Acyclic edge coloring of sparse graphs

Abstract

A proper edge coloring of a graph $G$ is called acyclic if there is no
bichromatic cycle in $G$. The acyclic chromatic index of $G$, denoted by
$\chi'_a(G)$, is the least number of colors $k$ such that $G$ has an acyclic
edge $k$-coloring. The maximum average degree of a graph $G$, denoted by
$\mad(G)$, is the maximum of the average degree of all subgraphs of $G$. In
this paper, it is proved that if $\mad(G)<4$, then
$\chi'_a(G)\leq{\Delta(G)+2}$; if $\mad(G)<3$, then
$\chi'_a(G)\leq{\Delta(G)+1}$. This implies that every triangle-free planar
graph $G$ is acyclically edge $(\Delta(G)+2)$-colorable.

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arXiv:1202.6129v1 [math.CO] 28 Feb 2012
Acyclic edge coloring of sparse graphs∗
Jianfeng Hou
Center for Discrete Mathematics, Fuzhou University,
Fujian, P. R. China, 350002
jfhou@fzu.edu.cn
Abstract
A proper edge coloring of a graph G is called acyclic if there is no bichromatic
cycle in G.The acyclic chromatic index of G, denoted by χ′a(G), is the least
number of colors k such that G has an acyclic edge k-coloring. The maximum
average degree of a graph G, denoted by mad(G), is the maximum of the average
degree of all subgraphs of G. In this paper, it is proved that if mad(G) < 4, then
χ′a(G) ≤ ∆(G) + 2; if mad(G) < 3, then χ′a(G) ≤ ∆(G) + 1. This implies that
every triangle-free planar graph G is acyclically edge (∆(G) + 2)-colorable.
Keywords: acyclic, coloring, average degree, critical
1 Introduction
In this paper, all considered graphs are finite, simple and undirected. We use V (G),
E(G), δ(G) and ∆(G) (or V,E,δ and ∆ for simple) to denote the vertex set, the edge set,
the minimum degree and the maximum degree of a graph G, respectively. For a vertex
v ∈ V (G), let N(v) denote the set of vertices adjacent to v and d(v) = |N(v)| denote
the degree of v. Let Nk(v) = {x ∈ N(v)|d(x) = k} and nk(v) = |Nk(v)|. A vertex of
degree k is called a k-vertex. We write a k+-vertex for a vertex of degree at least k, and
a k−-vertex for that of degree at most k. The girth of a graph G, denoted by g(G), is the
length of its shortest cycle.
As usual [k] stands for the set {1,2,...,k}.
A proper edge k-coloring of a graph G is a mapping φ from E(G) to the color set
[k] such that no pair of adjacent edges are colored with the same color. A proper edge
coloring of a graph G is called acyclic if there is no bichromatic cycle in G. In other
∗This work is supported by research grants NSFC (11001055) and NSFFP (2010J05004, 2011J06001).
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words, if the union of any two color classes induces a subgraph of G which is a forest.
The acyclic chromatic index of G, denoted by χ′
that G has an acyclic edge k-coloring.
Acyclic edge coloring has been widely studied over past twenty years. The first general
linear upper bound on χ′
χ′
the same method. In 2001, Alon, Sudakov and Zaks [2] stated the Acyclic Edge Coloring
Conjecture as follows.
a(G), is the least number of colors k such
a(G) was found by Alon et al. in [1]. Namely, they proved that
a(G) ≤ 64∆(G). This bound was improved to 16∆(G) by Molloy and Reed [12] using
Conjecture 1.1 χ′
a(G) ≤ ∆(G) + 2 for all graphs G.
Obviously, χ′
showed that if G is a connected graph with ∆(G) ≤ 3 different from K4and K3,3, then
χ′
for some special classes of graphs, including non-regular graphs with maximum degree at
most four [5], outerplanar graphs [11, 13], series-parallel graphs [10], planar graphs with
girth at least five [8, 10], graphs with large girth [2], and so on.
Recall that the maximum average degree of a graph G, denoted by mad(G), is the
maximum of the average degree of all of its subgraphs, i.e., mad(G) = maxH⊆G
Recently, Basavaraju and Chandran [6] consider the acyclic edge coloring of sparse graphs
and proved that if mad(G) < 4, then χ′
improved to ∆(G) + 2.
a(G) ≤ ∆(G) + 1 for graphs G with ∆(G) ≤ 2. Andersen et al. [4]
a(G) ≤ 4. Hence Conjecture 1.1 holds for ∆(G) ≤ 3. This conjecture was also verified
2|E(H)|
|V (H)|.
a(G) ≤ ∆(G) + 3. In this paper, the bound is
Theorem 1.1 Let G be a graph with mad(G) < 4. Then χ′
a(G) ≤ ∆(G) + 2.
Theorem 1.2 Let G be a graph with mad(G) < 3. Then χ′
a(G) ≤ ∆(G) + 1.
By an application of Euler’s formula, it is easy to see that every planar graph G with
girth g(G) satisfies mad(G) <
2g(G)
g(G)−2. Thus, we have the following corollaries.
Corollary 1.3 Every triangle-free planar graph is acyclic edge (∆(G) + 2)-colorable.
Corollary 1.4 Every planar graph with g(G) ≥ 6 is acyclic edge (∆(G) + 1)-colorable.
2Properties of k-critical graph
At first, we introduce some notations and definitions used in [7]. Let H be a subgraph
of G. Then an acyclic edge coloring φ of H is called a partial acyclic edge coloring of
G. Note that H can be G itself. Let φ : E(G) → [k] be a partial edge k-coloring of
G. For any vertex v ∈ V (G), let Fv(φ) = {φ(uv)|u ∈ NG(v)} and Cv(φ) = [k] − Fv(φ).
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For an edge uv ∈ E(G), we say the color φ(uv) appears on the vertex v and define
Suv(φ) = Fv(φ) − {φ(uv)}. Note that Suv(φ) need not be the same as Svu(φ).
Let α,β be two colors. An (α,β)-maximal bichromatic path with respect to a partial
edge coloring of G is maximal path consisting of edges that are colored α and β alter-
natingly. An (α,β,u,v)-maximal bichromatic path is an (α,β)-maximal bichromatic
path which starts at the vertex u with an edge colored α and ends at v. An (α,β,u,v)-
maximal bichromatic path which ends at the vertex v via an edge colored α is called an
(α,β,u,v)-critical path.
A graph G is called an acyclically edge k-critical graph if χ′
subgraph of G is acyclically edge k-colorable. Obviously, if G is an acyclically edge k-
critical graph with k > ∆(G), then ∆(G) ≥ 3. The following facts are obvious.
a(G) > k and any proper
Fact 2.1 Given a pair of color α and β of a proper edge coloring φ of G, there is at most
one (α,β)-maximal bichromatic path containing a particular vertex v, with respect to φ.
Fact 2.2 Let G be an acyclically edge k-critical graph, and uv be an edge of G. Then for
any acyclically edge k-coloring φ of G−uv, if Fu(φ)∩Fv(φ) = ∅, then d(u)+d(v) ≥ k+2.
If |Fu(φ)∩Fv(φ)| = t, say φ(uui) = φ(vvi) for i = 1,2,..,t, then
t ?
i=1d(vi)+d(u)+d(v) ≥
k + t + 2 and
t ?
i=1d(ui) + d(u) + d(v) ≥ k + t + 2.
In [9], Hou et al. considered the properties of acyclically edge k-critical graphs and
got the following lemmas.
Lemma 2.1 [9] Any acyclically edge k-critical graph is 2-connected.
Lemma 2.2 [9] Let G be an acyclically edge k-critical graph with k ≤ 2∆(G) − 2 and
v be a vertex of G adjacent to a 2-vertex. Then v is adjacent to at least k − ∆(G) + 1
vertices of degree at least k − ∆(G) + 2.
By above lemma, we have the following corollaries.
Corollary 2.1 Let v be a vertex of an acyclically edge (∆(G)+1)-critical graph G. Then
n2(v) ≤ ∆(G) − 2.
Corollary 2.2 Let v be a vertex of an acyclically edge (∆(G) + 2)-critical graph G. If
n2(v) ?= 0, then n2(v) + n3(v) ≤ ∆(G) − 3.
Now we consider the neighbors of 2-vertices in acyclically edge critical graphs.
Lemma 2.3 Let v be a 2-vertex of an acyclically edge k-critical graph G with k > ∆(G).
Then the neighbors of v are (k − ∆(G) + 3)+-vertices.
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Proof.
k−∆(G)+2. Let N(v) = {u,w} and N(u) = {v,u1,u2,...,ut}, where t ≤ k−∆(G)+1.
Then the graph G′= G−uv admits an acyclic edge k-coloring φ by the choice of G with
φ(uui) = i for 1 ≤ i ≤ t. Since d(u)+d(v) ≤ ∆(G)+2, we have φ(wv) ∈ Fu(φ) by Fact 2.2,
say φ(wv) = 1. Then for any t+1 ≤ i ≤ k, there is a (1,i,u,v)-critical path with respect
to φ through u1and w, since otherwise we can color uv with i properly with avoiding
bichromatic cycle, which is a contradiction to the choice of G. Thus d(u1) ≥ k − t + 1.
This implies that t = k − ∆(G) + 1 and Cu1(φ) = Cw(φ) = {2,3,...,t}. Recolor wv
with 2, by the same argument, there is a (2,i,u,v)-critical path through u2and w for
any t + 1 ≤ i ≤ k. Now exchange the colors on uu1and uu2, and color uv with t + 1.
The resulting coloring is an acyclic edge coloring of G using k colors by Fact 2.1, a
contradiction.
In [9], Hou et al. got the following lemma.
Suppose to the contrary that v has a neighbor u whose degree is at most
Lemma 2.4 [9] Let G be an acyclically edge k-critical graph with k ≥ ∆(G) + 2 and v
be a 3-vertex of G. Then the neighbors of v are (k − ∆(G) + 2)+-vertices.
Lemma 2.4 implies that the neighbors of a 3-vertex are 4+-vertices in an acyclically
edge (∆(G) + 2)-critical graph. Now we give the following lemma.
Lemma 2.5 Let v be a 3-vertex of an acyclically edge (∆(G) + 2)-critical graph G with
neighbors x,y,z. If d(x) = 4, then
1. both y and z are 5+-vertices.
2. one of vertices in {y,z}, say y, is adjacent to at least three 4+-vertices. The other
vertex z is adjacent to at least two 4+-vertices. Moreover, if d(z) = 5, then z is
adjacent to at least three 4+-vertices.
Proof.
Then for any cyclic edge k-coloring φ of G′= G − vx, Fv(φ) ∩ Fx(φ) ?= ∅ by Fact 2.2.
Now we show that there exists an cyclic edge k-coloring φ of G′= G − vx, such that
|Fv(φ)∩Fx(φ)| = 1. Otherwise, for any acyclic edge coloring ϕ of G′, |Fv(ϕ)∩Fx(ϕ)| = 2.
Assume that ϕ(xxi) = i for 1 ≤ i ≤ 3, ϕ(vy) = 1 and ϕ(vz) = 2. Let S (or T) denote the
set of colors such that for any i ≥ 4 and i ∈ S (or i ∈ T), there is a (1,i,v,x)-critical path
(or (2,i,v,x)-critical path) through x1and y (or x2and z). Then S ∪ T = {4,5,...,k}
by the choice of G. Assume that T ?= ∅.
If there exists a color α ∈ T such that α does not appear on y, then recolor vy with
α to get a partial edge-coloring φ, i.e., φ(vy) = α, and φ(e) = ϕ(e) for all other edges
e in G. Then φ is an acyclic edge coloring of G′with |Fv(φ) ∩ Fx(φ)| = 1 by Fact 2.1.
We may assume that N(x) = {v,x1,x2,x3}. Let k = ∆(G) + 2 for simplicity.
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Otherwise, T ⊆ Fy(ϕ). Similarly, S ⊆ Fz(ϕ). This implies that d(y) = d(z) = ∆(G) and
S ∩ T = ∅.
Exchange the colors on xy and xz. Then for any i ∈ S, there is a (2,i,v,x)-critical
path from through x2and y; for any j ∈ T, there is a (1,j,v,x)-critical path through
x1and z. This implies that Sxxi(ϕ) = S ∪ T for i = 1,2. If recolor vy with 3, then for
any i ∈ S, there is a (3,i,v,x)-critical path through x3and vy. If exchange the colors
on xx1and xx2, then for any i ∈ T, there is a (3,i,v,x)-critical path through x3and y.
We recolor vz with 3 and choose a color from T to color vx. It is easy to verify that the
resulting coloring is an acyclic edge coloring of G, a contradiction.
Now suppose that φ is an acyclic edge k-coloring of G′= G−vx with |Fv(φ)∩Fx(φ)| =
1. Let φ(xxi) = i for 1 ≤ i ≤ 3, φ(vy) = 1 and φ(vz) = 4. Then for any 5 ≤ i ≤ k, there
is a (1,i,v,x)-critical path through x1.
Claim 1. 4 ∈ Fy(φ)
Proof.
By contradiction, 4 / ∈ Fy(φ). Since |Cy(φ)| ≥ 2, {2,3} ∩ Cy(φ) ?= ∅, say
2 ∈ Cy(φ). Recolor vy with 2, the resulting coloring is also a acyclic edge coloring of G′.
Then there is a (2,i,v,x)-critical path through x2for any i ≥ 5. Note that {1,2} ⊆ Fz(φ),
since otherwise we can recolor vy with 4, vz with 1 or 2, and color xv with 5. This implies
that there is a color i0with 5 ≤ i0≤ k such that i0does not appear on z. Recolor vz
with i0and color xv with 4.
It follows from Claim 1 that d(y) = ∆(G) and Svy(φ) = {4,5,...,k}.
Claim 2. 1 ∈ Fz(φ).
Proof.
Otherwise, let φ′(vy) = 2, φ′(vz) = 1, and φ′(e) = φ(e) for the other edges e of
G′. Then φ′is also an acyclic edge coloring of G′. Note that coloring vx with i does not
create bichromatic cycle containing the edges vz and xx1by Fact 2.1 for any 4 ≤ i ≤ k.
Thus, there is a (2,i,v,x)-critical path for any 4 ≤ i ≤ k, since otherwise, we can color
vx with i. Similarly, let φ′(vy) = 3, φ′(vz) = 1, and φ′(e) = φ(e) for the edges e. Then
there is a (3,i,v,x)-critical path for any 4 ≤ i ≤ k with respect to φ′. Now we give an
acyclic edge coloring ϕ of G by letting ϕ(xx2) = φ(xx3), ϕ(xx3) = φ(xx2), ϕ(vy) = 2,
ϕ(vx) = 5, and ϕ(e) = φ(e) for the other edges e of G, a contradiction.
Claim 3. There is a (4,2,v,y)-critical path through z with respect to φ.
Proof.
Otherwise, recoloring vy with 2 results in a new acyclic edge coloring φ′of G′.
Then there is a (2,i,v,x)-critical path through x2for any i ∈ {5,...,k} with respect to
φ′. This implies that Cx2(φ) ⊆ {1,3,4}.
Now we show that there is a (1,4,v,x)-critical path through x1and y with respect
to φ. Otherwise, coloring vx with 4 and uncoloring vz results in an acyclic edge coloring
ϕ of G′′= G − vz. If 2 ∈ Cy(ϕ), then color vz with 2. Otherwise, there exists a color
i0∈ {5,6,...,k} such that i0∈ Cy(ϕ). Color vz with i0and get an acyclic edge coloring
of G.
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By the same argument above, there is a (2,4,v,x)-critical path through x2and y if
we recolor vy with 2. Let ϕ(xx1) = 2, ϕ(xx2) = 1, ϕ(vx) = 5, and ϕ(e) = φ(e) for the
other edges e of G. Then ϕ is an acyclic edge coloring of G, a contradiction.
As in the proof of Claim 3, there is a (3,4,v,y)-critical path through z.
{1,2,3,4} ⊆ Fz(φ) by Claims 2 and 3. This implies that at least two colors in {5,6,...,k}
such that neither appear on z, say 5, 6. Note that there is a (1,4,v,x)-critical path through
x1and y, since otherwise, recolor vz with 2 and color vx with 4. Recolor vz with 5 (or 6),
the resulting coloring φ′is also an acyclic edge coloring of G′. As in the proof of Claim
2, there is a (5,i,v,y)-critical path (or (6,i,v,y)-critical path) through z with respect to
φ′for any i ∈ {2,3}. Let φ(yyi) = i for i = 4,5,6 and φ(zzi) = i for i = 1,2,3. Then
{1,2,3,i} ⊆ Fyi(φ) for i = 4,5,6 and {4,5,6,i} ⊆ Fzi(φ) for i = 2,3. Thus y is adjacent
to at least three 4+-vertices and z is adjacent to at least two 4+-vertices.
Claim 4. d(y) ≥ 5
Proof.
Otherwise, ∆(G) = 4 and Fyi(φ) = {1,2,3,i} for any i = 4,5,6. Exchange the
colors on yy4and yy5, recolor vz with 6, and color vx with 5. It is easy to see that the
resulting coloring is an acyclic edge coloring of G, a contradiction.
Claim 5. d(z) ≥ 5
Proof.
Otherwise, d(z) = 4 and Fz(φ) = {1,2,3,4}. As in the proof of Claim 3, there
is a (2,i,v,z)-critical path (or (3,i,v,z)-critical path) for any 5 ≤ i ≤ k if recoloring vy
with 2 (or 3). This implies Fzi(φ) = {i,4,5,...,k}. Exchange the colors on zz1and zz2
and the argument in Claim 3 works.
Finally, we prove that if d(z) = 5, then n3(z) ≤ 2. Assume that N(z) = {z1,z2,z3,z7,v}
and Fz(φ) = {1,2,3,4,7}. Suppose that n3(v) ≥ 3, then d(z1) = d(z7) = 3 by Lemma
2.2, say N(zi) = {v,z′
and uncolor zz1. Then the resulting coloring is acyclic edge coloring ϕ of G′′= G − zz1
by Fact 2.1. By the choice of G, we have Fz1(ϕ) ∩ Fz(ϕ) ?= ∅.
We first consider the case that |Fz1(ϕ) ∩ Fz(ϕ)| = 1. If 7 / ∈ Fz1(ϕ), then coloring zz1
with a color from {4,5,6}\Fz1(ϕ). Otherwise, assume that ϕ(z1z′
Then for any 5 ≤ i ≤ k and i ?= 7, there is a (7,i,z,z1)-critical path through z7. This
implies that ∆(G) = 5 and Fz7(ϕ) = {5,6,7}. Note that 7 / ∈ Fz2(ϕ), since otherwise
recolor vz with 5, zz2 with 1, zz7with 4, and color zz1with 2. Similarly, 7 / ∈ Fz3(ϕ).
Then recolor zz2with 7, zz7with 4, and color zz1with 2.
Now we consider the case that |Fz1(ϕ)∩Fz(ϕ)| = 2. If 7 / ∈ Fz1(ϕ), then color zz1with
5. Otherwise, color zz1with a color from {4,5,6} \ Fz7(ϕ).
In any case, we can get an acyclic edge coloring of G, a contradiction.
In [5], Basavaraju and Chandran showed that if a connected graph G with n vertices
and m edges satisfies m ≤ 2n−1 and maximum degree ∆(G) ≤ 4, then χ′
result can be obtained immediately by Lemmas 2.3 and 2.5.
Thus
i,z′′
i} for i = 1,7. Recolor vy with 2, vz with 1, color vx with 4,
1) = 4 and ϕ(z1z′′
1) = 7.
a(G) ≤ 6. This
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Lemma 2.6 Let v be a t-vertex of an acyclically edge (∆(G) + 2)-critical graph G with
t ≥ 5. Then n2(v) ≤ t − 4. Moreover, if n2(v) = t − 4, then n3(v) = 0.
Proof.
5,6,...,t and k = ∆(G) + 2 for simplicity.
for 1 ≤ i ≤ 4. By contradiction, assume that d(x4) ≤ 3. We only consider the case that
d(x4) = 3, say N(x4) = {y′
graph G′= G − vxthas an acyclic edge k-coloring φ with φ(vxi) = i for 1 ≤ i ≤ t − 1.
Since d(v)+d(xi) ≤ ∆(G)+3 for i ≥ 4, we have φ(xtyt) ∈ {1,2,3} by Fact 2.2. Without
loss of generality, let φ(xtyt) = 1. Note that d(yt) = ∆(G) and Cyt(φ) = {2,3}, since oth-
erwise, recolor xtytwith a color i ≥ 4 not appearing on yt, and color xtytwith a coloring
appearing neither on v nor on xi. Moreover, for any t ≤ i ≤ k, there is a (1,i,v,xt)-critical
path through x1. Similarly, if recolor xtytwith 2 (or 3), then there is a (2,t,v,xt)-critical
path (or (2,i,v,xt)-critical path) through x2(or x3) for any t ≤ i ≤ k.
Claim 1. For any color i with 4 ≤ i ≤ t−1, there is a (1,i,v,xt)-critical path through
x1and ytwith respect to φ.
Proof.
Otherwise, there exists a color i0 ∈ {4,5,...,t − 1} such that there is no
(1,i0,v,xt)-critical path through x1 and yt. Recolor vxt with i0, uncolor vxi0and the
colors on the other edges unchange. We get an acyclic edge coloring ϕ of G′′= G − vxi0.
We first consider the case that i0≥ 5, say i0= 5. Since d(v) + d(x5) ≤ ∆(G) + 2, we
have φ(vxt) ∈ {1,2,3} by Fact 2.2. Color vx5with t and get an acyclic edge coloring of
G by Fact 2.1, a contradiction.
Now we consider the case that i0 = 4. Assume that Fx4(ϕ) = {α,β}. If Fx4(ϕ) ∩
{1,2,3} = ∅, we are done by Fact 2.1. Otherwise, let α = 1. For any i ≥ t, since there
is a (1,i,v,xt)-critical path, there is no (1,i,v,x4)-critical path. If β ∈ {2,3}, then color
vx4with t. If 5 ≤ β ≤ t − 1, then color vx4with a color appearing neither on v nor on
xβ. Otherwise, color vx4with a color appearing on neither v nor on x4.
Recolor xtytwith 2 (or 3), as in the proof of Claim 1, there is a (2,i,v,xt)-critical
path (or (3,i,v,xt)-critical path) through ytand x2or (x3). Thus Sxxi(φ) = {4,5,...,k}
for i = 1,2,3. For 1 ≤ i < j ≤ 3, let φijdenote the coloring that exchanging the colors
on vxiand vxj, and coloring vxtwith t with respect to φ.
Claim 2. There is a coloring in {φij: 1 ≤ i < j ≤ 3} such that there is no bichromatic
cycle containing vv4.
Proof.
If |Fv4(φ) ∩ {1,2,3}| ≤ 1, then there exist two colors i0,j0 with 1 ≤ i0 <
j0 ≤ 3 such that i0,j0neither appear on v4. The coloring φi0j0satisfies the condition.
Otherwise, assume that Fv4(φ) = {1,2,4}. If the coloring φ12has no bichromatic cycle
containing vv4, we are done. Otherwise, there is either a (1,4,v,x2)-critical through x4,
or a (2,4,v,x1)-critical path through x4with respect to φ12. In this case, the coloring φ13
has no bichromatic cycle containing vv4by Fact 2.1.
Suppose that v has neighbors x1,x2,...,xt with N(xi) = {v,yi} for i =
We only have to prove that d(xi) ≥ 4
4,y′′
4}. The case that d(x4) = 2 is the same, but easier. The
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We may assume the coloring ϕ = φ12is the coloring containing no bichromatic cycle
containing vv4. Let S = {xi: 5 ≤ i ≤ t − 1, there is a (i,1,v,x2)-critical path through
xiwith respect to ϕ}, and T = {xj: 5 ≤ j ≤ t − 1, there is a (j,2,v,x1)-critical path
through xj with respect to ϕ}. Then S ∪ T ?= ∅ by the choice of G. We may assume
that |S| ≤ 1, since otherwise, let S = {xi1,xi2,...,xit}, where t ≥ 2. We recolor vxip
with ϕ(vxip+1) for p = 1,2,...,t − 1, vxitwith ϕ(vxi1). Then the resulting coloring has
no bichromatic cycle containing vx2. Similarly, assume that |T| ≤ 1. We consider the
following three cases.
Case 1. |S| = 1 and |T| = 0.
Assume that S = {5}. Then ϕ(x5y5) = 1. We consider the colors not appearing on
y5. If there is a color i ≥ t such that i ∈ Cy5(ϕ), then recolor x5y5with i. If there is
a color 6 ≤ i ≤ t − 1 such that i ∈ Cy5(ϕ), then recolor x5y5with i, vx5with a color
appearing neither on v nor on xi. If 3 ∈ Cy5(ϕ), then recolor x5y5with 3, color vx5with
t + 1. Otherwise, 4 ∈ Cy5(ϕ). Then recolor x5y5with 4 and get a new coloring ϕ′of G.
If ϕ′is acyclic, we are done. Otherwise, recolor vx5with a color appearing neither on v
nor x4in ϕ′.
Case 2. |T| = 1 and |S| = 0.
This case is the same as Case 1.
Case 3. |S| = |T| = 1.
Assume that S = {5} and T = {6}. Then ϕ(x5y5) = 1 and ϕ(x6y6) = 2. If we
exchange the colors on vx5and vx6, then there is a (1,6,v,x5)-critical path through y5
and a (2,5,v,x6)-critical path through y6, since otherwise, the argument in Case 1 and
Case 2 works. If there exists a color α with α ≥ t or α = 1, such that α ∈ Cy6(ϕ), then
recolor x6y6with α and Case 1 works. If 3 ∈ Cy6(ϕ), we recolor x6y6with 3, exchange
the colors vx6and vxt, and then Case 1 works. Otherwise, there is a color 5 ≤ j0≤ t−1
such that j0∈ Cy6(ϕ). Recolor x6y6with j0in ϕ and the colors on other edges unchange.
Then we get a new coloring ϕ′of G. In ϕ′, if there is no bichromatic cycle containing vx6,
then Case 1 works. Otherwise, recoloring vx6with a color appearing neither on v nor on
xj0, and then Case 1 also works.
Lemma 2.7 Let v be a 5-vertex of an acyclically edge (∆(G)+2)-critical graph G. Then
n2(v) + n3(v) ≤ 3.
Proof.
If n2(v) > 0, we are done by Corollary 2.2. Suppose that n2(v) = 0 and k =
∆(G) + 2 for simplicity. By contradiction, let v be 5-vertex with neighbors v1,v2,...,v5
such that N(vi) = {v,xi,yi} for i = 2,3,4,5. The graph G′= G−vv5has an acyclic edge
k-coloring φ with φ(vvi) = i for i = 1,2,3,4. Then Fv(φ) ∩ Fv5(φ) ?= ∅ by Fact 2.2.
Case 1. |Fv(φ) ∩ Fv5(φ)| = 1.
Assume that φ(v5y5) = 5. We consider the color on v5x5.
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Subcase 1.1. φ(v5x5) ∈ {2,3,4}, say φ(v5x5) = 2.
For any i ≥ 6, there is a (2,i,v,v5)-critical path through v2and x5, since otherwise,
we can color vv5with i. This implies that ∆(G) = 5 and Fv2(φ) = {2,6,7}. This is also
a (5,i,v,v5)-critical path if we recolor vv2with 5 for any i ≥ 6.
Claim 1. 5 ∈ Fv5(φ).
Proof.
By contradiction, 5 / ∈ Fv5(φ). Firstly, we show that {3,4} ⊆ Fv5(φ). Otherwise,
assume that 3 / ∈ Fv5(φ). Since there is a (3,i,v,v5)-critical path for any i ∈ {6,7} if we
recolor v5x5with 3, Fv3(φ) = {3,6,7}. Exchange the colors on vv2and vv3, color vv5
with 6. The resulting coloring is an acyclic edge coloring of G, a contradiction. Thus
Fv5(φ) = {2,3,4,6,7}. Note that there is a (1,i,v,v5)-critical path through v1for any
i ∈ {6,7}, if we recolor v5x5with 1.
Secondly, we show that {1,2} ⊆ Fy5(φ). Otherwise, choose a color in {1,2} not
appearing on y5 to recolor v5y5, then recolor v5x5 with 5 and color vv5 with 6. Thus
Fy5(φ) = {1,2,5,6,7}.
Let φ′(v5y5) = 3, φ′(v5x5) = 5, and φ′(e) = φ(e) for other edges of G. Then φ′is an
acyclic edge coloring of G′. If there is no (3,i,v,v5)-critical path with respect to φ′for
some i ∈ {6,7}, then color vv5with i. Otherwise, exchange the colors on vv2and vv3,
color vv5with 6.
It follows from Claim 1 that {3,4} ∩ Cx5(φ) ?= ∅, say 3 ∈ Cx5(φ). Note that there is a
(5,3,v5,x5)-critical path, since otherwise, recoloring x5v5with 3 results in a new acyclic
edge coloring φ′of G′. If there is no (3,i,v,v5)-critical path with respect to φ′for some
i ∈ {6,7}, then color vv5with i. Otherwise, exchange the colors on vv2and vv3, color vv5
with 6.
Subcase 1.1.1 4 ∈ Cx5(φ).
By the same argument above, there is a (5,4,v5,x5)-critical path. Recolor x5v5with
3, v5y5with 2, and color vv5with 5.
Subcase 1.1.2 1 ∈ Cx5(φ).
If 1 ∈ Cy5(φ), then recolor v5y5with 1 and color vv5with 5. Otherwise, recolor x5v5
with 3, v5y5with 2, and color vv5with 5.
Subcase 1.2. φ(v5x5) = 1.
For any i ≥ 6, there is a (1,i,v,v5)-critical path through v1and x5. Note that for any
i ∈ {2,3,4} \ Fx5(φ), there is a (5,i,v5,x5)-critical path through y5, since otherwise, we
can recolor v5x5with i and the argument in Subcase 1.1 works. Thus 5 ∈ Fx5(φ), without
loss of generality, let d(x5) = ∆(G) and Fx5(φ) = {4,5,...,k}.
Color vv5with 2 and uncolor vv2, we get an acyclic edge coloring ϕ of G′′= G − vv2.
Then Fv(ϕ) ∩ Fv2(ϕ) ?= ∅ by Fact 2.2. We consider the following two cases.
Subcase 1.2.1 |Fv(ϕ) ∩ Fv2(ϕ)| = 1.
Assume that ϕ(v2x2) ∈ Fv(ϕ). If ϕ(v2x2) ∈ {3,4}, then the proof in Subcase 1.1
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works. Otherwise, ϕ(v2x2) = 1. We color vv2with a color from {6,7} \ Fv2(ϕ).
Subcase 1.2.2 |Fv(ϕ) ∩ Fv2(ϕ)| = 2.
Note that 1 / ∈ Fv2(ϕ), since otherwise, assume that ϕ(v2x2) = 1 and ϕ(v2y2) = i,
where i ∈ {3,4}. Choose a color from {5,6,7}\Fvi(ϕ) to color vv2. Thus Fv2(ϕ) = {3,4}.
For any i ≥ 5, there is either a (3,i,v,v2)-critical path through v3, or a (4,i,v,v2)-critical
path through v4. This implies that there exists a vertex in {v3,v4}, say v3, such that the
colors in Svv3(ϕ) are greater than or equal to 5, say Svv3(ϕ) = {5,6}. Recolor vv3with 7
and choose a color from {5,6} \ Fv4(ϕ) to color vv2, it is possible since 7 ∈ Fv3(ϕ).
Case 2. |Fv(φ) ∩ Fv5(φ)| = 2.
Then for any color i with 5 ≤ i ≤ k, there is either a (φ(v5x5),i,v,v5)-critical path, or a
(φ(v5y5), i,v,v5)-critical path. If there is a color i0≥ 5 such that i0not appearing on x5(or
y5), then we recolor x5v5(or y5v5) with i0and the argument in Case 1 works. Otherwise,
{5,...,k} ⊆ Fx5(φ)∩Fy5(φ). In this case, if there is a vertex vifor some i ∈ {2,3,4} such
that the colors in Fvvi(φ) are greater than or equal to 5, then recolor vviwith a color from
{4,5,...,k}\Fvvi(φ) and the argument in Case 1 works. Otherwise, 1 ∈ Fv5(φ). Assume
that φ(x5v5) = 1 and φ(y5v5) = 2. Since {5,6,...,k} ⊆ Fx5(φ), {3,4} ∩ Cx5(φ) ?= ∅, say
3 ∈ Cx5(φ). Then there is a (2,3,v5,x5)-critical path through y5, since otherwise, we can
recolor x5v5with 3 and choose a color from {4,5,...,k} \ (Fv2(φ) ∪ Fv2(φ)) to color vv5.
This implies that 4 / ∈ Fx5(φ). We can recolor recolor x5v5with 4 and choose a color from
{4,5,...,k} \ (Fv2(φ) ∪ Fv4(φ)) to color vv5.
In any case, we can get a acyclic edge coloring of G with ∆(G) + 2 colors, a contra-
diction.
3 Discharging
In this section, we give the proofs of Theorem 1.1 and 1.2 by discharging method.
3.1Proof of Theorem 1.1
Proof of Theorem 1.1. By contradiction, assume that G is an acyclically edge (∆(G)+
2)-critical graph, i.e., G is the minimum counterexample to the theorem in terms of the
number of edges. Then G is 2-connected by Lemma 2.1. It follows from Lemmas 2.3 and
2.5 that ∆(G) ≥ 4. Let us assign an initial charge of ω(v) = d(v) − 4 to each vertex
v ∈ V (G). It follows from mad(G) < 4 that
?
v∈V (G)
ω(v) < 0.
We shall design some discharging rules and redistribute weights according to them. Once
the discharging is finished, a new weight function ω′is produced. However, the total
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sum of weights is kept fixed when the discharging is in process. On the other hand, we
shall show that ω′(v) ≥ 0 for all v ∈ V (G). This leads to an obvious contradiction. The
discharging rules are defined as follows.
(R1) Every 2-vertex receives 1 from each neighbor.
(R2) Let v be a 3-vertex. If v is adjacent to a 4-vertex, then the other neighbors of v
are 5+-vertices by Lemma 2.5, and v receivers1
v receivers1
Now we show that the resultant charge function ω′(v) ≥ 0 for any v ∈ V (G).
If d(v) = 2, then ω(v) = −2 and the neighbors of v are 5+-vertices by Lemma 2.3,
each of which gives 1 to v, so ω′(v) = −2 + 2 × 1 = 0.
Suppose that d(v) = 3. Then ω(v) = −1 and the neighbor of v are 4+-vertices by
Lemma 2.4. If v is adjacent to a 4-vertex, then the other neighbors of v are 5+-vertices
by Lemma 2.5, so the other neighbors of v are 5+-vertices by Lemma 2.5 each of which
gives1
If d(v) = 4, then ω′(v) = ω(v) = 0.
Suppose that d(v) ≥ 5. Let u be the neighbor of v which has minimum degree in
N(v). If d(u) ≥ 4, then ω′(v) = ω(v) > 0. We first consider the case that d(u) = 2,
then n2(v) + n3(v) ≤ d(v) − 3 by Lemma 2.2. If n2(v) ≤ d(v) − 5, then ω′(v) = ω(v) −
n2(v) × 1 − n3(v) ×1
n2(v) = d(v)−4 and n3(v) = 0 by Lemma 2.6. Thus ω′(v) = ω(v)−n2(v)×1 = 0. At last
we consider the case that d(u) = 3. If u is adjacent to a 4-vertex, then n3(v) ≤ d(v) − 2
if d(v) ≥ 6, and n3(v) ≤ 2 if d(v) = 5 by Lemma 2.5. Thus ω′(v) ≥ ω(v) − 2 ×1
d(v) = 5; and ω′(v) ≥ ω(v)−(d(v)−2)×1
v. If d(v) = 5, then n3(v) ≤ 3 by Lemma 2.7, so ω′(v) ≥ ω(v) − 3 ×1
ω′(v) ≥ ω(v) − d(v) ×1
Thus ω′(v) ≥ 0 for any v ∈ V (G), a contradiction. This completes the proof.
2from each adjacent 5+-vertex. Otherwise,
3from each neighbor.
2to v, so ω′(v) = −1 + 2 ×1
2= 0. Otherwise, ω′(v) = −1 + 3 ×1
3= 0.
2≥ d(v) − 4 − n2(v) −
d(v)−3−n2(v)
2
=
d(v)−n2(v)−5
2
≥ 0. Otherwise,
2= 0 if
3from
2≥ 0 if d(v) ≥ 6. Otherwise, u receives1
3= 0. Otherwise,
3≥ 0.
3.2Proof of Theorem 1.2
Proof of Theorem 1.2. By contradiction, assume that G is an acyclically edge (∆(G)+
1)-critical graph. Then G is 2-connected by Lemma 2.1. Let us assign an initial charge
of ω(v) = d(v) − 3 to each vertex v ∈ V (G). It follows from mad(G) < 3 that
?
v∈V (G)
ω(v) < 0.
The discharging rule is defined as follows.
(R) Every 2-vertex receives1
Now we show that the resultant charge function ω′(v) ≥ 0 for any v ∈ V (G). If
d(v) = 2, then ω(v) = −1 and the neighbors of v are 4+-vertices by Lemma 2.3, each
2from each neighbor.
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of which gives
Otherwise, d(v) ≥ 4. By Corollary 2.1, n2(v) ≤ d(v)−2 and ω′(v) ≥ ω(v)−(d(v)−2)×1
d(v)
2− 2 ≥ 0.
Thus ω′(v) ≥ 0 for any v ∈ V (G), a contradiction. This completes the proof.
1
2to v, so ω′(v) = −1 + 2 ×1
2= 0. If d(v) = 3, then ω′(v) = ω(v) = 0.
2=
4 Open Problems
Since any cycle has acyclic chromatic index three, the bound ∆(G)+1 in Theorem 1.2 is
tight. It is easy to see that the graphs K4,K3,3have acyclic chromatic index five. So the
bounds mad < 3 in Theorem 1.2 and ∆(G) + 2 in Theorem 1.1 are tight.
Determining the acyclic chromatic index of a graph is a hard problem both from the-
oretical and algorithmic points of view. Even for complete graphs, the acyclic chromatic
index is still not determined exactly. It has been shown by Alon and Zaks [3] that deter-
mining whether χ′
following open problems.
a(G) ≤ 3 is NP-complete for an arbitrary graph G. Now we provide the
Problem 4.1 Find the necessary or sufficient conditions for a graph G with χ′
χ′(G), where χ′(G) is the chromatic index of G.
a(G) =
Problem 4.2 Determine the acyclic chromatic index of planar graphs.
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