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\lecture{7}{Feb 28, 2007}{Madhu Sudan}{Benjamin Rossman}
\paragraph{Today:} Communication Complexity
\begin{itemize}
\item
Definitions
\item
Connection to circuit complexity
\item
Basic results
\end{itemize}
The setting is the following. There are two players Alice and Bob,
separated by physical distance. They want jointly to compute
function $f(x,y)$ of inputs $x$ and $y$. Originally, only Alice
knows $x$ and only Bob knows $y$. Alice and Bob communicate by
sending each other bits, one at a time. Pictorially, we have
something like
\[
x\ \xrightarrow{\ \ \ \ \ }\ \boxed{\textup{Alice}}\
\begin{array}[t]{l}
\xrightarrow{\ \ b_1\ \ }\\
\xrightarrow{\ \ b_2\ \ }\\
\xleftarrow{\ \ b_3\ \ }\\
\xrightarrow{\ \ b_3\ \ }\\
\ \ \ \ \ \vdots\\
\xleftarrow{\ \ b_k\ \ }\\
\end{array}\
\boxed{\textup{Bob}}\ \xleftarrow{\ \ \ \ \ }\ y
\]
This goes on until Alice and Bob each have enough information to
determine the value $f(x,y)$ (possibly without knowing the other
party's complete input). At that point, the interaction ends. We
think of such an interaction as a cooperative ``game'' between Alice
and Bob, where the objective is for both parties to ascertain
$f(x,y)$ in the fewest possible rounds.
We ask the question: how many bits must Alice and Bob exchange so
that each of them can compute $f(x,y)$? For instance, Alice can
transmit $x$ to Bob, then Bob can compute $f(x,y)$ and send the
result back to Alice. Of course, we'd like to get away with
transmitting fewer bits. We don't care how long Alice and Bob spend
on their private computations; the only goal is minimizing the
length of the interaction, i.e., the number of bits passed between
Alice and Bob.
In a slightly more general framework, the combinatorial problem is
given by a relation $R(x,y,z)$. Here $x,y$ are the inputs and $z$ is
an admissible solution: given $x$ and $y$, we want $z$ such that
$(x,y,z) \in R$. Usually, we think of inputs and outputs as
sequences of bits, that is, $R \subseteq \{0,1\}^n \times \{0,1\}^n
\times \{0,1\}^m$. Also, it is common to consider $m \ll n$ ($m$
much smaller than $n$), since $z$ usually contains far less
information than the pair $(x,y)$. (For all inputs $x,y$, we will
assume there is at least one $z$ such that $(x,y,z) \in R$.)
Solutions to this combinatorial problem take the form of {\em
procotols}. At each stage in an interaction (game), a protocol $\pi$
specifies:
\begin{enumerate}
\item
whether the game ends, and
\item
if the game continues who transmits the next bit and what that
bit is.
\end{enumerate}
At the $k$th stage, Alice's action under a protocol $\pi$ must be
purely a function of the input $x$ and the history
$b_1,\dots,b_{k-1}$ of the interaction so far. (By ``Alice's
action'' is meant either: ``terminate with some output $z$'', ``send
bit 0/1 to Bob'', or ``wait for Bob to send the next bit''.)
Similarly, Bob's action under $\pi$ must be a function of $y$ and
$b_1,\dots,b_{k-1}$. For the protocol $\pi$ to be correct, Alice and
Bob should terminate at the same time (with both outputs in the set
$\{z : f(x,y,z) \in R\}$), and only one of them should send a bit in
each round so long as the game continues. This model was introduced
by [Yao $\sim$1980].
\begin{definition}
The {\em communication complexity} of a protocol $\pi$ is defined by
\[
CC(\pi) = \max_{x,y}\{\#\textup{ bits transmitted before $\pi$
terminates on inputs }x,y\}.
\]
\end{definition}
Communication complexity of relations and functions is defined as
follows. For a relation $R$, let
\[
CC(R) = \{CC(\pi) \mid \pi\textup{ is a protocol computing }R\}.
\]
For a function $f : \{0,1\}^n \times \{0,1\}^n \to \{0,1\}$, let
$
CC(f) = CC(R)$ where $R$ is the relation $
\{(x,y,f(x,y))\}.
$ We understand communication complexity far better for functions
than for relations.
We are also interested in communication complexity of partial
functions. A partial function is a function with range $\{0,1,?\}$
where $?$ denotes ``don't care''. This setup corresponds a {\em
promise problems} in which Alice and Bob must jointly compute a
function $f(x,y)$ with the promise that the input pair $(x,y)$ comes
from a specified set (if the promise is violated, then any output is
accepted). Communication complexity of a partial function $f :
\{0,1\}^n \times \{0,1\}^n \to \{0,1,?\}$ is defined by $CC(f) =
CC(R)$ where $R$ is the relation $\{(x,y,0) : f(x,y) \in \{0,?\}\}
\cup \{(x,y,1) : f(x,y) \in \{1,?\}\}$. Communication complexity of
partial functions is also very interesting: often lower bounds for
promise problems are better than for any completion you can choose.
Techniques for promise problems often don't extend to complete
functions.
This is the broad setting for communication complexity. Let's look
at upper and lower bounds. We know that $CC(R) \le n+m$ by a trivial
protocol: Alice sends $x$, Bob computes $z$ such that $R(x,y,z)$ and
sends $z$. We know linear lower bounds for some functions, but few
techniques for proving linear lower bounds in general.
\subsection*{Karchmer-Wigderson games}
Karchmer-Wigderson games involve a nonstandard way of associating a
relation $R_f$ with a function $f$. Consider a function $f :
\{0,1\}^n \to \{0,1\}$. The idea is look at pairs of inputs $x,y$
with the promise that $f(x) = 1$ and $f(y) = 0$. In particular, this
implies $x \ne y$, so there exists a bit where $x$ and $y$ differ
(i.e.\ $x_i \ne y_i$ for some $i \in \{1,\dots,n\}$. This leads us
to define a relation $R_f \subseteq \{0,1\}^n \times \{0,1\}^n
\times \{1,\dots,n\}$ by
\[
R_f = \{(x,y,i) \mid f(x) = 1,\ f(y) = 0,\ x_i \ne y_i\} \cup
\{(x,y,i) \mid f(x) = 0\textup{ or }f(y) = 1\}.
\]
Note that $R_f$ contains all triples $(x,y,i)$ where the promise
that $f(x)=1$ and $f(y)=0$ is violated. The relation $R_f$
correspond to a ``game'' in which Alice and Bob receive inputs $x$
and $y$ such that $f(x)=1$ and $f(y)=0$ and their objective is to
find an index $i$ such that $x_i \ne y_i$.
We now prove a remarkable theorem relating a standard complexity
measure of a function $f$ with the communication complexity of the
relation $R_f$. It is hard to see where the original intuition came
from, however the proof is nice and simple once you see it.
\begin{theorem}
$CC(R_f) = \textup{Depth}(f)$
\end{theorem}
Here $\textup{Depth}(f)$ is the circuit depth of $f$, that is, the
minimal depth of a circuit computing $f$ that has only 2-AND, 2-OR
and NOT gates.\\
\begin{proof}
We first show $CC(R_f) \le \textrm{Depth}(f)$ (easy direction). This
involves converting a circuit $\mathcal C$ computing $f$ into a
protocols. Wlog, $\textrm{Depth}(\mathcal C) = \textrm{Depth}(f)$.
Suppose the top (output) gate in $\mathcal C$ is an OR-gate with
subcircuits $\mathcal C_0$ and $\mathcal C_1$ feeding into it. Let
$f_0,f_1$ be the functions computed by $\mathcal C_0,\mathcal C_1$.
For $i=0,1$, we have $\textrm{Depth}(\mathcal C_i) \le
\textrm{Depth}(\mathcal C)-1$, so that by induction there exists a
protocol $\pi_i$ computing $f_i$ with $CC(\pi_i) \le
\textrm{Depth}(f)-1$. We now derive a protocol $\pi$ for $f$. We may
assume that Alice and Bob have inputs $x,y$ such that $f(x)=1$ and
$f(y)=0$ (since otherwise any output is okay). Clearly
$\max\{f_0(x),f_1(x)\}=1$ while $f_0(y)=f_1(y)=0$. In the first
round of the protocol, Alice can therefore send Bob a bit $b \in
\{0,1\}$ such that $f_b(x) = 1$ and $f_b(y) = 0$. Alice and Bob now
proceed according to protocol $\pi_b$.
The induction step is symmetric when the top (output) gate in
$\mathcal C$ is an AND-gate; this time Bob communicates a bit $b$
such that $f_b(x) = 1$ and $f_b(y) = 0$.
In other direction (just slightly more complicated), we convert
protocols to circuits. This argument requires KM games for partial
functions (in order to have a stronger induction hypothesis). We
will show that $CC(R_f) \ge \textup{Depth}(f)$ for every partial
function $f : \{0,1\}^n \to \{0,1,?\}$. Recall that $(x,y,i) \in
R_f$ iff either $\big[f(x)=1$ and $f(y)=0$ and $x_i \ne y_i\big]$ or
$f(x) \ne 0$ or $f(y) \ne 1$.
Let $\pi$ be a protocol computing $f$ (i.e.\ computing $R_f$), and
consider inputs $x,y \in \{0,1\}^n$ such that $f(x)=1$ and $f(y)=0$.
Suppose Alice sends the first bit under $\pi$. That is, she computes
$b(x) \in \{0,1\}$ and sends it to Bob. We claim there is a circuit
$\mathcal C$ computing $f$ of the form $\mathcal C_0$ OR $\mathcal
C_1$ where $\max\{\textup{depth}(\mathcal
C_0),\textup{depth}(\mathcal C_1)\} \le CC(R_f)-1$. (Had Bob
communicated the first bit, then we would have $\mathcal C_0$ AND
$\mathcal C_1$ instead.)
So what are circuits $\mathcal C_0$ and $\mathcal C_1$? %%%[[picture]]
In order to use the induction hypothesis, we define partial
functions $f_0,f_1 : \{0,1\}^n \to \{0,1,?\}$ by
\[
f_i(x) =
\left\{
\begin{array}{rl}
1&\textup{if }f(x)=1\textup{ and }b(x)=i,\\
0&\textup{if }f(x)=0,\\
?&\textup{otherwise.}
\end{array}\right.
\]
For $i=0,1$, protocol $\pi$ induces a protocol $\pi_i$ for computing
$f_i$: the protocol $\pi_i$ simply dictates what happens after Alice
sends the first bit $b(x)=i$. The longest communication under
$\pi_0$ or $\pi_1$ is precisely 1 less than the longest
communication under $\pi$, that is, $\max\{CC(\pi_0),CC(\pi_1)\} =
CC(\pi)-1$. We now apply the induction hypothesis to obtain circuits
$\mathcal C_i$ of depth $CC(\pi_i)$ computing $f_i$ for $i=0,1$. Let
$\mathcal C$ be the circuit $\mathcal C_0$ OR $\mathcal C_1$. We
claim that $\mathcal C$ computes $f$. This is easily seen from the
picture:\vspace*{-.5in}
\begin{center}
\includegraphics[bb=0 0 316 607, scale=.6, angle=270]{lect07-fig.eps}
\end{center}
The region where either $f_0=1$ or $f_1=1$ contains the region where
$f=1$, while the region where $f=0$ is contained by both regions
$f_0=0$ and $f_1=0$.
In the case where Bob sends the first bit, by a similar argument we
get a circuit $\mathcal C$ with an AND-gate on top.
\end{proof}
\subsubsection*{Application: Circuit Depth of PARITY}
Let's see an application of this theorem. It is known that the
smallest circuit computing PARITY using only 2-AND, 2-OR and NOT
gates has size $\ge n^2$. From this it follows that
$\textup{depth}(\oplus) \ge 2\log_2(n)$. However, we can prove this
directly using Theorem 1. It suffices to show that $CC(R_\oplus) \ge
2\log_2(n)$.
We will argue that, under any protocol computing PARITY, Alice must
send Bob at least $\log_2(n)$ bits and vice-versa, so that at least
$2\log_2(n)$ bits are transmitted in total. Why must Alice send Bob
at least $\log_2(n)$ bits? Let $x$ be uniformly distributed among
inputs in $\{0,1\}^n$ such that $\oplus(x) = 1$, and let $y = x+e_i$
where $i$ is uniform in $\{1,\dots,n\}$ and $e_i =
(0,\dots,0,1,0,\dots,0)$ with $1$ in the $i$th place. Clearly
$\oplus(y) = 0$. By the pigeonhole principle, Bob must receive $\ge
\log n$ bits from Alice in order to determine the unique coordinate
$i$ where $x$ and $y$ different, since for $k < \log_2(n)$ there are
more possibilities for $i \in \{1,\dots,n\}$ than possible sequences
of bits $b_1,\dots,b_k$ received from Alice.
This lower bound is almost the best known. Any lower bound greater
than $\Omega(\log_2(n))$ for $CC(R_\oplus)$ would be a real
breakthrough.
\subsection*{Communication Complexity of Functions}
Communication complexity of functions is better understand than for
relations or partial functions. For a function $f : \{0,1\}^n \to
\{0,1\}$, we clearly have $CC(f) \le n+1$. In fact,
$\mathrm{Pr}_f[CC(f) < n] \to 0$ as $n \to \infty$, so a random
function is very likely to have nearly the maximum possible
communication complexity. Explicit example of functions $f$ with
$CC(f) \ge n$ are known. We will describe a few.
\subsubsection*{Lower Bounds by Tiling}
Tiling is a technique for proving lower bounds on the communication
complexity of functions. Say we have a function $f : \{0,1\}^n
\times \{0,1\}^n \to \{0,1\}$ (it is useful to think of the $2^n
\times 2^n$ 0-1 matrix $M_f$) and inputs $x,y \in \{0,1\}^n$.
Consider some exchange of bits between Alice and Bob under a fixed
protocol $\pi$ computing $f$, say $A \xrightarrow{b_1} B$, $A
\xrightarrow{b_2} B$, $B \xrightarrow{b_3} A$, \ldots,
$B\xrightarrow{b_k} A$ (as in the earlier picture).
Let $\bar b = (b_1,b_2,b_3,\dots,b_k)$ be the transcript of this
interaction, and let $S_{\bar b} = \{(x,y) : \pi(x,y)$ has
interaction $\bar b\}$. What sets $S_{\bar b} \subseteq \{0,1\}^n
\times \{0,1\}^n$ can arise in this way?
\begin{claim}
For every $\bar b$, the set $S_{\bar b}$ is a ``rectangle''. That
is, there exist subsets $S_A \subseteq \{0,1\}^n$ and $S_B \subseteq
\{0,1\}^n$ such that $S_{\bar b} = S_A \times S_B$.
\end{claim}
\begin{proof-sketch}
Consider any two pairs $(x_1,y_1),(x_2,y_2) \in S_{\bar b}$.
Clearly, it suffices to show that $(x_1,y_2) \in S_{\bar b}$. To see
this, suppose Alice has input $x_1$ and Bob input $y_2$. We ask:
assuming $(x_1,y_2) \notin S_{\bar b}$, i.e.\ if $(x_1,y_2)$ leads
to some different interaction $\bar b'$, then when is the earliest
stage where $\bar b$ and $\bar b'$ can differ? Imagine that the
first $k-1$ stages of $\bar b$ and $\bar b'$ are identical, but the
$k$th stage is different. We immediately see that this is
impossible! Indeed, up until the $k$th stage, everything Alice sees
is consistent with the possibility that Bob has $y_1$ (since
$(x_1,y_1) \in S_{\bar b}$), and similarly, everything that Bob sees
is consistent with the possibility that Alice has $x_2$ (since
$(x_2,y_2) \in S_{\bar b}$). Because a protocol is, by definition, a
deterministic function of Alice and Bob's state of knowledge at each
stage, it follows that the $k$th stage of $\bar b'$ (i.e.\ who
transmit the $k$th bit and what that bit is) must the same as the
$k$th stage of $\bar b$. Therefore, $\bar b$ and $\bar b'$ cannot be
different, so we conclude $(x_1,y_2) \in S_{\bar b}$. (This type of
argument is called a {\em crossing sequence argument}: we show the
interaction histories $\bar b$ and $\bar b'$ can never cross.)
\end{proof-sketch}
Note that a protocol $\pi$ computing $f$ can terminate after an
interaction $\bar b$ if, and only if, both and Alice and Bob can be
certain that $f$ is {\em constant} on the sets $S_{\bar b}$. So the
goal of a protocol is, in some sense, to decompose the matrix $M_f$
into rectangles on which $f$ is constant. This is the idea behind
the result (proof omitted).
\begin{lemma}[Tiling Lemma]
Let $N_0$ (resp.\ $N_1$)be the minimal number of rectangles needed
to partition the set of zero entries (resp.\ one entries) in $M_f$.
Then $\max\{\log N_0,\log N_1\} \le CC(f)$.
\end{lemma}
As an application, consider the equality function $EQ(x,y)$ defined
as 1 if $x=y$ and 0 if $x \ne y$. $M_{EQ}$ is just the $2^n \times
2^n$ identity matrix. Clearly $N_1 = 2^n$. So by the Tiling Lemma,
we have $CC(EQ) \ge n$.
The next lemma, due to Yao, achieves a similar lower bound by means
of rank.
\begin{lemma}[Rank Lower Bound]
$\log\textup{rank}(M_f) \le CC(f)$ (where rank can be taken over any
field).
\end{lemma}
\begin{proof-idea}
$M_f$ is unique expressed as a sum of $N_1$ matrices having 1s at
entries lying in some ``rectangle'' in $\{1,\dots,2^n\} \times
\{1,\dots,2^n\}$ and 0s everywhere else, for example
\[
\begin{pmatrix}
1&0&1&1\\
0&0&0&0\\
0&0&0&0\\
1&0&1&1
\end{pmatrix}
\]
(corresponding to the rectangle $\{1,3,4\} \times \{1,4\}$ in the
case $n=2$). Every matrix of this form has rank 1 (over any field).
The result follows from the Tiling Lemma and the elementary linear
algebra inequality $\textup{rank}(A+B) \le \textup{rank}(A) +
\textup{rank}(B)$.
\end{proof-idea}
As an application, we get that the inner product function $IP(x,y) =
\sum_{i=1}^n x_iy_i$ mod 2 has communication complexity $\ge n$.
Indeed, this follows from the easily checked fact that $M_{IP}$ has
rank $\ge 2^n-1$. Thus, $CC(IP) \ge \lceil\log 2^n-1\rceil = n$ by
Lemma 5.
\end{document}