Average degrees of edge-chromatic critical graphs Yan Cao a ,Guantao Chen a , Suyun Jiang b * , Huiqing Liu c* , Fuliang Lu d* 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 Δ(G), d(G) and χ 0 (G) the maximum degree, the average degree and the chromatic index of G, respectively. A simple graph G is called edge-Δ-critical if χ 0 (G) = Δ(G) + 1 and χ 0 (H ) ≤ Δ(G) for every proper subgraph H of G. Vizing in 1968 conjectured that if G is edge-Δ-critical, then d(G) ≥ Δ(G) - 1+ 3 n . We show that for any edge-Δ-critical graph with Δ(G) ≥ 32 and any positive number c, ¯ d(G) ≥ q - 3q-2Δ(G)-2 3+f (c) , where q = min{ 2 √ 2(Δ(G)-1)-2 2 √ 2+1 , 3 4 Δ(G)-2} and f (c) = max{3 - c Δ(G)-q , 2}× (5c+2)Δ(G)-(6c+3)q+3c+2 cΔ(G) . Consequently, ¯ d(G) ≥ ( 0.69242Δ(G) - 0.1701 if Δ(G) ≥ 66, 0.68707Δ(G)+0.0136 if 32 ≤ Δ(G) ≤ 65. This result improves the best known bound 2 3 (Δ(G) + 2) obtained by Woodall in 2007 for Δ(G) ≥ 66. Additionally, Woodall constructed an infinite family of graphs showing his result cannot be improved by well-known Vizing’s Adjacency Lemma and other known edge-coloring techniques. To over come the barrier, we follow the recently developed recoloring technique of Tashkinov trees to expand Vizing fans technique 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 Δ(G) the maximum degree of G. An * Partially supported by NSFC of China (Nos. 11671232, 11571096, 61373019, 11671186). 1
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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.
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
3
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.
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.
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 +⌊
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
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
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,
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:
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 ∆-
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
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