Lecture 14 - 2025 / 5 / 27

Communication Complexity

Definition (Communication Complexity): $f(x, y)$, $x, y\in \{0, 1\}^*$, $f(\cdot) \in \{0, 1\}$. Alice has $x$ and Bob has $y$. The goal is to compute $f(x, y)$. Communication complexity is the number of bits Alice and Bob have to communicate in worst-case.

$$CC(f) \ge \log_2 \chi(f)$$

where $\chi(f) = \min_{\rm partition}R$, and $R := \#\rm chronomatic\ rectangles$.

Remark: $\log_2\chi(f)$ represents that Alice and Bob should both know the results, wich may actually differs 1 bit from the answer. But this is meaningless comparing to $n$.

Example: $EQ(x, y) = [x = y]$
$$\begin{matrix} \boxed{1} & 0 & 0 & 0\\ 0 & \boxed{1} & 0 & 0 \\ \hline 0 & 0 & \boxed{1} & 0 \\ 0 & 0 & 0 & \boxed{1} \end{matrix} \qquad \text{``Send 1 bit''}$$

Boxed elements can't be in the same chronomatic rectangles. $\chi(EQ) \ge 2^n$, so $CC(f) \ge n$

Example: $\text{NAND} (x, y) = \neg (x\land y)$
$$\begin{matrix} 1 & 1 & 1 & \boxed{1} \\ 1 & 0 & \boxed{1} & 0 \\ 1 & \boxed{1} & 0 & 0 \\ \boxed{1} & 0 & 0 & 0 \end{matrix}$$

Boxed elements can't be in the same chronomatic rectangles. $\chi(\text{NAND}) \ge 2^n$, so $CC(f) \ge n$

For $f(x, y)$, $x, y \in \{0, 1\} ^n$, let $M(f)_{2^n\times 2^n}$ be the matrix. What's the relation of $\text{rank}(M(f))$ and $\chi(f)$?

Recall that $\chi(f)$ just give a partition, decompositing $M(f)$ into some rank $1$ matrices. So $\text{rank}(M(f)) \le \chi(f)$ using $\text{rank}(A + B) \le \text{rank}(A) + \text{rank}(B)$.