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\newcommand{\TIME}[1]{\ensuremath{\mbox{TIME}(#1)}}
\newcommand{\SPACE}[1]{\ensuremath{\mbox{SPACE}(#1)}}
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\begin{document}
\lecture{4}{February 11, 1998}{Daniel A. Spielman}{Chris Peikert}
\section*{Polynomial Hierarchy}
We will make two equivalent set of definitions of some complexity
classes, and prove their equivalence. The first is presented here:
\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
$O(f(n))$. We also define
\[
\Sigma_k P = \bigcup_{i=1}^{\infty} \Sigma_k \TIME{n^i}.
\]
\end{definition}
Note that $NP = \Sigma_1 P$, since a NTM remains in an existential
state over its entire computation.
\begin{definition}
$\Pi_k \TIME{f(n)}$ and $\Pi_k P$ are defined similarly, but the ATM
must begin in a universal ($\forall$) state.
\end{definition}
Note that $\mbox{co-}NP = \Pi_1 P$, and in general it can be shown
directly that $\Pi_k P = \mbox{co-} \Sigma_k P$. For consistency, we
define the special cases $\Pi_0 P = \Sigma_0 P = P$. These
definitions are reasonable if we think of the deciders as using {\em
no} quantifiers at all, so that their execution is deterministic.
Note that, by definition, $\Sigma_k P \subseteq \Pi_{k+1} P$ and
$\Pi_k P \subseteq \Sigma_{k+1} P$, since a machine can simply
alternate quantifiers once at the beginning of its computation. This
prompts the following definition:
\begin{definition}
The polynomial hiearchy $PH = \bigcup_k \Sigma_k P = \bigcup_k \Pi_k P$.
\end{definition}
By the PSPACE-completeness of TQBF, it's easy to see that $\mbox{PH}
\subseteq \mbox{PSPACE}$, but it's an open question whether
$\mbox{PSPACE} \subseteq \mbox{PH}$.
There may be a temptation to say that $\mbox{PSPACE}(f(n)) \subseteq
PH$ because $TQBF \subseteq \Sigma_{f(n)} \TIME{f(n)}$,
but this reaoning is flawed because in PH the number of alternations {\em
may not} grow with $n$.
We now present an alternate set of definitions:
\begin{definition}
First, $\Sigma_0 P = \Pi_0 P = P$. Next,
\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}
It should be clear that $\Sigma_1 P = NP$. These definitions
generalize the notion of a ``certificate'' or ``witness'', which
should be familiar from the standard interpretation of $NP$.
\begin{theorem}
The two definitions of $\Sigma_k P$ are equivalent.
\end{theorem}
\begin{proof}
We can inductively ``unwind'' the second definition, if $k$ is even:
\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 \\
& & \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*}
If $k$ is odd, we get a similar definition. To show that definition 2
includes definition 4, we note that a $\Sigma_k P$ machine (from
definition 2) can solve the conditions for membership in $A$, because
the $y_i$ values are short enough to guess in polynomial time, and
because there are few enough of them.
To show that definition 4 includes definition 2, consider an ATM M,
beginning in an $\exists$ state, changing state at most $k-1$ times,
and running in polynomial time. Then we can produce a machine N with
inputs of the form $(x,y_1,y_2,\ldots,y_k)$. N will simulate M, but
it will use bits from $y_1$ in the first collection of existential
states, the bits from $y_2$ in the next collection of universal
states, etc. In other words, N simulates M by choosing an execution
path based on the first bit of $y_1$, then at M's next $\exists$ state
it branches on the second bit of $y_1$, etc. until M goes into a
$\forall$ state, at which point it uses the bits of $y_2$, and so on.
Then we have M accepts $x \iff \exists\; y_1 \forall\; y_2 \cdots
\forall\; y_k$ N accepts $(x,y_1,y_2,\ldots,y_k)$ so $B =$ L(N) $\in
\Sigma_k P$ (from definition 4). \end{proof}
Finally, we see another way of looking at $\Sigma_k P$ and $\Pi_k P$,
based on {\em oracle Turing Machines}.
\begin{definition}
An oracle TM (OTM) $M^?$ with oracle (a language) $A$ has an oracle
tape on which it can write a string $w$ and enter an ``oracle state''
and, in one step, return to state ``oracle yes'' if $w \in A$ or
``oracle no'' if $w \not\in A$.
\end{definition}
Note that $P^A = \{ L\; |\; \exists\; \mbox{a poly-time OTM } M^? \mbox{
that, on oracle A, accepts L} \}$. Consider that $P^{\mbox{3SAT}}
\supseteq NP$ (which is trivial to show), and also $P^{\mbox{3SAT}}
\supseteq \mbox{co-}NP$ since a machine can reject if the oracle
replies ``yes'', and accept otherwise.
Similarly, $NP^A = \{ L\; |\; \exists\; \mbox{a nondet poly-time OTM }
M^? \mbox{ that, on oracle A, accepts L} \}$. We will prove
that
that $NP^{\mbox{3SAT}} = \Sigma_2 P$. More generally, we define the
language $\exists\mbox{QBF}_k$, which is the language of true QBFs,
for which the first quantifier is $\exists$, and which alternate
quantifiers at most $k - 1$ times. Similarly, we will show that
$NP^{\exists\mbox{QBF}_k} = \Sigma_{k+1} P$.
Also, $\exists\mbox{QBF}_k$ is complete for $\Sigma_k P$, but the
proof will be done later (see the solutions to last year's problem
sets for a preview).
\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.'' (It's widely believed that PH does {\em not} collapse.)
\end{theorem}
\begin{proof}
We will use definition 4. Let $A \in \Sigma_{k+1} P \Rightarrow
\exists\; c > 0, B \in \Pi_k P$ such that $x \in A \iff \exists\; y:\;
|y| \leq |x|^c$ such that $(x,y) \in B$. If $\Pi_k P = \Sigma_k P,
\exists\; c' > 0, C \in \Pi_{k-1} P$ such that $x \in A \iff
\exists\; y:\; |y| \leq |x|^c$ 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}