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Adapted from “The story of set disjointness” by Arkadev Chattopadhyay and Toni Pitassi.
An $\Omega(\sqrt{n})$ lower bound on the randomized communication complexity of set disjointness.
Theorem
Let $f: 2^{[n]} \times 2^{[n]} \to \zo$ be the set disjointness function: that is, given any sets $A,B \subseteq [n]$, $f$ returns true if $A$ and $B$ are disjoint: $f(A,B) \coloneqq \One[A \cap B = \emptyset]$. Then the randomized communication complexity of $f$ is $\Omega(\sqrt{n})$.
More precisely, there is some constant probability of failure $\eps$ such that any randomized communication protocol that succeeds with probability $\geq 1-\eps$ on each input must use $\Omega(\sqrt{n})$ bits of communication.
Proof
Plan
By Yao’s minimax principle, it is enough to show a distribution over inputs on which any deterministic protocol needs $\Omega(\sqrt{n})$ communication in order to succed with probability $1-\eps$ on average. Here, we will consider the (product) distribution where both $A$ and $B$ are chosen uniformly at random in $\binom{[n]}{\sqrt{n}}$.
We’ll show that the corresponding $\binom{n}{\sqrt{n}} \times \binom{n}{\sqrt{n}}$ communication matrix has no large “nearly $0$-monochromatic” rectangles. That is, for any subsets $\CF, \CG \subseteq \binom{[n]}{\sqrt{n}}$, either
- $\CF$ or $\CG$ only represents a $2^{-\Omega(\sqrt{n})}$ fraction of $\binom{[n]}{\sqrt{n}}$;
- or a $\Omega(1)$ fraction of the pairs $(A,B) \in \CF \times \CG$ intersect.
Intuitively, we want to show that if $\CF$ and $\CG$ are large, then they must both be pretty equally spread over $[n]$, and if that’s the case, they must intersect a lot. However, there doesn’t seem to be a clean notion of “equally spread” that you can prove independently for $\CF$ and $\CG$, so instead we’ll:
- assume that and $\CF$ is large, and $\CF$ and $\CG$ don’t intersect much;
- find a small number of sets $A$ in $\CF$ that cover much of $[n]$ and yet don’t intersect $\CG$ too much;
- show that this limits the size of $\CG$.
Execution
Suppose that $|\CF| \geq 2^{-o(\sqrt{n})} \binom{n}{\sqrt{n}}$, and that only an $\epsilon$ fraction of the pairs $(A,B) \in \CF \times \CG$ intersect. Some sets $A \in \CF$ might be unusual in that they intersect with much more than an $\eps$ fraction of the sets in $\CG$, but at least half of them intersect with at most a $2\eps$ fraction of $\CG$. Let $\CF' \subseteq \CF$ be the family of such good sets; we have $|\CF'| \geq |\CF|/2 \geq 2^{-o(\sqrt{n})} \binom{n}{\sqrt{n}}$.
Then, find $k \leq O(\sqrt{n})$ sets $A_1, \ldots, A_k \in \CF'$ such that $|A_1 \cup \cdots \cup A_k| \geq n/3$. This is always possible given how big $\CF'$ is: Just choose those sets greedily to increase the size of the union as quickly as possible. If at some point $|A_1 \cup \cdots \cup A_i| < n/3$ and you can’t find some $A_{i+1} \in \CF'$ that extends the size of the union by at least $\sqrt{n}/2$, that would show that there are at most
\[\sum_{j=0}^{\sqrt{n}/2-1} \underbrace{\binom{n/3}{\sqrt{n}-j}}_\text{inside $A_1 \cup \cdots \cup A_i$}\quad\underbrace{\binom{2n/3}{j}}_\text{outside $A_1 \cup \cdots \cup A_i$} \leq O\left(\binom{n/3}{\sqrt{n}/2}\binom{2n/3}{\sqrt{n}/2}\right) \leq 2^{-\Omega(\sqrt{n})} \binom{n}{\sqrt{n}}\]
elements left in $\CF'$.
Now (using the same trick as earlier), the average set in $\CG$ intersects only a $2\eps$ fraction of the sets $A_1, \ldots, A_k$, so at least half of them intersect at most $4k\eps$ of them. Let $\CG' \subseteq \CG$ be the family of such good sets; we have $|\CG'| \geq |\CG|/2$.
Each set $B \in \CG$ can only intersect the union $A_1 \cup \cdots \cup A_k$ at $\leq 4k\eps$ points, and the rest of $B$ must lie entirely in the remaining $n-\Omega(n)$ elements of $[n]$. Therefore,
\[|\CG'| \leq \binom{k}{\leq 4k\eps} \binom{n - \Omega(n)}{\sqrt{n}}.\]
We can upper bound $\binom{n - \Omega(n)}{\sqrt{n}}$ by $2^{-\Omega(\sqrt{n})} \binom{n}{\sqrt{n}}$, and by choosing $\eps$ small enough, we can upper bound $\binom{k}{\leq 4k\eps}$ by $2^{c\sqrt{n}}$ for any $c>0$ we want, so we get
\[|\CG| \leq 2|\CG'| \leq 2^{-\Omega{(\sqrt{n})}} \binom{n}{\sqrt{n}},\]
as desired.