equitable list coloring and treewidth

Equitable List Coloringand Treewidth

Michael J. Pelsmajer

DEPARTMENT OF APPLIED MATHEMATICSILLINOIS INSTITUTE OF TECHNOLOGY

CHICAGO, ILLINOIS 60616E-mail: [email protected]

Received July 5, 2006; Revised September 15, 2008

Published online 6 April 2009 in Wiley InterScience (www.interscience.wiley.com).DOI 10.1002/jgt.20373

Abstract: Given lists of available colors assigned to the vertices of a graphG, a list coloring is a proper coloring ofG such that the color on each vertex ischosen from its list. If the lists all have size k, then a list coloring is equitableif each color appears on at most �|V(G)|/k� vertices. A graph is equitablyk-choosable if such a coloring exists whenever the lists all have size k.Kostochka, Pelsmajer, and West introduced this notion and conjecturedthat G is equitably k-choosable for k>�(G). We prove this for graphs oftreewidth w≤5 if also k≥3w−1. We also show that if G has treewidthw≥5, then G is equitably k-choosable for k≥max{�(G)+w−4,3w−1}. Asa corollary, if G is chordal, then G is equitably k-choosable for k≥3�(G)−4when �(G)>2. � 2009 Wiley Periodicals, Inc. J Graph Theory 61: 127–139, 2009

Keywords: graph coloring; equitable coloring; list coloring

1. INTRODUCTION

In many applications of graph coloring, it is desirable for the color classes to not bevery large. When scheduling jobs, for example, the number of jobs that can be run

Journal of Graph Theory� 2009 Wiley Periodicals, Inc.

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128 JOURNAL OF GRAPH THEORY

at the same time may be limited by the number of processors (see [12] for such anapplication). Equitable coloring is a well-known property (see [8] for a survey) thatrestricts the sizes of each color class. A proper vertex coloring of a graph is equitableif the sizes of the color classes differ by at most 1. For an equitable coloring of a graphG that uses k colors (an equitable k-coloring), each color class has at most �n(G)/k�vertices, where n(G) denotes the number of vertices of G.

Kostochka et al. introduced the list analogue of equitable coloring in [7]. A listassignment L for a graph G is a function that assigns to each vertex v∈V (G) a setL(v) of acceptable colors. An L-coloring of G is a proper vertex coloring such that forevery v∈V (G) the color on v belongs to L(v). A list assignment L for G is k-uniformif |L(v)|=k for all v∈V (G).

Given a k-uniform list assignment L for a graph G, we say that G is equitablyL-colorable if G has an L-coloring such that each color appears on at most �n(G)/k�vertices. A graph G is equitably list k-colorable or equitably k-choosable if G isequitably L-colorable whenever L is a k-uniform list assignment for G.

Hajnál and Szemerédi [6] proved that a graph G has an equitable k-coloring wheneverk>�(G), where �(G) denotes the maximum vertex degree in G; this answered aquestion of P. Erdos. Kostochka et al. [7] conjectured the list analogue of the Hajnál–Szemerédi Theorem [6]:

Conjecture 1. Every graph G is equitably k-choosable whenever k>�(G).

In this paper we consider equitable k-choosability for graphs of bounded treewidth.Treewidth is a well-known graph parameter (see [2] for a survey), in part because manyhard problems become theoretically feasible when restricted to graphs of boundedtreewidth. Forests are the graphs of treewidth at most 1; in [7] it was shown that forestsare equitably k-choosable for k≥1+�/2, where � is the maximum degree. It followseasily from Theorem 4 of [7] that graphs of treewidth 2 are equitably k-choosable fork≥max{�,5}. (Thus Conjecture 1 is true for all graphs of treewidth at most 2, sinceConjecture 1 has been verified for graphs of maximum degree at most 3 [9, 11].) Weprove similar results for graphs of arbitrary treewidth.

Theorem 2. Suppose that G is a graph of treewidth w and k≥3w−1. If w≤5 andk≥�(G)+1, then G is equitably k-choosable. If w≥5 and k≥�(G)+w−4, then Gis equitably k-choosable.

Note that for graphs G of small treewidth and relatively large maximum degree, thecondition k≥3w−1 is superfluous (when �(G)+1≥3w−1) and the bound �(G)+w−4 is close to Conjecture 1.

Conjecture 1 is known for interval graphs [7], so it is natural to ask whether someupper bound can be obtained for chordal graphs (see [13] for an introduction to thesegraph classes). Every graph G is equitably k-choosable for k≥�(G)(�(G)−1)/2+2[11]. For chordal graphs, Theorem 2 improves this quadratic bound to a linearbound.

Corollary 3. Let G be a chordal graph with maximum degree at most �. Then G isequitably k-choosable for k≥max{3�−4,�+1}.

Journal of Graph Theory DOI 10.1002/jgt

EQUITABLE LIST COLORING 129

Conjecture 1 has been proved for several classes of graphs in [7, 9–11], often usingan inductive approach which we again describe here. We use N (v) for {u : uv∈E(G)}and N [v] for N (v)∪{v}, and we say that the vertices of N (v) are the neighbors of v.

Consider a graph G with a k-uniform list assignment L , and let S be a set of kvertices in G. If G−S has an equitable L-coloring f that extends to an L-coloring f ′of G by giving the vertices in S different colors, then f ′ is an equitable L-coloringof G, since the extension augments each color class at most once. We give a simplesufficient condition for the existence of such an extension.

Lemma 4 (Kostochka et al. [7] and Pelsmajer [11]). Let G be a graph with a k-uniform list assignment L. Let S={x1, . . . , xk} be a set of k vertices in G, and supposethat G−S has an equitable L-coloring. If

|N (xi )−S|≤k−i (1)

for 1≤i≤k, then G has an equitable L-coloring such that k distinct colors are usedon S.

Proof. The vertices can be colored in order, such that the color of xi is distinct fromthe colors on (N (xi )−S)∪{x1, . . . , xi−1}. �

This will be our basic tool. In Section 2 we show how treewidth is used with Lemma 4by proving a weaker version of Theorem 2, and we use this to obtain Corollary 3. InSection 3 we refine this method to prove Theorem 2, including technical lemmas thatare potentially useful for strengthening Theorem 2 further. We finish by discussingthese possibilities and also whether these list colorings can be found quickly.

2. USING TREE DECOMPOSITIONS WITH LEMMA 4

We begin by giving some of the standard terminology and basic results on tree decom-positions. (For a more thorough introduction to tree decompositions, see the survey ofBodlaender [2] or Diestel [3].)

Definition 5. Let G be a graph, let T be a tree, and let W={Wt : t ∈V (T )} be afamily of subsets of the vertices of G indexed by the vertices of T . The pair (T,W) isa tree decomposition of G if it satisfies the following conditions:

1. every vertex of G is in some Wt ;2. every edge in G belongs to G[Wt ] for some t ∈T ;3. for every u∈V (G), the set T (u) induces a subtree of T , where T (u)={t ∈

V (T ) : u∈Wt }.The width of a tree decomposition is the maximum of |Wt |−1 over t ∈V (T ), and thetreewidth of G is the minimum width among all tree decompositions of G.

Every non-trivial tree has treewidth 1. In general, the smaller the size of the sets Wt ,the more closely G resembles a tree. The treewidth is thus a measure of how tree-like thegraph is. An important feature of a tree decomposition is that it transfers the separationproperties of its tree to the graph. We summarize some of these basic properties.



Remark 6. Let (T,W) be a tree decomposition of a graph G.

1. Definition 5(3) is equivalent to the following statement: If t2 is on the path fromt1 to t3 in T , then Wt1 ∩Wt3 ⊆Wt2 .

2. If t1 and t2 are adjacent vertices in T , then G−(Wt1 ∩Wt2 ) is the disjoint unionof two subgraphs; one subgraph is induced by

⋃x∈V (T1)Wx −Wt2 and the other is

induced by⋃

x∈V (T2)Wx −Wt1 , where T1 and T2 are the components of T − t1t2that contain t1 and t2, respectively.

3. If t is a vertex of T , and T1, . . . ,Tk are the components of T − t , then G−Wt isthe disjoint union of the k subgraphs G[Ui ], where Ui =

⋃x∈V (Ti )Wx −Wt .

A graph is chordal if it has no induced cycles of length greater than three. Whilechordal graphs have many interesting characterizations, we only need the fact that thetreewidth of a chordal graph plus 1 equals the maximum size of a clique in the graph(see [3, p. 326]).

A class of graphs G is hereditary if every induced subgraph of a graph G∈G is alsoin G. Thus, the class of chordal graphs is hereditary. Graphs of treewidth at most w

form a hereditary class of graphs, since a tree decomposition of a graph easily yieldsa valid tree decomposition for any of its subgraphs. Since graphs of bounded degree,intersections of hereditary classes of graphs, and chordal graphs are all hereditaryclasses of graphs, each of our results can be rephrased in the form, “for a certainhereditary class of graphs G and integer kG , each graph G∈G is equitably k-choosablefor all k≥kG .” Eroh [4] has shown that to prove such a result it suffices to showthat each G∈G is equitably kG-choosable; however, we do not make use of her workin this paper.

A graph G is w-degenerate if every subgraph of G contains a vertex of degree atmost w. It is not hard to show that a graph of treewidth at most w is w-degenerate(see [2, 3]). Clearly, w-degeneracy is potentially useful in conjunction with Lemma 4.With some additional work, this allowed Conjecture 1 to be verified for 2-degenerategraphs [7].

To apply Lemma 4 with fixed k, we need to construct a set of size k with fewneighbors outside of the set. The next lemma encapsulates everything else that wewill use from the theory of tree decompositions to construct the desired set. We usethe following notation: for a set of vertices S in a graph G, let N (S)={v∈V (G)−S : N (v)∩S �=∅} and N [S]= S∪N (S).

Lemma 7. Let G be a graph of treewidth at most w. Suppose that n(G)≥3w−1and w≥2. Then there exists U⊂V (G) such that |N (U )|≤w+1 and w≤|U |≤2(w−1).Moreover, if |N (U )|=w+1 then no vertex of U is adjacent to all of the verticesin N (U ).

Proof. Let (T,W) be a tree decomposition of width at most w. We may assumethat Ws �=Wt for each edge st ∈E(T ) (see [3, #13, p. 351] or use a “k-alternating treedecomposition” [5]). For each edge e=st of T , let Tst and Tts be the components ofT −e such that s∈V (Tst ) and t ∈V (Tts). Let Ust =

⋃x∈V (Tst )Wx −Wt and let Uts =⋃

x∈V (Tts )Wx −Ws . We will construct the desired set U such that N (U )⊆Wt for somet ∈V (T ).



Orient each edge st of T from s to t if |Uts |≤w−1. Let D be the set of directededges. If an edge st were oriented in both directions, then since V (G) is the disjointunion of Ust , Uts , and Ws∩Wt (by Remark 6(2)) and Wt �=Ws , we would have n(G)≤2(w−1)+w<3w−1. This contradiction shows that every edge of T is oriented in atmost one direction.

Claim. If rs∈D (i.e., rs is oriented from r to s) and st ∈E(T ), then st ∈D. SinceUsr =⋃

x∈V (Tsr )Wx −Wr and Tts ⊂Tsr , Usr contains⋃

x∈V (Tts )Wx −Wr , which equalsUts−(Wr−Ws). The setWr−Ws is contained inUst since r∈V (Tst ), so it does not inter-sect Uts by Remark 6(2). Therefore Uts⊆Usr , yielding |Uts |≤|Usr |≤w−1 and st∈D.

From the claim it follows that an endpoint of an undirected edge has no in-neighbors,and that a path between vertices of in-degree zero can contain only undirected edgesand vertices of in-degree zero. Also, there must be at least one vertex of in-degree zerobecause the sum of in-degrees equals |D| and |D|≤|E(T )|< |V (T )|. It follows thatthe vertices of in-degree zero and the undirected edges of T form a subtree T ′.

Let r be a leaf of T ′ if |V (T ′)|>1; otherwise, let r be the unique vertex of T ′. Lett1, . . . , tl be the out-neighbors of r . Let j be maximized such that

∑ ji=1 |Utir |≤w−1.

If j < l, let U =⋃ j+1i=1 Utir . Then |U |≥w and |U |≤ (w−1)+|Ut j+1r |. Since r t j+1∈D,

we have |Ut j+1r |≤w−1, giving us |U |≤2(w−1). According to Remark 6(3) (witht=r and Ui =Utir for i≤ l), N (Utir )⊆Wr for each i , so N (U )⊆Wr .

Otherwise j = l. Thus |⋃li=1Utir |≤w−1. If r is a leaf in T ′, let s be its neighbor

in T ′. Then⋃l

i=1Utir =Urs−Wr , and because sr �∈D we have |Urs |≥w. Hence wemay define U such that |U |=w and Urs−Wr ⊂U ⊆Urs . Also N (U )⊆Wr ∪N (Urs),so N (U )⊆Wr . If r is not a leaf then V (T ′)={r} and every neighbor of r in T is an out-neighbor, so

⋃li=1Utir =V (G)−Wr . However, since |Wr |≤w+1 we get n(G)≤2w,

contradicting n(G)≥3w−1 and w≥2.Since N (U )⊆Wr , |N (U )|≤w+1. Suppose that |N (U )|=w+1 and that some vertex

u∈U is adjacent to every vertex in N (U ). Observe that (T,W) is a valid tree decom-position for the graph formed by the union of G with a complete graph on Wr . Inthe new graph N (U )∪{u} is a clique of size w+2, which cannot exist in a graph oftreewidth at most w (see [3, p. 321], for example). �

The conclusion of Lemma 7 is still true if w≤n(G)<3w−1: simply let U be anyvertex set of size max{n(G)−w,w}. However, this is not useful to us. The condition|U |≥w is necessary because of the way that Lemma 7 will be applied. The otherbounds are best possible, as shown by the following construction.

For a path P , let Pw denote the wth power of P , meaning that vertices at distanceat most w in P are adjacent in Pw. Let P∗ =v∗

1 . . .v∗n be a path with n≥w+2, and

for j =1,2,3,4 let Pj be the path vj1 . . .v

j2w−1 and let S j ={v j

1 , . . . ,vjw}. Let G be

the union of Pw∗ , Pw1 , Pw

2 , Pw3 , Pw

4 with S1={v∗1 , . . . ,v

∗w}, S2={v∗

1 , . . . ,v∗w−1,v

∗w+1},

S3={v∗n−w+1, . . . ,v

∗n}, and S4={v∗

n−w,v∗n−w+2, . . . ,v

∗n}.

Let U be as in Lemma 7. N (U ) is a cut-set since n(G)−|N [U ]|≥ (5w−2)−(3w−1)>0. To separate two vertices on P∗ (or Pj for some j) a cut-set must containw consecutive vertices along P∗ (or Pj ). With this, it is not hard to check that the



N (U ) must be either S1∪S2 or S3∪S4, with U =V (Pw1 )∪V (Pw

2 )−V (Pw∗ ) or V (Pw3 )∪

V (Pw4 )−V (Pw∗ ). Then |N (U )|=w+1 and |U |=2(w−1).

For a tree decomposition of G of width w, let Wi∗ ={v∗i , . . . ,v

∗i+w} for 1≤ i≤n−w,

and for each j =1,2,3,4 let Wi j ={v ji , . . . ,v

ji+w} for 1≤ i≤w−1. The tree is formed

from the path 1∗ . . . (n−w)∗ and the paths 1 j . . . (w−1) j (for 1≤ j ≤4), and the edges111∗, 121∗, 13(n−w)∗, and 14(n−w)∗.

To demonstrate how Lemma 7 can be used together with Lemma 4, we first provea result weaker than Theorem 2, with a much simpler proof.

Proposition 8. Suppose G has treewidth w≥2. If k≥�(G)+w and k≥3w−1 thenG is equitably k-choosable.

Proof. Fix w≥2 and k≥3w−1; note that graphs G with treewidth at most w suchthat k≥�(G)+w form a hereditary class of graphs. Suppose that G is a smallest graphin this class that is not equitably k-choosable. Then G is a “minimal counterexample”in the sense that all of its induced subgraphs are equitably k-choosable. Let L be a k-uniform list assignment such that G has no equitable L-coloring. If n(G)≤k then colorall vertices with different colors from their lists. Thus we may assume that n(G)>k.

Apply Lemma 7 to obtain U . Since k−|U |+1≥k−2(w−1)+1≥w+2> |N (U )|,we may let N (U )={x1, . . . , x|N (U )|} and U ={xk−|U |+1, . . . , xk}. Since G is w-degenerate, we may repeatedly select and remove vertices from G−N [U ] with degreeat most w; we select xi for all |N (U )|< i<k−|U |+1 in this way. Let S={x1, . . . , xk}.

Now check that |N (xi )−S|≤k−i for 1≤ i≤k: For i≥k−|U |+1, it is satisfiedbecause N [U ]⊆ S. Since each element of N (U ) has a neighbor in U and hence inS, it is satisfied for each i≤|N (U )| if �(G)−1≤k−|N (U )|; this is true because|N (U )|≤w+1 and k≥�(G)+w. For |N (U )|< i≤k−|U |, the choice of xi implies|N (xi )−S|≤w, which suffices since w≤|U |≤k−i .

G−S is equitably L-colorable by minimality and the L-coloring extends to G byLemma 4, so we have a contradiction. �

Proposition 8 is strong enough to prove our result on chordal graphs.

Corollary 3. Let G be a chordal graph with maximum degree at most �. Then G isequitably k-choosable for k≥max{3�−4,�+1}.Proof. Fix k and � such that k≥max{3�−4,�+1}. The chordal graphs G such

that �(G)≤� form a hereditary class of graphs, so we may choose a minimal coun-terexample G in the sense that any induced subgraph of G is equitably k-choosable. LetL be a k-uniform list assignment such that G has no equitable L-coloring. If n(G)≤kthen color all vertices with different colors from their lists. Now assume that n(G)>k.

If G contains a (�+1)-clique, then it must induce a component of G. Labelits vertices {xk−�, . . . , xk}, and let x1, . . . , xk−�−1 be any other vertices. Then S={x1, . . . , xk} satisfies the condition in Lemma 4. G−S is equitably L-colorable by mini-mality, so the L-coloring extends to G, which is a contradiction. Now assume that Ghas no clique of size greater than �.

Let w be the treewidth of G. If w≤1 then G is a forest, so G is equitably k-choosableby Theorem 1.2 in [7]. Suppose that w≥2. Since G must contain a (w+1)-clique,



w≤�−1. Then 3�−4≥3w−1 and 3�−4≥�+2(w+1)−4≥�+w, so Proposition 8implies that G is equitably k-choosable. �

While max{3�−4,�+1}=3�−4 for �≥3, the �+1 is needed for �≤2 sincethere are graphs (odd cycles, K�+1) that are clearly not equitably �-choosable, sincethey are not even �-colorable.

3. REFINED METHOD FOR STRONGER RESULTS

We would like to improve k≥�+w to k≥�+1. The following technical lemma allowsus to apply Lemma 4 to more situations than Proposition 8, and we will use it later topush k≥�+w down to k≥�+1 for w≤5 and k≥�+w−4 for w≥5.

Lemma 9. Fix h≥0 and k=�+h with k≥3w−1. Let G be a w-degenerate graphwith �(G)≤� and let U ⊆V (G) such that w≤|U |≤2(w−1) and |N (U )|≤w+1.

If |N (U )|≤h+1 or if there are sets Z ⊆N (U ), {z j :2≤ j ≤|Z |−h}⊆Z , and {u j :0≤j < |N (U )−Z |}⊆U such that (i) |N (z j )∩(Z∪U )|≥ j and (ii) |N (u j )∩(N (U )−Z )|≤ j, then there exists S={x1, . . . , xk}⊆V (G) such that each |N (xi )−S|≤k−i .

Proof. We assign k distinct vertices to S as follows:

Case 1. |N (U )|≤h+1.

Let N (U )={x1, . . . , x|N (U )|} and U ={xk−|U |+1, . . . , xk} (which is okay because|N (U )|+|U |≤3w−1≤k). By w-degeneracy we may choose vertices xi from G−N [U ] for |N (U )|< i≤k−|U | such that |N (xi )−S|≤w; these xi are as desired becausek−i≥|U |≥w. For i≤|N (U )|, N (xi )∩U �=∅ and U ⊆ S, so |N (xi )−S|≤�−1=k−h−1≤k−|N (U )|≤k−i . For i>k−|U |, xi ∈U so N (xi )⊆N [U ], so |N (xi )−S|=0≤k−i .

Case 2. |N (U )|>h+1.

If |Z |≤h, then add vertices from N (U )−Z to Z until |Z |=h+1; note that (i) and(ii) are still satisfied. Thus, we may assume that |Z |≥h+1.

Let {x1, . . . , xh+1}= Z−{z j : 2≤ j ≤|Z |−h}, and let xi = zi−h for h+2≤ i≤|Z |.Since k−|U |≥3w−1−2(w−1)≥|N (U )|≥|Z | and |N (U )−Z |≤w≤|U |, we maylet {xi : k−|U |< i≤k−|N (U )−Z |}=U−{u j : 0≤ j< |N (U )−Z |}. For k−|N (U )−Z |< i≤k, let xi =uk−i . At this point, S contains Z∪U . By w-degeneracy we maychoose vertices xi from G−Z∪U for |Z |< i≤k−|U | such that |N (xi )−S|≤w; thesexi are as desired because k−i≥|U |≥w.

For each xi ∈ Z , |N (xi )−S|≤�−|N (xi )∩(Z∪U )|=k−(|N (xi )∩(Z∪U )|+h), andtherefore xi is as desired if |N (xi )∩(Z∪U )|≥ i−h. Since N (xi )∩U �=∅, this holds fori≤h+1. Otherwise h+2≤ i≤|Z |, so xi=zi−h . Then |N (xi )∩(Z∪U )|≥ i−h by (i).

For each xi ∈U , N (xi )−S⊆N (xi )∩(N (U )−Z ). Hence |N (xi )−S|≤k−i for k−i≥|N (U )−Z |. For k−i< |N (U )−Z |, xi =uk−i , so |N (xi )−S|≤k−i by (ii). �

Lemma 9 can always be applied if there are not too many edges from U to N (U ).The following technical lemma makes this precise.



Lemma 10. Suppose that h,k,�,w,G, and U are specified as in Lemma 9. Let �be the number of edges from U to N (U ).

If �≤w or |N (U )|≤ (�(h+1)−1)/(�−w), then Lemma 9 may be applied.If �≤h+w+1, then Lemma 9 applies unless h=0 and |N (U )|=w+1=�.

Proof. Assume that |N (U )|≥h+2, since otherwise Lemma 9 applies. It will sufficeto specify that N (U )−Z ⊆N (U ) and vertices u j that satisfy (ii) of Lemma 9 suchthat |N (U )−Z |=|N (U )|−h−1: then Z is defined implicitly with |Z |=h+1, so (i)is vacuously true. Note that |N (U )|−h−1≤w≤|U |.

Select the elements of {ui : 0≤ i< |N (U )|−h−1} in order: at the i th step, choose ui ∈U−{u j : 0≤ j< i} such that |N (ui )∩N (U )| is minimized. Then for 0<r ≤|N (U )|−h−1 we have |⋃0≤i<r N (ui )∩N (U )|≤∑

0≤i<r |N (ui )∩N (U )|≤��r/|U |�≤��r/w�,so |N (U )−⋃

0≤i<r N (ui )|≥|N (U )|−��r/w�. Briefly consider two cases separately.

Case 1. �≤w.

Then |N (U )−⋃0≤i<r N (ui )|≥|N (U )|−r . When r =|N (U )|−h−1, |N (U )|−r =

h+1≥1. Whenever r decreases by 1, |N (U )|−r increases by 1.

Case 2. �>w and |N (U )|≤ (�(h+1)−1)/(�−w).

Then for r =|N (U )|−h−1, |N (U )|−��r/w�=�(�/w)(h+1)−|N (U )|((�/w)−1)�≥�(�/w)(h+1)−(�(h+1)−1)/w�=�(1/w)�=1. Also, (|N (U )|−��(r−1)/w�)−(|N (U )|−��r/w�)=�(�r/w)�−�(�r/w)−(�/w)�≥1. Therefore |N (U )|−��r/w�≥ jfor r =|N (U )|−h− j with 1≤ j ≤|N (U )|−h−1.

In both cases |N (U )−⋃0≤i<r N (ui )|≥1 for r =|N (U )|−h−1 and, moreover,

|N (U )−⋃0≤i<r N (ui )|≥ j for 1≤ j ≤|N (U )|−h−1 and r =|N (U )|−h− j . There-

fore, as j advances from 1 to |N (U )|−h−1, we may select distinct verticesy j ∈N (U )−⋃

0≤i<r N (ui ), where r =|N (U )|−h− j . Let N (U )−Z ={y j : 1≤ j ≤|N (U )|−h−1}. Since each y j is not adjacent to any ui with 0≤ i< |N (U )|−h− j ,each ui is not adjacent to any y j for which j < |N (U )|−h−i . Therefore ui has atmost i neighbors in N (U )−Z , so (ii) is satisfied and Lemma 9 applies.

Finally, suppose that �≤h+w+1. Assume that �≥w+1, since otherwise thisreduces to Case 1. Then since |N (U )|≤�= (�(h+1)−1)/(�−w)−(�(h+1−�+w)−1)/(�−w), Case 2 applies unless �=h+w+1 and |N (U )|=�. However, in that casewe have w+1≥|N (U )|=h+w+1, which implies that h=0. �

In this paper we never apply Lemma 10 with h=0, but we chose to present it in aslightly stronger form because of the potential applications to future work.

Lemma 10 gathers together many cases where Lemma 9 can be applied with (i)satisfied vacuously. When there are many edges from U to N (U ), (i) can be satisfiedwith a larger set Z , which makes it easier to satisfy (ii). This is the basic idea whichallows us to finish the proof of the main theorem, although there is a fair amount ofdetail to work through. We begin with some small cases.

Lemma 11. If G has treewidth w∈{3,4}, k≥�(G)+1, and k≥3w−1, thenLemma 9 applies with h=1.



Proof. Use Lemma 7 to obtain U and let � be the number of edges from U toN (U ). Then |N (U )|≤w+1, w≤|U |≤2(w−1), and by Lemma 10 with h=1 we mayassume that �≥w+3.

Case 1. There are distinct z2, z3∈N (U ) such that z j has at least j neighbors in N [U ]for j =2,3.

If we let Z =N (U ) then (ii) is satisfied vacuously, and unless |Z |−1≥4, (i) is alsosatisfied. Thus we may assume that |N (U )|=5, which implies that w=4.

If z3 has a neighbor in N (U ) we may assume that it is z2, since each vertex of N (U )has a neighbor in U . It follows that we can select z∈N (U ) that avoids z2, z3, twoneighbors of z2, and three neighbors of z3. If z has at least four neighbors in N [U ],then (i) and (ii) are satisfied with Z =N (U ) and z4= z. Otherwise we can chooseu0∈U−N (z). Then letting Z =N (U )−{z} satisfies (i) and (ii).

Case 2. Either there is no vertex z3 in N (U ) with three neighbors in N [U ], or there issuch a vertex but there is no second vertex that can be z2. Hence �≤max{2|N (U )|, |U |+(|N (U )|−1)}.

Since �> |N (U )| we may let z2∈N (U ) be a vertex with at least two neighborsin U . Choose z∈N (U )−{z2}; then |N (z)|<3. Since |U |≥w≥3, we can choose u0∈U−N (z). Then letting Z =N (U )−{z}, (i) and (ii) are satisfied unless |N (U )|≥5.Therefore, assume that |N (U )|=5 and w=4.

First, suppose that there is a vertex z∈N (U ) with exactly one neighbor in N [U ] (inU ). Then z �= z2. Choose z′ ∈N (U )−{z, z2}. Since this is not Case 1, z′ has at most 2neighbors in N [U ]. Then because |U |≥w≥4, we may choose u0∈U−N (z)−N (z′)and u1∈U−N (z)−u0. Let Z =N (U )−{z, z′}. Then (i) is satisfied, and since z, z′ �∈N (u0) and z �∈N (u1), (ii) is satisfied.Otherwise every vertex of N (U ) has exactly 2 neighbors in N [U ]. Since �/|U |≤

104 =2.5, we can choose distinct u0,u1∈U such that |N (u0)∩N (U )|≤2 and |N (u0)∩N (U )|+|N (u1)∩N (U )|≤4. Then we can choose z∈N (U )−N (u0)−N (u1) and z′ ∈N (U )−N (u0)−z. Since z and z′ each has at most 1 neighbor in N (U ), we canalso choose z2∈N (U )−N [z]−N [z′]. Let Z =N (U )−{z, z′}; then (i) and (ii) aresatisfied. �

Lemma 12. Suppose G has treewidth w≥5. If k≥�(G)+w−4 and k≥3w−1 thenLemma 9 applies with h=w−4.

Proof. Get U from Lemma 7. By Lemma 9 we may assume that |N (U )|≥w−2,since h+1=w−3. By Lemma 10 we may assume that �≥2w−2, where � is thenumber of edges from U to N (U ). Since |N (U )|≤w+1<2w−2, some vertex inN (U ) has at least two neighbors in U . Note that |U |≥w≥5.

Case 1. |N (U )|=w+1 and there is a set {z2, z3, z4, z5}⊆N (U ) such that each z j hasat least j neighbors in N [U ], or

|N (U )|=w and there is a set {z2, z3, z4}⊆N (U ) such that each z j has at least jneighbors in N [U ], or



|N (U )|=w−1 and there is a set {z2, z3}⊆N (U ) such that each z j has at least jneighbors in N [U ], or

|N (U )|=w−2.

When |N (U )|=w−2, we may select z2∈N (U ) with at least 2 neighbors in U .Let Z =N (U ). Then (i) is satisfied. Since |N (U )−Z |=0, (ii) is satisfied vacuously.

Case 2. Not Case 1 and:|N (U )|=w+1 and there is a set {z2, z3, z4}⊆N (U ) such that each z j has at least

j neighbors in N [U ], or|N (U )|=w and there is a set {z2, z3}⊆N (U ) such that z j has at least j neighbors

in N [U ], or|N (U )|=w−1 (and some z2∈N (U ) has at least 2 neighbors in U ).

Suppose that we are able to choose another vertex z∈N (U ) such that |N (z2)∩N [U ]−{z}|≥2, |N (z3)∩N [U ]−{z}|≥3 if |N (U )|≥w, and |N (z4)∩N [U ]−{z}|≥4if |N (U )|=w+1. If also N (z)⊇U then this reduces to Case 1, because |U |≥5.Otherwise, let u0∈U−N (z) and let Z =N (U )−{z}. Then (i) and (ii) are satisfied.Now it suffices to find z as desired.

If |N (U )|=w−1, then since z2 has two neighbors inU , any z∈N (U )−{z2} suffices.Suppose that |N (U )|=w. If z3 has a neighbor in N (U ), we may let it to be z2 sinceevery vertex of N (U ) has a neighbor in U . Then, since |N (U )|>4 we can choose avertex in N (U ) to avoid z3 and 3 of its neighbors including z2; this defines z as desired.Otherwise z3 only has neighbors in U . Then we may choose z∈N (U ) to avoid z2, twoof its neighbors, and z3; this suffices.

Now suppose that |N (U )|=w+1. If a vertex of N (U ) has at least 3 neighbors inU , then we may assume that it is z3 or z4. In this case, since z3 and z4 each has atleast 1 neighbor in U and |N (U )|>5, we can choose z∈N (U ) to avoid z3 and 3 of itsneighbors and z4 and 4 of its neighbors. If z3 or z4 has a neighbor in N (U )−{z3, z4, z},we may assume that it is z2. This suffices. Otherwise, z can be chosen to avoid z2, z3, z4,and two neighbors of z2. This suffices.

Otherwise, each vertex of N (U ) has at most 2 neighbors in U . Then z3 has aneighbor in N (U ); assume that it is z2 (even if it is z4). Since |N (U )|−3≥2, we maylet z, z′ be distinct vertices in N (U ) that avoid z3 and 3 neighbors of z3 includingz2. Since |N (z)∩U |+|N (z′)∩U |≤4< |U |, we can choose u0∈U−N (z)−N (z′) andu1∈U−N (z)−u0. Let Z =N (U )−{z, z′}. Then |Z |−h=3 so (i) is satisfied. Sincez, z′ �∈N (u0) and z �∈N (u1), (ii) is satisfied.

Case 3. Not Case 1 or 2, |N (U )|=w+1, and there is a set {z2, z3}⊆N (U ) such thateach z j has at least j neighbors in N [U ].

Suppose that we find {z, z′}⊆N (U )−{z2, z3} such that z2 has at least 2neighbors in N [U ]−{z, z′}, z3 has at least 3 neighbors in N [U ]−{z, z′}, and|N (z)∩U |+|N (z′)∩U |≤4. Then (i) is satisfied with Z =N (U )−{z, z′}, and since|U |≥5 we can select u0∈U−N (z)−N (z′). We may assume that |N (z)∩U |≤|N (z′)∩U | and choose u1∈U−N (z)−u0; this satisfies (ii). Thus, it will suffice if weobtain such z, z′.



If each vertex of N (U ) has at most two neighbors in U , then N (z3)∩N (U ) isnonempty, and since every vertex of N (U ) has a neighbor in U we may assume thatz2∈N (z3). Then we may choose {z, z′}⊆N (U ) to avoid z3 and 3 of its neighborsincluding z2, because |N (U )|−3≥2. This suffices, so we may assume that some vertexof N (U ) has at least 3 neighbors in U (and that it is z3). If every other vertex of N (U )has at most two neighbors in U , we can choose distinct z, z′ to avoid z3, z2, and 2neighbors of z2, which suffices. Thus, assume that another vertex of N (U )−{z3} hasat least 3 neighbors in U , and that it is z2. If two vertices in N (U ) each has at mosttwo neighbors in U , let them be z, z′; this suffices.

Now we may assume that at most one vertex of N (U ) has less than three neighborsin U . Since this is not Case 2, every other vertex of N (U ) has exactly three neighborsin U (and none in N (U )). Then �≤3|N (U )|, so we can pick u0∈U to have at most3|N (U )|/|U | neighbors in N (U ). Since that is less than 4, we may choose distinctz, z′ ∈N (U )−N (u0). Then we can choose u1∈U−u0−N (z) because |U |≥5. LetZ =N (U )−{z, z′}, which satisfies (ii). We may then choose z3, z2∈ Z to satisfy (i).

Case 4. Not Case 1 or 2 or 3.

Then |N (U )| is w or w+1 and we cannot find z2, z3∈N (U ) such that eachz j has at least j neighbors in N [U ]. Then �≤max{2|N (U )|, |U |+(|N (U )|−1)}≤max{2(w+1),2(w−1)+w}≤3w−2. Select z2∈N (U ) so that |N (z2)∩U | is maxi-mized. Then z2 has at least two neighbors in U , and since this is not Case 2 or 3, everyother vertex of N (U ) has at most 2 neighbors in U . There is a vertex u0∈U with atmost 2 neighbors in N (U ), since otherwise �≥3|U |≥3w. Then we can define Z ⊂N (U )to contain z2 and N (u0)∩N (U ) such that |Z |=w−2, because w−2≥3. Since there areat most 2|N (U )−Z | edges from N (U )−Z toU−{u0} and �2|N (U )−Z |/|U−{u0}|�≤� 64�=1, there exists u1∈U−{u0} with at most one neighbor in N (U )−Z . Let z∈N (U )−Z . Since |U |≥5 we can choose u2∈U−{u0,u1}−N (z). Observe that (i) and(ii) are satisfied. �

The proof is finished in the usual way.

Theorem 2. Suppose that G is a graph of treewidth w and k≥3w−1. If w≤5 andk≥�(G)+1, then G is equitably k-choosable. If w≥5 and k≥�(G)+w−4, then Gis equitably k-choosable.

Proof. The cases w=1,2 follow from old results, as discussed in Section 1. Fixw≥3 and k≥3w−1. If w≤5, the graphs G such that k≥�(G)+1 form a hereditaryclass of graphs; if w≥5, the graphs G such that k≥�(G)+w−4 form a hereditaryclass of graphs. Thus, we may choose a minimal counterexample G in the sense thatany induced subgraph of G is equitably k-choosable. Let L be a k-uniform list assign-ment such that G has no equitable L-coloring. If n(G)≤k then we color all verticeswith different colors from their lists. Suppose now that n(G)>k. Then according toLemmas 11 and 12, Lemma 9 yields a set S={x1, . . . , xk} to which Lemma 4 applies.By minimality, G−S is equitably L-colorable, and the L-coloring extends to G, acontradiction. �



4. REMARKS AND CONCLUSIONS

We outline the algorithmic aspect of this paper. Let w be a constant. Then we can checkwhether an n-vertex graph has treewidth at most w and, if so, find a tree decompositionof width w in O(n)-time [1]. Then it is not too hard to find U and N (U ) as in Lemma 7in O(n)-time. Since |N [U ]| is bounded by the constant 3w−1 and the w-degeneratevertices can be found in O(n)-time using the structure of the tree decomposition, theset S of Lemma 4 can be constructed in O(n)-time. Hence it takes O(n2/k)-time toproduce an equitable L-coloring for a k-uniform list assignment L . However this isnot practical, since the constant factor associated with w is huge.

As far as improving the results here, it seems likely that with more extensive caseanalysis the w≤5 in Theorem 2 could be replaced by w≤6 or w≤7, and the �+w−4could be lowered to �+w−5 or �+w−6 (with w≥6 or w≥7). More interestingwould be to push �+w−4 down to �+o(w), or at least down to �+w� for some�<1, except for small values of w that would push the bound below �+1.

ACKNOWLEDGMENTS

The author would like to thank Douglas B. West and the anonymous referees for theirhelpful comments.

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