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\lecture{6}{February 23, 2005}{Madhu Sudan}{Mihai P\v{a}tra\c{s}cu}
\section{Communication Complexity}
Communication complexity is interesting from a complexity-theoretic
perspective because it offers ways to prove lower bounds in various
models of computation. For example, there are several known relations
to circuits, one of which we discuss in this lecture. This connection
would enable use to prove $\omega(\lg n)$ bounds on circuit depth if
we could prove a good lower bound on a certain comunication game.
Communication complexity was introduced by Yao \cite{yao}, and has
been studied intensively since then. Good references are the book by
Kushilevitz and Nisan \cite{kn} and a series of lecture notes by Ran
Raz \cite{raz}.
The most simple setup is a two-player communication game. We have two
players, Alice and Bob, who get inputs $x$ and $y$, respectively,
chosen from $\{0,1\}^n$. The two players want to compute a function
$f(x,y)$ by communicating a minimum number of bits. For now, let us
restrict the range of $f$ to be boolean. Observe the difference from
information theory: while information theory concentrates on sending a
message (the data), here we are not interested in the actual data,
just a function of the players' inputs. This allows the communication
complexity to be lower than sending the entire inputs in many cases.
Formally, the solution to a communication game is a protocol. The
protocol is a tuple of functions $(s, m_a, m_b, f_a, f_b)$:
\begin{itemize}
\item $f$ takes as argument the communication history and returns which
player speaks next, or that the protocol is over.
\item $m_a$ takes as arguments the communication history and Alice's
input, and decides the next bit sent by Alice; $m_b$ is similar for
Bob.
\item $f_a$ takes as arguments the communication history and Alice's
input, and decides the result of the protocol; $f_b$ is similar for
Bob. The protocol is correct if both $f_a$ and $f_b$ are equal to
$f(x,y)$ at the end of the protocol.
\end{itemize}
The complexity of the protocol is the maximum number of bits exchanged
over any pair of inputs $(x,y)$. The communication complexity of the
problem, $CC(f)$ is the minimum over all protocols for $f$ of the
complexity of the protocol.
The are many variations on this model, which are useful in different
setting. For example, one can consider: randomization (with zero or
bounded error), nondeterminism, message complexity (bound the number
of alternations between Alice speaking and Bob speaking), multiplayer
games (the most interesting variation being the number-on-the-forehead
model, where player $i$ gets all $x_j$'s except $x_i$) etc.
\section{Techniques for Communication Lower Bounds}
In the model which we defined, it is clear that $CC(f) \le n+1$ for
any $f$, because Alice can just send her input, and Bob can reply with
$f(x,y)$. Note that the model is nonuniform, and does not care about
computational power, so Bob can compute any function once he knows
enough information. Lower bounds of $n+1$ can be shown easily (see
below), even for very simple explicit functions. The interesting
questions are showing bounds for some specific problems, such as the
one we discuss in the next section, which relates to circuit
complexity.
Two of the most important techniques for communication lower bounds
originate in Yao's seminal paper \cite{yao}. The first one is the
fooling-set method. Consider a matrix $A^f$ of size $2^n \times 2^n$
where $A^f_{x,y} = f(x,y)$. The crucial observation is that
communication divides this matrix into disjoint \emph{rectangles}. A
rectangle is a set $X \times Y$, with $X \subset \{0,1\}^n, Y \subset
\{0,1\}^n$ (since $X,Y$ are arbitrary, we care about combinatorial
rectangles, not geometric rectangles). The claim is that for any
communication history, the set of inputs $(x,y)$ for which this
history materializes is a rectangle. Indeed, assume $(x_1, y_1)$ and
$(x_2, y_2)$ generate the same communication history. Now $(x_1, y_2)$
and $(x_2, y_1)$ must still generate the same history. This can be
seen by induction of the number of bits sent. Say Alice speaks next;
she decides her bit based on her input, and the communication history.
Both for $y_1$ and $y_2$ as Bob's input, the communication history so
far is the same (by the induction hypothesis), so she outputs the same
bit.
A correct protocol must announce $f(x,y)$ at the end. It follows that
at the end of the protocol, every communication history leads to a
monochromatic rectangle. Assume the rectangle contained both zeros and
ones. Then, there is either a row or a column containing both zeros
and ones. Say it's a row; then, if Alice gets the input corresponding
to that row, at the end of the communication she cannot tell whether
$f(x,y)$ is zero or one, because she cannot differentiate between
Bob's inputs leading to the rectangle.
Now consider the equality function. The matrix $A^f$ has zeros
everywhere except the diagonal. It is immediate that the number of
one-rectangles is $2^n$, because two ones cannot appear in the same
rectangle. Similarly, it is seen that there are at least $2^n$
zero-rectangles. Then, the communication complexity is $\ge n+1$,
because each new bit doubles the number of rectangles.
A second lower-bound technique takes an algebraic view of $A^f$. It
can be shown, using a similar argumentation as above, that $CC(f) \ge
\lg \rank(A^f)$. It is known that there exist $f$ for which $CC(f) =
\omega(\lg \rank(A^f))$. However, it is still an open question whether
$CC(f) = O(\poly(\lg \rank(A^f)))$.
\section{A Relation to Circuit Depth}
We now investigate an interesting connection between communication
complexity and circuit depth, which was described by Karchmer and
Wigderson \cite{kw}. For this, we need to generalize communication
complexity to consider relations. Consider a relation $R \subset X_A
\times X_B \times D$, where $X_A, X_B \subseteq \{0,1\}^n$ and $D$ is
arbitrary. At the beginning, Alice receives $x \in X_A$, and Bob
receives $y \in X_B$. At the end of the protocol, each player must
decide on a certain $i$ based on his input and the communication
history. It must be the case that both players decide the same $i$,
and that $(x,y,i) \in R$.
For a function $f: \{0,1\}^n \times \{0,1\}^n \to \{0,1\}$, we define
the following cannonical relation. Players receives inputs from $X_A =
f^{-1}(0)$ and $X_B = f^{-1}(1)$; we have $D = \{1,\dots,n\}$. The
players must decide the relation $R_f = \{ (x,y,i) \mid x \in X_A, y
\in X_B, x_i \ne y_i \}$. Clearly, if $f(x) \ne f(y)$, there must be
at least one bit where they differ; this is saying that the
communication problem is for the two players to find that bit.
\begin{theorem}
Let $d(f)$ be the circuit depth of $f$ on a circuit with basis
$(2-AND, 2-OR, NOT)$. Then $CC(R_f) = \Theta(d(f))$.
\end{theorem}
\begin{proof}
We first show $CC(R_f) \le d(f)$. The players know a circuit which
outputs $0$, respectively $1$, when applied to $x$ and $y$. Consider
the last gate of the circuit, and assume, for instance, that it is an
AND. Thus, the output of the circuit is $f(z) = f_0(z) \land
f_1(z)$. If $f(x) = 0$, at least one of $f_0(x), f_1(x)$ is zero;
since $f(x) = 1$, $f_0(y) = f_1(y) = 1$. Now Alice just outputs
$f_0(x)$. If this is zero, the players have a circuit of depth $d(f_0)
\le d(f)-1$ which differentiates $x$ and $y$; otherwise, $f_1$
differentiates between $x$ and $y$ and it also has depth $\le
d(f)-1$. Then, the conclusion follows by induction on the depth. The
base case is that a single bit is found (depth zero) which
differentiates $x$ and $y$.
We now show $CC(R_f) \ge d(f)$. This also works by induction on
$CC(R_f)$. Assume by induction that for any two disjoint sets $X_A,
X_B \subset \{0,1\}^n$ for which there is a $d$-bit protocol
``separating'' $X_A$ and $X_B$, there exists a depth $d$ protocol
implementing a function $g : \{0,1\}^n \to \{0,1\}$ with $g^{-1}(0)
\supseteq X_A, g^{-1}(1) \supseteq X_B$. The base case is $d=0$. Then,
Alice and Bob know a bit which is different for inputs with $f(x)=0$
vs $f(x)=1$. So the circuit can just look at that bit (this is fixed
nonuniformly), and $g$ is the bit or the bit negated, depending on
which way the separation goes.
We now prove the claim for $d+1$. Assume by symmetry that Alice
speaks, and her bit is $P(x)$ where $P : X_A \to \{0,1\}$. This bit
induces a partition of $X_A$ into $P^{-1}(0)$ and $P^{-1}(1)$. If the
first bit is zero, the remaining protocol has complexity $\le d$ and
it separates $P^{-1}(0)$ from $X_B$. If the first bit is one, we have
a protocol that separates $P^{-1}(1)$ from $X_B$. By induction, these
protocols can be converted into circuits $C_0$ and $C_1$. Observe that
if $x \in X_B$, both $C_0(x)$ and $C_1(x)$ output $1$, by
induction. If $x \in X_A$, either $x \in P^{-1}(0)$ or $x \in
P^{-1}(1)$. Then either $C_0(x)$ or $C_1(x)$ outputs zero. Then, if we
take $C = C_0 \land C_1$, it outputs one on $X_B$ and zero on $X_A$.
\end{proof}
\begin{thebibliography}{99}
\bibitem{kn} Eyal Kushilevitz and Noam Nisan: \emph{Communication
Complexity}, Cambridge University Press, 1996.
\bibitem{kw} Mauricio Karchmer, Avi Wigderson: \emph{Monotone Circuits
for Connectivity Require Super-logarithmic Depth}, STOC 1988: 539-550.
\bibitem{raz} Ran Raz: \emph{Circuit Complexity and Communication
Complexity}, lecture notes series from IAS summer school on Complexity
Theory. Park City Mathematical Series, Volume 10, 2000.
\bibitem{yao} Andrew Chi-Chih Yao: \emph{Some Complexity Questions
Related to Distributive Computing (Preliminary Report)}, STOC 1979:
209-213.
\end{thebibliography}
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