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Let $\alpha>0$ be a constant (can be sufficiently small if necessary) and $n$ be sufficiently large. What can we say about the cardinality of a family of subsets of $\{1,2,\ldots,n\}$, each of size $k$, such that any two of the subsets intersect in at most $\alpha k$ elements, when $k$ is of size linear in $n$?

I was able to find relevant results only when $k = O(\sqrt{n})$ (see Find a family $\mathcal{A} \subseteq {[n] \choose k}$ such that $|\mathcal{A}| \geq (n/2k)^h$ and $|A_i \cap A_j| \leq h-1$) and when $k$ is fixed ("On a Packing and Covering Problem", Rodl, 1985).

Any idea if results for other ranges of $k$ are available? I am more interested in constructions rather than upper bounds, since the context in Rodl's work anyway gives $\binom{n}{\alpha k}/\binom{k}{\alpha k}$ as a reasonable upper bound.

Any help appreciated!

Martin Sleziak
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DesmondMiles
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  • Does theorem 3 from https://www.khoury.northeastern.edu/home/viola/classes/gems-08/lectures/le3.pdf help? – Artur Riazanov Nov 03 '23 at 11:56
  • @ArturRiazanov not really, since $k = O(\log n)$ (this is $h$ in the notation of the theorem in the paper), while I need something like $k \sim cn$ for some constant $c$. – DesmondMiles Nov 03 '23 at 12:02
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    There is likely to be a big change in behaviour around $\alpha = k/n$, since random sets of size $k$ have expected intersection size $k^2/n$. Can you clarify whether $\alpha$ is bigger or smaller than $k/n$? – Sean Eberhard Nov 03 '23 at 12:33
  • Hmm, maybe $\alpha < k/n$ would fit better, but address any of the cases you wish. – DesmondMiles Nov 03 '23 at 12:36
  • A naive idea is to use the greedy choice to construct the required family. Can you estimate how small be a family constructed this way with the respect to the upper bound $\binom{n}{\alpha k}/\binom{k}{\alpha k}$? – Alex Ravsky Nov 05 '23 at 14:15
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    @AlexRavsky with a greedy approach the denominator would rather be a product of two binomial coefficients and not one (since you both have to choose which $\alpha k$ elements to overlap with when considering a bad set and also have to choose what happens with the other $k-\alpha k$ elements) – DesmondMiles Nov 05 '23 at 14:24
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    @DesmondMiles Do you know the standard Cauchy-Schwarz argument in this setting? – mathworker21 Nov 05 '23 at 22:23
  • @mathworker21 perhaps you mean Theorem 3.1 in https://arxiv.org/abs/2103.15850 ? I have seen it here and there indeed, but it does not give a lower bound for the number of sets, as it is not a constructive/probabilistic approach?? – DesmondMiles Nov 05 '23 at 22:59

1 Answers1

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Maybe the following simple construction can be useful for some choice of parameters, although I afraid it anyway provides too weak lower bound. Let $q$ be any natural power of any prime number, $\alpha=\frac 1{q}$, $\mathbb F_q$ be a field of order $q$, $s<r$ be any natural numbers, $n=|\mathbb F_q^r|=q^r$, $k=q^s$, and $\mathcal A$ be the family of all $s$-dimensional subspaces of $\mathbb F_q^r$. Then $$|\mathcal A|=\binom r s_q=\dfrac{(q^r-1)(q^r-q)\cdots(q^r-q^{s-1})}{(q^s-1)(q^s-q)\cdots(q^s-q^{s-1})},$$ see this answer, and the intersection of any distinct members of $\mathcal A$ has size at most $q^{s-1}=\alpha k$. In particular, when $s=r-1$ then $|\mathcal A|=\frac{q^r-1}{q-1}=\frac{n-1}{\alpha^{-1}-1}$.

Alex Ravsky
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