Sets with Small Intersection

Suppose that we want to construct subsets $S_1, \ldots, S_m \subseteq \{1,\ldots,n\}$ with the following properties:

$|S_i| \geq k$ for all
$|S_i \cap S_j| \leq 1$ for all $i \neq j$

The goal is to construct as large a family of such subsets as possible (i.e., to make as large as possible). If $k \geq 2\sqrt{n}$ , then up to constants it is not hard to show that the optimal number of sets is $\frac{n}{k}$ (that is, the trivial construction with all sets disjoint is essentially the best we can do).

Here I am interested in the case when $k \ll \sqrt{n}$ . In this case I claim that we can substantially outperform the trivial construction: we can take $m = \Omega(n^2 / k^3)$ . The proof is a very nice application of the asymmetric Lovasz Local Lemma. (Readers can refresh their memory here on what the asymmetric LLL says.)

Proof. We will take a randomized construction. For $i \in \{1,\ldots,n\}$ , $j \in \{1,\ldots,m\}$ , let $X_{i,a}$ be the event that $i \in S_a$ . We will take the $X_{i,a}$ to be independent each with probability $\frac{2k}{n}$ . Also define the events

$Y_{i,j,a,b} = I[i \in S_a \wedge j \in S_a \wedge i \in S_b \wedge j \in S_b]$

$Z_{a} = I[|S_a| < k]$

It suffices to show that with non-zero probability, all of the $Y_{i,j,a,b}$ and $Z_{a}$ are false. Note that each $Y_{i,j,a,b}$ depends on $Y_{i',j,a',b}, Y_{i',j,a,b'}, Y_{i,j',a',b}, Y_{i,j',a,b'}, Z_a, Z_b$ , and each Z_a depends on $Y_{i,j,a,b}$ . Thus each depends on at most 4nm other and other , and each depends on at most n^2m/2 of the . Also note that $P(Y_{i,j,a,b}) = (k/n)^4$ and $P(Z_a) \leq \exp(-k/4)$ (by the Chernoff bound). It thus suffices to find constants y, z such that

$(k/n)^4 \leq y(1-y)^{4nm}(1-z)^2$

$\exp(-k/4) \leq z(1-y)^{n^2m/2}$

We will guess $y = \frac{k}{n^2m}$ , $z = \frac{1}{2}$ , in which case the bottom inequality is approximately $\exp(-k/4) \leq \frac{1}{2}\exp(-k/2)$ (which is satisfied for large enough , and the top inequality is approximately $\frac{k^4}{n^4} \leq \frac{k}{4n^2m} \exp(-4k/n)$ , which is satisfied for $m \leq \frac{n^2}{4ek^3}$ (assuming $k \leq n/4$ ). Hence in particular we can indeed take $m = \Omega(n^2/k^3)$ , as claimed.