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\lecture{25}{04/22/2003}{Lautemann's Proof, Lemma on Fat and Thin Sets}
{Michael DeCoster, Vinod Ganapathy, and Charles Kahn}


\section{Approximate Counting}

We continue our discussion of approximate counting schemes.
For $T \subseteq \{0,1\}^n$, choose $k$ such that $1 \leq k \leq n \cdot m = l$.
We can repeat the process $m$ times, so define $S = T^m \subseteq \{0,1\}^{n \cdot m}$. \\

From the results shown in the last lecture, we know that if $|S| \leq \frac{2^k}{2}$, then
for $\forall x \in S$ with high probability there exists a hash function
$h_i$ that isolates $x$.
Conversely, if $|S| > k \cdot 2^k$, then no hash function $h_i$ isolates $x$. \\

There are only polynomially many $k$ to evaluate, so it's a question in $\Sigma_2^p$.  We claim that
there is a value of $k$ that represents the last ``flip" from a ``no" answer for $k-1$ to a ``yes"
answer (i.e., ``Yes, there is a hash function that isolates any $x$.") for all values at least $k$.

For $k = s + 1$,
\[
(s - \log s) \cdot 2^{s - \log s} < |S| \leq \frac{2^{s+1}}{2}
\]

Then $2^{k-2} < |S| \leq k \cdot 2^k$, so we can estimate $|S|$ with an error of approximate magnitude 
$4k$.  We estimate $|S| = 2^k$, which yields $|T| = 2^{k/m}$.

\[
\frac{1}{4} < \frac{|S|}{2^k} \leq k
\]

\[
(\frac{1}{4})^{\frac{1}{m}} < \frac{|S|^{\frac{1}{m}}}{2^\frac{k}{m}} \leq k^{\frac{1}{m}}
\]

This formulation gives us:
\[
k^{\frac{1}{m}} = e^{ \frac{1}{m} \log k} \leq e^{ \frac{1}{m} (\log m + \log n)} 
\]
\[
 = 1 + \frac{1}{m} (\log m + \log n)
 < 1 + \frac{1}{n^c}
\]

for arbitrarily large $c$, that is, $c$ as a polynomial in $n$. \\

Thus our process yields an estimate

$$ |T| = |S|^\frac{1}{m} \\
    \leq ( 2 ^\frac{k}{m} ) \cdot (1.00...0xxx...)
$$

where there are $O(\log n)$ zeroes after the decimal point, which, when multiplied by some power
of 2 gives us an approximation of the $O(\log n)$ leading bits.  This is Stockmeyer's approximate
counting scheme.  We can state this as a theorem. \\

\begin{theorem}
In $P^{\Sigma^p_2}$ we can generate $O(\log n)$ leading bits of any counting problem.
\end{theorem}

For instance, we can approximate $|S|$ for $S \subseteq \{0,1\}^n$, where 
we can decide whether $x \in S$ in deterministic polynomial time. \\

This can be done ``lower" in the polynomial hierarchy, and we'll use randomization
-- that is, a probabilistic algorithm -- to do so.  To achieve accuracy $c \log n$, 
choose a polynomial $m$ such that $l = n \cdot m$, randomly pick $h_1, h_2, \cdots, h_k :
\{0,1\}^l \mapsto \{0,1\}^k$ and determine if this random assignment works.

\[
\forall x \in S,  \exists h_i \forall y \in S, y \neq x [ h_i {\rm ~isolates~ } x ] 
\]

\[
\forall x \in S, \forall y_1,\ldots,y_k \in S [y_1 \neq x, \ldots, y_k \neq x]
    [ h_1 (y_1) \neq h_1 (x) \vee \cdots \vee  h_k (y_k) \neq h_k (x) ]
\]

Thus, it's a coNP question, so it can be decided in $BPP^{NP}$.  Note that the original
statement by Stockmeyer required $P ^ {\Sigma^p_2}$, and our revised approach can be
done in $BPP ^ {NP}$. \\

The algorithm can be made to err only in saying ``yes"; the last ``no" is at worst
an overestimate. \\

Lautemann's approach is simpler and more elegant, but not as versatile as Sipser's. \\



\section{Fat and Thin Sets}

\begin{definition}[Fat Set]
For set $S \subseteq \{0,1\}^n$, we say $S$ is $\frac{1}{m}$-fat if
$|S| > (1 - \frac{1}{m}) \cdot 2^n$.  
\end{definition}
\begin{definition}[Thin Set]
Set $S$ is $\frac{1}{m}$-thin if $S^c$ is fat.
\end{definition}

We can think of $\{0,1\}^n$ as an $n$-dimensional space, where we are interested
in the effect of ``shifting" a set $S$ by adding various $n$-dimensional vectors $u_i$.
As defined here, addition is the bitwise exclusive-or operation, where for any
$u,v \in \{0,1\}^n$, $(S \oplus v) \subseteq \{0,1\}^n$ and
$u \oplus v \oplus v = u$.\\

Before proving our lemmas about fat and thin sets, we state a simple lemma that aids in the proofs.\\

\begin{lemma}
$(E \oplus u)^c = E^c \oplus u$.
\end{lemma}

\begin{proof}
We'll show one direction of the proof.  We know $x = (x \oplus u) \oplus u$
for $\forall x \in E, \forall u \in \{0,1\}^n$.  Take $x \in (E \oplus u)^c$.  
We claim ($x \oplus u) \notin E$.  
If $(x \oplus u) \in E$, then $x = (x \oplus u) \oplus u$, which makes $(x \oplus u) \in E$,
which is a contradiction.
The other direction is proved in similar fashion.
\end{proof}

We will randomly and uniformly pick ``shift vectors" $u_1, u_2, \cdots, u_m \in \{0,1\}^n$.  
If $S$ is ``fat", then it's likely that for any ``point" in $\{0,1\}^n$,
some shifts will cover it.
We also can state that if $S$ is fat, then the intersection of the ``shifted"
sets $(S \oplus u_i)$ will be non-empty, that is, \\


\[
\mathop {\Pr }\limits_{u_1 , \cdots ,u_m } \left[ {\forall v \in \{ 0,1\} ^n ,\exists u_i ,1 \le i \le m,v \in S \oplus u_i } \right] \ge 1 - \frac{{2^n }}{{m^m }}
\]

Take $m > n$, so that $\frac{{2^n }}{{m^m }} \to 0$. \\


\[
\mathop {\Pr }\limits_{u_1 , \cdots ,u_m } \left[ {\bigcap\limits_{i = 1}^m {\left( {S \oplus u_i } \right) \ne \emptyset } } \right] = 1
\]

\[
 \left[ {\bigcap\limits_{i = 1}^m {\left( {S \oplus u_i } \right)} } \right]^c  = \bigcup\limits_{i = 1}^m {\left( {S \oplus u_i } \right)^c }  = \bigcup\limits_{i = 1}^m {\left( {S^c  \oplus u_i } \right)}
\]

\[
  \left| {\left[ {\bigcap\limits_{i = 1}^m {\left( {S \oplus u_i } \right)} } \right]^c } \right| < \frac{1}{m} \cdot m \cdot 2^n  = 2^n 
\]


\[
\mathop {\Pr }\limits_{u_1 , \cdots ,u_m } \left[ {\forall v \in \{ 0,1\} ^n ,\forall u_i ,1 \le i \le m,v \notin S \oplus u_i } \right] \le 2^n \left( {\mathop {\Pr }\limits_u \left[ {v \notin S \oplus u} \right]} \right)^m 
\]

\[
 v \notin S \oplus u \Rightarrow v \oplus u \notin S \Rightarrow u \notin S \oplus v 
\]

\[
  \left| {S \oplus v} \right| = \left| S \right| > \left( {1 - \frac{1}{m}} \right) \cdot 2^n
\]

So, if $S$ is thin, then the probability that the ``shifted" sets $(S \oplus u)$ cover all of the
points in $\{0,1\}^n$ is zero.  That is,

\[
\mathop {\Pr }\limits_{u_1 , \cdots ,u_m } \left[ {\forall v \in \{ 0,1\} ^n ,\forall u_i ,1 \le i \le m,v \notin S \oplus u_i } \right] = 0
\]

\[
\mathop {\Pr }\limits_{u_1 , \cdots ,u_m } \left[ {\bigcup\limits_{i = 1}^m {\left( {S \oplus u_i } \right) = \{ 0,1\} ^n } } \right] = 0
\]

The proof is the complement of the proof for ``fat" sets.  Also, for ``thin" sets, \\

\[
\mathop {\Pr }\limits_{u_1 , \cdots ,u_m } \left[ {\bigcap\limits_{i = 1}^m {\left( {S \oplus u_i } \right) \ne \emptyset } } \right] \le \frac{{2^n }}{{m^m }}
\]

\section{BPP}

\begin{theorem}
$BPP \subseteq \Sigma_2^P$.
\end{theorem}

If $x \in BPP$, the witness set $Y$ is either thin or fat in $\{0,1\}^m$, where
$m = $ poly $(n)$. For $L \in BPP$, $L = L(M)$: \\

If $x \in L$, then the witness set is fat in $\{0,1\}^m$ (for appropriate $m$, some
large polynomial in $n$), so \\

\[
\exists u_1 , \cdots ,u_m  ~s.t.~  \forall v,\bigvee\limits_{i = 1}^m {M\left( {x, u_i  \oplus v} \right)}  = 1
\]

Conversely, if $x \notin L$, the witness set is thin.



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