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New Bounds on the List-Chromatic Index of the Complete Graph andOther Simple Graphs

ROLAND HÄGGKVIST and JEANNETTE JANSSEN

Combinatorics, Probability and Computing / Volume 6 / Issue 03 / September 1997, pp 295 - 313DOI: null, Published online: 08 September 2000

Link to this article: http://journals.cambridge.org/abstract_S0963548397002927

How to cite this article:ROLAND HÄGGKVIST and JEANNETTE JANSSEN (1997). New Bounds on the List-Chromatic Index of the CompleteGraph and Other Simple Graphs. Combinatorics, Probability and Computing, 6, pp 295-313

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Combinatorics, Probability and Computing (1997) 6, 295–313. Printed in the United Kingdomc© 1997 Cambridge University Press

New Bounds on the List-Chromatic Index of the

Complete Graph and Other Simple Graphs

R O L A N D H A G G K V I S T1 and J E A N N E T T E J A N S S E N2†1Department of Mathematics, University of Umea,

S-901 87 Umea, Sweden

(e-mail: [email protected])2Department of Mathematics, London School of Economics,

Houghton Street, London WC2A 2AE, UK

(e-mail: [email protected])

Received 11 April 1995; revised 16 January 1996

In this paper we show that the list chromatic index of the complete graph Kn is at most n.

This proves the list-chromatic conjecture for complete graphs of odd order. We also prove

the asymptotic result that for a simple graph with maximum degree d the list chromatic

index exceeds d by at most O(d2/3√

log d).

1. Introduction

Many papers have been written concerning the list-chromatic index χ′`(G) of a graph G,

i.e. the smallest integer t such that if we assign an arbitrary list of t colours to the edges

of G there always exists an edge-colouring for which every edge receives a colour from its

list and every colour class is a matching. Some results from these papers shall be surveyed

below. Further references can be found in Chetwynd and Haggkvist [5] and Alon [1]. But

first let us mention some new results from the current paper.

Our main theorems are Theorems 1.1 and 1.2.

Theorem 1.1. For each n, χ′`(Kn) ≤ n.

Theorem 1.2. For all simple graphs G of maximal degree d, the list-chromatic index satisfies

χ′`(G) ≤ d+ O(d2/3√

log d).

Theorem 1.1 gives a bound for the list-chromatic index for a complete graph. This

bound is sharp for odd n. We thank Pavol Gvozdjak for carefully working through an

earlier version of our proof and supplying us with a more economical set of blocking out

values. We extend this result to a weaker bound of ∆ + 2k − 1 for any graph obtained

† This work was done while J. Janssen was visiting Umea University.

296 R. Haggkvist and J. Janssen

from a complete graph by replacing each vertex by an m-set and each edge by a k-regular

bipartite graph. By probabilistic reasoning we then obtain Theorem 1.2, which slightly

extends a result by Kahn [19], who obtains an o(d) term where we have O(d2/3√

log d).

It has been conjectured a number of times that the list-chromatic index of a graph or

multigraph equals the chromatic index (according to Kostochka (private communication),

Vizing formulated this conjecture for instance at a conference in Odessa in 1975). There

is no doubt that Vizing [24] was the first to propose this conjecture as well as the more

general question to study list-colourings (of vertices) for graphs, although these problems

took some time to seep through the iron curtain. It is equally clear that completely

independently from Vizing, a similar scenario was envisioned by Erdos, Rubin and

Taylor, whose paper on choosability (=(vertex) list colourings) [9] created a lot of interest

in the topic. The Erdos, Rubin and Taylor paper did not mention edge-colourings per se,

but they were certainly influenced by the question of whether or not the list-chromatic

index of Kn,n equals n. To demonstrate this we quote the acknowledgement in Erdos

Rubin and Taylor verbatim:

Acknowledgement

It got started when we tried to solve Jeff Dinitz’s problem. Jeffrey Dinitz is a mathematician at

Ohio State University. At the 10th S.E.Conference on Comb., Graph Theory, and Computing at Boca

Raton in April 1979 he posed the following problem.

Given a m x m array of m-sets, is it always possible to choose one from each set, keeping the chosen

elements distinct in every row, and distinct in every column?

To the best of our knowledge Jeff Dinitz problem remains unsolved for m ≥ 4.

The Dinitz problem was solved in 1994, when Galvin [10] proved the list-chromatic

conjecture for bipartite multigraphs, i.e. he proved that the list-chromatic index of every

bipartite multigraph is equal to its maximum degree. Galvin’s paper overshadowed earlier

work on the Dinitz problem by Haggkvist [11] (χ′`(Km,r) = m when 2m ≥ 7r), Alon and

Tarsi (verification of Dinitz’s conjecture for m = 4, 6) and Janssen [15, 16] (χ′`(Km,m) ≤ m+1

and χ′`(Km,r) = m when m > r).

Bollobas and Harris [3] were the first to obtain any results concerning the list chromatic

conjecture. They found, in part using probabilistic counting, that (here ∆ = ∆(G) denotes

the maximum degree in a graph G)

χ′`(G) ≤

2∆− 2 if ∆ ≥ 3

2∆− 3 if ∆ ≥ 37

2∆− 4 if ∆ ≥ 47

2∆− 5 if ∆ ≥ 56

and in general, for ∆ ≥ 3917,

χ′` ≤ 2∆−⌊

1

6∆− 1

3

√∆ log ∆

⌋Chetwynd and Haggkvist [5–7] found a deterministic way to obtain an upper bound of

95∆ for triangle free graphs. In his doctoral thesis, Hind [12] removed the restriction about

triangles and 2-cycles (he obtains this upper bound for loopless multigraphs – Theorem

4.5.1 in Hind [12]), and moreover, in collaboration with Bela Bollobas [4] improved the

New Bounds on the List-Chromatic Index 297

bounds to ( 74

+ o(1))∆ for simple graphs. For triangle free simple graphs Hind got the

upper bound 53∆ (Corollary 4.5.2 in Hind [12]). This ends one line of investigation.

A different route was taken by Kahn [19] who, building on work by Nick Pippenger

and Joel Spencer as well as Vojtek Rodl concerning the size of random matchings in

some families of hypergraphs, in a blockbuster approach managed to get asymptotically

sharp bounds on many problems concerning colourings of graphs, some 25 years old and

more. In particular, he showed that the list chromatic index for a simple graph is of the

order ∆ + o(∆) (other results with this method include, for instance, an n + o(n) bound

for the chromatic number of any graph which is the edge-disjoint union of at most n

complete graphs of order n, (conjectured to be n by Erdos, Lovasz and Faber)). Further

results using extensions of this method have recently been announced by Jeung Han Kim

[20], who shows that the choice number (= (vertex) list chromatic number) for a graph

of girth 5 is at most O(

∆log ∆

). Anders Johansson [17, 18], using yet another variation of

the method, has shown that the chromatic number of a triangle free graph is O(

∆log ∆

),

and that the choice number is o(∆).

Another breakthrough came with a result by Alon and Tarsi [2], which we describe

more in detail in the next section. They discovered some properties of the so-called graph

polynomial, which for a graph with vertices v1, v2, . . . , vn is defined as the product over

all edges vivj , i < j of the terms xi − xj . The resulting polynomial is, as any polynomial,

the sum of monomials, and it is shown that whenever the graph polynomial for G has

a nonzero coefficient for some monomial xd1

1 xd2

2 · · ·xdnn then if we assign a set Si of size

di + 1 to the vertex vi for every i, i = 1, 2, . . . , n there always exists a proper colouring of

the vertices of G such that no adjacent vertices in G receive the same colour and every

vertex receives a colour from its set. It must be emphasized that the graph polynomial is

usually too large to be displayed, and that no general fast method is known to decide if

some particular term in it has a non-zero coefficient. However, there are some examples

of families of graphs (examples are given later in the paper) where some (interesting)

coefficient in the graph polynomial can be calculated.

Some coefficients in the graph polynomial have a particularily striking interpretation.

When G is a t-regular graph on m edges, for instance, then the coefficient corresponding

to the monomial xt−11 xt−1

2 · · ·xt−1m in the graph polynomial of the line graph of G counts

the number of signed t-edge-colourings of G (see Noga Alon’s [1] survey paper at the

British Combinatorial Conference in Keele 1993). This immediately gives that uniquely

3-edge-chromatic graphs have list chromatic index 3. For t-regular planar multigraphs

every proper t-edge-colouring has the same sign [14, 23, 8], and therefore every t-regular

planar multigraph with a proper t-edge-colouring has list chromatic index t.In general, if G admits some orientation ~G for which each vertex has outdegree di,

then the coefficient for xd1

1 xd2

2 · · ·xdnn has modulus the number of even-edge subgraphs

of ~G with balanced in- and out-degrees at each vertex minus the number of odd-edge

such subgraphs (see Section 2). Consequently, if a graph admits some orientation with

out-degrees bounded by k and without odd cycles, then the choice number of this graph

is at most k+ 1. This implies, for instance, that bipartite simple planar graphs have choice

number 3 [2] (it is easy to see that planar bipartite graphs admit an orientation where

every vertex has out degree of at most 2). Recently, at the Erdos conference in Keszthely,


Margit Voigt [25] gave an example of a planar graph with choice number more than four,

and Carsten Thomassen [22] has announced a proof that the choice number of a planar

graph is at most five.

Lastly, we mention that Kostochka [21] has shown the list chromatic index for a simple

graph with girth at least 8∆(log ∆ + 1.1) to be at most ∆ + 1, and that the best published

bound for the list-chromatic index of a multigraph seems to be Hind’s bound of 1.8∆,

where ∆ denotes the maximum degree.

2. Orientations

In this paper we frequently use a result proven recently by Alon and Tarsi [2]. Their

main theorem establishes a relation between the number of odd and even orientations of

a graph, and the existence of S-legal vertex colourings of that graph. The statement of

the theorem requires some definitions.

Let G be a graph on a ordered vertex set V . An orientation D of G is a directed graph

that has the same set of vertices and edges as G. In a directed graph, an edge v ← w

such that v < w is called an inverted edge. An orientation D of G is called even or odd,

according to the parity of the number of inverted edges of D. An orientation of G is

said to obey a map ρ from the set of vertices V to the non-negative integers, if it has

the property that each vertex v ∈ V has out-degree (number of out-going edges) ρ(v).

DEG(ρ) is the number of even orientations of G that obey ρ, and DOG(ρ) is the number

of odd orientations with the same property. Let S = {Sv | v ∈ V } be a collection of sets.

An S-legal vertex colouring of G is a colouring of the vertices of G which assigns to each

vertex v in V an element from Sv , and which has the property that no vertices joined by

an edge are assigned the same colour.

Theorem 2.1. (Alon-Tarsi) Let G be a graph on an ordered vertex set V . Let S = {Sv | v ∈V } be a collection of sets. If there exists a map from the vertex set of G to the non-negative

integers ρ : V → N+ such that ρ(v) < |Sv| for all v ∈ V , and if

DEG(ρ) 6= DOG(ρ),

then G has an S-legal vertex colouring.

Since we are interested in edge colourings, we will apply the Alon–Tarsi theorem to

line graphs. An S-legal edge colouring of a graph G corresponds to an S-legal vertex

colouring of its line graph in the obvious way, and vice versa. All linegraphs share the

property that they are the edge-disjoint union of the cliques that correspond to the vertices

in the original graph (and such that every vertex in the line graph belongs to exactly two

cliques in the clique decomposition). The following proposition shows that in the case

where we know a particular clique decomposition of G, we can take this information

into account when we compute DEG(ρ)−DOG(ρ); it suffices to consider only orientations

where every clique in the decomposition is transitively oriented.

Definition. Let G = G1 ⊕ G2 ⊕ . . . ⊕ Gm, where each Gi is a complete graph. A clique

transitive orientation of G is an orientation of G that has the property that on each of the

particular cliques Gi of G the induced orientation is a transitive tournament. This implies


that the out-degrees given by a clique transitive orientation of G within a certain Gi (i.e.

the number of out-going edges from a vertex in Gi to other vertices in Gi) are exactly the

numbers 0, 1, . . . , k − 1, where k is the size of clique Gi.

It is important to note that the term clique transitive always refers to a particular clique

decomposition of G. Other cliques are not necessarily transitively oriented.

Note also that a transitive tournament is completely defined by its out-degrees, and

hence a clique transitive orientation of a graph that is the edge disjoint union of cliques

is completely defined by the out-degrees within the cliques.

Lemma 2.2. If G = Kn, the complete graph on n vertices, then for any map ρ : V (G)→ N,

DEG(ρ) 6= DOG(ρ) only when ρ(V (G)) = {0, 1, . . . , n − 1}. In this case, the only orientation

obeying ρ is a transitive tournament, and hence |DEG(ρ)− DOG(ρ)| = 1.

Proof. Let G = Kn with vertex set {v1, . . . , vn}. The graph polynomial fG of G, defined by

fG(x1, x2, . . . , xn) =∏vi∼vj

(xi − xj)

can be written as

fG(x1, x2, . . . , xn) =∏i<j

(xi − xj).

This expression is known to be the determinant of a Vandermonde matrix, which can be

alternatively expressed as

fG(x1, x2, . . . , xn) =∑σ∈Sn

(−1)sg(σ)xσ(1)−11 x

σ(2)−12 · · ·xσ(n)−1

n .

Writing out fG as the linear combination of monomials, each one of the terms corresponds

to an orientation of G in the following way: from each (xi−xj) in the product∏

vi∼vj ,i<j(xi−xj), either xi or xj has to be chosen. Choosing xi means directing the edge from vi to

vj , choosing xj gives the opposite direction. So only an inverted edge contributes a sign

inversion to the term. Using this correspondence, we can write fG in a different way:

fG(x1, x2, . . . , xn) =∑

D orient. of G

(−1)nr. of inv. edges in D xdD

1

1 xdD

2

2 . . . xdDnn ,

where dDi is the out-degree of vertex vi in orientation D. Since an even orientation

contributes a positive term, and an odd orientation a negative one,

fG(x1, x2, . . . , xn) =∑

ρ:V→N

(DEG(ρ)− DOG(ρ))xρ(v1)1 x

ρ(v2)2 . . . xρ(vn)

n .

The statement of the lemma now follows from comparison of the coefficients. It is easy

to check that the only orientation corresponding to the out-degrees 0, 1, . . . , n − 1 is the

transitive tournament where v → w precisely when ρ(v) > ρ(w).

Proposition 2.3. If G = G1 ⊕ G2 ⊕ · · · ⊕ Gn where each Gi is a complete graph, then for

each map ρ : V (G) → N, the number of even orientations obeying ρ which are not clique

transitive is equal to the number of odd such orientations.


Proof. Let G be as in the statement of the theorem. Then the graph polynomial of G can

be written as follows:

fG =∏vi∼vj

(xi − xj) =∏vi∼vjin G1

(xi − xj) · · ·∏vi∼vjin Gn

(xi − xj) = fG1fG2· · · fGn . (2.3a)

As shown in the proof of Lemma 2.2, the coefficient of xρ(v1)1 x

ρ(v2)2 · · ·xρ(vn)

n is DEG(ρ) −DOG(ρ). Equality (2.3a) then gives the relation

DEG(ρ)− DOG(ρ) =∑

ρ1 ,ρ2 ,...,ρn

(DOG1(ρ1)− DEG1

(ρ1)) · · · (DOGn(ρn)− DEGn(ρn)),

where the sum is taken over all maps ρi : V (Gi) → N that satisfy∑

i ρi(v) = ρ(v)

at each vertex v. Since for all ρi that do not correspond to a transitive tournament,

DEGi(ρi) − DOGi(ρi) = 0, by Lemma 2.2 we may take the sum to be over all clique

transitive orientations.

From Lemma 2.3, we can conclude that to use the Alon–Tarsi theorem, it is sufficient

to consider only clique transitive orientations. This reduces greatly the number of possible

orientations for any given map ρ. In this paper we will also use the technique of blocking

out certain values of possible out-degrees in a clique to further reduce the number of

orientations that obey a given map ρ. This technique is formally described by the following

definition and Proposition 2.4.

Definition. Let G be a graph with a given clique decomposition, and let ρ be a map from

the vertex set of G to the natural numbers. An orientation D of G obeying ρ with values

{b1, . . . , bm} blocked out in clique K is an orientation that can be embedded in an orientation

D of G, where G is the graph formed by extending clique K by vertices v1, . . . , vm, which has

the following properties:

1 D obeys ρ at each vertex of V (G).

2 For each i, vi has out-degree bi in D.

Note that D is clique transitive when D is clique transitive and vice versa. If D is clique

transitive, then the out-degrees in D of the vertices in K within the extended clique formed

by K and the adjoined vertices v1, . . . , vm take exactly the values {0, 1, . . . , |K|+ m+ 1} −{b1, . . . , bm}.

Proposition 2.4. Let G = H1 ⊕ H2 ⊕ · · · ⊕ Hm where each Hi is a complete graph. Let

B1, . . . , Bm be sets of non-negative integers. Let S = {Sv | v ∈ V (G)}, and ρ : V (G) → Nsuch that ρ(v) < |Sv| for all v ∈ V (G). If there exists a unique clique transitive orientation

D of G that obeys ρ with the values in Bi blocked out in clique Hi for each i, 1 ≤ i ≤ m,

then there exists an S-legal vertex colouring of G.

Proof. Let G, ρ, the sets Bi and S be as in the statement of the proposition. Suppose

that Bi = {bi1, . . . , bimi} for each i. Let G be the extension of G formed by extending each


clique Hi by vertices vi1, . . . , vimi . Define map ρ : V (G)→ N as follows:

ρ(v) =

{ρ(v) if v ∈ V (G)

bij if v = vij for some i, j.

Form S = S ∪ {Tij | 1 ≤ i ≤ m, 1 ≤ j ≤ mi}, where the Tij are arbitrary sets of size

|Tij | > bij . Now each clique transitive orientation of G that obeys ρ with the values in

Bi blocked out in clique Hi for each i, 1 ≤ i ≤ m, corresponds to a clique transitive

orientation of G obeying ρ. So if there is a unique such orientation of G, then there is a

unique clique transitive orientation of G obeying ρ, and we can conclude (using Lemma

2.3) that DEG(ρ) − DOG(ρ) = ±1. So by Theorem 2.1, there exists an S-legal vertex

colouring of G. But since G is a subgraph of G and S agrees with S on V (G), this implies

the existence of an S-legal vertex colouring of G.

3. The complete graph

In this section we prove our result on the list-chromatic index of the complete graph by

considering orientations of its line graph as described in the previous section.

Theorem 3.1. For each n, χ′`(Kn) ≤ n.

Proof. To prove this theorem we use Proposition 2.4. Suppose G = Kn has vertices labelled

{0, 1, . . . , n − 1}. Let H be the line graph of G. The vertex set of H can be indicated as

V (H) = {(i, j) | 0 ≤ i < j < n}. With this notation, vertices (i, j) and (i′, j ′) are joined by

an edge in H precisely when i = i′,i = j ′,j = i′ or j = j ′. H = C0 ⊕ C1 ⊕ . . .⊕ Cn−1, where

Ci is the clique of size n − 1 which corresponds to vertex vi of G. Define ρ : V (H) → Nas follows:

ρ(i, j) =

{n− 1 if i+ j ≥ n− 1

n− 2 otherwise.

Define values bi, 0 ≤ i < n:

bi =

n− 2− i if i <

⌊n2

⌋n− 1− i+

⌊n2

⌋if i ≥

⌊n2

⌋.

We will show that there exists a unique clique transitive orientation of H that obeys

ρ with value bi blocked out in clique Ci. Any vertex (i, j) of H lies at the intersection of

exactly two cliques, Ci and Cj , so in any orientation D of H obeying ρ the out-degrees of

this vertex within the cliques Ci and Cj , respectively, have to add up to ρ(i, j). We will give

our proof by describing an algorithm which assigns two out-degrees to each vertex v of H ,

one for each clique that contains the vertex, such that at each vertex v these out-degrees

add up to ρ(v), and for each i, clique Ci contains all out-degrees {0, 1, . . . , k}− {bi} exactly

once. Such an assignment uniquely determines a clique transitive orientation of H obeying

ρ and blocking values. If out-degree x in clique Ci is assigned to a vertex (i, j), then the

out-degree in Cj at this vertex such that both out-degrees add up to ρ(i, j) is called the

complementary degree at this vertex.


We will show that at each step of the algorithm, there is exactly one choice that

fulfills all requirements, and therefore we can conclude that the assignment, and hence the

corresponding orientation, is unique.

Algorithm 3.2

1 Set i = 0.

2 Set t = 0.

3 If t =⌊n2

⌋− i then go to step 6.

4 Assign degree n − i − 1 to a vertex in clique Ct+2i. The unique choice is at vertex

(t+ 2i, n− 1− t). The complementary degree at this vertex becomes i.

5 Set t = t+ 1. Go to step 3.

6 If t = n− 1− 2i then go to step 9.

7 Assign degree n − 1 − i to a vertex in clique Ct+2i+1. The unique choice is at vertex

(n− 1− t, t+ 2i+ 1). The complementary degree at this vertex becomes i.


9 If t = n− 1− i then go to step 12.

10 Assign degree n− 2− i to a vertex in clique Ct+2i−(n−1). The unique choice is at vertex

(t+ 2i− (n− 1), n− 1− t). The complementary degree at this vertex becomes i.


12 If n is even and i+ 1 ≥⌊n2

⌋then go to step 17.

13 If t > n− 1 then go to step 16.

14 Assign degree n−2− i to a vertex in clique Ct+2i+1−(n−1). The unique choice is at vertex

(n− 1− t, t+ 2i+ 1− (n− 1)). The complementary degree at this vertex becomes i.


16 Set i = i+ 1. If i <⌊n2

⌋then go to step 2.

17 End.

Note that the assignment given by the algorithm does not give a conflict with the blocked

out values, because for each i ≤⌊n2

⌋, out-degree n − 1 − i is assigned to all cliques Cj

except to Cj where j =⌊n2

⌋+ i, and bb n2c+i = n− 1− i, and out-degree n− 2− i is assigned

to all cliques except to Ci, and bi = n− 2− i, so the out-degrees are assigned to all cliques

except exactly those for which this out-degree is blocked out.

Before proving the uniqueness in steps 4, 7, 10 and 14 of the algorithm, we show

as an example the case where G = K6. Now H , the line graph of G, has vertex set

H = {(i, j) | 0 ≤ i < j < 6}. In the following we will, for reasons of clarity, represent H

as a 6 × 6 matrix. For all i, j, the 1 × 1 submatrix of H in row i and column j is called

cell (i, j) cells (i, j) and (j, i) are identified to represent vertex (i, j) of H . Cells (i, i) do not

represent any vertices. Hence both column i and row i represent clique Ci, and vertices

of H are connected exactly when the corresponding cells lie in the same row or column.

In this representation, the following matrix gives the values of ρ for our example; the

ρ-value of vertex (i, j) of H is given as entry (i, j) and entry (j, i) in this matrix. Since the

cells on the main diagonal of the matrix do not correspond to vertices of H , they are

used to show the blocked out values bi. So the entry (i, i) represents the value blocked out

in clique Ci.


4 4 4 4 54

3 4 4 5 54

4 2 5 5 54

4 5 5 5 54

5 5 5 4 54

5 5 5 5 35

C0 —

C1 —

C2 —

C3 —

C4 —

C5 —

C1 C2 C3 C4 C5C0

The following matrices show the horizontal and vertical degrees assigned after, respec-

tively, step 4 for i = 0 and t = 0, 1, 2 and step 7 for i = 0 and t = 3, 4. If integers km are

shown as entry (i, j) where i < j, this means that vertex (i, j) of H is assigned horizontal

degree k and vertical degree m. To be consistent with our representation, entry (j, i) is mk.

Thus the assigned out-degrees within a clique Ci can be found by looking at the leftmost

integers of the entries of row i, or the rightmost integers of the entries of column i. For

our assignment to be complying with the requirements, these integers should differ from

each other and from the value blocked out in Ci.

504

3 50

2 50

05 5

05 4

305

C0 —

C1 —

C2 —

C3 —

C4 —

C5 —

C1 C2 C3 C4 C5C0

504

3 50 05

2 50 05

05 5

50 50 4

50 305

C0 —

C1 —

C2 —

C3 —

C4 —

C5 —

C1 C2 C3 C4 C5C0

The next two matrices show the horizontal and vertical degrees assigned after, respec-

tively, steps 10 and 14 for i = 0 and t = 5 and steps 4 and 7 for i = 1 and t = 0, 1 and for

i = 1 and t = 2.

504

3 50

2 50

05 5

05 50 4

50 305

C0 —

C1 —

C2 —

C3 —

C4 —

C5 —

C1 C2 C3 C4 C5C0

04 504

3 50 05

2 50 05 41

05 5 41 14

05 50 14 4

50 14 41 305

C0 —

C1 —

C2 —

C3 —

C4 —

C5 —

C1 C2 C3 C4 C5C0

0540

04

05

40


The next matrix shows the assignment after step 10 and 14 for i = 1 and t = 3, 4, 5.

04 24 42 504

3 42 50 0540

24 2 50 05 4142

05 5 41 1424

05 50 41 4

50 41 14 305

C0 —

C1 —

C2 —

C3 —

C4 —

C5 —

C1 C2 C3 C4 C5C0

Finally, the following matrix shows the assignment after completion of the algorithm,

with next to it a schematic portrayal of the order in which the vertices recieved their

assignment of out-degrees. The reader is invited to check that this order is indeed as

described in the remark after Algorithm 3.2.

054

3 13 22 05

2 05

50 5

50 05 4

05 350

C0 —

C1 —

C2 —

C3 —

C4 —

C5 —

C1 C2 C3 C4 C5C0

4

3

2

5

4

3

C0 —

C1 —

C2 —

C3 —

C4 —

C5 —

C1 C2 C3 C4 C5C0

15

3113

2231

22

51 31 13 22

50

50 14

14 41

41 32

41 2314

Note that the out-degrees in clique Ci of the final assigment can be obtained by reading

the second digit of each entry in column i or the first digit of each entry in row i. It is

easy to check this way that the final assignment indeed has the property that each clique

Ci contains all out-degrees 0, 1, . . . , 5 except bi exactly once.

Remark. The previous example illustrates the pattern the algorithm follows in assigning

the out-degrees. If we understand diagonal i to mean those vertices (x, y) for which

x+y = i (mod n−1) (this corresponds to an antidiagonal in the matrix of the example),

then in each iteration the algorithm first assigns out-degrees to the vertices of diagonal 2i

with x+ y > n− 1 (in the example this means below the main antidiagonal) in ascending

order of x (rows in the example). Next the vertices of diagonal 2i+1 with x+y > n−1 are

filled in descending order of x (rows). Then the vertices of diagonal 2i with x+ y ≤ n− 1

(above the main antidiagonal) are considered in ascending x-value (row) order, and finally

the vertices of diagonal 2i+ 1 with x+ y ≤ n− 1 are filled in descending x-value order.

To prove the uniqueness in steps 4, 7, 10 and 14 of the algorithm, we will show that,

throughout the algorithm, the following predicate P holds.


P: Out-degrees 0, . . . , i− 1 have been assigned in all cliques. Out-degree i has been assigned

in cliques n− 1, . . . , n− t. Both complementary out-degrees have been assigned to all vertices

(x, y) with x + y = d (mod n − 1), where 0 ≤ d < 2i. Also, both complementary out-

degrees have been assigned to all vertices (x, y) (x < y) with x + y = 2i (mod n − 1)

and 2i ≤ x < t + 2i – below the main antidiagonal – or 0 ≤ x < t + 2i − (n − 1) – above

the main antidiagonal – and to all vertices (x, y) with x + y = 2i + 1 (mod n − 1) and⌊n2

⌋+ i+ 1 ≤ y < t+ 2i+ 1.

Suppose P holds before step 4 – this is trivially true when i = 0 and t = 0. Out-degree

n−1−i has to be assigned to a vertex (l, t+2i) or (t+2i, l) of clique Ct+2i. If l+t+2i < n−1

then the complementary degree at this vertex will be ρ(l, t+ 2i)− (n− 1− i) = i− 1. But

according to P, out-degree i − 1 has already been assigned in all cliques, so l + t + 2i ≥n − 1 and the complementary out-degree is i. Out-degree i has been assigned in cliques

n−1, . . . , n− t, according to P, so l ≤ n−1− t. Since l+ t+2i ≥ n−1 the representative of

l+t+2i when taken modulo n−1 is l+t+2i−(n−1). By P, all vertices (x, y) with x+y = d

(mod n− 1) where d < 2i already have assigned out-degrees, so l + t+ 2i− (n− 1) ≥ 2i,

which implies that l ≥ n−1− t. Hence the only possible assignment of out-degree n−1− iin clique Ct+2i is at vertex (t+ 2i, n− 1− t). It is easy to check that with this assignment

P again holds after t is augmented in step 5.

Suppose P holds before step 7. Out-degree n − 1 − i has to be assigned to a vertex

(l, t + 2i + 1) or (t + 2i + 1, l) of clique Ct+2i+1. Because out-degree i − 1 has already

been assigned in all cliques, the complementary degree at this vertex must be i, and thus

l + t+ 2i+ 1 ≥ n− 1. Out-degree i has been assigned in cliques n− 1, . . . , n− t according

to P, so l ≤ n − 1 − t. Since l + t + 2i ≥ n − 1, the representative of l + t + 2i when

taken modulo n − 1 is l + t + 2i + 1 − (n − 1). By P, all vertices (x, y) with x + y = d

(mod n− 1) with d < 2i already have assigned out-degrees, so l+ t+ 2i+ 1− (n− 1) ≥ 2i,

i.e. l ≥ n − 2 − t. If l = n − 2 − t then l + t + 2i + 1 = 2i (mod n − 1) and, since⌊n2

⌋− i ≤ t < n− 1− 2i, 2i ≤ l <

⌊n2

⌋+ i ≤ t+ 2i), so by P this vertex already has assigned

out-degrees. Hence the only possible assignment of out-degree n− 1− i in clique Ct+2i+1

is at vertex (n− 1− t, t+ 2i+ 1). P again holds after t is augmented in step 5.


(l, t+ 2i− (n− 1)) or (t+ 2i− (n− 1), l) of clique Ct+2i−(n−1). If t+ 2i− (n− 1) + l ≥ n− 1

then the representative of t+ 2i− (n− 1) + l modulo n− 1 is t+ 2i− (n− 1) + l− (n− 1) <

n − 1 − i + 2i + l − 2(n − 1) = i + l − (n − 1) ≤ i ≤ 2i. But according to P, all vertices

(x, y) with x + y = d (mod n − 1) where d < 2i already have assigned out-degrees, so

t+ 2i− (n− 1) + l < n− 1 and thus also t+ 2i− (n− 1) + l ≥ 2i, i.e. l ≥ n− 1− t. Since

t+ 2i− (n− 1) + l < n− 1, the complementary degree is n− 2− (n− 2− i) = i. Out-degree

i has been assigned in cliques n − 1, . . . , n − t so l ≤ n − 1 − t. Hence the only possible

assignment of out-degree n−2− i in clique Ct+2i−(n−1) is at vertex (t+2i− (n−1), n−1− t).P again holds after t is augmented in step 11.


(l, t+2i+1−(n−1)) or (t+2i+1−(n−1), l) of clique Ct+2i+1−(n−1). If t+2i+1−(n−1)+l ≥ n−1

then the representative of t+2i+1−(n−1)+l modulo n−1 is t+2i+1−(n−1)+l−(n−1) ≤n−1+2i+1+l−2(n−1) = 2i+1+l−(n−1) ≤ 2i+1. According to P, all vertices (x, y) with


x+y = d (mod n−1) where d < 2i or d = 2i (since 0 ≤ t+2i+1−(n−1) < t+2i) already

have assigned out-degrees. If equality holds, then l = t = n−1, so⌊n2

⌋+i+1 ≤ l < t+2i+1,

and hence the out-degrees of vertex (t + 2i + 1 − (n − 1), l) have already been assigned.

Hence t+2i−(n−1)+ l < n−1 and thus also t+2i+1−(n−1)+ l ≥ 2i, i.e. l ≥ n−1−t−1.

But if l = n− 1− t− 1 then t+ 2i+ 1− (n− 1) + l = 2i, and since n− 1− i ≤ t ≤ n− 1

so 0 ≤ n − 1 − t − 1 < t + 2i − (n − 1) this vertex already has assigned out-degrees.

Hence l ≥ n − 1 − t. Since t + 2i + 1 − (n − 1) + l < n − 1, the complementary degree is

n−2−(n−2−i) = i. Out-degree i has been assigned in cliques n−1, . . . , n−t so l ≤ n−1−t.Hence the only possible assignment of out-degree n − 2 − i in clique Ct+2i+1−(n−1) is at

vertex (n − 1 − t, t + 2i + 1 − (n − 1)). It can be checked that P again holds after t is

augmented in step 15, and after i is augmented in step 16.

So there exists a unique clique transitive orientation of H that obeys ρ with bi blocked

out in clique Ci for all i, and the maximal value of ρ is n− 1. Hence, by Proposition 2.4,

for each collection of sets S = {Sv| v ∈ V (H)} with |Sv| > n − 1, there exists an S-legal

vertex colouring of H , and hence a corresponding edge colouring of G. So χ′l(G) ≤ n.

The orientation obtained in the preceding proof will in the following be called the

designated orientation of Kn. This orientation will be represented by a function α :

V (H) → N ×N, where α(i, j) = (k, l) indicates that in the orientation, the out-degree of

vertex (i, j) within clique Ci is k, and the out-degree of vertex (i, j) within clique Cj is l.

Hence, for a certain value of n, α is defined as follows:

α(i, j) =

(i+ j + n− 1

2,i+ j − n+ 1

2

)if i+ j = n− 1 (mod 2) and i+ j ≥ n− 1(

i+ j − n+ 1

2,i+ j + n− 1

2

)if i+ j = n (mod 2) and i+ j ≥ n− 1(

n− 2− i+ j

2,i+ j

2

)if i+ j = 0 (mod 2) and i+ j < n− 1(

i+ j − 1

2, n− i+ j + 3

2

)if i+ j = 1 (mod 2)and i+ j < n− 1.

4. Edge compositions

To prepare for our result on the list chromatic index of simple graphs in general, we

describe in this section the construction of a class of graphs with list chromatic index close

to their maximum degree, and give a theorem on the list chromatic index of the union of

such graphs. In the next section we will then combine these results with a probabilistic

argument to express any simple graph as such a union.

Definition. The edge composition graph G〈H〉 is constructed from a graph G = (V , E) and

a family of graphs H = {He| e ∈ E}, each He with bipartition (W 1e ,W

2e ), where W 1

e and

W 2e are labelled and |W 1

e | = |W 2e | = m, by replacing each edge of G by the corresponding

bipartite graph of H and consequently each vertex of G by a vertex set of size m. In other

words, if W 1e = {w1

e,1, . . . , w1e,m} and W 2

e = {w2e,1, . . . , w

2e,m} for each e ∈ E, then G〈H〉 =

(V,E) can be characterized as follows:


Figure 1

• V =⋃v∈V Sv , where Sv = {sv,1, . . . , sv,m} for each v ∈ V ,

• (su,i, sv,j) ∈ E – u < v in some ordering imposed on V – precisely when (u, v) = e ∈ E, and

w1e,i ∼ w2

e,j in He.

Note that the maximal degree of G〈H〉 is at most the product of the maximal degree of

G and the largest degree that occurs in any of the He.

Figure 1 shows an example of the edge composition of K3 and a family of one-regular

bipartite graphs on four vertices.

Theorem 4.1. Let G = Kn〈H〉, where H is a family of one-regular graphs on 2k vertices.

Then χ′`(G) ≤ n = ∆(G) + 1.

Proof. We prove this theorem by methods similar to the ones used in the proof of

Theorem 1.1, including an adapted version of Algorithm 3.2.

Let the vertices of Kn be labelled 0, 1, . . . , n − 1. Let H = {Hij | 0 ≤ i < j < n}, where

Hij is the bipartite graph used to replace edge (i, j) in Kn. Let V (G) = {vi | 0 ≤ i < nk},where vin+k , 0 ≤ k < n, is the k-th vertex in the set of vertices that replaces vertex i of

Kn. Consider L, the line graph of G. As before, the vertex set of L can be represented

as V (L) = {(i, j)| i < j and vi ∼ vj in G}, and (i, j) ∼ (i′, j ′) in L exactly when i = i′,

j = j ′, i = j ′ or i′ = j. Note that L is the edge disjoint union of mn cliques Ci of size

n− 1, each one corresponding to a vertex of G. Let function ρ from the vertex set of the

line graph of Kn to N and blocking values bi be as in the proof of Theorem 1.1. Define

ρ : V (L)→ N and blocking values bi, 0 ≤ i < nm as follows:

ρ(im+ r, jm+ r′) = ρ(i, j) for all 0 ≤ r, r′ < m, 0 ≤ i, j < n

bim+r = bi for all 0 ≤ r < m, 0 ≤ i < n.

In other words, the out-degrees assigned by ρ to the vertex of the line graph corresponding

to edge (i, j) of Kn are now assigned by ρ to all vertices of L corresponding to edges

from Hij in Kn〈H〉.We will show that there exists a unique clique transitive orientation of L that obeys

ρ with value bi blocked out in clique Ci for all 0 ≤ i < nm. To do this, we describe how

to adapt Algorithm 3.2 so it finds an assignment of two out-degrees to each vertex of L


such that at each vertex v ∈ V (L) these out-degrees add up to ρ(v), and for each i, clique

Ci contains all out-degrees {0, 1, . . . , n} − {bi} exactly once. Such an assignment uniquely

determines a clique transitive orientation of L obeying ρ.

The only change we make to Algorithm 3.2 is that whenever (at step 4,7,10 or 14)

out-degree n − 1 − i or n − 2 − i is assigned to a vertex of clique Cs, we now do the

same for all cliques Csm, Csm+1, . . . , Csm+(m−1). Predicate P is hereby changed such that

what holds for clique Cs after a certain step of Algorithm 3.2 now holds for all cliques

Csm, Csm+1, . . . , Csm+(m−1) after the same step of the adapted algorithm. Suppose that at a

certain step in Algorithm 3.2 the objective is to assign out-degree n − 1 − i to a vertex

in clique Cs, and it follows from the predicate P that the only possible choice is at the

intersection of clique Cs with clique Ct. Then in the corresponding step in the adapted

algorithm the objective is to assign out-degree n− 1− i to a vertex in some clique Csm+r ,

0 ≤ r < m, and we can deduce by a completely analogous argument from the new

predicate that out-degree n − 1 − i has to be assigned to a vertex in some clique Ctm+r′ ,

0 ≤ r′ < m. But since Hst is a one-regular graph, there is only one of the cliques Ctm+r′ ,

0 ≤ r′ < m, that meets Csm+r . Hence there is a unique choice in each step of the algorithm,

and thus there exists a unique clique transitive orientation of L that obeys ρ with biblocked out in clique Ci for all i, and the maximal value of ρ is n− 1. So by Proposition

2.4 the list chromatic index of G is at most n.

By the designated orientation of Kn〈H〉 we denote, as for the complete graph, the

unique clique transitive orientation obeying map ρ and blocking values as defined in the

previous proof. This orientation is again represented by a map, say β :→ N ×N, where

V = {(i, j)| i < j, vi ∼ vj in Kn〈H〉} is the vertex set of the line graph of Kn〈H〉, and

β(i, j) = (l, r) indicates that in the orientation the out-degree of vertex (i, j) within clique

Ci is l, and the out-degree of (i, j) within clique Cj is r. For Kn〈H〉, β can be obtained

from α, the map that defines the designated orientation of Kn, as follows. For all i, j, r, r′,

0 ≤ i, j < n, 0 < r, r′ < m,

β(im+ r, jm+ r′) = α(i, j).

Next we give a theorem on the union of edge composition graphs. By the edge disjoint

union is meant here the union formed by identifying vertices, but duplicating edges

whenever they occur in more than one graph of the union. Hence, the edge disjoint union

of simple graphs is itself not necessarily a simple graph.

Theorem 4.2. Suppose G is a regular graph of degree d which is the edge disjoint union

of k graphs, each of which is an edge composition graph Kn〈Hi〉, where Hi is a family of

one-regular graphs on 2m vertices. Then χ′l(G) ≤ d+ 2k − 1.

Proof. Note that d = (n−1)k. Let G = G0⊕G1⊕ . . .⊕Gk−1, where for each 0 ≤ i < k, Gi =

Kn〈Hi〉, whereHi is a family of one-regular graphs on 2m vertices. LetL be the line graph

of G, with vertex set V (L) =⋃k−1s=0 Vs, where Vs = {(i, j, s)| i < j, (vi, vj) is an edge in Gs}.

Two vertices (i, j, s) and (i′, j ′, s′) are joined by an edge whenever i = i′, j = j ′, i = j ′ or

j = i′.

Again we define a map ρ : V (L) → N and some blocking sets and prove that there

exists a unique clique transitive orientation of L obeying ρ and with the values in the


corresponding blocking set blocked out in clique Ci. Map and blocking values are defined

as follows. Let D = nk − 1 when n is odd and D = (n + 1)k − 2 when n is even. For all

0 ≤ i, j < n and 0 ≤ r, r′ < m,

ρ(im+ r, jm+ r′, s) =

{D − s if i+ j ≥ n− 1

D − s− 1 if i+ j < n− 1.

For each i, t, 0 ≤ i < n, 0 ≤ t < m,

bs,in+t =

D − 1− i− s(⌊n

2

⌋+ 1)

if i <⌊n

2

⌋d− i+

⌊n2

⌋− s(⌊n

2

⌋+ 1)

if i ≥⌊n

2

⌋,

and when n is even, define additionally for each s < k− 1 and all i, t, 0 ≤ i < n, 0 ≤ t < m,

bs,in+t =

⌊n

2

⌋− 1 + s

⌊n2

⌋if i <

⌊n2

⌋D −

⌊n2

⌋− s(⌊n

2

⌋+ 1)

if i ≥⌊n

2

⌋.

Also, let Bl = {b0,l , b1,l , . . . , bk−1,l}, and when n is even, let Bl = {b0,l , b1,l , . . . , bk−2,l}. Next

we define a clique transitive orientation of L, and then prove that this orientation is in

fact the only clique transitive orientation obeying ρ and with values Bi (or Bi ∪ Bi when

n is even) blocked out in clique Ci for each i, 0 ≤ i < nm. Since the maximum value of ρ

is D ≤ nk + k − 2 = d+ 2k − 2, the result follows from Proposition 2.4.

For each s, let βs : Vs → N×N represent the designated orientation of Kn〈Hs〉. Now

define β : V (L)→ N×N as follows. If for some s, i, j, βs(i, j) = (x, y), then

β(i, j, s) =

(x+ s⌊n

2

⌋, D − s

(⌊n2

⌋+ 1)− (n− 1− y) if x ≤ y,(

D − s(⌊n

2

⌋+ 1)− (n− 1− x), y + s

⌊n2

⌋)if x > y.

It is straightforward to check that the orientation represented by β is clique

transitive and obeys ρ, and that each clique Ct, 0 ≤ t < nm, contains all out-degrees

in {0, 1, . . . , nk − 1} − Bi when n is odd, and {0, 1, . . . , (n + 1)k − 2} − (Bi ∪ Bi) when n is

even, exactly once. We will now demonstrate that this is indeed the only such orientation

of L, by proving the following statement.

P(q): Suppose out-degrees D, d− 1, . . . , D − q + 1 have been assigned according to β. Then

the only way to assign out-degree D− q in each clique Ct, 0 ≤ t < nm, for which D− q has

not been blocked out by Bt (or Bt ∪ Bt when n is even) such that this assignment obeys ρ,

is according to β.

Clearly, the validity of this statement for all q such that 0 ≤ q < k(⌊n2

⌋+ 1) implies

that β indeed represents the designated orientation of G and by Proposition 2.4 proves

the theorem.

Proof of P(q). First note that for each s, 0 ≤ s < k, the out-degrees assigned to the

vertices in Vs by β are exactly those out-degrees l for which s⌊n2

⌋≤ l < (s + 1)

⌊n2

⌋or

D−s(⌊n2

⌋+1) ≥ l > D−(s+1)(

⌊n2

⌋+1). Now suppose that out-degrees D,D−1, . . . , D−q+1


have been assigned to vertices of L according to β. Let s be such that s(⌊n2

⌋+ 1) ≤ q <

(s + 1)(⌊n2

⌋+ 1). The assignment then implies that all vertices in V0, V1, . . . , Vs−1 already

have assigned out-degrees. Hence out-degree D−q must be assigned to a vertex of⋃k−1i=s Vi.

The fact that out-degrees D,D − 1, . . . , D − q + 1 have been assigned according to β also

implies that out-degrees 0, 1, . . . , q − 2− s have been assigned in all cliques. If D − q were

to be assigned to a vertex of⋃k−1i=s+2 Vi, or in Vs+1 to a vertex (im+ r, jm+ r′, s+ 1) with

i+ j < n− 1, then in order for the assignment to obey ρ the complementary degree would

have to be at most q − 2 − s, an out-degree that is already taken in all cliques. Hence q

has to be assigned to a vertex of Vs or to a vertex (im+ r, jm+ r′, s+ 1) with the property

that 0 ≤ r, r′ < m and i+ j ≥ n− 1.

We can also deduce from the definition of β that the fact that out-degree D − q + 1

has been assigned in all cliques according to β implies that out-degree q − s− 1 has been

assigned in all cliques Ctm+r with n − 1 − (q − s(⌊n2

⌋+ 1)) < t < n. Suppose D − q is

assigned to a vertex (im+ r, jm+ r′, s+ 1) for which i+ j ≥ n−1. Then the complementary

out-degree is q− 1− s, and hence we get a conflict unless j ≤ n− 1− (q− s(⌊n2

⌋+ 1)), and

therefore i ≥ q − s(⌊n2

⌋+ 1) − 1. Hence for all cliques Cim+r with i < q − s(

⌊n2

⌋+ 1) − 1,

out-degree D − q is assigned to a vertex in Vs.

Now β restricted to Vs gives a one-to-one correspondence between assignments of out-

degrees s⌊n2

⌋, . . . , (s+1)

⌊n2

⌋−1 and D− (s+1)(

⌊n2

⌋+1)+1, . . . , D−s(

⌊n2

⌋+1) to the vertices

of Vs and assignments of out-degrees 0, 1, . . . , n− 1 to the vertex set of L(Kn〈Hs〉), the line

graph of Kn〈Hs〉, where both assignments correspond to clique transitive orientations.

Moreover, an assignment of those out-degrees to Vs obeys ρ and blocking values bs,iprecisely when the corresponding assignment in L(Kn〈Hs〉) obeys the map and blocking

values as defined in the proof of Theorem 4.1.

From the proof of Theorem 4.1 we know that whenever values n−1, n−2, . . . , n−1−(q−1−s

⌊n2

⌋) are assigned in all cliques according to βs, then the only way to assign out-degree

n− 1− (q− s⌊n2

⌋) is such that this assignment obeys the map and blocking values defined

in this proof, is according to βs. Hence, by the one-to-one correspondence, the only way

to assign out-degree D − q to vertices of Vs in cliques Cim+r with i < q − s(⌊n2

⌋+ 1) − 1

such that this assignment obeys ρ and the prescribed blocking values is according to β.

But when D − q has been assigned to all cliques Cim+r with i < q − s(⌊n2

⌋+ 1) − 1

according to β, then this implies that out-degree q− 1− s has been assigned to all cliques.

Hence D− q cannot be assigned to a vertex of Vs+1, and has to be assigned to a vertex of

Vs. By the one-to-one correspondence it then has to be assigned in all cliques Cim+r with

i ≥ q − s(⌊n2

⌋+ 1)− 1 according to β in order to obey ρ and blocking values.

5. Simple graphs

In this section we will combine some of the techniques introduced in the previous sections

with a probabilistic result to obtain a new bound on the list-chromatic index of simple

graphs. In order to prove this bound we need the following probabilistic lemma, which we

will then use to assert that every simple graph can be written as the union of sufficiently

few edge compositions of a complete graph and a family of one-regular graphs. This

lemma is a slight adaptation of Lemma 2.2 of Chetwynd and Haggkvist [7], where it is


proved using the Erdos & Lovasz local lemma and some standard bounds on the tail of

the binomial distribution. A similar lemma that gives slightly faster convergence can also

be found in Haggkvist and Thomason [13].

Lemma 5.1. (Chetwynd and Haggkvist) Let G be a simple graph of maximal degree d.

Then, for every integer p = 2k such that d60 log 3d

> 2k there exists a partition of the vertices

of G into p parts V1, . . . , Vp such that the degree d(v, Vi) from vertex v to any vertex in Vi

differs from its expected value d(v)/p by at most 3√

dp

log 3d.

Theorem 5.2. Let G be a simple graph of maximal degree d on n vertices. Then, for every

integer p = 2k such that d60 log 3d

> 2k , G can be embedded in a graph I∪ G, where G is the

edge disjoint union of s = d/p+ 3√

dp

log 3d graphs, each of which is the edge composition

of the complete graph Kp with a family of one-regular graphs, and I is a graph of maximum

degree at most s.

Proof. Let graph G and integer p be as in the statement of the theorem. By Lemma 5.1,

and the fact that d(v) ≤ d for all vertices v of G, there exists a partition of the vertex

set V of G into p parts, V1, . . . , Vp, such that d(v, Vi) ≤ s for each v ∈ V and all i. Let t

be the maximum size of any of the sets Vi. Form G′ by adding vertices (but no edges)

to G until each Vi has size t. Note that G′ also has the property that d(v, Vi) ≤ s for

all i and all v ∈ V . Label the vertices of G′ such that the vertices of Vi have indices

it, it+ 1, . . . , it+ (t− 1).

Let G′ij be the graph with vertex set Vi ∪ Vj which has as its edge set all edges {vi, vj}such that {vi, vj} is and edge of G′, and vi ∈ Vi, vj ∈ Vj . So G′ij is the subgraph of G′

that contains all edges between Vi and Vj . Each G′ij is a bipartite graph of maximum

degree at most s, and hence its edges can be properly coloured with s colours. Let

{ci | 0 ≤ i < s} be a set of colours, and use these to properly colour the edges of each

G′ij . Graphs G′b, b = 1, . . . , s, are the graphs which have as edge set all edges of colour

cb in G′, and for each b = 1, . . . , s, Gb is the graph obtained from G′b by embedding the

edges of colour cb in graph Vi,j in a perfect matching of Vi,j that preserves the bipartition.

This matching also has colour cb. Now each Gb is the edge composition Kp〈Hb〉, where

Hb = {Hbij | 0 ≤ i, j < p}, and Hb

ij is the perfect matching in Vij that contains colour cb.

Hence G =⋃b Gb is as in the statement of the theorem.

Let I be the graph with vertex set⋃i Vi, and as its edge set all edges {v, w} of G′ with

v, w ∈ Vi for some i. So I is the disconnected union of the induced subgraphs of G′ on

the vertex sets Vi. We have chosen the Vi such that d(v, Vi) ≤ s for all i and all v ∈ V ,

so the maximum degree of any of these induced subgraphs is at most s. Since I is their

disconnected union, I also has maximum degree at most s.

So the edges of G′ that go from one vertex set Vi to another can be embedded in the

family of Gb, and the edges that stay inside one of the Vi are present in I. Therefore, G′,

and thus also G, can be embedded in the union of I and G.

We now have developed all the tools necessary to prove Theorem 1.2, which we state

here in a slightly more specific form than given in the introduction.


Theorem 1.2. For all simple graphs G of maximal degree d, where d is such that d2/3 >

60 log 3d, the list-chromatic index satisfies χ′`(G) ≤ d+ 23d2/3√

log 3d.

Proof. Let G be a simple graph of maximal degree d. By Theorem 5.2, for every integer

p = 2k such that d60 log 3d

> 2k , G can be embedded in a graph I ∪ G, where where I has

maximum degree at most s, and G is the edge-disjoint union of s = d/p + 3√

dp

log 3d

graphs, each of which is the edge composition of the complete graph Kp with a family of

one-regular graphs. Note that G has degree ps.

By Theorem 4.2, χ′l(G) ≤ ps+ 3s− 1 = d+ 3√dp log 3d+ 3(d/p+ 3

√dp

log 3d)− 1. We

know that χ′`(I) ≤ 2s; if lists of size 2s are given to each edge, then the edges can be

list-coloured greedily. Now χ′`(G) ≤ χ′`(G) + 2s. Namely, if lists of size χ′`(G) + 2s are

assigned to all edges, then first I can be list-coloured, and then for each edge in G, all

colours occurring on edges of I adjacent to it (at most 2s) can be deleted from its list.

This then still leaves lists of size χ′`(G) to list-colour the edges of G.

So χ′l(G) ≤ ps+ 3s−1 + 2s = d+ 3√dp log 3d+ 5(d/p+ 3

√dp

log 3d)−1. This expression

is minimized if we take p to be the power of 2 closest to d1/3, in which case ps+ 5s− 1 =

d+ 3√d4/3 log 3d+ 5(d2/3 + 3

√d2/3 log 3d)− 1 ≤ 23d2/3

√log 3d.

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