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New Bounds on the List-Chromatic Index of the Complete Graph and
Other Simple Graphs
ROLAND HÄGGKVIST and JEANNETTE JANSSEN
Combinatorics, Probability and Computing / Volume 6 / Issue 03 / September 1997, pp 295 - 313
DOI: null, Published online: 08 September 2000
Link to this article: http://journals.cambridge.org/abstract_S0963548397002927
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ROLAND HÄGGKVIST and JEANNETTE JANSSEN (1997). New Bounds on the List-Chromatic Index of the Complete
Graph and Other Simple Graphs. Combinatorics, Probability and Computing, 6, pp 295-313
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Combinatorics, Probability and Computing (1997) 6, 295–313.
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New Bounds on the List-Chromatic Index of the
Complete Graph and Other Simple Graphs
R O L A N D H Ä G G K V I S T1 and J E A N N E T T E J A N S S E N2 †
1 Department
of Mathematics, University of Umeå,
S-901 87 Umeå, Sweden
(e-mail: [email protected])
2 Department 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
p 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 χ0` (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 Häggkvist [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, χ0` (Kn ) ≤ n.
Theorem 1.2. For all simple graphs G of maximal degree d, the list-chromatic index satisfies
√
χ0` (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 Umeå University.
296
R. Häggkvist 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 Erdős, Rubin and
Taylor, whose paper on choosability (=(vertex) list colourings) [9] created a lot of interest
in the topic. The Erdős, 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 Erdős
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 Häggkvist [11] (χ0` (Km,r ) = m when 2m ≥ 7r), Alon and
Tarsi (verification of Dinitz’s conjecture for m = 4, 6) and Janssen [15, 16] (χ0` (Km,m ) ≤ m+1
and χ0` (Km,r ) = m when m > r).
Bollobás 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)

2∆ − 2



2∆ − 3
χ0` (G) ≤

2∆ − 4


2∆ − 5
and in general, for ∆ ≥ 3917,
χ0` ≤ 2∆ −
if
if
if
if
∆≥3
∆ ≥ 37
∆ ≥ 47
∆ ≥ 56
1p
1
∆−
∆ log ∆
6
3
Chetwynd and Häggkvist [5–7] found a deterministic way to obtain an upper bound of
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 Bollobás [4] improved the
9
5∆
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 Rödl 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 Erdős, Lovász 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 vi vj , 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 xd11 xd22 · · · 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
t−1
t−1
to the monomial xt−1
1 x2 · · · xm 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 xd11 xd22 · · · 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 Erdős conference in Keszthely,
298
R. Häggkvist and J. Janssen
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
New Bounds on the List-Chromatic Index
299
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
Y
(xi − xj )
fG (x1 , x2 , . . . , xn ) =
vi ∼vj
can be written as
fG (x1 , x2 , . . . , xn ) =
Y
(xi − xj ).
i<j
This expression is known to be the determinant of a Vandermonde matrix, which can be
alternatively expressed as
X
fG (x1 , x2 , . . . , xn ) =
(−1)sg(σ) xσ(1)−1
xσ(2)−1
· · · xσ(n)−1
.
n
1
2
σ∈Sn
Writing out fG as the linear combination of monomials, each one of the terms corresponds
Q
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:
X
D
dD dD
fG (x1 , x2 , . . . , xn ) =
(−1)nr. of inv. edges in D x11 x22 . . . xdnn ,
D orient. of G
where dD
i 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,
X
(DEG (ρ) − DOG (ρ))x1ρ(v1 ) x2ρ(v2 ) . . . xnρ(vn ) .
fG (x1 , x2 , . . . , xn ) =
ρ:V →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.
300
R. Häggkvist and J. Janssen
Proof. Let G be as in the statement of the theorem. Then the graph polynomial of G can
be written as follows:
Y
Y
Y
fG =
(xi − xj ) =
(xi − xj ) · · ·
(xi − xj ) = fG1 fG2 · · · fGn .
(2.3a)
vi ∼vj
vi ∼vj
in G1
vi ∼vj
in Gn
As shown in the proof of Lemma 2.2, the coefficient of x1ρ(v1 ) x2ρ(v2 ) · · · xnρ(vn ) is DEG (ρ) −
DOG (ρ). Equality (2.3a) then gives the relation
X
DEG (ρ) − DOG (ρ) =
(DOG1 (ρ1 ) − DEG1 (ρ1 )) · · · (DOGn (ρn ) − DEGn (ρn )),
ρ1 ,ρ2 ,...,ρn
P i
where the sum is taken over all maps ρi : V (Gi ) → N that satisfy
i ρ (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) → N
such 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
New Bounds on the List-Chromatic Index
301
clique Hi by vertices vi1 , . . . , vimi . Define map ρ̄ : V (G) → N as follows:
ρ(v)
if v ∈ V (G)
ρ̄(v) =
bij
if v = vij
for some i, j.
Form S̄ = 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 ρ̄. So if there is a unique such orientation of G, then there is a
unique clique transitive orientation of G obeying ρ̄, and we can conclude (using Lemma
2.3) that DEḠ (ρ̄) − DOḠ (ρ̄) = ±1. So by Theorem 2.1, there exists an S̄-legal vertex
colouring of G. But since G is a subgraph of G and S̄ 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, χ0` (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 (i0 , j 0 ) are joined by
an edge in H precisely when i = i0 ,i = j 0 ,j = i0 or j = j 0 . 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) → N
as follows:
n − 1 if i + j ≥ n − 1
ρ(i, j) =
n − 2 otherwise.
Define values bi , 0 ≤ i < n:


 n−2−i
if
i<
jnk
2
jnk
jnk

 n−1−i+
if
i≥
.
2
2
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.
bi =
302
R. Häggkvist and J. Janssen
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.
8 Set t = t + 1. Go to step 6.
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.
11 Set t = t + 1. Go to step9.
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.
15 Set t = t + 1. Go tostep
13.
n
16 Set i = i + 1. If i < 2 then go to step 2.
17 End.
Note that the assignment given by the
algorithm does not give a conflict with the blocked
n
out values, because foreach
i
≤
2 , out-degree n − 1 − i is assigned to all cliques Cj
n
except to Cj where j = 2 + i, and bb n c+i = n − 1 − i, and out-degree n − 2 − i is assigned
2
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 .
303
New Bounds on the List-Chromatic Index
C0
C1
C2
C3
C4
C5
C0 —
4
4
4
4
4
5
C1 —
4
3
4
4
5
5
C2 —
4
4
2
5
5
5
C3 —
4
4
5
5
5
5
C4 —
4
5
5
5
4
5
C5 —
5
5
5
5
5
3
The following matrices show the horizontal and vertical degrees assigned after, respectively, 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 .
C0
C0 —
C1
C2
C3
C4
4
C1 —
C5
50
3
50
C0
C0 —
C1
C2
C3
4
C1 —
3
50
2
50
C2 —
2
50
C3 —
05
5
C3 —
05
5
05
4
C5 — 05
3
C5
50
C2 —
C4 —
C4
C4 —
50
C5 — 05
50
50
05
05
4
3
The next two matrices show the horizontal and vertical degrees assigned after, respectively, 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.
C0
C1
4
04
C1 — 40
3
C0 —
C2
C3
50
C2 —
2
50
C3 —
05
5
C4 —
05
C5 — 05
50
C4
50
C5
C0
C1
50
C0 —
4
04
05
C1 — 40
3
05
4
3
C2
C3
C4
C5
50
50
05
C2 —
2
50
05
41
C3 —
05
5
41
14
4
C4 —
05
50
14
C5 — 05
50
14
41
3
304
R. Häggkvist and J. Janssen
The next matrix shows the assignment after step 10 and 14 for i = 1 and t = 3, 4, 5.
C0
C1
C2
C3
4
04
24
42
C1 — 40
3
42
C2 — 42
24
2
C0 —
C3 — 24
C4
C5
50
50
05
50
05
41
05
5
41
14
4
C4 —
05
50
41
C5 — 05
50
41
14
3
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.
C0
C1
C2
C3
C4
C5
4
51
31
13
22
05
C0 —
C1 — 15
3
13
22
05
50
C1 —
C2 — 13
31
2
05
50
14
C2 —
C3 — 31
22
50
5
14
41
C3 —
C4 — 22
50
05
41
4
32
C4 —
C5 — 50
05
41
14
23
3
C5 —
C0 —
C0
C1
C2
C3
C4
C5
4
3
2
5
4
3
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.
New Bounds on the List-Chromatic Index
305
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 outdegrees 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
main antidiagonal – and to all vertices (x, y) with x + y = 2i + 1 (mod n − 1) and
the
n
2 + 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 − i
in 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,
l ≥ n − 2 − t. If l = n − 2−n t then l + t + 2i + 1 = 2i (mod n − 1) and, since
i.e.
n
2 − i ≤ t < n − 1 − 2i, 2i ≤ l < 2 + 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.
Suppose P holds before step 10. Out-degree n − 2 − i has to be assigned to a vertex
(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.
Suppose P holds before step 14. Out-degree n − 2 − i has to be assigned to a vertex
(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
306
R. Häggkvist and J. Janssen
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 χ0l (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:
 
 i + j + n − 1, i + j − n + 1
if i + j = n − 1 (mod 2) and i + j ≥ n − 1


2
2





i+j−n+1 i+j+n−1


,
if i + j = n (mod 2) and i + j ≥ n − 1

2
2
α(i, j) =
i+j i+j



,
if i + j = 0 (mod 2) and i + j < n − 1
n−2−


2
2




i+j+3
i+j−1



,n −
if i + j = 1 (mod 2)and i + j < n − 1.
2
2
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 GhHi is constructed from a graph G = (V , E) and
a family of graphs H = {He | e ∈ E}, each He with bipartition (We1 , We2 ), where We1 and
We2 are labelled and |We1 | = |We2 | = 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
1
1
2
2
, . . . , we,m
} and We2 = {we,1
, . . . , we,m
} for each e ∈ E, then GhHi =
words, if We1 = {we,1
(V, E) can be characterized as follows:
New Bounds on the List-Chromatic Index
307
Figure 1
S
• 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
1
2
∼ we,j
in He .
we,i
Note that the maximal degree of GhHi 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 hHi, where H is a family of one-regular graphs on 2k vertices.
Then χ0` (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) ∼ (i0 , j 0 ) in L exactly when i = i0 ,
j = j 0 , i = j 0 or i0 = 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
ρ̄ : V (L) → N and blocking values b̄i , 0 ≤ i < nm as follows:
ρ̄(im + r, jm + r0 ) = ρ(i, j)
b̄im+r = bi
for all
for all
0 ≤ r, r0 < m, 0 ≤ i, j < n
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 ρ̄ to all vertices of L corresponding to edges
from Hij in Kn hHi.
We will show that there exists a unique clique transitive orientation of L that obeys
ρ̄ with value b̄i 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
308
R. Häggkvist and J. Janssen
such that at each vertex v ∈ V (L) these out-degrees add up to ρ̄(v), and for each i, clique
Ci contains all out-degrees {0, 1, . . . , n} − {b̄i } exactly once. Such an assignment uniquely
determines a clique transitive orientation of L obeying ρ̄.
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+r0 ,
0 ≤ r0 < m. But since Hst is a one-regular graph, there is only one of the cliques Ctm+r0 ,
0 ≤ r0 < 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 ρ̄ with b̄i
blocked out in clique Ci for all i, and the maximal value of ρ̄ is n − 1. So by Proposition
2.4 the list chromatic index of G is at most n.
By the designated orientation of Kn hHi 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 hHi} is the vertex set of the line graph of Kn hHi, 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 hHi, β can be obtained
from α, the map that defines the designated orientation of Kn , as follows. For all i, j, r, r0 ,
0 ≤ i, j < n, 0 < r, r0 < m,
β(im + r, jm + r0 ) = 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 hHi i, where Hi is a family of
one-regular graphs on 2m vertices. Then χ0l (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 hHi i, where Hi is a family of one-regular graphs on 2m vertices. Let L be the line graph
S
of G, with vertex set V (L) = k−1
s=0 Vs , where Vs = {(i, j, s)| i < j, (vi , vj ) is an edge in Gs }.
0 0 0
Two vertices (i, j, s) and (i , j , s ) are joined by an edge whenever i = i0 , j = j 0 , i = j 0 or
j = i0 .
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
New Bounds on the List-Chromatic Index
309
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, r0 < m,
D−s
if i + j ≥ n − 1
0
ρ(im + r, jm + r , s) =
D−s−1
if i + j < n − 1.
For each i, t, 0 ≤ i < n, 0 ≤ t < m,
jnk
j n k

 D−1−i−s
+1
if
i<
2 k
j k
j
j n2 k
bs,in+t =
 d−i+ n −s n +1
if
i≥
,
2
2
2
and when n is even, define additionally for each s < k − 1 and all i, t, 0 ≤ i < n, 0 ≤ t < m,
jnk
jnk
 jnk

−1+s
if
i<
2 j k
2 k
j
j n2 k
b̄s,in+t =
 D− n −s n +1
if
i≥
.
2
2
2
Also, let Bl = {b0,l , b1,l , . . . , bk−1,l }, and when n is even, let B̄l = {b̄0,l , b̄1,l , . . . , b̄k−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 ∪ B̄i 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 hHs i. Now
define β : V (L) → N × N as follows. If for some s, i, j, βs (i, j) = (x, y), then
j n k
jnk

 (x + s
,D − s
+ 1 − (n − 1 − y)
if x ≤ y,
2 k
2
j
k
j
β(i, j, s) =
 D − s n + 1 − (n − 1 − x), y + s n
if x > y.
2
2
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 ∪ B̄i ) 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 ∪ B̄t 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
n
by
β
are
exactly
those
out-degrees
l
for
which
s
≤
l
<
(s + 1) n2 or
vertices
in
V
s
2
n
n
D−s( 2 +1) ≥ l > D−(s+1)( 2 +1). Now suppose that out-degrees D, D−1, . . . , D−q+1
310
R. Häggkvist and J. Janssen
n
have been
n assigned to vertices of L according to β. Let s be such that s( 2 + 1) ≤ q <
(s + 1)( 2 + 1). The assignment then implies that all vertices in V0 , V1 , . . . , Vs−1 already
S
have assigned out-degrees. Hence out-degree D−q must be assigned to a vertex of k−1
i=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
S
0
to be assigned to a vertex of k−1
i=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 + r0 , s + 1) with the property
that 0 ≤ r, r0 < 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 + r0 , 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 .
between
Now βrestricted
to V
ns gives a one-to-one correspondence
n
n assignments of outn
degrees s 2 , . . . , (s + 1) 2 − 1 and D − (s + 1)( 2 + 1) + 1, . . . , D − s( 2 + 1) to the vertices
of Vs and assignments of out-degrees 0, 1, . . . , n − 1 to the vertex set of L(Kn hH s i), the line
graph of Kn hH s i, where both assignments correspond to clique transitive orientations.
Moreover, an assignment of those out-degrees to Vs obeys ρ and blocking values bs,i
precisely when the corresponding assignment in L(Kn hH s i) 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 −
in all cliques according to βs , then the only way to assign out-degree
1 − s n2 ) are assigned
n
n − 1 − (q − s 2 ) 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
n to β.
But when D − q has been assigned to all cliques Cim+r with i < q − s( 2 + 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
correspondence it then has to be assigned in all cliques Cim+r with
Vs . By the one-to-one
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 Häggkvist [7], where it is
New Bounds on the List-Chromatic Index
311
proved using the Erdős & Lovász 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 Häggkvist and Thomason [13].
Lemma 5.1. (Chetwynd and Häggkvist) Let G be a simple graph of maximal degree d.
d
k
Then, for every integer p = 2k such that 60 log
3d > 2 there exists a partition of the vertices
of G into p parts V1 , . . . , Vp such that the degree d(v,
q Vi ) from vertex v to any vertex in Vi
differs from its expected value d(v)/p by at most 3
d
p
log 3d.
Theorem 5.2. Let G be a simple graph of maximal degree d on n vertices. Then, for every
d
2k , G can be embedded in a graph I ∪ Ḡ, where Ḡ is the
integer p = 2k such that 60 log
3d > q
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 G0 by adding vertices (but no edges)
to G until each Vi has size t. Note that G0 also has the property that d(v, Vi ) ≤ s for
all i and all v ∈ V . Label the vertices of G0 such that the vertices of Vi have indices
it, it + 1, . . . , it + (t − 1).
Let G0ij 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 G0 , and vi ∈ Vi , vj ∈ Vj . So G0ij is the subgraph of G0
that contains all edges between Vi and Vj . Each G0ij 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
G0ij . Graphs G0b , b = 1, . . . , s, are the graphs which have as edge set all edges of colour
cb in G0 , and for each b = 1, . . . , s, Ḡb is the graph obtained from G0b 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 Ḡb is the edge composition Kp hHb i, where
Hb = {Hijb | 0 ≤ i, j < p}, and Hijb is the perfect matching in Vij that contains colour cb .
S
Hence Ḡ = b Ḡb is as in the statement of the theorem.
S
Let I be the graph with vertex set i Vi , and as its edge set all edges {v, w} of G0 with
v, w ∈ Vi for some i. So I is the disconnected union of the induced subgraphs of G0 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 G0 that go from one vertex set Vi to another can be embedded in the
family of Ḡb , and the edges that stay inside one of the Vi are present in I. Therefore, G0 ,
and thus also G, can be embedded in the union of I and Ḡ.
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.
312
R. Häggkvist and J. Janssen
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 χ0` (G) ≤ d + 23d2/3 log 3d.
Proof. Let G be a simple graph of maximal degree d. By Theorem 5.2, for every integer
d
k
I has
p = 2k such that 60 log
3d > 2 , G can be embedded in a graph I ∪ Ḡ, where where
q
d
maximum degree at most s, and Ḡ is the edge-disjoint union of s = d/p + 3 p log 3d
graphs, each of which is the edge composition of the complete graph Kp with a family of
one-regular graphs. Note that Ḡ has degree ps.
q
√
By Theorem 4.2, χ0l (Ḡ) ≤ ps + 3s − 1 = d + 3 dp log 3d + 3(d/p + 3 dp log 3d) − 1. We
know that χ0` (I) ≤ 2s; if lists of size 2s are given to each edge, then the edges can be
list-coloured greedily. Now χ0` (G) ≤ χ0` (Ḡ) + 2s. Namely, if lists of size χ0` (Ḡ) + 2s are
assigned to all edges, then first I can be list-coloured, and then for each edge in Ḡ, 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 χ0` (Ḡ) to list-colour the edges
q of Ḡ.
√
0
So χl (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 bepthe power of 2 closest to d1/3 , in which case ps + 5s − 1 =
p
√
d + 3 d4/3 log 3d + 5(d2/3 + 3 d2/3 log 3d) − 1 ≤ 23d2/3 log 3d.
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