\documentclass{article}
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\newcommand{\TIME}[1]{\ensuremath{\mbox{TIME}(#1)}}
\newcommand{\SPACE}[1]{\ensuremath{\mbox{SPACE}(#1)}}
\newcommand{\setof}[1]{\ensuremath{\left\{ #1\right\}}}
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\newcommand{\NP}{\textrm{NP}}
\newcommand{\coNP}{\textrm{coNP}}
\newcommand{\Pclass}{\textrm{P}}

\begin{document}

\lecture{3}{February 13, 2000}{Shanghua Teng}{Hoeteck Wee}

\section*{Polynomial Hierarchy}

In this lecture, we introduce the polynomial hierarchy, which
generalizes the notions of P and NP. In particular, we provide
3 definitions of the polynomial hierarchy, which we claim
are equivalent.

\begin{definition}
Let $\Sigma_k \TIME{f(n)}$ be the class of languages $L$ accepted by
an Alternating Turing Machine (ATM) that begins in an existential
($\exists$) state, alternates between
existential and universal states at most $k-1$ times, and halts
(on all branches) within time $O(f(n))$. $\Pi_k \TIME{f(n)}$ is
defined similarly, except that the ATM begins in a universal ($\forall$)
state.

We also define
\begin{eqnarray*}
\Sigma_k P = \bigcup_{i=1}^{\infty} \Sigma_k \TIME{n^i}\\
\Pi_k P = \bigcup_{i=1}^{\infty} \Pi_k \TIME{n^i}
\end{eqnarray*}
\end{definition}

The first level of this hierarchy makes up the familiar complexity
classes, with $\Sigma_1 P = \NP$ and $\Pi_1 P = \coNP$. On
the other hand, {\it MIN-FORMULA} =
$\{\langle\phi\rangle \;|\; \phi \textrm{ is a minimal Boolean formula}\}
\in \Pi_2 P$.

\begin{definition}
$PH = \bigcup_{k=0}^{\infty} \Sigma_k P$.
\end{definition}

Note that $PH = \bigcup_{k=0}^{\infty} \Pi_k P$, since
$\Pi_k P \subset \Sigma_{k+1} P$ and $\Pi_k P \subset \Sigma_{k+1} P$.
In addition, we have $\NP \cup \coNP \subset PH \subset AP = PSPACE$.

\begin{definition}\label{def2}
First, $\Sigma_0 P = \Pi_0 P = P$; and for all $k \geq 1$,
\begin{eqnarray*}
\Sigma_k P & = & \{ A\; |\; \exists\; B \in \Pi_{k-1} P,\; c > 0
\mbox{ such that } x \in A \iff \exists\; y \mbox{ such that } (x,y) \in B
\mbox{ and } |y| \leq |x|^c \} \\
\Pi_k P & = & \{ A\; |\; \exists\; B \in \Sigma_{k-1} P,\; c > 0
\mbox{ such that } x \in A \iff \forall\; y \mbox{ such that } (x,y) \in B
\mbox{ and } |y| \leq |x|^c \}
\end{eqnarray*}
\end{definition}

\begin{definition}
An oracle is a language $A$. An oracle Turing machine $M^A$ is an ordinary
Turing machine with an extra tape called the oracle tape. Whenever $M$
writes a string on the oracle tape, it is informed whether that string
is a member of $A$ in a single computation step. Define $P^A$ to be the
class of languages decidable with a polynomial time oracle Turing machine
that uses oracle $A$. Define ${\NP}^A$ and ${\coNP}^A$ similarly. 
\end{definition}

\begin{definition}\label{oracledef}
First, $\Sigma_0 P = \Pi_0 P = P$; and for all $k \geq 1$,
\begin{eqnarray*}
&&\Sigma_k P = {\NP}^{\Sigma_{k-1}}\\
&&\Pi_k P = {\coNP}^{\Sigma_{k-1}}\\
&&\Delta_k P = P^{\Sigma_{k-1}}.
\end{eqnarray*}
\end{definition}

Note that $\Sigma_k P = \textrm{co}\Pi_k P$. It follows that
${\NP}^{\Sigma_{k-1}} = {\NP}^{\Pi_{k-1}}$, since we may check whether
a string is in $A \in \Pi_{k-1}P$ by asking the oracle tape whether it is
in $\textrm{co}A \in \Sigma_{k-1}P$ and then negating the outcome. By the
same argument, we have ${\coNP}^{\Sigma_{k-1}} = {\coNP}^{\Pi_{k-1}}$.

Next, observe that {\it MIN-COLOR} =
$\{\langle G, k \rangle \;|\; k$ is the minimum number of colors
needed to color the graph $G \} \in \Delta_2 P$. Checking
whether a $k$-coloring of $G$ exists can be done in NP, so we could
for each $k = 1,2,\ldots,n$ (where $n$ is the number of vertices
in $G$) query an oracle to check whether a $k$-coloring of $G$
exists. On the other hand, there is no known NP or coNP algorithm
for {\it MIN-COLOR}.

\begin{theorem}
The definitions of $\Sigma_k P$ in 1 and {\ref{def2}} are equivalent.
\end{theorem}

(The proof for this theorem is omitted; refer to the notes for the
next lecture.)

\begin{theorem}
The definitions of $\Sigma_k P$ and $\Pi_k P$ in {\ref{def2}} and
{\ref{oracledef}} are equivalent.
\end{theorem}

That {\ref{def2}} implies {\ref{oracledef}} is easy. For the converse,
our oracle Turing machine may make polynomially many queries to the
oracle tape nondeterministically, and intuitively, {\ref{def2}} allows
us to make only one such query, but note that we could use a different
$\Pi_k P$-language here. What we then do is to existentially guess the
computation history made by the oracle TM as well as the certificates
used by the queries to the oracle that returned ``yes''.
\vspace{8pt}

\begin{proof}
By induction on $k$. $k=0$ and $k=1$ are straight-forward. Note that
for each $k$, once we show that the 2 definitions of $\Sigma_k P$
equivalent, it follows that the 2 definitions of $\Pi_k P$ are equivalent,
because $\Pi_k P = \textrm{co}\Sigma_k P$ in both definitions.

Now, consider $k \geq 2$. That definition
{\ref{oracledef}} includes definition {\ref{def2}} is
easy: the nondeterministic machine on input $x$ guesses an appropriate
$y$ and makes a single query to check whether $(x,y) \in B$, where
$B \in \Sigma_{k-1} P$. 

Conversely, suppose $L \in \Sigma_k P$ according to definition
{\ref{oracledef}}, so $L$ can be decided by a polynomial-time NTM
$M^K$ using as an oracle a language $K \in \Sigma_{k-1}$, and by the
induction hypothesis, there exists a TM $C \in \Sigma_{k-2}$ and
an integer $c$ with the following property: $z \in K$ if and only
if there exists $w$ with $(z, w) \in C$ and $|w| \leq |z|^c$.

We may construct the TM $B$ in definition {\ref{def2}} as follows.
First, we know that $x \in L$ if and only if there exists a short
accepting computation of $M^K$ on $x$. Let $y$ be a string recording
such a computation of $M^K$, and $|y|$ is bounded by a polynomial
in $|x|$. Recall that $M^K$ is an oracle machine, so some of its
steps will be queries to $K$, and some of these queries will have
``yes'' answers and the remaining ones ``no'' answers. For each
``yes'' query $z_i$, our certificate $y$ also includes $z_i$'s own
certificate $w_i$ such that $(z_i, w_i) \in C$. This forms our
definition of $B$: $(x,y) \in B$ if and only if $y$ records an
accepting computation of $M^K$ on $x$, together with a certificate
$w_i$ for each ``yes'' query $z_i$ in the computation.

Next, we claim that checking whether $(x,y) \in B$ can be done in
$\Pi_{k-1}P$ (by the induction hypothesis, we may pick definition
{\ref{oracledef}} here). First, we need to check whether all steps of $M^K$
are legal, which can be done in deterministic polynomial time.
Then, we must check for polynomially many pairs $(z_i, w_i)$
whether $(z_i, w_i) \in C$, but this can be done in $\Pi_{k-2}P$,
which is in $\Pi_{k-1}P$. Finally, for all those ``no'' queries
$z_i'$, we must check that $z_i' \notin K$, or equivalently,
$z_i' \in \textrm{co}K$. However, $\textrm{co}K \in
\textrm{co}\Sigma_{k-1}P = \Pi_{k-1}P$. We may therefore collapse all
of these checks into a single $\Pi_{k-1}P$ computation.
\end{proof}

\begin{corollary}
The definitions of $\Pi_k P$ in {\ref{def2}} and {\ref{oracledef}}
are equivalent.
\end{corollary}

\begin{definition}
TQBF$^{\exists}_k$ = $\{ \langle \phi \rangle : \phi$ is
a satisfiable TQBF whose first quantifier is a $\exists$,
with $k-1$ alternations of the quantifiers$\}$.
TQBF$^{\forall}_k$ is defined similarly, except that
the first quantifier is a $\forall$.
\end{definition}

\begin{theorem}
For all $k \geq 1$, TQBF$^{\exists}_k$ is $\Sigma_kP$-complete.
\end{theorem}

\begin{proof}
We can inductively ``unwind'' definition {\ref{def2}}:
\begin{eqnarray*}
\Sigma_k P & = & \{ A\; |\; \exists\; B \in P, c_1, \ldots, c_k \ge 1
\mbox{ such that } x \in A \iff \\
& & \exists\; y_1 \quad |y_1| \leq |x|^{c_1} \\
& & \forall\; y_2 \quad |y_2| \leq (|x| + |y_1|)^{c_2} \\
& & \cdots \\
& & \exists/\forall\; y_k \quad |y_k| \leq \left(|x| + \sum_{i=1}^{k-1}
|y_i|\right)^{c_k} \\
& & \mbox{ such that } (x,y_1,y_2,\ldots,y_k) \in B \}
\end{eqnarray*}
(where the last quantifier depends on the parity of $k$).
It is then clear that TQBF$^{\exists}_k \in \Sigma_kP$
since we may evaluate a Boolean formula on a given input string in
polynomial time.

Conversely, given any $L \in \Sigma_k P$ in this form, we may in
polynomial time construct a Boolean formula that is equivalent to
$B \in P$ via Cook's Theorem, with variables that correspond to the
strings $y_1,y_2,\ldots,y_k$ so that we have an equivalent TQBF
with the same $k-1$ alternations of quantifiers. Hence, every language
$L \in \Sigma_k P$ is reducible to TQBF$^{\exists}_k$ in polynomial
time.
\end{proof}

By the same argument, we may show that TQBF$^{\forall}_k$ is
$\Pi_kP$-complete for $k \geq 1$.

\begin{theorem}
If $\Sigma_k P = \Pi_k P$ for some $k \ge 1$, then $\Sigma_j P =
\Sigma_k P\; \forall\; j > k$, in which case we say that ``the PH
collapses.''
\end{theorem}
 
\begin{proof}
It suffices to show that $\Sigma_k P = \Pi_k P$ implies
$\Sigma_{k+1} P = \Pi_{k+1} P$, from which the result follows by induction
on $j$.
We will use definition {\ref{def2}}.  Let $A \in \Sigma_{k+1} P \Rightarrow
\exists\; c > 0, B \in \Pi_k P$ satisfying: $x \in A \iff \exists\; y:\;
|y| \leq |x|^c$ such that $(x,y) \in B$.  If $\Pi_k P = \Sigma_k P,
\exists\; c' > 0, C \in \Pi_{k-1} P$ satisfying: $x \in A \iff
\exists\; y:\; |y| \leq |x|^c$ and $\exists\; y':\; |y'| \leq (|x| +
|y|)^{c'}$ such that $(x,y,y') \in C$.  But then we can just combine
$y$ and $y'$ into one string, which implies that $A \in \Sigma_k P$,
and the proof is complete.
\end{proof}

\end{document}