$\require{mathtools}
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% Greek
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\let\fi\phi % because it looks like an f
\let\phi\varphi % because it looks like a p
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% .. over and under
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% .. relations
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% .. miscer
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% (I needed to create my own because the MathJax version of \DeclarePairedDelimiter doesn't have \mathopen{} and that messes up the spacing)
% .. one-part
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% .... (use phantom to force at least the standard height of double bars)
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% ..... (use phantom to force at least the standard height of double bar)
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% .. dotted versions (is it equal in the limit?)
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% .. operators
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% .. analysis
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% Complexity classes
% .. classical
\newcommand{\Poly}{\mathsf{P}}
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\renewcommand{\L}{\mathsf{L}}
% .. probabilistic
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% .. circuits
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Adapted from scribe notes by Gabriel Cadamuro for Mark Braverman’s Fall 2011 “COS 597D: Information theory in computer science” class at Princeton.
Consider a function $f : X \times Y \to \zo$ (we’ll mostly think of $X = Y = \zo^n$). Suppose Alice has an input $x \in X$ and Bobo has an input $y \in Y$. How many bits do Alice and Bob need to communicate to each other in order to compute $f(x,y)$? Denote this as $D(f)$.
We can view a protocol as a binary tree, where each node is a function of only $x$ or only $y$, and the protocol takes the left or right branch based on the value of that function. Then $D(f)$ is the smallest depth of any such tree.
Each leaf in this tree corresponds to a (disjoint) monochromatic rectangle of $X \times Y$: a product set $S \times T \subseteq X \times Y$ such that:
- either $f(x,y) = 0$ for all $(x,y) \in S \times T$,
- or $f(x,y) = 1$ for all $(x,y) \in S \times T$.
Equivalently, we can think of the matrix $M_f$ such that $(M_f)_{x,y} \coloneqq f(x,y)$, and think of $0$ as “white” and $1$ as “black”. Then, monochromatic rectangles are submatrices (restrictions to certain rows and columns) where each cell has the same color.
If we can show that any partition of $M_f$ into monochromatic rectangles takes $\geq 2^t$ rectangles, then $D(f) \geq t$.
Fooling sets
One easy way to show that many monochromatic rectangles are needed is to exhibit a “fooling set”: a set $F \sse X \times Y$ such that no two cells in $F$ can be in the same rectangle.
Equality
Let $f(x,y) \coloneqq \One[x = y]$. Then $M_f$ is an identity matrix of dimension $2^n$.
Clearly, $M_f$ cannot be partitioned into fewer than $2^n$ monochromatic rectangles. In fact, there must at least be $2^n$ black rectangles. Indeed, we know that $(x,x)$ is black for all $x$, and on the other hand we can’t put two distinct black points $(x,x)$ and $(y,y)$ in the same rectangle: if they were, then $(x,y)$ would also be included, but $(x,y)$ is white.
Therefore, $D(f) \geq n$. In fact, we can improve this to $D(f) \geq n+1$ by observing that there must also be at least one white rectangle (if $n>1$, at least).
Hamming weight comparison
Let $f(x,y) \coloneqq \One[|x| \geq |y|]$ (where $|x|$ denotes the Hamming weight of $x$). Then if you group together inputs with the same Hamming weight, $M_f$ looks like a triangular matrix.
We’ll show that a partition of $M_f$ into monochromatic rectangles must have at least $n+1$ rectangles. First, define the inputs $x^{(i)} \ce 0^{n-i}1^i$ for $i=0,\ldots, n$. Clearly $|x^{(i)}| = i$. Now, we claim that for any $i<j$, $(x^{(i)},x^{(i)})$ and $(x^{(j)},x^{(j)})$ cannot be in the same rectangle. Indeed, they’re both black, whereas $(x^{(i)}, x^{(j)})$ is white, since $|x^{(i)}| < |x^{(j)}|$.
Therefore, $D(f) \geq \log(n+1)$.
Observe that unlike in the equality example, here we couldn’t have used the “other corner” $(x^{(j)},x^{(i)})$.
Set disjointness
This time, we’ll think of the inputs $x$ and $y$ instead as sets $A,B \subseteq [n]$. We define the set disjointness function to be true if $A$ and $B$ intersect: $f(A,B) \coloneqq \One[A \cap B \neq \emptyset]$.
We’ll show that there must be at least $2^n$ white rectangles, but we need to be a bit more clever than in the previous two cases. Concretely, we’ll consider white points of the form $(A, \comp{A})$ for some $A \sse [n]$: all pairs of sets that are “barely” disjoint. We claim that no two distinct points $(A, \comp{A})$ and $(B, \comp{B})$ can be in a rectangle together. To prove this, it suffices to show that either $A$ intersects $\comp{B}$, or $B$ intersects $\comp{A}$, which we do by picture: every element of $[n]$ must fit in this picture somewhere, so if the two highlighted regions below were both empty, then $A$ would be equal to $B$.
Thus $D(f) \geq n$.
An algebraic proof: inner product
Let $f(x,y) \coloneqq \One[x \cdot y \equiv 1 \pmod{2}]$.
This time, we’ll use a different technique: we’ll show that no white rectangle $S \times T$ can contain more than $2^n$ total cells, i.e. $|S \times T| \leq 2^n$. Since there are $2^{2n-1}$ white cells in total, this shows that at least $2^{n-1}$ rectangles are needed, and thus $D(f) \geq n-1$.
Take some white rectangle $S \times T$. Being white means that $S$ and $T$ are orthogonal in $\F_2^n$. I claim that we can always extend $S$ and $T$ to their spans. For example, for any $x,x' \in S$ and $y,y' \in T$, their sums are also orthogonal:
\[(x+x')\cdot(y+y') = x\cdot y+x\cdot y'+x'\cdot y+x'\cdot y' \equiv 0 \pmod{2}.\]
Therefore, we might as well assume that $S$ and $T$ are orthogonal subspaces of $\F_2^n$. But then $\dim(S) + \dim(T) \leq \dim(\F_2^n) = n$, and
\[
|S \times T| = |S| \cdot |T| = 2^{\dim(S)} \cdot 2^{\dim(T)} \leq 2^n.
\]
The rank method
In the equality and Hamming weight examples, the number of monochromatic rectangles needed matched the rank of the matrix exactly. Can we hope a general connection?
Yes, at least in one direction! If the black part of $M_f$ can be partitioned into $m$ rectangles, then it can also be obtained as the sum $M_f = R_1 + \cdots R_m$ of the “characteristic matrices” of those rectangles ($R_i$ is $1$ in the $i\th$ rectangle and $0$ elsewhere).
For any two matrices $A$ and $B$, $\rank(A+B) \leq \rank(A) + \rank(B)$, and the matrices $R_i$ all clearly have rank $1$, so $\rank(M_f) \leq m$. This shows that $D(f) \geq \log\rank(M_f)$.
The log-rank conjecture
An approximate converse might exist: the log-rank conjecture posits that $D(f) \leq \poly(\log\rank(M_f))$.
- The best known upper bound is $D(f) \leq \tilde{O}\p{\sqrt{\rank(M_f)}}$.
- There is some $f$ for which $D(f) \geq \tilde\Omega\p{(\log\rank(M_f))^2}$.