DISCRETE
MATHEMATICS
ELSEVIER
Discrete Mathematics 151 (1996) 155-160
On equitable coloring of bipartite graphs
K o - W e i Lih a'*, P o u - L i n W u b
"Institute of Mathematics, Academia Sinica, Taipei, Tmwan, ROC
bDepartment of Mathematics, Louisiana State University, Baton Rouge, LA 70803, USA
Received 3 December 1991; revised 1 December 1992
Abstract
If the vertices of a graph G are partitioned into k classes V~, I/2..... Vksuch that each V~is an
independent set and I1V~I- IV~[I~< 1 for all i#j, then G is said to be equitably colored with
k colors. The smallest integer n for which G can be equitably colored with n colors is called the
equitable chromatic number xc(G) of G. The Equitable Coloring Conjecture asserts that
zc(G) ~ A(G) for all connected graphs G except the complete graphs and the odd cycles. We
show that this conjecture is true for any connected bipartite graph G(X, Y). Furthermore, if
IXI = m/> n = IYI and the number of edges is less than Lm/(n + 1)J(m -- n) + 2m, then we can
establish an improved bound Xo(G) ~<Fm/(n + 1)-]+ 1.
1. Introduction
All graphs considered in this paper are finite, loopless, and without multiple edges.
If the vertices of a graph G are partitioned into k classes I/1, I/2..... Vk such that each V~
is an independent set and [[ V~[ - I Vj[[ ~< 1 for all i ~j, then G is said to be equitably
colored with k colors. The smallest integer n for which G can be equitably colored with
n colors is defined to be the equitable chromatic number of G, denoted by ;(e(G). It is
obvious that xe(G) is not less than the ordinary chromatic number x(G). This notion
of equitable colorability was introduced by Meyer [5]. Let degG(x), or deg(x) for
short, denote the degree of the vertex x in the graph G. Let A (G) be the largest degree
of any vertex in the graph G. Let [-x-] and [_xJ denote, respectively, the smallest integer
not less than x and the largest integer not greater than x. Meyer showed that a tree
T can be equitably colored with [-A(T)/2] + 1 colors. However, his proof was faulty. It
is reported in Guy [3, pp. 998-999] that Eggleton has remedied the defects. Eggleton
could prove that a tree T can be equitably colored with k colors, provided
* Corresponding author.
0012-365X/96/$15.00 © 1996--Elsevier Science B.V. All rights reserved
SSDI 0 0 1 2 - 3 6 5 X ( 9 4 ) 0 0 0 9 2 - W
156
K.-W. Lih, P.-L. Wu/Discrete Mathematics 151 (1996) 155-160
k > ~ A ( T ) / 2 q + 1. Meyer [5] also proposed the following conjecture which,
if true, is stronger than the well-known Brooks' theorem I-1] for the chromatic
number.
Equitable Coloring Conjecture. Let G be a connected graph. If G is neither a complete
graph nor an odd cycle, then
ze(G) ~< A ( G ) .
This conjecture has been verified for all graphs with 6 or fewer vertices. An earlier
results of Hajnal and Szemer6di [4-1 shows that a graph G can be equitably colored
with k colors ifk >/A(G) + 1. The statement of the conjecture is included in the survey
of Toft 1-6] as Problem 36.
2. Bipartite graphs
It will follow from our Theorems 1 and 2 that the Equitable Coloring Conjecture
holds for all connected bipartite graphs. We then derive a better bound for certain
bipartite graphs satisfying some restriction on the number of edges. We make
the convention that all bipartite graphs in this paper are assumed to have at least
one edge.
Theorem 1. Let G = G(X, Y) be a connected bipartite graph. I f G is different from any
complete bipartite graph K .... then G can be equitably colored with A(G) colors.
Proof. Let IX[ = m/> n = [Y[. Write A for A(G). If m > n , then X must contain
a vertex of degree less than A. Since G is connected, it follows that
n ~< (A - 2) + (m - 1)(A - 1) + 1. Hence r(m + n)/A] <<.m. On the other hand, we
have m <~ n(A - 1) + 1. It follows that [-(m + n - A + 1)/~q = l(m + n)/AJ <<.
k(n(A - 1) + 1 + n ) / A j = n. Since A >/2, both f ( m + n)/A] <<.m and L(m + n)/AJ <<,n
still hold when m = n.
It is easy to see the identity m + n = F ( m + n ) / A ] + F ( m + n - 1 ) / A
7+...+
~(m + n - A + 1)/A-~ If we can partition X into classes of sizes ~(m + n)/AT,
~(m + n - 1)/Aq ..... ~(m + n - k)/Aq for a certain k, 0 ~< k < A - 1, then we can
partition Y into classes of sizes r(m + n - k - 1)/dq ..... ~(m + n - ,t + 1)/~q. Thus
we have an equitable coloring with A colors. Otherwise, we partition X into classes of
sizes r(m + n)/~q, F(m + n - 1)/47 ..... F(m + n - k + 1)~AT, and s; partition Y into
classes of sizes ~ (m + n - k ) / A q ..... F(m + n - A + 2)/A-I,and t. The numbers s, t, and
k satisfy the conditions O < k < A - 1 ,
O<s<[(m+n-k)/A-],
and s + t =
~(m + n -- A + 1)/A~.
If the following claim is true, then S w T, regarded as a class, together with the other
classes constitute an equitable coloring of G with A colors and Theorem 1 is established.
K.-W. Lih, P.-L. W u / Discrete Mathematics 151 (1996) 155 160
157
C l a i m . T h e r e e x i s t S ~_ X a n d T ~ Y s u c h t h a t ISl = s, ITI = t, a n d S w T is an i n d e p e n dent set of vertices.
We
first
show
t h a t the t w o i n e q u a l i t i e s m - ( t ( A - 1 ) +
1)~<s-1
and
1 c a n n o t be t r u e s i m u l t a n e o u s l y . S u p p o s e t h a t t h e y could. Then,
b y a d d i n g t h e m t o g e t h e r , we h a v e m + n - (s + t ) ( A - 1) - 1 ~< s + t - 2. It follows
that
(m + n + 1)/A <<. s + t = F(m + n - A + 1)/A-] = L(m + n)/AJ<<. (m + n ) / A ,
a contradiction.
C a s e 1: Let m - (t(A - 1) + 1)/> s. W e use N ( V ) to d e n o t e the set of vertices which
are a d j a c e n t to at least o n e vertex in V. Since G is c o n n e c t e d , we can successively
n - s(A - - 1) ~< t -
c h o o s e d i s t i n c t vertices Yl,Y2 . . . . . y, in Y such t h a t IN(y,+ 1) c ~ N ( y l , y 2 . . . . . Yi)l >~ 1
for all 1 ~ < i < t .
Let T = { y a , Y 2 . . . . . y,}. T h e n I N ( T ) I ~ < t ( A - 1 ) + 1. Since
m - (t(A - - 1) + 1) ~> s, the set S can be selected a n d the c l a i m is true.
C a s e 2: Let m - (t(A - 1) + 1) ~< s - 1 a n d n - s ( A - 1) ~> t. O u r c l a i m is true if
we c a n find a s u b s e t S of X such t h a t ISI = s a n d IN(S)I ~< s ( d - 1). Since G is
c o n n e c t e d , we can successively c h o o s e d i s t i n c t vertices x l , x 2 . . . . . x~ in X such t h a t
I N ( x i + 1) n N ( x l , x 2 . . . . . xi) l >~ 1 for all 1 ~< i < s. U n l i k e case 1, we n o w c h o o s e x l to
be a vertex of d e g r e e less t h a n A if there is at least one such vertex. O t h e r w i s e we
c h o o s e a n a r b i t r a r y vertex as x l . Let S = { x l , x 2 . . . . . x , } .
If we s t a r t e d o u t with d e g ( x l ) < A, t h e n IN(S)I ~< (A - 2) + (s - I)(A - 1) + 1 =
s(A - - 1) a n d we are d o n e . O t h e r w i s e the g r a p h m u s t be a r e g u l a r g r a p h of d e g r e e
A a n d m = n. T h e trivial case is w h e n A = 2. T h u s G is an even cycle a n d ze(G) = 2. So
we let A >~ 3. If in the p r o c e s s of selecting the m e m b e r s of S we a c t u a l l y h a d
IN(xi+OnN(xl,x2
. . . . . xi)l / > 2 for s o m e i, l < < . i < s ,
t h e n it is easy to see
IN(S)I ~< s ( Z - - 1).
Now
suppose
. . . . . xi)l = 1
for
all
1 ~< i < s
and
IN(x)c~N(xl,x2
. . . . . x , _ l ) I <~ 1 for all x in X { x l , X z . . . . . x , - l } . In this case,
IN(S)I = s ( A - 1) + 1.
The
claim
is
true
if
the
stronger
inequality
n - (s(A - 1) + 1) t> t holds.
S u p p o s e , o n the c o n t r a r y , n - (s(A - 1) + 1) ~< t - 1. W e h a v e a l r e a d y a s s u m e d
t h a t n - (t(A - 1) + 1) = m - (t(A - 1) + 1) ~< s - 1. By a d d i n g these t w o i n e q u a l i ties t o g e t h e r , we get 2 n / A <~ s + t = 1 2 n / A I. T h i s implies t h a t A divides 2n. F r o m the
w a y of d e f i n i n g s a n d t, it follows t h a t s = t = n / A . N o w IX - { x l , x 2 . . . . . x~_ 1}1 =
n - s + 1 a n d IY - N ( x ~ , x 2 . . . . ,x~-~)l = n - (s - 1)(A - 1) - 1. W h e n we c o u n t the
n u m b e r of edges b e t w e e n X - { x l , x 2 , . . . , x ~ - l }
and Y-N(xl,x2
. . . . . x ~ - l ) in
t w o different ways, it c o m e s
out
both
/>(n-s
+ 1)(A - 1) a n d
~<
(n - (s - 1)(A - 1) - 1)A. Since s = n / A , we h a v e
IN(Xi+l)C~N(xl,x2
(n -- (s - - 1 ) ( A
--1)
- - l ) A /> (n -- s + l ) ( A -- 1 ) ,
A 2-nA-3A+3n-s+
d 3-nA
(A-n)(A
2-3A
2-3A
1 >10,
2+3nA-n+A
+ 1)>/0.
>~0,
158
K.-W. Lih, P.-L. Wu / Discrete Mathematics 151 (1996) 155-160
H o w e v e r the regular bipartite graph G is not a complete bipartite graph, hence A < n.
At the same time, d /> 3 implies that d 2 -- 3d + 1 is positive. It is impossible to have
(A -- n)(d 2 -- 3A + 1) >~ 0. Therefore n - (s(A -- 1) + 1) >/t and the claim is true. []
Theorem 2. The complete bipartite graph K U can be equitably colored with k colors if
and only if [n/[k/2 f] - [n/~k/2.]J <~ 1.
Proof. Case l: When k = 2p, it is clear that rnlLkl2Jq - Lnlrk121J = rnlp] - l_nlpd.
If there is an equitable coloring of Kn.n with k colors, then the size of any color class is
either [n/p-] or [n/pJ. It follows that [n/Lk/2]] - [n/[k/2]J ~< 1. Conversely, if the
inequality holds, then we partition X, respectively Y, into classes of sizes
In~p], [(n - 1)/p-] ..... L(n - p + 1)/p-].
Since
n = In~p] + ['(n - 1)/p-] + . . . +
[-(n - p + 1)/p], we have an equitable coloring with k colors.
Case 2: When k = 2p + 1, K~., can be equitably colored with k colors if and
only if X can be equitably colored with kl colors and Y with k2 colors such that
k, + k2 = k, k, < k2, and I n ~ k , ] - [nlk2J ~< 1. N o w [nlp]- ln/~p + 1)d
In~k1.]- [n/k2J. Hence the existence of an equitable coloring of K~.~ implies that
[n/Lk/2J]- [n/[k/E]J ~ 1. Conversely, if the inequality holds, then we partition
X into classes of sizes [n/p'],[(n- 1)/p] ..... [ ( n - p + 1)/p.] and Y into classes
of sizes [n/(p + 1).],[(n - 1)/(p + l).] ..... [(n - p)/(p + 1)]. Since we also have
n = [n/(p + 1)-] + [(n - 1)/(p + 1)'] + ... + [(n - p)/(p + 1)'], K~.n can be equitably
colored with k colors. []
Theorem 3. Let G(X,Y) be a connected bipartite graph with e edges. Suppose
IXl = m > / n = IYI a n d e < [m/(n + 1)J(m - n) + 2m. Then z,(G) <~Fm/(n + 1)] + 1.
Proof. Let q = [ m / ( n + 1)J. Hence m = q ( n + l ) + r
for some r such that
0~r<n+l.
Then[m/(n+l)]+l=[q+(r/(n+
1 ) ) ] + 1 which i s q + l
if r = 0
andisq+2if0<r<n+l.
If m = n, then z e ( G ) = 2 = [m/(n + 1)-] + 1. If m > n , then it is easy to see
xe(G) ~< q + 1 when r = 0 and xe(G) ~ q + 2 when r = n. So T h e o r e m 3 holds.
It remains to prove T h e o r e m 3 when m > n > r > 0. U n d e r these circumstances, we
have q / > 1 and we want to color G with q + 2 colors.
We first search for numbers k and t such that 0 ~< k < n, 0 ~< t ~< q + 1, and m + k
elements can be partitioned into t classes of size n - k and q + 1 - t classes of size
n-k+1.
Let k = [ ( n - r + l ) / ( q + 2 ) J ,
hence n - r + l = k ( q + 2 ) + t
for some
t such that 0 ~ < t ~ q + l .
Since 0 ~ < k ~ < ( n - r + l ) / ( q + 2 ) ~ < n / ( q + 2 ) ~ < n / 3 < n ,
the n u m b e r k falls into the desired range. Furthermore,
(m+k)-t(n-k)-(q+
l-t)(n-k
=q(n+ 1)+r+k-t(n-k)-(q+
= k ( q + 2) + t - ( n - r
+ l)=O.
Therefore k and t satisfy our requirements.
+ 1)
1-t)(n-k
+ 1)
K.-W. Lih, P.-L. Wu/Discrete Mathematics 151 (1996) 155-160
159
Case 1: S u p p o s e t h a t the n u m b e r k is equal to 0. W e m a y p a r t i t i o n X into
q + 1 - t classes of size n + 1 a n d t classes of size n. R e g a r d Y as a single class. W e
then c o l o r each class with a distinct color. T h u s G is e q u i t a b l y c o l o r e d with q + 2
colors.
Case 2: S u p p o s e that the n u m b e r k is different from 0. W e claim that there exist
A ~ X a n d B~_ Y s u c h t h a t IAI = n - 2k + 1, IBI = k, a n d A u B is an i n d e p e n d e n t set
of vertices o f size n - k + 1. W h e n t = 0, we c o l o r vertices in A u B with the same
color. A c c o r d i n g to the choices of k a n d t, we m a y then p a r t i t i o n X - A into q classes
of size n - k + 1. W h e n t > 0, we delete one element from A to get A' a n d c o l o r
vertices in A' u B with the same color. W e then p a r t i t i o n X - A' into q + 1 - t classes
of size n - k + 1 a n d t - 1 classes of size n - k. In b o t h cases, we always let Y - B be
a single class of size n - k. T h e n G is e q u i t a b l y c o l o r e d with q + 2 colors.
T o p r o v e the claim, let u = In~k] a n d hence n = uk + v for s o m e v such that
O<~v<k.
Let Y = {Yl,y2 . . . . . y,} be such that d e g ( y l ) / > deg(y2)/> . . . / > deg(y,). If v > 0,
then we define V = {Yl, Y2 . . . . . Yv}- Let w(S) d e n o t e the n u m b e r of edges which have at
least one end vertex in the set S. T h e n w(V) >i 2v if V c o n t a i n s no vertex of degree 1.
O n the o t h e r h a n d , Y - V c o n t a i n s only vertices o f degree 1 if V c o n t a i n s at least one
such vertex. In this case,
~o(V)=e-(n-v)l>(m+n-
1)-(n-v)
>~m-l+v>~n+v>2v.
M o r e o v e r , we j u s t let V = 0 if v = 0. T h u s in all cases we have o~(V) >1 2v.
N e x t we p a r t i t i o n Y - V into Y~, Y2 . . . . . Yu such t h a t I Yil = k for i = 1, 2 . . . . . u.
Subcase 1: If to(Y i ) < m - n + 2k for a certain Yi, then X c o n t a i n s at least
m - (m - n + 2k - 1) = n - 2k + 1 vertices which are i n d e p e n d e n t of Y, So we are
able to c h o o s e the required sets A a n d B.
Subcase 2: S u p p o s e ~o(Yi) >1 m - n + 2k for i = 1,2 . . . . . u. T h e n
e = ~o(V) + ~ ¢o(Yi) >1 2v + u ( m -
n + 2k)
i=1
= 2v + u ( m -
n) + 2 n -
2v = u ( m -
n) + 2 n ,
i.e., e >~ [.n/k](m - n) + 2n. H o w e v e r , [.n/k] >t Ln(q + 2)/(n - r + 1)] 1> q + 2. It follows t h a t e >f (q + 2)(m - n) + 2n = q(m - n) + 2m = [_m/(n + 1)J(m - n) + 2m. This
c o n t r a d i c t s the a s s u m p t i o n of T h e o r e m 3. []
R e m a r k . T h e b o u n d for xe(G) in T h e o r e m 3 is better t h a n d when e/> 2. This is
obvious whenn~m~<n+
1. S u p p o s e m > n +
1. T h e n n d t > e > / m + n - l .
Itfollows t h a t d >1 (m + n - 1)/n > (m + n + 1)/(n + 1). Since d is an integer, we have
A >~ V(m + n + 1)/(n + 1)-] = [m/(n + 1)-] + 1.
160
K.- W. Lih, P.- L. Wu / Discrete Mathematics 151 (1996) 155-160
3. Trees
A tree T can be viewed as a bipartite graph T(X, Y) with IXl = m and IY I = n.
The number of edges of T is exactly r e + n - 1 which is less than
Lm/(n + 1).J(m - n) + 2m. Theorem 3 implies the following.
Corollary. Regarding a tree T as a bipartite
zc(T) <~I-(ISl + IYI + 1)/(min{lXl, IYI} + 1)7.
graph
T(X, Y),
we
have
Remark. Recently, Bor-Liang Chen and the first author have found an exact formula
for xc(T) of a tree T. Their paper I-2] has appeared elsewhere.
Acknowledgements
The authors are grateful to Bor-Liang Chen and Shan-Pao Sun for fruitful discussions on this research.
References
[1]
[2]
I-3]
[4]
R.L. Brooks, On colouring the nodes of a network, Proc. Cambridge Philos. Soc. 37 (1941) 194-197.
B.-L. Chen and K.-W. Lih, Equitable coloring of trees, J. Combin. Theory Ser. B 61 (1994) 83-87.
R.K. Guy, Monthly research problems, 1969-1975, Amer. Math. Monthly 82 (1975) 995-1004.
A. Hajnal and E. Szemer6di, Proof of a conjecture of Erd6s, in: P. Erd6s, A. R6nyi and V.T. S6s, eds.,
Combinatorial Theory and Its Applications, Vol. II, Colloq. Math. Soc. Jfi.nos Bolyai 4 (NorthHolland, Amsterdam, 1970) 601-623.
[5] W. Meyer, Equitable coloring, Amer. Math. Monthly 80 (1973) 920-922.
F-6] B. Toft, 75 graph-colouring problems, in: R. Nelson and R.J. Wilson, eds., Graph Colourings
(Longman, Essex, 1990) 9-36.