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\lecture{5}{Feb 21, 2007}{Madhu Sudan}{Salman Abolfathe}

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\section{Overview}

\begin{itemize}
\item  Bounded depth circuits $=AC^0$.
\item  PARITY$\notin AC^0$.
\item  Switching lemma $DNF\rightarrow CNF$
\end{itemize}


\section{Introduction}

$AC^i$ is the class of languages recognized by a circuit of
polynomial size and depth $O(\log ^i n)$, that can use gates
$$\{\infty-AND, \infty-OR, NOT\}.$$
So for $AC^0$ circuits should have polynomial size and $O(1)$
depth. Loosely speaking, depth of a circuit shows the amount of
parallel time, and size is the amount of work needed for
computing.

Through this lecture we assume that the circuits are organized
into alternating levels of $AND$ and $OR$ gates. In fact, all
$NOT$ gates can be pushed to the first level, and since here we
use $\infty-AND$ and $\infty-OR$ gates instead of usual $AND$ and
$OR$ gates, we can combine consecutive $AND$ and $OR$ levels. So
each circuit by a constant blowup can be organized such that the
first level is $NOT$ gate, and the others are alternating $AND$
and $OR$. In a such circuit, we don't care about the first level
($NOT$ gates) and say the depth is number of $AND$ and $OR$
levels.


As an example the following language, for a constant $k$, is in
$AC^0$

$$T_{k,n}(x_1,x_2,\dots x_n)=\left\{\begin{array}{l}
1 \textrm{ if } \sum_i x_i\geq k\\
0 \textrm{ otherwise}
\end{array}\right.$$
In fact, $T_{k,n}(x_1,x_2,\dots x_n)=\bigvee_{A}\bigwedge_{x_i\in
A } x_i$, where $A$ ranges over all subsets of $\{x_1,x_2,\dots
x_n \}$ of size $k$.

The goal of this lecture is to show PARITY is not in $AC^0$. By
PARITY we mean
$$PARITY(x_1,x_2,\dots x_n)=\sum_i x_i\ (mod 2).$$

\section{Random Restriction}

In the last session we studied the idea of restriction of some
variables to get a bound on the size of a circuit that can
compute a certain language. Here we use another version of this
idea, called {\it random restriction}. Namely, if $x_1,x_2,\dots
x_n$ are variables, for each $i$, with probability $q$ we don't
restrict $x_i$, and restrict it as $x_i=0$ or $x_i=1$, each of
which with probability $\frac{1-q}{2}$.

$$x_i=\left\{\begin{array}{l}
x_i \textrm{ with prob. } q\\
0 \textrm{ with prob. } \frac{1-q}{2}\\
1 \textrm{ with prob. } \frac{1-q}{2}
\end{array}\right.$$
and we repeat it for each $i$ independently. Therefore, if we have
a function $f$ that can be computed by circuit $C$, after this
restriction, say $\rho(q)$, we get to the function $f\vert_\rho$
that can be computed by $C\vert_\rho$, which is hopefully is a
simpler circuit.

In our case, the PARITY problem, after restricting some variables
we get to the same problem, i.e. PARITY over remaining variables.

$$\textrm{ PARITY over}\ n\ \textrm{variables} \longrightarrow \textrm{PARITY over}\ qn\ \textrm{variables }.$$

\section{Switching Lemma}
Consider a depth $2$ circuit. Then it is either  a CNF or DNF
formula. By random restriction idea we want to convert a DNF
formula to a CNF formula. In general, if we want to write a DNF
as a CNF formula we may get to an exponential one (it is because
SAT is hard). But after random restriction we may get to a
polynomial size one. Indeed, if $q=0$ then we get the trivial
formula, and if $q=1$ we get the same function. So our goal is to
find the largest possible $q$ such that after restriction
$\rho(q)$, we get to a polynomial size CNF.

The following lemma, called Switching lemma, is first proved by
Furst, Saxe and Sipser in $1981$.

\begin{lemma}\label{switch} {\rm ({\bf Switching Lemma})} Let
$q=n^{-\frac{2}{3}}$, then for any DNF of polynomial size $P(n)$,
and $\delta=\frac{1}{poly(n)}$, after random restriction we get
to a CNF of size $C$ with probability $(1-\delta)$, where $C$ is
constant.
\end{lemma}

Before proving Switching lemma, let's use it to prove
PARITY$\notin AC^0$.

\begin{theorem} \label{parity} PARITY $\notin AC^0$.
\end{theorem}

\begin{proof} Suppose there is formula of size $s=poly(n)$ and depth $d$
that computes PARITY$_n$. By induction we can assume there is no
formula of size $poly(n)$ and depth $d-1$ that computes
PARITY$_n$. Now consider the DNF formulas in the first two levels
of this formula, and apply the Switching lemma on them. Since
number of these formulas is at most $s$, with probability
$1-s\delta$ each of DNF's will change to a CNF. Now by the same
idea as before, we have two consecutive AND levels and can combine
them. Therefore we get to a circuit that compute PARITY on
unrestricted variables, and has size $\leq O(s)$ and depth $d-1$,
which is a contradiction. For the base of induction it is not hard
to see that, if a circuit of depth $2$ computes PARITY then its
size is $O(2^n)$.

$\hfill$
\end{proof}

\section{Proof of Switching Lemma}

Let $f=T_1\vee T_2\vee \dots T_m$ be a formula that solves
PARITY. Call each $T_j$ a term. Suppose we restrict variables in
two stages

{\bf Stage 1}

Restrict variable with probability $\sqrt{q}$, i.e.

$$x_i=\left\{\begin{array}{l}
x_i \textrm{ with prob. } \sqrt{q}\\
0 \textrm{ with prob. } \frac{1-\sqrt{q}}{2}\\
1 \textrm{ with prob. } \frac{1-\sqrt{q}}{2}
\end{array}\right.$$


$$f \longrightarrow\ f\vert_{\rho_1}.$$

{\bf Stage 2}

Restrict variables in $f\vert_{\rho_1}$ with probability
$\sqrt{q}$, i.e.

$$x_i=\left\{\begin{array}{l}
x_i \textrm{ with prob. } \sqrt{q}\\
0 \textrm{ with prob. } \frac{1-\sqrt{q}}{2}\\
1 \textrm{ with prob. } \frac{1-\sqrt{q}}{2}
\end{array}\right.$$


$$f\vert_{\rho_1} \longrightarrow\ f\vert_{\rho_1\cup \rho_2}.$$

{\bf Claim 1} There is a constant $c$ such that all terms of
$f\vert_{\rho_1}$ are of size $\leq c$ (with high probability).

Consider two cases

\begin{itemize}
\item $T_i$ has a {\it large} size

Suppose $T_i$ consists of $k=\Omega(\log n)$ variables. Then
after restriction $T_i$ is non-zero iff non of the variables
restricted to $0$. Therefore $$Pr(T_i\neq 0)\leq
(\frac{1-\sqrt{q}}{2})^k\leq \frac{1}{poly}.$$

\item $T_i$ has{\it small} size

Suppose $T_i$ contains $k'=O(\log n)$ variables. Then the
probability that at least $c$ of them remain unrestricted is

$$Pr( c\ \textrm{variables\ remain\ unrestriced})\leq
{k'}^c(\sqrt{q})^c\leq \big(\frac{k'}{\sqrt{q}}\big)^c\leq
\frac{1}{poly}$$

\end{itemize}

Therefore {\bf Claim 1} is true.

{\bf Claim 2 }  Each term of $f\vert_{\rho_1\cup\rho_2}$ depends
on $b_c$ variables, again with high probability. Here $b_c$ is a
constant depends on $c$.

We prove it by induction on $c$. Consider two cases.

\begin{itemize}
\item There are $l=\Omega(\log n)$ terms $T_1, T_2,\dots T_l$ in $f\vert_{\rho_1}$ such
that they have disjoint variables.

In this case, the probability that $f\vert_{\rho_1\cup \rho_2}= 1$
is

$$Pr(f\vert_{\rho_1\cup \rho_2}= 1)\geq \log n\big( \frac{1-\sqrt{q}}{2}\big)^c\rightarrow 1$$
then with high probability $f\vert_{\rho_1\cup \rho_2}= 1$.

\item $T_1, T_2,\dots T_l$, where $l=O(\log n)$, are maximal disjoint terms in
$f\vert_{\rho_1}$, i.e. any other term has at least one variable
in $T_1, T_2,\dots T_l$.

Let $H$ be set of variables in $T_1, T_2,\dots T_l$, and $Y$ be
the remained variables, and assume after restriction $\rho_2$, $H$
changes to $H'$ and $Y$ to $Y'$. Since each $T_i$ has most $c$
variables (Claim 1) then $\#H\leq cl$. Therefore after the second
restriction, with high probability $\#H'\leq c'$, for some
constant $c'$.

Now set all variables in $H$, $0$ or $1$, then $f\vert_{\rho_1\cup
\rho_2\cup H'}$ depends just on $b_{c-1}$ variables. It is
because we assigned either $0$ or $1$ to each variable in $T_1,
T_2,\dots T_l$, and by the maximality assumption in $T_1,
T_2,\dots T_l$, each of the remained terms has at most $c-1$
variables. So by the induction assumption $f\vert_{\rho_1\cup
\rho_2\cup H'}$ has at most $b_{c-1}$ variables.

Set $b_c=c'+2^{c'}b_{c-1}$. Then $f\vert_{\rho_1\cup\rho_2}$
depends on at most $b_c$ variables. Because there are at most $c'$
variables in $H'$, and for each of $2^{c'}$ assignments of
variables in $H'$, $f\vert_{\rho_1\cup \rho_2 \cup H'}$ depends
on $b_{c-1}$ variables. Note that all of these arguments are true
with high probability.
\end{itemize}

So after two restrictions $\rho_1$ and $\rho_2$, with high
probability, we get to a formula that has constant number of
variables. Now use the distributive law to switch the order of
AND and OR gates. Since number of variables in constant, number
of gates after that is also constant.







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