On the average degree of edge chromatic critical …2 Coloring preliminaries and adjacency lemmas...

On the average degree of edge chromatic critical graphs

Yan Caoa and Guantao Chena,b

a Department of Mathematics and Statistics, Georgia State University, Atlanta, GA 30303b School of Mathematics and Statistics & Hubei Key Laboratory of Mathematical Sciences,

Central China Normal University, Wuhan 430079, P.R. China

Abstract

Let G be a simple graph, and let χ′(G) and ∆(G) denote the chromatic index and themaximum degree ofG, respectively. A graphG is a critical class two graph if χ′(G) = ∆(G)+1and χ′(H) ≤ ∆(G) for every proper subgraph H of G. Let d(G) denote the average degreeof G, i.e., d(G) = 2|E(G)|/|V (G)|. Vizing in 1968 conjectured that d(G) ≥ ∆(G)− 1 + 3

n if

G is a critical class two graph of order n. In this paper, we prove that d(G) ≥ 34∆(G) − 8

if G is a critical class two graph. Let δ(G) denote the minimum degree of G. We show thatthere exist two functions D and D0 such that for any ε ∈ (0, 1), if G is a critical class twograph with ∆(G) ≥ D(ε) and δ(G) ≥ D0(ε) then d(G) ≥ (1 − ε)∆(G). We will give twospecific functions satisfying the statement above in the paper. Moreover, we show that if Gis a critical class two graph and δ(G) ≥ (log ∆(G))

34 , then d(G) ≥ ∆(G)− o(∆(G)).

Keywords: edge coloring; critical class two graphs; Vizing’s average degree conjecture

1 Introduction

All graphs in this paper are finite, undirected and simple. We will generally follow Stiebitzet al. [12] for notation and terminology. An edge coloring of a graph G is to color the edges of Gsuch that each edge receives a color and adjacent edges, that is, distinct edges having a commonend, receive different colors. The minimum number of colors in such an edge coloring of G iscalled the chromatic index of G, written χ′(G). As proved by Holyer [6] the determination ofthe chromatic index is an NP-hard problem. On the other hand, the maximum degree ∆(G),that is, the maximum number of edges of G having a common end, is an obvious lower bound ofχ′(G). Vizing [13], and independently Gupta [11], proved that χ′(G) ≤ ∆(G) + 1 for any graphG. This leads to a natural classification of graphs. Following Fiorini and Wilson [4], a graph Gis said to be of class one if χ′(G) = ∆(G) and of class two if χ′(G) = ∆(G) + 1.

In investigating the classification problem, critical graphs are of particular interest. A graphG is called critical if χ′(H) < χ′(G) for every proper subgraph H of G. An edge e of G is calleda critical edge of G if χ′(G− e) < χ′(G). It is easy to see that a connected graph G is critical ifand only if every edge of G is critical. The only critical class one graph with maximum degree∆ is the star K1,∆. There is no critical class two graph with maximum degree ∆ = 1, and theonly critical class two graphs with maximum degree ∆ = 2 are the odd cycles. Critical graphsof class two have rather more structure than arbitrary graphs of class two, and it follows fromVizings theorem that every graph of class two contains a critical graph of class two with thesame maximum degree as a subgraph. These facts can be used when proving results in relationto the classification problem. In the remainder of this paper, a critical class two graph will becalled ∆-critical graph for convenience.

1

Let d(G) denote the average degree of a graph G. So 2|E(G)| = d(G) · n for any graph Gwith order n. The following is a well-known conjecture of Vizing proposed in 1968.

Conjecture 1. [Vizing’s Average Degree Conjecture] If G is a ∆-critical graph, then d(G) ≥∆(G)− 1 + 3

n ,

The conjecture has been verified for all graphs G with ∆(G) ≤ 6, see [7, 4, 8, 9]. Thelower bounds of d(G) for ∆-critical graphs have been heavily studied starting with the workof Fiorini [3] and Haile [5]. For long time, the best known result was due to Woodall [15],who showed that d(G) ≥ 2

3(∆(G) + 1) for every ∆-critical graph. His proof depends on theimprovement of Vizing’s Adjacency Lemma. Woodall also provided infinitely many examplesdemonstrating that his result cannot be improved by the use of his new adjacency lemmas andVizing’s Adjacency Lemma alone. Using the simple broom method developed in [2], Cao etal. [1] showed that d(G) ≥ 0.69242∆(G) − 0.1701 for every ∆-critical graph with ∆(G) ≥ 66,In this paper we develop a few new adjacency lemmas and obtain the following three results onVizing’s Average Degree Conjecture.

Theorem 1. If G is a ∆-critical graph, then d(G) ≥ 34∆(G)− 8.

Let δ(G) denote the minimum degree of graph G. For any constant ε ∈ (0, 1), let c0(ε) =†1−εε

£. Let c1(ε) = c0(ε)− 1. We will use c0, c1 instead of c0(ε), c1(ε) if ε is clear.

Theorem 2. There exist two functions D and D0 from (0, 1) to R such that for any positivereal number ε ∈ (0, 1), if G is a ∆-critical graph with ∆(G) ≥ D(ε) and δ(G) ≥ D0(ε), thend(G) ≥ (1− ε)∆(G).

In the proof, we show that Theorem 2 is true with the following two functions:

D0(ε) =

{2( 1

3ε)3 + 2

»( 1

3ε)6 − ( 1

3ε)3 if ε < 1

3 ,

2 if ε ≥ 13 ,

D(ε) = max

ßf(ε),

3c0 + 1

ρ2,N + c0

ε3

™,

where ρ = ε3

ε2+1, N = (c0 + 1)(1

ρ + 1)3c0+1 and f(ε) = 32/ε if ε > 1516 and f(ε) = 1

ε2(3c4

0 + 12c30 +

10c20 + 4c0 + 1) otherwise.

Consider the case when ε < 13 . Since c0 + 1 ≥ 1

ε > c0 =†

1−εε

£≥ 3 and 1

ρ = 1ε + 1

ε3, we have

D(ε) = N+c0ε3

by comparing the three functions above. To avoid unnecessary and complicated

calculation, we point out a rough upper bound of N+c0ε3

when ε < 13 which is ( 3

ε3)4ε .

Corollary 3. Let G be a ∆-critical graph with maximum degree ∆. If ∆ > 10972 and δ(G) ≥(log ∆)

34 , then d(G) ≥ ∆− 2∆

(log ∆)14.

Proof. Let ε = ( 12log ∆)

14 , since ∆ > 10972, 0 < ε < 1

3 . Then log ∆ = 12ε4≥ 4

ε log 3ε3≥ logD(ε), so

that ∆ ≥ D(ε). Also (log ∆)34 > 6

ε3> D0(ε). Thus the result follows from Theorem 2.

Theorem 2 will be proved in Section 3, assuming the truth of Lemmas 7 and 8, which areproved in Section 5. In a similar way, Theorem 1 will be proved in Section 4, assuming the truthof Lemmas 16 and 17, which are proved in Section 6.

2

2 Coloring preliminaries and adjacency lemmas

Let G be a graph. We denote by V (G) and E(G) the vertex set and the edge set of G,respectively. For a vertex v ∈ V (G), let N(v) = {w ∈ V (G) : vw ∈ E(G)} be the neighborhoodof v in G, and let d(v) = |N(v)| be the degree of v in G. Let N [v] = N(v) ∪ {v}. A vertexv ∈ V (G) is called a d-vertex if its degree is d. For a vertex set T , let N(T ) =

⋃v∈T N(v). For a

non-negative integer k, let [1, k] = {1, 2, . . . , k} and let Ck(G) denote the set of all edge coloringsof G with color set [1, k]. So ϕ ∈ Ck(G) if and only if ϕ assigns to each edge e ∈ E(G) a colorϕ(e) ∈ [1, k] such that ϕ(e) 6= ϕ(f) for any two adjacent edges e and f of G. Note that χ′(G) isthe smallest k such that Ck(G) 6= ∅.

In the reminder of this section, we always assume that G is a ∆-critical graph with maximumdegree ∆, xy ∈ E(G), and ϕ ∈ C∆(G− xy). We will re-state these assumptions in each lemmafor their completeness.

For a vertex v ∈ V (G), let ϕ(v) = {ϕ(vw) : w ∈ N(v)} and ϕ̄(v) = [1,∆]\ϕ(v). We call ϕ(v)the set of colors present at v and ϕ̄(v) the set of colors missing at v with respect to ϕ. For avertex set T , let ϕ̄(T ) =

⋃v∈T ϕ̄(v). A set X ⊆ V (G) is called elementary with respect to ϕ if

the sets ϕ̄(v) with v ∈ X are pairwise disjoint. For a color α ∈ [1,∆], let Eα denote the set ofedges e of G with ϕ(e) = α and call it a color class with respect to ϕ. Clearly, Eα is a matchingof G. So if α and β are two colors, then the spanning subgraph H of G with edge set Eα∪Eβ hasmaximum degree at most 2, so every component of H is either a path or an even cycle (whoseedges are colored alternately with α and β) and we refer to such a component as an (α, β)-chainof G with respect to ϕ. For a vertex v of G, let Pv(α, β, ϕ) denote the unique component ofH that contains the vertex v. Let ϕ′ = ϕ/Pv(α, β, ϕ) denote the mapping obtained from ϕ byswitching the colors α and β on the edges of Pv(α, β, ϕ). Then, clearly, ϕ′ ∈ C∆(G − xy) is anedge coloring of G − xy with color set [1,∆], too. This switching operation is called a Kempechange.

A multi-fan at x with respect to e = xy ∈ E(G) and ϕ ∈ C∆(G − e) is a sequenceF = (e1, y1, . . . , ep, yp) with p ≥ 1 consisting of edges e1, e2, . . . , ep and vertices x, y1, y2, . . . , ypsatisfying the following two conditions:

• The edges e1, e2, . . . , ep are distinct, e1 = e and ei ∈ xyi for i = 1, . . . , p.

• For every edge ei with 2 ≤ i ≤ p, there is a vertex yj with 1 ≤ j < i such that ϕ(ei) ∈ ϕ̄(yj).

Notice that multi-fan is slightly more general than Vizing-fan which requires j = i−1 in thesecond condition. The following lemma shows that a multi-fan is elementary. The proof can befound in the book [12].

Lemma 1. [Stiebitz, Scheide, Toft and Favrholdt [12]] Let G be a ∆-critical graph, e = xy ∈E(G) and ϕ ∈ Ck(G − e). Let F = (e1, y1, . . . , ep, yp) be a multi-fan at x with respect to e andϕ. Then the following statements hold:

(a) {x, y1, y2, . . . , yp} is elementary.(b) If α ∈ ϕ̄(x) and β ∈ ϕ̄(yi) for some i, then Px(α, β, ϕ) = Pyi(α, β, ϕ).

For a vertex v ∈ V (G) and a given positive number q, let

ϕbad(v) = ϕ̄(v) ∪ {ϕ(vv′) : v′ ∈ N(v) with d(v′) < q},

σq(x, y) = |{z ∈ N(y) \ {x} : d(z) ≥ q}|.

3

Lemma 2. [Vizing’s Adjacency Lemma [14]] If G is a ∆-critical graph, then σ∆(x, y) ≥ ∆(G)−d(x) + 1 for every xy ∈ E(G).

A Kierstead path with respect to e = xy and ϕ ∈ C∆(G−e) is a sequence K = (y0, e1, y1, . . . ,ep, yp) with p ≥ 1 consisting of edges e1, e2, . . . , ep and vertices y0, y1, . . . , yp satisfying thefollowing two conditions:

• The vertices y0, y1, . . . , yp are distinct, e1 = e and ei = yiyi−1 for 1 ≤ i ≤ p.

• For every edge ei with 2 ≤ i ≤ p, there is a vertex yj with 0 ≤ j < i such that ϕ(ei) ∈ ϕ̄(yj).

Notice that a Kierstead path with 3 vertices is a special multi-fan with center y = y1. For aKierstead path with 4 vertices, the following result was obtained.

Lemma 3. [Kostochka and Stiebitz [12], Luo and Zhao [10]] Let G be a ∆-critical graph,e1 = y0y1 be an critical edge and ϕ ∈ C∆(G− e1). If K = (y0, e1, y1, e2, y2, e3, y3) is a Kiersteadpath with respect to e1 and ϕ, then V (K) is elementary unless d(y1) = d(y2) = ∆(G), in whichcase, all colors in ϕ̄(y0), ϕ̄(y1), ϕ̄(y2) and ϕ̄(y3) are distinct except one possible common missingcolor in ϕ̄(y3) ∩ (ϕ̄(y0) ∪ ϕ̄(y1)).

A simple broom with respect to e = xy and ϕ ∈ C∆(G− e) is a sequence B = (y0, e1, y1, . . . ,ep, yp) with p ≥ 3 such that (y0, e1, y1, e1, y2, ei, yi) is a Kierstead path with respect to e and ϕwhenever 3 ≤ i ≤ p.

Lemma 4. [Chen, Chen and Zhao [2]] Let G be a ∆-critical graph, e1 = y0y1 ∈ E(G) andϕ ∈ C∆(G−e1) and let B = {y0, e1, y1, e2, y2, . . . , ep, yp} be a simple broom. If |ϕ̄(y0)∪ϕ̄(y1)| ≥ 4and min{d(y1), d(y2)} < ∆, then V (B) is elementary under ϕ.

By applying the lemma above, Cao et al. proved the following result.

Lemma 5. [Cao, Chen, Jiang, Liu, Lu [1]] Let G be a ∆-critical graph with maximum degree∆, xy ∈ E(G) and ϕ ∈ C∆(G − xy). For a positive number q with d(x) < q ≤ ∆ − 1, letZ∗q = {z ∈ N(x)\{y} : ϕ(xz) ∈ ϕ̄(y) and d(z) > q}. Then the following three inequalities hold.

|Z∗q | ≥ ∆− d(y) + 1−úd(x) + d(y)−∆− 2

∆− q

ü(1)∑

z∈Z∗q

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

For all z ∈ Z∗q ,

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

∆− q

ü(3)

Let α, β, and γ denote three arbitrary colors. For any edge incident to x with color α, weuse xα to denote the other end of it. Moreover, for any edge incident to xα with color β, we usexαβ to denote the other end of it. We use yγ to denote the other end of the edge incident withy with color γ. Since {x, y} is elementary, ϕ̄(x), ϕ̄(y) and ϕ(x) ∩ ϕ(y) form a partition of thecolor set [1,∆]. We divide ϕ(x) ∩ ϕ(y) into two sets of colors as the following:

Aϕ(x, y, q) = {α ∈ ϕ(x) ∩ ϕ(y) : d(yα) < q},

Bϕ(x, y, q) = {β ∈ ϕ(x) ∩ ϕ(y) : d(yβ) ≥ q}.

4

Let Sϕ(x, y, q) = Aϕ(x, y, q)∪ϕ̄(x)∪ϕ̄(y). We will use Aϕ(q), Bϕ(q) and Sϕ(q) in short if x, y areclear. Note that Sϕ(q)∩ϕ(x) = Aϕ(q)∪ϕ̄(y) since the set {x, y} is elementary and so ϕ̄(y) ⊆ ϕ(x).When q is clear, let NS(x) = {xα : α ∈ Sϕ(q) ∩ ϕ(x)} and NB(x) = {xβ : β ∈ Bϕ(q)}. ClearlyN(x) = NS(x) ∪NB(x) ∪ {y}.

In this paper, we will use the following operation very often.

(η, θ, δ)-switching: Let xy be an edge in a ∆-critical graph G and ϕ ∈ C∆(G − xy). Assumeη, θ, δ are three distinct colors such that η ∈ Aϕ(q), θ ∈ ϕ̄(y) and δ ∈ ϕ̄(yη) ∩ ϕ̄(x). Formϕ1 = ϕ/Pyη(δ, θ, ϕ), so that θ ∈ ϕ̄1(yη), and obtain ϕ2 by recoloring the edge yyη with θ. (Ifθ ∈ ϕ̄(yη) then ϕ1 = ϕ and the only edge recolored is yyη.)

Lemma 6. Suppose that an (η, θ, δ)-switching turns ϕ into ϕ1. Then ϕ1(e) = ϕ(e) for everyedge e incident with x or xθ, or incident with y except e = yyη. Moreover, Sϕ1(q) = Sϕ(q), andfor every edge e, ϕ1(e) ∈ Sϕ1(q) if and only if ϕ(e) ∈ Sϕ(q).

Proof. Since Px(δ, θ, ϕ) = Py(δ, θ, ϕ) (by Lemma 1), Pyη(δ, θ, ϕ) is disjoint from it. For everyedge e incident with x or y or xθ, ϕ1(e) = ϕ(e) unless e = yyη. It follows that δ ∈ ϕ̄(x) = ϕ̄1(x),η ∈ Aϕ(q)∩ ϕ̄1(y) and θ ∈ ϕ̄(y)∩Aϕ1(q), so that {δ, η, θ} ⊆ Sϕ(q)∩Sϕ1(q). Since edges of othercolors are unaffected by the switch, Sϕ1(q) = Sϕ(q) and the result follows.

We briefly explain how will we use this operation. For a color η ∈ Aϕ(q), we can show thatunder certain conditions there exist θ ∈ ϕ̄(y) and δ ∈ ϕ̄(yη)∩ ϕ̄(x) such that the switching aboveis valid. Then after the switching, η becomes to a missing color of y. This means in some sense,we may treat the colors in Aϕ(q) as same as the colors in ϕ̄(y).

3 Proof of Theorem 2

Theorem 2. There exist two functions D and D0 from (0, 1) to R such that for any positivereal number ε ∈ (0, 1), if G is a ∆-critical graph with ∆(G) ≥ D(ε) and δ(G) ≥ D0(ε), thend(G) ≥ (1− ε)∆(G).

In this section we will prove Theorem 2 based on Lemma 7 and Lemma 8. Let ε ∈ (0, 1),and let D(ε) and D0(ε) be defined as in Section 1. Let G be a ∆-critical graph with maximumdegree ∆ ≥ D(ε) and minimum degree δ(G) ≥ D0(ε). Let q = (1− ε)∆. For a vertex v ∈ V (G),let d<q(v) = |{w ∈ N(v) : d(w) < q}|. We will use the discharging method to prove the theoremand initially assign to each vertex x of G a charge M(x) = d(x) and redistribute the chargeaccording to the following rule:

• Rule of Discharge: Each vertex y with degree larger than q distributes d(y)− q equallyamong all neighbors of y with degree less than q.

Denote by M ′(x) the resulting charge on each vertex x. It is easy to see that Theorem 2 holdsif M ′(x) ≥ q for every x ∈ V (G) with d(x) ≥ D0(ε). We divide this statement into two claimsto prove it.Claim 3.1. For every x ∈ V (G) with d(x) > ε∆, M ′(x) ≥ q.Claim 3.2. For every x ∈ V (G) with D0(ε) ≤ d(x) ≤ ε∆, M ′(x) ≥ q.

3.1 Part I: Claim 3.1.

3.1.1 Lemmas (Part I)

Recall that ϕbad(v) = ϕ̄(v) ∪ {ϕ(vv′) : v′ ∈ N(v) with d(v′) < q} and c0 = c0(ε) =†

1−εε

£.

The following two lemmas will be used in the proof of Claim 3.1. Due to the length of the proofs,

5

they will be given in Section 5, which also contain the original definition of the function f .

Lemma 7. Let ε ∈ (0, 1), G be a ∆-critical graph with maximum degree ∆ ≥ f(ε), xy ∈ E(G)with d(x) < q = (1 − ε)∆ and ϕ ∈ C∆(G − xy). Then for any color α ∈ ϕ̄(y), there existsa set X ⊆ ϕbad(xα) ∩ ϕ(x)\{α} with |X| ≤ c0 such that |ϕbad(xβ) ∩ Sϕ(q)\{β}| < 3c0 for all

β ∈ ϕbad(xα) ∩ ϕ(x)\({α} ∪X).

Lemma 8. Let ε ∈ (0, 1), G be a ∆-critical graph with maximum degree ∆ ≥ f(ε), xy ∈ E(G)with d(x) < q = (1 − ε)∆ and ϕ ∈ C∆(G − xy). Then for any c0 + 2 colors ξ1, ξ2, . . . , ξc0+2 ∈Sϕ(q) ∩ ϕ(x), |⋂c0+2

i=1 ϕbad(xξi)\{ξi}| < 3c0 + 1.

Lemma 9. Let ρ ∈ (0, 1) and N1, N2,∆, l be positive integers such that ∆ ≥ N2/ρ2 and l >

(N1 − 1)(1ρ + 1)N2. Then for any l subsets A1, A2, . . . , Al of {1, 2, . . . ,∆} with |Ai| ≥ ρ∆ for

each i ∈ [1, l], there exist N1 sets Ai1 , Ai2 , . . . , AiN1such that |⋂N1

t=1Ait | ≥ N2.

Proof. We say a subset of {1, 2, . . . ,∆} is a N2-subset if it has size N2. There are(∆N2

)N2-

subsets in total, and each Ai contains at least(ρ∆N2

)of them. If one of the N2-subset is covered

by at least N1 Ai’s, we are done. Otherwise we have l(ρ∆N2

)≤ (N1 − 1)

(∆N2

). Since ρ < 1, if

0 ≤ i ≤ N2 − 1 < ρ2∆, then ρ∆−i∆−i >

ρ∆−ρ2∆∆−ρ2∆

= ρ1+ρ = (1

ρ + 1)−1. Hence(ρ∆N2

)> (1

ρ + 1)−N2(∆N2

).

Since l > (N1 − 1)(1ρ + 1)N2 , we have l

(ρ∆N2

)> (N1 − 1)

(∆N2

), which is a contradiction.

Corollary 10. Let ε, ρ ∈ (0, 1), G be a ∆-critical graph with maximum degree ∆ ≥ max{f(ε), (3c0

+1)/ρ2} and xy ∈ E(G) with d(x) < q = (1 − ε)∆. Then for any ϕ ∈ C∆(G − xy) there existsa set X ⊆ Sϕ(q) ∩ ϕ(x) with |X| ≤ (c0 + 1)(1

ρ + 1)3c0+1 such that |ϕbad(xξ)\{ξ}| < ρ∆ for all

ξ ∈ Sϕ(q) ∩ ϕ(x)\X.

Proof. Suppose on the contrary that there are l > (c0 + 1)(1ρ + 1)3c0+1 colors 1, 2, . . . , l in

Sϕ(q) ∩ ϕ(x) such that |ϕbad(xi)\{i}| ≥ ρ∆ for all i ∈ [1, l]. Let N1 = c0 + 2 and N2 =

3c0 + 1. Then, N1, N2, ρ,∆ and l satisfy the conditions of Lemma 9. Let Ai = ϕbad(xi)\{i}for i = 1, 2, . . . , l, by Lemma 9 there are c0 + 2 colors ξ1, ξ2, . . . , ξc0+2 in Sϕ(q) ∩ ϕ(x) such that

|⋂c0+2t=1 ϕbad(xξt)\{ξt}| ≥ 3c0 + 1. Since ∆ ≥ f(ε), we also have the conclusion of Lemma 8,

which is a contradiction.

3.1.2 Proof of Claim 3.1

Claim 3.1. For every x ∈ V (G) with d(x) > ε∆, M ′(x) ≥ q.

Proof. Recall M(x) = d(x) and q = (1 − ε)∆. If d(x) ≥ q = (1 − ε)∆, then M ′(x) = M(x) −d(x)−qd<q(x)d<q(x) = q.

We assume ε∆ ≤ d(x) < q. Choose y ∈ N(x) such that σq(x, y) is minimum over all neighborsof x. Let s = ∆ − σq(x, y). Then for each z ∈ N(x), d(z) > σq(x, z) ≥ σq(x, y) = ∆ − s. Nowwe consider two cases.

Case 1. s ≤ ε2∆ .In this case, for each z ∈ N(x), d(z) ≥ ∆ − s ≥ ∆ − ε2∆ > q and d(z) − σq(x, z) ≤

d(z) + s−∆ ≤ s. So we have:

6

M ′(x) >∑

z∈N(x)

d(z)− qd(z)− σq(x, z)

≥∑

z∈N(x)

∆− s− qs

= d(x)(∆− s− q

s) ≥ ε∆(

ε∆− ε2∆

ε2∆) = (1− ε)∆ = q

Case 2. s > ε2∆.For any y ∈ N(x) and ϕ ∈ C∆(G − xy), let Y (ϕ) = {v ∈ N(y) \ {x} : ϕ(vy) ∈ ϕ̄(x)} and

Cy(ϕ) = {ϕ(vy) : v ∈ Y (ϕ) and d(v) < q}. Since ϕ̄(x) ∩ ϕ̄(y) = ∅ by Lemma 1, ϕ̄(x) ⊆ ϕ(y),Y (ϕ) and Cy(ϕ) are well-defined.

The following properties will be used soon.

(1) |Cy(ϕ)| ≤ c0,

(2) |NS(x)| ≥ s− c0 ≥ ε2∆− c0.

Let ρ = ε3

ε2+1and N = (c0 + 1)(1

ρ + 1)3c0+1. Since ∆ ≥ D(ε) ≥ max{f(ε), (3c0 + 1)/ρ2},by Corollary 10, there is a subset X of NS(x) with |X| ≤ N such that |ϕbad(z)| ≤ ρ∆ for allz ∈ NS(x)\X.

(3) For all z ∈ NS(x)\X, d(z) ≥ ∆− |ϕbad(z)| > q and d(z)− σq(x, z) ≤ |ϕbad(z)|.

Now we prove the three statements above.For (1), since {x, y}∪Y (ϕ) is the vertex set of a multi-fan at y, it is elementary by Lemma 1.

Assume the contrary that |Cy(ϕ)| > c0, then ∆ ≥ |⋃yα∈Y (ϕ) ϕ̄(yα)| =∑yα∈Y (ϕ) |ϕ̄(yα)| ≥

(c0 + 1)ε∆ > ∆, giving a contradiction.For (2), recall Aϕ(x, y, q) = {α ∈ ϕ(x) ∩ ϕ(y) : d(yα) < q} and NS(x) = {xα : α ∈

Aϕ(q) ∪ ϕ̄(y)}. Then |Aϕ(q) ∪ ϕ̄(y) ∪ Cy(ϕ)| = |NS(x)| + |Cy(ϕ)| = ∆ − σq(x, y) = s. Since|Cy(ϕ)| ≤ c0 and s = |NS(x)|+ |Cy(ϕ)|, we have |NS(x)| ≥ s− c0 ≥ ε2∆− c0.

For (3), we have d(z) ≥ ∆−|ϕ̄(z)| ≥ ∆−|ϕbad(z)| ≥ ∆−ρ∆. Since ρ < 1ε , ∆−|ϕbad(z)| > q.

Notice that |ϕbad(z)| = ∆− σq(x, z). So d(z)− σq(x, z) ≤ |ϕbad(z)|.

By using these properties, we have

M ′(x) >∑

z∈NS(x)\X


≥∑

z∈NS(x)\X

∆− |ϕbad(z)| − q|ϕbad(z)|

≥ (ε2∆− c0 −N)(ε∆− ρ∆

ρ∆)

= ∆− c0 +N

ε2(Since ε−ρ

ρ = 1ε2

.)

≥ (1− ε)∆ = q. (Since ∆ ≥ D(ε) ≥ N+c0ε3

.)

7

3.2 Part II: Claim 3.2.

3.2.1 Lemmas (Part II)

In this subsection, we still assume that q = (1 − ε)∆ and consider the vertices with degreeat most ε∆. We point out that all the conclusions in Subsection 3.2 are true without assuming∆ ≥ D(ε).

If d(x) ≤ ε∆, then for any v ∈ V (G) with d(v) < q and any coloring ϕ ∈ C∆(G − xy),|ϕ̄(x)∩ ϕ̄(v)| ≥ 2. In this case we will use an operation called (η, θ)-switching instead of (η, θ, δ)-switching in some cases if pointing out δ is not necessary. We give the details here.

(η, θ)-switching: Let xy be an edge in a ∆-critical graph G and ϕ ∈ C∆(G − xy). For anyη ∈ Aϕ(q) and θ ∈ ϕ̄(y), since η ∈ Aϕ(q), yη exists and d(yη) < q, there exists a common missingcolor δ of x and yη. We do a (η, θ, δ)-switching in ϕ.

Notice that this operation is exactly the (η, θ, δ)-switching without pointing out δ, thus itsatisfies all the properties stated in Lemma 6. As we pointed out at the end of Section 2,this operation allows us to treat Aϕ(q) and ϕ̄(y) as the same set (in some sense). Recall thatSϕ(q) = Aϕ(q) ∪ ϕ̄(x) ∪ ϕ̄(y). In the reminder of this paper, we will write Aϕ(q) ∪ ϕ̄(y) asSϕ(q) ∩ ϕ(x) very often.

The next lemma is a direct corollary of a result proved by Woodall [15], it is actually Claim3.2, Claim 3.3 and Claim 3.6 in the proof of Lemma 2.4 in [15].

Lemma 11. Let G be a ∆-critical graph with maximum degree ∆, xy ∈ E(G) with d(x) ≤ ε∆and ϕ ∈ C∆(G − xy). Then for any two distinct colors α and β with α ∈ ϕ̄(y) and β ∈ Sϕ(q),xαβ exists and d(xαβ) ≥ ∆− d(x) + 2 > q.

Lemma 12. Let G be a ∆-critical graph with maximum degree ∆, xy ∈ E(G) with d(x) ≤ ε∆and ϕ ∈ C∆(G− xy). Then for any α ∈ Sϕ(q) ∩ ϕ(x), ϕbad(xα)\{α} ⊆ Bϕ(q).

Proof. Let β 6= α be a color in Sϕ(q). Recall that Sϕ(q)∪Bϕ(q) = [1,∆], we will prove that xαβexists and d(xαβ) ≥ q, which implies the lemma.

If α ∈ ϕ̄(y), we are done by Lemma 11. Assume that α ∈ Aϕ(q). Since xy is uncolored,there exists a color θ in ϕ̄(y). Now we do a (α, θ)-switching in ϕ to obtain a new coloring ϕ1 inwhich ϕ1(xxα) = α ∈ ϕ̄1(y).

It follows from Lemma 11 that Sϕ1(q) ⊆ ϕ1(xα). Now Lemma 6 implies that Sϕ(q) =Sϕ1(q) ⊆ ϕ(xα). In particular, xαβ exists. By Lemma 6, the color of xαxαβ under ϕ1 is still inSϕ(q) = Sϕ1(q), and by Lemma 11 again we have d(xαβ) ≥ ∆− d(x) + 2 > q.

Lemma 13. Let G be a ∆-critical graph with maximum degree ∆, xy ∈ E(G) with d(x) ≤ ε∆and ϕ ∈ C∆(G− xy). Let ξ ∈ Sϕ(q) ∩ ϕ(x). Then for any β ∈ ϕbad(xξ)\{ξ}, we have β ∈ ϕ(x)

and ϕbad(xβ)\{β} ⊆ Bϕ(q).

Proof. Since β ∈ ϕbad(xξ)\{ξ}, by Lemma 12 we have β ∈ Bϕ(q) ⊆ ϕ(x). Let α be a color inSϕ(q) = Aϕ(q) ∪ ϕ̄(x) ∪ ϕ̄(y). We will prove that xβα exists and d(xβα) ≥ q, which implies thelemma. We consider the three cases separately.

Case 1. α ∈ ϕ̄(x).We divide this case into two subcases.Subcase 1.1. ξ ∈ ϕ̄(y).In this case, we recolor the edge xy by ξ and leave xxξ uncolored to obtain a new coloring

ϕ1. In the new coloring, we have α ∈ ϕ̄1(x) ⊆ Sϕ1(x, xξ, q), β ∈ ϕbad(xξ)\{ξ} = Sϕ1(x, xξ, q).Thus by Lemma 12, xβα exists and d(xβα) ≥ ∆− d(x) + 2 > q.

8

Subcase 1.2. ξ ∈ Aϕ(q).In this case, there exists yξ ∈ N(y) such that ϕ(yyξ) = ξ and d(yξ) < q. Thus, there are

two colors in ϕ̄(x) ∩ ϕ̄(yξ). Choose t in ϕ̄(x) ∩ ϕ̄(yξ) such that t 6= α, choose ξ0 ∈ ϕ̄(y). Clearlyα, β /∈ {ξ, ξ0, t}, since ξ ∈ Aϕ(q), α ∈ ϕ̄(x) ⊆ ϕ(y), and β ∈ Bϕ(q) ⊆ ϕ(x) ∩ ϕ(y). Then we cando (ξ, ξ0, t)-switching in ϕ. By Lemma 6, this operation will not affect the edge xxξ and alsoother edges with color α, ξ or β except yyξ. After this operation, we are back to Subcase 1.1.

Case 2. α ∈ ϕ̄(y).Choose δ ∈ ϕ̄(x); clearly β /∈ {α, δ}. Assume first that at least one of α, δ is in ϕ̄(xβ).

Let ϕ1 = ϕ/Pxβ (α, δ, ϕ). This does not affect edges incident with x or y, since Px(α, δ, ϕ) =Py(α, δ, ϕ) by Lemma 1. Thus Bϕ1(q) = Bϕ(q), Sϕ1(q) = Sϕ(q), and δ ∈ ϕ̄1(x) ⊆ Sϕ1(q). Nowapply Case 1 with ϕ1, δ in place of ϕ, α, we have xβδ exists and d(xβδ) ≥ q under coloring ϕ1.Hence under coloring ϕ, xβα exists and d(xβα) ≥ q.

This argument fails if {α, δ} ⊆ ϕ(xβ). But then xβα exists and we may assume d(xβα) < q.Since d(x) < ε∆ = ∆− q, we could have chosen δ ∈ ϕ̄(x) ∩ ϕ̄(xβα), and the previous argumentwould then have worked.

Case 3. α ∈ Aϕ(q).Choose ξ0 ∈ ϕ̄(y) and do (α, ξ0)-switching of ϕ to give ϕ1. By Lemma 6, excepting yyα, this

operation does not affect edges incident with x or y or with color α or β, and so β ∈ Bϕ1(q) andξ ∈ Sϕ1(q) ∩ ϕ1(x) even when ξ0 = ξ. Now α ∈ ϕ̄1(y), and the result follows from Case 2 withϕ1 in place of ϕ.

Lemma 14. Let G be a ∆-critical graph with maximum degree ∆, xy ∈ E(G) with d(x) ≤ ε∆and ϕ ∈ C∆(G− xy). Then ϕbad(xξ1) ∩ ϕbad(xξ2) = ∅ for any two colors ξ1, ξ2 ∈ Sϕ(q) ∩ ϕ(x).

Proof. Let β ∈ ϕbad(xξ1). Then β ∈ Bϕ(q) by Lemma 12, so β 6= ξ1, ξ2. We will prove that xξ2βexists and d(xξ2β) ≥ q, which implies the lemma.

Choose ξ0 ∈ ϕ̄(y), we do (ξ1, ξ0)-switching in ϕ at y if ξ1 ∈ Aϕ(q). By Lemma 6, thisoperation will not affect the edges xxξ1 , xxξ2 and also other edges with color ξ1 or β exceptyyξ1 , ξ2 is still in Sϕ(q) ∩ ϕ(x) even when ξ0 = ξ2, thus we may assume that ξ1 ∈ ϕ̄(y) at thebeginning.

Now, we recolor the edge xy by ξ1 and leave xxξ1 uncolored to obtain a new coloring ϕ1.Notice that by Lemma 12, there exists xξ1ξ2 ∈ N(xξ1) such that d(xξ1ξ2) ≥ q. Thus in the

new coloring, ξ2 /∈ Sϕ1(x, xξ1 , q), β ∈ ϕbad(xξ1)\{ξ1} = Sϕ1(x, xξ1 , q), ϕ1(xy) = ξ1 ∈ ϕ̄1(xξ1) ⊆Sϕ1(x, xξ1 , q). Moreover, we have that ξ2 ∈ ϕbad

1 (y) because ξ2 ∈ Sϕ(x, y, q) ∩ ϕ(x). Thus byLemma 13, xξ2β exists and d(xξ2β) ≥ q. This completes the proof of Lemma 14.

Now combining Lemma 12 and Lemma 14, we have the following corollary.

Corollary 15. Let G be a ∆-critical graph with maximum degree ∆, xy ∈ E(G) with d(x) ≤ ε∆and ϕ ∈ C∆(G−xy). Then for any ξ ∈ Sϕ(q)∩ϕ(x), we have ϕbad(xξ)\{ξ} ⊆ Bϕ(q). Moreover,

ϕbad(xξ1) ∩ ϕbad(xξ2) = ∅ for any two colors ξ1, ξ2 ∈ Sϕ(q) ∩ ϕ(x).


Claim 3.2. For every x ∈ V (G) with D0(ε) ≤ d(x) ≤ ε∆, M ′(x) ≥ q.

Proof. For any v ∈ N(x), we have σ∆(x, v) ≥ ∆ − d(x) + 1 > (1 − ε)∆ = q by Lemma 2. Sod(v) ≥ σq(x, v) ≥ σ∆(x, v) ≥ q. Following the discharging rule, we have

M ′(x) ≥∑

v∈N(x)

d(v)− qd(v)− σq(x, v)

.

9

Choose y ∈ N(x) and ϕ ∈ C∆(G−xy) such that p := |Bϕ(q)| is minimum over all neighboursof x and all ∆-colorings of G − xy. Note that |Sϕ(q)| = ∆ − p and p = |Bϕ(q)| ≤ |Aϕ(q)| +|Bϕ(q)| = |ϕ(x)\ϕ̄(y)| = d(x) − 1 − |ϕ̄(y)| ≤ d(x) − 2. According to the discharging rule, a

vertex xξ ∈ Sϕ(q) ∩ ϕ(x) will distribute charge to x and xβ’s where β ∈ ϕbad(xξ)\ϕ̄(xξ). Let

Xϕ(q) =⋃ξ∈Sϕ(q)∩ϕ(x)(ϕ

bad(xξ)\(ϕ̄(xξ) ∪ {ξ})) and t = |Xϕ(q)|. Note that |ϕbad(xξ)\(ϕ̄(xξ) ∪{ξ})| = |{β ∈ ϕ(ξ)\{ξ} : d(xξβ) < q}| = d(xξ) − σq(x, xξ) − 1 denotes the number of vertices(except x) that receive charge from xξ. Now we claim the following statements and prove themone by one:

(1) Xϕ(q) ⊆ Bϕ(q),

(2) For any v ∈ N(x), σq(x, v) ≥ ∆− d(x) + 1 + p > q,

(3) For any v ∈ N(x), d(v)−qd(v)−σq(x,v) ≥

∆−q∆−σq(x,v) ,

(4) For any v ∈ NS(x), d(v) ≥ ∆− p+ t and

(5)∑

v∈NS(x)

1d(v)−σq(x,v) ≥

|NS(x)|2|NS(x)|+t = (d(x)−p−1)2

d(x)−p−1+t .

To prove (1), recall ϕbad(xξ)\(ϕ̄(xξ) ∪ {ξ}) ⊆ ϕbad(xξ)\{ξ} ⊆ Bϕ(q) by Corollary 15.

To prove (2), let v ∈ N(x) and ϕ′ ∈ C∆(G−xv). Then for any w ∈ N(v) with ϕ′(vw) ∈ ϕ̄′(x),{x, v, w} is elementary by Lemma 1, so d(w) ≥ |ϕ̄′(x)| = ∆ − d(x) + 1 > q. Thus σq(x, v) ≥|{w ∈ N(v)\{x} : ϕ′(vw) ∈ ϕ̄′(x)}|+ |Bϕ′(x, v, q)| ≥ |ϕ̄′(x)|+ |Bϕ(q)| ≥ ∆− d(x) + 1 + p > q.

To prove (3), let v ∈ N(x). Using the property above, we have d(v) ≥ σq(x, v) ≥ ∆− d(x) +

1 + p > q. Thus d(v)−qd(v)−σq(x,v) is a decreasing function of d(v). d(v)−q

d(v)−σq(x,v) ≥∆−q

∆−σq(x,v) holds.

To prove (4), for any xα ∈ NS(x), we have Sϕ(q) ⊆ ϕ(xα) by Lemma 12. For any β ∈ Xϕ(q),there exists ξ ∈ Sϕ(q) ∩ ϕ(x) such that xξβ exists and d(xξβ) < q. Obviously β ∈ ϕ(xα)if ξ = α and by Lemma 14, β ∈ ϕ(xα) if ξ 6= α. Thus, Sϕ(q) ∪ Xϕ(q) ⊆ ϕ(xα) and sod(xα) ≥ |Sϕ(q)|+ t = ∆− p+ t. The fourth statement holds.

To prove (5), let xα ∈ NS(x). Recall d(xα)−σq(x, xα)−1 = |ϕbad(xα)\(ϕ̄(xα)∪{α})|. Since

α is in Sϕ(q), not in Bϕ(q) and it follows from Corollary 15 that ϕbad(xξ)’s are disjoint for ξ ∈Sϕ(q)∩ϕ(x). Thus

∑xα∈NS(x)

(d(xα)−σq(x, xα)−1) = |⋃xα∈NS(x)

Äϕbad(xα)\(ϕ̄(xα) ∪ {α})

ä| = t.

Then by the Cauchy-Schwarz inequality, we have∑

v∈NS(x)

1d(v)−σq(x,v) ≥

|NS(x)|2|NS(x)|+t . Notice that

|NS(x)| = d(x) − |{y}| − |Bϕ(q)| = d(x) − p − 1, so |NS(x)|2|NS(x)|+t = (d(x)−p−1)2

d(x)−p−1+t . The last equality

holds.

Now by the properties above, we have

10

M ′(x) ≥ d(x) +∑

v∈NS(x)


+

Ñ ∑v∈NB(x)


é+

d(y)− qd(y)− σq(x, y)

≥∑

v∈NS(x)

∆− p+ t− qd(v)− σq(x, v)

+

Ñ ∑v∈NB(x)

∆− q∆− σq(x, v)

é+

∆− q∆− σq(x, y)

≥ (ε∆− p+ t)(d(x)− p− 1)2

d(x)− p− 1 + t+ (p+ 1)

ε∆

d(x)− p− 1.

Since d(x) ≤ ε∆, the first fraction above is a decreasing function of t. Since t ≤ p by (1), wehave:

M ′(x) ≥ ε∆

Ç(d(x)− p− 1)2

d(x)− 1+

p+ 1

d(x)− 1− p

å= ε∆

Ç(d(x)− p− 1)2

d(x)− 1+

d(x)

2(d(x)− 1− p)+

d(x)

2(d(x)− 1− p)− 1

å≥ ε∆

Ñ3

Çd(x)2

4(d(x)− 1)

å 13

− 1

é. (a+ b+ c ≥ 3(abc)

13 if a, b, c > 0.)

The number above is greater than q if ε ≥ 13 or ε < 1

3 and d(x) ≥ 2( 13ε)

3 +2»

( 13ε)

6 − ( 13ε)

3, we

briefly explain it. If ε ≥ 13 , then since d(x)2 ≥ 4(d(x) − 1), we have ε∆

Ç3(

d(x)2

4(d(x)−1)

) 13 − 1

å≥

ε∆(3− 1) ≥ 2ε∆ ≥ (1− ε)∆ = q. If ε < 13 , we solve the following inequality:

ε∆

Ñ3

Çd(x)2

4(d(x)− 1)

å 13

− 1

é≥ (1− ε)∆.

After simplifying, we have d(x)2 − 4( 13ε)

3d(x) + 4( 13ε)

3 ≥ 0. Then by the quadratic formula, weget what we want.

Claim 3.1 and Claim 3.2 together prove Theorem 2.

4 Proof of Theorem 1

Theorem 1. If G is a ∆-critical graph, then d(G) ≥ 34∆(G)− 8.

Let G be a ∆-critical graph with maximum degree ∆. In this section we assume ∆ ≥ 16since when ∆ < 16, 2

3∆ ≥ 34∆ − 8, Theorem 1 is implied by Woodall’s result [15]. We initially

assign to each vertex x ∈ V (G) a charge M(x) = d(x) and redistribute the charge according tothe following rule:

• Rule of Discharge (first step): Each vertex y with degree larger than 34∆−8 distributes

d(y)− (34∆− 8) equally among all its neighbors with degree less than 3

4∆− 8.

11

Denote by M ′(x) the resulting charge on each vertex x. We will prove the following two claims.Claim 4.1 For every x ∈ V (G) with d(x) > 1

4∆, M ′(x) ≥ 34∆− 8.

Claim 4.2 For every x ∈ V (G) with 4 ≤ d(x) ≤ 14∆, M ′(x) ≥ 3

4∆− 8.

The two claims above imply that M ′(x) ≥ 34∆ − 8 when d(x) ≥ 4. But for vertices with

degree 2 or 3, we only have the following result:

(∗) For every x ∈ V (G) with d(x) = 2, M ′(x) > 12∆; For every x ∈ V (G) with d(x) = 3,

M ′(x) > 58∆.

The proof of (∗) can be found in the proof of Claim 4.4 in Subsection 4.3. Obviously thisresult is not enough for proving Theorem 1. In order to deal with the vertices with degree 2 or3, we will adjust the discharging rule to complete the whole proof. Details of the last step willbe given in Subsection 4.3.

4.1 Part I: Claim 4.1.

4.1.1 Lemmas

The following lemma is a generalization of Lemma 12. We will use a separated section toprove it deal to its length. Recall that c1 = c1(ε) =

†1−εε

£− 1 for any ε ∈ (0, 1).

Lemma 16. Let ε ∈ (0, 1), q be a positive number, G be a ∆-critical graph with maximum degree∆, xy ∈ E(G) with d(x) < q and ϕ ∈ C∆(G − xy). If q ≤ min{(1 − ε)∆ − 2c1, ∆ − 6c1}, thenfor each z ∈ NS(x) except at most c1 vertices, there exists a coloring ϕ′ ∈ C∆(G− xy) such thatϕ′(xz) ∈ ϕ̄′(y) and for each ξ ∈ (Sϕ′(q) ∩ ϕ′(z))\{ϕ′(xz)} except at most 4c1 colors, d(x1ξ) ≥ qwhere 1 = ϕ′(xz).

The next lemma is a corollary of Lemma 16, we will explain why it can be implied byLemma 16 in Section 6 which also contain the proof of Lemma 16.

Lemma 17. Let ε ∈ (0, 1), q be a positive number, G be a ∆-critical graph with maximum degree∆, xy ∈ E(G) with d(x) < q and ϕ ∈ C∆(G−xy). If q ≤ min{(1−ε)∆−2c1, ∆−6c1}, then thereare at least ∆−σq(x, y)−2c1 vertices z ∈ N(x)\{y} such that σq(x, z) ≥ 2∆−d(x)−σq(x, y)−5c1.

Now let ε = 14 in Lemma 17. In this case c1 = 2. We have the following:

Corollary 18. Let G be a ∆-critical graph with maximum degree ∆, q ≤ min¶

34∆− 4,∆− 12

©be a positive number and xy ∈ E(G) with d(x) < q. Then there are at least ∆ − σq(x, y) − 4vertices z ∈ N(x) \ {y} such that σq(x, z) ≥ 2∆− d(x)− σq(x, y)− 10.

In general, for any x ∈ V (G) and positive number q with q ≤ ∆, we define the following twoparameters:

p(x, q) := miny∈N(x)

σq(x, y)− (∆− d(x) + 1) and

p′(x, q) := min{ p(x, q),úd(x)

2

ü− 6 }.

Lemma 19. Let G be a ∆-critical graph with maximum degree ∆, q ≤ min¶

34∆− 4,∆− 12

©be a positive number. Then for any x with d(x) < q, x has at least d(x)− p′(x, q)− 5 neighborsz for which σq(x, z) ≥ ∆− p′(x, q)− 11.

12

Proof. Let y ∈ N(x) such that p(x, q) = σq(x, y) − ∆ + d(x) − 1. If p′(x, q) = p(x, q), byCorollary 18, there are at least ∆−σq(x, y)−4 = ∆−(∆−d(x)+p(x, q)+1)−4 = d(x)−p(x, q)−5vertices z ∈ N(x) \ {y} such that σq(x, z) ≥ 2∆ − d(x) − 10 − σq(x, y) = ∆ − p(x, q) − 11. If

p′(x, q) =⌊d(x)

2

⌋− 6 < p(x, q), then for every z ∈ N(x), σq(x, z) > ∆− d(x) + 1 +

⌊d(x)

2

⌋− 6 ≥

∆−⌊d(x)

2

⌋− 6 = ∆− p′(x, q)− 12. So σq(x, z) ≥ ∆− p′(x, q)− 11.


In the reminder of Section 4, we fix the number q to be 34∆− 8 .

Claim 4.1. For every x ∈ V (G) with d(x) > 14∆, M ′(x) ≥ q.

Proof. Since M ′(x) ≥ q when d(x) ≥ q, we will consider the case when 14∆ < d(x) < q. Let

y ∈ N(x) such that d(y) is minimum over all neighbor of x, ϕ ∈ C∆(G − xy). Let Zq = {z ∈N(x) : d(z) > q}, Zy = {z ∈ N(x)\{y} : ϕ(xz) ∈ ϕ̄(y)} and Z∗q = Zq ∩ Zy. Clearly, for each

z ∈ Z∗q , x receives d(z)−qd(z)−σq(x,z) charge from it. Hence M ′(x) ≥ d(x) +

∑z∈Z∗q

d(z)−qd(z)−σq(x,z) . Since

the notation p′(x, q) will be used heavily in this proof, we let p′ = p′(x, q) for convenience.For any z ∈ Z∗q , the three inequalities below will be used soon.

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

(2) σq(x, z) ≥ 2∆− d(x)− d(y)− 1 and

(3) σq(x, z) ≥ 2∆− d(x)− d(y) if d(y) ≤ q.

We give the proof of these statements. (1) is exactly Lemma 5 (2). Recall that Lemma 5

also shows that σq(x, z) ≥ 2∆ − d(x) − d(y) + 1 − bd(x)+d(y)+d(z)−2∆−2∆−q c. Since d(x) < q and

q = 34∆ − 8, we have bd(x)+d(y)+d(z)−2∆−2

∆−q c ≤ 2, so σq(x, z) ≥ 2∆ − d(x) − d(y) − 1. Moreover,

if d(y) ≤ q, we further have bd(x)+d(y)+d(z)−2∆−2∆−q c ≤ 1 and σq(x, z) ≥ 2∆− d(x)− d(y).

Now we consider the following three cases to complete the proof.

Case 1. d(y) ≤ q.In this case, by (1) and (3) we have

∑z∈Z∗q

d(z)− qd(z)− σq(x, z))

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

∆− (2∆− d(x)− d(y))

=(∆− q)(∆− d(y) + 1) + 2

d(x) + d(y)−∆− 1.

Since ϕ̄(x) ∩ ϕ̄(y) = ∅ by Lemma 1, d(x) + d(y) ≥ ∆ + 2. Thus, d(x) ≥ ∆− q + 2 and

13

M ′(x) ≥ d(x) +(∆− q)(∆− d(y) + 1) + 2

d(x) + d(y)−∆− 1

≥ d(x) + q −∆ +(∆− q)(∆− q + 1) + 2

d(x) + q −∆− 1− q + ∆ (Since d(y) ≤ q.)

≥ 2»

(∆− q)(∆− q + 1) + 2 + ∆− q − 1

≥ 3(∆− q)− 1 > q.

Case 2. d(y) > q, 2(∆− q)− 10 < d(x) < q and p′ > ∆− q − 11.In this case, each neighbor of x has degree larger than q and Z∗q = Zy. So M ′(x) ≥

d(x) +∑z∈Zy

d(z)−qd(z)−σq(x,y) +

∑u∈N(x)\Zy

d(u)−qd(u)−σq(x,u) .

By (1) and (2) we have∑z∈Zy

d(z)− qd(z)− σq(x, y)

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

∆− (2∆− d(x)− d(y)− 1)

>(∆− d(y) + 1)(∆− q)d(x) + d(y)−∆ + 1

− 1.

Since d(x) > 2(∆− q)− 10 and p′ ≤ d(x)2 − 6, we have q ≥ ∆− d(x)

2 − 5 ≥ ∆− d(x) + 1 + p′.By the definition of p′, for each neighbor u of x we have σq(x, u) ≥ ∆ − d(x) + 1 + p′. Sinced(u) ≥ d(y) > q for each u ∈ N(x) and q ≥ ∆− d(x) + 1 + p′, we have

d(u)− qd(u)− σq(x, u)

≥ d(u)− qd(u)− (∆− d(x) + 1 + p′)

≥ d(y)− qd(y)− (∆− d(x) + 1 + p′)

.

Notice that |N(x)\Zy| ≥ d(x)− |ϕ̄(y)| = d(x)− (∆− d(y) + 1), we have

∑u∈N(x)\Zy


≥ |N(x) \ Zy| ·d(y)− q

d(y)− (∆− d(x) + 1 + p′)

≥ (d(x)− (∆− d(y) + 1))(d(y)− q)d(y)− (∆− d(x) + 1 + p′)

≥ (d(x) + d(y)−∆− 1))(d(y)− q)d(x) + d(y) + 2− 5

4∆

Where the last inequality follows from the condition that p′ > ∆− q − 11. Above all, we have:

M ′(x) ≥ d(x) +∑z∈Zy

d(z)− qd(z)− σq(x, y)

+∑

u∈N(x)\Zy


≥ d(x)− 1 +(∆− d(y) + 1)(∆− q)d(x) + d(y)−∆ + 1

+(d(x) + d(y)−∆− 1))(d(y)− q)

d(x) + d(y) + 2− 54∆

= d(x)− 1 +(∆− d(y) + 1)(∆− q)d(x) + d(y)−∆ + 1

+(1

4∆− 3))(d(y)− q)d(x) + d(y) + 2− 5

4∆+ d(y)− q

To prove this long expression is larger than q, we let A = d(y)− 34∆+8 and B = d(x)− 1

2∆−6be stepping-stones for the calculation. Easy to check that both A and B are positive andA < 1

4∆ + 8. Thus

14

M ′(x) ≥ B +1

2∆ + 6− 1 +

(14∆−A+ 9)(1

4∆ + 8)

A+B + 14∆− 1

+(1

4∆− 3)(A)

A+B+A

≥ 1

4∆ +A+B +

1

4∆ + (

1

4∆− 3)(

14∆−A

A+B + 14∆

+A

A+B) + 5

=1

4∆ +A+B +

1

4∆ + (

1

4∆− 3)

14∆(2A+B)

(A+B + 14∆)(A+B)

+ 5

≥ 1

4∆ +A+B +

1

4∆ + (

1

4∆− 3)

14∆

A+B + 14∆

+ 5

≥ 1

4∆ + 2

(1

4∆− 3)

1

4∆ + 5 ≥ 3

4∆− 1 > q

Case 3. d(y) > q and 14∆ < d(x) ≤ 2(∆− q)− 10 or p′ ≤ ∆− q − 11.

Notice that if d(x) ≤ 2(∆−q)−10, then p′ = min{ p(x, q),⌊d(x)

2

⌋−6 } ≤ d(x)

2 −6 ≤ ∆−q−11.

Thus in both cases, we have p′ ≤ ∆− q − 11.Since q = 3

4∆− 8 < 34∆− 4 and ∆ ≥ 16, we have ∆− q ≥ 12. Thus we may apply Lemma

19, x has at least d(x)− p′− 5 neighbors z for which σq(x, z) ≥ ∆− p′− 11. For such neighborsz, since q ≤ ∆− p′ − 11, we have


≥ d(z)− qd(z)− (∆− p′ − 11)

≥ ∆− qp′ + 11

.

Case 3.a q > ∆− d(x) + p′ + 1, i.e., p′ < d(x) + q −∆− 1.

In this case, d(x)−p′−5p′+11 > ∆−q−4

d(x)+q−∆+10 , thus

M ′(x) ≥ d(x) + (d(x)− p′ − 5)∆− qp′ + 11

≥ d(x) +∆− q − 4

d(x) + q −∆ + 10(∆− q)

= (d(x) + q −∆ + 10) +(∆− q)(∆− q − 4)

d(x) + q −∆ + 10− (q −∆ + 10)

≥ 2»

(∆− q)(∆− q − 4) + ∆− q − 10 ≥ 3(∆− q)− 18 > q.

Case 3.b q ≤ ∆− d(x) + p′ + 1.Following the definition of p′, for every u ∈ Zq, we have σq(x, u) ≥ ∆− d(x) + p′ + 1, which

in turn givesd(u)− q

d(u)− σq(x, u)≥ d(u)− qd(u)− (∆− d(x) + p′ + 1)

.

Since q ≤ ∆− d(x) + p′ + 1, we have d(u)−qd(u)−σq(x,u) ≥

∆−q∆−(∆−d(x)+p′+1) = ∆−q

d(x)−p′−1 . Thus,

M ′(x) ≥ d(x) + (d(x)− p′ − 5)∆− qp′ + 11

+ (p′ + 5)∆− q

d(x)− p′ − 1

15

Let A = d(x)− p′ − 5 ≥ d(x)2 + 1 and B = p′ + 11. Since A

B + BA ≥ 2 and A+B = d(x) + 6,

we have:

M ′(x) ≥ d(x) + (∆− q)(2− (B

A− B − 6

A+ 4))

= d(x) + (∆− q)(2− 4(A+B) + 2A

A(A+ 4))

= d(x) + (∆− q)(2− 4(d(x) + 6) + 2d(x)− 2p′ − 10

A(A+ 4))

≥ d(x) + (∆− q)(2− 6d(x) + 14

A(A+ 4))

Then since A ≥ d(x)2 + 1, we have

M ′(x) ≥ d(x) + (∆− q)(2− 6d(x) + 14

(d(x)2 + 1)(d(x)

2 + 5))

>∆

4+ (∆− q)(2−

32∆ + 14

(∆8 + 1)(∆

8 + 5)) (d(x) >

∆

4, function increasing.)

=∆

4+ (∆− q)(2−

96(∆ + 283 )

(∆ + 8)(∆ + 40))

>∆

4+ (∆− q)(2−

96(∆ + 283 )

(∆ + 10)(∆ + 32))

>∆

4+ (∆− q)(2− 96

∆ + 32) = q

This completes the proof of Case 3 and Claim 4.1.

4.2 Part II: Claim 4.2

Claim 4.2. For every x ∈ V (G) with 4 ≤ d(x) ≤ 14∆, M ′(x) ≥ q.

Proof. Let x ∈ V (G) with 4 ≤ d(x) ≤ 14∆. Since the proof of claim 3.2 does not contain

the assumption ∆ ≥ D(ε), we apply that claim here with ε = 14 . When ε = 1

4 , D0(ε) =

2( 13ε)

3 + 2»

( 13ε)

6 − ( 13ε)

3 is about 8.345. Thus we only need to prove this claim when d(x) is

in {4, 5, 6, 7, 8}. Choose y ∈ N(x) and ϕ ∈ C∆(G − xy) such that p = |Bϕ(x, y, q)| = |Bϕ(q)|is minimum over all neighbours of x and all ∆-colorings of G − xy. We will use the followingproperties to prove the claim:

(1) |Sϕ(q)| = ∆− p, p ≤ d(x)− 2 and |NS(x)| = d(x)− 1− p;

(2) For each v ∈ N(x), d(v) ≥ σq(x, v) ≥ ∆− d(x) + 1 + p and |ϕbad(v)| ≤ d(x)− 1− p;

(3) Let X ′ =⋃v∈NS(x)(ϕ

bad(v)\{ϕ(xv)}), then X ′ ⊆ Bϕ(q) and |X ′| = ∑v∈NS(x)(|ϕbad(v)|−1).

Consequently, |X ′| ≤ p;

16

(4) M ′(x) = d(x) +∑v∈N(x)

d(v)−q|ϕbad(v)|−|ϕ̄(v)| and

d(v)−q|ϕbad(v)|−|ϕ̄(v)| = ∆−q−|ϕ̄(v)|

|ϕbad(v)|−|ϕ̄(v)| ≥∆−q|ϕbad(v)| ≥

∆−qd(x)−1−p .

We then prove these properties one by one. For (1), |Sϕ(q)| = ∆ − |Bϕ(q)| = ∆ − p,p = |Bϕ(q)| ≤ |Aϕ(q)| + |Bϕ(q)| = |ϕ(x)\ϕ̄(y)| = d(x) − 1 − |ϕ̄(y)| ≤ d(x) − 2 and |NS(x)| =|Sϕ(q) ∩ ϕ(x)| = |ϕ(x)\Bϕ(q)| = d(x)− 1− p.

For (2), let v be a neighbor of x and ϕ′ ∈ C∆(G− xv). Since for any α ∈ ϕ̄′(x), there existsw ∈ N(v) such that ϕ′(vw) = α and d(w) ≥ |ϕ̄′(x)| = ∆− d(x) + 1 > q by Lemma 1. ({x, v, w}is elementary.) By the minimality of p, we have d(v) ≥ σq(x, v) ≥ |ϕ̄′(x)| + |Bϕ′(x, v, q)| =

∆−d(x)+1+p; Also, we have |Aϕ′(x, v, q)| = |{w ∈ N(v)\{x} : d(w) < q}|. So |ϕbad(v)| = |{w ∈N(v)\{x} : d(w) < q}|+ |ϕ̄(v)|+ |{ϕ(xz)}| = |Aϕ′(x, v, q)|+ |ϕ̄(v)|+ 1 = |Aϕ′(x, v, q)|+ |ϕ̄′(v)|.Hence |Bϕ′(x, v, q)| = |ϕ′(x)\(Aϕ′(x, v, q) ∪ ϕ̄′(v))| = |ϕ′(x)| − |ϕbad(v)|, |ϕbad(v)| = d(x)− 1−|Bϕ′(x, v, q)| ≤ d(x)− 1− p.

For (3), notice that X ′ ⊆ Bϕ(q) by Corollary 15 and ϕbad(v)’s are disjoint for v ∈ NS(x),

thus |X ′| = ∑v∈NS(x)(|ϕbad(v)| − 1).

For (4), since d(v) ≥ ∆ − d(x) + 1 + p > q for each v ∈ N(x) by (2), M ′(x) = d(x) +∑v∈N(x)

d(v)−q|ϕbad(v)|−|ϕ̄(v)| by the discharging rule. Also (2) implies |ϕbad(v)| ≤ d(v) − 1 − p, thus

the last inequality holds. Since d(x)−1−p < d(x) < ∆−q = 14∆+8, we have |ϕbad(v)| < ∆−q.

Hence the first inequality holds.

Now we start to prove M ′(x) ≥ q for each x with d(x) ∈ {4, 5, 6, 7, 8}.Case 0. p = 0 or p = d(x)− 2.Suppose p = 0. Then |Sϕ(q)| = ∆ by the first property, so Sϕ(q) contains all ∆ colors. Thus

by Lemma 12, for any v ∈ N(x)\{y}, d(y) ≥ |Sϕ(q)| = ∆ and ϕbad(v) = {ϕ(xv)}. Hence by

(4), M ′(x) ≥ (d(x)− 1) (∆−q)1 ≥ 3(∆− q) > q. Suppose p = d(x)− 2. Then for all v ∈ N(x), we

have σq(x, v) ≥ ∆ − d(x) + 1 + p = ∆ − 1 by (2). So x is the only vertex in N(v) with degree

less than q, |ϕbad(v)| = 1 and d(v) = ∆. Hence M ′(x) ≥ d(x)(∆− q) ≥ 4(∆− q) > q.

Case 1. d(x) = 4.In this case, p ≤ d(x)− 2 = 2. By using Case 0, we assume that p = 1. Then |NS(x)| = 2.Without loss of generality, assume that Sϕ(q) ∩ ϕ(x) = {ξ1, ξ2} and Bϕ(q) = {β}. So X ′ ⊆

{β}. If X ′ = ∅, then ϕbad(xξ1)\{ξ1} = ϕbad(xξ2)\{ξ2} = ∅. So M ′(x) ≥ ∑v∈NS(x)

∆−q|ϕbad(v)| +∑

v∈NB(x)∪{y}∆−q

d(x)−1−p ≥ 2(∆ − q) + 2(∆−q2 ) > q. Suppose X ′ = {β}, let’s say β ∈ ϕbad(xξ1).

Then ϕbad(xξ2)\{ξ2} = ∅. By Lemma 13 and Lemma 14, we have ϕbad(xβ)\{β} = ∅. Thus,

M ′(x) ≥∑v∈{ξ2,β}∆−q|ϕbad(v)| +

∑v∈{ξ1,y}

∆−qd(x)−1−p ≥ 2(∆− q) + 2(∆−q

2 ) > q. This proves Case 1.

Case 2. d(x) = 5.In this case, p ≤ d(x)− 2 = 3. By Case 0, we are done if p = 0 or 3.Suppose p = 1, then |NS(x)| = 3 by (1) and

∑v∈NS(x)(|ϕbad(v)| − 1) = |X ′| ≤ 1. Hence

M ′(x) ≥∑v∈NS(x)∆−q|ϕbad(v)| +

∑v∈NB(x)∪{y}

∆−qd(x)−1−p ≥

Ä2(∆− q) + ∆−q

2

ä+ 2(∆−q

3 ) > q.

Suppose p = 2, then for all v ∈ N(x), σq(x, v) ≥ ∆− d(x) + 1 + p = ∆− 2, so |ϕbad(v)| ≤ 2.If one of the neighbors of x has no neighbors other than x with degree less than q, we are donebecause M ′(x) ≥ 5+(∆−1−q)+4(∆−q

2 ) > q. Thus the only remaining case is: Each neighborsof x has exactly 1 neighbor other than x with degree less than q. Under these assumptions,|NS(x)| = 2 and |X ′| = 2. Let Sϕ(q) ∩ ϕ(x) = {ξ1, ξ2} and X ′ = Bϕ(q) = {β1, β2}. Notice that

17

ϕ̄(y) ⊆ Sϕ(q) ∩ ϕ(x) and |ϕ̄(y)| ≥ 1, we may assume ξ1 ∈ ϕ̄(y). Without loss of generality, byLemma 12, let’s say xξ1β1 , xξ2β2 have degrees less than q. By Lemma 13, the unique neighbor ofxβ1 with degree less than q is xβ1β2 ; The unique neighbor of xβ2 with degree less than q is xβ2β1 .Now, recolor xy by ξ1 and leave xxξ1 uncolored to obtain a new coloring ϕ1. Then in the new

coloring, β1 ∈ Sϕ1(x, xξ1 , q), xβ1β2 has degree less than q, i.e. β2 ∈ ϕbad1 (β1). By Lemma 13,

xβ2β1 should has degree at least q, giving a contradiction. This proves Case 2.

Case 3. d(x) = 6.In this case, p ≤ d(x)− 2 = 4. If p = 0 or 4, we are done by Case 0.If p = 1, then |NS(x)| = 4,

∑v∈NS(x)(|ϕbad(v)| − 1) = |X ′| ≤ 1. Thus


∑v∈NB(x)∪{y}

∆−qd(x)−1−p ≥

Ä3(∆− q) + ∆−q

2

ä+ 2(∆−q

4 ) > q.

If p = 2, then |NS(x)| = 3,∑v∈NS(x)(|ϕbad(v)| − 1) = |X ′| ≤ 2. Thus


∑v∈NB(x)∪{y}

∆−qd(x)−1−p ≥

Ä(∆− q) + 2(∆−q

2 )ä

+ 3(∆−q3 ) > q.

If p = 3, then M ′(x) ≥∑v∈N(x)∆−q

d(x)−1−p ≥ 6(∆−q2 ) > q. This proves Case 3.

Case 4. d(x) = 7.In this case, p ≤ d(x) − 2 = 5. Similarly, if p = 0 or p = 5 we are done. Also by similar

methods we used in Case 3, it is routine to check the cases when p = 1, 2 and 4.If p ≤ 2, then |NS(x)| ≥ 4,

∑v∈NS(x)(|ϕbad(v)| − 1) = |X ′| ≤ 2. Thus


∑v∈NB(x)∪{y}

∆−qd(x)−1−p ≥

Ä2(∆− q) + 2(∆−q

2 )ä

+ 3(∆−q5 ) > q.

If p = 4, then M ′(x) ≥∑v∈N(x)∆−q

d(x)−1−p ≥ 7(∆−q2 ) > q.

Now Suppose p = 3. Then |NS(x)| = 3, |X ′| ≤ 3 and for all v ∈ N(x), σq(x, v) ≥ ∆ −d(x) + 1 + p = ∆ − 3, so |ϕbad(v)| ≤ 3. If one vertex in NS(x) has no neighbors other than x

with degree less than q, then we have M ′(x) ≥Ä(∆− q) + (∆−q

2 ) + (∆−q3 )ä

+ 4(∆−q3 ) > q. So

we may assume that each vertex in NS(x) has exactly 1 neighbor other than x with degree lessthan q. Moreover, if one vertex in NB(x) has less than 2 neighbors other than x with degree

less than q, then we have M ′(x) ≥ 3(∆−q2 ) + 3(∆−q

3 ) + ∆−q2 > q. Therefore we may assume that

|NS(x)| = 3, |X ′| = 3, each vertex in NS(x) has exactly 1 neighbor other than x with degree lessthan q and each vertex in NB(x) has exactly 2 neighbors other than x with degree less than q.Let Sϕ(q) ∩ ϕ(x) = {ξ1, ξ2, ξ3} and X ′ = Bϕ(q) = {β1, β2, β3}. Notice that ϕ̄(y) ⊆ Sϕ(q) ∩ ϕ(x)and |ϕ̄(y)| ≥ 1, we may assume ξ1 ∈ ϕ̄(y). Without loss of generality, let’s say xξ1β1 has degreeless than q. By Lemma 13, the two neighbors of xβ1 with degree less than q are xβ1β2 , xβ1β3 ;The two neighbors of xβ2 with degree less than q are xβ2β1 , xβ2β3 . We don’t have to considerxβ3 . Now, recolor xy by ξ1 and leave xxξ1 uncolored to obtain a new coloring ϕ1. Then in thenew coloring, β1 ∈ Sϕ1(x, xξ1 , q), xβ1β2 has degree less than q, by Lemma 13, xβ2β1 should hasdegree at least q, giving a contradiction. This proves Case 4.

Case 5. d(x) = 8.In this case, p ≤ d(x) − 2 = 6. Actually all the cases can be verified by the same methods

we used in Case 3.If p ≤ 3, then |NS(x)| ≥ 4,

∑v∈NS(x)(|ϕbad(v)| − 1) = |X ′| ≤ 3. Thus


∑v∈NB(x)∪{y}

∆−qd(x)−1−p ≥

Ä(∆− q) + 3(∆−q

2 )ä

+ 4(∆−q6 ) > q.

If p = 4, then |NS(x)| = 3,∑v∈NS(x)(|ϕbad(v)| − 1) = |X ′| ≤ 4. Thus


∑v∈NB(x)∪{y}

∆−qd(x)−1−p ≥

Ä∆−q

3 + 2(∆−q2 )ä

+ 5(∆−q3 ) > q.

If p ≥ 5, then M ′(x) ≥∑v∈N(x)∆−q

d(x)−1−p ≥ 8(∆−q2 ) > q.

This completes the proof of Claim 4.2.

18

4.3 Part III: Vertices with degree two or three

4.3.1 Lemmas (Part III)

For the vertices with degree two or three, the neighbors of them are not enough to ensurewhat we want, hence we have to find more vertices other than the neighbors to help us attainthe requirement. Fortunately, we do have some candidates for 2-vertices.

Lemma 20. Let x be a 2-vertex in V (G), the following statements are true:

(1) For any z ∈ N(x), d(z) = ∆,

(2) For any z′ ∈ N(N(x))\{x}, d(z′) = ∆ and

(3) For any z′ ∈ N(N(x))\{x} and v ∈ N(z′)\{x}, d(v) ≥ ∆− 1.

Proof. Let z ∈ N(x), z′ ∈ N(z)\{x} and v ∈ N(z′)\{x}. We will prove that d(z) = d(z′) = ∆and d(v) ≥ ∆ − 1, which implies the lemma. Let ϕ ∈ C∆(G − xz). Since d(x) = 2, we have|ϕ̄(x)| = ∆− 1. Notice that |ϕ̄(z)| ≥ 1, |ϕ̄(z′)| ≥ 0 and ϕ̄(x)∩ ϕ̄(z) = ∅. So ∆ ≥ |ϕ̄(x)∪ ϕ̄(z)| ≥|ϕ̄(x)| + |ϕ̄(z)| ≥ ∆, the equality hold and |ϕ̄(z)| = 1. Thus all colors are in ϕ̄(x) ∪ ϕ̄(z),xzz′v form a Kierstead path when v 6= z. By Lemma 1 and 3, {x, z, z′} is elementary and|ϕ̄(v) ∩ (ϕ̄(x) ∪ ϕ̄(z) ∪ ϕ̄(z′))| ≤ 1 if v 6= z. Therefore |ϕ̄(z′)| ≤ |[1,∆]\(ϕ̄(x) ∪ ϕ̄(z))| = 0,d(z) = d(z′) = ∆. Moreover, if v 6= z, we have |ϕ̄(v)| ≤ 1, i.e. d(v) ≥ ∆− 1, which is also truewhen v = z.

Let L(x) = N (N(x)) \N [x]. Then for any x with d(x) = 2, each vertex in L(x) has degree∆ and all its neighbors have degrees at least ∆ − 1 by Lemma 20. Thus all vertices in L(x)are not used in the previous discharging rule. Now we consider all the vertices with degree 3.After applying the previous discharging rule, some 3-vertices may already have M ′ ≥ q, we callthese vertices type I 3-vertices and call the rest type II 3-vertices. Similar to the lemmaabove, for each type II 3-vertex x, we hope that many vertices in L(x) have large degrees and noneighbors of them have small degrees. We will prove the following lemma by using some resultsin previous sections.

Lemma 21. If x is a type II 3-vertex in V (G) with N(x) = {x1, x2, x3}. Then the followingtwo statements hold.

(1) Each xi has degree ∆ and there is only one vertex in L(x) with degree less than q .(2) There exists an i in {1, 2, 3} and a vertex w ∈ N(xi) such that for all v ∈ N(xi)\(N [x]∪

{w}) ⊆ L(x), d(v) ≥ ∆− 1 and d(v′) ≥ q for any v′ ∈ N(v).

Proof. We firstly prove (1). Choose xi ∈ N(x) such that σq(x, xi)−∆+d(x)−1 = pmin(x, q) = p,without loss of generality, let’s say it is x1. By the definition of p, we have p ≤ d(x) − 2 = 1.Let ϕ ∈ C∆(G − xx1) be a coloring such that ϕ(xx2) = 2, ϕ(xx3) = 3. If p = 1, then for allxi ∈ N(x), we have σq(x, xi) ≥ ∆− d(x) + 1 + p = ∆− 1, d(xi) = ∆ and |ϕbad(xi)| = 1. HenceM ′(x) = 3 + 3(∆ − q) > q, which contradict the fact that x is of type II. Thus p has to be 0.In this case, all the colors are contained in Sϕ(q), hence by Lemma 12, d(x2) = d(x3) = ∆ andx2, x3 have no neighbors other than x with degree less than q. Since x is of type II, x1 has aneighbor w 6= x with degree less than q, thus d(x1) ≥ ∆ − d(x) + 1 + |{x}| + |{w}| = ∆. Thisproves (1). Moreover, ϕ(x1w) cannot be a missing color of x by Lemma 1, so it has to be a colorin {2, 3}, without loss of generality, let’s say ϕ(x1w) = 2.

Now we prove (2). More specifically, we claim that under the setting in (1), i = 1 and the win (2) is just the w we used above. Choose an arbitrary vertex in N(x1)\{w, x, x2, x3} ⊆ L(x),assume the edge between this vertex and x1 is colored by α ∈ ϕ̄(x)∪ ϕ̄(x1) = {1, 3, 4, . . . ,∆} and

19

denote this vertex by v. By Lemma 2, we have d(v) ≥ ∆− 1. For any v′ ∈ N(v)\{x, x1, x2, x3},if ϕ(vv′) ∈ {1, 3, 4, . . . ,∆}, then d(v′) ≥ ∆ − 2 > q by Lemma 3. (This also implies that w isnot in N(v).) If ϕ(vv′) = 2, choose δ 6= α such that δ ∈ ϕ̄(x) ∩ ϕ̄(w), this is possible becaused(x) = 3 and d(w) < q. Then we do a (2, 3, δ)-switching in ϕ at x1 to obtain a new coloringϕ1. This will not affect x1v and vv′ by Lemma 6. Now we have 2 ∈ ϕ̄1(x1), by Lemma 3 again,d(v′) ≥ ∆− 2 > q. This proves (2).

4.3.2 Proof of Claim 4.3 and Claim 4.4

In this subsection, q is still 34∆− 8. We are finally able to complete our discharging rule as

the following:

• The First Step of Discharge: Each vertex y with degree larger than q distributesd(y)− q equally among all neighbors of y with degree less than q.

• The Second Step of Discharge: For all v ∈ V (G) with d(v) ≥ ∆− 1, if L(v) containsa 2-vertex or a 3-vertex of type II and all neighbors of v have degrees at least q, then vsends 1

4 to each 2-vertex in L(v) and 18 to each type II 3-vertex in L(v).

We still denote by M ′(x) the resulting charge on each vertex x. Now we prove the followingclaims.

Claim 4.3. Claim 4.1 and Claim 4.2 still hold under the new discharging rule.

Proof. Since the second step will only affect a vertex v when d(v) ≥ ∆ − 1, L(v) contains a2-vertex or a 3-vertex of type II and v has no neighbors with degree less than q, we only needto prove that M ′(v) is at least q for such v. Assume that x is a 2-vertex or a 3-vertex of typeII in L(v), then v ∈ L(x). By our assumption, we have that v has no neighbors with degree lessthan q, thus the first step of discharge will not affect v. Let z be a neighbor of v, by Lemma 20and Lemma 21 , if z is adjacent to a 2-vertex, then all other neighbors of z has degree ∆; If zis adjacent to a 3-vertex of type II, then z has at most 1 more neighbor with degree less thanq, this neighbor of z has degree at least 3 by Lemma 20. Thus in both cases, v sends at most 1

4

in total to the neighbors of z in the second step of discharge, M ′(v) ≥ ∆ − 1 − 14∆ > q. This

proves the claim.

Claim 4.4. For every x ∈ V (G) with d(x) = 2 or 3, M ′(x) ≥ q.

Proof. If d(x) = 2, by Lemma 20, all vertices in N(N(x))\{x} are ∆-vertices. So after the firststep of discharge, M ′(x) ≥ 2 + (∆− q) + (∆− q). By Lemma 20 again, there are at least ∆− 1distinct ∆-vertices in L(x) which have no neighbors with degree less than q, thus under the newdischarging rule, M ′(x) ≥ 2 + (∆− q) + (∆− q) + 1

4(∆− 1) > q.If x is a type I 3-vertex, we are done. If x is a type II 3-vertex, by Lemma 21, all its neighbors

are ∆-vertices and there is only one vertex in L(x) has degree less than q. So after the first stepof discharge, M ′(x) ≥ 3 + (∆− q) + (∆− q) + 1

2(∆− q). By Lemma 21 again, there are at least∆−4 distinct vertices in L(x) with degree at least ∆−1 and have no neighbors with degree lessthan q. (In Lemma 21, N(xi)\(N [x] ∪ {w}) has size ∆ − 4.) Thus under the new dischargingrule, M ′(x) ≥ 3 + (∆− q) + (∆− q) + 1

2(∆− q) + 18(∆− 4) > q. This proves the claim.

Now combining Claim 4.1, Claim 4.2, Claim 4.3 and Claim 4.4, we proved Theorem 1.

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5 Proofs of Lemma 7 and 8

In this section, let ε ∈ (0, 1) be a given real number and c0 =†

1−εε

£as before. For any two

positive integers m1 and m2, we define the functions f1, f2, f3 and f as below.

f1(m1,m2, ε) =1

ε2(m1 + c0m2),

f2(m1,m2, ε) = max{f1(m1,m2 + 1, ε),1

ε(m2 + (c0 + 1)m1 + 1)},

f3(ε) = (c0 + 1)(c0 + 3)(3c0) + c0 + 2,

f(ε) = f2(c0 + 1, f3(ε), ε).

We also assume that G is a ∆-critical graph with maximum degree ∆, q = (1 − ε)∆, xy ∈E(G) and d(x) < q.

Fact 1: Let v1, v2, . . . , vc0+1 be c0 + 1 vertices with degree less than q and ϕ ∈ C∆(G − xy).If |ϕ̄(vi) ∩ ϕ̄(x)| ≤ m1 for all i = 1, 2, . . . , c0 + 1 and ∆ ≥ f1(m1,m2, ε), then there exist twovertices vi, vj such that |(ϕ̄(vi) ∩ ϕ̄(vj))\ϕ̄(x)| ≥ m2.

Proof. Notice that a vertex with degree less than q has more than ε∆ missing colors. If Fact1 fails, then for each vi, there are at least ε∆ − m1 − c0m2 colors in ϕ̄(vi) that are not in(⋃j 6=i ϕ̄(vj)) ∪ ϕ̄(x). But then since c0 + 1 ≥ 1

ε and ∆ − d(x) > ε∆, there are at least (c0 +

1)(ε∆−m1 − c0m2) + ∆− d(x) + 1 > (c0 + 1)(ε∆− ε2∆) + ε∆ ≥ ∆ distinct colors, which is acontradiction.

Given a coloring ϕ ∈ C∆(G − xy), a color set C and a vertex set T ⊂ V (G), we call theKempe change ϕ→ ϕ/Pv(δ, η, ϕ) a S(C, T ) if v ∈ T , δ ∈ ϕ̄(x)\(C ∪ ϕ̄(T )), η ∈ ϕ̄(v)\(ϕ̄(x)∪C)and x /∈ V (Pv(δ, η, ϕ)).

Fact 2: Following the notation above, there exists a S(C, T ) if the conditions below hold:

• |T | = c0 + 1. For each v ∈ T , d(v) < q,

• |ϕ̄(v) ∩ ϕ̄(x)| ≤ m1 for each v ∈ T and ∆ ≥ f2(m1, |C|, ε).

Proof. Since ∆ > f2(m1, |C|, ε) ≥ f1(m1, |C|+ 1, ε), by Fact 1, there are two vertices u, v ∈ Tsuch that |(ϕ̄(u) ∩ ϕ̄(v))\ϕ̄(x)| ≥ |C|+ 1. So, there exists a color η ∈ (ϕ̄(u) ∩ ϕ̄(v))\(ϕ̄(x) ∪C).Since ∆ > f2(m1, |C|, ε) ≥ (|C|+ (c0 + 1)m1 + 1)/ε, |ϕ̄(x)| ≥ ∆− q = ε∆ ≥ |C|+ (c0 + 1)m1 + 1,we can find a color δ ∈ ϕ̄(x) such that δ /∈ C ∪ (ϕ̄(T ) ∩ ϕ̄(x)). Now, one of u, v is not onPx(η, δ, ϕ), say v. Then, ϕ′ = ϕ/Pv(η, δ, ϕ) is the desired switching.

The following observation follows directly from the definition of S(C, T ).

Fact 3. If ϕ → ϕ′ is a S(C, T ), then∑v∈T |ϕ̄′(v) ∩ ϕ̄′(x)| ≥ ∑

v∈T |ϕ̄(v) ∩ ϕ̄(x)| + 1 with δaccounting for the ’+1’, and edges with colors in C are unchanged.

Recall that Aϕ(q) = {α ∈ ϕ(x) ∩ ϕ(y) : d(yα) < q} and Sϕ(q) = Aϕ(q) ∪ ϕ̄(x) ∪ ϕ̄(y). Werestate Lemma 7 as the following for the proof purpose.Lemma 7. Let G, x and xy be defined as above. If ∆ ≥ f(ε), then for every ϕ ∈ C∆(G− xy)there does not exist a vertex z ∈ N(x)\{y} such that the following two statements hold.

21

A1: z ∈ N(x) and ϕ(xz) ∈ ϕ̄(y).

A2: There exists c0+1 colors β ∈ ϕbad(z)∩ϕ(x)\{ϕ(xz)} such that |ϕbad(xβ)∩Sϕ(q)\{β}| ≥ 3c0.

Proof. Suppose on the contrary there exist a coloring ϕ ∈ C∆(G− xy), a vertex z ∈ N(x) withϕ(xz) ∈ ϕ̄(y) and c0 + 1 colors β1, β2, . . . , βc0+1 satisfies A2. Without loss of generality, weassume ϕ(xz) = 1, that is, z = x1. We will show that the following statement holds.

(5.1). There exists a coloring π ∈ C∆(G − xy) such that π(xz) = ϕ(xz) = 1 ∈ π̄(y) and acolor β ∈ π(x) such that |π̄(x) ∩ πbad(xβ)| contains c0 colors ξ1, ξ2, . . ., ξc0 . Additionally, wehave either β ∈ π̄(z) or β ∈ π(z) and there exists a color δ1 ∈ (π̄(x1β) ∩ π̄(x))\{ξ1, ξ2, . . ., ξc0}.

We first prove Lemma 7 assuming (5.1) is true.Let π1 ∈ C∆(G− xz) be obtained from π by assigning the color 1 to edge xy and leaving xz

uncolored. We claim that we may assume β ∈ π̄1(z). Otherwise there exists δ1 ∈ π̄1(z1β)∩ π̄1(x)and δ1 /∈ {ξ1, ξ2, . . . , ξc0}. Then we can do a (β, 1, δ1)-switching at z in π1, β is missing at znow. Notice that the color of xzβ will not change by Lemma 6. Also the edges with colors in{ξ1, ξ2, . . . , ξc0} stay the same.

Now Under π1, {z, x, xβ} is a multi-fan, so it is elementary, which in turn shows xβ,ξ1 ,

xβ,ξ2 , . . . , xβ,ξc0 exist. For each i = 1, 2, . . . , c0, since ξi ∈ πbad1 (xβ). d(xβξi) < q. Under

π1, {z, x, xβ, xβξ1 , . . . , xβξc0} form a simple broom, so it is elementary by Lemma 4. Thus,

∆ > |π̄1(x)|+∑c0i=1 |π̄1(xβ,ξi)| > (c0 + 1)ε∆ ≥ ∆, giving a contradiction.

We now proceed with the proof of (5.1). For each color βi, let Ti ⊆ ϕbad(xβi) ∩ Sϕ(q)\{βi}with |Ti| = 3c0. Let W = (

⋃c0+1i=1 Ti)∪{1, β1, β2, . . . , βc0+1}. Clearly, |W | ≤ (c0 + 1)(3c0 + 1) + 1.

For any non-negative number m, we introduce the following two subsets of Aϕ(q).

TAϕ(m) = {η ∈ (∪c0+1i=1 Ti) ∩Aϕ(q) : |ϕ̄(yη) ∩ (ϕ̄(x) ∪ ϕ̄(y))| ≤ m};

TAoϕ(m) = {η ∈ (∪c0+1i=1 Ti) ∩Aϕ(q) : |ϕ̄(yη) ∩ (ϕ̄(x) ∪ ϕ̄(y))| = m}.

Then we letRϕ(m) =

⋃η∈TAoϕ(m+1)

ϕ̄(yη) ∩ (ϕ̄(x) ∪ ϕ̄(y)).

Notice that |Rϕ(m)| ≤ (m+ 1)|⋃c0+1i=1 Ti| ≤ (m+ 1)(c0 + 1)(3c0).

Claim 5.1. We may assume |TAϕ(c0 + 1)| ≤ c0.

Proof. Over all colorings satisfying A1 and A2, we assume ϕ satisfies the following additionalconditions: (a) |TAϕ(c0 + 1)| is minimum; (b) subject to (a),

∑η∈TAϕ(c0+1) |ϕ̄(yη) ∩ ϕ̄(x)| is

maximum. We claim |TAϕ(c0+1)| ≤ c0. Otherwise, there exists a set T = {yη : η ∈ TAϕ(c0+1)}with |T | = c0+1. Let Cϕ = W ∪Rϕ(c0+1). Since |Cϕ| ≤ |W |+(c0+2)|⋃c0+1

i=1 Ti| ≤ (c0+1)(3c0+1)+1+(c0 +2)(c0 +1)(3c0) = f3(ε) and ∆ ≥ f(ε), we have that ∆ ≥ f2(c0 +1, |Cϕ|, ε). Thus by

Fact 2, there is a S(Cϕ, T ): ϕ→ ϕ1. For any η ∈ (⋃c0+1i=1 Ti)∩Aϕ(q), if |ϕ̄(yη)∩(ϕ̄(x)∪ϕ̄(y))| >

c0 + 2, then |ϕ̄1(yη) ∩ (ϕ̄1(x) ∪ ϕ̄1(y))| > c0 + 1; If |ϕ̄(yη) ∩ (ϕ̄(x) ∪ ϕ̄(y))| ≤ c0 + 2, thenϕ̄(yη) ∩ (ϕ̄(x) ∪ ϕ̄(y)) ⊂ Rϕ(c0 + 1), the colors in this set are unchanged. Thus, we haveTAϕ1(c0 + 1) ⊆ TAϕ(c0 + 1). By the minimality of |TAϕ(c0 + 1)|, we have TAϕ1(c0 + 1) =TAϕ(c0 + 1). Then

∑η∈TAϕ1 (c0+1) |ϕ̄1(yη) ∩ ϕ̄1(x)| ≥ ∑η∈TAϕ(c0+1) |ϕ̄(yη) ∩ ϕ̄(x)| + 1 by Fact

3, which contradicts the maximality of∑η∈TAϕ(c0+1) |ϕ̄(yη) ∩ ϕ̄(x)|.

Claim 5.2. We may further assume that there exists a color in {β1, β2, . . . , βc0+1}, say β1, such

that either β1 ∈ ϕ̄(z) or (ϕ̄(x1β1) ∩ ϕ̄(x))\⋃c0+1i=1 Ti 6= ∅.

22

Proof. If Claim 5.2 fails for ϕ, then β1, β2, . . . , βc0+1 ∈ ϕ(z) and⋃c0+1i=1 ((ϕ̄(x1βi) ∩ ϕ̄(x))\⋃c0+1

i=1 Ti) = ∅. Since β1, β2, . . . , βc0+1 are in ϕbad(z), d(x1βi) < q for

i = 1, 2, . . . , c0 + 1. Let T = {x1β1 , x1β2 , . . . , x1βc0+1} and Cϕ = (⋃c0+1i=1 Ti) ∪ Rϕ(c0 + 1). Since

|Cϕ| ≤ (c0 + 1)(3c0) + (c0 + 2)(c0 + 1)(3c0) < f3(ε) and ∆ ≥ f(ε), we have that ∆ ≥ f2(c0 +

1, |Cϕ|, ε). Thus by Fact 2, let ϕ → ϕ1 be a S(Cϕ, T ). For any η ∈ (⋃c0+1i=1 Ti) ∩ Aϕ(q), if

|ϕ̄(yη)∩(ϕ̄(x)∪ϕ̄(y))| > c0+2, then |ϕ̄1(yη)∩(ϕ̄1(x)∪ϕ̄1(y))| > c0+1; If |ϕ̄(yη)∩(ϕ̄(x)∪ϕ̄(y))| =c0 + 2, then ϕ̄(yη) ∩ (ϕ̄(x) ∪ ϕ̄(y)) ⊂ Rϕ(c0 + 1), the colors in this set are unchanged. Thus,we have TAϕ1(c0 + 1) ⊆ TAϕ(c0 + 1), which in turn shows that Claim 5.1 holds for ϕ1 as well.

Moreover,∑c0+1i=1 |ϕ̄1(x1βi) ∪ ϕ̄1(x)| ≥∑c0+1

i=1 |ϕ̄(x1βi) ∪ ϕ̄(x)|+ 1 and the additional color is not

in⋃c0+1i=1 Ti ⊆ Cϕ. Hence, there exists βi such that (ϕ̄1(x1βi) ∩ ϕ̄1(x))\⋃c0+1

i=1 Ti 6= ∅.We assume that β1 is the color satisfies the claim above. Moreover, if β1 ∈ ϕ(z), we let

δ1 ∈ (ϕ̄(x1β1) ∩ ϕ̄(x))\⋃c0+1i=1 Ti.

For a sequence of colorings ϕ0, ϕ1, ϕ2, . . . , ϕi, . . . with ϕ0 = ϕ such that ϕi+1 = ϕi/Pi fori = 0, 1, . . . where Pi’s are (αi, γi)-chains not passing through x. (αi, γi 6= β1.) Let T1(ϕ0) = T1,we define T1(ϕi+1) for i = 0, 1, . . . as below:

T1(ϕi+1) =

T1(ϕi) ∪ {γi}\{αi} xβ1 ∈ V (Pi), αi ∈ T1(ϕi) and γi /∈ T1(ϕi),

T1(ϕi) ∪ {αi}\{γi} xβ1 ∈ V (Pi), γi ∈ T1(ϕi) and αi /∈ T1(ϕi),

T1(ϕi) Otherwise.

Easy to see that each T1(ϕi) is still in ϕbadi (xβ1) ∩ Sϕi(q)\{β1} if {αi, γi} ⊆ Sϕi−1(q) for

i = 1, 2, . . .. Let S1(ϕi) = T1(ϕi) ∩ ϕ̄i(x). Then

Claim 5.3. There exists a coloring π ∈ C∆(G − xy) obtained from ϕ through Kempe changessuch that T1(π) ∈ πbad(xβ1)∩Sπ(q)\{β1}, colors on edges incident with x unchanged and colorsin {β1, δ1} ∪ TAϕ(c0 + 1) stay unchanged, and the following three statements hold.

B1. π(xz) = ϕ(xz) = 1 and π(xz) ∈ π̄(y) or π(xz) ∈ Aπ(q)\TAπ(c0 + 1),

B2. |S1(π)| ≥ c0, and

B3. TAπ(c0 + 1) ⊆ TAϕ(c0 + 1).

We demonstrate how to use Claim 5.3 to prove (5.1) before proceed with its proof. Nowassume π satisfies Claim 5.3. Since π and ϕ agree on edges colored by β1 or δ1, we have β1 ∈ π(x)and there exists δ1 ∈ π̄(x1β1) ∩ π̄(x)\T1(π) if β1 /∈ π̄(z). Assume that S1(π) contains c0 colorsξ1, ξ2, . . ., ξc0 . Clearly δ1 /∈ {ξ1, ξ2, . . ., ξc0} since δ1 /∈ T1(π). {ξ1, ξ2, . . ., ξc0} ⊆ T1(π)∩ π̄(x) ⊆π̄(x) ∩ πbad(xβ) since T1(π) ∈ πbad(xβ1).

We claim that we may assume 1 = π(xz) ∈ π̄(y). Otherwise, 1 = π(xz) ∈ Aπ(q)\TAπ(c0+1),hence |π̄(y1) ∩ (π̄(x) ∪ π̄(y))| > c0 + 1. Choose θ ∈ π̄(y1) ∩ π̄(y) if π̄(y1) ∩ π̄(y) 6= ∅; Otherwiselet θ ∈ π̄(y) and choose δ ∈ π̄(y1) ∩ π̄(x) such that δ /∈ {ξ1, ξ2, . . . , ξc0} ∪ {δ1}. This is possiblesince |π̄(y1)∩ (π̄(x)∪ π̄(y))| > c0 + 1 and π̄(y1)∩ π̄(y) = ∅. Now we do a (1, θ, δ)-switching at yin π. After the switching, 1 = π(xz) is missing at y.

Now assume 1 = π(xz) ∈ π̄(y). We point out that the colors in {ξ1, ξ2, . . . , ξc0} ∪ {δ1} areunchanged in the switching above. By Lemma 6, xxβ stays unchanged as well. So we still have

δ1 ∈ π̄(x1β1)∩ π̄(x)\T1(π) and {ξ1, ξ2, . . . , ξc0} ⊆ πbad(xβ1)∩ π̄(x)\{β1}. Hence (5.1) holds.

Finally we give the proof of Claim 5.3.

Proof of Claim 5.3

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Proof. We firstly note that, besides the properties in the claim, using the Kempe changes wementioned in the proof of this claim will keep the following properties (assuming ϕ1 is the re-sulting coloring for the Kempe change):ϕ1(xz) = ϕ(xz) = 1; either β1 ∈ ϕ̄1(z) or β1 ∈ ϕ1(z) and there exists δ1 ∈ ϕ̄1(x1β1) ∩ϕ̄1(x)\⋃c0+1

i=1 Ti; T1(ϕ1) ⊆ ϕbad1 (xβ1) ∩ Sϕ1(q)\{β1} and |T1(ϕ1)| = 3c0.

Apparently ϕ satisfies all required conditions except B2. Hence we may let π ∈ C∆(G− xy)satisfying all required conditions in the claim except B2 with maximum value of |S1(π)|. Wewill show that |S1(π)| ≥ c0, which implies the claim.

Suppose otherwise |S1(π)| ≤ c0−1. Since |T1(π)| = |T1(ϕ)| = 3c0, by B3 and Claim 5.1, wecan find c0+1 colors τ1, τ2, . . . , τc0+1 in T1(π)\(S1(π)∪TAπ(c0+1)) = T1(π)\(π̄(x)∪TAπ(c0+1)).Recall that T1(π) ⊆ πbad(xβ1) ∩ Sπ(q)\{β1} ⊆ Sπ(q). We divide the proof into the followingcases.

Case 1. One of τ1, τ2, . . . , τc0+1, say τ1 is in π̄(xβ1).

Case 1.1. τ1 ∈ T1(π) ∩ π̄(y).Choose a color δ ∈ π̄(x)\(S1(π)∪ {δ1} ∪Rπ(c0 + 1)), it is possible since |π̄(x)| > ε∆ ≥ εf(ε)

and |S1(π) ∪ {δ1} ∪ Rπ(c0 + 1)| ≤ c0 + (c0 + 2)(c0 + 1)(3c0). Since Px(τ1, δ, π) = Py(τ1, δ, π),Pxβ1 (τ1, δ, π) is disjoint from x, y. Let π1 = π/Pxβ1 (τ1, δ, π), then color τ1 in T1(π) is replacedby δ . It is easy to see that under coloring π1 all previous assumptions and B1 are satisfied.Since both τ1 and δ are colors in π̄(x) ∪ π̄(y), TAπ1(c0 + 1) = TAπ(c0 + 1), so B3 holds. ButS1(π1) = S1(π) ∪ {t1} giving a contradiction to the maximality of |S1(π)|.

Case 1.2. τ1 ∈ T1(π) ∩Aπ(q)\TAπ(c0 + 1).In this case, we have |π̄(yτ1)∩ (π̄(x)∪ π̄(y))| > c0 + 1. If π̄(yτ1)∩ π̄(y) 6= ∅, let θ be a color in

it; If π̄(yτ1)∩ π̄(y) = ∅, let θ ∈ π̄(y) and choose δ ∈ π̄(yτ1)∩ π̄(x)\(S1(π)∪{δ1}). This is possiblesince |π̄(yτ1) ∩ (π̄(x) ∪ π̄(y))| > c0 + 1, π̄(yτ1) ∩ π̄(y) = ∅ and |S1(π)| ≤ c0 − 1. Then in bothcases, we can do a (τ1, θ, δ)-switching at y in π. (When θ ∈ π̄(yτ1) ∩ π̄(y), we do not need thecolor δ in the switching. ) Now τ1 is missing at y, we are back to Case 1.1. (Notice that evenwhen π(xz) ∈ π̄(y), the color of xz may not be missing at y after the switching if θ = π(xz) –this is the main reason why we replace A1 by a weaker condition B1. We also note that B3still holds since both θ and δ are colors in π̄(x) ∪ π̄(y), |π̄(yα) ∩ (π̄(x) ∪ π̄(y))| stays unchangedfor each yα ∈ TAπ(c0 + 1).)

Case 2. τ1, τ2, . . . , τc0+1 are in π(xβ1).As the previous method we used in Claim 5.1 for finding the coloring ϕ, we firstly adjust π

a little bit. Let T ′1 = {τ1, τ2, . . . , τc0+1}. For a given number m, let TA′π(m) = {τi ∈ T ′1∩Aπ(q) :|π̄(xβ1τi)∩ π̄(x)| ≤ m}. Now let π′ be a coloring satisfies all previous assumptions of π such that:(a) |TA′π′(c0 + 1)| is minimum; (b) subject to (a),

∑τi∈TA′π′ (c0+1) |π̄′(xβ1τi)∩ π̄′(x)| is maximum.

We then claim that |TA′π′(c0 + 1)| ≤ c0. If not, then T ′1 = TA′π′(c0 + 1). Let T ′ = {yτ : τ ∈T ′1}, R′π′(m) = ∪τi∈TA′π′ (m+1)(π̄

′(xβ1τi)∩ π̄′(x)) and C = T ′1 ∪R′π′(c0 + 1)∪Rπ′(c0 + 1). We have

|C| ≤ c0 +1+(c0 +2)(c0 +1)+(c0 +2)(c0 +1)(3c0) ≤ f3(ε) (since c0 ≥ 1). Since ∆ ≥ f(ε), thereexists a S(C, T ′) turns π′ to π1 by Fact 2. Easy to see that π1 satisfies all previous assumptionsand B1. Notice that the colors in Rπ′(c0 + 1) stay unchanged, so TAπ1(c0 + 1) ⊆ TAπ′(c0 + 1).Hence, B3 holds for π1. Moreover, since the colors in R′π′(c0+1) stay unchanged, TA′π1(c0+1) ⊆TA′π′(c0 + 1). By the minimality of |TA′π′(c0 + 1)|, we have TA′π1(c0 + 1) = TA′π′(c0 + 1).Then

∑τi∈TA′π1 (c0+1) |π̄1(xβ1τi)∩ π̄1(x)| ≥∑τi∈TA′π′ (c0+1) |π̄′(xβ1τi)∩ π̄′(x)|+ 1 by Fact 3, which

contradict the maximality of∑τi∈TA′π′ (c0+1) |π̄′(xβ1τi) ∩ π̄′(x)|.

24

Now we have |TA′π′(c0+1)| ≤ c0. Since |T ′1| = c0+1, we may assume that there exists τ1 ∈ T ′1such that |π̄′(xβ1τ1)∩π̄′(x)| > c0+1. Thus there exists a color δ ∈ π̄′(xβ1τ1)∩π̄′(x)\(S1(π′)∪{δ1}).

We first claim that we may assume τ1 ∈ π̄′(y). Suppose not, then τ1 ∈ Aπ′(q)\TAπ′(c0 + 1).So |π̄′(yτ1) ∩ (π̄′(x) ∪ π̄′(y))| > c0 + 1. If π̄′(yτ1) ∩ π̄′(y) 6= ∅, let θ be a color in it; If π̄′(yτ1) ∩π̄′(y) = ∅, let θ ∈ π̄′(y) and choose δ′ ∈ π̄′(yτ1) ∩ π̄′(x)\(S1(π′) ∪ {δ, δ1}). This is possible since|π̄′(yτ1)∩ (π̄′(x)∪ π̄′(y))| > c0 + 1, π̄′(yτ1)∩ π̄′(y) = ∅ and |S1(π′)| ≤ c0− 1. Then in both cases,we can do a (τ1, θ, δ

′)-switching at y in π′. (When θ ∈ π̄′(yτ1) ∩ π̄′(y), we do not need the colorδ in the switching. ) After the switching, τ1 is missing at y.

Now assume τ1 ∈ π̄′(y). Note that in the switching above, S1(π′) ∪ {δ, δ1} stay unchanged.Since δ ∈ π̄′(x) and τ1 ∈ π̄′(y), Px(τ1, δ, π

′) = Py(τ1, δ, π′) and Pxβ1 (τ1, δ, π

′) is disjoint from

x, y. Let π1 = π′/Pxβ1 (τ1, δ, π′). Clearly, π1 satisfies all previous assumptions and B1. Since

both τ1 and δ are colors in π̄′(x) ∪ π̄′(y), π1 satisfies B3. But S1(π1) = S1(π′) ∪ {τ1}, giving acontradiction to the maximality of |S1(π′)| = |S1(π)|. This completes the proof of Claim 5.3and Lemma 7.

By using Lemma 7, we prove Lemma 8.Lemma 8. Under the setting above, if ∆ ≥ f(ε), then for any c0 + 2 colors ξ1, ξ2, . . . , ξc0+2 ∈Sϕ(q) ∩ ϕ(x), |⋂c0+2

i=1 ϕbad(xξi)\{ξi}| < 3c0 + 1.

Proof. Suppose on the contrary, let S = {ξ1, ξ2, . . . , ξc0+2} ⊆ Sϕ(q) ∩ ϕ(x) be a color set such

that |⋂c0+2i=1 ϕbad(xξi)\{ξi}| ≥ 3c0 + 1. Let S′ = {ξ1, ξ2, . . . , ξc0+1}. Choose a color set T ′ ⊆⋂c0+2

i=1 ϕbad(xξi)\{ξi} with size 3c0 + 1.

Claim 5.4. We may assume that one of ξ1, ξ2, . . . , ξc0+1, say ξ1, has the following property:either ξ1 ∈ ϕ̄(y) or |ϕ̄(yξ1) ∩ (ϕ̄(x) ∪ ϕ̄(y))| > 3c0 + 1.

Proof. Suppose the contrary, which implies S′ ∩ ϕ̄(y) = ∅. For a constant m, let SAϕ(m) = {ξ :ξ ∈ S′ such that |ϕ̄(yξ)∩ (ϕ̄(x)∪ ϕ̄(y))| ≤ m}, R(ϕ) =

⋃ξ∈SAϕ(3c0+2) ϕ̄(yξ)∩ (ϕ̄(x)∪ ϕ̄(y)). Now

let ϕ1 be a coloring such that ξ1, ξ2, . . . , ξc0+2 ∈ Sϕ1(q) ∩ ϕ1(x) and T ′ ⊆ ⋂c0+2i=1 ϕbad

1 (xξi)\{ξi}with the following additional conditions: (a) |SAϕ1(3c0 + 1)| is minimum; (b) subject to (a),∑ξ∈SAϕ1 (3c0+1) |ϕ̄1(yξ) ∩ ϕ̄1(x)| is maximum.

We then claim that under the coloring ϕ1, |SAϕ1(3c0 + 1)| ≤ c0. Suppose not, thenSAϕ1(3c0+1) = S′. Let C = S∪T ′∪R(ϕ1) and T = {yξ : ξ ∈ S′}. Since |S| = c0+2, |T ′| = 3c0+1and |R(ϕ1)| ≤ (c0 + 1)(3c0 + 2), it is easy to check that |C| ≤ f3(ε). Then since ∆ ≥ f(ε), thereexists a S(C, T ) turns ϕ1 to ϕ2 by Fact 2. Notice that the colors in R(ϕ1) stay unchanged, soSAϕ2(3c0 + 1) ⊆ SAϕ1(3c0 + 1). By the minimality of SAϕ1(3c0 + 1), we have SAϕ2(3c0 + 1) =SAϕ1(3c0 + 1). Then

∑ξ∈SAϕ2 (3c0+1) |ϕ̄2(yξ) ∩ ϕ̄2(x)| ≥ ∑ξ∈SAϕ1 (3c0+1) |ϕ̄1(yξ) ∩ ϕ̄1(x)| + 1 by

Fact 3, which contradict the maximality of∑ξ∈SAϕ1 (3c0+1) |ϕ̄1(yξ) ∩ ϕ̄1(x)|.

By the discussion above, we have |SAϕ1(3c0 + 1)| ≤ c0. Since |S′| = c0 + 1, there exists acolor in S′, say ξ1, such that |ϕ̄1(yξ1) ∩ (ϕ̄1(x) ∪ ϕ̄1(y))| > 3c0 + 1. This proves the claim.

Now assume that under the coloring ϕ1, color ξ1 satisfies the claim above. We then claimthat we can find a color ξ ∈ S such that ξ ∈ ϕ̄1(y). Suppose not, we have S ∩ ϕ̄1(y) = ∅ and|ϕ̄1(yξ1) ∩ ϕ̄(x)| > 3c0 + 1. Now we choose δ ∈ ϕ̄1(yξ1) ∩ ϕ̄(x)\T ′, let θ ∈ ϕ̄1(y), we can do a(ξ1, θ, δ)-switching at y in ϕ1 to obtain ϕ2. Under the new coloring we have ξ1 ∈ ϕ̄2(y), which is

what we want. Notice that after this switching, T ′ may not be a subset of⋂c0+2i=1 ϕbad

2 (xξi)\{ξi}if θ ∈ T ′. But we still have T ′\{θ} ⊆ ⋂c0+2

i=1 ϕbad2 (xξi)\{ξi}.

By the discussion above, we assume that under ϕ1, ξ1 ∈ ϕ̄1(y) and |⋂c0+2i=1 ϕbad

1 (xξi)\{ξi}|≥ |T ′| − 1 = 3c0. Recolor xy by ξ1 and leave xxξ1 uncolored to obtain a new coloring ϕ2. Then

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ϕ2(xy) ∈ ϕ̄2(xξ1); S\{ξ1} is a subset of ϕbad2 (y) ∩ ϕ2(x) with size c0 + 1; For each ξ ∈ S\{ξ1},

|ϕbad2 (xξ) ∩ Sϕ2(x, xξ1 , q)\{s}| ≥ |(ϕbad

1 (xξ)\{ξ}) ∩ ϕbad1 (xξ1)| ≥ |⋂c0+2

i=1 ϕbad1 (xξi)\{ξi}| ≥ 3c0.

But this cannot happen by Lemma 7, which is a contradiction.

6 Proof of Lemma 16

Before the proof, we will discuss some basic properties first. Assume that ε ∈ (0, 1) and Gis a ∆-critical graph with maximum degree ∆. Let x ∈ V (G) with d(x) < (1 − ε)∆ = q andy ∈ N(x). In this section we don’t assume ∆ ≥ D(ε).

A neighbor z ∈ N(x)\{y} is called feasible if there exits a coloring ϕ ∈ C∆(G−xy) such thatϕ(xz) ∈ ϕ̄(y), and such a coloring ϕ is called z-feasible. Denote by Cz the set of all z-feasiblecolorings. For each ϕ ∈ Cz. let

YZ(ϕ) = {v ∈ N(y) \ {x} : ϕ(vy) ∈ ϕ̄(x) ∪ ϕ̄(z)},Ry(ϕ) = {ϕ(vy) : v ∈ YZ(ϕ) and d(v) < q},Z(ϕ) = {v ∈ N(z) \ {x} : ϕ(vz) ∈ ϕ̄(x) ∪ ϕ̄(y)},Rz(ϕ) = {ϕ(vz) : v ∈ Z(ϕ) and d(v) < q}.

In the remainder of the proof, we let Rz = Rz(ϕ) and Ry = Ry(ϕ) if the coloring ϕ is clear.

Recall that c1 = c0− 1 =†

1−εε

£− 1. When d(x) < q, we have for any ϕ ∈ C∆(G− xy), every

elementary vertex subset X with x ∈ X contains at most c1 other vertices with degree less thanq. Otherwise |⋃v∈X ϕ̄(v)| ≥ ε∆ + 1 + (c1 + 1)(ε∆) > ∆, giving a contradiction.

Now we claim that |Ry| ≤ c1 and |Rz| ≤ c1. Let z ∈ N(x) \ {y} be a feasible vertex andϕ ∈ Cz. By the definition of Z(ϕ), G[{x, y, z} ∪Z(ϕ)] forms a simple broom, so {x, y, z} ∪Z(ϕ)is elementary with respect to ϕ. Consequently, it contains at most c1 vertices other than x withdegree less than q. Thus, |Rz| ≤ c1. Similarly, if we color xy by ϕ(xz) and leave xz uncoloredto obtain a coloring ϕ′, then G[{x, y} ∪ YZ(ϕ′)] forms a simple broom, by the same reason, wehave |Ry(ϕ)| = |Ry(ϕ′)| ≤ c1. Let R = Rz ∪ Ry. A coloring ϕ ∈ Cz is called optimal if |R| ismaximum over all feasible colorings. Obviously, we have |R| ≤ 2c1.

Recall that Aϕ(q) ∪ Bϕ(q) = ϕ(x) ∩ ϕ(y) and ϕ̄(x) ∩ ϕ̄(y) = ∅. So [1,∆] = ϕ̄(x) ∪ ϕ̄(y) ∪(ϕ(x) ∩ ϕ(y)) = Sϕ(q) ∪ Bϕ(q). By Lemma 1, for a z-feasible coloring ϕ, ϕ̄(x), ϕ̄(y) and ϕ̄(z)are disjoint (notice that {x, y, z} is elementary), thus

[1,∆] = ϕ̄(x) ∪ ϕ̄(y) ∪ ϕ̄(z) ∪ (Aϕ(q) ∪Bϕ(q)\ϕ̄(z)), so

σq(x, y) = |ϕ(y)\(Aϕ(q) ∪Ry)| = |[1,∆]\(ϕ̄(y) ∪Aϕ(q) ∪Ry)|= |ϕ̄(x)|+ |ϕ̄(z)|+ |Bϕ(q)\ϕ̄(z)| − |Ry|.

By using this result, we further have

|Sϕ(q) ∩ ϕ(z)| = |ϕ(z)\Bϕ(q)| = ∆− |ϕ̄(z)| − |Bϕ(q)\ϕ̄(z)|= ∆− (σq(x, y)− |ϕ(x)|+ |Ry|) ≥ 2∆− d(x)− σq(x, y) + 1− c1 and

|NS(x)| = |ϕ̄(y) ∪Aϕ(q)| = |ϕ̄(y) ∪Aϕ(q) ∪Ry| − |Ry|= ∆− σq(x, y)− |Ry| ≥ ∆− σq(x, y)− c1.

By the two inequalities above, we show that Lemma 17 is a corollary of Lemma 16.Lemma 17. Let ε ∈ (0, 1), q be a positive number, G be a ∆-critical graph with maximumdegree ∆, xy ∈ E(G) with d(x) < q and ϕ ∈ C∆(G − xy). If q ≤ min{(1 − ε)∆ − 2c1,

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∆− 6c1}, then there are at least ∆− σq(x, y)− 2c1 vertices z ∈ N(x) \ {y} such that σq(x, z) ≥2∆− d(x)− σq(x, y)− 5c1.

Proof. By Lemma 16, there are at least |NS(x)| − c1 many vertices z ∈ NS(x) ⊆ N(x)\{y} anda z-feasible coloring ϕ′ for each z such that σq(x, z) ≥ |Sϕ′(q) ∩ ϕ′(z)| − 1− 4c1.

By the two inequalities we just proved, we have |NS(x)| − c1 ≥ ∆ − σq(x, y) − 2c1 and|Sϕ′(q) ∩ ϕ′(z)| − 1− 4c1 ≥ 2∆− d(x)− σq(x, y)− 5c1. Thus Lemma 17 holds.

Now we give the proof of Lemma 16.Lemma 16. Let ε ∈ (0, 1), q be a positive number, G be a ∆-critical graph with maximumdegree ∆, xy ∈ E(G) with d(x) < q and ϕ ∈ C∆(G− xy). If q ≤ min{(1− ε)∆− 2c1, ∆− 6c1},then for each z ∈ NS(x) except at most c1 vertices, there exists a coloring ϕ′ ∈ C∆(G − xy)such that ϕ′(xz) ∈ ϕ̄′(y) and for each ξ ∈ (Sϕ′(q) ∩ ϕ′(z))\{ϕ′(xz)} except at most 4c1 colors,d(x1ξ) ≥ q where 1 = ϕ′(xz).

Proof. When c1 = 0, i.e. ε ≥ 12 , this lemma is a corollary of Lemma 12. Thus we assume that

c1 ≥ 1. Lemma 16 follows the three statements below.

I. For any coloring ϕ ∈ C∆(G − xy) and any α ∈ Aϕ(q) except at most c1 colors, xα isfeasible. (Notice that by definition, xα is feasible if α ∈ (Sϕ(q) ∩ ϕ(x))\Aϕ(q) = ϕ̄(y).)

II. For any feasible vertex z ∈ N(x) and any z-feasible coloring ϕ, |Rz| ≤ c1.

III. For any feasible vertex z ∈ N(x), there exists a z-feasible optimal coloring ϕ (assumeϕ(xz) = 1) such that for all α ∈ Aϕ(q) ∩ ϕ(z) except at most 3c1 colors, d(x1α) ≥ q.(Together with II, we have for all α ∈ Sϕ(q)∩ϕ(z) except at most 4c1 colors, d(x1α) ≥ q.)

Since II has been proved at the beginning of this section, we give the proof of I and III.

6.1 Proof of I

Suppose on the contrary that there exist ϕ ∈ C∆(G− xy) and c1 + 1 colors α1, α2, . . . , αc1+1

in Aϕ(q) such that xα1 , xα2 , . . . , xαc1+1 are not feasible. We consider the following two cases:Case 1: One of yα1 , yα2 , . . . , yαc1+1 has a common missing color with x.

Assume that yα1 has a common missing color, say δ, with x. Let θ ∈ ϕ̄(y). Then we can doan (α1, θ, δ)-switching in ϕ at y to obtain a new coloring ϕ1 such that ϕ1(xxα1) = α1 ∈ ϕ1(y),which contradict the fact that xα1 is not feasible.

Case 2: None of yα1 , yα2 , . . . , yαc1+1 has a common missing color with x.As what we discussed before, every elementary vertex subset X with x ∈ X contains at most

c1 other vertices with degree less than q, thus {x, yα1 , yα2 , . . . , yαc1+1} is not elementary. Hencethere exist two vertices of yα1 , yα2 , . . . , yαc1+1 which have a common missing color. Assume

without loss of generality that η ∈ ϕ̄(yα1) ∩ ϕ̄(yα2). Choose δ ∈ ϕ̄(x), obviously δ 6= η. Then atleast one of yα1 , yα2 is not on the path Px(δ, η, ϕ), say yα1 . Now, let ϕ1 = ϕ\Pyα1 (δ, η, ϕ), thenδ ∈ ϕ̄1(yα1) ∩ ϕ̄1(x), we are back to Case 1.

6.2 Proof of III.

Let z ∈ N(x), ϕ be an optimal z-feasible coloring. Assume, without loss of generality,ϕ(xz) = 1 ∈ ϕ̄(y) and let T = {κ ∈ Aϕ(q) ∩ ϕ(z) : d(x1κ) < q}. Suppose on the contrary that|T | ≥ 3c1 + 1 under every optimal z-feasible colorings.

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

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Proof. We first show that z ∈ V (Px(η, κ, ϕ)). Otherwise, Pz(η, κ, ϕ) is disjoint with Px(η, κ, ϕ).Let ϕ′ = ϕ/Pz(η, κ, ϕ). Since 1 6= η, κ, ϕ′ is also feasible. Since colors in R are unchanged andd(x1κ) < q, Rz(ϕ

′) = Rz∪{η} and Ry(ϕ′) ⊇ Ry, giving a contradiction to the maximality of |R|.

Similarly, if we color xy by ϕ(xz) and leave xz uncolored to obtain a coloring ϕ′, then followthe same proof, we can easily verify that y ∈ V (Px(η, κ, ϕ)) as well.

Since |T | ≥ 3c1 + 1, there are 3c1 + 1 colors κ1, κ2, . . . , κ3c1+1 ∈ T . We let

VT = VT (ϕ) = {x1κ1 , x1κ2 , . . . , x1κ3c1+1} ∪ {yκ1 , yκ2 , . . . , yκ3c1+1},W = W (ϕ) = {u ∈ VT (ϕ) : ϕ̄(u) ∩ ϕ̄(x) ⊆ R},M = M(ϕ) = VT (ϕ)−W (ϕ) = {u ∈ VT (ϕ) : ϕ̄(u) ∩ ϕ̄(x) \R 6= ∅},ET = ET (ϕ) = {zx1κ1 , zx1κ2 , . . . , zx1κ3c1+1 , yyκ1 , yyκ2 , . . . , yyκ3c1+1},EW = EW (ϕ) = {e ∈ ET (ϕ) : e is incident to a vertex in W} and

EM = EM (ϕ) = ET (ϕ)− EW (ϕ) = {e ∈ ET (ϕ) : e is incident to a vertex in M}.

We assume that |EW | is minimum over all optimal z-feasible coloring ϕ and all sets of 3c0 +1colors in T . For each v ∈ M , pick a color αv ∈ ϕ̄(v) ∩ ϕ̄(x) \ R. Let CM = {αv : v ∈ M}.Clearly, |CM | ≤ |M |. Since {x1κ1 , x1κ2 , . . . , x1κ3c1+1} ∩ {yκ1 , yκ2 , . . . , yκ3c1+1} may not be empty,

so |EW |2 ≤ |W | ≤ |EW | and |EM |2 ≤ |M | ≤ |EM |.Claim B. The following can not happen: ϕ̄(x) \ (R ∪ CM ) 6= ∅ and there exist two vertices

u, v ∈W such that (ϕ̄(u) ∩ ϕ̄(v)) \R 6= ∅.

Proof. Suppose on the contrary. Let β ∈ ϕ̄(x) \ (R ∪CM ), u, v ∈W and α ∈ (ϕ̄(u) ∩ ϕ̄(v)) \R.Then by the definition of W , we have α ∈ ϕ(x), β ∈ ϕ(u) ∩ ϕ(v) and α 6= β. So both u and vare endpoints of (α, β)-chains, at least one of the two paths does not pass through x. Assumewithout loss of generality Pu(α, β, ϕ) is disjoint with Px(α, β, ϕ). We note that β ∈ ϕ(y) ∩ ϕ(z)since {x, y, z} is an elementary set.

We first consider the case of α = 1. In this case, Px(α, β, ϕ) = Py(α, β, ϕ) by Lemma 1. Sinceϕ(xz) = 1, z ∈ Px(α, β, ϕ). So, Pu(α, β, ϕ) ∩ {x, y, z} = ∅. Hence, coloring ϕ′ = ϕ/Pu(α, β, ϕ)is z-feasible, Ry(ϕ

′) = Ry, Rz(ϕ′) = Rz and T (ϕ′) = T . So, ϕ′ is also optimal, u ∈ M(ϕ′) and

|EW (ϕ′)| < |EW |, contradict the fact that |EW | is minimum.We now suppose α ∈ ϕ(x)\ (R∪{1}). So, both α and β are not in R∪{1}. Notice that even

if α is a color in T , Pu(α, β, ϕ) will not pass through zx1α or yyα because of Claim A. Thuswe may let ϕ′ = ϕ/Pu(α, β, ϕ). Then, ϕ′ is z-feasible, Ry(ϕ

′) = Ry and Rz(ϕ′) = Rz. Thus, ϕ′

is still an optimal coloring and β ∈ ϕ̄′(u). We have u ∈ M(ϕ′) and |EW (ϕ′)| < |EW |, giving acontradiction.

Claim C. There exist a color κ ∈ {κ1, κ2, . . . , κ3c0+1} and three distinct colors η, θ, λ suchthat η ∈ (ϕ̄(x)\R) ∩ ϕ̄(x1κ), θ ∈ (ϕ̄(x)\R) ∩ ϕ̄(yκ) and λ ∈ (ϕ̄(x) ∪ {1}\R) ∩ ϕ̄(yκ).

Before we give the proof of Claim C, we first show III assuming Claim C is true.Let κ, η, θ and λ be as stated in Claim C. If λ 6= 1, we consider coloring obtained from ϕ

by interchange colors 1 and λ for edges not on the path Px(1, λ, ϕ), and rename it as ϕ. So wemay assume 1 ∈ ϕ̄(yκ).

By Claim A, the paths Px(η, κ, ϕ) and Px(θ, κ, ϕ) both contain y, z. Since ϕ(yyκ) =ϕ(zx1κ) = κ, these two paths also contain yκ, x1κ. Since η ∈ ϕ̄(x1κ), we have x and x1κ

are the two endpoints of Px(η, κ, ϕ). So, η ∈ ϕ(y) ∩ ϕ(z) ∩ ϕ(yκ). Similarly, we have θ ∈ϕ(y) ∩ ϕ(z) ∩ ϕ(x1κ). We now consider the following sequence of colorings of G− xy.

Let ϕ1 be obtained from ϕ by assigning ϕ1(yyκ) = 1. Since 1 is missing at both y and yκ,ϕ1 ∈ C∆(G− xy) . Now κ is missing at y and yκ, η is still missing at x1κ. Since G is ∆-critical,

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Px(η, κ, ϕ1) = Py(η, κ, ϕ1); Furthermore, x1κ, yκ /∈ V (Px(η, κ, ϕ1)) since either η or κ is missingat these two vertices, which in turn shows that z /∈ V (Px(η, κ, ϕ1)) since ϕ1(zx1κ) = κ.

Let ϕ2 = ϕ1/Px(η, κ, ϕ1). We have κ ∈ ϕ̄2(x), η ∈ ϕ̄2(y) ∩ ϕ̄2(x1κ) and θ ∈ ϕ̄2(x) ∩ ϕ̄2(yκ).Then Px(η, θ, ϕ2) = Py(η, θ, ϕ2) contains neither yκ nor x1κ.

Let ϕ3 = ϕ2/Px(η, θ, ϕ2). Then κ ∈ ϕ̄3(x) and θ ∈ ϕ̄3(y) ∩ ϕ̄3(yκ).Let ϕ4 be obtained from ϕ3 by recoloring yyκ by θ. Then 1 ∈ ϕ̄4(y), ϕ4(xz) = 1, κ ∈ ϕ̄4(x),

ϕ4(zx1κ) = κ. Since ϕ4(xz) = 1 ∈ ϕ̄4(y), ϕ4 is z-feasible. Since η, θ, κ /∈ R, the colors in Rare unchanged during this sequence of re-colorings, so Ry(ϕ4) ⊇ Ry and Rz(ϕ4) ⊇ Rz. Sinceϕ4(zx1κ) = κ ∈ ϕ̄4(x) and d(x1κ) < q, we have κ = ϕ4(zx1κ) ∈ Rz(ϕ4). So, Rz(ϕ4) ⊇ Rz ∪ {κ}.We therefore have |Ry(ϕ4)|+ |Rz(ϕ4)| ≥ |Ry|+ |Rz|+ 1, giving a contradiction. This completesthe proof of III.

Finally we give the proof of Claim C.

Proof. We first note that the condition of d(x) and q gives

|ϕ̄(x)| = ∆− d(x) + 1 > ∆− q + 1 ≥ 6c1 + 1. (4)

Secondly, if there exist η, θ ∈ ϕ̄(x)\R such that η ∈ ϕ̄(x1κ) and θ ∈ ϕ̄(yκ), then η 6= θ;for otherwise, by Claim A, the path Px(η, κ, ϕ) contains three endpoints x, x1κ and yκ, acontradiction.

Now we start by showing that there exist c1 +1 colors γ1, γ2, . . . , γc1+1 ∈ {κ1, κ2, . . . , κ3c1+1}such that ϕ̄(x) ∩ ϕ̄(x1γi) 6= ∅ and ϕ̄(x) ∩ ϕ̄(yγi) 6= ∅ for i = 1, 2, . . . , c1 + 1.

Suppose not. Then |EM | ≤ 4c1 + 1 by the Pigeonhole Principle. Thus |EW | ≥ 2c1 + 1, whichin turn give |W | ≥ d2c1+1

2 e = c1 + 1 and |M | ≤ |EM | ≤ 4c1 + 1. Let w1, w2, . . . , wc1+1 ∈ W .By (4), |ϕ̄(x)| ≥ 6c1 + 2 > |R| + |M |. So ϕ̄(x)\(R ∪ CM ) 6= ∅. Recall |R| ≤ 2c1, d(x) < q andq ≤ (1− ε)∆− 2c1 , we have

c1+1∑i=1

|ϕ̄(wi) \R|+ |ϕ̄(x)| > (c1 + 1)(∆− q − 2c1) + ∆− d(x) + 1 > ∆.

So, there is a color α shared by at least two of these sets. Since w1, w2, . . . , wc1+1 ∈W , we mayassume that α ∈ (ϕ̄(w1)\R)∩ (ϕ̄(w2)\R)∩ϕ(x). Then by Claim B, this gives a contradiction.Thus we can assume that there exist c1 + 1 colors γ1, γ2, . . . , γc1+1 ∈ {κ1, κ2, . . . , κ3c1+1} suchthat ϕ̄(x) ∩ ϕ̄(x1γi) 6= ∅ and ϕ̄(x) ∩ ϕ̄(yγi) 6= ∅ for i = 1, 2, . . . , c1 + 1.

Let ηi ∈ ϕ̄(x)∩ ϕ̄(x1γi) and θi ∈ ϕ̄(x)∩ ϕ̄(yγi) for i = 1, 2, . . . , c1 + 1 (They don’t have to bedistinct). We now only need to show that additionally there exists another color λ ∈ ϕ̄(x)∪{1}\Rsuch that λ in ϕ̄(yγi) for some i ∈ {1, 2, . . . , c1 + 1}.

Suppose on the contrary that there is no such color λ. Then ϕ̄(yγn) ∩ ϕ̄(x)\R = {θi} fori = 1, 2, . . . , c1 + 1. Moreover, 1 /∈ ϕ̄(yγi). Since |R| ≤ 2c1, d(x) < q and q ≤ (1− ε)∆− 2c1, thefollowing inequalities hold.

c1+1∑i=1

|ϕ̄(yγi) \ (R ∪ {ηi})|+ |ϕ̄(x) ∪ (R ∪ {1})| > (c1 + 1)(∆− q − |R| − 1) + ∆− d(x) + 2 ≥ ∆

So, there is a color α in two of these sets. Since ϕ̄(yγi) ∩ ϕ̄(x)\R = {θi} and 1 /∈ ϕ̄(yγi) fori = 1, 2, . . . , c1 + 1, we may assume that α ∈ ϕ̄(yγ1) ∩ ϕ̄(yγ2) ∩ ϕ(x) \ (R ∪ {η1, θ1, η2, θ2, 1}).

Since |ϕ̄(x)| ≥ 6c1 + 2 ≥ |R| + 4c1 + 2 ≥ |R| + 6, ϕ̄(x) \ (R ∪ {η1, θ1, η2, θ2, 1}) 6= ∅, thereexists λ ∈ ϕ̄(x) \ (R ∪ {η1, θ1, η2, θ2, 1}). If α = λ we are done. Assume α 6= λ. Then, both yγ1and yγ2 are endpoints of (α, λ)-chains. Assume without loss of generality Pyγ1 (α, λ, ϕ) is disjoint

with Px(α, λ, ϕ). Let ϕ′ = ϕ/Pyγ1 (α, λ, ϕ). Since λ 6= η1, θ1, the previous assumptions still hold.

Now ϕ′, γ1, η1, θ1, λ satisfy the requirements. Claim C holds.

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7 Acknowledgement

We would like to thank Rong Luo and Yue Zhao for the helpful communication, and twoanonymous reviewers for their valuable comments on an earlier version of this paper.

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On the average degree of edge chromatic critical …2 Coloring preliminaries and adjacency lemmas...

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