ∆− d(z) for each z ∈ Zy, by (6), we get∑

z∈Zy

d(z) ≥ |Zy|∆− (d(x) + d(y)−∆− 2).

Since d(z) ≤ q for every z ∈ Zy − Z, we have∑z∈Z

(d(z)− q) ≥∑z∈Zy

(d(z)− q) ≥ |Zy|∆− (d(x) + d(y)−∆− 2)− |Zy|q.

Substituting |Zy| = ∆− d(y) + 1 in the above inequality, we get (4).

For each z ∈ Z, let U∗z = {u ∈ N(z)\{x} : ϕ(zu) ∈ ϕ̄(x) ∪ ϕ̄(y)} and Uz = {u ∈U∗z : d(u) > q}. Since ϕ(xz) ∈ ϕ̄(y), |U∗z | = 2∆ − d(x) − d(y) + 1 and {y, x, z} ∪ U∗zforms a simple broom. Since d(x) < q ≤ ∆ − 1, we have d(x) ≤ ∆ − 2. Thus |ϕ̄(x) ∪ϕ̄(y)| ≥ 4 and min{d(x), d(z)} = d(x) < ∆. By Lemma 7, {y, x, z} ∪ U∗z is elementary

with respect to ϕ. So∑

u∈U∗z|ϕ̄(u)| + |ϕ̄(x)| + |ϕ̄(y)| + |ϕ̄(z)| ≤ ∆, which in turn gives∑

u∈U∗z|ϕ̄(u)| ≤ d(x) + d(y) + d(z) − 2∆ − 2. Since d(u) ≤ q for every u ∈ U∗z − Uz,∑

u∈U∗z|ϕ̄(u)| ≥ (|U∗z | − |Uz|)(∆− q). So,

(|U∗z | − |Uz|)(∆− q) ≤ d(x) + d(y) + d(z)− 2∆− 2.

Solving the above inequality with |U∗z | = 2∆− d(x)− d(y) + 1, we get

σq(x, z) ≥ |Uz| ≥ 2∆− d(x)− d(y) + 1−⌊d(x) + d(y) + d(z)− 2∆− 2

∆− q

⌋.

So the inequality (5) holds.

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2.1 Proof of Lemma 4

Lemma 4. Let xy be an edge in a ∆-critical graph G and q be a positive number. If

∆(G)/2 < q ≤ ∆(G) − d(x)/2 − 2, then x has at least ∆(G) − σq(x, y) − 2 vertices

z ∈ N(x) \ {y} such that σq(x, z) ≥ 2∆(G)− d(x)− σq(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) ∈ ϕ̄(y), and such a coloring ϕ is called z-feasible. Denote by Cz the set of all

z-feasible colorings. For each feasible vertex z and each z-feasible coloring ϕ ∈ Cz, let

Z(ϕ) = {v ∈ N(z) \ {x} : ϕ(vz) ∈ ϕ̄(x) ∪ ϕ̄(y)},

Cz(ϕ) = {ϕ(vz) : v ∈ Z(ϕ) and d(v) < q} ⊆ ϕ̄(x) ∪ ϕ̄(y),

Y (ϕ) = {v ∈ N(y) \ {x} : ϕ(vy) ∈ ϕ̄(x) ∪ ϕ̄(z)}, and

Cy(ϕ) = {ϕ(vy) : v ∈ Y (ϕ) and d(v) < q} ⊆ ϕ̄(x) ∪ ϕ̄(z).

Note that Z(ϕ) and Y (ϕ) are vertex sets while Cz(ϕ) and Cy(ϕ) are color sets. For

each color α ∈ ϕ(z), let zα ∈ N(z) such that ϕ(zzα) = α. For each color β ∈ ϕ(y), let

yβ ∈ N(y) such that ϕ(yyβ) = β. Let

T (ϕ) = {α ∈ ϕ(x) ∩ ϕ(y) ∩ ϕ(z) : d(yα) < q and d(zα) < q}.

Since (ϕ(x)∩ϕ(y))∩ (ϕ̄(x)∪ ϕ̄(y)) = ∅ and (ϕ(x)∩ϕ(z))∩ (ϕ̄(x)∪ ϕ̄(z)) = ∅, we obtain

that T (ϕ) ∩ (Cz(ϕ) ∪ Cy(ϕ)) = ∅.Since G is ∆-critical and ϕ is z-feasible, {x, y, z} is elementary with respect to ϕ. So

ϕ̄(x), ϕ̄(y), ϕ̄(z) and ϕ(x) ∩ ϕ(y) ∩ ϕ(z) are mutually exclusive. It is not difficult to see

that|Z(ϕ)| = |ϕ̄(x) ∪ ϕ̄(y)| − 1 and |Y (ϕ)| = |ϕ̄(x) ∪ ϕ̄(z)|. (7)

Also,

ϕ̄(x) ∪ ϕ̄(y) ∪ ϕ̄(z) ∪ (ϕ(x) ∩ ϕ(y) ∩ ϕ(z)) = {1, 2, . . . ,∆}. (8)

Recall that σq(x, y) and σq(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

σq(x, y) + σq(x, z)

≥ |Y (ϕ)| − |Cy(ϕ)|+ |Z(ϕ)| − |Cz(ϕ)|+ |ϕ(x) ∩ ϕ(y) ∩ ϕ(z)| − |T (ϕ)|

= |ϕ̄(x) ∪ ϕ̄(z)|+ |ϕ̄(x) ∪ ϕ̄(y)| − 1 + |ϕ(x) ∩ ϕ(y) ∩ ϕ(z)| − |Cy(ϕ)| − |Cz(ϕ)| − |T (ϕ)|

= ∆ + |ϕ̄(x)| − 1− |Cy(ϕ)| − |Cz(ϕ)| − |T (ϕ)|

= 2∆− d(x)− |Cy(ϕ)| − |Cz(ϕ)| − |T (ϕ)|.

8

So, Lemma 4 follows from the three statements below.

I. For any ϕ ∈ Cz, |Cz(ϕ)| ≤ 1 and |Cy(ϕ)| ≤ 1;

II. there exists a ϕ ∈ Cz such that |T (ϕ)| ≤ 2; and

III. there are at least ∆− σq(x, y)− 2 feasible vertices z ∈ N(x) \ {y}.

For every z-feasible coloring ϕ ∈ C∆(G − xy), let ϕd ∈ C∆(G − xz) be obtained

from ϕ by assigning ϕd(xy) = ϕ(xz) and keeping all colors on other edges unchanged.

Clearly, ϕd is a y-feasible coloring and Z(ϕd) = Z(ϕ), Y (ϕd) = Y (ϕ), Cz(ϕd) = Cz(ϕ)

and Cy(ϕd) = Cy(ϕ). We call ϕd 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)/2− 2, we have

2(∆− q) + (∆− d(x) + 1)− 5 ≥ ∆. (†)

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. (∗)

Let z ∈ N(x) \ {y} be a feasible vertex and ϕ ∈ Cz. By the definition of Z(ϕ),

G[{y, x, z}∪Z(ϕ)] contains a simple broom, so {y, x, z}∪Z(ϕ) is elementary with respect

to ϕ. Thus, |Cz(ϕ)| ≤ 1 by (∗). By considering its dual ϕd, we have |Cy(ϕ)| = |Cy(ϕd)| ≤1. Hence, I holds. The proofs of II and III are much more complicated. Let R(ϕ) =

Cz(ϕ) ∪ Cy(ϕ). In the remainder of the proof, we let Z = Z(ϕ), Y = Y (ϕ), Cz = Cz(ϕ),

Cy = Cy(ϕ), R = R(ϕ) and T = T (ϕ) if the coloring ϕ is clear. Note that T ∩ R = ∅,ϕ(xz) /∈ T ∪R and |R| ≤ 2.

A coloring ϕ ∈ Cz is called optimal if |Cz(ϕ)|+ |Cy(ϕ)| is maximum over all z-feasible

colorings.

2.1.1 Proof of Statement II.

Suppose on the contrary that |T (ϕ)| ≥ 3 for every ϕ ∈ Cz. Let ϕ be an optimal

z-feasible coloring and assume, without loss of generality, ϕ(xz) = 1. Since χ′(G) > ∆

and 1 ∈ ϕ̄(y), it follows that

for every color i ∈ ϕ̄(x), Px(1, i, ϕ) = Py(1, i, ϕ). (‡)

Otherwise, an edge-∆-coloring of G can be gotten by Kempe change ϕ/Px(1, i, ϕ) and

then coloring xy with 1, a contradiction.

Claim A. For each i ∈ ϕ̄(x) \R and k ∈ T (ϕ), Px(i, k, ϕ) contains both y and z.

9

Proof. We first show that z ∈ V (Px(i, k, ϕ)). Otherwise, Pz(i, k, ϕ) is disjoint from

Px(i, k, ϕ). Let ϕ′ = ϕ/Pz(i, k, ϕ). Since 1 /∈ {i, k}, ϕ′ is also z-feasible. Since colors

in R are unchanged and d(zk) < q, Cz(ϕ′) = Cz(ϕ) ∪ {i} and Cy(ϕ

′) ⊇ Cy(ϕ), giving a

contradiction to the maximality of |Cy(ϕ)| + |Cz(ϕ)|. By considering the dual coloring

ϕd, we can verify that y ∈ V (Px(i, k, ϕ)).

Note that |T (ϕ)| ≥ 3. For each 3-element subsets S(ϕ) of T (ϕ), we may assume

S(ϕ) = {k1, k2, k3}, let

V (S(ϕ)) = {zk1 , zk2 , zk3} ∪ {yk1 , yk2 , yk3},

W (S(ϕ)) = {u ∈ V (S(ϕ)) : ϕ̄(u) ∩ ϕ̄(x) \R = ∅},

M(S(ϕ)) = {u ∈ V (S(ϕ)) : ϕ̄(u) ∩ ϕ̄(x) \R 6= ∅},

E(S(ϕ)) = {zzk1 , zzk2 , zzk3 , yyk1 , yyk2 , yyk3},

EW (S(ϕ)) = {e ∈ E(S(ϕ)) : e is incident to a vertex in W (S(ϕ))}, and

EM(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)) = EW (S(ϕ))∪EM(S(ϕ)). For con-

venience, we let W = W (S(ϕ)),M = M(S(ϕ)), EW = EW (S(ϕ)) and EM = EM(S(ϕ)) if

S(ϕ) is clear.

We assume that |EW (S(ϕ))| is minimum over all optimal z-feasible coloring ϕ and all

3-element subsets S(ϕ) of T (ϕ). For each v ∈ M , pick a color αv ∈ ϕ̄(v) ∩ ϕ̄(x) \ R. Let

CM = {αv : v ∈M}. Clearly, |CM | ≤ |M |. By (†) and the condition q > ∆/2, we have

|ϕ̄(x)| = ∆− d(x) + 1 ≥ ∆− 2(∆− q) + 5 > 5. (9)

Since |R| ≤ 2, we have

ϕ̄(x) \ (R ∪ CM) 6= ∅ if |CM | ≤ 3. (§)

Note that {zk1 , zk2 , zk3} ∩ {yk1 , yk2 , yk3} may be not empty, |EW |2≤ |W | ≤ |EW | and

|EM |2≤ |M | ≤ |EM |.

Claim B. Let u and v be two vertices of V (S(ϕ)). If ϕ̄(u)∩ ϕ̄(v)∩ϕ(x) \R 6= ∅, then

the following statements hold.

(i) If u, v ∈ W , then ϕ̄(x) \ (R ∪ CM) = ∅.(ii) There exists an optimal z-feasible coloring ϕ∗ with V (S(ϕ∗)) = V (S(ϕ)) such that

|EW (S(ϕ∗))| ≤ |EW | and {u, v} ∩M(S(ϕ∗)) 6= ∅.

Proof. Note that ϕ̄(u) ∩ ϕ̄(v) ∩ ϕ(x) \ R 6= ∅, we assume α ∈ ϕ̄(u) ∩ ϕ̄(v) ∩ ϕ(x) \ R. If

{u, v} ∩M 6= ∅, we are done (with ϕ∗ = ϕ). So suppose u, v ∈ W . Let β be an arbitrary

10

color in ϕ̄(x) \ (R ∪ CM) if this set is nonempty; otherwise, let β be a color in ϕ̄(x) \ R,

which is nonempty by (9) since |R| ≤ 2. 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

Pu(α, β, ϕ) is disjoint from Px(α, β, ϕ). Let ϕ∗ = ϕ/Pu(α, β, ϕ).

First, we note that T (ϕ∗) ⊇ T (ϕ). This is obvious if α /∈ T (ϕ), and if α ∈ T (ϕ)

it holds because that {z, y, zα, yα} ⊆ V (Px(α, β, ϕ)) by Claim A and so {z, y, zα, yα} ∩V (Pu(α, β, ϕ)) = ∅. So we can choose V (S(ϕ∗)) such that V (S(ϕ∗)) = V (S(ϕ)).

Next, we claim that ϕ∗ is an optimal z-feasible coloring. Note that ϕ∗(xz) = 1 ∈ϕ̄∗(y). This is obvious if α 6= 1 (since β 6= 1), and if α = 1 it holds because that

Px(α, β, ϕ) = Py(α, β, ϕ) by (‡) and so V (Pu(α, β, ϕ)) ∩ {x, y, z} = ∅. So ϕ∗ is z-feasible.

Since α, β /∈ R = Cy ∪ Cz, it follows that Cy ⊆ Cy(ϕ∗) and Cz ⊆ Cz(ϕ

∗). Since ϕ is

optimal, we have Cy(ϕ∗) = Cy, Cz(ϕ

∗) = Cz, R(ϕ∗) = R and so ϕ∗ is optimal.

Since α ∈ ϕ̄(u), it follows that β ∈ ϕ̄∗(u)∩ ϕ̄∗(x) \R(ϕ∗), so u ∈M(S(ϕ∗)) (that is, u

has moved from W to M(S(ϕ∗))). To avoid the contradiction that |EW (S(ϕ∗))| < |EW |,it is necessary that Pu(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 ϕ̄(w)∩ ϕ̄(x)\R, thus

ϕ̄∗(w)∩ϕ̄∗(x)\R(ϕ∗) = ∅ (that is, w has moved from M to W (S(ϕ∗))). But then we must

have chosen αw = β, so β ∈ CM . By the choice of β, we have ϕ̄(x) \ (R ∪CM) = ∅. Thus

(i) holds. By (§), we have |CM | ≥ 4. Since u, v ∈ W , |V (S(ϕ))| ≤ 6 and |CM | ≤ |M |,we have |V (S(ϕ))| = 6, |W | = 2 and |M | = 4. Thus |EW (S(ϕ∗))| = |EW |. Hence, (ii)

holds.

Claim C. There exist a color k ∈ {k1, k2, k3}, three distinct colors i, j, l and an optimal

z-feasible coloring ϕ∗ such that i ∈ ϕ̄∗(zk)∩ ϕ̄∗(x) \R(ϕ∗), j ∈ ϕ̄∗(yk)∩ ϕ̄∗(x) \R(ϕ∗) and

` ∈ (ϕ̄∗(zk) ∪ ϕ̄∗(yk)) ∩ (ϕ̄∗(x) ∪ {1} \R(ϕ∗)).

Proof. First we show that there exists a color k ∈ {k1, k2, k3} such that ϕ̄(zk)∩ϕ̄(x)\R 6= ∅and ϕ̄(yk)∩ ϕ̄(x)\R 6= ∅. If |EM | ≥ 4, by the definition of M and EM , it is not difficult to

see that the above statement holds. Thus we may assume that |EM | ≤ 3. So |EW | ≥ 3 and

then we have |W | ≥ 2 as |W | ≥ |EW |/2. Let u, v ∈ W . Since |CM | ≤ |M | ≤ |EM | ≤ 3,

we have ϕ̄(x) \ (R∪CM) 6= ∅ by (§). By Claim B (i), we have ϕ̄(u)∩ ϕ̄(v)∩ϕ(x) \R = ∅.On the other hand, by the definition of T (ϕ), we note that d(u) < q and d(v) < q. Since

|R| ≤ 2, by (†), we have

|ϕ̄(u) \R|+ |ϕ̄(v) \R|+ |ϕ̄(x)| > 2(∆− q − 2) + ∆− d(x) + 1 > ∆.

Moreover, by the definition of W and u, v ∈ W , we have (ϕ̄(u) \R)∩ ϕ̄(x) = (ϕ̄(v) \R)∩ϕ̄(x) = ∅. This implies that (ϕ̄(u) \R) ∩ (ϕ̄(v) \R) ∩ ϕ(x) 6= ∅, a contradiction.

11

Now we claim that ϕ̄(zk) ∩ ϕ̄(yk) ∩ ϕ̄(x) \ R = ∅. For otherwise, we assume i ∈ϕ̄(zk)∩ ϕ̄(yk)∩ ϕ̄(x) \R, then by Claim A, the path Px(i, k, ϕ) contains three endvertices

x, zk and yk, a contradiction.

Since ϕ̄(zk)∩ ϕ̄(yk)∩ ϕ̄(x) \R = ∅, ϕ̄(zk)∩ ϕ̄(x) \R 6= ∅ and ϕ̄(yk)∩ ϕ̄(x) \R 6= ∅, we

can find two distinct colors i, j such that i ∈ ϕ̄(zk)∩ ϕ̄(x) \R and j ∈ ϕ̄(yk)∩ ϕ̄(x) \R. If

there still exists another color ` ∈ (ϕ̄(zk)∪ ϕ̄(yk))∩ (ϕ̄(x)∪ {1} \R), then Claim C holds

with ϕ∗ = ϕ. Thus color ` does not exist, that is,

ϕ̄(zk) ∩ ϕ̄(x) \R = {i}, ϕ̄(yk) ∩ ϕ̄(x) \R = {j} and 1 /∈ ϕ̄(zk) ∪ ϕ̄(yk),

which implies that

((ϕ̄(zk) \ (R ∪ {i})) ∪ (ϕ̄(yk) \ (R ∪ {j}))) ∩ (ϕ̄(x) ∪R ∪ {1}) = ∅. (10)

By the definition of T (ϕ), we have d(zk) < q and d(yk) < q. Since |R| ≤ 2, by (†), we

have

|ϕ̄(zk)\(R∪{i})|+|ϕ̄(yk)\(R∪{j})|+|ϕ̄(x)∪(R∪{1})| > 2(∆−q−|R|−1)+∆−d(x)+2 ≥ ∆

So there exists a color α in the sets ϕ̄(zk) \ (R∪ {i}) and ϕ̄(yk) \ (R∪ {j}) by (10). Thus

α ∈ ϕ̄(zk) ∩ ϕ̄(yk) ∩ ϕ(x) \ (R ∪ {i, j, 1}).Since |R| ≤ 2, by (9), there exists a color β ∈ ϕ̄(x) \ (R ∪ {i, j}). By (10), we have

β ∈ ϕ(zk) ∩ ϕ(yk). We may assume that Pzk(α, β, ϕ) is disjoint from Px(α, β, ϕ). Let

ϕ∗ = ϕ/Pzk(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 ϕ̄∗(x) = ϕ̄(x).

Since β ∈ ϕ(zk), we have β ∈ ϕ̄∗(zk) ∩ ϕ̄∗(x) \ (R(ϕ∗) ∪ {i, j}). Moreover, we have

i ∈ ϕ̄∗(zk)∩ ϕ̄∗(x) \R(ϕ∗) and j ∈ ϕ̄∗(yk)∩ ϕ̄∗(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, ` and ϕ∗ be as stated in Claim C. If ` 6= 1, we consider coloring obtained

from ϕ∗ by interchange colors 1 and ` for edges not on the path Px(1, `, ϕ∗), and rename

it as ϕ∗. So we may assume 1 ∈ ϕ̄∗(yk) ∪ ϕ̄∗(zk).We first consider the case of 1 ∈ ϕ̄∗(yk). By Claim A, the paths Px(i, k, ϕ

∗) and

Px(j, k, ϕ∗) both contain y, z. Since ϕ∗(yyk) = ϕ∗(zzk) = k, these two paths also contain

yk, zk. Since i ∈ ϕ̄∗(zk), we have x and zk are the two endvertices of Px(i, k, ϕ∗). So,

i ∈ ϕ∗(y)∩ϕ∗(z)∩ϕ∗(yk). Similarly, we have j ∈ ϕ∗(y)∩ϕ∗(z)∩ϕ∗(zk). We now consider

the following sequence of colorings of G− xy.

Let ϕ1 be obtained from ϕ∗ by assigning ϕ1(yyk) = 1. Since 1 is missing at both y

and yk, ϕ1 is an edge-∆-coloring of G−xy. Now k is missing at y and yk, i is still missing

12

at zk. Since G is not ∆-colorable, Px(i, k, ϕ1) = Py(i, k, ϕ1); otherwise ϕ1/Py(i, k, ϕ1) can

be extended to an edge-∆-coloring of G, giving a contradiction. Furthermore, zk, yk /∈V (Px(i, k, ϕ1)) since either i or k is missing at these two vertices, which in turn shows

that z /∈ V (Px(i, k, ϕ1)) since ϕ1(zzk) = k.

Let ϕ2 = ϕ1/Px(i, k, ϕ1). We have k ∈ ϕ̄2(x), i ∈ ϕ̄2(y)∩ϕ̄2(zk) and j ∈ ϕ̄2(x)∩ϕ̄2(yk).

Since G is not edge-∆-colorable, we have Px(i, j, ϕ2) = Py(i, j, ϕ2), which contains neitheryk nor zk.

Let ϕ3 = ϕ2/Px(i, j, ϕ2). Then k ∈ ϕ̄3(x) and j ∈ ϕ̄3(y) ∩ ϕ̄3(yk).

Let ϕ4 be obtained from ϕ3 by recoloring yyk by j. Then 1 ∈ ϕ̄4(y), ϕ4(xz) = 1, k ∈ϕ̄4(x), ϕ4(zzk) = k. Since ϕ4(xz) = 1 ∈ ϕ̄4(y), ϕ4 is z-feasible. Since i, j, k /∈ R = Cy∪Cz,the colors in R are unchanged during this sequence of re-colorings, so Cy(ϕ4) ⊇ Cy and

Cz(ϕ4) ⊇ Cz. Since ϕ4(zzk) = k ∈ ϕ̄4(x) and d(zk) < q, we have k = ϕ4(zzk) ∈ Cz(ϕ4).

So, Cz(ϕ4) ⊇ Cz ∪ {k}. We therefore have |Cy(ϕ4)|+ |Cz(ϕ4)| ≥ |Cy|+ |Cz|+ 1, giving a

contradiction.

For the case of 1 ∈ ϕ̄∗(zk), we consider the dual coloring ϕd 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 |Cy| + |Cz|. This completes the

proof of Statement II.

2.1.2 Proof of Statement III.

For a coloring ϕ ∈ C∆(G − xy), let X(ϕ) = {z ∈ N(x) : ϕ(xz) ∈ ϕ̄(y)} and S(ϕ) =

{z ∈ N(x)\ (X(ϕ)∪{y}) : d(yϕ(xz)) < q}, where yj ∈ N(y) with ϕ(yyj) = 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 2.1. For any coloring ϕ ∈ C∆(G− xy), the following two statements hold.

a. |X(ϕ) ∪ S(ϕ)| ≥ ∆− σq(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) ∈ ϕ̄∗(y).

Proof. Let ϕ ∈ C∆(G− xy). Since G is ∆-critical, it is easy to see that ϕ̄(y) ⊆ ϕ(x) and

ϕ̄(x) ⊆ ϕ(y). To prove a, we divide ϕ(y) into two subsets:

ϕ(y,≥ q) = {i ∈ ϕ(y) : d(yi) ≥ q} and ϕ(y,< q) = {i ∈ ϕ(y) : d(yi) < q}.

Clearly, σq(x, y) = |ϕ(y,≥ q)| and |ϕ̄(y)|+ |ϕ(y,< q)| = ∆− σq(x, y). Also, |X(ϕ)| =|ϕ̄(y) ∩ ϕ(x)| = |ϕ̄(y)| since ϕ̄(y) ⊆ ϕ(x), and |S(ϕ)| = |ϕ(y,< q) ∩ ϕ(x)|. Since edge xy

13

and the edges incident to y with colors in ϕ̄(x) form a multi-fan F , the vertex set V (F )

is elementary with respect to ϕ. By (∗), V (F ) \ {x} contains at most one vertex with

degree less than q. Thus |ϕ̄(x) ∩ ϕ(y,< q)| ≤ 1 and so |S(ϕ)| ≥ |ϕ(y,< q)| − 1, and thisproves a.

To prove b, we show that for any two distinct vertices zk, z` ∈ S(ϕ), there is a coloring

ϕ∗ ∈ C∆(G − xy) such that at least one of ϕ∗(xzk) and ϕ∗(xz`) is in ϕ̄∗(y). We assume

ϕ(xzk) = k and ϕ(xz`) = `. Let yk, y` ∈ N(y) \ {x} such that ϕ(yyk) = k and ϕ(yy`) = `.

By the definition of S(ϕ), we have d(yk) < q and d(y`) < q. It follows from (†) that

|ϕ̄(x)|+ |ϕ̄(yk)|+ |ϕ̄(y`)| > ∆. (11)

Let z be an arbitrary vertex in X(ϕ), which exists since |X(ϕ)| = |ϕ̄(y)| > 0, and assume

ϕ(xz) = 1, so that (‡) holds. We claim that there exists a coloring ϕ′ ∈ C∆(G− xy) such

that ϕ′(xz) = 1 ∈ ϕ̄′(y), ϕ′(xzk) = ϕ′(yyk) = k, ϕ′(xz`) = ϕ′(yy`) = `, and

ϕ̄′(x) ∩ (ϕ̄′(yk) ∪ ϕ̄′(y`)) 6= ∅. (12)

Suppose (12) does not hold with ϕ′ = ϕ. Then ϕ̄(yk) ∪ ϕ̄(y`) ⊆ ϕ(x), and so by

(11) there exists a color r ∈ ϕ(x) ∩ ϕ̄(yk) ∩ ϕ̄(y`). Choose a color i ∈ ϕ̄(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, yk and y`, we may assume Pyk(i, r, ϕ) is disjoint from Px(i, r, ϕ), and also from

Py(i, r, ϕ) if r = 1, by (‡). Then (12) holds with ϕ′ = ϕ/Pyk(i, r, ϕ), since in this coloring

i is missing at both x and yk.

By (12), we may assume that there exists a color i ∈ ϕ̄′(x)∩ϕ̄′(yk). Applying (‡) to ϕ′,

we have Px(1, i, ϕ′) = Py(1, i, ϕ

′), so we have Pyk(1, i, ϕ′) is disjoint from Px(1, i, ϕ

′) and

Py(1, i, ϕ′). Thus in the coloring ϕ′′ = ϕ′/Pyk(1, i, ϕ

′), color 1 is missing at both y and yk.

Form ϕ∗ from ϕ′′ by changing the color of yyk from k to 1. Then ϕ∗(xzk) = k ∈ ϕ̄∗(y).

This proves b, and so completes the proofs of both Claim 2.1 and III.

3 Proof of Theorem 1

Let G be a ∆-critical graph with maximum degree ∆ ≥ 32. Let n = |V (G)| and m =

|E(G)|. Clearly, d(G) = 2m/n. Let q := min{2√

2(∆−1)−2

2√

2+1, 3

4∆−2}, that is, q = 2

√2(∆−1)−2

2√

2+1

if ∆ ≥ 66 and q = 34∆ − 2 if 32 ≤ ∆ ≤ 65. We initially assign to each vertex x of G a

charge M(x) = d(x) and redistribute the charge according to the following rule:

14

• Discharging Rule: each (> q)-vertex y distributes its surplus charge of d(y) − qequally among all (< q)-neighbors of y.

Denote by M ′(x) the resulting charge on each vertex x. Clearly,∑

x∈V (G) M′(x) =∑

x∈V (G)M(x) = 2m. Let X1 = {x ∈ V (G) : d(x) ≤ 3q − 2∆}. We show that

M ′(x) ≥ 2 + 2(∆ − q) for all vertices in X1 and M ′(x) ≥ q for all other vertices, which

gives d(G) ≥ q − (3q − 2∆− 2) |X1|n

(note that 3q − 2∆− 2 ≥ 0 since ∆ ≥ 32). We then

show that |X1|/n is small in order to complete our proof.

Since q = min{2√

2(∆−1)−2

2√

2+1, 3

4∆ − 2} and ∆ ≥ 32 > 12, we have ∆+2

2< q < 3∆

4. Thus

q > ∆− q + 2 > 3q − 2∆. Denote by d<q(y) the number of (< q)-neighbors of y.

Claim 3.1. If d(x) ≤ ∆− q + 2, then M ′(x) ≥ d(x) + 2(∆− q). Consequently, M ′(x) ≥d(x) + 2(∆− q) for each x ∈ X1.

Proof. Note that 3q− 2∆ < ∆− q+ 2 since q < 3∆4

, so the second statement in the claim

follows from the first. Also, d(x) ≤ ∆ − q + 2 < q since q > ∆+22

. Let y be an arbitrary

neighbor of x. Since 2∆− d(x)− d(y) + 2 ≥ ∆− d(x) + 2 ≥ q, we have σq(x, y) ≥ σ(x, y).

We will use lower bounds of σ(x, y) to estimate σq(x, y). By (1) and (2), we have

1 ≤ d<q(y) ≤ d(y)− σ(x, y) ≤ d(y)− (∆− d(x) + p(x) + 1). (13)

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). (14)

We first consider the case p(x) ≥ 1. In this case, by the hypothesis of Claim 3.1 and

(13), we have q ≤ ∆− d(x) + 2 ≤ ∆− d(x) + p(x) + 1 < d(y). Since d(y)−ad(y)−b with a ≤ b is

a decreasing function of d(y), for each y ∈ N(x), x receives charge at least

d(y)− qd(y)− (∆− d(x) + p(x) + 1)

≥ ∆− qd(x)− p(x)− 1

.

And there are at least d(x)− p(x)− 1 neighbors y of x giving x at least

d(y)− qd(y)− (∆− p(x)− 1)

≥ ∆− qp(x) + 1

,

15

where the inequality holds because q ≤ ∆−d(x)+2 ≤ ∆−p(x)−1 as 1 ≤ p(x) ≤ bd(x)2c−1,

which implies that d(x) ≥ 4 and p(x) ≤ d(x)− 3. Thus x receives at least

(d(x)− p(x)− 1)∆− qp(x) + 1

+ (p(x) + 1)∆− q

d(x)− p(x)− 1= (θ + θ−1)(∆− q) ≥ 2(∆− q),

where θ = d(x)−p(x)−1p(x)+1

. It follows that M ′(x) ≥M(x) + 2(∆− q) = d(x) + 2(∆− q).

We now consider the case p(x) = min{pmin(x), bd(x)2c−1} = 0. If d(x) = 2, then by (13)

for every neighbor y of x we have d<q(y) = 1 and d(y) = ∆, thus M ′(x) ≥M(x) + 2(∆−q) = d(x)+2(∆−q). If d(x) ≥ 3, then by (14) for at least d(x)−1 neighbors y of x, we have

d<q(y) = 1 and d(y) = ∆. Thus M ′(x) ≥M(x)+(d(x)−1)(∆−q) ≥ d(x)+2(∆−q).

Claim 3.2. For each x ∈ V (G)−X1, M′(x) ≥ q.

Proof. Let x ∈ V (G) − X1, i.e., d(x) > 3q − 2∆. If d(x) ≥ q, then M ′(x) = M(x) −d(x)−qd<q(x)

d<q(x) = q. If 3q − 2∆ < d(x) ≤ ∆ − q + 2, then by Claim 3.1, we have M ′(x) ≥

d(x) + 2(∆− q) > q. So we only need to consider the case ∆− q + 2 < d(x) < q.

Let y be a neighbor of x. Then there exists a coloring ϕ ∈ C∆(G − xy) as G is ∆-

critical. Let Zq = {z ∈ N(x) : d(z) > q}, Zy = {z ∈ N(x)\{y} : ϕ(xz) ∈ ϕ̄(y)} and

Z∗y = Zq ∩ Zy. 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 Zq and the discharging rule, for each

z ∈ Zq, x receives charge at least d(z)−qd(z)−σq(x,z) . Thus M ′(x) ≥ d(x) +

∑z∈Zq

d(z)−qd(z)−σq(x,z) .

Since ∆ ≥ 32 > 16, we have ∆− q > 6 and so ∆− q + 2 < 2(∆− q)− 4 < d(x). We

consider the following three cases to complete the proof.

Case 1. ∆− q + 2 < 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)− 2∆− 2 < 2q −∆ < 2(∆− q),

since q < 34∆. It follows from this and (5) in Lemma 8 that, for each vertex z ∈ Z∗y ,

σq(x, z) ≥ 2∆ − d(x) − d(y), and so d(z) − σq(x, z) ≤ d(x) + d(y) − ∆. Also, by (4) in

Lemma 8,∑

z∈Z∗y(d(z)− q) ≥ (∆− q)(∆− d(y) + 1) + 2− (d(x) + d(y)−∆). So, using

16

d(y) ≤ q in the next line gives

M ′(x) ≥ d(x) +∑z∈Z∗y

d(z)− qd(z)− σq(x, z)

≥ d(x) +(∆− q)(∆− d(y) + 1) + 2

d(x) + d(y)−∆− 1

≥ d(x) + q −∆ +(∆− q)(∆− q + 1) + 2

d(x) + q −∆+ ∆− q − 1

≥ 2√

(∆− q)(∆− q + 1) + 2 + ∆− q − 1

> 3(∆− q)− 1 > q,

where in the penultimate line we used a + ba≥ 2√b for all a, b > 0, and in the final line

that 3(∆− q) ≥ q + 8 since q ≤ 34∆− 2.

Case 2. 2(∆− q)− 4 < d(x) < q and d(y) > q for every y ∈ N(x).

Let y ∈ N(x) with minimum degree. Let u be an arbitrary vertex in N(x). Then

d(u) ≥ d(y) by the choice of y. By Lemma 1, we have σq(x, u) ≥ σ∆(x, u) ≥ ∆−d(x) + 1.

Since d(u) ≥ d(y) and q > ∆− d(x) + 1, we have

d(u)− qd(u)− σq(x, u)

≥ d(u)− qd(u)− (∆− d(x) + 1)

≥ d(y)− qd(y)− (∆− d(x) + 1)

,

where the inequality holds because d(u)−ad(u)−b with a > b > 0 is a increasing function of d(u).

Note that |N(x) \ Z∗y | ≥ |N(x) \ Zy| = d(x) − (∆ − d(y) + 1). Combining this with the

above inequality, we have

∑u∈N(x)\Z∗y

d(u)− qd(u)− σq(x, u)

≥ d(y)− q. (15)

For any vertex z, since d(y) ≤ ∆, d(z) ≤ ∆ and q < 34∆,

d(x) + d(y) + d(z)− 2∆− 2 < d(x) < q < 3(∆− q).

For each z ∈ Z∗y , by (5) in Lemma 8,

σq(x, z) ≥ 2∆− d(x)− d(y) + 1−⌊d(x) + d(y) + d(z)− 2∆− 2

∆− q

⌋≥ 2∆− d(x)− d(y)− 1,

17

and so d(z)− σq(x, z) ≤ d(x) + d(y)−∆ + 1. Combining this with (4) in Lemma 8, (15)

and using ∆− d(y) + 1 = d(x) + 2− (d(x) + d(y)−∆ + 1) in the next line gives

M ′(x) ≥ d(x) +∑z∈Z∗y

d(z)− qd(z)− σq(x, y)

+∑

u∈N(x)\Z∗y

d(u)− qd(u)− σq(x, u)

> d(x) +(∆− q)(∆− d(y) + 1)

d(x) + d(y)−∆ + 1− 1 + d(y)− q

= d(x) + d(y)−∆ + 1 +(∆− q)(d(x) + 2)

d(x) + d(y)−∆ + 1− 2

≥ 2√

(∆− q)(d(x) + 2)− 2

> 2√

2(∆− q − 1)− 2 ≥ q.

where in the penultimate line we used a + ba≥ 2√b for all a, b > 0, and in the final line

we used (∆ − q)(d(x) + 2) > 2(∆ − q − 1)2 since d(x) > 2(∆ − q) − 4 and we also used

q ≤ 2√

2(∆−1)−2

2√

2+1.

Case 3. ∆− q + 2 < d(x) ≤ 2(∆− 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{ pmin(x, q),

bd(x)2c−3 }, where pmin(x, q) = miny∈N(x) σq(x, y)−∆+d(x)−1. So σq(x, y) ≥ ∆−d(x)+

p′ + 1 for every y ∈ N(x).

Since d(x) ≤ 2(∆− q)− 4 and p′ ≤ bd(x)2c − 3, we have q ≤ ∆− d(x)

2− 2 ≤ ∆− p′− 5.

By Lemma 5, x has at least d(x)− p′ − 3 neighbors y for which σq(x, y) ≥ ∆− p′ − 5, so

for these neighbors y,

d(y)− qd(y)− σq(x, y)

≥ d(y)− qd(y)− (∆− p′ − 5)

≥ ∆− qp′ + 5

. (16)

18

If q > ∆− d(x) + p′ + 1, i.e., p′ < d(x) + q −∆− 1, then

M ′(x) ≥ d(x) + (d(x)− p′ − 3)∆− qp′ + 5

≥ d(x) +(∆− q − 2)(∆− q)d(x) + q −∆ + 4

= (d(x) + q −∆ + 4) +(∆− q)(∆− q − 2)

d(x) + q −∆ + 4− (q −∆ + 4)

≥ 2√

(∆− q)(∆− q − 2) + ∆− q − 4

≥ 3(∆− q)− 8 ≥ q. ( since q ≤ 34∆− 2)

We now consider the case q ≤ ∆ − d(x) + p′ + 1. Since σq(x, y) ≥ ∆ − d(x) + p′ + 1

for every y ∈ N(x),

d(y)− qd(y)− σq(x, y)

≥ d(y)− qd(y)− (∆− d(x) + p′ + 1)

≥ ∆− qd(x)− p′ − 1

. (17)

By (16), (17), the fact that d(x) > ∆−q+2 and the inequality θ+θ−1 ≥ 2 (θ = d(x)−p′−3p′+5

),

we have

M ′(x) ≥ d(x) + (d(x)− p′ − 3)∆− qp′ + 5

+ (p′ + 3)∆− q

d(x)− p′ − 1

> ∆− q + 2 + (∆− q)(d(x)− p′ − 3

p′ + 5+

p′ + 3

d(x)− p′ − 1

)≥ ∆− q + 2 + (∆− q)

(2−

( p′ + 5

d(x)− p′ − 3− p′ + 3

d(x)− p′ − 1

)),

Since d(x)− p′ − 1 > d(x)2

+ 2 and d(x)− p′ − 3 > d(x)2> ∆−q+2

2, we get

p′ + 5

d(x)− p′ − 3− p′ + 3

d(x)− p′ − 1<

2d(x) + 4

(∆−q+22

)(d(x)2

+ 2)=

8(d(x) + 2)

(∆− q + 2)(d(x) + 4).

Thus,

M ′(x) > ∆− q + 2 + (∆− q)(2− 8(d(x) + 2)

(∆− q + 2)(d(x) + 4))

≥ 3(∆− q)− 6 > q.

19

Recall that X1 = {x ∈ V (G) : d(x) ≤ 3q − 2∆} and let N(X1) = ∪x∈X1N(x).

Claim 3.3. d(y) > q for every y ∈ N(X1) and |N(X1)| ≥ 2|X1|.

Proof. For every y ∈ N(X1) and x ∈ X1 such that xy ∈ E(G), since G is ∆-critical and

q < 34∆, by Lemma 1, we have d(y) ≥ σ∆(x, y)+1 ≥ ∆−d(x)+2 ≥ ∆−(3q−2∆)+2 > q.

So X1 ∩N(X1) = ∅. Thus the vertices in N(X1) does not receive charges from any other

vertices. As the vertices in X1 receive charges only from the vertices in N(X1) and

d(y) ≤ ∆, we have∑x∈X1

M ′(x) +∑

y∈N(X1)

M ′(y) ≤∑x∈X1

M(x) +∑

y∈N(X1)

M(y) ≤∑x∈X1

d(x) + ∆|N(X1)|. (18)

Also, by Claims 3.1 and 3.2, we have M ′(x) ≥ d(x) + 2(∆ − q) for each x ∈ X1 and

M ′(y) ≥ q for each y ∈ N(X1). Thus we have∑x∈X1

M ′(x) +∑

y∈N(X1)

M ′(y) ≥∑x∈X1

d(x) + 2(∆− q)|X1|+ q|N(X1)|. (19)

Combining (18) with (19), we have |N(X1)| ≥ 2|X1|.

For each y ∈ N(X1), x ∈ N(y) ∩X1 and ϕ ∈ C∆(G− xy), let

Y (x, ϕ) = {w ∈ N(y) \ {x} : ϕ(yw) ∈ ϕ̄(x)}.

For any vertex w ∈ Y (x, ϕ), {x, y, w} is an elementary set with respect to ϕ. So d(w) ≥|ϕ̄(x)|+ |ϕ̄(y)| > ∆− (3q − 2∆) > 3q − 2∆ since q < 5

6∆. So we have Y (x, ϕ) ∩X1 = ∅.

LetY1(x, ϕ) = Y (x, ϕ) ∩N(X1) and Y2(x, ϕ) = Y (x, ϕ)−N(X1).

It is easy to see that Y1(x, ϕ)∪ Y2(x, ϕ) = Y (x, ϕ) and |Y (x, ϕ)| = |ϕ̄(x)| = ∆− d(x) + 1.

Claim 3.4. For each y ∈ N(X1) and x ∈ N(y) ∩X1, |Y2(x, ϕ)| ≥ ∆− 2d(x) + 3.

Proof. We will show the equivalent statement |Y1(x, ϕ)| ≤ d(x)− 2 by two subclaims.

Subclaim 3.4.1. For any w ∈ Y1(x, ϕ) and z ∈ N(w) ∩X1, we have ϕ(wz) ∈ ϕ(x) ∩ϕ(y).

Proof. Suppose on the contrary that ϕ(wz) ∈ ϕ̄(x) ∪ ϕ̄(y). So xywz is a Kierstead path.

By Lemma 6, we have |ϕ̄(z)∩ (ϕ̄(x)∪ ϕ̄(y))| ≤ 1, it follows that d(z) ≥ (∆− d(x) + 1) +

(∆− d(y) + 1)− 1 > 3q − 2∆, giving a contradiction to z ∈ X1.

20

For each color j ∈ ϕ(x) ∩ ϕ(y), let Yj = {w ∈ Y1(x, ϕ) : j ∈ ϕ(w)} and Zj = {z ∈X1 : there exists a vertex w ∈ Yj such that ϕ(wz) = j}. Clearly, for each z ∈ Zj, the

corresponding vertex w is unique. So the color ϕ(yw) is uniquely determined by z. By

Subclaim 3.4.1,∑

j∈ϕ(x)∩ϕ(y) |Zj| ≥ |Y1(x, ϕ)|. Since |ϕ(x)∩ϕ(y)| = ∆−|ϕ̄(x)|− |ϕ̄(y)| ≤d(x)− 2, to show that |Y1(x, ϕ)| ≤ d(x)− 2, we only need to prove that |Zj| ≤ 1 for each

j. Let |Zj| = t and Zj = {zα1 , . . . , zαt}, where for each zαi , yαi is the unique vertex in Yj

such that ϕ(yyαi) = αi and ϕ(yαizαi) = j. Clearly, αi ∈ ϕ̄(x) for each 1 ≤ i ≤ t.

Subclaim 3.4.2. For each color k ∈ ϕ̄(x), the followings hold.

(1) If k /∈ {α1, . . . , αt}, at least t− 1 vertices of Zj seeing the color k.

(2) If k ∈ {α1, . . . , αt}, at least t− 2 vertices of Zj seeing the color k.

Proof. First suppose that k /∈ {α1, . . . , αt}. We consider the path Px(j, k, ϕ), and let v

be the other end vertex of this path. We will show that the color k seen by each vertex

in Zj\{v}. For otherwise, we assume k /∈ ϕ(z) for some z ∈ Zj\{v}, say z = zα1 , then

Pzα1(j, k, ϕ) is disjoint from Px(j, k, ϕ), thus let ϕ′ = ϕ/Pzα1

(j, k, ϕ) be the Kempe change.

Clearly, we have ϕ′(yyα1) = ϕ(yyα1) ∈ ϕ̄′(x) and ϕ′(yα1zα1) = k ∈ ϕ̄′(x). Thus xyyα1zα1

is a Kierstead path. So by Lemma 6 we have |ϕ̄(zα1) ∩ (ϕ̄(x) ∪ ϕ̄(y))| ≤ 1, it follows that

d(zα1) ≥ (∆− d(x) + 1) + (∆− d(y) + 1)− 1 > 3q − 2∆ as d(x) ≤ 3q − 2∆ and q < 34∆,

this contradicts with the fact that zα1 ∈ X1.

Then suppose that k ∈ {α1, . . . , αt}. We may assume that k = αt. Clearly, k /∈{α1, . . . , αt−1}. Let Z ′j = Zj − {zk}. By (1), we have at least |Z ′j| − 1 vertices of Z ′j has

the color k, that is, at least t− 2 vertices of Zj has the color k.

Following Subclaim 3.4.2, we have

t∑i=1

d(zαi) ≥ (|ϕ̄(x)| − t)(t− 1) + t(t− 2).

On the other hand,t∑i=1

d(zαi) ≤ t(3q − 2∆).

Solving the inequality (∆− d(x) + 1− t)(t− 1) + t(t− 2) ≤ t(3q − 2∆), we get

t ≤ ∆− d(x) + 1

3∆− 3q − d(x)≤ 1 +

3q − 2∆ + 1

5∆− 6q.

Since ∆ > 4, q < 34∆ ≤ 7∆−1

9, which in turn gives 3q−2∆+1

5∆−6q< 1. So Claim 3.4 follows.

21

For a positive number c, let

Z1(c) = {z ∈ V (G)− (X1 ∪N(X1)) : d(z) ≥ ∆− c},

Z2(c) = {z ∈ V (G)− (X1 ∪N(X1)) : d(z) < ∆− c}.

Clearly, V (G) = X1 ∪N(X1) ∪ Z1(c) ∪ Z2(c) and n = |X1| + |N(X1)| + |Z1(c)| + |Z2(c)|as X1 ∩N(X1) = ∅.

Claim 3.5. |Z1(c)| > (5c+2)∆−(6c+3)q+3c+2c∆

|N(X1)|.

Proof. For each y ∈ N(X1), x ∈ N(y)∩X1 and ϕ ∈ C∆(G−xy), let Y<c = {z ∈ Y2(x, ϕ) :

d(z) < ∆ − c}. Since {x, y} ∪ Y (x, ϕ) is elementary and Y2(x, ϕ) ⊆ Y (x, ϕ), we have

∆ ≥ | ∪v∈{x,y}∪Y (x,ϕ) ϕ̄(v)| =∑

v∈{x,y}∪Y (x,ϕ) |ϕ̄(v)| > 2∆− d(x)− d(y) + 2 + c|Y<c|. Thus

|Y<c| < d(x)−2c

. So, by Claim 3.4, |Y2(x, ϕ) − Y<c| > ∆ − 2d(x) + 3 − d(x)−2c

, that is, for

each y ∈ N(X1) we have dZ1(c)(y) > ∆− 2d(x) + 3− d(x)−2c

. Hence,

(∆− 2d(x) + 3− d(x)− 2

c)|N(X1)| < |E(N(X1), Z1(c))| ≤ ∆|Z1(c)|,

where E(N(X1), Z1(c)) are the edges with one vertex in N(X1) and the other endvertex

in Z1(c). Since d(x) ≤ 3q − 2∆, solving the above inequality we have

|Z1(c)| > (5c+ 2)∆− (6c+ 3)q + 3c+ 2

c∆|N(X1)|.

By Claims 3.1 and 3.2, we have

M ′(x) ≥

{2 + 2(∆− q) x ∈ X1,

q x ∈ V (G)−X1.

And by the definitions of Z1(c) and Z2(c), we get the following two lower bounds of∑x∈V (G)M

′(x),

b1 = (2 + 2(∆− q))|X1|+ q|N(X1)|+ (∆− c)|Z1(c)|+ (3q − 2∆)|Z2(c)| and

b2 = (2 + 2(∆− q))|X1|+ (n− |X1|)q.

First we estimate the lower bound of b2. By Claims 3.3 and 3.5, we have

n ≥ |X1|+ |N(X1)|+ |Z1(c)| > (3 + f2(c))|X1|,

22

where f2(c) = 2 (5c+2)∆−(6c+3)q+3c+2c∆

. So we have |X1|n

< 13+f2(c)

. Hence, b2 = qn − (3q −

2∆− 2)|X1| > qn− 3q−2∆−23+f2(c)

n.

Now we divide into the following two cases to estimate the lower bound of∑

x∈V (G) M′(x).

First we consider the case ∆− c ≤ q. In this case, it is easy to see that b2 > b1. Thus∑x∈V (G)M

′(x) ≥ max{b1, b2} = b2 > qn− 3q−2∆−23+f2(c)

n.

Now we consider the case ∆− c > q. We claim that∑

x∈V (G) M′(x) ≥ qn− 3q−2∆−2

3+f1(c)n.

If b1 ≤ b2, then (∆ − c)|Z1(c)| + (3q − 2∆)|Z2(c)| ≤ q(|Z1(c)| + |Z2(c)|), that is

|Z2(c)| ≥ (12− c

2∆−2q)|Z1(c)|. So by Claims 3.3 and 3.5, we have |N(X1)|+|Z1(c)|+|Z2(c)| ≥

(2 + (3− c∆−q )

(5c+2)∆−(6c+3)q+3c+2c∆

)|X1|, which implies that

n = |X1|+ |N(X1)|+ |Z1(c)|+ |Z2(c)| ≥ (3 + f1(c))|X1|,

where f1(c) = (3 − c∆−q )

(5c+2)∆−(6c+3)q+3c+2c∆

. Then we have |X1|n≤ 1

3+f1(c). Hence,∑

x∈V (G)M′(x) ≥ max{b1, b2} = b2 = qn− (3q − 2∆− 2)|X1| > qn− 3q−2∆−2

3+f1(c)n.

Let

h(c) = (1

3 + f2(c)− 1

3 + f1(c))(3q − 2∆− 2)n.

It is easy to see that∑

x∈V (G) M′(x) ≥ qn− 3q−2∆−2

3+f1(c)n if b1 ≥ b2 + h(c). So we only need

to consider the case b2 < b1 < b2 + h(c). If |X1| ≤ n3+f1(c)

, then∑

x∈V (G) M′(x) ≥ b2 ≥

qn− 3q−2∆−23+f1(c)

n. Thus |X1| > n3+f1(c)

. Let θ = (5c+2)∆−(6c+3)q+3c+2c∆

. Then f1(c) = (3− c∆−q )θ.

By Claims 3.3 and 3.5, we have |Z1(c)| > 2θ|X1| and n > (3+2θ)|X1|+ |Z2(c)|. Note that

|X1| > n3+f1(c)

and f1(c) = (3− c∆−q )θ, we have |Z2(c)| < (1− c

∆−q )θ|X1|. We may assume

that |Z2(c)| = (1− c∆−q )θ|X1|−t for some positive number t. Then n > (3+f1(c))|X1|−t,

and so |X1| < n+t3+f1(c)

. It follows that∑x∈V (G)

M ′(x) ≥ max{b1, b2} = b1

= qn− (3q − 2∆− 2)|X1|+ (∆− c− q)|Z1(c)| − (2∆− 2q)|Z2(c)|

> qn− 3q − 2∆− 2

3 + f1(c)n− 3q − 2∆− 2

3 + f1(c)t+ 2(∆− c− q)θ|X1|

−(2∆− 2q)(1− c

∆− q)θ|X1|+ (2∆− 2q)t

= qn− 3q − 2∆− 2

3 + f1(c)n− 3q − 2∆− 2

3 + f1(c)t+ (2∆− 2q)t.

23

Since (2∆−2q)(3+f1(c))− (3q−2∆−2) > 3(2∆−2q)− (3q−2∆−2) = 8∆−9q+2 > 0,

we have∑

x∈V (G) M′(x) ≥ qn− 3q−2∆−2

3+f1(c)n.

Hence, we have

d̄(G) ≥ q − 3q − 2∆− 2

3 + f(c),

where

f(c) = max{3− c

∆− q, 2} × (5c+ 2)∆− (6c+ 3)q + 3c+ 2

c∆.

4 Proof of Corollary 1

LetG be a ∆-critical graph with maximum degree ∆ ≥ 32. Let q := min{2√

2(∆−1)−2

2√

2+1, 3

4∆−

2}. By Theorem 1, for any positive number c, we have

d̄(G) ≥

{q − 3q−2∆−2

3+f1(c)if ∆− q > c,

q − 3q−2∆−23+f2(c)

if ∆− q ≤ c.

where

f1(c) = (3− c

∆− q)× (5c+ 2)∆− (6c+ 3)q + 3c+ 2

c∆

and

f2(c) = 2× (5c+ 2)∆− (6c+ 3)q + 3c+ 2

c∆.

If ∆ ≥ 66, then q = min{2√

2(∆−1)−2

2√

2+1, 3

4∆ − 2} = 2

√2(∆−1)−2

2√

2+1, so ∆ − q = ∆+1

2√

2+1+ 1 ≥

672√

2+1+ 1 > 18.5. If ∆ ≤ 65, then q = 3

4∆− 2 and so ∆− q = ∆

4+ 2 ≤ 18.25. In order to

calculate conveniently, we set c = 18.5. Thus

d̄(G) ≥

{q − 3q−2∆−2

3+f1(c)if ∆ ≥ 66,

q − 3q−2∆−23+f2(c)

if 32 ≤ ∆ ≤ 65.

First we consider the case ∆ ≥ 66. In this case, q = 2√

2(∆−1)−2

2√

2+1. Let a1 = 2

√2

2√

2+1

and a2 = 2√

2+22√

2+1. So q = a1∆ − a2. Using q = a1∆ − a2 and f1(c) = (3 − c

∆−q ) ×

24

(5c+2)∆−(6c+3)q+3c+2c∆

in 3q−2∆−23+f1(c)

, we have

−3q − 2∆− 2

3 + f1(c)= − c∆(3q − 2∆− 2)

3c∆ + (3− c∆−q )((5c+ 2)∆− (6c+ 3)q + 3c+ 2)

= −(3a1 − 2)c∆2 − g1g2∆ + g1g2∆− (3a2 + 2)c∆

(18c+ 6− (18c+ 9)a1)∆ + g2

= g1∆ +(−g1g2 + 3a2c+ 2c)∆

(18c+ 6− (18c+ 9)a1)∆ + g2

,

where g1 = − (3a1−2)c18c+6−(18c+9)a1

and g2 = 9c+6+(18c+9)a2− (5c2+2c−(6c2+3c)a1)∆+(6c2+3c)a2+3c2+2c(1−a1)∆+a2

.

It is easy to see that g2 = 9c + 6 + (18c + 9)a2 − m1 + m1a2−(6c2+3c)a2−3c2+2c(1−a1)∆+a2

, where

m1 = −5c2+2c−(6c2+3c)a1

1−a1, and g2 is an increasing function of ∆. Since ∆ ≥ 66, g2 takes

minimum when ∆ = 66. Using c = 18.5, a1 = 2√

22√

2+1, a2 = 2

√2+2

2√

2+1and ∆ ≥ 66 in

g1, g2 and −g1g2 + 3a2c + 2c, we have −0.04637 ≤ g1 < 0, −275.7122 ≤ g2 < 0 and

−g1g2 + 3a2c + 2c > 94.212 > 0. Note that (18c + 6 − (18c + 9)a1)∆ + g2 > 0, we have(−g1g2+3a2c+2c)∆

(18c+6−(18c+9)a1)∆+g2> −g1g2+3a2c+2c

18c+6−(18c+9)a1. It follows that

−3q − 2∆− 2

3 + f1(c)≥ −0.04637∆ + 1.0912.

Hence

d̄(G) ≥ q − 0.04637∆ + 1.0998 ≥ 0.69242∆− 0.1701.

Now we consider the case 32 ≤ ∆ ≤ 65. In this case, we have q = 34∆ − 2. Using

q = 34∆− 2 and c = 18.5 in f2(c) and 3q−2∆−2

3+f2(c), we have f2(c) = 18∆+571

18.5∆and −3q−2∆−2

3+f2(c)=

−18.5∆2−592∆294∆+2284

= −18.5∆294

+( 18.5

294×2284+592)∆

294∆+2284. Thus

d̄(G) ≥ 3

4∆− 2− 18.5∆

294+

(18.5294× 2284 + 592)∆

294∆ + 2284≥ 101

147∆− 2 +

(18.5294× 2284 + 592)∆

294∆ + 2284.

It is not difficult to check that if 56 ≤ ∆ ≤ 65 then d̄(G) ≥ 23(∆ + 2), which improve

Woodall’s result in [15] when 56 ≤ ∆ ≤ 65. Clearly,( 18.5

294×2284+592)∆

294∆+2284=

18.5294×2284+592

294+ 2284∆

is an

increasing function of ∆, we have( 18.5

294×2284+592)∆

294∆+2284≥ 2.0136 as ∆ ≥ 32. Thus

d̄(G) ≥ 0.68707∆ + 0.0136.

25

Hence,

d̄(G) ≥

{0.69242∆− 0.1701 if ∆ ≥ 66,

0.68707∆ + 0.0136 if 32 ≤ ∆ ≤ 65.

This completes the proof of Corollary 1.

Remark. (i) Note that 0.69242∆− 0.1701 ≥ 23(∆ + 2) if ∆ ≥ 66 and d̄(G) ≥ 2

3(∆ + 2) if

56 ≤ ∆ ≤ 65, we improve Woodall’s result in [15] when ∆ ≥ 56.

(ii) If ∆ and c both are large enough such that ∆−q > c, we have d̄(G) ≥ q− 3q−2∆−23+f1(c)

≥

a1∆− a2 + g1∆ + (−g1g2+3a2c+2c)∆(18c+6−(18c+9)a1)∆+g2

. Note that g1 = − (3a1−2)c18c+6−(18c+9)a1

= − 3a1−2

18−18a1+6−9a1c

is an increasing function of c. Since c is large enough, the value of g1 will approximate to

− 3a1−218−18a1

and d̄(G) will approximate to 0.69277∆.

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