equitable list-coloring for graphs of maximum degree 3

Equitable List-Coloringfor Graphs of MaximumDegree 3

Michael J. PelsmajerDEPARTMENT OF APPLIED MATHEMATICS

ILLINOIS INSTITUTE OF TECHNOLOGY

CHICAGO, ILLINOIS 60616

E-mail: [email protected]

Received October 17, 2002; Revised November 25, 2003

Published online in Wiley InterScience (www.interscience.wiley.com).

DOI 10.1002/jgt.20011

Abstract: Given lists of available colors assigned to the vertices of a graphG, a list coloring is a proper coloring of G such that the color on each vertexis chosen from its list. If the lists all have size k , then a list coloring isequitable if each color appears on at most djV (G )j=ke vertices. A graph isequitably k -choosable if such a coloring exists whenever the lists all havesize k . Kostochka, Pelsmajer, and West introduced this notion andconjectured that G is equitably k -choosable for k > �(G). We prove thisfor �(G)¼ 3. We also show that every graph G is equitably k -choosablefor k � �(G)(�(G)�1)=2þ 2. � 2004 Wiley Periodicals, Inc. J Graph Theory 47: 1–8, 2004

Keywords: graph coloring; equitable coloring; list coloring

1. INTRODUCTION

In many applications of graph coloring, it is desirable that the color classes not be

very large. When scheduling jobs, for example, the number of jobs that can be run——————————————————

� 2004 Wiley Periodicals, Inc.

1

at the same time may be limited by the number of processors (see Ref. [6] for

such an application). Equitable coloring is a well known property (see Ref. [3] for

a survey) that restricts the sizes of each color class: A proper vertex coloring of a

graph is equitable if the sizes of the color classes differ by at most 1. For an

equitable k-coloring of a graph G, each color class has at most dnðGÞ=ke vertices,

where nðGÞ denotes the number of vertices of G.

Kostochka, Pelsmajer, and West introduced the list analogue of equitable

coloring in Ref. [2]. A list assignment L for a graph G assigns to each vertex

v 2 VðGÞ a set LðvÞ of acceptable colors. An L-coloring of G is a proper vertex

coloring such that for every v 2 VðGÞ the color on v belongs to LðvÞ. A list

assignment L for G is k-uniform if jLðvÞj ¼ k for all v 2 VðGÞ.Given a k-uniform list assignment L for a graph G, we say that G is equitably

L-colorable if G has an L-coloring such that each color appears on at most

dnðGÞ=ke vertices. A graph G is equitably list k-colorable or equitably k-

choosable if G is equitably L-colorable whenever L is a k-uniform list assignment

for G.

Hajnal and Szemeredi [1] proved that a graph G has an equitable k-coloring

whenever k > �ðGÞ, where �ðGÞ denotes the maximum vertex degree in G; this

answered a question of P. Erdos. Kostochka, Pelsmajer, and West [2] conjectured

the analogue of the Hajnal–Szemeredi Theorem [1]:

Conjecture 1.1. Every graph G is equitably k-choosable whenever k > �ðGÞ.

In this paper we prove Conjecture 1.1 for graphs with maximum degree at

most 3. (This has been proved independently and by a different method in

Ref. [4].)

The conjecture has been proved for several classes of graphs in Refs. [2,5],

often using an inductive approach which we will now describe. We use NðvÞ for

fu: uv 2 EðGÞg and N½v� for NðvÞ [ fvg, and we say that the vertices of NðvÞ are

the neighbors of v.

Consider a graph G with a k-uniform list assignment L, and let S ¼fx1; . . . ; xkg be a set of k vertices in G. If G� S has an equitable L-coloring f that

extends to an L-coloring f 0 of G by giving the vertices in S different colors, then f 0

is an equitable L-coloring of G, since the extension augments each color class at

most once. We give a simple sufficient condition for the existence of such an

extension (from Ref. [2], but stated here in a slightly stronger form).

Lemma 1.1. Let G be a graph with a k-uniform list assignment L. Let S ¼fx1; . . . ; xkg be a set of k vertices in G, and suppose that G� S has an equitable

L-coloring. If

jNGðxiÞ � Sj � k � i ð1Þ

for 1 � i � k, then G has an equitable L-coloring such that k distinct colors are

used on S.

2 JOURNAL OF GRAPH THEORY

Proof. Let G0 ¼ G� S. For 1 � i � k, let Gi ¼ G½VðGi�1Þ þ vi�, so that

G ¼ Gk. Let f0 be an equitable L-coloring of G0. We produce an L-coloring fiof Gi. For 1 � i � k, extend fi�1 to an L-coloring fi of Gi by giving xi a color in

LðxiÞ different from the colors that fi has used on neighbors of xi and on

the vertices x1; . . . ; xi�1. There are at most jNGðxiÞ � Sj þ ði� 1Þ forbidden

colors, so condition (1) guarantees that LðxiÞ contains a color of the desired type.

(See Fig. 1.) By construction, the colors used on S are distinct, and hence fk is an

equitable L-coloring of G. &

This will be our main tool. We now observe a general result in terms of

maximum degree.

Theorem 1.1. G is equitably k-choosable for k � �ðGÞð�ðGÞ�1Þ2

þ 2.

Proof. For fixed k, the graphs G such that k � �ðGÞð�ðGÞ�1Þ2

þ 2 form a

hereditary class of graph, so we may choose a minimal counterexample G in the

sense that any induced subgraph of G is equitably k-choosable. If G has at most k

vertices, then for any k-uniform list assignment L, we can color the vertices by

selecting f ðviÞ from LðviÞ � ff ðv1Þ; . . . ; f ðvi�1Þg. Therefore, we may assume that

nðGÞ > k. By Lemma 1.1, it suffices to find a set S ¼ fx1; . . . ; xkg that satisfies

condition (1). Any k-set suffices if G has no edges, so we may assume that

�ðGÞ � 1.

We will sequentially designate the vertices xk; . . . ; x1 in that order, according to

the following procedure. We begin with all vertices unlabeled. Let xk be a vertex

of maximum degree, and let xk�1; . . . ; xk��ðGÞ be its neighbors. We continue

indexing vertices in steps, for 1 � j � �ðGÞ � 2. The jth step: if there are more

than j unlabeled neighbors of xk�j, index all but j of the unlabeled neighbors of

xk�j using the next available indices; otherwise, do nothing. Note that after all the

steps, at most�ðGÞð�ðGÞ�1Þ

2þ 2 vertices have been indexed. Finish by indexing

arbitrary unlabeled vertices until a total of k vertices are indexed, then observe

that condition (1) holds for S. &

2. G IS EQUITABLY k-CHOOSABLE IF k >> ��(G)¼ 3

Theorem 1.1 immediately implies Conjecture 1.1 for graphs with maximum

degree 1 or 2 and also for graphs with maximum degree 3 when k � 5. Since

graphs with maximum degree 0 have no edges, for them Conjecture 1.1 follows

directly from Lemma 1.1. In this section, we prove Conjecture 1.1 for graphs with

FIGURE 1. An example of Lemma 1.1 with k ¼ 4 and S ¼ fx1; x2; x3; x4g.

EQUITABLE LIST-COLORING FOR GRAPHS 3

maximum degree at most 3 in the remaining case: we prove that graphs with

maximum degree 3 are equitably 4-choosable. Since K4 has maximum degree 3

but is not even 3-colorable, this result is sharp.

In the proof of our main theorem, nearly all of the cases reduce to variations of

Lemma 1.1. In each, we find an (ordered) list S of four vertices such that some

coloring of G� S can be properly extended so that S gets four distinct colors. The

most important variation is that we sometimes need two different colorings on

G� S to know that one of them extends to G.

Definition 2.1. If L is a list assignment on G, and f is an L-coloring of

a subgraph H � G, then for all v 2 VðGÞ, let Lf ðvÞ ¼ LðvÞ � ff ðuÞ:u 2 NðvÞ \ VðHÞg.

When considering potential extensions of f to an L-coloring of G (or

recolorings of a single vertex v), the restricted list Lf ðvÞ is the set of colors

available for use on v.

The following lemma is trivial.

Lemma 2.1. Suppose that S and S0 are sets of size at least two. There exists

c1; c2 2 S and c01; c02 2 S0 such that c1 6¼ c2; c01 6¼ c02; c1 6¼ c01, and c02 6¼ c02.

The next lemma will be used to begin the last argument in our main proof. It

will also make it easy to show that a minimal counterexample must be 3-regular.

Lemma 2.2. Let G be a graph with �ðGÞ � 3 and let L be a 4-uniform list

assignment of G. Suppose that every proper induced subgraph of G has an

equitable L-coloring.

If v; v0 2 VðGÞ are adjacent, and dðvÞ þ dðv0Þ � 4, then G has two equitable

L-colorings that agree on VðGÞ � v � v0 and differ on both v and v0.

Proof. If VðGÞ ¼ fv; v0; v00g, then choose f1ðv00Þ ¼ f2ðv00Þ 2 Lðv00Þ. Then we

can apply Lemma 2.1 with Lf1ðvÞ and Lf1ðv0Þ to obtain f1ðvÞ; f2ðvÞ 2 Lf1ðvÞ and

f1ðv0Þ; f2ðv0Þ 2 Lf1ðv0Þ such that f1ðvÞ 6¼ f2ðvÞ; f1ðv0Þ 6¼ f2ðv0Þ; f1ðvÞ 6¼ f1ðv0Þ, and

f2ðvÞ 6¼ f2ðv0Þ. Now f1 and f2 are the desired colorings.

The case nðGÞ ¼ 2 is easier, so we may assume that nðGÞ � 4.

We will find S ¼ ðx1; x2; x3; x4Þ, a list of distinct vertices in G such that

fx3; x4g ¼ fv; v0g; jNðx2Þ � Sj � 2; Nðx3Þ � S, and Nðx4Þ � S. From such a list

S, we can finish the proof as follows: let f be an equitable L-coloring of G� S.

Since jNðx1Þ � Sj � �ðGÞ � 3 and jNðx2Þ � Sj � 2, we may extend this L-

coloring to x1 and x2 such that f ðx1Þ 6¼ f ðx2Þ. Afterwards, jLf ðx3Þj and jLf ðx4Þj are

each at least 2, so Lemma 2.1 applies to further extend f in two ways to G. This

yields the two desired extensions, so it suffices to show that such a list S must

exist.

Suppose that both v and v0 have degree 2. Let x4 ¼ v0 and x3 ¼ v, and let x2 be

the other neighbor of x4. If x2 is not adjacent to x3, then let x1 be the other

neighbor of x3. Otherwise let x1 be any vertex outside fx2; x3; x4g.


Otherwise, we may assume that v0 has degree 1. Let x4 ¼ v0 and x3 ¼ v. If

dðx3Þ ¼ 3, then let x2 and x1 be its other neighbors. If dðx3Þ ¼ 2, then let x2 be its

other neighbor, and let x1 be any vertex outside fx2; x3; x4g. If dðx3Þ ¼ 1, then let

x1 and x2 be either two other adjacent vertices or two isolated vertices.

In each case, S ¼ ðx1; x2; x3; x4Þ has the desired properties. &

We now investigate the local structure of certain graphs which include any

possible counterexamples of minimum order. (See Fig. 2.)

Lemma 2.3. Let G be a graph with �ðGÞ ¼ 3, and let L be a 4-uniform list

assignment for G such that G is not equitably L-colorable. Suppose that for some

v 2 VðGÞ of degree 3, the graph G� N½v� has an equitable L-coloring f . Under

these conditions, each v0 2 NðvÞ has two neighbors u; u0 that are not in N½v�, suchthat f ðuÞ and f ðu0Þ are distinct colors in Lðv0Þ, and Lf ðv0Þ has size 2. Moreover, if

NðvÞ ¼ fv1; v2; v3g, then Lf ðv1Þ ¼ Lf ðv2Þ ¼ Lf ðv3Þ.Proof. Given v and f , we will attempt to extend f to an equitable L-coloring

of G using variations of Lemma 1.1—success gives us a contradiction.

Note that Lf ðviÞ ¼ LðviÞ � ff ðuÞ: u 2 NðviÞ � N½v�g for i 2 f1; 2; 3g. Since

jNðviÞ � N½v�j � 2; jLf ðviÞj � 4 � 2 ¼ 2 for i 2 f1; 2; 3g.

If there exists a color c in Lf ðv3Þ � Lf ðv2Þ or jLf ðv2Þj � 3 and c 2 Lf ðv3Þ, then

we can extend f to G as follows: choose f ðv1Þ 2 Lf ðv1Þ � c. Choose f ðv2Þ 2Lf ðv2Þ � f ðv1Þ � c. Let f ðv3Þ ¼ c. Choose f ðvÞ 2 LðvÞ � ff ðv1Þ; f ðv2Þ; f ðv3Þg.

Therefore, we may now assume that every Lf ðviÞ is the same set of size 2.

Let Lf ðviÞ ¼ fa; bg for i 2 f1; 2; 3g. Since jLðviÞ � Lf ðviÞj ¼ 2, each vi must

have two neighbors ui; u0i, besides v, such that LðviÞ ¼ fa; b; f ðuiÞ; f ðu0iÞg, for

i 2 f1; 2; 3g. &

Our plan is to show that for a counterexample G and some v 2 VðGÞ with

neighborhood fv1; v2; v3g, we can find two equitable L-colorings f and f 0 of

G� N½v� such that if every Lf ðviÞ is the same set of size 2, then the sets Lf 0 ðviÞcannot all be the same set of size 2. Then we will apply Lemma 2.3 to show

that one of these equitable L-colorings must extend to G, which will be a

contradiction.

Theorem 2.1. If �ðGÞ � 3, then G is equitably 4-choosable.

FIGURE 2. An equitable L-coloring of G�N [v] that cannot be extended to G.


Proof. By Theorem 1.1 we may assume that �ðGÞ ¼ 3. Since the class of

graphs with maximum degree at most 3 is hereditary, we may choose a minimal

counterexample G in the sense that any induced subgraph of G is equitably

4-choosable. Let L be a 4-uniform list assignment such that G has no equitable

L-coloring.

Suppose that G has an isolated vertex v. Let f be an equitable L-coloring of

G� v. Since nðG� vÞ < nðGÞ � 4dnðGÞ4e, there are fewer than four color classes

with dnðGÞ4e or more elements. Therefore some color in LðvÞ is available, and f

extends to an equitable L-coloring of G. Since this contradicts the choice of G and

L; G has no isolated vertex.

Since every proper induced subgraph of G is equitably L-colorable but G is

not, Lemma 2.2 must not be applicable. Hence if G has a vertex v of degree less

than 3, then dðvÞ ¼ 2, and each of its neighbors has degree 3. (Thus G is not

a forest.) Let NðvÞ ¼ fx2; x3g, and choose x1 2 Nðx3Þ � fv; x2g. Lemma 1.1

applies with S ¼ ðx1; x2; x3; vÞ. This produces a contradiction, so G is 3-regular.

Suppose that G has a triangle. Label its vertices x4; x3; x2, and let x1 be the

other neighbor of x4. Lemma 1.1 applies with S ¼ ðx1; x2; x3; x4Þ. This produces a

contradiction, so G is triangle-free.

Suppose that y1; y2; y3; y4 are the vertices in order along a 4-cycle in G (see

Fig. 3). Let Nðy2Þ ¼ fy1; y3; zg. Let f be an equitable L-coloring of G� N½y2�.Note that Lf ðy1Þ; Lf ðy3Þ; Lf ðzÞ, and Lf ðy4Þ each have size at least 2, since each

of these vertices has at most two colored neighbors. By Lemma 2.3 applied

with v ¼ y2; f ðy4Þ 2 Lðy1Þ \ Lðy3Þ. Color y1 and y3 with f ðy4Þ and recolor y4

with some � 2 Lf ðy4Þ � f ðy4Þ. Since y4 is adjacent to y1; f ðy4Þ 62 Lf ðy1Þ, and then

Lemma 2.3 implies that f ðy4Þ 62 Lf ðzÞ. Thus we can choose a color � from

Lf ðzÞ � f�; f ðy4Þg to color z. Coloring y2 from Lðy2Þ � ff ðy4Þ; �; �g completes

an equitable L-coloring of G. This is a contradiction, so G has no cycles of length

4.

Let C be a shortest cycle in G, and let y1; . . . ; yg be the vertices along C. We

now have g � 5. For 1 � i � g, let zi be the neighbor of yi that is not on C.

We claim that all zi are distinct. Suppose that zi ¼ zj for some i 6¼ j. Since the

longer yi; yj-path in C has length at least 3, we can shorten the cycle by replacing

that path with yiziyj. This contradicts the choice of C, so these vertices are

distinct. (See Fig. 4.)

Let m ¼ bg2c � 1. Thus g ¼ 2mþ 3 if g is odd, and g ¼ 2mþ 2 if g is even.

For 1 � i < m, let Si be the list ðz2i�1; z2i; y2i�1; y2iÞ. Let Sm be the list

FIGURE 3. A 4-cycle in G.


ðz2m; y2m�1; y2mþ1; y2mÞ. For 0 � k � m, let Gk ¼ G�S

kþ 1�i�m Si. Thus Gm ¼ G

and G0 ¼ G� fyi: 1 � i � 2mþ 1g � fzi: 1 � i � 2m� 2 or i ¼ 2mg.

If g is odd, then VðCÞ \ VðG0Þ ¼ fyg�1; ygg. These are adjacent vertices having

degree 2 in G0. By Lemma 2.2, we may let f1 and f2 be equitable L-colorings of

G0 that differ on yg�1, differ on yg, and are the same on all other vertices of G0. If

g is even, then VðCÞ \ VðG0Þ ¼ fygg. This vertex has only one neighbor in G0; it

is zg. By Lemma 2.2, we may let f1 and f2 be equitable L-colorings of G0 that

differ on yg, differ on zg, and are the same on all other vertices of G0. We will

extend f1 and f2 to L-colorings of Gk for 1 � k < m such that:

f1ðvÞ and f2ðvÞ are identical on VðGkÞ � VðCÞ;except that they differ on zg when g is even:

ð2Þ

Observe that (2) holds for f1; f2 on G0. To prove it for the other Gk, we will use

the following:

Claim 2.1. If g is even, then zg is not adjacent (in G) to any vertex ofS1� i�m Si. The only neighbor of zg in C is yg, and yg 62 [1� i�mSi. If zg is adjacent

to zj for some j < g, then the path yjzjzgyg yields a shorter cycle (contradicting the

choice of C) unless both yj; yg-paths along C have length at most 3. In this case,

the length of C is 6, and zj ¼ z3. However, if g ¼ 6, then 3 ¼ 2m� 1, and so

z3 62S

1�i�m Si. The claim is proved.

For 1 � k < m, given equitable L-colorings f1; f2 of Gk�1 that satisfy (2), we

will extend f1 and f2 to L-colorings of Gk satisfying (2) in which Sk gets four

distinct colors. (See Fig. 5.)

FIGURE 4. A shortest cycle with its neighbors, for g¼ 6.

FIGURE 5. The lists Si, with k ¼m� 1:


By Lemma 1.1 we may assign distinct colors �; �; �; � to z2k�1; z2k; y2k�1; y2k

(respectively) that extend f1 to an equitable L-coloring of Gk. Note that

� 2 Lf1ðz2k�1Þ and � 2 Lf1ðz2kÞ. Since f1 � f2 on all neighbors of z2k�1 and z2k in

Gk�1; Lf1ðz2k�1Þ ¼ Lf2ðz2k�1Þ and Lf1ðz2kÞ ¼ Lf2ðz2kÞ. Therefore, � 2 Lf2ðz2k�1Þand � 2 Lf2ðz2kÞ. Now since y2k�1 has only one neighbor in Gk�1, we may choose

�0 2 Lf2ðy2k�1Þ � f�; �g. Since all neighbors of y2k are in Sk, we can then choose

�0 2 Lf2ðy2kÞ � f�; �; �0g. Note that �; �; �0; �0 extends f2 as desired.

This produces two L-colorings of Gm�1 that differ on y2mþ2 and y2mþ3 if g is

odd, differ on y2mþ2 and z2mþ2 if g is even, and otherwise may only differ on

VðCÞ. The set of uncolored vertices is fz2m; y2m�1; y2mþ1; y2mg. The neighbors of

z2m in Gm�1 are colored the same way by f1 and f2, so Lf1ðz2mÞ ¼ Lf2ðz2mÞ. Exactly

one neighbor of y2mþ1 in Gm�1 is colored differently by f1 and f2 (namely y2mþ2),

so Lf1ðy2mþ1Þ 6¼ Lf2ðy2mþ1Þ. Therefore, if Lf1ðz2mÞ ¼ Lf1ðy2mþ1Þ, then Lf2ðz2mÞ 6¼Lf2ðy2mþ1Þ. By Lemma 2.3, at least one of these colorings extends to G, a

contradiction. &

ACKNOWLEDGMENTS

The author thank Professor Douglas B. West for greatly improving the exposition

of this paper, and Professor Alexandr V. Kostochka for simplifying the proof.

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[2] A. V. Kostochka, M. J. Pelsmajer, and D. B. West, A list analogue of equitable

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equitable list-coloring for graphs of maximum degree 3

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