Let $C_n^k$ be the $k$-th power of a cycle on $n$ vertices (i.e. the vertices
of $C_n^k$ are those of the $n$-cycle, and two vertices are connected by an
edge if their distance along the cycle is at most $k$). For each vertex draw
uniformly at random a subset of size $c$ from a base set $S$ of size $s=s(n)$.
In this paper we solve the problem of determining the asymptotic probability of
the existence of a proper colouring from the lists for all fixed values of
$c,k$, and growing $n$.Comment: 7 page

Krivelevich, Michael

Nachmias, Asaf

English

arXiv

AbstractLet Cnk be the kth power of a cycle on n vertices (i.e. the vertices of Cnk are those of the n-cycle, and two vertices are connected by an edge if their distance along the cycle is at most k). For each vertex draw uniformly at random a list of size c from a base set S of size s=s(n). In this paper we solve the problem of determining the asymptotic probability of the existence of a proper colouring from the random lists for all fixed values of c,k, and growing n

Elsevier - Publisher Connector 

Colouring powers of cycles from random lists 

European Journal of Combinatorics 25 (2004) 961–968
www.elsevier.com/locate/ejc
Colouring powers of cycles from random lists
Michael Krivelevich, Asaf Nachmias
Department of Mathematics, Raymond and Beverly Sackler Faculty of Exact Sciences, Tel Aviv University,
Tel Aviv 69978, Israel
Received 3 September 2003; received in revised form 27 December 2003; accepted 28 December 2003
Available online 16 January 2004
Abstract
Let Ckn be the kth power of a cycle on n vertices (i.e. the vertices of Ckn are those of the n-cycle,
and two vertices are connected by an edge if their distance along the cycle is at most k). For each
vertex draw uniformly at random a list of size c from a base set S of size s = s(n). In this paper we
solve the problem of determining the asymptotic probability of the existence of a proper colouring
from the random lists for all fixed values of c, k, and growing n.
© 2004 Elsevier Ltd. All rights reserved.
1. Introduction
Let G be a simple and undirected graph. Assign to each vertex x of G a set L(x) of
colours (positive integers). Such an assignment L of sets to vertices in G is referred to as a
colour scheme for G. An L-colouring of G is a mapping f of V (G) into the set of colours
such that f (x) ∈ L(x) for all x ∈ V (G) and f (x) = f (y) for each (x, y) ∈ E(G). If G
admits an L-colouring, then G is said to be L-colourable. In the case of L(x) = {1, . . . , k}
for all x ∈ V (G), we also use the terms k-colouring and k-colourable respectively. A graph
G is called k-choosable if G is L-colourable for every colour scheme L of G satisfying
|L(x)| = k for all x ∈ V (G). The chromatic number χ(G) (choice number ch(G)) of G is
the least integer k such that G is k-colourable (k-choosable).
The choosability concept was introduced, independently by Vizing [4] and by Erdo˝s,
Rubin and Taylor [2].
Denote by Ckn the kth power of a cycle on n vertices, i.e. V (Ckn ) = {0, . . . , n − 1} and
(i, j) ∈ E(G) if (i − j) mod n ∈ {−k, . . . ,−1, 1, . . . , k}. Equivalently, the vertices of
Ckn are those of the n-cycle, and two vertices are connected by an edge if their distance
E-mail addresses: krivelev@post.tau.ac.il. (M. Krivelevich), asafnach@post.tau.ac.il. (A. Nachmias).
0195-6698/$ - see front matter © 2004 Elsevier Ltd. All rights reserved.
doi:10.1016/j.ejc.2003.12.002
962 M. Krivelevich, A. Nachmias / European Journal of Combinatorics 25 (2004) 961–968
along the cycle is at most k. Let S be a set of colours of size s(n). For each vertex
of Ckn draw uniformly at random c colours from S to form a colour scheme L(c, k, s).
Denote by p(n, c, k, s(n)) the probability that Ckn is L(c, k, s)-colourable. The problem
of determining what is the minimum s(n) such that p(n, c, k, s(n)) = 1 − o(1) was
posed to us by Maurice Cochand. The problem originated in the chemical industry and
it is related to scheduling problems occurring in the production of colorants. The goal is to
get an estimate of the resources required for the production process. Basically, the vertices
represent jobs, the colours correspond to processors that perform the jobs, and the edges
represent technological restrictions on the processors’ assignments.
Of course, the same problem of colourability from randomly chosen lists can be posed
for other asymptotic families such as complete graphs Kn and complete bipartite graphs
Km,n as well. We plan to return to this question later.
In this paper we investigate the asymptotic behaviour of p(n) = p(n, c, k, s) for every
fixed c, k and n tending to infinity. We show that if c ≤ k,
p(n) =


o(1), s(n) = o(n1/c2),
e−(
k
c)t
−c2 (c!)c + o(1), s(n) ∼ tn1/c2 ,
1 − o(1), s(n) = ω(n1/c2).
On the other hand, if c = k + 1,
p(n) =
{
o(1), s(n) < c,
1 − o(1), s(n) > c,
and if c > k + 1
p(n) =
{
o(1), s(n) < c,
1 − o(1), s(n) ≥ c.
(The limit is easily seen not to exist for the case s = c = k + 1.)
We shall use the standard asymptotic notation and assumption. In particular we assume
that the parameter n is large enough whenever necessary. For two functions f (n) and g(n),
we write f = o(g) if limn→∞ f/g = 0, and f = ω(g) if g = o( f ). Also, f = O(g)
if there exists an absolute constant c > 0 such that f (n) < cg(n) for all large enough n;
f = Θ(g) if both f = O(g) and g = O( f ) hold; and f ∼ g if limn→∞ f/g = 1.
2. Case c ≤ k
For all results in this section we assume c ≤ k. We shall see that with probability tending
to 1 the graph becomes L-colourable at the same time cliques of size c + 1 with the same
list drawn for every vertex, vanish.
Proposition 2.1. If s(n) = o(n1/c2), then p(n) = o(1).
Proof. Every set {v1, . . . , vc+1} of c + 1 vertices of Ckn satisfies:
Pr[v1, . . . , vc+1 have the same list] =
(
s
c
)−c
.
M. Krivelevich, A. Nachmias / European Journal of Combinatorics 25 (2004) 961–968 963
Partition the first (c + 1) n
c+1	 vertices of V (Ckn ) into  nc+1	 disjoint sets of c + 1
consecutive (in the sense of the cycle) vertices. Clearly, the events “The set gets the same
list drawn for all its vertices” are mutually independent. Therefore, the probability that
none of those sets has the same list is:[
1 − 1(
s
c
)c
]n/(c+1)	
= o(1).
Hence with probability tending to 1, there exists a set of c + 1 consecutive vertices, which
forms a clique in Ckn , with the same list drawn for every vertex. Therefore, Ckn cannot be
coloured. 
Theorem 2.2. If s(n) = ω(n1/c2), then p(n) = 1 − o(1).
Proof. Fix with foresight a constant d satisfying d > c
2(c2+c−1)
c−1 . Order the vertices
v1, . . . , vn according to the cycle, with an arbitrary starting point. Call a colour scheme
L of Ckn good if it satisfies the following conditions:
1. There are no c + 1-cliques with the same list drawn for every vertex of the clique.
2. There exists a family of sets of k+1 consecutive vertices (in what follows: k+1-sets)
such that:
(a) The lists drawn for each of the sets do not intersect with any of the 2k lists of its
neighbours.
(b) The number of vertices between two such consecutive k +1-sets is no more than
n1/d .
3. Every set of at most n1/d consecutive vertices, U , has the following property: every
subset X ⊂ U , |X | ≥ c + 2, satisfies
|X | ≤
∣∣∣∣∣
⋃
x∈X
L(x)
∣∣∣∣∣ .
We prove the theorem in two steps. First we show that if L is good, then Ckn is L-colourable.
Secondly, we show that with probability tending to 1, L(c, k, s(n)) is good.
Assume L is good, then every set of at most n1/d consecutive vertices, W , can be
coloured from L. To see this construct a bipartite graph, HL ,W with sides W, A where
A = S and (v, a) ∈ E(H ) if a ∈ L(v). By Condition 3, |X | ≤ |N(X)| for all X of size
|X | ≥ c+2. Since the degree of every v ∈ W is c, it follows that |X |−|N(X)| ≤ 1 for every
X ⊂ W . By the defect version of Hall’s theorem (see [5]) there exists a matching in H
which saturates all but at most one vertex of W . This matching ensures us a legal choice of
colours for the saturated vertices. Furthermore, it is clear that if x ∈ W is the unsaturated
vertex, then x is in some X ⊂ W where |X | − |N(X)| = 1. This can only happen if
|X | = c + 1, meaning also that all the lists of X’s vertices are identical. By Condition 1, X
is not a clique. Colour two non-adjacent vertices of X with the same colour, if x is still not
coloured, colour it with the available colour. This clearly completes the matching and thus
completes an L-colouring of W .
964 M. Krivelevich, A. Nachmias / European Journal of Combinatorics 25 (2004) 961–968
Take a family of k + 1-sets as in Condition 2, find an L-colouring of every k + 1-set
and an L-colouring of every run of vertices between consecutive k + 1-sets. This is clearly
possible by the above argument and this completes an L-colouring of Ckn .
We show now that with probability tending to 1, L(c, k, s(n)) is good.
Condition 1. Let X be the random variable counting the number of c+1-cliques with the
same list drawn for every vertex. The number of cliques in Ckn is n
(k
c
) (choose the leftmost
vertex of the clique in n possible manners, then choose k vertices from its c neighbours to
the right in
(k
c
)
possible manners), so clearly:
E[X] = n
(
k
c
)(
s
c
)−c
= o(1),
so Pr[X ≥ 1] = o(1).
Condition 2. For a fixed k + 1-set, the probability that its lists do not intersect with its
neighbours’ lists is clearly more than:[(
s−2kc
c
)
(
s
c
)
]k+1
= 1 − Θ
(
1
s
)
;
hence, the chance that the k +1-set’s lists are not entirely disjoint from its neighbours’ lists
is at most Θ(1/s).
Partition the first (3k + 3) n3k+3	 vertices of V (Ckn ) into disjoint k + 1-sets, with distance
of 2k + 1 between them (this is a mere technicality to keep the events independent). So,
by the union bound, the event that there exists a run of at least n1/d consecutive vertices
in which every k + 1-set (in the former division) has failed to satisfy the list disjointness
condition occurs with probability at most:
n[Θ(1/s)]Θ(n1/d) = o(1).
Condition 3. We show that for a given set of n1/d	 consecutive vertices, W , the
probability that Condition 3 does not hold is o(1/n), thus with probability tending to 1,
every such W satisfies Condition 3. This obviously implies that, with probability tending
to 1, every set of at most n1/d vertices satisfies Condition 3. For a fixed W , construct a
bipartite graph HL ,W as before. The probability that Condition 3 does not hold is clearly:
Pr[∃X ⊂ W, |X | ≥ c + 2, |X | > |N(X)|] ≤
n1/d	∑
i=c+2
(n1/d	
i
)(
s
i − 1
)[(i−1
c
)
(
s
c
)
]i
.
The first term in the summation implies the choice of such X , the second implies the
choice of N(X) and the third is the probability that the neighbours of X are in N(X), i.e.
that the lists of the vertices of X were chosen from the i − 1 colours (at most) of N(X).
This is o( 1
n
):
n1/d 	∑
i=c+2
(n1/d	
i
)(
s
i − 1
)[(i−1
c
)
(
s
c
)
]i
≤
n1/d∑
i=c+2
(
en1/d
i
)i (
es
i − 1
)i−1 
(
e(i−1)
c
)c
(
s
c
)c


i
M. Krivelevich, A. Nachmias / European Journal of Combinatorics 25 (2004) 961–968 965
≤
n1/d∑
i=c+2
(
ec+2n1/d(i − 1)c−2
sc−1
)i
i − 1
es
≤ n
1/d
es
n1/d∑
i=c+2
(
ec+2n c−1d
sc−1
)i
≤ n
1/d
es
(
ec+2n
c−1
d
sc−1
)c+2
1 −
(
ec+2n
c−1
d
sc−1
)
= e(c+2)2(1 + o(1))n
c2+c−1
d
sc
2
sc−1
≤ e(c+2)2(1 + o(1))n
c2+c−1
d − c−1c2
sc
2
= o
(
1
n
)
. 
Theorem 2.3. For any constant t > 0, if s(n) ∼ tn1/c2 , then limn→∞ p(n) =
e−(
k
c)t
−c2 (c!)c
.
Proof. The same calculations as before show that if s(n) ∼ tn1/c2 , Conditions 2 and 3
hold with probability tending to 1. Thus, the problem reduces to calculating the asymptotic
probability of the appearance of a c + 1-clique with identical lists drawn for each vertex.
Using Brun’s Sieve (see, e.g. [1, Chapter 8]) we show that the number of such c+1-cliques
has an asymptotically Poisson distribution.
Let m = n(k
c
)
be the number of c + 1-cliques in Ckn , as explained previously. Let Bi ,
i = 1, . . . , m = n(k
c
)
, be the event that the i th c + 1-clique has the same list drawn for it.
For disjoint c + 1-cliques, it is clear that the corresponding Bi ’s are independent. Let Xi ,
i = 1, . . . , m, be Bi ’s indicator random variables, and let X = ∑i=1,...,m Xi denote the
random variable counting the number of c + 1-cliques having the same list drawn for its
vertices. We wish to estimate the probability that X = 0. Define:
µ =
(
k
c
)
t−c2(c!)c,
S(r) =
∑
i1,...,ir
Pr[Bi1 ∩ · · · ∩ Bir ],
where the sum is over all sets {i1, . . . , ir } ⊂ {1, . . . , m}. Now:
E[X] = S(1) =
m∑
i=1
Pr[Bi ] = n
(
k
c
)(
s
c
)−c
→ µ.
966 M. Krivelevich, A. Nachmias / European Journal of Combinatorics 25 (2004) 961–968
For r ≥ 2, divide the sum S(r) into two parts:
S(r) =
∗∑
i1,...,ir
Pr[Bi1 ∩ · · · ∩ Bir ] +
∗∗∑
i1,...,ir
Pr[Bi1 ∩ · · · ∩ Bir ],
where the first summation goes over all r -tuples of pairwise disjoint c + 1-cliques and the
second goes over all r -tuples of c + 1-cliques which are not pairwise disjoint. Note that
every c + 1-clique intersects with a constant number of other c + 1-cliques, therefore, the
number of terms in the first summation is
(
m
r
)− O(nr−1) and:
∗∑
i1,...,ir
Pr[Bi1 ∩ · · · ∩ Bir ] =
[(
m
r
)
− O(nr−1)
][(
s
c
)−cr]
→ µ
r
r ! .
To deal with the second summation, note that for every 1 ≤ i ≤ r − 1 there are Θ(ni )
manners to choose r c + 1-cliques such that the size of the maximal disjoint family is i .
The probability that every c + 1-clique in this maximal disjoint family has the same list
drawn is clearly
(
s
c
)−ci
. If i < r , then there exists at least one vertex of a clique not in
the maximal disjoint family whose list is identical to one of the lists of the cliques in the
maximal disjoint family. Therefore:
∗∗∑
i1,...,ir
Pr[Bi1 ∩ · · · ∩ Bir ] ≤
r−1∑
i=1
Θ(ni )
1(
s
c
)ci 1(s
c
) = o(1).
Thus S(r) → µr/r !, and by Brun’s Sieve, Pr[X = 0] → e−µ. It follows by the above
discussion that p(n) is also asymptotic to e−µ. 
3. Case c > k
For all results in this section we assume c > k.
A recent result of Prowse and Woodall [3] states that for all values of n and k,
ch(Ckn ) = χ(Ckn ). It is easy to see that when n ≥ k(k + 1), if k + 1 divides n then
χ(Ckn ) = k + 1, otherwise χ(Ckn ) = k + 2. Their result makes the following proposition
true:
Proposition 3.1. If s = s(n) satisfies s ≥ c ≥ k + 2, then p(n) = 1 − o(1).
Furthermore, it is trivial that if s = c = k + 1, no limit exists for the probability of a
legal colour assignment. We present an argument settling the case c = k + 1 and s > c,
which also implies Proposition 3.1 without the use of the result of Prowse and Woodall.
Theorem 3.2. 1. If c = k + 1 and s > c, then p(n) = 1 − o(1).
2. If c > k + 1 and s ≥ c, then p(n) = 1 − o(1).
Proof. If s = ω(1), then as shown before, for an arbitrary set of k + 1 consecutive
vertices, the probability that its lists intersect with its 2k neighbours lists is Θ(1/s). So,
with probability tending to 1, a fixed set of k+1 consecutive vertices will have lists disjoint
from its neighbours’ lists, which allows us to first colour the k + 1-set and then complete
M. Krivelevich, A. Nachmias / European Journal of Combinatorics 25 (2004) 961–968 967
the colouring greedily: colour by the order of the cycle, for each vertex choose a colour not
chosen by its preceding neighbours.
Consider now the case s = O(1). We prove the theorem for c = k + 1, s = k + 2.
This implies Theorem 3.2, since increasing c increases the probability of a colouring, and
in this case, the proof works with higher values of s as long as it stays bounded. Assume
then S = {1, . . . , k + 2}.
Given a set of 2[(k + 1)2 + (k + 1)] consecutive vertices, divide the set into the first and
last k + 1-sets, and into k + 1 pairs of disjoint k + 1-sets. Such a set will be called good if
it satisfies the following conditions:
1. Each vertex of the first and the last k + 1-set has the list {1, . . . , k + 1}.
2. For every 1 ≤ i ≤ k +1, each vertex of the second member of the i th pair has the list
{1, . . . , k + 1} and each vertex of the first member has the list {1, . . . , k + 2} − {i}.
The probability that a set of consecutive 2[(k + 1)2 + (k + 1)] vertices satisfies these
conditions is:(
1
k + 2
)2[(k+1)2+(k+1)]
.
After choosing  n2[(k+1)2+(k+1)] 	 disjoint consecutive sets in V (Ckn ) in the usual manner,
the probability that none of them has the above properties tends to 0 and therefore a good
set exists with probability tending to 1.
Assume we have a good set, colour Ckn in the following manner:
Colour the last k + 1-set with (1, . . . , k + 1) by that order. Continue colouring greedily
the vertices after the last k +1-set, by the order of the cycle (again, for each vertex choose a
colour not chosen by its preceding neighbours) until we complete colouring the first k + 1-
set (this can be done since every vertex after the first k + 1-set has exactly k coloured
neighbours). For each 1 ≤ i ≤ k + 1 colour as follows:
To the first i − 1 vertices of the first member of the i th pair, give colours (1, . . . , i − 1)
by that order. Give the i th vertex of that member the colour k + 2, and colour the rest of
the first member greedily. To the first i vertices of the second member of the i th pair, give
colours (1, . . . , i) by that order. Give the rest of the vertices the same colours as in the
matching vertices of the first member of the i th pair. It is easy to check that this colouring
is possible, and hence the theorem is proven. 
Acknowledgements
The authors wish to thank Maurice Cochand for posing the problem and for explaining
its motivation/industrial interpretation. The first author’s research was supported in part by
a USA–Israel BSF Grant and by a grant from the Israel Science Foundation.
References
[1] N. Alon, J.H. Spencer, The Probabilistic Method, second ed., Wiley, New York, 2000.
[2] P. Erdo˝s, A.L. Rubin, H. Taylor, Choosability in graphs, in: Proc. West Coast Conf. on Combinatorics, Graph
Theory and Computing, Congressus Numerantium XXVI, 1979, pp. 125–157.
968 M. Krivelevich, A. Nachmias / European Journal of Combinatorics 25 (2004) 961–968
[3] A. Prowse, D.R. Woodall, Choosability of powers of circuits, Graphs Combin. 19 (2003) 137–144.
[4] V.G. Vizing, Colouring the vertices of a graph with prescribed colours, Metody Diskretnogo Analiza Teorii
Kodov i Skhem (29) (1976) 3–10 (in Russian).
[5] D.B. West, Introduction to Graph Theory, second ed., Prentice Hall, 2001.


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Colouring powers of cycles from random lists

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