Colourings of the Cartesian Product of Graphs and Multiplicative Sidon Sets

Abstract

Let F be a family of connected bipartite graphs, each with at least three vertices. A proper vertex colouring of a graph G with no bichromatic subgraph in F is F-free. The F-free chromatic numberχ(G,F) of a graph G is the minimum number of colours in an F-free colouring of G. For appropriate choices of F, several well-known types of colourings fit into this framework, including acyclic colourings, star colourings, and distance-2 colourings. This paper studies F-free colourings of the cartesian product of graphs. Let H be the cartesian product of the graphs G1,G2,…,Gd. Our main result establishes an upper bound on the F-free chromatic number of H in terms of the maximum F-free chromatic number of the Gi and the following number-theoretic concept. A set S of natural numbers is k-multiplicative Sidon if ax=by implies a=b and x=y whenever x,y∈S and 1⩽a,b⩽k. Suppose that χ(Gi,F)⩽k and S is a k-multiplicative Sidon set of cardinality d. We prove that χ(H,F)⩽1+2k⋅maxS. We then prove that the maximum density of a k-multiplicative Sidon set is Θ(1/logk). It follows that χ(H,F)⩽O(dklogk).

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arXiv:math/0511262v2 [math.CO] 20 Oct 2008
COLOURINGS OF THE CARTESIAN PRODUCT OF GRAPHS
AND MULTIPLICATIVE SIDON SETS
ATTILA P´OR AND DAVID R. WOOD
Abstract. Let F be a family of connected bipartite graphs, each with at least three
vertices. A proper vertex colouring of a graph G with no bichromatic subgraph in
F is F-free. The F-free chromatic number χ(G,F) of a graph G is the minimum
number of colours in an F-free colouring of G. For appropriate choices of F, several
well-known types of colourings fit into this framework, including acyclic colourings,
star colourings, and distance-2 colourings. This paper studies F-free colourings of the
cartesian product of graphs.
Let H be the cartesian product of the graphs G1,G2,...,Gd. Our main result
establishes an upper bound on the F-free chromatic number of H in terms of the
maximum F-free chromatic number of the Gi and the following number-theoretic
concept. A set S of natural numbers is k-multiplicative Sidon if ax = by implies a = b
and x = y whenever x,y ∈ S and 1 ≤ a,b ≤ k. Suppose that χ(Gi,F) ≤ k and S is a
k-multiplicative Sidon set of cardinality d. We prove that χ(H,F) ≤ 1+2k·maxS. We
then prove that the maximum density of a k-multiplicative Sidon set is Θ(1/logk).
It follows that χ(H,F) ≤ O(dk logk).We illustrate the method with numerous
examples, some of which generalise or improve upon existing results in the literature.
1. Introduction
Sabidussi [50] proved that the chromatic number of the cartesian product of a set of
graphs equals the maximum chromatic number of a graph in the set. No such result is
known for more restrictive colourings (such as acyclic, star, and distance-2 colourings).
This paper investigates such colourings of cartesian products under a general model of
restriction, in which arbitrary bichromatic subgraphs are excluded. Our study leads to
a number-theoretic problem regarding multiplicative Sidon sets that is of independent
interest. This problem is then solved using a combination of number-theoretic and
graph-theoretic approaches.
Date: November 10, 2005. Revised: October 20, 2008.
2000 Mathematics Subject Classification. 05C15 (coloring of graphs and hypergraphs), 11N99 (mul-
tiplicative number theory).
Key words and phrases. graph, colouring, cartesian product, distance-2 colouring, acyclic colouring,
star colouring, L(p,1)-labelling, grid, toroidal grid, multiplicative Sidon set.
This is an amplified version of a paper to appear in Combinatorica. Research initiated at the Depart-
ment of Applied Mathematics and the Institute for Theoretical Computer Science, Charles University,
Prague, Czech Republic. Supported by project LN00A056 of the Ministry of Education of the Czech Re-
public, and by the European Union Research Training Network COMBSTRU (Combinatorial Structure
of Intractable Problems).
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Let G be a graph with vertex set V (G) and edge set E(G). (All graphs considered
are undirected, simple, and finite.) A colouring of G is a function c : V (G) → Z such
that c(v) ?= c(w) for every edge vw ∈ E(G). A colouring c with |{c(v) : v ∈ V (G)}| ≤ k
is a k-colouring. The chromatic number of G, denoted by χ(G), is the minimum integer
k for which there is a k-colouring of G.
Let F be a family of connected bipartite graphs, each with at least three vertices,
called a forbidden family. A colouring c of a graph G is F-free if it contains no bichro-
matic subgraph in F; that is, |{c(v) : v ∈ V (H)}| ≥ 3 for every subgraph H of G that is
isomorphic to a graph in F. The F-free chromatic number of G, denoted by χ(G,F), is
the minimum integer k for which there is an F-free k-colouring of G. When F = {H} is
a singleton, we write H-free instead of F-free, and refer to the H-free chromatic number
χ(G,H). The framework was introduced by Albertson et al. [7]; an even more general
model of restrictive graph colourings is considered by Neˇ setˇ ril and Ossona de Mendez
[44].
F-free colourings correspond to many well-studied types of colourings. Let Pnand
Cnrespectively be the path and cycle on n vertices. Let C := {Cn: n even}. Then
C-free colourings are the acyclic colourings [5, 6, 9, 10, 11, 12, 13, 14, 15, 16, 16, 17, 18,
23, 27, 30, 33, 37, 38, 41, 59, 60, 61]. Here each bichromatic subgraph is a forest. By
a further restriction we obtain the P4-free colourings, which are called star colourings,
since each bichromatic subgraph is a collection of disjoint stars [7, 9, 10, 18, 26, 43, 60].
A colouring is P3-free if and only if every pair of vertices at distance at most two receive
distinct colours (called a distance-2 colouring). That is, χ(G,P3) = χ(G2). Here Gkis
the k-th power of G, the graph with vertex set V (G), where two vertices are adjacent in
Gkwhenever they are at distance at most k in G. Often motivated by applications in
frequency assignment, colourings of graph powers has recently attracted much attention
[1, 2, 3, 4, 24, 39, 42, 57]. By definition,
χ(G) ≤ χ(G,C) ≤ χ(G,P4) ≤ χ(G,P3) .
Let G1and G2be graphs. The cartesian product of G1and G2, denoted by G1?G2,
is the graph with vertex set
V (G1?G2) := V (G1) × V (G2) := {(a,v) : a ∈ V (G1),v ∈ V (G2)} ,
where (a,v)(b,w) is an edge of G1?G2if and only if ab ∈ E(G1) and v = w, or a = b
and vw ∈ E(G2). Assuming isomorphic graphs are equal, the cartesian product is
associative, and G1?G2? ··· ?Gdis well-defined. Sabidussi [50] proved that
χ(G1?G2? ··· ?Gd) = max{χ(Gi) : 1 ≤ i ≤ d} .
This paper studies F-free colourings of cartesian products. The following upper
bound on the F-free chromatic number of a cartesian product is our main result. Here
and throughout the paper, γ = 0.5772... is Euler’s constant, and logarithms are base
e = 2.718... unless stated otherwise.
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Theorem 1. Let F be a forbidden family. Let G1,G2,...,Gd be graphs, each with
F-free chromatic number χ(Gi,F) ≤ k + 1. Then
χ(G1?G2? ··· ?Gd,F) ≤ 2k(kd − k + 1) + 1 .
Moreover, for all ǫ > 0 and for large d > d(k,ǫ),
χ(G1?G2? ··· ?Gd,F) ≤ 1 +2eγ
We actually prove a stronger result than Theorem 1 that is expressed in terms of
‘chromatic span’. This concept is introduced in Section 2. The key lemma of the paper,
which relates F-free colourings of a cartesian product to so-called k-multiplicative Sidon
sets, is proved in Section 3. In Section 4 we study k-multiplicative Sidon sets in their
own right. The obtained bounds establish our main colouring results. The remaining
sections contain numerous examples of the method, some of which generalise or improve
upon existing results in the literature. In particular, we consider distance-2 colourings
in Section 5, acyclic colourings in Section 6, and star colourings in Section 7. Finally,
in Section 8, we briefly discuss a generalisation of our method for L(p,1)-labellings.
1 − ǫdk logk .
2. Chromatic Span
Let c be a colouring of a graph G. The span of c is max{|c(v) − c(w)| : vw ∈ E(G)}.
(The number of colours is irrelevant.) The chromatic span of G, denoted by Λ(G), is the
minimum integer k for which there is a colouring of G with span k. Note that Λ(G) ≤ k
if and only if there is a homomorphism from G into Pk
Let [a,b] := [a,a+1,...,b] and [b] := [1,b] for all integers a ≤ b. We can assume that
the range of a k-colouring is [k]. Thus Λ(G) ≤ χ(G)−1 for every graph G. Conversely,
given a colouring c of G with span k, let c′(v) := c(v) mod (k + 1) for each vertex
v ∈ V (G). Then c′is a (k + 1)-colouring of G. Thus
Λ(G) = χ(G) − 1 .
This might suggest that chromatic span is pointless. Let the F-free chromatic span of
a graph G, denoted by Λ(G,F), be the minimum integer k for which there is an F-free
colouring of G with span k.
nfor some n.
Lemma 1. Let F be a forbidden family. For every graph G,
Λ(G,F) + 1 ≤ χ(G,F) ≤ 2 · Λ(G,F) + 1 .
Proof. Obviously Λ(G,F) ≤ χ(G,F) − 1. To prove that χ(G,F) ≤ 2 · Λ(G,F) + 1, let
c be an F-free colouring of G with span k := Λ(G,F). For every vertex v ∈ V (G), let
c′(v) := c(v) mod (2k + 1). Clearly c′is a (2k + 1)-colouring of G. For all i ∈ [0,2k],
let Vi := {v ∈ V (G) : c′(v) = i}, and for all j ∈ Z, let Vi,j := {v ∈ Vi : c(v) =
j(2k + 1) + i}. Thus the Vi’s are the colour classes of c′and the Vi,j’s are the colour
classes of c. For S,T ⊆ V (G) with S ∩ T = ∅, let G[S,T] be the subgraph of G with
vertex set S ∪ T and edge set {vw ∈ E(G) : v ∈ S,w ∈ T}. Consider two edges
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vw,xy ∈ E(G) with v,x ∈ Vi1,j1, w ∈ Vi2,j2, and y ∈ Vi2,j3. Since |c(v) − c(w)| ≤ k
and |c(x) − c(y)| ≤ k, we have j2 = j3. It follows that each bichromatic subgraph
of c′is the union of disjoint bichromatic subgraphs of c. In particular, G[Vi1,Vi2] =
∪{G[Vi1,j,Vi2,j] : j ∈ Z} or G[Vi1,Vi2] = ∪{G[Vi1,j,Vi2,j+1] : j ∈ Z}.
subgraph G[Vi1,j,Vi2,j] (or G[Vi1,j,Vi2,j+1] in the second case) is F-free, c′is F-free and
χ(G,F) ≤ 2 · Λ(G,F) + 1.
Lemma 1 cannot be improved in general, since it is easily seen that Λ(Pk
but χ(Pk
n,P3) = 2k + 1. Thus chromatic span is of interest when considering F-free
colourings. We prove the following result, which with Lemma 1, implies Theorem 1.
Since each
?
n,P3) = k
Theorem 2. Let F be a forbidden family. Let G1,G2,...,Gd be graphs, each with
F-free chromatic span Λ(Gi,F) ≤ k (which is implied if χ(Gi,F) ≤ k + 1). Then
Λ(G1?G2? ··· ?Gd,F) ≤ k(kd − k + 1) , and
χ(G1?G2? ··· ?Gd,F) ≤ 2k(kd − k + 1) + 1 .
Moreover, for all ǫ > 0 and for large d > d(k,ǫ),
Λ(G1?G2? ··· ?Gd,F) ≤
χ(G1?G2? ··· ?Gd,F) ≤ 1 +2eγ
eγ
1 − ǫdk logk, and
1 − ǫdklogk .
3. The Key Lemma
Our results depend upon the following number-theoretic concept (where N := {1,2,... }
and N0:= N ∪ {0}).
Definition 1. Let k ∈ N. A set A ⊆ N is k-multiplicative Sidon1if for all x,y ∈ A
and for all a,b ∈ [k], we have ax = by implies a = b and x = y. For brevity we write
k-multiplicative rather than k-multiplicative Sidon.
Consider a cartesian product˜G := G1?G2? ··· ?Gdto have vertex set
V (˜G) = {˜ v : ˜ v = (v1,v2,...,vd),vi∈ V (Gi),i ∈ [d]} ,
where ˜ v ˜ w ∈ E(˜G) if and only if viwi∈ E(Gi) for some i, and vj= wjfor all j ?= i; we
say that the edge ˜ v ˜ w is in dimension i.
Lemma 2. Let F be a forbidden family. Let G1,G2,...,Gdbe graphs, each with F-
free chromatic span Λ(Gi,F) ≤ k (which is implied if χ(Gi,F) ≤ k + 1). Let S :=
{s1,s2,...,sd} be a k-multiplicative set. Then
Λ(G1?G2? ··· ?Gd,F) ≤ k · maxS .
1Erd˝ os [19, 20, 21] defined a set A ⊆ N to be multiplicative Sidon if ab = cd implies {a,b} = {c,d}
for all a,b,c,d ∈ A; see [48, 49, 51]. Additive Sidon sets have been more widely studied; see the classical
papers [22, 53, 54] and the recent survey by O’Bryant [46].
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Proof. Let˜G := G1?G2? ··· ?Gd. For each i ∈ [d], let cibe an F-free colouring of
Giwith span k. For each vertex ˜ v ∈ V (˜G), let
c(˜ v) :=
?
i∈[d]
si· ci(vi) .
For every edge ˜ v ˜ w ∈ E(˜G) in dimension i,
(1)c( ˜ w) − c(˜ v) =

?
j∈[d]
sj· cj(wj)

−

?
j∈[d]
sj· cj(vj)

= si
?ci(wi) − ci(vi)?
.
Since 1 ≤ |ci(wi) − ci(vi)| ≤ k and si≥ 1, c is a colouring of˜G with span k · maxS.
Suppose, for the sake of contradiction, that c is not F-free.
bichromatic subgraph H of˜G that is isomorphic to some graph in F. First suppose
that all the edges of H have the same dimension i. By Equation (1), and since H is
connected, the edges {viwi: ˜ v ˜ w ∈ E(H)} induce a bichromatic subgraph of Gi that
is isomorphic to a graph in F, which is a contradiction. Thus not all the edges of H
are in the same dimension. Since H is connected and has at least three vertices, H
has two edges ˜ v˜ x and ˜ w˜ x with a common endpoint that are in distinct dimensions.
Say ˜ v˜ x is in dimension i and ˜ w˜ x is in dimension j ?= i.
c(˜ v) − c(˜ x) = c( ˜ w) − c(˜ x). By Equation (1),
si
?ci(vi) − ci(xi)?
Since cihas span k, we have 1 ≤ |ci(vi) − ci(xi)| ≤ k and 1 ≤ |cj(wj) − cj(xj)| ≤ k,
which implies that S is not k-multiplicative. This contradiction proves that c is an
F-free colouring of˜G.
That is, there is a
Since H is bichromatic,
= sj
?cj(wj) − cj(xj)?
.
?
4. k-Multiplicative Sidon Sets
Motivated by Lemma 2, in this section we study k-multiplicative sets. We measure
the ‘size’ of a k-multiplicative set by its density. The density of A ⊆ N is
δ(A) := lim
n→∞
|A ∩ [n]|
n
if the limit exists (otherwise the density is undefined). We say A ⊆ N is p-periodic if
x ∈ A if and only if x + p ∈ A for all x ∈ N. Observe that if A is p-periodic then
δ(A) =|A ∩ [p]|
(2)
p
.
The following theorem is our main result regarding k-multiplicative sets.
Theorem 3. For all k ∈ N, the maximum density of a k-multiplicative set is
Θ
?
1
logk
?
.
We start with a naive construction of a k-multiplicative set.
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Lemma 3. For all k ∈ N, the set Rk:= {x ∈ N : x ≡ 1 (mod k)} is k-multiplicative
and has density δ(Rk) = 1/k.
Proof. Suppose that ax = by for some x,y ∈ Rkand a,b ∈ [k]. Then x = pk + 1 and
y = qk + 1 for some p,q ∈ N. Thus (ap − bq)k = b − a. Since |b − a| ≤ k − 1, we have
a = b and ap = bq. Thus p = q and x = y. That is, Rkis k-multiplicative. Since Rkis
k-periodic, δ(Rk) = |Rk∩ [k]|/k = 1/k by Equation (2).
The lower and upper bounds in Theorem 3 are proved in Theorems 4 and 5, respec-
tively. Fix k ∈ N. Let Pk:= {p1,p2,...,pℓ} be the set of primes in [k]. Let
Πk :=
?
?
i∈[ℓ]
pi .
Every x ∈ N can be uniquely represented as
x = β∗(x)
?
i∈[ℓ]
pβi(x)
i
,
where βi(x) ∈ N0and β∗(x) is not divisible by pifor all i ∈ [ℓ]. That is, gcd(β∗(x),Πk) =
1. Let β(x) be the vector (β1(x),β2(x),...,βℓ(x)). For all x,y ∈ N,
(3)β(x · y) = β(x) + β(y) and β∗(x · y) = β∗(x) · β∗(y) .
Lemma 4. For all k ∈ N, if ax = by for some a,b ∈ [k] and x,y ∈ N, then β∗(x) =
β∗(y).
Proof. By Equation (3), we have β∗(a) · β∗(x) = β∗(b) · β∗(y). Since a,b ≤ k, we have
β∗(a) = β∗(b) = 1. Thus β∗(x) = β∗(y).
?
Theorem 4. For all k ∈ N, the set Sk:= {s ∈ N : gcd(s,Πk) = 1} is k-multiplicative
and has density
?
δ(Sk) =
?
i∈[ℓ]
1 −1
pi
?
∼
e−γ
logk.
Proof. Suppose that ax = by for some a,b ∈ [k] and x,y ∈ Sk. Thus β∗(x) = β∗(y) by
Lemma 4. Since gcd(x,Πk) = gcd(y,Πk) = 1, we have βi(x) = βi(y) = 0 for all i ∈ [ℓ].
Hence x = y, which implies that a = b, and Skis k-multiplicative.
It remains to compute the density of Sk. Let ϕ be Euler’s totient function, ϕ(x) :=
|{y ∈ [x] : gcd(x,y) = 1}|. If q1,q2,...,qrare the prime factors of x (with repetition),
then
?
i∈[r]
Observe that Skis Πk-periodic. By Equation (2),
ϕ(x) = x
?
1 −1
qi
?
.
δ(Sk) =|Sk∩ [Πk]|
Πk
=ϕ(Πk)
Πk
=
?
i∈[ℓ]
?
1 −1
pi
?
.
By Mertens’ Theorem (see [31]), δ(Sk) ∼ e−γ/logk; see Table 1.
?
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The following corollary is a straightforward consequence of Theorem 4.
Corollary 1. For all k ∈ N, ǫ > 0, and sufficiently large n > n(k,ǫ),
(1 − ǫ)n
eγlogk
≤ |Sk∩ [n]| ≤(1 + ǫ)n
eγlogk
.
Table 1. The first 15 elements of the set Skfor each k ≤ 30.
Sk
{1,3,5,7,9,11,13,15,17,19,21,23,25,27,29,... }
{1,5,7,11,13,17,19,23,25,29,31,35,37,41,43,... }
{1,7,11,13,17,19,23,29,31,37,41,43,47,49,53,... }
{1,11,13,17,19,23,29,31,37,41,43,47,53,59,61,... }
{1,13,17,19,23,29,31,37,41,43,47,53,59,61,67,... }
{1,17,19,23,29,31,37,41,43,47,53,59,61,67,71,... }
{1,19,23,29,31,37,41,43,47,53,59,61,67,71,73,... }
{1,23,29,31,37,41,43,47,53,59,61,67,71,73,79,... }
{1,29,31,37,41,43,47,53,59,61,67,71,73,79,83,... }
{1,31,37,41,43,47,53,59,61,67,71,73,79,83,89,... }
k
2
density
1/2
1/3
4/15
8/35
16/77
192/1001
3072/17017
55296/323323
110592/676039
442368/2800733
3,4
5,6
7,...,10
11,12
13,...,16
17,18
19,...,22
23,...,28
29,30
We can now prove Theorem 2.
Proof of Theorem 2. Lemma 3 implies that R := {ik+1 : i ∈ [0,d−1]} is k-multiplicative.
Since |R| = d and maxR = dk−k+1, by using R as a k-multiplicative set in Lemma 2,
we have Λ(G1?G2? ··· ?Gd,F) ≤ k(dk − k + 1). This proves the first part of the
theorem.
Let n be the minimum integer such that |Sk∩[n]| ≥ d. By Corollary 1, for d > d(k,ǫ),
max{Sk∩ [n]} ≤ n ≤
eγ
1 − ǫdlogk .
Using Sk∩ [n] as a k-multiplicative set in Lemma 2, we have
Λ(G1?G2? ··· ?Gd,F) ≤
eγ
1 − ǫdk logk .
The final claim in Theorem 2 follows from Lemma 1.
?
4.1. Proof of Optimality. We now prove that the lower bound in Theorem 4 is asymp-
totically optimal, which in turn completes the proof of Theorem 3.
Theorem 5. For all k ∈ N, ǫ > 0, and sufficiently large n > n(k,ǫ), every k-
multiplicative set A ⊆ [n] satisfies
|A| ≤(2 + ǫ)n
eγlogk+2n
4√k
= (2 + o(1))|Sk∩ [n]| =(2 + o(1))n
eγlogk
.
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To prove Theorem 5, we model k-multiplicative sets using graphs. Let Gn,kbe the
graph with vertex set V (Gn,k) := [n], where xy ∈ E(Gn,k) whenever ax = by for
some a,b ∈ [k]. Observe that a set A ⊆ [n] is k-multiplicative if and only if A is an
independent set of Gn,k. For each s ∈ Sk∩[n], let Gn,k,sbe the subgraph of Gn,kinduced
by Xn,k,s:= {x ∈ [n] : β∗(x) = s}.
Lemma 5. The connected components of Gn,kare {Gn,k,s: s ∈ Sk∩ [n]}.
Proof. If xy ∈ E(Gn,k), then β∗(x) = β∗(y) by Lemma 4, which implies that x,y ∈ Xn,k,s
for some s ∈ Sk∩ [n]. Thus distinct sets Xn,k,sand Xn,k,tare not joined by an edge
of Gn,k. It remains to prove that each subgraph Gn,k,sis connected. For each pair of
vertices x,y ∈ Xn,k,s, let
f(x,y) :=
?
i∈[ℓ]
|βi(x) − βi(y)|.
We claim that x and y are connected by a path of f(x,y) edges in Gn,k,s. The proof
is by induction on f(x,y). If f(x,y) = 0 then x = y (since β∗(x) = β∗(y) = s) and we
are done. Say f(x,y) > 0. Without loss of generality, βi(x) < βi(y) for some i. Let
z := pix. Then z ∈ Xn,k,sand xz is an edge of Gn,k,s. Moreover, βi(z) = βi(x) + 1,
which implies that f(z,y) = f(x,y) − 1. By induction, there is a path of f(z,y) edges
from z to y. Thus there is a path of f(z,y) + 1 = f(x,y) edges from x to y.
?
Lemma 6. Let Gn,k,s be a connected component of Gn,k with r vertices. Then the
min{k,r} smallest elements of Xn,k,sare {s,2s,3s,...,min{k,r} · s}, and they form a
clique of Gn,k,s.
Proof. Every element of Xn,k,sis a multiple of s and is at least s. Now is ∈ Xn,k,sfor
each i ∈ [min{k,r}]. Thus the min{k,r} smallest elements of Xn,k,sare {s,2s,3s,...,min{k,r}·
s}, which clearly form a clique of Gn,k,s.
For all x ∈ [n], let Nk(x) be the closed neighbourhood of x in Gn,k. That is, y ∈ Nk(x)
if and only if y ∈ [n] and ay = bx for some a,b ∈ [k].
?
Lemma 7. Let Gn,k,sbe a connected component of Gn,kwith at least k vertices. Then
|Nk(x)| ≥ ⌊√k⌋ for every x ∈ Xn,k,s.
Proof. By Lemma 6, the k smallest elements of Xn,k,sare {s,2s,3s,...,ks}, and they
form a clique of Gn,k,s. In particular, ks ≤ n.
Case (a). x ≤
and |Nk(x)| ≥ ⌊√k⌋.
Case (b). x >
p ≤ k. Thenax
suppose that there is no prime divisor p of x with
be the prime factors of x with duplication. Since x >
8
√ks: For each a ∈ [⌊√k⌋], we have ax ≤ ks ≤ n. Thus ax ∈ Nk(x)
√ks: First suppose that there is a prime p that divides x and
p∈ [x] for each a ∈ [p]. Thusax
√k ≤
p∈ Nk(x) and |Nk(x)| ≥ p ≥
√k ≤ p ≤ k. Let p1≤ p2≤ ··· ≤ pt
√k, for some ℓ ∈ [t], the integer
√k. Now

Page 9

q :=?
ax
i∈[ℓ]pi divides x and
q∈ Nk(x) and |Nk(x)| ≥ q ≥√k.
Proof of Theorem 5. Let k′:= ⌊√k⌋ and k′′:= ⌊√k′⌋. Note that k′′≥ 1 and k′′>
We proceed by studying the size of A within each connected component of the graph
Gn,k′. That is, we consider A as the union of the disjoint sets {A∩Xn,k′,s: s ∈ Sk′∩[n]}.
First consider s ∈ Sk′ ∩ [n] for which |Xn,k′,s| ≤ k′. By Lemma 6, Xn,k′,sis a clique
of Gn,k′. Since A is k-multiplicative, A is k′-multiplicative, and A is an independent set
of Gn,k′. Thus |A ∩ Xn,k′,s| ≤ 1. The set Sk′ ∩ [n] has exactly one element in Xn,k′,s.
Thus | ∪ {A ∩ Xn,k′,s: s ∈ Sk′ ∩ [n],|Xn,k′,s| ≤ k′}| ≤ |Sk′ ∩ [n]|. By Corollary 1,
???
Now consider s ∈ Sk′ ∩[n] for which |Xn,k′,s| > k′. We claim that Nk′(x)∩Nk′(y) = ∅
for distinct x,y ∈ A. Suppose that z ∈ Nk′(x) ∩ Nk′(y) for some x,y ∈ A. Then
a1x = b1z and a2y = b2z for some a1,a2,b1,b2∈ [k′]. Thus z = a1x/b1= a2y/b2and
(a1b2)x = (a2b1)y. Since a1b2,a2b1∈ [k] and A is k-multiplicative, x = y. This proves
the claim. Now Nk′(x) ⊆ Xn,k′,sfor each x ∈ Xn,k′,sby Lemma 5, and |Nk′(x)| ≥ k′′
by Lemma 7. Thus |A ∩ Xn,k′,s| · k′′≤ |Xn,k′,s|, and
???
Corollary 1 and Equations (4) and (5) imply that
√k ≤ q ≤ k. Thus
ax
q
∈ [x] for each a ∈ [q]. Thus
?
4√k/2.
(4)
?
{A ∩ Xn,k′,s: s ∈ Sk′ ∩ [n],|Xn,k′,s| ≤ k′}
??? ≤
(1 + ǫ)n
eγlogk′≤(2 + ǫ)n
eγlogk.
(5)
?
{A ∩ Xn,k′,s: s ∈ Sk′ ∩ [n],|Xn,k′,s| > k′}
??? ≤
n
k′′<
2n
4√k
.
|A| ≤(2 + ǫ)n
eγlogk+2n
4√k
≤(2 + o(1))n
eγlogk
= (2 + o(1))|Sk∩ [n]| .
?
4.2. An Improved Construction. While Sk∩ [n] is a k-multiplicative set whose
cardinality is within a constant factor of optimal, larger k-multiplicative sets in [n]
can be constructed. Recall that Pk= {p1,p2,...,pℓ} is the set of primes in [k]. Let
αi:= ⌊logpik⌋ + 1 for each pi∈ Pk. Define
Tk := {x ∈ N : βi(x) ≡ 0(mod αi), i ∈ [ℓ]} .
Lemma 8. For each k ∈ N, the set Tkis k-multiplicative.
Proof. Suppose that ax = by for some a,b ∈ [k] and x,y ∈ Tk. By Equation (3),
βi(a) + βi(x) = βi(b) + βi(y)
for all i ∈ [ℓ]. Now βi(x) ≡ βi(y) ≡ 0 (mod αi) since x,y ∈ Tk. Thus βi(a) ≡ βi(b)
(mod αi). Now pβi(a)
i
≤ a ≤ k. Thus βi(a) ≤ ⌊logpik⌋ = αi−1. Similarly βi(b) ≤ αi−1.
Hence βi(a) = βi(b) for all i ∈ [ℓ].
multiplicative.
9
Thus a = b and x = y.Therefore Tk is k-
?

Page 10

We now set out to determine the density of Tk. Observe that Sk= {x ∈ N : βi(x) =
0,i ∈ [ℓ]} ⊂ Tk. Thus (if it exists) the density of Tkis at least that of Sk.
Consider A,B ⊆ N with A ∩ B = ∅. If δ(A) and δ(B) exist, then δ(A ∪ B) =
δ(A) + δ(B). The following lemma extends this idea to an infinite union, where
δ(A) := sup
n→∞
|A ∩ [n]|
n
.
Lemma 9. Let A1,A2,... ⊆ N such that Ai∩Aj= ∅ whenever i ?= j. Suppose that for
each i ∈ N, δ(Ai) exists and δ(Ai) ≤ c · δ(Ai) for some constant c ≥ 1. Let A :=?
Then δ(A) =?
Proof. Let δ :=?
least integer such that
?
i>rǫ
Let nǫbe the minimum integer such that for all n > nǫand for all i ∈ [rǫ],
????
Let n > nǫ, X := A ∩ [n], Xi := X ∩ Ai and X∗:= ∪{Xi : i > rǫ}.
|Xi| < c · δ(Ai)n. Thus
|X∗| < cn
i>rǫ
Therefore
??????
<
?
i∈[rǫ]
iAi.
iδ(Ai).
iδ(Ai). Let ǫ > 0 be an arbitrary positive number. Let rǫbe the
δ(Ai) <ǫ
c
.
|Ai∩ [n]|
n
− δ(Ai)
????<
ǫ
rǫ
.
We have
?
δ(Ai) < ǫn .
????
|X|
n
− δ
????=

?
i∈[rǫ]
|Xi|
n
− δ(Ai)

+|X∗|
????+|X∗|
< ǫ
n
−
?
i>rǫ
δ(Ai)
??????
????
+ǫn
|Xi|
n
− δ(Ai)
n
+
?
i>rǫ
δ(Ai)
< rǫ
ǫ
rǫ
n+ǫ
c
?
2 +1
c
?
< 3ǫ .
This proves that δ(A) = δ.
?
Theorem 6. The set Tkis k-multiplicative with density
?
δ(Tk) = δ(Sk)
?
i∈[ℓ]
1 +
1
pαi
i− 1
?
=
?
i∈[ℓ]
?
1 −1
pi
??
1 +
1
pαi
i− 1
?
.
Proof. For all A ⊆ N and t ∈ N, let t · A := {ta : a ∈ A}. If δ(A) exists then
δ(t · A) =δ(A)
Now, for all v ∈ Nℓ

?
(6)
t
.
0, let
Sv
k:=
i∈[ℓ]
pviαi
i

· Sk .
10

Page 11

Note that Sv
Now Tk=?{Sv
k∩ Sw
k: v ∈ Nℓ
k= ∅ for distinct v,w ∈ Nℓ
0}. By Lemma 9,
0. For all v ∈ Nℓ
0we have
δ(Sv
δ(Sv
k)
k)=
δ(Sk)
δ(Sk).
δ(Tk) =
?
v∈Nℓ
0
δ(Sv
k) .
By Equation (6) with A = Skand t =?
ipviαi
i
,
δ(Tk) =
?
v∈Nℓ
0
δ(Sk)/
?
i∈[ℓ]
pviαi
i
.
Thus
δ(Tk) = δ(Sk)
?
v∈Nℓ
0
?
i∈[ℓ]
p−viαi
i
= δ(Sk)
?
i∈[ℓ]
pαi
i
i− 1= δ(Sk)
pαi
?
i∈[ℓ]
?
1 +
1
pαi
i− 1
?
.
The result follows by substituting the expression for δ(Sk) from Theorem 4; see Table 2.
?
Table 2. The first 15 elements of the set Tkfor each k ≤ 15.
Tk
{1,3,4,5,7,9,11,12,13,15,16,17,19,20,21,... }
{1,4,5,7,9,11,13,16,17,19,20,23,25,28,29,... }
{1,5,7,8,9,11,13,17,19,23,25,29,31,35,37,... }
5,6
{1,7,8,9,11,13,17,19,23,25,29,31,37,41,43,... }
7
{1,8,9,11,13,17,19,23,25,29,31,37,41,43,47,... }
8
{1,9,11,13,16,17,19,23,25,29,31,37,41,43,47,... }
9,10
{1,11,13,16,17,19,23,25,27,29,31,37,41,43,47,... }
11,12
{1,13,16,17,19,23,25,27,29,31,37,41,43,47,49,... }
13,14,15
{1,16,17,19,23,25,27,29,31,37,41,43,47,49,53,... }
k
2
3
4
density
2/3
1/2
3/7
5/14
5/16
7/24
7/26
77/312
11/48
We now show that δ(Tk) approaches δ(Sk) for large k.
Proposition 1. For all k ∈ N,
δ(Sk) < δ(Tk) = ck· δ(Sk) ,
for some constant ck→ 1 for large k.
Proof. By the Prime Number Theorem, ℓ ≤ O(k/logk). Thus
?
pαi
i− 1
≤ exp(O(1/log k)) → 1 .
ck =
?
i
1 +
1
?
<
?
i
?
1 +
1
k − 1
?
≤
?
1 +
1
k − 1
?O(k/logk)
?
11

Page 12

The case k = 2 was previously studied by Tamura [56] and Allouche et al. [8]. Observe
that T2= {22i(2j +1) : i,j ∈ N0}. Theorem 6 with k = 2 was proved by Allouche et al.
[8], who also proved that T2has the maximum density out of all 2-multiplicative sets.
Interesting relationships with the Thue-Morse sequence were also discovered.
Proposition 2 ([8]). The set T2is 2-multiplicative and has density 2/3. For all d ∈ N,
the d-th smallest element of T2is at most 3d/2 + O(logd).
Theorem 7. Let F be a forbidden family. Let G1,G2,...,Gd be graphs, each with
Λ(Gi,F) ≤ 2 or χ(Gi,F) ≤ 3. Let t be the d-th smallest element of T2. Then
Λ(G1?G2? ··· ?Gd,F) ≤ 2t ≤ 3d + O(logd) , and
χ(G1?G2? ··· ?Gd,F) ≤ 4t + 1 ≤ 6d + O(logd) .
Proof. By Lemma 1, χ(Gi,F) ≤ 3 implies Λ(Gi,F) ≤ 2. The result follows by applying
Lemma 2 with the d smallest elements of the 2-multiplicative set T2from Proposition 2.
?
5. P3-free Colourings
Recall that a colouring is P3-free if vertices at distance at most two receive distinct
colours. Let ∆(G) be the maximum degree of the graph G. Since a vertex and its
neighbours receive distinct colours in a P3-free colouring,
(7)χ(G,P3) ≥ ∆(G) + 1 .
Let Qd:= K2?K2? ··· ?K2be the d-dimensional hypercube. P3-free colourings
of Qd(and more generally, colourings of powers of Qd) have been extensively studied
[36, 45, 47, 55, 58, 63]. Wan [58] proved that
d + 1 ≤ χ(Qd,P3) ≤ 2⌈log2(d+1)⌉≤ 2d .
While our methods are not powerful enough to obtain the above upper bound, for grid
graphs we have the following result, which was first proved by Fertin et al. [24].
Example 1 ([24]). Every d-dimensional grid graph G := Pn1?Pn2? ··· ?Pndsatisfies
χ(G,P3) ≤ 2d + 1, with equality if every ni≥ 3.
Proof. The lower bound follows from Equation (7) since ∆(G) = 2d if every ni≥ 3.
Colour the i-th vertex in Pnby i. We obtain a P3-free colouring of Pnwith span 1.
Thus Λ(Pn,P3) = 1, and the upper bound follows from Theorem 2 with k = 1.
?
Example 1 highlights the utility of chromatic span. A weaker bound on χ(G,P3) is
obtained if the P3-free chromatic number, χ(Pn,P3) = 3, is used rather than the the
P3-free chromatic span, Λ(Pn,P3) = 1.
12

Page 13

Example 2. Let G be the d-dimensional graph G := P2
the d-th smallest element of T2. Then
n1?P2
n2? ··· ?P2
nd. Let t be
χ(G,P3) ≤ 4t + 1 ≤ 6d + O(logd) ,
and if each ni≥ 5 then χ(G,P3) ≥ 4d + 1.
Proof. Equation (7) implies the lower bound since ∆(G) = 4d if each ni≥ 5. Obviously
Λ(P2
n,P3) ≤ 2. Thus the upper bound follows from Theorem 7; see Table 3.
?
Table 3. Upper bound on χ(G,P3) for G := P2
G := Cn1?Cn2? ··· ?Cnd.
d1234
χ(G,P3) ≤
n1?P2
n2? ··· ?P2
ndor
5
29
6
37
7
45
8
49
9
53
10
61
11
65
12
69
13
77
14
81
15
85
...
...513 1721
Example 3. Let G be the graph Pk
χ(G,P3) ≥ k2, and if every ni≥ 2k + 1 then χ(G,P3) ≥ 2dk + 1. As an upper bound,
χ(G,P3) ≤ 2k(kd − k + 1) + 1 .
Moreover, for all ǫ > 0 and for large d > d(k,ǫ),
n1?Pk
n2? ··· ?Pk
nd. If there exists ni,nj≥ k then
χ(G,P3) ≤ 1 +2eγ
1 − ǫdk logk .
Proof. If ni,nj≥ k then G2contains a k2-vertex clique, and χ(G,P3) = χ(G2) ≥ k2.
The second lower bound follows from Equation (7) since ∆(G) = 2dk if every ni≥ 2k+1.
Obviously Λ(Pk
n,P3) ≤ k. Thus the upper bounds follow from Theorem 2.
?
Example 4. The d-dimensional toroidal grid G := Cn1?Cn2? ··· ?Cndsatisfies
2d + 1 ≤ χ(G,P3) ≤ 4t + 1 ≤ 6d + O(logd) ,
where t is the d-th smallest element of T2.
Proof. The lower bound follows from Equation (7) since G is 2d-regular. Say Cn =
(v1,v2,...,vn). By considering the vertex ordering
(v1,vn;v2,vn−1;...;vi,vn−i+1;...;v⌊n/2⌋,v⌈n/2⌉) ,
of Cn, we see that Cn⊂ P2
Fertin et al. [26] studied P4-free colourings of toroidal grids, and proved that the
minimum number of colours is at most 2d2+ d + 1, and at most 2d + 1 in the case
that 2d + 1 divides each ni. Thus Example 4 gives a linear upper bound on the P3-
free chromatic number of toroidal grids, where even for the weaker notion of P4-free
colourings, only a quadratic upper bound was previously known.
13
n. Thus the upper bound follows from Example 2.
?

Page 14

Example 5. Let G be the graph Ck
then χ(G,P3) ≥ k2, and if every ni≥ 2k + 1 then χ(G,P3) ≥ 2dk + 1. As an upper
bound,
n1?Ck
n2? ··· ?Ck
nd. If ni,nj≥ k for some i ?= j,
χ(G,P3) ≤ 4k(2kd − 2k + 1) + 1 .
Moreover, for all ǫ > 0 and for large d > d(k,ǫ),
χ(G,P3) ≤ 1 +4eγ
1 − ǫd · klog(2k) .
Proof. The lower bounds are the same as in Example 3. As proved in Example 4,
Cn⊂ P2
n. Thus Ck
n⊂ P2k
n, and the upper bound follows from Example 3.
?
6. Acyclic Colourings
Recall that a colouring with no bichromatic cycle is acyclic. The following elementary
lower bound is well known, where half the average degree of a graph G is denoted by
d(G) :=|E(G)|
|V (G)|
.
Lemma 10. Every graph G (with at least one edge) has acyclic chromatic number
χ(G,C) > d(G) + 1.
Proof. Say G has an acyclic k-colouring. Let nibe the number of vertices in the i-th
colour class. Let mi,jbe the number of edges between the i-th and j-th colour classes.
Thus mi,j≤ ni+ nj− 1. Hence
|E(G)| =
?
1≤i<j≤k
mi,j ≤
?
1≤i<j≤k
ni+nj−1 =
?
1≤i≤k
(k−1)ni −
?k
2
?
= (k−1)|V (G)|−
?k
2
?
.
Now k ≥ 2 since G has at least one edge.
d(G) + 1.
Thus k ≥ (|E(G)| + 1)/|V (G)| + 1 >
?
It is easily seen that a cartesian product satisfies
(8)
d(G1?G2? ··· ?Gd) =
?
i∈[d]
d(Gi) .
The following theorem, which was proved for paths by Fertin et al. [24], gives a special
case when a (k+1)-colouring can be obtained from a colouring with span k, rather than
the (2k + 1)-colouring guaranteed by Lemma 1.
Proposition 3. For all trees T1,T2,...,Td, the acyclic chromatic number
χ(T1?T2? ··· ?Td,C) ≤ d + 1 ,
with equality if every |V (Ti)| ≥ d.
14

Page 15

Proof. Let˜G := T1?T2? ··· ?Td. First we prove the lower bound. By Lemma 10 and
Equation (8), and since |V (Ti)| ≥ d,
|V (Ti)| − 1
|V (Ti)|
χ(˜G,C) > d(˜G) + 1 = 1 +
?
i∈[d]
= d + 1 −
?
i∈[d]
1
|V (Ti)|
≥ d .
Hence χ(˜G,C) ≥ d + 1.
Now we prove the upper bound. Root each tree Tiat some vertex ri. For each vertex
v ∈ V (Ti), let ci(v) be the distance between riand v in Ti. Then ciis a colouring of Ti
with span one. For each vertex ˜ v ∈ V (˜G), let
c(˜ v) :=
?
i∈[d]
i · ci(vi) .
For each edge ˜ v ˜ w ∈ E(˜G) in dimension i,


Thus c is a colouring of˜G with span d. Let c′(˜ v) := c(˜ v) mod (d + 1). Obviously c′is a
(d + 1)-colouring of˜G. We claim that c′is acyclic.
Consider each edge of Ti to be oriented away from the root ri. Orient each edge
˜ v ˜ w ∈ E(˜G) in dimension i according to the orientation of viwi. That is, orient ˜ v to ˜ w
so that ci(wi) − ci(vi) = 1. Clearly the orientation of˜G is acyclic.
Suppose that on the contrary there is a vertex ˜ v ∈ V (˜G) that has two incoming edges
˜ u˜ v and ˜ w˜ v for which c′(˜ u) = c′( ˜ w). Thus c(˜ u) ≡ c( ˜ w) (mod (d + 1)) and
c(˜ u) − c(˜ v) ≡ c( ˜ w) − c(˜ v)
Let i and j be the dimensions of ˜ u˜ v and ˜ w˜ v, respectively. By Equation (9),
(9)c( ˜ w) − c(˜ v) =
d
?
j=1
j · cj(wj)

−


d
?
j=1
j · cj(vj)

= i?ci(wi) − ci(vi)?
= ±i .
(mod (d + 1)) .
i(ci(ui) − ci(vi)) ≡ j(cj(wj) − cj(vj))
By the orientation of edges, ci(ui) − ci(vi) = 1 and cj(wj) − cj(vj) = 1. Thus i ≡ j
(mod (d + 1)), which implies that i = j. Hence ˜ u = ˜ w since vihas only one incoming
edge in Ti(from its parent). Thus every vertex of˜G has at most one incoming edge
in each bichromatic subgraph H (with respect to the colouring c′). Hence H has an
acyclic orientation with at most one incoming edge at each vertex. Therefore H is a
forest, and c′is the desired acyclic colouring of˜G.
(mod (d + 1)) .
?
7. P4-free Colourings
Recall that a colouring with no bichromatic P4is a star colouring.
Example 6. For trees T1,T2,...,Td, the star chromatic number
χ(T1?T2? ··· ?Td,P4) ≤ 2d + 1 .
15