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\section{Monotone Circuit Depth Lower Bounds}

Today we will finish the proof that
there are monotone functions that require
near linear monotone circuit depth.

Recall from last class that we will start with the
CNF search problem associated with Tseitin, and lift it
to obtain a lifted search problem.
From the theorem from last class, this implies that
there is a corresponding monotone function such that
the KW communication complexity of this monotone function
is large, if the communication complexity of the lifted
Tseitin search problem is large.

In particular, $Tseitin$ on a graph with $n$ vertices
has $n$ constraints, and we will select a gadget $g$ with $c=2$.
Thus, our monotone function will have $4n$ inputs
and its monotone depth will be at least the communication
complexity of $S(Tseitin(G))og^n$.

By using a different family of unsatisfiable CNF formulas,
it is also possible to obtain a function that is in monotone $P$
and that requires monotone circuits of depth $\sqrt n$;
however we will not do this here.


For the remainder of the lecture, we will prove that
for appropriate graphs $G$, the communication complexity of $S(Tseitin(G))og^n$ is
large.



\section{Block sensitivity, Critical Block Sensitivity, Decision Tree Depth}

Define block sensitivity, and critical block sensitivity.

Fix a search problem $S \subseteq \{0,1\}^n \times Q$.
An assignment $\alpha$ is said to be {\it critical} if it has
a unique feasible solution.

Let $f \subset S$ be a total function -- for each input $\alpha$,
$f$ picks out some feasible solution $f(\alpha)$ for $\alpha$.

The block sensitivity of $f$ at $\alpha$, $bs(f,\alpha)$,
is the maximal number $bs$ such that there are disjoint
blocks of coordinates $B_1, \ldots, B_{bs} \subset [n]$ satisfying
$f(\alpha) \neq f(\alpha^{B_i})$ for all $i$.
Here, $\alpha^{B_i}$ represents the assignment $\alpha$
where the block of variables $B_i$ has been flipped.

The block sensitivity of $f$, $bs(f)$ is then the maximal block
sensitivity over all assignments $\alpha$.
The block sensitivity of a search problem $S$ is just the
minimum over all total functions $f \subset S$ of the block sensitivity of $f$.

The critical block sensitivity of $S$, $bs_{crit}(S)$ is 
the minimum over all $f \subset S$ of the maximum
over all critical assignments $\alpha$, of $bs(f,\alpha)$.
So the critical block sensitivity is just like the block sensitivity
but we only look over critical assignments.


If a search problem has high critical block sensitivity,
then it has high block sensitivity, which in turn implies high
decision tree complexity.

It turns out that for the appropriate  choice of graph $G$,
that the Tseitin CNF formulas have very high critical block sensitivity.

\begin{theorem}
There are bounded-degree graphs
$G_n$ such that $bs_{crit}(S(Tseitin(G_n)) = \Omega(n/\log n)$.
\end{theorem}

\section{Defining the Gadget}

A "nice" two-party gadget $g$ will have some nice properties.
Let $f_1$ and $f_2$ be two-party functions.
We say that $f_1$ reduces to $f_2$, written $f_1 \leq f_2$ if
the communication matrix of $f_1$ appears as a submatrix of $f_2$.
This is equivalent to their being functions $\pi_A$ and $\pi_B$
such that $f_1(x,y) = f_2 (\pi_A(x), \pi_B(y))$ for all $x,y$.
In other words, if $f_1$ reduces to $f_2$, then from a
short communication complexity for $f_2$, Alice and Bob can solve
$f_1$. (Furthermore, the reduction is of the simple type where
Alice without communication maps her input $x$ to some $x'$
and similarly Bob without communication maps $y$ to $y'$
and then they run $f_2$ on $(x',y')$ and the answer returned
will be equal to $f(x,y)$.)

We want our gadget $g$ to have
three reducibility properties.
First, $\neg g$ is reducible to $g$.
Secondly the AND function is reducible to $g$.
Thirdly, $g$ is random self reducible.
A gadget $g$ satisfying these three properties will
be called {\it versatile}.

\begin{lemma}
The following 2-party function $g$ from $X \times Y \rightarrow \{0,1\}$
%$\{0,1\}^2 \times \{0,1\}^2 \rightarrow \{0,1\}$
is versatile.
Let $X=\{0,1,2,3\}$, $Y=\{0,1,2,3\}$.
Then $g(x,y) = 1$ if and only if $x+y (mod 4)$ equals $2$ or $3$.
\end{lemma}

\section{Communication Complexity Lower Bound for Lifted Tseitin via Set Disjointness}

We first give a warmup reduction that should help to explain the
main high level idea. Suppose that $f$ is a function (not a search problem)
with block sensitivity $bs$ and let $\alpha$ be an assignment that witnesses
this block sensitivity. Let the blocks be $B_1,\ldots,B_{bs}$.
Now consider $f o g^n$ where $g$ is some inner gadget such that
both $AND$ and $\neg$ are reducible to $g$ (that is, both of these
functions occur as subrectangles of the matrix for $g$).

The players, upon input $x,y$ to set disjointness, $|x|=|y|=bs$ want to
solve unique disjointness with a protocol for $f o g^n$.
They want to map input $x_i,y_i$ to values for block $B_i$.
If $x_i=y_i=1$ then they want to set the $i^{th}$ block to the value
specified by $\alpha$ and otherwise (if $AND(x_i,y_i)=0$) then they
want to toggle the values of the $i^{th}$ block.
All other indices that are not in a critical block will take
on a value consistent with $\alpha$.
They can do this as long as both AND and $\neg$ are reducible to $g$.
Now if $x,y$ are disjoint, then $g^n$ will be equal to $\alpha$,
but if $x,y$ are not disjoint and intersect uniquely at location $i$,
then $g^n$ will equal $\alpha^{B_i}$.
Since $f$ is sensitive on every block on $\alpha$, the protocol
for $f o g^n$ will either output $f(\alpha)$ and this happens whenever
$x,y$ are disjoint, or it will output $f(\alpha^{B_i})$ for some $i$
and this will happen whenever $x,y$ uniquely intersect at location $i$.
Thus Alice and Bob can determine whether or not $x,y$ intersect
from a protocol for $f o g^n$.

In this section, $S$ will be the search problem associated with
Tseitin formula over a fixed highly expanding graph on $O(n)$ vertices. By the above theorem,
we know that the critical block sensitivity of $S$ is very large -- $\Omega(n/\log n)$.

Let $S o g^n$ be the lifted search problem, where Alice gets
a binary vector $x$ of length $4n$, and similarly Bob gets a binary
vector $y$ of length $4n$, and their job is to solve the search
problem for $S$ on input $z_1 = g(x^1,y^1), \ldots, z_n=g(x^n,y^n)$,
where $g$ is the versatile gadget described in the previous section.

We want to show that if there is a short protocol for
$S o g^n$, then there is also a short protocol for unique set disjointness.

Let $\Pi$ be the short protocol for $S o g^n$.
We will record for each $\alpha \in \{0,1\}^n$ the most likely
feasible output of $\Pi$ on inputs $(x,y)$ that encode $\alpha$.
That is, for each $\alpha$, we define $\mu_{\alpha}$ to be
the uniform distribution on the set of preimages of $\alpha$, i.e.,
$\mu_{\alpha}$ is uniform over all $(x,y)$ such that $g^n(x,y)=\alpha$.
The most likely feasible solution output by $\Pi$ on inputs from
distribution $\mu_{\alpha}$ is now captured by a total function $f \subset S$.

Going back to our reduction, we want to show that from the short
protocol $\Pi$ for $S o g^n$ described above (and corresponding
total function $f \subset S$ described above), that Alice and Bob can
solve unique set disjointness.
To this aim, let $a=a_1,\ldots,a_m$ be Alice's input to unique set disjointness,
and let $b=b_1,\ldots,b_m$ be Bob's input to unique set disjointness.
We will eventually be setting $m$ equal to the block sensitivity of $S$,
so $m$ will be about $n/\log n$.


Since $S$ has critical block sensitivity $\Omega(n/\log n)$, there exists
some critical input $\gamma$ to $S$ that witnesses this high block sensitivity
with respect to $f$.
Let the blocks be denoted by $B_1,\ldots,B_{bs} \subset [n]$ be the sensitive
blocks with $f(\alpha^{B_i}) \neq f(\alpha)$.

\begin{lemma}
The protocol $\Pi$ can distinguish between
$\mu_{\alpha}$ and $\mu_{\alpha^{B_i}}$ in the sense that
if $(x,y)$ is chosen randomly from $\mu_{\alpha}$, then
the probability that $\Pi(x,y)=f(\alpha)$ is at least $1-\epsilon$,
but if $(x,y)$ is chosen randomly from $\mu_{\alpha^{B_i}}$, then
the probability that $\Pi(x,y)=f(\alpha)$ is at most 1/2.
\end{lemma}

\begin{proof}
Since $f(\alpha)$ is the most likely feasible solution output
by $\Pi$ over inputs consistent with $\alpha$, this solution must
be correct for these inputs.
On the other hand, the inputs consistent with $\alpha^{B_i}$ cannot
give this same answer because $f$ is sensitive to $B_i$ at $\alpha_i$.
Since the protocol is correct (most of the time), this means that
the protocol better give an answer other than $f(\alpha)$ a lot of the time.
\end{proof}


\subsection{The Reduction}


\begin{itemize}
\item On input $(a,b)=(a_1,\ldots,a_m, b_1,\ldots, b_m)$ to $UDISJ_m$,
we first take each pair $(a_i,b_i)$ through the reduction $AND \leq g$
to obtain instances $(a_1',b_1'), \ldots, (a_m',b_m')$ of $g$.
Note that if $UDISJ(a,b)=0$ then $g(a_i',b_i)=0$ for all $i$,
and if $UDISJ(a,b)=1$ then there is a unique $i$ with $g(a_i',b_i')=1$.

\item Next the instances $(a_i',b_i')$ are used to populate a vector
$(x,y)=(x_1,\ldots,x_n,y_1,\ldots,y_n)$ carrying $n$ instances of $g$ as follows.
Instance $(a_i',b_i')$ is plugged in for the coordinates $j \in B_i$
with the c opies corresponding to $\alpha_j=1$ flipped. That is,
we define for $j \in B_i$:

\begin{itemize}
\item if $\alpha_j=0$ then $(x_j,y_j) = (a_i',b_i')$;
\item if $\alpha_j=1$ then $(x_j,y_j) = (\pi_A(a_i'),\pi_B(b_i'))$ where $(\pi_A,\pi_B)$ is the reduction
$\neg g \leq g$.
\end{itemize}

For $j$ not in any of the blocks, we simply fix an arbitrary $(x_j,y_j) \in g^{-1}(\alpha_j)$.

We now have that if $UDISJ(a,b)=0$, then $g^n(x,y)=\alpha$,
and if $UDISJ(a,b)=1$ with $a_i=b_i=1$, then $g^n(x,y) = \alpha^{B_i}$.

\item Finally, we apply a random self reduction independently for each
component $(x_i,y_i)$ of $(x,y)$: this maps a $z$-input $(x_i,y_i)$ to a
uniformly random $z$-input $(x_i,y_i)$ from $\mu_{z}$.
The result is: if $UDISJ(a,b)=0$, then $(x,y)$ is a random vector
such that $g^n(x,y)=\alpha$, and if $UDISJ(a,b)=1$ with $a_i=b_i$
then $(x,y)$ is a random vector such that $g^n(x,y)=\alpha^{B_i}$.

By the above lemma, the protocol $\Pi$ distinguishes between
these 2 vectors, and thus Alice and Bob can determine whether or not $(a,b)$ are disjoint
with high probability.









\begin{figure}[t]%
%\ffigbox[.9\textwidth]
{\caption{The reduction mapping $(a,b)$ to a distribution $(x,y)$. In this example $bs=2$ and $n=7$. 
The critical input is $\alpha=1011010$ and the two sensitive blocks are $B_1=\{2,3,4\}$ and $B_2=\{6,7\}$. 
The input pair $(a_i,b_i)$, $i=1,2$, is plugged in for the block $B_i$.}
\label{fig:reduction}}
{\begin{lpic}[r(2cm)]{reduction(.35)}
\large
\lbl[t]{91,23;$(a_1,b_1)$}
\lbl[t]{224,23;$(a_2,b_2)$}
\lbl[c]{91,47;$\AND\leq g$}
\lbl[c]{224,47;$\AND\leq g$}
\lbl[c]{14,95;$(1,1)$}
\lbl[c]{166.7,95;$(0,0)$}
\lbl[c]{14,168;$g$}
\lbl[c]{52.2,168;$g$}
\lbl[c]{90.3,168;$g$}
\lbl[c]{128.5,168;$g$}
\lbl[c]{166.7,168;$g$}
\lbl[c]{204.3,168;$g$}
\lbl[c]{243,168;$g$}
\lbl[c]{128.5,228.5;\Large$S$}
\normalsize
\lbl[l]{270,130;{\bf-----~~Step 3}}
\lbl[l]{270,95;{\bf-----~~Step 2}}
\lbl[l]{270,47;{\bf-----~~Step 1}}
\end{lpic}}
\end{figure}














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