THE HARMONIOUS CHROMATIC NUMBER OF BOUNDED
DEGREE GRAPHS
KEITH EDWARDS
A
A harmonious colouring of a simple graph G is a proper vertex colouring such that each pair of colours
appears together on at most one edge. The harmonious chromatic number h(G) is the least number of colours
in such a colouring.
Let d be a fixed positive integer, and ε " 0. We show that there is a natural number M such that if G
is any graph with m & M edges and maximum degree at most d, then the harmonious chromatic number
h(G) satisfies
(2m)"/# % h(G) % (1­ε) (2m)"/#.
1. Introduction
A harmonious colouring of a simple graph G is a proper vertex colouring such that
each pair of colours appears together on at most one edge. Formally a harmonious
colouring is a function c from a colour set C to the set V(G) of vertices of G such that
for any edge e of G, with endpoints x, y say, c(x) 1 c( y), and for any pair of distinct
edges e, e«, with endpoints x, y and x«, y« respectively, ²c(x), c( y)´ 1 ²c(x«), c( y«)´. The
harmonious chromatic number h(G) is the least number of colours in such a colouring.
A survey on harmonious colourings is given by B. Wilson in [9].
If we have a harmonious colouring of G with k colours, then since each pair of
colours can appear on at most one edge, it is clear that the number of colour pairs,
k
namely
, must be at least m, the number of edges of G. This motivates the
2
following definition.
01
D. Let m be a positive integer. Then we define Q(m) to be the least
k
positive integer k such that
& m.
2
01
It is easily calculated that Q(m) ¯ 8"(1­(8m­1)"/#7, however for our purposes
#
it suffices to note that Q(m) is approximately (2m)"/#. We are interested in classes
of graphs for which h(G) is close to (2m)"/#, in particular, those classes Γ for which
h(G) ¯ (2m)"/#­o(m"/#).
It follows readily from the theorem of Wilson [10] on the packing of copies of a
graph into a complete graph, that if Γ(K ) consists of graphs which are disjoint unions
of a fixed graph K, then h(G) ¯ (2m)"/#­o(m"/#) for G ` Γ(K ), and this extends easily
Received 30 January 1995.
1991 Mathematics Subject Classification 05C15.
J. London Math. Soc. (2) 55 (1997) 435–447
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436
to classes with bounded component size. It is proved in [2] that the same is true for
the class Td consisting of trees of maximum degree at most d, and Γd, γ consisting of
graphs of maximum degree at most d and genus at most γ.
In this paper we extend all these results and show that if Γd consists of all graphs
of maximum degree at most d, then h(G) ¯ (2m)"/#­o(m"/#) for G ` Γd. The method
used is quite different from those used for the earlier results, and relies heavily on the
powerful Pippenger–Spencer theorem on the chromatic index of uniform hypergraphs
[8]. We also discuss the relationship of this result to Wilson’s Theorem, and offer an
alternative interpretation based on identifying vertices of graphs.
2. Regular graphs
In this section we show how to find asymptotically good colourings of regular
graphs. We start with some preliminaries.
D. An incomplete harmonious colouring of a graph G is a colouring of
a subset X of the vertices of G, such that the colouring is a harmonious colouring of
the graph induced by the vertices X.
D. A partial harmonious colouring of a graph G is an incomplete
harmonious colouring which satisfies the additional property that any two coloured
vertices of G which have a common neighbour must have distinct colours.
R. A partial harmonious colouring can always be extended to a
harmonious colouring ; this is not always possible with an incomplete colouring.
Hypergraphs
We now give some hypergraph notation and state the theorem of Pippenger and
Spencer.
Let G be a hypergraph ; then G is k-uniform if every edge contains k vertices. If
Š and w are vertices of G, then the degree of Š, denoted by degG(Š), is the number of
edges containing Š, and the codegree of Š and w, denoted by codegG(Š, w), is the
number of edges containing both Š and w. Let
D(G) ¯ max degG(Š),
`
v V(G)
d(G) ¯ min degG(Š),
`
v V(G)
C(G) ¯
max
v,w ` V(G),v1w
codegG(Š, w).
T (Pippenger–Spencer [8]). For eŠery k & 2 and δ " 0, there exist δ« " 0
and n such that if G is a k-uniform hypergraph on n(G) & n Šertices satisfying
!
!
d(G) & (1®δ«) D(G) and C(G) % δ«D(G), then the chromatic index χ(G) satisfies
χ(G) % (1­δ) D(G).
Probability Estimates
We now state two more theorems used to estimate probabilities. These can be
found, for example, in [5].
T (Azuma’s bounded differences inequality). Let X , … , Xn be inde"
pendent random Šariables with Xk taking Šalues in a set Ak for each k. Suppose that the
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437
(measurable) function f : 0 Ak ! R satisfies r f(x)®f(x«)r % ck wheneŠer the Šectors x
and x« differ only in the kth co-ordinate. Let Y be the random Šariable f [X , … , Xn].
"
Then for any t " 0, we haŠe
P [rY®Ex (Y )r & t] % 2 exp (®2t#}3 ck#).
T. Let X , … , Xn be independent random Šariables with 0 % Xk % 1 for
"
each k, let Xa ¯ (1}n) 3 Xk and let p ¯ Ex [Xa ]. Then for any 0 ! ε ! 1, we haŠe
P[Xa ®p & εp] % exp (®"ε#np),
$
P[Xa ®p %®εp] % exp (®"ε#np).
#
C 2.1. Let X , … , Xn be independent random Šariables with P[Xk ¯ 1]
"
¯ p and P[Xk ¯ 0] ¯ 1®p for each k. Then for any 0 ! ε ! 1, we haŠe
P[r3 Xk®npr & εnp] % 2 exp (®"ε#np).
$
Main lemma
We now introduce some notation and state and prove the main technical lemma.
N. Let G be a graph with vertex set V and edge set E. Let X, Y and Z
be subsets of V. We define the set of neighbours of X, denoted by N(X ), by
N(X ) ¯ ²Š ` V r (Š, w) ` E for some w ` X ´.
Define the edge set induced by X, denoted by E(X ), by
E(X ) ¯ ²(Š, w) ` E r Š, w ` X ´.
Define the edge set induced by X and Y, denoted by E(X, Y ), by
E(X, Y ) ¯ ²(Š, w) ` E r Š ` X, w ` Y ´.
Finally define the neighbourhood set of X and Y in Z, denoted by NZ(X, Y ), by
NZ(X, Y ) ¯ N(X )fN(Y )fZ.
N. Statements below of the form xj ¯ f(n)­O(z(n)) for j ¯ 1, … , J should be
taken to be uniform over j, that is, to mean that there is a constant α and an integer
N such that rxj®f(n)r % αz(n) for all n & N and for all j ¯ 1, … , J.
L 2.2. Let d be a fixed integer, and let η " 0. Then there is an integer N such
!
that if G is a d-regular graph with n Šertices and m (¯ "dn) edges, where n & N , then
#
!
we can find an incomplete harmonious colouring of G with at most (1­η) (2m)"/# colours
such that at least (1®η) n of the Šertices are coloured.
Proof. Let δ ¯ η}6, and let δ« be the corresponding real number, depending on
δ, given by the Pippenger–Spencer Theorem. We now choose an integer A satisfying
A & max ²16d %, d #}(δ«)#, 6}η, 9}η#, 2}δ«´. We proceed by a sequence of stages. At each
stage we shall colour approximately n}A of the vertices. We shall stop after Stage i
when A®i ¯ 8A"/#7 (thus leaving some vertices uncoloured). Hence we shall always
have A®i & A"/#.
Let k ¯ (2m)"/#, and V be the vertex set of G. We will ensure that at the start of
Stage i­1, the following conditions hold :
1. V is partitioned as V ¯ C e…eCieZ, where rZ r ¯ (A®i) n}A­O(n$/%), and
"
for each j ¯ 1, … , i, we have rCjr ¯ n}A­O(n$/%) ;
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2.
3.
4.
5.
6.
7.
each Cj is partitioned into colour classes Cj, , … , Cj,t , where tj ¯ k}A­O(n$/)) ;
"
j
rE(Cj)r ¯ 0 for each j ;
rE(Z )r ¯ (A®i) (A®i®1) m}A(A®1)­O(n$/%) ;
rE(Cj,r, Z )r ¯ (A®i) k}(A®1)­O(n$/)) for each j, r ;
rNZ(Cj,r, Cj «)r % kd}A­O(n$/)) for each j, r, and j « " j ;
for each j, no two distinct elements of Cj have a common neighbour in Z.
Note that before stage 1, that is, when i ¯ 0, all the above hold trivially.
Stage i­1. At stage i­1 we first choose a certain subset W of Z. Let Z « ¯ Z cW.
Then W and Z « will have the following properties :
1. rW r ¯ n}A­O(n$/%) and rZ «r ¯ (A®i®1) n}A­O(n$/%) ;
2. rE(Cj,r, W )r ¯ k}(A®1)­O(n$/)) for each j, r ;
3. rE(W )r ¯ 0 ;
4. rE(W, Z «)r ¯ 2(A®i®1) m}A(A®1)­O(n$/%) ;
5. rE(Z «)r ¯ (A®i®1) (A®i®2) m}A(A®1)­O(n$/%) ;
6. rNZ «(Cj,r, W )r % kd}A­O(n$/)) for each j, r ;
7. rNW(Cj,r, Cj «)r % kd}A$/#­O(n$/)) for each j, r, and j « " j.
To construct W, we first choose a subset B of Z by including each element,
independently, with probability pi, where pi is chosen so that pi(1®pi)d # ¯ 1}(A®i).
Note that this will always be possible since 1}(A®i) % 1}A"/# % 1}4d #, while an easy
calculation shows that f(x) ¯ x(1®x)d # has maximum value at least 1}4d # on (0, 1).
For each z ` Z, let the random variable Xz be the indicator function of the event
[z ` B]. Thus P[Xz ¯ 1] ¯ pi. We construct W from B as follows : choose an arbitrary
cyclic ordering of the vertices of Z. For each vertex Š of Z, construct the set T(Š) by
first including in T(Š) all vertices of Z which are at distances 1 or 2 from Š in the graph
induced by the vertices of Z. Then clearly rT(Š)r % d #. To ensure that each T(Š) has
exactly d # elements, add to T(Š) the first d #®rT(Š)r vertices after Š in the cyclic
ordering which are not already in T(Š). Note that for any Š« in Z, we have that Š« is
in T(Š) for at most 2d # vertices Š.
Now to obtain W, let R ¯ ²Š r T(Š)fB´ 1 W. Then let W ¯ BcR. It is clear that
P[Š ` W ] ¯ pi(1®pi)d # ¯ 1}(A®i) for any Š ` Z. We can establish, using Azuma’s
bounded differences inequality, that with probability very close to 1, the set W will
have each of the properties above. We do this as follows.
Let X be any subset of Z or of E(Z ), and suppose that rX r ¯ λ­O(λ$/%). Now
suppose that f is a function from subsets of Z to subsets of Z or E(Z ), and satisfies
the following conditions : (i) If S, S « are subsets of Z such that rS∆S «r % 1, then
r f(S ) ∆f(S «)r % d and (ii) the event [x ` f(T )] depends on at most d of the events
[z ` T ]. Let Y be the random variable rXff(W )r. Then we claim that
P[Y ¯ Ex (Y )­O(λ$/%)] & 1®2 exp (®Ω(λ"/#)).
To see this, first note that the random variables Xz are independent, and the random
variable Y is functionally dependent on them. Now a change in the value of one Xz,
corresponding to the removal from or addition to B of some vertex Š, results in a
change in W of at most 2d # elements. Hence the resulting change in f(W ) is of at most
2d $ elements.
Now for any x ` X, let Yx ¯ I[x ` f(W)], so that Y ¯ 3x ` X Yx. We know that the event
[z ` W ] depends on the event [Š ` B] only if Š ¯ z or if Š ` T(z). Hence [z ` W ] depends
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439
on at most d #­1 of the events [Š ` B]. It follows that [x ` f(W )] depends on at most
d(d #­1) of the events [Š ` B], and so Y depends on at most rX r d(d #­1) of the random
variables Xv. Furthermore, for those on which it may depend, the constant cv of
Azuma’s inequality is at most 2d $. Hence finally we have, by Azuma’s inequality, that
P[rY®Ex (Y )r & λ$/%] % 2 exp (®2λ$/#}rX r d(d #­1) 4d ') % 2 exp (®Ω(λ"/#)).
In a similar way we can show that if rX r % λ­O(λ$/%), and Y ¯ rXff(W )r then
P[Y % Ex (Y )­O(λ$/%)] & 1®2 exp (®Ω(λ"/#)).
Note that Ex (Y ) ¯ 3x ` X P[x ` f(W )].
We can now show that W has the required properties with high probability, as
follows.
1. rW r ¯ n}A­O(n$/%) and rZ« r ¯ (A®i®1) n}A­O(n$/%). For W, take X ¯ Z, so
that λ ¯ (A®i) n}A, and let f(S ) ¯ S. Then Y ¯ rW r and P[x ` f(W )] ¯ 1}(A®i),
so Ex (Y ) ¯ n}A­O(n$/%). Hence rW r ¯ n}A­O(n$/%) with probability at least
1®2 exp (®Ω(n"/#)). To deal with rZ «r, we simply note that rZ «r ¯ rZ r®rW r.
2. rE(Cj,r, W )r ¯ k}(A®1)­O(n$/)). Note that since any element of Z has at most
one neighbour in Cj,r, then rE(Cj,r, Z )r ¯ rN(Cj,r)fZ r. So take X ¯ N(Cj,r)fZ,
so that λ ¯ (A®i) k}(A®1), and let f(S ) ¯ S. Then Y ¯ rXfW r ¯ rE(Cj,r, W )r.
Now P[x ` f(W )] ¯ 1}(A®i), so Ex (Y ) ¯ k}(A®1)­O(n$/)). Hence Y ¯
k}(A®1)­O(n$/)) with probability at least 1®2 exp (®Ω(k"/#)).
3. rE(W )r ¯ 0. This follows immediately from the construction of W.
4. rE(W, Z «)r ¯ 2(A®i®1) m}A(A®1)­O(n$/%). Let X ¯ rE(Z )r, so that λ ¯
(A®i) (A®i®1) m}A(A®1), and let f(S ) ¯ E(S, Z®S ). Thus f(W ) ¯ E(W, Z «), so
that Y ¯ rE(W, Z «)r. Now let (u, Š) be any edge of E(Z ). Then by definition of W, if u is
in W then Š is in Z «, and vice versa, hence the probability that the edge is in E(W, Z «)
is 2}(A®i). Hence P[x ` f(W )] ¯ 2}(A®i). Thus Ex (Y ) ¯ 2(A®i®1) m}A(A®1), and
so Y ¯ 2(A®i®1) m}A(A®1)­O(n$/%) with probability at least 1®2 exp (®Ω(n"/#)).
5. rE(Z «)r ¯ (A®i®1) (A®i®2) m}A(A®1)­O(n$/%). This follows from the fact
that rE(Z «)r ¯ rE(Z )r®rE(W, Z «)r.
6. rNZ «(Cj,r, W)r % kd}A­O(n$/)). We have
NZ «(Cj,r, W ) ¯ N(Cj,r)fN(W )fZ « ¯ N(Cj,r)fZfN(W ).
So take X ¯ N(Cj,r)fZ. Now each element of Z has at most one neighbour in Cj,r,
hence we have λ ¯ rN(Cj,r)fZ r ¯ rE(Cj,r, Z )r ¯ (A®i) k}(A®1)­O(n$/)). Let
f(S ) ¯ N(S ). Then Y ¯ rNZ «(Cj,r, W )r, and P[x ` f(W )] % (d®1)}(A®i). Hence
Ex (Y ) % (d®1) k}(A®1) ! dk}A­O(n$/)), whence Y % kd}A­O(n$/)) with
probability at least 1®2 exp (®Ω(k"/#)).
7. rNW(Cj,r, Cj «)r % kd}A$/#­O(n$/)). Take X ¯ NZ(Cj,r, Cj «), so that λ ¯ dk}A,
and let f(S ) ¯ S. Then Y ¯ rNW(Cj,r, Cj «)r, and P[x ` f(W )] ¯ 1}(A®i). Then
Ex (Y ) ¯ kd}A(A®i)­(n$/)), whence Y % kd}A$/#­O(n$/)) with probability at least
1®2 exp (®Ω(k"/#)).
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 
It is clear that with probability near 1, provided that n is large enough, W has all
the above properties, hence we can certainly choose a W which does have all of these
properties.
Now in order to apply the Pippenger–Spencer Theorem, we need to give some
temporary ‘ colours ’ to any vertex in Z « which is the neighbour of a vertex in W. To
avoid confusion with the real colours, we shall refer to these temporary colours as
‘ shades ’. Let S ¯ N(W )fZ «. Note that, since no vertex in Z has two neighbours in
W, we have rS r ¯ rE(W, Z «)r ¯ (A®i®1) m}A(A®1)­O(n$/%). We first partition S as
follows : form a graph Γ with vertex set S by joining any vertices which have a
common neighbour in W. Since each vertex of W has degree at most d, it follows that
Γ has maximum degree at most d®1, and consequently the vertices can be partitioned
into d independent (in Γ) sets S , … , Sd. Let rSjr ¯ sj n. By moving some vertices if
"
necessary, we can assume that sj % dsj « for any j, j «, for if sj " dsj « some vertex of Sj
has no neighbour (in Γ) in S j! and so can be moved to S !j . Note that no two vertices
of any Sj have a common neighbour in W.
Now let sj A}d ¯ tj. Note that since sj % dsj « for any j, j «, and 3 sj is about
(A®i®1) d}2A(A®1), then the numbers sj are bounded above and below by
constants independent of j. Thus the same is true for the tj. We shall assign each
element of Sj independently to one of tj k shades, each with probability 1}tj k. (The sets
of shades for distinct Sj are disjoint.) Thus, for each j, we can define sj n random
variables Y j , Y j , … , Y js n such that P[Y jr ¯ α] ¯ 1}tj k for α ¯ 1, … , tj k. Furthermore
" #
j
all these random variables are independent. Now let C be some particular shade used
on Sj, and define the random variables Z jr by Z jr ¯ 1 if Y jr ¯ C and Z jr ¯ 0 otherwise.
Then clearly all the Z jr are independent. It follows that Ex (3r Z jr) ¯ sj n}tj k ¯ nd}Ak
¯ k}A and by Corollary 2.1
P [r3 Z jr®k}Ar & ηk}A] % 2 exp (®"η#k}A).
$
Hence taking η ¯ 1}n"/), we obtain
P [r3 Z jr®k}Ar & O(n$/))] % 2 exp (®Ω(n"/%)).
Thus we know that with high probability, each of the shades is assigned to
k}A­O(n$/)) vertices.
Now consider two of these shades, say C and C «, used on sets Sj and Sj «
respectively. We wish to determine the number N of vertices in W which are joined
to both C and C «. If j ¯ j « then there are none, so assume that j 1 j «. Now each vertex
of W has at most one neighbour in each of Sj and Sj « ; furthermore for distinct
elements of W these sets of neighbours are disjoint. For each vertex w of W, define
a random variable Xw by setting Xw ¯ 1 if w has a neighbour in both Sj and Sj « and
they are shaded with shades C and C « respectively, and Xw ¯ 0 otherwise. Then the
random variables Xw are independent, and for each P[Xw ¯ 1] ¯ 1}tj tj « k# % c}n for
some positive constant c independent of j, j «. Then
P[3 Xw & t] %
0rWt r1 (c}n) .
t
Using the (very crude) estimate rW r ! n, we obtain P[3 Xw & t] % (nt}t!) (c}n)t ¯
ct}t!. Taking t ¯ 8n"/%7, for example, we find that P[N & n"/%] % cn"/%}n"/%!. Similarly, if
we consider a new shade C and a colour Cj,r we can show that the number of common
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441
neighbours of these in W is O(n"/%) with probability very near to 1. It follows that the
choice of shades for the vertices in S can be made so that (i) each shade is used on
k}A­O(n$/)) vertices, and (ii) for any shade C, and colour or shade C «, we have
NW(C, C «) ¯ O(n"/%).
We now construct a d-uniform hypergraph H. The vertex set V(H ) is the set of
colours and shades. The edges are defined as follows. For each vertex w in W, the set
of colours and shades of the neighbours of w is an edge e(w). Note that these colours
and shades are all distinct, so that H is d-uniform as required. We shall call w the
parent vertex of the edge.
We have to consider the degrees and co-degrees of vertices in H. Firstly, the degree
of a vertex C in H (that is, a colour or shade C in G) is the number of vertices of W
which are adjacent to a vertex of C, which equals rE(C, W )r since each vertex in W is
adjacent to at most one in C. Now we have rE(C, W )r ¯ k}(A®1)­O(n$/)) or
k}A­O(n$/)) for any colour or shade C. Hence we have
d(H )}D(H ) ¯ (k}A­O(n$/)))}(k}(A®1)­O(n$/)))
¯ (A®1)}A­O(n−"/)) ¯ 1®1}A­O(n−"/)).
Hence since 1}A % δ«}2, provided that n is large enough we have
d(H ) & (1®δ«) D(H ).
Now for any vertices C, C « of H, we have
codeg (C, C «) ¯ rNW(C, C «)r % kd}A$/#­O(n$/))
for any C, C «. Since A & d #}(δ«)#, we have d % δ«A"/#, and so
C(H ) % kd}A$/#­O(n$/)) % kδ«}A­O(n$/))
¯ δ«k}(A®1)®δ«k}A(A®1)­O(n$/)) % δ«D(H ),
provided that n is large enough ; hence by the Pippenger–Spencer Theorem [8], we
have χ(H ) % (1­δ) D(H ). Thus we obtain a colouring of the edges of the hypergraph
H with at most (1­δ) D(H ) colours, so that any two non-disjoint edges receive
different colours. We now transfer the colour of each edge e(w) to its parent vertex.
Notice that if u and Š are two vertices of W with the same colour, then there cannot
be vertices x and y, with (u, x) and (Š, y) both edges of G, and x and y having the same
colour Cj,r. For then Cj,r ` e(u)fe(Š), and so u and Š receive distinct colours, a
contradiction. Thus, we have, roughly speaking, extended the harmonious colouring
to W using approximately k}A new colours.
We have to recolour a few of the newly coloured vertices to set up the correct
conditions for the next stage. We have that
rE(W, Z «)r ¯ 2(A®i®1) m}A(A®1)­O(n$/%).
This is equal to the sum over the colours C used for W of rE(C, Z «)r. Also we know
that each shade is used on k}A­O(n$/)) vertices, hence the number of shades is
[2(A®i®1) m}A(A®1)­O(n$/%)]}[k}A­O(n$/))]
which is (A®i®1) k}(A®1)­O(n$/)). Since rE(C, Z «)r is bounded by the number of
shades, we have rE(C, Z «)r ¯ (A®i®1) k}(A®1)­O(n$/)). Now let a be the maximum
value of rE(C, Z «)r, and let b ¯ :rE(W, Z «)r}a9. Then b ¯ k}A­O(n$/)). Now discard
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442
 
some of the colours of W to leave exactly b. We then recolour these vertices as follows :
consider them in some sequence, and for each, colour it with the colour C for which
rE(C, Z «)r is least. Since ab % rE(W, Z «)r, we know that the maximum value of
rE(C, Z «)r is now greater than a, and it follows that the difference between the
minimum and maximum values can be at most d. Suppose the maximum value is now
a«. Then
ba« & rE(W, Z «)r & b(a«®d )
and it follows easily that a« ¯ (A®i®1) k}(A®1)­O(n$/)) from which rE(C, Z «)r ¯
(A®i®1) k}(A®1)­O(n$/)). We now set Ci+ ¯ W, ti+ ¯ b and the sets Ci+ , … ,
"
"
""
Ci+ , t equal to the colour classes of the new colours.
" i+"
This completes step i­1. All the necessary conditions are satisfied.
It is clear that we can successively complete stage i for i ¯ 1, … , A®8A"/#7. Now
the number of colours used at stage i is k}A­O(n$/)), hence in total we have used at
most k­O(n$/)), which is at most (1­η) (2m)"/# provided that n is large enough.
Next we have to consider the vertices recoloured at each stage, after the colouring
of the hypergraph H. It is these vertices only which may cause the colouring not to
be harmonious. The number of colours used to colour H is at most (1­δ) D(H ),
where D(H ) ¯ k}(A®1)­O(n$/)). Hence the number of colours discarded at stage i
is D(H )®b, which is
(1­δ) k}(A®1)®k}A­O(n$/)) ¯ δk}(A®1)­k}A(A®1)­O(n$/)).
Now any colour used on H can be used on at most V(H )}d edges, and
rV(H )r ¯ ik}(A®1)­(A®i®1) k}(A®1)­O(n$/)) ¯ k­O(n$/)).
Hence each discarded colour is used on at most k}d­O(n$/)) vertices, so the total
number of vertices recoloured at stage i is at most
δk#}d(A®1)­k#}dA(A®1)­O(n(/)) ¯ δn}(A®1)­n}A(A®1)­O(n(/)).
Thus, since there are fewer than A®1 stages, in all at most δn­n}A­O(n(/)) vertices
are recoloured. Since δ ¯ η}6, and A & 6}η, this is less than ηn}2, provided that n is
large enough. We discard the colours of these vertices, so that these vertices become
uncoloured.
Finally we have to consider the vertices which are never coloured at all. After the
last stage i, where A®i ¯ 8A"/#7, we have rZ r ¯ (A®i) n}A­O(n$/%) ¯ n}A"/#­O(n$/%).
Since A"/# & 3}η, this is less than ηn}2, provided that n is large enough. Hence we have
coloured at least (1®η) n of the vertices with at most (1­η) (2m)"/# colours, as
required.
The next lemma shows how to extend the incomplete colouring to the whole
graph without using too many extra colours. Essentially this lemma appeared in [2,
Theorem 3.5] and [1, Theorem 2.1] ; we restate it here for convenience as a stand-alone
lemma.
L 2.3. Let d be a fixed integer, and let η " 0. Suppose that G is a graph with
maximum degree at most d, n Šertices and m edges, and G has an incomplete harmonious
colouring with k colours such that at least (1®η) n of the Šertices are coloured. Then
h(G) % k­12d #η"/#n"/#, proŠided that n is large enough.
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       
443
Proof. Denote the set of uncoloured vertices by S, so that rS r % ηn. Now set
N(S ) ¯ ²Š r (Š, x) ` E for some x ` S ´cS,
N #(S ) ¯ ²Š r (Š, x) ` E for some x ` N(S )´c(SeN(S )).
Note that rN(S )r % drS r % dηn, and rN #(S )r % (d®1) rN(S )r % d #ηn. We now extend
the harmonious colouring to S in three stages.
Stage 1. We ensure that no colour occurs more than 8η"/#n"/#7 times on N #(S ), as
follows. If colour c occurs more than 8η"/#n"/#7 times then pick 8η"/#n"/#7 of the vertices
with colour c and recolour them with a new colour not previously used for any vertex.
Repeat this until no colour occurs more than 8η"/#n"/#7 times. Now since each new
colour occurs exactly 8η"/#n"/#7 times, and rN #(S )r % d #ηn, then we have used at most
d #η"/#n"/# new colours.
Stage 2. Now recolour N(S ) with new colours not previously used. We shall
ensure that after this has been done, we have a partial harmonious colouring of G so
that in particular, for each Š ` S, no two neighbours of Š have the same colour. We
colour N(S ) sequentially. Order the vertices arbitrarily. We ensure that no colour set
on N(S ) ever has size greater than 8η"/#n"/#7. Thus when we colour an element Š ` N(S )
some colours are unavailable.
(a) The colours on the coloured neighbours of Š are unavailable. This excludes at
most d colours.
(b) If w ` N #(S )eN(S ) is adjacent to Š and is already coloured, and w« `
N #(S )eN(S ) is the same colour as w, then we cannot use the colour of any neighbour
of w«. There are at most d choices for w, then at most 8η"/#n"/#7 for w«, and then at most
d for a neighbour of w«. Hence this excludes at most d #8η"/#n"/#7 colours.
(c) We must ensure that no uncoloured vertex in N(S ) and no vertex in S gets two
neighbours in N(S ) of the same colour. This excludes at most d(d®1) colours.
(d) We must exclude any colour which has already occurred 8η"/#n"/#7 times on
N(S ) ; there are at most rN(S )r}8η"/#n"/#7 such colours. So we exclude at most dη"/#n"/#.
Hence in total at most d #8η"/#n"/#7­dη"/#n"/#­d(d®1)­d colours are excluded, so
we can colour N(S ) with at most d #8η"/#n"/#7­dη"/#n"/#­d #­1 colours.
Stage 3. We colour the elements of S sequentially, with brand new colours, never
allowing any colour set to become larger than 8η"/#n"/#7. For reasons as in (a) to (d)
above, we have to exclude at most d #8η"/#n"/#7­η"/#n"/#­d # colours when we colour
each vertex, so we can complete the colouring with at most d #8η"/#n"/#7­
η"/#n"/#­d #­1 new colours. Hence
h(G) % k­3d #8η"/#n"/#7­(d­1) η"/#n"/#­2d #­2.
It follows that if η"/#n"/# & 1, we have h(G) % k­12d #η"/#n"/#.
We can now give the main theorem for regular graphs.
T 2.4. Let d be a fixed integer, and let ε " 0. Then there is an integer
N(d, ε) such that if G is a d-regular graph with n Šertices and m (¯ "dn) edges, where
#
m & N, then
h(G) % (1­ε) (2m)"/#.
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 
444
Proof. Choose η so that η % ε}2 amd η % ε#}576d $. Then by Lemma 2.2,
provided that m is large enough, we can find an incomplete harmonious colouring of
G with at most (1­η) (2m)"/# colours, and with at most ηn vertices uncoloured. Then
by Lemma 2.3, provided that m, and hence n, is large enough, we can complete the
colouring with the addition of at most 12d #η"/#n"/# new colours. Hence
h(G) % (1­η) (2m)"/#­12d #η"/#n"/# % (1­ε}2) (2m)"/#­12d #εn"/#}24d $/#
% (1­ε}2) (2m)"/#­ε(2m)"/#}2 % (1­ε) (2m)"/#,
as required. It remains only to choose N(d, ε) appropriately.
3. Bounded degree graphs
We now extend the results of the previous section to graphs of bounded degree.
To do this we show how such a graph G can be turned into a regular graph H without
either decreasing the harmonious chromatic number or increasing the number of
edges too much.
Several authors have proved results of the form : for any graph G, with n vertices
and maximum degree at most d, we have h(G ) % Cd n"/#, where Cd is some
constant depending on d (see for example [2, 3, 4, 6]). For our purposes the most
convenient (though not the strongest) is the following result of McDiarmid and Luo
[6].
T.
For any nontriŠial graph G with maximum degree d, we haŠe
h(G ) % 2d(n®1)"/#.
T 3.1. Let d be a fixed integer, and let ε " 0. Then there is an integer
M such that if G is a graph with maximum degree at most d, and G has m edges,
!
where m & M , then
!
h(G ) % (1­ε) (2m)"/#.
Proof. Isolated vertices make no difference to the hypothesis or conclusion, so
assume that G has no isolated vertices. Let the number of vertices of G be n, and let
k ¯ 2dn"/#. Note that m & n}2. From the above theorem of McDiarmid and Luo, we
can find a harmonious colouring c : V(G ) ! ²1, … , k´. Define the degree sum σ(i) of
a colour i to be 3c(v)=i d(Š), where as usual d(Š) is the degree of Š in G. By recolouring
some vertices if necessary, we can ensure that for any colours i, j, we have σ(i) % dσ( j).
For suppose that σ(i) " dσ( j). Let the set of colours which occur adjacent to a vertex
of colour j be S, so that rSr ¯ σ( j). Then the number of vertices of colour i is at least
σ(i)}d " rS r, hence at least one vertex with colour i has no neighbour with a colour
from S. Hence we can recolour this vertex with colour j. We repeat this process until
σ(i) % dσ( j) for all pairs of colours. Let σmin be the minimum degree sum.
Now observe that 3ki= σ(i) ¯ 2m and σ(i) % dσmin for each i, hence 2m % kdσmin ¯
"
2d #n"/#σmin % 2d #(2m)"/# σmin from which we obtain σmin & (2m)"/#}2d #. Now choose
L to be the least even integer satisfying L & 16d}ε­2d. We consider each colour class,
and partition each into sets of vertices of total degree between L®2d and L®d. We
may be left with a residue of total degree less than L®2d, in this case we add at most
one of these extra vertices to each part. (There will certainly be enough parts provided
that m & M say.) Thus we obtain partitions of each colour class, with each part of
"
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445
total degree between L®2d and L. We now take each part, and identify its vertices
to form a single vertex of degree between L®2d and L. It is easy to see that we do
indeed obtain a graph G « by this process, that is, no loops or multiple edges are
formed. Furthermore, any harmonious colouring of G « immediately induces one of G
in the obvious way, hence h(G ) % h(G «).
We now form from G « a graph H which is regular of degree L. We do this as
follows. If G « has at least L vertices of degree less than L, add a new vertex and join
it to L such vertices. Repeat this until the set U of vertices of G « with degree less than
L satisfies rU r ! L. Now add L­1 new vertices Š , Š , … , ŠL to form a copy of KL+
"
! "
disjoint from G «. Let R be the set of vertices u of U for which L®d(u) is odd, and let
r ¯ rRr. Then r is necessarily even, for
3 L®d(u) ¯ 3 L®d(Š) ¯ LrV r®2m.
u`U
v`V
Since L is even, the result follows. Remove the edges (Š , Š ), … (Š , Šr/ ), and join Š
! "
! #
!
to r}2 of the elements of R, and each of Š , … , Šr/ to one of the remaining elements
"
#
of R. Now L®d(u) is even for each element of U. The vertices Š , … , ŠL induce a copy
"
of KL, which contains L®1 disjoint matchings of size L}2. For each element u of U,
select one of these matchings, remove (L®d(u))}2 of the edges, and join the
endpoints of these removed edges to u. This forms a graph H regular of degree L.
We must determine how many edges H has. Apart from the edges of KL+ , one end
"
of every new edge is a vertex of G «. Hence the number of new edges is at most
L­1
% 2d rV(G «)r­0
.
0L­1
1
2
2 1
3 L®dG «(Š)­
v ` V(G «)
But 2m & (L®2d ) rV(G «)r, hence since L & (16d}ε)­2d, the number of new edges is
at most
L­1
L­1
4dm}(L®2d )­
% εm}4­
.
2
2
0 1
0 1
Then if m is large enough, say m & M it follows that rE(H )r % (1­ε}2) m. Hence,
#
provided that m & N(L, ε}2), by Theorem 2.4 we have
h(H ) % (1­ε}2) (1­ε}2)"/# (2m)"/# % (1­ε) (2m)"/#.
Since h(G ) % h(H ), it remains only to choose M ¯ max ²M , M , N(L, ε}2)´. This
!
" #
completes the proof.
It is perhaps worth presenting an alternative formulation of this result. If we
identify two distinct vertices Š, w of a simple graph G, then we shall obtain a loop if
they are adjacent and a multiple edge if they have a common neighbour. However if
Š and w are at distance at least 3 in G, then we shall still have a simple graph after
identifying Š and w. Let us say that G can be collapsed to H if the graph H can be
obtained from G by a sequence of identifications of pairs of vertices distance at least
3 apart. It is easy to see that h(G ) is the smallest number of vertices in a graph H to
which G may be collapsed. Also, G can be collapsed to H if and only if H is a
detachment of G (see for example [7]). Define the density of a graph G as
rV(G )r
rE(G )r}
. Then Theorem 3.1 may be stated as follows.
2
0
1
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 
446
T 3.2. Let d be a fixed integer, and let ε " 0. Let G be a graph with
maximum degree at most d. Then if G has sufficiently many edges, it can be collapsed
to a graph of density at least 1®ε.
4. Locally injectiŠe harmonious colourings
Recall that in the Introduction we defined, for any graph K, the class Γ(K ) of
graphs consisting of a finite number of disjoint copies of K, and mentioned that, by
the theorem of Wilson [10], we have h(G ) ¯ (2m)"/#­o(m"/#) for G ` Γ(K ). (In fact
Wilson’s Theorem actually shows in some cases that h(G ) ¯ Q(m) ¯
8"(1­(8m­1)"/#7.) These colourings also have the added property of being one-to-one
#
on each copy of K, that is, within any copy of K, each vertex has a distinct colour.
In the situation considered in this paper of general bounded degree graphs, we clearly
cannot insist that the harmonious colourings are one-to-one, however a natural
generalisation of the property described above is that no two vertices joined by a short
path should have the same colour. This motivates the following definition.
D. Let r be a positive integer, and G a graph. A harmonious colouring
of G is r-locally injectiŠe if, whenever a pair Š, w of vertices of G is joined by a path
of length at most r, then Š and w have distinct colours.
R. By definition, any harmonious colouring is 2-locally injective.
By making small changes to the proofs of Lemma 2.2 and Lemma 2.3, we can
prove the following.
T 4.1. Let d, r be fixed integers, and let ε " 0. Then there is an integer M
!
such that if G is a graph with maximum degree at most d, and G has m edges, where
m & M , then there is an r-locally injectiŠe harmonious colouring of G with at most
!
(1­ε) (2m)"/# colours.
In the case covered by Wilson’s Theorem above, we can take r to be one greater
than the diameter of K to obtain a colouring which is one-to-one on each copy of K.
Thus Theorem 4.1 gives a weak form of the conclusions of Wilson’s Theorem, but
from considerably weaker hypotheses.
5. Concluding remarks
Let us recall the classes Γd of graphs of maximum degree d, and Γ(K ) of graphs
consisting of disjoint copies of a fixed graph K. We have seen that h(G ) ¯
(2m)"/#­o(m"/#) for G ` Γd. There are two natural ways in which this might be
strengthened, namely a strengthening of the conclusions or a weakening of the
conditions.
The first possibility is that the term o(m"/#) might be reduced, perhaps to a
constant. Let K be a d-regular graph, then the elements of Γ(K ) are of course also
d-regular. It follows from Wilson’s Theorem [10] that if G is a graph Γ(K ) with
C
m¯
edges, and d divides C®1, then h(G ) ¯ C ¯ Q(m), provided that m is large
2
01
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       
447
enough. It seems conceivable that such a result could hold for d-regular graphs in
general. More generally, it may be that for graphs in Γd, we have h(G ) % (2m)"/#­Cd,
where Cd is a constant.
On the other hand, it is natural to ask if the condition of bounded degree is
necessary. It appears that it may be, for Alon [3] proved by random methods that for
400 log (2n) % d % (n log (2n))"/#, there is a graph G with n vertices and maximum
degree at most d for which
h(G ) "
dn"/#
d "/#(2m)"/#
&
,
/
100 (log n)" # 100 (log n)"/#
so that h(G )}(2m)"/# ! 1 does not hold even for classes of graphs with quite slowly
growing degree.
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Department of Mathematics and Computer Science
University of Dundee
Dundee DD1 4HN
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