On K1,k-factorizations of a complete bipartite graph

Discrete Mathematics North-Holland

359

126 (1994) 359-364

On K1 , ,-factorizations of a complete bipartite graph Hong Wang Depariment of Mathematics and Statistics, The University of Calgary, Calgary, Alberta, Canada T2N IN4 Received 10 July 1990 Revised 30 October 1991

Abstract We present a necessary condition for a complete bipartite graph K,., to be K,.,-factorizable and whenever k is a prime number. These two a sufficient condition for K,,, to have a K,,,-factorization conditions provide Ushio’s necessary and sufficient condition for K,,, to have a K,,,-factorization.

1. Introduction

Let Km,, be a complete

bipartite

graph with two partite having m and n vertices, F of K,,, is called a spanning subgraph of K,,, if F contains It is clear that a graph with no isolated vertices is uniquely

respectively. A subgraph all the vertices of K,,,. decided by the set of its edges. So in this paper, we consider a graph with no isolated vertices to be a set of 2-element sets of its vertices. Let k be a positive integer, A K,,k-factor of K,,, is a spanning subgraph F of K,,. such that every component of of K,,, is a set of edge-disjoint K,,,-factors of F is a Kl,k. A K,,,-factorization K which partition the set of edges of K,,, . The graph K,,. is called K l,kif and faE&izable whenever it has a K,,k-factorization. Since K m,nis K1, ,-factorizable only if m= n [l], we will assume from now on that k > 1 holds. Ushio [2] gave a necessary and sufficient condition for K,,, to have a K 1, 2-factorization. In this paper, we prove the following two theorems, which include the result of [2] as a special case. Theorem 1.1. Let k( > l), m and n be positive integers. If K,,. has a KI,k-factorization and (kn-m)(km-n)=O then m< kn, n< km, kn-mrkm-n-O(mod(k2-1)) (modk(k’-l)(m+n)). Correspondence to: Hong Wang, Department Calgary, Alberta, Canada T2N 1N4. 0012-365X/94/$07.00 0 1994-Elsevier SSDI 0012-365X(92)00026-D

of Mathematics

and Statistics,

Science B.V. All rights reserved

The University

of Calgary,

360

H. Wang

Theorem 1.2. Let m and n be positive integers and k a prime m~kn,n~km,kn-m~km-n-O(mod(k2-l))and(kn-m)(km-n)~O(modk(k-1) (k2-

l)(m + n)). Then K,,.

number.

Assume

has a K,,k-factorization.

2. Proof of Theorem 1.1

Let X and Y be the two partites of K,,, with IX(=m and IYI=n. Let {Fl,FZ, . . . . F,} be a K,,,-factorization of K,,.. A vertex u of Fi is called a root of Ft if the degree of u in Fi is k. Let a and b be the numbers of roots of Fi contained in X and Y, respectively. Let c=(kn-m)(km-n)/k(k+ l)(m+n). Then we have a+kb=m,

ka+b=n.

(1)

Since 1Fi I= k (m + n)/(k + l), we have r=(k+l)mn/k(m+n)=a+b+c.

(2)

Thus, we get a =(kn -m)/(k2 - 1) and b =(km- n)/(k2 - 1). Since a, b and Y must be nonnegative integers, we have m < kn, n < km, (kn -m) z (km - n) = 0 (mod (k2 - 1)) and c is an integer. Let UEX. Suppose that there are only r’ Ft’s, each of which contains u as a root. Then kr’ + (r - r’) = n, i.e. (k - 1) r’ + a + b + c = ka + b. Therefore, c z 0 (mod (k - l)), i.e. (kn - m)(km - n) E 0 (mod k (k2 - l)(m + n)). This proves the theorem. 0 From the above proof, we obtain an obvious corollary as follows. Corollary 2.1. If K,,, is K,,, -factorizable, (k+l)mn/k(m+n)-m-n(mod(k-1)).

then (k + l)mn/k(m+

n) is an integer and

3. Proof of Theorem 1.2

The proof consists of the following lemmas. For any two integers x and y, we use gcd(x, y) to denote the greatest common divisor of x and y. Lemma 3.1. Let g, p and q be positive integers. If gcd(p, q) = 1 then gcd(pq, gp + q) = gcd (4, g). Proof. Obvious.

•!

Lemma 3.2. Zf K,, ,, has a K,,k-factorization, any positive integer s.

then K,,, ns has a Kl,k-factorization

Proof. We just outline how to construct a K,,,-factorization

for

of K,,,., and the reader can easily understand it. It is well known that the graph K,,, is K1, i-factorizable. Let

361

Klsk-factorizations of a complete bipartite graph (FtI l
replace every Gi (1
pairwise-disjoint and their union is K,,,.,. Since K,, n has a K,,,-factorization, it is as well. And consequently, K,,,,, has clear that the graphs Gi have K ,,,-factorizations 0 a K,,,-factorization. This proves the lemma. A corollary

of Lemma

3.2 is as follows.

Corollary 3.3. K,,ks is K I,k-factorizable for any positive integer s. Corollary 3.3 implies that we only need to treat the case m
d=Wp+q)4pq9

(4) m=Wp+q)(p+q)elpq,

(5)

n=(kp+q)(k2p+q)4pq,

(6)

a=Mkp+q)elpq,

(7)

b=q(kp+q)elpq.

(8)

Now we can establish

the following

lemma.

Recall that k is a prime number.

Lemma 3.4. Zf gcd (q, k2 ) = 1, then n=(kp+q)(k2p+q)s,

m=k(kp+q)(p+q)s, a=pk(kp+q)s,

b=q(kp+q)s,

for some positive integer s. Ifgcd(q, k2)=k, let q=kq,.

n=k(p+q,)(kp+q,)s, b=q,k(p+q,)s,

for some positive integer s. Ifgcd(q, k2)=k2, let q=k2q2. m=(p+kq2)(p+k2q2)s, a=p(p+kqz)s, for some positive integer s.

(10)

Then

m=k(p+qI)(p+kqI)sY a=pk(p+q,)s,

r=(p+q)(k2p+q)s

(9)

r=(p+kq,)(kp+q,)s

(11) (12)

Then n=k(p+kq,)(p+q,)s,

b=q,k(p+kq,),

r=(p+k2q2)(p+q2)s

(13) (14)

362

H. Wang

Proof. First

assume

that

gcd( q, k2) = 1 holds.

By

Lemma

3.1,

we

see

that

gcd(pq, p+q)= 1 and gcd(pq, k’p+q)=gcd(q, k2)= 1. Since r=(p+q)(k2p+q)e/pq is an integer, we see that e/pq must be an integer. Let s = e/pq. Then the equalities in (9) and (10) hold. that gcd(q, k2 ) = k holds. Then gcd(p, k) = 1 holds. It is easy to

Next we assume

see that r=(p+kqI)(kp+qI)e/pqI.

By Lemma

3.1, we see that gcd(pq,,p+kq,)=

gcd(p, k)= 1 and gcd(pq,, kp+q,)=gcd(q,, k)= 1. Therefore, ger. Let s = e/pq, . Then the equalities in (11) and (12) hold. Finally, easy

to

we assume see

that

that

gcd(q, k2)= k2 holds.

r=(p+k2q2)(p+q2)e/pq2.

Then By

e/pq,

must be an inte-

gcd(p, k2) = 1 holds.

Lemma

3.1,

we

see

It is that

gcd ( pq2, p + k2 q2 ) = gcd (p, k2) = 1 and gcd ( pq2, p + q2 ) = 1. Therefore, e/pq, must be an integer. Let s = e/pq,. Then the equalities in (13) and (14) hold. This proves the lemma. 0 Lemma 3.5. For n=(kp+q)(k2p+q).

any positive integers p and q, let Then K,,, has a KI,k-factorization.

Proof. Let a=pk(kp+q), b=q(kp+q), r=(p+q)(k2p+q), Let X and Y be the two partites of K,,, and set X={xi,jIl,

m=k(kp+q)(p+q)

r,=p+q

and

and r2=k2p+q.

l
We will construct a K,,,-factorization of K,,, . We remark in advance that the additions in the first subscripts of xi,j’s and yi,j’s are taken modulo r1 and r2 in r2}, respectively, and the additions in the second subscripts {1,2,..., r,)and{l,2,..., Of Xi,j(s and yi,j’s are taken modulo k(kp+q) and (kp+q) in {1,2, . . ..k(kp+q)} and kp + q}, respectively. (62, . . . . For each l
and for each 1 d id q, let Ep+i=(x

p+i,j+s(f)YpkZ+i,kp+j+i-lIl~j~kp+q,

lGf
We can perceive this construction in the following way. For each we identify xi,j as xj for all in{ 1,2, . . . . rl }. For each j~{l, 2, . . ..k(kp+q)}. $Z { 1,2, . . , kp + q}, we identify y,, I as ys for all eE { 1, 2, . . , r2 >. Then we would get aK Lu+,+q),Ckp+qJ. In this way, each of the graphs EL (1 < i
K,,,-factorizations

Let F= UFJFEi. c from XU Yonto

Then

F is a K ,,k-factor

the graph

of K,,,.

XU Yin such a way that a(xi,j)=Xi+1,j

each iE{l,2 ,..., rI} and eachjE{1,2 Fi, j={a’(X)[y’(y)lXEX,

363

of a complere bipartite graph

Define

a bijection

and a(yi,j)=yi+l,j.

For

,..., r2}, let Ye

Y, xY~F).

It is easy to show that the graphs Fi,j (l
3.6. For

n=k(p+q)(kp+q).

any

positive

Then K,,.

0 integers

p

and

q,

Proof. Let a=pk(p+q), b=qk(p+q), r=(p+kq)(kp+q), Let X and Y be the two partites of K,,, and set X={xi,jll
16 jdk(p+q)},

Y=(yi,jl

1 d j
ldiQr2,

m=k(p+q)(p+kq)

let

and

has a K,,k-factorization. r,=p+kq

and r2=kp+q.

We will construct a K,,,-factorization of K,,,. We remark in advance that the additions in the first subscripts of Xi,j’S and yi,j’S are taken modulo rl and r2 in (132, . . . . rl } and (1,2, . . . , r2 }, respectively, and the additions in the second subscripts of xi,j’s and yi,j’S are taken modulo k(p+q) in (1, 2, . . ..k(p+q)}. For each l
l
and each and let

lbh
l
Ep+i={Xt(i,h),jYkp+i,w(i,h)+jI

let t(i,h)=p+k(i-l)+h

1 d jGk(p+q),

and

w(i,h)=

l
We can perceive this construction in the following way. For each jE { 1,2, . . . , k (p + q) >, weidentifyxi,jasxjforall iE{l,2,...,r,). ForeachjE(1,2,...,k(p+q)},weidentify Y,,~ as yjfor all ee{l, 2, . . ..r2}. of the graphs Ei (1
Then we would get a KkCp+qI,kCp+q). In this way, each would consist of k edge-disjoint K1, i-factors of this

Ei (1
would

partition

the set of edges of this

K k(p+q),k(p+d~

Let F = u FzT’l”Ei.Then the graph F is a K 1, ,-factor of K,, n. Define a bijection CJfrom XV Y onto XU Yin such a way that a(xi,j)=Xi+l,j and o(yi,j)=yi+l,j. For each i~(l, 2, . . . . rI} and each jE{l, 2, . . . . r2}, let Fi,j=(o’(X)a’(y)(X~X,

yEY,

xyEF}.

It is easy to show that the graphs Fi,j (1 < i
H. Wang

364

Now it is clear that Theorem that when k=2, the conditions conclude

1.2 follows immediately of Theorems

our paper with the following

from Lemmas

1.1 and 1.2 coincide

3.2-3.6. Note

with each other. We

corollary.

Corollary 3.7 (Ushio [2]). Let m and n be positive integers. Then K,,, is KI,z-factorizable if and only if m<2n, n<2m, (2n-m)r(2m-n)=O(mod 3) and (2n-m)(2m-n)=O(mod 6(m + n)).

References [l]

J.A. Bondy and U.S.R. Murty, Graph Theory Basingstoke, 1976) 72-75. [Z] K. Ushio, P,-factorization of complete bipartite

with Applications graphs,

Discrete

(Macmillan Math.

Press,

London

72 (1988) 361-366.

and