Sets with Small Intersection

2 minute read


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

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

The goal is to construct as large a family of such subsets as possible (i.e., to make $m$ 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 $Y$ depends on at most $4nm$ other $Y$ and $2$ other $Z$, and each $Z$ depends on at most $n^2m/2$ of the $Y$. 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 $k$, 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.



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