Minimum resolvable coverings with small parallel classes

For prime power q, we determine the minimum number of parallel classes in a resolvable 2-(kq, L 1) covering for almost all values of k. @ 1999 Elsevier Science B.V. All rights reserved In September 1995 Jan van Haastrecht gave a dinner on the occasion of his retirement. He had invited 20 colleagues. The host and the guests were to be seated at three tables each of which had seven places. The dinner consisted of five courses. Jan wanted the guests to change places between the courses in such a way that everyone would meet everyone else (Jan included) at least once at some table. At that time no solution was known and so a few couples did not meet. A t-(r,k,)~) covering consists of a v-set V (of points) together with a collection o f k-subsets of V (called blocks) such that every t-subset of V is a subset of at least ,;. blocks. A covering is called resolvable if the collection of blocks admits a partitioning into classes (called parallel classes), such that each class consists of r"k disjoint blocks. Thus, the dinner problem just described can be rephrased as: does there exist a resolvable 2-(21,7, l ) covering with five parallel classes? We speak of a minimum resolvable t-(v,k,)~) covering i f there exists no resolvable t-(t',k, 2) covering with fewer parallel classes. In this note we consider minimum resolvable 2-(t~,k, 1) coverings with a given cardinality q = v/k of a parallel class. We shall show that if q is the order of an affine plane, the number o f parallel classes in such a covering equals q + 1 i f q divides k and q + 2 otherwise with just a small number of exceptions. In particular for q~<3 there are no exceptions, meaning that Van Haastrecht 's problem has a solution. Define r (q ,k ) to be the number of parallel classes in a minimum resolvable 2-(kq, k, 1) covering. It is well known (see [3]) that r(q, k ) ~ q + ( q 1 ) / ( k 1) with equality if and only i f there exists a resolvable 2-(kq, k, 1 ) design. In particular, r(q, 2) = 2q I E-mail: [email protected]. 0012-365X/99/$ see front matter @ 1999 Elsevier Science B.V. All rights reserved PlI: S001 2-365X(98)00247-7 394 W.H. Haemers / Discrete Mathematics 197/198 (1999) 393-396 for all q and r(q,q)=q + 1 if q is a prime power (or, more precisely, the order of an affine plane). Lemma 1. I f m is a positive integer then r(q, km) <<. r(q, k). Proof. Take a minimum resolvable 2-(kq, k, 1) covering and replace each of the kq points by an m-set. This gives a resolvable covering with block size km and the same number of classes as before. [] So r(q, qm)=q + 1 if there exists an affine plane of order q. The next result characterizes the case r( q,k )=q + 1. Theorem 1. The number of parallel classes in a minimum resolvable 2-(kq, k, 1 ) covering is at least q + 1. Equality holds if and only if q divides k and q is the order of an affine plane. Proof. Put r=r(q ,k) and let xi be the number of pairs that are covered exactly i times. Then clearly r ( ~ 2 ) ' £ qr(k)2 Xo : O, ~ Xi :ixi -= i