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\lecture{13}{March 16, 2000}{Dan Spielman}{Andrew Glenn}
In lecture 10, we introduced the class $\AC_0$ and showed that parity
$\not\in \AC_0$. This lecture presents a second proof of this
assertion, published by Furst, Saxe and Sipser (1984).
\section{Setup and Definitions}
\begin{definition}
A restriction is a function $\rho$ such that:
\[ \rho(x_i) = \left\{
\begin{array}{ll}
0,1 & (fixed)\\
\star & (free)
\end{array}
\right\}
\]
\end{definition}
$\\$
We designate the induced circuit obtained by applying the
restriction $\rho$ to the circuit $C$ as $C\mid_\rho$. To show parity
$\not\in \AC_0$, we will apply random restrictions as follows:
\[ \rho(x_i) = \left\{
\begin{array}{ll}
\star & \mbox{with probability } \frac{1}{\sqrt{n}}\\
0 & \mbox{with probability } \frac{1-\frac{1}{\sqrt{n}}}{2}\\
1 & \mbox{with probability } \frac{1-\frac{1}{\sqrt{n}}}{2}
\end{array}
\right\}
\]
$\\$
There are two ideas that motivate this proof.\\
Idea 1: If we apply a restriction to a parity circuit, we will get
parity, or parity on the remaining free variables.\\
Idea 2: If we apply two of these random restrictions on an $\AC_0$
circuit of depth $d$, we will end up with a circuit of depth $d-1$
with high probability and at least $\frac{n^\frac{1}{4}}{4}$ of the
variables will remain free. This results in a contradiction because
we will need exponential size in a depth two circuit for computing
parity. (After $k$ restrictions, we will have approximately
$n^{\frac{1}{4}^k}$ free variables.)
\section{Parity $\not\in \AC_0$}
We will consider a depth $d$ circuit which alternates AND and OR gates
at every level, and we will make use of the following fact.
\begin{fact}
Any function of $c$ bits can be expressed as an AND of $2^c$ ORs or an
OR of $2^c$ ANDs. This implies that switching from an AND of ORs to
an OR of ANDs can result in a blowup in circuit size of at most $2^c$,
which is constant size.
\end{fact}
\begin{lemma}
Let $c>0$, $G$ is an OR gate, and $n$ is large. Then, \\
$\Pr[G\mid_\rho \mbox{depends on more than c variables}] < n^{-\frac{c}{3}}$
\end{lemma}
\begin{proof-of-lemma}{3}\\
Case 1 (``fat case'' - the OR gate is large): $\abs{G}>c\lg{n}$\\
$\Pr[G\mid_\rho \mbox{is not fixed}] < ({\frac{1+\frac{1}{\sqrt{n}}}{2}})^{c\lg{n}} \sim \ {\frac{1}{2}}^{c\lg{n}} - n^{-c} < n^{-\frac{c}{3}}$
\\
\\
\\
Case 2 (``thin case''): $\abs{G} \leq c\lg{n}$\\
$\Pr[\mbox{there are $c$ free variables}] \leq {c\lg{n} \choose c} \cdot (\frac{1}{\sqrt{n}})^c \leq (c\lg{n})^c
\cdot (\frac{1}{\sqrt{n}})^c < n^{-\frac{c}{3}}$
\end{proof-of-lemma}
\begin{corollary}
Let $C$ be a circuit with $n^k$ gates. Then with high probability there exists a $\rho$ with $\frac{\sqrt{n}}{2}$ free variables such that all bottom-level gates of $C$ have fanin $\leq 4k$.
\end{corollary}
\begin{proof}
Apply lemma 3 with $c=4k$:\\
$\Pr[\mbox{any bottom gate depends on more than $c$ variables}] \leq n^k \cdot n^{\frac{-4k}{3}} \ll 1$\\
$\Pr[\mbox{there are fewer than $\frac{\sqrt{n}}{2}$ free variables}] \sim 2^{\bigO(-\sqrt{n})}$ (by Chernoff)
\end{proof}
\begin{lemma}
$\forall k, \forall c, \exists$ a constant $b_c$ such that for a sufficiently large $n$, if $G$ is an $AND$ of $OR$ gates each having fanin at most $c$, then:
$\Pr[G\mid_\rho \mbox{depends on more than $b_c$ variables}] < n^{-k}$
\end{lemma}
\begin{definition}
Let $M$ be a collection of the OR gates such that the ORs in the collection have disjoint sets of variables and every OR gate not in the collection $M$ overlaps with an OR gate in $M$.
\end{definition}
\begin{definition}
Let $V$ be the set of variables in the OR gates of $M$.
\end{definition}
\begin{proof-of-lemma}{7}\\
This proof is an induction on $c$, the maximum fanin of the OR gates.\\
Base case: If $c=1$, then the OR gates are unnecessary, and the circuit is really just an AND circuit. Thus, the base case holds by lemma 3.
\\
\\
Inductive step: Assume the lemma holds for $c-1$ to prove it for $c$.
$\\$
Case 1 (``Fat case''): $\abs{V}>a\lg{n}$\\
We wish to find the probability that $G$ is not fixed to 0.
Given $\frac{a}{c}\lg{n}$ variables in $M$, then
$\Pr[\mbox{a particular OR gate outputs 0}] = (\frac{1-\sqrt{n}}{2})^c$, and thus\\
$\Pr[\mbox{no OR gate outputs 0}] = \Pr[\mbox{G is not fixed}] =$\\
$ (1-(\frac{1-\sqrt{n}}{2})^c)^{\frac{a}{c}\lg{n}} \sim (1-2^{-c})^{\frac{a}{c}\lg{n}} \sim e^{-\frac{a}{c2^c}\lg{n}}$\\
Therefore, for the fat case, we need $a>2^c\cdot k$\\
\\
Case 2 (``Thin case''): $\abs{V} \leq a\lg{n}$\\
$\Pr[\mbox{V has i free variables}] \leq {a\lg{n} \choose i} \cdot (\frac{1}{\sqrt{n}})^i < n^{-\frac{i}{3}}$ (for sufficiently large n)\\
Therefore, we set $i = 4k$\\
Every OR gate not in $M$ shares an input with an OR gate in $M$. Thus, if we fix those $i$ free variables, then the other ORs will have a fanin of $c-1$. (We can then apply the inductive hypothesis to complete the proof.)\\
Finally, since there are only $2^i$ ways to fix these variables, $G$ will only depend on $i+2^ib_{c-1} = b_c$ variables.
\end{proof-of-lemma}
\end{document}