Haemers, W.H.

English

Research Papers in Economics

Minimum resolvable coverings with small
parallel classes
Willem H. Haemers
Tilburg University, Tilburg, The Netherlands
Abstract
For prime power q, we determine the minimum number of parallel
classes in a resolvable 2-(kq,k,1) covering for almost all values of k.
In September 1995 Jan van Haastrecht gave a dinner on the occasion of
his retirement. He had invited twenty colleagues. The host and the guests
were to be seated at three tables each of which had seven places. The diner
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 didn’t meet.
A t-(v,k,λ) covering consists of a v-set V (of points) together with a
collection of k-subsets of V (called blocks) such that every t-subset of V is a
subset of at least λ 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 v/k disjoint blocks. Thus the dinner problem just de-
scribed can be rephrased as: does there exist a resolvable 2-(21,7,1) covering
with ﬁve parallel classes? We speak of a minimum resolvable t-(v,k,λ) cover-
ing if there exists no resolvable t-(v,k,λ) covering with fewer parallel classes.
In this note we consider minimum resolvable 2-(v,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 of parallel classes in such a covering equals q+1
if 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
1problem 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 [1]) that r(q,k) ≥ q+(q−1)/(k−1)
with equality if and only if there exists a resolvable 2-(kq,k,1) design. In
particular r(q,2) = 2q − 1 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 If 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. 2
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
x0 = 0,
r X
i=1
xi =
 
kq
2
!
,
r X
i=1
ixi = qr
 
k
2
!
.
Let x be the number of times we can choose an unordered pair of points and
an unordered pair of blocks such that both points are in both blocks. On the
one hand
x =
r X
i=1
 
i
2
!
xi.
To ﬁnd a second expression for x, deﬁne σb,b′ to be the number of points in
the intersection of a block b and a block b′. Then
x =
1
2
X
b =b′
 
σb,b′
2
!
.
2Fix a parallel class C and a block b′  ∈ C. It is straightforward that
X
b∈C
 
σb,b′
2
!
≥ q
 
k/q
2
!
with equality if and only if σb,b′ = k/q for all b ∈ C. Hence
x ≥
r(qr − q)
2
  q
 
k/q
2
!
=
kr(r − 1)(k − q)
4
.
Combining our formulas for xi we obtain
0 ≥
r X
i=1
(i − 1)(i − r)xi =
r X
i=1
(2
 
i
2
!
xi − rixi + rxi) ≥
kr(r − 1)(k − q)
2
− qr
2
 
k
2
!
+ r
 
kq
2
!
=
k2r(q − 1)(q + 1 − r)
2
.
Therefore r ≥ q + 1. Suppose r = q + 1. Then we have equality everywhere,
which yields that xi = 0 if 2 ≤ i ≤ r − 1 and that any two blocks from
diﬀerent parallel classes meet in exactly k/q points. So the covering is just
an aﬃne plane wherein each point is replaced by a set of size k/q (as in
Lemma 1). 2
Theorem 2 Let q be the order of an aﬃne plane and let k be a positive
integer such that &
k
q
'
≤
2k
2q − 1
. (1)
Then there exists a resolvable 2-(kq,k,1) covering with q+2 parallel classes.
Proof. Deﬁne m = ⌈k/q⌉ and n = mq −k, then 0 ≤ n < q and (1) becomes
m ≥ 2n. Take a parallel class C of an aﬃne plane of order q and ﬁx n lines
in C. Replace (similar to the construction of Lemma 1) each point of these n
lines by an (m−1)-set and all the other points by an m-set. Then we obtain
a resolvable covering of kq points with q + 1 classes and all blocks, not com-
ing from a line of C have k points. However, from the blocks corresponding
to C, n have q(m − 1) points (the small blocks) and q − n have qm points
(the large blocks). From this ‘unbalanced’ covering we will make one with all
3block sizes equal by replacing the parallel class corresponding to C by two
new parallel classes with all block sizes equal to k in the following way. For
each large block, we ﬁx a set S of m points corresponding to one point of the
original aﬃne plane and, in addition, we take two disjoint subsets S1 and S2
of S, both of size n (we can do so because m ≥ 2n). For i = 1 and i = 2 we
make a new parallel class by moving the points of Si from each large block
to the small blocks such that all block sizes become k. It is easy to check
that this indeed produces gives a resolvable 2-(kq,k,1) covering with q + 2
parallel classes. 2
For a given prime power q, Theorem 1 and 2 give the value of r(q,k)
for almost all k. Indeed, for the inequality (1) to hold it suﬃces that k ≥
2(q − 1)2. In particular, for q = 2 there are no exceptions. Suppose q = 3.
Only for k = 2, 4 and 7 condition (1) is not satisﬁed. We know r(3,2) = 5
and, by Lemma 1 and Theorem 1, r(3,4) = 5. To make a covering with 5
classes if k = 7 (the dinner problem of the introduction) we ﬁrst give one for
k = 5 (diﬀerent from the one given by Theorem 2) in the table below.
1 2 3 A D 1 4 7 A B 1 5 9 A E 1 6 8 A F 1 4 5 A C
4 5 6 B F 2 5 8 C D 2 6 7 C F 2 4 9 D E 2 8 9 B E
7 8 9 C E 3 6 9 E F 3 4 8 B D 3 5 7 B C 3 6 7 D F
The point set is {1,...,9} ∪ {A,...,F} and the rows within each box rep-
resent the three blocks of a parallel class. The letters form a resolvable
2-(6,2,1) covering (a 1-factorization of the complete graph on six vertices)
and the digits in the ﬁrst four classes represent the aﬃne plane of order three.
Given these ingredients, it is an easy exercise to complete the covering by
hand. Next, by letting each letter represent a pair of points, we obtain the
desired covering with k = 7. In fact, replacing the digits by some m-set, and
the letters by some m′-set, provides coverings with ﬁve classes and q = 3 for
any k.
If q = 4 then (1) is satisﬁed for k  ∈ {2,3,5,6,9,10,13,17}. We know
r(4,2) = 7 and Lampken and Mills [2] showed that r(4,3) = 7. For the
remaining six values of k we don’t know the value of r(4,k). However, it can
be seen that r(4,k) ≤ 7. For k = 6, 9 and 10 this follows from Lemma 1 and
for k = 5, 13 and 17 it is a straightforward exercise.
If q = 5 we have r(5,2) = 9, r(5,3) = 7 (Kirkman’s schoolgirls problem),
r(5,4) = 7 (see [1] or [3]) and so, by Lemma 1, r(5,k) ≤ 7 if k is divisible
4by 3 or 4. Now Theorem 1 and 2 only leave k ∈ {7,11,13,17,22,26,31}
unsolved.
If q is not a prime power, little is known. An easy recursive construction
gives r(q1q2,k) ≤ r(q1,k)r(q2,k), but this bound is rather rough. For exam-
ple if q = 6 we get r(6,k) ≤ 12. But it is known that r(6,2) = 11, r(6,3) = 9
(see [1] or [2]) and that r(6,4) = 8 (see [1] or [3]). So, by Lemma 1, we can
easily do better for most values of k. Moreover, Theorem 1 gives r(6,k) ≥ 8
for all k, hence r(6,k) = 8 if k is divisible by 4. It seems more likely that
r(q,k) = q + 2 if q is not the order of an aﬃne plane, provided that k is big
enough (and maybe k ≥ q suﬃces).
Acknowledgement. The author wants to thank Jan van Haastrecht for the
posed problem and for the invitation to an excellent dinner.
References
[1] D.R. Stinson, Coverings, Chapter IV-8 in The CRC Handbook of Com-
binatorial Designs (C.J. Colbourn and J.H. Dinitz eds.), CRC Press,
1996, pp. 260-265.
[2] E.R. Lamken and W.H. Mills, Resolvable coverings, Congr. Numer. 96
(1993), 21-26.
[3] E.R, Lamken, W.H. Mills and R. Rees, Resolvable minimum covers with
quadruples, preprint.
Willem H. Haemers,
Tilburg University,
Dept. of Econometrics,
P.O. Box 90153,
5000 LE Tilburg,
The Netherlands.
e-mail: haemers@kub.nl
5

Minimum resolvable coverings with small parallel classes.

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