A coding problem for pairs of subsets

A coding problem for pairs of subsets Gyula O.H. Katona

Renyi Institute, Hungarian Academy of Sciences Budapest, Re altanoda u. 13{15, 1053 Hungary

 

X k

Let X be an n-element set,

denotes the family of all its k-element subsets.

 

The following problem was raised in a paper of Demetrovics, Sali and the

X k

author by motivations from database theory. Can we decompose

 

(unordered) pairs of disjoint pairs (with one exception if

n k

into

is odd) in such

a way that these pairs are not too close to each other in a certain sense? The answer was given in the following way.

Theorem j  k 1 Suppose that 1  k is an integer and n  n(k). n unordered pairs (A ; B ) of disjoint k-element subsets are 21 k i i ;; jAij = jBij = k) of an n-element set X such that

fj i \ j j j i \ j jg 

min

which implies

A

A

; B

B

fj i \ j j j i \ j jg 

min

A

B

; B

A

k 2

k 2

Then there (A i

\

B

6

(i = j )

i

=

(1)

6

(i = j )

(2)

by the unorderedness. The proof uses the following generalization of Dirac's theorem on the existence of Hamiltonian cycles.

Theorem 2 vertex set G0

and let

Let

j j V s

G0 = (V; E0 )

= N,

such that

and E0

\

E1 =

be the maximum degree of 2r

be simple graphs on the same . Let r be the minimum degree of

G1 = (V; E1 )

8s

2

;

G1 .

s

holds, then there is a Hamiltonian cycle in

Suppose, that

1 > N

G0

such that if

(a; b)

and

(c; d)

are

two vertex-disjoint edges of the cycle, then they do not form an alternating cycle with two edges of G1 .

Email address: [email protected] (Gyula Preprint submitted to Elsevier Science 233

O.H. Katona).

30 October 2003

It was noticed by Enomoto and the author that this is really a coding type problem. De ne the distance between two disjoint pairs fAi ; Bi g of k-element sets (that is, jAi j = jBi j = k; Ai \ Bi = ;(i = 1; 2)) in the following way. d

(fA1 ; B1 g; fA2 ; B2 g) = minfjA1

A2

j + j 1 2j j 1 2 j + j 1 2 jg (3) 2 g)  2 with equality i the four sets B

B

; A

B

B

A

It is easy to see that d(fA1 ; B1 g; fA2 ; B k are pairwise disjoint. It has been veri ed that (3) is really a distance, that is, it satis es the triangle inequality. We say that a set C of such pairs of disjoint subsets is a 2 (n; k; d){code if the distance of any two elements is at least d. Let C (n; k; d) be the maximum size of a 2 (n; k; d)-code. In fact, the following theorem was proved. Theorem j  k 3

are

1 2

n

unordered

; j j=j j= ; Ai

k

1  k is an integer and n  n(k). pairs (Ai ; Bi ) of disjoint k -element subsets

Suppose that

Bi

k)

of an

n-element

set

X

which implies

Aj

Bi

Bj

j \ j+j \ j Ai

Bj

(A i

\

Bi

=

such that

j \ j+j \ j Ai

Then there

Bi

Aj

k

(4)

k

(5)

by the unorderedness.

Observe, that conditions (4) and (5) are equivalent to the condition that the distance between the pairs is at least k. On the other hand they imply (1) and (2). Hence Theorem 3 implies Theorem 1. There is, however, one more condition here namely, a k-element subset may occur in at most one pair. Let C (n; k; d) denote the maximum size of such a 2 (n; k; d)-code that no two \codewords" (that is pairs of disjoint k-element sets) use the same k-element set. 0

Theorem 2 actually determines $

1 C (n; k; k ) = 2 0

n k

!% :

(6)

It is easy to see that C (n; k; k)  C (n; k; d) if d  k. Therefore (6) implies 0

0

$

1 C (n; k; d) = 2 0

n k

!%

(1  d  k):

Suppose that two pairs use the same k-element set: fA; B1 g; fA; B2 g. By the de nition of the distance we have d(fA; B1 g; fA; B2 g)  k. Therefore C

0

(n; k; d) = C (n; k; d)

234

holds for k < d. Hence C (n; k; d) is either determined or reduced to the determination of C (n; k; d). 0

The following upper estimate was proved by Brightwell and the author. Theorem 4

Let

d

 2k  n

C (n; k; d)

be integers. Then

 21 k(k

1)


(n 2k + d) d+1 1)


d d+1 2 e
k(k 1)


b 2 c

n (n

holds.

Bollobas, Leader and the author recently proved that the upper estimate is asymptotically sharp. Theorem 5

lim n!1

C (n; k; d)

1 = n2k d+1 2 k (k

1)


d 2

d+1

1

e
k (k

1)


b d+1 2 c

:

The proof uses a slight generalization of the Frankl-Rodl method worked out for achieving asymptotic packings and coverings. We strongly believe that for any given k and d there are in nitely many n with equality in the upper bound. A method towards this aim is exhibited which reduces the determination of C (n; k; 2k 1) to nding certain \dierence sets". The distance Æ (a; b) of two integers mod Æ (a; b)

= minfjb

m

(1  a; b  m) is de ned by

jj

a; b

jg

a+m :

(Imagine that the integers 1; 2; : : : ; m are listed around the circle clockwise uniformly. Then Æ (a; b) is the smaller distance around the circle from a to b.) Æ (a; b)  m 2 is trivial. Observe that b a  d c mod m implies Æ (a; b) = Æ (c; d). We say that the pair A = fa1 ; : : : ; ak g; B = disjoint sets is antagonistic mod m if (i) all the k(k

fb1; : : : ; bk g  f1; : : : ; mg of

1) integers Æ (ai ; aj )(i 6= j ) and Æ (bi ; bj )(i 6= j ) are dierent,

(ii) the k2 integers Æ (ai ; bj )(1  i; j  k) are all dierent and (iii) Æ (ai ; bj ) 6= m2 (1  i; j  k): If there is a pair of disjoint antagonistic k-element subsets mod 1  m must hold by (ii) and (iii).

235

m

then 2k2 +

Theorem 6

joint, antagonistic pair of k -element subsets

(

C n; k;

2k

1) =

S(

2k2 + 1; 2) mod 2k2 + 1 then

Suppose that there is Steiner family

(

1)

n n

2

2 k

n;

and a dis-

:

Unfortunately we found disjoint antagonistic pairs only for k = 1; 2; 3. Hence we have proved that the upper estimate is sharp for in nitely many n's for C (n; 2; 3) and C (n; 3; 5). The concept of antagonistic pairs is closely related to the concept of dierence sets. R.C. Bose had an essential role in developing the theory of dierence sets (see a paper of R.C. Bose and A. Chowla). Finally, Brightwell and the author considered the case of C (2k; k; d) for xed d and large k . The method of the construction was developed by Graham and Sloan. Let us show our result for the case d = 3. Theorem 7

1 2k 2(2k + 1) k

!



C

(2k; k; 3):

Note that the upper bound of Theorem 4 gives C

(2k; k; 3) 

236

1 2k 2 k

! :