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The following so-called "Eventown" problem is a classic chestnut:

The $n$ people of Eventown like to form clubs, subject to the following requirements:
(1) Each club has an even number of members.
(2) Each pair of clubs has even intersection.
(3) No two clubs have exactly the same membership.
What is the maximum number of clubs eventown can possibly have?

You can construct $2^{\lfloor n/2\rfloor}$ clubs by thinking of the town as consisting of $\lfloor n/2 \rfloor$ pairs of people (possibly with one person left out). Each club then consists of some subset of the pairs of people.

As it turns out, this is optimal, and there's a slick proof using linear algebra: To each club we associate a vector in $\mathbb{Z}_2^n$, where the vector is $1$ in position $i$ if person $i$ is in the club. The first two conditions then correspond to the statement that any two club vectors are orthogonal. So if $W$ is the subspace spanned by the club vectors, $W$ is contained in its orthogonal complement, so $\dim(W) \leq \lfloor n/2 \rfloor$, which then gives an upper bound on the number of vectors in $W$.

I presented this problem (along with the corresponding "Oddtown" problem) in our Math Club this week. In the discussion afterwards, someone had the following question: What happens if we replace "even" by "divisible by $3$" everywhere?

If you work over $\mathbb{Z}_3^n$ instead of $\mathbb{Z}_2^n$ (making each club vector $1$ in position $i$ if person $i$ is in the club, $0$ otherwise), the linear algebra still goes through, and an $n/2$ dimensional space still contains at most $2^{n/2}$ vectors from $\{0,1\}^n$, so you have the same upper bound as before.

On the other hand, the analogue of the lower bound construction (put people in groups of $3$, with each club consisting of some number of groups) now gives $2^{\lfloor n/3 \rfloor}$. So there's now a gap between what the two arguments give.

Which (if either) of the bounds is tight here? An equivalent way of stating it:

What is the maximal size of a collection $A_1, A_2, \dots, A_k$ of subsets of $\{1, 2, \dots, n\}$ such that both $|A_i|$ and $|A_i \cap A_j|$ are always divisible by $3$?

In this language it feels like this has to be a well-known problem in extremal set theory.

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As it turns out, this problem is open, and actually has been for some time.

Frankl and Odlyzko considered the more general question of how many subsets you can choose so that the intersection sizes are always divisible by some $\ell \geq 2$. Assuming the existence of a Hadamard matrix of order $4 \ell$ (which is known to exist for $\ell \leq 166$), they constructed a collection of $(8 \ell)^{\lfloor n/4 \ell \rfloor}$ sets having all intersections divisible by $\ell$.

In terms of my original question, this corresponds to giving a lower bound that is roughly $$24^{n/12} \approx (1.30)^n \approx 2^{0.38 n}$$ So in particular, the trivial "trios" construction is not optimal.

They also show that for $\ell \geq 5$, the $2^{n/2}$ bound from the linear algebra method is not optimal. However, I still don't know of any better boudn for the case $\ell=3$.